Fundamentals of Thermodynamics(Solutions)

10-1

CHAPTER 10 The correspondence between the new problem set and the previous 4th edition chapter 8 problem set. New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Old new new 1 2 3 5 6 7 9 15 16 10 11 49 47 13 14 38 52 8

New 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

Old new new 18 new 19 20 21 new new 23 24 new new 25 26 27 32 28 33 34

New 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Old 35 41 42 46 54 56 39 55 53 new 43 51 17 29 30 31 36 40 44 57

The problems that are labeled advanced starts at number 53. The English unit problems are: New 61 62 63 64 65 66 67 68 69 70

Old new 58 60 63 61 80 62 new 65 66

New 71 72 73 74 75 76 77 78 79 80

Old 67 new 68 new 69 70 72 new 79 84

New 81 82 83 84 85 86 87

Old 75 83 77 64 71 73 78

10-2 10.1

Calculate the reversible work and irreversibility for the process described in Problem 5.18, assuming that the heat transfer is with the surroundings at 20°C. C.V.: A + B. This is a control mass. Continuity equation: Energy:

m2 - (mA1 + mB1) = 0 ;

m2u2 - mA1uA1 - mB1uB1 = 1Q2 - 1W2

System: if VB ≥ 0 piston floats ⇒

PB = PB1 = const.

if VB = 0 then P2 < PB1 and v = VA/mtot see P-V diagram P State A1: Table B.1.1, x = 1 a vA1 = 1.694 m3/kg, uA1 = 2506.1 kJ/kg PB1

2

mA1 = VA/vA1 = 0.5903 kg

V2

State B1: Table B.1.2 sup. vapor vB1 = 1.0315 m3/kg, uB1 = 2965.5 kJ/kg mB1 = VB1/vB1 = 0.9695 kg => At (T2 , PB1)

m2 = mTOT = 1.56 kg

v2 = 0.7163 > v a = VA/mtot = 0.641 so VB2 > 0

so now state 2: P2 = PB1 = 300 kPa, T2 = 200 °C => u 2 = 2650.7 kJ/kg and V2 = m2 v2 = 1.56×0.7163 = 1.117 m3 (we could also have checked Ta at: 300 kPa, 0.641 m3/kg => T = 155 °C) ac 1W2 = 1Q 2

⌠ ⌡PBdVB = PB1(V2 - V1) = PB1(V2 - V1)tot = -264.82 kJ B

= m2u2 - mA1uA1 - mB1uB1 + 1W2 = -484.7 kJ

From the results above we have : sA1 = 7.3593, sB1 = 8.0329, s2 = 7.3115 kJ/kg K rev 1W2 =

To(S2 - S1) - (U2 - U1) + 1Q2(1 - To/TH) ac

= To(m2s2 - mA1sA1 - mB1sB1) + 1W2 - 1Q2To/TH = 293.15 (1.5598 × 7.3115 - 0.5903 × 7.3593 - 0.9695 × 8.0329) + (-264.82) - (-484.7) × 293.15 / 293.15 = -213.3 - 264.82 + 484.7 = 6.6 kJ 1I 2

rev

ac

= 1W2 - 1W2 = 6.6 - (-264.82) = 271.4 kJ

10-3 10.2

Calculate the reversible work and irreversibility for the process described in Problem 5.65, assuming that the heat transfer is with the surroundings at 20°C.

P 2

1

Linear spring gives 1W2

v

=⌠ ⌡PdV = 2(P1 + P2)(V2 - V1) 1

1Q 2

= m(u2 - u1) + 1W2 Equation of state: PV = mRT

State 1: V1 = mRT1/P1 = 2 x 0.18892 x 673.15 /500 = 0.5087 m3 State 2: V2 = mRT2/P2 = 2 x 0.18892 x 313.15 /300 = 0.3944 m3 1W2

1

= 2(500 + 300)(0.3944 - 0.5087) = -45.72 kJ

From Figure 5.10: Cp(Tavg) = 45/44 = 1.023 ⇒ Cv = 0.83 = C p - R For comparison the value from Table A.5 at 300 K is Cv = 0.653 kJ/kg K 1Q 2 rev 1W2 =

= mCv(T2 - T1) + 1W2 = 2 x 0.83(40 - 400) - 45.72 = -643.3 kJ To(S2 - S1) - (U2 - U1) + 1Q2(1 - To/TH) ac

= Tom(s2 - s1)+ 1W2 - 1Q2To/To ac

= Tom[CP ln(T2 / T1) − R ln(P2 / P1)] + 1W2 - 1Q2 = 293.15 x 2 [ 1.023 ln(313/673) - 0.1889 ln(300/500)] - 45.72 + 643.3 = -402.6 - 45.72 + 643.3 = 195.0 kJ 1I 2

10.3

rev

ac

= 1W2 - 1W2 = 195.0 - (-45.72) = 240.7 kJ

The compressor in a refrigerator takes refrigerant R-134a in at 100 kPa, −20°C and compresses it to 1 MPa, 40°C. With the room at 20°C find the minimum compressor work. Solution: C.V. Compressor out to ambient. Minimum work in is the reversible work. SSSF, 1 inlet and 2 exit energy Eq.:

wc = h1 - h2 + qrev

Entropy Eq.: s2 = s1 + ⌠ ⌡dq/T + sgen = s1 + qrev/T0 + 0 => qrev = T0(s2- s1) wc min = h1 - h2 + T0(s2- s1) = 387.22 - 420.25 + 293.15 × (1.7148 - 1.7665) = -48.19 kJ/kg

10-4 10.4

Calculate the reversible work out of the two-stage turbine shown in Problem 6.41, assuming the ambient is at 25°C. Compare this to the actual work which was found to be 18.08 MW. C.V. Turbine. SSSF, 1 inlet and 2 exits. Use Eq. 10.12 for each flow stream with q = 0 for adiabatic turbine. Table B.1: h1 = 3373.7, h2 = 2755.9, h3 = 251.4 + 0.9 x 2358.3 = 2373.9 s1 = 6.5966, s2 = 6.8382, s3 = 0.8319 + 0.9 x 7.0766 = 7.2009 . . . . . . . Wrev = T0(m2s2 + m3s3 - m1s1) - (m2h2 + m3h3 - m1h1) = 298.15(5 × 6.8382 + 15 × 7.2009 - 20 × 6.5966) −(5 × 2755.9 + 15 × 2373.9 - 20 × 3373.7) . . = 21.14 MW = Wac + Qrev = 3062.7 kW + 18084 kW

10.5

A household refrigerator has a freezer at TF and a cold space at TC from which energy is removed and rejected to the ambient at T A as shown in Fig. P10.5. . Assume that the rate of heat transfer from the cold space, QC, is the same as from . the freezer, QF, find an expression for the minimum power into the heat pump. . Evaluate this power when T A = 20°C, TC = 5°C, TF = −10°C, and QF = 3 kW. C.V. Refrigerator (heat pump), SSSF, no external flows except heat transfer. . . . . Energy Eq.: QF + Qc + W = −QA (amount rejected to ambient) Reversible gives minimum work in as from Eq. 10.1 or 10.9 on rate form. TA .  TA . .   293.15  293.15 W = QF 1 −  + Qc 1 −  = 31 −  + 31 −  263.15 278.15 TF  TC     = -0.504 kW

10.6

(negative so work goes in)

An air compressor takes air in at the state of the surroundings 100 kPa, 300 K. The air exits at 400 kPa, 200°C at the rate of 2 kg/s. Determine the minimum compressor work input. C.V. Compressor, SSSF, minimum work in is reversible work. ψ1 = 0 at ambient conditions s0 - s2 = sT°0 - sT°2 - R ln(P0/P2) = 6.86926 - 7.3303 - 0.287 ln(100/400) = -0.06317 ψ2 = h2 - h0 + T0(s0 - s2) = 475.79 - 300.473 + 300(s0 - s2) ψ2 = 156.365 kJ/kg . . . -WREV = m(ψ2-ψ1) = 312.73 kW = Wc

10-5 10.7

A supply of steam at 100 kPa, 150°C is needed in a hospital for cleaning purposes at a rate of 15 kg/s. A supply of steam at 150 kPa, 250°C is available from a boiler and tap water at 100 kPa, 15°C is also available. The two sources are then mixed in an SSSF mixing chamber to generate the desired state as output. Determine the rate of irreversibility of the mixing process. 1 3 2

B.1.1: h1 = 62.99 s1 = 0.2245 B.1.3: h2 = 2972.7 s2 = 7.8437 B.1.3: h3 = 2776.4 s3 = 7.6133

C.V. Mixing chamber, SSSF . . . Cont.: m1 + m2 = m3 . . . Energy: m1h1 + m2h2 = m3h3 . . . . Entropy: m1s1 + m2s2 + Sgen = m3s3

. . 2776.4 - 62.99 m2/m3 = (h3 - h1)/(h2 - h1) = = 0.9325 2972.7 - 62.99 . . m2 = 13.988 kg/s, m1 = 1.012 kg/s . . . . . I = T0Sgen = T0(m3s3 - m1s1 - m2s2) = 298.15 × (15 × 7.6133 - 1.012 × 0.2245 - 13.988 × 7.8437) = 1269 kW 10.8

Two flows of air both at 200 kPa of equal flow rates mix in an insulated mixing chamber. One flow is at 1500 K and the other is at 300 K. Find the irreversibility in the process per kilogram of air flowing out. 1 3 2

C.V. Mixing chamber . . . . Cont.: m1 + m2 = m3 = 2m1 . . . Energy: m1h1 + m1h2 = 2m1h3

Properties from Table A.7 h3 = (h1 + h2)/2 = (300.473 + 1635.8)/2 = 968.14 kJ/kg ° = 8.9867 ⇒ sT3 . . . . Entropy Eq.: m1s1 + m1s2 + Sgen = 2m1s3 ⇒ . . Sgen/2m1 = s3 - (s1 + s2)/2 = 8.9867 - (6.86926 - 8.61208)/2 = 1.246 kJ/kg K

10-6 10.9

A steam turbine receives steam at 6 MPa, 800°C. It has a heat loss of 49.7 kJ/kg and an isentropic efficiency of 90%. For an exit pressure of 15 kPa and surroundings at 20°C, find the actual work and the reversible work between the inlet and the exit. C.V. Reversible adiabatic turbine (isentropic) wT = hi - he,s ;

se,s = si = 7.6566 kJ/kg K,

hi = 4132.7 kJ/kg

xe,s = (7.6566 - 0.7548)/7.2536 = 0.9515, he,s = 225.91 + 0.9515×2373.14 = 2483.9 wT,s = 4132.7 - 2483.9 = 1648.79 kJ/kg C.V. Actual turbine wT,ac = ηwT,s = 1483.91 kJ/kg = hi - he,ac - qloss ⇒ he,ac = hi - qloss - wT,ac = 4132.7 - 49.7 - 1483.91 = 2599.1 Actual exit state: P,h ⇒

sat. vap.,

se,ac = 8.0085

C.V. Reversible process, work from Eq.10.12 qR = T0(se,ac - si) = 293.15 × (8.0085 - 7.6566) = 103.15

kJ kg

wR = hi - he,ac + qR = 4132.7 - 2599.1 + 103.16 = 1636.8 kJ/kg 10.10 A 2-kg piece of iron is heated from room temperature 25°C to 400°C by a heat source at 600°C. What is the irreversibility in the process? C.V. Iron out to 600°C source, which is a control mass. Energy Eq.:

mFe(u2 - u1) = 1Q2 - 1W2

Process: Constant pressure

=>

1W2

= PmFe(v2 - v1)

⇒ 1Q2 = mFe(h2 - h1) = mFeC(T2 - T1) 1Q2

= 2 × 0.42 × (400 - 25) = 315 kJ

mFe(s2 - s1) = 1Q2/Tres + 1S2 gen 1S2 gen

= mFe(s2 - s1) - 1Q2/Tres = mFeC ln (T2/T1) - 1Q2/Tres = 2 × 0.42 × ln

1I2

673.15 315 = 0.3233 kJ/K 298.15 873.15

= To (1S2 gen ) = 298.15 × 0.3233 = 96.4 kJ

10-7 10.11 A 2-kg/s flow of steam at 1 MPa, 700°C should be brought to 500°C by spraying in liquid water at 1 MPa, 20°C in an SSSF setup. Find the rate of irreversibility, assuming that surroundings are at 20°C. C.V. Mixing chamber, SSSF. State 1 is superheated vapor in, state 2 is compressed liquid in, and state 3 is flow out. No work or heat transfer. •

•

2

•

m3 = m1 + m2

Cont.:

•

Energy:

1

•

3

•

m3h3 = m1h1 + m2h2 •

Entropy:

•

•

•

m3s3 = m1s1 + m2s2 + Sgen

Table B.1.3: h1 = 3923.1, s1 = 8.2731,

h3 = 3478.5, s3 = 7.7622,

For state 2 interpolate between, saturated liquid 20°C table B.1.1 and, compressed liquid 5 MPa, 20°C from Table B.1.4: h2 = 84.9, s2 = 0.2964 •

•

x = m2/m1 = (h3 - h1)/(h2 - h3) = 0.13101 •

⇒ m2 = 2 × 0.131 = 0.262 kg/s ; •

•

•

•

•

m3 = 2 + 0.262 = 2.262 kg/s

Sgen = m3s3 - m1s1 - m2s2 = 0.9342 kW/K . . . . • I = Wrev - Wac = Wrev = ToSgen = 293.15 × 0.9342 = 273.9 kW

10-8 10.12 Fresh water can be produced from saltwater by evaporation and subsequent condensation. An example is shown in Fig. P10.12 where 150-kg/s saltwater, state 1, comes from the condenser in a large power plant. The water is throttled to the saturated pressure in the flash evaporator and the vapor, state 2, is then condensed by cooling with sea water. As the evaporation takes place below atmospheric pressure, pumps must bring the liquid water flows back up to P0. Assume that the saltwater has the same properties as pure water, the ambient is at 20°C and that there are no external heat transfers. With the states as shown in the table below find the irreversibility in the throttling valve and in the condenser. State 1 2 3 T [°C] 30 25 25 . . . C.V. Valve. m1 = mex = m2 + m3 Energy Eq.: h1 = 125.77 ,

h1 = he ;

4 --

Entropy Eq.:

s1 = 0.4369 ,

5 23

6 --

7 17

s1 + sgen = se

P2 = Psat(T2 = T3) = 3.169 kPa

he = h1 ⇒ xe = (125.77 - 104.87)/2442.3 = 0.008558 ⇒

se = 0.3673 + 0.008558 × 8.1905 = 0.4374

sgen = se - s1 = 0.4374 - 0.4369 = 0.000494 . . I = mT0sgen = 150 × 293.15 × 0.000494 = 21.72 kW C.V. Condenser. h2 = 2547.2 , s2 = 8.558 , h5 = 96.5 ,

. . m2 = (1 - xe)m1 = 148.716 kg/s

s5 = 0.3392

h7 = 71.37 , s7 = 0.2535 , h8 = 83.96 , s8 = 0.2966 . . . . m2h2 + m7h7 = m2h5 + m7h8 ⇒ . . 2547.2 - 96.5 kg m7 = m2 × (h2 - h5)/(h8 - h7) = 148.716 × = 28948 83.96 - 71.37 s . . . . . m2s2 + m7s7 + Sgen = m2s5 + m7s8 . . . . I = T0Sgen = T0 [ m2(s5 - s2) + m7(s8 - s7)] = 293.15[148.716(0.3392 - 8.558) + 28948(0.2966 - 0.2535)] = 293.15 × 25.392 = 7444 kW

8 20

10-9 10.13 An air compressor receives atmospheric air at T0 = 17°C, 100 kPa, and compresses it up to 1400 kPa. The compressor has an isentropic efficiency of 88% and it loses energy by heat transfer to the atmosphere as 10% of the isentropic work. Find the actual exit temperature and the reversible work. C.V. Compressor Isentropic: wc,in,s = he,s - hi ; Table A.7:

se,s = si

Pr,e,s = Pr,i × (Pe/Pi) = 0.9917 × 14 = 13.884 ⇒ he,s = 617.51

wc,in,s = 617.51 - 290.58 = 326.93 Actual: wc,in,ac = wc,in,s/ηc = 371.51 ;

qloss = 32.693

wc,in,ac + hi = he,ac + qloss => h e,ac = 290.58 + 371.51 - 32.693 = 629.4 => Te,ac = 621 K Reversible:

wrev = hi - he,ac + T0(se,ac - si) = 290.58 - 629.4 + 290.15 × (7.6121 - 6.8357) = -338.82 + 225.42 = -113.4 kJ/kg

Since qloss is also to the atmosphere it is not included as it will not be reversible. 10.14 A piston/cylinder has forces on the piston so it keeps constant pressure. It contains 2 kg of ammonia at 1 MPa, 40°C and is now heated to 100°C by a reversible heat engine that receives heat from a 200°C source. Find the work out of the heat engine. C.V. Ammonia plus heat engine Energy: mam(u2 - u1) = 1Q2,200 - WH.E. - 1W2,pist Entropy: mam(s2 - s1) = 1Q2/Tres + 0 =>

1Q2

= mam(s2 - s1)Tres

Process: P = const. ⇒ 1W2 = P(v2 - v1)mam WH.E. = mam(s2 - s1)Tres - mam(h2 - h1)

NH 3

⇑ QL H.E.

⇒W

⇑ QH 200 C

Table B.2.2: h1 = 1508.5, s1 = 5.1778, h2 = 1664.3, s2 = 5.6342 WH.E. = 2 × [(5.6342 - 5.1778)473.15 - (1664.3 - 1508.5)] = 120.3 kJ

10-10 10.15 Air flows through a constant pressure heating device, shown in Fig. P10.15. It is heated up in a reversible process with a work input of 200 kJ/kg air flowing. The device exchanges heat with the ambient at 300 K. The air enters at 300 K, 400 kPa. Assuming constant specific heat develop an expression for the exit temperature and solve for it. C.V. Total out to T0 s1 + qrev 0 /T 0 = s2

rev h1 + qrev = h2 0 -w

⇒ qrev 0 = T0(s2 - s1)

h2 - h1 = T0(s2 - s1) - wrev

(same as Eq. 10.12)

Constant Cp gives: Cp(T2 - T1) = T0Cp ln (T2/T1) + 200 T2 - T0ln(T2/T1) = T1 + 200/Cp T1 = 300 , Cp = 1.004 , T0 = 300 T2 - 300 ln (T2/300) = 300 + (200/1.004) = 499.3 At 600 K

LHS = 392 (too low)

At 800 K

Linear interpolation gives T2 = 790 K

LHS = 505.75

(LHS = 499.5 OK)

10.16 Air enters the turbocharger compressor (see Fig. P10.16), of an automotive engine at 100 kPa, 30°C, and exits at 170 kPa. The air is cooled by 50°C in an intercooler before entering the engine. The isentropic efficiency of the compressor is 75%. Determine the temperature of the air entering the engine and the irreversibility of the compression-cooling process. a) Compressor. First ideal which is reversible adiabatic, constant s: T2S = T1

P2

(P )

k-1 k

= 303.2

1

(170 ) 100

0.286

= 352.9 K

wS = CP0(T1 - T2S) = 1.004(303.2 - 352.9) = -49.9 w = wS/ηS = -49.9/0.75 = -66.5 kJ/kg = CP(T1 - T2) ⇒ T2 = 369.5 K T3(to engine) = T2 - ∆TINTERCOOLER = 369.5 - 50 = 319.5 K = 46.3°C b) Irreversibility from Eq.10.13 with rev. work from Eq.10.12, (q = 0 at TH) 319.4 170 kJ s3 - s1 = 1.004 ln  - 0.287 ln  = -0.1001 kg K 303.2 100 i = T(s3 - s1) - (h3 - h1) - w = T(s3 - s1) - CP(T3 - T1) - CP(T1 - T2) = 303.2(-0.1001) - 1.004(-50) = +19.8 kJ/kg

10-11 10.17 A car air-conditioning unit has a 0.5-kg aluminum storage cylinder that is sealed with a valve and it contains 2 L of refrigerant R-134a at 500 kPa and both are at room temperature 20°C. It is now installed in a car sitting outside where the whole system cools down to ambient temperature at −10°C. What is the irreversibility of this process? C.V. Aluminum and R-134a Energy Eq.: mAl(u2 - u1)Al + mR(u2 - u1)R = 1Q2 - 1W2 (1W2 = 0) Entropy Eq.:

mAL(s2 - s1)Al + mR(s2 - s1)R = 1Q2/T0 + 1S2 gen

(u2 - u1)Al = Cv,Al(T2 - T1) = 0.9(-10 - 20) = - 27 kJ/kg (s2 - s1)Al = Cp,Al ln(T2/T1) = 0.9 ln(263.15/293.15) = -0.09716 Table B.5.2: v1 = 0.04226, u1 = 411.65 - 500×0.04226 = 390.5 kJ/kg, s1 = 1.7342 kJ/kg K, v2 = v1 = 0.04226 & T2

=>

mR134a = V/v1 = 0.0473 kg x2 = (0.04226 - 0.000755)/0.09845 = 0.4216

u2 = 186.57 + 0.4216×185.7 = 264.9, s2 = 0.9507 + 0.4216×0.7812 = 1.2801 1Q2 1S2 gen 1I2

= 0.5 × (-27) + 0.0473(264.9 - 390.5) = - 19.44 kJ

= 0.5 (-0.09716) + 0.0473(1.2801 - 1.7342) +

19.44 = 0.003815 kJ/K 263.15

= T0 ( 1S2 gen ) = 263.15 × 0.003815 = 1.0 kJ

10.18 A steady combustion of natural gas yields 0.15 kg/s of products (having approximately the same properties as air) at 1100°C, 100 kPa. The products are passed through a heat exchanger and exit at 550°C. What is the maximum theoretical power output from a cyclic heat engine operating on the heat rejected from the combustion products, assuming that the ambient temperature is 20°C? PROD.e PROD.i Heat T i 1100 o o o Exchanger ( C) 1100 C 550 C . e ⇓ QH . 550 Heat ⇒ W Engine NET . ∆S ⇓ QL o

T L= 20 C

TL

. . QH = mPRODCP0(TL - Te)PROD = 0.15 × 1.004(1100 - 550) = 82.83 . . . ∆SL = - ∆Si-e = mPRODCP0 ln(Ti/Te)PROD = 0.15 × 1.004 ln(1373.15/823.15) = 0.077 kW/K . . QL = TL ∆SL = 293.15 × 0.077 = 22.57 kW . . . WNET = QH - QL = 82.83 - 22.57 = 60.26 kW

20

s

10-12 10.19 A counterflowing heat exchanger cools air at 600 K, 400 kPa to 320 K using a supply of water at 20°C, 200 kPa. The water flow rate is 0.1 kg/s and the air flow rate is 1 kg/s. Assume this can be done in a reversible process by the use of heat engines and neglect kinetic energy changes. Find the water exit temperature and the power out of the heat engine(s). 2

1 ⇓

H2O

⇒

AIR

·

W

• • • • m as 1 + mH2Os3 = mas 2 + mH2Os4

⇓

3

C.V. Total • • • • • m ah1 + mH2Oh3 = mah2 + mH2Oh4 + W Table A.7: h1 = 607.316,

4

° = 6.93413, Table A.7: h2 = 320.576, sT2

B.1.1:

(sgen = 0)

° = 7.57638 sT1

h3 = 83.96, s3 = 0.2966

• • s4 = (m a/mH2O)(s1 - s 2) + s3 = (1/0.1)(7.57638 - 6.93413) + 0.2966 = 6.7191

4: P4 = P3, s4 ⇒ Table B.1.2:

x4 = (6.7191-1.530)/5.597 = 0.9271,

h4 = 504.68 + 0.9271 × 2201.96 = 2546.1 kJ/kg, •

•

T4 = 120.20°C

•

W = ma(h1 - h2) + mH2O(h3 - h4) = 1(607.32 - 320.58) + 0.1(83.96 - 2546.1) = 40.53 kW 10.20 Water as saturated liquid at 200 kPa goes through a constant pressure heat exchanger as shown in Fig. P10.20. The heat input is supplied from a reversible heat pump extracting heat from the surroundings at 17°C. The water flow rate is 2 kg/min and the whole process is reversible, that is, there is no overall net entropy change. If the heat pump receives 40 kW of work find the water exit state and the increase in availability of the water. C.V. Heat exchanger + heat pump. . . . . . . . . . m1 = m2 = 2 kg/min, m1h1 + Q0 + Win = m1h2, m1s1 + Q/T0 = m1s2 . . Substitute Q0 into energy equation and divide by m1 h1 - T0s1 + win = h2 - T0s2 LHS = 504.7 - 290.15 × 1.5301 + 40×60/2 = 1260.7 kJ/kg State 2: P2 , h2 - T0s2 = 1260.7 kJ/kg At sat. vap. hg - T0sg = 638.8 so state 2 is superheated vapor at 200 kPa. At 600 C:

h2 - T0s2 = 3703.96 - 290.15 × 8.7769 = 1157.34 kJ/kg

At 700 C:

h2 - T0s2 = 3927.66 - 290.15 × 9.0194 = 1310.68 kJ/kg

Linear interpolation ⇒ T2 = 667°C ∆ψ = (h2 - T0s2) - (h1 - T0s1) = win = 1200 kJ/kg = 1260.7 - 504.7 + 290.15 × 1.5301 ≈ 1200 kJ/kg

10-13 10.21 Calculate the irreversibility for the process described in Problem 6.63, assuming that heat transfer is with the surroundings at 17°C. I = To Sgen so apply 2nd law out to To = 17 oC m2 s2 - m1s1 = misi + 1Q2 / To + 1S2gen ToSgen = To ( m2 s2 - m1s1 - mi si ) - 1Q2 m1 = 0.90 kg

mi = 3.082 kg m2 =3.982 kg ToSgen = I = To [ m1 (s2 - s1) + mi (s2 - si)] - 1Q2 = 290.15[0.9(Cp ln

350 400 350 400 - R ln ) + 3.082(Cpln - R ln )] 290 300 600 500

- ( - 819.2 kJ) = 290.15 (0.0956 - 1.4705) + 819.2 = 420.3 kJ 10.22 The high-temperature heat source for a cyclic heat engine is a SSSF heat exchanger where R-134a enters at 80°C, saturated vapor, and exits at 80°C, saturated liquid at a flow rate of 5 kg/s. Heat is rejected from the heat engine to a SSSF heat exchanger where air enters at 150 kPa and ambient temperature 20°C, and exits at 125 kPa, 70°C. The rate of irreversibility for the overall process is 175 kW. Calculate the mass flow rate of the air and the thermal efficiency of the heat engine. . C.V. R-134a Heat Exchanger, mR134a = 5 kg/s Inlet: T1 = 80oC, sat. vap. ‡ x1 = 1.0, Table B.5.1 h1 = hg = 429.189 kJ/kg, s1 = sg = 1.6862 kJ/kg-K Exit: T2 = 80oC, sat. liq. ‡ x2 = 0.0 h2 = hf = 322.794 kJ/kg, s2 = sf = 1.3849 kJ/kg-K C.V. Air Heat Exchanger, C p = 1.004 kJ/kg-K, R = 0.287 kJ/kg-K Inlet: T3 = 20oC, P3 = 150 kPa Exit: T4 = 70oC, P4 = 125 kPa . . . . 2nd Law: I = To Snet = mR134a (s2 - s1) + mair (s4 - s3) s4 - s3 = Cpln(T4/T3) - Rln(P4/P3) = 0.2103 kJ/kg-K . . . mair = [I - mR134a (s2 - s1)]/(s4 - s3) = 10.0 kg/s . . . . . 1st Law for each line: Q + mhin = mhex + W; W = 0 . . . . R-134a: 1Q2 = mR134a(h2 - h1) = -532 kW; QH = -1Q2 . . . . . Air: 3Q4 = mair (h4 - h3) = mair (T4 - T3) = 501.8 kW; QL = 3Q4 . . . . . Wnet = QH - QL = 30.2 kW; ηth = Wnet / QH = 0.057, or 5.7%

10-14 10.23 A control mass gives out 10 kJ of energy in the form of a. Electrical work from a battery b. Mechanical work from a spring c. Heat transfer at 500°C Find the change in availability of the control mass for each of the three cases. a) ∆φ = -Wel = -10 kJ b) ∆φ = -Wspring = -10 kJ  298.15 c) ∆φ = -[1 - (T0/TH)] Qout = -1  10 = -6.14 kJ  773.15 10.24 Calculate the availability of the water at the initial and final states of Problem 8.32, and the irreversibility of the process. 1: u1 = 83.94

s1= 0.2966

ac 1W2 =

2: u2 = 3124.3

s2= 7.7621

ac 1Q 2

0: uo = 104.86 ,

so= 0.3673

203 kJ

v1= 0.001

= 3243.4 v2= 0.354

vo= 0.001003 ,

TH = 873.15 K

φ = (u - Tos) - (uo - Toso) + Po( v - vo) φ1 = (83.94 - 298.15×0.2966) - (104.86 - 298.15×0.3673) + 100 (0.001002 - 0.001003) = 0.159 kJ/kg φ2 = (3124.3 - 298.15×7.7621) - (104.86 - 298.15×0.3673) + 100 (0.35411 - 0.001003) = 850 kJ/kg ac

ac

1I2 = m(φ1 - φ2) + [1-(T0/TH)]1Q2 - 1W2 + Po( V2 - V1) = -849.84 + (1 -

298.15 ) 3243.4 - 203 + 100 (0.3541 - 0.001) 873.15

= -849.84 + 2135.9 - 203 + 35.31 = 1118. kJ [(Sgen = 3.75 kJ/K ToSgen = 1118 kJ

so OK]

10.25 A steady stream of R-22 at ambient temperature, 10°C, and at 750 kPa enters a solar collector. The stream exits at 80°C, 700 kPa. Calculate the change in availability of the R-22 between these two states. R-22 i

e Solar

Collector

Inlet (T,P) Table B.4.1 (liquid) hi = 56.46 kJ/kg, si = 0.2173 kJ/kg K Exit (T,P) Table B.4.2 (sup. vap.) he = 305.91 kJ/kg, se = 1.0761 kJ/kg K

∆ψ ie = ψe - ψi = (he - hi) - T0(se - si) = (305.912 - 56.463) - 283.2(1.0761 - 0.2173) = 6.237 kJ/kg

10-15 10.26 Nitrogen flows in a pipe with velocity 300 m/s at 500 kPa, 300°C. What is its availability with respect to an ambient at 100 kPa, 20°C? ψ = h1 - h0 + (1/2)V12 - T0(s1 - s0) = Cp(T1 - T0) + (1/2)V12 - T0(Cp ln(T1/T0) - R ln(P1/P0)) = 1.0426(300 - 20) + (300 2/2000)  573.15 500 - 293.151.042 ln  - 0.2968ln  = 271.96 kJ/kg  293.15 100 10.27 A 10-kg iron disk brake on a car is initially at 10°C. Suddenly the brake pad hangs up, increasing the brake temperature by friction to 110°C while the car maintains constant speed. Find the change in availability of the disk and the energy depletion of the car’s gas tank due to this process alone. Assume that the engine has a thermal efficiency of 35%. All the friction work is turned into internal energy of the disk brake. m(u2 - u1) = 1Q2 - 1W2 ⇒ 1Q2 = mFeCFe(T2 - T1) 1Q2

= 10 × 0.45 × (110 - 10) = 450 kJ

Neglect the work to the surroundings at P 0 ∆φ = m(u2 - u1) - T0m(s2 - s1) 383.15 m(s2-s1) = mC ln(T2/T1) = 10 × 0.45 × ln   = 1.361 kJ/K 283.15 ∆φ = 450 - 283.15 × 1.361 = 64.63 kJ Wengine = ηthQgas = 1Q2 = Friction work Qgas = 1Q2/ηth = 450/0.35 = 1285.7 kJ

10-16 10.28 A 1 kg block of copper at 350°C is quenched in a 10 kg oil bath initially at ambient temperature of 20°C. Calculate the final uniform temperature (no heat transfer to/from ambient) and the change of availability of the system (copper and oil). C.V. Copper and oil. Cco = 0.42 Coil = 1.8 m2u2 - m1u1 = 1Q2 - 1W2 = 0 = mcoCco(T2 - T1)co + (mC)oil(T2 - T1)oil 1 × 0.42 ( T2 - 350) + 10 × 1.8 (T2 - 20) = 0 18.42 T2 = 507 => T = 27.5 C = 300.65 K For each mass copper and oil, we neglect work term (v = C) so Eq.10.22 is (φ2 - φ1) = u2 - u1 - To(s2 - s1) = mC [(T2 - T1) - Toln (T2 / T1) ] mcv(φ2 - φ1)cv + moil (φ2 - φ1)oil = 300.65 300.65 ] + 10 × 1.8 [7.5 - 293.15 ln ] 623.15 293.15 = - 45.713 + 1.698 = - 44.0 kJ

= 0.42 × [(-322.5) - 293.15 ln

10.29 Calculate the availability of the system (aluminum plus gas) at the initial and final states of Problem 8.74, and also the process irreversibility. State 1: T1 = 200 oC, v1 = V1/ m = 0.05 / 1.1186 = 0.0447 v2 = v1 × (2 / 1.5) × (298.15 / 473.15) = 0.03756 The metal does not change volume, so the combined is using Eq.10.22 as State 2:

φ1 = mgasφgas + mAlφAl = mgas[u1-uo-To(s1 - so)]cv + mgasPo(v1-vo) + mAl[u1-uo -To(s1-so)]Al T P = mgasCv (T1 - To) - mgasTo [Cp ln 1 - R ln 1 ] + mgasPo (v1 - vo) To Po + mAl [C (T1 - To) - ToC ln (T1/To) ]Al φ1 = 1.1186 [ 0.653(200-25) - 298.15 (0.842 ln

473.15 2000 - 0.18892 ln ) 298.15 100

+ 100 (0.0447 - 0.5633 ) ] + 4 × 0.90 [ 200 -25 - 298.15 ln

473.15 ] 298.15

= 128.88 + 134.3 = 263.2 kJ φ2 = 1.1186 [ 0.653(25 - 25) - 298.15 (0.842 ln

298.15 1500 - 0.18892 ln ) 298.15 100

+ 100 (0.03756 - 0.5633 ) ] + 4 × 0.9 [ 25 -25 - 298.15 ln

298.15 ] 298.15

= 111.82 + 0 = 111.82 kJ 1I2 = φ1 - φ2 + [1-(T0/TH)] 1Q2 - 1W2AC + Pom( V2 - V1) = 263.2 - 111.82 + 0 - (-14) + 100 × 1.1186 (0.03756 - 0.0447) = 164.58 kJ [(Sgen = 0.552

ToSgen = 164.58

so OK ]

10-17 10.30 Consider the springtime melting of ice in the mountains, which gives cold water running in a river at 2°C while the air temperature is 20°C. What is the availability of the water (SSSF) relative to the temperature of the ambient? ψ = h1 - h0 - T0(s1 - s0)

flow availability from Eq.10.19

Approximate both states as saturated liquid ψ = 8.392 - 83.96 - 293.15(0.03044 - 0.2966) = 2.457 kJ/kg Why is it positive? As the water is brought to 20°C it can be heated with qL from a heat engine using qH from atmosphere TH = T0 thus giving out work. 10.31 Refrigerant R-12 at 30°C, 0.75 MPa enters a SSSF device and exits at 30°C, 100 kPa. Assume the process is isothermal and reversible. Find the change in availability of the refrigerant. hi = 64.539, si = 0.2397, he = 209.866, se = 0.8482 ∆ψ = he - hi - T0(se - si) = 209.866 - 64.539 - 298.15(0.8482 - 0.2397) = -36.1 kJ/kg 10.32 A geothermal source provides 10 kg/s of hot water at 500 kPa, 150°C flowing into a flash evaporator that separates vapor and liquid at 200 kPa. Find the three fluxes of availability (inlet and two outlets) and the irreversibility rate. C.V. Flash evaporator chamber. SSSF with no work or heat transfer. . . . Cont. Eq.: m1 = m2 + m3 ; Vap. . . . 2 1 Energy Eq.: m1h1 = m2h2 + m3h3 Liq. . . . . 3 Entropy Eq.: m1s1 + Sgen = m2s2 + m3s3 B.1.1: B.1.2:

ho = 104.87, so = 0.3673, h1 = 632.18, s1= 1.8417 h2 = 2706.63, s2= 7.1271, h3 = 504.68, s3= 1.530 . . h -h h1 = xh2 + (1 - x) h3 => x = m2/m1 = 1 3 = 0.0579 h2 - h3 . . . . m2 = xm1 = 0.579 kg/s m3 = (1-x)m1 = 9.421 kg/s . Sgen = 0.579 × 7.1271 + 9.421 × 1.53 - 10 × 1.8417 = 0.124 kW/K

Flow availability Eq.10.22: ψ = (h - Tos) - (ho - Toso) = h - ho - To(s - so) ψ1 = 632.18 - 104.87 - 298.15 (1.8417 - 0.3673) = 87.72 ψ2 = 2706.63 - 104.87 - 298.15 (7.1271 - 0.3673) = 586.33 ψ3 = 504.68 - 104.87 - 298.15 (1.53 - 0.3673) = 53.15 . . . m1 ψ1 = 877.2 kW m2ψ2 = 339.5 kW m3ψ3 = 500.7 kW . . . . I = m1 ψ1 - m2 ψ2 - m3ψ3 = 37 kW

10-18 10.33 Air flows at 1500 K, 100 kPa through a constant pressure heat exchanger giving energy to a heat engine and comes out at 500 K. What is the constant temperature the same heat transfer should be delivered at to provide the same availability? C.V. Heat exchanger. SSSF, no work term. qout = h1 - h2 = 1635.8 - 503.36 = 1132.4 kJ/kg

Heat exch

1

T ∆ψ = (1 - o ) qout = ψ1 - ψ2 TH

.

⇓ QH H.E.

⇒

2 . W

. ⇓ QL

ψ1 - ψ2 = h1 - h2 - To ( s1 - s2 ) = 1132.4 - 298.15 (8.612 - 7.3869) = 1132.4 - 356.3 = 767.1 kJ/kg T 1 - o = (ψ1 - ψ2 ) / qout = 767.1 / 1132.4 = 0.6774 TH To = 0.3226 => TH = 924 K TH 10.34 A wooden bucket (2 kg) with 10 kg hot liquid water, both at 85°C, is lowered 400 m down into a mineshaft. What is the availability of the bucket and water with respect to the surface ambient at 20°C? C.V. Bucket and water. Both thermal availability and potential energy terms. v1 ≈ v0 for both wood and water so work to atm. is zero. Use constant heat capacity table A.3 for wood and table B.1.1 (sat. liq.) for water. From Eq.10.22 φ1 - φ0 = mwood[u1 - u0 - T0(s1- s0)] + mH2O[u1- u0- T0(s1- s0)] + mtotg(z1- z0) 273.15 + 85 ] + 10[ 355.82 - 83.94 293.15

= 2[1.26(85 - 20) - 293.15× 1.26 ln

- 293(1.1342 - 0.2966)] + 12 × 9.807 × (-400) /1000 = 15.85 + 263.38 - 47.07 = 232.2 kJ

10-19 10.35 A rigid container with volume 200 L is divided into two equal volumes by a partition. Both sides contains nitrogen, one side is at 2 MPa, 300°C, and the other at 1 MPa, 50°C. The partition ruptures, and the nitrogen comes to a uniform state at 100°C. Assuming the surroundings are at 25°C find the actual heat transfer and the irreversibility in the process. m2 = mA + mB =

PA1VA RTA1

+

PB1VB RTB1

=

2000×0.1 1000×0.1 + 0.2968×573.15 0.2968×323.15

= 1.176 + 1.043 = 2.219 kg P2 = m2RT2/Vtot = 2.219 × 0.2968 × 373.15/0.2 = 1228.8 kPa Energy Eq.: 1Q2

mA(u2 - u1)A + mB(u2 - u1)B = 1Q2 - 0/

= mACv(T2 - T1)A + mBCv(T2 - T1)B

= 1.176 × 0.745 × (100 - 300) + 1.043 × 0.745 × (100 - 50) = -136.4 kJ mA(s2 - s1)A + mB(s2 - s1)B = 1Q2/Tsur + 1Ss gen 373.15 1228.8 (s2 - s1)A = 1.042 × ln - 0.2968 × ln = -0.09356 573.15 2000 373.15 1228.8 (s2 - s1)B = 1.042 × ln - 0.2968 × ln = 0.0887 323.15 1000 1S2,gen

= 1.176× (-0.09356) + 1.043× 0.0887 + 136.4/298.15 = 0.4396 kJ/K

I = (T0)(1S2,gen) = 131.08 kJ 10.36 An air compressor is used to charge an initially empty 200-L tank with air up to 5 MPa. The air inlet to the compressor is at 100 kPa, 17°C and the compressor isentropic efficiency is 80%. Find the total compressor work and the change in availability of the air. C.V. Tank + compressor (constant inlet conditions) USUF, no heat transfer. Continuity: m2 - m1 = min ( m1 = 0 ) Energy: m2u2 = minhin - 1W2 Entropy:

m2s2 = minsin + 1S2 gen

Reversible compressor: 1S2 GEN = 0 ⇒ s2 = sin ⇒ Pr2 = Prin(P2/Pin) = 49.495 ⇒ T2,s = 855, u2,s = 637.2

⇒ 1w2,s = hin - u2,s = -346.77

Actual compressor: 1w2,AC = 1w2,s/ηc = -433.46 u2,AC = hin - 1w2,AC = 723.89 ⇒ T2,AC = 958 K ⇒ v2 = RT2/P2 = 0.055 State 2 u, P

m2 = V2/v2 = 3.636 ⇒ 1W2 = m2(1w2,AC) = -1576 kJ

m2(φ2 - φ1) = m2[u2 - u1 + P0(v2 - v1) - T0(s2 - s1)] = 3.636 [723.89 - 207.19 + 100(0.055 - 0.8323) - 290[8.0867 6.8352 - 0.287 ln(5000/100)] = 1460.3 kJ

10-20 10.37 Air enters a compressor at ambient conditions, 100 kPa, 300 K, and exits at 800 kPa. If the isentropic compressor efficiency is 85%, what is the second-law efficiency of the compressor process? 800 kPa T2s = 300(8)0.286 = 543.8 K -Ws = 1.004(543.8 - 300) = 244.6 kJ/kg 2 100 kPa 2s -Ws 244.6 -w = = = 287.8 K 0.85 ηs 300 K 1 -w 287.8 = 300 + = 586.8 K T 2 = T1 + s CP0 1.004 s2 - s1 = 1.004 ln(586.8/300) - 0.287 ln 8 = 0.07645 ψ2 - ψ1 = (h2 - h1) - T0(s2 - s1) = 287.8 - 300(0.07645) = 264.9 kJ/kg 2nd law efficiency: η2nd Law = (ψ2 - ψ1)/(-w) = 264.9/287.8 = 0.92 10.38 Steam enters a turbine at 25 MPa, 550°C and exits at 5 MPa, 325°C at a flow rate of 70 kg/s. Determine the total power output of the turbine, its isentropic efficiency and the second law efficiency. hi = 3335.6, si = 6.1765, he = 2996.5, se = 6.3289 Actual turbine: wT,ac = hi - he = 339.1 kJ/kg Isentropic turbine: se,s = si ⇒ he,s = 2906.6 wT,s = hi - he,s = 429 kJ/kg Rev. turbine: wrev = wT,ac + T0(se - si) = 339.1 + 45.44 = 384.54 kJ/kg ηT = wT,ac/wT,s = 339.1/429 = 0.79 ηII = wT,ac/wrev = 339.1/384.54 = 0.88 10.39 A compressor is used to bring saturated water vapor at 1 MPa up to 17.5 MPa, where the actual exit temperature is 650°C. Find the irreversibility and the second-law efficiency. Inlet state: Table B.1.2

hi = 2778.1,

si = 6.5864

Actual compressor Table B.1.3

h e,ac = 3693.9 , se,ac = 6.7356

Energy Eq. Actual compressor:

-wc,ac = he,ac - hi = 915.8 kJ/kg

From Eq.10.14: i = T0(se,ac - si) = 298.15 (6.7356 - 6.5864) = 44.48 kJ/kg From Eq.10.13:

wrev = i + w c,ac = -915.8 + 44.48 = -871.32

ηII = -wrev/wc,ac = 871.32/915.8 = 0.951

10-21 10.40 A flow of steam at 10 MPa, 550°C goes through a two-stage turbine. The pressure between the stages is 2 MPa and the second stage has an exit at 50 kPa. Assume both stages have an isentropic efficiency of 85%. Find the second law efficiencies for both stages of the turbine. 1

2

3

CV: T1, h1 = 3500.9 kJ/kg, s1 = 6.7561 kJ/kg K s2s = s1 ⇒ h2s = 3017.9 wT1,s = h1 - h2s = 483 kJ/kg

Isentropic T1

T2

Actual T1:

wT1,ac = ηT1 wT1,s = 410.55 = h 1 - h2ac

h2ac = h1 - wT1,ac = 3090.35,

s2ac = 6.8782

CV: T2, s3s = s2ac = 6.8782 ⇒ x3s = (6.8782-1.091)/6.5029 = 0.8899, h3s = 340.47 + 0.8899 × 2305.4 = 2392.2 kJ/kg wT2,s = h2ac - h3s = 698.15 ⇒

wT2,ac = ηT2 wT2,s = 593.4 kJ/kg

⇒ h3ac = 2496.9, x3ac = (2496.9 - 340.47)/2305.4 =0.9354, s3ac = 1.091 + 0.9354 × 6.5029 = 7.1736 kJ/kg K Actual T1:

iT1,ac = T0(s2ac-s1) = 298.15(6.8782 - 6.7561) = 36.4 kJ/kg

R

⇒ wT1 = wT1,ac + i = 447 kJ/kg,

R

ηII = wT1,ac/wT1 = 0.918

Actual T2: iT2,ac = T0(s3ac-s2ac) = 298.15(7.1736 - 6.8782) = 88.07 kJ/kg R

⇒ wT2 = wT2,ac + iT2,ac = 681.5,

R

ηII = wT2,ac/wT2 = 0.871

10.41 Consider the two-stage turbine in the previous problem as a single turbine from inlet to final actual exit and find its second-law efficiency. From solution to Problem 10.40 we have wT,ac = wT1,ac + wT2,ac = h1 - h3ac = 1004 kJ/kg The actual compared to the reversible turbine has, Eq.10.14, R i = qR = T0(s3 - si) = qR T1 + qT2 = 124.47 kJ/kg

wR = wT,ac + i = 1128.5 kJ/kg ηII = wT,ac/wR = 0.89

10-22 10.42 The simple steam power plant shown in Problem 6.39 has a turbine with given inlet and exit states. Find the availability at the turbine exit, state 6. Find the second law efficiency for the turbine, neglecting kinetic energy at state 5. h6 = 2393.2, s6 = 7.5501, h5 = 3404.3, s5 = 6.8953 ψ6 = h6 - h0 - T0(s6 - s0) = 2393.2 - 104.89 - 298.15(6.8953 - 0.3674) = 146.79 kJ/kg w rev = ψ5 - ψ6 = h5 - h6 - T0(s5 - s6) = 1206.3 kJ/kg. ηII = wac/wrev = 0.838

wac = h5 - h6 = 1011.1 ;

10.43 A compressor takes in saturated vapor R-134a at −20°C and delivers it at 30°C, 0.4 MPa. Assuming that the compression is adiabatic, find the isentropic efficiency and the second law efficiency. Table B.5

hi = 386.08, si = 1.7395, he,ac = 423.22, se,ac = 1.7895

Ideal exit: Pe, se,s = si ⇒ he,s = 408.51 Isentropic compressor wc,s = he,s - hi = 22.43 Actual compressor wc,ac = he,ac - hi = 37.14 Reversible between inlet and actual exit -wc,rev = hi - he,ac - T0(si - se,ac) = -37.14 - 298.15(1.7395 - 1.7895) = -22.23 ηTH,c = (wc,s/wc,ac) = (22.43/37.14) = 0.604 ηII = (wc,rev/wc,ac) = (22.23/37.14) = 0.599 10.44 Steam is supplied in a line at 3 MPa, 700°C. A turbine with an isentropic efficiency of 85% is connected to the line by a valve and it exhausts to the atmosphere at 100 kPa. If the steam is throttled down to 2 MPa before entering the turbine find the actual turbine specific work. Find the change in availability through the valve and the second law efficiency of the turbine. Take C.V. as valve and a C.V. as the turbine. Valve:

h2 = h1 = 3911.7 ,

s2 > s1 = 7.7571 ,

h2, P2 ⇒ s2 = 7.9425

ψ1 - ψ2 = h1−h2 −T0(s1-s2) = 0 -298.15(7.7571-7.9425) = 55.3 kJ/kg So some potential work is lost in the throttling process. Ideal turbine: s3 = s2 ⇒ h3s = 2929.13

wT,s = 982.57 kJ/kg

wT,ac = h2 - h3ac = ηwT,s = 835.2 kJ/kg h3ac = 3911.7 - 835.2 = 3076.5

⇒ s3ac = 8.219 kJ/kg K

wrev = h2 - h3ac - T0(s2 - s3ac) = 835.2 - 298.15(7.9425 - 8.219) = 917.63 kJ/kg

⇒

ηII = 835.2/917.63 = 0.91

10-23 10.45 The condenser in a refrigerator receives R-134a at 700 kPa, 50°C and it exits as saturated liquid at 25°C. The flowrate is 0.1 kg/s and the condenser has air flowing in at ambient 15°C and leaving at 35°C. Find the minimum flow rate of air and the heat exchanger second-law efficiency. C.V. Total heat exchanger.

4

3

AIR 2

1

R-134a

• • m a = m1 ×

⇒

• • • • m 1h1 + mah3 = m1h2 + mah4

h1 - h2 h4 - h3

436.89 - 234.59 = 1.007 kg/s 1.004(35 - 15)

= 0.1 ×

ψ1 - ψ2 = h1 - h2 - T0(s1 - s2) = 436.89 - 234.59 - 288.15(1.7919 - 1.1201) = 8.7208 ψ4 - ψ3 = h4 - h3 - T0(s4 - s3) = 1.004(35 - 15) - 288.15 × 1.004 × ln •

•

ηII = ma(ψ4 - ψ3)/m1(ψ1 - ψ2) =

308.15 = +0.666 288.15

1.007(0.666) = 0.77 0.1(8.7208)

10.46 A piston/cylinder arrangement has a load on the piston so it maintains constant pressure. It contains 1 kg of steam at 500 kPa, 50% quality. Heat from a reservoir at 700°C brings the steam to 600°C. Find the second-law efficiency for this process. Note that no formula is given for this particular case so determine a reasonable expression for it. 1: Table B.1.2

P 1 , x1

⇒

v1 = 0.001093 + 0.5×0.3738 = 0.188,

h1 = 640.21 + 0.5×2108.47 = 1694.5 , s1 = 1.8606 + 0.5×4.9606 = 4.341 2: P2 = P1,T2 ⇒ v2 = 0.8041, h2 = 3701.7 , s2 = 8.3521 m(u2 - u1) = 1Q2 - 1W2 = 1Q2 - P(V2 - V1) 1Q2 1W2

= m(u2 - u1) + Pm(v2 - v1) = m(h2 - h1) = 2007.2 kJ = Pm(v2 - v1) = 308.05 kJ

1W2 to atm =

P0m(v2 - v1) = 61.61 kJ

Useful work out = 1W2 - 1W2 to atm = 246.44  298.15 ∆φ reservoir = (1 - T0/Tres)1Q2 = 1  2007.2 = 1392.2  973.15 ηII = Wnet/∆φ = 0.177

10-24 10.47 Air flows into a heat engine at ambient conditions 100 kPa, 300 K, as shown in Fig. P10.47. Energy is supplied as 1200 kJ per kg air from a 1500 K source and in some part of the process a heat transfer loss of 300 kJ/kg air happens at 750 K. The air leaves the engine at 100 kPa, 800 K. Find the first and the second law efficiencies. C.V. Engine out to reservoirs hi + q1500 = q750 + he + w wac = 300.473 + 1200 - 300 - 822.202 = 378.27 kJ/kg ηTH = w/q1500 = 0.3152 For second law efficiency also a q to/from ambient si + (q1500/TH) + (q0/T0) = (q750/Tm) + se q0 = T0(se - si) + (T0/Tm)q750 - (T0/TH)q1500  100 300 = 3007.88514 - 6.86925 - 0.287 ln  + 300 100 750  -(300/1500) 1200 = 184.764 kJ/kg wrev = hi - he + q1500 - q750 + q0 = wac + q0 = 563.03 kJ/kg ηII = wac/wrev = 378.27/563.03 = 0.672 10.48 Consider the high-pressure closed feedwater heater in the nuclear power plant described in Problem 6.42. Determine its second-law efficiency. For this case with no work the second law efficiency is from Eq. 10.25: • • ηII = m 16(ψ18 - ψ16)/m17(ψ17 - ψ15)

Properties (taken from computer software): h15 = 585

h16 = 565

h17 = 2593

h18 = 688

s15 = 1.728

s16 = 1.6603

s17 = 6.1918

s18 = 1.954

ψ18 - ψ16 = h18 - h16 - T0(s18 - s16) = 35.433 ψ17 - ψ15 = h17 - h15 - T0(s17 - s15) = 677.12 ηII = (75.6 × 35.433)/(4.662 × 677.12) = 0.85

10-25 10.49 Consider a gasoline engine for a car as an SSSF device where air and fuel enters at the surrounding conditions 25°C, 100 kPa and leaves the engine exhaust manifold at 1000 K, 100 kPa as products assumed to be air. The engine cooling system removes 750 kJ/kg air through the engine to the ambient. For the analysis take the fuel as air where the extra energy of 2200 kJ/kg of air released in the combustion process, is added as heat transfer from a 1800 K reservoir. Find the work out of the engine, the irreversibility per kilogram of air, and the first- and second-law efficiencies. 2

1 AIR

C.V. Total out to reservoirs • • • • • m ah1 + Q H = mah2 + W + Q out • • • • • m as 1 + Q H/TH + S gen = mas s + Q out/T 0 ° = 6.8631 h1 = 298.61 sT1

⇒ W· · · ⇓ Qout ⇑ QH

T0

TH

h2 = 1046.22

° = 8.1349 sT1

•

• wac = W/m a = h1 - h2 + qH - qout = 298.6 - 1046.22 + 2200 - 750 = 702.4 kJ/kg

ηTH = w/qH = 702.4/2200 = 0.319 sgen = s2 - s1 +

qout qH 750 2200 = 8.1349 - 6.8631 + = 2.565 kJ/kg K 298.15 1800 T0 TH

itot = (T0)sgen = 764.8 kJ/kg For reversible case have

sgen = 0 and qR0 from T0, no qout

qR 0,in = T0(s2 - s1) - (T0/TH)qH = 14.78 kJ/kg wrev = h1 - h2 + qH + qR 0,in = wac + itot = 1467.2 kJ/kg ηII = wac/wrev = 0.479 10.50 Air enters a steady-flow turbine at 1600 K and exhausts to the atmosphere at 1000 K. The second law efficiency is 85%. What is the turbine inlet pressure? C.V.: Turbine, exits to atmosphere so assume Pe = 100 kPa o

Inlet: Ti = 1600 K, Table A.7: hi = 1757.3 kJ/kg, si = 8.1349 kJ/kg K o

Exit: Te = 1000 K, he = 1046.2 kJ/kg, se = 8.6905 kJ/kg K 1st Law: q + hi = he + w; q = 0 => w = (hi - he) = 711.1 kJ/kg 2nd Law: ψi - ψe = w/η2ndLaw = 711.1/0.85 = 836.6 kJ/kg ψi - ψe = (hi - he) - To(si - se) = 836.6 kJ/kg hi - he = w = 711.1 kJ/kg, assume To = 25oC ‡ si - se = 0.4209 kJ/kg-K o

o

si - se = se - si - R ln(Pi/Pe) = 0.4209 Pi = 3003 kPa

=> Pe/Pi = 30.03;

10-26 10.51 Consider the two turbines in Problem 9.72. What is the second-law efficiency for the combined system? From the solution to Problem 9.72 we have • • s1 = 6.2828, s2 = 6.7240, s5 = 6.5233, m T1/mtot = 0.6848 • • m T1/mtot = 0.3152

=>

• • wac = W net/mtot = 301.775 kJ/kg

C.V. Total system out to T0 • •  •  rev rev rev • m ⇒ wqc + qrev T1/mtoth1 + mT2/mtoth2 + q0 = h5 + w 0 =w • •  •  rev • m T1/mtots 1 + mT2/mtots 2 + q0 = s5

⇒ qrev 0 = 298.15[6.5233 - 0.6848 × 6.2825 - 0.3152 × 6.724] = 30.242 kJ/kg wrev = wac + qrev 0 = 301.775 + 30.242 = 332 kJ/kg ηII = wac/wrev = 301.775/332 = 0.909 10.52 Air in a piston/cylinder arrangement is at 110 kPa, 25°C, with a volume of 50 L. It goes through a reversible polytropic process to a final state of 700 kPa, 500 K, and exchanges heat with the ambient at 25°C through a reversible device. Find the total work (including the external device) and the heat transfer from the ambient. C.V. Total out to ambient ma(u2 - u1) = 1Q2 - 1W2,tot ,

ma(s2 - s1) = 1Q2/T0

ma = 110 × 0.05/0.287 × 298.15 = 0.0643 kg 1Q2

= T0ma(s2 - s1) = 298.15 × 0.0643[7.3869 - 6.8631 - 0.287 ln (700/110)] = -0.14 kJ

1W2,tot

= 1Q2 - ma(u2 - u1) = -0.14 - 0.0643 × (359.844 - 213.037) = -9.58 kJ

10-27 Advanced Problems 10.53 Refrigerant-22 is flowing in a pipeline at 10°C, 600 kPa, with a velocity of 200 m/s, at a steady flowrate of 0.1 kg/s. It is desired to decelerate the fluid and increase its pressure by installing a diffuser in the line (a diffuser is basically the opposite of a nozzle in this respect). The R-22 exits the diffuser at 30°C, with a velocity of 100 m/s. It may be assumed that the diffuser process is SSSF, polytropic, and internally reversible. Determine the diffuser exit pressure and the rate of irreversibility for the process. C.V. Diffuser out to T0, Int. Rev. flow ⇒ sgen R-22 = 0/ • • 2 2 m 1 = m2 , h1 + (1/2)V1 + q = h2 + (1/2)V2

s1 + ∫ dq/T + sgen = s2 = s1 + q/T0 + sgen Rev. Polytropic Process:

w12 = 0/ = - ∫ v dP + (V12 - V22)/2

n

1

2

2

1

LHS = ∫ v dP = n-1(P2v2 - P1v1) = 2(V1 - V2) = 2(2002 - 1002)/1000 = 15 kJ/kg P1v1n = P2v2n ⇒ n = ln(P2/P1)/ln(v1/v2) Inlet state: v1 = 0.04018, s1 = 0.9295, h1 = 255.1 Exit state: T2,V2, ? ⇒ only P2 unknown.

⇒

v2 = v(P2), n = n(P2)

Trial and error on P2 to give LHS = 15: P2 = 1 MPa ⇒ v2 = 0.0246,

1000 0.04018 n = ln 600  / ln 0.0246  = 1.0412    

LHS = (1.0412/0.0412)(1000 × 0.0246 - 600 × 0.04018) = 12.43 LHS too small so P2 > 1.0 MPa Assume sat. vapor @ 30°C, P2 = 1.1919 MPa v2 = 0.01974, LHS =

1.1919 0.04018 n = ln 0.6  /ln0.01974  = 0.9657    

0.9657 (1191.9 × 0.01974 - 600 × 0.04018) = 16.33 > 15 -0.0343

P2 = 1.1 MPa ⇒ v2 = 0.021867, n = 0.9963, LHS = 14.596 Interpolate to get the final state: P2 = 1.121 MPa, v2 = 0.021353, n = 0.9887, LHS = 15.01 OK s2 = 0.8967 h2 = 260.56 Find q from energy equation (Assume T0 = T1 = 10°C) 1

q = h2 - h1 + 2(V22 - V12) = 260.56 - 255.1 - 15 = -9.54 kJ/kg . I• = m[T0(s2 - s1) - q ]= 0.1[283.15(0.8967 - 0.9295) + 9.54] = 0.0253 kW

10-28 10.54 A piston/cylinder contains ammonia at −20°C, quality 80%, and a volume of 10 L. A force is now applied to the piston so it compresses the ammonia in an adiabatic process to a volume of 5 L, where the piston is locked. Now heat transfer with the ambient takes place so the ammonia reaches the temperature of the ambient at 20°C. Find the work and heat transfer. If it is done in a reversible process, how much work and heat transfer would be involved? P Process 1 → 2 → 3 1: v1 = 0.49926 m3/kg, s1 = 4.5658 kJ/kg K u1 = 1152.25 - 94.9 = 1057.4 kJ/kg m = V1/v1 = 0.010/0.49926 = 0.02 kg V V2 V1 Assume reversible adiabatic compression m(u2 - u1) = - 1W2 ;

m(s2 - s1) = 0, v2 = v1V2/V1 = 0.2496

2: s2 = s1, v1 ⇒ Trial and error on T2. For 0°C, v2 ⇒ s = 4.6925,

For -2°C, v2 ⇒ s = 4.431

⇒ T2 = -1°C, x = 0.832, u2 = 1125 kJ/kg 1W2

= -m(u2 - u1) = - 1.345 kJ, (1Q2 = 0)

3: T3, v3 = v2 ⇒ P3 = 537 kPa, u3 = 1351.6, s3 = 5.382 2Q3

= m(u3 - u2) = 0.02(1351.6 - 1125) = 4.532 kJ

Qrev = mT0(s3 - s1) = 0.02×293.15(5.382 - 4.5658) = 4.785 kJ Wrev = Qrev - m(u3 - u1) = 4.785 - 0.02(1351.6-1057.4) = -1.1 kJ

10-29 10.55 Consider the irreversible process in Problem 8.34. Assume that the process could be done reversibly by adding heat engines/pumps between tanks A and B and the cylinder. The total system is insulated, so there is no heat transfer to or from the ambient. Find the final state, the work given out to the piston and the total work to or from the heat engines/pumps. C.V. Water mA + mB + heat engines. No Qexternal, only 1W2,cyl + WHE m2 = mA1 + mB1 = 6 kg, m2u2 - mA1uA1 - mB1uB1 = -1W2,cyl - WHE m2s2 - mA1sA1 - mB1sB1 = 0/ + 0/ vA1 = 0.06283 uA1 = 3448.5 sA1 = 7.3476 VA = 0.2513 m3 vB1 = 0.09053 uB1 = 2843.7 sB1 = 6.7428 VB = 0.1811 m3 m2s2 = 4×7.3476 + 2×6.7428 = 42.876 ⇒ s2 = 7.146 kJ/kg K If P2 < Plift = 1.4 MPa then V2’ = VA + VB = 0.4324 m3 , v2’ = 0.07207 m3/kg (Plift , s2) ⇒ v2 = 0.20135 ⇒ V2 = 1.208 m3 > V2’

OK

⇒ P2 = Plift = 1.4 MPa u2 = 2874.2 kJ/kg 1W2,cyl =

Plift(V2 - VA - VB) = 1400×(1.208 - 0.4324) = 1085.84 kJ

WHE = mA1uA1 + mB1uB1 - m2u2 - 1W2,cyl = 4 × 3447.8 + 2 × 2843.7 - 6 × 2874.2 - 1085.84 = 1147.6 kJ

10-30 10.56 Water in a piston/cylinder is at 100 kPa, 34°C, shown in Fig. P10.56. The cylinder has stops mounted so Vmin = 0.01 m3 and Vmax = 0.5 m3. The piston is loaded with a mass and outside P0, so a pressure inside of 5 MPa will float it. Heat of 15000 kJ from a 400°C source is added. Find the total change in availability of the water and the total irreversibility. P 1a 2

1

1b v

CV water plus cyl. wall out to reservoir m2 = m1 = m, m(u2- u1) = 1Q2 - 1W2 m(s2 - s1) = 1Q2/Tres + 1S2 gen System eq: States must be on the 3 lines

Here Peq = 5 MPa, so since P1 < Peq ⇒ V1 = Vmin = 0.01 m3 1: v1 = 0.0010056 m3/kg, u1 = 142.49 kJ/kg, s1 = 0.4916 kJ/kg K ⇒ m = V1/v1 = 9.9443 kg Heat added gives q = Q/m = 1508.4 kJ/kg so check if we go from 1 past 1a (Peq,v1) and past 1b (Peq,vmax) 1a: T = 40°C, u1a = 166.95

⇒

1q1a

1b: T = 337.5°C, u1b= 2780.94 1q1b =

= u1a - u1 = 24.5 kJ/kg

⇒ as 1w1b = Peq(v1b - v1) = 246.4

h1b - u1 - Peqv1 = 2884.8

The state is seen to fall between 1a and 1b so P2 = Peq. 1w2

= Peq(v2 - v1), u2 - u1 = 1q2 - 1w2

⇒

u2 + Peqv2 = h2 = 1q2 + u1 + Peqv1 = 1655.9 < hg(P2) => T2 = 264°C x2 = (1655.9 - 1154.21)/1640.12 = 0.3059, v2 = 0.001286 + 0.3059 × 0.03815 = 0.01295, s2 = 2.9201 + 0.3059 × 3.0532 = 3.8542 The non-flow availability change of the water (Eq.10.22) m(φ2 - φ1) = m[u2 - u1 - T0(s2 - s1) + P0(v2 - v1)] = 9.944[1591.15 - 142.49 - 298.15(3.8542 - 0.4916) + 100(0.01295 - 0.0010056)] = 9.944 × 447.3 = 4448 kJ I = T(1S2 gen) = T [m(s2 - s1) - 1Q2/Tres ]= 298.15 [ 9.944(3.8542 - 0.4916) - 15000/673.15] = 3326 kJ (Piston and atm. have an increase in availability)

10-31 10.57 A rock bed consists of 6000 kg granite and is at 70°C. A small house with lumped mass of 12000 kg wood and 1000 kg iron is at 15°C. They are now brought to a uniform final temperature with no external heat transfer. a. For a reversible process, find the final temperature and the work done in the process. b. If they are connected thermally by circulating water between the rock bed and the house, find the final temperature and the irreversibility of the process, assuming that surroundings are at 15°C. a) For a reversible process a heat engine is installed between the rockbed and the house. Take C.V. Total mrock(u2 - u1) + mwood(u2 - u1) + mFe(u2 - u1) = -1W2 mrock(s2 - s1) + mwood(s2 - s1) + mFe(s2 - s1) = 0/ (mC)rockln

T2 T1

+ (mC)woodln

T2 T1

+ (mC)Feln

T2 T1

= 0/

6000 × 0.89 ln (T2/343.15) + 12000 × 1.26 ln (T2/288.15) + 1000 × 0.46 ln (T2/288.15) = /0 =>

T2 = 301.3 K

-1W2 = 6000 × 0.89(301.3 - 343.15) + (12000 × 1.26 + 460)(301.3 - 288.15) ⇒ 1W2 = 18602 kJ b) No work, no Q irreversible process. (mC)rock(T2 - 70) + (mCwood + mCFe)(T2 - 15) = 0/ T2 = 29.0°C = 302.2 K Sgen = ∑mi(s2 - s1)i = 5340 ln 1I2

302.2 302.2 + 15580 ln = 63.13 kJ/K 343.15 288.15

= (T0)1S2,gen = 288.15 × 63.13 = 18191 kJ

10-32 10.58 Consider the heat engine in Problem 10.47. The exit temperature was given as 800 K, but what are the theoretical limits for this temperature? Find the lowest and the highest, assuming that the heat transfers are as given. For an exit temperature that is the average of the highest and lowest possible, find the firstand second-law efficiencies for the heat engine. The lowest exhaust temperature will occur when the maxumum amount of work is delivered which is a reversible process. Assume no other heat transfers then 2nd law:

si + qH/TH + 0/ = se + qm/Tm se - si = qH/TH - qm/Tm = s°Te - s°Ti - R ln(Pe/Pi)

s°Te = s°Ti + R ln(Pe/Pi) + qH/TH - qm/Tm = 6.86926 + 0.287 ln(100/100) + 1200/1500 - 300/750 = 7.26926 ⇒ Te,min = 446 K The maximum exhaust temperature occurs with no work out hi + qH = qm + he ⇒ he = 300.473 + 1200 - 300 = 1200.5 Te,max = 1134 K Te = (1/2)(Te,min + Te,max) = 790 K see solution for 10.47 to follow calculations he = 811.23,

s°Te = 7.8713

wac = 300.47 + 1200 - 300 - 811.23 = 389.24, q0 = 300[7.8713 - 6.86926 - 0.287 ln wrev = wac + q0 = 569.85,

ηTH = wac/qH = 0.32

100 300 1200 + ] = 180.61 kJ/kg 100 750 1500

ηII = wac/wrev = 389.24/569.85 = 0.683

10-33 10.59 Air in a piston/cylinder arrangement, shown in Fig. P10.59, is at 200 kPa, 300 K with a volume of 0.5 m3. If the piston is at the stops, the volume is 1 m3 and a pressure of 400 kPa is required to raise the piston. The air is then heated from the initial state to 1500 K by a 1900 K reservoir. Find the total irreversibility in the process assuming surroundings are at 20°C. m2 = m1 = m = P1V1/RT1

P 2 Pstop

1

= 200 × 0.5/0.287 × 300 = 1.161 kg m(u2 - u1) = 1Q2 - 1W2 1W 2

P1

Vstop

V1

⌠ dQ/T + 1S2 gen m(s2 - s1) = ⌡ Process: P = P0 + α(V-V0)

if V ≤ Vstop

Information: Pstop = P0 + α(Vstop-V0)

Eq. of state ⇒ Tstop = T1PstopVstop/P1V1 = 1200 < T2 So the piston will hit the stops => V 2 = Vstop 1

1W2 = 2(P 1 1Q2 =

1

+ Pstop)(Vstop- V1) = 2(200 + 400)(1 - 0.5) = 150 kJ

M(u2 - u1) + 1W2 = 1.161(1205.25 - 214.36) + 150 = 1301 kJ o

o

s2 - s1 = sT2 - sT1 -R ln(P2/P1) = 8.6121 - 6.8693 - 0.287 ln 2.5 = 1.48 kJ/kg K Take control volume as total out to reservoir at T RES 1S2 gen tot = 1I2

m(s2 - s2) - 1Q2/TRES = 1.034 kJ/K

= T0(1S2 gen) = 293.15 × 1.034 = 303 kJ

10-34 10.60 Consider two rigid containers each of volume 1 m 3 containing air at 100 kPa, 400 K. An internally reversible Carnot heat pump is then thermally connected between them so it heats one up and cools the other down. In order to transfer heat at a reasonable rate, the temperature difference between the working substance inside the heat pump and the air in the containers is set to 20°C. The process stops when the air in the coldest tank reaches 300 K. Find the final temperature of the air that is heated up, the work input to the heat pump, and the overall second-law efficiency.

A

⇐ H.P. ⇐ QA QB ⇑W

The high and the low temperatures in the heat pump are TA+20 and TB-20, respectively. Since TA and TB change during the process,

B

H.P.

the coefficient of performance changes, and so it must be integrated. dUA = mAduA = mACvdTA = dQA dUB = mBduB = mBcvdTB = -dQB Carnot heat pump: ⇒

dQA dQB

=

TA + 20 TB - 20

=

-mACvdTA -dTA = mBCvdTB dTB

(dTA)/(TA + 20) = - (dTB)/(TB - 20)

⌠ dTA TA2 + 20 ⌠ dTB TB2 - 20  ⇒ = ln = - = -ln TA1 + 20 ⌡TB - 20 TB1 - 20 ⌡TA + 20 TA1 = TB1 = 400 K, TB2 = 300 K ⇒ ln

TA2 + 20 420

= -ln

300 - 20 ⇒ TA2 = 550 K 380

mA = mB = P1V1/RT1 = (100 × 1)/(0.287 × 400) = 0.871 kg WH.P. = QA - QB = mCv(TA2 - TA1) + mCv(TB2 - TB1) = 0.871 × 0.717 (550 - 400 + 300 - 400) = 31.2 kJ Second law efficiency is total increase in air availability over the work input. ∆φ = mCv(T2 - T1) + P0(V2 - V1) - T0(s2 - s1) ∆φ A = QA - T0m(s2 - s1)A

∆φB = -QB - T0m(s2 - s1)B

∆φ A + ∆φ B = WH.P. - mT0[ (s2 - s1)A + (s2 - s1)B] To find entropies we need the pressures (P2/P1) = T2/T1 for both A and B s2 - s1 = Cpln(T2/T1) - R ln(T2/T1) = Cvln(T2/T1) ∆φ tot = WH.P. - mT0Cv[ln(T2/T1)A + ln(T2/T1)B]  550 300 = 31.2 - 0.871 × 298.15 × 0.7175 × ln + ln  = 25.47 400  400 ηII = ∆φ tot/WH.P. = 25.47/31.2 = 0.816

10-35

ENGLISH UNIT PROBLEMS 10.61ECalculate the reversible work and irreversibility for the process described in Problem 5.122, assuming that the heat transfer is with the surroundings at 68 F. P

Linear spring gives

1

1 =⌠ ⌡PdV = (P1 + P2)(V2 - V1) 2 1Q 2 = m(u2 - u1) + 1W2 Equation of state: PV = mRT

2

1W2

v

State 1: V1 = mRT1/P1 =

4 × 35.1 × (750 + 460) = 16.85 ft3 70 × 144

State 2: V2 = mRT2/P2 =

4 × 35.1 × (75 + 460) = 11.59 ft3 45 × 144

1W2

1

= 2(70 + 45)(11.59 – 16.85) x144/778 = -55.98 Btu

From Table C.7 Cp(Tavg) = [(6927-0)/(1200-537)]/M = 10.45/44.01 = 0.2347 Btu/lbm R ⇒ CV = Cp – R = 0.2375 – 35.10/778 = 0.1924 1Q 2 rev 1W2 =

= mCV(T2 - T1) + 1W2 = 4x 0.1924(75 - 750) - 55.98 = -575.46 Btu ac

To(S2 - S1) - (U2 - U1) + 1Q2(1 - To/TH) = Tom(s2 - s1)+ 1W2 - 1Q2To/To ac

= Tom[CP ln(T2 / T1) − R ln(P2 / P1)] + 1W2 - 1Q2 = 527.7 x 4 [ 0.2347 ln(535/1210) - 0.0451 ln(45/70)] - 55.98 + 575.46 = -362.24 - 55.98 + 575.46 = 157.2 Btu 1I 2

rev

ac

= 1W2 - 1W2 = 157.2 - (-55.98) = 213.2 Btu

10.62EThe compressor in a refrigerator takes refrigerant R-134a in at 15 lbf/in.2, 0 F and compresses it to 125 lbf/in.2, 100 F. With the room at 70 F find the reversible heat transfer and the minimum compressor work. C.V. Compressor out to ambient. Minimum work in is the reversible work. SSSF, 1 inlet and 2 exit energy Eq.:

wc = h1 - h2 + qrev

Entropy Eq.: s2 = s1 + ∫ dq/T + sgen = s1 + qrev/T0 + 0 => qrev = T0(s2- s1) qrev = 529.67×(0.41262 - 0.42288) = -5.43 Btu/lbm wc min = h1 - h2 + T0(s2- s1) = 167.193 - 181.059 - 5.43 = -19.3 Btu/lbm

10-36 10.63EA supply of steam at 14.7 lbf/in.2, 320 F is needed in a hospital for cleaning purposes at a rate of 30 lbm/s. A supply of steam at 20 lbf/in. 2, 500 F is available from a boiler and tap water at 14.7 lbf/in.2, 60 F is also available. The two sources are then mixed in an SSSF mixing chamber to generate the desired state as output. Determine the rate of irreversibility of the mixing process. 1 3

C.V. Mixing chamber . . . m1 + m2 = m3 . . . m1h1 + m2h2 = m3h3 . . . . m1s1 + m2s2 + Sgen = m3s3

2

C.8.1 h1 = 28.08, s1 = 0.05555 C.8.2 h2 = 1286.8, s2 = 1.8919 C.8.2 h3 = 1202.1 , s3 = 1.828

. . 1202.1 - 28.08 m2/m3 = (h3 - h1)/(h2 - h1) = = 0.9327 1286.8 - 28.08 . . m2 = 27.981 lbm/s , m1 = 2.019 lbm/s . . . . . I = T0Sgen = T0(m3s3 - m1s1 - m2s2) = 536.67 × (30 × 1.828 - 2.019 × 0.05555 - 27.981 × 1.8919) = 961 Btu/s 10.64EA 4-lbm piece of iron is heated from room temperature 77 F to 750 F by a heat source at 1100 F. What is the irreversibility in the process? C.V. Iron out to 1100 F source, which is a control mass. Energy Eq.:

mFe(u2 - u1) = 1Q2 - 1W2

Process: Constant pressure

=>

1W2

= PmFe(v2 - v1)

⇒ 1Q2 = mFe(h2 - h1) = mFeC(T2 - T1) 1Q2

= 4 × 0.107 × (750 - 77) = 288.04 Btu

mFe(s2 - s1) = 1Q2/Tres + 1S2 gen 1S2 gen

= mFe(s2 - s1) - 1Q2/Tres = mFeC ln (T2/T1) - 1Q2/Tres = 4 × 0.107 × ln

1I2

1209.67 288.04 = 0.163 Btu/R 536.67 1559.67

= To (1S2 gen ) = 536.67 × 0.163 = 87.57 Btu

10-37 10.65EFresh water can be produced from saltwater by evaporation and subsequent condensation. An example is shown in Fig. P10.12 where 300-lbm/s saltwater, state 1, comes from the condenser in a large power plant. The water is throttled to the saturated pressure in the flash evaporator and the vapor, state 2, is then condensed by cooling with sea water. As the evaporation takes place below atmospheric pressure, pumps must bring the liquid water flows back up to P0. Assume that the saltwater has the same properties as pure water, the ambient is at 68 F, and that there are no external heat transfers. With the states as shown in the table below find the irreversibility in the throttling valve and in the condenser. State 1 T [F] 86

2 77

3 77

4 --

. . . C.V. Valve: m1 = mex = m2 + m3, Energy Eq.:

h1 = he ;

5 74

Entropy Eq.:

6 --

7 63

8 68

si + sgen = se

h1 = 54.08 si = 0.1043 P2 = Psat(T2 = T3) = 0.4641 he = h1 ⇒ xe = (54.08 - 45.08)/1050.0 = 0.008571 ⇒ se = 0.08769 + 0.008571 × 1.9565 = 0.1045 sgen = se - si = 0.1045 - 0.1043 = 0.0002 . . I = mT0sgen = 300 × 528 × 0.0002 = 31.68 Btu/s C.V. Condenser. h2 = 1095.1 , s2 = 2.044 ,

. . m2 = (1 - xe)mi = 297.44 lbm/s

h5 = 42.09 , s5 = 0.0821 h7 = 31.08 , s7 = 0.0613 , h8 = 36.09 , s8 = 0.0708 . . . . Energy Eq.: m2h2 + m7h7 = m2h5 + m7h8 ⇒ . . 1095.1 - 42.09 lbm m7 = m2 × (h2 - h5)/(h8 - h7) = 297.44 = 62516 36.09 - 31.08 s . . . . . Entropy Eq.: m2s2 + m7s7 + Sgen = m2s5 + m7s8 . . . . I = T0Sgen = T0[m2(s5 - s2) + m7(s8 - s7)] = 528[297.44(0.0821 - 2.044) + 62516(0.0708 - 0.0613)] = 528 × 10.354 = 5467 Btu/s

10-38 10.66EAir flows through a constant pressure heating device as shown in Fig. P10.15. It is heated up in a reversible process with a work input of 85 Btu/lbm air flowing. The device exchanges heat with the ambient at 540 R. The air enters at 540 R, 60 lbf/in.2. Assuming constant specific heat develop an expression for the exit temperature and solve for it. C.V. Total out to T0

Energy Eq.:

Entropy Eq.:

s1 + qrev 0 /T 0 = s2

rev h1 + qrev = h2 0 -w

⇒ qrev 0 = T0(s2 - s1)

h2 - h1 = T0(s2 - s1) - wrev Constant Cp gives: Cp(T2 - T1) = T0Cp ln (T2/T1) + 85 T2 - T0ln(T2/T1) = T1 + 85/Cp T1 = 540

Cp = 0.24

T0 = 540

T2 - 540 ln (T2/540) = 540 + (85/0.24) = 894.17 At 1400 R

LHS = 885.56 (too low)

Interpolate to get

T2 = 1414 R

At 1420 R (LHS = 894.19

LHS = 897.9

OK)

10.67EAir enters the turbocharger compressor of an automotive engine at 14.7 lbf/in2, 90 F, and exits at 25 lbf/in2, as shown in Fig. P10.16. The air is cooled by 90 F in an intercooler before entering the engine. The isentropic efficiency of the compressor is 75%. Determine the temperature of the air entering the engine and the irreversibility of the compression-cooling process. 2

3 Cooler

⇐.

Comp.

-Wc

1

⇓-Q. c

T3 = T2 - 90 F, P2 = 25 lbf/in2 ηs COMP = 0.75, P1 = 14.7 lbf/in2 T1 = 90 F = 550 R T2s = T1

Air

P2

(P )

k-1 k

= 550

1

25 (14.7 )

0.286

= 640.2 R

-ws = CP0(T2s - T1) = 0.24(640.2 - 550) = 21.65 -w = - ws/ηs = 21.65/0.75 = 28.87 = CP0(T2 - T1) = 0.24(T2 - 550) =>

T2 = 670.3 R

⇒ T3 = 670.3 - 90 = 580.3 R

b) Irreversibility from Eq.10.13 with rev. work from Eq.10.12, (q = 0 at TH) s3 - s1 = 0.24 ln

580.3 53.34 25 ln = -0.0235 Btu/lbm R 550 778 14.7

i = T(s3 - s1) - (h3 - h1) - w = T(s3 - s1) - CP(T3 - T1) - CP(T1 - T2) = 550(-0.0235) - 0.24(-90) = +8.7 Btu/lbm

10-39 10.68ECalculate the irreversibility for the process described in Problem 6.101, assuming that the heat transfer is with the surroundings at 61 F. I = To Sgen

so apply 2nd law out to To = 61 F

m2 s2 - m1s1 = misi + 1Q2 / To + 1S2gen ToSgen = To ( m2 s2 - m1s1 - mi si ) - 1Q2 m1 = 2.104 lbm

mi = 7.152 lbm Use from table C.4: Cp = 0.24

m2 =9.256 lbm R = 53.34 / 778 = 0.06856

ToSgen = I = To [ m1 (s2 - s1) + mi (s2 - si)] - 1Q2 = 520.7[2.104(Cp ln

630 60 630 60 - R ln ) +7.152(Cpln - R ln )] 519.7 45 1100 75

- ( - 868.9) = 520.7 (0.05569 - 0.8473) + 868.9 = 456.7 Btu 10.69EA control mass gives out 1000 Btu of energy in the form of a. Electrical work from a battery b. Mechanical work from a spring c. Heat transfer at 700 F Find the change in availability of the control mass for each of the three cases. a) ∆φ = -Wel = -1000 Btu b) ∆φ = -Wspring = -1000 Btu  T0  537  c) ∆φ = -1 -  = Qout = -1  1000 = -537 Btu  1160  TH 10.70EA steady stream of R-22 at ambient temperature, 50 F, and at 110 lbf/in. 2 enters a solar collector. The stream exits at 180 F, 100 lbf/in.2. Calculate the change in availability of the R-22 between these two states. R-22 i

e Solar

50 F 110 lbf/in2 hi = 24.275 si = 0.0519

Collector

180 F 100 lbf/in2 he = 132.29 se = 0.2586

∆ψie = ψe - ψi = (he - hi) - T0(se - si) = (132.29 - 24.275) -510(0.2586 - 0.0519) = 2.6 Btu/lbm

10-40 10.71EA 20-lbm iron disk brake on a car is at 50 F. Suddenly the brake pad hangs up, increasing the brake temperature by friction to 230 F while the car maintains constant speed. Find the change in availability of the disk and the energy depletion of the car’s gas tank due to this process alone. Assume that the engine has a thermal efficiency of 35%. All the friction work is turned into internal energy of the disk brake. m(u2 - u1) = 1Q2 - 1W2 ⇒ 1Q2 = mFeCFe(T2 - T1) 1Q2

= 20 × 0.107 × (230 - 50) = 385.2 Btu

Neglect the work to the surroundings at P 0 690 m(s2 - s1) = mC ln(T2 / T1) = 20 × 0.107 × ln 510 = 0.6469  

∆φ = m(u2 - u1) - T0m(s2 - s1) = 385.2 - 510 × 0.6469 = 55.28 Btu Wengine = ηthQgas = 1Q2 = Friction work Qgas = 1Q2/ηth = 385.2/0.35 = 1100 Btu 10.72ECalculate the availability of the system (aluminum plus gas) at the initial and final states of Problem 8.113, and also the irreversibility. State 1: T1 = 400 F v1 = 2/2.862 = 0.6988 P1 = 300 psi Ideal gas v2 = v1(300 / 220)(537 / 860) = 0.595 ; vo = 8.904 = RTo / Po The metal does not change volume so the terms as Eq.10.22 are added φ1 = mgasφgas + mAlφAl T P = mgasCv (T1 - To) - mgasTo [Cp ln 1 - R ln 1 ] + mgasPo (v1 - vo) To Po + mAl [C (T1 - To) - ToC ln (T1 / To) ]Al 860 35.1 300 φ1 = 2.862[0.156(400-77) - 537(0.201 ln ln ) 537 778 14.7 144 860 + 14.7 (0.6988 - 8.904 )( ) ] + 8 × 0.21 [ 400 -77 - 537 ln ] 778 537 = 143.96 + 117.78 = 261.74 Btu 220 537 35.1 ln ) φ2 = 2.862[ 0.156(77 - 77) - 537 (0.201 ln 14.7 537 778 144 537 + 14.7(0.595 - 8.904 )( )] + 8 × 0.21 [ 77 -77 - 537 ln ] 778 537 = 122.91 + 0 = 122.91 Btu 1I2 = φ1 - φ2 + (1 - To/TH )1Q2 - 1W2AC + Pom( V2 - V1) = 261.74 - 122.91 + 0 - (-14.29) + 14.7 × 2.862 ×

144 778

( 0.595 - 0.6988)

= 152.3 Btu [(Sgen = 0.2837 Btu/R

ToSgen = 152.3

so OK ]

10-41 10.73EConsider the springtime melting of ice in the mountains, which gives cold water running in a river at 34 F while the air temperature is 68 F. What is the availability of the water (SSSF) relative to the temperature of the ambient? ψ = h1 - h0 - T0(s1 - s0)

flow availability from Eq.10.19

Approximate both states as saturated liquid ψ = 1.9973 - 36.088 - 527.67 × (0.00405 - 0.07081) = 1.136 Btu/lbm Why is it positive? As the water is brought to 68 F it can be heated with qL from a heat engine using qH from atmosphere TH = T0 thus giving out work. 10.74EA geothermal source provides 20 lbm/s of hot water at 80 lbf/in. 2, 300 F flowing into a flash evaporator that separates vapor and liquid at 30 lbf/in.2. Find the three fluxes of availability (inlet and two outlets) and the irreversibility rate. C.V. Flash evaporator chamber. SSSF with no work or heat transfer. . . . Cont. Eq.: m1 = m2 + m3 ; Vap. . . . 2 1 Energy Eq.: m1h1 = m2h2 + m3h3 Liq. . . . . 3 Entropy Eq.: m1s1 + Sgen = m2s2 + m3s3 C.8.1:

ho = 45.08,

so = 0.08769, h1 = 269.73, s1= 0.4372 C.8.1: h2 = 1164.3, s2= 1.6997, h3 = 218.9, s3= 0.36815 . . h -h h1 = xh2 + (1 - x) h3 => x = m2/m1 = 1 3 = 0.05376 h2 - h3 . . . . m3 = (1-x)m1 = 18.925 lbm/s m2 = xm1 = 1.075 lbm/s . Sgen = 0.579 × 7.1271 + 9.421 × 1.53 - 10 × 1.8417 = 0.124 kW/K Flow availability Eq.10.22: ψ = (h - Tos) - (ho - Toso) = h - ho - To(s - so) ψ1 = 269.73 - 45.08 - 537 × (0.4372 - 0.08769) = 36.963 ψ2 = 1164.3 - 45.08 - 537 × (1.6997 - 0.08769) = 253.57 ψ3 = 218.9 - -45.08 - 537 × (0.36815 - 0.08769) = 23.21 . . . m1 ψ1 = 739.3 Btu/s m2ψ2 = 272.6 Btu/s m3ψ3 = 439.3 Btu/s . . . . I = m1 ψ1 - m2 ψ2 - m3ψ3 = 27.4 Btu/s

10-42 10.75EA wood bucket (4 lbm) with 20 lbm hot liquid water, both at 180 F, is lowered 1300 ft down into a mineshaft. What is the availability of the bucket and water with respect to the surface ambient at 70 F? v1 ≈ v0 for both wood and water φ1 - φ0 = mwood[u1 - u0 -T0(s1 - s0)] + mH2O[u1 - u0 -T0(s1 - s0)] + mtotg(z1 -z0) = 4[0.3(180 - 70) - 0.3 × 530 ln

640 ] + 20[147.76 - 38.09 530

- 530(0.263 - 0.074)] + 24 × (g/gc) × (-1300 / 778) = 12.05 + 199.3 - 40.1 = 171.25 Btu 10.76EAn air compressor is used to charge an initially empty 7-ft3 tank with air up to 750 lbf/in.2. The air inlet to the compressor is at 14.7 lbf/in.2, 60 F and the compressor isentropic efficiency is 80%. Find the total compressor work and the change in energy of the air. C.V. Tank + compressor (constant inlet conditions) Continuity: Entropy:

m2 - 0 = min

Energy: m2u2 = minhin - 1W2

m2s2 = minsin + 1S2 GEN

To use isentropic efficiency we must calc. ideal device Reversible compressor: 1S2 GEN = 0 ⇒ s2 = sin ⇒ Pr2 = Prin(P2/Pin) = 0.9767 × (750/14.7) = 49.832 ⇒ T2,s = 1541 R u2,s = 274.49 Btu/lbm ⇒ 1w2,s = hin - u2,s = 124.38 - 274.49 = -150.11 Actual compressor: 1w2,AC = 1w2,s/ηc = -187.64 Btu/lbm u2,AC = hin - 1w2,AC = 312 ⇒ T2,AC = 1729 R Final state 2 u, P ⇒ v2 = RT2/P2 = 0.854 ft3/lbm m2 = V2/v2 = 8.2 lbm ⇒ 1W2 = m2(1W2,AC) = -1539 Btu m2(φ2 - φ1) = m2 [u2 - u1 + P0(v2 - v1) - T0(s2 - s1) ] 144 - 520 1.9311 778

[

= 8.2 312 - 88.733 + 14.7(0.854 - 13.103) - 1.63074 -

53.34 750 ln 778 14.7

(

)] = 8.2 × 173.94 = 1426.3 Btu

10-43 10.77EA compressor is used to bring saturated water vapor at 150 lbf/in. 2 up to 2500 lbf/in.2, where the actual exit temperature is 1200 F. Find the irreversibility and the second law efficiency. Inlet state: Table C.8.1

hi = 1194.9 Btu/lbm, si = 1.5704 Btu/lbm R

Actual compressor C.8.2: he = 1587.7 Btu/lbm, se = 1.6101 Btu/lbm R Energy Eq. actual compressor:

-wc,ac = he - hi = 392.8 Btu/lbm

Eq.10.14: i = T0(se - si) = 536.67 × (1.6101 - 1.5704) = 21.3 Btu/lbm Eq.10.13:

wrev = i + w c,ac = 21.3 + 1194.9 - 1587.7 = -371.5 Btu/lbm

ηII = -wrev/(-wc,ac) = 371.5 / 392.8 = 0.946 10.78EThe simple steam power plant in Problem 6.91, shown in Fig P6.39 has a turbine with given inlet and exit states. Find the availability at the turbine exit, state 6. Find the second law efficiency for the turbine, neglecting kinetic energy at state 5. Turbine from 6.91 C.8:

h6 = 1028.7 ho = 45.08

s6= 1.8053

h5 = 1455.6

s5= 1.6408

so= 0.08769

ψ6 = h6 - ho - To ( s6 - so ) = 61.26 Btu/lbm wrev = ψ5 - ψ6 = h5 - h6 - To( s5 - s6 ) = 515.2 Btu/lbm wAC = h5 - h6 = 426.9 ηII = wac / wrev = 426.9 / 515.2 = 0.829 10.79ESteam is supplied in a line at 450 lbf/in.2, 1200 F. A turbine with an isentropic efficiency of 85% is connected to the line by a valve and it exhausts to the atmosphere at 14.7 lbf/in.2. If the steam is throttled down to 300 lbf/in.2 before entering the turbine find the actual turbine specific work. Find the change in availability through the valve and the second law efficiency of the turbine. C.V. Valve: h2 = h1 = 1630.8, s2 > s1, h2, P2 ⇒ s2 = 1.86348 Ideal turbine:

s3 = s2 ⇒ h3s = 1230.88

Actual turbine:

wT,ac = ηTwT,s = 339.93 Btu/lbm

wT,s = h2-h3s = 399.92

ψ2 - ψ1 = h2 - h1 - T0(s2 - s1) = -536.67(1.86348 - 1.8192) = -23.76 Btu/lbm h3ac = h2 - wTac = 1290.87 ⇒ s3ac = 1.9299 wrev = ψ2 - ψ3 = 1630.8 - 1290.87 - 536.67(1.86348 - 1.9299) = 375.58 ηII = wac/wrev = 339.93/375.58 = 0.905

10-44 10.80EA piston/cylinder arrangement has a load on the piston so it maintains constant pressure. It contains 1 lbm of steam at 80 lbf/in.2, 50% quality. Heat from a reservoir at 1300 F brings the steam to 1000 F. Find the second-law efficiency for this process. Note that no formula is given for this particular case, so determine a reasonable expression for it. 1: P1, x1 ⇒ v1 = 2.7458 h1 = 732.905 s1 = 1.0374 2: P2 = P1,T2 ⇒ v2 = 10.831 h2 = 1532.6 s2 = 1.9453 m(u2 - u1) = 1Q2 - 1W2 = 1Q2 - P(V2 - V1) 1Q2 1W2

= m(u2 - u1) + Pm(v2 - v1) = m(h2 - h1) = 799.7 Btu = Pm(v2 - v1) = 119.72 Btu

1W2 to atm =

P0m(v2 - v1) = 22 Btu

Useful work out = 1W2 - 1W2 to atm = 119.72 - 22 = 97.72 Btu  T0   536.67  ∆φ reservoir = 1  1Q2 = 1  799.7 = 556 Btu  1759.67  Tres nII = Wnet/∆φ = 0.176 10.81EAir flows into a heat engine at ambient conditions 14.7 lbf/in.2, 540 R, as shown in Fig. P10.47. Energy is supplied as 540 Btu per lbm air from a 2700 R source and in some part of the process a heat transfer loss of 135 Btu per lbm air happens at 1350 R. The air leaves the engine at 14.7 lbf/in.2, 1440 R. Find the first- and the second-law efficiencies. C.V. Engine out to reservoirs hi + qH = qL + he + w wac = 129.18 + 540 - 135 - 353.483 = 180.7 ηTH = w/qH = 180.7/540 = 0.335 For second law efficiency also a q to/from ambient si + (qH/TH) + (q0/T0) = (qloss/Tm) + se q0 = T0[se - si + (qloss/Tm) - (qH/TH)]  135 540  = 5401.88243 - 1.6398 +  = 77.02 1350 2700  wrev = hi - he + qH - qloss + q0 = wac + q0 = 257.7 ηII = wac/wrev = 180.7/257.7 = 0.70

10-45 10.82EConsider a gasoline engine for a car as an SSSF device where air and fuel enters at the surrounding conditions 77 F, 14.7 lbf/in.2 and leaves the engine exhaust manifold at 1800 R, 14.7 lbf/in.2 as products assumed to be air. The engine cooling system removes 320 Btu/lbm air through the engine to the ambient. For the analysis take the fuel as air where the extra energy of 950 Btu/lbm of air released in the combustion process, is added as heat transfer from a 3240 R reservoir. Find the work out of the engine, the irreversibility per pound-mass of air, and the first- and second-law efficiencies. 2 C.V. Total out to reservoirs 1 • • • • • m ah1 + Q H = mah2 + W + Q out AIR • • • • • · ⇒ W m as 1 + Q H/TH + S gen = mas s + Q out/T 0 · · ° = 1.63831 h1 = 128.381 sT1 Q Q ⇓ out T0

⇑ H TH

h2 = 449.794

° = 1.94209 sT1

•

• wac = W/m a = h1 - h2 + qH - qout = 128.38 - 449.794 + 950 - 320 = 308.6 Btu/lbm

ηTH = w/qH = 308.6/950 = 0.325 itot = (T0)sgen = T0(s2 - s1) + qout- qHT0/TH 536.67 = 536.67(1.94209 - 1.63831) + 320 - 950  3240  = 325.67 Btu/lbm  

For reversible case have

sgen = 0 and qR0 from T0, no qout

qR 0,in = T0(s2 - s1) - (T0/TH)qH = itot - qout = 5.67 Btu/lbm wrev = h1 - h2 + qH + qR 0,in = wac + itot = 634.3 Btu/lbm ηII = wac/wrev = 0.486 10.83EConsider the two turbines in Problem 9.119, shown in Fig. P9.72. What is the second-law efficiency of the combined system? From the solution to 9.119 we have • • s1 = 1.499 s2 = 1.582 s5 = 1.5651 m T1/mtot = 0.525 • • m T1/mtot = 0.475

• • wac = W net/mtot = 146.317 Btu/lbm

C.V. Total system out to T0 rev • •  •  rev rev • m ⇒ wqc + qrev 0 =w T1/mtoth1 + mT2/mtoth2 + q0 = h5 + w rev • •  •  • m T1/mtots 1 + mT2/mtots 2 + q0 = s5

⇒ qrev 0 = 536.67[1.5651 - 0.525 × 1.499 - 0.475 × 1.582] = 14.316 Btu/lbm wrev = wac + qrev 0 = 146.317 + 14.316 = 160.63 ηII = wac/wrev = 146.317/160.63 = 0.911

10-46 10.84E(Adv.) Refrigerant-22 is flowing in a pipeline at 40 F, 80 lbf/in.2, with a velocity of 650 ft/s, at a steady flowrate of 0.2 lbm/s. It is desired to decelerate the fluid and increase its pressure by installing a diffuser in the line (a diffuser is basically the opposite of a nozzle in this respect). The R-22 exits the diffuser at 80 F, with a velocity of 160 ft/s. It may be assumed that the diffuser process is SSSF, polytropic, and internally reversible. Determine the diffuser exit pressure and the rate of irreversibility for the process. Note: The exit velocity should be 320 ft/s at 80 F C.V. Diffuser out to T0, Int. Rev. flow ⇒ sgen R-22 = 0/ 2 2 • • m 1 = m2 , h1 + (1/2)V1 + q = h2 + (1/2)V2

s1 + ∫ dq/T + sgen = s2 = s1 + q/T0 + sgen Rev. Process: w12 = 0/ = -∫ v dP + (V12 - V22)/2 Polytropic process: n

LHS = ∫ v dP = (P v - P1v1) = (V12 - V22)/2 n-1 2 2 = (1/2)(6502 - 3202)/32.174 = 4974.5 ft•lbf P1v1n = P2v2n ⇒ n = ln(P2/P1)/ln(v1/v2) Inlet state: v1 = 0.68782, s1 = 0.2211, h1 = 108.347 Exit state: T2,V2, ? ⇒ only P2 unknown.

⇒ v2 = v(P2), n = n(P2)

Trial and error on P 2 to give LHS = 4974.5 ft•lbf: P2 = 125 psi ⇒ v2 = 0.46219 n = ln(125/80)/ln(0.68782/0.46219) = 1.12257 1.12257

LHS = 0.12257 (125 × 0.462187 - 80 × 0.68782) × 144 = 3624 P2 = 150 psi ⇒ v2 = 0.370548, n = 1.01627, LHS = 5006 Interpolate to get P2 = 149.4 psi. Take 150 psi as OK. P2 = 150 psi ⇒ v2 = 0.370548, h2 = 111.556 s2 = 0.21484 Find q from energy equation (Assume T0 = T1 = 40 F) q = h2 - h1 + (V22 - V12)/2 = 111.556 - 108.347 - (4974.5/778) = -3.185 Btu/lbm • i = •I/m = T0(s2 - s1) - q = 499.67(0.21484 - 0.2211) + 3.185 = 0.057 Btu/lbm

10-47 10.85E(Adv.) Water in a piston/cylinder is at 14.7 lbf/in.2, 90 F, as shown in Fig. P10.56. The cylinder has stops mounted so that Vmin = 0.36 ft3 and Vmax = 18 ft3. The piston is loaded with a mass and outside P0, so a pressure inside of 700 lbf/in.2 will float it. Heat of 14000 Btu from a 750 F source is added. Find the total change in availability of the water and the total irreversibility. CV water plus cyl. wall out to reservoir m2 = m1 = m, m(u2- u1) = 1Q2 - 1W2 m(s2 - s1) = 1Q2/Tres + 1S2 gen System eq: States must be on the 3 lines

P 1a 2

1

1b v

Here Peq = 700 lbf/in2, since P1 < Peq => V1 = Vmin = 0.36 ft3 1: v1 = 0.016099, u1 = 58.07, ⇒ m = V1/v1 = 22.362 lbm Heat added gives q = Q/m = 626.062 Btu/lbm so check if we go from 1 past 1a (Peq,v1) and past 1b (Peq,vmax) 1a: T ≈ 100 F, u1a ≅ 67.8

⇒

1b: T ≈ 600 F, u1b= 1180

=>

1q1a

= u1a - u1 ≅ 9.73

1q1b =

h1b - u1 - Peqv1 = 1221.8

The state is seen to fall between 1a and 1b so P2 = Peq. 1w2

= Peq(v2 - v1), u2 - u1 = 1q2 - 1w2

⇒

u2 + Peqv2 = h2 = 1q2 + u1 + Peqv1 = 686.22 < hg(P2) x2 = 0.3132, v2 = 0.3043, s2 = 0.9044 T2 = 467 F, 1S2 gen

V2 = 6.805

= m(s2 - s1) - 1Q2/Tres = 22.362(0.9044 - 0.11165) - 14000/1210 = 6.157 Btu/R

1I2

= T0(1S2,gen) = 3304 Btu

The non-flow availability change of the water, Eq.10.22 φ2 - φ1 = u2 - u1 + P0(v2 - v1) - T0(s2 - s1) = 646.794 - 58.07 + 14.7(0.3043 - 0.016099) × - 536.67(0.9044 - 0.11165) = 164.06 Btu/lbm m∆φ = 3669 Btu

144 778

10-48 10.86EA rock bed consists of 12000 lbm granite and is at 160 F. A small house with lumped mass of 24000 lbm wood and 2000 lbm iron is at 60 F. They are now brought to a uniform final temperature with no external heat transfer. a. For a reversible process, find the final temperature and the work done in the process. b. If they are connected thermally by circulating water between the rock bed and the house, find the final temperature and the irreversibility of the process assuming that surroundings are at 60 F. a) For a reversible process a heat engine is installed between the rockbed and the house. Take C.V. Total mrock(u2 - u1) + mwood(u2 - u1) + mFe(u2 - u1) = -1W2 mrock(s2 - s1) + mwood(s2 - s1) + mFe(s2 - s1) = 0/ (mC)rockln

T2 T1

+ (mC)woodln

T2 T1

+ (mC)Feln

T2 T1

= 0/

12000 × 0.243 ln (T2/619.67) + 24000 × 0.42 ln (T2/519.67) + 2000 × 0.107 ln (T2/519.67) = /0 ⇒ T2 = 540.3 R -1W2 = 12000 × 0.243(540.3 - 619.67) + (24000 × 0.42 + 314)(540.3 - 519.67) ⇒ 1W2 = 19078 Btu b) No work, no Q irreversible process. (mC)rock(T2 - 160) + (mCwood + mCFe)(T2 - 60) = 0/ T2 = 82 F = 541.7 R Sgen = ∑mi(s2 - s1)i = 2916 ln 1I2

541.7 541.7 + 10294 ln = 35.26 Btu/R 619.67 519.67

= T0Sgen = 519.67 × 35.26 = 18324 Btu

10-49

10.87EAir in a piston/cylinder arrangement, shown in Fig. P10.59, is at 30 lbf/in.2, 540 R with a volume of 20 ft3. If the piston is at the stops the volume is 40 ft3 and a pressure of 60 lbf/in.2 is required. The air is then heated from the initial state to 2700 R by a 3400 R reservoir. Find the total irreversibility in the process assuming surroundings are at 70 F. m2 = m1 = m = P1V1/RT1

P 2 Pstop

= 30 × 20 × 144/(53.34 × 540) = 3.0 lbm m(u2 - u1) = 1Q2 - 1W2

1

1W 2

P1

Vstop

V1

⌠ dQ/T + 1S2 gen m(s2 - s1) = ⌡ Process: P = P0 + α(V-V0) if V ≤ Vstop Information: Pstop = P0 + α(Vstop-V0)

Eq. of state ⇒ Tstop = T1PstopVstop/P1V1 = 2160 < T2 So the piston hits the stops => V 2 = Vstop => P2 = 2.5 P1 1

1W2 = 2(P 1 1Q2 =

1

+ Pstop)(Vstop- V1) = 2(30 + 60)(40 - 20) = 166.6 Btu

m(u2 - u1) + 1W2 = 3(518.165 - 92.16) + 166.6 = 1444.6 Btu o

o

s2 - s1 = sT2 - sT1 -R ln(P2/P1) = 2.0561 - 1.6398 - 53.34 × ln (2.5)/778 = 0.3535 Btu/lbm R Take control volume as total out to reservoir at T RES 1S2 gen tot = 1I2

m(s2 - s2) - 1Q2/TRES = 0.6356 Btu/R

= T0(1S2 gen) = 530 × 0.6356 = 337 Btu

11-1

CHAPTER 11 The correspondence between the new problem set and the previous 4th edition chapter 9 problem set. New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 94 95 96 97 98 99

Old New 2 New 3 4a 4b New 5 New 6 7 8 1 13 New 14 15 mod 16 17 New 18 New New 19 mod 20 mod 22 New 23 mod 24 a,b 24 c,d mod New New New 75 76 77 78

New 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 100 101 102 103 104 105

Old 25 New 26 new 27 New 28 New 30 31 mod 32 New New 33 34 35 36 37 38 New 39 40 41 42 43 44 mod 44 mod 45 46 New 47 48 49 79 80 81 82 83 85

New 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 106 107 108 109

Old 50 51 52 53 New 54 55 56 57 mod 58 mod New New 59 mod 60 61 mod New New New 62 63 New 64 65 66 67 68 New 69 New 70 71 87 88 89 91

11-2

The problems that are labeled advanced are: New

Old

New

Old

New

Old

110

9

113

84

116

92

111

21

114

86

112

New

115

90

The English unit problems are: New

Old

New

Old

New

Old

117

New

130

106

143

New

118

94

131

107

144

118

119

95 a

132

108

145

119

120

95 b

133

109

146

New

121

93

134

110

147

121

122

99

135

111

148

122

123

100

136

112

149

123

124

98

137

113

150

124

125

101 mod

138

114

151

125

126

102 mod

139

115 mod

152

127

127

103

140

116 mod

153

128

128

New

141

117 mod

154

130

129

105

142

New

155

96

11-3 11.1

A steam power plant as shown in Fig. 11.3 operating in a Rankine cycle has saturated vapor at 3.5 MPa leaving the boiler. The turbine exhausts to the condenser operating at 10 kPa. Find the specific work and heat transfer in each of the ideal components and the cycle efficiency. C.V. Pump Rev adiabatic -wP = h2 - h1 ; s2 = s1 since incompressible it is easier to find work as -w p = ∫ v dP = v1 (P2 - P1) = 0.00101 (3500 - 10) = 3.525 kJ/kg => h 2 = h1 - wp = 191.81 + 3.525 = 195.33 C.V. Boiler : qh = h3 - h2 = 2803.43 - 195.33 = 2608.1 kJ/kg C.V. Turbine : wt = h3 - h4 ; s4 = s3 s4 = s3 = 6.1252 = 0.6492 + x4 (7.501) => x4 = 0.73 => h 4 = 191.81 + 0.73 (2392.82) = 1938.57 wt = 2803.43 - 1938.57 = 864.9 kJ/kg C.V. Condenser : qL = h4 - h1 = 1938.57 - 191.81 = 1746.8 kJ/kg ηcycle = wnet / qH = (wt + wp) / qH = (864.9 - 3.5) / 2608.1 = 0.33

11.2

Consider a solar-energy-powered ideal Rankine cycle that uses water as the working fluid. Saturated vapor leaves the solar collector at 175°C, and the condenser pressure is 10 kPa. Determine the thermal efficiency of this cycle. H2O ideal Rankine cycle

T

T3 = 175°C ⇒ P3 = PG 175°C = 892 kPa CV: turbine s4 = s3 = 6.6256 = 0.6493 + x4 × 7.5009 x4 = 0.797 h4 = 191.83 + 0.797 × 2392.8 = 2098.3 wT = h3 - h4 = 2773.6 - 2098.3 = 675.3 kJ/kg -wP = v1(P2 - P1) = 0.00101(892 - 10) = 0.89 wNET = wT + wP = 675.3 - 0.89 = 674.4 kJ/kg h2 = h1 - wP = 191.83 + 0.89 = 192.72 qH = h3 - h2 = 2773.6 - 192.72 = 2580.9 kJ/kg ηTH = wNET/qH = 674.4/2580.9 = 0.261

3 2 1

4 s

11-4 11.3

A utility runs a Rankine cycle with a water boiler at 3.5 MPa and the cycle has the highest and lowest temperatures of 450°C and 45°C respectively. Find the plant efficiency and the efficiency of a Carnot cycle with the same temperatures. Solution: 1: 45oC , x = 0

=> h1 = 188.42 , v1 = 0.00101 , Psat = 9.6 kPa 3: 3.5 MPa , 450oC => h 3 = 3337.2 , s3 = 7.0051

C.V. Pump Rev adiabatic -wP = h2 - h1 ; s2 = s1 since incompressible it is easier to find work as -wp = ∫ v dP = v1 (P2 - P1) = 0.00101 (3500 - 9.6) = 3.525 => h 2 = h1 - wp = 188.42 + 3.525 = 191.95 C.V. Boiler : qh = h3 - h2 = 3337.2 - 191.95 = 3145.3 C.V. Turbine : wt = h3 - h4 ; s4 = s3 s4 = s3 = 7.0051 = 0.6386 + x4 (7.5261)

=>

x4 = 0.8459 => h 4 = 188.42 + 0.8459 (2394.77) = 2214.2

wt = 3337.2 - 2214.2 = 1123 kJ/kg C.V. Condenser : qL = h4 - h1 = 2214.2 - 188.42 = 2025.78 kJ/kg ηcycle = wnet / qH = (wt + wp) / qH = (1123 - 3.5) / 3145.3 = 0.356 273.15 + 45 = 0.56 ηcarnot = 1 - TL / TH = 1 273.15 + 450 11.4

A steam power plant operating in an ideal Rankine cycle has a high pressure of 5 MPa and a low pressure of 15 kPa. The turbine exhaust state should have a quality of at least 95% and the turbine power generated should be 7.5 MW. Find the necessary boiler exit temperature and the total mass flow rate. C.V. Turbine

wT = h3 - h4; s4 = s3

4: 15 kPa, x4 = 0.95 => s4 = 7.6458 , h4 = 2480.4 3: s3 = s4, P3 ⇒ h3 = 4036.7, T3 = 758°C wT = h3 - h4 = 4036.7 - 2480.4 = 1556.3 . . m = WT/wT = 7.5 × 1000/1556.3 = 4.82 kg/s

11-5 11.5

A supply of geothermal hot water is to be used as the energy source in an ideal Rankine cycle, with R-134a as the cycle working fluid. Saturated vapor R-134a leaves the boiler at a temperature of 85°C, and the condenser temperature is 40°C. Calculate the thermal efficiency of this cycle. CV: Pump (use R-134a Table B.5)

T

2

⌠ vdP ≈ v1(P2-P1) -wP = h2 - h1 = ⌡

3

D

1

2 1

= 0.000873(2926.2 - 1017.0) = 1.67 kJ/kg

o

85 C o

4

h2 = h1 - wP = 256.54 + 1.67 = 258.21 kJ/kg

40 C

CV: Boiler

s

qH = h3 - h2 = 428.10 - 258.21 = 169.89 kJ/kg CV: Turbine s4 = s3 = 1.6782 = 1.1909 + x4 × 0.5214 =>

x 4 = 0.9346

h4 = 256.54 + 0.9346 × 163.28 = 409.14 Energy Eq.:

wT = h3 - h4 = 428.1 - 409.14 = 18.96 kJ/kg

wNET = wT + wP = 18.96 - 1.67 = 17.29 kJ/kg ηTH = wNET/qH = 17.29/169.89 = 0.102 11.6

Do Problem 11.5 with R-22 as the working fluid. Same T-s diagram and analysis as in problem 11.5 with R-22 T CV: Pump (use R-22 Table B.4) 2

⌠ vdP ≈ v1(P2-P1) -wP = h2 - h1 = ⌡

D

1

= 0.000884(4037 - 1534) = 2.21 kJ/kg

2 1

3

o

85 C o

4

h2 = h1 - wP = 94.27 + 2.21 = 96.48

40 C s

CV: Boiler:

qH = h3 - h2 = 253.69 - 96.48 = 157.21 kJ/kg

CV: Turbine s4 = s3 = 0.7918 = 0.3417 + x4 × 0.5329, => h4 = 94.27 + 0.8446 × 166.88 = 235.22 wT = h3 - h4 = 253.69 - 235.22 = 18.47 kJ/kg ηTH = wNET/qH = (18.47 - 2.21)/157.21 = 0.1034

x4 = 0.8446

11-6 11.7

Do Problem 11.5 with ammonia as the working fluid. Solution: T

CV: Pump (use Ammonia Table B.2) 2 -w = h - h = ⌠ ⌡ vdP = v (P -P ) P

2

1

1

1

2

1

= 0.001725(4608.6 - 1554.9) = 5.27 kJ/kg h2 = h1 - wP = 371.43 + 5.27 = 376.7 kJ/kg

D 2 1

3

o

85 C o

4

40 C

CV: Boiler

s

qH = h3 - h2 = 1447.8 - 376.7 = 1071.1 kJ/kg CV: Turbine s4 = s3 = 4.3901 = 1.3574 + x4 × 3.5083 h4 = 371.43 + 0.8643 × 1098.8 = 1321.13 Energy Eq.: wT = h3 - h4 = 1447.8 - 1321.13 = 126.67 kJ/kg wNET = wT + wP = 126.67 - 5.27 = 121.4 kJ/kg ηTH = wNET/qH = 121.4/1071.1 = 0.113

=>

x4 = 0.8643

11-7 11.8

Consider the ammonia Rankine-cycle power plant shown in Fig. P11.8, a plant that was designed to operate in a location where the ocean water temperature is 25°C near the surface and 5°C at some greater depth. a. Determine the turbine power output and the pump power input for the cycle. b. Determine the mass flow rate of water through each heat exchanger. c. What is the thermal efficiency of this power plant? a) Turbine s2S = s1 = 5.0863 = 0.8779 + x2S × 4.3269 x2S = 0.9726 h2S = 227.08 + 0.9726 × 1225.09 = 1418.6 wST = h1 - h2S = 1460.29 - 1418.6 = 41.69 wT = ηS wST = 0.80 × 41.69 = 33.35 kJ/kg

T

1

3

2s

. . WT = mwT = 1000 × 33.35 = 33 350 kW Pump: wSP ≈ -v3(P4 - P3) = -0.0016(857 - 615) = -0.387 wP = wSP/ηS = -0.387/0.80 = -0.484 kJ/kg . . WP = mwP = 1000(-0.484) = -484 kW b) h2 = h1 - wT = 1460.29 - 33.35 = 1426.94 . Qto low T H2O = 1000(1426.94 - 227.08) = 1.1999×106 kW 1.1999×106 . mlow T H2O = = 142 840 kg/s 29.38 - 20.98 h4 = h3 - wP = 227.08 + 0.48 = 227.56 . Qfrom high T H2O = 1000(1460.29 - 227.56) = 1.2327×106 kW 1.2327×106 . mlow T H2O = = 147 280 kg/s 104.87 - 96.50 c)

. . 33 350 - 484 = 0.027 ηTH = WNET/QH = 1.2327×106

o

20 C

4 4s 2

o

10 C s

11-8 11.9

Do Problem 11.8 with R-134a as the working fluid in the Rankine cycle. Solution: a) Turbine s2S = s1 = 1.7183 = 1.0485 + x2S × 0.6733

T

x2S = 0.9948 h2S = 213.58 + 0.9948 × 190.65 = 403.24 wST = h1 - h2S = 409.84 - 403.24 = 6.6 kJ/kg wT = ηS wST = 0.80 × 6.6 = 5.28 kJ/kg

1

o

20 C

4 4s 3

2s

2

. . WT = mwT = 5280 kW Pump: wSP ≈ -v3(P4 - P3) = -0.000794(572.8 - 415.8) = -0.125 kJ/kg . . WP = mwP = -156 kW wP = wSP/ηS = -0.156 => b) h2 = h1 - wT = 409.84 - 5.28 = 404.56 . Qto low T H2O = 1000(404.56 - 213.58) = 190980 kW . mlow T H2O =

190980 = 22736 kg/s 29.38 - 20.98

h4 = h3 - wP = 213.58 + 0.156 = 213.74 . Qfrom high T H2O = 1000(409.84 - 213.74) = 196100 kW . mlow T H2O =

196100 = 23429 kg/s 104.87 - 96.50

. . 5280 - 156 = 0.026 c) ηTH = WNET/QH = 196100

o

10 C s

11-9 11.10 Consider the boiler in Problem 11.5 where the geothermal hot water brings the R134a to saturated vapor. Assume a counter flowing heat exchanger arrangement. The geothermal water temperature should be equal to or greater than the R-134a temperature at any location inside the heat exchanger. The point with the smallest temperature difference between the source and the working fluid is called the pinch point. If 2 kg/s of geothermal water is available at 95°C, what is the maximum power output of this cycle for R-134a as the working fluid? (hint: split the heat exchanger C.V. into two so the pinch point with ∆T = 0, T = 85°C appears) 2 kg/s of water is available at 95 oC for the boiler. The restrictive factor is the boiling temperature of 85° C. Therefore, break the process up from 2-3 into two parts as shown in the diagram.

liquid

2

LIQUID HEATER

R-134a

R-134a

BOILER

sat liq . -QBC at 85 oC

liquid

3 sat. vap

D

. -QAB

B

C

H2O out

85o C

liq H2O at 85 oC

A

liquid H2O 95 oC

Write the enrgy equation for the first section A-B and D-3: . . -QAB = mH2O(hA - hB) = 2(397.94 - 355.88) = 84.12 kW . . = mR134A(428.1 - 332.65) ⇒ mR134A = 0.8813 kg/s To be sure that the boiling temp. is the restrictive factor, calculate TC from the energy equation for the remaining section: . -QAC = 0.8813(332.65 - 258.21) = 65.60 kW = 2(355.88 - h C) ⇒ hC = 323.1, TC = 77.2°C > T2 OK CV Pump:

-wP = v1(P2-P1) = 0.000873(2926.2 - 1017.0) = 1.67 kJ/kg

CV: Turbine s4 = s3 = 1.6782 = 1.1909 + x4 × 0.5214 =>

x 4 = 0.9346

h4 = 256.54 + 0.9346 × 163.28 = 409.14 Energy Eq.:

wT = h3 - h4 = 428.1 - 409.14 = 18.96 kJ/kg

Cycle: wNET = wT + wP = 18.96 - 1.67 = 17.29 kJ/kg . . WNET = mR134AwNET = 0.8813 × 17.29 = 15.24 kW

11-10 11.11 Do the previous problem with R-22 as the working fluid. A flow with 2 kg/s of water is available at 95 oC for the boiler. The restrictive factor is the boiling temperature of 85oC. Therefore, break the process up from 2-3 into two parts as shown in the diagram.

2

liquid

LIQUID HEATER

R-22

3 sat. vap

D

sat liq . -QBC at 85 oC

H2O out

85o C

. -QAB

B

C

liquid

R-22

BOILER

A

liquid H2O 95 oC

liq H2O at 85 oC

. . -QAB = mH2O(hA - hB) = 2(397.94 - 355.88) = 84.12 kW . . = mR-22(253.69 - 165.09) ⇒ mR-22 = 0.949 kg/s To verify that TD = T3 is the restrictive factor, find TC. . -QAC = 0.949(165.09 - 96.48) = 65.11 = 2.0(355.88 - hC) hC = 323.32 ⇒ TC = 77.2 oC OK . From Problem 11.6 then: Wnet = 0.949 × (18.47 - 2.21) = 15.43 kW 11.12 The power plant in Problem 11.1 is modified to have a superheater section following the boiler so the steam leaves the super heater at 3.5 MPa, 400°C. Find the specific work and heat transfer in each of the ideal components and the cycle efficiency. C.V. Tubine:

Energy:

wT,s = h3 - h4s;

Entropy:

s4s = s3 = 6.8404

⇒ x4s = (6.8404 - 0.6492)/7.501 = 0.8254 ; h4s = 191.81 + 0.8254 × 2392.82 = 2166.8 kJ/kg wT,s = 3222.24 - 2166.8 = 1055.4 kJ/kg ⌡v dP = v1(P2 - P1) = 0.00101(3500 - 10) = 3.525 kJ/kg C.V. Pump: -wP = ⌠ ⇒ h2 = h1 - wP = 191.81 + 3.525 = 195.34 C.V. Condenser: -qC = h4 - h1 = 2166.8 - 191.81 = 1975 C.V. Boiler:

qH = h3 - h2 = 3222.24 - 195.34 = 3027

ηCYCLE = wNET/qH = (1055.4 - 3.53)/3027 = 0.3475

11-11 11.13 A steam power plant has a steam generator exit at 4 MPa, 500°C and a condenser exit temperature of 45°C. Assume all components are ideal and find the cycle efficiency and the specific work and heat transfer in the components. T

From the Rankine cycle we have the states: 1: 45°C x = 0/ , v1 = 0.00101 , h1 = 188.45

3

2: 4 MPa 3: 4 MPa, 500°C , h3 = 3445.3 , s3 = 7.0901

2 4

1

4: Psat(45°C) = 9.593 s

C.V. Turbine: s4 = s3 ⇒ x4 = (7.0901 - 0.6386)/7.5261 = 0.8572, h4 = 188.42 + 0.8572 × 2394.77 = 2241.3 wT = h3 - h4 = 3445.3 - 2241.3 = 1204 kJ/kg C.V. Pump: -wP = v1(P2 - P1) = 0.00101(4000 - 9.6) = 4.03 kJ/kg -wP = h2 - h1 ⇒ h2 = 188.42 + 4.03 = 192.45 kJ/kg C.V. Boiler:

qH = h3 - h2 = 3445.3 - 192.45 = 3252.8 kJ/kg

C.V. Condenser:

qL,out = h4 - h1 = 2241.3 - 188.42 = 2052.9 kJ/kg

ηTH = wnet/qH = (wT + wP)/qH = (1204 - 4.03)/3252.8 = 0.369 11.14 Consider an ideal Rankine cycle using water with a high-pressure side of the cycle at a supercritical pressure. Such a cycle has a potential advantage of minimizing local temperature differences between the fluids in the steam generator, such as the instance in which the high-temperature energy source is the hot exhaust gas from a gas-turbine engine. Calculate the thermal efficiency of the cycle if the state entering the turbine is 25 MPa, 500°C, and the condenser pressure is 5 kPa. What is the steam quality at the turbine exit? s4 = s3 = 5.9592 = 0.4764 + x4 × 7.9187

3 25 MPa o 500 C

T

x4 = 0.6924 Very low for a turbine exhaust s1 = 0.4764 , h1 = 137.82 h4 = 1816 , h3 = 3162.4 s2 = s1 => h2 = 162.8 wNET = h3 - h4 - (h2 - h1) = 1321.4 qH = h3 - h2 = 2999.6, η = wNET/qH = 0.44

5 kPa

2 1

4 s

11-12 11.15 Steam enters the turbine of a power plant at 5 MPa and 400°C, and exhausts to the condenser at 10 kPa. The turbine produces a power output of 20 000 kW with an isentropic efficiency of 85%. What is the mass flow rate of steam around the cycle and the rate of heat rejection in the condenser? Find the thermal efficiency of the power plant and how does this compare with a Carnot cycle. . Solution: WT = 20 000 kW and ηTs = 85 % State 1 : T1 = 400oC , P1 = 5 MPa , superheated h1 = 3195.6 kJ/kg , s1 = 6.6458 kJ/kgK State 3 : P3 = P2 = 10 kPa , sat liq , x3 = 0 T3 = 45.8oC , h3 = hf = 191.8 kJ/kg , v3 = vf = 0.00101 m3/kg C.V Turbine : 1st Law:

qT + h1 = h2 + wT ;

qT = 0

wT = h1 - h2 , Assume Turbine is isentropic s2s = s1 = 6.6458 kJ/kgK , s2s = sf + x2s sfg , solve for x2s = 0.7994 h2s = hf + x2shfg = 1091.0 kJ/kg wTs = h1 - h2s = 1091 kJ/kg , wT = ηTswTs = 927.3 kJ/kg . . WT m= = 21.568 kg/s , h2 = h1 - wT = 2268.3 kJ/kg wT qc + h2 = h3 + wc ; wc = 0 . . qc = h3 - h2 = -2076.5 kJ/kg , Qc = m qc = -44786 kW

C.V. Condenser: 1st Law :

C.V. Pump: Assume Isentropic wps = - ∫ v dP = -v3(P4 - P3) = -5.04 kJ/kg 1st Law : qp + h3 = h4 + wp ; qp = 0 h4 = h3 - wp = 196.8 kJ/kg C.V Boiler : 1st Law : qB + h4 = h1 + wB ; wB = 0 qB = h1 - h4 = 2998.8 kJ/kg (or qB = wT + wB - q c) wnet = wT + wP = 922.3 kJ/kg ηth = wnet / qB = 0.307 Carnot cycle : TH = T1 = 400oC , TL = T3 = 45.8oC ηth =

T H - TL TH

= 0.526

11-13 11.16 Consider an ideal steam reheat cycle where steam enters the high-pressure turbine at 3.5 MPa, 400°C, and then expands to 0.8 MPa. It is then reheated to 400°C and expands to 10 kPa in the low-pressure turbine. Calculate the cycle thermal efficiency and the moisture content of the steam leaving the low-pressure turbine. T

P3 = 3.5 MPa, T3 = T5 = 400 oC

3.5 MPa o 3 5 400 C

P4 = P5 = 0.8 MPa, P6 = 10 kPa

4

2

From solution 11.12: -wP = 3.525 kJ/kg,

1

h2 = 195.34

10 kPa 6

h3 = 3222.2, s3 = 6.8404

s

qH1 = h3-h2 = 3222.3-195.35 = 3026.9 s4 = s3 => h4 = 2851.6 ; h5 = 3267.1 qH = qH1 + h5 - h4 = 3026.9 +3267.1 - 2851.6 = 3442.4 kJ/kg s6 = s5 = 7.5715 => x6 = (7.5715-0.6492)/7.501 = 0.9229 ; h6 = 2400 wT,tot = h3- h4 + h5 - h6 = 3222.2 - 2851.6 + 3267.1 - 2400 = 1237.8 kJ/kg ηCYCLE = (1237.8 - 3.525)/3442.4 = 0.3586 11.17 The reheat pressure effect the operating variables and thus turbine performance. Repeat Problem 11.16 twice, using 0.6 and 1.0 MPa for the reheat pressure. 3.5 MPa o 3 5 400 C

T

P3 = 3.5 MPa, T3 = T5 = 400 oC P4 = P5 = 0.6 MPa, P6 = 10 kPa -wP = 3.525 kJ/kg, h3 = 3222.2,

4

2

From solution 11.12:

1

h2 = 195.34

10 kPa 6

s3 = 6.8404

s

qH1 = h3-h2 = 3222.2-195.34 = 3026.9 wT,tot = h3- h4 + h5 - h6,

qH = qH1 + h5 - h4,

s4 = s3,

s6 = s5

For P4=P5 = 1 MPa: s5 = 7.465, state 4 is superheated vapor For P4=P5 = 0.6 MPa: s5 = 7.7078, state 4 is superheated vapor x6 0.9087 0.9410

P4=P5 h4 1 2900.7 0.6 2793.2

h5 3263.9 3270.3

h6 2366 2443.5

Notice the very small changes in efficiency.

wT 1219.5 1255.9

qH 3390.1 3504.0

ηCYCLE 0.3587 0.3574

11-14 11.18 The effect of a number of reheat stages on the ideal steam reheat cycle is to be studied. Repeat Problem 11.16 using two reheat stages, one stage at 1.2 MPa and the second at 0.2 MPa, instead of the single reheat stage at 0.8 MPa. P4 = P5= 1.2 MPa, P6 = P7 = 0.2 MPa

3.5 MPa o 3 5 7 400 C

T

From solution to 11.12, -wP = 3.525 kJ/kg, h2 = 195.34

4

3: h3 = 3222.2, s3 = 6.8404 4: P4, s4 = s3 ⇒ sup. vap. h4 = 2940.9

2 1

5: h5 = 3260.7 , s5 = 7.3773

10 kPa 6 8 s

6: P6, s6 = s5 ⇒ sup. vap. h6 = 2811.2 7: h7 = 3276.5, s7 = 8.2217

8: P8, s8 = s7 ⇒ sup. vap. h8 = 2607.9

wT = (h3 - h4) + (h5 - h6) + (h7 - h8) = 281.4 + 449.5 + 668.6 = 1399.5 kJ/kg qH = (h3 - h2) + (h5 - h4) + (h7 - h6) = 3027 + 319.8 + 465.3 = 3812.1 kJ/kg ηTH = (wT + wP)/qH = (1399.5 - 3.53)/3812.1 = 0.366 11.19 A closed feedwater heater in a regenerative steam power cycle heats 20 kg/s of water from 100°C, 20 MPa to 250°C, 20 MPa. The extraction steam from the turbine enters the heater at 4 MPa, 275°C, and leaves as saturated liquid. What is the required mass flow rate of the extraction steam? 3 2

From table B.1 1

4

B.1.4: 100°C, 20 MPa

h1 = 434.06

B.1.4: 250°C, 20 MPa

h2 = 1086.75

B.1.3: 4 MPa, 275°C

h3 = 2886.2

B.1.2: 4 MPa, sat. liq.

h4 = 1087.31

C.V. Feedwater Heater . . . . Energy Eq.: m1h1 + m3h3 = m1h2 + m3h4 . . m3 = m1(h1-h2)/(h4-h3) = 7.257 kg/s

11-15 11.20 An open feedwater heater in a regenerative steam power cycle receives 20 kg/s of water at 100°C, 2 MPa. The extraction steam from the turbine enters the heater at 2 MPa, 275°C, and all the feedwater leaves as saturated liquid. What is the required mass flow rate of the extraction steam? Solution: 2

C.V Feedwater heater . . . m1 + m2 = m3 . . . . . m1h1 + m2h2 = m3h3 = (m1 + m2) h3

3

1

Table B.1.1 and Table B.1.4 at 100 C interpolate to 2 MPa: h1 = 420.4 Table B.1.3: h2 = 2963, Table B.1.2: h3 = 908.8 . . h3 - h1 908.8 - 420.4 m2 = m1 = 20 × = 4.755 kg/s 2963 - 908.8 h2 - h3 Remark: h1 was interpolated between sat liq and compressed liquid at 5 MPa both at 100oC. The saturated liquid value 419.02 could have been used with very small error. 11.21 A power plant with one closed feedwater heater has a condenser temperature of 45°C, a maximum pressure of 5 MPa, and boiler exit temperature of 900°C. Extraction steam at 1 MPa to the feedwater heater condenses and is pumped up to the 5 MPa feedwater line where all the water goes to the boiler at 200°C. Find the fraction of extraction steam flow and the two specific pump work inputs. From turbine s1 = 0.6387, h1 = 188.45 v1 = 0.00101 s4 = 2.1387, h4 = 762.81 T6 => h 6 = 853.9

6

s3 = sIN = 7.9593,

Pump 1

5 2 7

Pump 2 C.V. Turbine:

3

T3 = 573.8,

4 h3 = 3640.58

C.V. P1: wP1 = -v1(P2 - P1) = -5.04 ⇒ h2 = h1 - wP1 = 193.49 C.V. P2: wP2 = -v4(P7 - P4) = -4.508 ⇒ h7 = h4 - wP2 = 767.31 . . . . . . . C.V. Tot: m6 = m1 + m3 , m6/m1 = 1 + x, m3/m1 = x (1 + x)h6 = xh3 - xwP2 - wP1 + h1 = xh3 - xwP2 + h2 . . x = (h6 -h2)/(h3 + wP2 - h6) = 0.2364, m3/m6 = 0.1912

1

From condenser

11-16 11.22 A power plant with one open feedwater heater has a condenser temperature of 45°C, a maximum pressure of 5 MPa, and boiler exit temperature of 900°C. Extraction steam at 1 MPa to the feedwater heater is mixed with the feedwater line so the exit is saturated liquid into the second pump. Find the fraction of extraction steam flow and the two specific pump work inputs. Solution:

From turbine To boiler

State out of boiler 5

6

Pump 1

1

h5 = 4378.82 , s5 = 7.9593 C.V. Turbine: s7 = s6 = s5 P6 , s6

=>

h6 = 3640.6,

4

3

Pump 2

2

T6 = 574oC C.V Pump P1 -wP1 = h2 - h1 = v1(P2 - P1) = 0.00101(1000 - 9.6) = 1.0 kJ/kg => h 2 = h1 - wP1 = 188.42 + 1.0 = 189.42 kJ/kg . . C.V. Feedwater heater: Call m6 / mtot = x Energy Eq.: x=

(1 - x) h2 + x h6 = 1 h3

h3 - h2 h6 - h2

=

762.79 - 189.42 = 0.1661 3640.6 - 189.42

C.V Pump P2 -wP2 = h4 - h3 = v3(P4 - P3) = 0.001127(5000 - 1000) = 4.5 kJ/kg

From condenser

11-17 11.23 A steam power plant operates with a boiler output of 20 kg/s steam at 2 MPa, 600°C. The condenser operates at 50°C dumping energy to a river that has an average temperature of 20°C. There is one open feedwater heater with extraction from the turbine at 600 kPa and its exit is saturated liquid. Find the mass flow rate of the extraction flow. If the river water should not be heated more than 5°C how much water should be pumped from the river to the heat exchanger (condenser)? Solution: The setup is as shown in Fig. 11.10,

Condenser:

1: 50oC sat liq. v1 = 0.001012 , h1 = 209.31 2: 600 kPa

To river

7 Ex turbine From

s 2 = s1

3: 600 kPa

sat liq.

5: P , T

h3 = hf = 670.54

h5 = 3690.1

To pump 1

s5 = 7.7023

1

6: 600 kPa s6 = s5 => h 6 = 3270.0 CV P1 wP1 = -v1(P2 - P1) = -0.001012 (600 - 12.35) = -0.595 h2 = h1 - wP1 = 209.9 C.V FWH x h6 + (1 -x) h2 = h3 x=

h3 - h2 h6 - h2

=

670.54 - 209.9 = 0.1505 3270.0 - 209.9

CV Turbine s 7 = s6 = s5

=>

x 7 = 0.9493 ,

h7 = 2471.17 kJ/kg

qL = h7 - h1 = 2471.17 - 209.31 = 2261.86 kJ/kg . . QL = (1 - x) mqL = 0.85 × 20 × 2261.86 = 38429 kW . . = mH2O ∆hH2O = m (20.93) = 38429 kW . m = 1836 kg/s CV Condenser + Heat Exchanger

. QL . . 0 = mH2O (s7 - s1) (1 - x) + Sg TL

. 38429 Sg = - 20 × 0.85 × (7.7023 - 0.7037) = 12.184 kW/K 293.15

river

11-18 11.24 Consider an ideal steam regenerative cycle in which steam enters the turbine at 3.5 MPa, 400°C, and exhausts to the condenser at 10 kPa. Steam is extracted from the turbine at 0.8 MPa for an open feedwater heater. The feedwater leaves the heater as saturated liquid. The appropriate pumps are used for the water leaving the condenser and the feedwater heater. Calculate the thermal efficiency of the cycle and the net work per kilogram of steam. Solution: C.V. Turbine

2nd Law

s7 = s6 = s5 = 6.8404 kJ/kg K

P6 , s6 => h6 = 2851 (sup vap) C.V Pump P1 -wP1 = h2 - h1 = v1(P2 - P1) = 0.00101(800 - 10) = 0.798 kJ/kg => h 2 = h1 - wP1 = 191.81 + 0.798 = 192.61 . . Call m6 / mtot = x (1 - x) h2 + x h6 = 1 h3 x=

h3 - h2 h6 - h2

=

721.1 - 192.61 = 0.1988 2851 - 192.61

C.V Pump P2 -wP2 = h4 - h3 = v3(P4 - P3) = 0.001115(3500 - 800) = 3.01 h4 = h3 - wP2 = 721.1 + 3.01 = 724.1 CV Boiler qH = h5 - w4 = 3222.2 + 724.1 = 2498.1 CV Turbine

. . WT / m5 = h5 - h6 + (1 - x) (h6 - h7) s7 = s6 = s5 = 6.8404

=>

x7 =

6.8404 - 0.6492 7.501

x7 = 0.8254 => h7 = 191.81 + x7 2392.82 = 2166.8 wT = 3222.2 - 2851 + (1 - 0.1988) ( 2851 - 2166.8) = 919.38 wnet = wT + (1 - x) wP1 + wP2 = 919.38 + (1 - 0.1968)(-0.798) - 3.01 = 915.7 kJ/kg ηcycle = wnet / qH = 915.7 / 2498.1 = 0.366

11-19 11.25 Repeat Problem 11.24, but assume a closed instead of an open feedwater heater. A single pump is used to pump the water leaving the condenser up to the boiler pressure of 3.5 MPa. Condensate from the feedwater heater is drained through a trap to the condenser. Solution: C.V. Turbine, 2nd law:

BOILER

s4 = s5 = s6 = 6.8404 kJ/kg K

4 3

TURBINE.

h4 = 3222.24 , h5 = 2851 FW HTR

=> x 6 = (6.8404 - 0.6492)/7.501

5 6

= 0.8254

Trap 7

h6 = 191.81 + x6 2392.82 = 2166.8

2

P

1

COND.

Assume feedwater heater exit at the T of the condensing steam C.V Pump -wP = h2 - h1 = v1(P2 - P1) = 0.00101(3500 - 10) = 3.525 h2 = h1 - wP = 191.81 + 3.525 = 195.3 T3 = Tsat (P5) = 170.43 , h3 = hf = h7 = 721.1 C.V FWH . . m5 / m3 = x , x=

h3 - h2 h5 - hf 800

Energy Eq.: =

h2 + x h5 = h3 + h7 x

721.1 - 195.3 = 0.2469 2851 - 721.1

wT = h4 - h5 + (1 - x)(h5 - h6) = 3222.2 - 2851 + 0.7531 ( 2851 - 2166.8) = 886.51 wnet = wT + wP = 886.51 - 3.53 = 883 qH = h4 - h3 = 3222.24 - 721.1 = 2501.1 ηcycle = wnet / qH = 883 / 2501.1 = 0.353

11-20 11.26 A steam power plant has high and low pressures of 25 MPa and 10 kPa, and one open feedwater heater operating at 1 MPa with the exit as saturated liquid. The maximum temperature is 800°C and the turbine has a total power output of 5 MW. Find the fraction of the flow for extraction to the feedwater and the total condenser heat transfer rate. The physical components and the T-s diagram is as shown in Fig. 11.10 in the main text for one open feedwater heater. The same state numbering is used. From the Steam Tables: h5 = 4047.1, s5 = 6.9345, h1 = 191.83, v1 = 0.00101 h3 = 762.8 v3 = 0.001127 Pump P1: -wP1 = -v1(P2 - P1) = 0.00101 × 990 = 1 kJ/kg h2 = h1 - wP1 = 192.83 kJ/kg Turbine 5-6: s6 = s5 ⇒ h6 = 2948 kJ/kg wT56 = h5 - h6 = 1099.1 kJ/kg . . . . . Feedwater Heater (mTOT = m5): xm5h6 + (1 - x)m5h2 = m5h3 ⇒ x=

h3 - h2 h6 - h2

=

762.8 - 192.83 = 0.2069 2948 - 192.83

To get state 7 into condenser consider turbine. s7 = s6 = s5 ⇒ x7 = (6.9345 - 0.6493)/7.5009 = 0.83793 h7 = 191.81 + 0.83793 × 2392.82 = 2196.8 kJ/kg Find specific turbine work to get total flow rate . . . . . WT = mTOTh5 - xmTOTh6 - (1 - x)mTOTh7 = mTOT × 1694.9 . mTOT = 5000/1694.9 = 2.95 kg/s . . QL = mTOT (1-x) (h7-h1) = 2.95 × 0.7931(2196.8-191.83) = 4691 kW

11-21 11.27 Do Problem 11.26 with a closed feedwater heater instead of an open and a drip pump to add the extraction flow to the feed water line at 25 MPa. Assume the temperature is 175°C after the drip pump flow is added to the line. One main pump brings the water to 25 MPa from the condenser. Solution: v1 = 0.00101 , h1 = 191.83

From condenser

From turbine 6

T4 = 175oC ; h4 = 741.16

3

4

h6 = 4047.08 , s6 = 6.9345 h6a = hf 1MPa = 762.79 ,

2

6b

v6a = 0.001127

Pump 2

Pump 1

6a

C.V Pump 1 -wP1 = h2 - h1 = v1(P2 - P1) = 0.00101(25000 - 10) = 25.24 kJ/kg => h 2 = h1 - wP1 = 191.83 + 25.24 = 217.07 C.V Pump 2 -wP2 = h6b - h6a = v6a(P6b - P6a) = 0.001127(25000 - 1000) = 27.05 kJ/kg Turbine section 1:

s 6 = s5

P6 = 1 MPa

=> h6 = 2948 kJ/kg

C.V FWH + P2 . . m6 / m4 = x x h6 + (1 - x) h2 + x (-wP2)= h4 x= Turbine:

h4 - h 2 h6 - h2 - wP2

=

741.16 - 217.07 = 0.19 2948 - 217.07 + 27.05

s7 = s6 = s5 & P7 = 10 kPa

=> x 7 =

1

6.9345 - 0.6493 = 0.83793 7.5009

. . WT = m5 [ h5 - h6 + (1 - x) (h6 - h7) ] . . = m5 [ 4047.08 - 2948 + (2948 - 2196.8)] = m5 1707.6 . . . WT = 5000 kW = m5 1707.6 => m5 = 2.928 kg/s . . QL = m5(1 - x) (h7 - h1) = 2.928 × 0.81 (2196.8 - 191.83) = 4755 kW

11-22 11.28 Consider an ideal combined reheat and regenerative cycle in which steam enters the high-pressure turbine at 3.5 MPa, 400°C, and is extracted to an open feedwater heater at 0.8 MPa with exit as saturated liquid. The remainder of the steam is reheated to 400°C at this pressure, 0.8 MPa, and is fed to the lowpressure turbine. The condenser pressure is 10 kPa. Calculate the thermal efficiency of the cycle and the net work per kilogram of steam. 5: h5 = 3222.24,

s5 = 6.8404

7: h7 = 3267.07,

s7 = 7.5715

7 5

T1

3: h3 = hf = 721.1 C.V. T1 s 5 = s6

6

1-x

=> h6 = 2851

8

x

wT1 = h5 - h6 = 3222.2 - 2851 = 371.2

HTR

4

C.V. Pump 1 -wP1 = h2 - h1 = v1(P2 - P1)

T2

1-x

COND.

2 3

P

1

P

= 0.00101(800 - 10) = 0.798 => h 2 = h1 - wP1 = 191.81 + 0.798 = 192.61

T

400 oC

5 7

C.V. FWH x h6 + (1 - x) h2 = h3 h3 - h2

721.1 - 192.61 x= = = 0.1988 h6 - h2 2851 - 192.61

4 6 2 3 1

10 kPa

8 s

C.V. Pump 2 -wP2 = h4 - h3 = v3(P4 - P3) = 0.001115(3500 - 800) = 3.01 => h 4 = h3 - wP2 = 721.1 + 3.01 = 724.11 qH = h5 - h4 + (1 - x)(h7 - h6 ) = 2498.13 + 333.35 = 2831.5 C.V. Turbine 2 s 7 = s8

=> x8 = (7.5715 - 0.6492)/7.501 = 0.92285

h8 = hf + x8 hfg = 191.81 + 0.92285 × 2392.82 = 2400.0 wT2 = h7 - h8 = 3267.07 - 2400.02 = 867.05 wnet = wT1 + (1 - x) wT2 + (1 - x) wP1 + wP2 = 371.2 + 694.7 - 0.64 - 3.01 = 1062.3 kJ/kg ηcycle = wnet / qH = 1062.25 / 2831.5 = 0.375

11-23 11.29 An ideal steam power plant is designed to operate on the combined reheat and regenerative cycle and to produce a net power output of 10 MW. Steam enters the high-pressure turbine at 8 MPa, 550°C, and is expanded to 0.6 MPa, at which pressure some of the steam is fed to an open feedwater heater, and the remainder is reheated to 550°C. The reheated steam is then expanded in the low-pressure turbine to 10 kPa. Determine the steam flow rate to the high-pressure turbine and the power required to drive each of the pumps. 7 a) T 5 5 550 oC 7 HI P LOW P T1 T2 4 6 6 10 kPa 8 2 3 6a 1 8 COND. HTR s 4 2

3

P

P

1

b) -wP12 = 0.00101(600 - 10) = 0.6 kJ/kg h2 = h1 - wP12 = 191.8 + 0.6 = 192.4 -wP34 = 0.00101(8000 - 600) = 8.1 kJ/kg h4 = h3 - wP34 = 670.6 + 8.1 = 678.7 ; s6 = s5 = 6.8778 ⇒ T6 = 182.32 oC

h5 = 3521.0, h6 = 2810.0,

h7 = 3591.9, s8 = s7 = 8.1348 = 0.6493 + x8 ×7.5009 ⇒ x8 = 0.9979 h8 = 191.83 + 0.9979 × 2392.8 = 2579.7 CV: heater Cont: m6a + m2 = m3 = 1 kg, 1st law: m6a =

m6ah6 + m2h2 = m3h3

670.6 - 192.4 = 0.1827, m2 = m7 = 1 - m6a = 0.8173 2810.0 - 192.4

CV: turbine wT = (h5 - h6) + (1 - m6a)(h7 - h8) = 3521 - 2810 + 0.8173(3591.9 - 2579.7) = 1538.2 CV: pumps wP = m2wP12 + m4wP34 = 0.8214×(-0.6) + 1×(-8.1) = -8.6 kJ/kg wN = 1538.2 - 8.6 = 1529.6 kJ/kg (m5) . . m5 = WN/wN = 10000/1529.6 = 6.53 kg/s

11-24 11.30 The low pressure turbine in a reheat and regenerative cycle receives 10 kg/s steam at 600 kPa, 550°C. The turbine exhausts to a condenser operating at 10 kPa. The condenser cooling water temperature is restricted to a maximum of 10°C increase so what is the needed flow rate of the cooling water? The steam velocity in the turbine-condenser connecting pipe is restricted to a maximum of 100 m/s, what is the diameter of the connecting pipe? Solution: 7 a) T 5 550 oC 5 7 HI P LOW P T1 T2 4 6 6 10 kPa 8 2 3 6a ∆T COND. 1 8 HTR s 4 2 1 V8 = 100 m/s, ∆TCOOL H2O = 10 oC 3 P P h7 = 3591.83, s7 = 8.13465 CV T2

=> s8 = s7 = 8.13465 = 0.6492 + x8 × 7.501 =>

x8 = 0.99793

v8 = vf + x8 vfg = 0.00101 + x8 × 14.67254 = 14.643 h8 = hf + x8 hfg = 191.81 + x8 × 2392.82 = 2579.7 CV CONDENSER: . . QL = m(h8- h1) = 10(2579.7 - 191.81) = 23879 kW ∆ hH2O = h30 - h20 = 125.77 - 83.94 = 41.83 (=CP liq(∆T) = 4.184 × 10) . . mH2O = QL/∆ hH2O = 23879 / 41.83 = 571 kg/s Turbine - condenser connecting pipe: . . m = ρAV = AV/v => A = mv/V = 10 × 14.643/100 = 1.464 m2 D = (4A/π)1/2 = 1.37 m

11-25 11.31 A steam power plant has a high pressure of 5 MPa and maintains 50°C in the condenser. The boiler exit temperature is 600°C. All the components are ideal except the turbine which has an actual exit state of saturated vapor at 50°C. Find the cycle efficiency with the actual turbine and the turbine isentropic efficiency. Simple Rankine cycle. Boiler exit:

h3 = 3666.5 ,

s3 = 7.2588

4s: 50°C, s = s3 => x = (7.2588 - 0.7037)/7.3725 = 0.88913,

Ideal Turbine:

h4s = 209.31 + 0.88913 × 2382.75 = 2327.88 =>

wTs = h3 - h4s = 1338.62

Condenser exit:

h1 = 209.31 ,

Actual turbine:

wTac = h3 - h4ac = 1074.4

Actual turbine exit: h4ac = hg = 2592.1

ηT = wTac / wTs = 0.803: Isentropic Efficiency Pump:

-wP = v1( P2 - P1) = 0.001012(5000-12.35) = 5.05 kJ/kg h2 = h1 - wP = 209.31 + 5.05 = 214.36 kJ/kg qH = h3 - h2 = 3666.5 - 214.36 = 3452.14 kJ/kg ηcycle = (wTac + wP) / qH = 0.31: Cycle Efficiency

11.32 A steam power cycle has a high pressure of 3.5 MPa and a condenser exit temperature of 45°C. The turbine efficiency is 85%, and other cycle components are ideal. If the boiler superheats to 800°C, find the cycle thermal efficiency. Basic Rankine cycle as shown in Figure 11.3 in the main text. C.V. Turb.: wT = h3 - h4, s4 = s3 + sT,GEN Ideal: s4 = s3 = 7.9134 =>

h4S = 2503.2,

wT,S = 1640.44

Actual: wT,AC = η × wT,S = 1394.38 C.V. Pump:

-wP = ∫ v dP ≈ v1(P2 - P1) = 3.52 kJ/kg

C.V. Boiler:

qH = h3 - h2 = h3 - h1 + wP = 3951.7

η = (wT,AC + wP)/qH = (1394.4 - 3.5)/3951.7 = 0.352

11-26 11.33 A steam power plant operates with with a high pressure of 5 MPa and has a boiler exit temperature of of 600°C receiving heat from a 700°C source. The ambient at 20°C provides cooling for the condenser so it can maintain 45°C inside. All the components are ideal except for the turbine which has an exit state with a quality of 97%. Find the work and heat transfer in all components per kg water and the turbine isentropic efficiency. Find the rate of entropy generation per kg water in the boiler/heat source setup. Take CV around each component (all SSSF) in standard Rankine Cycle. 1: v = 0.00101; h = 188.42, s = 0.6386 (saturated liquid at 45°C). 3: h = 3666.5, s = 7.2588 superheated vapor 4ac: h = 188.42 + 0.97 × 2394.8 = 2511.4 kJ/kg CV Turbine: no heat transfer q = 0 wac = h3 - h4ac = 3666.5 - 2511.4 = 1155.1 kJ/kg Ideal turbine: s4 = s3 = 7.2588 =>

x4s = 0.88,

h4s = 2295

ws = h3 - h4s = 3666.5 - 2295 = 1371.5, Eff = wac / ws = 1155.1 / 1371.5 = 0.842 CV Condenser: no shaft work w = 0 qOut = h4ac - h1 = 2511.4 - 188.42 = 2323 kJ/kg CV Pump: no heat transfer, q = 0 incompressible flow so v = constant -w = v(P2- P1) = 0.00101(5000-9.59) = 5.04 kJ/kg CV Boiler: no shaft work, w = 0 qH = h3 - h2 = h3 - h1 + wP = 3666.5 - 188.42 -5.04 = 3473 kJ/kg s2 + (qH/ TH) + sGen = s3 and s2 = s1 (from pump analysis) sGen = 7.2588 - 0.6386 - 3473/(700+273) = 3.05 kJ/kg K

11-27 11.34 Repeat Problem 11.26 assuming the turbine has an isentropic efficiency of 85%. The physical components and the T-s diagram is as shown in Fig. 11.10 in the main text for one open feedwater heater. The same state numbering is used. From the Steam Tables: h5 = 4047.1, s5 = 6.9345,

h1 = 191.83, v1 = 0.00101

h3 = 762.8 , v3 = 0.001127 Pump P1: -wP1 = -v1(P2 - P1) = 0.00101 × 990 = 1 kJ/kg h2 = h1 - wP1 = 192.83 kJ/kg Turbine 5-6: s6 = s5 ⇒ h6 = 2948 kJ/kg wT56,s = h5 - h6 = 1099.1 kJ/kg ⇒ wT56,AC = 1099.1 × 0.85 = 934.24 wT56,AC = h5 - h6AC ⇒ h6AC = h5 - wT56,AC = 4047.1 - 934.24 = 3112.86 . . . . . Feedwater Heater (mTOT = m5): xm5h6AC + (1 - x)m5h2 = m5h3 ⇒

x=

h3 - h2 h6 - h2

=

762.8 - 192.83 = 0.1952 3112.86 - 192.83

To get the turbine work apply the efficiency to the whole turbine. (i.e. the first section should be slightly different). s7s = s6s = s5 ⇒ x7s = 0.83793 ,

h7s = 2196.8

wT57,s = h5 - h7s = 4047.1 - 2196.8 = 1850.3 wT57,AC = wT57,sηT = 1572.74 = h5 - h7AC => h 7AC = 2474.4 Find specific turbine work to get total flow rate . WT . 5000 mTOT = = = 3.453 kg/s xwT56 + (1-x)wT57 0.1952×934.24 + 0.8048×1572.74 . . QL = mTOT(1 - x)(h7 - h1) = 3.453 × 0.8048(2474.4 - 191.83) = 6343 kW

11-28 11.35 Steam leaves a power plant steam generator at 3.5 MPa, 400°C, and enters the turbine at 3.4 MPa, 375°C. The isentropic turbine efficiency is 88%, and the turbine exhaust pressure is 10 kPa. Condensate leaves the condenser and enters the pump at 35°C, 10 kPa. The isentropic pump efficiency is 80%, and the discharge pressure is 3.7 MPa. The feedwater enters the steam generator at 3.6 MPa, 30°C. Calculate the the thermal efficiency of the cycle and the entropy generation for the process in the line between the steam generator exit and the turbine inlet, assuming an ambient temperature of 25°C. 2

T

1

TURBINE.

ST. GEN. 6

η = 0.88 5

4

5 5S 6 4

3

COND.

3.5 MPa 3.4 MPa 1 400 oC 375 oC 2 10 kPa 3S 3 s

P 1: h1 = 3222.3,

s1 = 6.8405 ,

2:

h2 = 3165.7,

s2 = 6.7675

3s: s3S = s2 ⇒ x3S = 0.8157, h3S = 2143.6 wT,S = h2 - h3S = 3165.7 - 2143.6 = 1022.1 wT,AC = ηwT,S = 899.4 kJ/kg,

3ac:

h3 = h2 - wT,AC = 2266.3

-wP,S = vf(P5 - P4) = 0.001006(3700 - 10) = 3.7 kJ/kg -wP,AC = -wP,S/ηP = 4.6 kJ/kg qH = h1 - h6 = 3222.3 - 129.0 = 3093.3 kJ/kg η = wNET/qH = (899.4 - 4.6)/3093.3 = 0.289 C.V. Line from 1 to 2: Energy Eq.:

w = 0/ ,

q = h2 - h1 = 3165.7 - 3222.3 = - 56.6 kJ/kg

Entropy Eq.: s1 + sgen + q/T0 = s2

=>

sgen = s2 - s1 -q/T0 = 6.7675 - 6.8405 - (-56.6/298.15) = 0.117 kJ/kg K

11-29 11.36 For the steam power plant described in Problem 11.1, assume the isentropic efficiencies of the turbine and pump are 85% and 80%, respectively. Find the component specific work and heat transfers and the cycle efficiency. CV Pump, Rev & Adiabatic: -wPs = h2s - h1 = v1(P2 - P1) = 0.00101(3500 - 10) = 3.525 ;

s2s = s1

-wPac = -wPs / ηP = 3.525/0.8 = 4.406 = h2a - h1 h2a = -wPac + h1 = 4.406 + 191.81 = 196.2 kJ/kg CV Boiler:

qH = h3 - h2a = 2803.43 - 196.2 = 2607.2 kJ/kg

CV Turbine:

wTs = 2803.43 - 1938.57 = 864.9 kJ/kg wTac = wTs × ηT = 735.2 = h3 - h4a h4a = h3 - wTac = 2803.43 - 735.2 = 2068.2

CV Condensor:

qL = h4a - h1 = 2068.2 - 191.81 = 1876.4

ηcycle = (wTac + wPac) / qH = (735.2 - 4.41) / 2607.2 = 0.28 This compares to 0.33 for the ideal case. 11.37 A small steam power plant has a boiler exit of 3 MPa, 400°C while it maintains 50 kPa in the condenser. All the components are ideal except the turbine which has an isentropic efficiency of 80% and it should deliver a shaft power of 9.0 MW to an electric generator. Find the specific turbine work , the needed flow rate of steam and the cycle efficiency. CV Turbine (Ideal): s4s = s3 = 6.9211, x4s = (6.9211 - 1.091)/6.5029 = 0.8965 h4s = 2407.35, h3 = 3230.8

=> wTs = h3 - h4s = 823.45

CV Turbine (Actual): wTac = ηT × wTs = 658.76 = h3 - h4ac, => h 4ac = 2572 . . m = W / wTac = 9000/658.76 = 13.66 kg/s C.V. Pump: -wP = h2 - h1 = v1(P2 - P1) = 0.00103 (3000 - 50) = 3.04 kJ/kg => h 2 = h1 -wP = 340.47 + 3.04 = 343.51 kJ/kg C.V. Boiler:

qH = h3 - h2 = 3230.8 - 343.51 = 2887.3 kJ/kg

ηcycle = (wTac + wP) / qH = (658.76 - 3.04) / 2887.3 = 0.227

11-30 11.38 In a particular reheat-cycle power plant, steam enters the high-pressure turbine at 5 MPa, 450°C and expands to 0.5 MPa, after which it is reheated to 450°C. The steam is then expanded through the low-pressure turbine to 7.5 kPa. Liquid water leaves the condenser at 30°C, is pumped to 5 MPa, and then returned to the steam generator. Each turbine is adiabatic with an isentropic efficiency of 87% and the pump efficiency is 82%. If the total power output of the turbines is 10 MW, determine the mass flow rate of steam, the pump power input and the thermal efficiency of the power plant. 5

5 MPa

T

2 2S 1

3

0.5 MPa o 3 450 C 5 7.5 kPa 4S 4

LP TURBINE.

HP

4

ηST1= ηST2= 0.87

6S 6 2

s

P

COND.

η SP= 0.82

a) s4S = s3 = 6.8185 = 1.8606 + x4S × 4.9606

1

=>

x 4S = 0.999

h4S = 640.21 + 0.999 × 2108.5 = 2746.6 wT1,S = h3 - h4S = 3316.1 - 2746.6 = 569.5 kJ/kg wT1 = ηT1,S × wT1,S = 0.87 × 569.5 = 495.5 kJ/kg h4ac = 3316.1 - 495.5 = 2820.6 s6S = s5 = 7.9406 = 0.5764 + x6S × 7.675

⇒

x6S = 0.9595

h6S = 168.79 + 0.9595 × 2406 = 2477.3 kJ/kg wT2,S = h5 - h6S = 3377.9 - 2477.3 = 900.6 wT2 = 0.87 × 900.6 = 783.5 wT1+T2 = 783.5 + 495.5 = 1279 kJ/kg . . m = WT/wT1+T2 = 10000/1279 = 7.82 kg/s b)

-wP,S = (0.001004)(5000 - 7.5) = 5.01 kJ/kg -wP = -wSP/ηSP = 5.01/0.82 = 6.11 kJ/kg . . WP = wPm = -7.82 × 6.11 = -47.8 kW

c)

qH = (h3 - h2) + (h5 - h4) = 3316.1 - 130.2 + (3377.9 - 2820.6) = 3743.2 kJ/kg wN = 1279.0 - 6.11 = 1272.9 kJ/kg ηTH = wN/qH = 1272.9/3743.2 = 0.34

6

11-31

11.39 A supercritical steam power plant has a high pressure of 30 MPa and an exit condenser temperature of 50°C. The maximum temperature in the boiler is 1000°C and the turbine exhaust is saturated vapor There is one open feedwater heater receiving extraction from the turbine at 1MPa, and its exit is saturated liquid flowing to pump 2. The isentropic efficiency for the first section and the overall turbine are both 88.5%. Find the ratio of the extraction mass flow to total flow into turbine. What is the boiler inlet temperature with and without the feedwater heater?

Basically a Rankine Cycle 1: 50 °C, 12.35 kPa, h = 209.31, s = 0.7037 2: 30 MPa 3: 30 MPa, 1000 °C, h = 4554.7, s = 7.2867 4AC: 50°C, x = 1, h = 2592.1

3

T 2b 2 1a

1b 1

30 MPa 1000 C 1 MPa 3b 3a 50 C

4s 4ac s

a) C.V. Turbine Ideal: s4S = s3 ⇒ x4S = 0.8929, h4S = 2336.8 =>

wT,S = h3 - h4S = 2217.86

Actual: wT,AC = h3-h4AC = 1962.6, η = wT,AC/wT,S = 0.885 b)

P2 1b

2b

m tot 3b m1

1a

P1

1

1b: Sat liq. 179.91°C, h = 762.81 3a: 1 MPa, s = s3 -> h3a = 3149.09, T3a = 345.96 -> wT1s = 1405.6 3b: 1 MPa, wT1ac = ηwT1s = 1243.96 wT1ac = h3-h3b => h 3b = 3310.74 1a: wP1 = -v1(P1a-P1) ≈ -1 h1a = h1 - wP1 = 210.31

. . . . C.V. Feedwater Heater: mTOTh1b = m1h3b + (mTOT - m1)h1a . . ⇒ m1/mTOT = x = (h1b - h1a)/(h3b - h1a) = 0.178 . . . . c) C.V. Turbine: (mTOT)3 = (m1)3b + (mTOT - m1)4AC . . . . . _T=m W TOTh3 - m1h3b - (mTOT - m1)h4AC = 25 MW = mTOTwT . wT = h3-xh3b - (1-x)h4AC = 1834.7 => mTOT = 13.63 kg/s d) C.V. No FWH, Pump Ideal:

-wP = h2S - h1, s2S = s1

Steam table ⇒ h2S = 240.1, T2S = 51.2°C 1 FWH, CV: P2. s2b = s1b = 2.1386 => T 2b = 183.9°C

11-32 11.40 In one type of nuclear power plant, heat is transferred in the nuclear reactor to liquid sodium. The liquid sodium is then pumped through a heat exchanger where heat is transferred to boiling water. Saturated vapor steam at 5 MPa exits this heat exchanger and is then superheated to 600°C in an external gas-fired superheater. The steam enters the turbine, which has one (open-type) feedwater extraction at 0.4 MPa. The isentropic turbine efficiency is 87%, and the condenser pressure is 7.5 kPa. Determine the heat transfer in the reactor and in the superheater to produce a net power output of 1 MW. 5 MPa T 6 6 600o C SUP. HT.

TURBINE.

Q

0.4 MPa

7

5

8

3 4

2 3 1

HTR.

REACT.

P

5 7s

4

COND. 2

7

7.5 kPa

8s 8

s . WNET = 1 MW , ηST = 0.87

1

P -wP12 = 0.001008(400 - 7.5) = 0.4 kJ/kg h2 = h1 - wP12 = 168.8 + 0.4 = 169.2

-wP34 = 0.001084(5000 - 400) = 5.0 kJ/kg h4 = h3 - wP34 = 604.7 + 5.0 = 609.7 s7S = s6 = 7.2589, P7=0.4 MPa => T7S= 221.2 oC, h7S = 2904.5 h6 - h7 = ηST(h6 - h7S) ⇒ 3666.5 - h7 = 0.87(3666.5 - 2904.5) = 662.9 ⇒ h7 = 3003.6 s8S = s6 = 7.2589 = 0.5764 + x8S × 7.6750

x8S = 0.8707

h8S = 168.8 + 0.8707 × 2406.0 = 2263.7 h6 - h8 = ηST(h6 - h8S) ⇒ 3666.5 - h8 = 0.87(3666.5 - 2263.7) = 1220.4 ⇒ h8 = 2446.1 CV: heater cont: m2 + m7 = m3 = 1.0 kg,

Energy Eq.:

m2h2 + m7h7 = m3h3

m7 = (604.7-169.2)/(3003.6-169.2) = 0.1536 CV: turbine wT = (h6 - h7) + (1 - m7)(h7 - h8) = 3666.5-3003.6 + 0.8464(3003.6-2446.1) = 1134.8 kJ/kg

11-33 CV: pumps wP = m1wP12 + m3wP34 = 0.8464(-0.4) + 1(-5.0) = -5.3 kJ/kg . wNET = 1134.8 - 5.3 = 1129.5 => m = 1000/1129.5 = 0.885 kg/s CV: reactor . . QREACT = m(h5 - h4) = 0.885(2794.3 - 609.7) = 1933 kW CV: superheater . QSUP = 0.885(h6 - h5) = 0.885(3666.5 - 2794.3) = 746 kW 11.41 A cogenerating steam power plant, as in Fig. 11.17, operates with a boiler output of 25 kg/s steam at 7 MPa, 500°C. The condenser operates at 7.5 kPa and the process heat is extracted as 5 kg/s from the turbine at 500 kPa, state 6 and after use is returned as saturated liquid at 100 kPa, state 8. Assume all components are ideal and find the temperature after pump 1, the total turbine output and the total process heat transfer. Pump 1:

Inlet state is saturated liquid: h1 = 168.79, v1 = 0.001008

wP1 = -∫ v dP = -v1 ( P2 - P1) = -0.001008(7000-7.5) = -7.05 kJ/kg - wP1 = h2 - h1 => h2 = h1 - wP1 = 175.84, T5 = 42°C Turbine:

h5 = 3410.3, s5 = 6.7975

P6, s6 = s5 => x6 = 0.9952, h6 = 2738.6 P7, s7 = s5 => x7 = 0.8106, h7 = 2119.0 . . . WT = m5 ( h5 - h6) + 0.80m5 ( h6 - h7) = 25 (3410.3 - 2738.6) + 20 (2738.6 - 2119) = 16792.5 + 12392 = 29.185 MW . . Qproc = m6(h6 - h8) = 5(2738.6 - 417.46) = 11.606 MW

11-34 11.42 A 10 kg/s steady supply of saturated-vapor steam at 500 kPa is required for drying a wood pulp slurry in a paper mill. It is decided to supply this steam by cogeneration, that is, the steam supply will be the exhaust from a steam turbine. Water at 20°C, 100 kPa, is pumped to a pressure of 5 MPa and then fed to a steam generator. It may be assumed that the isentropic efficiency of the pump is 75%, and that of the turbine is 85%. What is the steam temperature exiting the steam generator? What is the additional heat transfer rate to the steam generator beyond what would have been required to produce only the desired steam supply? What is the difference in net power? State 3: P3 = 5 MPa State 4: P4 = 500 kPa, sat. vap. -> x4 = 1.0, T4 = 151.9°C h4 = hg = 2748.7 kJ/kg, s4 = sg = 6.8212 kJ/kg-K State 1: T1 = 20°C, P1 = 100 kPa h1 = hf = 83.94 kJ/kg, v1 = vf = 0.001002 m3/kg State 2: P2 = 5 MPa (a) C.V.: Turbine, Assume reversible & adiabatic, s3 = s4s s3 = s4s = sf + x4s sfg = 1.8606 + x4s4.9606 h4s = hf + x4s hfg = 640.2 + x4s2108.5 ηts =

wt

=

h 3 - h4

= 0.85 wts h3 - h4s Trial and Error to find T3 @ P3 = 5 MPa Assume T3 = 400°C -> s3 = 6.6458 kJ/kg-K, h3 = 3195.6 kJ/kg s4s = s3 -> x4s = 0.9646,

h4s = hf + x4s hfg = 2674.1 kJ/kg

h3 - h 4

= 0.857 ≈ ηts; T3 = 400°C h3 - h4s (b) With Cogeneration; C.V. Pump wPs = -∫ vdP = -v1( P2- P1) = -4.91 kJ/kg wPw = wPs / ηPs = -6.55 kJ/kg st 1 Law: q + h1 = h2 + w; q = 0; h2 = h1 - wPw = 90.5 kJ/kg

C.V. Steam Generator: qw = h3 - h2 = 3105.1 kJ/kg Without Cogeneration; C.V. Pump: wPs = -v1( P4- P1) = -0.4 kJ/kg wPw/o = wPs / ηPs = -0.53 kJ/kg; h2 = h1 - wPw/o = 84.5 kJ/kg C.V. Steam Generator: qw/o = h4- h2 = 2664.2 kJ/kg . Additional Heat Transfer: qw - qw/o = 440.9 kJ/kg; Qextra = 4409 kW Difference in Net Power: w diff = (wt + wPw) - wPw/o, wt = h3 - h4 = 446.9 kJ/kg . wdiff = 440.9 kJ/kg, Wdiff = 4409 kW

11-35 11.43 An industrial application has the following steam requirement: one 10-kg/s stream at a pressure of 0.5 MPa and one 5-kg/s stream at 1.4 MPa (both saturated or slightly superheated vapor). It is obtained by cogeneration, whereby a highpressure boiler supplies steam at 10 MPa, 500°C to a turbine. The required amount is withdrawn at 1.4 MPa, and the remainder is expanded in the lowpressure end of the turbine to 0.5 MPa providing the second required steam flow. Assuming both turbine sections have an isentropic efficiency of 85%, determine the following. a. The power output of the turbine and the heat transfer rate in the boiler. b. Compute the rates needed were the steam generated in a low-pressure boiler without cogeneration. Assume that for each, 20°C liquid water is pumped to the required pressure and fed to a boiler. o

BOILER. 2

. -W P

3

. QH

P 1

10 MPa, 500 C

ηs = 0.85

o

20 C

. W HPT

HP TURB. 4

MPa {1.4 5 kg/s

H 2O IN

ηs = 0.85

LP TURB. 5

. W LPT MPa {0.5 10 kg/s

a) high-pressure turbine s4S = s3 = 6.5966 ⇒ T4S = 219.9 oC, h4S = 2852.6 wS HPT = h3 - h4S = 3373.7 - 2852.6 = 521.1 wHPT = ηSwS HPT = 0.85 × 521.1 = 442.9 h4 = h3 - w = 3373.7-442.9 = 2930.8 ⇒ T4 = 251.6°C, s4 = 6.7533 low-pressure turbine s5S = s4 = 6.7533 = 1.8607 + x5S × 4.9606, x5S = 0.9863 h5S = 640.23 + 0.9863 × 2108.5 = 2719.8 wS LPT = h4 - h5S = 2930.8 - 2719.8 = 211.0 wLPT = ηSwS LPT = 0.85 × 211.0 = 179.4 kJ/kg

STEAM

STEAM

11-36 h5 = h4 - w = 2930.8 - 179.4 = 2751.4 > hG OK . WTURB = 15 × 442.9 + 10 × 179.4 = 8438 kW . b) -wP = 15[0.001002(10000 - 2.3)] = 150.3 kW h2 = h1 - wP = 83.96 + 10.02 = 94.0 . . QH = m1(h3 - h2) = 15(3373.7 - 94.0) = 49196 kW This is to be compared to the amount of heat required to supply 5 kg/s of 1.4 MPa sat. vap. plus 10 kg/s of 0.5 MPa sat. vap. from 20 oC water. 1

5 kg/s o 20 C

2

3

P

5 kg/s SAT. VAP. AT 1.4 MPa

. Q 56 10 kg/s o 20 C

4

5

P . -W P2

6

10 kg/s SAT. VAP. AT 0.5 MPa

h2 = h1 - wP = 83.96 - 0.001002(1400 - 2.3) = 85.4 . . Q = m1(h3 - h2) = 5(2790.0 - 85.4) = 13523 kW 2 3 . -WP1 = 5 × 14.0 = 7 kW h5 = h4 - wP2 = 83.96 + 0.001002(500 - 2.3) = 84.5 . . Q = m4(h6 - h5) = 10(2748.7 - 84.5) = 26642 kW 5 6 . -WP2 = 10 × 0.5 = 5 kW . Total QH = 13523 + 26642 = 40165 kW

11-37 11.44 In a cogenerating steam power plant the turbine receives steam from a highpressure steam drum and a low-pressure steam drum as shown in Fig. P11.44. The condenser is made as two closed heat exchangers used to heat water running in a separate loop for district heating. The high-temperature heater adds 30 MW and the low-temperature heaters adds 31 MW to the district heating water flow. Find the power cogenerated by the turbine and the temperature in the return line to the deaerator. Assume a reversible turbine . . . . . m1h1 + m2h2 = m3h3 + m4h4 + WT . . . m1s1 + m2s2 = mTOTsmix h1 = 3445.9

2

.

WT

1

Turbine

s1 = 6.9108

h2 = 2855.4 s2 = 7.0592 . mTOT = 27 kg/s ⇒ sMIX = 6.9383

3

s3 = sMIX ⇒ h3 = 2632.4 x3 = 0.966 s4 = sMIX ⇒ h4 = 2413.5 x4 = 0.899 . WT = 22 × 3445.9 + 5 × 2855.4 - 13 × 2632.4 - 14 × 2413.5 = 22077 kW = 22 MW . . District heating line QTOT = m(h95 - h60) = 60935 kW OK, this matches close enough . . . . C.V. Both heaters: m3h3 + m4h4 - QTOT = mTOThEX 13 × 2632.4 - 14 × 2413.5 - 60935 = 7075.2 = 27 × hEX hEX = 262 ≈ hf ⇒ TEX = 62.5°C

4

11-38 11.45 A boiler delivers steam at 10 MPa, 550°C to a two-stage turbine as shown in Fig. 11.17. After the first stage, 25% of the steam is extracted at 1.4 MPa for a process application and returned at 1 MPa, 90°C to the feedwater line. The remainder of the steam continues through the low-pressure turbine stage, which exhausts to the condenser at 10 kPa. One pump brings the feedwater to 1 MPa and a second pump brings it to 10 MPa. Assume the first and second stages in the steam turbine have isentropic efficiencies of 85% and 80% and that both pumps are ideal. If the process application requires 5 MW of power, how much power can then be cogenerated by the turbine? h5 = 3500.9, s5 = 6.7567

5

4s: s6S = s5 ⇒ h6S = 2932.1

T1

wT1,S = h5 - h6S = 568.8

6

⇒ wT1,AC = 483.5

7

h6AC = h5 - wT1,AC = 3017.4

4

6ac: P6, h6AC ⇒ s6AC = 6.9129 7s: s7S = s6AC ⇒ h7S = 2189.9 wT2,S = h6AC - h7S = 827.5

wT2,AC = 622 = h 6AC - h7AC

T2

Boiler

P2

8

Process heat 5 MW

3

2 P1

⇒

h7AC = 2355.4

1

8: h8 = 377.6 qPROC = h6AC - h8 = 2639.8 . . . m6 = Q/qPROC = 5000/2639.8 = 1.894 kg/s = 0.25 mTOT . . . . . ⇒ mTOT = m5 = 7.576 kg/s, m7 = m5 - m6 = 5.682 kg/s . . . . WT = m5h5 - m6h6AC - m7h7AC = 7424 kW

C

11-39 11.46 Consider an ideal air-standard Brayton cycle in which the air into the compressor is at 100 kPa, 20°C, and the pressure ratio across the compressor is 12:1. The maximum temperature in the cycle is 1100°C, and the air flow rate is 10 kg/s. Assume constant specific heat for the air, value from Table A.5. Determine the compressor work, the turbine work, and the thermal efficiency of the cycle. P

T P

2

P1 = 100 kPa

s

T1 = 20 oC

3 s

s

P 2

1

3

P

s

4

P2

4

P1

P

1 s

v

= 12

T3 = 1100 oC . m = 10 kg/s

k-1 P2 k a) T2 = T1  = 293.2(12)0.286 = 596.8 K P1

-wC = -1w2 = CP0(T2 - T1) = 1.004(596.8 - 293.2) = 304.8 kJ/kg k-1 P4 k  1 0.286 T4 = T3  = 1373.2  = 674.7 K 12 P3

wT = CP0(T3 - T4) = 1.004(1373.2 - 674.7) = 701.3 kJ/kg . . . . WC = mwC = -3048 kW, WT = mwT = 7013 kW qH = CP0(T3 - T2) = 1.004(1373.2 - 596.8) = 779.5 kJ/kg ηTH = wNET/qH = (701.3 - 304.8)/779.5 = 0.509 b)

v4 = RT4/P4 =

0.287 × 674.7 = 1.9364 m3/kg 100

v2 = RT2/P2 =

0.287 × 596.8 = 0.1427 1200

mep =

wNET v4 - v2

=

396.5 = 221 kPa 1.9364 - 0.1427

Too low for a reciprocating machine (compared with corresponding values for Otto and Diesel cycles.)

11-40 11.47 Repeat Problem 11.46, but assume variable specific heat for the air, table A.7. a) From A.7: s 2 = s1

h1 = 293.6,

Pr1 = 1.0286

⇒ Pr2 = Pr1 × (P2/P1) = 1.0286 × 12 = 12.343

⇒ T2 = 590 K, h2 = 597.3 kJ/kg -wC = -1w2 = h2 - h1 = 597.3 - 293.6 = 303.7 kJ/kg From A.7:

h3 = 1483.3,

Pr3 = 333.59

s4 = s3 ⇒ Pr4 = Pr3 × (P4/P3) = 333.59 × (1/12) = 27.8 ⇒ T4 = 735 K, h4 = 751.2 kJ/kg wT = h3 - h4 = 1483.2 - 751.2 = 732 kJ/kg . . . . ⇒ WC = mwC = -3037 kW, WT = mwT = 7320 kW qH = h3 - h2 = 1483.2 - 597.3 = 885.9 kJ/kg wNET = 732 - 303.7 = 428.3 kJ/kg ηTH = wNET/qH = 428.3/885.9 = 0.483 v4 = (0.287 × 735)/100 = 2.1095 m3/kg

b)

v2 = (0.287 × 590)/1200 = 0.1411 mep = 428.3/(2.1095 - 0.1411) = 217.6 kPa

11.48 An ideal regenerator is incorporated into the ideal air-standard Brayton cycle of Problem 11.46. Find the thermal efficiency of the cycle with this modification. T

3

Problem 11.46 + ideal regen. From 11.46: wT = 701.3, wC = -304.8 kJ/kg

s x 2 s

wNET = 396.5 kJ/kg

4 y

Ideal regen.:

TX = T4 = 674.7 K

qH = h3 - hX = 1.004(1373.2 - 674.7)

1 s

= 701.3 kJ/kg = wT

ηTH = wNET/qH = 396.5/701.3 = 0.565

11-41 11.49 A Brayton cycle inlet is at 300 K, 100 kPa and the combustion adds 670 kJ/kg. The maximum temperature is 1200 K due to material considerations. What is the maximum allowed compression ratio? For this calculate the net work and cycle efficiency assuming variable specific heat for the air, table A.7. Combustion: h3 = h2 + qH; 2w3 = 0 and Tmax = T3 = 1200 K h2 = h3 - qH = 1277.8 - 670 = 607.8 T2 ≈ 600 K; Pr2 = 13.0923 ; T1 = 300 K; Pr1 = 1.1146 Ideal Compression: P2 / P1 = Pr2 / Pr1 = 11.75 Ideal Expansion:

Pr4 = Pr3 / (P3 / P4) = 191.174 / 11.75 = 16.27

T4 ≈ 636 K, h4 = 645.7

linear interpolation

wT = h3 - h4 = 1277.8 - 645.7 = 632.1 -wC = h2 - h1 = 607.8 - 300.47 = 307.3 wnet = wT + wC = 632.1 - 307.3 = 324.8 η = wnet / qH = 324.8 / 670 = 0.485 11.50 A large stationary Brayton cycle gas-turbine power plant delivers a power output of 100 MW to an electric generator. The minimum temperature in the cycle is 300 K, and the maximum temperature is 1600 K. The minimum pressure in the cycle is 100 kPa, and the compressor pressure ratio is 14 to 1. Calculate the power output of the turbine. What fraction of the turbine output is required to drive the compressor? What is the thermal efficiency of the cycle? T

3

T1 = 300 K, P2/P1 = 14, T3 = 1600 K a) Assume const CP0: ⇒ T2 = T1(P2/P1

k-1 )k

s

s2 = s1 2

= 300(14)

0.286

= 638.1 K

-wC = -w12 = h2 - h1 = CP0(T2 - T1)

s 1

= 1.004 (638.1 - 300) = 339.5 kJ/kg Also, s4 = s3 → T4 = T3(P4/P3

k-1 )k

4

s

= 1600 (1/14)0.286 = 752.2 K

wT = w34 = h3 − h4 = CP0(T3 − T4) = 1.004 (1600 − 752.2) = 851.2 kJ/kg wNET = 851.2 - 339.5 = 511.7 kJ/kg . . m = WNET/wNET = 100000/511.7 = 195.4 kg/s . . WT = mwT = 195.4 × 851.2 = 166.32 MW -wC/wT = 339.5/851.2 = 0.399 b) qH = CP0(T3 - T2) = 1.004 (1600 - 638.1) = 965.7 kJ/kg ηTH = wNET/qH = 511.7/965.7 = 0.530

11-42 11.51 Repeat Problem 11.50, but assume that the compressor has an isentropic efficiency of 85% and the turbine an isentropic efficiency of 88%. Same as problem 11.50 , except ηSC = 0.85 & ηST = 0.88 P1 = 100 kPa, T1 = 300 K, P2 = 1400 kPa, T3 = 1600 K a) From solution 11.50: T2S = 638.1 K,

wSC = -339.5 kJ/kg

⇒ -wC = -wSC/ηSC = 339.5/0.85 = 399.4 kJ/kg = CP0(T2-T1) ⇒ T2 = T1 - wc/CP0 = 300 +

399.4 = 697.8 K 1.004

From solution 11.50: T4S = 752.2 K, wST = 851.2 kJ/kg ⇒ wT = ηST wST = 0.88 × 851.2 = 749.1 kJ/kg = C P0(T3-T4) ⇒ T4 = T3 - wT/CP0 = 1600 -

749.1 = 853.9 K 1.004

wNET = 749.1 - 399.4 = 349.7 kJ/kg . . m = WNET/wNET = 100000/349.7 = 286.0 kg/s . . wT = mwT = 286.0×749.1 = 214.2 MW -wC/wT = 399.4/749.1 = 0.533 b)

qH = CP0(T3 - T2) = 1.004(1600 - 697.8) = 905.8 kJ/kg ηTH = wN/qH = 349.7/905.8 = 0.386

11.52 Repeat Problem 11.51, but include a regenerator with 75% efficiency in the cycle. T

3 x' 2s 2

4

x

⇒ TX = 814.9 K

4s 1 s

b)

Same as 11.51, but with a regenerator hX - h2 TX - T2 TX - 697.8 = = ηREG = 0.75 = T4 - T2 853.9 - 697.8 ' hX - h2 a) Turbine and compressor work not affected by regenerator.

qH = CP0(T3 - TX) = 1.004(1600 - 814.9) = 788.2 kJ/kg ηTH = wNET/qH = 349.7/788.2 = 0.444

11-43 11.53 A gas turbine with air as the working fluid has two ideal turbine sections, as shown in Fig. P11.53, the first of which drives the ideal compressor, with the second producing the power output. The compressor input is at 290 K, 100 kPa, and the exit is at 450 kPa. A fraction of flow, x, bypasses the burner and the rest (1 − x) goes through the burner where 1200 kJ/kg is added by combustion. The two flows then mix before entering the first turbine and continue through the second turbine, with exhaust at 100 kPa. If the mixing should result in a temperature of 1000 K into the first turbine find the fraction x. Find the required pressure and temperature into the second turbine and its specific power output. C.V.Comp.: -wC = h2 - h1; s2 = s1 Pr2 = Pr1(P2/P1) = 0.9899(450/100) = 4.4545, Tr2 = 445 h2 = 446.74,

-wC = 446.74 - 290.43 = 156.3

C.V.Burner: h3 = h2 + qH = 446.74 + 1200 = 1646.74 kJ/kg ⇒ T3 = 1509 K C.V.Mixing chamber: (1 - x)h3 + xh2 = hMIX = 1046.22 kJ/kg h3 - hMIX

1646.74 - 1046.22 = 0.50 1646.74 - 446.74 h3 - h2 . . . WT1 = WC,in ⇒ wT1 = -wC = 156.3 = h 3 - h4 x=

=

h4 = 1046.22 - 156.3 = 889.9 ⇒ T4 = 861 K P4 = (Pr4/PrMIX)PMIX = (51/91.65) × 450 = 250.4 kPa s4 = s5 ⇒ Pr5 = Pr4(P5/P4) = 51(100/250.4) = 20.367 h5 = 688.2 T5 = 676 K wT2 = h4 - h5 = 889.9 - 688.2 = 201.7 kJ/kg

11-44 11.54 The gas-turbine cycle shown in Fig. P11.54 is used as an automotive engine. In the first turbine, the gas expands to pressure P , just low enough for this turbine 5 to drive the compressor. The gas is then expanded through the second turbine connected to the drive wheels. The data for the engine are shown in the figure and assume that all processes are ideal. Determine the intermediate pressure P , the 5 net specific work output of the engine, and the mass flow rate through the engine. Find also the air temperature entering the burner T , and the thermal efficiency of 3 the engine. k-1 P2 k a) s2 = s1 ⇒ T2 = T1  = 300(6)0.286 = 500.8 K P1

-wC = -w12 = CP0(T2 - T1) = 1.004(500.8 - 300) = 201.6 kJ/kg wT1 = -wC = 201.6 = C P0(T4 - T5) = 1.004(1600 - T5) ⇒ T5 = 1399.2 K k-1 T5 k 1399.23.5 s5 = s4 ⇒ P5 = P4  = 600  = 375 kPa  1600  T4 k-1 P6 k 1000.286 b) s6 = s5 ⇒ T6 =T5  = 1399.2  = 958.8 K 375 P5

wT2 = CP0(T5 - T6) = 1.004(1399.2 - 958.8) = 442.2 kJ/kg . . m = WNET/wT2 = 150/442.2 = 0.339 kg/s c) Ideal regenerator ⇒ T3 = T6 = 958.8 K qH = CP0(T4 - T3) = 1.004(1600 - 958.8) = 643.8 kJ/kg ηTH = wNET/qH = 442.2/643.8 = 0.687

11-45 11.55 Repeat Problem 11.54, but assume that the compressor has an efficiency of 82%, that both turbines have efficiencies of 87%, and that the regenerator efficiency is 70%. k-1 P2 k a) From solution 11.54: T2 = T1  = 300(6)0.286 = 500.8 K P1

-wC = -w12 = CP0(T2 - T1) = 1.004(500.8 - 300) = 201.6 kJ/kg -wC = -wSC/ηSC = 201.6/0.82 = 245.8 kJ/kg = wT1 = CP0(T4 - T5) = 1.004(1600 - T5) ⇒ T5 = 1355.2 K wST1 = wT1/ηST1 = 245.8/0.87 = 282.5 = CP0(T4 - T5S) = 1.004(1600 - T5S) s5S = s4 ⇒ P5 = P4(T5S/T4

k k-1 )

= 600(

⇒

T5S = 1318.6 K

1318.6 3.5 ) = 304.9 kPa 1600

b) P6 = 100 kPa, s6S = s5 k-1 P6 k  100 0.286 T6S = T5  = 1355.2  = 985.2K 304.9 P5

wST2 = CP0(T5-T6S) = 1.004(1355.2- 985.2) = 371.5 kJ/kg wT2 = ηST2 × wST2 = 0.87 × 371.5 = 323.2 kJ/kg 323.2 = CP0(T5-T6) = 1.004(1355.2 -T6) ⇒ T6 = 1033.3K . . m = WNET/wNET = 150/323.2 = 0.464 kg/s c)

wC = 245.8 = C P0(T2 - T1) = 1.004(T2 – 300) ⇒ T2 = 544.8 K ηREG =

h3 - h2 h6 - h2

=

T 3 - T2

T3 - 544.8

= = 0.7 T6 - T2 1033.3 - 544.8

⇒ T3 = 886.8 K qH = CP0(T4 - T3) = 1.004(1600 – 886.8) = 716 kJ/kg ηTH = wNET/qH = 323.2/716 = 0.451

11-46 11.56 Repeat the questions in Problem 11.54 when we assume that friction causes pressure drops in the burner and on both sides of the regenerator. In each case, the pressure drop is estimated to be 2% of the inlet pressure to that component of the system, so P3 = 588 kPa, P4 = 0.98 P3 and P6 = 102 kPa. k-1

P2 k a) From solution 11.54: T2 = T1  = 300(6)0.286 = 500.8 K P1 -wC = -w12 = CP0(T2 - T1) = 1.004(500.8 - 300) = 201.6 kJ/kg P3 = 0.98 × 600 = 588 kPa, P4 = 0.98 × 588 = 576.2 kPa s5 = s4 ⇒ P5 = P4(T5S/T4

k k-1 )

= 576.2(

1399.2 3.5 ) = 360.4 kPa 1600

b) P6 = 100/0.98 = 102 kPa, s6S = s5 k-1

P6 k  102 0.286 T6 = T5  = 1399.2  = 975.2K 292.8 P5 wST2 = CP0(T5-T6) = 1.004(1399.2 - 975.2) = 425.7 kJ/kg . . m = WNET/wNET = 150/425.7 = 0.352 kg/s c) T3 = T6 = 975.2 K qH = CP0(T4 - T3) = 1.004 (1600 - 975.2) = 627.3 kJ/kg ηTH = wNET/qH = 425.7/627.3 = 0.678

11-47 11.57 Consider an ideal gas-turbine cycle with two stages of compression and two stages of expansion. The pressure ratio across each compressor stage and each turbine stage is 8 to 1. The pressure at the entrance to the first compressor is 100 kPa, the temperature entering each compressor is 20°C, and the temperature entering each turbine is 1100°C. An ideal regenerator is also incorporated into the cycle. Determine the compressor work, the turbine work, and the thermal efficiency of the cycle. 10

REG

I.C. CC

1

5 2

COMP

9 6

4

COMP

TURB

TURB

7

8

CC

P2/P1 = P4/P3 = P6/P7 = P8/P9 = 8.0

T

8

6

P1 = 100 kPa T1 = T3 = 20 oC, T6 = T8 = 1100 oC

5 7

Assume const. specific heat s2 = s1 and s4 = s3

4

k-1

P2 k ⇒ T4 = T2 = T1  = 293.2(8)0.286 = 531.4 K P1

3

2

9

10

1 s

Total -wC = 2 × (-w12) = 2CP0(T2 - T1) = 2 × 1.004(531.4 - 293.2) = 478.1 kJ/kg k-1

P7 10.286 Also s6 = s7 and s8 = s9: ⇒ T7 = T9 = T6  k = 1373.2  = 757.6 K 8 P6 Total wT = 2 × w67 = 2CP0(T6 - T7) = 2 × 1.004(1373.2 - 756.7) = 1235.5 kJ/kg wNET = 1235.5 - 478.1 = 757.4 kJ/kg Ideal regenerator: T5 = T9, T10 = T4 ⇒ qH = (h6 - h5) + (h8 - h7) = 2CP0(T6 - T5) = 2 × 1.004(1373.2 - 757.6) = 1235.5 kJ/kg ηTH = wNET/qH = 757.4/1235.5 = 0.613

11-48 11.58 Repeat Problem 11.57, but assume that each compressor stage and each turbine stage has an isentropic efficiency of 85%. Also assume that the regenerator has an efficiency of 70%. T T1 = T3 = 20 oC, T6 = T8 = 1100 oC 8 6 P2/P1 = P4/P3 = P6/P7 = P8P9 = 8.0

9 5

From solution 11.57: T4S = T2S = 531.4 K, wSC = 478.1 kJ/kg

4 4S

T7S = T9S = 757.6 K, wST = 1235.5 kJ/kg ⇒ -wC = -wSC/ηSC = 478.1/0.85 = 562.5 kJ/kg

7S

2S

3

1

wT = ηST wST = 0.85 × 1235.5 = 1050.2 kJ/kg w67 = w89 = 1050.2/2 = 525.1 kJ/kg 281.3 = 573.5 K 1.004

T7 = T9 = T6 - (+w67/CP0) = 1373.2 -

525.1 = 849.9 K 1.004

ηREG =

h5 - h4 h9 - h4

=

T 5 - T4 T9 - T 4

=

T5 - 573.5 849.9 - 573.5

= 0.7 ⇒ T5 = 767 K

qH = CP0(T6 - T5) + CP0(T8 - T7) = 1.004(1373.2 - 767.0) + 1.004(1373.2 - 849.9) = 1133.5 kJ/kg wNET = wT + wC = 1050.2 - 562.5 = 487.7 ηTH = wNET/qH = 487.7/1133.5 = 0.430

9S

2

-w12 = -w34 = 562.5/2 = 281.3 kJ/kg

T2 = T4 = T1 + (-w12/CP0) = 293.2 +

7

s

11-49 11.59 A gas turbine cycle has two stages of compression, with an intercooler between the stages. Air enters the first stage at 100 kPa, 300 K. The pressure ratio across each compressor stage is 5 to 1, and each stage has an isentropic efficiency of 82%. Air exits the intercooler at 330 K. The maximum cycle temperature is 1500 K, and the cycle has a single turbine stage with an isentropic efficiency of 86%. The cycle also includes a regenerator with an efficiency of 80%. Calculate the temperature at the exit of each compressor stage, the second-law efficiency of the turbine and the cycle thermal efficiency. State 1: P1 = 100 kPa, T1 = 300 K

State 7: P7 = Po = 100 kPa

State 3: T3 = 330 K; State 6: T6 = 1500 K, P6 = P4 P2 = 5 P1 = 500 kPa;

P4 = 5 P3 = 2500 kPa

Ideal compression T2s = T1 (P2/P1)(k-1)/k = 475.4 K st 1 Law: q + hi = he + w; q = 0

=>

wc1 = h1 - h2 = CP (T1 - T2)

wc1 s = CP (T1 - T2s) = -176.0 kJ/kg,

wc1 = wc1 s/ η = -214.6

T2 = T1 - wc1/CP = 513.9 K T4s = T3 (P4/P3)(k-1)/k = 475.4 K wc2 s = CP (T3 - T4s) = -193.6 kJ/kg;

wc2 = -236.1 kJ/kg

T4 = T3 - wc2 / CP = 565.2 K Ideal Turbine (reversible and adiabatic) T7s = T6(P7/P6)(k-1)/k = 597.4 K => w Ts = CP (T6 - T7s) = 905.8 kJ/kg st 1 Law Turbine:

q + h6 = h7 + w;

q=0

wT = h6 - h7 = CP (T6 - T7) = ηTs wTs = 0.86 × 905.8 = 779.0 kJ/kg T7 = T6 - wT/ CP = 1500 - 779/1.004 = 723.7 K s6 - s7 = CP ln

T6 T7

- R ln

P6 P7

= -0.1925 kJ/kg K

ψ6 - ψ7 = (h6 - h7) - To(s6 - s7) = 779.0 - 298.15(-0.1925) = 836.8 kJ/kg η2nd Law = d)

wT ψ6-ψ7

ηth = qH / wnet ;

= 779.0 / 836.8 = 0.931 wnet = wT + wc1 + wc2 = 328.3 kJ/kg

st 1 Law Combustor: q + hi = he + w;

w=0

qc = h6 - h5 = CP (T6 - T5) Regenerator: ηreg =

T 5 - T4 T7 - T4

= 0.8

qH = qc = 810.7 kJ/kg;

->

T5 = 692.1 K

ηth = 0.405

11-50 11.60 A two-stage air compressor has an intercooler between the two stages as shown in Fig. P11.60. The inlet state is 100 kPa, 290 K, and the final exit pressure is 1.6 MPa. Assume that the constant pressure intercooler cools the air to the inlet temperature, T3 = T1. It can be shown, see Problem 9.130, that the optimal pressure, P2 = (P1P4)1/2, for minimum total compressor work. Find the specific compressor works and the intercooler heat transfer for the optimal P2. Optimal intercooler pressure P2 =

100 × 1600 = 400 kPa

1: h1 = 290.43, Pr1 = 0.9899 C.V.: C1 -wC1 = h2 - h1, s2 = s1 ⇒ Pr2 = Pr1(P2/P1) = 3.9596, T2 = 430, h2 = 431.95 -wC1 = 431.95 - 290.43 = 141.5 kJ/kg C.V.Cooler: T3 = T1 ⇒ h3 = h1 qOUT = h2 - h3 = h2 - h1 = -wC1 = 141.5 kJ/kg C.V.: C2 T3 = T1, s4 = s3 ⇒ Pr4 = Pr3(P4/P3) = Pr1(P2/P1) = 3.9596, T4 = T2 Thus we get -wC2 = -wC1 = 141.5 kJ/kg 11.61 Repeat Problem 11.60 when the intercooler brings the air to T = 320 K. The 3 corrected formula for the optimal pressure is P =[ P P (T /T )n/(n-1)]1/2 see 2

1 4

3

1

Problem 9.131, where n is the exponent in the assumed polytropic process. The polytropic process has n = k (isentropic) so n/(n - 1) = 1.4/0.4 = 3.5 P2 = 400

(320/290)3.5 = 475.2 kPa

C.V. C1: s2 = s1 ⇒ Pr2 = Pr1(P2/P1) = 0.9899(475.2/100) = 4.704 ⇒ T2 = 452 K, h2 = 453.75 -wC1 = h2 - h1 = 453.75 - 290.43 = 163.3 kJ/kg C.V. Cooler:

qOUT = h2 - h3 = 453.75 - 320.576 = 133.2 kJ/kg

C.V. C2: s4 = s3 ⇒ Pr4 = Pr3(P4/P3) = 1.3972(1600/475.2) = 4.704 ⇒

T4 = T2 = 452 K,

h4 = 453.75

-wC2 = h4 - h3 = 453.75 - 320.576 = 133.2 kJ/kg

11-51 11.62 Consider an ideal air-standard Ericsson cycle that has an ideal regenerator as shown in Fig. P11.62. The high pressure is 1 MPa and the cycle efficiency is 70%. Heat is rejected in the cycle at a temperature of 300 K, and the cycle pressure at the beginning of the isothermal compression process is 100 kPa. Determine the high temperature, the compressor work, and the turbine work per kilogram of air. T P P2 = P3 = 1 MPa P

2

3

3

T 4

P1 = 100 kPa

T

T

P 1

P

q = -4q1 (ideal reg.)

2 3

P

4

⇒ qH = 3q4 & wT = qH

2 T

T1 = T2 = 300 K

1 s

v

rp = P2/P1 = 10

ηTH = ηCARNOT TH. = 1 - TL/TH = 0.7 ⇒ T3 = T4 = TH = 1000 K P2 1000 ⌡v dP = RT1ln  = 0.287 × 300 × ln  = 198.25 qL = -wC = ⌠  100  P1 ⌠v dP = -RT ln(P /P ) = 660.8 kJ/kg wT = qH = -⌡ 3 4 3 11.63 An air-standard Ericsson cycle has an ideal regenerator. Heat is supplied at 1000°C and heat is rejected at 20°C. Pressure at the beginning of the isothermal compression process is 70 kPa. The heat added is 600 kJ/kg. Find the compressor work, the turbine work, and the cycle efficiency. See the cycle diagrams in solution to 11.62: T3 = T4 = 1273.15 K, P1 = 70 kPa,

T1 = T2 = 293.15 K,

Ideal regenerator: 2q3 = -4q1, wT = qH = 600 kJ/kg ηTH = ηCARNOT = 1 - 293.15/1273.15 = 0.7697 wNET = ηTHqH = 0.7697 × 600 = 461.82 qL = -wC = 600 - 461.82 = 138.2

qH = 600 kJ/kg

11-52 11.64 Consider an ideal air-standard cycle for a gas-turbine, jet propulsion unit, such as that shown in Fig. 11.27. The pressure and temperature entering the compressor are 90 kPa, 290 K. The pressure ratio across the compressor is 14 to 1, and the turbine inlet temperature is 1500 K. When the air leaves the turbine, it enters the nozzle and expands to 90 kPa. Determine the pressure at the nozzle inlet and the velocity of the air leaving the nozzle.

2

T

3

3

BURN P

s4 s 5

2

COMP

TURB

P = 90 kPa

5 1

1

4

s

NOZ

C.V. Comp.: s2 = s1 ⇒ Pr2 = Pr1 × (P2/P1) = 0.9899 × 14 = 13.8586 h2 = 617.2, T2 = 609 K -wC = h2 - h1 = 617.2 - 290.43 = 326.8 kJ/kg C.V. Turb.: wT = h3 - h4 = -wC ⇒ h4 = h3 + wC = 1635.8 - 326.8 = 1309

⇒ Pr4 = 209.1, T4 = 1227

P4 = Pr4 × (P3/Pr3) = 209.1(1260/483.155) = 545.3 kPa C.V. Nozzle: s2 = s1 ⇒ Pr5 = Pr4 × (P5/P4) = 209.1(90/545.3) = 34.51 => T5 = 778 K,

h5 = 798.1 kJ/kg

(1/2)V25 = h4 - h5 = 510.9 ⇒

V5 =

2 × 1000 × 510.9 = 1010.8 m/s

11-53 11.65 The turbine in a jet engine receives air at 1250 K, 1.5 MPa. It exhausts to a nozzle at 250 kPa, which in turn exhausts to the atmosphere at 100 kPa. The isentropic efficiency of the turbine is 85% and the nozzle efficiency is 95%. Find the nozzle inlet temperature and the nozzle exit velocity. Assume negligible kinetic energy out of the turbine. C.V. Turb.: hi = 1336.7, Pri = 226, se = si ⇒ Pre = Pri × (Pe/Pi) = 226(250/1500) = 37.667 Te = 796, he = 817.9, wT,s = 1336.7 - 817.9 = 518.8 wT,AC = wT,s × ηT = 441 = he,AC - hi ⇒ he,AC = 895.7 ⇒ Te,AC = 866 K, Pre,AC = 52.21 C.V. Nozzle: (1/2)V2e = hi - he; se = si ⇒ Pre = Pri × (Pe/Pi) = 52.21(100/250) = 20.884 ⇒ Te,s = 681 K, he,s = 693.1 kJ/kg 2 = hi - he,s = 895.7 - 693.1 = 202.6 kJ/kg (1/2)Ve,s 2 2 = (1/2)Ve,s × ηNOZ = 192.47 kJ/kg (1/2)Ve,AC

Ve,AC =

2 × 1000 × 192.47 = 620 m/s

11-54 11.66 Repeat Problem 11.64, but assume that the isentropic compressor efficiency is 87%, the isentropic turbine efficiency is 89%, and the isentropic nozzle efficiency is 96%. Same as 11.64 except ηSC = 0.87, ηST = 0.89, ηSN = 0.96

T

Assume const. specific heat P1 = P5 = 90 kPa, T1 = 290 K

3

4 5

2S

2

5S

P2/P1 = 14, T3 = 1500 K From solution 11.64: T2S = 609 K, wSC = 326.8 kJ/kg

1

s

⇒ wC = wSC/ηSC = 326.8/0.87 = 375.6 kJ/kg wT = wC = 375.6 = h 3 - h4 ⇒ h4 = 1635.8 - 375.6 = 1260.2, T4 = 1185 K wST = wT/ηST = 375.6/0.89 = 422.0 kJ/kg = 1635.8 - h4s ⇒ h4s = 1213.8, T4s = 1145, Pr4s = 158.06 s4S= s3 ⇒ P4 = P3 Pr4s/Pr3 = 1260×158.06/483.155 = 412 kPa s5S = s4 ⇒ Pr5 = Pr4 P5/P4 = 181.6×90/412 = 39.67 ⇒ T5S = 807, h5S = 829.8, 2 /2000 = h4 − h5S = 1260.2 - 829.8 = 430.4 V5S 2 /2000 = 0.96 × 430.4 = 413.2 kJ/kg => V25/2000 = ηSN V5s

⇒ V5 = 2000*413.2 = 909 m/s

11-55 11.67 Consider an air standard jet engine cycle operating in a 280K, 100 kPa environment. The compressor requires a shaft power input of 4000 kW. Air enters the turbine state 3 at 1600 K, 2 MPa, at the rate of 9 kg/s, and the isentropic efficiency of the turbine is 85%. Determine the pressure and temperature entering the nozzle at state 4. If the nozzle efficiency is 95%, determine the temperature and velocity exiting the nozzle at state 5. . . . C.V. Shaft: WT = m(h3 - h4) = WC . . h3 - h4 = WC / m = 4000/9 = 444.4 h4 = 1757.3 – 444.4 = 1135.8 wTa = wC = 444.4

⇒ wTs = wTa / η = 522.82 = h 3 - h4s

h4s = 1234.5 ⇒ Pr = 168.28,

T4s ≈ 1163 K

P4 = (Pr4 / Pr3) P3 = (168.28/ 634.967)2000 = 530 kPa 4a: 530 kPa, h = 1312.85, T ≈ 1230 K, Pr4 = 211.92 5: 100 kPa ⇒ Pr5s = Pr4(100/530) = 40 h5s = 830.95

2

⇒ 0.5V5s = h4a - h5s = 481.9 kJ/kg

2

2

0.5V5a = η(0.5V5s) = 457.808 2

⇒ V5a= 957 m/s

h5a = h4 - 0.5V5a = 1312.85 – 457.808 = 855 ⇒

T5a ≈ 830 K

11-56 11.68 A jet aircraft is flying at an altitude of 4900 m, where the ambient pressure is approximately 55 kPa and the ambient temperature is −18°C. The velocity of the aircraft is 280 m/s, the pressure ratio across the compressor is 14:1 and the cycle maximum temperature is 1450 K. Assume the inlet flow goes through a diffuser to zero relative velocity at state 1. Find the temperature and pressure at state 1 and the velocity (relative to the aircraft) of the air leaving the engine at 55 kPa. 3

T

4

Ambient TX = -18oC = 255.2 K, PX = 55 kPa = P5

5

also VX = 280 m/s

2 1

Assume that the air at this state is reversibly decelerated to zero velocity and then enters the comp. at 1.

P amb X

P2/P1 = 14 & T3 = 1450 K

s

2

T 1 = TX +

VX 2 × 1000

= 255.2 +

(280)2 = 294.3 K 2 × 1000 × 1.0035

k

 T1k-1 294.33.5  = 90.5 kPa P1 = PX  = 55 255.2 TX Then, T2 = T1 (P2/P1

k-1 )k

= 294.3(14)0.286 = 626.0 K

-wC = -w12 = CP0(T2-T1) = 1.004(1450 - T4)

⇒ T4 = 1118.3 K

P3 = P2 = 14 × 90.5 = 1267 kPa P4 =

k k-1 P3 (T4/T3)

T5 = T4 (P5/P4 V25 2 × 1000

k-1 )k

= 1267(1118.3/1450)3.5 = 510 kPa = 1118.3(55/510)

0.286

= 591.5 K

= CP0(T4 - T5) = 1.004(1118.3 - 591.5) = 528.7 kJ/kg

⇒ V5 = 1028 m/s

11-57 11.69 Air flows into a gasoline engine at 95 kPa, 300 K. The air is then compressed with a volumetric compression ratio of 8;1. In the combustion process 1300 kJ/kg of energy is released as the fuel burns. Find the temperature and pressure after combustion. Compression 1 to 2: s2 = s1 ⇒ vr2 = vr1/8 = 179.49/8 = 22.436, T2 = 673 K, u2 = 491.5, Pr2 = 20 P2 = Pr2(P1/Pr1) = 20(95/1.1146) = 1705 kPa Compression 2 to 3: u3 = u2 + qH = 491.5 + 1300 = 1791.5 T3 = 2118 K P3 = P2 × (T3/T2) = 1705(2118/673) = 5366 kPa 11.70 A gasoline engine has a volumetric compression ratio of 9. The state before compression is 290 K, 90 kPa, and the peak cycle temperature is 1800 K. Find the pressure after expansion, the cycle net work and the cycle efficiency using properties from Table A.7. Use table A.7 and interpolation. Compression 1 to 2: s2 = s1 ⇒ vr2 = vr1(v2/v1) vr2 = 196.37/9 = 21.819 ⇒ T2 ≅ 680 K,

Pr2 ≅ 20.784,

P2 = P1(Pr2/Pr1) = 90(20.784/0.995) = 1880 kPa w = u1 - u2 = 207.19 - 496.94 = -289.75 kJ/kg

1 2

Combustion 2 to 3: qH = u3 - u2 = 1486.33 - 496.94 = 989.39 kJ/kg P3 = P2(T3/T2) = 1880(1800/680) = 4976 kPa Expansion 3 to 4: s4 = s3 ⇒ vr4 = vr3 × 9 = 1.143 × 9 = 10.278 ⇒ T4 = 889 K, Pr4 = 57.773, u4 = 665.8 P4 = P3(Pr4/Pr3) = 4976(57.773/1051) = 273.5 kPa w = u3 - u4 = 1486.33 - 665.8 = 820.5 kJ/kg

3 4

wNET = 3w4 + 1w2 = 820.5 - 289.75 = 530.8 kJ/kg η = wNET/qH = 530.8/989.39 = 0.536

u2 = 496.94

11-58 11.71 To approximate an actual spark-ignition engine consider an air-standard Otto cycle that has a heat addition of 1800 kJ/kg of air, a compression ratio of 7, and a pressure and temperature at the beginning of the compression process of 90 kPa, 10°C. Assuming constant specific heat, with the value from Table A.5, determine the maximum pressure and temperature of the cycle, the thermal efficiency of the cycle and the mean effective pressure. P

T

3 s

v

v

4 v 1

2 s

2 s

3

qH = 1800 kJ

s

v1/v2 = 7

4

P1 = 90 kPa T1 = 10 oC

v

1 v

s

k

a) P2 = P1(v1/v2) = 90(7)1.4 = 1372 kPa T2 = T1(P2/P1)(v2/v1)= 283.15 × 15.245 ×

1 = 616.6 K 7

T3 = T2 + qH/CV0 = 616.6 + 1800/0.717 = 3127 K P3 = P2T3/T2= 1372 × 3127 / 616.6 = 6958 kPa b)

ηTH = 1 - T1/T2 = 1 - 283.15/616.5 = 0.541

c)

wNET = ηTH × qH = 0.544 × 1800 = 979.2 kJ/kg v1 = RT1/P1 = (0.287 × 283.2)/90 = 0.9029 v2 = (1/7) v1 = 0.1290 m3/kg mep =

wNET v1-v2

=

979.2 = 1265 kPa 0.9029 - 0.129

11-59 11.72 Repeat Problem 11.71, but assume variable specific heat. The ideal gas air tables, Table A.7, are recommended for this calculation (and the specific heat from Fig. 5.10 at high temperature). Table A.7 is used with interpolation. a)

T1 = 283.15 K,

u1 = 202.3, vr1 = 207.94

s2 = s1 ⇒ vr2 = vr1(v2/v1) = 207.94(1/7) = 29.705 ⇒ T2 = 606.7 K, u2 = 440.2 => u3 = 440.2 + 1800 = 2240.2 =>

-w12 = u2 - u1 = 237.9,

T3 = 2575.5 K ,

vr3 = 0.3402

P3 = 90 × 7 × 2575.5 / 283.15 = 5730 kPa b)

vr4 = vr3 × 7 = 2.3814 3w 4

Ë

=>

T 4 = 1437 K;

u4 = 1147

= u3 - u4 = 2240.2 - 1147 = 1093.2 wnet = 1093.2 - 237.9 = 855.3 kJ/kg ηTH = wnet / qH = 855.3 / 1800 = 0.475

(c)

mep = 855.3 / (0.9029 - 0.129) = 1105 kPa

11.73 A gasoline engine takes air in at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which the temperature is 2050 K. Use the cold air properties (i.e. constant heat capacities at 300 K) and find the compression ratio, the compression specific work and the highest pressure in the cycle. Standard Otto Cycle T3 = 2050 K u2 = u3 - qH T2 = T3 - qH / Cvo = 2050 - 1000 / 0.717 = 655.3 K P2 = P1(T2 / T1)k/(k-1) = 90(655.3/290) 3.5 = 1561 kPa CR = v1 / v2 = (T2 / T1)1/(k-1) = (655.3 / 290) 2.5 = 7.67 -1w2 = u2 - u1 = Cvo( T2 - T1) = 0.717(655.3 - 290) = 262 kJ / kg P3 = P2T3 / T2 = 1561 × 2050 / 655.3 = 4883 kPa

11-60 11.74 Answer the same three questions for the previous problem, but use variable heat capacities (use table A.7). T3 = 2050 K u3 = 1725.714 u2 = u3 - qH = 1725.714 - 1000 = 725.714 ⇒ T2 = 960.56

vr2 = 8.216

vr1 = 195.361 =>

v1 / v2 = vr1 / vr2 = 195.361 / 8.216 = 23.78

-1w2 = u2 - u1 = 725.714 - 207.19 = 518.5 kJ/kg P3=P2T3 / T2 = P1( T3 / T1)( v1 / v3) = 90×(2050 / 290)×23.78 = 15129 kPa

11.75 When methanol produced from coal is considered as an alternative fuel to gasoline for automotive engines, it is recognized that the engine can be designed with a higher compression ratio, say 10 instead of 7, but that the energy release with combustion for a stoichiometric mixture with air is slightly smaller, about 1700 kJ/kg. Repeat Problem 11.71 using these values. same as 11.71 except v1/v2 = 10 & qH = 1700 kJ k

a) P2 = P1(v1/v2) = 90(10)1.4 = 2260.7 kPa k-1

T2 = T1(v1/v2)

= 283.15(10)0.4 = 711.2 K

T3 = T2 + qH / Cvo = 711.2 + 1700 / 0.7165 = 3084 K P3 = P2(T3 / T2) = 2260.7×3084 / 711.2 = 9803 kPa ηTH = 1 - T1/T2 = 1 - 283.15/711.2 = 0.602 wnet = ηTH × qH = 0.6 × 1700 = 1023.4 kJ/kg v1 = RT1/P1 = 0.287×283.15/90= 0.9029, v2 = v1/10= 0.0903 mep = wNET/(v1 - v2)= 1023.4 / (0.9029 - 0.0903) = 1255 kPa

11-61 11.76 It is found experimentally that the power stroke expansion in an internal combustion engine can be approximated with a polytropic process with a value of the polytropic exponent n somewhat larger than the specific heat ratio k. Repeat Problem 11.71 but assume that the expansion process is reversible and polytropic (instead of the isentropic expansion in the Otto cycle) with n equal to 1.50. See solution to 11.71 except for process 3 to 4. T3 = 3127 K, P3 = 6.958 MPa v3 = RT3/P3 = v2 = 0.129 m3/kg, v4 = v1 = 0.9029 Process:

Pv1.5 = constant.

P4 = P3(v3/v4)1.5 = 6958 (1/7)1.5 = 375.7 kPa T4 = T3(v3/v4)0.5 = 3127(1/7)0.5 = 1181.9 k w =⌠ ⌡Pdv = 1-1.4(T2 - T1) = -0.4 (606.6 -283.15)= -239.3 R

0.287

1 2

w =⌠ ⌡Pdv = R(T4 - T3)/(1 - 1.5)

3 4

= -0.287(1181.9-3127)/0.5 = 1116.5 wNET = 1116.5 - 239.3 = 877.2 ηCYCLE = wNET/qH = 877.2/1800 = 0.487 mep = wNET/(v1 - v2) = 877.2/(0.9029 - 0.129) = 1133 kPa Note a smaller wNET, ηCYCLE, mep compared to an ideal cycle.

11-62 11.77 In the Otto cycle all the heat transfer qH occurs at constant volume. It is more realistic to assume that part of qH occurs after the piston has started its downward motion in the expansion stroke. Therefore, consider a cycle identical to the Otto cycle, except that the first two-thirds of the total qH occurs at constant volume and the last one-third occurs at constant pressure. Assume that the total qH is 2100 kJ/kg, that the state at the beginning of the compression process is 90 kPa, 20°C, and that the compression ratio is 9. Calculate the maximum pressure and temperature and the thermal efficiency of this cycle. Compare the results with those of a conventional Otto cycle having the same given variables. P

3

T

4

3 s

P1 = 90 kPa, T1 = 20oC

s

rV = v1/v2 = 7

v

5

2

4

2 s

s 1

5

= 1400 kJ/kg;

1 s

v

b)

a) q23 = (2/3) × 2100

v

q34 = 2100/3 = 700

P2 = P1(v1/v2)k = 90(9)1.4 = 1951 kPa T2 = T1(v1/v2)k-1 = 293.15(9)0.4 = 706 K T3 = T2 + q23/CV0 = 706 + 1400/0.717 = 2660 K P3 = P2T3/T2 = 1951(2660/706) = 7350.8 kPa = P4 T4 = T3 + q34/CP0 = 2660 + 700/1.004 = 3357 K P4 T1 7350.8 293.15 = = × = × = 7.131 90 3357 v4 v4 P1 T4

v5

v1

T5 = T4(v4/v5)k-1 = 3357(1/7.131)0.4 = 1530 K qL = CV0(T5-T1) = 0.717(1530 - 293.15) = 886.2 kJ/kg ηTH = 1 - qL/qH = 1 - 886.2/2100 = 0.578 Std. Otto Cycle:

ηTH = 1 - (9)-0.4 = 0.585, small difference

11-63 11.78 A diesel engine has a compression ratio of 20:1 with an inlet of 95 kPa, 290 K, state 1, with volume 0.5 L. The maximum cycle temperature is 1800 K. Find the maximum pressure, the net specific work and the thermal efficiency. P2 = 95×(20) 1.4 = 6297.5 kPa T2 = T1(v1 / v2)k-1 = 290×200.4 = 961 K P3 = P2(T3 / T2) = 6297.5 × (1800 / 961) = 11796 kPa -1w2 = u2 - u1 ≈ Cvo( T2 - T1) = 0.717(961 - 290) = 481.1 kJ / kg T4 = T3( v3 / v4)0.4 = 1800(1 / 20) 0.4 = 543 K 3w 4

= u3 - u4 ≈ Cvo(T3 - T2) = 0.717(1800 - 543) = 901.3 kJ / kg

wnet = 3w4 + 1w2 = 901.3 - 481.1 = 420.2 kJ / kg η = wnet / qH = 1- (1 / 20) 0.4 = 0.698

( = 420.2 / 0.717(1800 - 961))

11.79 A diesel engine has a bore of 0.1 m, a stroke of 0.11 m and a compression ratio of 19:1 running at 2000 RPM (revolutions per minute). Each cycle takes two revolutions and has a mean effective pressure of 1400 kPa. With a total of 6 cylinders find the engine power in kW and horsepower, hp. Power from mean effective pressure. mep = wnet / (vmax - vmin) -> wnet = mep(vmax - vmin) ∆V = πBore2 × 0.25 × S = π × 0.12 × 0.25 × 0.11 = 0.000864 m3 W = mep(Vmax - Vmin) = 1400× π × 0.25 × 0.12 × 0.11 = 1.2096 Work per Cylinders per Power Stroke . W = W × n × RPM × 0.5 (cycles / min)×(min / 60 s)×(kJ / cycle) = 1.2096 × 6 × 2000 × 0.5 × (1/60) = 121 kW = 162 hp

11-64 11.80 At the beginning of compression in a diesel cycle T = 300 K, P = 200 kPa and after combustion (heat addition) is complete T = 1500 K and P = 7.0 MPa. Find the compression ratio, the thermal efficiency and the mean effective pressure. P2 = P3 = 7000 kPa => v1 / v2 = (P2/P11/ ) k = (7000 / 200)0.7143 = 12.67 T2 = T1(P2 / P1)(k-1) / k = 300(7000 / 200) 0.2857 = 828.4 v3 / v2 = T3 / T2 = 1500 / 828.4 = 1.81 v4 / v3 = v1 / v3 = (v1 / v2)( v2 / v3) = 12.67 / 1.81 = 7 T4 = T3(v3 / v4)k-1 = (1500 / 7) 0.4 = 688.7 qL = Cvo(T4 - T1) = 0.717(688.7 - 300) = 278.5 qH = h3 - h2 ≈ Cpo(T3 - T2) = 1.004(1500 - 828.4) = 674 η = 1 - qL / qH = 1- 278.5 / 674 = 0.587 wnet = qnet = 674 - 278.5 = 395.5 vmax = v1 = R T1 / P1 = 0.287×300 / 200 = 0.4305 m3 / kg vmin = vmax / (v1 / v2) = 0.4305 / 12.67 = 0.034 m3 / kg mep = wnet / (vmax - vmin) = 395.5 / (0.4305 - 0.034) = 997 kPa 11.81 Consider an ideal air-standard diesel cycle in which the state before the compression process is 95 kPa, 290 K, and the compression ratio is 20. Find the maximum temperature (by iteration) in the cycle to have a thermal efficiency of 60%? P1 = 95 kPa,

Diesel cycle:

k-1

T2 = T1(v1/v2)

T1 = 290 K,

v1/v2 = 20, ηTH = 0.6

= 290(20)0.4 = 961.2 K

v1 = 0.287 × 290/95 = 0.876 = v4,

v2 = 0.876 / 20 = 0.0438

v3 = v2(T3/T2) = 0.043883(T3/961.2) = 0.0000456 T3 k-1

T3 = T4 (v4/v3)

ηTH = 0.60 = 1 -

0.876 =( )0.4 ⇒ T4 = 0.019345 T1.4 3 0.0000456 T3 T 4 - T1 k(T3 - T2)

=1-

0.019345 × T1.4 - 290 3 1.4(T3 - 961.2)

⇒ 0.019345 × T1.4 - 0.56 × T3 + 248.272 = 0 3 3050: LHS = +1.06 T3 = 3040 K

3040: LHS = -0.036,

11-65 11.82 Consider an ideal Stirling-cycle engine in which the state at the beginning of the isothermal compression process is 100 kPa, 25°C, the compression ratio is 6, and the maximum temperature in the cycle is 1100°C. Calculate the maximum cycle pressure and the thermal efficiency of the cycle with and without regenerators. T

P 3

T 4

3 T

v

4

2

v

v T

P1 = 100 kPa

v

v1/v2 = 6

2

1

Ideal Stirling cycle T1 = T2 = 25 oC

T

1 s

v

T3 = T4 = 1100 oC

a) T1 = T2 ⇒ P2 = P1(v1/v2) = 100 × 6 = 600 kPa V2 = V3 ⇒ P3 = P2T3/T2 = 600×1373.2/298.2 = 2763 kPa b) w34 = q34 = RT3 ln(v4/v3) = 0.287×1373.2×ln6 = 706.1 kJ/kg q23 = CV0(T3 - T2) = 0.7165(1100 - 25) = 770.2 kJ/kg w12 = q12 = -RT1 ln(v1/v2) = -0.287× 298.2 ln(6) = -153.3 kJ/kg wNET = 706.1 - 153.3 = 552.8 kJ/kg ηNO REGEN =

552.8 = 0.374, 706.1+770.2

ηWITH REGEN =

552.8 = 0.783 706.1

11.83 An air-standard Stirling cycle uses helium as the working fluid. The isothermal compression brings helium from 100 kPa, 37°C to 600 kPa. The expansion takes place at 1200 K and there is no regenerator. Find the work and heat transfer in all of the 4 processes per kg helium and the thermal cycle efficiency. Helium table A.5:

R = 2.077 ,

Compression/expansion:

Cvo = 3.1156

v4 / v3 = v1 / v2 = P2 / P1 = 600 / 100 = 6

1 -> 2 -1w2 = -q12 = ∫ P dv = R T1ln(v1 / v2) = RT1ln (P2 /P1) = 2.077 × 310 × ln6 = 1153.7 kJ/kg 2 -> 3 : 2w3 = 0; 3 -> 4:

q23 = Cvo(T3 - T2) = 3.1156(1200 - 310) = 2773 kJ/kg

v4 w = q = R T ln = 2.077×1200 ln6 = 4465.8 kJ/kg 3 4 34 3 v 3

4 -> 1 4w1 = 0;

q41 = Cvo(T4 - T1) = -2773 kJ/kg

ηcycle = (1w2+3w4)/(q23+q34) = (-1153.7+4465.8) /(2773 +4465.8) = 0.458

11-66 11.84 Consider an ideal air-standard Stirling cycle with an ideal regenerator. The minimum pressure and temperature in the cycle are 100 kPa, 25°C, the compression ratio is 10, and the maximum temperature in the cycle is 1000°C. Analyze each of the four processes in this cycle for work and heat transfer, and determine the overall performance of the engine. Ideal Stirling cycle diagram as in solution 11.82, with P1 = 100 kPa,

T1 = T2 = 25oC,

From 1-2 at const T:

v1/v2 = 10,

T3 = T4 = 1000oC

w = 1q2 = T1(s2 - s1)

1 2

= -RT1ln(v1/v2) = -0.287 × 298.2 × ln(10) = -197.1 kJ/kg From 2-3 at const V:

w = 0/

2 3

q23 = CV0(T3 - T2) = 0.7165(1000 - 25) = 698.6 From 3-4 at const T; = +RT3 × ln

w = 3q4 = T3(s4 - s3)

3 4

v4 v3

From 4-1 at const V;

= 0.287 × 1237.2 × ln(10) = 841.4 kJ/kg w = 0/

4 1

q41 = CV0(T1 - T4) = 0.7165(25 - 1000) = -698.6 kJ/kg wNET = -197.1 + 0 + 841.4 + 0 = 644.3 kJ/kg Since q23 is supplied by -q41 (regenerator) qh = q34 = 841.4 kJ/kg, NOTE:

ηTH =

qH = q34 = RT3 × ln(10),

wNET qH

=

644.3 = 0.766 841.4

qL = -q12 = RT1 × ln(10)

ηTH = (qH - qL)/qH = (T3 - T1)/T3 = (975/1273.2) = 0.766 = Carnot efficiency

11-67 11.85 The air-standard Carnot cycle was not shown in the text; show the T–s diagram for this cycle. In an air-standard Carnot cycle the low temperature is 280 K and the efficiency is 60%. If the pressure before compression and after heat rejection is 100 kPa, find the high temperature and the pressure just before heat addition. η = 0.6 = 1 - TH/TL ⇒ TH = TL/0.4 = 700 K

T H

T 2

3

P1 = 100 kPa T L

1

P2 = P1(TH/TL)k-1 = 2.47 MPa

1

4 s

[or P2 = P1(Pr2/Pr1) = 2.645 MPa]

11.86 Air in a piston/cylinder goes through a Carnot cycle in which T = 26.8°C and the L total cycle efficiency is η = 2/3. Find T , the specific work and volume ratio in H the adiabatic expansion for constant C , C . Repeat the calculation for variable p v heat capacities. Carnot cycle: Same T-s diagram as in previous problem. η = 1 - TL/TH = 2/3 ⇒ TH = 3 × TL = 3 × 300 = 900 K Adiabatic expansion 3 to 4: w = (P4v4 - P3v3/(1 - k) =

3 4

Pvk = constant R (T - T ) = u3 - u4 1-k 4 3

= Cv(T3 - T4) = 0.7165(900 - 300) = 429.9 kJ/kg v4/v3 = (T3/T4)1/(k - 1) = 32.5 = 15.6 For variable Cp, Cv we get, TH = 3 × TL = 900 K w = u3 - u4 = 674.824 - 214.364 = 460.5 kJ/kg

3 4

v4/v3 = vr4/vr3 = 179.49/9.9169 = 18.1

11-68 11.87 Consider an ideal refrigeration cycle that has a condenser temperature of 45°C and an evaporator temperature of −15°C. Determine the coefficient of performance of this refrigerator for the working fluids R-12 and R-22. T

Ideal Ref. Cycle Tcond = 45 oC = T3 Tevap = -15 oC

2 3

4

1 s

h1, kJ/kg

R-12 180.97

R-22 244.13

s 2 = s1

0.7051

0.9505

P2, MPa

1.0843

1.729

T2, oC

54.7

74.4

212.63

289.26

h3=h4, kJ/kg

79.71

100.98

wC = h2-h1

31.66

45.13

qL = h1-h4

101.26

143.15

β =qL/wC

3.198

3.172

h2, kJ/kg

11.88 The environmentally safe refrigerant R-134a is one of the replacements for R-12 in refrigeration systems. Repeat Problem 11.87 using R-134a and compare the result with that for R-12. T h1 = 389.2, s2 = s1 = 1.7354 h3 = 264.11, P3 = P2 = 1.16 MPa At 1 MPa, T2 = 45.9, h2 = 426.8 At 1.2 MPa,T2 = 53.3, h2 = 430.7 ⇒ T2 = 51.8, h2 = 429.9

2 o

45 C o

-15 C

-wC = h2 - h1 = 429.9 - 389.2 = 40.7 kJ/kg qL = h1 - h4 = h1 - h3 = 389.2 - 264.11 = 125.1 kJ/kg β = qL/(-wC) = 125.1/40.7 = 3.07

3

4

1 s

11-69 11.89 A refrigerator using R-22 is powered by a small natural gas fired heat engine with a thermal efficiency of 25%. The R-22 condenses at 40°C and it evaporates at 20°C and the cycle is standard. Find the two specific heat transfers in the refrigeration cycle. What is the overall coefficient of performance as QL/Q1? Evaporator: Inlet State is sat. liq-vap with

h4 = h3 =94.272 kJ / kg

The exit state is sat. vap. with h = 242.055 kJ / kg qL = h1 - h4 = h1 - h3 = 147.78 kJ / kg Compressor: Inlet State 1 and Exit State 2 about 1.6 MPa wC = h2 - h1 ; 2:

T2 ≈ 70°C

s2 = s1 = 0.9593 kJ / kg K h2 = 287.17 kJ / kg

wC = h2 - h1 = 45.11 kJ / kg Condenser: Brings it to sat liq at state 3 qH = h2 - h3 = 287.17 - 94.272 = 192.9 kJ / kg Overall Refrigerator: β = qL / wC = 147.78 / 45.11 = 3.276 Heat Engine:

. . . . WHE = ηHEQ1 = WC = QL / β . . QL / Q1 = ηβ = 0.25×3.276 = 0.819

11.90 A refrigerator with R-12 as the working fluid has a minimum temperature of −10°C and a maximum pressure of 1 MPa. Assume an ideal refrigeration cycle as in Fig. 11.32. Find the specific heat transfer from the cold space and that to the hot space, and the coefficient of performance. h1 = 183.058, s1 = 0.7014, h3 = 76.155 s2 = s1 & P2 ⇒ h2 ≈ 210.16 qL = h1 - h4 = h1 - h3 = 183.058 - 76.155 = 106.9 kJ/kg qH = h2 - h3 = 210.16 - 76.155 = 134 kJ/kg b = qL/(-wc) = qL/(qH - qL) = 3.945

11-70 11.91 A refrigerator in a meat warehouse must keep a low temperature of -15°C and the outside temperature is 20°C. It uses R-12 as the refrigerant which must remove 5 kW from the cold space. Find the flow rate of the R-12 needed assuming a standard vapor compression refrigeration cycle with a condenser at 20°C. Basic refrigeration cycle:

T1 = T4 = -15°C,

T3 = 20°C

Table B.3: h4 = h3 = 54.27 ; h1 = hg = 180.974 . . . QL = mR-12 × qL = mR-12(h1 - h4) qL = 180.974 - 54.874 = 126.1 kJ/kg . mR-12 = 5.0 / 126.1 = 0.03965 kg / s

11.92 A refrigerator with R-12 as the working fluid has a minimum temperature of −10°C and a maximum pressure of 1 MPa. The actual adiabatic compressor exit temperature is 60°C. Assume no pressure loss in the heat exchangers. Find the specific heat transfer from the cold space and that to the hot space, the coefficient of performance and the isentropic efficiency of the compressor. State 1: Inlet to compressor, sat. vap. -10°C, h1 = 183.058, s1 = 0.7014 State 2: Actual compressor exit, h2AC = 217.81 State 3: Exit condenser, sat. liq. 1MPa, h3 = 76.155 State 4: Exit valve, h4 = h3 C.V. Evaporator: qL = h1 - h4 = h1 - h3 = 106.9 kJ/kg C.V. Ideal Compressor: wC,S = h2,S - h1, s2,S = s1 State 2s: T2,S = 49.66, h2,S = 209.9, wC,S = 26.842 C.V. Actual Compressor: wC = h2,AC - h1 = 34.752 kJ/kg β = qL/wC = 3.076, ηC = wC,S/wC = 0.7724 C.V. Condenser: qH = h2,AC - h3 = 141.66 kJ/kg

11-71 11.93 Consider an ideal heat pump that has a condenser temperature of 50°C and an evaporator temperature of 0°C. Determine the coefficient of performance of this heat pump for the working fluids R-12, R-22, and ammonia. T

2

Ideal Heat Pump: Tcond = 50 oC = T3

3

Tevap = 0 oC 4

h1, kJ/kg s 2 = s1 P2, MPa T2, oC h2, kJ/kg h3=h4, kJ/kg wC = h2-h1 qH = h2-h3

1 s

R-12

R-22

NH3

187.53 0.6965 1.2193 56.7

249.95 0.9269 1.9423 72.2

1442.32 5.3313 2.0333 115.6

211.95 84.94 24.42 127.01

284.25 107.85 34.3 176.4

1672.84 421.58 230.52 1251.26

5.201

5.143

5.428

β′ =qH/wC

11.94 The air conditioner in a car uses R-134a and the compressor power input is 1.5 kW bringing the R-134a from 201.7 kPa to 1200 kPa by compression. The cold space is a heat exchanger that cools atmospheric air from the outside down to 10°C and blows it into the car. What is the mass flow rate of the R-134a and what is the low temperature heat transfer rate. How much is the mass flow rate of air at 10°C? Standard Refrigeration Cycle Table B.5

h1 = 392.285;

s1 = 1.7319 ;

h4 = h3 = 266

C.V. Compressor (assume ideal) . . m1 = m2 wC = h2 - h1; s2 = s1 + sgen P2, s = s1 -> h2 = 429.5 -> wC = 37.2 . . . m wC = WC -> m = 1.5 / 37.2 = 0.0403 kg /s C.V. Evaporator . . QL = m(h1 - h4) = 0.0405(392.28 - 266) = 5.21 kW C.V. Air Cooler . . . mair∆hair = QL ≈ mairCp∆T . . mair = QL / (Cp∆T) = 5.21 / (1.004×20) = 0.26 kg / s

11-72 11.95 A refrigerator using R-134a is located in a 20°C room. Consider the cycle to be ideal, except that the compressor is neither adiabatic nor reversible. Saturated vapor at -20°C enters the compressor, and the R-134a exits the compressor at 50°C. The condenser temperature is 40°C. The mass flow rate of refrigerant around the cycle is 0.2 kg/s, and the coefficient of performance is measured and found to be 2.3. Find the power input to the compressor and the rate of entropy generation in the compressor process. Table B.5: P2 = P3 = Psat 40C = 1017 kPa, s2 ≈ 1.7472 , h2 ≈ 430.87 ;

h4 = h3 = 256.54 kJ/kg

s1 = 1.7395 , h1 = 386.08

β = qL / wC -> wC = qL / β = (h1- h4) / β = (386.08 - 256.54) / 2.3 = 56.32 . . WC = m wC = 11.26 kW C.V. Compressor h1 + wC + q = h 2 -> qin = h2 - h1 - wC = 430.87 - 386.08 - 56.32 = -11.53

i.e. a heat loss

s1 + ∫ dQ/T + sgen = s2 sgen = s2 - s1 - q / To = 1.7472 - 1.7395 + (11.53 / 293.15) = 0.047 . . Sgen = m sgen = 0.2 × 0.047 = 0.0094 kW / K 11.96 A small heat pump unit is used to heat water for a hot-water supply. Assume that the unit uses R-22 and operates on the ideal refrigeration cycle. The evaporator temperature is 15°C and the condenser temperature is 60°C. If the amount of hot water needed is 0.1 kg/s, determine the amount of energy saved by using the heat pump instead of directly heating the water from 15 to 60°C. T Ideal R-22 heat pump o o T1 = 15 C, T3 = 60 C 2 h1 = 255.02, s2 = s1 = 0.9062

3

P2 = P3 = 2.427 MPa, h3 = 122.18 T2 = 78.4 oC, h2 = 282.86

4

wC = h2 - h1 = 27.84 qH = h2 - h3 = 160.68 To heat 0.1 kg/s of water from 15 oC to 60 oC, . . QH2O = m(∆h) = 0.1(251.11 - 62.98) = 18.81 kW Using the heat pump . . WIN = QH2O(wC/qH) = 18.81(27.84/160.68) = 3.26 kW a saving of 15.55 kW

1 s

11-73 11.97 The refrigerant R-22 is used as the working fluid in a conventional heat pump cycle. Saturated vapor enters the compressor of this unit at 10°C; its exit temperature from the compressor is measured and found to be 85°C. If the isentropic efficiency of the compressor is estimated to be 70%, what is the coefficient of performance of the heat pump? T R-22 heat pump: 2 o TEVAP = 10 C, ηS COMP = 0.70 2S

o

3

T2 = 85 C Isentropic compressor: s2S = s1 = 0.9129

4

but P2 unknown. Trial & error.

1 s

Assume P2 = 2.11 MPa At P2,s2S: T2S = 72.1oC, h2S = 281.59, At P2,T2 : h2 = 293.78 calculate ηS COMP = (h2S - h1)/(h2 - h1) =

281.59 - 253.42 = 0.698 ≈ 0.70 293.28 - 253.42

OK ⇒ P2 = 2.11 MPa = P3 ⇒ T3 = 53.7 oC, h3 = 112.99, wC = h2 - h1 = 40.36 qH = h2 - h3 = 180.79,

β′ = qH/wC = 4.48

11-74 11.98 In an actual refrigeration cycle using R-12 as the working fluid, the refrigerant flow rate is 0.05 kg/s. Vapor enters the compressor at 150 kPa, −10°C, and leaves at 1.2 MPa, 75°C. The power input to the compressor is measured and found be 2.4 kW. The refrigerant enters the expansion valve at 1.15 MPa, 40°C, and leaves the evaporator at 175 kPa, −15°C. Determine the entropy generation in the compression process, the refrigeration capacity and the coefficient of performance for this cycle. Actual ref. cycle 2 T P1 = 0.15 MPa, P2 = 1.2 MPa 1: comp. inlet

T1 = -10 oC, T2 = 75 oC

3

P3 = 1.15 MPa, P5 = 0.175 MPa

4

5

T3 = 40 oC, T5 = -15 oC 5: evap. exit . . WCOMP = 2.4 kW in, m = 0.05 kg/s Table B.3

1 s

h1 = 184.619, s1 = 0.7318, h2 = 226.543, s2 = 0.7404

CV Compressor: h1 + qCOMP = h2 + wCOMP ;

s1 + ∫ dq/T + sgen = s2

wCOMP = -2.4/0.05 = -48.0 kJ/kg qCOMP = h2 + wCOMP - h1 = 226.5 - 48.0 - 184.6 = -6.1 kJ/kg sgen = s2 - s1 - q / To = 0.7404 - 0.7318 + 6.1/298.15 = 0.029 kJ / kg K b)

qL = h5 - h4 = 181.024 - 74.527 = 106.5 kJ/kg . . ⇒ QL = mqL = 0.05 × 106.5 = 5.325 kW

c)

β = qL/-wCOMP = 106.5/48.0 = 2.219

11.99 Consider a small ammonia absorption refrigeration cycle that is powered by solar energy and is to be used as an air conditioner. Saturated vapor ammonia leaves the generator at 50°C, and saturated vapor leaves the evaporator at 10°C. If 7000 kJ of heat is required in the generator (solar collector) per kilogram of ammonia vapor generated, determine the overall performance of this system. T

NH3 absorption cycle: o sat. vapor at 50 oC exits the generator 50 C o sat. vapor at 10 C exits the evaporator qH = qGEN = 7000 kJ/kg NH3 out of gen. 10 oC

GEN. EXIT

1

2

EVAP EXIT

qL = h2 - h1 = hG 10oC - hF 50oC = 1452.2 - 421.6 = 1030.6 kJ/kg ⇒ qL/qH = 1030.6/7000 = 0.147

s

11-75 11.100 The performance of an ammonia absorption cycle refrigerator is to be compared with that of a similar vapor-compression system. Consider an absorption system having an evaporator temperature of −10°C and a condenser temperature of 50°C. The generator temperature in this system is 150°C. In this cycle 0.42 kJ is transferred to the ammonia in the evaporator for each kilojoule transferred from the high-temperature source to the ammonia solution in the generator. To make the comparison, assume that a reservoir is available at 150°C, and that heat is transferred from this reservoir to a reversible engine that rejects heat to the surroundings at 25°C. This work is then used to drive an ideal vapor-compression system with ammonia as the refrigerant. Compare the amount of refrigeration that can be achieved per kilojoule from the high-temperature source with the 0.42 kJ that can be achieved in the absorption system. o

T

T H = 50 C o

T' H = 150 C

QH

Q' H = 1 kJ WC

REV. H.E.

2

3 3

COND. 2

1

4

COMP.

s

Q L'

o

T1 = -10 C

EVAP. 1

o

T L'= 25 C

QL

4

o

T L = -10 C

h1 = 1430.8 , s1 = 5.4673 h4 = h3 = 421.48

298.2 ′ ′ ηTH = 1 - TL/TH = 1 = 0.295 423.2

For the rev. heat engine:

′ ⇒ WC = ηTH Q = 0.295 kJ H For the NH3 refrig. cycle: s2 = s1 = 5.4673

P2 = P3 = 2033 kPa , =>

T2 ≈ 135°C

Use 2000 kPa Table h2 ≈ 1724

wC = h2 - h1 = 1724 - 1430.8 = 293.2 qL = h1 - h4 = 1430.8 - 421.48 = 1009.3 β = qL/wC = 1009.3 / 293.2 = 3.44 ⇒ QL = βwC = 3.44 × 0.295 = 1.015 kJ based on assumption of ideal heat engine & refrig. cycle.

11-76 11.101 A heat exchanger is incorporated into an ideal air-standard refrigeration cycle, as shown in Fig. P11.101. It may be assumed that both the compression and the expansion are reversible adiabatic processes in this ideal case. Determine the coefficient of performance for the cycle. q 6

T

L

2 3

1

2

qH

5

3

1

4 6

4

COMP

5

EXP

s

T1 = T3 = 15 oC = 288.2 K, P1 = 100 kPa, P2 = 1.4 MPa T4 = T6 = -50 oC = 223.2 K, s2 = s1 ⇒ T2 = T1(P2/P1

k-1 )k

= 288.2(1400/100)0.286 = 613 K

wC = -w12 = CP0(T2 - T1) = 1.0035(613 - 288.2) = 326 kJ/kg k-1

s5 = s4 ⇒ T5= T4(P5/P4) k = 223.2(100/1400)

0.286

= 104.9 K

wE = CP0(T4 - T5) = 1.0035(223.2 - 104.9) = 118.7 kJ/kg wNET = 118.7 - 326.0 = -207.3 kJ/kg qL = CP0(T6 - T5) = 1.0035(223.2 - 104.9) = 118.7 kJ/kg β = qL/wNET = 118.7/207.3 = 0.573 11.102 Repeat Problem 11.101, but assume an isentropic efficiency of 75% for both the compressor and the expander. 2 From solution 11.101 : T2S = 613 K, wSC = 326 kJ/kg 2S T T5S = 104.9 K, wSE = 118.7 ⇒ wC = wSC / ηSC = 326/0.75 = 434.6 kJ/kg wE = ηSE × wSE = 0.75 × 118.7 = 89.0 kJ/kg = CP0(T4-T5) = 1.004(223.2 -T5) ⇒ T5 = 134.5 K wNET = 89.0 - 434.6 = -345.6 kJ/kg qL = CP0(T6 - T5) = 1.004(223.2 - 134.5) = 89.0 kJ/kg β = qL/(-wNET) = 89.0/345.6 = 0.258

3 1

4 6 5S

5

s

11-77 11.103 Repeat Problems 11.101 and 11.102, but assume that helium is the cycle working fluid instead of air. Discuss the significance of the results. a) Problem 11.101 , except for Helium 14000.40  100 0.40 T2 = 288.2  = 828.2 K, T5 = 223.2  = 77.7 K  100  1400 wC = -w12 = 5.1926(828.2 - 288.2) = 2804.1 kJ/kg wE = 5.1926(223.2 - 77.7) = 755.5 kJ/kg wNET = 755.5 - 2804.1 = -2048.6 kJ/kg qL = 5.1926(223.2 - 77.7) = 755.5 kJ/kg β = 755.5/2048.6 = 0.369 b) Problem 11.102 , except for Helium From part a): T2S = 828.2 K, wSC = 2804.1 T5S = 77.7 K, wSE = 755.5 ⇒ wC = wSC/ηSC = 2804.1/0.75 = 3738.8 kJ/kg wE = ηSE × wSE = 0.75 × 755.5 = 566.6 kJ/kg wNET = 566.6 - 3738.8 = -3172.2 kJ/kg wE = 566.6 = 5.1926(223.2 - T5) ⇒ T5 = 114.1 K qL = 5.1926(223.2 - 114.1) = 566.6 kJ/kg β = 566.6/3172.2 = 0.179

11-78 11.104 A binary system power plant uses mercury for the high-temperature cycle and water for the low-temperature cycle, as shown in Fig. 11.39. The temperatures and pressures are shown in the corresponding T–s diagram. The maximum temperature in the steam cycle is where the steam leaves the superheater at point 4 where it is 500°C. Determine the ratio of the mass flow rate of mercury to the mass flow rate of water in the heat exchanger that condenses mercury and boils the water and the thermal efficiency of this ideal cycle. The following saturation properties for mercury are known Tg, °C

P, MPa 0.04 1.60

hf, kJ/kg hg, kJ/kg sf kJ/kgK

309 562

42.21 75.37

335.64 364.04

0.1034 0.1498

sg, kJ/kgK 0.6073 0.4954

a) For the mercury cycle: sd = sc = 0.4954 = 0.1034 + xd × 0.5039, xd = 0.7779 hb = ha - wP HG ≈ ha ( since vF is very small) qH = hc - ha = 364.04 - 42.21 = 321.83 kJ/kg qL = hd - ha = 270.48 - 42.21 = 228.27 kJ/kg For the steam cycle: s5 = s4 = 7.0097 = 0.6493 + x5 × 7.5009, x5 = 0.8480 h5 = 191.83 + 0.848 × 2392.8 = 2220.8 -wP ≈ v1(P2 - P1) = 0.00101(4688 - 10) = 4.7 kJ/kg h2 = h1 - wP = 191.8 + 4.7 = 196.5 qH (from Hg) = h3 - h2 = 2769.9 - 196.5 = 2600.4 qH (ext. source) = h4 - h3 = 3437.4 - 2796.9 = 640.5 CV: Hg condenser - H2O boiler: mHg/mH2O = b)

1st law: mHg(hd - ha) = mH2O(h3 - h2)

2796.9 - 196.5 = 11.392 270.48 - 42.21

qH TOTAL = (mHg/mH2O)(hc - hb) + (h4 - h3) (for 1 kg H2O) = 11.392 × 321.83 + 640.5 = 4306.8 kJ

All qL is from the H2O condenser: qL = h5 - h1 = 2220.8 - 191.8 = 2029.0 kJ wNET = qH - qL = 4306.8 - 2029.0 = 2277.8 kJ ηTH = wNET/qH = 2277.8/4306.8 = 0.529

11-79 11.105 A Rankine steam power plant should operate with a high pressure of 3 MPa, a low pressure of 10 kPa, and the boiler exit temperature should be 500°C. The available high-temperature source is the exhaust of 175 kg/s air at 600°C from a gas turbine. If the boiler operates as a counterflowing heat exchanger where the temperature difference at the pinch point is 20°C, find the maximum water mass flow rate possible and the air exit temperature. T -wP = h2 - h1 = v1(P2 - P1) = 0.00101(3000 - 10) = 3.02 h2 = h1 - wP = 191.83 + 3.02 = 194.85

3 2a

h3 = 3456.5, hair,in = 903.16

2

State 2a: T2a = TSAT = 233.9 °C

1

h2a = 1008.42

s

hair(T2a + 20) = 531.28 Air temperature should be 253.9°C at the point where the water is at state 2a. C.V. Section 2a-3, . i-a . mH O(h3 - h2a) = mair(hi - ha)

e

a i HEAT EXCH

2

2a

3

2

. 903.16 - 531.28 mH O = 175 = 26.584 kg/s 3456.5 - 1008.42 2 . . Take C.V. Total: mH O(h3 - h2) = mair(hi - he) 2 . . ⇒ he = hi - mH O(h3 - h2)/mair 2

= 903.6 - 26.584(3456.5 - 194.85)/175 = 408.13 ⇒ Te = 406.7 K = 133.6 °C, Te > T2 = 46.5 °C OK. 11.106 A simple Rankine cycle with R-22 as the working fluid is to be used as a bottoming cycle for an electrical generating facility driven by the exhaust gas from a Diesel engine as the high temperature energy source in the R-22 boiler. Diesel inlet conditions are 100 kPa, 20°C, the compression ratio is 20, and the maximum temperature in the cycle is 2800°C. Saturated vapor R-22 leaves the bottoming cycle boiler at 110°C, and the condenser temperature is 30°C. The power output of the Diesel engine is 1 MW. Assuming ideal cycles throughout, determine a. The flow rate required in the diesel engine. b. The power output of the bottoming cycle, assuming that the diesel exhaust is cooled to 200°C in the R-22 boiler.

11-80 T

T

3 s

v

7

4

2 s

v

1

DIESEL AIR-STD CYCLE

IDEAL R-12 RANKINE BOTTOMING CYCLE

6 5

8

s

s

P1 = 100 kPa, T1 = 20 oC, T7 = 110 oC, x7 = 1.0 . v1/v2 = 20, T3 = 2800oC, T5 = T8 = 30oC, WDIESEL = 1.0 MW a)

T2 = T1(v1/v2)k-1 = 293.2(20)0.4 = 971.8 K P2 = P1(v1/v2)k = 100(20)1.4 = 6629 kPa qH = CP0(T3 - T2) = 1.004(3073.2 - 971.8) = 2109.8 kJ/kg v1 =

0.287 × 293.2 0.8415 = 0.8415, v2 = = 0.04208 20 100

v3 = v2(T3/T2) = 0.04208(3073.2/971.8) = 0.13307 v3k-1 0.133 070.4  = 1469.6 K T4 = T3  = 3073.2  0.8415  v4 qL = 0.717(293.2 - 1469.6) = -843.5 wNET = 2109.8 - 843.5 = 1266.3 kJ/kg . . mAIR = WNET/wNET = 1000/1266.3 = 0.79 kg/s b)

s8 = s7 = 0.60758 = 0.2399 + x8 × 0.4454, x8 = 0.8255 h8 = 64.59 + 0.8255 × 135.03 = 176.1 wT = h7 - h8 = 198.0 - 176.1 = 21.9 kJ/kg -wP = v5(P6 - P5) = 0.000774(3978.5 - 744.9) = 2.50 h6 = h5 - wP = 64.6 + 2.5 = 67.1 qH = h7 - h6 = 198.0 - 67.1 = 130.9 kJ/kg . QH available from Diesel exhaust cooled to 200 oC: . QH = 0.79 × 0.717(1469.6 - 473.2) = 564 kW . . ⇒ mR-12 = QH/qH = 564/130.9 = 4.309 kg/s . WR-12 = 4.309(21.9 - 2.5) = 83.6 kW

11-81 11.107 For a cryogenic experiment heat should be removed from a space at 75 K to a reservoir at 180 K. A heat pump is designed to use nitrogen and methane in a cascade arrangement (see Fig. 11.41), where the high temperature of the nitrogen condensation is at 10 K higher than the low-temperature evaporation of the methane. The two other phase changes take place at the listed reservoir temperatures. Find the saturation temperatures in the heat exchanger between the two cycles that gives the best coefficient of performance for the overall system. The nitrogen cycle is the bottom cycle and the methane cycle is the top cycle. Both std. refrigeration cycles. THm = 180 K = T 3m ,

TLN = 75 K = T 4N = T1N

TLm = T4m = T1m = T3N - 10, Trial and error on T3N or TLm. For each cycle we have, -wC = h2 - h1, s2 = s1,

-qH = h2 - h3,

qL = h1 - h4 = h1 - h3

Nitrogen: T4 = T1 = 75 K ⇒ h1 = 74.867 s1 = 5.4609 T3 120 115 110

N2 a) b) c)

h3 -17.605 -34.308 -48.446

P2 2.5125 1.9388 1.4672

h2 202.96 188.35 173.88

-wc 128.1 113.5 99.0

-qH 220.57 222.66 222.33

qL 92.47 109.18 123.31

-qH 540.8 581.6 630.2

qL 221.5 212.7 203.4

Methane: T3 = 180 K ⇒ h3 = -0.5 P2 = 3.28655 MPa CH4 a) b) c)

T4 110 105 100

h1 221 212.2 202.9

s1 9.548 9.691 9.851

h2 540.3 581.1 629.7

-wc 319.3 368.9 426.8

The heat exchanger that connects the cycles transfers a Q . . . . . . QHn = qHn mn = QLm = qLm mm => mm/mn = qHn/qLm The overall unit then has . . . . . QL 75 K = mn qLn ; Wtot in = - (mnwcn + mmwcm) . . . . β = QL 75 K/Wtot in = qLn/[-wcn -(mm/mn)wcm] Case a) b) c)

. . mm/mn 0.996 1.047 1.093

. . wcn+(mm/mn)wcm 446.06 499.65 565.49

β 0.207 0.219 0.218

A maximum coeff. of performance is between case b) and c).

11-82 11.108 A cascade system is composed of two ideal refrigeration cycles, as shown in Fig. 11.41. The high-temperature cycle uses R-22. Saturated liquid leaves the condenser at 40°C, and saturated vapor leaves the heat exchanger at −20°C. The low-temperature cycle uses a different refrigerant, R-23 (Fig. G.3 or the software). Saturated vapor leaves the evaporator at −80°C, and saturated liquid leaves the heat exchanger at −10°C. Calculate the ratio of the mass flow rates through the two cycles and the coefficient of performance of the system. COND

2'

3'

T3' = 40o C x '= 0.0 3

C

R-22

o

T1' = -20 C 1' x1'= 1.0

4'

2

3

T3 = -10o C x = 0.0 3

T1 = -80o C 1 x1 = 1.0

R-23

C

4

EVAP

T

T 2'

2

3'

3

4'

1

4

1' s

1′ 2′ 3′ 4′

T,oC -20 71 40 -20

s

P

h

s

0.245 1.534 1.534

242.1 289.0 94.3 94.3

0.9593 0.9593

1 2 3 4

T,oC -80 50 -10 -80

P

h

s

0.12 1.90 1.90 0.12

330 405 185 185

1.76 1.76

h′1 - h′4 242.1 - 94.3 . . a) m/m′ = = = 0.672 405 - 185 h2 - h3 . . b) QL/m = h1 - h4 = 330 - 185 = 145 . . . . - WTOT/m = (h2 - h1) + (m′/m)(h′2 - h′1) = (405 - 330) + (1/0.672)(289 - 242.1) = 144.8 . β = QL/-WTOT = 145/144.8 = 1.0

11-83 11.109 Consider an ideal dual-loop heat-powered refrigeration cycle using R-12 as the working fluid, as shown in Fig. P11.109. Saturated vapor at 105°C leaves the boiler and expands in the turbine to the condenser pressure. Saturated vapor at − 15°C leaves the evaporator and is compressed to the condenser pressure. The ratio of the flows through the two loops is such that the turbine produces just enough power to drive the compressor. The two exiting streams mix together and enter the condenser. Saturated liquid leaving the condenser at 45°C is then separated into two streams in the necessary proportions. Determine the ratio of mass flow rate through the power loop to that through the refrigeration loop. Find also the performance of the cycle, in terms of the ratio Q /Q L H. T

1 TURB.

COMP.

7

2

6 BOIL.

COND.

5

3 P

E V A P .

. QL

6 5

2

3

7 4

4

1 s

o

T3 = 45 C sat. liq. T1 = -15 oC sat. vap.

P5 = P6 = 3.6509 MPa; T6 = 105 oC sat. vap.; P2 = P3 = P7 = 1.0843 MPa h1 = 180.846; h3 = h4 = 76.647; h6 = 206.099 s7 = s6 = 0.6319 = 0.2875 + x7 × 0.3931; x7 = 0.8761 h7 = 79.647 + 0.8761 × 125.074 = 189.226 s2 = s1 = 0.7046, P2 => T2 = 54.7 oC, h2 = 212.401 a) CV: turbine + compressor . . . . . . . . Cont: m1 = m2, m6 = m7 ; 1st law: m1h1 + m6h6 = m2h2 + m7h7 . . m6/m1 = (212.401 - 180.846)/(206.099 - 189.226) = 1.870 b) CV: pump -wP = v3(P5 - P3) = 0.000811(3651 - 1084) = 2.082 kJ/kg h5 = h3 - wP = 81.729 . . CV: evaporator ⇒ QL = m1(h1 - h4) . . CV: boiler ⇒ QH = m6(h6 - h5) . . QL m1(h1 - h4) 180.846 - 79.647 = = 0.435 β= . = . QH m6(h6 - h5) 1.87(206.099 - 81.729)

11-84 11.110 Find the availability of the water at all four states in the Rankine cycle described in Problem 11.12. Assume that the high-temperature source is 500°C and the lowtemperature reservoir is at 25°C. Determine the flow of availability in or out of the reservoirs per kilogram of steam flowing in the cycle. What is the overall cycle second law efficiency? Reference State: 100 kPa, 25°C, s0 = 0.3674, h0 = 104.89 ψ1 = h1 - h0 - T0(s1 - s0) = 191.83 - 104.89 - 298.15(0.6493 - 0.3674) = 2.89 ψ2 = 195.35 - 104.89 - 298.15(0.6493 - 0.3674) = ψ1 + 3.525 = 6.42 ψ3 = 3222.3 - 104.89 - 298.15(6.8405 - 0.3674) = 1187.5 ψ4 = ψ3 - wT,s = 131.96 ∆ψ H = (1 - T0/TH)qH = 0.6144 × 3027 = 1859.7 ∆ψ L = (1 - T0/T0)qC = 0/ ηII = wNET/∆ψH = (1055.5 - 3.53)/1859.7 = 0.5657 Notice— TH > T3, TL < T4 = T1 so cycle is externally irreversible. Both qH and qC over finite ∆T.

11-85 11.111 The effect of a number of open feedwater heaters on the thermal efficiency of an ideal cycle is to be studied. Steam leaves the steam generator at 20 MPa, 600°C, and the cycle has a condenser pressure of 10 kPa. Determine the thermal efficiency for each of the following cases. A: No feedwater heater. B: One feedwater heater operating at 1 MPa. C: Two feedwater heaters, one operating at 3 MPa and the other at 0.2 MPa. a) no feed water heater 3

2

wP = ⌠ ⌡ vdP 1

≈ 0.00101(20000 - 10) = 20.2 kJ/kg h2 = h1 + wP = 191.8 + 20.2 = 212.0

ST. GEN.

TURBINE.

4

COND.

s4 = s3 = 6.5048

2

= 0.6493 + x 4 × 7.5009

1

P

x4 = 0.78064 h4 = 191.83 + 0.780 64 × 2392.8

20 MPa o 3 600 C

T

= 2059.7 wT = h3 - h4 = 3537.6 - 2059.7 = 1477.9 kJ/kg wN = wT - wP = 1477.9 - 20.2 = 1457.7

10 kPa

2

qH = h3 - h2 = 3537.6 - 212.0 = 3325.6

4 1 s

ηTH =

wN qH

=

1457.7 = 0.438 3325.6

b) one feedwater heater wP12 = 0.00101(1000 - 10) = 1.0 kJ/kg h2 = h1 + wP12 = 191.8 + 1.0 = 192.8

5

ST. GEN.

TURBINE. 6

wP34 = 0.001127 (20000- 1000)

HTR.

= 21.4 kJ/kg h4 = h3 + wP34 = 762.8 + 21.4 = 784.2 s6 = s5 = 6.5048 = 2.1387 + x6 × 4.4478

7

3

COND. 1

P 4

2

P

11-86 x6 = 0.9816

20 MPa

T

o

h6 = 762.8 + 0.9816 × 2015.3 = 2741.1

1 MPa

CV: heater const: m3 = m6 + m2 = 1.0 kg

4

1st law: m6h6 + m2h2 = m3h3 m6 =

600 C

5

6

2 3 1

762.8 - 192.8 = 0.2237 2741.1 - 192.8

10 kPa

7 s

m2 = 0.7763, h7 = 2059.7 ( = h4 of part a) ) CV: turbine

wT = (h5 - h6) + m2(h6 - h7)

= (3537.6 - 2741.1) + 0.7763(2741.1 - 2059.7) = 1325.5 kJ/kg CV: pumps wP = m1wP12 + m3wP34 = 0.7763(1.0) + 1(21.4) = 22.2 kJ/kg wN = 1325.5 - 22.2 = 1303.3 kJ/kg CV: steam generator qH = h5 - h4 = 3537.6 - 784.2 = 2753.4 kJ/kg ηTH = wN/qH = 1303.3/2753.4 = 0.473 c) two feedwater heaters wP12 = 0.00101 × (200 - 10) = 0.2 kJ/kg h2 = wP12 + h1

7

= 191.8 + 0.2 = 192.0 wP34 = 0.001061 ×

8

9 10

HP HTR

(3000 - 200) = 3.0 kJ/kg h4 = h3 + wP34 = 504.7 + 3.0 = 507.7

TURBINE.

ST. GEN.

LP HTR 1

5

3

P 6

COND.

P 4

P 2

11-87 T

wP56 = 0.001217(20000 - 3000)

80 MPa

o

600 C

7

= 20.7 kJ/kg h6 = h5 + wP56 = 1008.4 + 20.7 = 1029.1

3 MPa 6

s8 = s7 = 6.5048 T8 = 293.2 oC  at P8 = 3 MPa  h8 = 2974.8 s9 = s8 = 6.5048 = 1.5301 + x9 × 5.5970

4 5 2 3 1

8 0.2 MPa 10 kPa 9 10 s

x9 = 0.8888 => h9 = 504.7 + 0.888 × 2201.9 = 2461.8 CV: high pressure heater cont:

m5 = m4 + m8 = 1.0 kg ;

m8 =

1st law: m5h5 = m4h4 + m8h8

1008.4 - 507.7 = 0.2030 2974.8 - 507.7

m4 = 0.7970

CV: low pressure heater cont:

m9 + m2 = m3 = m4 ;

m9 =

1st law: m9h9 + m2h2 = m3h3

0.7970(504.7 - 192.0) = 0.1098 2461.8 - 192.0

m2 = 0.7970 - 0.1098 = 0.6872 CV: turbine wT = (h7 - h8) + (1 - m8)(h8 - h9) + (1 - m8 - m9)(h9 - h10) = (3537.6 - 2974.8) + 0.797(2974.8 - 2461.8) + 0.6872(2461.8 - 2059.7) = 1248.0 kJ/kg CV: pumps wP = m1wP12 + m3wP34 + m5wP56 = 0.6872(0.2) + 0.797(3.0) + 1(20.7) = 23.2 kJ/kg wN = 1248.0 - 23.2 = 1224.8 kJ/kg CV: steam generator qH = h7 - h6 = 3537.6 - 1029.1 = 2508.5 kJ/kg ηTH = wN/qH = 1224.8/2508.5 = 0.488

11-88 11.112 Find the availability of the water at all the states in the steam power plant described in Problem 11.36. Assume the heat source in the boiler is at 600°C and the low-temperature reservoir is at 25°C. Give the second law efficiency of all the components. From solution to 11.1 and 11.36 : States

1 - 2a - 3 - 4a - 1

1:

hf = 191.81,

3:

h3 = 2803.43,

s3 = 6.1252

4a:

h4a = 2068.2,

s4a = 6.5313,

ψ = h - ho - To(s - so)

sf = 0.6492

2a: h2a = 196.2,

s2a ≈ 0.6628

x4a = 0.7842

ho = 104.89, so = 0.3674

ψ1= 191.81 - 104.89 - 298.15(0.6492 - 0.3674) = 2.90 ψ2a = 196.2 - 104.89 - 298.15(0.6628 - 0.3674) = 3.24 ψ3 = 2803.43 - 104.89 - 298.15(6.1252 - 0.3674) = 981.85 ψ4a = 2068.2 - 104.89 - 298.15(6.5313 - 0.3674) = 125.54 ηII Pump = (ψ2a - ψ1) / wp ac = (3.24 - 2.9) / 4.406 = 0.077 ηII Boiler = (ψ3 - ψ2a) / [(1- To/TH) qH] = (981.85 - 3.24) / [0.658×2607.2] = 0.57 ηII Turbine = wT ac / (ψ3 - ψ4a) = 735.2 / (981.85 - 125.54) = 0.859 ηII Cond = ∆ψamb / (ψ4a - ψ1) = 0

11-89 11.113 The power plant shown in Fig. 11.40 combines a gas-turbine cycle and a steamturbine cycle. The following data are known for the gas-turbine cycle. Air enters the compressor at 100 kPa, 25°C, the compressor pressure ratio is 14, and the isentropic compressor efficiency is 87%; the heater input rate is 60 MW; the turbine inlet temperature is 1250°C, the exhaust pressure is 100 kPa, and the isentropic turbine efficiency is 87%; the cycle exhaust temperature from the heat exchanger is 200°C. The following data are known for the steam-turbine cycle. The pump inlet state is saturated liquid at 10 kPa, the pump exit pressure is 12.5 MPa, and the isentropic pump efficiency is 85%; turbine inlet temperature is 500°C and the isentropic turbine efficiency is 87%. Determine a. The mass flow rate of air in the gas-turbine cycle. b. The mass flow rate of water in the steam cycle. c. The overall thermal efficiency of the combined cycle. .

QH = 60 MW P2 /P1 = 14

HEAT 2

3

T3 = 1250o C

η = 0.87 SC

.

GAS TURB

COMP

W NET CT P4 = 100 kPa

AIR 5

1

4

ST

P 1= 100 kPa T1 = 25o C T5 = 200 oC HEAT EXCH

o

7 6

SP

T7 = 500 C

P6 = P7 = 12.5 MPa

.

9

P

WST

STEAM TURB

H 2O

η = 0.85

η = 0.87

ηST= 0.87 8

COND

a) From Air Tables, A.7: Pr1 = 1.0913,

P8 = P9 = 10 kPa h1 = 298.66,

h5 = 475.84

s2 = s1 ⇒ Pr2S = Pr1(P2/P1) = 1.0913 × 14 = 15.2782 T2S = 629 K, h2S = 634.48 wSC = h1 - h2S = 298.66 - 634.48 = -335.82 kJ/kg wC = wSC/ηSC = -335.82/0.87 = -386 = h1 - h2 At T3 = 1523.2 K: Pr3 = 515.493, h3 = 1663.91

⇒ h2 = 684.66

11-90 . . mAIR = QH/(h3 - h2) = b)

60 000 = 61.27 kg/s 1663.91 - 684.66

Pr4S = Pr3(P4/P3) = 515.493(1/14) = 36.8209 => T4S = 791 K, h4S = 812.68 wST = h3 - h4S = 1663.91 - 812.68 = 851.23 wT = ηST × wST = 0.87 × 851.23 = 740.57 = h 3 - h4 => h4 = 923.34 kJ/kg Steam cycle: -wSP ≈ 0.00101(12500 - 10) = 12.615 -wP = - wSP/ηSP = 12.615/0.85 = 14.84 h6 = h9 - wP = 191.83 + 14.84 = 206.67 At 12.5 MPa, 500 oC: h7 = 3341.7, s7 = 6.4617 h4 - h5 923.34 - 475.84 . = m = 61.27 = 8.746 kg/s O AIR h - h 3341.7 - 206.67 7 6 2

. mH

c) s8S = s7 = 6.4617 = 0.6492 + x8S × 7.501, x8S = 0.7749 h8S = 191.81 + 0.7749 × 2392.8 = 2046.0 wST = h7 - h8S = 3341.7 - 2046.0 = 1295.7 wT = ηST × wST = 0.87 × 1295.7 = 1127.3 kJ/kg . . . WNET = m(wT+wC)AIR +m(wT+wP)H O 2

= 61.27(740.57 - 386.0) + 8.746(1127.3 - 14.84) = 21725 + 9730 = 31455 kW = 31.455 MW . . ηTH = WNET/QH = 31.455/60 = 0.524

11-91 11.114 For Problem 11.105, determine the change of availability of the water flow and that of the air flow. Use these to determine a second law efficiency for the boiler heat exchanger. From solution to 11.105 : . mH2O = 26.584 kg/s, h2 = 194.85, s2 = 0.6587 h3 = 3456.5, s3 = 7.2338, s° = 7.9820, s°Te = 7.1762 Ti hi = 903.16, he = 408.13 ψ3 - ψ2 = h3 - h2 - T0(s3 - s2) = 1301.28 kJ/kg ψi - ψe = hi - he - T0(s°Ti - s°Te) = 254.78 kJ/kg . (ψ3 - ψ2)mH2O 1301.28 × 26.584 . ηII = = = 0.776 (ψi - ψe)mair 254.78 × 175

11-92 11.115 One means of improving the performance of a refrigeration system that operates over a wide temperature range is to use a two-stage compressor. Consider an ideal refrigeration system of this type that uses R-12 as the working fluid, as shown in Fig. P11.115. Saturated liquid leaves the condenser at 40°C and is throttled to −20°C. The liquid and vapor at this temperature are separated, and the liquid is throttled to the evaporator temperature, −70°C. Vapor leaving the evaporator is compressed to the saturation pressure corresponding to −20°C, after which it is mixed with the vapor leaving the flash chamber. It may be assumed that both the flash chamber and the mixing chamber are well insulated to prevent heat transfer from the ambient. Vapor leaving the mixing chamber is compressed in the second stage of the compressor to the saturation pressure corresponding to the condenser temperature, 40°C. Determine a. The coefficient of performance of the system. b. The coefficient of performance of a simple ideal refrigeration cycle operating over the same condenser and evaporator ranges as those of the two-stage compressor unit studied in this problem. ROOM . +QH COND

5

T 5 1

SAT.LIQ. o 40 C

40 C o

-20 C

COMP. ST.2 4

o

1 6 2

o

SAT.VAP. o -20 C FLASH MIX.CHAM

-70 C

2

6 SAT.LIQ. o

9

-20 C

COMP. ST.1 EVAP .

R-12 ref. with 2-stage compression

7

-QL

SAT.VAP. o -70 C

9 8 s

CHAMBER

3

8

7

4 3

COLD SPACE

CV: expansion valve, upper loop h2 = h1 = 74.527 = 17.8 + x2 × 160.81; x2 = 0.353 m3 = x2m2 = x2m1 = 0.353 kg ( for m1=1 kg) m6 = m1 - m3 = 0.647 kg CV: expansion valve, lower loop h7 = h6 = 17.8 = -26.1 + x7 × 181.64, x7 = 0.242

11-93 QL = m3(h8 - h7) = 0.647(155.536 - 17.8) qL = 89.1 kJ/kg-m1 CV: 1st stage compressor s8 = s9 = 0.7744, P9 = PSAT -20 oC= 0.1509 MPa ⇒ T9 = 9 oC, h9 = 196.3 kJ/kg CV: mixing chamber (assume const. press. mixing) 1st law: m6h9 + m3h3 = m1h4 or h4 = 0.647 × 196.3 + 0.353 × 178.61 = 190.06 kJ/kg h4, P4 = 0.1509 MPa ⇒ T4 = -1.0 oC, s4 = 0.7515 CV: 2nd stage compressor P4 = 0.1509 MPa = P9 = P3 P5 = Psat 40oC = 0.9607 MPa, s5 = s4 ⇒ T5 = 70oC, h5 = 225.8 CV: condenser 1st law: -qH = h1 - h5 = 74.527 - 225.8 = -151.27 β 2 stage = qL/(qH - qL) = 89.1/(151.27 - 89.1) = 1.433 b) 1 stage compression h3 = h4 = 74.527

T

h1 = 155.536 qL = h1 - h4 = 81.0 kJ/kg s1 = s2 = 0.7744   P2 = 0.9607 MPa

2 40o C

3

o

-70 C

4

⇒ T2 = 80.9 oC, h2 = 234.0 qH = h2 - h3 = 234.0 - 74.527 = 159.47 kJ/kg β 1 stage = qL/(qH - qL) = 81.0/(159.47 - 81.0) = 1.032

1 s

11-94 11.116 A jet ejector, a device with no moving parts, functions as the equivalent of a coupled turbine-compressor unit (see Problems 9.82 and 9.90). Thus, the turbinecompressor in the dual-loop cycle of Fig. P11.109 could be replaced by a jet ejector. The primary stream of the jet ejector enters from the boiler, the secondary stream enters from the evaporator, and the discharge flows to the condenser. Alternatively, a jet ejector may be used with water as the working fluid. The purpose of the device is to chill water, usually for an air-conditioning system. In this application the physical setup is as shown in Fig. P11.116. Using the data given on the diagram, evaluate the performance of this cycle in terms of the ratio Q /Q . L H a. Assume an ideal cycle. b. Assume an ejector efficiency of 20% (see Problem 9.90). T

2

VAP o 150 C

2 JET EJECT.

BOIL. . QH

3 11

30 C HP P.

10

1

3

4

9 5,10 8 7 6

VAP o 10 C

COND. o

11

1'

2' 1 s

4

o

5

FLASH CH.

6

9 o

LIQ o 10 C

7

20 C

CHILL . QL

8

LP P.

T1 = T7 = 10 C T2 = 150 oC T4 = 30 oC T9 = 20 oC Assume T5 = T10

(from mixing streams 4 & 9). P3 = P4 = P5 = P8 = P9 = P10= PG 30 oC = 4.246 kPa P11 = P2 = PG 150oC = 475.8 kPa, P1 = P6 = P7 = PG 10oC = 1.2276 kPa . . . . . . . . Cont: m1 + m9 = m5 + m10, m5 = m6 = m7 + m1 . . . . . . . . m7 = m8 = m9, m10 = m11 = m2, m3 = m4 . . . a) m1 + m2 = m3; ideal jet ejector s′1 = s1 & s′2 = s2 (1' & 2' at P3 = P4) . . then, m1(h′1 - h1) = m2(h2 - h′2) From s′2 = s2 = 0.4369 + x′2 × 8.0164; x′2 = 0.7985 h′2 = 125.79 + 0.7985 × 2430.5 = 2066.5

11-95 From s′1 = s1 = 8.9008 ⇒ T′1 = 112 °C, h′1 = 2710.4 2746.5 - 2066.5 . . ⇒ m1/m2 = = 3.5677 2710.4 - 2519.8 Also h4 = 125.79, h7 = 42.01, h9 = 83.96 Mixing of streams 4 & 9 ⇒ 5 & 10: . . . . . . (m1 + m2)h4 + m7h9 = (m7 + m1 + m2)h5 = 10 . . . . Flash chamber (since h6 = h5) : (m7+m1)h5 = 10 = m1h1 + m7h1 . ⇒ using the primary stream m2 = 1 kg/s: . . 4.5677 × 125.79 + m7 × 83.96 = (m7 + 4.5677)h5 . . & (m7 + 3.5677)h5 = 3.5677 × 2519.8 + m7 × 42.01 . Solving, m7 = 202.627 & h5 = 84.88 LP pump: -wLP P = 0.0010(4.246 - 1.2276) = 0.003 h8 = h7 - wLP P = 42.01 + 0.003 = 42.01 . . . Chiller: QL = m7(h9-h8) = 202.627(83.96 - 42.01) = 8500 kW (for m2 = 1) HP pump: -wHP P = 0.001002(475.8 - 4.246) = 0.47 h11 = h10 - wHP P = 84.88 + 0.47 = 85.35 . . Boiler: Q11 = m11(h2 - h11) = 1(2746.5 - 85.35) = 2661.1 kW . . ⇒ QL/QH = 8500/2661.1 = 3.194 . . . . b) Jet eject. eff. = (m1/m2)ACT/(m1/m2)IDEAL = 0.20 ⇒

(m. 1/m. 2)ACT = 0.2 × 3.5677 = 0.7135

. . . using m2 = 1 kg/s: 1.7135 × 125.79 + m7 × 83.96 = (m7 + 1.7135)h5 . . & (m7 + 0.7135)h5 = 0.7135 × 2519.8 + m7 × 42.01 . Solving, m7 = 39.762 & h5 = h10 = 85.69 . Then, QL = 39.762(83.96 - 42.01) = 1668 kW h11 = 85.69 + 0.47 = 86.16 . QH = 1(2746.5 - 86.16) = 2660.3 kW . . & QL/QH = 1668/2660.3 = 0.627

11-96

English Unit Problems 11.117E A steam power plant, as shown in Fig. 11.3, operating in a Rankine cycle has saturated vapor at 600 lbf/in.2 leaving the boiler. The turbine exhausts to the condenser operating at 2 lbf/in.2. Find the specific work and heat transfer in each of the ideal components and the cycle efficiency. 1: h1 = 93.73, v1 = 0.01623, 3: h3 = hg = 1204.06, s3 = sg = 1.4464 144 = 1.8 Btu/lbm C.V. Pump: -wP = ∫ v dP = v1(P2 - P1) = 0.01623(600 - 2) 778 h2 = h1 - wP = 93.73 + 1.8 = 95.81 Btu/lbm C.V. Boiler: qH = h3 - h2 = 1204.06 - 95.53 = 1108.53 Btu/lbm C.V. Tubine: wT = h3 - h4, s4 = s3 s4 = s3 = 1.4464 = 0.1744 + x4 × 1.7461

=> x 4 = 0.7285,

h4 = 93.73 + 0.7285 × 1022.2 = 838.42 wT = 1204.06 - 838.42 = 365.63 Btu/lbm ηCYCLE = (wT + wP)/qH = (365.63 - 1.8)/1108.53 = 0.33 C.V. Condenser:

qL = h4 - h1 = 838.42 - 93.73 = 744.69 Btu/lbm

11.118E Consider a solar-energy-powered ideal Rankine cycle that uses water as the working fluid. Saturated vapor leaves the solar collector at 350 F, and the condenser pressure is 1 lbf/in. 2. Determine the thermal efficiency of this cycle. T

H2O ideal Rankine cycle CV: turbine s4 = s3 = 1.5793 = 0.13266 + x4 × 1.8453 x4 = 0.784

3 2 1

4

h4 = 69.74 + 0.784 × 1036.0 = 881.9 wT = h3 - h4 = 1193.1 - 881.9 = 311.2 Btu/lbm -wP = ∫ vdP ≈ v1(P2 - P1) = 0.016136(67 - 1) 144 / 778 = 0.2 ⇒ wNET = wT + wP = 311.2 - 0.2 = 311 h2 = h1 - wP = 69.7 + 0.2 = 69.9 Btu/lbm qH = h3 - h2 = 1193.1 - 69.9 = 1123.2 Btu/lbm ηTH = wNET/qH = 311/1123.2 = 0.277

s

11-97 11.119E A supply of geothermal hot water is to be used as the energy source in an ideal Rankine cycle, with R-134a as the cycle working fluid. Saturated vapor R-134a leaves the boiler at a temperature of 180 F, and the condenser temperature is 100 F. Calculate the thermal efficiency of this cycle. T

a) From the R-134a tables, h1 = 108.86 v1 = 0.01387

D

P1 = 138.93 P2 = P3 = 400.4 h3 = 184.36 s3 = 0.402

2 1

3

4

CV. Pump: -wP = v1(P2 - P1) = 0.01387(400.4-138.93)

180 F 100 F s

144 = 0.671 Btu/lbm = h2 - h1 778

h2 = h1 - wP = 108.86 + 0.671 = 109.53 CV: Turbine s4 = s3 ⇒ x4 = (0.402 - 0.2819)/0.1272 = 0.9442 h4 = 176.08, wT = h3 - h4 = 8.276 btu/lbm CV: Boiler qH = h3 - h2 = 184.36 - 109.53 = 74.83 Btu/lbm ηTH = (wT + w12)/qH = (8.276 - 0.671)/74.83 = 0.102 11.120E Do Problem 11.119 with R-22 as the working fluid. Same diagram as in problem 11.119, now from the R-22 tables, h1 = 39.267, v1 = 0.01404, P1 = 210.6, P2 = P3 = 554.8, h3 = 110.07, s3 = 0.1913 CV: Pump

144 = 0.894 Btu/lbm -wP = v1(P2-P1) = -0.01404 (554.8-210.6) 778 h2 = h1 - wP = 39.267 + 0.894 = 40.16

CV: Turbine s4 = s3 ⇒ x4 = (0.1913 - 0.07942)/0.13014 = 0.9442 h4 = 101.885, wT = h3 - h4 = 8.185 Btu/lbm CV: Boiler qH = h3 - h2 = 110.07 - 40.16 = 69.91 Btu/lbm ηTH = (wT + w12)/qH = (8.185 - 0.894)/157.21 = 0.104

11-98 11.121E The power plant in Problem 11.117 is modified to have a superheater section following the boiler so the steam leaves the super heater at 600 lbf/in.2, 700 F. Find the specific work and heat transfer in each of the ideal components and the cycle efficiency. h3 = 1350.6, s3 = 1.5871, h1 = 94.01, v1 = 0.01623 ⌠v dP = v (P - P ) C.V. Pump: -wP = ⌡ 2 2 1 = 0.01623(600 - 2)(144 / 778) = 1.8 Btu/lbm h2 = h1 - wP = 95.81 Btu/lbm C.V. Boiler: qH = h3 - h2 = 1350.6 - 95.81 = 1254.79 Btu/lbm C.V. Tubine: wT = h3 - h4, s4 = s3 ⇒ x4 = 0.8093, h4 = 921.23 wT = 1350.6 - 921.23 = 429.37 Btu/lbm ηCYCLE = (wT + wP)/qH = (429.37 - 1.8)/1254.79 = 0.341 C.V. Condenser: qL = h4 - h1 = 921.23 - 94.01 = 827.22 Btu/lbm

11.122E Consider a simple ideal Rankine cycle using water at a supercritical pressure. Such a cycle has a potential advantage of minimizing local temperature differences between the fluids in the steam generator, such as the instance in which the high-temperature energy source is the hot exhaust gas from a gasturbine engine. Calculate the thermal efficiency of the cycle if the state entering the turbine is 3500 lbf/in.2, 1100 F, and the condenser pressure is 1 lbf/in.2. What is the steam quality at the turbine exit? s4 = s3 = 1.5193 = 0.1326 + x4×1.8453 x4 = 0.7515

3500 lbf/in2 3 1100 F

T

very low for a turbine exhaust h4 = 848.3, h1 = 69.73, h3 = 1496 s2 = s1 = 0.1326 => h2 = 80.13 qH = h3 - h2 = 1415.9 Btu/lbm

1 lbf/in2 2 1

4 s

wnet = h3 - h4 - (h2 - h1) = 637.3 Btu/lbm η = wnet/qH = 637.3/1415.9 = 0.45

11-99 11.123E Consider an ideal steam reheat cycle in which the steam enters the highpressure turbine at 600 lbf/in.2, 700 F, and then expands to 120 lbf/in.2. It is then reheated to 700 F and expands to 2 lbf/in.2 in the low-pressure turbine. Calculate the thermal efficiency of the cycle and the moisture content of the steam leaving the low-pressure turbine. Basic cycle as in 11.121 plus reheat. From solution 11.121: -wP = 1.8 Btu/lbm, h2 = 95.81

T 3 5

h3 = 1350.6, s3 = 1.5871

2

qH1 = h3-h2 = 1350.6-95.81 = 1254.79

4 1

6

CV T1: wT1 = h3 - h4; s4 = s3 => x4 = 0.9986, h4 = 1190 Btu/lbm

s

CV T2: wT2 = h5 - h6; s6 = s5 = 1.7825, h5 = 1378.2 => x 6 = 0.9213, h6 = 1035.7, wT2 = 1378.2 - 1035.7 = 342.5 wT,tot = wT1 + wT2 = 160.6 + 342.5 = 503.1 Btu/lbm qH = qH1 + h5 - h4 = 1254.79 + 1378.2 - 1190 = 1443 Btu/lbm ηCYCLE = wT,tot/qH = 503.1/1443 = 0.349

11.124E A closed feedwater heater in a regenerative steam power cycle heats 40 lbm/s of water from 200 F, 2000 lbf/in.2 to 450 F, 2000 lbf/in.2. The extraction steam from the turbine enters the heater at 500 lbf/in.2, 550 F and leaves as saturated liquid. What is the required mass flow rate of the extraction steam? 3 1

2

From the steam tables: h1 = 172.6 h2 = 431.14 all h3 = 1266.6 h4 = 449.5 Btu/lbm

4

C.V. Feedwater Heater . . m3 = m1(h1 - h2)/(h4 - h3) = 40

lbm 172.6 - 431.14 = 12.656 s 449.5 - 1266.6

11-100 11.125E Consider an ideal steam regenerative cycle in which steam enters the turbine at 600 lbf/in.2, 700 F, and exhausts to the condenser at 2 lbf/in.2. Steam is extracted from the turbine at 120 lbf/in. 2 for an open feedwater heater. The feedwater leaves the heater as saturated liquid. The appropriate pumps are used for the water leaving the condenser and the feedwater heater. Calculate the thermal efficiency of the cycle and the net work per pound-mass of steam.

5

From solution 11.121: h5 = 1350.6, s5 = 1.5871 h7 = 921.23, h1 = 94.01 2nd law: s6 = s5 =>

TURBINE.

ST. GEN.

6 x6 = 0.999

7

4 FW HTR

h6 = 1190 Btu/lbm P

Also: h3 = 312.66

COND. P

2

1

3

T wP12 = 0.01623(120 - 2)144/778 = 0.35

600 psi 5

wP34 = 0.0179(600 - 120)144/778 = 1.59 ⇒ h2 = 94.36,

120 psi

4 3

2

h4 = 314.25

6 7

1 CV: feedwater heater cont : m3 = m6 + m2 = 1 lbm,

2 psi s

1st law: m6h6 + m2h2 = m3h3

m6(1190) + (1 - m6)(94.36) = 312.66 ⇒ m6 = 0.1992, m2 = 0.8008 CV: turbine wT = (h5 - h6) + (1 - m6)(h6 - h7) = (1350.6 - 1190) + 0.8008(1190 - 921.2) = 375.9 Btu/lbm CV: pumps wP = m2wP12 + m3wP34 = 0.8008 × 0.35 + 1 × 1.59 = 1.87 Btu/lbm wNET = wT - wP = 375.9 - 1.87 = 374 Btu/lbm CV: steam generator qH = h5 - h4 = 1350.6 - 314.25 = 1036.35 Btu/lbm ηTH = wNET/qH = 374/1036.35 = 0.361

11-101 11.126E Consider an ideal combined reheat and regenerative cycle in which steam enters the high-pressure turbine at 500 lbf/in.2, 700 F, and is extracted to an open feedwater heater at 120 lbf/in.2 with exit as saturated liquid. The remainder of the steam is reheated to 700 F at this pressure, 120 lbf/in.2, and is fed to the lowpressure turbine. The condenser pressure is 2 lbf/in.2. Calculate the thermal efficiency of the cycle and the net work per pound-mass of steam.

5: h5 = 1356.66,

s5 = 1.6112

7: h7 = 1378.17,

s7 = 1.7825

7 5

T1

3: h3 = hf = 312.59, v3 = 0.01788 s 5 = s6

6

1-x

C.V. T1

= 146.9 Btu/lbm

8

x

=> h6 = 1209.76

wT1 = h5 - h6 = 1356.66 - 1209.76

T2

4

HTR

P

1-x

COND.

2 3

1

P

C.V. Pump 1 -wP1 = h2 - h1 = v1(P2 - P1) = 0.01623(120 - 2) = 0.354 => h 2 = h1 - wP1 = 93.73 + 0.354 = 94.08 Btu/lbm

T

700 F

C.V. FWH x h6 + (1 - x) h2 = h3

5 7

4

h3 - h2

312.59 - 94.08 x= = = 0.1958 1209.76 - 94.08 h6 - h2 C.V. Pump 2

6

2 psi

2 3 1

8 s

-wP2 = h4 - h3 = v3(P4 - P3) = 0.01788(500 - 120)(144/778) = 1.26 Btu/lbm => h 4 = h3 - wP2 = 312.59 + 1.26 = 313.85 Btu/lbm qH = h5 - h4 + (1 - x)(h7 - h6 ) = 1042.81 + 135.43 = 1178.2 Btu/lbm C.V. Turbine 2 s 7 = s8

=> x8 = (1.7825 - 0.1744)/1.746 = 0.921

h8 = hf + x8 hfg = 93.73 + 0.921 × 1022.2 = 1035.2 wT2 = h7 - h8 = 1378.17 - 1035.2 = 342.97 wnet = wT1 + (1 - x) wT2 + (1 - x) wP1 + wP2 = 146.9 + 275.8 - 0.285 - 1.26 = 421.15 kJ/kg ηcycle = wnet / qH = 421.15 / 1178.2 = 0.357

11-102 11.127E A steam power cycle has a high pressure of 500 lbf/in. 2 and a condenser exit temperature of 110 F. The turbine efficiency is 85%, and other cycle components are ideal. If the boiler superheats to 1400 F, find the cycle thermal efficiency. C.V. Turb.: wT = h3 - h4, s4 = s3 + sT,GEN Ideal: s4S = s3 = 1.8704 => x4S = 0.9519, h4S = 1059.7 Btu/lbm => wT,S = 1741 - 1059.7 = 681.3 Btu/lbm Actual:

wT,AC = η × wT,S = 579.1 Btu/lbm

C.V. Pump: -wP = ⌠ ⌡vdP ≈ v1(P2 - P1) = 0.016166(500-1.28)144/778 = 1.49 Btu/lbm C.V. Boiler:

qH = h3 - h2 = h3 - h1 + wP = 1661.5 Btu/lbm

C.V. Condenser:

qL = h4ac - h1 = 1161.9 - 78.01 = 1083.9

η = (wT,AC - wP)/qH = (579.1 - 1.49)/1661.5 = 0.348

11.128E The steam power cycle in Problem 11.117 has an isentropic efficiency of the turbine of 85% and that for the pump it is 80%. Find the cycle efficiency and the specific work and heat transfer in the components. States numbered as in fig 9.3 of text. CV Pump: -wP,S = v1(P2-P1) = 0.01623(600 - 2)144/778 = 1.8 Btu/lbm ⇒ -wP,AC = 1.8/0.8 = 2.245 Btu/lbm h2 = h1 - wP,AC = 93.73 + 2.245 = 95.975 CV Turbine: wT,S = h3 - h4s ,

s4 = s3 = 1.4464 Btu/lbm R

⇒ wT,S = 1204.06 - 838.42 = 365.64 Btu/lbm wT,AC = h3 - h4AC = 365.64 × 0.85 = 310.78 ⇒ h4AC = 893.3 CV Boiler:

qH = h3 - h2 = 1204.06 - 95.975 = 1108.1 Btu/lbm

qL = h4AC - h1 = 893.3 - 93.73 = 799.5 Btu/lbm ηCYCLE = (wT + wP)/qH = (310.78 - 2.245)/1108.1 = 0.278 Compared to (365.64-1.8)/1108.5 = 0.33 in the ideal case.

11-103 11.129E Steam leaves a power plant steam generator at 500 lbf/in.2, 650 F, and enters the turbine at 490 lbf/in.2, 625 F. The isentropic turbine efficiency is 88%, and the turbine exhaust pressure is 1.7 lbf/in.2. Condensate leaves the condenser and enters the pump at 110 F, 1.7 lbf/in.2. The isentropic pump efficiency is 80%, and the discharge pressure is 520 lbf/in.2. The feedwater enters the steam generator at 510 lbf/in.2, 100 F. Calculate the thermal efficiency of the cycle and the entropy generation of the flow in the line between the steam generator exit and the turbine inlet, assuming an ambient temperature of 77 F. 2

T

1

500 lbf/in

2

1 650 F 625 F

TURBINE.

ST. GEN.

ηsT =

2

0.88

5 5s 6 4

3

6

COND. 5

3s 3 s

ηST = 0.88, ηSP = 0.80

4

P

1.7 lbf/in

h1 = 1328.0, h2 = 1314.0 s3S = s2 = 1.5752 = 0.16483 + x3S×1.7686

=>

x 3S = 0.79745

h3S = 88.1 + 0.797 45×1025.4 = 905.8 wST = h2 - h3S = 1314.0 - 905.8 = 408.2 wT = ηSTwST = 0.88×408.2 = 359.2 Btu/lbm h3 = h2 - wT = 1314.0 - 359.2 = 954.8 -wSP = 0.016166(520-1.7)

144 = 1.55 Btu/lbm 778

-wp = -wSP/ηSP = 1.55/0.80 = 1.94 Btu/lbm qH = h1 - h6 = 1328.0 - 68.1 = 1259.9 Btu/lbm ηTH = wNET/qH = (359.2 - 1.94)/1259.9 = 0.284 C.V. Line from 1 to 2: Energy Eq.:

w = 0/ ,

q = h2 - h1 = 1314 - 1328 = - 14 Btu/lbm

Entropy Eq.: s1 + sgen + q/T0 = s2

=>

sgen = s2 - s1 -q/T0 = 1.5752 - 1.586 - (-14/536.7) = 0.0153 Btu/lbm R

2

11-104 11.130E In one type of nuclear power plant, heat is transferred in the nuclear reactor to liquid sodium. The liquid sodium is then pumped through a heat exchanger where heat is transferred to boiling water. Saturated vapor steam at 700 lbf/in.2 exits this heat exchanger and is then superheated to 1100 F in an external gas-fired superheater. The steam enters the turbine, which has one (open-type) feedwater extraction at 60 lbf/in.2. The isentropic turbine efficiency is 87%, and the condenser pressure is 1 lbf/in.2. Determine the heat transfer in the reactor and in the superheater to produce a net power output of 1000 Btu/s. 700 lbf/in 2 6 SUP. HT.

60 lbf/in 2

7

5

8

HTR. 3

4

1100 F

TURBINE.

Q REACT.

6

T

P

COND. 2

1

P

4

5 7s

2 3 1

7

1 lbf/in 2

8s 8

s . WNET = 1000 Btu/s, ηST = 0.87

-wP12 = 0.016136(60 - 1)144/778 = 0.18 Btu/lbm h2 = h1 - wP12 = 69.73 + 0.18 = 69.91 Btu/lbm -wP34 = 0.017378(700 - 60)144/778 = 2.06 Btu/lbm h4 = h3 - wP34 = 262.24 + 2.06 = 264.3 Btu/lbm s7S=s6 = 1.7682, P7

=> T7S = 500.8 F,

h7S = 1283.4

h7 = h6 - ηST(h6 - h7S) = 1625.8 - 0.87(1625.8 - 1283.4) = 1327.9 s8S = s6 = 1.7682 = 0.13264 + x8S × 1.8453 =>

x 8S = 0.8863

h8S = 69.73 + 0.8863 × 1036 = 987.9 Btu/lbm h8 = h6 - ηST(h6 - h8S) = 1625.8 - 0.87(1625.8 - 987.9) = 1070.8 CV: heater: cont: m2 + m7 = m3 = 1.0 lbm, 1st law: m2h2 + m7h7 = m3h3 m7 = (262.24-69.91) / (1327.9-69.91) = 0.1529 CV: turbine: wT = (h6 - h7) + (1 - m7)(h7 - h8) = 1625.8-1327.9 + 0.8471(1327.9-1070.8) = 515.7 Btu/lbm CV pumps: wP = m1wP12 + m3wP34 = -(0.8471×0.18 + 1×2.06) = -2.2 Btu/lbm . wNET = 515.7 - 2.2 = 513.5 Btu/lbm => m = 1000/513.5 = 1.947 lbm/s . . CV: reactor QREACT = m(h5-h4) = 1.947(1202 - 264.3) = 1825.7 Btu/s . . CV: superheater QSUP = m(h6 - h5) = 1.947(1625.8 - 1202) = 825 Btu/s

11-105 11.131E A boiler delivers steam at 1500 lbf/in.2, 1000 F to a two-stage turbine as shown in Fig. 11.17. After the first stage, 25% of the steam is extracted at 200 lbf/in.2 for a process application and returned at 150 lbf/in. 2, 190 F to the feedwater line. The remainder of the steam continues through the low-pressure turbine stage, which exhausts to the condenser at 2 lbf/in. 2. One pump brings the feedwater to 150 lbf/in.2 and a second pump brings it to 1500 lbf/in.2. Assume the first and second stages in the steam turbine have isentropic efficiencies of 85% and 80% and that both pumps are ideal. If the process application requires 5000 Btu/s of power, how much power can then be cogenerated by the turbine? 3: h3 = 1490.3, s3 = 1.6001 4s: s4S = s3 ⇒ h4S = 1246.6

3

wT1,S = h3 - h4S = 243.7

T2

T1

⇒ wT1,AC = 207.15 Btu/lbm

B 4

h4AC = h3 - wT1,AC = 1283.16 4ac: P4, h4AC ⇒ s4AC = 1.6384 5s: s5S = s4AC ⇒ h5S = 951.3 wT2,S = h4AC - h5S = 331.9

5 2 P2 7 1

Proc. 5000 Btu/s 6 P1

wT2,AC = 265.5 = h 4AC - h5AC ⇒ h5AC = 1017.7 Btu/lbm 7: h7 = 158.02 ; qPROC = h4AC - h7 = 1125.1 Btu/lbm . . . m4 = Q/qPROC = 5000/1125.1 = 4.444 lbm/s = 0.25 mTOT . . . . . ⇒ mTOT = m3 = 17.776 lbm/s, m5 = m3 - m4 = 13.332 lbm/s . . . . WT = m3h3 - m4h4AC - m5h5AC = 7221 Btu/s

C

11-106 11.132E A large stationary Brayton cycle gas-turbine power plant delivers a power output of 100000 hp to an electric generator. The minimum temperature in the cycle is 540 R, and the maximum temperature is 2900 R. The minimum pressure in the cycle is 1 atm, and the compressor pressure ratio is 14 to 1. Calculate the power output of the turbine, the fraction of the turbine output required to drive the compressor and the thermal efficiency of the cycle? Brayton: . wNET = 100 000 hp

3

T

s

P1 = 1 atm, T1 = 540 R P2/P1 = 14, T3 = 2900 R

2 s

a) Assume const CP0:

1

4

s 2 = s1

s

P2

(P )

→ T2 = T 1

k-1 k

= 540(14)0.286 = 1148.6 R

1

-wC = -w12 = h2 - h1 = CP0(T2-T1) = 0.24(1148.6-540) = 146.1 Btu/lbm P4

(P )

As s4 = s3 → T4 = T3

k-1 k

= 2900(

3

1 0.286 ) = 1363.3 R 14

wT = w34 = h3 - h4 = CP0(T3-T4) = 0.24(2900-1363.3) = 368.8 Btu/lbm wNET = wT + wC = 368.8 - 146.1 = 222.7 Btu/lbm . . m = WNET/wNET = 100 000×2544/222.7 = 1 142 344 lbm/h . . WT = mwT = 165 600 hp, -wC/wT = 0.396 b)

qH = CP0(T3-T2) = 0.24(2900-1148.6) = 420.3 Btu/lbm ηTH = wNET/qH = 222.7/420.3 = 0.530

11-107 11.133E An ideal regenerator is incorporated into the ideal air-standard Brayton cycle of Problem 11.132. Calculate the cycle thermal efficiency with this modification. T

Problem 9.108 + ideal regen., where : wT = 368.8, wC = 146.1 Btu/lbm

3 s

wNET = 222.7 Btu/lbm

x 2

Ideal regen.: TX = T4 = 1363.3 R

4 y

s

qH = h3 - hX = 0.24(2900 - 1363.3)

1

= 368.8 Btu/lbm = w T ηTH = wNET/qH = 222.7/368.8 = 0.604

s

11.134E Consider an ideal gas-turbine cycle with two stages of compression and two stages of expansion. The pressure ratio across each compressor stage and each turbine stage is 8 to 1. The pressure at the entrance to the first compressor is 14 lbf/in.2, the temperature entering each compressor is 70 F, and the temperature entering each turbine is 2000 F. An ideal regenerator is also incorporated into the cycle. Determine the compressor work, the turbine work, and the thermal efficiency of the cycle. 10

REG

I.C. CC

1

9

5 2

6

4

COMP

COMP

TURB

TURB

7

8

CC

P2/P1 = P4/P3 = P6/P7 = P8/P9 = 8.0

T

8

6

P1 = 14 lbf/in2 T1 = T3 = 70 F, T6 = T8 = 2000 F Assume const. specific heat s2 = s1 and s4 = s3 T4 = T2 = T1(P2/P1

k-1 )k

= 529.67(8)0.2857 = 959.4 R

5 7 4 3

2

9

10

1 s

Total compressorwork -wC = 2 ×(-w12) = 2CP0(T2 - T1) = 2 × 0.24(959.4 - 529.67) = 206.3 Btu/lbm

11-108 Also s6 = s7 and s8 = s9 k-1

P7 k 10.2857 ⇒ T7 = T9 = T6  = 2459.67  = 1357.9 R 8 P6 Total turbinework wT = 2× w67 = 2CP0(T6 - T7) = 2 × 0.24(2459.67 - 1357.9) = 528.85 Btu/lbm wNET = 528.85 - 206.3 = 322.55 Btu/lbm Ideal regenerator: T5 = T9, T10 = T4 ⇒ qH = (h6 - h5) + (h8 - h7) = 2CP0(T6 - T5) = 2 × 0.24(2459.67 - 1357.9) = wT = 528.85 Btu/lbm ηTH = wNET/qH = 322.55/528.85 = 0.61

11.135E Repeat Problem 11.134, but assume that each compressor stage and each turbine stage has an isentropic efficiency of 85%. Also assume that the regenerator has an efficiency of 70%. T4S = T2S = 959.4 R, -wCS = 206.3

T

8

6

T7S = T9S = 1357.9 R, wTS = 528.85 ⇒

9 5

-wC = -wSC/ηSC = 242.7 Btu/lbm

4

-w12 = -w34 = 242.7/2 = 121.35

4S

T2 = T4 = T1 + (-w12/CP0)

3

7S

7 9S

2S 2 1

= 529.67 + 121.35/0.24 = 1035.3 R

s

wT = ηT wTS = 449.5 T7 = T9 = T6 - (+w67/CP0) = 2459.67 - 449.5/2×0.24 = 1523 R ηREG =

h5 - h4 h9 - h4

=

T 5 - T4

T5 - 1035.3

= = 0.7 T9 - T4 1523 - 1035.3

⇒ T5 = 1376.7 R

qH = CP0(T6 - T5) + CP0(T8 - T7) = 0.24(2459.67 - 1376.7) + 0.24(2459.67 - 1523) = 484.7 Btu/lbm wNET = wT + wC = 449.5 - 242.7 = 206.8 Btu/lbm ηTH = wNET/qH = 206.8/484.7 = 0.427

11-109 11.136E An air-standard Ericsson cycle has an ideal regenerator as shown in Fig. P11.62. Heat is supplied at 1800 F and heat is rejected at 68 F. Pressure at the beginning of the isothermal compression process is 10 lbf/in.2. The heat added is 275 Btu/lbm. Find the compressor work, the turbine work, and the cycle efficiency. T P T4 = T3 = 1800 F P

2

3

3

T 4

P1 = 10 lbf/in2

T

T

P 1

P

q = -4q1 (ideal reg.)

P

4

2 3

⇒ qH = 3q4 &

2 T

T1 = T2 = 68 F = 527.7R

1

v

s

wT = qH = 275 Btu/lbm

ηTH = ηCARNOT TH. = 1 - TL/TH = 1 - 527.7/2349.7 = 0.775 wnet = ηTH qH = 0.775 × 275 = 213.13 Btu/lbm qL = -wC = 275 - 213.13 = 61.88 Btu/lbm 11.137E The turbine in a jet engine receives air at 2200 R, 220 lbf/in.2. It exhausts to a nozzle at 35 lbf/in.2, which in turn exhausts to the atmosphere at 14.7 lbf/in.2. The isentropic efficiency of the turbine is 85% and the nozzle efficiency is 95%. Find the nozzle inlet temperature and the nozzle exit velocity. Assume negligible kinetic energy out of the turbine. C.V. Turb.: hi = 560.588, Pri = 206.092, se = si ⇒ Pre = Pri × (Pe/Pi) = 206.092(35/220) = 32.787 Te = 1381, he = 338.13, wT,s = 560.588 - 338.13 = 222.46 wT,AC = wT,s × ηT = 189.09 = he,AC - hi ⇒ he,AC = 371.5 ⇒ Te,AC = 1508.4 R, Pre,AC = 45.917 C.V. Nozzle: (1/2)V2e = hi - he; se = si ⇒ Pre = Pri × (Pe/Pi) = 45.917(14.7/35) = 19.285 ⇒ Te,s = 1199.6 R, he,s = 291.3 Btu/lbm 2 = hi - he,s = 371.5 - 291.3 = 80.2 Btu/lbm (1/2)Ve,s 2 2 = (1/2)Ve,s × ηNOZ = 76.19 Btu/lbm (1/2)Ve,AC

Ve,AC =

2 × 25037 × 76.19 = 1953 ft/s

11-110 11.138E Air flows into a gasoline engine at 14 lbf/in.2, 540 R. The air is then compressed with a volumetric compression ratio of 8;1. In the combustion process 560 Btu/lbm of energy is released as the fuel burns. Find the temperature and pressure after combustion. ⇒ vr2 = vr1/8 = 179.49/8 = 22.436

Compression 1 to 2: s2 = s1

T2 = 1212 R, u2 = 211.31 Btu/lbm, Pr2 = 20 P2 = Pr2(P1/Pr1) = 20(14/1.1146) = 251.2 lbf/in2 Compression 2 to 3: u3 = u2 + qH = 211.31 + 560 = 771.3, T3 = 3817 R P3 = P2 × (T3/T2) = 251.2(3817/1212) = 791 lbf/in2 11.139E To approximate an actual spark-ignition engine consider an air-standard Otto cycle that has a heat addition of 800 Btu/lbm of air, a compression ratio of 7, and a pressure and temperature at the beginning of the compression process of 13 lbf/in.2, 50 F. Assuming constant specific heat, with the value from Table C.4, determine the maximum pressure and temperature of the cycle, the thermal efficiency of the cycle and the mean effective pressure. P

3

qH = 800 Btu

s

v1/v2 = 7

4

P1 = 13 lbf/in2, T1 = 50 F

T

3 s

v

v

4 v 1

2 s

2 s 1 v

a)

v1 = RT1/P1

v

s

= 53.34×510/13×144 = 14.532 ft3/lbm v2 = v1/7 = 2.076 ft3/lbm

k

P2 = P1(v1/v2) = 13(7)1.4 = 198.2 lbf/in2 k-1

T2 = T1(v1/v2)

= 510(7)0.4 = 1110.7 R

T3 = T2 + qH/CV0 = 1110.7 + 800/0.171 = 5789 R P3 = P2T3/T2= 198.2 × 5789/1110.7 = 1033 lbf/in2 b)

ηTH = 1 - (T1/T2) = 1 - 510/1110.7 = 0.541

c)

wNET = ηTH × qH = 0.541 × 800 = 432.8 Btu mep =

wNET v1-v2

=

432.8×778 = 188 lbf/in2 (14.532-2.076)×144

11-111 11.140E

In the Otto cycle all the heat transfer qH occurs at constant volume. It is

more realistic to assume that part of qH occurs after the piston has started its downwards motion in the expansion stroke. Therefore consider a cycle identical to the Otto cycle, except that the first two-thirds of the total qH occurs at constant volume and the last one-third occurs at constant pressure. Assume the total qH is 700 Btu/lbm, that the state at the beginning of the compression process is 13 lbf/in.2, 68 F, and that the compression ratio is 9. Calculate the maximum pressure and temperature and the thermal efficiency of this cycle. Compare the results with those of a conventional Otto cycle having the same given variables. P

3

T

4

3 s

P1 = 13, T1 = 527.67 R

s

rV = v1/v2 = 7

5

2 Btu q23 = ×700 = 466.7 3 lbm 1 Btu q34 = ×700 = 233.3 3 lbm

v

5

2

4

2 s

s 1

v

1 s

v k

P2 = P1(v1/v2) = 13(9)1.4 = 281.8 lbf/in2 k-1

T2 = T1(v1/v2)

= 527.67(9)0.4 = 1270.7 R

T3 = T2 + q23/CV0 = 1270.7 + 466.7/0.171 = 4000 R P3 = P2(T3/T2) = 281.8 × 4000/1270.7 = 887.1 lbf/in2 = P4 T4 = T3 + q34/CP0 = 4000 + 233.3/0.24 = 4972 R v5 v4

=

v1 v4

= (P4/P1) × (T1/T4) = k-1

T5 = T4(v4/v5)

88.1 527.67 × = 7.242 13 4972

= 4972(1/7.242)

0.4

= 2252 R

qL = CV0(T5-T1) = 0.171(2252 - 527.67) = 294.9 Btu/lbm ηTH = 1 - qL/qH = 1 - 294.9/700 = 0.579 Std Otto cycle: ηTH = 1 - (9)-0.4 = 0.585

11-112 11.141E It is found experimentally that the power stroke expansion in an internal combustion engine can be approximated with a polytropic process with a value of the polytropic exponent n somewhat larger than the specific heat ratio k. Repeat Problem 11.139 but assume the expansion process is reversible and polytropic (instead of the isentropic expansion in the Otto cycle) with n equal to 1.50. See solution to 11.139 :

T3 = 5789 R,

P3 = 1033 lbf/in2

v3 = RT3/P3 = v2 = 2.076 ft3/lbm, v4 = v1 = 14.532 except for process 3 to 4:

Pv1.5 = constant.

P4 = P3(v3/v4)1.5 = 1033(1/7)1.5 = 55.78 lbf/in2 T4 = T3(v3/v4)0.5 = 5789(1/7)0.5 = 2188 R 1w2

= ∫ P dv = R(T2 - T1)/(1 - 1.4) = -53.34(1110.7 - 510)/(0.4×778) = -102.96 Btu/lbm

3w4

= ∫ P dv = R(T4 - T3)/(1 - 1.5) = -53.34(2188 - 5789)/(0.5×778) = 493.8 Btu/lbm

wNET = 493.8 - 102.96 = 390.84 Btu/lbm ηCYCLE = wNET/qH = 390.84/700 = 0.488 mep = wNET/(v1-v2) = 390.84×778/(14.532 - 2.076) = 169.5 lbf/in2 Notice a smaller wNET, ηCYCLE, mep compared to ideal cycle.

11.142E A diesel engine has a bore of 4 in., a stroke of 4.3 in. and a compression ratio of 19:1 running at 2000 RPM (revolutions per minute). Each cycle takes two revolutions and has a mean effective pressure of 200 lbf/in. 2. With a total of 6 cylinders find the engine power in Btu/s and horsepower, hp. Power from mean effective pressure. mep = wnet / (vmax - vmin)

->

wnet = mep(vmax - vmin)

∆V = πBore2 × 0.25 × S = π × 42 × 0.25 × 4.3 = 54.035 in3 W = mep(Vmax - Vmin) = 200×54.035/(12×778) = 1.1575 Btu Work per Cylinders per Power Stroke . W = W×n×RPM×0.5 (cycles / min)×(min / 60 s)×(Btu / cycle) = 1.1575×6×2000×0.5×(1/60) = 115.75 Btu/s = 115.75 × 3600/2544.43 hp = 164 hp

11-113 11.143E At the beginning of compression in a diesel cycle T = 540 R, P = 30 2 lbf/in. and the state after combustion (heat addition) is 2600 R and 1000 lbf/in.2. Find the compression ratio, the thermal efficiency and the mean effective pressure. P2 = P3 = 1000 lbf/in2 T2 = T1(P2/P1)(k-1)/k = 540(1000/30) 0.2857 = 1470.6 v1/v2 = (P2/P1)1/k = (1000/30)0.7143 = 12.24 v3/v2 = T3/T2 = 2600/1470.6 = 1.768 v4/v3 = v1/v3 = (v1/v2)(v2/v3) = 12.24/1.768 = 6.923 T4 = T3(v3/v4)k-1 = 2600*6.923-0.4 = 1199 qL = CV(T4 - T1) = 0.171(1199-540) = 112.7 qH = h3 - h2 = CP (T3 - T2) = 0.24(2600 - 1470.6) = 271.1 η = 1 - qL/qH = 1 - 112.7 / 271.1 = 0.5843 wnet = qnet = 271.1 - 112.7 = 158.4 vmax = v1 = RT1/P1 = 53.34 × 540/(30 × 144) = 6.6675 vmin = vmax(v1/v2) = 6.6675 / 12.24 = 0.545 mep = [158.4/(6.6675 - 0.545)] × (778/144) = 139.8 lbf/in2

11-114 11.144E Consider an ideal air-standard diesel cycle where the state before the compression process is 14 lbf/in.2, 63 F and the compression ratio is 20. Find the maximum temperature(by iteration) in the cycle to have a thermal efficiency of 60%. Diesel cycle: P1= 14, T1 = 522.67 R, v1/v2 = 20, ηTH = 0.60 k-1

T2 = T1(v1/v2) v1 =

= 522.67(20)0.4 = 1732.4 R

53.34×522.67 13.829 = 13.829, v2 = = 0.6915 20 14×144

v3 = v2×T3/T2 = 0.6915×T3/1732.4 = 0.000399 T3 T3 T4

=

v4

13.829 (v ) =(0.000399 ) T k-1

3

3

ηTH = 0.60 = 1 ⇒

0.4

T4-T1 k(T3-T2)

1.4

→ T4 = 0.0153 T3 1.4

=1-

0.0153 T3 -522.67 1.4(T3-1732.4)

1.4

0.0153 T3 - 0.56 T3 + 447.5 = 0

5100R: LHS = -35.54,

5500R: LHS = 5.04,

Linear interpolation, T3 = 5455 R

5450R: LHS = -0.5

11-115

11.145E Consider an ideal Stirling-cycle engine in which the pressure and temperature at the beginning of the isothermal compression process are 14.7 lbf/in.2, 80 F, the compression ratio is 6, and the maximum temperature in the cycle is 2000 F. Calculate the maximum pressure in the cycle and the thermal efficiency of the cycle with and without regenerators. T

P 3

T 4

3 T

v 2

P1 = 14.7 lbf/in2

4

v

v T

Ideal Stirling cycle T1 = T2 = 80 F v1

v

v2

2

1

T

1 s

v

=6

T3 = T4 = 2000 F

T1 = T2 → P2 = P1× v1/v2 = 14.7×6 = 88.2 V2 = V3 → P3 = P2× T3/T2 = 88.2×

2460 = 401.8 lbf/in2 540

w34 = q34 = RT3 ln (v4/v3) = (53.34/778) × 2460 ln 6 = 302.2 Btu/lbm q23 = CV0(T3-T2) = 0.171(2460-540) = 328.3 Btu/lbm w12 = q12 = -RT1 ln

v1 v2

=-

53.34 ×540 ln 6 = -66.3 Btu/lbm 778

wNET = 302.2 - 66.3 = 235.9 Btu/lbm ηNO REGEN =

235.9 = 0.374, 302.2+328.3

ηWITH REGEN =

235.9 = 0.781 302.2

11-116

11.146E An ideal air-standard Stirling cycle uses helium as working fluid. The 2 2 isothermal compression brings the helium from 15 lbf/in. , 70 F to 90 lbf/in. . The exspansion takes place at 2100 R and there is no regenerator. Find the work and heat transfer in all four processes per lbm helium and the cycle efficiency. Substance helium C.4:

R = 386 ft-lbf/lbmR

CV = 0.753

v4/v3 = v1/v2 = P2/P1 = 90/15 = 6 1 -> 2:

-1w2 = -1q2 = ∫ P dV = 386 × 530 × ln(6)/778 = 471.15 Btu/lbm

2 -> 3:

2w 3

= 0; 2q3 = CP (T3 - T2) = 0.753(2100 - 530) = 1182.2

3 -> 4:

3w 4

= 3q4 = RT3ln(v4/v3) = 386 × 2100 × ln(6)/778 = 1866.8

4 -> 1:

4w 1

= 0;

4q 1

= CP (T4 - T1) = -1182.2

ηCycle = wnet/ qH = (-471.15 + 1866.0) / (1182.2 + 1866.8) = 0.458

11.147E The air-standard Carnot cycle was not shown in the text; show the T–s diagram for this cycle. In an air-standard Carnot cycle the low temperature is 500 R and the efficiency is 60%. If the pressure before compression and after heat rejection is 14.7 lbf/in.2, find the high temperature and the pressure just before heat addition. η = 0.6 = 1 - TH/TL ⇒ TH = TL/0.4 = 500/0.4 = 1250 R P2 = P1(TH/TL

1 k-1 )

= 363.2 lbf/in

= 14.7(

1250 3.5 ) 500

T H T L

T 2

3

1

2

[or P2 = P1(Pr2/Pr1) = 14.7 × 22.48 / 0.8515 = 388 lbf/in2]

4 s

11-117 11.148E Air in a piston/cylinder goes through a Carnot cycle in which TL = 80.3 F and the total cycle efficiency is η = 2/3. Find TH, the specific work and volume ratio in the adiabatic expansion for constant Cp, Cv. Repeat the calculation for variable heat capacities. Carnot cycle: Same process diagram as in previous problem. η = 1 - TL/TH = 2/3 ⇒ TH = 3 × TL = 3 × 540 = 1620 R Pvk = constant

Adiabatic expansion 3 to 4:

w = (P4v4 - P3v3)/(1 - k) = [R/(1-k)](T4 - T3) = u3 - u4

3 4

= Cv(T3 - T4) = 0.171(1620 - 540) = 184.68 Btu/lbm v4/v3 = (T3/T4)1/(k - 1) = 32.5 = 15.6 For variable Cp, Cv we get, TH = 3 × TL = 1620 R w = u3 - u4 = 290.13 - 92.16 = 197.97 Btu/lbm

3 4

v4/v3 = vr4/vr3 = 179.49/9.9289 = 18.1

11.149E Consider an ideal refrigeration cycle that has a condenser temperature of 110 F and an evaporator temperature of 5 F. Determine the coefficient of performance of this refrigerator for the working fluids R-12 and R-22. T

Ideal Ref. Cycle Tcond = 110 F = T3

2 3

Tevap = 5 F

4

Use Table C.10 for R-22 Use computer table for R-12

1 s

h1, Btu/lbm

R-12 77.803

R-22 104.954

s 2 = s1

0.16843

0.22705

P2, lbf/in2

151.11

241.04

T2 , F

127.29

161.87

h2, Btu/lbm

91.107

123.904

h3=h4, Btu/lbm

33.531

42.446

13.3

18.95

44.27

62.51

3.33

3.30

-wC = h2-h1 qL = h1-h4 β =qL/(-wC)

11-118 11.150E The environmentally safe refrigerant R-134a is one of the replacements for R-12 in refrigeration systems. Repeat Problem 11.149 using R-134a and compare the result with that for R-12. T

Ideal Ref. Cycle Tcond = 110 F

2 3

Tevap = 5 F h1 = 167.325 Btu/lbm

1

4

s2 = s1 = 0.4145 Btu/lbm R

s

P2 = 161.124 lbf/in2, T2 = 122.2 F,

h2 = 184.44 Btu/lbm

h3=h4 = 112.455 Btu/lbm -wC = h2-h1 = 17.115 Btu/lbm ; qL = h1-h4 = 54.87 Btu/lbm β =qL/(-wC) = 3.206

11.151E Consider an ideal heat pump that has a condenser temperature of 120 F and an evaporator temperature of 30 F. Determine the coefficient of performance of this heat pump for the working fluids R-12, R-22, and ammonia. T 2

Ideal Heat Pump Tcond = 120 F

3

Tevap = 30 F Use Table C.9 for NH3 Use Table C.10 for R-22 Use computer table for R-12 R-12 h1, Btu/lbm 80.42

1

4

s

R-22 107.28

NH3 619.58

s 2 = s1

0.1665

0.2218

1.2769

P2, lbf/in2

172.3

274.6

286.5

T2 , F

132.2

160.4

239.4

h2, Btu/lbm

91.0

122.17

719.5

h3=h4, Btu/lbm

36.011

45.71

178.83

-wC = h2-h1

10.58

14.89

99.92

qH = h2-h3

54.995

76.46

540.67

5.198

5.135

5.411

β′ =qH/(-wC)

11-119 11.152E The refrigerant R-22 is used as the working fluid in a conventional heat pump cycle. Saturated vapor enters the compressor of this unit at 50 F; its exit temperature from the compressor is measured and found to be 185 F. If the isentropic efficiency of the compressor is estimated to be 70%, what is the coefficient of performance of the heat pump? T 2

R-22 heat pump: T2 = 185 F

2S

TEVAP = 50 F, ηS COMP = 0.70

3

Isentropic compressor: s2S = s1 = 0.2180

1

4

but P2 unknown. Trial & error.

s

Assume P2 = 307 lbf/in2 At P2,s2S: T2S = 162 F, h2S = 121.07, At P2,T2: h2 = 126.24 calculate ηS COMP=

h2S - h1 h2 - h1

=

121.07 - 108.95 = 0.701 ≈ 0.70 126.24 - 108.95

OK ⇒ P2 = 307 lbf/in2 = P3 ⇒ T3 = 128.8 F, h3 = 48.66 -wC = h2 - h1 = 17.29,

qH = h2 - h3 = 77.58,

β′ = qH/(-wC) = 4.49

11.153E Consider a small ammonia absorption refrigeration cycle that is powered by solar energy and is to be used as an air conditioner. Saturated vapor ammonia leaves the generator at 120 F, and saturated vapor leaves the evaporator at 50 F. If 3000 Btu of heat is required in the generator (solar collector) per pound-mass of ammonia vapor generated, determine the overall performance of this system. NH3 absorption cycle: sat. vapor at 120 F exits the generator. Sat. vapor at 50 F exits the evaporator qH = qGEN = 3000 Btu/lbm NH3

T GEN. EXIT

120F 50 F

out of generator.

1

2

EVAP EXIT s

qL = h2 - h1 = hG 50 F - hF 120 F = 624.28 - 178.79 = 445.49 Btu/lbm ⇒ qL/qH = 445.49/3000 = 0.1485

11-120 11.154E Consider an ideal dual-loop heat-powered refrigeration cycle using R-12 as the working fluid, as shown in Fig. P11.109. Saturated vapor at 220 F leaves the boiler and expands in the turbine to the condenser pressure. Saturated vapor at 0 F leaves the evaporator and is compressed to the condenser pressure. The ratio of the flows through the two loops is such that the turbine produces just enough power to drive the compressor. The two exiting streams mix together and enter the condenser. Saturated liquid leaving the condenser at 110 F is then separated into two streams in the necessary proportions. Determine the ratio of mass flow rate through the power loop to that through the refrigeration loop. Find also the performance of the cycle, in terms of the ratio QL /QH. T 1 TURB.

COMP.

7

2

6 BOIL.

COND.

5

E V A P .

. QL

6 5

2

3

7 4

1

4

s

3 P

T F

P 2

h Btu/lbm

s Btu/lbm R

Computer tables for properties.

77.271

168.88

P2=P3=PSAT at 110 F

168.88

P5=P6=PSAT at 220 F

0.067 45

s2=s1=0.168 88

1

0

lbf/in 23.849

2

-

151.11

3

110

151.11

33.531

4

0

23.849

33.531

5 6

220

524.43 524.43

7

110

151.11

89.036

-wP = 0.0129(524.4 - 151.1)

h2=91.277 0.067 45 0.151 49 0.151 49

144 = 0.894 778

h5 = 33.531 + 0.894 = 34.425 Btu/lbm (1-x7) =

0.162 79 - 0.151 49 0.011 30 = = 0.1187 0.095 34 0.095 34

h7 = 87.844 - 0.1187(54.313) = 81.397 Btu/lbm

Pump work: -wP = h5-h3 ≈ v5(P5-P3)

11-121 CV: turbine + compressor . . . . Continuity Eq.: m1 = m2, m6 = m7 . . . . Energy Eq.: m1h1 + m6h6 = m2h2 + m7h7 89.036-81.397 7.639 . . m1/m6 = = = 0.545, 91.277-77.271 14.006

. . m6/m1 = 1.833

-wP = v3(P5-P3), h5 = h3 - wP . . . . CV evaporator: QL = m1(h1-h4), CV boiler: QH = m6(h6-h5) . m1(h1-h4) . . 77.271-33.531 = = 0.436 ⇒ β = QL/QH = . m6(h6-h5) 1.833(89.036-34.425) CV: pump:

11.155E (Adv.) Find the availability of the water at all four states in the Rankine cycle described in Problem 11.121. Assume the high-temperature source is 900 F and the low-temperature reservoir is at 65 F. Determine the flow of availability in or out of the reservoirs per pound-mass of steam flowing in the cycle. What is the overall cycle second law efficiency? Ref. state 14.7 lbf/in2, 77°F, h0 = 45.08 Btu/lbm,

s0 = 0.08774 Btu/lbm R

From solution to 11.121: s1 = 0.17497, h1 = 94.01,

s2 = 0.175 = s1, h2 = 95.81,

s4 = s3 = 1.5871 Btu/lbm R

h3 = 1350.6,

h4 = 921.23

ψ = h - h0 - T0(s - s0) ψ1 = 94.01 - 45.08 - 536.67(0.17497 - 0.08774) = 2.116 ψ2 = 3.92, ψ3 = 500.86, ψ4 = 71.49 Btu/lbm ∆ψ H = (1 - T0/TH)qH = 0.6054 × 1254.79 = 759.65 Btu/lbm ∆ψ L = (1 - T0/T0)qC = 0/ ηII = wNET/∆ψH = (429.37 - 1.8)/759.65 = 0.563 Notice TH > T3, TL < T4 = T1, so cycle is externally irreversible. Both qH and qC over finite ∆T.

12-1

CHAPTER 12 The correspondence between the new problem set and the previous 4th edition chapter 11 problem set. New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Old New New New 1 New 3 new new new new new new 5 new 4 new new new 6 7 new new 12 8a mod 8b mod 10 new 11 13 new

New 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Old 15 new new 16 17 new new 20 new new 24 new 26 new 23 32 40 37 46 28 new new 39 new new new 35 new 27 21

New 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

The problems that are labeled advanced start at number 74.

Old new 22 25 38 50 new 43 44 47 48 49 52 41 9 14 33 mod 34 mod 36 42 51

12-2

The English unit problems are: New

Old

81 82 83 84 85 86 87 88 89

new new new new new 77 78 79 82

New 90 91 92 93 94 95 96 97 98 99

Old 80 84 86 85 mod 87 91 93 94 88 97

New 100 101 102 103

Old 96 99 83 98

12-3 12.1

A gas mixture at 20°C, 125 kPa is 50% N2, 30% H2O and 20% O2 on a mole basis. Find the mass fractions, the mixture gas constant and the volume for 5 kg of mixture. From Eq. 12.3: ci = yi Mi/ ∑ yjMj MMIX = ∑ yjMj = 0.5×28.013 + 0.3×18.015 + 0.2×31.999 = 14.0065 + 5.4045 + 6.3998 = 25.811 cN2 = 14.0065 / 25.811 = 0.5427, cH2O = 5.4045 / 25.811 = 0.2094 cO2 = 6.3998 / 25.811 = 0.2479, sums to 1 OK − RMIX = R/MMIX = 8.3145 / 25.811 = 0.3221 kJ/kg K V = mRMIX T/P = 5×0.3221×393.15/125 = 5.065 m3

12.2

A 100 m3 storage tank with fuel gases is at 20°C, 100 kPa containing a mixture of acetylene C2H2, propane C3H8 and butane C4H10. A test shows the partial pressure of the C2H2 is 15 kPa and that of C3H8 is 65 kPa. How much mass is there of each component? Assume ideal gases, then the ratio of partial to total pressure is the mole fraction, y = P/Ptot yC2H2 = 15/100 = 0.15, yC3H8 = 65/100 = 0.65, yC4H10 = 20/100 = 0.20 − ntot = PV/RT = 100 × 100 / (8.31451 × 293.15) = 4.1027 kmoles mC2H2 = (nM)C2H2 = yC2H2ntotMC2H2 = 0.15×4.1027×26.038 = 16.024 kg mC3H8 = (nM)C3H8 = yC3H8ntotMC3H8 = 0.65×4.1027×44.097 = 117.597 kg mC4H10 = (nM)C4H10 = yC4H10ntotMC4H10 = 0.20×4.1027×58.124 = 47.693 kg

12.3

A mixture of 60% N2, 30% Ar and 10% O2 on a mole basis is in a cylinder at 250kPa, 310 K and volume 0.5 m3. Find the mass fractions and the mass of argon. From Eq. 12.3:

ci = yi Mi/ ∑ yjMj

MMIX = ∑ yjMj = 0.6×28.013 + 0.3×39.948 + 0.1×31.999 = 31.992 cN2 = (0.6×28.013) / 31.992 = 0.5254, cAr = (0.3×39.948) / 31.992 = 0.3746 cO2 = (0.1×31.999) / 31.992 = 0.1, sums to 1 OK − RMIX = R/MMIX = 8.3145 / 31.992 = 0.260 kJ/kg K mMIX = PV/(RMIX T) = 250×0.5 / 0.26×310 = 1.551 kg mAr = cAr × mMIX = 0.3746×1.551 = 0.581 kg

12-4 12.4

A carbureted internal combustion engine is converted to run on methane gas (natural gas). The air-fuel ratio in the cylinder is to be 20 to 1 on a mass basis. How many moles of oxygen per mole of methane are there in the cylinder? The mass ratio mAIR/mCH = 20, so relate mass and mole 4 nAIR nCH4 →

12.5

=

(mm )× M AIR

CH4

nO2 nCH4

CH4/MAIR

n = m/M

= 20× 16.04/28.97 = 11.0735

= 0.21×11.0735 = 2.325 mole O2/mole CH4

Weighing of masses gives a mixture at 60°C, 225 kPa with 0.5 kg O2, 1.5 kg N2 and 0.5 kg CH4. Find the partial pressures of each component, the mixture specific volume (mass basis), mixture molecular weight and the total volume. From Eq. 12.4: yi = (mi /Mi) / ∑ mj/Mj ntot = ∑ mj/Mj = (0.5/31.999) + (1.5/28.013) + (0.5/16.04) = 0.015625 + 0.053546 + 0.031172 = 0.100343 yO2 = 0.015625/0.100343 = 0.1557,

yN2 = 0.053546/0.100343 = 0.5336,

yCH4 = 0.031172/0.100343 = 0.3107 PO2 = yO2 Ptot = 0.1557×225 = 35 kPa, PN2 = yN2 Ptot = 0.5336×225 = 120 kPa PCH4 = yCH4 Ptot = 0.3107×225 = 70 kPa − Vtot = ntot RT/P = 0.100343 × 8.31451 × 333.15 / 225 = 1.235 m3 v = Vtot/mtot = 1.235 / (0.5 + 1.5 + 0.5) = 0.494 m3/kg MMIX = ∑ yjMj = mtot/ntot = 2.5/0.100343 = 24.914

12-5 12.6

At a certain point in a coal gasification process, a sample of the gas is taken and stored in a 1-L cylinder. An analysis of the mixture yields the following results: Component H2 CO CO2 N2 Percent by volume 25 40 15 20 Determine the mass fractions and total mass in the cylinder at 100 kPa, 20°C. How much heat transfer must be transferred to heat the sample at constant volume from the initial state to 100°C? Volume fractions same as mole fractions so From Eq. 12.3: ci = yi Mi/ ∑ yjMj yi

× Mi

= kgi/kg

= ci

kmoli/kmol

× kgi/kmoli

= kgi/kmol

= kgi/kg

H2

0.25

× 2.016

= 0.504/23.9121

= 0.0211

CO CO2

0.40 0.15

× 28.01 × 44.01

= 11.204/23.9121 = 6.6015/23.9121

= 0.4685 = 0.2761

N2

0.20

× 28.013

= 5.6026/23.9121

= 0.2343

MMIX = 23.9121 − RMIX = R/MMIX = 8.3145/23.9121 = 0.34771 kJ/kg/K m = PV/RT = 100×10-3/0.34771× 293.15 = 9.81×10-4 kg CV0 MIX = ∑ ci×CV0 i = 0.0211×10.0849 + 0.4685×0.7445 + 0.2761×0.6529 + 0.2343×0.7448 = 0.9164 kJ/kg K Q 12 = ∆U12 = mCV0(T2-T1) = 9.81×10-4× 0.9164×(100-20) = 0.0719 kJ 12.7

A pipe, cross sectional area 0.1 m2, carries a flow of 75% O2 and 25% N2 by mole with a velocity of 25 m/s at 200 kPa, 290 K. To install and operate a mass flow meter it is necessary to know the mixture density and the gas constant. What are they? What mass flow rate should the meter then show? From Eq. 12.3: ci = yi Mi/ ∑ yjMj MMIX = ∑ yjMj = 0.75×31.999 + 0.25×28.013 = 31.0025 − RMIX = R/MMIX = 8.3145 / 31.0025 = 0.2682 kJ/kg K v = RMIX T/P = 0.2682× 290 / 200 = 0.38889 m3/kg ρ = 1/v = 2.5714 kg/m3 . m = ρAV = 2.5714×0.1×25 = 6.429 kg/s

12-6 12.8

A pipe flows 0.05 kmole a second mixture with mole fractions of 40% CO2 and 60% N2 at 400 kPa, 300 K. Heating tape is wrapped around a section of pipe with insulation added and 2 kW electrical power is heating the pipe flow. Find the mixture exit temperature. C.V. Pipe heating section. Assume no heat loss to the outside, ideal gases. . . - .− . Energy Eq.: Q = m(he − hi) = n(he − hi) = nCP mix(Te − Ti) − − CP mix = ∑ yi Ci = 0.4 × 37.056 + 0.6 × 29.189 = 32.336 kJ/kmole . .− Te = Ti + Q / nCP mix = 300 + 2/(0.05 × 32.336) = 301.2 K

12.9

A rigid insulated vessel contains 0.4 kmol of oxygen at 200 kPa, 280 K separated by a membrane from 0.6 kmol carbon dioxide at 400 kPa, 360 K. The membrane is removed and the mixture comes to a uniform state. Find the final temperature and pressure of the mixture. C.V. Total vessel. Control mass with two different initial states. Mass:

n = nO2 + nCO2 = 0.4 + 0.6 = 1.0 kmole

Process: V = constant (rigid)

=> W = 0, insulated => Q = 0 − − Energy: U2 - U1 = 0 - 0 = n O2 CV O2(T2 - T1 O2) + nCO2CV CO2(T2 - T1 CO2) Initial state from ideal gas Table A.5 − − CV O2 = 0.6618×31.999 = 21.177, CV CO2 = 0.6529×44.01 = 28.73 kJ/kmole − O2 : VO2 = nRT1/P = 0.4 × 8.3145 × 280/200 = 4.656 m3, − CO2 : VCO2 = nRT1/P = 0.6 × 8.3145 × 360/400 = 4.49 m3 − − − − nO2 CV O2 T2 + nCO2CV CO2 T2 = nO2 CV O2 T1 O2 + nCO2CV CO2 T1 CO2 (8.4708 + 17.238 ) T2 = 2371.8 + 6205.68 = 8577.48 kJ => T2 = 333.6 K − P2 = nRT2/V = 1 × 8.3145 × 333.6 / 9.146 = 303.3 kPa 12.10 An insulated gas turbine receives a mixture of 10% CO 2, 10% H2O and 80% N2 on a mole basis at 1000 K, 500 kPa. The volume flow rate is 2 m 3/s and its exhaust is at 700 K, 100 kPa. Find the power output in kW using constant specific heat from A.5 at 300 K. C.V. Turbine, SSSF, 1 inlet, 1 exit flow with an ideal gas mixture, q = 0. . . - .− . Energy Eq.: WT = m(hi − he) = n(hi − he) = nCP mix(Ti − Te) . . − − PV = nRT => n = PV / RT = 500×2/(8.3145×1000) = 0.1203 kmole/s − − CP mix = ∑ yi Ci = 0.1 × 44.01 × 0.842 + 0.1 × 18.015 × 1.872 + 0.8 × 28.013 × 1.042 = 30.43 kJ/kmol K . WT = 0.1203 × 30.43 (1000 − 700) = 1098 kW

12-7 12.11 Solve Problem 12.10 using the values of enthalpy from Table A.8. C.V. Turbine, SSSF, 1 inlet, 1 exit flow with an ideal gas mixture, q = 0. . . - . . - Energy Eq.: WT = m(hi − he) = n(hi − he) = n [ ∑yj (hi − he)j ] . . − − PV = nRT => n = PV / RT = 500×2/(8.3145×1000) = 0.1203 kmole/s . WT = 0.1203[0.1(33397 - 17754) + 0.1(26000 - 14190) + 0.8(21463 - 11937)] = 1247 kW 12.12 Consider Problem 12.10 and find the value for the mixture heat capacity, mole basis and the mixture ratio of specific heats, kmix, both estimated at 850 K from values (differences) of h in Table A.8. With these values make an estimate for the reversible adiabatic exit temperature of the turbine at 100 kPa. − We will find the individual heat capacities by : CP i = (h900 -h800)/(900 - 800) − − CP CO2 = (28030 - 22806)/100 = 52.24; CP H2O = (21937 - 18002)/100 = 39.35 − CP N2 = (18223 - 15046)/100 = 31.77 − − CP mix = ∑ yi CP i = 0.1 × 52.24 + 0.1 × 39.35 + 0.8 × 31.77 = 34.575 kJ/kmol − − − − − CV mix = CP mix - R = 34.575 - 8.3145 = 26.26, k = CP mix/CV mix = 1.3166 Rev. ad. turbine Te =

=> s = constant. Assume const. avg. heat capacities:

k-1 Ti(Pe/Pi) k =

0.3166

1000 (100/500)1.3166 = 1000 (0.2)0.2405 = 679 K

12.13 The gas mixture from Problem 12.6 is compressed in a reversible adiabatic process from the initial state in the sample cylinder to a volume of 0.2 L. Determine the final temperature of the mixture and the work done during the process. CP0 MIX = CV0 MIX + RMIX = 0.9164 + 0.34771 = 1.2641 → k = CP0/CV0 = 1.2641/0.9164 = 1.379

(VV )

→ T2 = T 1

1 k-1 2

= 293.15

(0.21 )

0.3805

= 539.5 K

W12 = - ∆U12 = -mCV0(T2-T1) = - 9.81×10-4 × 0.9164×(539.5-293.15) = - 0.22 kJ

12-8 12.14 Three SSSF flows are mixed in an adiabatic chamber at 150 kPa. Flow one is 2 kg/s of O2 at 340 K, flow two is 4 kg/s of N2 at 280 K and flow three is 3 kg/s of CO2 at 310 K. All flows are at 150 kPa the same as the total exit pressure. Find the exit temperature and the rate of entropy generation in the process. C.V. Mixing chamber, no heat transfer, no work. . . . . Continuity: m1 + m2 + m3 = m4 . . . . Energy: m1h1 + m2h2 + m3h3 = m4h4 . . . . . Entropy: m1s1 + m2s2 + m3s3 + Sgen = m4s4

1 O2 N2 CO2

2 3

4 mix

Assume ideal gases and since T is close to 300 K use heat capacity from A.5 . . . m1CP O2(T1 - T4) + m2CP N2(T2 - T4) + m3CP CO2(T3 - T4)= 0 2 × 0.922 × 340 + 4 × 1.042 × 280 + 3 × 0.842 × 310 = (2 × 0.922 + 4 × 1.042 + 3 × 0.842 ) T4 =>

2577.06 = 8.538 T 4

=>

T4 = 301.83 K

State 4 is a mixture so the component exit pressure is the partial pressure. For each component se − si = CP ln(Te / Ti) − R ln(Pe / Pi) and the pressure ratio is

Pe / Pi = y P4 / Pi = y for each.

n=∑

m 2 4 3 = + + = 0.0625 + 0.1428 + 0.06817 = 0.2735 M 32 28.013 44.01

0.0625 0.1428 0.06817 = 0.2285, yN2 = = 0.5222, yCO2 = = 0.2493 yO2 = 0.2735 0.2735 0.2735 . . . . Sgen = m1(s4 - s1) + m2(s4 - s2) + m3(s4 - s3) =

2 [ 0.922 ln(301.83/340) - 0.2598 ln(0.2285)] + 4 [ 1.042 ln(301.83/280) - 0.2968 ln(0.5222)] + 3 [ 0.842 ln(301.83/310) - 0.1889 ln(0.2493)]

= 0.5475 + 1.084 + 0.2399 = 1.871 kW/K

12-9 12.15 Carbon dioxide gas at 320 K is mixed with nitrogen at 280 K in a SSSF insulated mixing chamber. Both flows are at 100 kPa and the mole ratio of carbon dioxide to nitrogen is 2;1. Find the exit temperature and the total entropy generation per mole of the exit mixture. . . CV mixing chamber, SSSF. The inlet ratio is nCO = 2 nN and assume no 2 2 external heat transfer, no work involved. . . . . Continuity: nCO + 2nN = nex = 3nN ; 2 2 2 . . Energy Eq.: nN (hN + 2hCO ) = 3nN hmix ex 2 2 2 2 - Take 300K as reference and write h = h300 + CPmix(T-300). CP N (Ti N -300) + 2CP CO (Ti CO -300) = 3CP mix(Tmix ex-300) 2 2 2 2 CP mix = ∑ yiCP i = (29.178 + 2 37.05)/3 = 34.42 3CP mixTmix ex = CP N2Ti N2 + 2CP CO2Ti CO2 = 31882 Tmix ex = 308.7 K; Pex N = Ptot/3; Pex CO = 2Ptot/3 2 2 . . ... - . - Sgen = nexsex-(ns)iCO -(ns)iN = nN (se-si)N + 2nN (se-si)CO 2 2 2 2 2 2 Tex − Tex . . − Sgen/3nN2 = [CPN2ln - Rln yN2 + 2CPCO2ln - 2 Rln yCO2]/3 TiN2 TiCO2 = [2.8485 + 9.1343 - 2.6607+6.742]/3 = 5.35 kJ/kmol mix K 12.16 A piston/cylinder contains 0.5 kg argon and 0.5 kg hydrogen at 300 K, 100 kPa. The mixture is compressed in an adiabatic process to 400 kPa by an external force on the piston. Find the final temperature, the work and the heat transfer in the process. C.V. Mixture in cylinder. Control mass with adiabatic process: 1Q2 = 0 Cont.Eq.:

m 2 = m1 = m ;

Entropy Eq.:

Energy Eq.:

m(u2 − u1) = − 1W2

m(s2 − s1) = ∫ dQ/T + 1S2 gen = 0 + 0

assume reversible.

Rmix = ∑ ciRi = 0.5 × 0.2081 + 0.5 × 4.1243 = 2.1662 CP mix = ∑ ciCPi = 0.5 × 0.52 + 0.5 × 14.209 = 7.3645 CV = CP - R = 7.3645-2.1662 = 5.1983,

k = CP / CV = 1.417

s2 = s1 => T2 = T1(P2/P1)(k-1)/k = 300 (400/100)0.2943 = 451 K 1W2

= m(u1 − u2) = mCV (T1 − T2) = 1×5.1983(300 - 451) = -784.9 kJ

12-10 12.17 Natural gas as a mixture of 75% methane and 25% ethane by volume is flowing to a compressor at 17°C, 100 kPa. The reversible adiabatic compressor brings the flow to 250 kPa. Find the exit temperature and the needed work per kg flow. C.V. Compressor. SSSF, adiabatic q = 0, reversible sgen = 0 Energy Eq.:

-w = hex - hin ;

Entropy Eq.:

Process: reversible => sgen = 0 =>

si + sgen = si = se

se = si

Assume ideal gas mixture and constant heat capacity, so we need k and CP CP = ∑ yi CPi = 0.75 × 2.254 × 16.043 + 0.25 × 1.766 × 30.07 = 40.3966 Mmix = 0.75 × 16.043 + 0.25 × 30.07 = 19.55 − Cp = CP/ Mmix = 2.066 , R = R/ Mmix = 8.31451 / 19.55 = 0.4253 Cv = Cp - R = 1.6407 ,

k = Cp/ Cv = 1.259

Te = Ti (Pe/ Pi)(k-1)/k = 290 (250/100) 0.2057 = 350.2 K wc in = CP (Te- Ti) = 2.066 (350.2 – 290) = 124.3 kJ/kg 12.18 Take Problem 12.15 with inlet temperature of 1400 K for the carbon dioxide and 300 K for the nitrogen. First estimate the exit temperature with the specific heats from Table A.5 and use this to start iterations using A.8 to find the exit temperature. . . CV mixing chamber, SSSF. The inlet ratio is nCO = 2 nN and assume no 2 2 external heat transfer, no work involved. CP CO2 = 44.01 × 0.842 = 37.06 CP N2 = 28.013 × 1.042 = 29.189 . . . . Cont. Eq.: 0 = Σnin - Σnex; Energy Eq.: 0 = Σnin hin - Σnex hex . . 0 = 2nN2 CP CO2(Tin- Tex) CO2 + nN2 CP N2 (Tin- Tex) N2 0 = 2 × 37.06 × (1400-Tex) + 29.189 × (300-Tex) 0 = 103768 + 8756.7 – 103.309 Tex _ Tex = 1089 K . . . From Table A.8: Σnin hin = nN2 [2 × 55895 + 1 × 54] = nN2 × 111844 . . . @ 1000K : Σnex hex = nN2 [2 × 33397 + 21463] = nN2 × 88257 . . . @ 1100K : Σnex hex = nN2 [2 × 38885 + 24760] = nN2 × 102530 . . . @ 1200K : Σnex hex = nN2 [2 × 44473 + 28109] = nN2 × 117055 Now linear interpolation between 1100 K and 1200 K Tex = 1100 + 100 ×

111844-102530 = 1164 K 117055-102530

12-11 12.19 A mixture of 60% helium and 40% nitrogen by volume enters a turbine at 1 MPa, 800 K at a rate of 2 kg/s. The adiabatic turbine has an exit pressure of 100 kPa and an isentropic efficiency of 85%. Find the turbine work. Assume ideal gas mixture and take CV as turbine. wT s = hi - hes, ses = si, Tes = Ti(Pe/Pi)(k-1)/k CP mix = 0.6× 5.1926× 4.003 + 0.4× 1.0416× 28.013 = 24.143 − (k-1)/k = R/CP mix = 8.31451/24.143 = 0.3444 Mmix= 0.6×4.003 + 0.4×28.013 = 13.607, CP=CP/Mmix= 1.774 Tes = 800(100/1000)0.3444 = 362 K,

wTs = CP(Ti-Tes) = 777

. wT ac = ηwTs = 660.5 kJ/kg; W = mwTs = 1321 kW 12.20 A mixture of 50% carbon dioxide and 50% water by mass is brought from 1500 K, 1 MPa to 500 K, 200 kPa in a polytropic process through a SSSF device. Find the necessary heat transfer and work involved using values from Table A.5. Process Pvn = constant

leading to

n ln(v2/v1) = ln(P1/P2); v = RT/P n = ln(1000/200)/ln(500 × 1000/200 × 1500) =3.1507 Rmix = ciRi = 0.5 × 0.18892 + 0.5 × 0.46152 = 0.32522 ∑

CP mix = ciCPi = 0.5 × 0.8418 + 0.5 × 1.8723 = 1.357 ∑

⌠vdP = - n (P v -P v )= - nR (T -T ) = 476.4 kJ/kg w = -⌡ n-1 e e i i n-1 e i q = he-hi + w = CP(Te-Ti) + w = -880.6 kJ/kg 12.21 Solve Problem 12.20 using specific heats CP = ∆h/∆T, from Table A.8 at 1000 K. 38885-28030 = 1.223 (1100-900)×44.01 30190-21937 CP H2O = = 2.2906 (1100-900)×18.015

CP CO2 =

CP mix = Σ ciCPi = 0.5 × 1.223 + 0.5 × 2.2906 = 1.7618 Rmix = Σ ciRi = 0.5 × 0.18892 + 0.5 × 0.46152 = 0.32522 Process Pvn = C

=>

n = ln(P1/P2) / ln(v2/v1) = 3.1507

⌠vdP = - n (P v -P v )= - nR (T -T ) = 476.4 kJ/kg w = -⌡ n-1 e e i i n-1 e i q = he-hi + w = 1.7618(500-1500) + 476.4 = -1285.4 kJ/kg

12-12 12.22 A 50/50 (by mole) gas mixture of methane CH 4 and ethylene C2H4 is contained in a cylinder/piston at the initial state 480 kPa, 330 K, 1.05 m3. The piston is now moved, compressing the mixture in a reversible, polytropic process to the final state 260 K, 0.03 m3. Calculate the final pressure, the polytropic exponent, the work and heat transfer and net entropy change for the process. Ideal gas mixture: CH4, C2H4, 50% each by mol => yCH4 = yC2H4 = 0.5 − n = P1V1/RT1 = 480 × 1.05 / (8.31451 × 330) = 0.18369 kmol

State 1:

Cv mix = ∑ yi Mi Cvi = 31.477 kJ/kmol-K V1 T2 = 13236 kPa State 2: T2 = 260 K, V2 = 0.03 m3, Ideal gas => P2 = P1 V T 2

1

V1n-1 T2 V1 =   , ln = (n-1) ln => n = 0.933 T1 V2 T1 V2 1 => 1W2 = ∫ P dv = 1-n(P 2V2- P1V1) = -1595.7 kJ - Energy Eq.: 1Q2 = n(u2- u1) + 1W2 = n Cv mix (T2- T1) + 1W2

Process: PVn = constant,

T2

= 0.18369 × 31.477(260-330) -1595.7 = -2000.4 kJ - s2- s1 = Cv mix ln (T2 / T1) + R ln (V2 / V1) = -37.065 kJ/kmol-K 2nd Law:

∆ Snet = n(s-2- -s1) - 1Q2 /To

with

T0 = 260 K

∆Snet = 0.18369 × (-37.065) - (-2000.4)/260 = 0.8854 kJ/K 12.23 A mixture of 2 kg oxygen and 2 kg of argon is in an insulated piston cylinder arrangement at 100 kPa, 300 K. The piston now compresses the mixture to half its initial volume. Find the final pressure, temperature and the piston work. C.V. Mixture. Control mass, boundary work and no Q, assume reversible. Energy Eq.: u2 - u1 = 1q2 - 1w2 = - 1w2 ; Entropy Eq.: s2-s1 = 0 + 0 = 0 Process: constant s

=>

Pvk = constant, v2 = v1/2,

Assume ideal gases (T1 >> TC ) and use kmix and Cv mix for properties. Rmix = Σ ciRi = 0.5 × 0.25983 + 0.5 × 0.20813 = 0.234 kJ/kg mix CPmix = Σ ciCPi = 0.5 × 0.9216 + 0.5 × 0.5203 = 0.721 kJ/kg mix Cvmix = CPmix -Rmix = 0.487, kmix = CPmix/Cvmix = 1.4805 P2 = P1(v1/v2)k = P1(2)k = 100(2)1.4805 = 279 kPa T2 = T1(v1/v2)k-1 = T1(2)k-1 = 300(2)0.4805 = 418.6 K 1W2

= mtot (u1-u2) = mtot Cv(T1-T2) = 4×0.487 (300 - 418.6) = -231 kJ

12-13 12.24 Two insulated tanks A and B are connected by a valve. Tank A has a volume of 1 m3 and initially contains argon at 300 kPa, 10°C. Tank B has a volume of 2 m3 and initially contains ethane at 200 kPa, 50°C. The valve is opened and remains open until the resulting gas mixture comes to a uniform state. Determine the final pressure and temperature. C.V. Tanks A + B. Control mass no W, no Q. Energy Eq.: U2-U1 = 0 = nArCV0(T2-TA1) + nC H CVO(T2-TB1) 2 6 − nAr = PA1VA/RTA1 = (300 × 1) / (8.31451 × 283.15) = 0.1274 kmol − nC2H6 = PB1VB/RTB1 = (200 × 2) / (8.31451 × 323.15) = 0.1489 kmol Continuity Eq.: Energy Eq.:

n2 = nAr + nC

2H6

= 0.2763 kmol

0.1274 × 39.948 × 0.3122 (T2 - 283.2) + 0.1489 × 30.07 × 1.4897 (T2 - 323.2) = 0

Solving, T2 = 315.5 K − P2 = n2RT2/(VA+VB) = 0.2763×8.3145×315.5/3 = 242 kPa

12-14 12.25 Reconsider the Problem 12.24, but let the tanks have a small amount of heat transfer so the final mixture is at 400 K. Find the final pressure, the heat transfer and the entropy change for the process. C.V. Both tanks. Control mass with mixing and heating of two ideal gases. − nAr = PA1VA/RTA1 =

300×1 = 0.1274 kmol 8.3145×283.2

− nC2H6 = PB1VB/RTB1 = Continuity Eq.: Energy Eq.:

200×2 = 0.1489 kmol 8.3145×323.2

n2 = nAr + nC

2H6

= 0.2763 kmol

U2-U1 = nArCV0(T2-TA1) + nC H CVO(T2-TB1) = 1Q2 2 6

− P2 = n2RT2/(VA+VB) = 0.2763×8.3145×400 / 3 = 306.3 kPa 1Q 2

= 0.1274×39.948×0.3122(400 - 283.15)

+ 0.1489×30.07×1.4897(400 - 323.15) = 698.3 kJ ∆SSYS = nAr∆SAr + nC2H6∆SC2H6 ∆SSURR = -1Q2/TSURR; yAr = 0.1274/0.2763 = 0.4611 T2 − yArP2 ∆SAr = CP Ar ln - R ln TA1 PA1 = 39.948×0.5203 ln

0.4611×306.3 400 - 8.3145 ln 283.15 300

= 13.445 kJ/kmol K ∆SC

T2 − yC2H6P2 = CC2H6 ln - R ln 2H6 TB1 PB1 = 30.07×1.7662 ln

0.5389×306.3 400 - 8.3145 ln 323.15 200

= 12.9270 kJ/kmol K Assume the surroundings are at 400 K (it heats the gas) ∆SNET = nAr∆SAr + nC2H6∆SC2H6 + ∆SSURR = 0.1274×13.445 + 0.1489×12.9270 - 698.3/400 = 1.892 kJ/K

12-15 12.26 A piston/cylinder contains helium at 110 kPa at ambient temperature 20°C, and initial volume of 20 L as shown in Fig. P12.26. The stops are mounted to give a maximum volume of 25 L and the nitrogen line conditions are 300 kPa, 30°C. The valve is now opened which allows nitrogen to flow in and mix with the helium. The valve is closed when the pressure inside reaches 200 kPa, at which point the temperature inside is 40°C. Is this process consistent with the second law of thermodynamics? P1 = 110 kPa, T1 = 20 oC, V1 = 20 L, Vmax = 25 L = V2 P2 = 200 kPa, T2 = 40 oC, Pi = 300 kPa, Ti = 30 oC Const. P to stops, then const. V = Vmax Qcv = U2 - U1 + Wcv - nihi, Wcv = P1(V2 - V1) = n2h2 - n1h1 - nihi - (P2 - P1)V2 = nA(hA2 - hAi) + nB(hB2 - hB1) - (P2 - P1)V2 nB = n1 = P1V1/RT1 = 110×0.02/8.3145×293.2 = 0.0009 kmol n2 = nA + nB = P2V2/RT2 = 200×0.025/8.3145×313.2 = 0.00192 kmol, nA = n2 - nB = 0.00102 kmol Qcv = 0.00102×28.013×1.0416(40-30) + 0.0009×4.003×5.1926(40-20) - (200 - 110) 0.025 = 0.298 + 0.374 - 2.25 = - 1.578 kJ Sgen = n2s2 - n1s1 - nisi - Qcv/T0 = nA(sA2 - sAi) + nB(sB2 - sB1) - Qcv/T0 yA2 = 0.00102/0.00192 = 0.5313, yB2 = 0.4687 313.2 - 0.5313*200 sA2 - sAi = 29.178 ln - R ln = 9.5763 303.2 300 313.2 - 0.4687*200 - R ln = 2.7015 sB2 - sB1 = 20.786 ln 293.2 110 Sgen = 0.00102×9.5763 + 0.0009×2.7015 + 1.578/293.2 = 0.0176 kJ/K > 0

Satisfies 2nd law.

12-16 12.27 Repeat Problem 12.17 for an isentropic compressor efficiency of 82%. C.V. Compressor. SSSF, adiabatic q = 0, reversible sgen = 0 Energy Eq.:

-w = hex - hin ;

Entropy Eq.:

si + sgen = si = se

Process: For ideal compressor reversible => sgen = 0 =>

se = si

Assume ideal gas mixture and constant heat capacity, so we need k and CP CP = ∑ yi CPi = 0.75 × 2.254 × 16.043 + 0.25 × 1.766 × 30.07 = 40.3966 Mmix = 0.75 × 16.043 + 0.25 × 30.07 = 19.55 − Cp = CP/ Mmix = 2.066 , R = R/ Mmix = 8.31451 / 19.55 = 0.4253 Cv = Cp - R = 1.6407 ,

k = Cp/ Cv = 1.259

Te = Ti (Pe/ Pi)(k-1)/k = 290 (250/100) 0.2057 = 350.2 K wc in = CP (Te- Ti) = 2.066 (350.2 – 290) = 124.3 kJ/kg wc actual = wc in/η = 124.3/0.82 = 151.6 kJ/kg = Cp (Te actual - Ti) => Te actual = T + w c actual/CP = 290 + 151.6 / 2.066 = 363.4 K 12.28 A spherical balloon has an initial diameter of 1 m and contains argon gas at 200 kPa, 40°C. The balloon is connected by a valve to a 500-L rigid tank containing carbon dioxide at 100 kPa, 100°C. The valve is opened, and eventually the balloon and tank reach a uniform state in which the pressure is 185 kPa. The balloon pressure is directly proportional to its diameter. Take the balloon and tank as a control volume, and calculate the final temperature and the heat transfer for the process. PA1VA1 π 200×0.5236 A = = 1.606 kg VA1 = 13 = 0.5236, mA1 = RTA1 6 0.208 13×313.2 mB1 = PB1VB1/RTB1 = 100×0.50/0.18892×373.2 = 0.709 kg -1/3

(PP ) = 0.5236(185 ) 200

-1/3

P2VA2 = PA1VA1 → VA2 = VA1

B

CO 2

2: Uniform ideal gas mix. :

2

3

3

= 0.4144 m3

A1

P2(VA2+VB) = (mARA+mBRB)T2

T2 = 185(0.4144+0.50) / (1.606×0.20813 + 0.709×0.18892) = 361.3 K W12 =

P2VA2-PA1VA1 1-(-1/3)

=

185×0.4144-200×0.5236 = -21.0 kJ (4/3)

Q = mACV0A(T2-TA1) + mBCV0B(T2-TB1) + W12 = 1.606×0.3122(361.3-313.2) + 0.709×0.6529(361.3-373.2) - 21.0 = 18.6 - 21.0 = -2.4 kJ

12-17 12.29 A large SSSF air separation plant takes in ambient air (79% N2, 21% O2 by volume) at 100 kPa, 20°C, at a rate of 1 kmol/s. It discharges a stream of pure O2 gas at 200 kPa, 100°C, and a stream of pure N 2 gas at 100 kPa, 20°C. The plant operates on an electrical power input of 2000 kW. Calculate the net rate of entropy change for the process. Air 79 % N2 21 % O2 P1 = 100 kPa

1

P2 = 200 kPa

2 pure O 2

T2 = 100 oC

pure N 2 3

T1 = 20 oC . n1 = 1 kmol/s

P3 = 100 kPa

. -WIN = 2000 kW

T3 = 20 oC

. . . . . . . Sgen = Σ ni∆s-i - QCV/T0 = (n2-s2 + n3-s3 - n1-s1) - QCV/T0 . . . . . . QCV = Σ n∆hi + WCV = nO2CP0 O2(T2-T1) + nN2CP0 N2(T3-T1) + WCV = 0.21 ×32 × 0.9216 ×(100-20) + 0 - 2000 = -1505 kW .

[

373.2

200

Σ ni∆s-i = 0.21 29.49 ln 293.2 - 8.3145 ln 21

]

+ 0.79 [0 - 8.3145 ln (100/79)]= -3.9894 kW/K . Sgen = 1505/293.2 - 3.9894 = 1.1436 kW/K

12-18 12.30 An insulated vertical cylinder is fitted with a frictionless constant loaded piston of cross sectional area 0.1 m 2 and the initial cylinder height of 1.0 m. The cylinder contains methane gas at 300 K, 150 kPa, and also inside is a 5-L capsule containing neon gas at 300 K, 500 kPa. The capsule now breaks, and the two gases mix together in a constant pressure process. What is the final temperature, final cylinder height and the net entropy change for the process. Ap = 0.1 m2, h = 1.0 m =>

Vtot = Va1 + Vb1 = 0.1 m3

Methane: M = 16.04 kg/kmol, Cp = 2.2537 kJ/kg-K, R = 0.51835 kJ/kg-K Neon: M = 20.183 kg/kmol, Cp = 1.0299 kJ/kg-K, R = 0.41195 kJ/kg-K State 1: Methane, Tal = 300 K, Pa1 = 150 kPa, Va1 = Vtot- Vb1 = 0.095 m3 Neon, Tbl = 300 K, Pb1 = 500 kPa, Vb1 = 5 L ma = mb =

PalVal

= 0.0916 kg, na =

RaTal PblVbl RbTbl

= 0.0202 kg, nb =

ma Ma mb Mb

= 0.00571 kmol

= 0.001 kmol

State 2: Mix, P2mix = Pa1 = 150 kPa Energy Eq: 1Q2 = ma(ua2-ua1) + mb( ub2 - ub1) + 1W2;

1Q2

=0

= ∫ P dV = P2(V2-V1) tot; P2V2 = mtotRmixT2 = (maRa + mbRb)T2

1W2

Assume Constant Specific Heat 0 = maCpa(T2-Ta1) + mbCpb(T2-Tb1 + (maRa + mbRb)T2 - P2V1 tot Solving for T2 = 293.3 K V2 =

(maRa + mbRb)T2 P2

Entropy Eq.:

ya =

na na+nb

= 0.1087 m3,

h2 =

V2 Ap

∆Snet = ma(sa2- sa1) + mb(sb2-sb1) = 0.851,

= 1.087 m 1Q 2

T0

yb = 1 - ya = 0.149

T2 yaP2 = 0.02503 kJ/kg-K sa2 - sa1 = Cpa ln - Raln Ta1 Pa1 T2 ybP2 - Rb ln = 1.2535 kJ/kg-K sb2 - sb1 = Cpb ln Tb1 Pb1 ∆Snet = 0.0276 kJ/K

;

1Q2

=0

12-19 12.31 The only known sources of helium are the atmosphere (mole fraction approximately 5 × 10−6) and natural gas. A large unit is being constructed to separate 100 m3/s of natural gas, assumed to be 0.001 He mole fraction and 0.999 CH 4. The gas enters the unit at 150 kPa, 10°C. Pure helium exits at 100 kPa, 20°C, and pure methane exits at 150 kPa, 30°C. Any heat transfer is with the surroundings at 20°C. Is an electrical power input of 3000 kW sufficient to drive this unit? P2 = 100 kPa 1 2 He 0.999 CH4 T = 20 oC 2

0.001 He at 150 kPa, 10 oC . V1 = 100 m3/s

CH 4 3

P3 = 140 kPa

. -WCV = 3000 kW

T3 = 30 oC

. . n1 = P1V1/RT1 = 150×100/(8.3145×283.2) = 6.37 kmol/s . . . => n2 = 0.001; n1 = 0.006 37; n3 = 6.3636 CP CH4= 16.04×2.2537 = 36.149 CP He= 4.003×5.1926 = 20.786, . . . . . - . . . QCV = n2h2 + n3h3 - n1h1 + WCV = n2CP0 He(T2-T1) + n3CP0 CH (T3-T1) + WCV 4 = 0.00637×20.786(20-10)+6.3636×36.149(30-10)+(-3000) = +1600 kW . . . . . Sgen = n2-s2 + n3-s3 - n1-s1 - QCV/T0

[

= 0.00637 20.786 ln

[

293.2 100 - 8.3145 ln 283.2 0.001×150

+ 6.3636 36.149 ln

]

303.2 140 - 8.3145 ln 283.2 0.999×150

] - 1600/293.2

= +13.498 kW/K > 0 12.32 A steady flow of 0.01 kmol/s of 50% carbon dioxide and 50% water at 1200K and 200 kPa is used in a heat exchanger where 300 kW is extracted from the flow. Find the flow exit temperature and the rate of change of entropy using Table A.8. C.V. Heat exchanger, SSSF, 1 inlet, 1 exit, no work. Continuity Eq.: yCO2 = yH2O = 0.5 . - . . . . Energy Eq.: Q = m(he − hi) = n(he − hi) => he = hi + Q/n Inlet state: Table A.8 hi = 0.5 × 44473 + 0.5 × 34506 = 39489.5 kJ/kmol . . Exit state: he = hi + Q/n = 39489.5 - 300/0.01 = 9489.5 kJ/kmol Trial and error for T with h values from Table A.8 @500 K he = 0.5(8305 + 6922) = 7613.5 @600 K he = 0.5(12906 + 10499) = 11702.5 Interpolate to have the right h:

T = 545.9 K

12-20 12.33 An insulated rigid 2 m3 tank A contains CO2 gas at 200°C, 1MPa. An uninsulated rigid 1 m3 tank B contains ethane, C2H6, gas at 200 kPa, room temperature 20°C. The two are connected by a one-way check valve that will allow gas from A to B, but not from B to A. The valve is opened and gas flows from A to B until the pressure in B reaches 500 kPa and the valve is closed. The mixture in B is kept at room temperature due to heat transfer. Find the total number of moles and the ethane mole fraction at the final state in B. Find the final temperature and pressure in tank A and the heat transfer to/from tank B. Tank A: VA = 2 m3, state A1 : CO2, TA1 = 200°C = 473.2 K, PA1 = 1 MPa C v0 CO2 = 0.653 × 44.01 = 28.74, C P0 CO2 = 0.842 × 44.01 = 37.06 Tank B: VB = 1 m3, state B1: C2H6, TB1 = 20°C = 293.2 K, PB1 = 200 kPa Slow Flow A to B to PB2 = 500 kPa and assume TB2 = TB1 = T0 PB1VB = nB1RTB1 , yC2H6 B2 = nB1 =

PB1VB RTB1

nB2 mix =

nB1 nB2

=

=

PB1 PB2

PB2VB = nB2 mix RTB2 =

200 = 0.400 500

200 × 1 = 0.08204 kmol R × 293.2

PB2VB RTB2

=

500 × 1 = 0.2051 kmol R × 293.2

nCO2 B2 = 0.2051 – 0.08201 = 0.12306 kmol PA1VA

1000×2 = 0.50833 kmol ; nA2 = nA1 - nCO2 B2 = 0.38527 kmol RTA1 R×473.2 = 0 = n u- n u+ n hC.V. A: All CO Energy Eq.: Q nA1 =

=

2

CV A

A2

A2

A1

A1

ave

ave

0 = nA2 C v0 CO2TA2 - nA1 C v0 CO2TA1 + nave C P0 CO2 ( TA1 + TA2)/2 0 = 28.74( 0.38527× TA2 - 0.50833×473.2) + 0.12306 × 37.06(473.2 + TA2)/2 => PA2 =

TA2 = 436.9 K

nA2 RTA2 VA

=

0.38527 × R × 436.9 = 700 kPa 2

C.V. B: QCV B + nBi h Bi ave = (nu-) B2 - (nu-) B1 = (nu-)CO2 B2 + (nu-)C2H6 B2 - (nu-)C2H6 B1 QCV B = 0.12306 × 28.74 × 293.2 + 0 – 0.12306 × 37.06 (473.2 + 436.9)/2 = -1037.8 kJ

12-21 12.34 A tank has two sides initially separated by a diaphragm. Side A contains 1 kg of water and side B contains 1.2 kg of air, both at 20°C, 100 kPa. The diaphragm is now broken and the whole tank is heated to 600°C by a 700°C reservoir. Find the final total pressure, heat transfer and total entropy generation. C.V. Total tank out to reservoir. Energy Eq.:

U2-U1 = ma(u2-u1)a + mv(u2-u1)v = 1Q2

Entropy Eq.: S2-S1 = ma(s2-s1)a + mv(s2-s1)v = 1Q2/Tres + Sgen Volume:

V2 = VA + VB = mvvv1 + mava1 = 0.001 + 1.009 = 1.01 m3

vv2 = V2/mv = 1.01, T2 => P2v = 400 kPa va2 = V2/ma = 0.8417, T2 => P2a = mRT2/V2 = 297.7 kPa P2tot = P2v + P2a = 697.7 kPa Water table B.1:

u1 = 83.95, u2 = 3300, s1 = 0.2966, s2 = 8.4558

Air table A.7: u1 = 293, u2 = 652.3, sT1 = 2.492, sT2 = 3.628 From energy equation we have 1Q2

= 1(3300 - 83.95) + 1.2(652.3 - 293) = 3647.2 kJ

From the entropy equation we have Sgen = 1(8.4558 - 0.2966) + 1.2[3.628 - 2.492 - 0.287×ln(301.6/100)] - 3647.2 / 973.2 = 5.4 kJ/K

12-22 12.35 A 0.2 m3 insulated, rigid vessel is divided into two equal parts A and B by an insulated partition, as shown in Fig. P12.35. The partition will support a pressure difference of 400 kPa before breaking. Side A contains methane and side B contains carbon dioxide. Both sides are initially at 1 MPa, 30°C. A valve on side B is opened, and carbon dioxide flows out. The carbon dioxide that remains in B is assumed to undergo a reversible adiabatic expansion while there is flow out. Eventually the partition breaks, and the valve is closed. Calculate the net entropy change for the process that begins when the valve is closed. A CH 4

∆PMAX = 400 kPa,

B CO 2

PA1 = PB1 = 1 MPa

VA1 = VB1 = 0.1 m3 TA1 = TB1 = 30 oC = 303.2 K

CO2 inside B: sB2 = sB1 to PB2 = 600 kPa (PA2 = 1000 kPa) 0.289

600 1.289 = 270.4 K For CO2, k = 1.289 => TB2 = 303.2 1000 nB2 = PB2VB2/RTB2 = 600×0.1/8.3145×270.4 = 0.026 688

(

)

nA2 = nA1 = 1000×0.1/8.3145×303.2 = 0.039 668 kmol Q23 = 0 = n 3u-3 - ni2u-i2 + 0 = n A2CV0 A(T3-TA2) + nB2CV0 B(T3-TB2) = 0 ∑

i

0.039 668×16.04×1.736(T3-303.2) + 0.026 688×44.01×0.653(T3-270.4) = 0 Solve

T3 = 289.8 K

P3 = 0.066356×8.3145×289.8/0.2 = 799.4 kPa PA3 = 0.5978×799.4 = 477.9 kPa ,

PB3 = P3 - PA3 = 321.5 kPa

-s - -s = 16.04×2.254 ln(289.8) - 8.3145 ln 477.9 = 4.505 kJ/kmol K A3 A2 303.2 1000 -s - -s = 44.01×0.842 ln(289.8) - 8.3145 ln 321.5 = 7.7546 kJ/kmol K B3 B2 270.4 600 ∆SNET = 0.039668×4.505 + 0.026688×7.7546 = +0.3857 kJ/K

12-23 12.36 Consider 100 m3 of atmospheric air which is an air–water vapor mixture at 100 kPa, 15°C, and 40% relative humidity. Find the mass of water and the humidity ratio. What is the dew point of the mixture? Air-vap P = 100 kPa, T= 15 oC, φ = 40% Pg = Psat15 = 1.705 kPa mv =

P vV RvT

=

=>

Pv = φ Pg = 0.4×1.705 = 0.682 kPa

0.682×100 = 0.513 kg 0.461×288.15

Pa = Ptot- Pv1 = 100 – 0.682 = 99.32 ma = w1 =

P aV RaT mv ma

=

=

99.32×100 = 120.1 kg 0.287×288.15

0.513 = 0.0043 120.1

Tdew is T when Pv = Pg = 0.682 kPa;

T = 1.4 oC

12.37 The products of combustion are flowing through a SSSF heat exchanger with 12% CO2, 13% H2O and 75% N2 on a volume basis at the rate 0.1 kg/s and 100 kPa. What is the dew-point temperature? If the mixture is cooled 10°C below the dewpoint temperature, how long will it take to collect 10 kg of liquid water? yH

2O

= 0.13; PH2O = 0.13×100 = 13 kPa, TDEW = 50.95 oC

Cool to 40.95 oC → PG = 7.805 kPa yH

2O

nH

= 7.805/100 = nH2O(v)/(nH2O(v) + 0.87)

2O(v)

= 0.07365 per kmol mix in

→ nLIQ = 0.13 - 0.07365 = 0.05635 MMIX IN = 0.12×44.01 + 0.13×18.015 + 0.75×28.013 = 28.63 kg/kmol . . nMIX IN = mTOTAL/MMIX IN = 0.1/28.63 = 0.003493 kmol/s . nLIQ COND = 0.003 493×0.05635 = 0.000 197 kmol/s . or mLIQ COND = 0.000 197 × 18.015 = 0.003 55 kg/s For 10 kg, takes ~ 47 minutes

12-24 12.38 A new high-efficiency home heating system includes an air-to-air heat exchanger which uses energy from outgoing stale air to heat the fresh incoming air. If the outside ambient temperature is −10°C and the relative humidity is 30%, how much water will have to be added to the incoming air, if it flows in at the rate of 1 m3/s and must eventually be conditioned to 20°C and 40% relative humidity? Outside ambient air:

PV1 = φ1PG1 = 0.30×0.2602 = 0.078 kPa

Assuming P1 = P2 = 100 kPa, => PA1 = 100 - 0.078 = 99.922 kPa . PA1V1 99.922×1 0.078 . mA = = = 1.3228 kg/s w1 = 0.622× = 0.00049 99.922 RAT1 0.287×263.2 Conditioned to :

T2 = 20 oC, φ2 = 0.40

PV2 = φ2PG2 = 0.40×2.339 = 0.9356 =>

w2 = 0.622×

0.9356 = 0.00587 99.064

. . mLIQ IN = mA(w2-w1) = 1.3228(0.00587 - 0.00049) = 0.00712 kg/s = 25.6 kg/h 12.39 Consider 100 m3 of atmospheric air at 100 kPa, 25°C, and 80% relative humidity. Assume this is brought into a basement room where it cools to 15°C, 100 kPa. How much liquid water will condense out? Pg = Psat25 = 3.169 kPa => Pv = φ Pg = 0.8 × 3.169 = 2.535 kPa P vV 2.535×100 mv1 = = = 1.844 kg RvT 0.461×298.15 Pg15°C = 1.705 < Pv1 => State 2 is saturated φ2 = 100% , Pv2 = Pg2 = 1.705 P vV 1.705×100 mv2 = = = 1.2835 kg RvT 0.461×288.15 mliq = mv1 - mv2 = 1.844 – 1.2835 = 0.56 kg 12.40 A flow of 2 kg/s completely dry air at T1, 100 kPa is cooled down to 10°C by spraying liquid water at 10°C, 100 kPa into it so it becomes saturated moist air at 10°C. The process is SSSF with no external heat transfer or work. Find the exit moist air humidity ratio and the flow rate of liquid water. Find also the dry air inlet temperature T1. 2: saturated

Pv = Pg = 1.2276 kPa

and

hfg (10°C)= 2477.7 kJ/kg

w2 = 0.622 × 1.2276/ (100 - 1.2276) = 0.00773 C.V. Box . . . . . ma + mliq = m a(1 + w2) => mliq = w2 ma = 0.0155kg/s . . . ma ha1 - mliq hf = ma (ha2 + w2 hg2) ha1 - ha2 = Cpa (T1- T2) = w2 hg2- w2 hf = w2 hfg = 19.15 =>

T1 = 29.1°C

12-25 12.41 A piston/cylinder has 100 kg of saturated moist air at 100 kPa, 5°C. If it is heated to 45°C in an isobaric process, find 1Q2 and the final relative humidity. If it is compressed from the initial state to 200 kPa in an isothermal process, find the mass of water condensing. Energy Eq.:

m(u2-u1) = Q12 - W12,

Initial state 1: φ1 = 100%, h˜1 = 39 kJ/kg dry air, w1 = 0.054 ma = mtot/(1+ w1) = 99.463 kg, mv1 = w1ma = 0.537 kg Case a: P = constant => W12 = mP(v2-v1) => Q12 = m(u2-u1) + W12 = m(h2-h1) = ma(h˜2 - h˜1) State 2: w2 = w1, T2 => h˜2 = 79 , φ2 = 9.75% Q12 = 99.463 (79 - 39) = 3979 kJ. Case b: T = constant & φ2 = 100% => Pv = Pg = 0.8721 kPa w2 = 0.622 Pv2/Pa2 = 0.622 Pg2/(Ptot2-Pg2) = 0.002697 mv2 = w2 ma = 0.268 kg, mliq = mv1 - mv2 = 0.269 kg 12.42 A flow moist air at 100 kPa, 40°C, 40% relative humidity is cooled to 15°C in a constant pressure SSSF device. Find the humidity ratio of the inlet and the exit flow, and the heat transfer in the device per kg dry air. . . . C.V. Cooler. mv1 = mliq + mv2 Psychrometric chart: State 2: T < Tdew = 23°C => φ2 = 100% . . . . ~ mv2/ma = ω2 = 0.0107 , h˜2 = 62 mv1/ma = ω1 = 0.018, h1 = 106; . . mliq/ma = ω1- ω2 = 0.0073 , h˜f = 62.98 . . . . m q- = m h˜ - m h - m h˜ => a

out

a 1

liq f

a

2

q-out = h˜1 - (ω1- ω2) hf - h˜2 = 106 – 0.0073 × 62.98 – 62 = 43.54 kJ/kg-dry air Tables:

Pg1 = 7.384 kPa

ω1 = 0.0189

Pv1 = 2.954

Pv2 = 1.705 kPa = Pg2 => ω2 = 0.0108 hv1 = 2574.3 ,

hv2 = 2528.9 ,

hf = 62.98

q-out = CP (T1 - T2) + ω1hv1 - ω2 hv2 - (ω1- ω2) hf = 0.717(40 - 15) + 0.0189 × 2574.3 - 0.0108 × 2528.9 - 0.0073 × 62.98 = 38.8 kJ/kg dry air

12-26 12.43 Ambient moist air enters a steady-flow air-conditioning unit at 102 kPa, 30°C, with a 60% relative humidity. The volume flow rate entering the unit is 100 L/s. The moist air leaves the unit at 95 kPa, 15°C, with a relative humidity of 100%. Liquid condensate also leaves the unit at 15°C. Determine the rate of heat transfer for this process. State 1:

PV1 = φ1PG1 = 0.60×4.246 = 2.5476

w1 = 0.622 × 2.5476/(102 - 2.5476) = 0.01593 . PA1V1 99.45×0.1 . = = 0.1143 kg/s mA = RAT1 0.287×303.2 PV2 = PG2 = 1.705, w2 = 0.622 × 1.705/(95 - 1.705) = 0.01137 . . . . . . Energy Eq.: QCV + mAhA1 + mV1hV1 = mAhA2 + mV2hA2 + m3hL3 . . QCV/mA = CP0A(T2-T1) + w2hV2 - w1hV1 + (w1-w2)hL3 = 1.004(15-30) + 0.01137×2528.9 - 0.01593×2556.2 + 0.00456×63.0 = -26.732 kJ/kg air . Q CV = 0.1143(-26.73) = -3.055 kW 12.44 A steady supply of 1.0 m3/s air at 25°C, 100 kPa, 50% relative humidity is needed to heat a building in the winter. The outdoor ambient is at 10°C, 100 kPa, 50% relative humidity. What are the required liquid water input and heat transfer rates for this purpose? Air: R = 0.287 kJ/kg K, Cp = 1.004 kJ/kg-K State 1: T1 = 10°C,

φ1 = 50%, P1 = 100 kPa

Pg1= 1.2276 kPa,

Pv1= φ1Pg1 = 0.6138 kPa,

Pa1 = P1- Pv1 = 99.39 kPa => ω1 = 0.622 Pv1/Pa1 = 0.003841 . State 2: T2 = 25°C, P2 = 100 kPa, φ2 = 50%, V a2 = 1 m3/s Pg2 = 3.169 kPa,

Pv2= φ2Pg2 = 1.5845 kPa,

P = P2 - Pv2 = 98.415 kPa, ω2 = 0.622 Pv2/Pa2 = 0.010014 . a2 . m a2 = Pa2 V a2/RTa2 = 98.415 × 1/(0.287 × 298.15) = 1.15 kg/s Steam tables B.1.1:

hv1 = 2519.7 kJ/kg,

hv2 = 2547.2 kJ/kg

State 3: Assume: Liq. Water at T3 = 25°C, hf3 = 104.9 kJ/kg . . . . . Conservation of Mass: ma1 = ma2, mf3 = mv2 - mv1 . . mf3= ma2(ω2 - ω1) = 1.15 × 0.006173 = 0.0071 kg/s . . . . . . 1stLaw: Q + ma1ha1 + mv1hv1 + mf3hf3 = ma2ha2 +mv2hv2 . . mf3 . Q . = Cp(T2- T1) + ω2hv2 - ω1hv1 - . hf3 => Q = 34.76 kW ma ma

12-27 12.45 Consider a 500-L rigid tank containing an air–water vapor mixture at 100 kPa, 35°C, with a 70% relative humidity. The system is cooled until the water just begins to condense. Determine the final temperature in the tank and the heat transfer for the process. Pv1 = φPG1 = 0.7×5.628 = 3.9396 kPa Since mv = const & V = const & also Pv = PG2: PG2 = Pv1× T2/T1 = 3.9396× T2/308.2 = 0.01278 T2 Assume T2 = 30oC:

0.01278×303.2 = 3.875 =/ 4.246 = PG 30C

Assume T2 = 25oC:

0.01278×298.2 = 3.811 =/ 3.169 = PG 25C

interpolating → T2 = 28.2 oC w2 = w1 = 0.622

3.9396 = 0.025 51 (100-3.9369)

ma = Pa1V/RaT1 = (100-3.94)×0.5/(0.287×308.2) = 0.543 kg 1st law: Q12 = U2-U1 = ma(ua2-ua1) + mv(uv2-uv1) = 0.717(28.2 - 35) + 0.02551 (2414.2 - 2423.4) = -5.11 kJ/kg → Q12 = 0.543(-5.11) = -2.77 kJ 12.46 Air in a piston/cylinder is at 35°C, 100 kPa and a relative humidity of 80%. It is now compressed to a pressure of 500 kPa in a constant temperature process. Find the final relative and specific humidity and the volume ratio V2/V1. Check to see if the second state is saturated or not. First assume no water is condensed 1: w1 = 0.029

2: w2 = 0.622 Pv2/(P2-Pv2) w2 = w1 => Pv2 = 22.568 > Pg = 5.628 kPa Conclusion is state 2 is saturated φ2 = 100%, w2 = 0.622 Pg/(P2-Pg) = 0.00699 To get the volume ratio, write the ideal gas law for the vapor phases V2 = Va2 + Vv2 + Vf2 = (maRa + mv2Rv)T/P2 + mliqvf V1 = Va1 + Vv1 = (maRa + mv1Rv)T/P1 Take the ratio and devide through with maRaT/P2 to get V2/V1 = (P1/P2)[1 + 0.622w2 + (w1-w2)P2vf/RaT] / [1+0.622 w1] = 0.1973 The liquid contribution is nearly zero (=0.000126) in the numerator.

12-28 12.47 A 300-L rigid vessel initially contains moist air at 150 kPa, 40°C, with a relative humidity of 10%. A supply line connected to this vessel by a valve carries steam at 600 kPa, 200°C. The valve is opened, and steam flows into the vessel until the relative humidity of the resultant moist air mixture is 90%. Then the valve is closed. Sufficient heat is transferred from the vessel so the temperature remains at 40°C during the process. Determine the heat transfer for the process, the mass of steam entering the vessel, and the final pressure inside the vessel. i

Pv1 = φ1PG1 = 0.1×7.384 = 0.7384 kPa

H 2O

Pv2 = 0.9×7.384 = 6.6456 kPa Pa2 = Pa1 = 150 - 0.738 = 149.262 kPa AIR + H 2O

w1 = 0.622 ×

0.7384 = 0.003 08 149.262

w2 = 0.622 ×

6.6456 = 0.0277 149.262

ma = 149.262×0.3/(0.287×313.2) = 0.5 kg P2 = 149.262 + 6.6456 = 155.9 kPa mvi = 0.5(0.0277 - 0.00308) = 0.0123 kg uv1 = uv2 ≈ uG at 40 oC and

ua2 = ua1

QCV = ma(ua2 - ua1) + mv2uv2 - mv1uv1 - mvihi = mvi(uG at T - hi) = 0.0123(2430.1 - 2850.1) = -5.15 kJ

12-29 12.48 A combination air cooler and dehumidification unit receives outside ambient air at 35°C, 100 kPa, 90% relative humidity. The moist air is first cooled to a low temperature T2 to condense the proper amount of water, assume all the liquid leaves at T2. The moist air is then heated and leaves the unit at 20°C, 100 kPa, relative humidity 30% with volume flow rate of 0.01 m3/s. Find the temperature T2, the mass of liquid per kilogram of dry air and the overall heat transfer rate.

MIX IN

1

2

COOL

MIX OUT

HEAT 2'

. -Q C LIQ OUT

CV . QH

a) Pv1 = φ1PG1 = 0.9 × 5.628 = 5.0652 kPa w1 = 0.622 ×

5.0652 = 0.033 19 100-5.0652

Pv3 = φ3PG3 = 0.3 × 2.339 = 0.7017 kPa 0.7017 w2 = w3 = 0.622 × = 0.0044 100-0.7017 . . mLIQ 2′/ma = w1 - w2 = 0.033 19 - 0.0044 = 0.028 79 kg/kg air PG2 = Pv3 = 0.7017 kPa

→

T2 = 1.7 oC

b) For a C.V. around the entire unit . . . QCV = QH + QC Net heat transfer, 1st law: . . . . QCV/ma = (ha3-ha1) + w3hv3 - w1hv1 + mL2′ hL2′/ma = 1.004(20-35) + 0.0044×2538.1 - 0.033 19×2565.3 + 0.028 79×7.28 = -88.82 kJ/kg air . Pa3V3 (100-0.7017)×0.01 . ma = = = 0.0118 kg/s RaT3 0.287×293.2 . QCV = 0.0118(-88.82) = -1.05 kW

12-30 12.49 A rigid container, 10 m3 in volume, contains moist air at 45°C, 100 kPa, φ = 40%. The container is now cooled to 5°C. Neglect the volume of any liquid that might be present and find the final mass of water vapor, final total pressure and the heat transfer. CV container.

m 2 = m1 ;

m2u2 - m1u1 = 1Q2

1: 45°C, φ = 40% => w1 = 0.0236 ,

Tdew = 27.7°C

Final state T2 < Tdew so condensation, φ2 = 100% Pv1 = 0.4 Pg = 0.4*9.593 = 3.837 kPa,

Pa1 = Ptot - Pv1 = 97.51 kPa

ma = Pa1V/RT1 = 10.679 kg, mv1 = w1 ma = 0.252 kg Pv2 = Pg2 = 0.8721 kPa, Pa2 = Pa1T2/T1 = 85.25 kPa P2 = Pa2 + Pv2 = 86.12 kPa mv2 = Pv2V/RvT2 = 0.06794 kg

(=V/vg= 0.06797 steam table)

mf2 = mv1 - mv2 = 0.184 kg The heat transfer from the energy equation becomes 1Q 2

= ma(u2-u1)a + mv2ug2 + mf2uf2 - mv1ug1 = ma Cv(T2-T1) + mv2 2382.3 + mf2 20.97 - mv1 2436.8 = -306.06 + 161.853 + 3.858 - 614.07 = - 754.4 kJ

12.50 A saturated air-water vapor mixture at 20 oC, 100 kPa, is contained in a 5-m3 closed tank in equilibrium with 1 kg of liquid water. The tank is heated to 80oC. Is there any liquid water in the final state? Find the heat transfer for the process. a) Since VLIQ = mLIQvF ≈ 0.001 m3, VGAS ≈ V

AIR +

VAP LIQ

ma =

Pa1V RaT1

At state 2:

φ1 = 1.00 → Pv1 = PG1 = 2.339 kPa

Q12

w1 = 0.622× 2.339 /(100 - 2.339) = 0.0149 =

97.661×4.999 = 5.802 kg 0.287×293.2 Pa2 = 97.661×

=>

mv1 = w1ma = 0.086 kg

353.2 4.999 × = 117.623 kPa 293.2 5

wMAX 2 = 0.622 × 47.39 / 117.623 = 0.2506 But

w2 ACTUAL =

0.086 + 1.0 = 0.1872 < wMAX 2 5.802

→ No liquid at 2

Q12 = ma (ua2-ua1) + mv2 uv2 - mv1uv1 - mliq 1uliq 1 = 5.802 × 0.717(80 - 20) + 1.086 × 2482.2 - 0.086 × 2402.9 - 1 × 84.0 = 249.6 + 2695.7 - 206.65 - 84 = 2655 kJ

12-31 12.51 An air-water vapor mixture enters a steady flow heater humidifier unit at state 1: 10°C, 10% relative humidity, at the rate of 1 m3/s. A second air-vapor stream enters the unit at state 2: 20°C, 20% relative humidity, at the rate of 2 m3/s. Liquid water enters at state 3: 10°C, at the rate of 400 kg per hour. A single airvapor flow exits the unit at state 4: 40°C. Calculate the relative humidity of the exit flow and the rate of heat transfer to the unit. Assume: P = 100 kPa . State 1: T1 = 10°C, φ1 = 10%, Va1 = 1 m3/s Pv1= φ1Pg1 = 0.1228 kPa, Pa1 = P - Pv1 = 99.877 kPa . Pv1 Pa1Va1 . ω1 = 0.622 = 0.000765, ma1 = = 1.2288 kg/s Pa1 RTa1 . . mv1 = ω1m a1 = 0.00094 kg/s, hv1 = hg1 = 2519.7 kJ/kg . State 2: T2 = 20°C, φ2 = 20%, Va2 = 2 m3/s Pg1= 1.2276 kPa,

Pv2= φ Pg2 = 0.4677 kPa, Pa2 = P - Pv2 = 99.532 kPa . Pv2 Pa2 V a2 . ω2 = 0.622 = 0.002923, m a2 = = 2.3656 kg/s Pa2 RTa2 . . mv2 = ω2ma2 = 0.00691 kg/s, hv2 = hg2 = 2538.1 kJ/kg . State 3: Liquid. T3 = 10°C, mf3 = 400 kg/hr = 0.1111 kg/s, hf3 = 42 kJ/kg Pg2 = 2.3385 kPa,

State 4: T4 = 40°C

. . . m a4 = m a2 + m a1 = 3.5944, . . . . mv4 = mv1 + mv2 + mf3 = 0.11896 kg/s

Continuity Eq. air:

Continuity Eq. water: . m v4 Pv4 ω4 = . = 0.0331 = 0.622 m a4 P-Pv4 Pg4 = 7.384 kPa,

φ4 =

Pv4 Pg4

→ Pv4 = 5.052 kPa

= 0.684, hv4 = hg4 = 2574.3 kJ/kg

. . . . . . . . 1stLaw: Q + ma1ha1 + mv1hv1 + ma2ha2 +mv2hv2 + mf3hf3 = ma4ha4 +mv4hv4 . Q = 1.004(3.5944 × 40 - 1.2288 × 10 - 2.3656 × 20) + 0.11896 × 2574.3 - 0.00094 × 2519.7 - 0.00691 × 2538.1 - 0.1111 × 42.0 = 366 kW

12-32 12.52 In a hot and dry climate, air enters an air-conditioner unit at 100 kPa, 40°C, and 5% relative humidity, at the steady rate of 1.0 m3/s. Liquid water at 20°C is sprayed into the air in the AC unit at the rate 20 kg/hour, and heat is rejected from the unit at at the rate 20 kW. The exit pressure is 100 kPa. What are the exit temperature and relative humidity? . State 1: T1 = 40°C, P1 = 100 kPa, φ1 = 5%, Va1 = 1 m3/s Pg1= 7.3837 kPa, Pv1= φ1Pg1 = 0.369 kPa, Pa1 = P- Pv1 = 99.63 kPa . Pv1 Pa1Va1 . ω1 = 0.622 = 0.0023, ma1 = = 1.108 kg/s, hv1 = 2574.3 kJ/kg Pa1 RTa1 . State 2 : Liq. Water. 20°C, mf2 = 20 kg/hr = 0.00556 kg/s, hf2 = 83.9 kJ/kg . . . . . Conservation of Mass: ma1 = ma3, mv1 + m12 = mv3 . . ω3 = (mf2 / ma1) + ω1 = ( 0.00556/1.108 ) + 0.0023 = 0.0073 State 3 : P3 = 100 kPa and Pv3 = P3ω3/(0.622 + ω3) = 1.16 kPa . . . . . . . 1stLaw: Q + ma1ha1 + mv1hv1 + mf2hf2 = ma3ha3 +mv3hv3; Q = - 20 kW . . . (ha3-ha1) + ω3hv3 = ω1hv1 + (mf2hf2 - Q )/ma1 = 24.39; Unknowns: ha3, hv3 Trial and Error for T3; T3 = 301.6 K, Pg3 = 3.901 kPa , φ3 =

Pv3 Pg3

= 0.297

12.53 A water-filled reactor of 1 m3 is at 20 MPa, 360°C and located inside an insulated containment room of 100 m3 that contains air at 100 kPa and 25°C. Due to a failure the reactor ruptures and the water fills the containment room. Find the final pressure. CV Total container. mv(u2-u1) + ma(u2-u1) = 1Q2 - 1W2 = 0 Initial water: Initial air:

v1 = 0.0018226, u1 = 1702.8, mv = V/v = 548.67 kg ma = PV/RT = 100×99/0.287×298.2 = 115.7 kg

Substitute into energy equation 548.67 (u2 - 1702.8) + 115.7×0.717 (T2 - 25) = 0 u2 + 0.1511 T2 = 1706.6 Trial and error 2-phase T = 150 LHS = 1546

&

v2 = V2/mv = 0.18226 m3/kg

(T guess,

v2 => x2 => u 2 => LHS) T = 160 LHS = 1820.2

T = 155 LHS = 1678.1 => T = 156 C LHS = 1705.7 OK x2 = 0.5372, Psat = 557.5 kPa Pa2 = Pa1V1T2/V2T1 = 100× 99 × 429.15 / (100×298.15) = 142.5 kPa =>

P2 = Pa2 + Psat = 700 kPa.

12-33 12.54 Use the psychrometric chart to find the missing property of: φ, ω, Twet, Tdry a. Tdry = 25°C, φ =80%

b. Tdry =15°C, φ =100%

c.Tdry = 20°C, and ω = 0.008

d. Tdry = 25°C, Twet = 23°C

a. 25°C, φ = 80%

=> ω = 0.016; Twet = 22.3°C

b. 15°C, φ = 100%

=> ω = 0.0106; Twet = 15°C

c. 20°C, ω = 0.008

=> φ = 57%; Twet = 14.4°C

d. 25°C, Twet = 23°C

=>

ω = 0.017; φ = 86%

12.55 Use the psychrometric chart to find the missing property of: φ, ω, Twet, Tdry a. φ = 50%, ω = 0.012

b. Twet =15°C, φ = 60%.

c. ω = 0.008 and Twet =17°C

d. Tdry = 10°C, ω = 0.006

a. φ = 50%, ω = 0.012

=> Tdry = 23.5°C, Twet = 20.6 °C

b. Twet = 15°C, φ =60%

=> Tdry = 20.2°C, ω = 0.0086

c. ω = 0.008, Twet =17°C => Tdry = 27.2°C, φ = 37% d. Tdry= 10°C, ω = 0.006 => φ = 80%, Twet = 8.2°C

12.56 For each of the states in Problem 12.55 find the dew point temperature. The dew point is the state with the same humidity ratio (abs humidity ω) and completely saturated φ = 100%. From psychrometric chart: a.

Tdew = 16.8°C

c. Tdew = 10.9°C

b.

Tdew = 12°C

d. Tdew = 6.5°C

Finding the solution from the tables is done for cases a,c and d as Eq.12.21 solve:

Pv = Pg = ωPtot /[ω + 0.622] = Psat(Tdew) in B.1.1

For case b use energy Eq. 12.23 to find ω1 first from Tad sat = Twet.

12-34 12.57 One means of air-conditioning hot summer air is by evaporative cooling, which is a process similar to the SSSF adiabatic saturation process. Consider outdoor ambient air at 35°C, 100 kPa, 30% relative humidity. What is the maximum amount of cooling that can be achieved by such a technique? What disadvantage is there to this approach? Solve the problem using a first law analysis and repeat it using the psychrometric chart, Fig. F.5. Cooled

Ambient Air 1

2

Air

P1 = P2 = 100 kPa T1 = 35 oC, φ1 = 30% Pv1 = φ1Pg1 = 0.30×5.628 = 1.6884

3 Liquid

ω1 = 0.622×1.6884/98.31 = 0.01068

For adiabatic saturation (Max. cooling is for φ2 = 1), 1st law, Eq.12.23 ω1 (hv1 - hf2) = Cp(T2 - T1) + ω2 hfg2 φ2 = 1

&

ω2 = 0.622 × PG2/(P2 - PG2)

Only one unknown: T2 . Trial and error on energy equation: CpT2 + ω2 hfg2 + ω1 hf2 = CpT1 + ω1hv1 = 1.004 × 35 + 0.01068 × 2565.3 = 62.537 T2 = 20 oC: PG2 = 2.339, hf2 = 83.94 , hfg2 = 2454.12 => ω2 = 0.622 × 2.339/ 97.661 = 0.0149 LHS = 1.004 × 20 + 0.0149 × 2454.1 + 0.01068 × 83.94 = 57.543 T2 = 25 oC: PG2 = 3.169, hf2 = 104.87 , hfg2 = 2442.3 => ω2 = 0.622 × 3.169/ 96.831 = 0.02036 LHS = 1.004 × 25 + 0.02036 × 2442.3 + 0.01068 × 104.87 = 75.945 linear interpolation: b)

T2 = 21.4 oC

chart F.5 : Ad. sat. T ≈ WetBulbTemperature ≈ 21.5 oC

12-35 12.58 Use the formulas and the steam tables to find the missing property of: φ, ω, and Tdry, total pressure is 100 kPa; repeat the answers using the psychrometric chart a. φ = 50%, ω = 0.010

b. Twet =15°C, φ = 50%

c. Tdry = 25°C, Twet = 21°C

a. From Eq.12.21 Pv = P ω /(0.622 + ω) = 100 × 0.01/0.632 = 1.582 kPa => T = 25°C

From Eq.12.18 Pg = Pv/φ = 1.582/0.5 = 3.165 kPa

b. Assume Twet is adiabatic saturation T and use energy Eq.12.23 At 15°C: Pg = 1.705 => ω = 0.622 × 1.705/(100 - 1.705) = 0.01079 LHS = ω1 (hv1 - hf2) + CpT1 = RHS = CpT2 + ω2 hfg2 RHS = 1.004×15 + 0.01079 × 2465.93 = 41.667 kJ/kg ω1 = 0.622 φPg/(100 - φPg)

where Pg is at T1.

LHS25C = 49.98, LHS20C = 38.3

=>

Trial and error.

T = 21.4°C, ω1 = 0.008

c. At 21°C:: Pg = 2.505 => ω2 = 0.622 × 2.505/(100 - 2.505) = 0.016 hf2 = 88.126

and

hfg2 = 2451.76 kJ/kg,

hv1 = 2547.17

From Eq.12.23: ω1 = [Cp(T2-T1) + ω2 hfg2 ]/(hv1 - hf2) = 0.0143 Pv = P ω /(0.622 + ω) = 2.247,

φ = 2.247/3.169 = 0.71

Using the psychrometric chart F.5: a: Tdry = 25.3 °C

b. Tdry = 21.6°C, ω = 0.008

c: ω = 0.0141, φ = 71-72%

12.59 Compare the weather two places where it is cloudy and breezy. At beach A it is 20°C, 103.5 kPa, relative humidity 90% and beach B has 25°C, 99 kPa, relative humidity 20%. Suppose you just took a swim and came out of the water. Where would you feel more comfortable and why? Your skin being wet and air is flowing over it will feel Twet. With the small difference in pressure from 100 kPa use the psychrometric chart. A: 20°C, φ = 90%

=> Twet = 18.7°C

B: 25°C, φ = 20%

=> Twet = 12.3°C

At beach A it is comfortable, at B it feels chilly.

12-36 12.60 Ambient air at 100 kPa, 30°C, 40% relative humidity goes through a constant pressure heat exchanger in a SSSF process. In one case it is heated to 45°C and in another case it is cooled until it reaches saturation. For both cases find the exit relative humidity and the amount of heat transfer per kilogram dry air. . . . . CV heat exchanger: mAi = mAe, mvi = mve, we = wi (ha + whv)i + q = (h a + whv)e = h˜e, Using the psychrometric chart:

i:

q = h˜e - h˜i

wi = 0.0104, h˜i = 76

Case I) e: Te = 45 oC, we = wi => h˜e = 92, φe = 17%, q = 92-76 = 16 kJ/kg dry air Case II) e: w e = wi, φe = 100% => h˜e = 61, Te = 14.5oC q = 61-76 = -15 kJ/kg dry air 12.61 A flow, 0.2 kg/s dry air, of moist air at 40°C, 50% relative humidity flows from the outside state 1 down into a basement where it cools to 16°C, state 2. Then it flows up to the living room where it is heated to 25°C, state 3. Find the dew point for state 1, any amount of liquid that may appear, the heat transfer that takes place in the basement and the relative humidity in the living room at state 3. Solve using psychrometric chart: a) Tdew = 27.2 (w = w1, φ = 100%)

w1 = 0.0232, h˜1 = 118.2

b) T2 < Tdew so we have φ2 = 100% liquid water appear in the basement. => w2 = 0.0114 h˜2 = 64.4 and from steam tbl. hf = 67.17 . . mliq = mair (w1-w2) = 0.2(0.0232-0.0114) = 0.00236 kg/s . . . c) Energy equation: mair h˜1 = mliq hf + mair h˜2 + Qout Qout = 0.2[118.2 - 64.4 - 0.0118×67.17] = 10.6 kW d)

w3 = w2 = 0.0114 & 25°C

=>

φ3 = 58%.

If you solve by the formula's and the tables the numbers are: Pg40 = 7.384 ; Pv1 = 0.5×7.384 = 3.692 kPa w1 = 0.622 × 3.692 / (100 - 3.692) = 0.02384 Pv1 = Pg (Tdew) => Tdew 1 = 27.5 °C 2: φ = 100%, Pv2 = Pg2 = 1.832 kPa, w2 = 0.622×1.832/98.168 = 0.0116 . . mliq = mair (w1-w2) = 0.2×0.01223 = 0.00245 kg/s 3:

w3 = w2 => Pv3 = Pv2 = 1.832 & Pg3 = 3.169 φ3 = Pv/Pg = 1.832/3.169 = 57.8%

12-37 12.62 A flow of air at 5°C, φ = 90%, is brought into a house, where it is conditioned to 25°C, 60% relative humidity. This is done in a SSSF process with a combined heater-evaporator where any liquid water is at 10°C. Find any flow of liquid, and the necessary heat transfer, both per kilogram dry air flowing. Find the dew point for the final mixture. CV heater and evaporator. Use psychrometric chart. Inlet: w1 = 0.0048, h˜1 = 37.5, hf = 42.01 Exit: w2 = 0.0118, h˜2 = 75, Tdew = 16.5°C From these numbers we see that water and heat must be added. Continuity eq. and energy equation give . . mLIQ IN/mA = w2 - w1 = 0.007 kg/kg dry air q = h˜2 - h˜1 - (w2-w1)hf = 37.3 kJ/kg dry air

12.63 Atmospheric air at 35°C, relative humidity of 10%, is too warm and also too dry. An air conditioner should deliver air at 21°C and 50% relative humidity in the amount of 3600 m3 per hour. Sketch a setup to accomplish this, find any amount of liquid (at 20°C) that is needed or discarded and any heat transfer. CV air conditioner. Check from psychrometric chart. Inlet: w1 = 0.00305, h˜mix,1 = 62.8, hf,20 = 83.96 w2 = 0.0076, h˜mix,2 = 60.2 Water must be added. Continuity and energy equations . . . . . . . mA(1+w1) + mliq = mA(1+w2); mAh˜1mix + mliqhf + QCV = mAh˜2mix . . mA = PaV/RT = (101.325 - 1.25)1/0.287×294.15 = 1.1845 kg/s . . mliq = mA(w2 - w1) = 0.00539 kg/s = 19.4 kg/h . QCV = 1.1845(60.2 - 62.8 - 0.00455×83.96) = - 3.53 kW Exit:

Liquid water

Cooler Exit

Inlet

12-38 12.64 In a car’s defrost/defog system atmospheric air, 21°C, relative humidity 80%, is taken in and cooled such that liquid water drips out. The now dryer air is heated to 41°C and then blown onto the windshield, where it should have a maximum of 10% relative humidity to remove water from the windshield. Find the dew point of the atmospheric air, specific humidity of air onto the windshield, the lowest temperature and the specific heat transfer in the cooler. Solve using the psychrometric chart Air inlet: T = 21°C, φ = 80%

=> w1 = 0.0124, Tdew = 17.3°C

Air exit: T = 41°C, φ = 10%

=> w3 = 0.0044, Tdew = 1.9°C To remove enough water we must cool to the exit T dew, followed by heating to Tex. The enthalpies are h˜1 = 72, h˜2 = 32.5, hf(1.9°C) = 8 CV cooler: . . mliq /mair = w1-w3 = 0.0124 - 0.0044 = 0.008 kg liq/kg air . q = Q/mair = h˜2 + (w1 - w3) hf - h˜1 = 32.5 + 0.008×8 - 72 = -39.4 kJ/kg dry air If the steam and air tables are used the numbers are 1: Pg1 = 2.505, Pv1 = 2.004 => w1 = 0.01259 hg1 = 2539.9, ha1 = 294.3 => h˜1 = 326.3 3: Pg3 = 7.826, Pv3 = 0.783 => w3 = 0.00486 2: wg3 = w3 => T2 = T3dew = 3.3°C, hf2 = 13.77 hg2 = 2507.4, ha2 = 276.56 => h˜2 = 288.75 . . mliq /mair = 0.00773, q = 288.75 + 0.00773× 13.77 - 326.3 = -37.45 kJ/kg air 12.65 Two moist air streams with 85% relative humidity, both flowing at a rate of 0.1 kg/s of dry air are mixed in a SSSF setup. One inlet flowstream is at 32.5°C and the other at 16°C. Find the exit relative humidity. 2: w2 = 0.0094, h˜2 = 60 1: w1 = 0.0266, h˜1 = 120 CV mixing chamber. Continuity Eq. water: Energy Eq.:

. . . mair w1 + mair w2 = 2mair wex; . . . mair h˜1 + mair h˜2 = 2mair h˜ex

wex = (w1 + w2)/2 = 0.018 ; ˜hex = (h˜1 + h˜2)/2 = 90 exit: wex, h˜ex => Tex = 24.5°C, φ = 94%

12-39 12.66 A flow of moist air at 21°C, 60% relative humidity should be produced from mixing of two different moist air flows. Flow 1 is at 10°C, relative humidity 80% and flow 2 is at 32°C and has Twet = 27°C. The mixing chamber can be followed by a heater or a cooler. No liquid water is added and P = 100 kPa. Find the two controls one is the ratio of the two mass flow rates ma1/ma2 and the other is the heat transfer in the heater/cooler per kg dry air. w1 = 0.006 , h˜ 1 = 45 2: w2 = 0.0208 , h˜ 2 = 105 4: w4 = 0.0091 , h˜ 4 = 64 , Tdew 4 = 12.5°C C.V : Total Setup . . . . ma1 w1 + ma2 w2 = (ma1 + ma2) w4 . . . . . ma1 h˜ 1 + ma2 h˜ 2 + Qa1 = (ma1 + ma2) h˜ 4 w4- w2 . . x = ma1/ma2 => x w1 + w2 = (1+x) w4 => x = = 3.773 w -w 1:

1

4

. . . q˜ = Qa1/(ma1 + ma2) = h˜ 4 - [x/(1+x)] h˜ 1 - [1/(1+x)] h˜2 = 64 – 0.7905 × 45 − 0.2095 × 105 = 6.43 kJ/kg-dry air 12.67 Consider two states of atmospheric air. (1) 35°C, Twet = 18°C and (2) 26.5°C, φ = 60%. Suggest a system of devices that will allow air in a SSSF process to change from (1) to (2) and from (2) to (1). Heaters, coolers (de)humidifiers, liquid traps etc. are available and any liquid/solid flowing is assumed to be at the lowest temperature seen in the process. Find the specific and relative humidity for state 1, dew point for state 2 and the heat transfer per kilogram dry air in each component in the systems. Use the psychrometric chart 1: w1= 0.006, h˜1 = 70.5, φ1 = 18%, Tdew = 6.5°C, h˜dew = 42 2: w2= 0.013, h˜2 = 79.4, φ2 = 60%, Tdew = 18°C, h˜dew = 71 Since w2 > w1 water must be added in process I to II and removed in the process II to I. Water can only be removed by cooling below dew point temperature so I to II: Adiab. sat I to Dew,II, then heater from Dew,II to II II to I: Cool to Dew,I then heat Dew,I to I The first one can be done because Tdew II = Tad sat I I to II: q = h˜II - h˜dewII = 79.4 - 71 = 8.4 kJ/kg air II to I: qcool = h˜II - h˜dewI - (w2-w1)hf(at TdewI) = 79.4 - 0.007 × 27.29 = 37.2 kJ/kg air qheat = h˜I - h˜dewI = 70.5 - 42 = 28.5 kJ/kg air

12-40 12.68 An insulated tank has an air inlet, ω1 = 0.0084, and an outlet, T2 = 22°C, φ2 = 90% both at 100 kPa. A third line sprays 0.25 kg/s of water at 80°C, 100 kPa. For a SSSF operation find the outlet specific humidity, the mass flow rate of air needed and the required air inlet temperature, T 1. Take CV tank in SSSF. Continuity and energy equations are: . . . . . . Continuity: m3 + ma w1 = ma w2; Energy Eq.: m3hf + ma h˜1 = ma h˜2 All state properties are known except T1. From the psychrometric chart we get State 2: w2 = 0.015, h˜2 = 79.5 State 3: hf = 334.91 (steam tbl) . . ma = m3/(w2-w1) = 0.25/(0.015-0.0084) = 37.88 kg/s h˜1 = h˜2 - (w2-w1)hf = 79.5 - 0.0066 × 334.91 = 77.3 Chart (w1, h˜1) =>

T1 = 36.5°C

12.69 You have just washed your hair and now blow dry it in a room with 23°C, φ = 60%, (1). The dryer, 500 W, heats the air to 49°C, (2), blows it through your hair where the air becomes saturated (3), and then flows on to hit a window where it cools to 15°C (4). Find the relative humidity at state 2, the heat transfer per kilogram of dry air in the dryer, the air flow rate, and the amount of water condensed on the window, if any. The blowdryer heats the air at constant specific humidity to 2 and it then goes through an adiabatic saturation process to state 3, finally cooling to 4. 1: 23°C, 60% rel hum => w1 = 0.0104, h˜1 = 69 2: w2 = w1, T2 => φ2 = 15%, h˜2 = 95 CV. 1 to 2: w 2 = w1; CV. 2 to 3:

q = h˜2 - h˜1 = 95 - 69 = 26 kJ/kg dry air . ma = Q/q = 0.5/26 = 0.01923 kg/s . . . . . w3 - w2 = mliq /ma ; ma h˜2 + mliq hf = ma h˜3

3: φ = 100% => T3 = Twet,2 = 24.8°C, w3 = 0.0198 4: φ = 100%, T4 => w4 = 0.01065 . . mliq = (w3-w4)ma = (0.0198-0.01065)×0.01923 = 0.176 g/s If the steam tables and formula's are used then we get hg1 = 2543.5, hg2 = 2590.3, Pg1 = 2.837, Pv1 = 1.7022, Pg2 = 11.8, w1 = 0.01077, w2 = w1 , Pv2 = Pv1 φ2 = Pv2/Pg2 = 14.4%, hf3 = 114, Trial and error for adiabatic saturation temperature. T3 = 25°C, w3 = 0.02, Pv4 =Pg4 = 1.705 kPa, w4 = 0.622×1.705/(100-1.705) = 0.0108

12-41 12.70 A water-cooling tower for a power plant cools 45°C liquid water by evaporation. The tower receives air at 19.5°C, φ = 30%, 100 kPa that is blown through/over the water such that it leaves the tower at 25°C, φ = 70%. The remaining liquid water flows back to the condenser at 30°C having given off 1 MW. Find the mass flow rate of air, and the amount of water that evaporates. CV Total cooling tower, SSSF. . . Water in air: win + mevap/ma = wex . . . . . Energy: ma h˜in + m1 h45 = ma h˜ex + (m1 - mevap) h30 Inlet: 19.5°C, 30% rel hum => win = 0.0041, h˜in = 50 Exit : 25°C, 70% rel hum => wex = 0.0138, h˜ex = 80 Take the two water flow difference to mean the 1 MW . . . . Q = m1 h45 - (m1 - mevap) h30 = 1 MW . . . ma(h˜ex-h˜in) = ma(80-50) = 1000 kW => ma = 33.33 kg/s . . mevap = (wex - win) ma = 0.0097 × 33.33 = 0.323 kg/s The needed make-up water flow could be added to give a slightly different meaning to the 1 MW. 12.71 An indoor pool evaporates 1.512 kg/h of water, which is removed by a dehumidifier to maintain 21°C, φ = 70% in the room. The dehumidifier, shown in Fig. P12.71, is a refrigeration cycle in which air flowing over the evaporator cools such that liquid water drops out, and the air continues flowing over the condenser. For an air flow rate of 0.1 kg/s the unit requires 1.4 kW input to a motor driving a fan and the compressor and it has a coefficient of performance, β = QL/ Wc = 2.0. Find the state of the air as it returns to the room and the compressor work input. The unit must remove 1.512 kg/h liquid to keep steady state in the room. As water condenses out state 2 is saturated. 1: 21 °C, 70% => w1 = 0.0108, h˜1 = 68.5 . . . . CV 1 to 2: mliq = ma(w1 - w2) => w2 = w1 - mliq /ma qL = h˜1 - h˜2 - (w1 - w2) hf2 w2 = 0.0108 - 1.512/3600×0.1 = 0.0066 2: w2, 100% => T2 = 8°C, h˜2 =45, hf2 = 33.6 qL = 68.5 - 45 - 0.0042×33.6 = 23.36 kJ/kg dry air . . CV Total system : h˜3 = h˜1 + Wel/ma - (w1-w2) hf = 68.5 + 14 - 0.14 = 82.36 kJ/kg dry air 3: w3 = w2, h˜3 => T3 = 46°C, φ3 = 11-12% . . Wc = ma qL/ β = 1.165 kW

12-42 12.72 To refresh air in a room, a counterflow heat exchanger, see Fig. P12.72, is mounted in the wall, drawing in outside air at 0.5°C, 80% relative humidity and pushing out room air, 40°C, 50% relative humidity. Assume an exchange of 3 kg/min dry air in a SSSF device, and also that the room air exits the heat exchanger to the atmosphere at 23°C. Find the net amount of water removed from the room, any liquid flow in the heat exchanger and (T, φ) for the fresh air entering the room. State 1: w1 = 0.0232, h˜1 = 119.2, Tdew,1 = 27°C The room air is cooled to 23°C < Tdew1 so liquid will form in the exit flow channel and state 2 is saturated. 2: 23°C, φ = 100% => w2 = 0.0178, h˜2 = 88, hf2 = 96.52 3: 0.5°C, φ = 80% => w3 = 0.0032, h˜3 = 29.2 CV 1 to 2: . . mliq,2 = ma (w1 - w2) = 3 (0.0232 - 0.0178) = 0.0162 kg/min . . . CV room: mv,out = ma (w1 - w4) = ma (w1 - w3) = 3(0.0232-0.0032) = 0.06 kg/min ma(h˜4 - h˜3) = ma(h˜1 - h˜2) - mliqhf2

CV Heat exchanger: h˜4 = h˜3 + h˜1 - h˜2 - (w1-w2) hf2 = 29.2 + 119.2 - 88 - 0.0054×96.52 = 59.9 kJ/kg dry air 4: w4 = w3, h˜4 => T4 = 32.5°C, φ = 12%

12-43 12.73 Steam power plants often utilize large cooling towers to cool the condenser cooling water so it can be recirculated; see Fig. P12.73. The process is essentially evaporative adiabatic cooling, in which part of the water is lost and must therefore be replenished. Consider the setup shown in Fig. P12.73, in which 1000 kg/s of warm water at 32°C from the condenser enters the top of the cooling tower and the cooled water leaves the bottom at 20°C. The moist ambient air enters the bottom at 100 kPa, dry bulb temperature of 18°C and a wet bulb temperature of 10°C. The moist air leaves the tower at 95 kPa, 30°C, and relative humidity of 85%. Determine the required mass flow rate of dry air, and the fraction of the incoming water that evaporates and is lost. Air + vap. P4 = 95 kPa 4 1 T1 = 32 oC T4 = 30 oC . φ4 = 0.85 m1 = 1000 kg/s LIQ H 2O P2 = 100 kPa T2 = 18 oC

T3 = 20 oC

3

WBT2 = 10 oC

2

Air + vap. 2'

2 o

18 C w2′ = 0.622× LIQ IN w2 =

(ha2′-ha2) + w2′hFG2′ hv2 - hF 2′

=

φ2′ = 1.0

T2′ = 10 oC,

1.2276 = 0.00773 100-1.2276

1.0035(10-18) + 0.00773×2477.7 = 0.00446 2534.4 - 42.0

3.609 Pv4 = 0.85 × 4.246 = 3.609, w4 = 0.622 × = 0.02456 95-3.609 . . . . . . . m1 + mv2 = m3 + mv4 Cons. mass: ma2 = ma4 = ma, . . . . . or m3 = m1 + ma(w2-w4) and set r = m1/ma . . . . . . 1st law: m1h1 + maha2 + mv2hv2 = m3h3 + maha4 + mv4hv4 or

r h1 + (ha2-ha4) + w2hv2 = (r + w2 - w4)h3 + w4hv4

r(h1-h3) = ha4 - ha2 + w4hv4 - w2hv2 - (w4-w2)h3 r(134.15-83.96) = 1.004(30-18) + 0.024 56×2556.3 - 0.004 46×2534.4 - 0.0201×83.96 . . . r = m1/ma = 1.232 → ma = 811.7 kg/s . . . m3 = m1 + ma(w2-w4) = 1000 - 811.7×0.0201 = 983.7 . . ∆m/m1 = 0.0163

12-44 12.74 A semipermeable membrane is used for the partial removal of oxygen from air that is blown through a grain elevator storage facility. Ambient air (79% nitrogen, 21% oxygen on a mole basis) is compressed to an appropriate pressure, cooled to ambient temperature 25°C, and then fed through a bundle of hollow polymer fibers that selectively absorb oxygen, so the mixture leaving at 120 kPa, 25°C, contains only 5% oxygen. The absorbed oxygen is bled off through the fiber walls at 40 kPa, 25°C, to a vacuum pump. Assume the process to be reversible and adiabatic and determine the minimum inlet air pressure to the fiber bundle. A = N2 ; 0.79 A + 0.21 B P1 = ?

1

B = O2 2

3

0.79 A + 0.0416 B (0.79/0.95)

MIX 5 % O2

P2 = 120 kPa P3 = 40 kPa All T = 25 oC

0.1684 B

Let sA1 = sB1 = 0 at T = 25 oC & P1 -s MIX 1 = 0 + 0 - y A1R ln yA1 - yB1R ln yB1 -s MIX 2 = 0 + 0 - R ln (P 2/P1) - yA2R ln yA2 - yB2R ln yB2 Pure B: -s3 = 0 - R ln (P3/P1) . . . For n -s = n -s + n -s 1 1

2 2

3 3

[

R -0.8316 ln (P2/P1) - 0.79 ln 0.95 - 0.0416 ln 0.05

]

- 0.1684 ln (P3/P1) + 0.79 ln 0.79 + 0.21 ln 0.21 = 0 0.8316 ln (P2/P1) + 0.1684 ln (P3/P1) = -0.3488 + 4.6025 - ln P1 = -0.3488 P1 min = 141 kPa For P1 > P1 min →

. ∆S > 0

12-45 12.75 A 100-L insulated tank contains N2 gas at 200 kPa and ambient temperature 25° C. The tank is connected by a valve to a supply line flowing CO2 at 1.2 MPa, 90° C. A mixture of 50% N2, 50% CO2 by mole should be obtained by opening the valve and allowing CO2 flow in to an appropriate pressure is reached and close the valve. What is the pressure? The tank eventually cools to ambient temperature. Find the net entropy change for the overall process. CO 2 i V = 100 L, P1 = 200 kPa, T1 = T0 = 25 oC

N

Pi = 1.2 MPa, Ti = 90 oC At state 2: yN2 = yCO2 = 0.50 n2 CO2 = n2 N2 = n1 N2 = P1V/RT1

2

= 200*0.1/8.3145*298.2 = 0.00807 kmol n2 = 0.01614 kmol 1st law: nihi = n2u-2 - n1u-1 , for const specific heats niCPoiTi = (niCVoi+n1CVo1)T2 - n1CVo1T1 But ni = n1 → CPoiTi = CVoiT2 + CVo1(T2-T1) 44.01×0.842×363.2 = 44.01×0.653 T2 + 28.013×0.745(T2-298.2) T2 = 396.7 K P2 = n2RT2/V = 0.01614×8.3145×396.7/0.1 = 532 kPa Cool to T3 = T0 = 298.2 K P3 = P2× T3/T2 = 532× 298.2/396.7 = 400 kPa Q23 = n2CVo2(T3-T2) = 0.016 14(0.5×28.013×0.745 + 0.5×44.01×0.653)(298.2-396.7) = -39.4 kJ ∆SNET = n3-s3 - n1-s1 - ni-si - Q23/T0 = n [(s- ) - -s ] + n [(s- ) - s- ] - Q /T i

CO2 3

i

1

N2 3

= 0.00807(44.01×0.8418 ln

1

23

0

0.5×400 298.2 - 8.3145 ln ) 363.2 1200

+ 0.00807×( - 8.3145 ln

0.5×400 -39.4 )298.2 200

= +0.0613 + 0 + 0.1321 = +0.1934 kJ/K

12-46 12.76 A cylinder/piston loaded with a linear spring contains saturated moist air at 120 kPa, 0.1 m3 volume and also 0.01 kg of liquid water, all at ambient temperature 20°C. The piston area is 0.2 m2, and the spring constant is 20 kN/m. This cylinder is attached by a valve to a line flowing dry air at 800 kPa, 80°C. The valve is opened, and air flows into the cylinder until the pressure reaches 200 kPa, at which point the temperature is 40°C. Determine the relative humidity at the final state, the mass of air entering the cylinder and the work done during the process. P1 = 120 kPa, T1 = 20 oC = T0, V1 = 0.1 m3, mLIQ 1 = 0.01 kg, AP = 0.2 m2, ks = 20 kN/m Pi = 800 kPa P2 = 200 kPa

A+V

Ti = 80 oC

i

T2 = 40 oC 2

P2 = P1 + (ks/Ap)(V2-V1) 200 = 120 + (20/0.22)(V2-0.1) → V2 = 0.26 m3

DRY AIR

φ1 = 1.0 mv1 =

(or

Pv1V1 RvT1

=

w1 = 0.622×2.339/117.66 = 0.012 36) 2.339×0.1 = 0.001 73 ( = w1mA1 ) 0.461 52×293.2

Assume no liquid at state 2 mv2 = mv1 + mL1 = 0.01173 kg → Pv2 =

mvRvT2 0.0011 73×0.461 52×313.2 = = 6.521 kPa V2 0.26

a) φ2 =

6.521 = 0.883 7.384

b) mA1 = mA2 =

PA1V1 RAT1

=

117.66×0.1 = 0.1398 0.287×293.2

193.479×0.26 = 0.5596 0.287×313.2

mAi = mA2 - mA1 = 0.4198 kg 1 1 c) WCV = ⌠ ⌡ PdV = (P1+P2)(V2-V1) = (120+200)(0.26-0.1) = 25.6 kJ 2 2

12-47 12.77 Consider the previous problem and additionally determine the heat transfer. Show that the process does not violate the second law. a) QCV = mA2hA2 - mA1hA1 - mAihAi + mv2hv2 -mv1hv1 - mL1hL1 - P2V2 + P1V1 + WCV = 0.5596×1.004×313.2 - 0.1398×1.004×293.2 - 0.4198×1.004×353.2 + 0.001173×2574.3 - 0.00173×2538.4 – 0.01 ×83.9 - 200×0.26 + 120×0.1 + 25.6 = -3.48 kJ b) ∆SCV = S2 - S1 = mA2sA2 - mA1sA1 + mvsv2 - mv1sv1 - mL1sL1 ∆SSURR = - QCV/T0 - mAisAi ∆SNET = mA2sA2 - mA1sA1 - mAisAi + mvsv2- mv1sv1 - mL1sL1 - QCV/T0 = mA2(sA2-sAi) + mA1(sAi-sA1) + mvsv2- mv1sv1 - mL1sL1 - QCV/T0 sA2 - sAi = 1.004 ln

313.2 193.479 - 0.287 ln = +0.2866 353.2 800

sAi - sA1 = 1.004 ln

313.2 800 - 0.287 ln = -0.3633 293.2 117.66

sv2 = sG2 - Rv ln φ2 = 8.2569 - 0.46152 ln 0.883 = 8.3143 sv1 = sG1 - Rv ln φ1 = 8.6671 – 0 = 8.6671 sL1 = sF1 = 0.2966 → ∆SNET = 0.5596 (+0.2868) + 0.1398 (-0.3633) + 0.01173×8.3143 – 0.00173×8.6671 – 0.01×0.2966 + 3.48/293.2 = +0.201 kJ/K

12-48 12.78 The air-conditioning by evaporative cooling in Problem 12.57 is modified by adding a dehumidification process before the water spray cooling process. This dehumidification is achieved as shown in Fig. P12.78 by using a desiccant material, which absorbs water on one side of a rotating drum heat exchanger. The desiccant is regenerated by heating on the other side of the drum to drive the water out. The pressure is 100 kPa everywhere and other properties are on the diagram. Calculate the relative humidity of the cool air supplied to the room at state 4, and the heat transfer per unit mass of air that needs to be supplied to the heater unit. States as noted on Fig. P12.78, text page 466. At state 1, 35 oC: PV1 = φ1PG1 = 0.30×5.628 = 1.6884 w1 = 0.622×1.6884/98.31 = 0.010 68 At T3 = 25 oC:

w3 = w2 = w1/2 = 0.00534

Evaporative cooling process to state 4, where T4 = 20oC As in Eq. 12.23:

w3(hV3-hL4) = CP0A(T4-T3) + w4hFG4

0.005 34(2547.2-83.9) = 1.004(20-25) + w4×2454.2 w4 = 0.0074 = 0.622×PV4/(100-PV4) PV4 = 1.176 kPa, φ4 = 1.176/2.339 = 0.503 At T5 = 25 oC, w5 = w4 = 0.0074 Evaporative cooling process to state 6, where T6 = 20oC w5(hV5-hL6) = CP0A(T6-T5) + w6hFG6 0.0074(2547.2-83.9) = 1.004(20-25) + w6×2454.2 For adiabatic heat exchanger, . . . . . mA2 = mA3 = mA6 = mA7 = mA,

=>

w6 = 0.009 47

Also w2 = w3, w6 = w7

hA2 + w2hV2 + hA6 + w6hV6 = hA3 + w3hV3 + hA7 + w7hV7 or CP0AT7 + w6(hV7-hV6) = CP0A(T2+T6-T3) + w2(hV2-hV3) 1.004 T7 + 0.009 47(hV7 - 2538.1) = 1.004(60 + 20 - 25) + 0.005 34(2609.6-2547.2) = 55.526 By trial and error, T7 = 54.7 oC, hV7 = 2600.3 For the heater 7-8, w8 = w7, . . Q/mA = CP0A(T8-T7) + w7(hV8-hV7) = 1.004(80-54.7) + 0.009 47(2643.7-2600.3) = 25.8 kJ/kg dry air

12-49 12.79 A vertical cylinder is fitted with a piston held in place by a pin, as shown in Fig. P12.79. The initial volume is 200 L and the cylinder contains moist air at 100 kPa, 25°C, with wet-bulb temperature of 15°C. The pin is removed, and at the same time a valve on the bottom of the cylinder is opened, allowing the mixture to flow out. A cylinder pressure of 150 kPa is required to balance the piston. The valve is closed when the cylinder volume reaches 100 L, at which point the temperature is that of the surroundings, 15°C. a. Is there any liquid water in the cylinder at the final state? b. Calculate the heat transfer to the cylinder during the process. c. Take a control volume around the cylinder, calculate the entropy change of the control volume and that of the surroundings. From Fig. F.5, at T1 = 25 oC & WBT1 = 15 oC a) w1 = 0.006 67 for saturation at state 2, where P2 = PEXT = 150 kPa, T2 = 15 oC; Max PV2 = PG2 = 1.705 kPa wMAX 2 = 0.622×1.705/(150-1.705) = 0.00715 > w 1 → not saturated at 2, no liquid b) w1 = 0.006 67 = 0.622×

PV1 100-PV1

→ PV1 = 1.061 kPa

mA1 =

(100-1.061)×0.2 = 0.2312 kg, mV1 = w1mA1 = 0.00154 kg 0.287×298.2

mA2 =

(150-1.592)×0.1 = 0.1794 kg, mV2 = w2mA2 = 0.0012 kg 0.287×288.2

WCV = ⌠ ⌡ PEXTdV = PEXT(V2-V1) = 150(0.1-0.2) = -15 kJ QCV = mA2uA2 - mA1uA1 + mV2uV2 - mV1uV1 + (mAEhAE AVE + mVEhVE AVE) + WCV = 0.1794×0.717×288.2 - 0.2312×0.717×298.2 + 0.0012×2396.0 - 0.00154×2409.8 + 0.0518×1.004(298.2 +288.2)/2 + 0.00034(2547.2+2528.9)/2 - 15.0 = -12.1 kJ c) ∆SNET = mA2(sA2- sAE AVE) + mA1(sAE AVE - sA1) + mV2(sV2-sVE AVE) + mV1(sVE AVE - sV1) - QCV/T0

12-50 Assume that the mixture exiting is throttled across the value to 100 kPa and then discharged. Therefore, since the composition is constant. PAE = PA1 = 98.94 kPa,

PVE = PV1 = 1.061 kPa

Also, PA2 = 148.41 kPa & PV2 = 1.592 kPa Using TAE ≈ TVE ≈ 293.2 K & T0 = 15 oC = 288.2 K ∆SNET = 0.1794( 1.004 ln

288.2 148.425 - 0.287 ln ) 293.2 98.94

+ 0.2312( 1.004 ln

293.2 -0) 298.2

+ 0.0012( 1.8723 ln

288.2 1.575 - 0.461 52 ln ) 293.2 1.05

+ 0.001 54( 1.8723 ln = +0.0138 kJ/K

293.2 - 0 ) − (−12.1)/288.2 298.2

12-51 12.80 Ambient air is at a condition of 100 kPa, 35°C, 50% relative humidity. A steady stream of air at 100 kPa, 23°C, 70% relative humidity, is to be produced by first cooling one stream to an appropriate temperature to condense out the proper amount of water and then mix this stream adiabatically with the second one at ambient conditions. What is the ratio of the two flow rates? To what temperature must the first stream be cooled? P1 = P2 = 100 kPa 1 5

2

T1 = T2 = 35 oC

MIX COOL

. -Q COOL

3

4

. -Q MIX= 0

LIQ H 2O

φ1 = φ2 = 0.50, φ4 = 1.0 P5 = 100, T5 = 23 oC φ5 = 0.70

Pv1 = Pv2 = 0.5×5.628 = 2.814 kPa => w1 = w2 = 0.622× Pv5 = 0.7×2.837 = 1.9859 kPa => w5 = 0.622× C.V.: Mixing chamber:

Call the mass flow ratio

2.814 = 0.0180 100-2.814

1.9859 = 0.0126 100-1.9859 r = ma2/ma1

cons. mass:

w1 + r w4 =(1+ r)w5

Energy Eq.:

ha1 + w1hv1 + rha4 + rw4hv4 = (1+r)h a5 + (1+r)w5hv5

→ 0.018 + rw4 = (1+r) 0.0126 or

ma2

0.018-0.0126 = , r= 0.0126-w4 ma1

with

w4 = 0.622 ×

PG4 100-PG4

1.004×308.2 + 0.018×2565.3 + r×1.004×T4 + r w4hv4 = (1+r)×1.004 × 296.2 + (1+r)×0.0126×2543.6 or

[

]

r 1.004×T4 + w4hG4 - 329.3 + 26.2 = 0

Assume T4 = 5 oC

→ PG4 = 0.8721, hG4 = 2510.5

w4 = 0.622×0.8721/(100-0.8721) = 0.0055 r = ma2/ma1 =

0.018-0.0126 = 0.7606 0.0126-0.0055

0.7606[1.004×278.2 + 0.0055×2510.5 - 329.6] + 26.2 = -1.42 ≈ 0

OK

12-52

ENGLISH UNIT PROBLEMS 12.81EA gas mixture at 70 F, 18 lbf/in.2 is 50% N2, 30% H2O and 20% O2 on a mole basis. Find the mass fractions, the mixture gas constant and the volume for 10 lbm of mixture. From Eq. 12.3: ci = yi Mi/ ∑ yjMj MMIX = ∑ yjMj = 0.5 × 28.013 + 0.3 × 18.015 + 0.2 × 31.999 = 14.0065 + 5.4045 + 6.3998 = 25.811 cN2 = 14.0065 / 25.811 = 0.5427, cH2O = 5.4045 / 25.811 = 0.2094 cO2 = 6.3998 / 25.811 = 0.2479, sums to 1 OK − RMIX = R/MMIX = 1545.36 / 25.811 = 59.87 lbf ft/lbm R V = mRMIX T/P = 10 × 59.87 × 710 / 18 × 144 = 164 ft3 12.82EWeighing of masses gives a mixture at 80 F, 35 lbf/in.2 with 1 lbm O2, 3 lbm N2 and 1 lbm CH4. Find the partial pressures of each component, the mixture specific volume (mass basis), mixture molecular weight and the total volume. From Eq. 12.4: yi = (mi /Mi) / ∑ mj/Mj ntot = ∑ mj/Mj = (1/31.999) + (3/28.013) + (1/16.04) = 0.031251 + 0.107093 + 0.062344 = 0.200688 yO2 = 0.031251/0.200688 = 0.1557,

yN2 = 0.107093/0.200688 = 0.5336,

yCH4 = 0.062344/0.200688 = 0.3107 PO2 = yO2 Ptot = 0.1557 × 35 = 5.45 lbf/in.2, PN2 = yN2 Ptot = 0.5336 × 35 = 18.676 lbf/in.2, PCH4 = yCH4 Ptot = 0.3107 × 35 = 10.875 lbf/in.2 − Vtot = ntot RT/P = 0.200688 × 1545 × 539.7 / (35 × 144) = 33.2 ft3 v = Vtot/mtot = 33.2 / (1 + 3 + 1) = 6.64 ft3/lbm MMIX = ∑ yjMj = mtot/ntot = 5/0.200688 = 24.914

12-53 12.83EA pipe flows 0.15 lbmol a second mixture with molefractions of 40% CO2 and 60% N2 at 60 lbf/in.2, 540 R. Heating tape is wrapped around a section of pipe with insulation added and 2 Btu/s electrical power is heating the pipe flow. Find the mixture exit temperature. C.V. Pipe heating section. Assume no heat loss to the outside, ideal gases. . . - .− . Energy Eq.: Q = m(he − hi) = n(he − hi) = nCP mix(Te − Ti) − − CP mix = ∑ yi Ci = 0.4 × 8.846 + 0.6 × 6.975 = 7.723 Btu/lbmole . .− Te = Ti + Q / nCP mix = 540 + 2/(0.15 × 7.723) = 541.7 R 12.84EAn insulated gas turbine receives a mixture of 10% CO 2, 10% H2O and 80% N2 on a mole basis at 1800 R, 75 lbf/in.2. The volume flow rate is 70 ft3/s and its exhaust is at 1300 R, 15 lbf/in.2. Find the power output in Btu/s using constant specific heat from C.4 at 540 R. C.V. Turbine, SSSF, 1 inlet, 1 exit flow with an ideal gas mixture, q = 0. . . - .− . Energy Eq.: WT = m(hi − he) = n(hi − he) = nCP mix(Ti − Te) . . − − PV = nRT => n = PV / RT = 75 × 144 × 70 / (1545.4 × 1800) = 0.272 lbmole/s − − CP mix = ∑yi Ci = 0.1 × 44.01 × 0.201 + 0.1 × 18.015 × 0.447 + 0.8 × 28.013 × 0.249 = 7.27 Btu/lbmol R . WT = 0.272 × 72.7 × (1800 − 1300) = 988.7 Btu/s

12.85ESolve Problem 12.84 using the values of enthalpy from Table C.7 C.V. Turbine, SSSF, 1 inlet, 1 exit flow with an ideal gas mixture, q = 0. . . - . Energy Eq.: WT = m(hi − he) = n(hi − he) . . − − PV = nRT => n = PV / RT = 75 × 144 × 70 / (1545.4 × 1800) = 0.272 lbmole/s . WT = 0.272 × [0.1(14358 − 8121) + 0.1(11178 − 6468.5) + 0.8(9227 − 5431)] = 1123.7 Btu/s

12-54 12.86ECarbon dioxide gas at 580 R is mixed with nitrogen at 500 R in a SSSF insulated mixing chamber. Both flows are at 14.7 lbf/in.2 and the mole ratio of carbon dioxide to nitrogen is 2;1. Find the exit temperature and the total entropy generation per mole of the exit mixture. . . CV mixing chamber, SSSF. The inlet ratio is nCO2 = 2 nN2 and assume no external heat transfer, no work involved. . . . . . . nCO + 2nN = nex = 3nN ; nN (hN + 2hCO ) = 3nN hmix ex 2

2

2

2

2

2

2

- Take 540R as reference and write h = h540 + CPmix(T-540). CP N (Ti N -540) + 2CP CO (Ti CO -540) = 3CP mix(Tmix ex -540) 2

2

2

2

CP mix = ∑yiCP i = (29.178 + 2×37.05)/3 = 8.2718 3CP mixTmix ex = CP N Ti N + 2CP CO Ti CO = 13837 2

2

2

2

Tmix ex = 557.6 R; Pex N = Ptot/3; Pex CO = 2Ptot/3 2

2

. . ... - . - Sgen = nexsex-(ns)iCO - (ns)iN = nN (se - si)N + 2nN (se - si)CO 2

2

2

2

2

2

Tex Tex . . Sgen/nN = CPN ln - Rln yN + 2CPCO ln - 2 Rln yCO 2 2 TiN 2 2 TiCO 2 2

2

= 0.7575 + 2.1817 - 0.7038 + 1.6104 = 3.846 Btu/lbmol N2 R 12.87EA mixture of 60% helium and 40% nitrogen by volume enters a turbine at 150 lbf/in.2, 1500 R at a rate of 4 lbm/s. The adiabatic turbine has an exit pressure of 15 lbf/in.2 and an isentropic efficiency of 85%. Find the turbine work. Assume ideal gas mixture and take CV as turbine. wT s = hi-hes,

ses = si,

Tes = Ti(Pe/Pi)(k-1)/k

CP mix = 0.6× 1.25× 4.003 + 0.4× 0.248× 28.013 = 5.7811 - (k-1)/k = R/CP mix = 1545/(5.7811×778) = 0.3435 Mmix= 0.6×4.003 + 0.4×28.013 = 13.607, CP=CP/Mmix= 0.4249 Tes= 1500(15/150)0.3435 = 680 R, wTs = CP(Ti-Tes) = 348.4 . . wT ac = wTs = 296.1 Btu/lbm; W = mwTs = 1184 Btu/s

12-55 12.88EA mixture of 50% carbon dioxide and 50% water by mass is brought from 2800 R, 150 lbf/in.2 to 900 R, 30 lbf/in.2 in a polytropic process through a SSSF device. Find the necessary heat transfer and work involved using values from C.4. Process Pvn = constant

leading to

n ln(v2/v1) = ln(P1/P2); v = RT/P n = ln(150/30)/ln(900 × 150/30 × 2800) =3.3922 Rmix = ΣciRi = (0.5 × 35.1 + 0.5 × 85.76)/778 = 0.07767 CP mix = ΣciCPi = 0.5 × 0.203 + 0.5 × 0.445 = 0.324 ⌠vdP = - n (P v -P v ) = - nR (T -T ) = 209.3 Btu w = -⌡ n-1 e e i i n-1 e i lbm q = he-hi + w = CP(Te-Ti) + w = -406.3 Btu/lbm

12.89EA mixture of 4 lbm oxygen and 4 lbm of argon is in an insulated piston cylinder arrangement at 14.7 lbf/in.2, 540 R. The piston now compresses the mixture to half its initial volume. Find the final pressure, temperature and the piston work. Since T1 >> TC assume ideal gases. u2-u1 = 1q2 - 1w2 = - 1w2 ;

s2-s1 = 0

Pvk = constant, v2 = v1/2 P2 = P1(v1/v2)k = P1(2)k; T2 = T1(v1/v2)k-1 = T1(2)k-1 Find kmix to get P2,T2 and Cv mix for u2-u1 Rmix = ΣciRi = (0.5 × 48.28 + 0.5 × 38.68)/778 = 0.055887 CPmix = ΣciCPi = 0.5 × 0.219 + 0.5 × 0.1253 = 0.17215 Cvmix = CPmix -Rmix = 0.11626, kmix = CPmix/Cvmix = 1.4807 P2 = 14.7(2)1.4805= 41.03 lbf/in2, T2 = 540*20.4805= 753.5 R w = u1-u2= Cv(T1-T2)= 0.11626(540-753.5) = -24.82 Btu/lbm

1 2 1

W2 = mtot 1w2 = 8 (-24.82) = -198.6 Btu

12-56 12.90ETwo insulated tanks A and B are connected by a valve. Tank A has a volume of 30 ft3 and initially contains argon at 50 lbf/in.2, 50 F. Tank B has a volume of 60 ft3 and initially contains ethane at 30 lbf/in.2, 120 F. The valve is opened and remains open until the resulting gas mixture comes to a uniform state. Find the final pressure and temperature and the entropy change for the process. a) 1st law: U2-U1 = 0 = nArCV0(T2-TA1) + nC H CVO(T2-TB1) 2 6

50×144×30 nAr = PA1VA/RTA1 = = 0.2743 lbmol 1545×509.7 30×144×60 = P V /R T = = 0.2894 lbmol H B1 B B1 1545×579.7 2 6

nC

n2 = nAr + nC

H 2 6

= 0.5637 lbmol

0.2743×39.948×0.0756(T2-509.7) + 0.2894×30.07×0.361(T2-509.7) = 0 Solving, T2 = 565.1 R 0.5637×1545×565.1 = 38 lbf/in2 P2 = n2RT2/(VA+VB) = 90×144 b) ∆SSURR = 0 → ∆SNET = ∆SSYS = nAr∆SAr + nC H ∆SC 2 6

H 2 6

yAr = 0.2743/0.5637 = 0.4866 T2 - yArP2 ∆SAr = CP Ar ln - R ln TA1 PA1 = 39.948×0.1253 ln

565.1 1545 0.4866×38 ln = 2.4919 Btu/lbmol R 509.7 778 50

T2 - yC2H6P2 - R ln ∆SC2H6 = CC2H6 ln TB1 PB1 = 30.07×0.427 ln

565.1 1545 0.5134×38 ln 579.7 778 30

= 0.5270 Btu/lbmol R ∆SNET = 0.2743×2.4919 + 0.2894×0.5270 = 0.836 Btu/R

12-57 12.91EA large SSSF air separation plant takes in ambient air (79% N2, 21% O2 by volume) at 14.7 lbf/in.2, 70 F, at a rate of 2 lb mol/s. It discharges a stream of pure O2 gas at 30 lbf/in.2, 200 F, and a stream of pure N2 gas at 14.7 lbf/in.2, 70 F. The plant operates on an electrical power input of 2000 kW. Calculate the net rate of entropy change for the process. Air 79 % N2

1

2 pure O 2

21 % O2

pure N 2

P1 = 14.7

3

P2 = 30 T2 = 200 F P3 = 14.7

. T1 = 70 F T3 = 70 F -WIN = 2000 kW . n1 = 2 lbmol/s . . QCV . QCV . dSNET . . =+ ni∆s-i = + (n2-s2 + n3-s3 - n1-s1) dt T0 i T0 . . . . . . QCV = Σn∆hi + WCV = nO CP0 O (T2-T1) + nN CP0 N (T3-T1) + WCV ∑

2

2

2

2

= 0.21×2× [32×0.213×(200-70)] + 0 - 2000×3412/3600 = +382.6 - 1895.6 = -1513 Btu/s 660 1545 30 . Σni∆s-i = 0.21×2 32×0.219 ln ln 530 778 0.21×14.7

[

[

+ 0.79×2 0 -

1545 14.7 ln 778 0.79×14.7

]

= -1.9906 Btu/R s dSNET dt

=+

1513 - 1.9906 = 0.864 Btu/R s 530

]

12-58 12.92EA tank has two sides initially separated by a diaphragm. Side A contains 2 lbm of water and side B contains 2.4 lbm of air, both at 68 F, 14.7 lbf/in.2. The diaphragm is now broken and the whole tank is heated to 1100 F by a 1300 F reservoir. Find the final total pressure, heat transfer, and total entropy generation. U2-U1 = ma(u2-u1)a + mv(u2-u1)v = 1Q2 S2-S1 = ma(s2-s1)a + mv(s2-s1)v = ⌠ ⌡ 1Q2/T + Sgen V2 = VA + VB = mvvv1 + mava1 = 0.0321 + 31.911 = 31.944 ft3 vv2 = V2/mv = 15.9718, T2 => P2v = 58.7 lbf/in2 va2 = V2/ma = 13.3098, T2 => P2a = mRT2/V2 = 43.415 lbf/in2 P2tot = P2v + P2a = 102 lbf/in2 Water: u1 = 36.08, u2 = 1414.3, s1 = 0.0708, s2 = 2.011 Air: u1 = 90.05, u2 = 278.23, sT1 = 1.6342, sT2 = 1.9036 Q = 2(1414.3 - 36.08) + 2.4(278.23 - 90.05) = 3208 Btu

1 2

Sgen = 2(2.011-0.0708) + 2.4[1.9036 - 1.6342 - (53.34/778)×ln(43.415/14.7)] - 3208/1760 = 3.8804 + 0.4684 - 1.823 = 2.526 Btu/R 12.93EConsider a volume of 2000 ft3 that contains an air-water vapor mixture at 14.7 lbf/in.2, 60 F, and 40% relative humidity. Find the mass of water and the humidity ratio. What is the dew point of the mixture? Air-vap P = 14.7 lbf/in.2, T= 60 F, φ = 40% Pg = Psat60 = 0.256 lbf/in.2 Pv = φ Pg = 0.4 × 0.256 = 0.1024 lbf/in.2 mv1 =

P vV RvT

=

0.1024 × 144 × 2000 = 0.661 lbm 85.76 × 520

Pa = Ptot- Pv1 = 14.7 – 0.1024 = 14.598 ma = w1 =

P aV RaT mv ma

=

=

14.598 × 144 × 2000 = 151.576 lbm 53.34 ×⊇520

0.661 = 0.00436 151.576

Tdew is T when Pg(Tdew) = 0.1024;

T = 35.5 F

12-59 12.94EConsider a 10-ft3 rigid tank containing an air-water vapor mixture at 14.7 lbf/in.2, 90 F, with a 70% relative humidity. The system is cooled until the water just begins to condense. Determine the final temperature in the tank and the heat transfer for the process. Pv1 = φPG1 = 0.7 × 0.6988 = 0.489 lbf/in2 Since mv = const & V = const & also Pv = PG2: PG2 = Pv1×T2/T1 = 0.489×T2/549.7 For T2 = 80 F:

0.489×539.7/549.7 = 0.4801 =/ 0.5073 ( = PG at 80 F )

For T2 = 70 F:

0.489×529.7/549.7 = 0.4712 =/ 0.3632 ( = PG at 70 F )

interpolating → T2 = 78.0 F w2 = w1 = 0.622 ma =

Pa1V RaT1

=

0.489 = 0.0214 (14.7-0.489)

14.211×144×10 = 0.698 lbm 53.34×549.7

1st law: Q12 = U2-U1 = ma(ua2-ua1) + mv(uv2-uv1) = 0.698[0.171(78 - 90) + 0.0214(1036.3 - 1040.2)] = 0.698(-2.135 Btu/lbm air) = -1.49 Btu 12.95EAir in a piston/cylinder is at 95 F, 15 lbf/in.2 and a relative humidity of 80%. It is now compressed to a pressure of 75 lbf/in.2 in a constant temperature process. Find the final relative and specific humidity and the volume ratio V2/V1. Check if the second state is saturated or not. First assume no water is condensed 1:

Pv1= φ1PG1 = 0.66,

w1 = 0.622×0.66/14.34 = 0.0286

2: w2 = 0.622 Pv2/(P2-Pv2) = w1 => Pv2 = 3.297 > Pg = 0.825 lbf/in2 Conclusion is state 2 is saturated φ2 = 100%, w2 = 0.622 Pg/(P2-Pg) = 0.00692 To get the volume ratio, write the ideal gas law for the vapor phases V2 = Va2 + Vv2 + Vf2 = (maRa + mv2Rv)T/P2 + mliqvf V1 = Va1 + Vv1 = (maRa + mv1Rv)T/P1 Take the ratio and devide through with maRaT/P2 to get V2/V1 = (P1/P2)[1 + 0.622w2 + (w1-w2)P2vf/RaT]/(1+0.622 w1) = 0.1974 The liquid contribution is nearly zero (=0.000127) in the numerator.

12-60 12.96EA 10-ft3 rigid vessel initially contains moist air at 20 lbf/in.2, 100 F, with a relative humidity of 10%. A supply line connected to this vessel by a valve carries steam at 100 lbf/in.2, 400 F. The valve is opened, and steam flows into the vessel until the relative humidity of the resultant moist air mixture is 90%. Then the valve is closed. Sufficient heat is transferred from the vessel so the temperature remains at 100 F during the process. Determine the heat transfer for the process, the mass of steam entering the vessel, and the final pressure inside the vessel. i Air-vap mix: P1 = 20 lbf/in2, T1 = 560 R H 2O

φ1 = 0.10, T2 = 560 R, φ2 = 0.90 Pv1 = φ1PG1 = 0.1×0.9503 = 0.095 lbf/in2 AIR + H 2O

Pv2 = 0.9×0.9503 = 0.8553 lbf/in2 Pa2 = Pa1 = P1 - Pv1 = 20 - 0.095 = 19.905 w1 = 0.622×0.095/19.905 = 0.002 96

w2 = 0.622×0.8553/19.905 = 0.026 64 w=

mv ma

→ mvi = ma(w2-w1), ma =

19.905×144×10 = 0.96 lbm 53.34×560

P2 = 19.905 + 0.855 = 20.76 lbf/in2 mvi = 0.96(0.02664 - 0.00296) = 0.0227 lbm CV: vessel QCV = ma(ua2-ua1) + mv2uv2 - mv1uv1 - mvihi uv ≈ uG at T → uv1 = uv2 = uG at 100 F, ua2 = ua1 → QCV = mvi(uG at T - hi) = 0.0227(1043.5-1227.5) = -4.18 Btu

12-61 12.97EA water-filled reactor of 50 ft3 is at 2000 lbf/in.2, 550 F and located inside an insulated containment room of 5000 ft3 that has air at 1 atm. and 77 F. Due to a failure the reactor ruptures and the water fills the containment room. Find the final pressure. CV Total container.

Energy:

mv(u2-u1) + ma(u2-u1) = 1Q2 - 1W2 = 0

Initial water: v1 = 0.021407, u1 = 539.24, Initial air:

mv = V/v = 2335.7 lbm

ma = PV/RT = 14.7×4950×144/53.34×536.67 = 366.04 lbm

Substitute into energy equation 2335.7 (u2-539.24) + 366.04×0.171 (T2-77) = 0 u2 + 0.0268 T2 = 541.3

&

v2 = V2/mv = 2.1407 ft3/lbm

Trial and error 2-phase (T guess, v2 => x2 => u 2 => LHS) T = 300 LHS = 550.789

T = 290 LHS = 506.05

T2 = 298 F, x2 = 0.3198, Psat = 65 lbf/in2, LHS=541.5 OK Pa2 = Pa1V1T2/V2T1 = 14.7×4950×757.7/5000×536.67 = 20.55 lbf/in2 => P2 = Pa2 + Psat = 85.55 lbf/in2 12.98EAtmospheric air at 95 F, relative humidity of 10%, is too warm and also too dry. An air conditioner should deliver air at 70 F and 50% relative humidity in the amount of 3600 ft3 per hour. Sketch a setup to accomplish this, find any amount of liquid (at 68 F) that is needed or discarded and any heat transfer. CV air conditioner. Check from psychrometric chart, inlet 1, exit 2. In: w1 = 0.0034, h˜mix,1 = 27, hf,68 = 36.08, Ex: w2 = 0.0078, h˜mix,2 = 25.4 Water must be added. Continuity and energy equations . . . . . . . mA(1 + w1) + mliq = mA(1 + w2) & mAh1mix + mliqhf + QCV = mAh2mix . . mtot = PVtot /RT = 14.7×3600×144/53.34×529.67 = 270 lbm/h . . mA = mtot/(1+w2) = 267.91 lbm/h . . mliq = mA(w2 - w1) = 267.91(0.0078 - 0.0034) = 1.179 lbm/h . QCV = 267.91(25.4-27-0.0044×36.08) = - 471.2 Btu/h If the properties are calculated from the tables then Pg1 = 0.8246, hg1 = 1102.9, Pv1 = φ1 Pg1 = 0.08246,

Pg2 = 0.36324,

hg2 = 1092

Pv2 = φ2 Pg2 = 0.1816

w1 = 0.622Pv1/(Ptot -Pv1) = 0.0035, w2 = 0.00778 h˜2-h˜1 = 0.24(70-95) + 0.00778×1092 - 0.0035×1102.9 = -1.364 . mliq = 267.91(0.00778 - 0.0035) = 1.147 lbm/h . QCV = 267.91(-1.364 - 0.00428×36.08) = - 406.8 Btu/h

12-62 12.99ETwo moist air streams with 85% relative humidity, both flowing at a rate of 0.2 lbm/s of dry air are mixed in a SSSF setup. One inlet flowstream is at 90 F and the other at 61 F. Find the exit relative humidity. CV mixing chamber. . . . ma w1 + ma w2 = 2ma wex;

. . . ma h1 + ma h2 = 2ma hex

1: w1 = 0.0262, h1 = 50.4 2: w2 = 0.0097, h2 = 25.2 wex = (w1 + w2)/2 = 0.018 ; hex = (h1 + h2)/2 = 37.8 Exit: wex, hex => Tex = 75 F,

φ = 95%

12.100E An indoor pool evaporates 3 lbm/h of water, which is removed by a dehumidifier to maintain 70 F, Φ = 70% in the room. The dehumidifier is a refrigeration cycle in which air flowing over the evaporator cools such that liquid water drops out, and the air continues flowing over the condenser, as shown in Fig. P12.71. For an air flow rate of 0.2 lbm/s the unit requires 1.2 Btu/s input to a motor driving a fan and the compressor and it has a coefficient of performance, β = Q_L /W_c = 2.0. Find the state of the air after the evaporator, T2, ω2, Φ2 and the heat rejected. Find the state of the air as it returns to the room and the compressor work input. The unit must remove 3 lbm/h liquid to keep steady state in the room. As water condenses out state 2 is saturated. 1: 70 F, 70% => w1 = 0.011, h1 = 28.8 . . . . CV 1 to 2: mliq = ma(w1 - w2) => w2 = w1 - mliq /ma qL = h1 - h2 - (w1 - w2) hf2 w2 = 0.011 - 3/(3600 × 0.2) = 0.0068 2: w2, 100% => T2 = 47 F, h˜2 = 18.8, hf2 = 15.04 qL = 28.8 - 18.8 - 0.0042×15.04 = 9.94 Btu/lbm dry air . . CV Total system : h˜3 = h˜1 + Wel/ma - (w1-w2) hf = 28.8 + 1.2/0.2 - 0.0632 = 34.74 Btu/lbm dry air 3: w3 = w2, h3 => T3 = 112 F, φ3 = 12% . . Wc = ma qL/ β = 1 Btu/s

12-63 12.101E To refresh air in a room, a counterflow heat exchanger is mounted in the wall, as shown in Fig. P12.72. It draws in outside air at 33 F, 80% relative humidity and draws room air, 104 F, 50% relative humidity, out. Assume an exchange of 6 lbm/min dry air in a SSSF device, and also that the room air exits the heat exchanger to the atmosphere at 72 F. Find the net amount of water removed from room, any liquid flow in the heat exchanger and (T, φ) for the fresh air entering the room. State 1: w1 = 0.0236, h1 = 51, Tdew,1 = 81.5 F The room air is cooled to 72 F < Tdew1 so liquid will form in the exit flow channel and state 2 is saturated. 2: 72 F, φ = 100% => w2 = 0.017, h2 = 36, hf2 = 40.09 3: 33 F, φ = 80% => w3 = 0.0031, h3 = 11.3 . . CV 1 to 2: mliq,2 = ma (w1 - w2) = 6 (0.0236 - 0.017) = 0.04 lbm/min . . . CV room: mv,out = ma (w1 - w4) = ma (w1 - w3) = 6(0.0236-0.0031) = 0.123 lbm/min . . . CV Heat exchanger: ma(h˜4 - h˜3) = ma(h˜1 - h˜2) - mliq hf2 h4 = h3 + h1 - h2 - (w1-w2) hf2 = 11.3 + 51 - 36 - 0.0066 × 40.09 = 26 Btu/lbm dry air 4: w4 = w3, h4 => T4 = 94 F, φ = 9%

12-64 12.102E A 4-ft3 insulated tank contains nitrogen gas at 30 lbf/in.2 and ambient temperature 77 F. The tank is connected by a valve to a supply line flowing carbon dioxide at 180 lbf/in.2, 190 F. A mixture of 50 mole percent nitrogen and 50 mole percent carbon dioxide is to be obtained by opening the valve and allowing flow into the tank until an appropriate pressure is reached and the valve is closed. What is the pressure? The tank eventually cools to ambient temperature. Calculate the net entropy change for the overall process. V = 4 ft3, P1 = 30 lbf/in2, T1 = T0 = 77 F

CO 2 i

Pi = 180 lbf/in.2, Ti = 190 F

N

n2 CO

2

2

At state 2: yN = yCO = 0.50 2 2 = n2 N = n1 N = P1V/RT1 2

2

= 30×4×144/(1545×536.67) = 0.02084 lbmol n2 = 0.04168 lbmol 1st law: nihi = n2u-2 - n1u-1 , for const specific heats niCPoiTi = (niCVoi+n1CVo1)T2 - n1CVo1T1 But ni = n1

→ CPoiTi = CVoiT2 + CVo1(T2-T1)

44.01×0.201×649.67 = 44.01×0.156 T2 + 28.013×0.178(T2-536.67) T2 = 710.9 R P2 = n2RT2/V = 0.04168×1545×710.9/4×144 = 79.48 lbf/in2 Cool to T3 = T0 = 77 F = 536.67 R P3 = P2× T3/T2 = 79.48× 536.67/710.9 = 60 lbf/in2 Q23 = n2CVo2(T3-T2) = 0.04168(0.5×28.013×0.178 + 0.5×44.01×0.156)(536.67-710.9) = - 43.0 Btu ∆SNET = n3-s3 - n1-s1 - ni-si - Q23/T0 = ni[(s-CO )3 - -si] + n1[(s-N )3 - s-1] - Q23/T0 2

2

= 0.02084(44.01×0.201 ln + 0.02084×( - 1.98589 ln

0.5×60 536.67 - 1.98589 ln ) 649.67 180 0.5×60 −43.0 )− 30 536.67

= +0.03893 + 0 + 0.0801 = +0.119 Btu/R

12-65 12.103E Ambient air is at a condition of 14.7 lbf/in.2, 95 F, 50% relative humidity. A steady stream of air at 14.7 lbf/in.2, 73 F, 70% relative humidity, is to be produced by first cooling one stream to an appropriate temperature to condense out the proper amount of water and then mix this stream adiabatically with the second one at ambient conditions. What is the ratio of the two flow rates? To what temperature must the first stream be cooled? 1 P1 = P2 = P5 = 14.7 lbf/in2 5 2

T1 = T2 = 95 F

MIX COOL

. -Q COOL

3

φ1 = φ2 = 0.50, φ4 = 1.0

. -Q MIX= 0

4

T5 = 73 F, φ5 = 0.70

LIQ H 2O

Pv1 = Pv2 = 0.5×0.8246 = 0.4123 w1 = w2 = 0.622×

0.4123 = 0.0179 14.7-0.4123

Pv5 = 0.7×0.4064 = 0.2845

=>

w5 = 0.622×

0.2845 = 0.0123 14.7-0.2845

MIX: Call the mass flow ratio r = ma2/ma1 Conservation of water mass:

w1 + r w4 =(1+ r)w5

Energy Eq.: ha1 + w1hv1 + rha4 + rw4hv4 = (1+r)h a5 + (1+r)w5hv5 → or

r=

0.0179 + rw4 = (1+r) 0.0123

0.0179-0.0123 , with 0.0123-w4

w4 = 0.622×

PG4 14.7-PG4

0.24×555 + 0.0179×1107.2 + r×0.24×T4 + rw4hv4 = (1+r)×0.24×533 + (1+r)×0.0123×1093.3

[

Assume T4 = 40 F w4 = 0.622× ma2 ma1

]

r 0.24×T4 + w4hG4 - 141.4 + 11.66 = 0

or

=

→ PG4 = 0.121 66, hG4 = 1078.9

0.121 66 = 0.0052 14.7-0.121 66

0.0179-0.0123 = 0.7887 0.0123-0.0052

0.7887[0.24×500 + 0.0052×1078.9 - 141.4] + 11.66 = -0.29 ≈ 0 => T4 = 40 F

OK

13-1

CHAPTER 13 The correspondence between the new problem set and the previous 4th edition chapter 10 problem set: New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Old 1 10 2 3 4 5 6 7 9 12 13 14 15 16 17 18 19 21 22 23 24

New 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

Old 25 new 26 mod 27 28 new 30 31 mod new 35 36 37 39 new 40 new 41 38 45 44 47

New 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61

Old 48 new 49 new new 51 53 55 56 61 63 new 65 66 new 67 68 new new

Old 52 70 mod 57 58 62

New 72 73 74

Old new new new

The problems that are labeled advanced are: New 62 63 64 65 66

Old 8 29 20 46 50

New 67 68 69 70 71

13-2

The English-unit problems are: New 75 76 77 78 79 80 81 82

Old 74 75 76 78 79 80 82 83

New 83 84 85 86 87 88 89 90

Old 84 81 86 mod 87 88 89 90 91

New 91 92 93 94 95 96 97

Old 92 93 96 97 98 99 101

mod indicates a modification from the previous problem that changes the solution but otherwise is the same type problem.

13-3

The following table gives the values for the compressibility, enthalpy departure and the entropy departure along the saturated liquid-vapor boundary. These are used for all the problems using generalized charts as the figures are very difficult to read accurately (consistently) along the saturated liquid line. It is suggested that the instructor hands out copies of this page or let the students use the computer for homework solutions.

Tr

Pr

Zf

Zg

d(h/RT)f

d(h/RT)g

d(s/R)f

d(s/R)g

0.96

0.78

0.14

0.54

3.65

1.39

3.45

1.10

0.94

0.69

0.12

0.59

3.81

1.19

3.74

0.94

0.92

0.61

0.10

0.64

3.95

1.03

4.00

0.82

0.90

0.53

0.09

0.67

4.07

0.90

4.25

0.72

0.88

0.46

0.08

0.70

4.17

0.78

4.49

0.64

0.86

0.40

0.07

0.73

4.26

0.69

4.73

0.57

0.84

0.35

0.06

0.76

4.35

0.60

4.97

0.50

0.82

0.30

0.05

0.79

4.43

0.52

5.22

0.45

0.80

0.25

0.04

0.81

4.51

0.46

5.46

0.39

0.78

0.21

0.035

0.83

4.58

0.40

5.72

0.35

0.76

0.18

0.03

0.85

4.65

0.34

5.98

0.31

0.74

0.15

0.025

0.87

4.72

0.29

6.26

0.27

0.72

0.12

0.02

0.88

4.79

0.25

6.54

0.23

0.70

0.10

0.017

0.90

4.85

0.21

6.83

0.20

0.68

0.08

0.014

0.91

4.92

0.18

7.14

0.17

0.66

0.06

0.01

0.92

4.98

0.15

7.47

0.15

0.64

0.05

0.009

0.94

5.04

0.12

7.81

0.12

0.60

0.03

0.005

0.95

5.16

0.08

8.56

0.08

0.58

0.02

0.004

0.96

5.22

0.06

8.97

0.07

0.54

0.01

0.002

0.98

5.34

0.03

9.87

0.04

0.52

0.0007

0.0014

0.98

5.41

0.02

10.38

0.03

13-4

13.1

A special application requires R-12 at −140°C. It is known that the triple-point temperature is −157°C. Find the pressure and specific volume of the saturated vapor at the required condition. The lowest temperature in Table B.3 for R-12 is -90oC, so it must be extended to -140oC using the Clapeyron Eq. 13.7 integrated as in example 13.1 Table B.3:

at T1 = -90oC = 183.2 K,

P1 = 2.8 kPa.

8.3145 = 0.068 76 120.914 P hfg (T - T1) 189.748 (133.2 - 183.2) ln = = = -5.6543 P1 R T × T 0.068 76 133.2 × 183.2

R=

1

P = 2.8 exp(-5.6543) = 0.0098 kPa

13-5

13.2

In a Carnot heat engine, the heat addition changes the working fluid from saturated liquid to saturated vapor at T, P. The heat rejection process occurs at lower temperature and pressure (T − ∆T), (P − ∆P). The cycle takes place in a piston cylinder arrangement where the work is boundary work. Apply both the first and second law with simple approximations for the integral equal to work. Then show that the relation between ∆P and ∆T results in the Clapeyron equation in the limit ∆T → dT. T 1

T T −∆T

P

P

2

P-∆P

qH = TsFG;

4

s

sfg at T

qL = (T-∆T)sFG ;

2

T

P-∆P

3

4

1

P

3

vfg at T

T-∆T v

wNET = qH - qL = ∆TsFG

Problem similar to development in section 13.1 for shaft work, here boundary movement work, w = ⌠ ⌡ Pdv 3

4

wNET = P(v2-v1) + ⌠ ⌡ Pdv + (P-∆P)(v4-v3) + ⌠ ⌡ Pdv 2

1

Approximating, 3

∆P ⌠ ⌡ Pdv ≈ (P - ) (v3-v2); 2 2

4

⌠ ⌡ Pdv ≈ 1

& collecting terms, v2+v3 v1+v4 wNET ≈ ∆P[( ) -( )] 2 2 (the smaller the ∆P, the better the approximation) ⇒

sFG ∆P ≈ v +v v +v ∆T ( 22 3) - ( 12 4)

In the limit as ∆T → 0:

v3 → v2 = vG ,

lim ∆P dPSAT sFG & ∆T→0 = = dT vFG ∆T

v4 → v1 = vF

∆P

(P - 2 ) (v1-v4)

13-6

13.3

Ice (solid water) at −3°C, 100 kPa is compressed isothermally until it becomes liquid. Find the required pressure. Water, triple point T = 0.01oC , P = 0.6113 kPa Table B.1.1: vF = 0.001, hF = 0.01 kJ/kg, Tabel B.1.5:

vI = 0.001 0908, hI = -333.4 kJ/kg dPIF

Clapeyron

dT ∆P ≈

dPIF dT

=

hF-hI (vF-vI)T

=

333.4 = -13 442 -0.0000908×273.16

∆T = -13 442(-3 - 0.01) = 40 460 kPa

P = Ptp + ∆P = 40 461 kPa

13.4

Calculate the values hFG and sFG for nitrogen at 70 K and at 110 K from the Clapeyron equation, using the necessary pressure and specific volume values from Table B.6.1. dPG hFG sFG = = Clapeyron equation Eq.13.7: dT TvFG vFG For N2 at 70 K, using values for PG from Table B.6 at 75 K and 65 K, and also vFG at 70 K, ∆PG

hFG ≈ T(vG-vF)

∆Τ

= 70(0.525 015)

(76.1-17.41 ) = 215.7 kJ/kg (207.8) 75-65

sFG = hFG/T = 3.081 kJ/kg K (2.97) Comparison not very close because PG not linear function of T. Using 71 K & 69 K, hFG = 70(0.525 015)

(44.56-33.24 ) = 208.0 71-69

At 110 K, hFG ≈ 110(0.014 342) sFG =

(1938.8-1084.2 ) = 134.82 kJ/kg (134.17) 115-105

134.82 = 1.226 kJ/kg K (1.22) 110

13-7

13.5

Using thermodynamic data for water from Tables B.1.1 and B.1.5, estimate the freezing temperature of liquid water at a pressure of 30 Mpa. P

30 MPa

H 2O

dPIF dT

=

hIF TvIF

≈ const

At the triple point, vIF = vF - vI = 0.001 000 - 0.001 090 8

T.P.

= -0.000 090 8 m3/kg

T

hIF = hF - hI = 0.01 - (-333.40) = 333.41 kJ/kg dPIF dT

=

333.41 = -13 442 kPa/K 273.16(-0.000 090 8)

⇒ at P = 30 MPa, T ≈ 0.01 +

13.6

(30 000-0.6) = -2.2 oC (-13 442)

Helium boils at 4.22 K at atmospheric pressure, 101.3 kPa, with hFG = 83.3 kJ/kmol. By pumping a vacuum over liquid helium, the pressure can be lowered and it may then boil at a lower temperature. Estimate the necessary pressure to produce a boiling temperature of 1 K and one of 0.5 K. Helium at 4.22 K: P1 = 0.1013 MPa, hFG = 83.3 kJ/kmol dPSAT dT

=

hFG TvFG

≈

hFGPSAT RT2

⇒

ln

P2 P1

=

hFG 1 1 [ − ] R T1 T2

For T2 = 1.0 K: ln

P2

101.3 For T2 = 0.5 K: ln

P2 101.3

=

83.3 1 1 [ − ] 8.3145 4.22 1.0

=

83.3 1 1 [ − ] 8.3145 4.22 0.5

=> P2 = 0.048 kPa = 48 Pa

P2 = 2.1601×10-6 kPa = 2.1601×10-3 Pa

13-8

13.7

A certain refrigerant vapor enters an SSSF constant pressure condenser at 150 kPa, 70°C, at a rate of 1.5 kg/s, and it exits as saturated liquid. Calculate the rate of heat transfer from the condenser. It may be assumed that the vapor is an ideal gas, and also that at saturation, vf ! vg. The following quantities are known for this refrigerant: ln Pg = 8.15 - 1000/T CP = 0.7 kJ/kg K with pressure in kPa and temperature in K. The molecular weight is 100. Refrigerant: State 1 T1 = 70oC P1 = 150 kPa State 2 P2 = 150 kPa

x2 = 1.0

State 3

P3 = 150 kPa x3 = 0.0

ln (150) = 8.15 - 1000/T2 => T2 = 318.5 K = 45.3oC = T3 q13 = h3 -h1 = (h3 -h2) + (h2 -h1) = - hFG T3 + CP0(T2 -T1) dPG dT

=

hFG TvFG

d ln PG dT

,

dPG

RT vFG ≈ vG = , PG

dT

= PG

d ln PG dT

=

hFG RT2

PG

= +1000/T2 = hFG/RT2

hFG = 1000 × R = 1000 × 8.3145/100 = 83.15 q13 = -83.15 + 0.7(45.3 - 70) = -100.44 kJ/kg . QCOND = 1.5(-100.44) = -150.6 kW

13.8

A container has a double wall where the wall cavity is filled with carbon dioxide at room temperature and pressure. When the container is filled with a cryogenic liquid at 100 K the carbon dioxide will freeze so the wall cavity has a mixture of solid and vapor carbon dioxide at the sublimation pressure. Assume that we do not have data for CO2 at 100 K, but it is known that at −90°C: Psat = 38.1 kPa, hIG = 574.5 kJ/kg. Estimate the pressure in the wall cavity at 100 K. For CO2 space: at T1 = -90 oC = 183.2 K , P1 = 38.1 kPa, hIG = 574.5 kJ/kg For T2 = TcO2 = 100 K: Clapeyron ln

P2 P1

=

hIG R

dPSUB dT

=

hIG TvIG

≈

hIGPSUB RT2

1 1 574.5 1 1 [183.2 − ] = [ − ] = -13.81 100 0.188 92 183.2 100

or P2 = P1 × 1.005×10-6

⇒ P2 = 3.83×10-5 kPa = 3.83×10-2 Pa

13-9

13.9

Small solid particles formed in combustion should be investigated. We would like to know the sublimation pressure as a function of temperature. The only information available is T, hFG for boiling at 101.3 kPa and T, hIF for melting at 101.3 kPa. Develop a procedure that will allow a determination of the sublimation pressure, Psat(T). P

TNBP = normal boiling pt T.

Solid

TNMP = normal melting pt T.

101.3 kPa PTP

TTP = triple point T.

Liquid Vap.

T TP T

1) TTP ≈ TNMP

T T NBP

NMP

TTP P TP

⌠h  FG ⌠ ⌡ (1/PSAT) dPSAT ≈ ⌡ RT2 dT

2)

0.1013 MPa

TNMP

Since hFG ≈ const ≈ hFG NBP we ca perform the integral over temperature ln

PTP 0.1013

≈

hFG NBP R

[T 1

-

NBP

1 ] TTP

→

get PTP

3) hIG at TP = hG - hI = (hG - hF) + (hF - hI) ≈ hFG NBP + hIF NMP Assume hIG ≈ const. again we can evaluate the integral

ln

PSUB PTP

P SUB

T

⌠h h  IG dT ≈ IG[ 1 − 1 ] = ⌠ (1/P ) dP ≈ ⌡ SUB SUB R TTP T ⌡ RT2 P TP

or PSUB = fn(T)

TTP

13-10

13.10 Derive expressions for ( ∂T/∂v)u and for (∂h/∂s)v that do not contain the properties h, u, or s. ∂P P - T( )v ∂T ∂u ∂u ∂T (∂v )u = - (∂v)T (∂T)v = C v

/

(see Eqs. 13.33 and 13.34)

∂h ∂P ∂T As dh = Tds + vdP => ( )v = T + v( )v = T - v( )s ∂s ∂s ∂v ∂P T( )v ∂T ∂s ∂s ∂T (∂v )s = - (∂v)T (∂T)v = - C v

/

But

∂h

(Eq.13.20)

(Eq.13.22)

∂P

(∂s )v = T + vT ( ) C ∂T v

⇒

v

13.11 Derive expressions for ( ∂h/∂v)T and for (∂h/∂T )v that do not contain the properties h, u, or s. Find

(∂h∂v)T and (∂h∂T)v

dh = Tds + vdP ⇒

and use Eq.13.22

(∂h∂v)T = T(∂s∂v)T + v(∂P ) ∂v T

=T

∂P + v (∂P ) ( ) ∂T v ∂v T

Also for the second first derivative use Eq.13.28 ∂s ∂P = C + v (∂h∂T)v = T(∂T )v + v(∂P ) ( ) v ∂T v ∂T v

13.12 Develop an expression for the variation in temperature with pressure in a constant entropy process, (∂T/∂P)s, that only includes the properties P–v–T and the specific heat, Cp. ∂s

∂T

(∂P)s = ∂s

(∂P)T ∂s

(∂T)P

∂v

∂v ) ∂T P T ∂v == ( ) (CP/T) CP ∂T P -(

{(∂P)T = -(∂T)P, Maxwell relation Eq. 13.23 and the other is Eq.13.27}

13-11

13.13 Determine the volume expansivity, α P, and the isothermal compressibility, β T, for water at 20°C, 5 MPa and at 300°C, and 15 MPa using the steam tables. Water at 20oC, 5 MPa (compressed liquid) 1 ∂v 1 ∆v ≈ ( )P ( ) P v ∂T v ∆T

αP =

Estimate by finite difference.

Using values at 0oC, 20oC and 40oC,

αP ≈

1 0.001 0056 - 0.000 9977 = 0.000 1976/oC 0.000 9995 40 - 0

βT = -

1 ∂v 1 ∆v ≈ ( ) ( ) v ∂P T v ∆P T

Using values at saturation, 5 MPa and 10 MPa,

βT ≈ -

1 0.000 9972 - 0.001 0022 = 0.000 50 MPa-1 0.000 9995 10 - 0.0023

Water at 300oC, 15 MPa (compressed liquid)

αP ≈

1 0.001 4724 - 0.001 3084 = 0.002 977/oC 0.001 377 320 - 280

βT ≈ -

1 0.001 3596 - 0.001 3972 = 0.002 731 MPa-1 0.001 377 20 - 10

13.14 Sound waves propagate through a media as pressure waves that cause the media to go through isentropic compression and expansion processes. The speed of sound c is defined by c 2 = (∂P/∂ ρ )s and it can be related to the adiabatic compressibility, which for liquid ethanol at 20°C is 940 µm2/N. Find the speed of sound at this temperature. c2 =

∂P = −v (∂P ) ( ) = ∂ρ s ∂v s

1

2

1 ∂v ρ v ∂P s

-

From Table A.4 for ethanol, ⇒

c=

(940×101

-12

) ×783

( )

=

1 β sρ

ρ = 783 kg/m3 1/2

= 1166 m/s

13-12

13.15 Consider the speed of sound as defined in Problem 13.14. Calculate the speed of sound for liquid water at 20°C, 2.5 MPa and for water vapor at 200°C, 300 kPa using the steam tables. c2 =

From problem 13.14 :

∂P = -v (∂P ) ( ) ∂ρ s ∂v s 2

Liquid water at 20oC, 2.5 MPa, assume ∆P ≈ (∂P ) ( ) ∂v s ∆v

T

Using saturated liquid at 20oC and compressed liquid at 20oC, 5 MPa, 9995 5-0.0023 (0.001 002+0.000 ) ( ) = 2.002×10 2 0.000 9995-0.001 002 2

c2 = -

=>

6

c = 1415 m/s

Superheated vapor water at 200oC, 300 kPa v = 0.7163, s = 7.3115 At P = 200 kPa & s = 7.3115: T = 157oC,

v = 0.9766

At P = 400 kPa & s = 7.3115: T = 233.8oC, v = 0.5754 c2 = -(0.7163)2 =>

0.400-0.200 (0.5754-0.9766 ) = 0.2558×10

6

c = 506 m/s

13.16 Find the speed of sound for air at 20°C, 100 kPa using the definition in Problem 13.14 and relations for polytropic processes in ideal gases. From problem 13.14 :

c2 =

∂P = -v (∂P ) ( ) ∂ρ s ∂v s 2

For ideal gas and isentropic process, Pvk = const P = Cv-k ⇒

∂P = -kCv-k-1 = -kPv-1 ∂v

c2 = -v2(-kPv-1) = kPv = kRT

c = kRT =

1.4×0.287×293.15×1000 = 343.2 m/s

13-13

13.17 A cylinder fitted with a piston contains liquid methanol at 20°C, 100 kPa and volume 10 L. The piston is moved, compressing the methanol to 20 MPa at constant temperature. Calculate the work required for this process. The isothermal compressibility of liquid methanol at 20°C is 1220 (µm)2/N. 2

2

⌠ ∂v ⌠ w12 = ⌠ ⌡ Pdv =  ⌡ P ∂P T dPT = -⌡ vβ T PdPT

( )

1

1

For v ≈ constant & βT ≈ constant the integral can be evaluated w12 = -

vβ T 2

(P

2 2

2

- P1

)

For liquid methanol, from Table A.4: ρ = 787 m3/kg V1 = 10 L, W12 =

m = 0.01×787 = 7.87 kg

0.01×1220 (20)2 - (0.1)2 2

[

] = 2440 J = 2.44 kJ

13.18 A piston/cylinder contains 5 kg of butane gas at 500 K, 5 MPa. The butane expands in a reversible polytropic process with polytropic exponent, n = 1.05, until the final pressure is 3 MPa. Determine the final temperature and the work done during the process. C4H10 m = 5 kg T1 = 500 K P1 = 5 MPa Rev. polytropic process (n=1.05), P2 = 3 MPa Tr1 =

500 5 = 1.176, Pr1 = = 1.316 425.2 3.8

V1 = n P1V1

From Fig. D.1:

Z1 = 0.68

5×0.68×0.14304×500 = 0.0486 m3 5000 =

Z2Tr2 =

n P2V2

=>

P2V2 mRTC

=

V2 = 0.0486

1 1.05 (5/3)

= 0.0791 m3

3000×0.0791 = 0.7803 5×0.143 04×425.2

at Pr2 = 3/3.8 = 0.789 Trial & error: Tr2 = 1.062, Z2 = 0.735 2

W12 = ⌠ ⌡ PdV = 1

P2V2 - P1V1 1-n

=

=>

T2 = 451.6 K

3000×0.0791 - 5000×0.0486 = 114 kJ 1 - 1.05

13-14

13.19 Show that the two expressions for the Joule–Thomson coefficient µJ given by Eq. 13.54 are valid. µj = ( Also µj =

∂T ∂h = ) ( ) ∂P h ∂P T

/(∂h∂T)P = [T(∂v∂T)P - v]/C

v=

ZRT , P

P

∂v

(∂T)P =

RT ∂Z [ZRT + ( ) - v]/C P P ∂T P 2

= P

ZR RT ∂Z + ( ) P P ∂T P

RT2 ∂Z ( ) CPP ∂T P

13.20 A 200-L rigid tank contains propane at 9 MPa, 280°C. The propane is then allowed to cool to 50°C as heat is transferred with the surroundings. Determine the quality at the final state and the mass of liquid in the tank, using the generalized compressibility chart, Fig. D.1. Propane C3H8: V = 0.2 m3, P1 = 9 MPa, T1 = 280oC = 553.2 K cool to T2 = 50 oC = 323.2 K From Table A.2:

TC = 369.8 K,

PC = 4.25 MPa

9 553.2 = 2.118, Tr1 = = 1.496 From Fig. D.1: Z1 = 0.825 4.25 369.8 Z1RT1 0.825×0.188 55×553.2 v2 = v1 = = = 0.00956 P1 9 000

Pr1 =

From Fig. D.1 at Tr2 = 0.874, PG2 = 0.45 × 4250 = 1912 kPa vG2 = 0.71 × 0.188 55 × 323.2/1912 = 0.02263 vF2 = 0.075 ×0.188 55× 323.2/1912 = 0.00239 0.00956 = 0.002 39 + x 2(0.02263 - 0.00239) mLIQ 2 = (1-0.354)×0.2/0.00956 = 13.51 kg

=>

x 2 = 0.354

13-15

13.21 A rigid tank contains 5 kg of ethylene at 3 MPa, 30°C. It is cooled until the ethylene reaches the saturated vapor curve. What is the final temperature? V = const m = 5 kg P1 = 3 MPa T1 = 30 oC = 303.2 K

C2 H4 T

cool to x2 = 1.0

1

Pr1 =

2

Fig. D.1:

v Pr2 = Pr1

Z2Tr2 Z1Tr1

Trial & error: Tr2 ZG2 0.866

0.72

3 = 0.595, 5.04

= 0.595

Pr2

Tr1 =

303.2 = 1.074 282.4

Z1 = 0.82

ZG2Tr2 0.82×1.074

= 0.6756 ZG2Tr2

Pr2 CALC

0.42

0.421

~ OK

=> T2 = 244.6 K

13.22 Two uninsulated tanks of equal volume are connected by a valve. One tank contains a gas at a moderate pressure P 1, and the other tank is evacuated. The valve is opened and remains open for a long time. Is the final pressure P2 greater than, equal to, or less than P1/2? VA = VB ⇒ V2 = 2V1, T2 = T1 = T V1 Z2 mRT 1 Z2 = = 2 Z1 P1 V2 Z1 mRT

P2

If T < TB, Z2 > Z1

A GAS

P2

1 ⇒ > P1 2

If T > TB, Z2 < Z1 ⇒

P2 P1

Z 1 1.0

<

1 2

B EVAC.

T > TB

2 2 P2

T < TB 1 P1 P

13-16

13.23 Show that van der Waals equation can be written as a cubic equation in the compressibility factor involving the reduced pressure and reduced temperature as 27 Pr2 27 Pr  Z– Z3 – ( + 1 ) Z2 +  =0 8Tr 64 T2 512 Tr 3 r  Pr

van der Waals equation, Eq.13.55: 2 2 27 R TC a= 64 PC

P=

RT a v-b v2

b=

RTC 8PC

v2(v-b) multiply equation by P Get:

v3 - (b +

Multiply by Get:

RT 2 a ab )v +( )v- =0 P P P

Pv P3 3 3 and substitute Z = RT R T

bP aP abP2 Z3 – ( + 1) Z2 + ( 2 2) Z – ( 3 3) = 0 RT RT R T

Substitute for a and b, get: 27 Pr  27 Pr2 Z– Z3 – ( + 1 ) Z2 +  =0 8Tr 64 T2 512 Tr 3 r  Pr

Where

Pr =

P , Pc

Tr =

T Tc

13-17

13.24 Develop expressions for isothermal changes in enthalpy and in entropy for both van der Waals equation and Redlich-Kwong equation of state. RT a van der Waals equation of state: P= − v-b v2 R = (∂P ) ∂T v v-b RT RT a -P= (∂u∂v)T = T(∂P ) − + v v-b v-b v ∂T

2

2

2

⌠ ∂P ⌠a 1 1  P]dv = (u2-u1)T =  [T dv = a( − ) 2 ⌡ v1 v2 ∂T v ⌡v

( )

1

1

1 1 (h2-h1)T = (u2-u1)T + P2v2 - P1v1 = P2v2 − P1v1 + a( − ) v1 v2 2

2

v2-b ⌠ ∂P ⌠ R  dv = dv = R ln (s2-s1)T =  ⌡ ∂T v ⌡ v-b v1-b

( )

( )

1

1

Redlich-Kwong equation of state: R a = (∂P ) + v v-b 2v(v+b)T ∂T

P=

RT a v-b v(v+b)T1/2

3/2

2

v2+b v2 ⌠ 3a -3a (u2-u1)T =  dv = ln[( )(v +b)] v2 ⌡ 2v(v+b)T1/2 2bT1/2 1 1

(h2-h1)T = P2v2 - P1v1 -

3a ln 2bT1/2

[(vv+b)(vv+b)] 2

2

2

1

2

⌠ R a/2 (s2-s1)T =  + ⌡ v-b v(v+b)T3/2 dv

[

]

1

= R ln

(vv -b-b)- 2bTa ln[(vv+b)(vv+b)] 2 1

2

3/2

1

2

1

13-18

13.25 Determine the reduced Boyle temperature as predicted by an equation of state (the experimentally observed value for most substances is about 2.5), using the van der Waals equation and the Redlich–Kwong equation. Note: It is helpful to use Eqs. 13.47 and 13.48 in addition to Eq. 13.46 lim ∂Z ( ) =0 P→0 ∂P T

The Boyle temp. is that T at which

RT lim Z-1 1 lim lim ∂Z But P→0( )T = P→0 = ( v- ) P→0 P-0 RT P ∂P van der Waals:

P=

multiply by v-b =

RT a − v-b v2

v-b , get P

RT a(v-b) or P Pv2

v-

RT a(1-b/v) =b− P Pv

&

a(1-0) lim ∂Z RT × P→0( )T = b − =0 RT ∂P

or

TBoyle =

Redlich-Kwong:

only at TBoyle

a 27 = T = 3.375 TC Rb 8 C P=

RT a − v-b v(v+b)T1/2

as in the first part, get RT a(1-b/v) v=b− P Pv(1+b/v)T1/2 a(1-0) lim ∂Z & RT × P→0( )T = b − =0 ∂P Pv(1+0)T1/2 5/2

or

3/2 TBoyle =

TBoyle = (

only at TBoyle

2 PC a 0.427 48 R TC = × Rb RPC 0.08 664 R TC

0.427 48 2/3 ) TC = 2.9 TC 0.086 64

13-19

13.26 Consider a straight line connecting the point P = 0, Z = 1 to the critical point P = PC, Z = ZC on a Z versus P compressibility diagram. This straight line will be tangent to one particular isotherm at low pressure. The experimentally determined value is about 0.8 TC. Determine what value of reduced temperature is predicted by an equation of state, using the van der Waals equation and the Redlich–Kwong equation. See also note for Problem 13.25.

slope =

ZC - 1

1.0 Z

PC - 0

lim ∂Z ( ) for T = T′ P→0 ∂P T From solution 13.25

C.P.

ZC

But also equals

0

PC P

1 lim RT lim Z-1 lim ∂Z = ( ) = ( v − ) P→0 ∂P T P→0 P-0 RT P→0 P VDW: using solution 13.25:

ZC - 1 1 a lim ∂Z = ( ) = [ b − ] P→0 ∂P T PC RT′ RT′

1-Z

( P C)(RT′)2 + bRT′ − a = 0

or

C

2

3 Substituting ZC = , 8

2 27 R TC a= , 64 PC

b=

2

40 T′r + 8 T′r − 27 = 0

RTC 8PC

solving,

Redlich-Kwong: using solution 13.25, ZC-1 1 a lim ∂Z = ( ) = [ b ] or P→0 ∂P T PC RT′ RT′ 3/2

′ Tr = 0.727

1-Z

( P C)R2T′ 5/2 + bRT′ 3/2 - a = 0 C

5/2

Substitute get

1 ZC = , 3

a = 0.42748

R2TC PC

,

b = 0.08664

2 5/2 3/2 T′r + 0.086 64 T′r − 0.427 48 = 0 3 solving,

T′r = 0.787

RTC PC

13-20

13.27 Determine the 2nd virial coefficient B(T) using the van der Waals equation and the Redlich–Kwong equation of state. Find also its value at the critical point (the − experimentally observed value is about -0.34 RTC/PC). − RT − lim where Eq.13.47: α = −v From Eq.13.51: B(T) = - P→0 α P − − RT a v -b van der Waals: P = − - −2 which we can multiply by , get P v -b v − − − -b) R T a(v RT a(1-b/v− ) − − v -b= − −2 or v − =b− P P Pv− Pv Taking the limit for P -> 0 then we get : − RTC 1 27 TC − B(T) = b − a/RT = ( − ) PC 8 64 T where a,b are from Eq.13.59. At T = TC then we have − − RTC 19 RTC B(TC) = ( - ) = −0.297 64 PC PC For Redlich Kwong the result becomes − a(1- b/v− ) RT − =b− v− − ) T1/2 P Pv− (1 + b/v

=>

Now substitute Eqs.13.61 and 13.62 for a and b, − RTC TC3/2 B(T) = 0.08664 - 0.42748   T PC

[

a B(T) = b − − 3/2 RT

]

and evaluated at TC it becomes − − RTC RTC B(TC) = 0.08664 - 0.42748 = −0.341 PC PC

[

]

13-21

13.28 One early attempt to improve on the van der Waals equation of state was an expression of the form RT a P= v-b v2T Solve for the constants a, b, and v using the same procedure as for the van der c Waals equation. From the equation of state take the first two derivatives of P with v: ∂P RT 2a (∂v )T = - (v-b) and 2+ 3 vT

∂ 2P

2RT 6a (∂v2)T = - (v-b) 3- 4 vT

Since both these derivatives are zero at the critical point: RT 2a 2RT 6a and =0 2+ 3 =0 (v-b) v T (v-b)3 v4T RTC a − 2 Also, PC = vC-b v T C C solving these three equations: 3

vC = 3b,

2 27 R TC a= , 64 PC

b=

RTC 8PC

13.29 Use the equation of state from the previous problem and determine the Boyle temperature. RT a P= − 2 v-b v T v-b RT a(1-b/v) Multiplying by gives: v − b = − P P PvT Using solution from 13.25 for TBoyle: RT a(1-0) a ( v − ) = b − = b − = 0 at TBoyle P→0 P RT×T RT2 lim

3

or

TBoyle =

a = Rb

2 27 R TC 1 8PC = 64 PC R RTC

27 T 8 C

13-22

13.30 Calculate the difference in internal energy of the ideal-gas value and the real-gas value for carbon dioxide at the state 20°C, 1 MPa, as determined using the virial equation of state, including second virial coefficient terms. For carbon dioxide we have: B = -0.128 m3/kmol, T(dB/dT) = 0.266 m3/kmol, both at 20°C. virial eq.:

P=

⌠v u-u = - ⌡∞

[

*

∂P

RT BRT + 2 ; v v

R

RT dB (∂T) v = v + BR 2 + 2 (dT ) v v

⌠v RT2 dB ∂P dB (∂T) v - P dv = - ⌡∞ v2 (dT ) dv = - RT T( ) v dT

]

[

]

[

]

Solution of virial equation (quadratic formula): − − RT 8.3145×293.15 − 1 RT − v= 1 + 1 + 4BP/RT where: = = 2.43737 2 P 1000 P

[

]

− 1 v = × 2.43737 2

[1 +

1 + 4(-0.128)/2.43737

] = 2.3018 m /kmol 3

Using the minus-sign root of the quadratic formula results in a compressibility factor < 0.5, which is not consistent with such a truncated equation of state. u-u* =

-8.3145 × 293.15 0.266 = - 281.7 kJ/kmol 2.3018

[

]

13.31 Refrigerant-123, dichlorotrifluoroethane, which is currently under development as a potential replacement for environmentally hazardous refrigerants, undergoes an isothermal SSSF process in which the R-123 enters a heat exchanger as saturated liquid at 40°C and exits at 100 kPa. Calculate the heat transfer per kilogram of R123, using the generalized charts, Fig. D.2 R-123: M = 152.93, TC = 456.9 K, PC = 3.67 MPa 1

T1 = T2 = 40 oC, x1 = 0 P2 = 100 kPa

2

Heat Exchanger SSSF

From D.1: P1 = 0.084×3670 = 308 kPa

Tr1 = Tr2 = 313.2/456.9 = 0.685, From Fig. D.2:

Pr1 = 0.084,

Pr2 = 0.1/3.67 = 0.027

(h* − h)1/RTC = 4.9

P2 < P1 with no work done, so process is irreversibel. Energy Eq.:

q + h1 = h2,

From Fig. D.2:

Entropy Eq.:

s1 + ∫ dq/T + sgen = s2, sgen > 0

(h*-h)2/RTC = 0.056

h2 - h1 = 8.3145 × 456.9 [-0.056 + 0 + 4.901]/152.93= 120.4 kJ/kg

13-23

13.32 Calculate the heat transfer during the process described in Problem 13.18. From solution 13.18, V1 = 0.0486 m3, V2 = 0.0791 m3, Tr1 = 1.176, Pr1 = 1.316 ; From Fig. D.2: *

W12 = 114 kJ

Tr2 = 1.062, Pr2 = 0.789, T2 = 451.6 K

(h*- h)1 = 1.30 RTC ,

(h*- h)2 = 0.90 RTC

*

h2 - h1 = 1.7164(451.6 - 500) = -83.1 kJ/kg 8.3145×425.2 (-0.90 + 1.30) = -58.8 kJ/kg 58.124 U2 - U1 = m(h2 - h1) - P2V2 + P1V1

h2 - h1 = -83.1 +

= 5(-58.8) - 3000×0.0791 + 5000×0.0486 = -288.3 kJ Q12 = U2 - U1 + W12 = -174.3 kJ

13.33 Saturated vapor R-22 at 30°C is throttled to 200 kPa in an SSSF process. Calculate the exit temperature assuming no changes in the kinetic energy, using the generalized charts, Fig. D.2 and the R-22 tables, Table B.4 R-22 throttling process 1st law:

*

*

*

*

h2-h1 = (h2-h2) + (h2-h1) + (h1-h1) = 0

a) Generalized Chart, Fig. D.2, R = 8.31451/86.469 = 0.096156 303.2 * = 0.821 => (h1-h1) = 0.096156 × 369.3 (0.53) = 18.82 Tr1 = 369.3 For CP0, use h values from Table B.4 at low pressure. CP0 ≈ 278.115 - 271.594) / (30 - 20) = 0.6521 kJ/kg K *

Substituting: (h2-h2) + 0.6521(T2-30) + 18.82 = 0 at Pr2 = 200/4970 = 0.040 Assume T2 = 5.0 oC => Tr2 =278.2/369.3 = 0.753 *

(h2-h2) = RT × 0.07 = 0.096156 × 369.3 (0.07) = 2.49 Substituting :

-2.49 + 0.6521(5.0-30) + 18.82 = -0.03 ≈ 0

⇒ T2 = 5.0 oC b)

R-22 tables, B.4: at T1 = 30 oC, x1 = 1.0

=> h1 = 259.12

h2 = h1 = 259.12 , P2 = 0.2 MPa => T2 = 4.7 oC

13-24

13.34 250-L tank contains propane at 30°C, 90% quality. The tank is heated to 300°C. Calculate the heat transfer during the process. V = 250 L = 0.25 m3 T1 = 30 oC = 303.2 K, x1 = 0.90

C3 H8 T

2

Heat to T2 = 300 oC = 573.2 K 1 v

M = 44.094, TC = 369.8 K, PC = 4.25 MPa R = 0.188 55, CP0 = 1.6794

Tr1 = 0.82 → Fig. D.1: Z1 = (1- x1) Zf1 + x1 Zg1 = 0.1 × 0.05 + 0.9 × 0.785 = 0.711 *

Fig D.2: SAT

Pr

m=

h1-h1 RTc SAT

= 0.30 P1

= 0.1 × 4.43 + 0.9 × 0.52 = 0.911 = 1.275 MPa

1275×0.25 = 7.842 kg 0.711×0.188 55×303.2

Pr2 =

7.842×Z2×0.188 55×573.2

at Tr2 = 1.55

0.25×4250

=

Z2 1.254

Trial and error on Pr2

Pr2 = 0.743 => P2 = 3.158 MPa, Z2 = 0.94 , (h*- h)2 = 0.35 RTC *

*

(h2-h1) = 1.6794(300-30)

= 453.4 kJ/kg

*

(h1-h1) = 0.911×0.188 55×369.8 = 63.5 kJ/kg *

(h2-h2) = 0.35×0.188 55×369.8 = 24.4 kJ/kg Q12 = m(h2-h1) - (P2-P1)V = 7.842(-24.4+453.4+63.5) - (3158-1275)×0.25 = +3862 - 471 = 3391 kJ

13-25

13.35 The new refrigerant fluid R-123 (see Table A.2) is used in a refrigeration system that operates in the ideal refrigeration cycle, except the compressor is neither reversible nor adiabatic. Saturated vapor at -26.5°C enters the compressor and superheated vapor exits at 65°C. Heat is rejected from the compressor as 1 kW and the R-123 flow rate is 0.1 kg/s. Saturated liquid exits the condenser at 37.5°C. Specific heat for R-123 is Cp =0.6 kJ/kg. Find the coefficient of performance. R-123: Tc = 456.9 K, Pc = 3.67 MPa, M = 152.93 kg/kmol, R = 0.05438 kJ/kg K State 1: T1 = -26.5oC = 246.7 K, sat vap., x1 = 1.0 Tr1 = 0.54, Fig D.1, Pr1 = 0.01, P1 = Pr1Pc = 37 kPa *

Fig. D.2,

h1-h1 = 0.03 RTC = 0.8 kJ/kg State 2: T2 = 65oC = 338.2 K State 3: T = 37.5oC = 310.7 K, sat. liq., x = 0 3

3

Tr3 = 0.68, Fig. D.1: Pr3 = 0.08, P3 = Pr3Pc = 294 kPa P2 = P3 = 294 kPa, Pr2 = 0.080, Tr2 = 0.74, *

Fig. D.2:

h2-h2 = 0.25 RTC = 6.2 kJ/kg *

h3-h3 = 4.92 RTC = 122.2 kJ/kg State 4: T4 = T1 = 246.7 K, h4 = h3 1st Law Evaporator: qL + h4 = h1 + w; w = 0, *

qL = h1 - h3 = (h1 − h1) *

*

*

h4 = h3

*

+ (h1 − h3) + (h3 − h3)

*

h1 − h3 = Cp(T1 - T3) = -38.4 kJ/kg, qL = -0.8 – 38.4 + 122.2 = 83.0 kJ/kg . . 1st Law Compressor: q + h1 = h2 + wc; Q = -1.0 kW, m = 0.1 kg/s *

wc = h1 - h2 + q; h1 - h2 = (h1 − h1) *

*

*

h1 − h2 = CP (T1 - T2) = -54.9 kJ/kg, wc = -0.8 –54.9 + 6.2 – 10.0 = -59.5 kJ/kg β = qL/wc = 83.0/59.5 = 1.395

*

*

+ (h1 − h2) + (h2 − h2)

13-26

13.36 A cylinder contains ethylene, C2H4, at 1.536 MPa, −13°C. It is now compressed in a reversible isobaric (constant P) process to saturated liquid. Find the specific work and heat transfer. Ethylene C2H4 P1 = 1.536 MPa = P2 , T1 = -13oC = 260.2 K State 2: saturated liquid, x2 = 0.0 Tr1 =

260.2 1.536 = 0.921 Pr1 = Pr2 = = 0.305 282.4 5.04 *

From Figs. D.1, D.2: Z1 = 0.85 , (h1-h1)/RTc = 0.40 Z1RT1 0.85×0.29637×260.2 v1 = = = 0.042675 P1 1536 *

(h1-h1) = 0.296 37×282.4×0.40 = 33.5 From Figs. D.1, D.2: T2 = 0.824×282.4 = 232.7 K *

Z2 = 0.05 , (h2-h2)/RTc = 4.42 v2 =

Z2RT2 0.05×0.29637×232.7 = = 0.002245 P2 1536

*

(h2-h2) = 0.296 37×282.4×4.42 = 369.9 *

*

(h2-h1) = CP0(T2-T1) = 1.5482(232.7-260.2) = -42.6 w12 = ⌠ ⌡ Pdv = P(v2-v1) = 1536(0.002 245-0.042 675) = -62.1 kJ/kg q12 = (u2-u1) + w12 = (h2-h1) = -369.9 - 42.6 + 33.5 = -379 kJ/kg

13-27

13.37 A piston/cylinder initially contains propane at T = -7°C, quality 50%, and volume 10L. A valve connecting the cylinder to a line flowing nitrogen gas at T = 20°C, P = 1 MPa is opened and nitrogen flows in. When the valve is closed the cylinder contains a gas mixture of 50% nitrogen, 50% propane on a mole basis at T = 20°C, P = 500 kPa. What is the cylinder volume at the final state and how much heat transfer took place? State 1: Propane, T1 = -7oC, x1 = 0.5, V1 = 10 L Tcp = 369.8 K, Pcp = 4.25 kPa, Cpp = 1.679 kJ/kg-K, M = 44.097 kg/kmol Fig. D.1: Tr1 = 0.72, Pr1 = 0.12, P1 = Pr1Pc = 510 kPa Fig. D.1: Zf1 = 0.020, Zg1 = 0.88, Z1 = (1 - x1)Zf + x1Zg = 0.45 − n1 = P1V1/(Z1RT1) = 510 × 0.01/(0.45 × 8.3145 × 266.2) = 0.00512 kmol −* − − −* −* − h 1 = h 1o + CPP(T1 - To) + (h 1 - h 1 ) ; h 1o = 0, −* − − −* − − (h 1-h 1)f /RTc = 4.79, (h 1-h 1)g /RTc = 0.25 −* − −* − −* − h 1 - h 1 = (1 - x1) (h 1 - h 1)f + x1 (h 1 - h 1)g = 7748 kJ/kmol − h 1 = 0 + 1.679 × 44.094(-7 - 20) - 7748 = -9747 kJ/kmol Inlet: Nitrogen, Ti = 20oC, Pi = 1.0 MPa, Tcn = 126.2 K, Pcn = 3.39 MPa, Cpn = 1.042 kJ/kg-K, M = 28.013 kg/kmol −* − Tri = 2.323, Pri = 0.295, h i -h i = 0.06 × 8.3145 × 126.2 = 62.96 kJ/kmol − −* − − −* −* h i = h io + CPn (Ti - To) + (h i - h i ) ; h io = 0, Ti - To = 0 State 2: 50% Propane, 50% Nitrogen by mol, T2 = 20oC, P2 = 500 kPa Tcmix = ∑yiTci = 248 K, Pcmix = ∑yiPci = 3.82 MPa −* − − Tr2 = 1.182, Pr2 = 0.131, Z2 = 0.97, (h 2-h 2)/RTc = 0.06 − −* − − −* −* h 2 = h 2o + CPmix(T2 - To) + (h 2 - h 2 ) ;; h 2o = 0, T2 - To = 0 − a) ni = n1 => n 2 = n1 + ni = 0.1024, V2 = n2Z2RT2/P2 = 0.0484 m3 b) 1st Law: Qcv + nihi = n2u-2 - n21u-21 + Wcv; u- = h - PvWcv = (P1 + P2)(V2 - V1)/2 = 19.88 kJ Qcv = n2h2 - n1h1 - nihi - P2V2 + P1V1 + Wcv − − h i = -62.96 kJ/kmol, h 2 = -123.7 kJ/kmol, Qcv = 50.03 kJ

13-28

13.38 An ordinary lighter is nearly full of liquid propane with a small amount of vapor, the volume is 5 cm3 and temperature is 23°C. The propane is now discharged slowly such that heat transfer keeps the propane and valve flow at 23°C. Find the initial pressure and mass of propane and the total heat transfer to empty the lighter. Propane C3H8

T1 = 23oC = 296.2 K = constant,

V1 = 5 cm3 = 5×10-6 m3,

x1 = 0.0

Tr1 = 296.2/369.8 = 0.804

From Figs. D.1 and D.2, P1 = PG T1 = 0.25×4.25 = 1.063 MPa,

Z1 = 0.04

*

(h1-h1) = 0.188 55×369.8×4.51 = 314.5 m1 =

P1V1

1063×5×10-6 = = 0.00238 kg Z1RT1 0.04×0.188 55×296.2

State 2: Assume vapor at 100 kPa, 23oC Therefore, m2 much smaller than m1 ( ∼ 9.0 × 10-6 kg) QCV = m2u2 - m1u1 + mehe = m2h2 - m1h1 - (P2-P1)V + (m1-m2)he = m2(h2-he) + m1(he-h1) - (P2-P1)V (he - h1) = 0 + 0 + 314.5 QCV = ≈ 0 + 0.00238(314.5) - (100-1063)×5×10-6 = 0.753 kJ

13-29

13.39 An uninsulated piston/cylinder contains propene, C3H6, at ambient temperature, 19°C, with a quality of 50% and a volume of 10 L. The propene now expands very slowly until the pressure in the cylinder drops to 460 kPa. Calculate the mass of propene, the work and heat transfer for this process. Propene C3H6:

T1 = 19oC = 292.2 K,

x1 = 0.50,

From Fig. D.1:

Tr1 = 292.2/364.9 = 0.80,

Pr1 = Pr sat = 0.25,

P1 = 0.25 × 4.6 = 1.15 MPa

Z1 = 0.5 × 0.04 + 0.5 × 0.805 = 0.4225

From D.1: m=

P1V1 Z1RT1

V1 = 10 L

=

1150×0.010 = 0.471 kg 0.4225×0.197 58×292.2

Assume reversible and isothermal process (slow, no friction, not insulated) Q12 = m(u2-u1) + W12 2

2

W12 = ⌠ ⌡ PdV (cannot integrate);

Q12 = ⌠ ⌡ TdS = Tm(s2-s1)

1

1

From Figs. D.2 and D.3: *

h1-h1 = 0.19758 × 364.9(0.5 × 4.51 + 0.5 × 0.46) = 179.2 *

(s1-s1) = 0.197 58 (0.5 × 5.46 + 0.5 × 0.39) = 0.5779 *

*

(h2-h1) = 0 and

* *

(s2-s1) = 0 - 0.197 58 ln

460 = + 0.1829 1161

At Tr2 = 0.80, Pr2 = 0.10, from D.1, D.2 and D.3, Z2 = 0.93 *

(h2-h2) = 0.197 58 × 364.9 × 0.16 = 11.5 *

(s2-s2) = 0.197 58 × 0.13 = 0.0257 (s2-s1) = -0.0257 + 0.1829 + 0.5779 = 0.7351 q12 = 292.2 × 0.7351 = 214.8 kJ/kg

=>

Q12 = m q12 = 101.2 kJ

(h2-h1) = -11.5 + 0 + 179.2 = 167.7 u2-u1 = (h2-h1) + RT(Z1-Z2) = 167.7 + 0.197 58×292.2(0.4225-0.93) = 138.4 w12 = q12 - (u2-u1) = 214.8 - 138.4 = 76.4 kJ/kg W12 = m w12 = 36.0 kJ

13-30

13.40 A 200-L rigid tank contains propane at 400 K, 3.5 MPa. A valve is opened, and propane flows out until half the initial mass has escaped, at which point the valve is closed. During this process the mass remaining inside the tank expands according to the relation Pv1.4 = constant. Calculate the heat transfer to the tank during the process. C3H8: V = 200 L, T1 = 400 K, P1 = 3.5 MPa Flow out to m2 = m1/2 ; Pv1.4 = const inside Tr1 =

400 3.5 = 1.082, Pr1 = = 0.824 Fig D.1: Z1 = 0.74 369.8 4.25

0.74×0.188 55×400 = 0.01594, v2 = 2v1 = 0.03188 3500 0.2 1 m1 = = 12.55 kg, m2 = m1 = 6.275 kg, 0.015 94 2

v1 =

P 2 = P1

(vv )

1 1.4

=

2

3500 = 1326 kPa 21.4

Trial & error: saturated with 1.326 Pr2 = = 0.312  T2 = 0.826×369.8 = 305.5 K & 4.25  P2v2 = Z2RT2  Z2 = 1326×0.03188 = 0.734 0.188 55×305.5 Z2 = ZF2 + x2(ZG2 - ZF2) = 0.734 = 0.05 + x2(0.78-0.05)

=>

x2 = 0.937

*

(h1-h1) = 0.188 55×369.8(0.9) = 62.8 *

*

(h2-h1) = 1.6794(305.5-400) = -158.7 *

[

*

(h2-h2) = (h2-hF2) - x2hFG2 = 0.188 55×369.8 4.41 - 0.937(4.41-0.55) = 55.3 1st law: QCV = m2h2 - m1h1 + (P1-P2)V + mehe AVE *

Let h1 = 0 then *

*

*

*

h1 = 0 + (h1-h1) = -62.8 *

h2 = h1 + (h2-h1) + (h2-h2) = 0 - 158.7 – 55.3 = -214.0 he AVE = (h1+h2)/2 = -138.4 QCV = 6.275(-214.0) - 12.55(-62.8) + (3500-1326)×0.2 + 6.275(-138.4) = -981.4 kJ

]

13-31

13.41 A newly developed compound is being considered for use as the working fluid in a small Rankine-cycle power plant driven by a supply of waste heat. Assume the cycle is ideal, with saturated vapor at 200°C entering the turbine and saturated liquid at 20°C exiting the condenser. The only properties known for this compound are molecular weight of 80 kg/kmol, ideal gas heat capacity CPO= 0.80 kJ/kg K and TC= 500 K, PC= 5 MPa. Calculate the work input, per kilogram, to the pump and the cycle thermal efficiency. T1 = 200oC = 473.2 K, x1 = 1.0

1 . WT

Turbine . QH

Ht. Exch

T3 = 20oC = 293.2 K, x3 = 0.0 Properties known: M = 80, CPO = 0.8 kJ/kg K

2

TC = 500 K, PC = 5.0 MPa

Cond

Tr1 = P 4

. -W P

473.2 293.2 = 0.946 , Tr3 = = 0.586 500 500

3

From Fig. D.1, Pr1 = 0.72, P1 = 0.72×5 = 3.6 MPa = P4 Pr3 = 0.023, P3 = 0.115 MPa = P2 , ZF3 = 0.004 vF3 =

ZF3RT3 P3

=

0.004×8.3145×293.2 = 0.00106 115×80

4

⌡ vdP ≈ vF3(P4 -P3) = -0.00106(3600-115) = -3.7 kJ/kg wP = - ⌠ 3

qH + h4 = h1 , but h3 = h4 + wP =>

q H = (h1-h3) + wP

From Fig. D.2: *

(h1-h1) = 0.103 93×500×1.25 = 64.9 *

(h3-h3) = 0.103 93×500×5.2 = 270.2 *

*

(h1-h3) = CP0(T1-T3) = 0.80(200-20) = 144.0 (h1-h3) = -64.9 + 144.0 + 270.2 = 349.3

13-32

qH = 349.3 + (-3.7) = 345.6 kJ/kg *

*

*

*

Turbine, (s2 - s1) = 0 = -(s2 - s2)+(s2 - s1) + (s1 - s1) From Fig. D.3, *

(s1-s1) = 0.10393×0.99 = 0.1029 * *

(s2-s1) = 0.80 ln

293.2 115 - 0.103 93 ln = -0.0250 473.2 3600

Substituting, *

*

s2-s2 = +0.1029 - 0.0250 = 0.0779 = (s2-sF2) - x2sFG2 0.0779 = 0.103 93×8.85 - x2×0.103 93(8.85-0.06) *

*

(h2-h2) = (h2-hF2) - x2hFG2 From Fig. D.2, hFG2 = 0.10393×500(5.2-0.07) = 266.6 *

(h2-h2) = 270.2 -0.922 × 266.6 = 25.0 wT = (h1-h2) = -64.9 + 144.0 + 25.0 = 104.1 kJ/kg ηTH =

wNET qH

=

104.1-3.7 = 0.29 345.6

=> x 2 = 0.922

13-33

13.42 A geothermal power plant on the Raft river uses isobutane as the working fluid. The fluid enters the reversible adiabatic turbine, as shown in Fig. P13.42, at 160°C, 5.475 MPa and the condenser exit condition is saturated liquid at 33°C. Isobutane has the properties Tc= 408.14 K, Pc= 3.65 MPa, CP0= 1.664 kJ/kg K and ratio of specific heats k = 1.094 with a molecular weight as 58.124. Find the specific turbine work and the specific pump work. Turbine inlet: T1 = 160oC , P1 = 5.475 MPa Condenser exit: T3 = 33oC , x3 = 0.0, Tr3 = 306.2 / 408.1 = 0.75 From Fig. D.1: => P2 = P3 = 0.16 × 3.65 = 0.584 MPa

Pr3 = 0.16, Z3 = 0.03

Tr1 = 433.2 / 408.1 = 1.061,

Pr1 = 5.475 / 3.65 = 1.50

From Fig. D.2 & D.3: *

(h1-h1) = 0.143 05×408.1×2.84 = 165.8 *

(s1-s1) = 0.143 05×2.15 = 0.3076 * *

(s2-s1) = 1.664 ln *

306.2 0.584 - 0.143 05 ln = -0.2572 433.2 5.475

*

(s2-s2) = (s2-sF2) - x2sFG2 = 0.143 05×6.12 - x2×0.143 05(6.12-0.29) = 0.8755 - x2×0.8340 (s2-s1) = 0 = -0.8755 + x2×0.8340 - 0.2572 + 0.3076 *

=>

x2 = 0.99

*

(h2-h1) = CP0(T2-T1) = 1.664(306.2 - 433.2) = -211.3 From Fig. D.2:, *

*

(h2-h2) = (h2-hF2) - x2hFG2 = 0.143 05×408.1[4.69-0.99(4.69-0.32)] = 273.8 − 0.99 × 255.1 = 21.3 Turbine:

wT = (h1-h2) = -165.8 + 211.3 + 21.3 = 66.8 kJ/kg

Pump:

vF3 =

ZF3RT3 P3

=

0.03×0.143 05×306.2 = 0.00225 584

wP = - ∫ v dP ≈ vF3(P4 -P3) = -0.00225 (5475-584) = -11.0 kJ/kg

13-34

13.43 Carbon dioxide collected from a fermentation process at 5°C, 100 kPa should be brought to 243 K, 4 MPa in an SSSF process. Find the minimum amount of work required and the heat transfer. What devices are needed to accomplish this change of state? 278.2 100 Tri = = 0.915 Pri = = 0.0136 304.1 7380 From D.2 and D.3 : (h*-h) /RTC = 0.02, (s*-s)ri/R = 0.01 ri

Tre =

243 4 = 0.80, Pre = = 0.542 304.1 7.38

From D.2 and D.3: *

(h*-h) /RTC = 4.5 re

*

*

(s*-s)re/R = 4.74

*

(hi-he) = - (hi -hi) + (hi -he) + (he-he) = - 0.188 92×304.1×0.01 + 0.8418(278.2-243) + 0.188 92×304.1×4.5 *

* *

= 287.6 kJ/kg

*

(si-se) = - (si -si) + (si -se) + (se-se) = - 0.188 92×0.01 + 0.8418 ln(278.2/243) - 0.188 92 ln(0.1/4) + 0.188 92×4.74 = 1.7044 kJ/kg K wrev = (hi-he) -T0(si-se) = 287.6 - 278.2(1.7044) = -186.6 kJ/kg qrev = (he-hi) + wrev = -287.6 -186.6 = -474.2 kJ/kg

13-35

13.44 An insulated cylinder fitted with a frictionless piston contains saturated-vapor carbon dioxide at 0oC, at which point the cylinder volume is 20 L. The external force on the piston is now slowly decreased, allowing the carbon dioxide to expand until the temperature reaches - 30 oC. Calculate the work done by the CO2during this process. CO2: Tc = 304.1 K, Pc = 7.38 MPa, Cp = 0.842 kJ/kg-K, R = 0.18892 kJ/kg K State 1: T1 = 0oC, sat. vap., x1 = 1.0, V1 = 20 L Tr1 = 0.9, P1 = Pr1Pc = 0.53 × 7380 = 3911 kPa, Z1 = Zg = 0.67 *

(h1 − h1)g

P 1V 1 * = 0.9 RTC, (s1 − s1)g/R = 0.72, m = = 2.262 kg Z1RT1

State 2: T2 = -30oC Tr2 = 0.8, P2 = Pr2Pc = 0.25 × 7380 = 1845 kPa 2nd Law:

∆Snet =m(s2 − s1) − 1Q2/T ; *

s2 - s1 = (s2 − s2) *

*

s2 − s1 = CP ln

T2 T1

*

*

1Q 2

− R ln

*

*

*

P2

*

P1

= 0.044 kJ/kg-K, s1 − s1 = 0.136 kJ/kg-K

*

*

(s2 − s2)f = 5.46 R, (s2 − s2)g = 0.39 R

*

*

(s2 − s2) = (1-x2)(s2 − s2)f + x2 (s2 − s2)g 1Q 2

∆Snet = 0

+ (s2 − s1) + (s1 − s1) = 0

s2 - s2 = 0.180 kJ/kg K,

1st Law:

= 0,

‡ x2 = 0.889

= m(u2 − u1) + 1W2 ; 1Q2 = 0,

u = h - Pv

Z2 = (1 - x2)Zf + x2Zg = 0.111 × 0.04 + 0.889 × 0.81 = 0.725; *

(h2 - h1) = (h2 − h2) *

*

*

*

+ (h2 − h1) + (h1 − h1)

*

*

h2 − h1 = Cp(T2 - T1) = -25.3 kJ/kg, (h1 − h1) = 51.7 kJ/kg *

(h2 − h2)f

= 4.51 RTC ,

*

*

*

(h2 − h2)g = 0.46 RTC *

(h2 − h2) = (1 - x2)(h2 − h2)f + x2 (h2 − h2)g =

52.2 kJ/kg

h2 - h1 = -52.2 – 25.3 + 51.7 = -25.8 kJ/kg u2 - u1 = (h2 - h1) - Z2RT2 + Z1RT1 = -25.8 – 0.725 × 0.18892 × 243.2 + 0.67 × 0.18892 × 273.2 = -24.5 kJ/kg 1W2 = 55.4 kJ

13-36

13.45 An evacuated 100-L rigid tank is connected to a line flowing R-142b gas, chlorodifluoroethane, at 2 MPa, 100°C. The valve is opened, allowing the gas to flow into the tank for a period of time and then it is closed. Eventually, the tank cools to ambient temperature, 20°C, at which point it contains 50% liquid, 50% vapor, by volume. Calculate the quality at the final state and the heat transfer for the process. The ideal-gas specific heat of R-142b is C = 0.787 kJ/kg K. Rigid tank V = 100 L, m1 = 0 Line: R-142b CH3CClF2 M = 100.495, TC = 410.3 K, PC = 4.25 MPa, CP0 = 0.787 kJ/kg K − R = R/M = 8.31451 / 100.495 = 0.082 73 kJ/kg K Line Pi = 2 MPa, Ti = 100 oC, Flow in to T2 = T0 = 20oC VLIQ 2 = VVAP 2 = 50 L Continuity: mi = m2 ; Energy:

QCV + mihi = m2u2 = m2h2 - P2V

From D.2 at i: Pri = 2 / 4.25 = 0.471,

Tri = 373.15 / 410.3 = 0.91

*

(hi -hi) = 0.082 73×410.3×0.72 = 24.4 *

*

(h2-hi ) = CP0(T2-Ti) = 0.787(20-100) = -63.0 From D.2: Tr2 =

293.2 = 0.715 => P2 = 0.115×4250 = 489 kPa 410.3

sat. liq.: ZF = 0.02, (h*-hF) = RTC×4.85 = 164.6 sat. vap.: ZG = 0.88, (h*-hG) = RTC×0.25 = 8.5 mLIQ 2 = mVAP 2 =

P2VLIQ 2 ZFRT2

=

P2VVAP 2 ZGRT2

489×0.050 = 50.4 kg 0.02×0.082 73×293.2

= 1.15 kg,

m2 = 51.55 kg

x2 = mVAP 2/m2 = 0.0223 *

*

*

(h2-h2) = (1-x2)(h2-hF2) + x2(h2-hG2) = 0.9777 × 164.6 + 0.0223 × 8.5 = 161.1 QCV = m2(h2-hi) - P2V = 51.55(-161.1-63.0+24.4) - 489×0.10 = -10 343 kJ

13-37

13.46 A cylinder fitted with a movable piston contains propane, initially at 67oC and 50 % quality, at which point the volume is 2 L. The piston has a cross-sectional area of 0.2 m2. The external force on the piston is now gradually reduced to a final value of 85 kN, during which process the propane expands to ambient temperature, 4oC. Any heat transfer to the propane during this process comes from a constant-temperature reservoir at 67oC, while any heat transfer from the propane goes to the ambient. It is claimed that the propane does 30 kJ of work during the process. Does this violate the second law? +Q from Tres = 67oC C3 H8

-Q to Environment To = 4oC

Fext

Fext 2 = 85 kN

Propane: Tc= 369.8 K, Pc = 4.25 MPa, R = 0.18855 kJ/kg K, Cp = 1.679 kJ/kg K State 1: T1 = 67oC = 340.2 K, x1 = 0.5, V1 = 2.0 L Tr1 = 0.92, Fig D.1, Pr1 = 0.61, P1 = Pr1Pc = 2.592 MPa Zf1 = 0.10, Zg1 = 0.64, Z1 = (1 - x1)Zf1 + x1Zg1 = 0.37 P 1V 1 * * m= = 0.218 kg, (h1 − h1)f = 3.95 RTc , (h1 − h1)g = 1.03 RTc Z1RT1 *

*

(s1 − s1)f = 4.0 R , (s1 − s1)g = 0.82 R State 2: T2 = 4oC = 277.2 K, Fext 2 = 85 kN sat

sat

Tr2 = 0.75, P2 = Pr2 Pc = 0.165 × 4250 = 701 kPa P2 = Fext 2/Ap = 425 kPa, Pr2 = 0.10, Z2 = 0.92,

sat

P2 < P2 ‡ State 2 is a vapor V2 = mZ2RT2/P2 = 0.0247 m3

*

h2 − h2 = 0.18 RTc =12.6 kJ/kg, 1st Law: 1Q2 = m(u2 - u1) + 1W2; 1Q 2

*

s2 − s2 = 0.16 R = 0.0302 kJ/kg K 1W2

= 30 kJ, u = h - Pv

= m(h2 - h1) - P2V2 + P1V1 + 1W2 *

(h2 - h1) = (h2 − h2) *

*

*

*

+ (h2 − h1) + (h1 − h1) *

*

(h1 − h1) = (1 - x1)(h1 − h1)f + x1 (h1 − h1)g = *

173.6 kJ/kg

*

h2 − h1 = Cp(T2 - T1) = -105.8 kJ/kg 1Q 2

= 0.218 (-12.6 - 105.8 + 173.6) - 425×0.0247 + 2592×0.002 + 30 = 36.7 kJ

13-38

1Q 2 ∆Snet = m(s2 − s1) − ; Tres = 67oC = 340.2 K T

2nd Law:

*

s2 - s1 = (s2 − s2) *

*

*

*

+ (s2 − s1) + (s1 − s1) *

*

s1 − s1 = (1 - x1)(s1 − s1)f + x1 (s1 − s1)g = 0.4544 kJ/kg-K *

*

s2 − s1 = Cpln

T2 T1

- R ln

P2 P1

= -0.0030 kJ/kg K

∆Snet = 0.218 (-0.0302-0.0030+0.4544) – 36.7/340.2 = -0.0161 kJ/K; ∆Snet < 0 Process is Impossible

13.47 Consider the following equation of state, expressed in terms of reduced pressure and temperature: Pr 6 ( 1- 2) Z=1+ 14 Tr Tr What does this equation predict for enthalpy departure from the ideal gas value at the state Pr = 0.4, Tr = 0.9 ? What does this equation predict for the reduced Boyle temperature? Pr Pv 6 a) Z= =1+ ( 1 - 2) RT 14 Tr Tr 3

6Tc2 RT RTc v= + (1 - T2 ) ; P 14Pc

∂v  R 12RTc   = + ∂T p P 14PcT3 3

RTc 18RTc ∂v  v-T  = ∂T p 14Pc 14PcT3 P

RTc ∂v  18 h - h* = ⌠ ⌡ [v - T   ] dP = ( 1 − 2) Pr = 0.606 RTc 14 ∂T p Tr 0 ∂Z 1 6 b)   = ( 1 - 2) 14P ∂P T Tr c Tr

(1 - T62) = 0 r

‡ Tr =

=> 6 = 2.45

Lim ∂Z P→0 ∂P T = 0 at Tboyle

13-39

13.48 Saturated liquid ethane at 2.44 MPa enters (SSSF) a heat exchanger and is brought to 611 K at constant pressure, after which it enters a reversible adiabatic turbine where it expands to 100 kPa. Find the heat transfer in the heat exchanger, the turbine exit temperature and turbine work. From D.2, Pr1 = 2.44/4.88 = 0.50 , Tr1 = 0.89, T1 = 0.89×305.4 = 271.8 K *

(h1-h1) = 0.2765×305.4×4.12 = 347.9 *

*

(h2-h1) = 1.766 (611 - 271.8) = 599.0 Pr2 = 0.50 , Tr2 = 611/305.4 = 2.00 From D.2:

*

(h2-h2) = RTc × 0.14 = 0.2765×305.4×0.14 = 11.8

q = (h2-h1) = -11.8 + 599.0 + 347.9 = 935.1 kJ/kg From D.3, *

(s2-s2) = 0.2765×0.05 = 0.0138 * *

(s3-s2) = 1.766 ln

T3

- 0.2765 ln

611 Assume T3 = 368 K , Tr3 = 1.205

100 2440

at Pr3 = 0.020 * *

(s3-s2) = -0.8954 + 0.8833 = -0.0121 From D.3, *

(s3-s3) = 0.2765×0.01 = 0.0028 (s3-s2) = -0.0028 - 0.0121 + 0.0138 ≈ 0

ΟΚ

Therefore, T3 = 368 K From D.2, *

(h3-h3) = 0.2765×305.4×0.01 = 0.8 w = (h2-h3) = -11.8 + 1.766 (611 - 368) + 0.8 = 418.1 kJ/kg

13-40

13.49 A flow of oxygen at 230 K, 5 MPa is throttled to 100 kPa in an SSSF process. Find the exit temperature and the entropy generation. Tri =

230 5 = 1.488, Pri = = 0.992, 154.6 5.04

Pre =

0.1 = 0.02 5.04

*

From D.2:

(hi -hi) = 0.2598×154.6×0.50 = 20.1 *

First law:

*

*

*

(he-hi) = 0 = - (h e-he) + (he-hi ) + (hi -hi) *

Assume Te = 208 K , Tre = 1.345:

*

(he-hi ) = 0.9216(208 - 230) = -20.3

*

From D.2:

(he-he) = 0.2598×154.6×0.01 = 0.4 (he-hi) = -0.4 -20.3 + 20.1 ≈ 0 OK

Check first law

=>

Te = 208 K

From D.3, *

(si -si) = 0.2598×0.25 = 0.0649 * *

(se-si ) = 0.9216 ln

and

*

(se-se) = 0.2598×0.01 = 0.0026

208 0.1 - 0.2598 ln = 0.9238 230 5

(se-si) = -0.0026 + 0.9238 + 0.0649 = 0.9861 kJ/kg K

13.50 A cylinder contains ethylene, C2H4, at 1.536 MPa, −13°C. It is now compressed isothermally in a reversible process to 5.12 MPa. Find the specific work and heat transfer. Ethylene C2H4 P1 = 1.536 MPa , T2 = T1 = -13oC = 260.2 K Tr2 = Tr1 = 260.2 / 282.4 = 0.921 , Pr1 = 1.536 / 5.04 = 0.305 From D.1, D.2 and D.3:

Z1 = 0.85

*

(h1-h1) = 0.2964×282.4×0.40 = 33.5 and From D.1, D.2 and D.3:

Z2 = 0.17 ,

Pr2 = 5.12/5.04 = 1.016 (comp. liquid)

*

(h2-h2) = 0.2964×282.4×4.0 = 334.8 and Ideal gas:

*

*

(h2-h1) = 0

and

*

(s1-s1) = 0.2964×0.30 = 0.0889 *

(s2-s2) = 0.2964×3.6 = 1.067

* *

(s2-s1) = 0 - 0.2964 ln

5.12 = -0.3568 1.536

q12 = T(s2-s1) = 260.2(-1.067 - 0.3568 + 0.0889) = -347.3 kJ/kg (h2-h1) = -334.8 + 0 + 33.5 = -301.3 (u2-u1) = (h2-h1) - RT(Z2-Z1) = -301.3 - 0.2964×260.2(0.17-0.85) = -248.9 w12 = q12 - (u2-u1) = -347.3 + 248.9 = -98.4 kJ/kg

13-41

13.51 A control mass of 10 kg butane gas initially at 80°C, 500 kPa, is compressed in a reversible isothermal process to one-fifth of its initial volume. What is the heat transfer in the process? T1 = 80 oC, P1 = 500 kPa

Butane C4H10: m = 10 kg,

Compressed, reversible T = const, to V2 = V1/5 353.2 500 = 0.831, Pr1 = = 0.132 425.2 3800

Tr1 =

From D.1 and D.3:

Z1 = 0.92,

*

(s1- s1) = 0.143×0.16 = 0.0230

Z1RT1 0.92×0.143×353.2 v1 = = = 0.09296 m3/kg P1 500 v2 = v1/5 = 0.01859 m3/kg At Tr2 = Tr1 = 0.831 From D.1: PG = 0.325×3800 = 1235 kPa sat. liq.: ZF = 0.05,

(s*-sF) = R×5.08 = 0.7266 (s*-sG) = R×0.475 = 0.0680

sat. vap.: ZG = 0.775, Therefore vF =

0.05×0.143×353.2 = 0.00205 1235

vG =

0.775×0.143×353.2 = 0.0317 1235

Since vF < v2 < vG → x2 = (v2-vF)/(vG-vF) = 0.5578 *

*

*

(s2-s2) = (1-x2)(s2-sF2) + x2(s2-sG2) = 0.4422×0.7266 + 0.5578×0.0680 = 0.3592 * *

(s2-s1) = CP0 ln (T2/T1) - R ln (P2/P1) = 0 - 0.143 ln (1235/500) = -0.1293 (s2-s1) = -0.3592 - 0.1293 + 0.0230 = -0.4655 Q12 = Tm(s2-s1) = 353.2×10(-0.4655) = -1644 kJ

13-42

13.52 An uninsulated compressor delivers ethylene, C2H4, to a pipe, D = 10 cm, at 10.24 MPa, 94°C and velocity 30 m/s. The ethylene enters the compressor at 6.4 MPa, 20.5°C and the work input required is 300 kJ/kg. Find the mass flow rate, the total heat transfer and entropy generation, assuming the surroundings are at 25°C. 293.7 6.4 = 1.040 , Pri = = 1.270 282.4 5.04 From D.2 and D.3, Tri = *

(hi -hi) = 0.296 37×282.4×2.65 = 221.8 *

(si -si) = 0.296 37×2.08 = 0.6164 367.2 10.24 = 1.30 , Pre = = 2.032 => From D.1: 282.4 5.04 ZeRTe 0.69×0.296 37×367.2 ve = = = 0.0073 m3/kg Pe 10 240

Tre =

π 2 Ae = De = 0.007 85 m2 4

=>

Ze = 0.69

. AeVe 0.007 85×30 m= = = 32.26 kg/s ve 0.0073

From D.2 and D.3, *

(he-he) = 0.296 37×282.4×1.6 = 133.9 *

(se-se) = 0.296 37×0.90 = 0.2667 *

*

(he-hi ) = 1.5482(367.2-293.7) = 113.8 * *

(se-si ) = 1.5482 ln

367.2 10.24 - 0.296 37 ln = 0.2065 293.7 6.4

(he-hi) = -133.9 + 113.8 + 221.8 = 201.7 kJ/kg (se-si) = -0.2667 + 0.2065 + 0.6164 = 0.5562 kJ/kg K First law: 302 - 300 = -97.9 kJ/kg q = (he-hi) + KEe + w = 201.7 + 2×1000 . . Qcv = mq = 32.26(-97.9) = -3158 kW . Q . . cv 3158 Sgen = − + m(se - si) = + + 32.26(0.5562) = 28.53 kW/K 298.2 T0

13-43

13.53 A distributor of bottled propane, C3H8, needs to bring propane from 350 K, 100 kPa to saturated liquid at 290 K in an SSSF process. If this should be accomplished in a reversible . .setup given the surroundings at 300 K, find the ratio of the volume flow rates Vin/Vout, the heat transfer and the work involved in the process. From Table A.2: Tri =

350 = 0.946 , 369.8

Pri =

0.1 = 0.024 4.25

From D.1, D.2 and D.3, Zi = 0.99 *

(hi -hi) = 0.1886×369.8×0.03 = 2.1 *

(si -si) = 0.1886×0.02 = 0.0038 290 = 0.784, 369.8 From D.1, D.2 and D.3, Tre =

Pre = 0.22 , Pe = 0.22×4.25 = 0.935 MPa

and

Ze = 0.036

*

(he-he) = 0.1886×369.8×4.57 = 318.6 *

(se-se) = 0.1886×5.66 = 1.0672 *

*

(he-hi ) = 1.679(290 - 350) = -100.8 * *

(se-si )

= 1.679 ln

290 0.935 - 0.1886 ln = -0.7373 350 0.1

(he-hi) = -318.6 - 100.8 + 2.1 = -417.3 (se-si) = -1.0672 - 0.7373 + 0.0038 = -1.8007 . ZiTi/Pi Vin 0.99 350 0.935 . = = × × = 310.3 0.1 Vout ZeTe/Pe 0.036 290 wrev = (hi-he) -T0(si-se) = 417.3 - 300(1.8007) = -122.9 kJ/kg qrev = (he-hi) + wrev = -417.3 –122.9 = -540.2 kJ/kg

13-44

13.54 Saturated-liquid ethane at T1 = 14°C is throttled into a SSSF mixing chamber at the rate of 0.25 kmol/s. Argon gas at T2 = 25°C, P2 = 800 kPa, enters the chamber at the rate of 0.75 kmol/s. Heat is transferred to the chamber from a heat source at a constant temperature of 150 oC at a rate such that a gas mixture exits the chamber at T3 = 120oC, P3 = 800 kPa. Find the rate of heat transfer and the rate of entropy generation. . Argon, Ta2 = 25oC, P2 = 800 kPa, n2 = 0.75 kmol/s Tca = 150 K, Pca = 4.87 MPa, Ma = 39.948 kg/kmol, Cpa = 0.52 kJ/kg K ha3 - ha2 = MaCpa(T3 - Ta2) = 1973.4 kJ/kmol . Inlet: Ethane, Tb1 = 14oC, sat. liq., xb1 = 0, n1 = 0.25 kmol/s Tcb = 305.4 K, Pcb = 4.88 MPa, Mb = 30.07 kg/kmol, Cpb = 1.766 kJ/kg-K Tr1 = 0.94, Pb1 = Pr1Pcb = 0.69 × 4880 = 3367 kPa ∗ − − −∗ − h b1 − h b1 = 3.81 RTcb = 9674.5 kJ/kmol, -sb1 − -sb1 = 3.74 R = 31.1 −∗ −∗ h b3 - h b1 = MbCpb(T3 - Tb1) = 5629.6 kJ/kmol Exit: Mix, T3 = 120oC, P3 = 800 kPa consider this an ideal gas mixture. . . . . . . SSSF Energy Eq.: n1hb1 + n2ha2 +Q = n3h3 = n1hb3 + n2ha3 . . . Q = n1(hb3 - hb1) + n2(ha3 - ha2) = 0.25 (5629.6 + 9674.5) + 0.75(1973.4) = 5306 kW . . . . Entropy Eq.: Sgen = n1(s-b3 − -sb1) + n2(s-a3 − -sa2) - Q/TH ; . . . . ya = n2/ntot = 0.75; yb = n1/ntot = 0.25 T yaP3 -s − -s = M C ln 3 - − R ln = 8.14 kJ/kmol-K a3 a2 a pa T P a2

TH = 150oC

a2

T ybP3 ∗ -s − -s = M C ln 3 - − R ln + -sb1 − -sb1 = b3 b1 b pb T Pb1 b1 = 40.172 + 31.1 = 71.27 kJ/kmol K . Sgen = 0.25 × 71.27 + 0.75 × 8.14 - 5306 / 423 = 11.38 kW/K

13-45

13.55 One kilogram per second water enters a solar collector at 40°C and exits at 190°C, as shown in Fig. P13.55. The hot water is sprayed into a direct-contact heat exchanger (no mixing of the two fluids) used to boil the liquid butane. Pure saturated-vapor butane exits at the top at 80°C and is fed to the turbine. If the butane condenser temperature is 30°C and the turbine and pump isentropic efficiencies are each 80%, determine the net power output of the cycle. H2O cycle: solar energy input raises 1 kg/s of liquid H2O from 40oC to 190oC. Therefore, corresponding heat input to the butane in the heat exchanger is . . QH = m(hF 190 C-hF 40 C)H2O = 1(807.62-167.57) = 640.05 kW C4H10 cycle

1 . WT

Turbine . QH

ηST = ηSP = 0.80

Ht. Exch

2 Cond P

4

Tr3 =

T1 = 80 oC, x1 = 1.0 ; T3 = 30 oC, x3 = 0.0 353.2 = 0.831 425.2 From D.1, D.2 and D.3: P1 = 0.325×3800 = 1235 kPa

Tr1 =

*

. -W P

3

(h1-h1) = 0.143 04×425.2×0.56 = 34.1 *

(s1-s1) = 0.143 04×0.475 = 0.0680

303.2 = 0.713 425.2

From D.1, D.2 and D.3: P3 = 0.113×3800 = 429 kPa sat. liq.: (h*-hF) = RTC×4.81 = 292.5 ;

(s*-sF) = R×6.64 = 0.950

sat. vap.: (h*-hG) = RTC×0.235 = 14.3 ;

(s*-sG) = R×0.22 = 0.031

Because of the combination of properties of C 4H10 (particularly the large CP0 /R), s1 is larger than sG at T3. To demonstrate, * *

(s1-sG3) = 1.7164 ln

353.2 1235 - 0.143 04 ln = 0.1107 303.2 429

(s1-sG3) = -0.0680 + 0.1107 + 0.031 = +0.0737 kJ/kg K

13-46

T

so that T2S will be > T3, as shown in the T-s

1

diagram. A number of other heavy hydrocarbons also exhibit this behavior. Assume T2S = 315 K, Tr2S = 0.741

2s 2

3

s From D.2 and D.3: *

(h2S-h2S) = RTC×0.21 = 12.8 * *

(s1-s2S) = 1.7164 ln

and

*

(s2S-s2S) = R×0.19 = 0.027

353.2 1235 - 0.143 04 ln = +0.0453 315 429

(s1-s2S) = -0.0680 + 0.0453 + 0.027 ≈ 0 ⇒ T2S = 315 K *

*

(h1-h2S) = 1.7164(353.2-315) = 65.6 wST = h1-h2S = -34.1 + 65.6 + 12.8= 44.3 kJ/kg wT = ηS×wST = 0.80×44.3 = 35.4 kJ/kg At state 3, v3 =

0.019×0.143 04×303.2 = 0.001 92 m3/kg 429

-wSP ≈ v3(P4-P3) = 0.001 92(1235-429) = 1.55 kJ/kg -wP =

-wSP ηSP

=

1.55 = 1.94 kJ/kg 0.8

wNET = wT + wP = 35.4 - 1.94= 33.46 kJ/kg For the heat exchanger, . . QH = 640.05 = mC4H10(h1-h4) But

h1-h4 = h1-h3+wP *

*

*

*

h1-h3 = (h1-h1) + (h1-h3) + (h3-h3) = -34.1 + 1.716(80 - 30) + 292.5 = 344.2 kJ/kg Therefore, 640.05 . mC4H10 = = 1.87 kg/s 344.2-1.94 . . WNET = mC4H10wNET = 1.87 × 33.46 = 62.57 kW

13-47

13.56 A line with a steady supply of octane, C8H18, is at 400°C, 3 MPa. What is your best estimate for the availability in an SSSF setup where changes in potential and kinetic energies may be neglected? Availability of Octane at 3 = 1.205, 2.49 From D.2 and D.3, Pri =

Tri =

Ti = 400 oC, Pi = 3 MPa 673.2 = 1.184 568.8

*

(h1-h1) = 0.072 79×568.8×1.13 = 46.8 ;

*

(s1-s1) = 0.072 79×0.69 = 0.05

Exit state in equilibrium with the surroundings, assume T0 = 298.2 K, P0 = 100 kPa 298.2 0.1 = 0.524 , Pr0 = = 0.040 568.8 2.49 From D.2 and D.3, Tr0 = *

(h0-h0) = RTC×5.4 = 223.6 *

and

*

(s0-s0) = R×10.37 = 0.755

*

(hi -h0) = 1.7113(673.2-298.2) = 641.7 * *

(si -s0) = 1.7113 ln

673.2 3 - 0.072 79 ln = 1.1459 298.2 0.1

(hi-h0) = -46.8 + 641.7 + 223.6 = 818.5 (si-s0) = -0.05 + 1.1459 + 0.755 = 1.8509 ϕi = wrev = (hi-h0) - T0(si-s0) = 818.5 - 298.2(1.8509) = 266.6 kJ/kg

13-48

13.57 A piston/cylinder contains ethane gas, initially at 500 kPa, 100 L, and at ambient temperature, 0°C. The piston is now moved, compressing the ethane until it is at 20°C, with a quality of 50%. The work required is 25% more than would have been required for a reversible polytropic process between the same initial and final states. Calculate the heat transfer and the net entropy change for the process. Ethane: Tc = 305.4 K, Pc = 4.88 MPa, R = 0.2765 kJ/kg-K, Cp = 1.766 kJ/kg K State 1: Tr1 = 0.895, Pr1 = 0.102 ‡ Z1 = 0.95 v1 = Z1RT1/P1 = 0.1435 m3/kg, m1 = V1/v1 = 0.697 kg *

*

(h1 − h1) = 0.13RTc = 11.0 kJ/kg, (s1 − s1) = 0.09 R = 0.025 kJ/kg K State 2: T2 = 20oC, x2 = 0.5, 1W2 = 1.25Wrev Tr2 = 0.96, Pr2 = 0.78, P2 = Pr2Pc = 3806 kPa Zf2 = 0.14, Zg2 = 0.54, Z2 = (1 - x2)Zf + x2Zg = 0.34 *

(h2 − h2) = (1 - x2) 3.65 RTc + x2 (1.39 RTc) = 212.8 kJ/kg *

(s2 − s2) = (1 - x2) 3.45 R + x2 × 1.10 R = 0.629 kJ/kg K v2 = Z2RT2/P2 = 0.0072 m3/kg, n

n

P1V1 = P2V2 ,

ln

Wrev = ∫ P dV =

P2 P1

= n ln

V2 = mv2 = 0.005 m3

V1 V2

‡

n = 0.6783

P 2V 2 - P 1V 1

= -96.3 kJ, 1W2 = 1.25Wrev = -120.4 kJ 1-n a) 1st Law: 1Q2 = m(u2 - u1) + 1W2; u = h - Pv u2 - u1 = (h2 - h1) - (P2v2 - P1v1) *

h2 - h1 = = (h2 − h2) *

*

*

*

+ (h2 − h1) + (h1 − h1)

*

(h2 − h1) = Cp(T2 - T1) = 35.3 kJ/kg, h2 - h1 = -212.8 +35.3 + 11.0 = -166.5 kJ/kg, u2 - u1 = -122.2 kJ/kg 1Q2 = 0.697(-122.2) - 120.4 = -205.6 kJ b) 2nd Law: ∆Snet = m(s2 - s1) - 1Q2 /To; To = 0oC *

s2 - s1 = (s2 − s2) *

*

*

*

+ (s2 − s1) + (s1 − s1)

*

(s2 − s1) = Cp ln(T2 / T1) − R ln(P2 / P1) = -0.436 kJ/kg K, 205.6 ∆Snet = 0.697(-0.629 - 0.436 + 0.025) + = 0.028 kJ/K 273.2

13-49

13.58 The environmentally safe refrigerant R-152a (see Problem 13.65) is to be evaluated as the working fluid for a heat pump system that will heat winter households in two different climates. In the colder climate the cycle evaporator temperature is −20°C, and the more moderate climate the evaporator temperature is 0°C. In both climates the cycle condenser temperature is 30°C. For this study assume all processes are ideal. Determine the cycle coefficient of performance for the two climates. R152a difluoroethane. From 13.65: CP0 = 0.996 Ideal Heat Pump TH = 30 oC M = 66.05, R = 0.125 88, T C = 386.4 K, PC = 4.52 MPa

From A.2:

CASE I: T1 = -20 oC T

303.2 = 0.785 386.4 Pr3 = Pr2 = 0.22 =>

2

Tr3 =

3 1

4

Sat.liq.:

v

CASE I)

CASE II: T1 = 0 oC

P3 = P2 = 994 kPa

*

h3 - h3 = 4.56×RTC = 221.8

T1 = -20 oC = 253.2 K, Tr1 = 0.655, Pr1 = 0.058 → P1 = 262 kPa

*

h1 - h1 = 0.14×RTC = 6.8

and

*

s1 - s1 = 0.14×R = 0.0176

Assume T2 = 307 K, Tr2 = 0.795 given Pr2 = 0.22 *

From D.2, D.3: s2 - s2 = 0.34×R = 0.0428 ; *

*

s2 - s1 = 0.996 ln

*

h2 - h2 = 0.40×RTc = 19.5

307 994 - 0.125 88 ln = 0.0241 253.2 262

s2 - s1 = -0.0428 + 0.0241 + 0.0176 = -0.001 ≈ 0 OK ⇒ h2 - h1 = -19.5 + 0.996(307-253.2) + 6.8 = 40.9 h2 - h3 = -19.5 + 0.996(307-303.2) + 221.8 = 206.1 β=

qH wIN

=

h2 - h3 h2 - h1

=

206.1 = 5.04 40.9

13-50

case II) T1 = 0 oC = 273.2 K, Tr1 = 0.707 => Pr1 = 0.106, P1 = 479 kPa *

h1 - h1 = 0.22×RTC = 10.7

*

and

s1 - s1 = 0.21×R = 0.0264

Assume T2 = 305 K, Tr2 = 0.789 *

s2 - s2 = 0.35×R = 0.0441 *

*

s2 - s1 = 0.996 ln

*

and

h2 - h2 = 0.38×RTC = 18.5

305.0 994 - 0.125 88 ln = 0.0178 273.2 479

s2 - s1 = -0.0441 + 0.0178 + 0.0264 = 0.0001 ≈ 0 OK h2 - h1 = -18.5 + 0.996(305.0-273.2) + 10.7 = 23.9 h2 - h3 = -18.5 + 0.996(305.0-303.2) + 221.8 = 205.1 β=

h2 - h3 h2 - h1

=

205.1 = 8.58 23.9

13.59 Repeat the calculation for the coefficient of performance of the heat pump in the two climates as described in Problem 13.58 using R-12 as the working fluid. Compare the two results. Problem the same as 13.58 , exept working fluid is R-12 For R-12: At T3 = 30 oC: CASE I)

T1 = -20 oC,

P3 = P2 = 0.745 MPa, h3 = 64.6 h1 = 178.7,

s1 = 0.7087

 T2 = 39.8 oC At s2 = s1 & P2 →   h2 = 206.8 ⇒β= CASE II:

h2 - h3 h2 - h1

=

T1 = 0oC,

At s2 = s1 & P2 ⇒

206.8 - 64.6 142.2 = = 5.06 206.8 - 178.7 28.1

β=

h1 = 187.5,

s1 = 0.6965

→ T = 34.6oC , h = 203.0

203.0-64.6 138.4 = = 8.93 203.0-187.5 15.5

13-51

13.60 One kmol/s of saturated liquid methane, CH4, at 1 MPa and 2 kmol/s of ethane, C2H6, at 250°C, 1 MPa are fed to a mixing chamber with the resultant mixture exiting at 50°C, 1 MPa. Assume that Kay’s rule applies to the mixture and determine the heat transfer in the process. Control volume the mixing chamber, inlet CH4 is 1, inlet C2H6 is 2 and the exit state is 3. Energy equation is . . . . QCV = n3 h3 - n1 h1 - n2 h2 Select the ideal gas reference temperature to be T3 and use the generalized charts for all three states. Pr1 = Prsat = 1/4.60 = 0.2174 => Trsat = 0.783, T1 = 0.783 × 190.4 = 149.1 K,

∆h1 = 4.57

Pr2 = 1/4.88 = 0.205, Tr2 = 523/305.4 = 1.713 h1 = C1(T1 - T3) - ∆h1 RTc = 36.15(149.1 - 323.2) - 4.57 × 8.3145 × 190.4 = -13528 kJ/kmol h2 = C2(T2 - T3) - ∆h2 RTc = 53.11(250 - 50) - 0.08 × 8.3145 × 305.4 = 10419 kJ/kmol Tcmix = (1 × 190.4 + 2 × 305.4)/3 = 267.1 K Pcmix = (1 × 4.60 + 2 × 4.88)/3 = 4.79 MPa Tr3 = 323.2/267.1 = 1.21 , Pr3 = 1/4.79 = 0.21 h3 = 0 - 0.15 × 267.1 × 8.3145 = - 333 kJ/kmol . QCV = 3(-333) - 1(-13528) - 2(10419) = - 8309 kW

13.61 Consider the following reference state conditions: the entropy of real saturated liquid methane at −100°C is to be taken as 100 kJ/kmol K, and the entropy of hypothetical ideal gas ethane at −100°C is to be taken as 200 kJ/kmol K. Calculate the entropy per kmol of a real gas mixture of 50% methane, 50% ethane (mole basis) at 20°C, 4 MPa, in terms of the specified reference state values, and assuming Kay’s rule for the real mixture behavior. CH : T = -100 oC, -s = 100 kJ/kmol K 4

0

LIQ 0

* C2H6: T0 = -100 oC, P0 = 1 MPa, -s0 = 200 kJ/kmol K

Also for CH4: TC = 190.4 K, PC = 4.60 MPa For a 50% mixture: Tcmix = 0.5 × 190.4 + 0.5 × 305.4 = 247.9 K Pcmix = 0.5 × 4.60 + 0.5 × 4.88 = 4.74 MPa

13-52

IG MIX at T0(=-100 oC), P0(=1 MPa): CH4: Tr0 = 0.91 ,

PG = 0.57 × 4.60 = 2.622 MPa

-* -s* 0 CH4 = s LIQ 0 P + (s -s LIQ)at P - R ln (P 0/PG) G

G

= 100 + 4.01×8.3145 - 8.3145 ln (1/2.622) = 141.36 -s* = 0.5×141.36 + 0.5×200 - 8.3145(0.5 ln 0.5 + 0.5 ln 0.5) = 176.44 0 MIX CP0 MIX = 0.5×16.04×2.254 + 0.5×30.07×1.766 = 44.629 293.2 4 -s* = 176.44 + 44.629 ln 8.3145 ln = 188.41 kJ/kmol K TP MIX 173.2 1 For the mixture at T,P: Tr = 1.183, Pr = 0.844 -s* - -sTP MIX = 0.4363×8.3145 = 3.63 kJ/kmol K TP MIX Therefore, -s = 188.41 - 3.63 = 184.78 kJ/kmol K TP MIX An alternative is to form the ideal gas mixture at T,P instead of at T0,P0 : T P -s - *-s- ) + C -s* = + (s ln R ln LIQ 0 LIQ P0 CH4 TP CH4 T0 PG P ,T G 0

at P , T G 0

= 100 + 33.34 + 16.04×2.254 ln

293.2 4 - 8.3145 ln 173.2 2.6

= 100 + 33.34 + 19.03 - 3.53 = 148.84 293.2 4 -s* = 200 + 30.07×1.766 ln - 8.3145 ln TP C2H6 173.2 1 = 200 + 27.96 - 11.53 = 216.43 -s* TP MIX = 0.5×148.84 + 0.5×216.43 - 8.3145(0.5 ln 0.5 + 0.5 ln 0.5) = 188.41 -s = 188.41 - 3.63 = 184.78 kJ/kmol K TP MIX

13-53

Advanced Problems 13.62 An experiment is conducted at −100°C inside a rigid sealed tank containing liquid R-22 with a small amount of vapor at the top. When the experiment is done the container and the R-22 warms up to room temperature of 20°C. What is the pressure inside the tank during the experiment? If the pressure at room temperature should not exceed 1 MPa, what is the maximum percent of liquid by volume that can be used during the experiment? T R-22 tables Go to -70 oC a) For hFG ≈ const & 2

vFG ≈ vG ≈ RT/PG hFG 1 1 ln ≈ PG0 R T0 T1 PG1

[

]

o

20 C o

-100 C 1

extrapolating from -70 oC

v

(Table B.4.1) to TAVE = -85 oC, hFG ≈ 256.5 8.3145 = 0.096 15 86.469 For T0 = 203.2 K & T1 = 173.2 K

Also R =

PG1

256.5 (20.5)= 0.096 [ 1 - 1 ] 15 203.2 173.2

ln

PG1 = 2.107 kPa b) Extrapolating vF from -70 oC to T1 = -100 oC vF1 ≈ 0.000 634 Also vG1 ≈ RT1/PG1 =

0.096 15×173.2 = 7.9037 2.107

Since v1 = v2 ≈ vF2 = 0.000 824 0.000 824 = 0.000 634 + x 1×7.9031 => x 1 = 2.404×10-5 VLIQ 1 m

= (1-x1)vF1 = 0.000 634,

% LIQ, by vol. =

VVAP 1 m

= x1vG1 = 0.000 190

0.000 634 ×100 = 76.9 % 0.000 824

13-54

13.63 Determine the low-pressure Joule–Thomson inversion temperature from the condition in Eq. 13.54 as predicted by an equation of state, using the van der Waals equation and the Redlich–Kwong equation. ∂v T( ) P - v ∂T ∂T µ j = ( )n = , But CP ∂P

a) van der Waal VDW:

∂P

∂v

(∂T)P = -

∂v (∂T)P = -

1 CP

[

RT

2a

[- (v-b)2 + v3] v2 and then rearranging, v2

1 b -RT( )( )+ 2a 1-b/v 1-b/v 1 2 2a ) -v RT( 1-b/v

]

lim

Then or

∂P

(∂v )T

R (v-b )

Substitute into µj eq'n & multiply by

µj =

(∂T)v

µ =

P→0 j

TINV =

1 2a [-b + RT ]= 0 CP

only at TINV

2a 27 = T = 6.75 TC Rb 4 C 1

∂P R 2a = (∂T)v v-b + v(v+b)T3/2

b) R-K: ∂P

-RT

a(2v+b) (∂v )T = (v-b)2 + v2(v+b) 2 1/2 T Substituting as in part a) 1

µj =

1 CP

[

2a bRT (1-b/v)2 (1+b/v)2 T1/2 RT a(2+b/v) (1-b/v)2 v(1+b/v)2 T1/2

]

then lim

µ =

P→0 j

or

1 5 a [ b]= 0 only at TINV CP 2 RT3/2

TINV = (

5a 2/3 ) = 5.3 TC 2Rb

13-55

13.64 Suppose the following information is available for a given pure substance: vx Known: T PSAT = fn(T) 5 4 6

3 1

2

vapor P = fn(T,v) values of vF, PC, & TC vapor CV at vx

s

Outline the procedure that should be followed to develop a table of thermodynamic properties comparable to Tables 1,2 and 3 of the steam tables. 1) from vapor pressure curve, get P SAT at each T. 2) vF values are known. 3) reference state: T0 & sat. liquid. Set

h0 = 0, s0 = 0, u0 = 0 - P0v0

4) from eq'n of state, find v1 = vG at T0

then vFG(T0) = vG(T0) - vF(T0)

5) from clapeyron eq'n, using slope of vapor pressure curve & vFG, find hFG(T0) then h1 = hG(T0) = 0 + h FG(T0) 6) u1 = uG(T0) = h1 - P1v1; uFG(T0) = u1 - u0 hFG(T0) 7) sFG(T0) = ; s1 = sG(T0) = 0 + sFG(T0) T0 8) find P2 from eq'n of state ( T2 = T0, v2 = vx ) 2

⌠ ∂P 9) u2 - u1 = [T( )v - P]dvT ⌡ ∂T 1

evaluate from eq'n of state then h2 = u2 + P2v2 2

⌠ ∂P 10) s2 - s1 =  ( )v dvT, find s2. ⌡ ∂T 1 3

11) u3 - u2 = ⌠ ⌡ Cvx dT, find u3. 2

12) find P3 from eq'n of state ( T3, v3 = vx ) 13) h3 = u3 + P3v3

13-56

3

14) s3 - s2 = ⌠ ⌡ (Cvx/T) dT, find s3 2

15) at any point 4 (T4=T3) given by desired P4, find v4 from eq'n of state. 16) find P5 = PSAT at T5 = T3 from vap. pressure. 17) find v5 = vG(T5) from eq'n of state. 18) find u5, h5, s5 as in steps 9 & 10. 19) find hFG(T5) from Clapeyron eq'n as in step 5. 20) h6 = hF(T5) = h5 - hFG(T5) 21) sFG(T5) =

hFG(T5) , s6 = sF(T5) = sF - sFG(T5) T5

22) u6 = h6 - P6v6, uFG(T5) = u5 - u6 23) pick a different T (instead of T3) and repeat steps 11 to 22. For T ⇒ T5 steps 16 to 22 are eliminated.

13-57

13.65 The refrigerant R-152a, difluoroethane, is tested by the following procedure. A 10-L evacuated tank is connected to a line flowing saturated-vapor R-152a at 40°C. The valve is then opened, and the fluid flows in rapidly, so that the process is essentially adiabatic. The valve is to be closed when the pressure reaches a certain value P2, and the tank will then be disconnected from the line. After a period of time, the temperature inside the tank will return to ambient temperature, 25°C, through heat transfer with the surroundings. At this time, the pressure inside the tank must be 500 kPa. What is the pressure P2 at which the valve should be closed during the filling process? The ideal gas specific heat of R-152a is CP0 = 0.996 kJ/kg K. R-152a CHF2CH3 : A.2:

M = 66.05, TC = 386.4 K, PC = 4.52 MPa, T3 = T0 = 25oC, P3 = 500 kPa, R = R/M = 8.3145/66.05 = 0.12588 Tr3 = 298.2/386.4 = 0.772,

Pr3 = 500/4520 = 0.111

From D.1 and D.2 at 3: Z3 = 0.92, ⇒

m3 = m2 = mi =

Filling process: or

P 3V Z3RT3

Energy Eq.:

=

(h*-h)3 = 0.19 RTC

500×0.010 = 0.145 kg 0.92×0.125 88×298.2 hi = u2 = h2 - Z2RT2

*

*

(h2-h2) + CP0(T2-Ti) + (hi -hi) - P2V/m2 = 0

From D.2 with Tri = 313.2/386.4 = 0.811, *

(hi -hi) = 0.125 88×386.4×0.49 = 23.8 ;

Pi = 0.276×4520 = 1248 kPa

Assume P2 = 575 kPa, Pr2 = 0.127 Now assume T2 = 339 K, Tr2 = 0.877 => ⇒

From D.1: Z2 = 0.93

Z2T2 0.93×339 Z3T3 0.92×298.2 = = 0.5483 ≈ = = 0.5487 P2 P3 575 500

⇒ T2 = 339 K is the correct T2 for the assumed P2 of 575 kPa. Now check the 1st law to see if 575 kPa is the correct P 2. From D.2,

*

h2-h2 = 0.125 88×386.4×0.17 = 8.3

Substituting into 1st law, -8.3 + 0.996(339-313.2) + 23.8 -

575×0.010 = +1.5 ≈ 0 0.1456

⇒ P2 = 575 kPa (Note: for P2 = 580 kPa, T2 = 342 K,

1st law sum = +4.2)

13-58

13.66 An insulated cylinder has a piston loaded with a linear spring (spring constant of 600 kN/m) and held by a pin, as shown in Fig. P13.66. The cylinder crosssectional area is 0.2 m2, the initial volume is 0.1 m3, and it contains carbon dioxide at 2.5 MPa, 0°C. The piston mass and outside atmosphere add a force per unit area of 250 kPa, and the spring force would be zero at a cylinder volume of 0.05 m3. Now the pin is pulled out; what is the final pressure inside the cylinder? AP = 0.2 m3, VFs = 0 = 0.05 m3 , ks = 600 kN/m P0 = 250 kPa, P1 = 2.5 MPa, V1 = 0.1 m3, T1 = 0oC R = 0.1889, C = 0.842 ks Any point: PEXT = P0 + 2 (V-V0) Ap 600 Pin pulled: PEXT = 250 + 2 (0.1-0.05) = 1000 kPa 0.2 But P1 = 2500 kPa

P0

CO2

Q = 0

⇒ Irr. process, expansion to 1000 < P2 < 2500

At 2: P2 = 250 +

600 (V -0.05) = 15 000 V2 - 500 0.22 2

2

1 W12 = ⌠ ⌡ PEXTdV = 2(PEXT 1 + PEXT 2)(V2-V1) 1

Q12 = 0 = m(h2-h1) - P2V2 + P1V1 + W12 *

*

h2-h1 = (h2-h2) + CP0(T2-T1) + (h1-h1) Tr1 =

273.2 2.5 = 0.898, Pr1 = = 0.339 , 304.1 7.38

Also P1V1 = mZ1RT1 From D.2,

m=

2500×0.1 = 5.91 kg 0.82×0.188 92×273.2

*

(h1-h1) = 0.49×0.1889×304.1 = 28.2

Assume P2 = 1655 kPa Z2Tr2 =

=>

From Fig. D.1, Z1 = 0.82

P2V2 mRTC

=

=>

V2 =

1655×0.1437 = 0.7004 5.91×0.188 92×304.1

At Tr2 = 0.824 and Pr2 = 0.224 => T2 = 250.6 K

1655+500 = 0.1437 m3 15 000

Z2 = 0.850 , ZTr = 0.7004 ~ OK

13-59

*

(h2-h2) = 0.37 × 0.1889 × 304.1 = 21.3 *

*

(h2-h1) = 0.842(250.6 - 273.2) = -19.0 W12 = 0.5 (1000 + 1655) (0.1437-0.10) = 58.0 kJ Q12 = 5.91(-21.3-19.0+28.2) - 1655×0.1437 + 250 + 58.0 = -1.3 ≈ 0 OK (Note: for P2 = 1660 kPa, Q12 = +2.7) P2 = 1655 kPa

13.67 A 10- m3 storage tank contains methane at low temperature. The pressure inside is 700 kPa, and the tank contains 25% liquid and 75% vapor, on a volume basis. The tank warms very slowly because heat is transferred from the ambient. a. What is the temperature of the methane when the pressure reaches 10 MPa? b. Calculate the heat transferred in the process, using the generalized charts. c. Repeat parts (a) and (b), using the methane tables, Table B.7. Discuss the differences in the results. CH4: V = 10 m3, P1 = 700 kPa VLIQ 1 = 2.5 m3, VVAP 1 = 7.5 m3 0.70 10 = 0.152, Pr2 = = 2.174 4.60 4.60 From D.1: ZF1 = 0.025, ZG1 = 0.87 &

a) Pr1 =

T1 = 0.74 × 190.4 = 140.9 K vF1 =

0.025×0.518 35×140.9 = 0.00261 700

0.87×0.518 35×140.9 = 0.0908 700 2.5 7.5 mLIQ 1 = = 957.9 kg, mVAP 1 = = 82.6 kg 0.00261 0.0908 vG1 =

Total m = 1040.3 kg v2 = v1 =

Z2×0.518 35×190.4×Tr2 V 10 = = 0.00961 = m 1040.5 10 000

or Z2Tr2 = 0.9737 at Pr2 = 2.174 By trial and error

13-60

Tr2 = 1.334 & Z2 = 0.73, T2 = 1.334×190.4 = 254.0 K b) 1st law: Q12 = m(u2-u1) = m(h2-h1) - V(P2-P1) Using D.2 & x1 = *

82.6 = 0.0794 1040.5

*

(h1-h1) = (h1-hF1) - x1hFG1

[

= 0.518 35×190.4 4.72-0.0794(4.72-0.29) *

] = 431.1

*

(h2-h1) = 2.2537(254.0-140.9) = 254.9 *

(h2-h2) = 0.518 35×190.4(1.47) = 145.1 (h2-h1) = -145.1 + 254.9 + 431.1 = 540.9 kJ/kg Q12 = 1040.5(540.9) - 10(10 000-700) = 469 806 kJ c) Using Table B.7 for CH4 T1 = TSAT 1 = 141.7 K,

vF1 = 0.002 675,

uF1 = -178.47

vG1 = 0.090 45 , uG1 = 199.84 mLIQ 1 =

2.5 7.5 = 934.6, mVAP 1 = = 82.9 0.002 675 0.090 45

Total mass

m = 1017.5 kg

and

v2 =

10 = 0.009 828 m3/kg 1017.5

 T2 = 259.1 K At v2 & P2 = 10 MPa →   u2 = 296.11 Q12 = m(u2-u1) = 1017.5×296.11 - 934.6(-178.47) - 82.9(199.84) = 451 523 kJ

13-61

13.68 Calculate the difference in entropy of the ideal-gas value and the real-gas value for carbon dioxide at the state 20°C, 1 MPa, as determined using the virial equation of state. Use numerical values given in Problem 13.30. CO2 at T = 20oC, P = 1 MPa RT/P*

RT/P*

⌠ ∂P * sP* - sP =  ⌡ ∂T v dv ; ID Gas,

⌠R * P sP* - sP =  dv = R ln * ⌡v P

( )

v(P)

v(P) RT/P* *

Therefore, at P:

sP - sP = -R ln

⌠ ∂P P  + dv P* ⌡ ∂T v

( )

v(P)

virial:

P=

R BR RT dB = + (∂P ) + ( ) v v v ∂T v dT

RT BRT + 2 and v v

2

2

Integrating, P RT dB - sP = -R ln * + R ln * + R B + T dT P P v RT dB 1 = R ln + B+T Pv dT v Using values for CO2 from solution 13.30,

[

* sP

[

(

( )]( ( )) ]

1 P* v RT

)

1 -s* - -s = 8.3145 ln 2.437 37 + -0.128 + 0.266 P P 2.3018 2.3018 = 0.9743 kJ/kmol K

[

(

)

]

13.69 Carbon dioxide gas enters a turbine at 5 MPa, 100°C, and exits at 1 MPa. If the isentropic efficiency of the turbine is 75%, determine the exit temperature and the second-law efficiency. CO2 turbine: ηS = w/wS = 0.75 inlet: T1 = 100oC, P1 = 5 MPa, exhaust: P2 = 1 MPa 5 373.2 1 = 0.678, Tr1 = = 1.227, Pr2 = = 0.136 7.38 304.1 7.38 From D.2 and D.3,

a) Pr1 = *

(h1-h1) = 0.188 92×304.1×0.52 = 29.9 *

(s1-s1) = 0.188 92×0.30 = 0.0567

13-62

Assume T2S = 253 K, Tr2S = 0.832 From D.2 and D.3:

*

(h2S-h2S) = RTC×0.20 = 11.5

*

(s2S-s2S) = R×0.17 = 0.0321 *

*

(s2S-s1) = 0.8418 ln

253 1 - 0.188 92 ln = -0.0232 373.2 5

(s2S-s1) = -0.0321 - 0.0232 + 0.0567 ≈ 0 ⇒ T2S = 253 K *

*

(h2S-h1) = 0.8418(253-373.2) = -101.2 wS = (h1-h2S) = -29.9 + 101.2 + 11.5 = 82.8 kJ/kg *

*

*

*

w = ηS×wS = 0.75×82.8 = 62.1 kJ/kg = (h 1-h1) + (h1-h2) + (h2-h2) Assume T2 = 275 K, Tr2 = 0.904 *

*

(h1-h2) = 0.8418(373.2-275) = 82.7 From D.2 and D.3, *

(h2-h2) = RTC×0.17 = 9.8 ;

*

(s2-s2) = R×0.13 = 0.0245

Substituting, w = -29.9 + 82.7 + 9.8 = 62.7 ≈ 62.1 ⇒ T2 = 275 K * *

b) (s2-s1) = 0.8418 ln

275 1 - 0.188 92 ln = +0.0470 373.2 5

(s2-s1) = -0.0245 + 0.0470 + 0.0567 = +0.0792 Assuming T0 = 25 oC, (ϕ1-ϕ2) = (h1 - h2) - T0(s1 - s2) = 62.1 + 298.2(0.0792) = 85.7 kJ/kg η2nd Law =

w 62.1 = = 0.725 ϕ1-ϕ2 85.7

13-63

13.70 A 4- m3 uninsulated storage tank, initially evacuated, is connected to a line flowing ethane gas at 10 MPa, 100°C. The valve is opened, and ethane flows into the tank for a period of time, after which the valve is closed. Eventually, the whole system cools to ambient temperature, 0°C, at which time the it contains one-fourth liquid and three-fourths vapor, by volume. For the overall process, calculate the heat transfer from the tank and the net change of entropy. Rigid tank V = 4 m3, m1 = 0 Line: C2H6 at Pi = 10 MPa, Ti = 100 oC Flow in, then cool to M = 30.07,

T2 = T0 = 0 oC, VLIQ 2 = 1 m3 & VVAP 2 = 3 m3

R = 0.2765, CP0 = 1.766

10 = 2.049, 4.88 From D.2 and D.3, Pri =

Tri =

373.2 = 1.225 305.4

*

(hi -hi) = 0.2765×305.4×2.0 = 168.9 and Tr2 =

*

(si -si) = 0.2765×1.22 = 0.3373

273.2 = 0.895 305.4

From D.1, D.2 and D.3,

P2 = PG = 0.51×4880 = 2489 kPa

sat. liq.: ZF = 0.087 ; (h*-hF) = RTC×4.09 = 345.4 ; (s*-sF) = R×4.3 = 1.189 sat. vap. : ZG = 0.68 ; (h*-hG) = RTC×0.87 = 73.5 ; (s*-sG) = R×0.70 = 0.193 mLIQ 2 =

2489×1 = 378.7 kg 0.087×0.2765×273.2

mVAP 2 =

2489×3 = 145.4 kg 0.68×0.2765×273.2

m2 = 524.1 kg

=>

x2 =

145.4 = 0.277 524.1

1st law: *

*

*

*

QCV = m2u2 - mihi = m2(h2-hi) - P2V = m2[(h2-h2) + (h2-hi ) + (hi -hi)]- P2V *

*

(h2-hi ) = 1.7662(0-100) = -176.6 *

*

*

(h2-h2) = (1-x2)(h2-hF2) + x2(h2-hG2) = 0.723 × 345.4 + 0.277 × 73.5 = 270.1 QCV = 524.1[-270.1 - 176.6 + 168.9]- 2489 × 4 = -155 551 kJ

13-64

∆SNET = m2(s2-si) - QCV/T0 *

* *

*

(s2-si) = (s2-s2) + (s2-si ) + (si -si) * *

(s2-si ) = 1.7662 ln *

273.2 2.489 - 0.2765 ln = -0.1664 373.2 10

*

*

(s2-s2) = (1-x2)(s2-sF2) + x2(s2-sG2) = 0.723 ×1.189 + 0.277 × 0.193 = 0.9131 (s2-si) = -0.9131 - 0.1664 + 0.3373 = -0.7422 ∆SNET = 524.1(-0.7422) -

-155 551 = 180.4 kJ/K 273.2

13.71 The environmentally safe refrigerant R-142b (see Problem 13.45) is to be evaluated as the working fluid in a portable, closed-cycle power plant, as shown in Fig. P13.71. The air-cooled condenser temperature is fixed at 50°C, and the maximum cycle temperature is fixed at 180°C, because of concerns about thermal stability. The isentropic efficiency of the expansion engine is estimated to be 80%, and the minimum allowable quality of the fluid exiting the expansion engine is 90%. Calculate the heat transfer from the condenser, assuming saturated liquid at the exit. Determine the maximum cycle pressure, based on the specifications listed T 1 R-142b CH3CClF2 M = 100.495, TC = 410.3 K, PC = 4.25 MPa 4

CP0 = 0.787 kJ/kg K

3

R = 0.082 73 kJ/kg K

2S2

s

T1 = 180 oC, T2 = T3 = 50 oC, x2 = 0.90, ηS EXP ENG = 0.80 a) Tr2 =

323.2 = 0.788 410.3

From D.1, D.2 and D.3: P2 = P3 = 0.23×4250 = 978 kPa sat. liq.: (h*-hF) = RTC×4.55 = 154.4 ;

(s*-sF) = R×5.62 = 0.4649

sat. vap.: (h*-hG) = RTC×0.42 = 14.3 ;

(s*-sG) = R×0.37 = 0.0306

Condenser, 1st law qCOND = h3 - h2 = hF - (hF + x2hFG) = -x2hFG = -0.90(154.4 - 14.3) = -126.1 kJ/kg

13-65

ηS EXP ENGINE =

b) expansion engine, efficiency

w12 w12S

1st law & 2nd law, ideal engine:

w12S = h1 - h2S,

s2S - s1 = 0

1st law & 2nd law, real engine:

w12 = h1 - h2,

s2 - s1 > 0

*

*

*

*

(h1-h2) = (h1-h2S) = 0.787(180-50) = +102.3 * * (s1-s2)

=

* * (s1-s2S)

P1 453.2 = 0.787 ln - 0.08273 ln 323.2 978

= 0.266 05 - 0.08273 ln *

P1 978

*

*

(h2-h2) = (1-x2)(h2-hF2) + x2S(h2-hG2) = 0.1 × 154.4 + 0.9 × 14.3 = 28.3 *

*

*

(h2S-h2S) = (1-x2S)(h2S-hF2) + x2S(h2S-hG2) = 154.4 - x2S×140.1 *

*

*

(s2S-s2S) = (1-x2S)(s2S-sF2) + x2S(s2S-sG2) = 0.4649 - x2S×0.4343 Substituting, * (s1-s1)

+ 0.266 05 - 0.082 73 ln

P1

+ 0.4649 - x2S × 0.4343 = 0 (Eq.1)

978

Also *

(h1-h1) + 102.3 + 28.3 * (h1-h1)

+ 102.3 + 154.4 - x2S×140.1

= 0.80 (Eq.2)

P1(kPa) 453.2 , Tr1 = = 1.105 Both Pr1 = 410.3 4250 Trial & Error solution: Assume P1 = 9300 kPa, Pr1 = 2.188 From D.2 and D.3: *

(h1-h1) = RTC×3.05 = 103.5 and

*

(s1-s1) = R×2.10 = 0.1737

Substituting into eq. 1, -0.1737 + 0.730 75 - 0.1863 - x2S×0.4343 = 0

=>

Substituting into eq. 2, 27.1 -103.4 + 130.5 = = 0.806 ≈ 0.8 -103.4 + 256.7 - 0.8537 ×⊇140.1 33.6 ⇒ P1 = 9300 kPa

x2S = 0.8537

13-66

13.72 The refrigerant fluid R-21 (see Table A.2) is to be used as the working fluid in a solar energy powered Rankine-cycle type power plant. Saturated liquid R-21 enters the pump, state 1, at 25oC, and saturated vapor enters the turbine, state 3, at 88oC. For R-21 : CP0 = 0.582 kJ/kg K and find the boiler heat transfer q23 and the thermal efficiency of the cycle. R-21: Tc = 451.6 K, Pc = 5.18 MPa, M = 102.925 kg/kmol, R = 0.08079 kJ/kg-K State 1: T1 = 25oC, x1 = 0 From D.1 and D.2: Tr1 = 0.66, Pr1 = 0.06, P1 = Pr1Pc = 311 kPa Z1 = 0.01 ,

*

(h1 − h1)f = 4.98 RTc ,

v1 = Z1RT1/P1 = 0.000775 m3/kg

State 3: T3 = 88oC, x3 = 1.0 Tr3 = 0.8, Pr3 = 0.25, P3 = Pr3Pc = 1295 kPa *

*

(h3 − h3)g = 0.46 RTc , (s3 − s3)g = 0.39 R => qH = h3 - h2 a) 1st Law Boiler: qH + h2 = h3 + w; w = 0 1st Law Pump: q + h1 = h2 + wp; q = 0 wp = -∫ v dP = -v1(P2 - P1) = -0.000775(1295 - 311) = -0.76 kJ/kg *

h3 - h1 = (h3 − h3) *

*

*

*

+ (h3 − h1) + (h1 − h1)

*

h3 − h1 = CP (T3 - T1) = 0.582(88 - 25) = 36.7 kJ/kg, h3 - h1 = 36.7 + RTc( -0.46 + 4.98) = 201.6 kJ/kg h2 = h1 - wp;

qH = h3 - h1 + wp = 201.6 - 0.76 = 200.8 kJ/kg

b) ηth = wnet/qH; wnet = wt + wp 1st Law Turbine: q + h3 = h4 + wt; q = 0 *

wt = h3 - h4 = (h3 − h3) *

*

*

*

h3 − h4 = Cp(T3 - T4) = 36.7 kJ/kg; *

*

+ (h3 − h4) + (h4 − h4)

*

T4 = T1

*

h4 − h4 = (1 - x4)(h4 − h4)f + x4 (h4 − h4)g; x4 = ? 2nd Law Turbine:

∆Snet = m(s3 - s4) - Q/T; *

s3 - s4 = (s3 − s3) *

*

*

*

Assume Isentropic

*

+ (s3 − s4) + (s4 − s4) = 0

T3

(s3 − s4) = CP ln T

4

− R ln

P3 P4

= -0.0033 kJ/kg K

13-67

*

s4 − s4 = 0.0348 kJ/kg-K, *

*

*

s4 − s4 = (1 - x4)(s4 − s4)f + x4 (s4 − s4)g = 7.47 R - x4(7.47 - 0.15) R Tr4 = Tr1 = 0.66,

*

*

(s4 − s4)f = 7.47 R, (s4 − s4)g = 0.15 R

0.0348 = 7.47 R - x4(7.47 - 0.15) R *

(h4 − h4)f = 4.98 RTc , *

h4 − h4 = 11.5,

=>

x 4 = 0.962

*

(h4 − h4)g = 0.15 RTc

wt = h3 - h4 = 31.5 kJ/kg;

wnet = 30.7 kJ/kg

ηth = wnet/qH = 0.153

13.73 A cylinder/piston contains a gas mixture, 50% CO2 and 50% C2H6 (mole basis) at 700 kPa, 35°C, at which point the cylinder volume is 5 L. The mixture is now compressed to 5.5 MPa in a reversible isothermal process. Calculate the heat transfer and work for the process, using the following model for the gas mixture: a. Ideal gas mixture. b. Kay’s rule and the generalized charts. c. The van der Waals equation of state. a) Ideal gas mixture U2 - U1 = mCv mix(T2 - T1) = 0 Q12 = W12 = ⌠ ⌡ P dV = P1V1 ln(V2/V1) = - P1V1 ln(P2/P1) = - 700 × 0.005 ln(5500/700) = -7.71 kJ b) Kay's rule Tcmix = 0.5 × 304.1 + 0.5 × 305.4 = 304.75 K Pcmix = 0.5 × 7.38 + 0.5 × 4.88 = 6.13 MPa Tr1 = 308.15/304.75 = 1.011, P r1 = 0.7/6.13 = 0.1142 Z1 = 0.96, ∆h1 = 0.12, ∆s1 = 0.08 n = P1V1/Z1R T1 =

700*0.005 = 0.00142 kmol 0.962*8.3145*308.15

Tr2 = Tr1 , Pr2 = 5.5/6.13 = 0.897, Z2 = 0.58, ∆h2 = 1.35, ∆s2 = 1.0 - - h2 - h1 = (h2 - h1) - R Tc(∆h2 - ∆h1) = 0 - 8.3145 × 304.75(1.35 - 0.12) = - 3117 - u-2 - u-1 = h2 - h1 + RT(Z1 - Z2) = - 3117

13-68

+ 8.3145 × 308.15(0.96 - 0.58) = -2143 kJ/kmol Q12 = nT(s-2 - -s1)T = 0.00142 × 308.15 × 8.3145[ 0 - ln(5.5/0.7) -1.0 W12 = Q12

+ 0.08 ] = - 10.85 kJ - n(u- - u- ) = -10.85 - 0.00142(-2143) = - 7.81 kJ 2

1

c) van der waal's equation For CO2 : b = R Tc/8Pc = 8.3145 × 304.1/8 × 7380 = 0.04282 a = 27 Pc b2 = 27 × 7380 × 0.042822 = 365.45 For C2H6 : b = R Tc/8Pc = 8.3145 × 305.4/8 × 4880 = 0.06504 a = 27 Pc b2 = 27 × 4880 × 0.065042 = 557.41 amix = (0.5 365.45 + 0.5 557.41)2 = 456.384 bmix = 0.5 × 0.04282 + 0.5 × 0.06504 = 0.05393 8.3145*308.2 456.384 - - 2 - 700 = 0 v-1 - 0.05393 v1 By trial and error: v- = 3.5329 m3/kmol 1

8.3145*308.2 456.384 - - 2 - 5500 = 0 v-2 - 0.05393 v2 By trial and error: v-2 = 0.2815 m3/kmol n = V1/v-1 = 0.005/3.5329 = 0.00142 v-2 - b Q12 = nT(s2 - s1)T = n R T ln v -b 1

= 0.00142 × 8.3145 × 308.2 ln

0.2815 - 0.05392 = - 9.93 kJ 3.5329 - 0.05392

U2-U1 = 0.00142 × 456.39(3.5329-1 - 0.2815-1) = -2.12 kJ Q12 = U2-U1 + W12 => W12 = -9.93 -(-2.12) = -7.81 kJ

13-69

13.74 A gas mixture of a known composition is frequently required for different purposes, e.g., in the calibration of gas analyzers. It is desired to prepare a gas mixture of 80% ethylene and 20% carbon dioxide (mole basis) at 10 MPa, 25°C in an uninsulated, rigid 50-L tank. The tank is initially to contain CO2 at 25°C and some pressure P1. The valve to a line flowing C2H4 at 25°C, 10 MPa, is now opened slightly, and remains open until the tank reaches 10 MPa, at which point the temperature can be assumed to be 25°C. Assume that the gas mixture so prepared can be represented by Kay’s rule and the generalized charts. Given the desired final state, what is the initial pressure of the carbon dioxide, P1? Determine the heat transfer and the net entropy change for the process of charging ethylene into the tank. Pi =10 MPa A A = C2H4, B = CO2 T1 = 25 oC P2 = 10 MPa, T2 = 25 oC yA2 = 0.8, yB2 = 0.2

o

Ti = 25 C V=0.05 m 3

a) Mixture at 2 : PC2 = 0.8 × 5.04 + 0.2 × 7.38 = 5.508 MPa TC2 = 0.8 × 282.4 + 0.2 × 304.1 = 286.7 K Tr2 = 298.15/286.7 = 1.040; Pr2 = 10/5.508 = 1.816 D.1 : n2 =

Z2 = 0.32

P 2V 10 000×0.05 = 0.6302 kmol - = Z2RT2 0.32×8.3145×298.2

nA2 = ni = 0.8 n2 = 0.5042 kmol C2H4 nB2 = n1 = 0.2 n2 = 0.1260 kmol CO2 298.2 = 0.981 304.1 n1ZB1RT1 0.126 ZB1× 8.3145×298.2 Pr1 = = = 0.8466 ZB1 PCBV 7380×0.05 Tr1 =

By trial & error: Pr1 = 0.618 & ZB1 = 0.73

`

⇒ P1 = 0.618 × 7.38 = 4.56 MPa b) 1st law: QCV + nihi = n2u-2 - n1u-1 = n2h2 - n1h1 - (P2-P1)V - -* - -* - -* or QCV = n2(h2-h2) - n1(h1-h1) - ni(hi-hi ) - (P2-P1)V

B

13-70

-* -* -* (since Ti = T1 = T2, hi = h1 = h2) -* (h1-h1) = 0.83×8.3145×304.1 = 2099 kJ/kmol -* (h2-h2) = 3.40×8.3145×286.7 = 8105 kJ/kmol Tri =

298.2 10 = 1.056, Pri = = 1.984 282.4 5.04

-* (hi -hi) = 3.35×8.3145×282.4 = 7866 kJ/kmol QCV = 0.6302(-8105) - 0.126(-2099) - 0.5042(-7866) - (10 000-4560)×0.05 = -1149 kJ ∆SCV = n2-s2 - n1-s1 , ∆SSURR = - QCV/T0 - ni-si ∆SNET = n2-s2 - n1-s1 - QCV/T0 - ni-si * * Let -sA0 = -sB0 = 0 at T0 = 25 oC, P0 = 0.1 MPa *

Then -sMIX 0 = -8.3145 (0.8 ln 0.8 + 0.2 ln 0.2) = 4.161 kJ/kmol K -s = -s* + (s-* -s-* ) + (s- -s-* ) B0 P1 T1 P0 T0 B 1 1 P1 T1 B = 0 + (0-8.3145 ln

4.56 ) - 0.60 × 8.3145 = -36.75 kJ/kmol K 0.1

-s = -s* + (s-* -s-* ) + (s- -s-* ) A0 Pi Ti P0 T0 A i i Pi Ti A = 0 + (0-8.3145 ln

10 ) - 2.44×8.3145 = -58.58 kJ/kmol K 0.1

-s = -s* -* -* - -* MIX 0 + (s P2 T2-s P0 T0)MIX + (s 2-s P2 T2)MIX 2 = 4.161 + (0-8.3145 ln

10 ) - 2.551×8.3145 = -55.34 kJ/kmol K 0.1

∆SNET = 0.6302(-55.33) - 0.126(-36.75) - 0.5042(-58.58) + 1149/298.2 = +3.15 kJ/K

13-71

English Unit Problems 13.75EA special application requires R-22 at −150 F. It is known that the triple-point temperature is less than −150 F. Find the pressure and specific volume of the saturated vapor at the required condition. The lowest temperature in Table C.10 for R-22 is -100 F, so it must be extended to -150 F using the Clapeyron eqn. At T1 = -100 F = 359.7 R , P1 = 2.398 lbf/in.2

and

R=

1.9859 = 0.022 97 Btu/lbm R 86.469

P hfg (T-T1) 107.9 (309.7-359.7) ln = = = -2.1084 P1 R T × T 0.022 97 309.7 × 359.7 1 P = 0.2912 lbf/in.2 vg =

RT 0.022 97 × 778 × 309.7 = = 132 ft3/lbm Pg 0.2912 × 144

13.76EIce (solid water) at 27 F, 1 atm is compressed isothermally until it becomes liquid. Find the required pressure. Water, triple point T = 32.02 F = 491.69 R, P = 0.088 67 lbf/in.2 vF = 0.016 022 vI = 0.017 473 hF = 0.00 dPIF dT ∆P ≈

=

hI = -143.34

hF-hI (vF-vI)T

dPIF dT

=

143.34×778.2 = -1085.8 -0.001 451×491.69×144

∆T = -1085.8(27-32.02) = 5450.7 lbf/in.2

P = Ptp + ∆P = 5451 lbf/in.2

13-72

13.77E Using thermodynamic data for water from Tables C.8.1 and C.8.3, estimate the freezing temperature of liquid water at a pressure of 5000 lbf/in.2. P

H 2O T.P.

dPIF dT

=

hIF TvIF

≈ constant

At the triple point, vIF = vF - vI = 0.016 022 - 0.017 473

T

= -0.001 451 ft3/lbm hIF = hF - hI = 0.0 - (-143.34) = 143.34 Btu/lbm dPIF dT

=

143.34 778.2 × = -1085.8 lbf/in.2 R 491.69(-0.001 451) 144

⇒ at P = 5000 lbf/in2, T ≈ 32.02 +

(5000-0.09) = 27.4 F (-1085.8)

13.78E Determine the volume expansivity, α P, and the isothermal compressibility, β T , for water at 50 F, 500 lbf/in.2 and at 500 F, 1500 lbf/in.2 using the steam tables. Water at 50 F, 500 lbf/in.2 (compressed liquid) 1 ∂v 1 ∆v ≈ ( ) ( ) v ∂T P v ∆T P

αP =

Using values at 32 F, 50 F and 100 F αP ≈

1 0.016 106 - 0.015 994 = 0.000 103 F-1 0.015 998 100 - 32

βT = -

1 ∂v 1 ∆v ≈ ( ) ( ) v ∂P T v ∆P T

Using values at saturation, 500 and 1000 lbf/in.2 1 0.015 971 - 0.016 024 βT ≈ = 0.000 0033 in.2/lbf 0.015 998 1000 - 0.178 Water at 500 F, 1500 lbf/in.2 (compressed liquid) 1 0.021 579 - 0.019 264 αP ≈ = 0.001 143 F-1 0.020 245 550 - 450 βT ≈ -

1 0.020 139 - 0.020 357 = 0.000 0108 in.2/lbf 0.020 245 2000 - 1000

13-73

13.79E Sound waves propagate through a media as pressure waves that causes the media to go through isentropic compression and expansion processes. The speed of sound c is defined by c2 = (∂P/∂ρ)s and it can be related to the adiabatic compressibility, which for liquid ethanol at 70 F is 6.4 in.2/lbf. Find the speed of sound at this temperature. ∂P = -v (∂P ) ( )= ∂ρ s ∂v s

c2 =

1

2

1 ∂v ρ v ∂P s

-

( )

32.174×144 (6.4×10 ) ×48.9

1/2

-6

1 β sρ

ρ = 48.9 lbm/ft3

From Table C.3 for ethanol, ⇒c=

=

= 3848 ft/s

13.80E Consider the speed of sound as defined in Problem 13.79. Calculate the speed of sound for liquid water at 50 F, 250 lbf/in.2 and for water vapor at 400 F, 80 lbf/in.2 using the steam tables. From problem 13.79 :

∂P = -v (∂P ) ( ) ∂ρ s ∂v s

c2 =

2

Liquid water at 50 F, 250 lbf/in.2 Assume

∆P ≈( ) (∂P ) s ∂v ∆v

T

Using saturated liquid at 50 F and compressed liquid at 50 F, 500 lbf/in.2, (500-0.18)×144×32.174 (0.016024+0.015998 ) ( ) = 22.832×10 2 0.015998-0.016024

c2 = -

2

c = 4778 ft/s Superheated vapor water at 400 F, 80 lbf/in.2 v = 6.217, s = 1.6790 At P = 60 lbf/in.2 & s = 1.6790: T = 343.8 F, v = 7.7471 At P = 100 lbf/in.2 & s = 1.6790: T = 446.2 F, v = 5.2394 c2 = -(6.217)2 c = 1690 ft/s

((100-60)×144×32.174 ) = 2.856×10 5.2394-7.7471

6

6

13-74

13.81E A cylinder fitted with a piston contains liquid methanol at 70 F, 15 lbf/in.2 and volume 1 ft3. The piston is moved, compressing the methanol to 3000 lbf/in.2 at constant temperature. Calculate the work required for this process. The isothermal compressibility of liquid methanol at 70 F is 8.3 × 10−3 in.2/lbf. 2

2

⌠ ∂v ⌠ w12 = ⌠ ⌡ Pdv =  ⌡ P ∂P T dPT = -⌡ vβ T PdPT

( )

1

1

For v ≈ const & β T ≈ const.

=>

w 12 = -

vβ T

(P 2

2 2

2

- P1

)

For liquid methanol, from Table C.3 : ρ = 49.1 lbm/ft3 m = 1.0 × 49.1 = 49.1 lbm

V1 = 1.0 ft3, W12 =

1.0×0.0083 (3000)2 - (15)2 ×144 = 37 349 ft lbf = 48.0 Btu 2

[

]

13.82E A piston/cylinder contains 10 lbm of butane gas at 900 R, 750 lbf/in.2. The butane expands in a reversible polytropic process with polytropic exponent, n = 1.05, until the final pressure is 450 lbf/in.2. Determine the final temperature and the work done during the process. C4H10 ,

P1 = 750 lbf/in.2

m = 10 lbm , T1 = 900 R ,

Rev. polytropic process (n=1.05), P2 = 450 lbf/in.2 Tr1 =

900 750 = 1.176, Pr1 = = 1.361 765.4 551 V1 = n P1V1

10 × 0.67 × 26.58 × 900 750 × 144 =

Z2Tr2 =

n P2V2

750 → V2 = 1.484 450

P2V2 mRTC

Z1 = 0.67

= 1.484 ft3

( )

=

=> From Fig. D.1 :

1 1.05

450 × 144 × 2.414 10 × 26.58 × 765.4

= 2.414 ft3 = 0.7688

at Pr2 = 450/551 = 0.817 Trial & error: 2

W12 = ⌠ ⌡ PdV = 1

Tr2 = 1.068, P2V2 - P1V1 1-n

=

Z2 = 0.72

=>

T2 = 817.4 R

750 × 1.484 (450 × 2.4141 --1.05 ) × 144 = 98.8 Btu 778

13-75

13.83E A 7-ft3 rigid tank contains propane at 1300 lbf/in.2, 540 F. The propane is then allowed to cool to 120 F as heat is transferred with the surroundings. Determine the quality at the final state and the mass of liquid in the tank, using the generalized compressibility charts. Propane C3H8: V = 7.0 ft3, P1 = 1300 lbf/in.2, T1 = 540 F = 1000 R cool to T2 = 120 F = 580 R From Table C.1 : TC = 665.6 R, PC = 616 lbf/in.2 1300 1000 = 2.110, Tr1 = = 1.502 616 665.6 From D.1: Z1 = 0.83 Pr1 =

Z1RT1 0.83 × 35.04 × 1000 v2 = v1 = = P1 1300 × 144

= 0.1554 ft3/lbm

From D.1 at Tr2 = 0.871, saturated => PG2 = 0.43 × 616 = 265 lbf/in.2 vG2 =

0.715 × 35.04 × 580 265 × 144

= 0.3808

vF2 =

0.075 × 35.04 × 580 265 × 144

= 0.0399

0.1554 = 0.0399 + x 2(0.3781-0.0399) mLIQ 2 = (1 - 0.3388)

7.0 = 29.8 lbm 0.1554

=>

x2 = 0.3388

13-76

13.84E A rigid tank contains 5 lbm of ethylene at 450 lbf/in.2, 90 F. It is cooled until the ethylene reaches the saturated vapor curve. What is the final temperature? C2H4 m = 5 lbm C 2H 4

P1 = 450 lbf/in2, T1 = 90 F = 249.7 R

T

1

Pr1 =

450 = 0.616, 731

Fig. D.1

2

Tr1 =

549.7 = 1.082 508.3

⇒ Z1 = 0.82

v

Pr2 = Pr1

Z2Tr2 Z1Tr1

= 0.616

ZG2Tr2 0.82 ×⊇1.082

Trial & error: Tr2 ZG2

Pr2

Pr2 CALC

0.871 0.715

0.43

0.432

= 0.6943 ZG2Tr2

~ OK

=>

T2 = 442.7 R

13.85E Calculate the difference in internal energy of the ideal-gas value and the real-gas value for carbon dioxide at the state 70 F, 150 lbf/in. 2, as determined using the virial equation of state. For carbon dioxide at 70 F, B = -2.036 ft3/lb mol , T(dB/dT) = 4.236 ft3/lb mol CO2 at 70 F, 150 lbf/in2

Solution: virial:

P=

R BR RT dB = (∂P ) + + ( ) ∂T v v v v dT

RT BRT + 2 ; v v

2

2

⌠∞ ⌠∞ RT2 dB ∂P RT2 dB   - P dv = u-u = - ⌡ T ⌡v v2 dT dv = - v dT ∂T v v *

[( ) ]

( )

B = -2.036 ft3/lbmol

T

(dB )= 4.236 ft /lbmol dT 3

-v = 1 RT 1 + 1+4BP/R T 2 P RT 1545×529.7 But = = 37.8883 P 150×144

[

]

[

v- = 0.5×37.8883 1 +

1+4(-2.036)/37.8883

] = 35.7294 ft /lbmol

-1.9859×529.7 4.236 = -123.9 Btu/lbmol u--u-* = 35.7294

3

13-77

13.86E Calculate the heat transfer during the process described in Problem 13.81. From solution 13.82, V1 = 1.473 ft3, Tr1 = 1.176,

V2 = 2.396 ft3,

Pr1 = 1.361, *

From D.1:

(hRT-h) C

*

W12 = 98.8 Btu

Tr2 = 1.068,

Pr2 = 0.817, T2 = 817.4 R

*

= 1.36, 1

(hRT-h) C

2

= 0.95

*

h2 - h1 = 0.415 (817.4 - 900) = -34.3 Btu/lbm 26.58×765.4 (-0.95 + 1.36) = -23.6 Btu/lbm 778 U2-U1 = m(h2-h1) - P2V2 + P1V1

h2 - h1 = -34.3 +

450×144×2.414 750×144×1.484 + = -231.1 Btu 778 778 Q12 = U2-U1 + W12 = -132.3 Btu = 10(-23.6) -

13-78

13.87E Saturated vapor R-22 at 90 F is throttled to 30 lbf/in.2 in a SSSF process. Calculate the exit temperature assuming no changes in the kinetic energy, using the generalized charts, Fig. D.2 and repeat using the R-22 tables, Table C.10. R-22 throttling process T1 = 90 F, x1 = 1.00, P2 = 30 lbf/in.2 *

Energy Eq.:

*

*

*

h2-h1 = (h2-h2) + (h2-h1) + (h1-h1) = 0

Generalized charts,

Tr1 =

549.7 = 0.827 664.7

1.9859×664.7 (0.55) = 8.40 86.469 To get CP0, use h values from Table C.10 at low pressure. *

From D.2:

(h1-h1) =

CP0 ≈

121.867-118.724 = 0.1572 100-80

Substituting into energy Eq.: at Pr2 =

*

(h2-h2) + 0.1572 (T2 - 30) + 8.40 = 0

30 = 0.042 721

Assume T2 = 43.4 F = 503.1 R => Tr2 =

503.4 = 0.757 664.7

1.9859×664.7 (0.07) = 1.07 86.469 Substituting, *

(h2-h2) =

-1.07 + 0.1572(43.4 - 90) + 8.40 = 0.005 ≈ 0 ⇒ T2 = 43.4 F b) R-22 tables, C.10:

T1 = 90 F, x1 = 1.0

=> h 1 = 111.62

h2 = h1 = 111.62 , P2 = 30 lbf/in.2 => T2 = 42.1 F

13-79

13.88E A 10-ft3 tank contains propane at 90 F, 90% quality. The tank is heated to 600 F. Calculate the heat transfer during the process. V = 10 ft3 T1 = 90 F = 549.7 R, x1 = 0.90

C 3H 8 T

Heat to T2 = 600 F = 1059.7 R

2

M = 44.094, TC = 665.6 R 1

PC = 616 lbf/in.2

v

R = 35.04, CP0 = 0.407 Tr1 = 0.826 Figs. D.1 and D.2 → Z1 = 0.1 × 0.053 + 0.9 × 0.78 = 0.707, *

h1-h1 RTc SAT

Pr

m=

= 0.1 × 4.4 + 0.9 × 0.55 = 0.935 = 0.31 ;

SAT

P1

= 0.31 × 616 = 191 lbf/in.2

191 × 144 × 10 0.707 × 35.04 × 549.7

Pr2 =

= 20.2 lbm

20.2 × Z 2 × 35.04 × 1059.7 616 × 144 × 10

=

Z2 1.183

at Tr2 = 1.592

 Pr2 = 0.79 Trial & error:   Z2 = 0.94 *

P2 = 490 lbf/in.2 *

h2-h2 RTc

= 0.36

*

(h2-h1) = 0.407(600-90) = 207.6 Btu/lbm *

(h1-h1) = 0.935 × 35.04 × 665.9/ 778 = 28.0 *

(h2-h2) = 0.36 × 35.04 × 665.9/ 778 = 10.8 Q12 = m(h2-h1) - (P2-P1)V = 20.2 (-10.8 + 207.6 + 28.0) - (490 - 191) × = +4541 - 553 = 3988 Btu

144 × 60 778

13-80

13.89E A newly developed compound is being considered for use as the working fluid in a small Rankine-cycle power plant driven by a supply of waste heat. Assume the cycle is ideal, with saturated vapor at 400 F entering the turbine and saturated liquid at 70 F exiting the condenser. The only properties known for this compound are molecular weight of 80 lbm/lbmol, ideal gas heat capacity Cpo = 0.20 Btu/lbm R and Tc = 900 R, Pc = 750 lbf/in.2. Calculate the work input, per lbm, to the pump and the cycle thermal efficiency. T1 = 400 F = 860 R

1 . WT

Turbine . QH

x1 = 1.0 T3 = 70 F = 530 R

Ht. Exch

2 Cond

Properties known: M = 80, CPO = 0.2 Btu/lbm R

3

TC = 900 R, PC = 750 lbf/in.2

P 4

. -W P

x3 = 0.0

Tr1 =

860 530 = 0.956, Tr3 = = 0.589 900 900

From D.1: Pr1 = 0.76, P1 = 0.76 × 750 = 570 lbf/in.2 = P4 Pr3 = 0.025, P3 = 19 lbf/in.2 = P2, ZF3 = 0.0045 vF3 = ZF3RT3/P3 =

0.0045 × 1545 × 530 = 0.0168 19 × 144 × 80

4

⌠ vdP ≈ v (P -P ) = -0.0168 (570 - 19) × wP = - ⌡ F3 4 3 3

144 = -1.71 Btu/lbm 778

qH + h4 = h1 , but h3 = h4 + wP => q H = (h1-h3) + wP From D.2, *

(h1-h1) = (1.9859/80) × 900 × 1.34 = 30.0 *

(h3-h3) = (1.9859/80) × 900 × 5.2 = 116.1 *

*

(h1-h3) = CP0(T1-T3) = 0.2(400-70) = 66.0 (h1-h3) = -30.0 + 66.0 + 116.1 = 152.1 qH = 152.1 + (-1.71) = 150.4 Btu/lbm

13-81

*

* *

*

Turbine, (s2-s1) = 0 = -(s2-s2)+(s2-s1)+(s1-s1) From D.3, *

(s1-s1) = (1.9859/80) × 1.06 = 0.0263 * *

(s2-s1) = 0.20 ln

530 19 - 0.024 82 ln = -0.0124 860 570

Substituting, *

*

(s2-s2) = - 0.0124 + 0.0263 = 0.0139 = (s2-sF2) - x2sFG2 0.0139 = 0.024 82 × 8.77 - x2 × 0.024 82 (8.77 - 0.075) => x 2 = 0.9444 *

*

(h2-h2) = (h2-hF2) - x2hFG2 From D.2, hfg2 = 0.024 82 × 900 (5.2 − 0.07) = 114.6 *

(h2-h2) = 116.1 - 0.9444 × 114.6 = 7.9 wT = (h1-h2) = -30.0 + 66.0 + 7.9 = 43.9 Btu/lbm ηTH =

wNET qH

=

43.9 - 1.7 = 0.281 150.4

13-82

13.90E A 7-ft3 rigid tank contains propane at 730 R, 500 lbf/in.2. A valve is opened, and propane flows out until half the initial mass has escaped, at which point the valve is closed. During this process the mass remaining inside the tank expands according to the relation Pv1.4 = constant. Calculate the heat transfer to the tank during the process. C3H8: V = 7.0 ft3, T1 = 400 K, P1 = 500 lbf/in.2 Pv1.4 = const inside

Flow out to m2 = m1/2; Tr1 =

730 500 = 1.097, Pr1 = = 0.812 665.6 616

0.76 × 35.04 × 730 = 0.270, 500 × 144 m1 = 7.0/0.270 = 25.92 lbm, v1 =

P 2 = P1

v1

(v )

1.4

=

2

=> From D.1:

Z1 = 0.76

v2 = 2v1 = 0.54 m2 = m1/2 = 12.96 lbm,

500 = 189.5 lbf/in.2 21.4

Trial & error: saturated with 189.5 Pr2 = = 0.308  T2 = 0.825 × 665.6 = 549.4 R & 616  P2v2 = Z2RT2  Z2 = 189.5 × 144 × 0.54 = 0.764 35.04 × 549.4 Z2 = ZF2 + x2(ZG2 - ZF2) 0.764 = 0.052 + x 2(0.78 - 0.052) =>

x2 = 0.978

*

(h1-h1) = 35.04 × 665.6 × 0.85 / 778 = 25.5 *

*

(h2-h1) = 0.407 (549.4 - 730) = -73.4 *

*

(h2-h2) = (h2-hF2) - x2hFG2 = 18.7 1st law:

[4.41 - 0.978(4.41 - 0.54)] =

QCV = m2h2 - m1h1 + (P1-P2)V + mehe AVE

*

Let h1 = 0 then: *

35.04 × 665.6 778

*

*

*

h1 = 0 + (h1-h1) = -25.5 *

h2 = h1 + (h2-h1) + (h2-h2) = 0 - 73.4 - 18.7 = -92.1 he AVE = (h1 + h2)/2 = -58.8 QCV = 12.96 (-92.1) - 25.92(-25.5) + (500-189.5)×7.0× = -892 Btu

144 + 12.96(-58.8) 778

13-83

13.91E A geothermal power plant on the Raft river uses isobutane as the working fluid as shown in Fig. P13.42. The fluid enters the reversible adiabatic turbine at 320 F, 805 lbf/in.2 and the condenser exit condition is saturated liquid at 91 F. Isobutane has the properties Tc = 734.65 R, Pc = 537 lbf/in.2, Cpo = 0.3974 Btu/lbm R and ratio of specific heats k = 1.094 with a molecular weight as 58.124. Find the specific turbine work and the specific pump work. R = 26.58 ft lbf/lbm R = 0.034 166 Btu/lbm R Turbine inlet: T1 = 320 F , P1 = 805 lbf/in.2 Condenser exit: T3 = 91 F , x3 = 0.0 ;

Tr3 = 550.7 / 734.7 = 0.75

Pr3 = 0.165, Z3 = 0.0275

From D.1 :

P2 = P3 = 0.165 × 537 = 88.6 lbf/in.2 Tr1 = 779.7 / 734.7 = 1.061,

Pr1 = 805 / 537 = 1.499

From D.2 and D.3, *

(h1-h1) = 0.034 166 × 734.7 × 2.85 = 71.5 *

(s1-s1) = 0.034 166 × 2.15 = 0.0735 * *

(s2-s1) = 0.3974 ln *

550.7 88.6 - 0.034 166 ln = -0.0628 779.7 805

*

(s2-s2) = (s2-sF2) - x2sFG2 = 0.034 166 × 6.12 - x2× 0.034 166(6.12 - 0.29) = 0.2090 - x2× 0.1992 (s2-s1) = 0 = -0.2090 + x2 × 0.1992-0.0628 + 0.0735 *

=>

x2 = 0.9955

*

(h2-h1) = CP0(T2-T1) = 0.3974(550.7-779.7) = -91.0 From D.2, *

*

(h2-h2) = (h2-hF2) - x2hFG2 = 0.034 166×734.7[4.69 - 0.9955(4.69 - 0.32)] = 117.7 − 0.9955 × 109.7 = 8.5 Turbine: Pump

wT = (h1-h2) = -71.5 + 91.0 + 8.5 = 28.0 Btu/lbm vF3 = 4

ZF3RT3 P3

=

0.0275 ×⊇26.58 × 550.7 = 0.031 55 88.6 ×⊇144

144 wP = - ⌠ ⌡ vdP ≈ vF3(P4 -P3) = -0.031 55(805-88.6) × = -4.2 Btu/lbm 778 3

13-84

13.92E Carbon dioxide collected from a fermentation process at 40 F, 15 lbf/in.2 should be brought to 438 R, 590 lbf/in.2 in a SSSF process. Find the minimum amount of work required and the heat transfer. What devices are needed to accomplish this change of state? 35.1 R= = 0.045 12 Btu/lbm R 778 Tri =

500 = 0.913 , 547.4

438 = 0.80, 547.4

From D.2 and D.3: *

15 = 0.014 1070

h*-h (RT ) = 0.02, C

From D.2 and D.3: Tre =

Pri =

Pre =

590 = 0.551 1070

*

he-he = 4.50 RTc, *

*

s*-s ( R ) = 0.01 R

*

se-se = 4.70 R

*

(hi-he) = - (hi -hi) + (hi -he) + (he-he) = - 0.045 12 × 547.4 × 0.02 + 0.203(500 - 438) + 0.045 12 × 547.4 × 4.50 = 123.2 Btu/lbm *

* *

*

(si-se) = - (si -si) + (si -se) + (se-se) = - 0.045 12×0.01 + 0.203 ln

15 500 - 0.045 12 ln + 0.045 12×4.70 590 438

= 0.4042 Btu/lbm R wrev = (hi-he) -T0(si-se) = 123.2 - 500(0.4042) = -78.4 Btu/lbm qrev = (he-hi) + wrev = -123.2 - 78.9 = -202.1 Btu/lbm

13-85

13.93E A control mass of 10 lbm butane gas initially at 180 F, 75 lbf/in.2, is compressed in a reversible isothermal process to one-fifth of its initial volume. What is the heat transfer in the process? T1 = 180 F, P1 = 75 lbf/in.2

Butane C4H10: m = 10 lbm,

Compressed, reversible T = const, to V2 = V1/5 Tr1 =

640 75 = 0.836, Pr1 = = 0.136 => From D.1 and D.3: Z 1 = 0.92 765.4 551 *

(s1-s1) = 0.16 ×

26.58 = 0.0055 778

Z1RT1 0.92 ×⊇26.58 ×⊇640 v1 = = = 1.449 ft3/lbm P1 75 ×⊇144 v2 = v1/5 = 0.2898 ft3/lbm At Tr2 = Tr1 = 0.836 From D.1 and D.3: sat. liq.: sat. vap.:

PG = 0.34 × 551 =187 lbf/in.2 (s*-sF) = R × 5.02 = 0.1715

ZF = 0.058 ;

(s*-sG) = R × 0.49 = 0.0167

ZG = 0.765 ;

Therefore vF =

0.058 ×⊇26.58 ×⊇640 = 0.0366 187 ×⊇144

vG =

0.77 ×⊇26.58 ×⊇640 = 0.4864 187 ×⊇144

Since vF < v2 < vG → two-phase *

*

x2 =

v2-vF vG-vF

= 0.563

*

(s2-s2) = (1-x2)(s2-sF2) + x2(s2-sG2) = 0.437 × 0.1715 + 0.563 × 0.0167 = 0.0843 * * (s2-s1)

= CP0 ln

T2 T1

- R ln

P2 P1

=0-

26.58 187 × ln = -0.0312 778 75

(s2-s1) = -0.0843 - 0.0312 + 0.0055 = -0.110 Btu/lbm R Q12 = Tm(s2-s1) = 640×10(-0.110) = -704 Btu

13-86

13.94E A cylinder contains ethylene, C2H4, at 222.6 lbf/in.2, 8 F. It is now compressed isothermally in a reversible process to 742 lbf/in.2. Find the specific work and heat transfer. Ethylene C2H4 P1 = 222.6 lbf/in.2, T2 = T1 = 8 F = 467.7 R State 2: P2 = 742 lbf/in.2 R = 55.07 ft lbf/lbm R = 0.070 78 Btu/lbm R 467.7 222.6 Tr2 = Tr1 = = 0.920 Pr1 = = 0.305 508.3 731 From D.1, D.2 and D.3:

Z1 = 0.85 ,

h*-h RTC

( )

1

= 0.40

*

(h1-h1) = 0.070 78 × 508.3 × 0.40 = 14.4 *

(s1-s1) = 0.070 78 × 0.30 = 0.0212 742 = 1.015 (comp. liquid) Pr2 = 731 From D.1, D.2 and D.3: Z2 = 0.17 *

(h2-h2) = 0.070 78 × 508.3 × 4.0 = 143.9 *

(s2-s2) = 0.070 78 × 3.6 = 0.2548 *

*

(h2-h1) = 0 * *

(s2-s1) = 0 - 0.070 78 ln

742 = -0.0852 222.6

q12 = T(s2-s1) = 467.7(-0.2548 - 0.0852 + 0.0212) = -149.1 Btu/lbm (h2-h1) = -143.9 + 0 + 14.4 = -129.5 (u2-u1) = (h2-h1) - RT(Z2-Z1) = -129.5 - 0.070 78 × 467.7 (0.17 - 0.85) = -107.0 w12 = q12 - (u2-u1) = -149.1 + 107.0 = -42.1 Btu/lbm

13-87

13.95E A cylinder contains ethylene, C 2H4, at 222.6 lbf/in.2, 8 F. It is now compressed in a reversible isobaric (constant P) process to saturated liquid. Find the specific work and heat transfer. Ethylene C2H4 P1 = 222.6 lbf/in.2 = P2, T1 = 8 F = 467.7 R State 2: saturated liquid, x2 = 0.0 R = 55.07 ft lbf/lbm R = 0.070 78 Btu/lbm R 467.7 222.6 = 0.920 Pr1 = Pr2 = = 0.305 Tr1 = 508.3 731 From D.1 and D.2:

v1 =

Z1RT1 P1

=

Z1 = 0.85 ,

h*-h RTC

( )

1

= 0.40

0.85 × 55.07 × 467.7 = 0.683 222.6 × 144

*

(h1-h1) = 0.070 78 × 508.3 × 0.40 = 14.4 From D.1 and D.2: Z2 = 0.05 , v2 =

Z2RT2 P2

h*-h RTC

T2 = 0.822 × 508.3 = 417.8 R

( ) =

2

= 4.42

0.05 × 55.07 × 417.8 = 0.035 89 222.6 × 144

*

(h2-h2) = 0.070 78 × 508.3 × 4.42 = 159.0 *

*

(h2-h1) = CP0(T2-T1) = 0.411(417.8 - 467.7) = -20.5 2

144 w12 = ⌠ ⌡ Pdv = P(v2-v1) = 222.6(0.035 89 - 0.683) × = -26.7 Btu/lbm 778 1

q12 = (u2 - u1) + w12 = (h2-h1) = -159.0 - 20.5 + 14.4 = -165.1 Btu/lbm

13-88

13.96E A distributor of bottled propane, C3H8, needs to bring propane from 630 R, 14.7 lbf/in.2 to saturated liquid at 520 R in a SSSF process. If this should be accomplished in a reversible setup given the surroundings at 540 R, find the ratio of the volume flow rates V.in / V.out, the heat transfer and the work involved in the process. R = 35.04/778 = 0.045 04 Btu/lbm R 630 14.7 Tri = = 0.947 Pri = = 0.024 665.6 616 From D.1, D.2 and D.3 : Zi = 0.99 *

(hi -hi) = 0.045 03 × 665.6 × 0.03 = 0.9 *

(si -si) = 0.045 04 × 0.02 = 0.0009 Tre = 520/665.6 = 0.781, From D.1, D.2 and D.3 : Pre = = 0.21 , Pe = 0.21 × 616 = 129 lbf/in.2 Ze = 0.035 *

(he-he) = 0.045 04 × 665.6 × 4.58 = 137.3 *

(se-se) = 0.045 04 × 5.72 = 0.2576 *

*

(he-hi ) = 0.407 (520 - 630) = -44.8 * *

(se-si ) = 0.407 ln

520 132 - 0.045 04 ln = -0.1770 630 14.7

(he-hi) = -137.3 - 44.8 + 0.9 = -181.2 (se-si) = -0.2576 - 0.1759 + 0.0009 = -0.4326 . ZiTi/Pi Vin 0.99 630 129 . = = × × = 300.7 Vout ZeTe/Pe 0.035 520 14.7 wrev = (hi-he) -T0(si-se) = 181.2 - 540(0.4326) = -52.4 Btu/lbm qrev = (he-hi) + wrev = -181.2 - 52.4 = -233.6 Btu/lbm

13-89

13.97E A line with a steady supply of octane, C8H18, is at 750 F, 440 lbf/in.2. What is your best estimate for the availability in an SSSF setup where changes in potential and kinetic energies may be neglected? Availability of Octane at

Ti = 750 F, Pi = 440 lbf/in.2

R = 13.53 ft lbf/lbm R = 0.017 39 Btu/lbm R 440 1209.7 Pri = = 1.219, Tri = = 1.182 361 1023.8 From D.2 and D.3: *

(h1-h1) = 0.017 39 × 1023.8 × 1.15 = 20.5 *

(s1-s1) = 0.017 39 × 0.71 = 0.0123 Exit state in equilibrium with the surroundings Assume T0 = 77 F, P0 = 14.7 lbf/in.2 536.7 14.7 = 0.524 , Pr0 = = 0.041 1023.8 361 From D.2 and D.3: Tr0 = *

(h0-h0) = RTC × 5.41 = 96.3 *

and

*

(s0-s0) = R × 10.38 = 0.1805

*

(hi -h0) = 0.409(1209.7 - 536.7) = 275.3 * *

(si -s0) = 0.409 ln

1209.7 440 - 0.017 39 ln = 0.2733 536.7 14.7

(hi-h0) = -20.5 + 275.3 + 96.3 = 351.1 (si-s0) = -0.0123 + 0.2733 + 0.1805 = 0.4415 ψi = wrev = (hi-h0) - T0(si-s0) = 351.1 - 536.7 (0.4415) = 114.1 Btu/lbm

14-1

CHAPTER 14 The correspondence between the new problem set and the previous 4th edition chapter 12 problem set. New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Old new 3 new new 1 2 6 4 5 10 7 8 9 11 12 13 new 15 16 18 19 new 20 21 22 23 24 25 26 27

New 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Old 28 new 29 30 31 32 33 34 36 37 39 40 mod 42 17 43 44 46 mod 49 50 47 48 52 new 56 57 new 58 59 67 65

New 61 62 63 64 65 66 67 68 69 70 71 72 73 74

The problems that are labeled advanced start at number 66.

Old new 66 69 70 71 14 38 41 54 60 61 62 68 new

14-2

The English unit problems are: New

Old

75 76 77 78 79

72 73 74 76 77

New 80 81 82 83 84 85 86 87 88 89

Old 78 79 80 81 82 83 84 85 86 87

New 90 91 92 93 94 95 96 97 98

Old 88 89 90 92 mod 96 97 98 100 101

14-3 14.1

Calculate the theoretical air-fuel ratio on a mass and mole basis for the combustion of ethanol, C2H5OH. Reaction Eq.: C2H5OH + νO2(O2 + 3.76N2) => aCO 2+ bH2O + cN2 Balance C: 2 = a

Balance H: 6 = 2b

=> b = 3 => νO2 = 3

Balance O: 1 + 2νO2 = 2a + b = 4 + 3 = 7 (air/fuel)mol = νO2(1 + 3.76)/1 = 3 × 4.76 = 14.28 (air/fuel)mass = (νO2MO2 + νN2 MN2)/MFuel

= (3×31.999 + 11.28×28.013)/46.069 = 8.943

14.2

In a combustion process with decane, C10H22, and air, the dry product mole fractions are 86.9% N2, 1.163% O2, 10.975% CO2 and 0.954% CO. Find the equivalence ratio and the percent theoretical air of the reactants. x C10H22 + φ νO (O2 + 3.76 N2) → a H2O + b CO2 + c CO + d N2 + e O2 2

Stoichiometric combustion: φ = 1, c = 0, e = 0,

C balance: b = 10x

H balance: a = 22x/2 = 11x, O balance: 2 νO = a + 2b = 11x + 20x = 31x 2

νO = 15.5x, νN = 58.28x 2

=> (A/F)s = (νO

2

Actual combustion: C balance:

d = 86.9

2

→ φ νO × 3.76 = 86.9 2

10x = 10.975 + 0.954 = 11.929

+ν

N2

)/x = 73.78

→ φ νO = 23.11 2

=> x = 1.1929

(A/F)ac = φ ν O × 4.76/1.1929 = 92.22 2

φ = (A/F)ac / (A/F)s = 92.22 / 73.78 = 1.25

14.3

or 125%

A certain fuel oil has the composition C10H22. If this fuel is burned with 150% theoretical air, what is the composition of the products of combustion? C10H22 + φ νO (O2 + 3.76 N2) → a H2O + b CO2 + c N2 + d O2 2

Stoichiometric combustion: φ = 1, d = 0, H balance: a = 22/2 = 11, => νO2 = 15.5

O balance: 2 νO = a + 2b = 11 + 20 = 31

Actual case:

H balance: a = 11, C balance: O2 balance:

C balance: b = 10 2

φ = 1.5 => νO2 = 1.5×15.5 = 23.25 b = 10,

N balance: c = 23.25×3.76 = 87.42

d = 23.25 - 10 - 11/2 = 7.75 (excess oxygen)

14-4 14.4

Natural gas B from Table 14.2 is burned with 20% excess air. Determine the composition of the products. The reaction equation (stoich. complete comb.) with the fuel composition is: 60.1 CH4 + 14.8 C2H6 + 13.4 C3H8 + 4.2 C4H10 + 7.5 N2 + νO2 (O2 + 3.76 N2) →

a H2O + b CO2 + c N2

C balance:

60.1 + 2×14.8 + 3×13.4 + 4×4.2 = b = 146.7

H balance:

4×60.1 + 6×14.8 + 8×13.4 + 10×4.2 = 2a = 478.4 => a = 239.2

O balance:

2 νO2 = a + 2b = 239.2 + 2×146.7

N2 balance:

7.5 + 3.76 νO2 = c = 1008.8

20% excess air: νO2 = 1.2×266.3 = 319.56

so now more O2 and N2

Extra oxygen: d = 319.56 - 266.3 = 53.26,

c = 7.5 + 3.76×319.56 = 1209

239.2 H2O + 146.7 CO2 + 1209 N2 + 53.26 O2

Products: 14.5

=> νO2 = 266.3

A Pennsylvania coal contains 74.2% C, 5.1% H, 6.7% O, (dry basis, mass percent) plus ash and small percentages of N and S. This coal is fed into a gasifier along with oxygen and steam, as shown in Fig. P14.5. The exiting product gas composition is measured on a mole basis to: 39.9% CO, 30.8% H2, 11.4% CO2, 16.4% H2O plus small percentages of CH4, N2, and H2S. How many kilograms of coal are required to produce 100 kmol of product gas? How much oxygen and steam are required? Number of kmol per 100 kg coal: C : n = 74.2/12.01 = 6.178

H2: n = 5.1/2.016 = 2.530

O2: n = 6.7/31.999 = 0.209 (6.178 C + 2.53 H2 + 0.209 O2)x + y H2O + z O2 in and 39.9 CO + 30.8 H2 + 11.4 CO2 + 16.4 H2O out in 100 kmol of mix out →

C balance:

6.178 x = 39.9 + 11.4

H2 balance:

2.53×8.304 + y = 30.8 + 16.4

O2 balance: 0.209 × 8.304 +

→ y = 26.191

26.191 39.9 16.4 +z= + 11.4 + → z = 24.719 2 2 2

Therefore, for 100 kmol of mixture out require:

x = 8.304

830.4 kg of coal 26.191 kmol of steam 24.719 kmol of oxygen

14-5 14.6

Repeat Problem 14.5 for a certain Utah coal that contains, according to the coal analysis, 68.2% C, 4.8% H, 15.7% O on a mass basis. The exiting product gas contains 30.9% CO, 26.7% H2, 15.9% CO2 and 25.7% H2O on a mole basis. Number of kmol per 100 kg coal: C : 68.2/12.01 = 5.679

H2: 4.8/2.016 = 2.381

O2: 15.7/32.00 = 0.491 (5.679 C + 2.381 H2 + 0.491 O2)x + y H2O + z O2 in 30.9 CO + 26.7 H2 + 15.9 CO2 + 25.7 H2O out in 100 kmol of mix out →

C:

5.679x = 30.9 + 15.9

x = 8.241

H 2:

2.381 × 8.241 + y = 26.7 + 25.7

O 2:

0.491 × 8.241 +

→ y = 32.778

32.778 30.9 25.7 +z= + 15.9 + 2 2 2

→

z = 23.765

Therefore, for 100 kmol of mixture out, require: 824.1 kg of coal 32.778 kmol of steam 23.765 kmol of oxygen 14.7

A sample of pine bark has the following ultimate analysis on a dry basis, percent by mass: 5.6% H, 53.4% C, 0.1% S, 0.1% N, 37.9% O and 2.9% ash. This bark will be used as a fuel by burning it with 100% theoretical air in a furnace. Determine the air–fuel ratio on a mass basis. Converting the Bark Analysis from a mass basis: Substance

S

H2

C

O2

N2

c/M =

0.1/32

5.6/2

53.4/12

37.9/32

0.1/28

kmol / 100 kg coal

0.003

2.80

4.45

1.184

0.004

Product

SO2

H 2O

CO2

oxygen required

0.003

1.40

4.45

--

--

Combustion requires: 0.003 + 1.40 + 4.45 = 5.853 kmol O2 there is in the bark 1.184 kmol O2 so the net from air is 4.669 kmol O2 AF = (4.669 + 4.669 × 3.76) ×

28.97 kg air = 6.44 100 kg bark

14-6 14.8

Liquid propane is burned with dry air. A volumetric analysis of the products of combustion yields the following volume percent composition on a dry basis: 8.6% CO2, 0.6% CO, 7.2% O2 and 83.6% N2. Determine the percent of theoretical air used in this combustion process. a C3H8 + b O2 + c N2 → 8.6 CO2 + 0.6 CO + d H2O + 7.2 O2 + 83.6 N2 C balance:

3a = 8.6 + 0.6 = 9.2 ⇒ a = 3.067

H2 balance:

4a = d ⇒ d = 12.267

N2 balance:

c = 83.6

O2 balance:

b = 8.6 +

Air-Fuel ratio =

0.6 12.267 + + 7.2 = 22.234 2 2

22.234 + 83.6 = 34.51 3.067

Theoretical: C3H8 + 5 O2 + 18.8 N2 → 3 CO2 + 4 H2O + 18.8 N2 ⇒ theo. A-F ratio = % theoretical air =

14.9

5 + 18.8 = 23.8 1

34.51 × 100 % = 145 % 23.8

A fuel, CxHy, is burned with dry air and the product composition is measured on a dry basis to be: 9.6% CO2, 7.3% O2 and 83.1% N2. Find the fuel composition (x/y) and the percent theoretical air used. νFuCxHy + νO O2 + 3.76νO N2 → 9.6CO2 + 7.3O2 + 83.1N2 + νH OH2O 2

2

2

N2 balance: 3.76νO = 83.1 ⇒ νO = 22.101 2

2

1

O2 balance: νO = 9.6 + 7.3 + 2 νH

2O

2

H balance: νFu y = 2 νH

2O

⇒ νH

2O

= 10.402

= 20.804

C balance: νFu x = 9.6 Fuel composition ratio = x/y = 9.6/20.804 = 0.461 Theoretical air =

νO AC 2

=

22.101

νO stoich 9.6 + 1 × 29.804 2 4

= 1.493

14-7 14.10 Many coals from the western United States have a high moisture content. Consider the following sample of Wyoming coal, for which the ultimate analysis on an as-received basis is, by mass: Component Moisture H C S N O Ash % mass 28.9 3.5 48.6 0.5 0.7 12.0 5.8 This coal is burned in the steam generator of a large power plant with 150% theoretical air. Determine the air–fuel ratio on a mass basis. Converting from mass analysis: Substance

S

H2

C

O2

N2

c/M =

0.5/32

3.5/2

4.86/12

12/32

0.7/28

kmol / 100 kg coal

0.0156

1.75

4.05

0.375

0.025

Product

SO2

H 2O

CO2

oxygen required

0.0156

0.875

4.05

--

--

Combustion requires then oxygen as: 0.0156 + 0.875 + 4.05 = 4.9406 The coal does include 0.375 O2 so only 4.5656 O2 from air/100 kg coal AF = 1.5 × (4.5656 + 4.5656 × 3.76) × 28.97/100 = 9.444 kg air/kg coal 14.11 Pentane is burned with 120% theoretical air in a constant pressure process at 100 kPa. The products are cooled to ambient temperature, 20°C. How much mass of water is condensed per kilogram of fuel? Repeat the answer, assuming that the air used in the combustion has a relative humidity of 90%. C5H12 + 1.2 × 8 (O2 + 3.76 N2) → 5 CO2 + 6 H2O + 0.96 O2 + 36.1 N2 Products cooled to 20oC, 100 kPa, so for H2O at 20°C: Pg = 2.339 kPa yH2O MAX = Pg/P = Therefore,

nH2O MAX 2.339 = => n H2O MAX = 1.007 < n H2O 100 nH2O MAX + 42.06

nH2O VAP = 1.007,

mH2O LIQ =

nH2O LIQ = 6 - 1.007 = 4.993

4.993 × 18.015 = 1.247 kg/kg fuel 72.151

Pv1 = 0.9 × 2.339 = 2.105 kPa nH2O IN = 0.013375 ×

=>

w1 = 0.622 ×

2.105 = 0.013375 97.895

28.97 × (9.6 + 36.1) = 0.983 kmol 18.015

nH2O OUT = 0.983 + 6 = 6.983 => nH2O LIQ = 6.983 - 1.007 = 5.976 kmol nH2O LIQ =

5.976 × 18.015 = 1.492 kg/kg fuel 72.151

14-8 14.12 The coal gasifier in an integrated gasification combined cycle (IGCC) power plant produces a gas mixture with the following volumetric percent composition: Product

CH4

% vol.

0.3

H

2 29.6

CO

CO2

41.0

10.0

N

2 0.8

H 2O

H 2S

NH3

17.0

1.1

0.2

This gas is cooled to 40°C, 3 MPa, and the H2S and NH are removed in 3 water scrubbers. Assuming that the resulting mixture, which is sent to the combustors, is saturated with water, determine the mixture composition and the theoretical air–fuel ratio in the combustors. CH4 H2 CO CO2 N2 n 0.3 yH2O =

29.6 nV

, nV+81.7

where

41.0

10.0

0.8

81.7

nV = number of moles of water vapor

Cool to 40°C PG = 7.384, P = 3000 kPa nV 7.384 = yH2O MAX = 3000 nV+81.7 a) Mixture composition: CH4 H2 0.3 kmol

29.6

→

nV = 0.2016

CO

CO2

N2

H2O(v)

41.0

10.0

0.8

0.2016

81.9016 kmol (from 100 kmol of the original gas mixture) 0.3 CH4 + 0.6 O2 → 0.3 CO2 + 0.6 H2O 29.6 H2 + 14.8 O2 → 29.6 H2O 41 CO + 20.5 O2 → 41 CO2 ⇒ Number of moles of O2 = 0.6 + 14.8 + 20.5 = 35.9 Number of moles of air = 35.9 + 3.76 × 35.9 (N2) Air Fuel ratio = 28.97(35.9 + 3.76(35.9)) 0.3(16) + 29.6(2) + 41(28) + 10(44) + 0.8(28) + 0.2016(18) = 2.95 kg air/kg fuel

14-9 14.13 The hot exhaust gas from an internal combustion engine is analyzed and found to have the following percent composition on a volumetric basis at the engine exhaust manifold. 10% CO2, 2% CO, 13% H2O, 3% O2 and 72% N2. This gas is fed to an exhaust gas reactor and mixed with a certain amount of air to eliminate the carbon monoxide, as shown in Fig. P14.13. It has been determined that a mole fraction of 10% oxygen in the mixture at state 3 will ensure that no CO remains. What must the ratio of flows be entering the reactor? 1 Exh. gas

Exhaust gas at state 1: CO2 10 %, H2O 13%,

Reactor

CO 2%, O2 3%, N2 72% CO = 0 %, O2 = 10 %

Exhaust gas at state 3:

3 gas

2 Air

⇒ 0.02 CO + x O2 + 3.76x N2 → 0.02 CO2 + (x-0.01) O2 + 3.76x N2 At 3: CO2 = 0.10 + 0.02 = 0.12,

H2O = 0.13

O2 = (x-0.01) + 0.03 + x + 0.02

N2 = 0.72 + 3.76x

or nTOT = 0.12 + 0.13 + x + 0.02 + 0.72 + 3.76x = 0.99 + 4.76x yO2 = 0.16 = or

x+0.02 0.99+4.76x

air 2 Exh. Gas 1

=

→

x = 0.151

4.76x kmol air = 0.718 1 kmol Exh. gas

14.14 Methanol, CH3OH, is burned with 200% theoretical air in an engine and the products are brought to 100 kPa, 30°C. How much water is condensed per kilogram of fuel? CH3OH + νO {O2 + 3.76 N2}→ CO2 + 2 H2O + 3.76 νO N2 2

Stoichiometric νO

2

2S

= 1.5 ⇒ νO

2 AC

=3

Actual products: CO2 + 2 H2O + 1.5 O2 +11.28 N2 Psat(30°C) = 4.246 kPa ⇒ yH2O = 0.04246 =

νH2O 1 + νH2O + 1.5 + 11.28

⇒ νH2O = 0.611 ⇒ ∆ν H2O cond = 2 - 0.611 = 1.389 MFu = 32.042

∆MH2O 1.389 × 18 kg H2O = = 0.781 kg fuel MFu 32.042

out

14-10 14.15 The output gas mixture of a certain air–blown coal gasifier has the composition of producer gas as listed in Table 14.2. Consider the combustion of this gas with 120% theoretical air at 100 kPa pressure. Determine the dew point of the products and find how many kilograms of water will be condensed per kilogram of fuel if the products are cooled 10°C below the dew-point temperature. {3 CH4 + 14 H2 + 50.9 N2 + 0.6 O2 + 27 CO + 4.5 CO2} + 31.1 O2 + 116.9 N2 → 34.5 CO2 + 20 H2O + 5.2 O2 + 167.8 N2 Products: yH2O = yH2O MAX = PG/100 = ⇒ PG = 8.79 kPa →

20 34.5 + 20 + 5.2 + 167.8

TDEW PT = 43.2°C

At T = 33.2°C, P G = 5.13 kPa yH2O = mH2O LIQ =

nH2O 5.13 = 100 nH2O+34.5+5.2+167.8

→

nH2O = 11.22

8.78(18) 3(16) +14(2) +50.9(28) +0.6(32) +27(28) +4.5(44)

= 0.0639 kg/kg fuel

14.16 Pentene, C5H10 is burned with pure oxygen in an SSSF process. The products at one point are brought to 700 K and used in a heat exchanger, where they are cooled to 25°C. Find the specific heat transfer in the heat exchanger. C5H10 + νO O2 → 5 CO2 + 5 H2O ⇒ νO = 7.5 2

2

. . 5 nF hCO + 5 nF hH 2

. . -° + Q = 5 n F hf CO O

2

2

. + (5 - x) nF h°liq H

2O

Find x: yH

2O max

=

Pg(25°) Ptot

= 0.0313 =

. + (x) nF h°vap H

2O

x ⇒ x = 0.1614 5+x

Out of the 5 H2O only 0.1614 still vapor. . Q -° -° -° -° ⋅n = -5 ∆hCO2,700 + (5-x)(hf liq - hf vap - ∆h700) + x(hf vap - hf vap - ∆h700) F = -5(17761) + 4.84(-44011-14184) - 0.16(14184) = -372738 kJ/kmol Fu

14-11 14.17 Butane gas and 200% theoretical air, both at 25C, enter a SSSF combustor. The products of combustion exits at 1000 K. Calculate the heat transfer from the combustor per kmol of butane burned. C4H10 + (1/φ) νO2(O2 + 3.76 N2) → a CO2 + b H2O + c N2 + d O2 First we need to find the stoichiometric air ( φ = 1, d = 0 ) C balance: 4 = a,

H balance: 10 = 2b =>

b=5

O balance: 2νO2 = 2a + b = 8 + 5 = 13 => νO2 = 6.5 Now we can do the actual air: (1/φ) = 2 => νO2 = 2 × 6.5 = 13 N balance: c = 3.76 νO2 = 48.88, O balance: d = 13 - 6.5 = 6.5 o

o

o

q = HR - HP = HR - HP - ∆HP = M(-HRP) - ∆HP o

Table 14.3: HRP = -45714, The rest of the values are from Table A.8 − − − − ∆h CO2 = 33397, ∆h N2 = 21463, ∆h O2 = 22703, ∆h H2O = 26000 kJ/kmol q = 58.124(45714) - (4 × 33397 + 5 × 26000 + 48.88 × 21463 + 6.5 × 22703) = 2 657 080 - 1 460 269 = 1 196 811 kJ/kmol butane 14.18 Liquid pentane is burned with dry air and the products are measured on a dry basis as: 10.1% CO2, 0.2% CO, 5.9% O2 remainder N2. Find the enthalpy of formation for the fuel and the actual equivalence ratio. νFuC5H12 + νO O2 + 3.76 νO N2 → 2

2

x H2O + 10.1 CO2 + 0.2 CO + 5.9 O2 + 83.8 N2 Balance of C: 5 νFu = 10.1 + 0.2 ⇒ νFu = 2.06 Balance of H: 12 νFu = 2 x ⇒ x = 6 νFu = 12.36 Balance of O: 2 νO = x + 20.2 + 0.2 + 2 × 5.9 ⇒ νO = 22.28 2

2

Balance of N: 2 × 3.76 νO = 83.8 × 2 ⇒ νO = 22.287 ⇒ OK 2

2

νO for 1 kmol fuel = 10.816 2

φ = 1, C5H12 + 8 O2 + 8 × 3.76 N2 → 6 H2O + 5 CO2 + 30.08 N2 ° ° ° -° -° -° HRP = HP - HR = 6 hf H2O + 5 hf CO2 - hf fuel °

14.3: HRP = 44983 × 72.151

⇒

φ = νO2 stoich/νO2 AC = 8/10.816 = 0.74

-° hf fuel = -172 998 kJ/kmol

14-12 14.19 A rigid vessel initially contains 2 kmol of carbon and 2 kmol of oxygen at 25°C, 200 kPa. Combustion occurs, and the resulting products consist of 1 kmol of carbon dioxide, 1 kmol of carbon monoxide, and excess oxygen at a temperature of 1000 K. Determine the final pressure in the vessel and the heat transfer from the vessel during the process. 1

2 C + 2 O2 → 1 CO2 + 1 CO + 2 O2 Process P 2 = P1 ×

V = constant, n2T2 n1T1

= 200 ×

C: solid, n1(GAS) = 2, n2(GAS) = 2.5

2.5 × 1000 = 838.4 kPa 2 × 298.2

H1 = 0 H2 = 1(-393522 + 33397) + 1(-110527 + 21686) + (1/2)(0 + 22703) = -437615 kJ Q = (U2 - U1) = (H2 - H1) - n2RT2 + n1RT1 1 2 = (-437615 - 0) - 8.3145(2.5 × 1000 - 2 × 298.2) = -453442 kJ 14.20 In a test of rocket propellant performance, liquid hydrazine (N2H4) at 100 kPa, 25°C, and oxygen gas at 100 kPa, 25°C, are fed to a combustion chamber in the ratio of 0.5 kg O2/kg N2H4. The heat transfer from the chamber to the surroundings is estimated to be 100 kJ/kg N2H4. Determine the temperature of the products exiting the chamber. Assume that only H2O, H2, and N2 are present. The enthalpy of formation of liquid hydrazine is +50417 kJ/kmol. 1 o

Liq. N2H4: 100 kPa, 25 C Gas O2: 100 kPa, 25oC

Products

2

. . . . mO2/mN2H4 = 0.5 = 32nO2/32nN2H4 Energy Eq.:

3

Comb. Chamber

and

. . Q/mN2H4 = -100 kJ/kg

QCV = HP - HR = -100 × 32.045 = -3205 kJ/kmol fuel

1

1 N2H4 + 2 O2 → H2O + H2 + N2 1

HR = 1(50417) + 2(0) = 50417 kJ HP = -241826 + ∆hH2O + ∆hH2 + ∆hN2 ∆HP = ∆hH2O + ∆hH2 + ∆hN2 = 241826 + 50417 -3205 = 289038 Table A.8 : 2800 K ∆HP = 282141 Interpolate to get

3000 K ∆HP = 307988

TP = 2854 K

14-13 14.21 Repeat the previous problem, but assume that saturated-liquid oxygen at 90 K is used instead of 25°C oxygen gas in the combustion process. Use the generalized charts to determine the properties of liquid oxygen. Problem same as 14.20, except oxygen enters at 2 as saturated liquid at 90 K. . . . . . . mO2/mN2H4 = 0.5 = 32nO2/32nN2H4 and Q/mN2H4 = -100 kJ/kg Energy Eq.:

QCV = HP - HR = -100 × 32.045 = -3205 kJ/kmol fuel 1

1 N2H4 + 2 O2 → H2O + H2 + N2

Reaction equation:

At 90 K, Tr2 = 90/154.6 = 0.582 ⇒

~ ∆hf = 5.2

-* Figure D.2, (h - h) = 8.3145 × 154.6 × 5.2 = 6684 kJ/kmol ∆hAT 2 = -6684 + 0.922 × 32(90 - 298.15) = -12825 kJ/kmol 1

HR = 50417 + 2(0 - 12825) = 44005 kJ, 1st law: ∆hP = ∆hH

2O

HP° = -241826

+ ∆hH + ∆hN = Qcv + HR - H°P = 282626 2

2

From Table A.8, ∆HP 2800K = 282141, ∆HP 3000K = 307988 Therefore, T P = 2804 K 14.22 The combustion of heptane C7H16 takes place in a SSSF burner where fuel and air are added as gases at P , T . The mixture has 125% theoretical air and the o o products are going through a heat exchanger where they are cooled to 600 K.Find the heat transfer from the heat exchanger per kmol of heptane burned. The reaction equation for stiochiometric ratio is: C7H16 + vO2 (O2 + 3.76 N2)

=> 7CO2 + 8 H2O + vO2 × 3.76 N2

So the balance (C and H was done in equation) of oxygen gives vO2 = 7 + 4 = 11, and actual one is 11×1.25 = 13.75. Now the actual reaction equation is: C7H16 + 13.75 O2 + 51.7 N2

=> 7CO2 + 8 H2O + 51.7 N2+ 2.75 O2

To find the heat transfer take control volume as combustion chamber and heat exchanger Q = HPo + ∆HP - HRo = HRPo + ∆HP Q = 100.205(-44922) + 7(12906) + 8(10499) + 51.7(8894) + 2.75(9245) HR + Q = HP =>

= -4 501 409 + 90342 + 83922 + 459 819.8 + 25423.75 = - 3 841 831 kJ/kmol fuel

14-14 14.23 Ethene, C2H4, and propane, C3H8, in a 1;1 mole ratio as gases are burned with 120% theoretical air in a gas turbine. Fuel is added at 25°C, 1 MPa and the air comes from the atmosphere, 25°C, 100 kPa through a compressor to 1 MPa and mixed with the fuel. The turbine work is such that the exit temperature is 800 K with an exit pressure of 100 kPa. Find the mixture temperature before combustion, and also the work, assuming an adiabatic turbine. φ=1 C2H4 + C3H8 + 8 O2 + 30.08 N2 → 5 CO2 + 6 H2O + 30.08 N2 φ = 1.2 C2H4 + C3H8 + 9.6 O2 + 36.096 N2 → 5 CO2 + 6 H2O + 1.6 O2 + 36.096 N2 45.696 kmol air per 2 kmol fuel C.V. Compressor (air flow) Energy Eq.:

wc = h2 - h1

Entropy Eq.:

s2 = s1

⇒

Pr = Pr × P2/P1 = 13.573 ⇒ T2 air = 570.8 K 2

1

wc = 576.44 - 298.34 = 278.1 kJ/kg = 8056.6 kJ/kmol air C.V. Mixing Chanber (no change in composition) . . . nairhair in + nFu1h1 in + nFu2h2 in = (SAME)exit CP F1 + CP F2Texit - T0 = 45.696 CP airT2 air - Texit C2H4: CP F1 = 43.43, C3H8: CP F2 = 74.06, CP air = 29.07 45.696CP airT2 + CP F1 + CP F2T0 Texit = = 548.7 K CP F1 + CP F2 + 45.696 CP air 6

Dew Point Products: yH

2O

PH

2O

= 5 + 6 + 1.6 + 36.096 = 0.1232

= yH OPtot = 123.2 kPa ⇒ Tdew = 105.5°C 2

C.V. Turb. + combustor + mixer + compressor (no Q) wnet

= Hin - Hout = HR - HP 800

(800°K out so no liquid H2O)

-° -° = hf C2H4 + hf C3H8 - 5 hCO - 6 hH O - 1.6 hO - 36.096 hN 2

2

kJ

= 2 576 541 2 kmol Fu wT= wnet + wcomp = 2 944 695

kJ 2 kmol Fu

2

2

14-15 14.24 One alternative to using petroleum or natural gas as fuels is ethanol (C 2H5OH), which is commonly produced from grain by fermentation. Consider a combustion process in which liquid ethanol is burned with 120% theoretical air in an SSSF process. The reactants enter the combustion chamber at 25°C, and the products exit at 60°C, 100 kPa. Calculate the heat transfer per kilomole of ethanol, using the enthalpy of formation of ethanol gas plus the generalized charts. C2H5OH + 1.2 × 3 (O2 + 3.76 N2) → 2CO2 + 3H2O + 0.6O2 + 13.54N2 -0 Fuel: hf = -235000 kJ/kmol for IG yH2O MIX =

Products at 60°C, 100 kPa

nV MAX 19.94 = => n V MAX = 4.0 > 3 ⇒ No liq. 100 nV MAX+2+0.6+13.54 Tr = 298.2 / 513.9 = 0.58

⇒

∆hf = 5.23

from Figure D.2

-* (h - h)LIQ = 8.3145 × 513.9 × 5.23 = 22347 kJ/kmol HR = 1(-235000 - 22347) + 0 + 0 = -257 347 kJ HP = 2(-393522 + 1327) + 3(-241826 + 1178) + 0.6(0+1032) + 13.54(0+1020) = -1 491 904 kJ QCV = HP - HR = -1 234 557 kJ 14.25 Another alternative to using petroleum or natural gas as fuels is methanol, CH3OH, which can be produced from coal. Both methanol and ethanol have been used in automotive engines. Repeat the previous problem using liquid methanol as the fuel instead of ethanol. CH3OH + 1.2 × 1.5 (O2 + 3.76 N2) → 1 CO2 + 2 H2O + 0.3 O2 + 6.77 N2 React at 25 oC, Prod at 60 oC = 333.2 K, 100 kPa yH2O MAX =

nV MAX 19.94 = => n V MAX = 2.0 > 2 ⇒ No liq. 100 nV MAX+1+0.3+6.77

-o CH3OH: hf = -201 300 kJ, TC = 512.6 K ~ Tr = 298.15 / 512.6 = 0.582 ⇒ ∆hf = 5.22 Figure D.2 - (h*-h)LIQ = 8.3145 × 512.6 × 5.22 = 22 248 HR = 1 hLIQ = -201 300 - 22 248 = -223 548 kJ HP = 1(-393 522 + 1327) + 2(-241 826 + 1178) + 0.3(1032) + 6.77(1020) = -866 276 kJ Q = HP - HR = -642 728 kJ

14-16 14.26 Another alternative fuel to be seriously considered is hydrogen. It can be produced from water by various techniques that are under extensive study. Its biggest problem at the present time is cost, storage, and safety. Repeat Problem 14.24 using hydrogen gas as the fuel instead of ethanol. H2 + 1.2 × 0.5 O2 + 1.2 × 3.76 × 0.5 N2 → 1 H2O + 0.1 O2 + 2.256 N2 Products at 60°C, 100 kPa yH2O MAX =

nV MAX 19.94 = 100 nV MAX+0.1+2.256

Solving, nV MAX = 0.587 < 1 => nV = 0.587, nLIQ = 0.413 HR = 0 + 0 + 0 = 0 HP = 0.413(-285830 + 18.015(251.1 - 104)) + 0.587(-241826 + 1178) + 0.1(0 + 1032) + 2.256(0 + 1020) = -255 816 kJ Q CV = HP - HR = -255 816 kJ 14.27 Hydrogen peroxide, H2O2, enters a gas generator at 25°C, 500 kPa at the rate of 0.1 kg/s and is decomposed to steam and oxygen exiting at 800 K, 500 kPa. The resulting mixture is expanded through a turbine to atmospheric pressure, 100 kPa, as shown in Fig. P14.27. Determine the power output of the turbine, and the heat transfer rate in the gas generator. The enthalpy of formation of liquid H2O2 is − 187 583 kJ/kmol. . mH2O2 1 0.1 . H2O2 → H2O + O2 nH2O2 = = = 0.00294 kmol/s 2 34.015 M . . nMIX = nH2O2× 1.5 = 0.00441 kmol/s 2 1 CP0 MIX = × 1.872 × 18.015 + × 0.922 × 31.999 = 32.317 3

3

CV0 MIX = 32.317 - 8.3145 = 24.0 => kMIX = 32.317/24.0 = 1.3464 CV: turbine. Assume reversible → s3 = s2 P3

k-1 k

100 0.2573 T3 = T2( ) = 800( ) = 528.8 K 500 P2 w = CP0(T2 - T3) = 32.317(800 - 528.8) = 8765 kJ/kmol . WCV = 0.00441 × 8765 = 38.66 kW CV: Gas Generator . H1 = 0.00294(-187 583 + 0) = -551.49 . H2 = 0.00294(-241 826 + 18002) + 0.00147(0 + 15836) = -634.76 . . . QCV = H2 - H1 = -634.76 + 551.49 = -83.27 kW

14-17 14.28 In a new high-efficiency furnace, natural gas, assumed to be 90% methane and 10% ethane (by volume) and 110% theoretical air each enter at 25°C, 100 kPa, and the products (assumed to be 100% gaseous) exit the furnace at 40°C, 100 kPa. What is the heat transfer for this process? Compare this to an older furnace where the products exit at 250°C, 100 kPa. 0.90CH4 + 0.10C2 H6

Prod. 40 oC 100 kPa

Furnace o

25 C 110% Air

0.9 CH4 + 0.1 C2H6 + 1.1 × 2.15 O2 + 3.76 × 2.365 N2 → 1.1 CO2 + 2.1 H2O + 0.215 O2 + 8.892 N2 HR = 0.9(-74873) + 0.1(-84740) = -75860 kJ HP = 1.1(-393 522 + 562) + 2.1(-241 826 + 504) + 0.215(441) + 8.892(437) = -935052 kJ assuming all gas QCV = HP - HR = -859 192 kJ/kmol fuel b) TP = 250 oC HP = 1.1(-393 522 + 9346) + 2.1(-241 826 + 7740) + 0.215(6808) + 8.892(6597) = -854 050 kJ QCV = HP - HR = -778 190 kJ/kmol fuel

14.29 Repeat the previous problem, but take into account the actual phase behavior of the products exiting the furnace. Same as 14.28, except check products for saturation at 40 oC, 100 kPa yV MAX =

nV MAX 7.384 = => 100 nV MAX+10.207

nV = 0.814,

Solving, nV MAX = 0.814

nLIQ = 2.1 - 0.814 = 1.286

HLIQ = 1.286[-285 830 + 18.015(167.6 - 104.9)] = -366125 kJ HGAS = 1.1(-393 522 + 562) + 0.814(-241826 + 504 ) + 0.215(441) + 8.892(437) = -624711 kJ QCV = HP - HR = (-366 125 - 624 711) + 75860 = -914 976 kJ/kmol fuel b) no liquid, answer the same as 14.28.

14-18 14.30 Methane, CH4, is burned in an SSSF process with two different oxidizers: Case Pure oxygen, O2 and case B: A mixture of O2 + x Ar. The reactants are

A:

supplied at T0, P0 and the products should for both cases be at 1800 K. Find the required equivalence ratio in case (A) and the amount of Argon, x, for a stoichiometric ratio in case (B). a) Stoichiometric has ν = 2, actual has: CH4 + νO2 → CO2 + 2H2O + (ν - 2)O2 °

°

°

HR = HP + ∆HP 1800 => ∆HP 1800= - HRP = 50010 × 16.04 = 802 160 ∆h = 79432, ∆ hH O= 62693, ∆ hO = 51674 CO 2

2

2

∆HP 1800= 101470 + ν 51674 = 802160 ⇒

ν = 13.56,

φ = 6.78

CH4 + 2 O2 + 2x Ar → CO2 + 2H2O + 2x Ar

b)

∆HP 1800= 79432 + 2 × 62693 + 2x × 0.5203 × 39.948(1800 - 298) = 204818 + x62438 ⇒

802160 = 204818 + x 62438

x = 9.567

14.31 Butane gas at 25°C is mixed with 150% theoretical air at 600 K and is burned in an adiabatic SSSF combustor. What is the temperature of the products exiting the combustor? o

25 C GAS 150% Air 600 K

C4 H10

Adiab. Comb.

Prod. at Tp

QCV = 0 C4H10 + 1.5×6.5 (O2 + 3.76 N2 )→ 4 CO2 + 5 H2O + 3.25 O2 + 36.66 N2 Reactants: hAIR = 9.75(9245) + 36.66(8894) = 416193 kJ ; -o hC4H10 = hf IG = -126 200 kJ

=>

H R = +289 993 kJ

-* -* -* -* HP = 4(-393522 + ∆hCO2) + 5(-241826 + ∆hH2O) + 3.25 ∆hO2 + 36.66 ∆hN2 Energy Eq.:

HP - HR = 0

-* -* -* -* 4 ∆hCO2 + 5 ∆hH2O + 3.25 ∆hO2 + 36.66 ∆hN2 = 3 073 211 Trial and Error:

LHS2000 K = 2 980 000,

Linear interpolation to match RHS

LHS2200 K = 3 369 866

=> TP = 2048 K

14-19 14.32 In a rocket, hydrogen is burned with air, both reactants supplied as gases at Po, To. The combustion is adiabatic and the mixture is stoichiometeric (100% theoretical air). Find the products dew point and the adiabatic flame temperature (~2500 K). The reaction equation is: H2 + vO2 (O2 + 3.76 N2) => H2O + 3.76 vO2 N2 The balance of hydrogen is done, now for oxygen we need v O2 = 0.5 and thus we have 1.88 for nitrogen. yv = 1/(1+1.88) = 0.3472 => Pv = 101.325 × 0.3472 = 35.18 kPa = Pg Table B.1.2: Tdew = 72.6 C. HR = HP => 0 = -241826 + ∆hwater + 1.88 ∆hnitrogen Find now from table A.13 the two enthalpy terms At 2400 K : ∆HP = 93741 + 1.88 × 70640 = 226544 At 2600 K : ∆HP = 104520 + 1.88 × 77963 = 251090.4 Then interpolate to hit 241826 to give

T = 2524.5 K

14.33 Liquid butane at 25°C is mixed with 150% theoretical air at 600 K and is burned in an adiabatic SSSF combustor. Use the generalized charts for the liquid fuel and find the temperature of the products exiting the combustor. o TC = 425.2 K 25 C LIQ. C4 H10 Prod. Adiab. TR = 0.701 150% Air Comb. at Tp 600 K QCV = 0 -o hC4H10 = hf IG + (hLIQ - h*)

see Fig. D.2

= -126200 + (-4.85 × 8.3145 × 425.2) = -143 346 kJ C4H10 + 1.5 × 6.5 O2 + 3.76 × 9.75 N2 → 4 CO2 + 5 H2O + 3.25 O2 + 36.66 N2 hAIR = 9.75(9245) + 36.66(8894) = 416 193 kJ ⇒

HR = 416 193- 143 346 = +272 847 kJ

-* -* -* -* HP = 4(-393522 + ∆hCO2) + 5(-241826 + ∆hH2O) + 3.25 ∆hO2 + 36.66 ∆hN2 Energy Eq.:

HP - HR = 0

-* -* -* -* 4 ∆hCO2 + 5 ∆hH2O + 3.25 ∆hO2 + 36.66 ∆hN2 = 3 056 065 Trial and Error:

LHS2000 K = 2 980 000,

Linear interpolation to match RHS

LHS2200 K = 3 369 866

=> TP = 2039 K

14-20 14.34 A stoichiometric mixture of benzene, C6H6, and air is mixed from the reactants flowing at 25°C, 100 kPa. Find the adiabatic flame temperature. What is the error if constant specific heat at T0 for the products from Table A.5 are used? C6H6 + νO O2 + 3.76 νO N2 → 6CO2 + 3H2O + 3.76 νO N2 2

2

2

νO = 6 + 3/2 = 7.5 ⇒ νN = 28.2 2

2

°

°

°

HP=HP + ∆HP = HR = HR ⇒ ∆HP = -HRP = 40576 × 78.114 = 3169554 ∆HP 2600K= 3280600,

∆HP 2400K= 2968000

⇒

TAD= 2529 K

∑νiC-Pi = 6 × 37.048 + 3 × 33.73 + 28.2 × 29.178 = 1146.3 ∆T = ∆HP/∑νiCPi = 2765

⇒

TAD= 3063 K, 21% high

14.35 Liquid n-butane at T0, is sprayed into a gas turbine with primary air flowing at 1.0 MPa, 400 K in a stoichiometric ratio. After complete combustion, the products are at the adiabatic flame temperature, which is too high, so secondary air at 1.0 MPa, 400 K is added, with the resulting mixture being at 1400 K. Show that Tad > 1400 K and find the ratio of secondary to primary air flow. C.V. Combustion Chamber. C4H10 + 6.5 O2 + 6.5 × 3.76 N2 → 5 H2O + 4 CO2 + 24.44 N2

Primary Air

Fuel Combustion Chamber

Air

Secondary

To turbine Mixing

TAD

1400 K

Energy Eq.: Hair + Hfuel = HR = HP °

°

HP + ∆HP = HR + ∆HR

⇒

°

°

∆HP = HR + ∆HR - HP

∆HP = 45344 × 58.124 + 6.5(3.76 × 2971 + 3027) = 2727861 ∆HP 1400 = 5.43491 + 4.55895 + 24.44 × 34936 = 1294871 < ∆HP Try TAD > 1400: ∆HP = 2658263 @2400 K, C.V. Mixing Ch.

∆HP = 2940312 @2600 K

Air Second: νO2 sO2 + 3.76 N2

∆ΗP + νO2 second ∆Hair = ∆HP 1400 + νO2 second ∆Hair 1400 ⇒ νO2 second=

∆HP - ∆HP 1400 ∆Hair 1400 - ∆Hair 400

=

ratio = νO2 sec/νO2 prim = 9.3/6.5 = 1.43

1432990 = 9.3 168317 - 14198

14-21 14.36 Consider the gas mixture fed to the combustors in the integrated gasification combined cycle power plant, as described in Problem 14.12. If the adiabatic flame temperature should be limited to 1500 K, what percent theoretical air should be used in the combustors? Product

CH4

% vol.

0.3

H

2 29.6

CO

CO2

41.0

10.0

N

2 0.8

H 2O

H 2S

NH3

17.0

1.1

0.2

Mixture may be saturated with water so the gases are ( H 2S and NH3 out) CH4 H2 CO CO2 N2 n 0.3

29.6

41.0

10.0

0.8

81.7

yV MAX = 7.384/3000 = nV/(nV + 81.7) Solving, nV = 0.2 kmol, rest condensed

{0.3 CH4 + 29.6 H2 + 41.0 CO + 10.0 CO2 + 0.8 N2 + 0.2 H2O + 35.9x O2 + 3.76 × 35.9x N2}→ 51.3 CO2 + 30.4 H2O + 35.9(x - 1) O2 + (135.0x + 0.8) N2 For the fuel gas mixture, nCP0 MIX = 0.3 × 16.04 × 2.2537 + 29.6 × 2.016 × 14.2091 + 41.0 × 28.01 × 1.0413 + 10.0 × 44.01 × 0.8418 + 0.8 × 28.013 × 1.0416 + 0.2 × 18.015 × 1.8723 = 2455.157 -0 nhf MIX = 0.3(-74873) + 29.6(0) + 41.0(-110527) + 10.0(-393522) + 0.8(0) + 0.2(-241826) = -8537654 kJ At 40°C, for the fuel mixture: HMIX = -8537654 + 2455.157(40 - 25) = -8500827 kJ Assume air enters at 25°C: hAIR = 0 Products at 1500 K: HP = 51.3(-393522 + 61705) + 30.4(-241826 + 48149) + 35.9(x - 1)(0 + 40600) + (135x + 0.8)(0 + 38405) = -24336806 + 6642215x 1st law: HP = HR = HMIX x=

+24336809 - 8500827 = 2.384 6642215

or

238 % theo. air

14-22 14.37 Acetylene gas at 25°C, 100 kPa is fed to the head of a cutting torch. Calculate the adiabatic flame temperature if the acetylene is burned with a. 100% theoretical air at 25°C. b. 100% theoretical oxygen at 25°C. C2H4 + 2.5 O2 + 2.5 × 3.76 N2 → 2 CO2 + 1 H2O + 9.4 N2

a)

-o HR = hf C2H2 = +226 731 kJ -* -* -* HP = 2(-393 522 + ∆hCO2) + 1(-241 826 + ∆hH2O) + 9.4 ∆hN2 -* -* -* 2 ∆hCO2 + 1 ∆hH2O + 9.4 ∆hN2 = 1 255 601 kJ

QCV = HP - HR = 0 ⇒

Trial and Error A.8: LHS2800 = 1 198 369, Linear interpolation: TPROD = 2909 K

LHS3000 = 1 303 775

C2H2 + 2.5 O2 → 2 CO2 + H2O

b)

HR = +226 731 kJ ; ⇒

-* -* HP = 2(-393 522 + ∆hCO2) + 1(-241 826 + ∆hH2O)

-* -* 2 ∆hCO2 + 1 ∆hH2O = 1 255 601 2 × 343 782 + 302 295 = 989 859

At 6000 K (limit of A.8)

2 × 337 288 + 296 243 = 970 819

At 5900 K or 19040/100 K change

Difference, extrapolating

TPROD ≈ 6000 + 265 742/190.40 ≈ 7400 K 14.38 Ethene, C2H4, burns with 150% theoretical air in an SSSF constant-pressure process with reactants entering at P0, T0. Find the adiabatic flame temperature. C2H4 + 3(O2 + 3.76N2) → 2CO2 + 2H2O + 11.28N2 C2H4 + 4.5(O2 + 3.76N2) → 2CO2 + 2H2O + 1.5 O2 + 16.92N2 yH

2O

= 2/(2 + 2 + 1.5 + 16.92) = 0.0892 => Pv = 9.041 ⇒ TDEW = 43.8°C

° HP = HP + 2∆hCO + 2∆hH 2

° -° HR = hf Fu

2O

+ 1.5∆hO + 16.92∆hN 2

°

2

°

∆HP + HP = HR °

kJ

⇒ ∆HP = -HRP = 28.054 × 47158 = 1322970.5 kmol Fu Initial guess based on N2 from A.8 ∆HP(2000) = 1366982,

T1 = 2100 K

∆HP(1900) = 1278398

=> TAD ≅ 1950 K

14-23 14.39 Solid carbon is burned with stoichiometric air in an SSSF process. The reactants at T0, P0 are heated in a preheater to T2 = 500 K as shown in Fig. P14.39, with the energy given by the product gases before flowing to a second heat exchanger, which they leave at T0. Find the temperature of the products T4, and the heat transfer per kmol of fuel (4 to 5) in the second heat exchanger. Control volume: Total minus last heat exchanger. C + O2 + 3.76N2 → CO2 + 3.76N2 Energy Eq.: ° ° HR = HR = HP = HP + ∆HP = hf CO + ∆hCO + 3.76∆hN 3

hf CO = -393 522, ∆HP

3

2

=381 436, ∆HP

3 2400

2

2

⇒ T3 = Tad.flame = 2461 K Control volume: Total. Then energy equation: ° ° − HR + Q = HP kJ − -° -° Q = HRP = hf CO2 - 0 = -393 522 kmol fuel

= 401 339

3 2500

2

14-24 14.40 A study is to be made using liquid ammonia as the fuel in a gas-turbine engine. Consider the compression and combustion processes of this engine. a. Air enters the compressor at 100 kPa, 25°C, and is compressed to 1600 kPa, where the isentropic compressor efficiency is 87%. Determine the exit temperature and the work input per kilomole. b. Two kilomoles of liquid ammonia at 25°C and x times theoretical air from the compressor enter the combustion chamber. What is x if the adiabatic flame temperature is to be fixed at 1600 K? Air P1 = 100 kPa

1

2

P2 = 1600 kPa

COMP.

T1 = 25 oC

ηS COMP = 0.87

-W

a) ideal compressor process (adiabatic reversible): P2

s2S = s1 ⇒ T2S = T1( ) P1

k-1 k

= 298.2(

1600 0.286 ) = 659 K 100

-wS = CP0(T2S - T1) = 1.004(659 - 298.2) = 362.2 Real process: -w = -wS/ηS = 362.2/0.87 = 416.3 kJ/kg T2 = T1 - w/CP0 = 298.2 + 416.3/1.004 = 713 K Also -w = 416.3 × 28.97 = 12060 kJ/kmol b) o

2 liq NH3, 25 C

COMP

Air 100 kPa, 25 oC

-W

COMB. CHAMBER

Prod. PP = 1600 kPa TP = 1600 K

-Q = 0

2 NH3 + 1.5x O2 + 5.64x N2 → 3 H2O + 1.5(x - 1) O2 + (5.64x + 1) N2 Using Tables A.16 and A.2, hNH3 = -45720 + 17.031(298.36 - 1530.04) = -66697 kJ/kmol HR = 2(-66697) + 0 = -133394 kJ -W = 12060 × 7.14x = 86108 x kJ HP = 3(-241826 + 52907) + 1.5(x - 1)(0 + 44267) + (5.64x + 1)(0 + 41904) = 302739x - 591254 1st law: HR = HP + W = -133394 = 302739 x - 591254 - 86108 x =>

x = 2.11

14-25 14.41 A closed, insulated container is charged with a stoichiometric ratio of oxygen and hydrogen at 25°C and 150 kPa. After combustion, liquid water at 25°C is sprayed in such that the final temperature is 1200 K. What is the final pressure? 1

H2 + 2O2 → H2O   P: 1 H2O + xiH2O ° 3U2 - U1 = xihi = xihf liq = 1 + xiHP - HR - 1 + xiRTP + 2RTR °

HR = φ, HP = -241826 + 34506 = -207320, hf liq= -285830 Substitute xi( -285830 + 207320 + 8.3145 × 1200) = 3 -207320 - 8.3145 1200 - 2×298.15= -213579

xi = 3.116 P1V1 = nRRT1, P2V1 = npRTp P2 =

P11 + xiTP 3 2 T1

=

150 × 4.116 × 1200 3 2×

= 1657 kPa

298.15

14.42 Wet biomass waste from a food-processing plant is fed to a catalytic reactor, where in an SSSF process it is converted into a low-energy fuel gas suitable for firing the processing plant boilers. The fuel gas has a composition of 50% methane, 45% carbon dioxide, and 5% hydrogen on a volumetric basis. Determine the lower heating value of this fuel gas mixture per unit volume. For 1 kmol fuel gas, 0.5 CH4 + 0.45 CO2 + 0.05 H2 + 1.025 O2 → (0.5 + 0.45) CO2 + 1.05 H2O The lower heating value is with water vapor in the products. Since the 0.45 CO2 cancels, hRP = 0.5(-393522) + 1.05(-241826) - 0.5(-74873) - 0.05(0) = -413242 kJ/kmol fuel gas With

n 100 = P/RT = = 0.04033 kmol/m3 V 8.3145 × 298.2

LHV = +413242 × 0.04033 = 16 666 kJ/m3

14-26 14.43 Determine the lower heating value of the gas generated from coal as described in Problem 14.12. Do not include the components removed by the water scrubbers. The gas from problem 14.12 is saturated with water vapor. Lower heating value LHV has water as vapor. °

°

°

LHV = -HRP = HP - HR Only CH4, H2 and CO contributes. From 14.12 the gas mixture after the scrubbers has ∑νi = 81.9 of composition: 0.3CH4 + 29.6H2 + 41CO + 10CO2 + 0.8N2 + 0.2016H2O -° -° -° LHV = -[0.3HRPCH4 + 29.6HRPH2 + 41HRPCO]/81.9 = -[0.3(-50010 × 16.043) + 29.6(-241826) + 41(-393522 + 110527)]/81.9 kJ

= 232009 kmol gas

14.44 Propylbenzene, C9H12, is listed in Table 14.3, but not in table A.9. No molecular weight is listed in the book. Find the molecular weight, the enthalpy of formation for the liquid fuel and the enthalpy of evaporation. ^ = 9 × 12.011 + 12 × 1.0079 = 120.194 Hfg = 384 kJ/kg M C9H12 + 12 O2 → 9 CO2 + 6 H2O -° -° -° -° -° -° HRP = νihfi - hfFu ⇒ hfFu = νihfi - hRP ∑

∑

P

P

-° -° -° ^ (-41219) hfFu = 9hfCO2 + 6hfH2O g + M liq

Fu H2O vap

= -38383.5 kJ/kmol

14.45 Determine the higher heating value of the sample Wyoming coal as specified in Problem 14.10. -0 Note: For SO2 , hf = -296842 kJ/kmol From solution 14.10 , for 100 kg of coal there are 4.05 kmol CO2, 1.75 kmol H2O, 0.0156 kmol SO2 in the products. Therefore, hRP0 = 4.05(-393522) + 1.75(-285830) + 0.0156(-296842) = -2098597 kJ/100 kg coal So that HHV = +20986 kJ/kg coal

14-27 14.46 Consider natural gas A and natural gas D, both of which are listed in Table 14.2. Calculate the enthalpy of combustion of each gas at 25°C, assuming that the products include vapor water. Repeat the answer for liquid water in the products. Natural Gas A 0.939 CH4 + 0.036 C2H6 + 0.012 C3H8 + 0.013 C4H10 + 2.1485 O2 + 3.76 × 2.1485 N2 → 1.099CO2 + 2.099 H2O + 8.0784 N2 HR = 0.939(-74878) + 0.036(-84740) + 0.012(-103900) + 0.013(-126200) = -76244 kJ a) vapor H2O HP = 1.099(-393522) + 2.099(-241826) = -940074 hRP = HP - HR = -863830 kJ/kmol b) Liq. H2O HP = 1.099(-393522) + 2.099(-285830) = -1032438 hRP = -956194 kJ/kmol Natural Gas D: 0.543 CH4 + 0.163 C2H6 + 0.162 C3H8 + 0.074 C4H10 + 0.058 N2 + O2 + N2

→

1.651 CO2 + 2.593 H2O + N2

HR = 0.543(-74873) + 0.163(-84740) + 0.162(-130900) + 0.074(-126200) = -80639 kJ a) vapor H2O HP = 1.651(-393522) + 2.593(-241826) = -1276760 kJ hRP = -1196121 kJ/kmol b) Liq. H2O HP = 1.651(-393522) + 2.593(-285830) = -1390862 kJ hRP = -1310223 kJ/kmol

14-28 14.47 Blast furnace gas in a steel mill is available at 250°C to be burned for the generation of steam. The composition of this gas is, on a volumetric basis, Component CH4 H2 CO CO2 N2 H2O Percent by volume 0.1 2.4 23.3 14.4 56.4 3.4 Find the lower heating value (kJ/m3) of this gas at 250°C and ambient pressure. Of the six components in the gas mixture, only the first 3 contribute to the heating value. These are, per kmol of mixture: 0.024 H2,

0.001 CH4,

0.233 CO

For these components, 0.024 H2 + 0.001 CH4 + 0.233 CO + 0.1305 O2 → 0.026 H2O + 0.234 CO2 The remainder need not be included in the calculation, as the contributions to reactants and products cancel. For the lower HV(water as vapor) at 250°C hRP = 0.026(-241826+7742) + 0.234(-393522+9348) - 0.024(0+6558) - 0.001(-74873+2.254×16.04(250-25)) - 0.233(-110527+6625) – kJ

0.1305(0+6810) = -72 573 kmol fuel

− v-0 = R To/Po = 8.3145 × 523.2/100 = 43.5015 m3/kmol LHV = +72573 / 43.5015 = 1668 kJ/m3 14.48 The enthalpy of formation of magnesium oxide, MgO(s), is −601827 kJ/kmol at 25°C. The melting point of magnesium oxide is approximately 3000 K, and the increase in enthalpy between 298 and 3000 K is 128449 kJ/kmol. The enthalpy of sublimation at 3000 K is estimated at 418000 kJ/kmol, and the specific heat of magnesium oxide vapor above 3000 K is estimated at 37.24 kJ/kmol K. a. Determine the enthalpy of combustion per kilogram of magnesium. b. Estimate the adiabatic flame temperature when magnesium is burned with theoretical oxygen. a)

1 Mg + O2 → MgO(s) 2 ∆hCOMB = ∆hCOMB/M = h°f /M = -601827/24.32 = -24746 kJ/kg

b) assume TR = 25°C and also that T P > 3000 K, 1st law: QCV = HP - HR = 0,

(MgO = vapor phase)

but HR = 0

⇒ HP = h°f + (h3000 - h298)SOL + ∆hSUB + CP VAP(TP - 3000) = -601827 + 128449 + 418000 + 37.24(TP - 3000) = 0 Solving, TP = 4487 K

14-29 14.49 A rigid container is charged with butene, C4H8, and air in a stoichiometric ratio at P0, T0. The charge burns in a short time with no heat transfer to state 2. The products then cool with time to 1200 K, state 3. Find the final pressure, P3, the total heat transfer, 1Q3, and the temperature immediately after combustion, T2. The reaction equation is, having used C and H atom balances: C4H8 + νO O2 + 3.76 N2 → 4 CO2 + 4 H2O + 3.76 νO N2  2 2 Counting now the oxygen atoms we get νO = 6. 2

C.V. analysis gives: U2 - U1 = Q - W = Q = H2 - H1 - P2V2 + P1V1 = H2 - H1 - Rn2T2 - n1T1 °

°

°

°

^ HRP + ∆H H2 - H1 = HP 1200 - HR = HP - HR + ∆HP = M P = -2542590 + 950055 = -1592535 ^ = 56.108 Where M

n1= 1 + 6 × 4.76 = 29.56,

and

n2 = 4 + 4 + 6 × 3.76 = 30.56, Τable A.8 at 1200 K:

∆hCO = 44473, 2

∆hH O=34506, 2

∆hN =28109. 2

Now solving for the heat transfer: kJ

Q = -1592535 - 8.3145(30.56 × 1200 - 29.56 × 298.15) = -1824164 kmol fuel To get the pressure, assume ideal gases: n2RT2 n2T2 P2 = = P1 = 421.6 kPa V2 n1T1 Before heat transfer takes place we have constant U so: U1 - U1 = 0 = H1 - H1 - n2RT1 + n1RT1 a

a

a

°

Now split the enthalpy H1 = HP + ∆HPT1 and arrange things with the a  a unknowns on LHS and knowns on RHS: ° ∆HP- n2RT = HR - HP - n1RT1= 2 542 590 - 73278 = 2 469 312 Trial and error leads to: LHS (3000 K) = 3 209 254 - 30.56 × 8.31451 × 3000 = 2 446 980 LHS (3000 K) = 3 471 331 - 30.56 × 8.31451 × 3200 = 2 658 238 linear interpolation

T = 3021 K

14-30 14.50 In an engine a mixture of liquid octane and ethanol, mole ratio 9;1, and stoichiometric air are taken in at T0, P0. In the engine the enthalpy of combustion is used so that 30% goes out as work, 30% goes out as heat loss and the rest goes out the exhaust. Find the work and heat transfer per kilogram of fuel mixture and also the exhaust temperature. 0.9 C8H18 + 0.1 C2H5OH + 11.55 O2 + 43.428 N2 → 8.4 H2O + 7.4 CO2 + 43.428 N2 For 0.9 octane + 0.1 ethanol ° ° kJ -° HRP mix= 0.9HRP C8H18 + 0.1HRP C2H5OH = -4690690.3 kmol

^ ^ + 0.1 M ^ = 107.414 M = 0.9 M mix oct alc Energy:

-° -° hin + qin = hex + ωex = hex + ∆hex + ωex

-° -° -° hex - hin = HRP mix ⇒

-° ωex + ∆hex - qin = -HRP mix

kJ kJ -° ωex= -qin = 0.3-HRP= 1407207 kmol = 13101 kg Fu kJ -° ∆hprod = ∆hex = 0.4-HRP= 1 876 276 kmol Fu

∆hprod = 8.4 ∆hH

2O

+ 7.4∆hCO + 43.428 ∆hN 2

2

⇒ satisfied for T = 1216 K 14.51 Consider the same situation as in the previous problem. Find the dew point temperature of the products. If the products in the exhaust are cooled to 10°C, find the mass of water condensed per kilogram of fuel mixture. Reaction equation with 0.9 octane and 0.1 ethanol is 0.9 C8H18 + 0.1 C2H5OH + 11.55 O2 + 43.428 N2 → 8.4 H2O + 7.4 CO2 + 43.428 N2 yH

= O

2

PH

2O

8.4 = 0.1418 8.4 + 7.4 + 43.428 ⇒ Tdew= 52.9 °C

= yH OPtot = 14.3 kPa 2

10 °C ⇒ PH

2O

= 1.2276 ⇒

yH

2O

= 0.012276 = kmol

x x + 7.4 + 43.428

⇒ x = 0.6317 ⇒ ∆ν H O= -7.77 kmol Fu mix 2

-∆ν H O × 18.015 mH

2

= O cond

2

107.414

kmol

= 1.303 kmol Fu mix

14-31 14.52 Calculate the irreversibility for the process described in Problem 14.19. From solution 14.19, Reactants: 2 C(s) + 2 O2 at 25°C, 200 kPa 1 Product mixture: 1 CO2 + 1 CO + O2 at 1000 K, 838.4 kPa 2 & 1Q2 = -453442 kJ Reactants: SR = 2(5.740) + 2(205.148 - 8.31451 ln

200 ) = 410.250 kJ/K 100

Products: ni

yi

-s° i

S-i

yiP

-Rln

P0

CO2

1.0

0.40

269.299

-10.061

259.238

CO O2

1.0 0.5

0.40 0.20

234.538 243.579

-10.061 -4.298

224.477 239.281

SP = 1.0(259.238) + 1.0(224.477) + 0.5(239.281) = 603.355 kJ/K I = T0(SP - SR) - 1Q2 = 298.15(603.355 - 410.250) - (-453442) = +511 016 kJ

14-32

14.53 Pentane gas at 25°C, 150 kPa enters an insulated SSSF combustion chamber. Sufficient excess air to hold the combustion products temperature to 1800 K enters separately at 500 K, 150 kPa. Calculate the percent theoretical air required and the irreversibility of the process per kmol of pentane burned. C5H12 + 8XO2 + 30.08XN2 ‡ 5CO2 + 6H2O + 8(X-1)O2 + 30.08XN2 1st Law: Qcv + HR = HP + Wcv; Wcv = 0 , Qcv = 0 => HR = HP −o − Reactants: C5H12 : h f from A.9 and ∆h 500 for O2 and N2 from A.8 −o − − HR =(h f )C5H12 + 8X ∆h O2 + 30.08X ∆h N2 = -146 500 + 8X 6086 + 30.08 X 5911 = 226 491 X - 146 500 −o − −o − − − HP = 5(h f + ∆h )CO2 + 6(h f + ∆h )H2O + 8(X-1) ∆h O2 + 30.08 X ∆h N2 = 5(-393522 + 79432) + 6(-241826 + 62693) + 8(X-1) 51674 + 30.08 X 48979 = 1 886 680 X - 3 058 640 Energy Eq. solve for X;

=>

X = 1.754

o b) Reactants: Pi = 150 kPa, Po = 100 kPa, −s f

ni

yi

−s o / −s o f 500

− yiPi - R ln Po

Si

C5H12

1

1

348.945

-3.371

352.316

O2

8X

0.21

220.693

9.604

211.089

N2

30.08 X

0.79

206.74

1.411

208.151

kJ kmol K

SR = ∑ niSi = 14410.249 kJ/K Products: Pe = 150 kPa, Po = 100 kPa ni

yi

−s o 1800

− yiPe - R ln Po

Si

CO2

5

0.0716

302.969

18.550

321.519

H2O

6

0.086

259.452

17.027

276.479

O2

8(X-1)

0.0864

264.797

16.988

281.785

N2

30.08X

0.756

248.304

-1.045

247.259

SP = ∑ niSi = 17732.073 kJ/K; I = To(SP - SR) = 298.15(17 732.07 - 14 410.25) = 990 MJ

kJ kmol K

14-33 14.54 Consider the combustion of methanol, CH3OH, with 25% excess air. The combustion products are passed through a heat exchanger and exit at 200 kPa, 40°C. Calculate the absolute entropy of the products exiting the heat exchanger per kilomole of methanol burned, using appropriate reference states as needed. CH3OH + 1.25 × 1.5 (O2 + 3.76 N2) → CO2 + 2 H2O + 0.38 O2 + 7.05 N2 Products exit at 40 oC, 200 kPa, check for saturation: PG

nV MAX 7.384 yV MAX = = = 200 P nV MAX + 1 + 0.38 + 7.05 nV = nV MAX = 0.323 Gas mixture: ni

nLIQ = 1.677

yi

-s° i

- yiP -Rln P0

Si

CO2

1.0

0.1143

215.633

+12.270

227.903

H 2O

0.323

0.0369

190.485

+21.671

212.156

O2

0.38

0.0434

206.592

+20.322

226.914

N2

7.05

0.8054

193.039

-3.964

189.075

SGAS MIX =

∑ n S- = 1715.64 kJ/K i i

-s = 69.950 + 18.015(0.5725 - 0.3674) = 73.645 LIQ SLIQ = 1.677 × 73.645 = 123.50 kJ/K SPROD = 1715.64 + 123.50 = 1839.14 kJ/K

14-34 14.55 The turbine in Problem 14.23 is adiabatic. Is it reversible, irreversible, or impossible? Inlet to the turbine is the exit from the mixing of air and fuel at 1 MPa. From solution to 14.23, we have: CP C H = 43.43, CP C H = 74.06, Tturbine,in = 548.7 K 2 2

3 8

C2H4 + C3H8 + 9.6 O2 + 36.096 N2 → 5 CO2 +6 H2O + 1.6 O2 + 36.096 N2 ⌠dQ Sex - Sin =  + Sgen = Sgen ≥ φ ⌡T Inlet: 1 MPa, 548.7 K

-° SFu = Si + CP Fu lnT/T0 -s° i - yiP -Rln P0

Si

ni

yi

C2H4

1

0.02097

245.82

12.989

258.809

C3H8

1

0.02097

315.09

12.989

328.079

O2

9.6

0.2013

223.497

-5.816

217.681

N2

36.096

0.7568

209.388

-16.828

192.56

Sin = 258.809 + 328.079 + 9.6 × 217.681 + 36.096 × 192.56 = 9627.3 ni

yi

-s° i

- yiP -Rln P0

S-i

CO2

5

0.1027

257.496

18.925

276.421

H 2O

6

0.1232

223.826

17.409

241.235

O2

1.6

0.0329

235.92

28.399

264.319

N2

36.096

0.7413

221.016

2.489

223.505

Sex

= 5 × 276.421 + 6 × 241.235 1.6 × 264.319 kJ

+ 36.096 × 223.505 = 11320 2kmol Fu K Sgen

kJ

= Sex - Sin = 1693 2kmol Fu K > 0

Possible, but one should check the state after combustion to account for generation by combustion alone and then the turbine expansion separately.

14-35 14.56 Saturated liquid butane enters an insulated constant pressure combustion chamber at 25°C, and x times theoretical oxygen gas enters at the same P and T. The combustion products exit at 3400 K. With complete combustion find x. What is the pressure at the chamber exit? and what is the irreversibility of the process? Butane: T1 = To = 25oC, sat liq., x1 = 0, Tc = 425.2 K, Pc = 3.8 MPa Fig. D.1: Tr1 = 0.7, Pr1 = 0.1, P1 = Pr1Pc = 380 kPa − * −  − − −* − h 1 − h 1 f = 4.85 RTc , (s 1 - s 1)f = 6.8 R Oxygen: T2 = To = 25oC, X - Theoretical O2 Products: T3 = 3400 K, Assumes complete combustion

Figs. D.2 and D.3:

C4H10 + 6.5XO2 ‡ 4CO2 + 5H2O + 6.5(X-1)O2 1st Law: Q

cv + HR = HP + Wcv; Qcv = 0, Wcv = 0 −o − HR = n(h f + ∆h )C4H10 = 1(-126200 + -17146) = -143346 kJ

Products: CO2

−o − n(h f + ∆h )CO2 = 4(-393522 + 177836) = -862744 kJ

H2O

−o − n(h f + ∆h )H2O = 5(-241826 + 149073) = -463765 kJ

O2

−o − n(h f + ∆h )O2 = 6.5(X-1)(0 + 114101) = (X-1)741656.5 kJ

HP =

−o

−

∑ni (h f + ∆h )i = 741656.5X - 2068165.5

HP = HR solve for X;

X = 2.594

Assume that the exit pressure equals the inlet pressure: −s C

Pe = Pi = 380 kPa

P1 P1 −s o - − − * - −s ) ] ; −s = [−s o - − = [ R ln (s R ln ] 1 f O2 f 1 4H10 Po Po

SR = SC4H10 + SO2 = [306.647 - 11.10 - 56.539] + [205.48 - 11.10] × 6.5 × 2.594 = 3516.45 kJ/K Products: ni

yi

−s o i

− yiPe − R ln Po

Si

CO2

4

0.2065

341.988

2.016

344.004

H2O

5

0.2582

293.550

0.158

293.708

O2

10.368

0.5353

289.499

-5.904

283.595

SP = ∑ niSi = 5784.87 kJ/K; I = To(SP - SR) = 298.15 (5784.87 - 3516.45) = 676329 kJ

kJ kmol K

14-36 14.57 An inventor claims to have built a device that will take 0.001 kg/s of water from the faucet at 10°C, 100 kPa, and produce separate streams of hydrogen and oxygen gas, each at 400 K, 175 kPa. It is stated that this device operates in a 25°C room on 10-kW electrical power input. How do you evaluate this claim? Liq H2O

H2 gas

10oC, 100 kPa 0.001 kg/s

O2 gas . -WCV = 10 kW

 each at  400 K  175 kPa T0 = 25 oC

1

H 2O → H 2 + 2 O 2 1

Hi - He = [-285830 + 18.015(42.01 - 104.89)] - 2961 - 2 (3027) = -291437 kJ (Si - Se) = [69.950 + 18.015(0.151 - 0.3674)] - (139.219 - 8.3145 ln 1.75) 1

- 2 (213.873 - 8.3145 ln 1.75) = -173.124 WREV = (Hi - He) - T0(Si - Se) = -291437 - 298.15(-173.124) = -239820 kJ . WREV = (0.001/18.015)(-239820) = -13.31 kW . . . I = WREV - WCV = -13.31 - (-10) < 0 Impossible 14.58 Two kilomoles of ammonia are burned in an SSSF process with x kmol of oxygen. The products, consisting of H2O, N2, and the excess O2, exit at 200°C, 7 MPa. a. Calculate x if half the water in the products is condensed. b. Calculate the absolute entropy of the products at the exit conditions. 2NH3 + xO2 → 3H2O + N2 + (x - 1.5)O2 Products at 200 oC, 7 MPa with a) yH2O VAP = PG/P =

nH2O LIQ = nH2O VAP = 1.5

1.5538 1.5 = 7 1.5 + 1 + x - 1.5

b) SPROD = SGAS MIX + SH2O LIQ

=>

x = 5.757

Gas mixture:

ni

yi

-s° i

-Rln(yiP/P0)

S-i

H 2O

1.5

0.222

204.595

-22.810

181.785

O2

4.257

0.630

218.985

-31.482

187.503

N2

1.0

0.148

205.110

-19.439

185.671

SGAS MIX = 1.5(181.785) + 4.257(187.503) + 1.0(185.67) = 1256.55 kJ/K SH2O LIQ = 1.5[69.950 + 18.015(2.3223 - 0.3674)] = 157.75 kJ/K SPROD = 1256.55 + 157.75 = 1414.3 kJ/K

14-37 14.59 Consider the SSSF combustion of propane at 25°C with air at 400 K. The products exit the combustion chamber at 1200 K. It may be assumed that the combustion efficiency is 90%, and that 95% of the carbon in the propane burns to form carbon dioxide; the remaining 5% forms carbon monoxide. Determine the ideal fuel–air ratio and the heat transfer from the combustion chamber. Ideal combustion process, assumed adiabatic, excess air to keep 1200 K out. C3H8 + 5x O2 + 18.8x N2 → 3 CO2 + 4 H2O + 5(x - 1) O2 + 18.8x N2 HR = -103900 +5x(0 + 3027) + 18.8x(0 + 2971) = -103900 + 70990x HP = 3(-393522 + 44473) + 4(-241826 + 34506) + 5(x - 1)(0 + 29761) + 18.8x(0 + 28109) = -2025232 + 677254x 1st law: HP - HR = 0

Solving, x = 3.169

FAIDEAL = 1/(23.8 × 3.169) = 0.01326 b)

FAACTUAL = 0.01326/0.90 = 0.01473 C3H8 + 14.26 O2 + 53.62 N2 → 2.85 CO2 + 0.15 CO + 4 H2O + 9.335 O2 + 53.62 N2 HR = -103900 + 14.26(0 + 3027) + 53.62(0 + 2971) = +98570 kJ HP = 2.85(-393522 + 44473) + 0.15(-110527 + 28427) + 4(-241826 + 34506) + 9.335(0 + 29761) + 53.62(0 + 28109) = -51361 kJ QCV = HP - HR = -149931 kJ

14-38 14.60 Graphite, C, at P0, T0 is burned with air coming in at P0, 500 K in a ratio so the products exit at P0, 1200 K. Find the equivalence ratio, the percent theoretical air, and the total irreversibility. C + φb(O2 + 3.76 N2) → CO2 + ( φ - 1) O2 + 3.76 φ N2 °

°

HP = HR ⇒ ∆HP 1200 - ∆HR = HR - HP 44473 + ( φ - 1) 29761 + 3.76φ28109 - φ( 6086 + 3.76×5911) = 0 - (-393522) ⇒ φ = 3.536 Sgen = SP - SR =

∑ν-s

°

- R ln( y) 

P-R

R: yO = 0.21, yN = 0.79 2

2

P: yO = 0.1507, yN = 0.79, yCO = 0.0593 2

2

2

°

SP = 279.39 + 2.536 × 250.011 + 13.295 × 234.227 = 4027.5 °

SR = 5.74 + 3.536(220.693 + 3.76 × 206.74) = 3534.8 For the pressure correction the term with the nitrogen drops out (same y). R

∑-ν ln(y) = 8.3145(2.8235 + 1.8927 - 1.5606) = 26.236 P-R

Sgen = 4027.5 - 3534.8 + 26.236 = 518.94 kJ

I = T0 Sgen = 154721 kmol C

14-39 14.61 A gasoline engine is converted to run on propane. Assume the propane enters the engine at 25°C, at the rate 40 kg/h. Only 90% theoretical air enters at 25°C such that 90% of the C burns to form CO2, and 10% of the C burns to form CO. The combustion products also include H2O, H2 and N2, exit the exhaust at 1000 K. Heat loss from the engine (primarily to the cooling water) is 120 kW. What is the power output of the engine? What is the thermal efficiency? . Propane: T = 25oC, m = 40 kg/hr, M = 44.094 kg/kmol 1

Air: T2 = 25oC, 90% theoretical Air produces 90% CO 2, 10% CO Products: T3 = 1000 K, CO2, CO, H2O, H2, N2 C3H8 + 4.5O2 + 16.92N2 ‡ 2.7 CO2 + 0.3CO + 3.3H2O + 0.7H2 + 16.92N2 . . nC3H8 = m/(M×3600) = 0.000252 kmol/s . . . 1st Law: Q + H = H + W ; Q = -120 kW R

P

−o HR = nC3H8 h f = -103 900 kJ Products: CO2

-

−o − nCO2(h f + ∆h ) = 2.7(-393522 + 33397) = -972337.5 kJ

CO

-

−o − nCO(h f + ∆h ) = 0.3(-110527 + 21686) = -26652 kJ

H 2O

-

−o − nH2O(h f + ∆h ) = 3.3(-241826 + 26000) = -712226 kJ

H2

-

−o − nH2(h f + ∆h ) = 0.7(0 + 20663) = 14464.1 kJ

N2

-

−o − nN2(h f + ∆h ) = 16.92(0 + 21463) = 363154 kJ

−o − HP = ∑ni (h f + ∆h )i = -1 333 598 kJ . . . W = Q + n(HR - HP) = 189.9 kW C3H8: Table 14.3 HRPo = -50343 kJ/kg . . HHV = nC3H8 M(-HRPo) = 559.4 kW . . ηth = W/HHV = 0.339

14-40 14.62 A small air-cooled gasoline engine is tested, and the output is found to be 1.0 kW. The temperature of the products is measured to 600 K. The products are analyzed on a dry volumetric basis, with the result: 11.4% CO2, 2.9% CO, 1.6% O2 and 84.1% N2. The fuel may be considered to be liquid octane. The fuel and air enter the engine at 25°C, and the flow rate of fuel to the engine is 1.5 × 10-4 kg/s. Determine the rate of heat transfer from the engine and its thermal efficiency. a C8H18 + b O2 + 3.76b N2 → 11.4 CO2 + 2.9 CO + c H2O + 1.6 O2 + 84.1 N2 84.1 1 = 22.37, a = (11.4 + 2.9) = 1.788 3.76 8 c = 9a = 16.088 C8H18 + 12.5 O2 + 47.1 N2

b=

→ 6.38 CO2 + 1.62 CO + 9 H2O + 0.89 O2 + 47.1 N2 -0 HR = hf C8H18 = -250105 kJ HP = 6.38(-393522 + 15788) + 1.62(-110527 + 10781) + 9(-241826 + 12700) + 0.89(0 + 11187) + 47.1(0 + 10712) = -4119174 kJ H P - HR = -4119174 - (-250105) = -3869069 kJ . . HP - HR = (0.00015/114.23)(-3869069) = -5.081 kW . QCV = -5.081 + 1.0 = -4.081 kW . QH = 0.00015(47893) = 7.184 kW . . ηTH = WNET/QH = 1.0/7.184 = 0.139 14.63 A gasoline engine uses liquid octane and air, both supplied at P , T , in a 0 0 stoichiometric ratio. The products (complete combustion) flow out of the exhaust valve at 1100 K. Assume that the heat loss carried away by the cooling water, at 100°C, is equal to the work output. Find the efficiency of the engine expressed as (work/lower heating value) and the second law efficiency. C8H18 + νO O2 + 3.76 N2 → 8 CO2 + 9 H2O + 47 N2  2 2 νO = 16 + 9 ⇒ νO = 12.5 2

2

kJ

kJ

LHV = 44425 kg fuel ⇒ LHV = 5.07476×106 kmol fuel ∆HP 1100 = 8 × 38885 + 9 × 30190 + 47 × 24760 = 1746510 C.V. Total engine

14-41 Hin = Hex + W + Qloss = Hex + 2 W °

⇒ 2 W = Hin - Hex= HR - Hν= -HRP + ∆HR- ∆HP 1100 = 5.07476×106 + 0 - 1746510 = 3328250 kJ

W = 1.664×106 kmol fuel 1.664×106 W ηth = = = 0.328 LHV 5.07476×106 Find entropies in and out: -° inlet: SFu = 360.575 1 SO° = 205.148 - 8.3145 ln = 218.12 4.76 2 3.76 = 193.57 SN° = 191.609 - 8.3145 ln 4.76 2 -° = 360.575 + 12.5 × 218.12 + 47 × 193.57 = 12185 Sin 8 - ° exit: SCO = 275.528 - 8.3145 ln = 292.82 64 2 9 SH° O = 236.732 - 8.3145 ln = 253.04 64 2 47 SN° = 231.314 - 8.1345 ln = 233.88 64 2 -° = 8 × 292.82 + 9 × 253.04 + 47 × 233.88 = 15612 Sex Assume the same Q loss out to 100°C reservoir in the reversible case and rev

compute Q0 : rev Sin + Q0 /T0 = Sex + Qloss/Tres rev

Q0

= T0( Sex - Sin) + Qloss T0/Tres = 298.15(15612 - 12185) + 1.664×106 × 298.15/373.15 kJ

= 2.351×106 kmol fuel rev

Hin + Q0 = Hex + Wrev + Qloss rev

rev

kJ

⇒ Wrev = Hin - Hex - Qloss + Q0 = Wac + Q0 = 4.015×106 kmol fuel ηII = Wac/ Wrev =1.664×106/4.015×106 = 0.414

14-42 14.64 In Example 14.16, a basic hydrogen–oxygen fuel cell reaction was analyzed at 25°C, 100 kPa. Repeat this calculation, assuming that the fuel cell operates on air at 25°C, 100 kPa, instead of on pure oxygen at this state. Anode: 2 H2

→ 4 e- + 4 H+

Cathode: 4 H+ + 4 e- → 1 O2 + 2 H2O Overall: 2 H2 + 1 O2 → 2 H2O Example 14.16: ∆G25°C = -474283 kJ (for pure O2) For PO2 = 0.21 × 0.1: SO = 205.148 – 8.3145 ln0.21 = 218.124 2

∆S = 2(69.950) – 2(130.678) – 1(218.124) = -339.58 kJ/K ∆G25°C = -571660 – 298.15(-339.58) = -470414 kJ E° = +470414/(96487 × 4) = 1.219 V

14-43 14.65 Consider a methane-oxygen fuel cell in which the reaction at the anode is CH4 + 2H2O CO2 + 8e- + 8H+ The electrons produced by the reaction flow through the external load, and the positive ions migrate through the electrolyte to the cathode, where the reaction is 8 e- + 8 H+ + 2 O2 4 H2O a. Calculate the reversible work and the reversible EMS for the fuel cell operating at 25°C, 100 kPa. b. Repeat part (a), but assume that the fuel cell operates at 600 K instead of at room temperature. CH4 + 2H2O → CO2 + 8e- + 8H+ and 8e- + 8H+ + 2CO2 → 4H2O Overall

CH4 + 2O2 → CO2 + 2H2O

a) 25 oC assume all liquid H2O and all comp. at 100 kPa 0

∆H25 C = -393522 + 2(-285830) – (-74873) – 0 = -890 309 kJ 0

∆S25 C = 213.795 + 2(69.950) – 186.251 – 2(205.148) = - 242.852 kJ/K 0

∆G25 C = -890309 – 298.15(-242.852) = - 817 903 kJ Wrev = -∆G0 = +817903 kJ -∆G0 +817903 0 = = 1.06 V E = 96485 × 8 96485 × 8 b) 600 K assume all gas H2O and all comp. at 100 kPa 0

∆H600 K = 1(-393522 + 12906) + 2(-241826 + 10499) - 2(0 + 9245) - 1[-74873 + 16.04 × 2.2537(600 - 298.2)] = -797797 kJ 0

∆S600 K = 1(243.284) + 2(213.051) - 1(186.251 + 16.04 × 2.2537 ln

600 ) - 2(226.450) 298.2

= 4.9606 kJ/K 0

0

0

∆G600 K = ∆H600 K - T∆S600 K = -797797 - 600(4.9606) = -800773 kJ Wrev = +800773 kJ

E0=

+800773 = 1.037 V 96485 × 8

14-44 Advanced Problems 14.66 A gas mixture of 50% ethane and 50% propane by volume enters a combustion chamber at 350 K, 10 MPa. Determine the enthalpy per kilomole of this mixture relative to the thermochemical base of enthalpy using Kay’s rule. -* hMIX O = 0.5(-84740) + 0.5(-103900) = -94320 kJ/kmol CP0 MIX = 0.5 × 30.07 × 1.7662 + 0.5 × 44.097 × 1.67 = 63.583 -* -* h350 - h298 = 63.583(350 - 298.2) = 3294 kJ/kmol Kay’s rule:

TC MIX = 0.5 × 305.4 + 0.5 × 369.8 = 337.6 K PC MIX = 0.5 × 4.88 + 0.5 × 4.25 = 4.565 MPa

Tr = 350/337.6 = 1.037,

P r = 10/4.565 = 2.19

-* From Fig. D.2: h - h = 8.3145 × 337.6 × 3.53 = 9909 kJ/kmol hMIX 350K,10MPa = -94320 + 3294 - 9909 = -100935 kJ/kmol

14-45 14.67 A mixture of 80% ethane and 20% methane on a mole basis is throttled from 10 MPa, 65°C, to 100 kPa and is fed to a combustion chamber where it undergoes complete combustion with air, which enters at 100 kPa, 600 K. The amount of air is such that the products of combustion exit at 100 kPa, 1200 K. Assume that the combustion process is adiabatic and that all components behave as ideal gases except the fuel mixture, which behaves according to the generalized charts, with Kay’s rule for the pseudocritical constants. Determine the percentage of theoretical air used in the process and the dew-point temperature of the products. Reaction equation: -0 Fuel mix: hf FUEL = 0.2(-74873) + 0.8(-84740) = -82767 kJ/kmol CP0 FUEL = 0.2 × 2.2537 × 16.04 + 0.8 × 1.7662 × 30.07 = 49.718 -* ∆hFUEL = 49.718(65 - 25) = 1989 kJ/kmol TCA = 305.4 K, TCB = 190.4 K ⇒ Tc mix=282.4 K PCA = 4.88, PCB=4.60 ⇒ Pc mix= 4.824 MPa Tr = 338.2/282.4 = 1.198, P r = 10/4.824 = 2.073 - (h* - h)FUEL IN = 8.31451 × 282.4 × 2.18 = 5119 kJ ⇒ hFUEL IN = -82767 + 1989 - 5119 = -85897 kmol

1st law: 1.8(-393522 + 44473) + 2.8(-241826 + 34506) + 3.2(x - 1)(29761) + (12.03x)(28109) + 85897 - (3.2x)(9245) - (12.03x)(8894) = 0 a)

x = 4.104 or 410.4 %

b) nP = 1.8 + 2.8 + 3.2(4.104 - 1) + 12.03 × 4.104 = 63.904 yH2O = 2.8/63.904 = PV/100 ;

PV = 4.38 kPa, T = 30.5°C

14-46 14.68 Gaseous propane mixes with air, both supplied at 500 K, 0.1 MPa. The mixture goes into a combustion chamber and products of combustion exit at 1300 K, 0.1 MPa. The products analyzed on a dry basis are 11.42% CO2, 0.79% CO, 2.68% O2, and 85.11% N2 on a volume basis. Find the equivalence ratio and the heat transfer per kmol of fuel. C3H8 + α O2 + 3.76 α N2 → β CO2 + γ H2O + 3.76 α N2 β = 3, γ = 4, α = β +

γ =5 2

( A/F) S = 4.76α = 23.8

φ = ( A/F) /( A/F) S = 1.1123

% Theoretical Air = 111.23 % °

hP = hP + ∑νj∆h(1300 K) °

q = hP - hR = hP - hR + ∑νj∆h(1300 K) = -∆H + ∑νj∆h(1300 K) kJ

∑νj∆h(1300 K) = 983230.7 kmol Fu kJ

q = -1182480 kmol Fu

14-47 14.69 A closed rigid container is charged with propene, C H , and 150% theoretical air 3 6 at 100 kPa, 298 K. The mixture is ignited and burns with complete combustion. Heat is transferred to a reservoir at 500 K so the final temperature of the products is 700 K. Find the final pressure, the heat transfer per kmole fuel and the total entropy generated per kmol fuel in the process. C3H6 + νO O2 + 3.76 N2 → 3 CO2 + 3 H2O + x N2  2 2 νO = 6 + 3 = 9

Oxygen, O2 , balance: Actual Combustion:

2

φ = 1.5 ⇒ νO

2 ac

⇒

νO = 4.5 2

= 1.5 × 4.5 = 6.75

C3H6 + 6.75 O2 + 25.38 N2 → 3 CO2 + 3 H2O + 25.38 N2 + 2.25 O2 P 2 = P1

npT2 nRT1

= 100 ×

33.63 × 700 = 238.3 kPa 33.13 × 298.15 kJ

∆HP 700 = 3×17754 + 3×14190 + 25.38×11937 + 2.25×12499 = 426916 kmol fuel U2 - U1 = 1Q2 - 0 = H2 - H1 - n2R  T2 + n1R  T1 °

Q = HRP + ∆HP 700 - nPR  T2 + n1R  T1

1 2

= -45780 × 42.081 + 426916 - 33.63 × 8.3145 × 700 kJ

+ 33.13 × 8.3145 × 298.15 = -1.613×106 kmol fuel -s° Reactants: ni yi -Rln(yiP/P0) i

S-i

C3H8

1.0

0.0302

267.066

29.104

296.17

O2

6.75

0.2037

205.143

13.228

218.376

N2

25.38 0.7661

191.609

2.216

189.393 kJ

S1= 296.17 + 6.75 × 218.376 + 25.38 × 189.393 = 6577 kmol fuel K -s° Products: ni yi -Rln(yiP/P0) S-i i CO2

3

0.0892

250.752

+12.875

263.627

H 2O

3

0.0892

218.739

+12.875

231.614

O2

2.25

0.0669

231.465

+15.266

246.731

N2

25.38

0.7547

216.865

- 4.88

211.985

S2 = 3(263.627 + 231.614) + 2.25 × 246.731 + 25.38 × 211.985 = 7421 kJ/kmol fuel K 1.613×106 kJ S = S S Q /T = 7421 6577 + = 4070 kmol fuel K 1 2 gen 2 1 1 2 res 500

14-48 14.70 Consider one cylinder of a spark-ignition, internal-combustion engine. Before the compression stroke, the cylinder is filled with a mixture of air and methane. Assume that 110% theoretical air has been used, that the state before compression is 100 kPa, 25°C. The compression ratio of the engine is 9 to 1. a. Determine the pressure and temperature after compression, assuming a reversible adiabatic process. b. Assume that complete combustion takes place while the piston is at top dead center (at minimum volume) in an adiabatic process. Determine the temperature and pressure after combustion, and the increase in entropy during the combustion process. c. What is the irreversibility for this process? 1 CH4 + 1.1 × 2 O2 + 3.76 × 2.2 N2 → 1 CO2 + 2 H2O + 0.2 O2 + 2 H2O P1 = 100 kPa, T1 = 298.2 K,

V2/V1 = 1/8, Rev. Ad. s2 = s1

Assume T2 ~ 650 K → TAVE ~ 475 K Table A.6: CP0 CH4 = 44.887, CP0 O2 = 30.890, CP0 N2 = 29.415 CP0 MIX = (1 × 44.887 + 2.2 × 30.890 + 8.27 × 29.415)/11.47 = 31.047 - CV0 MIX = CP0 - R = 22.732, k = CP0/CV0 = 1.366 a) T2 = T1(V1/V2)k-1 = 298.2 (9)0.366 = 666.4 K (avg OK) P2 = P1(V1/V2)k = 100 (9)1.366 = 2011 kPa b) comb. 2-3 const. vol., Q = 0 Q = 0 = (H3 - H2) - R(n3T3 - n2T2) 2 3 -0 H2 = 1 hf CH4 + n2 CP0 MIX (T2 - T1) H2 = -74873 + 11.47 × 31.047(666.4 -298.2) = +56246 kJ -* -* -* -* H3 = 1(-393522 + ∆hCO2) + 2(-241826 + ∆hH2O) + 0.2 ∆hO2 + 8.27 ∆hN2 Substituting, -* -* -* -* 1 ∆hCO2 + 2 ∆hH2O + 0.2 ∆hO2 + 8.27 ∆hN2 - 95.366 T3 - 869868 = 0 Trial & error: T3 = 2907 K 1 × 147072 + 2 × 121377 + 0.2 × 94315 + 8.27 × 89274 - 95.366 × 2907 - 869868 ≈ 0 OK P 3 = P2

n3T3 n2T2

= P2

T3 T2

= 2011 ×

2907 = 8772 kPa 666.4

14-49

c) state 1 REAC

ni

yi

-s° i

- R ln(yiP/P0)

S-i

CH4

1

0.0872

186.251

+20.283

206.534

O2

2.2

0.1918

205.148

+13.730

218.878

N2

8.27

0.7210

191.609

+2.720

194.329

11.47 S2 = S1 = ∑ niSi = 2295.17 kJ/K state 3 PROD

ni

yi

-s° i

- R ln(yiP/P0)

S-i

CO2

1

0.0872

332.213

-16.916

315.297

H 2O

2

0.1744

284.753

-22.680

262.073

O2

0.2

0.0174

283.213

-3.516

279.697

N2

8.27

0.7210

265.726

-34.480

231.246

11.47 S3 = ∑ niSi = 2807.79 kJ/K I = T0(S3 - S2) = 298.2(2807.79 - 2295.17) = 152860 kJ

14-50 14.71 Consider the combustion process described in Problem 14.67. a. Calculate the absolute entropy of the fuel mixture before it is throttled into the combustion chamber. b. Calculate the irreversibility for the overall process. From solution 14.67 , fuel mixture 0.8 C2H6 + 0.2 CH4 at 65°C, 10 MPa CP0 FUEL = 49.718 kJ/kmol K. Using Kay’s rule: Tr1 = 1.198, Pr1 = 2.073 and x = 410.4 % theoretical air or 13.13 O2 + 49.36 N2

in at 600 K, 100 kPa

and 1.8CO2 + 2.8H2O + 9.93O2 + 49.36N2

out at 100 kPa, 1200 K

*

a) -s0 FUEL = 0.2(186.251) + 0.8(229.597) - 8.3145(0.2 ln 0.2 + 0.8 ln 0.8) = 225.088 *

∆sTP = 49.718 ln

338.2 10 - 8.3145 ln = -32.031 298.2 0.1 * (s- -s-)FUEL = 1.37 × 8.3145 = 11.391

From Fig. D.3:

-s = 225.088 - 32.031 - 11.391 = 181.66 kJ/kmol K FUEL b) Air at 600 K, 100 kPa ni

yi

-s° i

-Rln(yiP/P0)

S-i

O2

13.13

0.21

226.45

+12.976

239.426

N2

49.36

0.79

212.177

+1.96

214.137

SAIR = ∑ niSi = 13713.47 kJ/K SR = 181.66 + 13713.47 = 13895.1 kJ/K Products at 1200 K, 100 kPa PROD ni yi

-so i

-Rln(yiP/P0)

S-i

CO2

1.8

0.0282

279.390

+29.669

309.059

H 2O

2.8

0.0438

240.485

+26.008

266.493

O2

9.93

0.1554

250.011

+15.479

265.490

N2

49.36

0.7726

234.227

+2.145

236.372

SP = ∑ niSi = 15606.1 kJ/K I = T0(SP - SR) - QCV = 298.15(15606.1 - 13895.1) + 0 = 510132 kJ

14-51 14.72 Liquid acetylene, C2H2, is stored in a high-pressure storage tank at ambient temperature, 25°C. The liquid is fed to an insulated combustor/steam boiler at the steady rate of 1 kg/s, along with 140% theoretical oxygen, O2, which enters at 500 K, as shown in Fig. P14.72. The combustion products exit the unit at 500 kPa, 350 K. Liquid water enters the boiler at 10°C, at the rate of 15 kg/s, and superheated steam exits at 200 kPa. a.Calculate the absolute entropy, per kmol, of liquid acetylene at the storage tank state. b. Determine the phase(s) of the combustion products exiting the combustor boiler unit, and the amount of each, if more than one. c. Determine the temperature of the steam at the boiler exit. -° a) C2H2: SIG 25°C = 200.958 TR = 298.2/308.3 = 0.967 => From Fig. D.1:

PR = 0.82

1

P1 = 0.82 × 6.14 = 5.03 MPa, Sliq T

1 P1

= S°T

0 P0

1

- ( S* - S) 1 = 3.33R = 27.687

- + ∆T - R lnP1/P° + ( S - S*) P

kJ

1 T1

= 140.695 kmol K

b) 1 C2H2 + 1.4 × 2.5 O2 → 2 CO2 + 1 H2O + 1 O2 H1 = 226731 + (-3.56 × R × 308.3) = 217605 kJ H2 = 3.5(0 + 6086) = 21301 kJ Products T3 = 350 K = 76.8°C ⇒ PG = 41.8 kPa yV max =

PG P

=

nV max 41.8 = 0.0836 = ⇒ nV max = 0.2737 = n V gas mix 500 nV max + 2 + 1

⇒ nliq = 1 - 0.2737 = 0.7263 Gas Mix = 2 CO2 + 0.2737 H2O + 1 O2 c) Hliq = 0.7263(-285830 + 18.015(321.5 - 104.9)) = -204764 kJ 3

Hgas mix = 2(-393522 + 2036) + 0.2737(-241826 + 1756) + 1541 3

= -847138 kJ H3 = Hliq + Hgas mix = -204 764 -847 138 = -1 051 902 kJ 3

3

H3 - H1 - H2 = -1 290 808 kJ . . . . or H3 - H1 - H2 = -1290808/26.038 = -49574 kW = mH Oh4 - h5 2 h5 = 42.01 +

49574 = 3346.9 ⇒ T5 = 433.4 °C 15

14-52 14.73 Natural gas (approximate it as methane) at a ratio of 0.3 kg/s is burned with 250% theoretical air in a combustor at 1 MPa where the reactants are supplied at T0. Steam at 1 MPa, 450°C at a rate of 2.5 kg/s is added to the products before they enter an adiabatic turbine with an exhaust pressure of 150 kPa. Determine the turbine inlet temperature and the turbine work assuming the turbine is reversible. CH4 + νO O2 + 3.76 N2 → CO2 + 2 H2O + 7.52 N2  2 2 νO = 2 + 2 ⇒ ν O = 2 2

Actual νO = 2 × 2.5 = 5

=>

2

2

CH4 + 5 O2 + 18.8 N2 → CO2 + 2 H2O + 3 O2 + 18.8 N2 C.V. combustor and mixing chamber HR + nH OhH O in = HP ex 2

2

. . nH2O mH2OMFu 2.5 × 16.043 kmol steam nH O = . = . = = 7.421 kmol fuel 2 nFu mFuMH2O 0.3 × 18.015 Energy equation becomes nH Ohex - hinH O + ∆hCO + 2∆hH O + 3∆hO + 18.8∆hN ex  2 2 2 2 2 2 °

= -HRP = 50010 × 16.043 = 802 310 hex - hinH O = ∆hH O ex - 15072.5, so then: 2 2 kJ ∆hCO2 + 9.421∆hH2O + 3∆hO2 + 18.8∆hN2ex = 914163 kmol fuel

Trial and error on Tex Tex = 1000 K ⇒ LHS = 749956 ; Tex = 1200 K ⇒ LHS = 987286

Tex = 1100 K ⇒ LHS = 867429

Tex ≅ 1139 K = Tin turbine If air then Tex turbine ≈ 700 K and Tavg≈ 920 K. Find CP mix between 900 and 1000K. From Table A.8: =>

53.67 + 9.421(40.63) + 3(34.62) + 18.8(32.4) CP mix = ∑ niCPi/∑ ni = 32.221 = 35.673 kJ/kmol K CV mix = CP mix - R = 27.3587, kmix= 1.304 Tex turbine = 1139 (150 / 1000)0.2331= 732 K ∆H732 = 19370.6 + 9.421(15410) + 3(13567) + 18.8(12932) = 448371 kJ

wT = Hin - Hex = ∆Hin - ∆Hex = 914163 - 448371 = 465792 kmol fuel . . • ^ WT = nFuwT = m Fuw T/MFu = (0.3 × 465792)16.043 = 8710 kW

14-53 14.74 Liquid hexane enters a combustion chamber at 31°C, 200 kPa, at the rate 1 kmol/s 200% theoretical air enters separately at 500 K, 200 kPa, and the combustion products exit at 1000 K, 200 kPa. The specific heat of ideal gas hexane is C = p 143 kJ/kmol K. Hexane:

Tc = 507.5 K, Pc = 3010 kPa

Tr1 = 0.6, Fig. D.1: Prg = 0.028, Pg1 = Pr1Pc = 84.47 kPa −* − − − * Figs D.2 and D.3: (h 1 - h 1) f = 5.16 RTc , (s− 1 - −s 1) f = 8.56 R Air: T2 = 500 K, P2 = 200 kPa, 200% theoretical air Products: T3 = 1000 K, P3 = 200 kPa − −o −* − −* −* −* − a) h C6H14 = h f - (h 1 - h 1)f + (h 1 - h 0) + (h 0 - h 0) −* − −* −* − −o h 0 - h 0 = 0 , h 1 - h 0 = CP (T1 - To) = 858 kJ/kmol, hf = -167300 kJ/kmol −* − − h 1 - h 1 = 5.16 × 8.3145 × 507.5 = 21773 kJ/kmol, h C6H14 = -188215 kJ/kmol T1 − P1 o * − −s −s = + C ln - R ln + (−s 1 - −s 1) C6H14 P To To Po T1 − P1 o − −s + C ln - R ln = 387.979 + 2.85 – 5.763 = 385.066 kJ/kmol-K P To To Po −s * - −s = 8.56×8.3145 = 71.172 kJ/kmol-K, −s 1 C6H14 = 313.894 kJ/kmol-K 1 b) C6H14 + 19O2 + 71.44N2 ‡ 6CO2 + 7H2O + 9.5O2 + 71.44N2 T3 Tc prod = ∑yiTci = 179.3 K, Tr3 = = 5.58 ‡ Ideal Gas Tc prod c) 1st Law: Q + HR = HP + W;

W=0

=>

Q = HP - H R

− − − HR = (h )C6H14 + 19∆h O2 + 71.44 ∆h N2 = -188 215 + 19 6086 + 71.44 5911 = 349701 kJ/kmol fuel −o − −o − −o − −o − HP = 6(h f + ∆h )CO2 = 7(h f + ∆h )H2O + 9.5 (h f + ∆h )O2 + 71.44(h f + ∆h )N2 CO2

-

(−h of + ∆h−) = (-393522 + 33397) = -360125 kJ/kmol

H2O

-

(−h of + ∆h−) = (-241826 + 26000) = -215826 kJ/kmol

O2

-

(−h of + ∆h−) = (0 + 22703) = 22703 kJ/kmol

14-54

N2

(−h of + ∆h−) = (0 + 21463) = 21463 kJ/kmol

-

. HP = -1922537 kJ; Q = -2272238 kW . . . d) I = To n (SP - SR) - Q; To = 25oC

o o − yO2P2 − yN2P2 SR = (−s )C6H14 + 19 (−s 500 - Rln )O2 + 71.44(s− 500 - Rln ) Po Po N2

(−s )C6H14 = 313.894 kJ/kmol K, (−s 500)O2 = 220.693 kJ/kmol K o

(−s 500)N2 = 206.740 kJ/kmol K, o

. SR = 19141.9 kW/K

yO2 = 0.21, yN2 = 0.79

Products: ni

yi

−s o i

Si

− yiPe − R ln Po

(kJ/kmol-K)

CO2

6

0.0639

269.299

17.105

286.404

H2O

7

0.0745

232.739

15.829

248.568

O2

9.5

0.1011

243.579

13.291

256.87

N2

71.44 0.7605 228.171 -3.487 SP = ∑ ni−s i = 21950.1 kJ/K; . . . I = To n (SP - SR) - Q = 3109628 kW

224.684

14-55

English unit problems 14.75E

Pentane is burned with 120% theoretical air in a constant pressure process at 14.7 lbf/in2. The products are cooled to ambient temperature, 70 F. How much mass of water is condensed per pound-mass of fuel? Repeat the answer, assuming that the air used in the combustion has a relative humidity of 90%. C5H12 + 1.2 × 8 (O2 + 3.76 N2) →

5 CO2 + 6 H2O + 0.96 O2 + 36.1 N2

Products cooled to 70 F, 14.7 lbf/in2 a) for H2O at 70 F: PG = 0.3632 lbf/in2 PG

nH2O MAX 0.3632 yH2O MAX = = = 14.7 P nH2O MAX + 42.06 Solving, nH2O MAX = 1.066 < n H2O Therefore, nH2O VAP = 1.066, nH2O LIQ = 6 - 1.066 = 4.934 mH2O LIQ =

4.934 × 18.015 = 1.232 lbm/lbm fuel 72.151

b) Pv1 = 0.9 × 0.3632 = 0.3269 lbf/in2 w1 = 0.622 ×

0.3269 = 0.014 147 14.373

nH2O IN = 0.014147 ×

28.97 × (9.6 + 36.1) = 1.040 lbmol 18.015

nH2O OUT = 1.04 + 6 = 7.04 nH2O LIQ = 7.04 - 1.066 = 5.974 lb mol nH2O LIQ =

5.974 × 18.015 = 1.492 lbm/lbm fuel 72.151

14-56 14.76E The output gas mixture of a certain air-blown coal gasifier has the composition of producer gas as listed in Table 14.2. 2 Consider the combustion of this gas with 120% theoretical air at 14.7 lbf/in. pressure. Find the dew point of the products and the mass of water condensed per pound-mass of fuel if the products are cooled 20 F below the dew point temperature? a) {3 CH4 + 14 H2 + 50.9 N2 + 0.6 O2 + 27 CO + 4.5 CO2} + 31.1 O2 + 116.9 N2 → 34.5 CO2 + 20 H2O + 5.2 O2 + 167.8 N2 Products: yH2O = yH2O MAX =

PG 14.7

=

20 34.5 + 20 + 5.2 + 167.8

⇒ PG = 1.2923 lbf/in2 → TDEW PT = 110.4 F b) At T = 90.4 F, PG = 0.7089 lbf/in2 nH2O 0.7089 yH2O = = 14.7 nH2O + 34.5 + 5.2 + 167.8

=>

n H2O = 10.51

⇒ nH2O LIQ = 20 - 10.51 = 9.49 lb mol mH2O LIQ =

9.49(18) 3(16)+14(2)+50.9(28)+0.6(32)+27(28)+4.5(44)

= 0.069 lbm/lbm fuel 14.77E Pentene, C5H10 is burned with pure oxygen in an SSSF process. The products at one point are brought to 1300 R and used in a heat exchanger, where they are cooled to 77 F. Find the specific heat transfer in the heat exchanger. C5H10 + νO O2 → 5 CO2 + 5 H2O, stoichiometric: νO = 7.5 2

2

Heat exchanger in at 1300 R, out at 77 F, so some water will condense. 5 H2O → (5 - x)H2Oliq + x H2Ovap yH

2Omax

=

Pg 77 Ptot

=

0.464 x = 0.03158 = ⇒ x = 0.163 14.696 5+x

Q⋅ q= . = 5hex - hinCO + 5hex - hinH O - (5 -x)hfg H O nfuel 2 2 vap 2 Btu

= -164340 lb mol fuel

14-57 14.78E A rigid vessel initially contains 2 pound mole of carbon and 2 pound mole of oxygen at 77 F, 30 lbf/in. 2. Combustion occurs, and the resulting products consist of 1 pound mole of carbon dioxide, 1 pound mole of carbon monoxide, and excess oxygen at a temperature of 1800 R. Determine the final pressure in the vessel and the heat transfer from the vessel during the process. 1 2 C + 2 O2 → 1 CO2 + 1 CO + O2 2 V = constant, C: solid, n1(GAS) = 2, n2(GAS) = 2.5 P 2 = P1 ×

n2T2 n1T1

= 30 ×

2.5 × 1800 lbf = 125.8 2 in 2 × 536.7

H1 = 0 1

H2 = 1(-169184 + 14358) + 1(-47518 + 9323) + 2(0 + 9761) = -188141 Btu Q = (U2-U1) = (H2-H1) - n2RT2 + n1RT1

1 2

= (-188141 - 0) - 1.98589(2.5 × 1800 - 2 × 536.67) = -194945 Btu 14.79E In a test of rocket propellant performance, liquid hydrazine (N2H4) at 14.7 lbf/in.2, 77 F, and oxygen gas at 14.7 lbf/in.2, 77 F, are fed to a combustion chamber in the ratio of 0.5 lbm O2/lbm N2H4. The heat transfer from the chamber to the surroundings is estimated to be 45 Btu/lbm N2H4. Determine the temperature of the products exiting the chamber. Assume that only H2O, H2, and N2 are present. The enthalpy of formation of liquid hydrazine is +21647 Btu/lb mole. 1

N2H4

3

Comb. Chamber

O2

Products

2 1

N2H4 + 2 O2 → H2O + H2 + N2 ⋅ ≅ (m ⋅ × 32)/(m ⋅ × 32) = 0.5 ; n⋅ O /n Fu O Fu 2

2

. ⋅ = 45 QCV/m Fu

Btu

QCV = -45 × 32.045 = -1442 lb mol fu . C.V. combustion chamber: n⋅ Fuh1 + n⋅ O h2 + QCV = n⋅ toth3 ⇒

2

- or - H1 + H2 + QCV = HP

3

°

=>

°

°

HR + QCV = HP + ∆HP

3

°

Btu

∆HP = HR - HP + QCV = 21647 + 103966 - 1442 = 124171 lb mol fuel 3

Trial and error on T3:

T3 = 5000 ⇒ ∆HP = 120071,

T3 = 5200 ⇒ ∆HP = 126224

⇒ T3 = 5133 R

14-58 14.80E Repeat the previous problem, but assume that saturated-liquid oxygen at 170 R is used instead of 77 F oxygen gas in the combustion process. Use the generalized charts to determine the properties of liquid oxygen. Problem the same as 14.79, except oxygen enters at 2 as saturated liquid at 170 R. At 170 R, Tr2 =

170 − = 0.61 ⇒ ∆h f = 5.1 278.6

From Fig. D.2: -* (h - h) = 1.98589 × 278.6 × 5.1 = 2822 Btu/lbmol °

°

∆HP = HR + ∆HR - HP + QCV = 21647 - 0.5(2822) 3

+ 0.5(0.219)(170 - 536.67)(32) + 103966 - 1442 = 121475 With ∆HP

3 5000 R

= 120071, ∆HP

⇒ T3 = 5045 R

3 5200 R

= 126224

14-59 14.81E Ethene, C2H4, and propane, C3H 8 , in a 1;1 mole ratio as gases are burned with 120% theoretical air in a gas turbine. Fuel is added at 77 F, 150 lbf/in.2 and the air comes from the atmosphere, 77 F, 15 lbf/in.2 through a compressor to 150 lbf/in.2 and mixed with the fuel. The turbine work is such that the exit temperature is 1500 R with an exit pressure of 14.7 lbf/in. 2. Find the mixture temperature before combustion, and also the work, assuming an adiabatic turbine. C2H4 + C3H8 + νO2(O2+3.76N2) → 5CO2 + 6H2O + νN2N2 φ = 1 ⇒ νO2 = 8

φ = 1.2 ⇒ νO2 = 9.6

so we have 45.696 lbmol air per 2 lbmol fuel C2H4 + C3H8 + 9.6(O2 + 3.76N2) → 5CO2 + 6H2O + 1.6 O2 + 36.096N2 C.V. Compressor (air flow) wc,in = h2 - h1

s2 = s1 ⇒ Pr1 = 1.0907

Pr2 = Pr1P2/P1 = 10.907 ⇒ T2 air = 1027.3 R wc,in = 247.81 - 128.38 = 119.53 Btu/lbm = 3462.4 Btu/lbmol air C.V. Mixing chamber • • • nairhair in + nfu1hfu1 + nfu2hfu2 = (same)exit (CPF1 + CPF2)(Tex - T0) = 45.696 CP air(T2 air - Tex) CPF1 = 11.53, CPF2 = 17.95, CP air = 6.953 45.696CP airT2 + (CPF1 + CPF2)T0 = 985.6 R = Tin combust Tex = CPF1 + CPF2 + 45.696CP air Turbine work: take C.V. total and subtract compressor work. Wtotal = Hin - Hout = HR - HP,1500 = h°f F1 + h°f F2 - 5hCO2 - 6hH2O - 36.096hN2 - 1.6hO2 = 22557 + (-44669) - 5(10557 - 169184) - 6(8306 - 103966) - 36.096 × 6925 - 1.6 × 7297.5 = 1083342 Btu/2 lbmol Fuel wT = wtot + wc,in = 1083342 + 3462.4 × 45.696 = 1 241 560 Btu/2 lbmol fuel

14-60 14.82E One alternative to using petroleum or natural gas as fuels is ethanol (C2H5OH), which is commonly produced from grain by fermentation. Consider a combustion process in which liquid ethanol is burned with 120% theoretical air in an SSSF process. The reactants enter the combustion chamber at 77 F, and the products exit at 140 F, 14.7 lbf/in.2. Calculate the heat transfer per pound mole of ethanol, using the enthalpy of formation of ethanol gas plus the generalized tables or charts. C2H5OH + 1.2 × 3(O2 + 3.76N2) → 2CO2 + 3H2O + 0.6O2 + 13.54N2 Products at 140 F, 14.7 lbf/in2, yH2O = 2.892/14.7 = nv/(2 + 0.6 + nv + 13.54) nv = 3.953 > 3 ⇒ no condensation h°f = -101032 Btu/lbmol as gas Tr = 536.67/925 = 0.58 ⇒ D.2:

∆h/RTc = 5.23

HR = -101032 - 5.23 × 1.98589 × 925 + 0 + 0 = -110639 Btu/ lbmol fuel HP = 2(-169184 + 570) + 3(-47518 + 506.5) + 0.6(443.7) + 13.54(438.4) = -472060 Btu/ lbmol fuel QCV = HP - HR = -361421 Btu/lbmol fuel

14-61 14.83E Hydrogen peroxide, H2O2, enters a gas generator at 77 F, 75 lbf/in. 2 at the rate of 0.2 lbm/s and is decomposed to steam and oxygen exiting at 1500 R, 75 lbf/in.2. The resulting mixture is expanded through a turbine to atmospheric pressure, 14.7 lbf/in.2, as shown in Fig. P14.27. Determine the power output of the turbine, and the heat transfer rate in the gas generator. The enthalpy of formation of liquid H2 O2 is −80541 Btu/lb mol. H2O2 •

→H

1

2

O + 2 O2

•

nFu = mFu/MFu = 0.2/34.015 = 0.00588 lbmol/s •

•

nex,mix = 1.5 × nFu = 0.00882 lbmol/s 2 1 Cp mix = 3 × 0.445 × 18.015 + 3 × 0.219 × 31.999 = 7.6804 Cv mix = Cp mix - 1.98588 = 5.6945 ; kmix = Cp mix/Cv mix = 1.3487

Reversible turbine T3 = T2 × (P3/P2)(k-1)/k = 1500 × (14.7/75)0.2585 = 984.3 R - w = Cp(T2 - T3) = 7.6804(1500 - 984.3) = 3960.8 Btu/lbmol • • WCV = nmix × w = 0.00882 × 3960.8 = 34.9 Btu/s

C.V. Gas generator •

•

•

QCV = H2 - H1 = 0.00588 × (-103966 + 8306) + 0.00294(7297.5) - 0.00588(-80541) = -67.45 Btu/s

14-62 14.84E In a new high-efficiency furnace, natural gas, assumed to be 90% methane and 10% ethane (by volume) and 110% theoretical air each enter at 77 F, 14.7 2 lbf/in. , and2 the products (assumed to be 100% gaseous) exit the furnace at 100 F, 14.7 lbf/in. . What is the heat transfer for this process? 2 Compare this to an older furnace where the products exit at 450 F, 14.7 lbf/in. . 0.90CH 4+ 0.10C H 2 6

Prod. Furnace

77 F 110% Air

100 F 2 14.7 lbf/in

HR = 0.9(-32190) + 0.1(-36432) = -32614 Btu 0.9 CH4 + 0.1 C2H6 + 1.1 × 2.15 O2 + 3.76 × 2.365 N2 → 1.1 CO2 + 2.1 H2O + 0.215 O2 + 8.892 N2 a) TP = 100 F HP = 1.1(-169184 + 206) + 2.1(-103966 + 185) + 0.215(162) + 8.892(160) = -402360 Btu, assuming all gas QCV = HP - HR = -369746 Btu/lb mol fuel b) TP = 450 F HP = 1.1(-169184 + 3674) + 2.1(-103966 + 3057) + 0.215(2688) + 8.892(2610) = -370184 Btu QCV = HP - HR = -337570 Btu/lb mol fuel 14.85E Repeat the previous problem, but take into account the actual phase behavior of the products exiting the furnace. Same as 14.84 , except possible condensation. a) 100 F, 14.7 lbf/in2 yv max = 0.9503 / 14.7 = nv max/[nv max + 10.207] nv max = 0.705 ⇒ nv = 0.705; nliq = 2.1 - 0.705 = 1.395 Hliq = 1.395[-122885 + 18.015(68.05 - 45.09)] = -170847 Hgas = 1.1(-169184 + 206) + 0.705(-103966 + 185) + 0.215(162) + 8.892(160) = -257584 HP = Hliq + Hgas = -428431 QCV = HP - HR = -395817 Btu/lbmol fuel b) Older furnace has no condensation, so same as in 14.84.

14-63 14.86E Methane, CH , is burned in an SSSF process with two different oxidizers: A. Pure oxygen, 4O2 and B a mixture of O2 + x Ar. The reactants are supplied at T , 0 P and the products in are at 3200 R both cases. Find the required equivalence 0 ratio in case A and the amount of Argon, x, for a stoichiometric ratio in case B. a) CH4 + νO2 → CO2 + 2H2O + (ν - 2)O2 νs = 2 for stoichiometric mixture. HP 3200 = HR° = HP° + ∆HP 3200 ∆hCO2 = 33579, ∆hH2O = 26479, ∆hO2 = 21860 ° = (50010/2.326) × 16.04 = 344867 ∆HP = HR° - H°P = -HRP = 33579 + 2 × 26479 + (ν - 2)21860 = 42817 + ν × 21860 ν = 13.8175 ⇒ φ = ν/νs = 6.91 b) CH4 + 2O2 + 2xAr → CO2 + 2H2O + 2xAr ∆HP = 33579 + 2 × 26479 + 2 × 0.1253 × 39.948 × (3200 - 536.67) = 86537 + x × 26662.5 = 344867 ⇒ x = 9.689

14.87E Butane gas at 77 F is mixed with 150% theoretical air at 1000 R and is burned in an adiabatic SSSF combustor. What is the temperature of the products exiting the combustor? C4H10 + 1.5 × 6.5(O2 + 3.76N2) → 4CO2 + 5H2O + 3.25 O2 + 36.66N2 HR = HR° + ∆Hair,in HP = HP° + 4∆hCO + 5∆hH 2

2O

+ 3.25∆hO + 36.66∆hN 2

2

HP = HR ⇒ ∆HP = HR° - H°P + ∆Hair,in 45714 ° + ∆H ∆HP = -HRP = 2.326 × 58.124 + 9.75 × 3366 + 36.66 × 3251 air,in

= 1294339 Btu/lbmol fuel = 4∆hCO2 + 5∆hH2O + 3.25∆hO2 + 36.66∆hN2 at Tad ∆HP,3600R = 1281185

∆HP,3800R = 1374068

Tad = 3628 R

14-64 14.88E Liquid n-butane at T , is sprayed into a gas turbine with primary air flowing at 2 150 lbf/in. , 700 R 0in a stoichiometric ratio. After complete combustion, the products are at the 2 adiabatic flame temperature, which is too high so secondary air at 150 lbf/in. , 700 R is added, with the resulting mixture being at 2500 R. Show that T > 2500 R and find the ratio of secondary to primary air flow. ad C4H10 + 6.5(O2 + 3.76N2) → 4CO2 + 5H2O + 24.44N2 2nd.air

Fuel Primary air

Tad MIXING

COMBUSTOR

2500 R

C.V. Combustor HR = Hair + HFu = HP = HP° + ∆HP = HR° + ∆HR ° + ∆H ∆HP = HR° − H°P + ∆HR = -HRP R = 45344 × 58.124/2.326 + 6.5(1158 + 3.76 × 1138) = 1168433 Btu/lbmol fuel ∆HP,2500R = 4 × 23755 + 5 × 18478 + 24.44 × 14855 = 550466 ∆HP > ∆HP,2500R ⇒ Tad > 2500 R (If iteration Tad≅ 4400 R) C.V. Mixing chamber ∆HP + νO2 2nd∆Hair,700 = ∆HP,2500R + νO2 2nd∆Hair,2500R νO2 2nd =

∆HP - ∆HP,2500 ∆Hair,2500 - ∆Hair,700

=

1168433 - 550466 = 9.344 71571 - 5437

Ratio = νO2 2nd/νO2 Prim. = 9.344/6.5 = 1.44

14-65 2 14.89E Acetylene gas at 77 F, 14.7 lbf/in. is fed to the head of a cutting torch. Calculate the adiabatic flame temperature if the acetylene is burned with 100% theoretical air at 77 F. Repeat the answer for 100% theoretical oxygen at 77 F. a) C2H4 + 2.5 O2 + 2.5 × 3.76 N2 → 2 CO2 + 1 H2O + 9.4 N2 -o HR = hf C2H2 = 97477 Btu -* -* -* HP = 2(-169184 + ∆hCO2) + 1(-103966 + ∆hH2O) + 9.4 ∆hN2 QCV = HP - HR = 0 -* -* -* ⇒ 2 ∆hCO2 + 1 ∆hH2O + 9.4 ∆hN2 = 539811 Btu Trial and Error: TPROD = 5236 R 2 × 147196 + 121488 + 9.4 × 89348 = 1255751 OK b) C2H2 + 2.5 O2 → 2 CO2 + H2O HR = 97477 Btu -* -* HP = 2(-169184 + ∆hCO2) + 1(-103966 + ∆hH2O) -* -* ⇒ 2 ∆hCO2 + 1 ∆hH2O = 539811 At 10000 R (limit of C.7): 2 × 135426 + 118440 = 389292 At 9500 R: 2 × 127734 + 111289 = 366757 or 4507/100 R change, TPROD ≈ 10000 +

Difference, extrapolating

150519 ≈ 13340 R 45.07

14.90E Ethene, C2H 4 , burns with 150% theoretical air in an SSSF constant-pressure process with reactants entering at P , T . Find the adiabatic flame temperature. 0 0 C2H4 + 1.5 × 3(O2 + 3.76N2) → 2CO2 + 2H2O + 1.5O2 + 16.92N2 HP = HR = HP° + ∆HP = HR° ⇒ ° = 28.054 × 47158/2.326 = 568775 ∆HP = HR° - H°P = -HRP ∆HP = 2∆hCO2 + 2∆hH2O + 1.5∆hO2 + 16.92∆hN2 Trial and error on Tad... ∆HP,3400R = 545437 Tad = 3510 R

∆HP,3600R = 587736

14-66 14.91E Solid carbon is burned with stoichiometric air in an SSSF process, as shown in Fig. P14.39. The reactants at T , P are heated in a preheater to T = 900 R with 0 0before flowing to a second heat 2 exchanger, the energy given by the products which they leave at T . Find the temperature of the products T , and the heat 0 (4 to 5) in the second heat exchanger. 4 transfer per lb mol of fuel a) Following the flow we have: Inlet T1, after preheater T2, after mixing and combustion chamber T3, after preheater T4, after last heat exchanger T5 = T1. b) Products out of preheater T 4. Control volume: Total minus last heat exchanger. C + O2 + 3.76N2 → CO2 + 3.76N2 Energy Eq.: ° ° HR = HR = HP = HP + ∆HP = hf CO + ∆hCO + 3.76∆hN 3

h°f CO = -169184, ∆HP

3

3 4400

2

2

= 167764, ∆HP

3 4600

2

2

= 177277

⇒ T3 = Tad.flame = 4430 R c) Control volume total. Then energy equation: ° ° HR + q- = HP Btu ° = h-°f CO - 0 = -169184 lbmol fuel q- = HRP 2

14.92E A closed, insulated container is charged with a stoichiometric ratio of oxygen and 2 hydrogen at 77 F and 20 lbf/in. . After combustion, liquid water at 77 F is sprayed in such that the final temperature is 2100 R. What is the final pressure? 1

H2 + 2O2 → H2O ;

P: 1 H2O + xiH2O

3-° U2 - U1 = xihi = xihf liq = 1 + xiHP - HR - 1 + xiRTP + 2RTR

HR = φ, HP = -103966 + 14218.5 = -89747.5, h°f liq= -122885 Substitute xi(-122885 + 89747.5 + 1.98588 × 2100) 3

= -89747.5 - 1.98588(2100 - 2 × 536.67)= -92319.2 xi = 3.187 P1V1 = nRRT1, P2V2 = npRTp ⇒ P2 =

P1(1 + xi)TP 1.5 T1

=

20(4.187)(2100) = 218.5 lbf/in2 1.5(536.67)

14-67 14.93E Blast furnace gas in a steel mill is available at 500 F to be burned for the generation of steam. The composition of this gas is, on a volumetric basis, Component CH4 H2 CO CO2 N2 H2O Percent by volume 0.1 2.4 23.3 14.4 56.4 3.4 3 Find the lower heating value (Btu/ft ) of this gas at 500 F and P . 0 Of the six components in the gas mixture, only the first 3 contribute to the heating value. These are, per lb mol of mixture: 0.024 H2, 0.001 CH4, 0.233 CO For these components, 0.024 H2 + 0.001 CH4 + 0.233 CO + 0.1305 O2 → 0.026 H2 + 0.234 CO2 The remainder need not be included in the calculation, as the contributions to reactants and products cancel. For the lower HV(water vapor) at 500 F hRP = 0.026(-103966 + 3488) + 0.234(-169184 + 4229) - 0.024(0 + 2101) - 0.001(-32190 + 0.538 × 16.04(500-77)) - 0.233(-47518 + 2981) - 0.1305(0 + 3069) = -31257 Btu/lb mol fuel RT0 1545 × 536.7 -v = = = 391.47 ft3/lb mol 0 P0 14.7 × 144 LHV = 31 680 / 391.47 = 79.85 Btu/ft3

14-68 14.94E Two pound moles of ammonia are burned in an SSSF process with x lb mol of oxygen. The2 products, consisting of H2O, N2, and the excess O2, exit at 400 F, 1000 lbf/in. . a. Calculate x if half the water in the products is condensed. b. Calculate the absolute entropy of the products at the exit conditions. 2NH3 + xO2 → 3H2O + N2 + (x - 1.5)O2 Products at 400 F, 1000 lbf/in2 with nH2O LIQ= nH2O VAP= 1.5 a) yH2O VAP =

PG P

=

247.1 1.5 = 1000 1.5 + 1 + x - 1.5

x = 5.070 b) SPROD = SGAS MIX + SH2O LIQ Gas mixture: ni

yi

-s° i

yiP

-Rln

Si

P0

H 2O

1.5

0.2471

48.939

-5.604

43.335

O2

3.57

0.5881

52.366

-7.326

45.040

N2

1.0

0.1648

49.049

-4.800

44.249

SGAS MIX = 1.5(43.335) + 3.57(45.040) + 1.0(44.249) = 270.04 Btu/R SH2O LIQ = 1.5[16.707 + 18.015(0.5647 - 0.0877)] = 37.95 Btu/R SPROD = 270.04 + 37.95 = 307.99 Btu/R

14-69 14.95E Graphite, C, at P , T is burned with air coming in at P , 900 R in a ratio so the 0 R. Find the equivalence ratio, the 0 percent theoretical air products exit at P0 , 2200 0 and the total irreversibility. C + φ(O2 + 3.76N2) → CO2 + (φ - 1)O2 + 3.76φN2 ° HP = HR ⇒ ∆HP,2200 - ∆HR = HR° - H°P = -HRP 19659 + (φ - 1)13136 + 3.76 × φ × 12407 - φ(2616 + 3.76 × 2541) = 0 - (-169184) ⇒ 6523 + φ × 47616.2 = 169184 ⇒ φ = 3.416 or 342 % theoretical air Equivalence ratio = 1/φ = 0.293 S gen = sP - sR =

∑ν (s° - R-ln y ) - ∑ν (s° - R-ln y ) i

i

i

i

P

R: yO = 0.21 2

i

i

R

yN = 0.79 2

P: yO = 0.1507 2

yN = 0.79 2

yCO = 0.0593

Sgen = 66.952 + 5.610 + 2.416(59.844 + 3.758) + 3.76 × 3.416(56.066 + 0.468) - 1.371 - 3.416(52.686 + 3.099 + 3.76(49.353 + 0.468)) = 120.5 Btu/lbmol C × R I = T0Sgen = 64677 Btu/lbmol C

2

14-70 14.96E A small air-cooled gasoline engine is tested, and the output is found to be 2.0 hp. The temperature of the products is measured and found to be 730 F. The products are analyzed on a dry volumetric basis, with the following result 11.4% CO 2, 2.9% CO, 1.6% O2 and 84.1% N2 . The fuel may be considered to be liquid octane. The fuel and air enter the engine at 77 F, and the flow rate of fuel to the engine is 1.8 lbm/h. Determine the rate of heat transfer from the engine and its thermal efficiency. a C8H18 + b O2 + 3.76b N2 → 11.4 CO2 + 2.9 CO + c H2O + 1.6 O2 + 84.1 N2 b = 84.1/3.76 = 22.37,

a = (1/8)(11.4 + 2.9) = 1.788,

C8H18 + 12.5 O2 + 47.1 N2 → 6.38 CO2 + 1.62 CO + 9 H2O + 0.89 O2 + 47.1 N2 -° a) HR = hf C8H18 = -107526 Btu HP = 6.38(-169184 + 6807) + 1.62(-47518 + 4647) + 9(-103966 + 5475) + 0.89(0+4822) + 47.1(0 + 4617) = -1770092 Btu HP-HR = -1770092 - (-107526) = -1662566 Btu . . 1.8 HP-HR = (-1662566) = -26198 Btu/h 114.23 . QCV = -26198 + 2.0(2544) = -21110 Btu/h . b) QH = 1.8 × 20590 = 37062 . WNET = 2.0 × 2544 = 5088 ;

ηTH =

5088 = 0.137 37062

c = 9a = 16.088

14-71 14.97E A gasoline engine uses liquid octane and air, both supplied at P , T , in a 0 stoichiometric ratio. The products (complete combustion) flow out of 0the exhaust valve at 2000 R. Assume that the heat loss carried away by the cooling water, at 200 F, is equal to the work output. Find the efficiency of the engine expressed as (work/lower heating value) and the second law efficiency. C8H18 + 12.5(O2 + 3.76N2) → 8CO2 + 9H2O + 47N2 LHV = 44425 × 114.232/2.326 = 2181753 Btu/lbmol fuel ∆HP,2000 = 8 × 16982 + 9 × 13183 + 47 × 10804 = 643470 C.V. Total engine: Hin = Hex + W + Qloss = Hex + 2W W = (Hin - Hex)/2 = (HR - HP)/2 = (-H°RP - ∆HP,2000)/2 = (2181753 - 643470)/2 = 769142 Btu/lbmol fuel ηTH = W/LHV = 769142/2181753 = 0.353 For 2nd law efficiency we must find reversible work Sin = -sfuel + 12.5(s-O + 3.76s-N ) 2

2

= 86.122 + 12.5 [48.973 - 1.98589 ln (1/4.76)] + 47[45.739 - 1.98589 ln (3.76/4.76)] = 2908.8 Btu/lbmol fuel × R Sex = 8sCO + 9sH 2

2O

+ 47s-N = 8[65.677 - 1.98589 ln (8/64)] 2

+ 9[56.619 - 1.98589 ln (9/64)] + 47[55.302 - 1.98589 ln (47/64)] = 3731.1 Btu/lbmol fuel × R Assume the same Qloss out to 200 F = 659.67 R reservoir and compute Qrev : 0 /T0 = Sex + Qloss/Tres Sin + Qrev 0 Qrev = T0(Sex - Sin) + QlossT0/Tres 0 = 536.67(3731.1 - 2908.8) + 769142*536.67/659.67 = 1067034 Btu/lbmol fuel Wrev = Hin - Hex - Qloss + Qrev = Wac + Qrev 0 0 = 769142 + 1067034 = 1836176 ηII = Wac/Wrev = 769142/1836176 = 0.419

14-72 14.98EIn Example 14.16, a basic hydrogen–oxygen fuel cell reaction was analyzed at 25°C, 100 kPa. Repeat this calculation, assuming that the fuel cell operates on air at 77 F, 14.7 lbf/in.2, instead of on pure oxygen at this state. Anode: 2 H2

→ 4 e- + 4 H+

Cathode: 4 H+ + 4 e- → 1 O2 + 2 H2O Overall: 2 H2 + 1 O2 → 2 H2O Example 14.16 : ∆G25°C = -474283 kJ/kmol Or ∆G77 F = -203904 Btu/lbmol PO = yO2 × P = 0.21 × 14.7 = 3.087 lbf/in2 2

-s = 48.973 – 1.98589 ln 0.21 = 52.072 O 2

∆S = 2(16.707) – 2(31.186) – 1(52.072) = - 81.03 Btu/R ∆H = 2(-122885) – 2(0) – 1(0) = -245770 Btu ∆G77 F = -245770 – 536.67(-81.03) = 202 284 Btu E° = -∆G/N0e ne = 202284 × 2.326/(96485 × 4) = 1.219 V

15-1

CHAPTER 15 The correspondence between the new problem set and the previous 4th edition chapter 13 problem set. New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Old 1 mod 2 mod new 11 new 12 new 13 14 new 16 new 17 new 15 new 18 20 new 21

New 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

Old 22 19 23 24 25 27 28 new 29 30 new new 31 32 33 35 34 new 36 37 mod

New 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57

Old 38 39 40 41 42 43 45 47 49 mod 50 51 52 new new new 44 48

The problems that are labeled advanced start at number 53.

The English unit problems are: New 58 59 60 61 62 63

Old 53 new 58 59 60 61

New 64 65 66 67 68 69

Old 62 63 64 new new 65

New 70 71 72 73 74 75

Old 67 68 69 70 71 72

15-2 15.1

Carbon dioxide at 15 MPa is injected into the top of a 5-km deep well in connection with an enhanced oil-recovery process. The fluid column standing in the well is at a uniform temperature of 40°C. What is the pressure at the bottom of the well assuming ideal gas behavior?

Z1

(Z1-Z2) = 5000 m, P1 = 15 MPa

1

T = 40 oC = const

CO 2 Z2

Equilibrium at constant T

2

-wREV = 0 = ∆g + ∆PE = RT ln (P2/P1) + g(Z2-Z1) = 0 ln (P2/P1) =

9.807×5000 = 0.8287 1000×0.188 92×313.2

P2 = 15 exp(0.8287) = 34.36 MPa 15.2

Consider a 2-km-deep gas well containing a gas mixture of methane and ethane at a uniform temperature of 30°C. The pressure at the top of the well is 14 MPa, and the composition on a mole basis is 90% methane, 10% ethane. Determine the pressure and composition at the bottom of the well, assuming an ideal gas mixture. Z1 (Z1-Z2) = 2000 m, Let A = CH4, B = C2H6 1 GAS MIX A+B 2

P1 = 14 MPa, yA1 = 0.90, yB1 = 0.10

T = 30 oC = const Z2 From section 15.1, for A to be at equilibrium between 1 and 2: WREV = 0 = n A(GA1-GA2) + nAMAg(Z1-Z2) Similarly, for B: WREV = 0 = n B(GB1-GB2) + nBMBg(Z1-Z2) Using eq. 15.10 for A: RT ln (PA2/PA1) = MAg(Z1-Z2) with a similar expression for B. Now, ideal gas mixture, Substituting:

ln

yA2P2 yA1P1

=

MAg(Z1-Z2) RT

ln (yA2P2) = ln(0.9×14) +

and

ln

PA1 = yA1P, etc.

yB2P2 yB1P1

=

16.04×9.807(2000) = 2.6585 1000×8.3145×303.2

=> y A2P2 = 14.2748 ln (yB2P2) = ln(0.1×14) + =>

30.07×9.807(2000) = 0.570 43 1000×8.3145×303.2

y B2P2 = (1-yA2)P2 = 1.76903

Solving: P2 = 16.044 MPa &

yA2 = 0.8897

MBg(Z1-Z2) RT

15-3 15.3

Using the same assumptions as those in developing Eq. d in Example 15.1: Develop an expression for pressure at the bottom of a deep column of liquid in terms of the isothermal compressibility, β T. For liquid water at 20°C, β T=0.0005 [1/MPa]. Use the result of the first question to estimate the pressure in the Pacific ocean at the depth of 3 km. d gT = v° (1-β TP) dPT

d gT + g dz = 0

v° (1-β TP) dPT + g dz = 0 and integrate ⌠P (1-β P) dP = + g ⌠+Hdz T T ⌡P 0 v° ⌡0 P (1 -

⌠ ⌡dz ⌡v°(1-β TP) dPT = - g ⌠

1 g P - P0 - β T [P2 - P02] = H 2 v°

=>

1 1 g β P) = P0 - β T P02 + H 2 T 2 v°

v° = vf 20°C = 0.001002; H = 3000 m , g = 9.80665 m/s2; βT = 0.0005 1/MPa 1

1

P (1 - 2 × 0.0005P) = 0.101 - 2 × 0.0005 × 0.1012 + [9.80665 × 3000/0.001002] × 10-6 = 29.462 MPa,

which is close to P

P = 29.682 MPa

15.4

Calculate the equilibrium constant for the reaction O2 <=> 2O at temperatures of 298 K and 6000 K. Reaction

O2

<=>

2O

At 25 oC (298.15 K): -0 -0 ∆H0 = 2hf O - 1hf O2 = 2(249 170) - 1(0) = 498 340 kJ/kmol 0 0 ∆S0 = 2s-O - 1s-O2 = 2(161.059) - 1(205.148) = 116.97 kJ/kmol K

∆G0 = ∆H0 - T∆S0 = 498 340 - 298.15×116.97 = 463 465 kJ/kmol ∆G0 463 465 = -186.961 ln K = - - = RT 8.3145×298.15 At 6000 K: ∆H0 = 2(249 170+121 264) - (0+224 210) = 516 658 kJ/kmol ∆S0 = 2(224.597) -1(313.457) = 135.737 kJ/kmol K ∆G0 = 516 658 - 6000×135.737 = -297 764 kJ/kmol ln K =

+297 764 = +5.969 8.3145×6000

15-4 15.5

Calculate the equilibrium constant for the reaction H2 <=> 2H at a temperature of 2000 K, using properties from Table A.8. Compare the result with the value listed in Table A.10. ∆hH = 52942

sH = 188.419

−o hf = 0

∆hH = 35375

sH = 154.279

−o h f = 217999

2

2

∆G0 = ∆H - T∆S = HRHS - HLHS – T (S0RHS - S0LHS) = 2 × (35375+217999) – 52943 – 2000(2×154.279-182.419) = 213528 ln K = -∆G0/RT = -213528 / (8.3145×2000) = -12.8407 Table A.10 ln K = -12.841 OK

15.6

Plot to scale the values of lnK versus 1/T for the reaction 2CO2 <=> 2CO + O 2. Write an equation for lnK as a function of temperature. 2 CO2 ⇔ 2 CO + 1 O2 T(K)

104 ×

2000 2400 2800 3200 3600

5.000 4.167 3.571 3.125 2.778

1 T

For the range below ~ 5000 K, ln K ≈ A + B/T Using values at 2000 K & 5000 K A = 19.5056 B = -65 543 K

ln K

T(K)

-13.266 -7.715 -3.781 -0.853 1.408

4000 4500 5000 5500 6000

1 T 2.500 2.222 2.000 1.818 1.667

ln K

104 ×

8

3.204 4.985 6.397 7.542 8.488

almost

4 linear

0 -4 -8 -12 0

1

2

3

4

5

4 1 10 x _ T

15-5 15.7

Calculate the equilibrium constant for the reaction:2CO <=> 2CO + O at 3000 2 2 K using values from Table A.8 and compare the result to Table A.10. ∆hCO = 93504

-0 hf CO = -110527

sCO = 273.607

∆hCO = 152853

-0 hf CO = -393522

sCO = 334.17

∆hO = 98013

sO = 284.466

2

2

2

2

2

∆G0 = ∆H - T∆S = 2 HCO + HO – 2 HCO - T (2sCO + sO - 2sCO ) 2

2

2

2

= 2 (93504 – 110527) + 98013 + 0 – 2(152853-393522) -3000(2×273.607 + 284.466 - 2×334.17) = 55285 ln K = -∆G0/RT = -55285/ (8.31451×3000) = -2.2164 Table A.10 ln K = -2.217 OK

15.8

Pure oxygen is heated from 25°C to 3200 K in an SSSF process at a constant pressure of 200 kPa. Find the exit composition and the heat transfer. The only reaction will be the dissociation of the oxygen O2 ⇔ 2O ; K(3200) = 0.046467 Look at initially 1 mol Oxygen and shift reaction with x nO2 = 1 - x; nO = 2x; ntot = 1 + x; yi = ni/ntot y2O P 4x2 1 + x 8x2 2-1 K= ( ) = 2= y0 Po (1 + x)2 1 - x 1 - x2 2

K/8 ⇒ x = 0.07599; y0 = 0.859; y0 = 0.141 1 + K/8 2 q = n0 exh0 ex + n0exhOex - h0 in = (1 + x)(y 0 h0 + y0hO) - 0 x2 =

2

2

h0 = 106022; 2

2

hO = 249170 + 60767

2

2

⇒

q = q/32 = 4532 kJ/kg (=3316.5 if no reaction)

q = 145015 kJ/kmol O2

15-6 15.9

Pure oxygen is heated from 25°C, 100 kPa to 3200 K in a constant volume container. Find the final pressure, composition, and the heat transfer. As oxygen is heated it dissociates O2 ⇔ 2O

ln Keq = -3.069

C. V. Heater: U2 - U1 = 1Q2 = H2 - H1 - P2v + P1v - Per mole O2: 1q2 = h2 - h1 + R(T1 - (n2/n1)T2) Shift x in reaction 1 to have final composition: (1 - x)O 2 + 2xO n1 = 1 yO

22

n2 = 1 - x + 2x = 1 + x

= (1 - x)/(1 + x) ; yO2 = 2x/(1 + x)

Ideal gas and V2 = V1 ⇒ P2 = P1n2T2/n1T1 Keq = e ⇒

-3.069

⇒ P2/Po = (1 + x)T2/T1

y2O P2 2x 2 1 + x 1 + x T2 = ( )=( ) ( )( )( ) 1+x 1-x 1 y02 Po T1

4x2 T1 -3.069 = e = 0.00433 ⇒ x=0.0324 1 - x T2

(nO )2 = 0.9676, (nO)2 = 0.0648, n2 = 1.0324 2 q = 0.9676(106022) + 0.0648(249170 + 60767) - Ø 1 2 + 8.3145(298.15 - 1.0324(3200)) = 97681 kJ/kmolO2 yO 2=0.9676/1.0324 = 0.937; yO2=0.0648/1.0324 = 0.0628 2

15.10 Nitrogen gas, N2 , is heated to 4000 K, 10 kPa. What fraction of the N2 is dissociated to N at this state? N2 <=> 2 N Initial

1

0

Change

-x

2x

Equil.

1-x

2x

yN2 =

@ T = 4000 K, lnK = -12.671 K = 3.14x10-6

ntot = 1 - x + 2x = 1 + x

2x 1-x , yN = 1+x 1+x

2

yN  P 2-1 K=   ; => yN2 Po nN2 = 0.99911,

3.14x10 -6 =

nN = 0.00177,

4x2  10    1 - x2 100

yN2 = 0.9982,

=> x = 0.000886 yN = 0.0018

15-7 15.11 Hydrogen gas is heated from room temperature to 4000 K, 500 kPa, at which state the diatomic species has partially dissociated to the monatomic form. Determine the equilibrium composition at this state. H2 ⇔ 2H

Equil.

-x +2x

nH2 = 1 - x nH = 0 + 2x n =1+ x

(2x)2 P 2-1 K= ( ) (1-x)(1+x) P0

at 4000 K:

2.545 x2 = 0.127 25 = 4×(500/100) 1-x2

ln K = 0.934 => K = 2.545

Solving,

x = 0.3360

nH2 = 0.664, nH = 0.672, ntot = 1.336 yH2 = 0.497, yH = 0.503

15.12 Consider the chemical equilibrium involving H2O, H2, CO, and CO2, and no other substances. Show that the equilibrium constant at any temperature can be found using values from Table A.10 only. The reaction between the species is (1)

1 CO2 + 1 H2 ⇔ 1 CO + 1 H2O

From A.10 we find: (2) 2 H2O ⇔ 2 H2 + 1 O2,

(3) 2 CO2 ⇔ 2 CO + 1 O2

Form (1) from (2) and (3) as: (1) = From A.10

1 1 (3) - (2) 2 2 ln K

=>

ln K = (ln K3 - ln K2)/2

at any T is listed for reactions (2) and (3)

15-8 15.13 One kilomole Ar and one kilomole O2 is heated up at a constant pressure of 100 kPa to 3200 K, where it comes to equilibrium. Find the final mole fractions for Ar, O2, and O. The only equilibrium reaction listed in the book is dissociation of O2. So assuming that we find in Table A.10:

ln(K) = -3.072

Ar + O2 ⇒ Ar + (1 - x)O2 + 2x O The atom balance already shown in above equation can also be done as Species

Ar

O2

Start

1

0

Change

0

-x

2x

Total

1

1-x

2x

1

The total number of moles is

O

ntot = 1 + 1-x + 2x = 2 + x

so

yAr = 1/(2 + x); yO = 1 - x/(2 + x); yO = 2x/(2 + x) 2 and the definition of the equilibrium constant (Ptot = Po) becomes K=e

-3.072

y2O

4x2 = 0.04633 = = y02 (2 + x)(1 - x)

The equation to solve becomes from the last expression (K + 4)x2 + Kx - 2K = 0 If that is solved we get x = -0.0057 ± 0.1514 = 0.1457; yO = 0.1358;

y02 = 0.3981;

x must be positive yAr = 0.4661

15-9 15.14 A piston/cylinder contains 0.1 kmol hydrogen and 0.1 kmol Ar gas at 25°C, 200 kPa. It is heated up in a constant pressure process so the mole fraction of atomic hydrogen is 10%. Find the final temperature and the heat transfer needed. When gas is heated up H2 splits partly into H as H2 ⇔ 2H Component

and the gas is diluted with Ar H2

Ar

H

Initial

0.1

0.1

0

Shift

-x

0

2x

Final

0.1-x 0.1

2x

yH = 0.1 = 2x/(0.2+x) ⇒ 2x = 0.02+0.1x

Total = 0.2 + x ⇒

x = 0.010526

⇒ ntot = 0.21053 yH2 = 0.425 = [(0.1-x)/(0.2+x)];

yAr = 1 – rest = 0.475

Do the equilibrium constant: 2

K(T) =

yH yH2

0.01 (PP0)2-1 = (0.425 ) × (200 ) = 0.047059 100

ln (K) = -3.056 so from Table A.10 interpolate to get T = 3110 K To do the energy eq., we look up the enthalpies in Table A.8 at 3110K hH2 = 92829.1; hH = 217999 + 58447.4 = 276445.4 (= hf + ∆h) hAr = CP(3110–298.15) = 20.7863 × (3110-298.13) = 58447.9 (same as ∆h for H) Now get the total number of moles to get nH = 0.021053;

nH2 = ntot ×

1-x = 0.08947; 2+x

nAr = 0.1

Since pressure is constant W = P∆V and Q becomes differences in h Q = n∆h = 0.08947 × 92829.1 – 0 + 0.021053 × 276446.4 – 0 + 0.1 × 58447.9 = 19970 kJ

15-10 15.15 Air (assumed to be 79% nitrogen and 21% oxygen) is heated in an SSSF process at a constant pressure of 100 kPa, and some NO is formed. At what temperature will the mole fraction of N.O be 0.001? 0.79 N2 + 0.21 O2 heated at 100 kPa, forms NO N2 + O2 ⇔ 2 NO

nN2 = 0.79 - x

-x -x

nO2 = 0.21 - x

+2x

nNO = 0

+ 2x

ntot = 1.0 At exit, yNO = 0.001 =

2x 1.0

⇒

x = 0.0005

⇒ nN2 = 0.7895, nO2 = 0.2095 2

K=

yNO yN2yO2

-6

10 (PP0)0 = 0.7895×0.2095 = 6.046×10-6

From Table A.10,

T = 1444 K

or ln K = -12.016

15-11 15.16 Saturated liquid butane enters an insulated constant pressure combustion chamber at 25°C, and x times theoretical oxygen gas enters at the same pressure and temperature. The combustion products exit at 3400 K. Assuming that the products are a chemical equilibrium gas mixture that includes CO, what is x? Butane: T1 = 25oC, sat. liq., x1 = 0, Tc = 425.2 K, Pc = 3.8 Mpa Tr1 = 0.7, Figs. D.1and D.2, Pr1 = 0.10, P1 = Pr1Pc = 380 kPa − * −  h 1 − h 1f = 4.85 RTc Oxygen: T = 25oC, X * theoretical air

Products: T3 = 3400 K C4H10 + 6.5XO2 => ?CO 2 + 5H2O + ?O2 2

2CO2 <=> 2CO + O2 Initial

4

0

6.5(X-1)

Change

-2a

2a

a

Equil.

4-2a

2a

6.5(X-1) + a

nCO = 4 - 2a, nCO = 2a, nO = 6.5(X-1) + a, nH2O = 5, ntot = 2.5 + a + 6.5X 2 2 2a 4 - 2a 6.5(x - 1) + a yCO = , yCO2 = , yO2 = 2.5 + a +6.5X 2.5 + a +6.5X 2.5 + a +6.5X 2

yCOyO P12+1-2  a 2 6.5X - 6.5 + a P1 K= 2 2  =     2-a 6.5X - 2.5 + a Po Po   y CO 2

@ T3 = 3400 K,

ln(K) = 0.346,

K = 1.4134

 a 2 6.5X - 6.5 + a 1.4134 =     (3.76) 2-a 6.5X - 2.5 + a

Equation 1.

Need a second equation: 1st Law: Qcv + HR = HP + Wcv; Qcv = 0, Wcv = 0 −o − HR = (h f + ∆h )C H = (-126200 - 17146) = -143346 kJ 4 10

Products @ 3400 K: −o − −o − −o − −o − HP = n(h f + ∆h )CO2 + n(h f + ∆h )CO + n(h f + ∆h )O2 + n(h f + ∆h )H2O = (4 - 2a)(-393522 + 177836) + 2a(-110527 + 108440) + [6.5(X - 1) + a](0 + 114101) + 5(-241826 + 149073) = -463765 kJ/kmol H P = HR

=>

1924820 = 541299a + 741656.5 X

Two equations and two unknowns, solve for X and a. a ≅ 0.87, X ≅ 1.96

Equation 2.

15-12 15.17 The combustion products from burning pentane, C5H12, with pure oxygen in a stoichiometric ratio exists at 2400 K, 100 kPa. Consider the dissociation of only CO2 and find the equilibrium mole fraction of CO. C5H12 + 8 O2 → 5 CO2 + 6 H2O 2 CO2 ⇔ 2 CO + 1 O2

At 2400K, ln K = -7.715 K = 4.461x10-4

Initial Change Equil.

5 -2z 5-2z

0 +2z 2z

0 +z z

Assuming P = Po = 0.1 MPa, and ntot = 5 + z + 6 = 11 + z y2COyO K=

2

y2CO

2

P  2z 2  z  ( 0) =    (1) = 4.461x10-4 ; P 5 - 2z 11 + z

Trial & Error (compute LHS for various values of z): nCO = 4.418; 2

nCO = 0.582;

nO = 0.291 2

z = 0.291

=>

yCO = 0.0515

15.18 Find the equilibrium constant for the reaction 2NO + O ⇔ 2NO from the 2 2 elementary reactions in Table A.10 to answer which of the nitrogen oxides, NO or NO , is the more stable at ambient conditions? What about at 2000 K? 2 2 NO + O2 ⇔ 2 NO2 (1) But N2 + O2 ⇔ 2 NO (2) N2 + 2 O2 ⇔ 2 NO2 (3) Reaction 1 = Reaction 3 - Reaction 2 ⇒

0

0

0

∆G1 = ∆G3 - ∆G2

At 25 oC, from Table A.10: or

=>

ln K1 = ln K3 - ln K2

ln K1 = -41.355 - (-69.868) = +28.513

K1 = 2.416×1012

an extremely large number, which means reaction 1 tends to go very strongly from left to right. At 2000 K:

ln K1 = -19.136 - (-7.825) = -11.311

or K1 = 1.224×10-5

meaning that reaction 1 tends to go quite strongly from right to left.

15-13 15.19 Methane in equilibrium with carbon and hydrogen as: CH4 ⇔ C + 2H2. has lnK = -0.3362 at 800 K. For a mixture at 100 kPa, 800 K find the equilibrium mole fractions of all components (CH 4, C, and H2 neglect hydrogen dissociation).Redo the molefractions for a mixture state of 200 kPa, 800 K. CH4 ⇔ C + 2H2

ln K = - 0.3362

nCH = 1-x; nC = x; nH = 2x; ntot = 1+2x 4 2 x   4x  1+2x ×   (1+2x)  100  = ×    1-x   100  1+2x  2

y2 yC H2

yCH

4

 P 1+2-1 = K = e -0.3362 P   o

2

2

x3 0.71448 = = = 0.17862 2 4 (1+2x) (1-x) Solve by trial and error or successive substitutions x = 0.5

LHS = 0.0625

x = 0.6

LHS = 0.11157

x = 0.7

LHS = 0.198495

x ≈ 0.67

LHS = 0.16645

Interpolate

LHS = 0.17787 ⇒ x = 0.682

x = 0.6814

 1-x  yCH4 =   = 0.1345 1+2x

 x  yC =   = 0.2885 1+2x

 2x  yH2 =   = 0.577 1+2x  P  200 P  = 100 = 2  o  

⇒

x3 0.71448 = = 0.044655 2 (1+2x) (1-x) 4 × 22

x = 0.4

LHS = 0.03292

x = 0.45

x = 0.44

LHS = 0.04304

x = 0.446

 1-x   = 0.293; yCH4 =  1+2x

LHS = 0.045895

 x   2x  yC =   = 0.236;yH =   = 0.471 2 1+2x 1+2x

15-14 15.20 A mixture of 1 kmol carbon dioxide, 2 kmol carbon monoxide, and 2 kmol oxygen, at 25°C, 150 kPa, is heated in a constant pressure SSSF process to 3000 K. Assuming that only these same substances are present in the exiting chemical equilibrium mixture, determine the composition of that mixture. initial mix: Equil. mix: CO2, CO, O2 at Const. 1 CO2, 2 CO, T = 3000 K, Pressure

2 O2

P = 150 kPa Q

Reaction

2 CO2

initial change equil.

1 -2x (1-2x)

⇔ 2 CO

+

2 +2x (2+2x)

O2 2 +x (2+x)

From A.10 at 3000 K: K = 0.1090 1

For each n > 0 ⇒ -1 < x < +2 2

yCOyO2 P 1 1+x 2 2+x 150 K= 2 (P0) = 4(1-2x ) (5+x)(100) y CO2

or

1+x (1-2x ) (2+x )= 0.018 167, 5+x 2

nCO2 = 2.042 nCO = 0.958

Trial & error: x = -0.521

y = 0.4559 nO2 = 1.479   CO2 = 0.2139  y nTOT = 4.479   CO  yO2 = 0.3302

15.21 Repeat the previous problem for an initial mixture that also includes 2 kmol of nitrogen, which does not dissociate during the process. Same as Prob. 15.20 except also 2 kmol of inert N2 Equil.: nCO2 = (1-2x), nCO = (2+2x), n O2 = (2+x), n N2 = 2 2

yCOyO2 P 1 1+x 2 2+x 150 (P0) = 4(1-2x ) (7+x)(100) K= 2 y CO2

or (

1+x 2 2+x ) ( )= 0.018167 Trial & error: x = -0.464 1-2x 7+x

nCO2 = 1.928 nCO = 1.072

  yCO2 = 0.295   y = 0.164 nTOT = 6.536   CO

nO2 = 1.536 nN2 = 2.0

yO2 = 0.235 yN2 = 0.306

15-15 15.22 Complete combustion of hydrogen and pure oxygen in a stoichiometric ratio at P , T to form water would result in a computed adiabatic flame temperature of 0 0 4990 K for an SSSF setup. a. How should the adiabatic flame temperature be found if the equilibrium reaction 2H + O ⇔ H O is considered? Disregard all other possible reactions 2 2 2 (dissociations) and show the final equation(s) to be solved. b. Find the equilibrium composition at 3800 K, again disregarding all other reactions. c. Which other reactions should be considered and which components will be present in the final mixture? 2H2 + O2 ⇔ 2H2O

a)

Species

HP = HR = HoP + ∆HP = HoR = Ø

y2H Keq =

2O

P

( ) y2H yO P0 2 2

-1

,

H2

O2

H 2O

Initial

2

1

Ø

Shift

-2x

-x

2x

Final

2-2x

1-x

2x

ntot = 2-2x + 1-x + 2x = 3-x

Hp = (2-2x)∆hH + (1-x)∆hO + 2x(hofH2O + ∆hH2O) = Ø 2

2

4x2 (3-x)2 3-x x2(3-x) = = Keq(T) Keq = (3-x)2 (2-2x)2 1-x (1-x)3 hofH2O = -241826; ∆hH (T), ∆hO (T), ∆hH O(T) 2

2

(1)

(2)

2

Trial and Error (solve for x,T) using Eqs. (1) and (2). yO = 0.15; 2

b) At 3800 K

yH = 0.29; yH 2

2O

= 0.56]

Keq = e1.906 (Reaction is times -1 of table)

x2(3-x)(1-x)-3 = e1.906 = 6.726 ⇒ x ≅ 0.5306 yH

2

= O

2x 1-x 2-2x = 0.43; yO = = 0.19; yH = = 0.38 3-x 3-x 2 3-x 2

c) Other possible reactions from table A.10 H2 ↔ 2 H

O2 ↔ 2 O

2 H2O ↔ H2 + 2 OH

15-16 15.23 Gasification of char (primarily carbon) with steam following coal pyrolysis yields a gas mixture of 1 kmol CO and 1 kmol H2. We wish to upgrade the hydrogen content of this syngas fuel mixture, so it is fed to an appropriate catalytic reactor along with 1 kmol of H 2O. Exiting the reactor is a chemical equilibrium gas mixture of CO, H 2, H2O, and CO2 at 600 K, 500 kPa. Determine the equilibrium composition. Note: see Problem 15.12. 1 CO + 1 H 2

Chem. Equil. Mix CO, H2, H2O, CO2

1 H 2O

600 K 500 kPa

(1) 1 CO + 1 H2O ⇔ 1 CO2 + 1 H2 -x

-x

+x

+x

(2) 2 H2O ⇔ 2 H2 + 1 O2 (1) =

(3) 2 CO2 ⇔ 2 CO + 1 O2

1 1 (2) - (3) 2 2

ln K1 =

1 [-85.79-(-92.49)]= +3.35, 2

K1 = 28.503

Equilibrium: nCO = 1-x,

∑ n = 3, 28.503 =

nH2O = 1-x,

nCO2 = 0 + x,

n H2 = 1 + x

yCO2yH2 P yCO2yH2 0 K= ( ) =y y yCOyH2O P0 CO H2O x(1+x) → x = 0.7794 (1-x)2 n y CO 0.2206 0.0735 H2O 0.2206 0.0735

% 7.35 7.35

CO2

0.7794

0.2598

26.0

H2

1.7794

0.5932

59.3

15-17 15.24 A gas mixture of 1 kmol carbon monoxide, 1 kmol nitrogen, and 1 kmol oxygen at 25°C, 150 kPa, is heated in a constant pressure SSSF process. The exit mixture can be assumed to be in chemical equilibrium with CO 2, CO, O2, and N2 present. The mole fraction of CO2 at this point is 0.176. Calculate the heat transfer for the process. initial mix: Equil. mix: Const. 1 CO, 1 O2, CO2, CO, O2, N2 Pressure

1 N2

yCO2 = 0.176 Q

reaction

2 CO2

initial change equil.

0 +2x 2x

yCO2 = 0.176 =

2x 3-x

⇔

2 CO

P = 150 kPa +

1 -2x (1-2x) ⇒

O2

also,

1 -x (1-x)

N2 1 0 1

x = 0.242 65

nCO2 = 0.4853 nO2 = 0.7574  yCO2 = 0.176 yO2 = 0.2747  nCO = 0.5147 nN2 = 1   yCO = 0.1867 2

yCOyO2 P 1 0.18672×0.2747 150 K= 2 (P0) = 0.1762 (100) = 0.4635 y CO2

From A.10, TPROD = 3213 K From A.9, HR = -110 527 kJ HP = 0.4853(-393 522 + 166 134) + 0.5147(-110 527 + 101 447) + 0.7574(0 + 106 545) + 1(0 + 100 617) = +66 284 kJ QCV = HP - HR = 66 284 - (-110 527) = +176 811 kJ

15-18 15.25 A rigid container initially contains 2 kmol of carbon monoxide and 2 kmol of oxygen at 25°C, 100 kPa. The content is then heated to 3000 K at which point an equilibrium mixture of CO2, CO, and O2 exists. Disregard other possible species and determine the final pressure, the equilibrium composition and the heat transfer for the process. 2 CO + 2 O2 ⇔ 2 CO2 + O2 Initial 2

2

Species:

CO

-2x+x

Final 2-2x

2-x

2x : ntot = 2-2x + 2-x + 2x = 4-x

yO =

2-x ; 4-x

2-2x ; 4-x

CO2

0

Shift -2x

yCO =

O2

2

2x

yCO = 2

2x 4-x

U2 - U1 = 1Q2 = H2 - H1 - P2v + P1v = (2 - 2x)hCO 2 + (2 - x)hO

22

+ 2xhCO - 2hofCO - 2hofO 2

2

2

- R(4 - x)T2 + 4RT1 Keq = e2.217 =

y2CO

P

2 2 -1 2 (P ) y0 yCO o 2

=

4x2 4-x 4T1 4(1-x)2 2-x (4-x)T2

x 2 1 1 T2 2.217 ⇒ ( ) = e = 23.092 1-x 2-x 4 T1 x = 0.8382;

yCO = 0.102;

yO = 0.368; 2

yCO = 0.53 2

P2 = P1(4-x)T2/4T1 = 100(3.1618)(3000/4)(298.15) = 795.4 kPa Q = 0.3236(-110527 + 93504) + 1.1618(98013) + 1.6764(-393522

1 2

+ 152853) - 2(-110527) - 2(Ø) + 8.3145(4(298.15) - 3000(3.1618)) = -142991 kJ

15-19 15.26 One approach to using hydrocarbon fuels in a fuel cell is to “reform” the hydrocarbon to obtain hydrogen, which is then fed to the fuel cell. As a part of the analysis of such a procedure, consider the reforming section a. Determine the equilibrium constant for this reaction at a temperature of 800 K. b. One kilomole each of methane and water are fed to a catalytic reformer. A mixture of CH4, H 2O, H2, and CO exits in chemical equilibrium at 800 K, 100 kPa; determine the equilibrium composition of this mixture. For CH4, use CP0 at ave. temp., 550 K. Table A.6, CP0 = 49.316 kJ/kmol K -0 -0 a) h800 K = hf + CP0∆T = -74 873 + 49.316(800-298.2) = -50 126 kJ/kmol 800 -s0 800 K = 186.251 + 49.316 ln 298.2 = 234.918 kJ/kmol K CH4 + H2O ⇔ 3H2 + CO

For 0

∆H800 K = 3(0+14 681) + 1(-110 527+15 174) - 1(-50 126) - 1(-241 826+18 002) = +222 640 kJ 0

∆S800 K = 3(159.554) + 1(227.277) - 1(234.918) - 1(223.826) = +247.195 kJ/K ∆G0 = ∆H0 - T∆S0 = 222 640 - 800(247.195) = +24 884 kJ ∆G0 -24 884 ln K = - - = = -3.7411 => K = 0.0237 RT 8.3144×800 b) CH4 +H2O + CO ⇔ 3 H2 initial change equil.

1 -x (1-x)

1 -x (1-x)

0 +3x 3x

0 +x x

nTOTAL = 2 + 2x 3

K=

yH2yCO

P 2 (3x)3x 100 2 ( ) = ( ) yCH4yH2O P0 (1-x)(1-x)(2+2x)2 100

or

x x (1-x ) (1+x )

or

x2 = 1-x2

2

2

=

4×0.0237 = 0.003 51 27×1

0.003 51 = 0.059 25

 y y = 0.7095 y = 0.2365  y  = 2.473 

nCH4 = 0.7635

Solving,

nH2O = 0.7635

CH4

= 0.3087

nH2

H2O

= 0.3087

nCO

nTOT

H2

= 0.2870

CO

= 0.0956

x = 0.2365

15-20 15.27 In a test of a gas-turbine combustor, saturated-liquid methane at 115 K is to be burned with excess air to hold the adiabatic flame temperature to 1600 K. It is assumed that the products consist of a mixture of CO , H O, N , O , and NO in 2 2 2 2 chemical equilibrium. Determine the percent excess air used in the combustion, and the percentage of NO in the products. CH4 + 2x O2 + 7.52x N2 → 1 CO2 + 2 H2O + (2x-2) O2 + 7.52x N2 Then

N2 + O2 = 2 NO

init ch. equil.

7.52x 2x-2 0 -a -a +2a (7.52x-a) (2x-2-a) 2a

Also CO2 H2O 1 0 1

2 0 2

nTOT = 1 + 9.52x 1600 K: ln K = -10.55, K = 2.628×10-5 2

2

yNO

yNO P 0 4a2 -5 2.628×10 K = ( ) = y y = (7.52x-a)(2x-2-a) yN2yO2 P0 N2 O2 From A.9 and B.7, HR = 1[-74 873 + 16.043(-274.7-624.1)]+ 0 + 0 = -89 292 kJ (Air assumed 25 oC) HP = 1(-393 522 + 67 569) + 2(-241 826 + 52 907) + (7.52x-a)(41 904) + (2x-2-a)(44 267) + 2a(90 291 + 43 319) = -792 325 + 403 652 x + 181 049 a Assume a ~ 0, then from

HP - HR = 0 → x = 1.7417

Subst. 2.628×10-5 a2 = , (13.098-a)(1.483-a) 4

get a ≈ 0.0113

Use this a in 1st law x=

703 042 - 181 049×0.0113 = 1.7366 403 652 -5

2.628×10 a2 = , a = 0.0112 ⇒ (13.059-a)(1.4732-a) 4 % excess air = 73.7 % % NO =

2×0.0112×100 = 0.128 % 1+9.52×1.7366

⇒

x = 1.7366

15-21 15.28 The van't Hoff equation o

∆H d ln K = − 2 dTPo RT relates the chemical equilibrium constant K to the enthalpy of reaction ∆Ho. From the value of K in Table A.10 for the dissociation of hydrogen at 2000 K and the value of ∆Ho calculated from Table A.8 at 2000 K use van’t Hoff equation to predict the constant at 2400 K. H2 ⇔ 2H ∆H° = 2 × (35375+217999) – 52942 = 453806 lnK2000 = -12.841; Assume ∆H° is constant and integrate the Van’t Hoff equation 2000 ∆H° 1 1 ⌠ - 2 lnK2400 - lnK2000 = ⌡(∆H°/RT )dT = - - ( ) R T2400 T2000 2400 1

lnK2400 = lnK2000 + ∆H° (

-

1

)/R

T2400 T2000

= -12.841 + 453806 (

6-5 ) / 8.31451 = -12.841 + 4.548 12000

= -8.293 Table A.10 lists –8.280 (∆H° not exactly constant)

15-22 15.29 Catalytic gas generators are frequently used to decompose a liquid, providing a desired gas mixture (spacecraft control systems, fuel cell gas supply, and so forth). Consider feeding pure liquid hydrazine, N H , to a gas generator, from 2 4 which exits a gas mixture of N , H , and NH in chemical equilibrium at 100°C, 2 2 3 350 kPa. Calculate the mole fractions of the species in the equilibrium mixture. Initially, 2 N2H4 → 1 N2 + 1 H2 + 2 NH3 then,

N2

initial change equil.

+

1 -x (1-x)

K=

3

yN2yH2

(PP0)

-2

⇔

1 -3x (1-3x)

2

yNH3

3 H2

=

2 NH3 2 +2x (2+2x)

nTOTAL = (4-2x)

(2+2x)2(4-2x)2 350 -2 ( ) (1-x)(1-3x)3 100

At 100 oC = 373.2 K, for NH3 use A.5

CP0 = 17.03×2.130 = 36.276

-0 hNH3 = -45 720 + 36.276(373.2-298.2) = -42 999 kJ/kmol -s0 = 192.572 + 36.276 ln 373.2 = 200.71 kJ/kmol K NH3 298.2 Using A.8, 0

∆H100 C = 2(-42 999) - 1(0+2188) - 3(0+2179) = -94 723 kJ 0

∆S100 C = 2(200.711) - 1(198.155) - 3(137.196) = -208.321 kJ/K 0

∆G100 C = ∆H0 - T∆S0 = -94 723 - 373.2(-208.321) = -16 978 kJ ∆G0 +16 978 ln K = - - = = 5.4716 RT 8.3145×373.2

=>

Therefore,

[

(1+x)(2-x) (1-3x)

]

2

237.84×3.52 1 = = 182.096 (1-x)(1-3x) 16

By trial and error, x = 0.226 y = 0.2181 nN2 = 0.774 nNH3 = 2.452   N2  y = 0.0908 nH2 = 0.322 nTOT = 3.518   H2  yNH3 = 0.6911

K = 237.84

15-23 15.30 Acetylene gas at 25°C is burned with 140% theoretical air, which enters the burner at 25°C, 100 kPa, 80% relative humidity. The combustion products form a mixture of CO2, H 2O, N2, O2, and NO in chemical equilibrium at 2200 K, 100 kPa. This mixture is then cooled to 1000 K very rapidly, so that the composition does not change. Determine the mole fraction of NO in the products and the heat transfer for the overall process. C2H2 + 3.5 O2 + 13.16 N2 + water → 2 CO2 + 1 H2O + 1 O2 + 13.16 N2 + water water: PV = 0.8×3.169 = 2.535 kPa nV = nA PV/PA = (3.5+13.16) 2.535/97.465 = 0.433 So, total H2O in products is 1.433. a) reaction: N2 + O2 <-> 2 NO change : -x

-x

+2x

at 2200 K, from A.10: Equil. products:

K = 0.001 074

nCO2 = 2,

nH2O = 1.433, nO2 = 1-x,

nN2 = 13.16-x, nNO = 0+2x,

n TOT = 17.593

(2x)2 = 0.001 074 => x = 0.0576 K= (1-x)(13.16-x) yNO =

2×0.0576 = 0.006 55 17.593

b) Final products (same composition) at 1000 K HR = 1(226 731) + 0.433(-241 826) = 122 020 kJ HP = 2(-393 522 + 33 397) + 1.433(-241 826+26 000) + 0.9424(0+22 703) + 13.1024(0+21 463) + 0.1152(90 291+22 229) = -713 954 kJ QCV = HP - HR = -835 974 kJ

15-24 15.31 The equilibrium reaction as: CH4 ⇔ C + 2H2. has ln K = -0.3362 at 800 K and lnK = -4.607 at 600 K. By noting the relation of K to temperature show how you would interpolate ln K in (1/T) to find K at 700 K and compare that to a linear interpolation. A.10:

ln K = - 0.3362 at 800K

ln K = -4.607 at 600K

1 1 800 -1 700 800 700 lnK700 = lnK800 + × (-4.607+0.3362) = -0.3362 + ×(-4.2708) 1 1 800 -1 600 800 600 = -2.1665 Linear interpolation: lnK700 = lnK600 + = -4.607 +

700 - 600 (lnK800 - lnK600) 800 -600 1 (-0.3362 + 4.607) = -2.4716 2

o 15.32 Use the information in Problem 15.31 to estimate the enthalpy of reaction, ∆H , at 700 K using Van’t Hoff equation with finite differences for the derivatives. - 2 dlnK = [∆H°/RT ]dT

or

solve for ∆H°

- 2 dlnK - 2 ∆lnK ∆H° = RT = RT dT ∆T 2

= 8.31451 × 700 ×

-0.3362 + 4.607 = 86998 kJ/kmol 800 - 600

[Remark: compare this to A.8 values + A.5, A.9, ∆H° = HC + 2HH - HCH = 0.61 × 12 × (700-298) + 2 × 11730 2

4

– 2.254 × 16.04 × (700-298) - (-74873) = 86739 ]

15-25 15.33 A step in the production of a synthetic liquid fuel from organic waste matter is the following conversion process: 1 kmol of ethylene gas (converted from the waste) at 25°C, 5 MPa, and 2 kmol of steam at 300°C, 5 MPa, enter a catalytic reactor. An ideal gas mixture of ethanol, ethylene, and water in chemical equilibrium leaves the reactor at 700 K, 5 MPa. Determine the composition of the mixture and the heat transfer for the reactor. IG chem. equil. o 1 C H mixture 25 C, 5 MPa 2 4 C2H5OH, C2H4, H2O 2 H 2O 700 K, 5 MPa 300 oC, 5 MPa A.6 at ~ 500 K: 1 C2H4 + 1 H2O ⇔ 1 C2H5OH init 1 2 0 CP0 C2H4 = 62.3 ch. equil.

-x (1-x)

-x (2-x)

+x x

0

a) ∆H700 K = 1(-235 000 + 115(700-298.2)) - 1(+52 467 + 62.3(700-298.2)) - 1(-241 826 + 14 190) = -38 656 kJ 0

∆S700 K = 1(282.444 + 115 ln

700 700 ) - 1(219.330 + 62.3 ln ) - 1(218.739) 298.2 298.2

= -110.655 kJ/K 0

∆G700 K = ∆H0 - T∆S0 = +38 803 kJ -∆G0 ln K = - = -6.667 RT ⇒

=>

K = 0.001 272 =

yC2H5OH

(PP )

yC2H4yH2O

-1

0

5.0 (1-xx )(3-x ) = 0.001272× = 0.0636 2-x 0.1

By trial and error: x = 0.0404 => C 2H5OH: n = 0.0404, y = 0.01371 C2H4: n = 0.9596, y = 0.3242, b) Reactants:

H2O: n = 1.9596 , y = 0.6621

C2H4:

Tr = 298.2/282.4 = 1.056, P r = 5/5.04 = 0.992 - A.15: (h*-h) = 1.30×8.3145×282.4 = 3062 kJ ⇒ HC2H4 = 1(+52 467 - 3062) = +49 405 kJ

H2O, LIQ Ref. + St. Table: HH2O = 2(-285830 + 18.015(2924.5-104.9)) = -470 070 kJ HPROD = 0.0404(-235 000 + 115(700-298.2)) + 0.9596(+52 467 + 62.3(700-298.2)) + 1.9596(-241 826 + 14 190) = -379 335 kJ QCV = HP - HR = +41 330 kJ

15-26 15.34 Methane at 25°C, 100 kPa, is burned with 200% theoretical oxygen at 400 K, 100 kPa, in an adiabatic SSSF process, and the products of combustion exit at 100 kPa. Assume that the only significant dissociation reaction in the products is that of carbon dioxide going to carbon monoxide and oxygen. Determine the equilibrium composition of the products and also their temperature at the combustor exit. CH4 + 4O2 → CO2 + 2H2O + 2O2

Combustion:

Dissociation: 2 CO2 ⇔ initial change equil.

1 -2x 1-2x

2 CO + O2

,H2O

0 +2x 2x

2 0 2

2 +x 2+x

inert

nTOT = 5+x

2

yCOyO2 P x 2 2+x P (P0)= (0.5-x ) (5+x)(P0) Equil. Eq'n: K = 2 y CO2

or

x K (0.5-x ) (2+x ) = 5+x (P/P0)

(1)

2

1st law: HP - HR = 0 (1-2x)(-393 522 + ∆hCO2) + 2x(-110 527 + ∆hCO) + 2(-241 826 + ∆hH2O) + (2+x)∆hO2 - 1(-74 873) - 4(3027) = 0 or (1-2x)∆hCO2 + 2x∆hCO + 2∆hH2O + (2+x)∆hO2 + 565 990x - 814 409 = 0 Assume TP = 3256 K. From A.10:

K = 0.6053

Solving (1) by trial & error, x = 0.2712 Substituting x and the ∆h values from A.8 (at 3256 K) into (2) 0.4576×168 821 + 0.5424×103 054 + 2×140 914 + 2.2712×108 278 + 565 990×0.2712 - 814 409 ≈ 0 OK TP = 3256 K & x = 0.2712 nCO2 = 0.4576, nCO = 0.5424, nH2O = 2.0, nO2 = 2.2712

15-27 15.35 Calculate the irreversibility for the adiabatic combustion process described in the previous problem. From solution of Prob. 15.34 , it is found that the product mixture consists of 0.4576 CO2, 0.5424 CO, 2.0 H2O & 2.2712 O2 at 3256 K, 100 kPa. The reactants include 1 CH4 at 25 oC, 100 kPa and 4 O2 at 400 K, 100 kPa. Reactants: SR = 1(186.251) + 4(213.873) = 1041.74 kJ/K Products: ni

yi

-s0 i

- yiP -R ln P0

-* Si

CO2

0.4576

0.0868

339.278

+20.322

359.600

CO H 2O

0.5424 2.0

0.1029 0.3794

276.660 291.099

+18.907 +8.058

295.567 299.157

O2

2.2712

0.4309

287.749

+7.000

294.749

SP = 0.4576(359.600) + 0.5424(295.567) + 2.0(299.157) + 2.2712(294.749) = 1592.62 kJ/K I = T0(SP-SR) - QCV = 298.15(1592.62 - 1041.74) - 0 = 164 245 kJ

15.36 In rich (too much fuel) combustion the excess fuel may be broken down to give H2 and CO may form. In the products at 1200 K, 200 kPa the reaction called the water gas reaction may take place: CO2 + H2 ⇔ H2O + CO Find the equilibrium constant for this reaction from the elementary reactions. (1)

1 CO2 + 1 H2 ⇔ 1 CO + 1 H2O

(2)

2 H 2O ⇔ 2 H 2 + 1 O 2,

(1) =

1 1 (3) - (2) 2 2

=>

(3)

2 CO2 ⇔ 2 CO + 1 O2

ln K = (ln K3 - ln K2)/2

From A.10 at 1200 K for reactions (2) and (3) ln K2 = - 36.363, ln K3 = - 35.736 => ln K = 0.3135

=> K = 1.3682

15-28 15.37 An important step in the manufacture of chemical fertilizer is the production of ammonia, according to the reaction: N 2 + 3H2 ⇔ 2NH3 a. Calculate the equilibrium constant for this reaction at 150°C. b. For an initial composition of 25% nitrogen, 75% hydrogen, on a mole basis, calculate the equilibrium composition at 150°C, 5 MPa. 1N2 + 3H2 <=> 2NH3 at 150 oC -o hNH3 150 C = -45 720 + 2.13×17.031(150-25) = -41 186 423.2 -so NH3 150 C = 192.572 + 2.13×17.031 ln 298.2 = 205.272 o

∆H150 C = 2(-41 186) - 1(0+3649) - 3(0+3636) = -96 929 kJ 0

∆S150 C = 2(205.272) - 1(201.829) - 3(140.860) = -213.865 kJ/K 0

∆G150 C = -96 929 - 423.2(-213.865) = -6421 kJ/kmol ln K =

+6421 = 1.8248, K = 6.202 8.3144×423.2

b) nNH3 = 2x, nN2 = 1-x, nH2 = 3-3x 2

K=

yNH3 3

yN2yH2

P P0

( )

2

-2

(2x)222(2-x)2 P = 33(1-x)4 P0

( )

2

27 5 2 × 6.202 × ( ) = 26165 16 0.1

or

(1-xx ) (2-x ) 1-x

or

(1-xx )(2-x ) = 161.755 1-x

→ Trial & Error: x = 0.9215

=

-2

NH3

n 1.843

y 0.8544

N2

0.0785

0.0364

H2

0.2355

0.1092

15-29 15.38 A space heating unit in Alaska uses propane combustion is the heat supply. Liquid propane comes from an outside tank at -44°C and the air supply is also taken in from the outside at -44°C. The airflow regulator is misadjusted, such that only 90% of the theoretical air enters the combustion chamber resulting in incomplete combustion. The products exit at 1000 K as a chemical equilibrium gas mixture including only CO2, CO, H2O, H2, and N2. Find the composition of the products. Propane: Liquid, T1 = -44oC = 229.2 K Air: T2 = -44oC = 229.2 K, 90% Theoretical Air Products: T3 = 1000 K, CO2, CO, H2O, H2, N2 Theoretical Air: C3H8 + 5O2 + 18.8N2 => 3CO2 + 4H2O + 18.8N2 90% Theoretical Air: C3H8 + 4.5O2 + 16.92N2 => aCO2 + bCO + cH2O + dH2 + 16.92N2 Carbon:

a+b=3

Oxygen:

2a + b + c = 9 Where: 2 ≤ a ≤ 3

Hydrogen: c + d = 4 CO + H2O

Reaction:

↔ CO2

+ H2

Initial:

b

c

a

d

Change:

-x

-x

x

x

Equil:

b-x

c-x

a+x

d+x

Chose an Initial guess such as: a = 2, b = 1, c = 4, d = 0 Note: A different initial choice of constants will produce a different value for x, but will result in the same number of moles for each product. nCO2 = 2 + x, n CO = 1 - x, nH2O = 4 - x, nH2 = x, nN2 = 16.92 The reaction can be broken down into two known reactions to find K (1)

2CO2 ↔ 2CO + O2 @ 1000 K ln(K1) = -47.052

(2)

2H2O ↔ 2H2 + O2 @ 1000 K ln(K2) = -46.321

For the overall reaction: lnK = (ln(K2) - ln(K1))/2 = 0.3655; K = 1.4412 K=

yCO2yH2  P 1+1-1-1 yCO2yH2 (2 + x)x = = 1.4412 = P  yCOyH2O  o yCOyH2O (1 − 4)(4 − x) =>

x = 0.6462

nCO2 = 2.6462

nCO = 0.3538

nH2O = 3.3538

nH2 = 0.6462

nN2 = 16.92

15-30 15.39 One kilomole of carbon dioxide, CO2 , and 1 kmol of hydrogen, H2 at room temperature, 200 kPa is heated to 1200 K at 200 kPa. Use the water gas reaction, see problem 15.36, to determine the mole fraction of CO. Neglect dissociations of H2 and O2. 1 CO2 + 1 H2 ⇔ 1 CO + 1 H2O Initial

1

1

0

0

Shift

-x

-x

+x

+x

Total

1-x

1-x

x

x;

ntot = 2

yH2O = yCO = x/2, yH2 = yCO2 = (1-x)/2 From solution to problem 15.36, K = 1.3682 (x/2)(x/2) x2 =K= 1-x 1-x (1-x)2 ( )( ) 2 2

=>

x = 1.1697 1-x

x = 1.1697 / 2.1697 = 0.5391 yH2O = yCO = x/2 = 0.27,

yH2 = yCO2 = (1-x)/2 = 0.23

15-31

15.40 Consider the production of a synthetic fuel (methanol) from coal. A gas mixture of 50% CO and 50% H leaves a coal gasifier at 500 K, 1 MPa, and enters a 2 catalytic converter. A gas mixture of methanol, CO and H2 in chemical equilibrium with the reaction: CO + 2H 2 <=> CH3OH leaves the converter at the same temperature and pressure, where it is known that ln K = -5.119. a. Calculate the equilibrium composition of the mixture leaving the converter. b. Would it be more desirable to operate the converter at ambient pressure? 1 CO 1 H2

a)

Equil. Mix

CONVERTER

CH3OH, CO, H2 500 K, 1 MPa

2 H2 ⇔

CH3OH

1

1

0

change

-x

-2x

+x

equil.

(1-x)

(1-2x)

x

Reaction:

CO

initial

K=

+

yCH3OH P -2 x 2-2x 2 P -2 2 ( 0) = (1-x)(1-2x) ( 0) P y y P

=>

CO H2

ln K = -5.119,

K = 0.005 98

x(1-x) 0.005 98 1 2 = (0.1) = 0.1495 4 (1-2x)2 nCH3OH = x = 0.1045, yCH3OH

x(1-x) K P 2 = ( ) (1-2x)2 4 P0

=>

x = 0.1045

n CO = 1-x = 0.8955,

= 0.0583,

yCO = 0.5000,

b) For P = 0.1 MPa x(1-x) 0.005 98 0.1 2 = (0.1) = 0.001 495 4 (1-2x)2 x is much smaller (~ 0.0015) not good

n H2 = 1 - 2x = 0.791 yH2 = 0.4417

15-32 15.41 Consider the following coal gasifier proposed for supplying a syngas fuel to a gas turbine power plant. Fifty kilograms per second of dry coal (represented as 48 kg C plus 2 kg H) enter the gasifier, along with 4.76 kmol/s of air and 2 kmol/s of steam. The output stream from this unit is a gas mixture containing H2, CO, N2, CH4, and CO2 in chemical equilibrium at 900 K, 1 MPa. a. Set up the reaction and equilibrium equation(s) for this system, and calculate the appropriate equilibrium constant(s). b. Determine the composition of the gas mixture leaving the gasifier. a) Entering the gasifier:

4 C + 1 H2 + 1 O2 + 3.76 N2 + 2 H2O

Since the chem. equil. outlet mixture contains no C, O2 or H2O, we must first consider “preliminary” reaction (or reactions) to eliminate those substances in terms of substances that are assumed to be present at equilibrium. One possibility is 4 C + 1 O2 + 2 H2O → 4 CO + 2 H2 such that the "initial" composition for the equilibrium reaction is 4 CO + 3 H2 + 3.76 N2 (or convert equal amounts of CO and H2 to half of CH4 and CO2 - also present at equilibrium. The final answer will be the same.) reaction

2 CO

initial change equil.

4 -2x (4-2x)

+

2 H2 3 -2x (3-2x)

CH4 0 +x x

+

CO2 0 +x x

also N2 3.76 0 3.76

nTOT = 10.76 - 2x For CH4 at 600 K (formula in Table A.6), CP0 = 52.22 At 900 K -0 hCH4 = -74 873 + 52.22(900 - 298.2) = -43 446 kJ/kmol -s0 = 186.251 + 52.22 ln (900 / 298.2) = 243.936 kJ/kmol K CH4 (The integrated-equation values are -43 656 and 240.259) 0

∆H900 K = 1(-43 447) + 1(-393 522 + 28 030) - 2(-110 527 + 18 397) - 2(0 + 17 657) = -259 993 kJ 0

∆S900 K = 1(243.935) + 1(263.646) - 2(231.074) - 2(163.060) = -280.687 kJ/K 0

∆G900 K = -259 993 - 900(-280.687) = -7375 kJ

15-33

ln K =

b)

+7375 = 0.9856, 8.3145×900

K=

K = 2.679

yCH4yCO2 P -2 x×x×(10.76-2x)2 P -2 (P0) = (4-2x)2(3-2x)2 (P0) 2 2 y y CO H2

or

x(10.76-2x) P 1 = 0 K= 2.679 = 16.368 (4-2x)(3-2x) P 0.1

By trial & error,

x = 1.2781

nCO = 1.444,

nH2 = 0.444, nCH4 = nCO2 = 1.278,

nN2 = 3.76

yCO = 0.176,

yH2 = 0.054, yCH4 = yCO2 = 0.156,

yN2 = 0.458

15-34 15.42 Ethane is burned with 150% theoretical air in a gas turbine combustor. The products exiting consist of a mixture of CO , H O, O , N , and NO in chemical 2 2 2 2 equilibrium at 1800 K, 1 MPa. Determine the mole fraction of NO in the products. Is it reasonable to ignore CO in the products? Combustion: C2H6 + 5.25 O2 + 19.74 N2 → 2 CO2 + 3 H2O + 1.75 O2 + 19.74 N2 Products at 1800 K, 1 MPa Equilibrium mixture: CO2, H2O, O2, N2, NO +

N2 initial change equil.

19.74 -x 19.74-x

O2

2 NO

1.75 -x 1.75-x

0 +2x 2x

Equil. comp. nCO2 = 2, nO2 = 1.75-x, nNO = 2x , nH2O = 3, nN2 = 19.74-x 2

yNO

P 0 4x2 ( ) = (19.74-x)(1.75-x) K = 1.192×10 = yN2yO2 P0 -4

Solving, x = 0.031 75 yNO =

2×0.031 75 = 0.0024 26.49 2 CO2 ⇔

b) initial

2

2 CO +

O2

0

0

change -2a

+2a

+2x

equil. 2-2a

2a

2x

2

yCOyO2 P 1 2a 2 1.75-x+a 1 -8 K = 4.194×10 = 2 (P0) =(2-2a ) ( 26.49+a )×0.1 y CO2

This equation should be solved simultaneously with the equation solved in part a) (modified to include the unknown a). Since x was found to be small and also a will be very small, the two are practically independent. Therefore, use the value x = 0.031 75 in the equation above, and solve for a. a 75+a 0.1 (1-a ) (1.75-0.031 ) = ( )×4.194×10-8 26.49+a 1.0 2

Solving, a = 0.000 254 or yCO = 1.92×10-5 negligible for most applications.

15-35 15.43 One kilomole of liquid oxygen, O2, at 93 K, and x kmol of gaseous hydrogen, H , 2 at 25°C, are fed to an SSSF combustion chamber. x is greater than 2, such that there is excess hydrogen for the combustion process. There is a heat loss from the chamber of 1000 kJ per kmol of reactants. Products exit the chamber at chemical equilibrium at 3800 K, 400 kPa, and are assumed to include only H2O, H2, and O. a. Determine the equilibrium composition of the products and also x, the amount of H2 entering the combustion chamber. b. Should another substance(s) have been included in part (a) as being present in the products? Justify your answer. x H2 + 1 O2 → 2 H2O + (x-2) H2 (1) 1 H2O ⇔ 1 H2 + 1 O shift

-a

+a

Equil 2-a

x-2+a

+a

and a > 0

a

a < 2 and

(2) 2 H2O ⇔ 2 H2 + 1 O2 (3)

ln K2 = -1.906

1 O2 ⇔ 2 O

ln K3 = -0.017

ln K1 = 0.5( ln K2 + ln K3 ) = -0.9615 Equil.:

ntot = x + a

=>

K1 = 0.3823

K1 (x-2+a)a 0.3823 = = = 0.95575 1 (2-a)(x+a) 4 (P/Po)

1st law: Q + HR = HP , Q = (1+x)(-1000) kJ -* Table A.8: ∆hIG = -5980 kJ/kmol [or = 0.922 × 32(93 - 298.2) = - 6054 kJ/kmol ] Fig. D.2: Tr = 93/154.6 = 0.601, ∆hf = -5.16×R×154.6 = -6633 -* HR = x(0 + 0) + 1(0 + ∆hIG + ∆hf) = 1(-5980 - 6633) = - 12613 kJ HP = (2-a)(-241 826 + 171 981) + (x-2+a)(0 + 119077) + a(249170 + 73424) = 119077 x + 511516 a - 377844 = Q + HR = -1000 - 1000 x - 12613 Rearrange eq. to:

x + 4.2599 a = 3.03331

Substitute it into the equilibrium eq.: Solve a = 0.198, LHS = 0.09547, yH2O =

2-a = 0.755, x+a

yH2 =

Other substances and reactions: H2 <=> 2 H,

(1.03331 + 5.2599 a) a = 0.095575 (2-a)(3.03331-3.2599 a) x = 2.1898

x-2+a = 0.162, x+a

yO =

a = 0.083 x+a

2 H2O <=> H2 + 2 OH,

: ln K = 0.201,

O2 <=> 2 O, :

All are significant as K's are of order 1.

ln K = -0.984, ln K = -0.017

15-36 15.44 Butane is burned with 200% theoretical air, and the products of combustion, an equilibrium mixture containing only CO , H O, O , N , NO, and NO , exit from 2 2 2 2 2 the combustion chamber at 1400 K, 2 MPa. Determine the equilibrium composition at this state. Combustion: C4H10 + 13 O2 + 48.9 N2 → 4 CO2 + 5 H2O + 6.5 O2 + 48.9 N2 Dissociation: N2 + O2 ⇔ 2 NO

1)

change -a

-a

N2 + 2O2 ⇔ 2 NO2

2)

+2a

change

-b

-2b

At equilibrium: nN2 = 48.9-a-b nH2O = 5

nNO = 2a

nCO2 = 4

nNO2 = 2b

nO2 = 6.5-a-2b

+2b

nTOT = 64.4-b At 1400 K, from A.10: K1 =

K1 = 3.761×10-6, K2 = 9.026×10-10

(2a)2 ; (48.9-a-b)(6.5-a-2b)

K2 =

(2b)2(64.4-b) P -1 ( ) 2 (6.5-a-2b) (48.9-a-b) P0

As K1 and K2 are both very small, with K2 << K1, the unknowns a & b will both be very small, with b << a. From the equilibrium eq.s, for a first trial a~

1 2

1 K1×48.9×6.5 ~ 0.0173 ;b ~ ×6.5 2

2 48.9 K 2× × ~ 0.000 38 0.1 64.4

Then by trial & error, 3.761×10-6 a2 = = 0.940 25×10-6 (48.9-a-b)(6.5-a-2b) 4 2

b (64.4-b) = (6.5-a-2b)2(48.9-a-b)

9.026×10-10×( 4

2 ) 0.1

= 45.13×10-10

Solving, a = 0.017 27, b = 0.000 379 nCO2 = 4 ,

nH2O = 5 ,

yCO2 = 0.062 11 , yH2O = 0.077 64 , nNO = 0.034 54 ,

nNO2 = 0.000 76

yNO = 0.000 55 ,

yNO2 = 0.000 01

nN2 = 48.882 ,

nO2 = 6.482 ,

yN2 = 0.759 04 , yO2 = 0.100 65

15-37 15.45 A mixture of 1 kmol water and 1 kmol oxygen at 400 K is heated to 3000 K, 200 kPa, in an SSSF process. Determine the equilibrium composition at the outlet of the heat exchanger, assuming that the mixture consists of H2O, H2, O2, and OH. Reactions and equilibrium eq'ns the same as in example 15.8 (but different initial composition). At equil.: nH2O = 1-2a-2b, nH2 = 2a+b, nO2 = 1+a nOH = 2b,

nTOT = 2+a+b

Since T = 3000 K is the same, the two equilibrium constants are the same: K1 = 0.002 062,

K2 = 0.002 893

The two equilibrium equations are K1 =

2a+b 1+a P ( ); 1-2a-2b 2+a+b P0

K2 =

2 P 2a+b 2b ( ) ( ) 2+a+b 1-2a-2b P0

which must be solved simultaneously for a & b. If solving manually, it simplifies the solution to divide the first by the second, which leaves a quadratic equation in a & b - can solve for one in terms of the other using the quadratic formula (with the root that gives all positive moles). This reduces the problem to solving one equation in one unknown, by trial & error. Solving =>

b = 0.116,

a = -0.038

=>

nH2O = 0.844, nH2 = 0.0398, nO2 = 0.962, nOH = 0.232, nTOT = 2.0778 yH2O = 0.4062, yH2 = 0.0191, yO2 = 0.4630, yOH = 0.1117

15-38 15.46 One kilomole of air (assumed to be 78% nitrogen, 21% oxygen, and 1% argon) at room temperature is heated to 4000 K, 200 kPa. Find the equilibrium composition at this state, assuming that only N2, O2, NO, O, and Ar are present. 1 kmol air (0.78 N2, 0.21 O2, 0.01 Ar) heated to 4000 K, 200 kPa. Equil.: 1)

N2 + O2 ⇔ 2 NO

change -a -a

nN2 = 0.78-a

+2a

nO2 = 0.21-a-b nAr = 0.01

2)

O2 ⇔ 2 O

nO = 2b

change -b +2b

nNO = 2a ntot = 1+b

4a2 200 0 ( ) K1 = 0.0895 = (0.78-a)(0.21-a-b) 100 4b2 200 K2 = 2.221 = ( ) (1+b)(0.21-a-b) 100 Divide 1st eq'n by 2nd and solve for a as function(b), using K1 P X = ( 0)= 0.0806 K2 P Get Xb2 a= [-1+ 2(1+b)

1+

4×0.78(1+b) Xb2

(1)

]

Also K2 b2 = = 0.277 63 (1+b)(0.21-a-b) 4(P/P0) Assume

b = 0.1280

From (1), get a = 0.0296

Then, check a & b in (2) ⇒ OK Therefore, nN2 = 0.7504 nO = 0.2560 yN2 = 0.6652 yO = 0.2269 nO2 = 0.0524 nNO = 0.0592 yO2 = 0.0465 yNO = 0.0525 nAr = 0.01

yAr = 0.0089

(2)

15-39 15.47 Acetylene gas and x times theoretical air (x > 1) at room temperature and 500 kPa are burned at constant pressure in an adiabatic SSSF process. The flame temperature is 2600 K, and the combustion products are assumed to consist of N 2, O2, CO2, H2O, CO, and NO. Determine the value of x. Combustion: C2H2 + 2.5x O2 + 9.4x N2 → 2 CO2 + H2O + 2.5(x-1)O2 + 9.4xN2 Eq. products 2600 K, 500 kPa: N2, O2, CO2, H2O, CO & NO 2 Reactions: 1)

2 CO2 ⇔ 2 CO + O2

change -2a Equil. Comp.:

+2a

+a

nN2 = 9.4x-b ,

nO2 = 2.5x - 2.5 + a - b ,

N2 + O2 ⇔ 2 NO

2)

change -b nH2O = 1 ,

-b

+2b

nCO = 2a ,

nNO = 2b

nCO2 = 2 - 2a , nTOT = 11.9x + 0.5 + a

At 2600 K, from A.10: K1 = 3.721×10-3,

K2 = 4.913×10-3

K1

3.721×10-3 a 2 2.5x-2.5+a-b = = ( ) (11.9x+0.5+a ) 1-a 5 (P/P0) (2b)2 K2 = 4.913×10 = (9.4-b)(2.5x-2.5+a-b) -3

Also, from the 1st law: HP - HR = 0 where HR = 1(+226 731) + 0 + 0 = +226 731 kJ HP = (9.4x-b)(0+77 963) + (2.5x-2.5+a-b)(0+82 225) + (2-2a)(-393 522+128 074) + 1(-241 826+104 520) + 2a(-110 527+78 679) + 2b(90 291+80 034) Substituting, EQ1:

988 415x + 549 425a + 180 462b - 1 100 496 = 0

which results in a set of 3 equations in the 3 unknowns x, a, b. Assume x = 1.07 Then EQ2:

3.721×10-3 a 2 0.175+a-b =( ) ( ) 1-a 13.233+a 5

4.913×10-3 b2 EQ3: = 4 (10.058-b)(0.175+a+b) Solving, a = 0.1595,

b = 0.0585

Then checking in EQ1,

988 415×1.07 + 549 425×0.1595 + 180 462×0.0585 - 1 100 496 ≈ 0 Therefore, x = 1.07

15-40 15.48 One kilomole of water vapor at 100 kPa, 400 K, is heated to 3000 K in a constant pressure SSSF process. Determine the final composition, assuming that H 2O, H2, H, O2, and OH are present at equilibrium. Reactions: 1)

2 H 2O ⇔ 2 H 2 + O2

change -2a 3)

+2a

2)

2 H2O ⇔ H2 + 2 OH

+a change

-2b

+b

+2b

H2 ⇔ 2 H

change -c

+2c

At equilibrium (3000 K, 100 kPa) nH2O = 1-2a-2b nO2 = a

nH =

2c

nH2 =

nTOT =

1+a+b+c

2a+b-c

nOH =

2b

K1

2.062×10-3 2a+b-c 2 a = = ( ) ( ) 0 1-2a-2b 1+a+b+c 1 (P/P ) K2 (P/P0)

=

2 2.893×10-3 2a+b-c 2b =( )( ) 1+a+b+c 1-2a-2b 1

K3

2.496×10-2 (2a)2 = = 1 (2a+b-c)(1+a+b+c) (P/P0) These three equations must be solved simultaneously for a, b & c: a = 0.0622, and

b = 0.0570,

c = 0.0327

nH2O = 0.7616 yH2O = 0.6611 nH2 = 0.1487 yH2 = 0.1291 nO2 = 0.0622 yO2 = 0.0540 nOH = 0.1140 yOH = 0.0990 nH = 0.0654 yH = 0.0568

15-41 15.49 Operation of an MHD converter requires an electrically conducting gas. It is proposed to use helium gas “seeded” with 1.0 mole percent cesium, as shown in Fig. P15.49. The cesium is partly ionized (Cs ⇔ Cs` + e−) by heating the mixture to 1800 K, 1 MPa, in a nuclear reactor to provide free electrons. No helium is ionized in this process, so that the mixture entering the converter consists of He, Cs, Cs`, and e−. Determine the mole fraction of electrons in the mixture at 1800 K, where ln K = 1.402 for the cesium ionization reaction described. +

Reaction: CS ⇔ CS + e-, init

0.01

change

-x

Also He

0

0

0.99

+x

+x

0

x

x

Equil (0.01-x)

ln K = 1.402 =>

0.99 ;

total: 1 + x

K=

ye-yCs+ P x = 0 0.01-x yCs P

or

x x (0.01-x )(1+x )= 4.0633 / (1/0.1) = 0.40633

( )(

Quadratic equation:

K = 4.0633

x )(1+x )(PP ) 0

x = 0.009767

⇒

ye- = x/(1+x) = 0.00967

15.50 One kilomole of argon gas at room temperature is heated to 20000 K, 100 kPa. Assume that the plasma in this condition consists of an equilibrium mixture of Ar, Ar`, Ar``, and e− according to the simultaneous reactions 1) Ar ⇔ Ar+ + e-

2) Ar+ ⇔ Ar++ + e-

The ionization equilibrium constants for these reactions at 20000 K have been calculated from spectroscopic data as ln K1 = 3.11 and ln K2 = -4.92. Determine the equilibrium composition of the plasma. 1) Ar ⇔ Ar+ + e-

2) Ar+ ⇔ Ar++ + e-

ch. -a

ch. -b

+a

+a

+b

+b

Equil. Comp.: nAr = 1-a, nAr+ = a-b, nAr++ = b, ne- = a+b, nTOT = 1+a+b K1 =

yAr+ ye- P (a-b)(a+b) = (1) = 22.421 0 (1-a)(1+a+b) yAr P

K2 =

yAr++ye- P b(a+b) = (1) = 0.0073 0 (a-b)(1+a+b) yAr+ P

( )

( )

By trial & error:

a = 0.978 57,

b = 0.014 13

nAr = 0.02143, nAr+ = 0.96444, nAr++ = 0.01413, ne- = a0.9927 yAr = 0.0107,

yAr+ = 0.484,

yAr++ = 0.0071,

ye- = 0.4982

15-42 15.51 Plot to scale the equilibrium composition of nitrogen at 10 kPa over the temperature range 5000 K to 15000 K, assuming that N 2 , N, N` , and e − are present. For the ionization reaction N ⇔ N` + e−, the ionization equilibrium constant K has been calculated from spectroscopic data as T [K] 10000 12 000 14 000 16 000 100K 6.26 10-2 1.51 15.1 92 1) N2 ⇔ 2N

2) N ⇔ N+ + e-

change -a +2a change -b +b +b Equil. Comp.: nN2 = 1-a, nN = 2a-b, nN+ = b,

ne- = b

2

yN

(2a-b) (PP )= (1-a)(1+a+b) (PP )

EQ1:

K1 =

EQ2:

yN+ ye- P b2 P K2 = = 0 yN P (2a-b)(1+a+b) P0

yN2

0

0

( )

For T < 10 000 K:

( )

b ~ 0 so neglect EQ2:

To extrapolate K1 above 6000 K:

⇒ K1 =

ln K1 ≈ 16.845 -

4a2 10 ( ) (1-a2) 100

118 260 T

(from values at 5000 K & 6000 K) T(K) 5000 6000 7000 8000 10000

K1 0.0011 0.0570 0.9519 7.866 151.26

For T > 10 000 K: T(K)

a

a ≈ 1.0

K2

0.0524 0.3532 0.8391 0.9755 0.9987

=> b

yN 0.0996 0.5220 0.9125 0.9876 0.9993

yN2 0.9004 0.4780 0.0875 0.0124 0.0007

b2 10 b2 K2 = ( ) = (4-b2) 0.1 (2-b)(2+b) 100 yN

yN2

10 000

6.26×10-4

0.1577

0.8538

0.0731

12 000

1.51×10-2 0.151 0.92

0.7244

0.4862

0.2659

1.5512 1.8994

0.1264 0.0258

0.4368 0.4871

14 000 16 000

Note that b ≈ 0 is not a very good approximation in the vicinity of 10 000 K. In this region, it would be better to solve the original set simultaneously for a & b. The answer would be approximately the same.

15-43 15.52 Hydrides are rare earth metals, M, that have the ability to react with hydrogen to form a different substance MH with a release of energy. The hydrogen can then x be released, the reaction reversed, by heat addition to the MH . In this reaction x only the hydrogen is a gas so the formula developed for the chemical equilibrium is inappropriate. Show that the proper expression to be used instead of Eq. 15.34 is ln (PH2/Po) = ∆G0/RT when the reaction is scaled to 1 kmol of H2. 1 M + x H2 <=> MH x 2 At equilibrium GP = GR , assume g of the solid is a function of T only. 0 0 -0 g-MHx = hMHx - Ts-MHx = g-MHx ,

0 0 -0 g-M = hM - Ts-M = g-M

0 0 -0 g-H2 = hH2 - Ts-H2 + RT ln(PH2/Po) = g-H2 + RT ln(PH2/Po)

GP = GR:

1 0 1 0 g-MHx = g-M + x g-H2 = g-M + x[g-H2 + RT ln(PH2/Po)] 2 2

0 0 0 0 0 ∆G0 = g-MHx - g-M - x g-H2/2 = g-MHx - g-M

Scale to 1 mole of hydrogen ~ 0 = (g-0 - g-0 )/(x/2) = R ∆G T ln(PH2/Po) MHx M which is the desired result.

15-44

Advanced Problems 15.53 Repeat Problem 15.1 using the generalized charts, instead of ideal gas behavior. (Z1-Z2) = 5000 m, P1 = 15 MPa Z1

1

T = 40 oC = const 313.2 15 Tr = = 1.03, Pr1 = = 2.033 304.1 7.38 Equilibrium: -wREV = 0 = ∆g + ∆PE

CO 2 Z2

2

g2 - g1 = h2 - h1 -T(s2 - s1) = g(Z1 - Z2) =

9.807×5000 = 49.04 kJ/kg 1000

From Figures D.2 and D.3, *

*

h1 - h1 = RTc× 3.54 = 203.4 kJ/kg ; s1 - s1 = R × 2.61 = 0.4931 kJ/kg K *

*

h2 - h 1 = 0 ;

*

*

s2 - s1 = 0 - R ln(P2 /P1) = 0.18892 ln( P2 /15)

Trial and error. Assume P2 = 55 MPa (Pr2 = 55/7.38 = 7.45) *

*

h2 - h2 = RTc× 3.60 = 206.8 kJ/kg ; s2 - s2 = R × 2.14 = 0.4043 kJ/kg K ∆g = -206.8 + 0 + 203.4 - 313.2[-0.4043 - 0.18892 ln(55/15) + 0.4931] = 45.7 Too low so assume P2 = 60 MPa (Pr2 = 60/7.38 = 8.13) *

*

h2 - h2 = RTc× 3.57 = 205.1 kJ/kg ; s2 - s2 = R × 2.11 = 0.3986 kJ/kg K ∆g = -205.1 + 0 + 203.4 - 313.2[-0.3986 - 0.18892 ln(60/15) + 0.4931] = 50.7 Make linear interpolation

⇒

P2 = 58 MPa

15.54 Derive the van’t Hoff equation given in problem 15.28, using Eqs.15.12 and 15.15. Note: the d(g-/T) at constant P for each component can be expressed using the relations in Eqs. 13.18 and 13.19. 0 0 0 0 Eq. 15.12: ∆G0 = vC g-C + vD g-D - vA g-A - vB g-B

Eq. 15.15: lnK = ∆G0/RT

Eq. 13.19:

∆G0 = ∆Η° - T ∆S0

1 dlnK d ∆G0 1 dG0 ∆G0 dG0 0 =( )=+ = [∆G – T ] dT dT R-T RT dT R-T2 R-T2 dT 1 dg0 0 = - 2 [∆G + T ∆S ] used Eq.13.19 = - -s dT RT 1 = - 2 ∆Η0 RT

15-45 15.55 A coal gasifier produces a mixture of 1 CO and 2H2 that is then fed to a catalytic converter to produce methane. A chemical-equilibrium gas mixture containing CH4, CO, H2, and H2O exits the reactor at 600 K, 600 kPa. Determine the mole fraction of methane in the mixture. CO

+ 3H2 ↔ CH4 + H2O

Initial

1

2

0

0

Change

-x

-3x

x

x

Equil.

1-x

2-3x

x

x

n = (1 - x) + (2 - 3x) + x + x = 3 - 2x yCH4yH2O  P (1+1-1-3) = -2 x2  P -2 K= =     3 P (1-x)(2-3x)3 Po yCOyH2  o − lnK = - ∆Go/R; ∆Go = ∆Ho - T∆So −o − −o − HP = nCH4 [h f + CP(T - To)] + nH2O (h f + ∆h ) = [-74873 + 2.254×16.04(600 - 298.15)] + (-241826 + 10499) = -295290 −o − HR = nCO (h f + ∆h )

+ nH2 (−h f + ∆h−) = 1(-110527 + 8942) + 3(0 + 8799) o

= -75188 kJ o

∆Η600 = HP - HR = -295290 - (-75188) = -220102 kJ o o − (s− T)CH4 = −s To + CPln(T/To) = 186.251 + 2.254×16.04 ln(600/298.2) = 211.549 o (s− T )H2O = 213.051 kJ/kmol-K;

SP = 424.6 kJ/K

o o (s− T)CO = 218.321 kJ/kmol-K, (s− T)H2 = 151.078 kJ/kmol-K o o o o o ∆S600 = SP - SR = (n−s T)CH4 + (n−s T )H2O - (ns− T)CO - (ns− T)H2

= (211.549 + 213.051) - (218.321 + 3 × 151.078) = -246.955 kJ/K ∆Go = ∆Ho - T∆So = -220 102 - 600(-246.955) = -71929 kJ, K = 1.827×106

lnK = -(-71915)/(8.31451×600) = 14.418

=>

Solve for x,

yCH4 = 0.4

x = 0.6667,

ntot = 1.6667,

15-46 15.56 Dry air is heated from 25°C to 4000 K in a 100-kPa constant-pressure process. List the possible reactions that may take place and determine the equilibrium composition. Find the required heat transfer. Air assumed to be 21% oxygen and 79% nitrogen by volume. From the elementary reactions we have at 4000 K (A.10) 2

(1) O2 <=> 2 O

K1 = 2.221 = y O/yO2

(2) N2 <=> 2 N

K2 = 3.141 × 10-6 = yN/yN2

2

2

(3) N2 + O2 <=> 2 NO

K3 = 0.08955 = yNO/yN2 yO2

Call the shifts a,b,c respectively so we get nO2 = 0.21-a-c, nO = 2a, nN2 = 0.79-b-c, nN = 2b, nNO = 2c, ntot = 1+a+b From which the molefractions are formed and substituted into the three equilibrium equations. The result is 2

K1 = 2.221 = y O/yO2 = 4a2/[(1+a+b)(0.21-a-c)] 2

K2 = 3.141 × 10-6 = yN/yN2 = 4b2/[(1+a+b)(0.79-b-c)] 2

K3 = 0.08955 = y NO/yN2 yO2 = 4c2/[(0.79-b-c)(0.21-a-c)] which gives 3 eqs. for the unknowns (a,b,c). Trial and error assume b = c = 0 solve for a from K 1 then for c from K3 and finally given the (a,c) solve for b from K2. The order chosen according to expected magnitude K1>K3>K2 a = 0.15, b = 0.000832, c = 0.0244 => nO2 = 0.0356, nO = 0.3, nN2 = 0.765, nN = 0.00167, nNO = 0.049 Q = Hex - Hin = nO2∆hO2 + nN2∆hN2 + nO(hfO + ∆hO) + nN(hfN + ∆hN) + nNO(hfNO + ∆hNO) - 0 = 0.0356 × 138705 + 0.765 × 130027 + 0.3(249170 + 77675) + 0.00167(472680 + 77532) + 0.049(90291 + 132671) = 214 306 kJ/kmol air [If no reac. Q = nO2∆hO2 + nN2∆hN2 = 131 849 kJ/kmol air]

15-47 15.57 Methane is burned with theoretical oxygen in an SSSF process, and the products exit the combustion chamber at 3200 K, 700 kPa. Calculate the equilibrium composition at this state, assuming that only CO , CO, H O, H , O , and OH are 2 2 2 2 present. CH4 + 2 O2 → CO2 + 2 H2O

Combustion: Dissociation reactions:

2 H 2O ⇔ 2 H 2 + O2

1)

change -2a

+2a +a

2)

2 H2O ⇔ H2 + 2 OH

change -2b

+b +2b

2 CO2 ⇔ 2 CO + O2

3)

change -2c

+2c +c

At equilibrium: NH2O = 2-2a-2b

nO2 =

a+c

nCO2 =

1-2c

NH2 =

nOH =

2b

nCO =

2c

nTOT =

3+a+b+c

2a+b

Products at 3200 K, 700 kPa K1 = 0.007 328 = (

2a+b 2 a+c 700 ) ( )( ) 2-2a-2b 3+a+b+c 100

K2 = 0.012 265 = (

2 2b 2a+b 700 ) ( )( ) 2-2a-2b 3+a+b+c 100

K3 = 0.426 135 = (

2c 2 ) ( a+c )(700) 1-2c 3+a+b+c 100

These 3 equations must be solved simultaneously for a, b, & c. If solving by hand divide the first equation by the second, and solve for c = fn(a,b). This reduces the solution to 2 equations in 2 unknowns. Solving, a = 0.024, b = 0.1455, c = 0.236 Substance:

H 2O

H2

O2

OH

CO2

CO

n

1.661

0.1935

0.260

0.291

0.528

0.472

y

0.4877

0.0568

0.0764

0.0855

0.1550

0.1386

15-48

ENGLISH UNIT PROBLEMS 15.58ECarbon dioxide at 2200 lbf/in.2 is injected into the top of a 3-mi deep well in connection with an enhanced oil recovery process. The fluid column standing in the well is at a uniform temperature of 100 F. What is the pressure at the bottom of the well assuming ideal gas behavior?

Z1

(Z1-Z2) = 3 miles = 15 840 ft

1

P1 = 2200 lbf/in2, T = 100 F = const

CO 2 Z2

Equilibrium and ideal gas beahvior

2

-wREV = 0 = ∆g + ∆PE = RT ln (P2/P1) + g(Z2-Z1) = 0

ln (P2/P1) =

32.2×15 840 = 0.8063 32.2×35.1×559.7

P2 = 2200 exp(0.8063) = 4927 lbf/in2

15.59E Calculate the equilibrium constant for the reaction O2 <=> 2O at temperatures of 537 R and 10 000 R. Find the change in Gibbs function at the two T’s from Table C.7: ∆H0 = 2(249 170+121 264) - (0+224 210) = 516 658 kJ/kmol ∆S0 = 2(224.597) -1(313.457) = 135.737 kJ/kmol K ∆G0 = 516 658 - 6000×135.737 = -297 764 kJ/kmol ln K = 537 R:

+297 764 = +5.969 8.3145×6000

-0 -0 ∆H0 = 2hf O - 1hf O2 = 2 × 107 124 = 214 248 ; 0 0 ∆S0 = 2s-O - 1s-O2 = 2 × 38.442 - 48.973 = 27.911

∆G0 = ∆H0 - T∆S0 = 214 248 - 537 × 27.911 = 199 260 ∆G0 +199 260 - = - 185.85 ln K = - - = RT 1.98589×537 -0 -0 10 000 R: ∆H0 = 2hf O - 1hf O2 = 2 (107 124 + 47897) - 87997 = 222 045; 0 0 ∆S0 = 2s-O - 1s-O2 = 2 × 53.210 - 74.034 = 32.386

∆G0 = ∆H0 - T∆S0 = 222 045 - 10 000 × 32.386 = -101 815 ln K = - ∆G0/RT = 101 815 / (1.98589 × 10 000) = +5.127

15-49 15.60E Pure oxygen is heated from 77 F to 5300 F in an SSSF process at a constant pressure of 30 lbf/in.2. Find the exit composition and the heat transfer. The only reaction will be the dissociation of the oxygen O2 ⇔ 2O ;

K(5300 F) = K(3200 K) = 0.046467

Look at initially 1 mol Oxygen and shift the above reaction with x nO2 = 1 - x; nO = 2x; ntot = 1 + x; yi = ni/ntot y2O P 4x2 1 + x 8x2 K= ( )2-1 = 2 = y02 Po (1 + x)2 1 - x 1 - x2 K/8 ⇒ x = 0.07599; y02 = 0.859; y0 = 0.141 1 + K/8 q = n02exh02ex + n0exhOex - h02in = (1 + x)(y 02h02 + y0hO) - 0 h02 = 45 581; hO = 107124 + 26125 = 133 249 q = 1.076(0.859 × 45581 + 0.141 × 133249) = 62345 Btu/lbmol O2 q = q/32 = 1948 Btu/lbm (=1424 if no dissociation) x2 =

15.61E Pure oxygen is heated from 77 F, 14.7 lbf/in.2 to 5300 F in a constant volume container. Find the final pressure, composition, and the heat transfer As oxygen is heated it dissociates O2 ⇔ 2O C. V. Heater: Per mole O2:

ln Keq = -3.069

U2 - U1 = 1Q2 = H2 - H1 - P2v + P1v - 1q2 = h2 - h1 + R(T1 - (n2/n1)T2)

Shift x in reaction final composition: n1 = 1

(1 - x)O2 + 2xO

n2 = 1 - x + 2x = 1 + x

yO2 = (1 - x)/(1 + x) ;

yO = 2x/(1 + x)

Ideal gas and V2 = V1 ⇒ P2 = P1n2T2/n1T1 ⇒ P2/Po = (1 + x)T2/T1 Keq = e ⇒

-3.069

y2O P2 2x 2 1 + x 1 + x T2 = ( )=( ) ( )( )( ) 1+x 1-x 1 y02 Po T1

4x2 T1 -3.069 = e = 0.00433 ⇒ x=0.0324 1 - x T2

(nO2)2 = 0.9676, (nO)2 = 0.0648, n2 = 1.0324 1q2 = 0.9676(45581) + 0.0648(107124 + 26125) - Ø + 1.98589(536.67 - 1.0324 × 5760) = 41996 Btu/lbmol O2 yO2=0.9676 / 1.0324 = 0.937;

yO =0.0648/1.0324 = 0.0628

15-50 15.62E Air (assumed to be 79% nitrogen and 21% oxygen) is heated in an SSSF process at a constant pressure of 14.7 lbf/in.2, and some NO is formed. At what temperature will the mole fraction of NO be 0.001? heated at 14.7 lbf/in2, forms NO

0.79N2 + 0.21O2

At exit, yNO = 0.001 N2 + O2 ⇔ 2 NO

nN2 = 0.79 - x

-x

nO2 = 0.21 - x

-x

+2x

nNO = 0

+ 2x

n = 1.0 2x ⇒ 1.0

yNO = 0.001 =

x = 0.0005

2

K=

yNO

⇒

nN2 = 0.7895,

nO2 = 0.2095

-6

yN2yO2

10 (PP0)0 = 0.7895×0.2095 = 6.046×10-6

From Table A.10,

or

ln K = -12.016

T = 1444 K = 2600 R

15.63E The combustion products from burning pentane, C5H12, with pure oxygen in a stoichiometric ratio exists at 4400 R. Consider the dissociation of only CO2 and find the equilibrium mole fraction of CO. C5H12 + 8 O2 → 5 CO2 + 6 H2O 2 CO2 ⇔ 2 CO + 1 O2

At 4400 R, ln K = -7.226 K = 7.272x10-4

Initial 5 Change Equil. 5-2z

0 -2z 2z

Assuming P = Po = 0.1 MPa, y2COyO K=

P 2z 2 z ( 0) = ( )( )(1); T & E : z = 0.2673 5 - 2z 5 + z P

2

y2CO 2

nCO = 4.4654; 2

nCO = 0.5346;

yCO = 0.1015

nO = 0.2673 2

0 +2z z

+z

15-51 15.64E A gas mixture of 1 pound mol carbon monoxide, 1 pound mol nitrogen, and 1 pound mol oxygen at 77 F, 20 lbf/in.2, is heated in a constant pressure SSSF process. The exit mixture can be assumed to be in chemical equilibrium with CO2, CO, O2, and N2 present. The mole fraction of CO2 at this point is 0.176. Calculate the heat transfer for the process. initial mix: 1 CO, 1 O2,

Equil. mix: CO2, CO, O2, N2

Const. Pressure

1 N2

yCO2 = 0.176 Q

reaction

2 CO2⇔

initial change equil.

0 +2x 2x

yCO2 = 0.176 =

2x 3-x

2 CO

P = 20 lbf/in2 +

1 -2x (1-2x) ⇒

O2

also,

1 -x (1-x)

N2 1 0 1

x = 0.242 65

y = 0.176 nCO2 = 0.4853 nO2 = 0.757 35  CO2   yCO = 0.186 67 nCO = 0.5147 nN2 = 1  y = 0.274 67  O2 2

yCOyO2 P 1 0.186 672×0.274 67 20 (P0) = (14.504)= 0.426 07 K= 2 0.1762 y CO2

Since Table A.10 corresponds to a pressure P0 of 100 kPa, which is 14.504 lbf/in2. Then, from A.10, TPROD = 3200 K = 5760 R HR = -47 518 Btu HP = 0.4853(-169 184+71 075) + 0.5147(-47 518+43 406) + 0.757 35(0+45 581) + 1(0+43 050) = +27 842 Btu QCV = HP - HR = 27 842 - (-47 518) = +75 360 Btu

15-52 15.65EIn a test of a gas-turbine combustor, saturated-liquid methane at 210 R is to be burned with excess air to hold the adiabatic flame temperature to 2880 R. It is assumed that the products consist of a mixture of CO2, H2O, N2, O2, and NO in chemical equilibrium. Determine the percent excess air used in the combustion, and the percentage of NO in the products. CH4 + 2x O2 + 7.52x N2 → 1 CO2 + 2 H2O + (2x-2) O2 + 7.52x N2 Then

N2

init ch. equil.

+

O2 ⇔ 2 NO

7.52x 2x-2 -a -a (7.52x-a) (2x-2-a)

Also CO2 H2O

0 +2a 2a

1 0 1

2 0 2

nTOT = 1 + 9.52x 2880 R:

ln K = -10.55,

K = 2.628×10-5

2

2

yNO

yNO P 0 4a2 2.628×10-5 K = ( ) = = yN2yO2 P0 yN2yO2 (7.52x-a)(2x-2-a) HR = 1[-32 190 + (-1854-4300)]+ 0 + 0 (assume 77 F) = -38 344 Btu HP = 1(-169 184 + 29 049) + 2(-103 966 + 22 746) + (7.52x-a)(18 015) + (2x-2-a)(19 031) + 2a(38 818 + 18 624) = -340 639 + 173 535 x + 77 838 a Assume

a ~ 0, then from

→ x = 1.742

HP - HR = 0

Subst. 2.628×10-5 a2 = , (13.1-a)(1.484-a) 4

a ≈ 0.01125

get

Use this a in 1st law x=

302 295 - 77 838×0.01125 = 1.737 173 535 -5

⇒ ⇒

2.628×10 a2 = , a = 0.0112 (13.062-a)(1.474-a) 4 x = 1.737 % NO =

% excess air = 73.7 % 2×0.0112×100 = 0.128 % 1+9.52×1.737

15-53 15.66E Acetylene gas at 77 F is burned with 140% theoretical air, which enters the burner at 77 F, 14.7 lbf/in.2, 80% relative humidity. The combustion products form a mixture of CO2, H2O, N2, O2, and NO in chemical equilibrium at 3500 F, 14.7 lbf/in.2. This mixture is then cooled to 1340 F very rapidly, so that the composition does not change. Determine the mole fraction of NO in the products and the heat transfer for the overall process. C2H2 + 3.5 O2 + 13.16 N2 + water → 2 CO2 + 1 H2O + 1 O2 + 13.16 N2 + water Water:

PV = 0.8×0.46 = 0.368 lbf/in2

PV 0.368 nV = nA× = (3.5+13.16)× = 0.428 14.332 PA So, total H2O in products is : 1 + nV = 1.428. a)

reaction:

N2 + O2 ⇔ 2 NO

change :

-x

-x

+2x

at 3500F = 3960 R (=2200 K), from A.10:

K = 0.001 074

Equilibrium products: nCO2 = 2,

nH2O = 1.428, nO2 = 1-x,

nN2 = 13.16-x, K=

nNO = 0+2x,

n TOT = 17.588

(2x)2 = 0.001 074 (1-x)(13.16-x)

By trial and error, yNO =

x = 0.0576

2×0.0576 = 0.006 55 17.588

b) Final products (same composition) at 1340 F = 1800 R HR = 1(97 476) + 0.428(-103 966) = 52 979 Btu HP = 2(-169 184+14 358) + 1.428(-103 966+11 178) + 0.9424(0+9761) + 13.1024(0+9227) + 0.1152(38 818+9557) = -306 486 Btu QCV = HP - HR = -359 465 Btu

15-54

15.67E The equilibrium reaction with methane as CH4 ⇔ C + 2H2 has ln K = -0.3362 at 1440 R and ln K = -4.607 at 1080 R. By noting the relation of K to temperature, show how you would interpolate lnK in (1/T) to find K at 1260 R and compare that to a linear interpolation. ln K = -0.3362 at 1440 R

ln K = -4.607 at 1080 R

1 1 1260 1440 lnK1260 = lnK1440 + × (-4.607+0.3362) 1 1 1080 1440 1440 -1 1260 = -0.3362 + × (-4.2708) = -2.1665 1440 -1 1080 Linear interpolation: lnK1260 = lnK1080 + = -4.607 +

1260 - 1080 (lnK1440 - lnK1080) 1440 - 1080

1 (-0.3362 + 4.607) = -2.4716 2

o 15.68E Use the information in problem 15.67 to estimate the enthalpy of reaction, ∆H , at 1260 R using the van’t Hoff equation (see problem 15.28) with finite differences for the derivatives. - 2 dlnK = [∆H°/RT ]dT or

solve for ∆H°

- 2 dlnK - 2 ∆lnK ∆H° = RT = RT dT ∆T 2

= 1.98589 × 1260 ×

-0.3362 + 4.607 = 37403 Btu/lb mol 1440 - 1080

[Remark: compare this to C.7 values + C.4, C.12, ∆H° = HC + 2HH - HCH = 0.146 × 12 × (1260-537) + 2 × 5044 2

4

– 0.538 × 16.043 × (1260-537) -(-32190) = 37304 ]

15-55 15.69E An important step in the manufacture of chemical fertilizer is the production of ammonia, according to the reaction: N 2 + 3H2 ⇔ 2NH3 a. Calculate the equilibrium constant for this reaction at 300 F. b. For an initial composition of 25% nitrogen, 75% hydrogen, on a mole basis, calculate the equilibrium composition at 300 F, 750 lbf/in.2. 1N2 + 3H2 <=> 2NH3 at 300 F -o a) hNH3 300 F = -19 656 + 0.509×17.031(300-77) = -17723 760 -so NH3 300 F = 45.969 + 0.509×17.031 ln 537 = 48.980 o

∆H300 F = 2(-17723) - 1(0 + 1557) - 3(0 + 1552) = -41 659 Btu 0

∆S300 F = 2(48.98) - 1(48.164) - 3(33.60) = -51.0 Btu/R 0

∆G300 F = -41 659 - 760(-51.0) = -2899 Btu ln K = b)

+2899 = 1.9208, 1.98589×760

nNH3 = 2x,

nN2 = 1-x,

2

K=

yNH3 3

yN2yH2

(PP )

-2

0

2

=

2

(1-xx ) (2-x ) 1-x

or

(1-xx )(2-x )= 173.16 1-x

→ Trial & Error: x = 0.9242

nH2 = 3-3x

(2x)222(2-x)2 P 33(1-x)4 P0

or

=

K = 6.826

( )

-2

27 750 2 ×6.826× ( ) = 29985 16 14.7

NH3

n 1.848

y 0.8591

N2

0.0758

0.0352

H2

0.2273

0.1057

15-56 15.70E Ethane is burned with 150% theoretical air in a gas turbine combustor. The products exiting consist of a mixture of CO2, H2O, O2, N2, and NO in chemical equilibrium at 2800 F, 150 lbf/in.2. Determine the mole fraction of NO in the products. Is it reasonable to ignore CO in the products? Combustion: C2H6 + 5.25 O2 + 19.74 N2 → 2 CO2 + 3 H2O + 1.75 O2 + 19.74 N2 Products at 2800 F, 150 lbf/in2 a) Equilibrium mixture: CO2, H2O, O2, N2, NO +

N2 initial change equil.

O2

19.74 -x 19.74-x

⇔

1.75 -x 1.75-x

2 NO 0 +2x 2x

Equil. comp. nCO2 = 2, nH2O = 3, nO2 = 1.75-x, nN2 = 19.74-x, nNO = 2x 2

yNO

P 0 4x2 ( ) = (19.74-x)(1.75-x) K = 1.283×10 = yN2yO2 P0 -4

Solving,

x = 0.032 95

yNO =

2×0.032 95 = 0.002 49 26.49

b) 2 CO2 initial change equil.

⇔

2 CO

2 -2a 2-2a

0 +2a 2a

+

O2 0 +2x 2x

2

yCOyO2 P 1 2a 2 1.75-x+a 150 -8 K = 5.259×10 = 2 (P0) = (2-2a ) ( 26.49+a )(14.504 ) y CO2

Since Table A.10 corresponds to a pressure P0 of 100 kPa, which is 14.504 lbf/in2. This equation should be solved simultaneously with the equation solved in part a) (modified to include the unknown a). Since x was found to be small and also a will be very small, the two are practically independent. Therefore, use the value x = 0.032 95 in the equation above, and solve for a. a 95+a 14.504 (1-a ) (1.75-0.032 ) =( )×5.259×10-8 26.49+a 150 2

Solving, a = 0.000 28 or yCO = 2.1×10-5 negligible for most applications.

15-57 15.71E One pound mole of air (assumed to be 78% nitrogen, 21% oxygen, and 1% argon) at room temperature is heated to 7200 R, 30 lbf/in.2. Find the equilibrium composition at this state, assuming that only N2, O2, NO, O, and Ar are present. 1 lbmol air (0.78 N2, 0.21 O2, 0.01 Ar) heated to 7200 R, 30 lbf/in2. N2 + O2 ⇔ 2 NO

1)

change -a

-a

+2a

2) change

Equil.: nAr =

nN2 =

0.78-a

nO2 =

0.21-a-b

O2 ⇔ 2 O

nO =

-b

+2b

0.01

nNO =

2a

2b

n =

1+b

4a2 30 0 ( ) K1 = 0.0895 = (0.78-a)(0.21-a-b) 14.504 4b2 30 K2 = 2.221 = ( ) 14.504 (1+b)(0.21-a-b) Divide 1st eq'n by 2nd and solve for a as function(b), using K1 P X= ( ) = 0.083 35 K2 P0 Get Xb2 a= [-1 + 2(1+b) Also

1+

4×0.78(1+b) Xb2

(1)

]

K2 b2 = = 0.268 44 (1+b)(0.21-a-b) 4(P/P0) Assume b = 0.1269

(2)

From (1), get a = 0.0299

Then, check a & b in (2) ⇒ OK Therefore, Subst.

N2

O2

Ar

O

NO

n

0.7501

0.0532

0.01

0.2538

0.0598

y

0.6656

0.0472

0.0089

0.2252

0.0531

15-58 15.72E Dry air is heated from 77 F to 7200 R in a 14.7 lbf/in.2 constant-pressure process. List the possible reactions that may take place and determine the equilibrium composition. Find the required heat transfer. Air assumed to be 21% oxygen and 79% nitrogen by volume. From the elementary reactions at 4000 K = 7200 R (A.10): 2

(1) O2 <=> 2 O

K1 = 2.221 = y O/yO2

(2) N2 <=> 2 N

K2 = 3.141 × 10-6 = yN/yN2

2

(3) N2 + O2 <=> 2 NO

2

K3 = 0.08955 = yNO/yN2 yO2

Call the shifts a,b,c respectively so we get nO2 = 0.21-a-c, nNO = 2c,

nO = 2a,

nN2 = 0.79-b-c,

nN = 2b,

ntot = 1+a+b

From which the molefractions are formed and substituted into the three equilibrium equations. The result is corrected for 1 atm = 14.7 lbf/in2 = 101.325 kPa versus the tables 100 kPa 2

K1 = 2.1511 = y O/yO2 = 4a2/[(1+a+b)(0.21-a-c)] 2

K2 = 3.042×10-6 = yN/yN2 = 4b2/[(1+a+b)(0.79-b-c)] 2

K3 = 0.08955 = y NO/yN2 yO2 = 4c2/[(0.79-b-c)(0.21-a-c)] which give 3 eqs. for the unknowns (a,b,c). Trial and error assume b = c = 0 solve for a from K 1 then for c from K3 and finally given the (a,c) solve for b from K2. The order chosen according to expected magnitude K1>K3>K2 a = 0.15, b = 0.000832, c = 0.0244 => nO2 = 0.0356, nO = 0.3, nN2 = 0.765, nN = 0.00167, nNO = 0.049 Q = Hex - Hin = nO2∆hO2 + nN2∆hN2 + nO(hfO + ∆hO) + nN(hfN + ∆hN) + nNO(hfNO + ∆hNO) - 0 = 0.0356×59632 + 0.765×55902 + 0.3(107124 + 33394) + 0.00167(203216 + 33333) + 0.049(38818 + 57038) = 92 135 Btu/lbmol air [If no reaction: Q = nO2∆hO2 + nN2∆hN2 = 56 685 Btu/lbmol air]

15-59 15.73E Acetylene gas and x times theoretical air (x > 1) at room temperature and 75 lbf/in.2 are burned at constant pressure in an adiabatic SSSF process. The flame temperature is 4600 R, and the combustion products are assumed to consist of N2, O2, CO2, H2O, CO, and NO. Determine the value of x. Combustion: C2H2 + 2.5x O2 + 9.4x N2 → 2 CO2 + H2O + 2.5(x-1)O2 + 9.4x N2 Eq. products 4600 R, 75 lbf/in2: N2, O2, CO2, H2O, CO, NO 2 Reactions: 1)

2 CO2 ⇔ 2 CO + O2

change -2a

+2a

+a

2) change

N2 + O2 ⇔ 2 NO -b

-b

Equil. Comp.: 9.4x-b nN2 =

nCO2 =

2-2a

nO2 =

2.5x-2.5+a-b

nH2O =

1

nCO =

2a

nNO =

2b

nTOT =

11.9x+0.5+a

+2b

At 4600 R, from A.17: K1 = 2.359×10-3, K2 = 4.249×10-3 K1

2.359×10-3 a 2 2.5x-2.5+a-b = 4.622×10-4 =( ) ( ) 0 = 1-a 11.9x+0.5+a 5.103 (P/P ) (2b)2 K2 = 4.249×10 = (9.4-b)(2.5x-2.5+a-b) -3

Also, from the 1st law: HP - HR = 0 where HR = 1(+97 477) + 0 + 0 = +97 497 Btu HP = (9.4x-b)(0+32 817) + (2.5x-2.5+a-b)(0+34 605) + (2-2a)(-169 184+53 885) + 1(-103 966+43 899) + 2a(-47 518+33 122) + 2b(38 818+31 161) Substituting, 394 992 × x + 236 411 × a + 72 536 × b - 377 178 = 97477 which results in a set of 3 equations in the 3 unknowns x,a,b. Trial and error solution from the last eq. and the ones for K1 and K2. The result is x = 1.12 , a = 0.1182, b = 0.05963

15-60 15.74E One pound mole of water vapor at 14.7 lbf/in. 2, 720 R, is heated to 5400 R in a constant pressure SSSF process. Determine the final composition, assuming that H2O, H2, H, O2, and OH are present at equilibrium. Reactions: 1)

2 H 2O ⇔ 2 H 2 + O2

change -2a 3)

+2a

+a

2)

2 H2O ⇔ H2+ 2 OH

change

-2b

+b

H2 ⇔ 2 H

change -c +2c At equilibrium (5400 R, 14.7 lbf/in2) nH2O = 1-2a-2b nOH =

2b

nH2 =

2a+b-c

nH =

2c

nO2 =

a

nTOT =

1+a+b+c

K1

2.062×10-3 2a+b-c 2 a = = ( ) ( ) 1-2a-2b 1+a+b+c 1.03 (P/P0) K2

2 2.893×10-3 2a+b-c 2b =( )( ) 0 = 1+a+b+c 1-2a-2b 1.03 (P/P )

K3

2.496×10-2 (2a)2 = = 1.03 (2a+b-c)(1+a+b+c) (P/P0) These three equations must be solved simultaneously for a, b & c: and

a = 0.0622, b = 0.0570, c = 0.0327 nH2O = 0.7616 yH2O = 0.6611 nH2 = 0.1487 yH2 = 0.1291 nO2 = 0.0622 yO2 = 0.0540

+2b

15-61 15.75E Methane is burned with theoretical oxygen in an SSSF process, and the products exit the combustion chamber at 5300 F, 100 lbf/in.2. Calculate the equilibrium composition at this state, assuming that only CO2, CO, H2O, H2, O2, and OH are present. Combustion: 1)

Dissociation reactions: 2 H 2O ⇔ 2 H 2 + O2

change -2a 2)

At equilibrium: nH2O =

+2a +a

2 H2O ⇔ H2 + 2 OH

change -2b 3)

CH4 + 2 O2 → CO2 + 2 H2O

+b +2b

2 CO2 ⇔ 2 CO + O2

change -2c

+2c +c

2-2a-2b

nH2 =

2a+b

nO2 =

a+c

nOH =

2b

nCO2 =

1-2c

nCO =

2c

nTOT =

3+a+b+c

Products at 5300F, 100 lbf/in2 K1 = 0.007 328 = (

2a+b 2 ) ( a+c )( 100 ) 2-2a-2b 3+a+b+c 14.504

K2 = 0.012 265 = (

2 2b 2a+b 100 ) ( )( ) 2-2a-2b 3+a+b+c 14.504

K3 = 0.426 135 = (

2c 2 a+c 100 ) ( )( ) 1-2c 3+a+b+c 14.504

These 3 equations must be solved simultaneously for a, b, & c. If solving by hand divide the first equation by the second, and solve for c = fn(a,b). This reduces the solution to 2 equations in 2 unknowns. Solving, a = 0.0245, b = 0.1460, c = 0.2365 Substance:

H2O

H2

O2

OH

CO2

CO

n

1.659

0.195

0.260

0.292

0.527

0.473

y

0.4871

0.0573

0.0763

0.0857

0.1547

0.1389

16-1

CHAPTER 16 Notice that most of the solutions are done using the computer tables, which includes the steam tables, air table, compressible flow table and the normal shock table. This significantly reduces the amount of time it will take to solve a problem, so this should be considered in problem assignments and exams.

Changes of problems from the 4th edition Chapter 14 to the new Chapter 16 are: New 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 New

Old 1 new 2 new new 3 new new 5 6 7 new new new new 8 9 mod skip a) 10 11 12 Old

New 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 New

Old 13 14 new 15 16 17 18 19 20 21 22 23 24 25 26 new 27 28 29 new Old

New 41 42 43 44E 45E 46E 47E 48E 49E 50E 51E 52E 53E 54E 55E 56E 57E

Old 30 31 4 39 38 new new 40 new new new 43 44 45 46 47 new

New

Old

16-2 16.1

Steam leaves a nozzle with a pressure of 500 kPa, a temperature of 350°C, and a velocity of 250 m/s. What is the isentropic stagnation pressure and temperature?

h0 = h1 + V21/2 = 3167.7 + 2502/2000 = 3198.4 kJ/kg s0 = s1 = 7.6329 kJ/kg K Computer software: (ho, so) ⇒ To = 365°° C, Po = 556 kPa 16.2

An object from space enters the earth’s upper atmosphere at 5 kPa, 100 K with a relative velocity of 2000 m/s or more . Estimate the object`s surface temperature.

ho1 - h1 = V21/2 = 20002/2000 = 2000 kJ/kg ho1 = h1 + 2000 = 100 + 2000 = 2100 kJ/kg

=>

T = 1875 K

The value for h from ideal gas table A.7 was estimated since the lowest T in the table is 200 K.

16.3

The products of combustion of a jet engine leave the engine with a velocity relative to the plane of 400 m/s, a temperature of 480°C, and a pressure of 75 kPa. Assuming that k = 1.32, Cp = 1.15 kJ/kg K for the products, determine the stagnation pressure and temperature of the products relative to the airplane.

ho1 - h1 = V21/2 = 4002/2000 = 80 kJ/kg To1 - T1 = (ho1 - h1)/Cp = 80/1.15 = 69.6 K To1 = 480 + 273.15 + 69.6 = 823 K Po1 = P1(To1/T1)k/(k-1) = 75(823/753.15)4.125 = 108 kPa

16.4

A meteorite melts and burn up at temperatures of 3000 K. If it hits air at 5 kPa, 50 K how high a velocity should it have to experience such a temperature? Assume we have a stagnation T = 3000 K h1 + V21/2 = hstagn.

Use table A.7, hstagn. = 3525.36, h1 = 50

V21/2 = 3525.36 – 50 = 3475.4 kJ/kg V1 =

2 × 3475.4 × 1000 = 2636 m/s

(remember convert to J/kg = m2/s2)

16-3 16.5

I drive down the highway at 110 km/h on a day with 25°C, 101.3 kPa. I put my hand , cross sectional area 0.01 m2, flat out the window. What is the force on my hand and what temperature do I feel? The air stagnates on the hand surface :

h1 + V21/2 = hstagn.

Use constant heat capacity Tstagn. = T1 +

V21/2

0.5 × 1102 × (1000/3600)2 = 25 + = 25.465°°C Cp 1004

Assume a reversible adiabatic compression Pstagn. = P1 (Tstagn./T1)k/(k-1) = 101.3 (298.615/298.15)3.5 = 101.85 kPa

16.6

Air leaves a compressor in a pipe with a stagnation temperature and pressure of 150°C, 300 kPa, and a velocity of 125 m/s. The pipe has a cross-sectional area of 0.02 m2. Determine the static temperature and pressure and the mass flow rate. ho1 - h1 = V21/2 = 1252/2000 = 7.8125 kJ/kg To1 - T1 = (ho1 - h1)/Cp = 7.8125/1.004 = 7.8 K T1 = To1 - ∆T = 150 - 7.8 = 142.2 °C = 415.4 K P1 = Po1(T1/To1)k/(k-1) = 300(415.4/423.15)3.5 = 281 kPa AV P1AV1 281.2(0.02)(125) . m = ρAV = = = = 5.9 kg/s v 0.287(415.4) RT1

16.7

A stagnation pressure of 108 kPa is measured for an air flow where the pressure is 100 kPa and 20°C in the approach flow. What is the incomming velocity? Assume a reversible adiabatic compression 108 0.2857 ) = 299.667 K 100

To1 = T1 × (Po1/P1)(k-1)/k = 293.15 × (

V21/2 = ho1 - h1 = Cp (To1 - T1) = 6.543 V1 =

2 × 6.543 × 1000 = 114.4 m/s

16-4 16.8

A jet engine receives a flow of 150 m/s air at 75 kPa, 5°C across an area of 0.6 m 2 with an exit flow at 450 m/s, 75 kPa, 600 K. Find the mass flow rate and thrust. . m = ρAV; ideal gas ρ = P/RT 75 ) × 0.6 × 150 = 0.9395 × 0.6 × 150 0.287 × 278.15

. m = (P/RT)AV = (

= 84.555 kg/s . Fnet = m (Vex-Vin) = 84.555 × (450-150) = 25367 N

16.9

A water cannon sprays 1 kg/s liquid water at a velocity of 100 m/s horizontally out from a nozzle. It is driven by a pump that receives the water from a tank at 15°C, 100 kPa. Neglect elevation differences and the kinetic energy of the water flow in the pump and hose to the nozzle. Find the nozzle exit area, the required pressure out of the pump and the horizontal force needed to hold the cannon. 0.001001 . . m = ρAV = AV/v ⇒ A= mv/V = 1 = 1.0 × 10-5 m2 100 . . . 2 _ W = mwp = mv(Pex - Pin) = mVex /2 p 2 Pex = Pin + Vex /2v = 100 + 1002/2 × 1000 × 0.001 = 150 kPa

. F = mVex = 1 × 100 = 100 N

16.10 An irrigation pump takes water from a lake and discharges it through a nozzle as shown in Fig. P16.10. At the pump exit the pressure is 700 kPa, and the temperature is 20°C. The nozzle is located 10 m above the pump and the atmospheric pressure is 100 kPa. Assuming reversible flow through the system determine the velocity of the water leaving the nozzle. Assume we can neglect kinetic energy in the pipe in and out of the pump. Incompressible flow so Bernoulli's equation applies (V1 ≅ V2 ≅ V3 ≅ 0) v(P3 - P2)+ (V23 - V22)/2 + g(Z3 - Z2) = 0 P 3 = P2 -

g(Z3 - Z2) v

= 700 -

9.807(10) = 602 kPa 1000(0.001002)

V24/2 = v(P3 - P4) ⇒ V4 =

2v(P3 - P4) =

2 × 0.001002 × 502.1 × 1000

= 31.72 m/s

16-5 16.11 A water turbine using nozzles is located at the bottom of Hoover Dam 175 m below the surface of Lake Mead. The water enters the nozzles at a stagnation pressure corresponding to the column of water above it minus 20% due to friction. The temperature is 15°C and the water leaves at standard atmospheric pressure. If the flow through the nozzle is reversible and adiabatic, determine the velocity and kinetic energy per kilogram of water leaving the nozzle. ∆P = ρg∆Z =

g∆Z 9.807 × 175 = = 1714.5 kPa v 0.001001 × 1000

∆Pac = 0.8 ∆P = 1371.6 kPa 2 /2 ⇒ Vex = v∆P = Vex

Vex =

2v∆P

2 × 0.001001 × 1000 × 1371.6 = 62.4 m/s

2 Vex /2 = v∆P = 1.373 kJ/kg

16.12 A water tower on a farm holds 1 m3 liquid water at 20°C, 100 kPa in a tank on top of a 5 m tall tower. A pipe leads to the ground level with a tap that can open a 1.5 cm diameter hole. Neglect friction and pipe losses and estimate the time it will take to empty the tank for water. Bernoulli Equation: Pe = Pi; Vi = 0; Ze = 0; Zi = H V2e /2 = gZi ⇒

Ve =

2gZ =

2 × 9.807 × 5 = 9.9 m/s

. m = ρAVe = AVe/v = ∆m/∆t ∆m = V/v; A = πD2/4 = π × 0.0152 / 4 = 1.77 × 10-4 m2 ⇒ ∆t =

∆t = ∆mv/AVe = V/AVe 1 = 571.6 sec = 9.53 min 1.77 × 10 -4 × 9.9

16-6 16.13 Find the speed of sound for air at 100 kPa at the two temperatures 0°C and 30°C. Repeat the answer for carbon dioxide and argon gases. From eq. 16.28 we have c0 = kRT = c30 =

1.4 × 0.287 × 273.15 × 1000 = 331 m/s

1.4 × 0.287 × 303.15 × 1000 = 349 m/s

For Carbon Dioxide: R = 0.1889, k = 1.289 c0 =

1.289 × 0.1889 × 273.15 × 1000 = 257.9 m/s

c30 =

1.289 × 0.1889 × 303.15 × 1000 = 271.7 m/s

For Argon: R = 0.2081, k = 1.667 c0 =

1.667 × 0.2081 × 273.15 × 1000 = 307.8 m/s

c30 =

1.667 × 0.2081 × 303.15 × 1000 = 324.3 m/s

16.14 If the sound of thunder is heard 5 seconds after seing the lightning and the weather is 20°C how far away is the lightning taking place? The sound travels with the speed of sound in air (ideal gas). L = c × t = kRT × t =

1.4 × 0.287 × 293.15 × 1000 × 5 = 1716 m

16.15 Estimate the speed of sound for steam directly from Eq. 16.28 and the steam tables for a state of 6 MPa, 400°C. Use table values at 5 and 7 MPa at the same entropy as the wanted state. Eq. 16.28 is then done by finite difference. Find also the answer for the speed of sound assuming steam is an ideal gas. c2 = (

δP ∆P ) = ( ) δρ s ∆ρ s

State 6 MPa, 400°C ⇒ s = 6.5407 7 MPa, s ⇒ v = 0.04205; ρ = 1/v = 23.777 5 MPa, s ⇒ v = 0.05467; ρ = 1/v = 18.2909 c2 =

7000 - 5000 = 364.56 × 1000 ⇒ c = 603.8 m/s 23.777 - 18.2909

Cp =

14190 - 10499 = 36.91; 100

k = Cp/Cv = 1.2908; c = kRT =

Cv = Cp - R = 36.91 - 8.31451 = 28.595

R = 0.4615

1.2908 × 0.4615 × 673.15 × 1000 = 633.2 m/s

16-7 16.16 A convergent-divergent nozzle has a throat diameter of 0.05 m and an exit diameter of 0.1 m. The inlet stagnation state is 500 kPa, 500 K. Find the back pressure that will lead to the maximum possible flow rate and the mass flow rate for three different gases as: air; hydrogen or carbon dioxide.

There is a maximum possible flow when M=1 at the throat, T* =

2 2 k 2 1 To; P* = Po ( )k-1; ρ* =ρ o ( )k-1 k+1 k+1 k+1

. m = ρ *A*V = ρ*A*c = P*A* k/RT* A* = πD2/4 = 0.001963 m2 ρ*

. m

k

T*

P*

a)

1.400

416.7

264.1

448.2

2.209

1.944

b)

1.409

415.1

263.4

1704.5

0.154

0.515

c)

1.289

436.9

273.9

348.9

3.318

2.273

c

AE/A* = (DE/D*)2 = 4. There are 2 possible solutions corresponding to points c and d in Fig. 16.13 and Fig. 16.17. For these we have

ME

PE/Po

ME

PE/Po

a)

0.1466

0.985

2.940

0.0298

b)

0.1464

0.985

2.956

0.0293

c)

0.1483

0.986

2.757

0.0367

PB = PE ≅ 0.985 × 500 = 492.5 kPa all cases point c PB = PE = a) 14.9 b) 14.65 c) 18.35 kPa, point d

16-8 16.17 Air is expanded in a nozzle from 2 MPa, 600 K, to 200 kPa. The mass flow rate through the nozzle is 5 kg/s. Assume the flow is reversible and adiabatic and determine the throat and exit areas for the nozzle.  2 k P * = Po  k-1 k+1

Mach #

= 2 × 0.5283 = 1.056 MPa

Velocity Area

Density

T* = To × 2/(k+1) = 600 × 0.8333 = 500 K v* = RT*/P* = 0.287 × 500/1056

2.0 MPa

P

0.2 MPa

= 0.1359 m3/kg

kRT* = 1.4 × 1000 × 0.287 × 500 = 448.2 m/s . A* = mv*/c* = 5 × 0.1359/448.2 = 0.00152 m2 c* =

P2/Po = 200/2000 = 0.1 ⇒ M*2 = 1.701 = V2/c* V2 = 1.701 × 448.2 = 762.4 m/s T2 = 500 × 0.5176 = 258.8 K v2 = RT2/P2 = 0.287 × 258.8/200 = 0.3714 m3/kg . A2 = mv2/V2 = 5 × 0.3714/762.4 = 0.002435 m2 16.18 Consider the nozzle of Problem 16.17 and determine what back pressure will cause a normal shock to stand in the exit plane of the nozzle. This is case g in Fig. 16.17. What is the mass flow rate under these conditions? PE/Po = 200/2000 = 0.1 ; ME = 2.1591 = Mx My = 0.5529 ; Py/Px = 5.275 Py = 5.275 × 200 = 1055 kPa . m = 5 kg/s same as in Problem 16.9

16-9 16.19 At what Mach number will the normal shock occur in the nozzle of Problem 16.17 if the back pressure is 1.4 MPa? (trial and error on Mx) Relate the inlet and exit conditions to the shock conditions with reversible flow before and after the shock. It becomes trial and error. Assume Mx = 1.8 ⇒

My = 0.6165 ;

Poy/Pox = 0.8127

AE/A*x = A2/A* = 0.002435/0.001516 = 1.6062 Ax/A*x = 1.439 ; Ax/A*y = 1.1694 AE/A*y = (AE/A*x)(Ax/A*y)/(Ax/A*x) =

1.6062(1.1694) = 1.3053 1.439

⇒ ME = 0.5189 ; PE/Poy = 0.8323 PE = (PE/Poy)(Poy/Pox)Pox = 0.8323 × 0.8127 × 2000 = 1353 kPa So select the mach number a little less Mx = 1.7 ⇒ My = 0.64055 ; Poy/Pox = 0.85573 Ax/A*x = 1.3376 ; Ax/A*y = 1.1446 AE/A*y = (AE/A*x)(Ax/A*y)/(Ax/A*x) =

1.6062(1.1446) = 1.3744 1.3376

⇒ ME = 0.482 ; PE/Poy = 0.853 PE = (PE/Poy)(Poy/Pox)Pox = 0.853 × 0.85573 × 2000 = 1459.9 kPa Now interpolate between the two Mx = 1.756 ⇒ My = 0.6266 ; Poy/Pox = 0.832 Ax/A*x = 1.3926 ; Ax/A*y = 1.1586 AE/A*y = 1.6062 × 1.1586/1.3926 = 1.3363 ⇒ ME = 0.5 ; PE/Poy = 0.843 PE = 0.843 × 0.832 × 2000 = 1402.7 kPa OK

16-10 16.20 Consider the nozzle of Problem 16.17. What back pressure will be required to cause subsonic flow throughout the entire nozzle with M = 1 at the throat? AE/A* = 0.002435/0.001516 = 1.6062 ME = 0.395 ;

PE/Po = 0.898

PE = 0.898 × 2000 = 1796 kPa

16.21 Determine the mass flow rate through the nozzle of Problem 16.17 for a back pressure of 1.9 MPa. PE/Pox = 1.9/2.0 = 0.95 ⇒ ME = 0.268 TE = (T/To)To = 0.9854 × 600 = 591.2 K cE =

kRTE =

1.4 × 1000 × 0.287 × 591.2 = 487.4 m/s

VE = MEcE = 0.268 × 487.4 = 130.6 m/s vE = RT/P = 0.287 × 591.2/1900 = 0.0893 m3/kg . m = AEVE/vE = 0.002435 × 130.6/0.0893 = 3.561 kg/s

16.22 At what Mach number will the normal shock occur in the nozzle of Problem 16.16 flowing with air if the back pressure is halfway between the pressures at c and d in Fig. 16.17? First find the two pressure that will give exit at c and d. See solution to 16.8 a) 2

AE/A* = (DE/D*) = 4 ⇒

PE = 492.5 (c) 16.9 (d)

PE = (492.5 + 14.9)/2 = 253.7 kPa Assume ME = 2.4 ⇒ My = 0.5231 ; Poy/Pox = 0.54015 Ax/A*x = 2.4031 ; Ax/A*y = 1.298 AE/A*y = (AE/A*x)(Ax/A*y) / (Ax/A*x) = 4 × 1.298/2.4031 = 2.1605 ⇒ ME = 0.2807 ; PE/Poy = 0.94675 PE = (PE/Poy)(Poy/Pox)Pox = 0.94675 × 0.54015 × 500 = 255.7 kPa Repeat if Mx = 2.5 ⇒ PE = 233.8 kPa Interpolate to match the desired pressure =>

Mx = 2.41

16-11 16.23 A convergent nozzle has minimum area of 0.1 m2 and receives air at 175 kPa, 1000 K flowing with 100 m/s. What is the back pressure that will produce the maximum flow rate and find that flow rate? P* 2 k =( )k-1 = 0.528 Po k+1

Critical Pressure Ratio

Find Po: h0 = h1 + V21/2 = 1046.22 + 1002/2000 = 1051.22 T0 = Ti + 4.4 = 1004.4 from table A.7 P0 = Pi (T0/Ti)k/(k-1) = 175 × (1004.4/1000)3.5 = 177.71 P* = 0.528 Po = 0.528 × 177.71 = 93.83 kPa T* = 0.8333 To = 836.97 K ρ* =

P* 93.83 = = 0.3906 RT* 0.287 × 836.97

V = c = kRT* = 1.4 × 1000 × 0.287 × 836.97 = 579.9 m/s . m = ρAV = 0.3906 × 0.1 × 579.9 = 22.65 kg/s 16.24 A convergent-divergent nozzle has a throat area of 100 mm 2 and an exit area of 175 mm2. The inlet flow is helium at a total pressure of 1 MPa, stagnation temperature of 375 K. What is the back pressure that will give sonic condition at the throat, but subsonic everywhere else? AE/A* = 175/100 = 1.75 ; kHe = 1.667 ME = 0.348 ; PE/PO = 0.906 PE = 0.906 × 1000 = 906 kPa

16-12 16.25 A nozzle is designed assuming reversible adiabatic flow with an exit Mach number of 2.6 while flowing air with a stagnation pressure and temperature of 2 MPa and 150°C, respectively. The mass flow rate is 5 kg/s, and k may be assumed to be 1.40 and constant. a. Determine the exit pressure, temperature, and area, and the throat area. b. Suppose that the back pressure at the nozzle exit is raised to 1.4 MPa, and that the flow remains isentropic except for a normal shock wave. Determine the exit Mach number and temperature, and the mass flow rate through the nozzle. (a)

From Table A.11:

ME = 2.6

PE = 2.0 × 0.05012 = 0.1002 MPa T* = 423.15 × 0.8333 = 352.7 K P* = 2.0 × 0.5283 = 1.057 MPa c* =

1.4 × 1000 × 0.287 × 352.7 = 376.5 m/s

v* = 0.287 × 352.7/1057 = 0.0958 m3/kg A* = 5 × 0.0958/376.5 = 1.272 × 10-3m2 AE = 1.272 × 10-3 × 2.896 = 3.68 × 10-3m2 TE = 423.15 × 0.42517 = 179.9 K Assume Mx = 2 then My = 0.57735, Poy/Pox = 0.72088, AE/A*x = 2.896 Ax/A*x = 1.6875, Ax/A*y = 1.2225, AE/A*y = 2.896 × 1.2225/1.6875 = 2.098 ⇒ ME = 0.293, PE/Poy = 0.94171 PE = 0.94171 × 0.72088 × 2.0 = 1.357 MPa, OK TE = 0.98298 × 423.15 = 416 K,

. m = 5 kg/s

16-13 16.26 A jet plane travels through the air with a speed of 1000 km/h at an altitude of 6 km, where the pressure is 40 kPa and the temperature is −12°C. Consider the inlet diffuser of the engine where air leaves with a velocity of 100 m/s. Determine the pressure and temperature leaving the diffuser, and the ratio of inlet to exit area of the diffuser, assuming the flow to be reversible and adiabatic. v 1 = 0.287 × 261.15/40 = 1.874

V = 1000 km/h = 277.8 m/s, h1 = 261.48, Pr1 = 0.6862

ho1 = 261.48 + 277.8 2/2000 = 300.07 ⇒ To1 = 299.7 K, Pro1 = 1.1107 Po1 = 40 × 1.1107/0.6862 = 64.74 kPa h2 = 300.07 - 1002/2000 = 295.07

⇒

T2 = 294.7 K, Pr2 = 1.0462

P2 = 64.74 × 1.0462/1.1107 = 61 kPa v2 = 0.287 × 294.7/61 = 1.386 A1/A2 = (1.874/1.386)(100/277.8) = 0.487 16.27 A 1-m3 insulated tank contains air at 1 MPa, 560 K. The tank is now discharged through a small convergent nozzle to the atmosphere at 100 kPa. The nozzle has an exit area of 2 × 10−5 m2. a. Find the initial mass flow rate out of the tank. b. Find the mass flow rate when half the mass has been discharged. c. Find the mass of air in the tank and the mass flow rate out of the tank when the nozzle flow changes to become subsonic. a. The back pressure ratio: PB/Po1 = 100/1000 = 0.1 < (P*/Po)crit = 0.5283 so the initial flow is choked with the maximum possible flow rate. ME = 1 ; PE = 0.5283 × 1000 = 528.3 kPa TE = T* = 0.8333 × 560 = 466.7 K VE = c =

kRT* =

1.4 × 1000 × 0.287 × 466.7 = 433 m/s

vE = RT*/PE = 0.287 × 466.7/528.3 = 0.2535 m3/kg

16-14 . m1 = AVE/vE = 2 × 10-5 × 433/0.2535 = 0.0342 kg/s b. The initial mass is m1 = P1V/RT1 = 1000 × 1/0.287 × 560 = 6.222 kg with a mass at state 2 as m2 = m1/2 = 3.111 kg. Assume an adiabatic reversible expansion of the mass that remains in the tank. P2 = P1(v1/v2)k = 100 × 0.51.4 = 378.9 kPa T2 = T1(v1/v2)k-1 = 560 × 0.50.4 = 424 K The pressure ratio is still less than critical and the flow thus choked. PB/Po2 = 100/378.9 = 0.264 < (P*/Po)crit ME = 1 ; PE = 0.5283 × 378.9 = 200.2 kPa TE = T* = 0.8333 × 424 = 353.7 K VE = c =

kRT* =

1.4 × 1000 × 0.287 × 353.7 = 377 m/s -5

2×10 (377)(200.2) . = 0.0149 kg/s m2 = AVEPE/RTE = 0.287(353.7) c. The flow changes to subsonic when the pressure ratio reaches critical. PB/Po3 = 0.5283 ⇒ Po3 = 189.3 kPa v1/v3 = (Po3/P1)1/k = (189.3/1000)0.7143 = 0.3046 m3 = m1v1/v3 = 1.895 kg T3 = T1(v1/v3)k-1 = 560 × 0.30460.4 = 348 K PE = PB = 100 kPa ; ME = 1 TE = 0.8333 × 348 = 290 K ; VE = -5

kRTE = 341.4 m/s

2×10 (341.4)(100) . = 0.0082 kg/s m3 = AVEPE/RTE = 0.287(290)

16-15 16.28 A 1-m3 uninsulated tank contains air at 1 MPa, 560 K. The tank is now discharged through a small convergent nozzle to the atmosphere at 100 kPa while heat transfer from some source keeps the air temperature in the tank at 560 K. The nozzle has an exit area of 2 × 10−5 m2. a. Find the initial mass flow rate out of the tank. b. Find the mass flow rate when half the mass has been discharged. c. Find the mass of air in the tank and the mass flow rate out of the tank when the nozzle flow changes to become subsonic. a.

Same solution as in 16.27 a)

b.

From solution 16.27 b) we have m2 = m1/2 = 3.111 kg P2 = P1/2 = 500 kPa ; T2 = T1 ;

PB/P2 = 100/500 = 0.2 < (P*/Po)crit

The flow is choked and the velocity is: VE = c =

kRT* = 433 m/s from 16.27 a)

PE = 0.5283 × 500 = 264.2 kPa ; ME = 1 -5

2×10 (433)(264.2) . = 0.01708 kg/s m2 = AVEPE/RTE = 0.287(466.7) c. Flow changes to subsonic when the pressure ratio reaches critical. PB/Po = 0.5283 ; P3 = 189.3 kPa m3 = m1P3/P1 = 1.178 kg ; T3 = T1 ⇒ VE = 433 m/s -5

2×10 (433)(189.3) . = 0.01224 kg/s m3 = AVEPE/RTE = 0.287(466.7)

16.29 The products of combustion enter a convergent nozzle of a jet engine at a total pressure of 125 kPa, and a total temperature of 650°C. The atmospheric pressure is 45 kPa and the flow is adiabatic with a rate of 25 kg/s. Determine the exit area of the nozzle. The critical pressure: Pcrit = P2 = 125 × 0.5283 = 66 kPa > Pamb The flow is then choked. V 2 = c2 =

T2 = 923.15 × 0.8333 = 769.3 K

1.4 × 1000 × 0.287 × 769.3 = 556 m/s

v2 = 0.287 × 769.3/66 = 3.3453 A2 = 25 × 3.3453/556 = 0.1504 m2

16-16 16.30 Air is expanded in a nozzle from 700 kPa, 200°C, to 150 kPa in a nozzle having an efficiency of 90%. The mass flow rate is 4 kg/s. Determine the exit area of the nozzle, the exit velocity, and the increase of entropy per kilogram of air. Compare these results with those of a reversible adiabatic nozzle. T2s = T1(P2/P1)(k-1)/k = 473.2 (150/700)0.286 = 304.6 K V2s2 = 2 × 1000 × 1.0035(473.2 - 304.6) = 338400 V22 = 0.9 × 338400 ⇒ V2 = 552 m/s h2 + V22/2 = h1 ⇒ T2 = T1 - V22/2Cp T2 = 473.2 - 5522/(2 × 1000 × 1.004) = 321.4 K ; v2 = 0.287 × 321.4/150 = 0.6149 m3/kg A2 = 4 × 0.6149/552 = 0.00446 m2 = 4460 mm2 321.4 150 s2 - s1 = 1.0035 ln  - 0.287 ln  = 0.0539 kJ/kg K 473.2   700

16.31 Repeat Problem 16.26 assuming a diffuser efficiency of 80%. Same as problem 16.26, except ηD = 0.80. We thus have from 16.26 h3 - h1 ho1 - h1

=

h3 - 261.48 300.07 - 261.48

h 01 02

= 0.8

3 2

⇒ h3 = 292.35, Pr3 = 1.0129

1

Po2 = P3 = 40 × 1.0129/0.6862 = 59.04 kPa

s

Pro2 = Pro1 = 1.1107 h2 = 300.07 - 1002/2000 = 295.07

⇒ T2 = 294.7 K,

P2 = 59.04 × 1.0462/1.1107 = 55.6 kPa v2 = 0.287 × 294.7/55.6 = 1.521 A1/A2 = (1.874/1.521)(100/277.8) = 0.444

Pr2 = 1.0462

16-17 16.32 Consider the diffuser of a supersonic aircraft flying at M = 1.4 at such an altitude that the temperature is −20°C, and the atmospheric pressure is 50 kPa. Consider two possible ways in which the diffuser might operate, and for each case calculate the throat area required for a flow of 50 kg/s. a. The diffuser operates as reversible adiabatic with subsonic exit velocity. b. A normal shock stands at the entrance to the diffuser. Except for the normal shock the flow is reversible and adiabatic, and the exit velocity is subsonic. This is shown in Fig. P16.32. a. Assume a convergent-divergent diffuser with M = 1 at the throat. Relate the inlet state to the sonic state P1/Po = 0.31424 ; P*/Po1 = 0.5283 P* = c* =

0.5283 0.8333 50 = 84 kPa ; T* = ; 253.2 = 293.7 K 0.31424 0.71839 kRT* =

1.4 × 1000 × 0.287 × 293.7 = 343.5 m/s

v* = RT*/P* = 0.287 × 293.7/84 = 1.0035 . A* = mv*/c* = 50 × 1.0035/343.5 = 0.1461 m2 b.

Across the shock we have My = 0.7397 ; Py = 50 × 2.12 = 106 kPa ; Ty = 253.2 × 1.2547 = 317.7 K P* =

0.5283 × 106 = 80.6 kPa 0.6952

T* =

0.8333 × 317.7 = 293.7 K, c* = 343.5 m/s 0.9011

v* = 0.287 × 293.7/80.6 = 1.0458 A* = 50 × 1.0458/343.5 = 0.1522 m2

16-18 16.33 Air enters a diffuser with a velocity of 200 m/s, a static pressure of 70 kPa, and a temperature of −6°C. The velocity leaving the diffuser is 60 m/s and the static pressure at the diffuser exit is 80 kPa. Determine the static temperature at the diffuser exit and the diffuser efficiency. Compare the stagnation pressures at the inlet and the exit. To1 = T1 + V21/2Cp = 267.15 + 2002/(2000 × 1.004) = 287.1 K To2 = To1 ⇒ T2 = To2 - V22/2Cp = 287.1 - 602/(2000 × 1.004) = 285.3 K To1 - T1

k-1 Po1 - P1 = k P1

⇒

Po1 - P1 = 18.25 ⇒ Po1 = 88.3 kPa

k - 1 Po2 - P2 = k P2

⇒

Po2 - P2 = 1.77 ⇒ Po2 = 81.8 kPa

T1 To2 - T2 T2 ex

Ts = T1 (Po2/P1)k-1/k = 267.15 × 1.0454 = 279.3 K ex

ηD =

Ts - T1 To1 - T1

=

279.3 - 267.15 = 0.608 287.1 - 267.15

16.34 Steam at a pressure of 1 MPa and temperature of 400°C expands in a nozzle to a pressure of 200 kPa. The nozzle efficiency is 90% and the mass flow rate is 10 kg/s. Determine the nozzle exit area and the exit velocity. First do the ideal reversible adiabatic nozzle s2s= s1= 7.4651 ⇒ T2s = 190.4°C ; h2s = 2851, h1= 3263.9 Now the actual nozzle can be calculated h1 - h2ac = ηD(h1 - h2s) = 0.9(3263.9 - 2851) = 371.6 h2ac = 2892.3 , T2 = 210.9°C, v2 = 1.1062 V2 = 2000(3263.9 - 2892.3) = 862 m/s . A2 = mv2/V2 = 10 × 1.1062/862 = 0.01283 m2

16-19 16.35 Steam at 800 kPa, 350°C flows through a convergent-divergent nozzle that has a throat area of 350 mm2. The pressure at the exit plane is 150 kPa and the exit velocity is 800 m/s. The flow from the nozzle entrance to the throat is reversible and adiabatic. Determine the exit area of the nozzle, the overall nozzle efficiency, and the entropy generation in the process. ho1 = 3161.7, so1 = 7.4089 P*/Po1 = (2/(k+1))k/(k-1) = 0.54099 ⇒ P* = 432.7 kPa At *: (P*,s* = so1) ⇒ h* = 2999.3, v* = 0.5687 ∆ h = V2/2 ⇒ V* = 2000(3161.7-2999.3) = 569.9 m/s . m = AV*/v* = 350 × 10-6 × 569.9/0.5687 = 0.3507 kg/s he = ho1 - V2e /2 = 3161.7 - 8002/2 × 1000 = 2841.7 Exit: Pe, he: ve = 1.395, se =7.576 . Ae = mve/Ve = 0.3507 × 1.395/800 = 6.115 × 10-4 m2 sgen = se - so1 = 7.576 - 7.4089 = 0.167 kJ/kg K

16.36 Air at 150 kPa, 290 K expands to the atmosphere at 100 kPa through a convergent nozzle with exit area of 0.01 m2. Assume an ideal nozzle. What is the percent error in mass flow rate if the flow is assumed incompressible? Pe k-1 Te = Ti ( ) k = 258.28 K Pi V2e /2 = hi - he = Cp (Ti - Te) = 1.004 (290 - 258.28) = 31.83 Ve = 252.3 m/s;

ve =

RTe Pe

=

0.287 × 258.28 = 0.7412 100

0.01 × 252.3 . m = AVe / ve = = 3.4 kg/s 0.7413 Incompressible Flow: vi = RT/P = 0.287 × 290/150 = 0.55487 m3/kg V2e /2 = v ∆P = vi (Pi - Pe) = 0.55487 (150 - 100) = 27.74 kJ/kg . => Ve = 235 m/s => m = AVe / vi = 0.01 × 235 / 0.55487 = 4.23 kg/s . mincompressible 4.23 . = = 1.25 about 25% overestimation. 3.4 mcompressible

16-20 16.37 A sharp-edged orifice is used to measure the flow of air in a pipe. The pipe diameter is 100 mm and the diameter of the orifice is 25 mm. Upstream of the orifice, the absolute pressure is 150 kPa and the temperature is 35°C. The pressure drop across the orifice is 15 kPa, and the coefficient of discharge is 0.62. Determine the mass flow rate in the pipeline. k-1∆P 0.4 15 ∆T = Ti  = 308.15 × × = 8.8 1.4 150  k  Pi vi = RTi/Pi = 0.5896 Pe = 135 kPa, Te = 299.35, ve = 0.6364 . . mi = me ⇒ Vi = Ve(De/Di)2 vi/ve = 0.0579 hi - he = V2e (1 - 0.05792)/2 = Cp(Ti - Te) e

Vs =

2 × 1000 × 1.0035 × 8.8/(1 - 0.0579)2 = 133.1 m/s . m = 0.62 (π/4) (0.025)(2133.1/0.6364) = 0.06365 kg/s

16.38 A critical nozzle is used for the accurate measurement of the flow rate of air. Exhaust from a car engine is diluted with air so its temperature is 50°C at a total pressure of 100 kPa. It flows through the nozzle with throat area of 700 mm2 by suction from a blower. Find the needed suction pressure that will lead to critical flow in the nozzle, the mass flow rate and the blower work, assuming the blower exit is at atmospheric pressure 100 kPa. P* = 0.5283 Po = 52.83 kPa, T* = 0.8333 To = 269.3 K v* = RT*/P* = 0.287 × 269.3/52.83 = 1.463 c* = kRT* = 1.4 × 1000 × 0.287 × 269.3 = 328.9 m/s . m = Ac*/v* = 700 × 10-6 × 328.9/1.463 = 0.157 kg/s

16-21 16.39 A convergent nozzle is used to measure the flow of air to an engine. The atmosphere is at 100 kPa, 25°C. The nozzle used has a minimum area of 2000 mm2 and the coefficient of discharge is 0.95. A pressure difference across the nozzle is measured to 2.5 kPa. Find the mass flow rate assuming incompressible flow. Also find the mass flow rate assuming compressible adiabatic flow. Assume Vi ≅ 0,

vi = RTi/Pi = 0.287 × 298.15/100 = 0.8557

Ve,s2/2 = hi - he,s = vi(Pi - Pe) = 2.1393 kJ/kg Ve,s =

2 × 1000 × 2.1393 = 65.41 m/s

. ms = AVe,s/vi = 2000 × 10-6 × 65.41/0.8557 = 0.153 kg/s . . ma = CDms = 0.1454 kg/s Te,s = Ti (Pe/Pi)(k-1)/k = 298.15(97.5/100)0.2857 = 296 K ∆h = Cp∆T = 1.0035 × 2.15 = 2.1575 = Ve,s2/2 Ve,s =

2 × 1000 × 2.1575 = 65.69 m/s

ve,s = 0.287 × 296/97.5 = 0.8713 . ms = AVe,s/ve,s = 2000 × 10-6 × 65.69/0.8713 = 0.1508 kg/s . . ma = CDms = 0.1433 kg/s

16-22 16.40 A convergent nozzle with exit diameter of 2 cm has an air inlet flow of 20°C, 101 kPa (stagnation conditions). The nozzle has an isentropic efficiency of 95% and the pressure drop is measured to 50 cm water column. Find the mass flow rate assuming compressible adiabatic flow. Repeat calculation for incompressible flow. Convert ∆P to kPa: ∆P = 50 cm H2O = 0.5 × 9.8064 = 4.903 kPa T0 = 20°C = 293.15 K

P0 = 101 kPa

Assume inlet Vi = 0

Pe = P0 - ∆P = 101 - 4.903 = 96.097 kPa

Pe k-1 96.097 0.2857 Te = T0 ( ) k = 293.15 ×( ) = 289.01 101 P0 V2e /2 = hi - he = Cp (Ti - Te) = 1.004 × (293.15 - 289.01) = 4.1545 kJ/kg = 4254.5 J/kg Ve 2ac/2 = η Ve2s/2 = 0.95 × 4154.5 = 3946.78 Te ac = Ti ρe ac =

Pe RTp

Ve 2ac/2 Cp =

= 293.15 -

=> Ve = 91.15 m/s ⇒ Ve ac = 88.85 m/s

3.9468 = 289.2 1.0035

96.097 = 1.158 kg/m3 0.287 × 289.2

π . m = ρAV = 1.158 × × 0.022 × 88.85 = 0.0323 kg/s 4

16.41 Steam at 600 kPa, 300°C is fed to a set of convergent nozzles in a steam turbine. The total nozzle exit area is 0.005 m 2 and they have a discharge coefficient of 0.94. The mass flow rate should be estimated from the measurement of the pressure drop across the nozzles, which is measured to be 200 kPa. Determine the mass flow rate. se,s = si = 7.3724 kJ/kg K ;

(Pe, se,s) ⇒ he,s = 2961 kJ/kg

ve,s = 0.5932, hi = 3061.6 Ve,s =

2 × 1000(3061.6 - 2961) = 448.55 m/s

. m = 0.005 × 448.55/0.5932 = 3.781 kg/s

16-23 16.42 The coefficient of discharge of a sharp-edged orifice is determined at one set of conditions by use of an accurately calibrated gasometer. The orifice has a diameter of 20 mm and the pipe diameter is 50 mm. The absolute upstream pressure is 200 kPa and the pressure drop across the orifice is 82 mm of mercury. The temperature of the air entering the orifice is 25°C and the mass flow rate measured with the gasometer is 2.4 kg/min. What is the coefficient of discharge of the orifice at these conditions? ∆P = 82 × 101.325/760 = 10.93 kPa k-1 0.4 ∆T = Ti   ∆P/Pi = 298.15 × × 10.93/200 = 4.66 1.4 k  vi = RTi/Pi = 0.4278, ve = RTe/Pe = 0.4455 Vi = VeAevi/Aive = 0.1536 Ve (V2e - V2i )/2 = V2e (1 - 0.15362)/2 = hi - he = Cp∆T Ve =

2 × 1000 × 1.004 × 4.66/(1 - 0.15362) = 97.9 m/s

π . m = AeVe/ve = × 0.022 × 97.9/0.4455 = 0.069 kg/s 4 CD = 2.4/60 × 0.069 = 0.58

16-24 16.43 (Adv.) Atmospheric air is at 20°C, 100 kPa with zero velocity. An adiabatic reversible compressor takes atmospheric air in through a pipe with cross-sectional area of 0.1 m2 at a rate of 1 kg/s. It is compressed up to a measured stagnation pressure of 500 kPa and leaves through a pipe with cross-sectional area of 0.01 m2. What is the required compressor work and the air velocity, static pressure and temperature in the exit pipeline? C.V. compressor out to standing air and exit to stagnation point. . . . _ m ho1 + W = m(h + V2/2)ex = mho,ex c . . o mso1 = msex ⇒ Pr,o,ex = Pr,o1 (Pst,ex/Po1) = 1.028(500/100) = 5.14 ⇒ To,ex = 463,

ho,ex = 465.38, ho1 = 209.45

. _ = m (ho,ex - ho1) = 1(465.38 - 209.45) = 255.9 kW W c Pex = Po,ex(Tex/To,ex)k/(k-1)

2 Tex = To,ex - Vex /2Cp

. m = 1 kg/s = (ρAV)ex = PexAVex/RTex . Now select 1 unknown amongst Pex, Tex, Vex and write the continuity eq. m and solve the nonlinear equation. Say, use Tex then Vex =

2Cp(To,ex - Tex)

o . m = 1 kg/s = Pex(Tex/To,ex)k/k-1A

2Cp(To,ex - Tex)/RTex

solve for Tex/To,ex (close to 1) Tex = 462.6 K ⇒ Vex= 28.3 m/s, Pex = 498.6 kPa

16-25

English Unit Problems 16.44E Steam leaves a nozzle with a velocity of 800 ft/s. The stagnation pressure is 100 lbf/in2, and the stagnation temperature is 500 F. What is the static pressure and temperature? h1 = ho1 -

V21/2gc=

8002 Btu 1279.1 × 778 = 1266.3 lbm 2 × 32.174

s1 = s0 = 1.7085 Btu/lbm R (h, s) Computer table ⇒ P1 = 88 lbf/in.2, T = 466 F

16.45E Air leaves the compressor of a jet engine at a temperature of 300 F, a pressure of 45 lbf/in2, and a velocity of 400 ft/s. Determine the isentropic stagnation temperature and pressure. ho1 - h1 = V21/2gc = 4002/2 × 32.174 × 778 = 3.2 Btu/lbm To1 - T - 1 = (ho1 - h1)/Cp = 3.2/0.24 = 13.3 To1 = T + ∆T = 300 + 13.3 = 313.3 F = 773 R k

Po1 = P1 To1/T1k-1 = 45(773/759.67)3.5 = 47.82 lbf/in2  

16.46E A meteorite melts and burn up at temperatures of 5500 R. If it hits air at 0.75 lbf/in.2, 90 R how high a velocity should it have to reach such temperature? Assume we have a stagnation T = 5500 R h1 + V21/2 = hstagn. Extrapolating from table C.6,

hstagn. = 1546.5, h1 = 21.4

V21/2 = 1546.5 – 21.4 = 1525.1 Btu/lbm V1 =

2 × 32.174 × 778 × 1525.1 = 8738 ft/s

16-26 16.47E A jet engine receives a flow of 500 ft/s air at 10 lbf/in.2, 40 F inlet area of 7 ft2 with an exit at 1500 ft/s, 10 lbf/in.2, 1100 R. Find the mass flow rate and thrust. . m = ρAV; ideal gas ρ = P/RT . m = (P/RT)AV =

10 × 144 × 7 × 500 = 189.1 lbm/s 53.34 × 499.7

. Fnet = m (Vex - Vin) = 189.1 × (1500 - 500) / 32.174 = 5877 lbf 16.48E A water turbine using nozzles is located at the bottom of Hoover Dam 575 ft below the surface of Lake Mead. The water enters the nozzles at a stagnation pressure corresponding to the column of water above it minus 20% due to friction. The temperature is 60 F and the water leaves at standard atmospheric pressure. If the flow through the nozzle is reversible and adiabatic, determine the velocity and kinetic energy per kilogram of water leaving the nozzle. ∆P =

ρg∆Z = g(∆Z/v)/gc = 575/(0.016035 × 144) = 249 lbf/in.2 gc

∆Pac = 0.8∆P = 199.2 lbf/in.2 Vex =

2v∆ P =

2g∆ Z =

and Bernoulli

v∆P = Vex2/2

2× 32.174× 575 = 192.4 ft/s

2 Vex /2 = v∆P = g∆Z/gc = 575/778 = 0.739 Btu/lbm

16.49E Find the speed of sound for air at 15 lbf/in.2, at the two temperatures of 32 F and 90 F. Repeat the answer for carbon dioxide and argon gases. From eq. 16.28 we have c32 = c90 =

kRT =

1.4 × 32.174 × 53.34 × 491.7 = 1087 ft/s

1.4 × 32.174 × 53.34 × 549.7 = 1149 ft/s

For Carbon Dioxide: R = 35.1, k = 1.289 c32 =

1.289 × 32.174 × 35.1 × 491.7 = 846 ft/s

c90 =

1.289 × 32.174 × 35.1 × 549.7 = 894.5 ft/s

For Argon: R = 38.68, k = 1.667 c32 =

1.667 × 32.174 × 38.68 × 491.7 = 1010 ft/s

c90 =

1.667 × 32.174 × 38.68 × 549.7 = 1068 ft/s

16-27 16.50E Air is expanded in a nozzle from 300 lbf/in.2, 1100 R to 30 lbf/in. 2. The mass flow rate through the nozzle is 10 lbm/s. Assume the flow is reversible and adiabatic and determine the throat and exit areas for the nozzle.  2 k P * = Po  k-1 k+1

Mach #

Velocity Area

Density

= 300 × 0.5283 = 158.5 lbf/in.2. T* = To× 2/(k+1) = 1100 × 0.8333 = 916.6 R v* = RT*/P* = 53.34 × 916.6/(158.5 × 144)

300 psia

P

30 psia

= 2.1421 ft3/lbm

kRT* = 1.4 × 32.174 × 53.34 × 916.6 = 1484 ft/s . A* = mv*/c* = 10 × 2.1421/1484 = 0.0144 ft2 c* =

P2/Po = 30/300 = 0.1

Table A.11 ⇒

M*2 = 1.701 = V2/c*

V2 = 1.701 × 1484 = 2524 ft/s T2 = 916.6 × 0.5176 = 474.4 R v2 = RT2/P2 = 53.34 × 474.4/(30 × 144) = 5.8579 ft3/lbm . A2 = mv2/V2 = 10 × 5.8579 / 2524 = 0.0232 ft2

16-28 16.51E A convergent nozzle has a minimum area of 1 ft2 and receives air at 25 lbf/in.2, 1800 R flowing with 330 ft/s. What is the back pressure that will produce the maximum flow rate and find that flow rate? P* 2 k =( )k-1 = 0.528 Po k+1 Find Po:

Critical Pressure Ratio

Cp = (463.445 - 449.794)/50 = 0.273

h0 = h1 + V21/2 T0 = 1800 +

from table C.6

⇒ T0 = Ti + V2/2Cp

3302/2 = 1807.97 => T * = 0.8333 To = 1506.6 R 32.174 × 778 × 0.273

P0 = Pi (T0/Ti)k/(k-1) = 25 × (1807.97/1800)3.5 = 25.39 P* = 0.528 Po = 0.528 × 25.39 = 13.406 lbf/in2 13.406 × 144 P* ρ = = = 0.024 RT* 53.34 × 1506.6 *

V = c = kRT* = 1.4 × 53.34 × 1506.6 × 32.174 = 1902.6 ft/s . m = ρAV = 0.024 × 1 × 1902.6 = 45.66 lbm/s

16.52E A jet plane travels through the air with a speed of 600 mi/h at an altitude of 20000 ft, where the pressure is 5.75 lbf/in. 2 and the temperature is 25 F. Consider the diffuser of the engine where air leaves at with a velocity of 300 ft/s. Determine the pressure and temperature leaving the diffuser, and the ratio of inlet to exit area of the diffuser, assuming the flow to be reversible and adiabatic. V = 600 mi/h = 880 ft/s v1 = 53.34 × 484.67/(5.75 × 144) = 31.223 , h1 = 115.91, Pr1 = 0.7637 ho1 = 115.91 + 880 2/(2 × 32.174 × 778) = 131.38 Btu/lbm Pro1 = 1.1822 ⇒ Po1 =

5.75 × 1.1822 = 8.9 lbf/in.2 0.7637

h2 = 131.38 - 3002/(2 × 32.174 × 778) = 129.58 T2 = 542 R, Pr2 = 1.1267 => P2 = 8.9 × 1.1267/1.1822 = 8.48 lbf/in.2 v2 = 53.34 × 542/(8.482 × 144) = 23.67 ft3/lbm A1/A2 = (31.223/23.67)(300/880) = 0.45

16-29 16.53E The products of combustion enter a nozzle of a jet engine at a total pressure of 18 lbf/in.2, and a total temperature of 1200 F. The atmospheric pressure is 6.75 lbf/in.2. The nozzle is convergent, and the mass flow rate is 50 lbm/s. Assume the flow is adiabatic. Determine the exit area of the nozzle. Pcrit = P2 = 18 × 0.5283 = 9.5 lbf/in.2 > Pamb The flow is then choked. T2 = 1660 × 0.8333 = 1382 R V 2 = c2 =

1.4 × 32.174 × 53.34 × 1382 = 1822 ft/s

v2 = 53.34 × 1382/9.5 × 144 = 53.9 A2 = 50 × 53.9/1822 = 1.479 ft2

16.54E Repeat Problem 16.52 assuming a diffuser efficiency of 80%. From solution to 16.52 h1 = 115.91, ho1 = 131.38, Pr1 = 0.7637, v1 = 31.223 h2 = 129.58, Pr2 = 1.1267, Pro1 = 1.1822 ηD = (h3 - h1)/(ho1 - h1) = 0.8

⇒

h3 = 128.29, Pr3 = 1.088

Po2=P3 = 5.75 × 1.088/0.7637 = 8.192 lbf/in.2 P2 = Po2Pr2/Pro1 = 8.192 × 1.1267/1.1822 = 7.807 lbf/in.2 T2 = 542 R ⇒ v2 =

53.34 × 542 × 144 = 25.716 ft3/lbm 7.807

A1/A2 = v1V2/v2V1 = 31.223 × 300/(25.716 × 880) = 0.414

16-30 16.55E A 50-ft3 uninsulated tank contains air at 150 lbf/in.2, 1000 R. The tank is now discharged through a small convergent nozzle to the atmosphere at 14.7 lbf/in.2 while heat transfer from some source keeps the air temperature in the tank at 1000 R. The nozzle has an exit area of 2 × 10−4 ft2. a. Find the initial mass flow rate out of the tank. b. Find the mass flow rate when half the mass has been discharged. c. Find the mass of air in the tank and the mass flow rate out of the tank when the nozzle flow changes to become subsonic. PB/Po = 14.7/150 = 0.098 < (P*/Po)crit = 0.5283 a. The flow is choked, max possible flow rate ME =1 ;

PE = 0.5283 × 150 = 79.245 lbf/in.2

TE = T* = 0.8333 × 1000 = 833.3 R VE = c =

kRT* =

1.4 × 53.34 × 833.3 × 32.174 = 1415 ft/s

vE = RT*/PE = 53.34 × 833.3/(79.245 × 144) = 3.895 ft3/lbm . Mass flow rate is : m1 = AVE/vE = 2 × 10-4 × 1415/3.895 = 0.0727 lbm/s b. m1 = P1V/RT1 = 150 × 50 × 144/53.34 × 1000 = 20.247 lbm m2 = m1/2 = 10.124 lbm,

P2=P1/2 = 75 lbf/in.2 ;

T2 = T1

PB/P2= 14.7/75 = 0.196 < (P*/Po)crit The flow is choked and the velocity is the same as in 1) PE = 0.5283 × 75 = 39.623 lbf/in.2 ; ME =1 2 × 10-4 × 1415 × 39.623 × 144 . = 0.0303 lbm/s m2 = AVEPE/RTE = 53.34 × 1000 c. Flow changes to subsonic when the pressure ratio reaches critical. PB/Po = 0.5283 P3 = 27.825 lbf/in.2 m3 = m1P3/P1 = 3.756 lbm ; T3 = T1 ⇒ VE = 1415 ft/s 2 × 10-4 × 1415 × 27.825 × 144 . = 0.02125 lbm/s m3 = AVEPE/RTE = 53.34 × 1000

16-31 16.56E Air enters a diffuser with a velocity of 600 ft/s, a static pressure of 10 lbf/in.2, and a temperature of 20 F. The velocity leaving the diffuser is 200 ft/s and the static pressure at the diffuser exit is 11.7 lbf/in. 2 . Determine the static temperature at the diffuser exit and the diffuser efficiency. Compare the stagnation pressures at the inlet and the exit. To1 = T1 + To2 = To1

V21 2gcCp

= 480 + 6002/(2 × 32.174 × 778 × 0.24) = 510 R

⇒

T2 = To2 - V22/2Cp = 510 - 2002/(2 × 32.174 × 0.24 × 778) = 506.7 R To2 - T2 T2

=

k-1 Po2 - P2 ⇒ Po2 - P2 = 0.267 ⇒ Po2 = 11.97 lbf/in.2 k P2

Tex,s = T1 (Po2/P1)(k-1)/k = 480 × 1.0528 = 505.3 R ηD =

Tex,s - T1 To1 - T1

=

505.3 - 480 = 0.844 51 - 480

16-32 16.57E A convergent nozzle with exit diameter of 1 in. has an air inlet flow of 68 F, 14.7 lbf/in.2 (stagnation conditions). The nozzle has an isentropic efficiency of 95% and the pressure drop is measured to 20 in. water column. Find the mass flow rate assuming compressible adiabatic flow. Repeat calculation for incompressible flow. Convert ∆P to lbf/in2 ∆P = 20 in H2O = 20 × 0.03613 = 0.7226 lbf/in2 T0 = 68 F = 527.7 R

P0 = 14.7 lbf/in2

Assume inlet Vi = 0

Pe = P0 - ∆P = 14.7 - 0.7226 = 13.977 lbf/in2

Pe k-1 13.977 0.2857 Te = T0 ( ) k = 527.7 ×( ) = 520.15 14.7 P0 V2e /2 = hi - he = Cp (Ti - Te) = 0.24 × (527.7 - 520.15) = 1.812 Btu/lbm Ve 2ac/2 = η V2e /2 = 0.95 × 1.812 = 1.7214 Btu/lbm ⇒ Ve ac =

2 × 32.174 × 1.7214 × 778 = 293.6 ft/s

Te ac = Ti -

Ve 2ac/2

ρe ac =

Pe RTe ac

Cp =

= 527.7 -

1.7214 = 520.53 0.24

13.977 × 144 = 0.07249 lbm/ft3 53.34 × 520.53

π 1 . m = ρAV = 0.07249 × × ( )2 × 293.6 = 0.116 lbm/s 4 12 P0

14.7 × 144 = 0.0752 lbm/ft3 53.34 × 527.7

Incompressible:

ρi =

V2e /2 = vi (Pi - Pe) =

∆P 0.7226 × 144 = = 1.7785 Btu/lbm ρi 0.0752 × 778

RT0

=

Ve 2ac/2 = η V2e /2 = 0.95 × 1.7785 = 1.6896 Btu/lbm ⇒ Ve ac =

2 × 32.174 × 1.6896 × 778 = 290.84 ft/s

π . 1 m = ρAV = 0.0752 × × ( )2 × 290.84 = 0.119 lbm/s 4 12