Transport Phenomena

= 0), thus:

If the water flow is doubled, the pressure drop which is proportional to the momentum (dp .- ¢ .v ,..., v 2 ) is increased by a factor 4. Thus 4>,. ("'
vdPL dt

=0

and

d(pv) . V dt = 0 respecttvely

31

whereas the energy balance becomes: V

d(pLcpT) dt

= tPn

1 -

tPn

2

= (Xl(~- T)A 1

-

(XiT- TaJA:z

During boiling the balances become: mass: Vd:L

=

momentum:

Vdt =

-p,vA3 = -
0

=IF

where the force in the y-direction is given by FY = Mg the x-direction by Fx = 4>mvx; and:

+ mvy and the force in

(where, of course, H 1 - 4>H 2 - 4>mllHv = 0) if we neglect all other forms of energy because they·are so small. Problem 7

A macromomentum balance over the entire droplet will provide the answer we are looking for. Because the flow velocities are very low we can neglect the momen tum of the fluid and our momentum balance becomes a force balance:

the forces acting are the weight fo rce of the droplet and the surface tension of the liquid in contact with the capillary as shown in the drawing. Thus the force balance becomes: ~ow,

'L F

n 3 pg = 0 = -d

6

2nra

32

t I

and thus : d=

F=21TfCT

ru;:;

\}pg =

4·4mm

Problem 11

This problem is easily solved by application of the law of conservation of matter. The complication here is that the composition (Y) of the vapour phase changes as the distillation proceeds, because of the changing composition (X) of the residue. Apparently, the right thing to do therefore is to set up an unsteadystate material balance for a short time interval dt and to integrate the result found over the total process time te . The mass balance (equation 1.1) for this case becomes (no flow into the system, no production of methanol but variable contents of the evaporator): d(NX) dN dt = -4>molY = dt Y

where the symbols have the meaning given in the figure. After partial differentia-

tion we can write:

d(NX) = NdX dt dt

fi

't'mo1 .L

+ XdN dt

dN =- = constant d/

Mole fraction CH30H in vapour= Y

N kmole mole fraction CH 3 0H =X ott =0, x0 , No

33 and combinin~ these two equations we obtain: dX dt

dN dt

dN dt

N - = - Y- - X = -

mol

(Y- X)

or: dX

lPmol

Y- X.

dt

N

which yields after integration between t

dN

= -

= 0, x 0

N

and t, x:

N= No exp ( s:, y d: X) This equation gives the relation between the total number of moles in the evaporator and the methanol content of this residue. We are, however, interested in the average composition of the distillate < Y) which is given by : ( Y)

= f t.0 ¢ mol Y dt

=

-ft.. d(NX) 0

No - N

¢mol te

=

-No N 0 -N

{xex

P

=

- (NX)I'•·x.. O.Xo

No - N

(Jx

dX )} ~- x~ = x 0 Y-X . Xo

X

0 -

Xeexp(J~~dX/Y-

1-exp(J~~ dX/Y-X)

X)

This is the desired result ; all that remains is for us to evaluate graphically the integral, as shown in the figure. We find for the area under the curve :

f

x~ dX

Xa

= 1·17

Y- X

xand thus :

( Y) =

0·45 - 0·05 exp (- 1·17) 0·435 = - - = 63·1 per cent 1 - ex p (- 1·17) 0·69

34

The amount of distillate is found as: N0

-

N = N 0 {1 - exp

(J:., x)} yd:

= 69 kmol

We see that sometimes the solution of a simple mass balance can be quite involved. There is, however, a much simpler way to solve this problem approximately by carrying out the distillation stepwise. We start with 100 kmol of a mixture containing a mole fraction of X 0 = 0-45 methanol. Let us distil off so much distillate that X drops t~ X 1 = 0-35. Apparently,in this range the average mole fraction of methanol in the vapour phase will be Y0 = 0·73 and thus we can find the total number of moles distilled off from a simple mass balance as follows: xl

= NoXo- (No- N)Yo = 0-35 = 100 N 0 - (N 0 - N) ·

X

0·45- 0·73(No- N) 100 - (N 0 - N)

We find N 0 - N = 26·3 kmol, containing 19·2 kmol of methanol. Now we repeat this exercise as stated in the following table until we reach the desired methanol fraction of X = 0·05 in the liquid phase. Total number Mole fraction of Mole fraction of Number ofkmol Numberofkmol of kmol in pot methanol in pot methanol in vapour distilled off of methanol N X Y N0 - N distilled off 100 73-7 56-1 43·1 31·5

0·45- 0·35 0·35- 0-25 0·25 - 0-15 0·15- 0-05

0·73 0·67 0-58 0·42

Total

26·3 17-6 13·0 11·6

19·2 11·8

68·5

43·5

7·6 4-9

Thus we find that 68·5 kmol were distilled off, containing 43·5 kmol of methanol, i.e. 63·5 per cent. There is a residue of 31·5 kmol, containing 45 - 43·5 = 1·5 kmol of methanol, which is near enough to the desired 5 mol per cent. The reader will realize that the principal considerations in this approximate solution are the same as for the exact solution given before. Instead of infinitesimal steps dX or dt which made integration necessary, we just used bigger steps which enabled us to obtain an approximate analytical solution a fter a reasonable amount of additions and multiplications. Problem 12

J ohn and the burnt-down factory

The flow rate of air through the plant ¢v can be found by means of an energy balance (equation !.6). Under steady-state conditions dEtfdt = 0 and realizing that the change in heat content of the air pc P L\ Tis much bigger than the changes

35 in the other forms of energy (equation 1.5), we can write :

dEt V dt

= 4>vPCp ll.T +¢A- ¢H = 0

Now, no mechanjcaJ energy is added, ¢A = 0 and the amount of heat supplied equals ,. is the rate of steam loss. Thus : _

A. t.y

"

-

¢,. /l.H ., _ 70 X 10 3 • 2·3 X 106 pcPAT 24. 3600.1·2 x 10 3 . 30

_ -

/

3 52 ms ·

If we assume that the air in the building is well mixed, the hexane concentration in the building will equal the concentration in the exit stream. A material balance then yields with cin = 0 for the steady state:

de

V- = - ¢ c dt v

+r=

0

where r equals the rate of hexane loss. T hus we find: r

9000

c= - = = 0·002 kgjm 3 ¢v 24 . 3600 . 52 and, regarding the hexan e vapour as an ideal gas, we find :

86 22·4

p = - = 3-83 kg/m 3

and thus c = 0·052 per cent by volume, which is indeed well below the explosion limit.

CHAPTER II

Flow Phenomena In this chapter the transport of momentum is studied and we·will discuss the prediction of flow resistance in pipe systems and of the resistance due to obstacles placed in a flow field. The analysis of pipe flow is divided into two sections, laminar and turbulent flow, the difference being that in the first case shear stresses in the fluid are predictable from velocity gradients (deformation rates) but in ·the latter they are not. This means that the 'theory' of turbulent pipe flow relies strongly on empirical knowledge; however, this knowledge can be made generally applicable by scientific reasoning. In stationary flow through pipes of a uniform cross-section the momentum balance degenerates to a balance between shear forces and pressure forces, but this is no longer the case in flow systems in which the cross-section available for flow changes. H ence, in that case a more general and practical approach will be n eeded. A class of flow problems, in which the enery dissipation due to shearing is small in comparison to the amount of mechanical energy which is transformed, e.g. from kinetic energy into potential energy, will be dealt with separately. This will lead us to the design and use of flow meters. F low around obstacles will not only be extended to flow through beds of particles but also to stirrers. This brings us to the problem of mixing and of the non-uniform distribution of residence time of the fluid elements which pass through a continuous flow system. In all hydrodynamic theory we will make use of the three fundamental conservation laws which were introduced in chapter I. The idea of conservation of mass, momentum and energy provides us with five relationships for a threedimensional flow field, from which the velocity distribution (v,.., v,. and vz) and the temperature and pressure distributions can be obtained in principle, as soon as : these quantities are defined at the boundaries of the system and relationships between shear stresses and deformation rates are known. The whole problem in using the concepts of conservation of mass, energy and momentum in predicting flow fields and flow resistances is to define the control ,·olume in such a way that it facilitates the analysis and to make clever guesses about the order of magnitude of the several terms in the balances in order to

37

make them amendable for an algebraic treatment. Hence, although the whole theory on transport phenomena does not go beyond the concepts which are already introduced in chapter I, the professional engineer needs considerable experience before he is able to tum these concepts into practical use. Therefore~ this chapter has been written. It treats a number of problems in an elementary way in order to illustrate the practical application of the basic principles and it provides a number of elementary solutions which can be the starting point to the solution of more complex problems.

ll.I. Laminar Oow John looked at the spoon in his hand. Syrup dripped from it into his porridge at a decreasing rate. H e thought about writing a criminal story in which someone was imprisoned until the very last drop of syrup would leave the spoon. John had a feeling that this would last very long, months, perhaps,. or years. Being a scientific story writer and working efficient ly, he read the following introduction and found out that the thickness of the syrup layer on his spoon decreased about 100 times in 25 minutes. Consequently, he thought, this layer would diminish to molecular dimensions in about 3 x 105 years-a long sentence indeed.

Dtlring laminar flow, fluid elements move along parallel streamlines. Consider, for example, the flow of a liquid film (e.g. syrup) on a vertical plane surface. Here the weight of the fluid constitutes the driving force, resulting (if the viscosity of the fluid is not too low) in laminar flow. At the wall, flow velocity will be zero and there will be a velocity gradient over the thickness of the film. Thus, adjacent fluid elements will have different flow velocities, causing shear forces between the fluid elements. A great number of laminar flow problems can be solved with the aid of the balances discussed in the foregoing chapter if, for the fluids used, the relation between shear forces and the velocity gradient is known. This relation reads for unidirectional flo.w of Newtonian fluids : r

yx

d(pvx) = - vdy --

(II.la)

and if the fluids are non-compressible: dvx

"[ y x = -11dy

(Il.Ib)

is the shear force in the x-direction in a plane xz normal to they-axis. This relationship is shown, together with two other possible relations between ryx and the velocity gradient, in Figure 11. 1. To solve a flow pro blem we start with a momentum balance which, for channels with constant cross-section, takes the form of a force balance, from which we find a relation between the shear force at any place and the pressure drop. No assumption about the type of liquid or flow has been made until this ryx

38

-dvx l dy

-dv1 1dy

Newton

Binghom

Ostwald

Tyx

Figure D.l

: - K

dv )"-I .::..:.!. dv ( _dy£ dy

Relation between shear force and velocity gradient

point. This relation is always valid for any kind of flow and for any type of rheology. Using the shear force-velocity gradient relation for the fluid concerned (e.g. equation ILl a for Newtonian liquids) we obtain after integration an expression for the velocity distribution in laminar flow. From this velocity profile the mean velocity and the volumetric flow rate can be determined as a function of the driving force. We will illustrate this procedure in the following paragraphs. II.l .l . Stationary laminar flow between two fiat horizontal plates

Consider a liquid flowing between two parallel plates, as shown in Figure II.2. As the control volume we choose the volume between the plane y = 0 (centre of flow field) and the plane y = constant between the two cross-sections '1' and '2'. Under steady-state conditions, the momentum balance over this control volume is given by equation (I. 7) as: d(pvx)

V

dt

=0 =

tPv,inPV::c.in -

1Jv.outPVx.out

+ L Fx

Now, since the cross-section available for flow is constant and the liquid is incompressible, the amount of momentum transported by the liquid into and out of the control volume is the same :

----

i'l' Figure 11.2 Flow between two flat plates

39

and we find that the momentum balance reduces to a force balance I Fx = 0. This is the case in all situations where compressibility of the liquid can be neglected and where the flow channel has a constant cross-section. The forces in the x-direction per unit of width acting on the control volume are the following if, at the two cross-sections, the pressures are p 1 and p 2 respectively: pressure force at ' 1':

F

'dt h =

Wl

F

at '2'. x · width

-

p1 y

(positive, because it acts in the positive x-direction)

- .PzY

(acts in the negative x-direction)

shear force

(negative, because it acts in the negative x-direction)

We can now set up our force balance and we find :

~ Fx Width

=0=

PtY - PzY -

T1

xL

or (II.2) Thus we see that the shear force is maximal at the wall (y = + d/2) and zero in the centre (y = 0). Applying equation II.lb we obtain for a Newtonian noncompressible fluid flowing in a two-dimensional slit of constant width : dvx = - Pt dy

- Pz -y 11L

Integrating with the boundary conditions vx = 0 at y = zero at the wall) w·e ~nd the velocity distribution : v, =

+ d/2 (flow velocity is

p,;;t(~- y2)

(II.3)

We notice that the velocity profile is parabolic, the maximum velocity being (at y = 0): .

_ Pt - P2d2 8rfL

~max -

The flow rate per unit width is found as: lf>v = width

f

+ df 2

-d;z

v dy = Pl - P2 d3 x 1211L

= d( v)

(II.4)

------------------------------------------------ ----40

and we see that the maximum velocity is given by timax

= f(v)

In Figure Il.3 the shear force and velocity distribution found for laminar flow in a narrow slit are shown. The reader is asked to work out on his own the

X

I

I I

Pt-P2

vmox

= 3/2

L

Figure 0.3 Shear force and velocity distribution for laminar flow in a narrow slit

solution to a similar problem of a film falling on a vertical plane surface {problem 2 on page 52). We could also have solved the flow problem just discussed by means of the general micromomentum balance developed in paragraph 1.3. Using equation (!.25) and realizing that there is no flow velocity in the y- and z-directions (v1 = 0, vz = 0), no velocity gradient in the x- and z-directions (dvxf dx = 0, dvjdz = 0), no shear force in the x- or z-directions (d-rxxfdx = 0, dtzjdz = 0) and that the plates are horizontal (thus pgx = 0), we find for steady-state conditions (dv:.:f dt = 0):

or, after integration : r,x

= - ~~y + constant

which is identical with equation (11.2) if we realize that there is no shear force at the centre of the channel (at y = 0) and hence that the integration constant must be zero. II.l.2. Stationary laminar flow through a horizontal circular tube

To solve this problem we can use the same approach as in paragraph II.l.l. As a control volume we choose a cylindrical plug of length L and radius r as shown in Figure 11.4. The forces acting on the control volume are again pressure forces acting on the cross-sections and a shear force acting at the cylindrical surface in the negative x-direction.

41

L

·{}+. ' I

I

I

-

1

P2

xz -x.

P•

Figure 11.4

X

Flow through a circular tube

W'riting down the force balance for steady-state conditions we obtain : Fx = 0

I

= nr2 p 1

-

nr 2 p 2

r~.x2nr(x 2

-

-

x 1)

or:

= !:_ P1

r

= ~( _ dp)

- P2

2 x2 - xl

rx

2

(II.S)

dx

This relation between shear stress and pressure drop is generally valid (there has been no assumption about the type of flow), as was the case with equation (II.2). The shear stress distribution is linear : r,.x = 0 at the centre (where thus the velocity gradient is also zero) and maximum at the wall. The velocity distribution fo r laminar flow of a Newtonian liquid is fou nd by eliminating tr.x in equation (11.5) by equation (II.lb) and by integrating the differential equation obtained with vx = 0 at the wall (r = R ). The result fo und IS :

dp) (R 2· = -1(~-

vx

4~

r2)

dx

(11.6)

Thus, again, the velocity distribution is parabolic, with maximum flow velocity at r = 0: Vmax

dp) .

1 (

= 4'7 - dx R

2

The flow rate at a given' pressure gradient is found from th e velocity distribution as:

f

4

2nrv x dr = -nR o 8rr R

(

dp) -dx - = nR 2 (v)

(II. 7)

Combining equations (Il.6) and (11.7) we find for the velocity distribution :

(

r2)

v = 2(v) 1 - x

R2

= -R

2 (

4'1

dp) ( r2) 1 -dx R

--

and we see that in the case of flow through a circular tube : Vmax

= 2(v)

2

(11.8)

42 Equation (11.7) is the Hagen-Poiseuille relation which (with some corrections) is used by various methods for the measurement of viscosities of Newtonian liquids. In Figure II.5 the shear stress and the velocity distribution found are illustrated.

I I I

I I I

r (PJ-P2l

Figure 11.5

2 · -L-

vmax

= 2

Shear fo~ce and velocity distribution for laminar flow

in a circular pipe

The velocity gradient at the wall can be found by differentiating equation (11.8) and by determining the value of the first derivative for r = R. We obtain:

~; ,~. = - 4~) =

2: (- ~~)

= -

~

· (II.9)

where rw is the shear stress at the wall. We will need this result later on, when we develop equations to predict pressure drop during flow of liquids in pipes. The reader may check that equation (II.5)" could also have been obtained from the appropriate microbalance, equation (1.30), and that equation (11.6) can be derived from equation (1.37):

I 1.1.3. Stationary laminar flow through a horizontal annulus In many cases it is not possible to define directly a macroscopic momentum balance as used in the foregoing two examples. In such cases a microscopic momentum balance must be developed. Consider the annulus shown in Figure II.6. As a control volume we choose a streamline with radius r, thickness dr and length L. We can then set up the following force balanc.e over the control volume:

I

Fr

= 0 = 2nr dr p~

+ rrx2nrL!, -

- 2nr drp 2

r,.x2nrLI.+ dr

Dividing by 2nL dr we obtain: Pt - P2 r= :....:..-...:....;:;; L

rr,xlr

rr,xlr +dr -

. dr

The right-hand side of this equation is the mathematical definition of the first derivative of n rx, so we can write: d(rr,.x)

--'---'-=

dr

=

P1

-

L

P2

r

43

I

---.1

-- rr

b

!

r

I

I I

I I

I I

I

:

1

·+--- ------ -:--- -- ·

- - - L - -I

1

...-x

1

P1

P2

Figure 11.6 Laminar flow in an annulus and integrating this equation we obtain : r rx

=Pt-P2r + C1 2L r

(11.10)

Applying Newton's law (eq1:1ation Il.lb) and integrating again we obtain:

(11.11) In this equation the constants C 1 and C 2 still have to be evaluated. We can do so by applying the boundary conditions vx = 0 at r = a and at r = b. We then find for the shear stress distribution (from equB:tio!lll.lO) :

r

· Pt - Pz

u

=

2L

(-ba) z

/ 1/ -

2

(II.12)

!

r

b

2ln -

and for the velocit~ dis_tribution : , =

vx

Pt - Pz bl 1 _

4~L

(~)

2

+

b

1

(WIn (~) In(~) -

(IL 13)

b

whereas the fl ow rate is given by:

I - (~)• -[I - (~) ']' b

lnm

(11.14)

44

In Figure Il.7 the shear stress and the velocity distribution for laminar flow in an annulus are shown. The reader may check that, indeed: r,.x = 0

at

and: vx

= vmax at the same value of r2 jb2

For a = 0, equations (II.l2), (II.13) and (II.14) reduce to the corresponding relations for flow through a circular tube. For a very narrow annulus, thus for a = b - d (d « a) and r = a + y (y « a), the corre-sponding relations for flow in a narrow slit between two flat plates are obtained (problem 1).

+ Trx

-

vx

Figure ll.7 Shear stress and velocity distribution for laminar flow in a horizontal annulus

The shear stress and the velocity distribution for flow in a circular pipe can, of course, also be found fro m equation (11.11) by applying the proper boundary conditions Yx = 0 at r = Rand vx ¥= oo at r = 0. The last boundary condition leads to C 1 = 0, which we would have also found from equation (II.lO) by realizing that at r = 0, r,.x ¥= oo. The reader may check that equation (11.10) can also be derived from the proper micromomentum balance, equation (1.30). Coming back to laminar flow in an annulus it is interesting to compare the shear force at the inner and outer wall. From equation (II.l2) we find the force at the wall as : at inner wall : Fa =

-~aL-rwla =

I

-na(p 1

-

P2)

45 The sum of the two forces is then (ra.x acts in the negative x-direction !) :

Fa

+ Fb = -2naLrwla + 2nbLrwlb = n(b2

-

a2 )(pt - p 2 )

i.e. the total shear force at both walls equals the pressure force on the crosssection of the annulus, as should be the case. The ratio of the forces at the inner and outer wall is:

b 2 /a2

-

1

- - -- -1 2ln (bja)

- arax

b2/a2 - 1 21n (bi a)

-------~~~--

2

b ja

2 -

This relation is shown graphically in Figure II.8. We see that for the limiting case bja ...,. 1 the two forces become equal. The ratio of the force on the inner cylinder to the total force is given by: Fa

-arax -arax + btbx

1

----=--- - - - --'---- - - - - -

Fa + Fb

2ln (bja)

1 (bj a) 2

-

1

This relation is also shown in Figure II.8. If the ratio of the surfaces of the outer to the inner cylinder is 5:1, then 27 per cent of the total force still acts on the smaller cylinder ! Considerations of this kind are of importance during the extrusion of, fo r example, plastic pipes.

bl a-

Figure 11.8

Forces acting on the walls of an annulus

11.1.4. Flow caused by moving surfaces

The foregoing sections have dealt with the fl ow of liquids caused by a pressure gradient over a length of pipe. Flow can also be caused by moving a surface through a liquid.

46 Let us consider what would happen if, in an infinitely long horizontal annulus filled with a Newtonian liquid, the inner tubes were moving in the x-direction . (Figure II.6) at a constant velocity v0 • Since there is no pressure gradient induced from the outside, the force balance under stationary conditions becomes for this case:

I

Fx = 0 = r,_x1nrLI,. - r:,.~nrLI,.+dr

from which we find : d(rr,.x) =

dr

0

Integration, application of Newton's law (equation II.16), second integration

and determination of the two constants of integration via the boundary conditions (r = b, vx = 0 and r = a, vx = v0 ) leads to : r

YfVo

rx

=-

1

(11.15)

r In (b/a)

and : vx

In (bj r) = Vo ln (b/a)

(11.16)

The first relation shows that the forces on both surfaces are equal, as should be the case: 1tYfVo

Fa = nar:,.aL = Fb = nbr:,~ = In (b/a) L

In Figure 11.9 both relations are shown graphically. The reader may check that for bj a-+- 1 both expressions reduce to the proper relations for two parallel flat plates (see also problem 4).

Figure ll.9 Horizontal annulus with moving inner

cylinder A different situation occurs if the inner cylinder is not moving in the xdirection but is rotating at a rotational speed n (s - 1 ). This case can easily be treated by applying cylindrical coordinates (Figure II.lO).

47

Figure ll.lO Annulus with inner

moving cylinder

A momentum balance leads to : F = 0 = 2nrLn:,81,

-

2nrLn,elr+dr

from which follows: 2

d(r r..e) = 0 dr Integration of this equation yields, after application of Newton's law, which read s in cylindrical coordinates for non-compressible fluids:

and second integration:

v8

cl

=--

2w

c2

-r 11

and, applying the boundary conditions v 8 = v 0 at r = a, v 8 find (using v 0 = na) : .-

=0

at r = b, we

(II.17) and: 2na 2b2ry r,e = (bl - a2)r2

(II.18)

As should be the case, the torque at the inner and at the outer wall Fr = 2nr-r,rL is the same. The shear stress and the velocity distribution are shown in Figure II. ll.

48

o~----------------------~ a r b

Figure 11.11 Shear stress and velocity distribution in an annulus with inner moving cylinder .

The above situation is present, for example, in friction bearings. A similar situation occurs in a rotary viscosimeter. Here the inner cylinder is immobile and the outer cylinder rotates. The annulus is filled with liquid and the force on the inner cylinder is measured (e.g. by means of the torque of a wire). The reader may check that the viscosity of the liquid in the annulus is given by: F b2 - a 2 '1 = - -t -...,..--,-4rtnL a 2 b2

if Fr is the torque, L the height of the cylinder and if effects caused by the bottom of the annulus are neglected. The reader is asked (problem 4) to solve by himself the analogous simple problem of two parallel moving plates. Il.l .5. Flow through pipes with other cross-sections The calculation of velocity profiles for pipes with non-circular cross-sections is difficult but possible. Boussinesq developed analytical solutions for the velocity profiles for Newtonian liquids in pipes with elliptic, triangular and square cross-section. These results can be summarized by defining a flow coefficient M 0 for which the volumetric flow rate of flow through a slit with the same smallest dimension has to be corrected. For flow through two parallel plates we have developed equation (1I.4), which reads, if B represents the width of the slit : tip 3 ¢ v == 121JL Bd Thus, for other cross-sections the flow rate is given by :

¢v

=

llp 3 12'1L Bd M o

Values for the correction factor M 0 can be found from Figure II.12.

(II.19)

49

Rectangle

0 ·1

•!2:{ 8-----1

Isosceles triangle

d/8

Figure 11.12 The flow coefficient lvf 0 for stationary flow of Newtonian liquids in cross-sectjons of various geometries

II.J.6. Non-stationary flow

Until now we have concentrated on stationary (steady-state) situations, i.e. on situations which were independent of the elapsed time. ln this section we will consider what happens in the time between starting a flow and reaching the stationary state (transient behaviour) and how long it takes to obtain a steady-state situation. Consider the situation drawn in Figure II.13. From t = 0 a big fiat horizontal plate is moved with a constant velocity v0 . At t = 0 the liquid is stagnant (va:: = 0). At t > 0 momentum transport in the + y-direction takes place, resulting in a velocity distribution which is not only a function of the x-coordinate but also of time.

50 <

/aQ

<

Figure D.t3 Horizontal flat plate in infinite viscous liquid

To analyse this problem we select a control volume parallel to the plate and of height dy. The micromomentum balance over this' height dy then reads:

d(pvJ d dt

y =

I

Tyxy- r,xly+dy

=-

dr1 "' dy dy

and, after applying·· Newton's law (equation 11.16) we obtain :

dvX d 2 vX - = v- dt dy 2 The same equation could have been obtained from the general micromomentum balance, equation (130). This differential equation is analogous to the equations describing the penetration of heat and mass: dT d2 T dt = a dx 2

and

de dt =

d2c

[]l dx 2

. respectively

which will be treated in more detail in chapters III and IV. There, also, the exact solution of this differential equation will be discussed. In Figure II.14 the velocity distribution for the system of Figure 11.13 is shown· for two short times. As can be seen, the distribution can be approximated by straight lines which cut the abscissa at y = foi. This distance foi is called the penetration depth, i.e. the distance over which momentum has penetrated. This solu tion is applicable as long as the penetration depth is smaller than the depth of the liquid. We can now estimate the velocity gradient as:

or, since

V 00

= 0:

d v"' Vo17 = -1]-=--

r w

dy

Fvi

(II.20)

51

;-;;-;-;

j;;t(

X

Figure 11.14 Velocity distribution for system of Figure II.13

A somewhat different situation occurs if, instead of a moving plate, liquid with an initial velocity v0 flows along a fixed plate. This situation will be dis-

cussed in paragraph II.5. 1 (see Figure II.37). We can now estimate the distance needed for laminar pipe flow to develop its velocity distribution completely (Figure 11.15). If we move with the liquid and assume that the velocity distribution will be developed completely when the penetration depth at x = xs is equal to the distance R to the centre of the channel, we can write: R

~ foj = V~v (v) Xs 1l

Thus, after a distance of approximately: x ~ (v) Rz ~ 0·1 cf>v s

rrv

v

the velocity profile is fully developed.

Figure II. 15

Development of velocity profiles in tubes

(11.21)

52 JI .l.7. Problems 1. Show that equations (II.12) and (U.13) reduce to equations (II.5) and (11.6) if af b goes to zero. Show that these equations reduce to the corresponding expressions fo r flow in a slit if aj b goes to 1.

2. (a) Develop equations for the shear stress distribution, the velocity profile and the volumetric flow rate for a liquid film flowing down on a vertical surface. Use the notation given in Figure II.16 (width of plate = W).

Figure 11.16 Laminar film on a vertical plate

g

Answer :

'txy

v

y

=

- pg(d- x)

= gp (dx 11

-

1x2 )

cPv = (v ) d = pgd3

w

)'

311

(b) On a vertical plane {width B = 1m) 1200 kg/ h water flows downwards. What is the mean velocity, the film thickness and the maximum velocity? Answer :


0·725 m/ s d = 0·46 X 10- 3 m Vm a)l = 1·09 m/s

*3. In a rotary viscosimeter filled with paint (a Bingham liquid) the shear velocities are measured as a function of shear foree : Shear force (N/m 2 ) 15 18

21

Shear velocity (s- 1 ) 6 12 18

53 If this paint is smeared on a vertical wall, how thick can the layer be before the paint starts tc drip (p = 1200 kgfm 3 )?

Answer : d < 10- 3 m

4. The slit (width 1 m m) between two horizontal flat plates is filled with a viscous Newtonian liquid ('1 = 100 Nsjm 2 ). If the surface of the plates is 10 - 2 m 2 , which force has to be applied under steady-state conditions to one plate in order to move it parallel to the other p late at a velocity of 1 crnfs? Calculate and draw the shear force and the velocity distribution between the plates. After what time is the stationary state reached? Answer :

F = 10 N,t~3x10 - 6 s

5. The 'viscosity' of a non-Newtonian liquid is approxima tely given by :

dvx' dr

YJ = YJo

where '7o is a constant. Find the volumetric flow rate of this liquid in a horizontal pipe (diameter 2R) under laminar and isothermal conditions.

'

Answer :

v =

2n ,

{!;P

7 Ri ...j~

6. Viscous oil ('7 = 0·1 Nsjm 2 ) is pumped through a horizontal pipe (diameter 1 em, length 10m) into an open vessel. The absolute pressure just behind the pump is 3 x 10 5 N j m 2 . {a) Determine the volumetric flow rate of t he liquid. (b) What would be the flow rate if a pipe with rectangular cross-section (0·5 x 2·0 em) was used? Answer : (a)

4>v = 4-9

x 10- 5 m 3 js

(b) ¢ ., = 3-47 x lo-s m3 /s

*7. Check John 's calculations about the syrup dripping from the spoon, reported at the beginning of this paragraph. Assume:

YJ = 1 000~,

spoon 5 em

p

= 2000 kgjm 3 , initial thickness 5 mm, height of

Comments on problems Problem 3

For a liquid with Bingham characteristics the following relation between shear force and velocity gradient is valid if we use the coordinates shown in the small drawing overleaf:

54 B

The size of the yield value t 0 for this paint can be determined from the experimental results as t 0 = 12 N j m 2 • A force balance will lead to the maximum possible layer thickness before dripping takes place :

pghdB < -r 0 hB and thus: to

d
Problem 7 John and his spoon We are dealing with a situation where a liquid film is flowing down from a vertical surface. An expression for t:f>v was found in problem 2 of this section:

cP = pgd3W = WHdd v

Rearrangement leads to: d=d

L=do

dt

3~

dd pg sr =r ---dl d3 r=O

3~H

and after integration:

1[ 1 1] pgt d~ - d 2 = J17H

-2 or:

d -=

55

Spoon ( face on)

Spoon ( in profi l e) \

-w- l

\ \

':-do I

J

H

I I

J For long times : 2pgtd~ _:_:::__..::.. » 1

3t7H

and we can simplify the above relation to:

d/3riH

do="'>/~ For djd 0 = 0·01 we find t = 1500 s = 25 minutes. The period to diminish the syrup layer to molecular dimensions is found by substitution of d 5 x 10- 10 d = 5 x 10- 3 0

=

10 - 7

as 15 x 1012 s.

11.2. Turbulent flow John stood at the edge of a circular sewer channel of 4 m diameter and contemplated that the dead body had been found floating in the channel 7·2 km downstream, in a fisher village, at 11.45 p.m. The suspect had a watertight alibi for the time before 9.15 p.m. The channel was now half-filled and a piece of wood passed John at a speed of 1 mjs. One of the sewage plant operators had informed him that until the time che body was found the water height had been less than onequarter of the diameter of the channel. Th inking about one of his college courses, of which you will .find a brief outline below, he made a quick calculation and ordered to sec the suspected man free.

11.2.1. Turbulent flow in pipes In a well-known experiment in which he injected a filament of dye into a fluid, Reynolds demonstrated that above a certain flow velocity in a pipe the flow changes from laminar into turbulent. In this case velocity fluctuations occur with components perpendicular to the main direction of the fiow.

56

The turbulence of the flow field makes the effective viscosity (from a macroscopic viewpoint) larger than the molecular viscosity 17, because neighbouring layers of different mean velocity not only transmit momentum to each other by molecule interaction but also by cross-currents (eddies). These cross-currents decrease in intensity the nearer they are to the wall. Because of this varying effective viscosity (which can be supposed to be composed of the t7 and the 'eddy' viscosity pE) the velocity distribution differs from that in laminar flow. The eddy viscosity just introduced is, of course, not a physical property of the liquid but is dependent on the place and on the type of flow. The dimension of the eddy viscosity pE is kg/ms ; thus the term E has the same dimension as the kinematic viscosity v (m2 js). Figure Il.17 shows an example of the velocity profile in a pipe with a circular

cross-section for laminar (parabolic) and turbulent flow (more flattened). r Vmox"' 2
>

I

I

Vmox

I

•

i
= I· I to I · 2 < v > 1 •

>

I

I

I

I

I

I I

I I

X

---r--I I

I

I ----r--··

I I

I

I

I

I

I I

Figure 11.17

Laminar and turbulent velocity profiles in a pipe

If we want to know something about the frictional force exerted on the wall by a liquid flowing at a velocity (vx), the shear force -rw at the wall must be known. Since for incompressible Newtonian liquids : !w = r[

dv.x dr r=R

holds (no turbulence at the wall, pEat wall = 0), we must know the velocity distribution in order to calculate -rw . For turbulent flow this is not possible by momentum balances, because with turbulence their solution presents insurmountable difficulties. However, on the basis of physical intuition it can be posed that : -rw = f (p, rJ, (vx), D, geometry)

from which, with dimensional analysis, it follows that the quotient of -rw and p(vx) 2 for geometrically similar pipes only depends on the Reynolds number (= p( vx ) D/tt). This is usually written as: (1 1.22)

57

With this equation the constant f, the so-called friction factor is defined: on account of the above it is a function of the geometry and of Re. For turbulent flow this function can be obtained only from experiments on friction losses in pipes and channels. Conversely, these friction losses are, in practice, calculated using this friction factor f, the value of which is well known for widely divergent conditions (see paragraph Il.4.). · 2 Equation (II.22) can be read as follows: the quotient of -rw and p Recr ~ 2000. This also applies to straight, non-circular pipes if, in the Re number, as the characteristic length the hydraulic diameter is introduced, which is defined as four times the surface area of the cross-section divided by the wet circumference {see, further, the next section and check that for a circular pipe the hydraulic diameter equals D). TheRe number can be interpreted as the ratio of the momentum transport by convection (,.., pv2 ) and the momentum transport by diffusion (,.., qvf D). Apparently, in turbulent flow the transport by convection, i.e. by eddies, prevails. In a free-falling liquid film these eddies already occur at lower velocities; film flow becomes turbulent at aRe number ~ 1000. During flow through a curved pipe (e.g. a spiral; radius of spiral R) the centripetal forces stabilize the flow and the change to turbulence does not occur until higher Renumbers have been attained. Empirically it was found for the critical Renumber :

Re" =

2o,ooo(~r , 32

·

ror

~ ~ to'

(11.23)

The fact that turbulence decreases towards the wall has led to the concept of the laminar boundary layer. No definite thickness can be assigned to such a layer because the turbulence changes gradually. For the qualitative insight :t can be useful to scliematize the flow field by a laminar boundary layer, .with thickness ~h in which the velocity gradient is constant, and an adjacent turbulent ~ow region, with the mean velocity in which the velocity gradient is ;>ractically zero. The shear force at the wall is then given by: ( vx) rw='f'/T

(II.24)

and the thickness (J h can be estimated with the help of equation (11.22) as soon as the friction factor is known. With the aid of the laminar boundary layer thickness b1, we can give a new mterpretation on the physical meaning of theRe number. The ratio of the tube

58 diameter and thickness of the laminar boundary layer is found to be : D c5h

=

Ihw 1'/(V),)

= f Dp(v;;,)

= f Re

211

(II.25)

2

This means that for pipe flow at Re = 10s (f = 0·0045) the distance from the wall at which turbulent momentum transport overrides the transport by viscosity is c511 ~ 4 x 10- 3 D. A velocity distribution as given above is physically impossible, but it can serve as a rough model for tre~ting heat and mass transfers in turbulent flow. In the case of heat transfer a layer of thickness c5T is assumed over which the complete temperature drop occurs. Outside this layer the turbulent eddies disperse the heat so efficiently that a uniform temperature exists in the core of the flow. Within the layer eddies are assumed to be absent. This layer of thickness {JT is called the thermal boundary layer analogous to the hydrodynamic boundary layer (thickness oh) treated above. It will be clear 'that both models do not contradict each other as long as c5T ~ c5 11 . The question now arises as to whether a relationship exists between the thicknesses of the hydrodynamic and the thermal boundary layers. Some insight into this problem can be gained from the fact that the eddy diffusivity E increases with the third power of the distance from the pipe wall : E = C1 y3

From the definition of viscous transfer if:

o

11 ,

(II.26)

this means that eddy momentum transfer overrides

pE = pC 1 y 3 ~ 11 . or

E

= C 1y3

~ 11/P

=v

Hence: {II.27)

The diffusivity of heat by eddies is also described by equation (ll.26), whereas the diffusivity of heat by conduction only is given by a= J..jpcP (m 2/s). Hence, the distance c5T from the pipe wall at which heat transfer by eddies overtakes conductive heat transfer is given by a reasoning similar to that a pplied above to be : (II.28)

H ence, the sought relationship between (ll.27) and (11.28) and is found to be:

~.!!. = (~) t Or

a

or and oh is obtained from equations =

Prt

(II.29)

where Pr is named the Prandtl number. For gases Pr ~ 1 and therefore c51, ~ oT, for viscous liquids P r ~ 1000 and o11 ~ lOoT, and for water at room temperature Pr ~ 7 and c511 ~ 2c5r.

59 The reader can imagine that mass transfer between a wall and turbulent pipe How can be depicted by using a concentration boundary layer of thickness beThe ratio between this thickness and the thickness of the hyd rodynamic boundary layer will then be (see also paragraph IV.3.3):

b: = (v)t = set

{>

II)

(11.30)

where Sc is the Schmidt number. For gases Sc ~ 1 and ~, ~ l>c and for liquids Sc ~ 1000 and b" ~ lObe. We see that in these cases the thickness of the hydrodynamic boundary layer is greater than those of the respective thermal and concentration boundary layers, b11 > bT, We have .already stated at the beginning of this chapter that it is impossible to derive theoretically a relation for the velocity distribution during turbulent Bow. We can, however, construct a relation between velocity and the friction factor and see whether we can determine the coefficients in this relation by experimental results. Analogously to the devdopment of equation (II.22), we assume that the mean velocity is a function of the shear stress at the wall, the specific gravity and the kinematic viscosity of the fluid and of the pipe diameter ; so:

oc.

Via dimensional analysis we find:

~=~{~ ~} =C{~~r

(II.31)

which can be rearranged to : rw = f = (C)-2/(l +n>{-v-}2n!<1+ n> p(v) 2 2 (v)D

(11.32)

Equation (11.31) indicates that in the turbulent region also the friction factor will be a function bf the Reynolds number only. Results of friction factor measurements in smooth pipes have been collected a nd correlated by Blasius as:

_ 4f = 0·316 Re - 0.25

This correlation is discussed in more detail in the next section. At the moment

we can use it to determine the constant n in equation (II.32). Via : 2n - - =0·25 1+ n we find n = l Repetition of the foregoing procedure for the flow velocity in the x -direction at any place r yields, if we realize that vx must be zero at 2r = D, instead of equation (II.32}:

(11.33)

60

Thus, maximum flow velocity occurs in the centre of the tube at r can write :

(l _2r)-+ D

_2_ = Vmax

=

0 and we (II.34)

In order to find the ratio between mean and maximum velocity we can calculate ( v) from equation (II.34) as :

= _1_ JD/2v

( v)

D/ 2

0

dr =

Vmax

D/2

X

JD/2(1 -

2r) -+ dr D

0

Solving this equation we find: (v)

-

7

=-

8

Vmax

= 0·875

-

compared with (v)/ vmax =!found for laminar flow in circular tubes. The velocity distribution (equation II.34) gives a rather good description of the actual situation, as illustrated in Figure 11.18, where measured values of (vfvmu? are plotted against (1 - 2r/D). ,CY 0 ,..t

I

c;/

0

~ 0\

'o

/

><

0

\

~0

E

"'

'"'

'o

v6 0

i

>

~

5l

' r;\

I

/

0 ·5

0 _ . 1-2r/D

0 ·5

~

Figure ll.18 Measured velocity distribution for turbulent flow

I 1.2.2. Pressure drop in straight channels

At the end of paragraph 11.1.2 we saw that the pressure drop in a pipe system can only be predicted as a function of the flow rate if the shear force at the wall is known. For laminar flow this quantity can be predicted directl y because the relation between -rxy and dvy/dx (e.g. according to Newton, equation 11.1) characteristic of the fluid offers sufficient additional information. For turbulent flow we lack this additional information. In order to be able to predict the flow rate for turbulent flow on the basis of a given pressure drop (or, inversely, the pressure drop for a given flow rate), use is made of graphs and tables of earlier measuring data in which numerical

61

values can be found of the dimensionless groups that are important to this problem. On the basis of dimensional considerations these results can then be used for solving every new but analogous problem. We shall deal in succession with the fiow through a straight channel, the fiow through fittings and the frictionless fiow. Since in what follows we only speak of a mean velocity and not of velocity distributions, the microbalances are of no use to us in this chapter. The quantities which we shall now come across belong to balances over an entire channel or channel system (i.e. they are macroscopic quantities to be used in macrobalances). If the shear stress on the wall of a horizontal straight channel (cross-section A, circumference S) is known, we find for the stationary state the momentum balance (see also paragraph 11.1.2 and Figure Il.4) :

L F.x =

0

= p1A

- p 2 A - twS(x 2

-

x 1)

and, using equation (II.22), we find for the pressure drop : Pt - P2 =

tw

S(x 2

-

A

x 1)

=

Jl < ) 2 S(x 2 2P vx

-

A

x 1)

(11.35)

From this equation, which is named after Fanning, it is clear that A/S is the characteristic dimension of the channel cross-section, so that it is obvious to call A/ S the hydraulic diameter* of the channel. If the numerical factor 4 is introduced in the above relation, which is done for historical reasons, fS/ A can be written as 4fS/4A and it is still general practice to state the value of 4f instead of the value off The problem of predicting the pressure drop in straight channels can thus be reduced to predicting friction factors. We shall now collect data on the friction factors for Newtonian flow in circular pipes. The Hagen- Poiseuille formula (II.7) valid in the laminar field yields, together with equation (11.22), for the friction factor : '7 64 (II.36) 64 4f = p(v)D = -Re In the turbulent region the friction factor is considerably less dependent on Re, notably proportional to- .Re- 0 · 25-0. As a consequence, the pressure drop is proportional to (v) 1 . 75- 2 , so that for a given pipe the pressure drop increases :nuch more strongly with ( v) than in the laminar field, but the viscosity of the 9ow has less influence. For pipes with a smooth wall the experimentally found relation proposed b y Blasius:

4f =

0·316 Re- o·:z.s

(4000 < Re < 105 )

(11.37)

gives a good description of the dependency off on Re. • The quotient A /S is also called the hydraulic radius, with the consequence that the hydraulic diameter is four times the hydraulic radius!!!

0 ·06 ---~,----------------------10 · 025

" "'

----_::~------------------~0·01

1

"-..

-----"'-.::::...,o:::-----------------J5xl0-~

'

'-

0

u

.E c:

.Q

..

u

·.:

j

........

'

----------~·"~~----------~to- 3

..

.......... ~--------=.:..:a...-:. ,-----~5x 10- 4

Laminar Turbulent (circular . cross- section! only l Roughness of wall x .( m l

LL

.........

Cost iron Wood,concrete

...........

2xto-6 40xlo- 6

Drown tube

Steel tube Galvanized iron

..........

:-------=.::::::.....:----t2x 10- 4 :::-----....::::~ to- 4

t50xlo-6 0·3 x10-3 0 ·2-2ltl0-3

--.::--._-------15x 10-5

0·007~~-L---~-L-L--~~L---L--L-~-L~-L---~-L~--~~~--~-~~-L~~--~

56

s

103

2

3 4

G

s 104

2

3 4

6

a

to5

2

3 4

6

8 106

2

3 4

6

8 107

2

Re= <~ 4A II

S

Figure 11.19 Friction factor for flow in tubes ~~~~~~~-

'

-~--

----

-

--~~~~~~~~~-

-

~~~~~-

63 In the case of turbulent flow the roughness of the wall surface also influences the friction losses. When the mean height x of these irregularities (roughness) becomes of the same order of magnitude as the thickness of the laminar boundary region (either in the case of great roughness or at high Renumbers), part of the resistance will be caused by direct momentum transmission between the turbulent core of the flow and the protuberances of the wall. In extreme cases (very great roughness or high Re) this causes a pressure loss proportional to the kinetic energy of the flow, so that the friction factor no longer depends on Re. The wall roughness is allowed for by the dimensionless quantity i / D = relative roughness. Consequently, the friction factor is in general a function of Re and i j D.

In Figure II.l9 the relation between 4f and Re is shown for various values of the relative roughness. It appears that in the turbulent region two parts can be distinguished : at high Re numbers, the so-called completely turbulent region where f only depends on the roughness, and at lower Re, a turbulent transition region where both Re and roughness play a part. The mean roughness of a number of materials is also given in Figure 1119. For pipes with non-circular cross-section the friction factor in the laminar region is proportional to 1/ Re. The proportionality constant depends on the form of the cross-section and is known for a number of cases (see paragraph 11.1.5 and figure II.12). If the flow is turbulent, approximately the same values of the friction factor apply as in tubes with circular cross-section if, in the Fanning equation (II.35) and in the Reynolds number, use is made of the hydraulic diameter. Hence the advantage of using the hydraulic diameter is that (only for turbulent flow) with one graph for the relation between the friction factor and Re the pressure drop

F low si tuation

8

a -0E}r CJe w

- =-=rH

go~ JH

Cir;cu lo r pipe

Concentri c pipe or sl it

Open channel

Open channel

~v2D ~8

=

R ect a ngu la r p i pe

r? P

A

Hydraul ic d iameter o, 4A ! S

Ha lf - f i lled

L iqu id film

r. 4

D 0 2 -01

= 28

2 WB

W+ B 4WH W+2 H

2H

12 0

48

Figure Il.20 Hydraulic diameter of various channels

o2

2 0 z> -.,. ( DzI

4

WB

WH

J.. Hz 2

.!!:.oz 8

a.,. o

64 can be calculated for all forms of the cross-section and for all Newtonian fluids (i.e. accurately for circular cross-sections and approximately for other forms). A number of frequently occurring hydraulic diameters is shown in Figure II.20, together with the cross sections available for flow. The pressure drop in spirals is, for the reasons stated in paragraph 11.2.1, higher than in a straight tube. For a rough estimation of pressure drop of turbulent flow of waterlike liquids, the empirical relation :

llp,ph~ =

dp,.n;&b
X (

1 + 3·74

~)

(II.38)

can be used (D = tube diameter, 2R = diameter of spiral). In practice, the chemical engineer often has to determine which flow rate can be attained at a given pressure drop. Thus he has to solve the relation (11.35) :

L tlp =4ftp(v) 2 D for ( v), which can be done by a trial and error procedure. Rearrangement of equation (!1.35) yields: JRe2 = 2llppD3 4Lyt2

(11.39)

For a given problem the right-hand side of this equation is a known quantity ; thusfRe 2 is known. In Figure 11.21 the relation betweenfRe 2 andRe is shown for pipes of differing roughness. With the help of this graph the Re number can be found and consequently the volumetric flow rate. Now that withfthe pressure drop in the straight horizontal channel is known, it is obvious that the energy dissipation in the channel is also fixed. From the energy equations {1.5) and (I.6) it follows that the frictional beat released for each transported unit of mass in this case equals (p 1 - p 2 )fp. The energy necessary for pumping the fluid is then :

¢A =

lPm (pl ~ P2) = lf>v(pl

- P2) = 4>m; (v)2 S(x2; X1)

(11.40)

This means that the dissipated mechanical energy per unit of volume equals the undergone pressure difference. Thus, for laminar flow 4> AI¢ v is proportional to (v) and for turbulent flow to (v) 1 ' 7 5 •

I 1.2.3. Pressure drop in pipe systems The calculations for pressure drop in straight pipes cannot easily be extended to piping systems with fittings (valves, bends, constrictions, etc.). The reason is that the forces which the piping system (now not in one but in three directions) exerts on the flow are position-dependent, which makes the three momentum balances too complicated for mathematical treatment. It was Bernoulli who in principle indicated how this problem could be solved By scalar multiplication

65

1

10 6

1....-...1...-.A..LM'-L.J.___.J~.L_..L..L..L..--L---"'--L-,.u_---L...--L-L.oU--__..J

10 3

104

10 5

10 6

Re--

10 7

Figure ll.ll The relation ofJRe2 andRe for turbulent flow

of the three momentum balances with the corresponding flow velocities he obtained, after addition of the three equations, a relation between the various forms of mechanical energy in a flow field Whereas the terms of the momentum balance have the dimension of a force or momentum flow, multiplication by the velocity leads to an equation between energy flows. In the stationary state for part of a pipe between cross-sections 1 and 2 Bernoulli's equation reads: 0= -

{f ~

dp + g(h, - hi ) + :l{(v,) 2

-

(vi ) 2 )} 1/>m + 1/> A

-

Er,l/>m

(11.41)

Err in this equation is the amount of mechanical energy which per unit of mass is converted into heat owing to internal friction. This constant is always positive because the dissipation is a result of statistical transport, which according to thermodynamics is attended by degradation of energy (entropy production).

66 Hence we may conclude from equation (II.41 ) that owing to friction the total mechanical potential (the first four terms of the equation) decreases in the direction of the flow, so that here we can speak of the law of 'non-conservation' of mechanical energy. The readers should realize that the Bernoulli equation is not a new balance but rather a summary of the three momentum balances. If the amount of energy added to the system per unit of mass 4>A = 0 and the energy loss due to friction is negligible, equation (II.41) can be written for any point in the flow field as : 1

- dp + g dh + v dv p

(11.42)

= 0

or, also, if at the same time the specific gravity is constant:

!?.p + gh + tv 2

= constant along streamline

.

(11.43)

These are two forms of the well-known equation of Bernoulli, which can also be derived in a simple way by using the momentum balance for a frictionless fluid (Newton's first law). The more general equation (11.41), which plays a very important role in the technical flow theory, is furt her referred to as the extended Bernoulli equation. On account of what we know about the pressure drop and the energy dissipation in a straight channel we can relate the constant Err for such a channel (length L) to the friction factor and the mean velocity (v) in that channel (equation II.40): 1

Err= f2.(v)

2SL

A ·

(11.44)

Local pressure losses are mainly due to a sudden increase in the cross-section of the pipe. According to Bernoulli's law in such a diverging flow the velocity downstream decreases and the pressure increases. This pressure gradient may cause the fluid near the wall to flow upstream. This results in systematic eddies (not the same a~ turbulence!). In these eddies part of the kinetic energy is destroyed by internal friction. Only if the diameter is carefully widened (e.g. via a diffuser with a vertical angle of at most 8°) do these eddies not occur. The dissipation per unit of mass in a fitting is described with a friction loss factor K w which is defined as : (II.45) In practical cases K w is only a function of local geometry and is for not too low Renumbers independent of Re. For (v) the mean velocity in the pipe on the downstream side of the local resistance is usually taken. To calculate the resistance of a pipe having various pieces of different diameter, bends, valves and other local resistances, both equations (II.44) and (II.45) should be introduced in the extended Bernoulli equation (11.41), via

67 the lost frictional energy Err. The latter constant integrated over the entire pipe then becomes :

Err =~ (ft(v)

2

I

s:).+ 2;:

(K...,t (v) 2 )j

(11.46)

J

I

where the first sum is taken over all pieces of straight pipe and the second over all local resistances. In air conditioning channels, often short bends have to be made owing to lack of space. However, the K..., values then become unpermissibly high so that frequently guide vanes are built in to lower these values (avoidance of eddies, streamlines). Empirical correlations between K..., values ·and the geometry of the pipe system are known for a great number of situations. A few K..., values are listed in Table IL L In some cases it is possible to derive K ..., values theoretically. Consider the situation shown in Figure II.22. For the calculation of the energy dissipation three equations are available, viz.:

I I I

---- :--------------·-·+·- --1

I

1 l

I

I I

..

I

Figure 11.22

Flow in a channel with changing cross-section

the mass balance :

= PVzAz + ¢mvt -

¢m = pv1A1

the momentum balance :

0

=

the Bernoulli tquatioQ (II.41) : 0 = -

P2A 2 - ¢mv2 +F...,

u: ; f f

P1A1

+

g dh

+

v dv} t/>m

+ A - Errm The latter equation simplifies for horizontal pipes, no energy input and incom;>ressible fluids to: Err=

Pt - Pz

p

1 + 1((v l) 2 - (vz) 2 )

using the above three balances it is possible to estimate the amount of dissipated !nergy Err if we can find an expression for the force the liquid exerts on the ~·aUF w· We will illustrate the procedure to be applied in four examples.

68 Table II.l Some values of the friction Joss factor

K• • k{

~~

I -

2 }

rY =1 <1001 10° 120°130°1 40° 150• ,600,70°,80° 190• I

A2

0

=

k

fr = Kw

0 · 17 0-41 0 ·71 0-90 1·03 1·12 1·13

1-10 1·05

to• l 2o• 130° 140° !so• I so• jmo 1800 1

=0 ·16 0·20 0·24 0·28_ 0·31 0·32 0 ·34 0 ·35 ..:

Az

Kw

__ ,

=(

A2 A I

) 2

0·2

A,

Ao

A2

A1 > 20A 2

~

~~----:-1--

0 · 3 0·4

0 ·5

Ao A, • 0 · 1 0·2 0·3 0·4 0 ·5 Kw ,.

232

20

51

9 ·6

5 ·3

0 ·6

0·7

0 ·6

0 ·7

3 ·1

I ·9

20. 140• 16o· lao• 19o. j'oo. lt2o•

0·8

0 ·8

1'40° Kw = OD5 0 ·14 0 ·36 0-74 0 ·98 I ·26 1·86 2-43

..J- =

K

w

=

{

0·131

+

3·5}

0·163 ( -R0 )

,j-goo

0 ·9

0 ·9 1· 0

1·2 0 ·73 0·4 8

j

69 Kw (referring to downstream velocity for Re > 10

,

;

15°

5

)

45°

3()0

60°

I

900

K,= (}02 10 ·II 10 · 26 10 ·50 1 1·20

2 - - - Kw-0 ·75

- - Kw =-

.m2 Kw2

+·- s·---~

Gate valve

tPm2 .ml K.,

=

Kw2

:

OJ6

0 ·2

0

1· 0

0 ·8

= 0 ·95 0 ·88 0 ·89 0·95 1·10 1·28

-

-0 08 -oo 0 ·07 0 ·21 0 ·35

600

7QO

II

4

1·5

80° 185° 0 ·52 0 ·24

17

5·5

1·6

0 ·29 0 ·05

20°

30°

400 150°

K," 751

118

33

Kw = 4 86

206

53

~:

0

' Fraction closed

i/8 2/8 3 / 814/815/8 6 / 8

Kw

=0·05 0·07 0 · 26 0 · 81 2 · I

Globe 110lve

Open

K.

Sieve plate

«

.-

...

:

= 4

0·61 0·51 0 ·2 0 ·7

I ·5

Fraction- free space i

Kw

Cyclone

Kw

Water meter

K,., = 6 to 12

10 to 20

2 ·0

5 ·5

17

7181 981

70 Example 1. Straight tube

Here the momentum balance reduces to (A 1 = A 2 , (v 1 ) p 1A I

-

P2A 2

+

= ( v2 ) ) :

Fw = 0

Since the liquid exerts on the wall a shear force only : F..., = - t w- JSL

- fi pv2 SL

=

we find : 2SL

t

I 2 PV A

Pt - P2 =

The Bernoulli equation reads under these conditio!!.s ((v 1 )

= ( v2 )):

_ P1 - P2 Ef r - : p as we found earlier. Substituting for p 1

momentum balance, we finally find : Err

p2

-

t

= f zv

as should be the case. Extending the straight pipes:

the expression developed from the

2SL

A

defini~ion

of the Kw value to the case of

it follows that for s traight pipes : K w

=JSL A

Example 2. Sudden expansion

In the case of sudden expansions eddy currents will occur, leading to extra energy losses. The momentum balance for this case (see Figure 11.23) between 18

'I I

I

i2 I I I

Figure 11.23 Sudden expansion of pipe

71

Pt Al - PzA2

+

+ F..., = 0

mvl - mv2

If we neglect wall friction, F..., is given by

Fw = (A2 - At)PB where p8 is the pressure just behind the expansion in plane 'B'. This pressure will be p 1 < p8 < p2 . Only experimental results can show which pressure bas :o be used. -\. Let us assume p 8 ~ p 2 , thus Fw = p 2 (A 2 - A 1 ). Then we find from the momentum balance :

Pt - Pz

=

- c/>m(v1 - Vz)

A1

2

Az(A2 - 1)

= - pv2- -

A1 A 1

We see that the momentum balance predicts correctly that stituting p 1 - p 2 in the Bernoulli equation: E rr -_ Pt - P2 p

+ li 2 2\vl

-

p2

>

p 1 . Sub-

2)

Vz

we find after some sorting and applying of the momentum balance :

This is evidently an impossible solution (Err must be positive, otherwise we would gain energy), thus our estimate of p8 must have been wrong. B. Repeating the exercise assuming that PB ~ p 1 , we find:

and, with the BernouHi equation, finally:

Thus, the resistance factor is given by : K

...,

=

A~ ) _::_1

( At

2

which fits very well with experimental results.

72 Example 3. Gradual expansion

For the case shown in Figure II.24 we can roughly estimate (neglecting friction losses): F = Pt w

+ 2

P2(A _ A ) 1

2

I I

12

I I

I

I

'

I

I I

I I

I I

1

I

- - - +-·-----------·---+--· ·--

Figure 11.24 Gradual expansion of pipe

where (p 1 find :

+ p2 )/2

is the average pressure. With the momentum balance we

and, applying the Bernoulli equation, we get : E fr

_ P1 -

Now (A 1 + A2 )/2 is the mean cross-section, where the mean flow velocity (v 1 + v2 )/2 is present. Thus: A..

v1

_

'+'m -

p

+ v2 2

A1

+ A2 2

Substituting 4>m in the Bernoulli equation we find Err = 0-there is no energy loss. Err has been proved experimentally to be negligible indeed if 8/ 2 < 8°. In practical cases of smoothly diverging flow, only small friction losses occur and a practical value of the friction loss factor K w is 0-05. Let us now reverse the reasoning. How great is Fw? Knowing that Err = 0 we can derive an expression for Fw. This force acts at a mean cross-section A, where the flow velocity is given by {v 1 + v2 )/2. A mass balance shows:

75 'lwater 11 • •

'latr

Answer:

= 10- 3 Ns/rn 2

Pwater =

= 1·6 x 10-. 6 Nsj m 2

Pair

103 kgjrn3

= 1·2 kg(m3

133 m js

2. An air heater consists of a rectangular box in which a bundle of heating pipes has been fitted perpendicular to the direction of flow. In a scale model of this heater the pressure drop has been measured as a function of the air velocity related to the empty cross-section. If in the actual air heater the diameter of the pipes is twice as large. what is the ratio between the pressure drops over the two apparatus at the same Re number (related to the pipe diameter and the said velocity)?

Answer : dp actual : t:.p scale = 1 : 4 3. Through a horizontal smooth tube there is a turbulent flow of liquid. By what factor does the pressure drop increase if the flow rate is trebled? What should the diameter of a new tube be to obtain the original pressure drop at the trebled flow rate?

Answer :

pressure increase by a factor of 6·8; D 2

=

1·50D1

4. Develop an expression for the friction factor for laminar flow in a falling film and for flow between

Answer:

f

6

;=

Re

h. ~~ .fiat

(film);

f

plates.

12 = Re

(plates)

*5. Through a smooth horizontal tube 1 kg/s water is pressed (Re ~ 2 x 104 ). The pressure difference over the tube is 2 x 10 5 N j m 2 • What energy Pis needed and what is the temperature rise .1 T of the water if there is no heat exchange with th:e environment? What will be the values for P and aT: (a) if the water fiow_is doubled? (b) if 1 kg/ s W(\ter is pressed through a tube 1·5 times as narrow? Answer : 200 W, 0-~8oC (a) 1280 W, O·l6°C (b) i400 W , 0·34°C 1

6. Calculate the pressure drop over a capillary flow meter with a smooth wall if the gas velocity is 5 m/ s. 50 mj s and 100 mj s respectively. 5 capillary length 10 em 1Jair = 2 x 10- Ns/ m2 3 Pair= 1·2 kg/m capillary diameter 1 mm

Answer:

~p

= 3·21

x 102 N j m 2

6·5 x 10 3 N / m 2 2·1 x 104 N j rn 2

76 *7. Through a steel tube of 60 m length 120m 3 /h of water is pumped against a height difference of 25 m. \Vhat will be the difference in pressure between the beginning and end of the tube if: (a) the tube is of circular cross-section (D = 10 em)? (b) the tube has a square cross-section (10 x lO em)? (c) the water flows through an annulus (D0 = 14 em, Di = 10 em)? Answer :

(a) 3·4 x 105 N j m 2 (b) 3·1 x 10 5 N /m 2 (c) 5·2 x 10 5 N j m 2

8. In a factory there is an 8 inch pipe from a trough of a rotating filter to a hollander which is situated 3·75 m lower. Cellulose slurry which flows over the brim of the trough is returned to the system through this pipe. The pipe is 15 m long and has two short 90°. bends and one open gate valve. What is the flow rate ofthe slurry in the pipe.and what is the capacity? The viscosity of the slurry is 5 x 10- 3 kg/ m s (at 20°C, that is five times as viscous as water) and the density is 1200 kg/m3 • Answer :

4·5 m j s; 520m 3 fh

9. A vertical pipe (1 inch diameter, 2m long, sm ooth) contains one rotameter, three right-angled bends for connecting the pipe to the ro~ameter and two valves. The Kw value of these valve~ when open, is 6. The rotameter, which has approximately the diameter of the pipe, contains a float of 25 x to- 3 kg (p = 2600 kg/m 3 ). Calculate the pressure drop over this pipe if water flows through it at an upward velocity of 1 m/s. Answer :

.1p

= 28,060 N j m 2

*10. When combustion gases enter a chimney the pressure is 250 N / m 2 below atmospheric pressure. The chimney is made of smooth s teel plates which are bent to an internal radius of 1·68 m. 15 tjh combustion gases are discharged. The mean temperature of the gases is 260°C, the outside temperature is 21 oc and the barometer reading is 1000 mbar. What is the height H of the chimney? The density of the burned gas at 25°C is p = 1·27 kgfmJ and rt = 15·6 x 10 - 6 kgjms at 260°C . Answer :

H = 47 m

11 . Through a pipe with a 5 em internal diameter flows 241/s of water. An obstacle in the pipe shows a flow resistance of 100 mm water column. (a) What is the force the flow exerts on the obstacle? (b) H ow much energy is lost by friction? Answer : (a) = 2 N (b) = 1·96 w

77 *12. A strong fireman sends up a jet of water of 2 m 3 jmin at an angle of 45°. The maximum height he can reach is 20m. What is the force with which he is pressed against the ground as a result of spouting?

Answer :

66 7 N

13. A garden hose (D i = 1 ern) of smooth, inelastic material, 20m long, is connected to a tap in the ground ; the resistance value of the tap related to the downstream velocity is K .,., = 7-5. The pressure before the tap is 2·65 x 10 5 N j m 2 • A nozzle (D i = 1/ 4 em) can be connected to the spouting end of the hose, in this case Kw = 0·1 related to the velocity in the nozzle. The spouting end is kept 1·5 m above the ground. Calculate the velocity vat which water of l8°C leaves the garden hose: (a) without a nozzle. (b) with a nozzle. Neglect the water velocity before the tap.

Answer : (a) 2·3 m js (b) 14·7 m js 14. At least how high can a pole vaulter jump if he can run at a speed of 10 m j s?

Answer : 5·0 m 15. On the Mons St. Nicolas (Luxemburg) lies an artificial lake with a capacity of 7-6 X 10 6 m 3 . For twelve hours (mainly during the night) the lake is filled with water from the Our, 500 m lower. During the next twelve hours turbines on the Our are driven with water from the lake in order to generate electric power. The water supply from the lake to the turbines takes place through two parallel circular concrete tubes, 650 m long and 6 m in diameter. (a) What power could this station deliver, apart from losses, if it is in continuous operation for 12 out of 24 hours? (b) What are the .fri~on losses expressed as a percentage of the theoretically possible power? Let the resistance value for the entry and exit losses, each related to the velocity in the tubes, be 2·5. (c) Calculate that, if the pump delivery and the turbine efficiency are 75 per cent, this way of energy storage makes the electric power 1·8 times as expensive : the engineering works as such may be considered as being fully written off.

Answer : (a) 550 MW ; (b) 0-65 per cent •16. Show that, in the case discussed at the beginning of this paragraph, John was right in letting the suspected man go.

78 Comments on problems Problem 3

(a) An increase in throughput will affect, amongst others, the Re number and therefore also the friction factor. (See Figure Il.19.) (b) Here again you should not forget the change in f due to the changing Re number. Problem 5

This problem has been included to show you bow widely Figure 11.19 can be used, provided that you base your Re number on the proper hydraulic diameter. Don't forget the hydrostatic head pressure! Problem 10

A sketch of the chimney with an expression for tlie pressures at the bottom and top will help you to find an expression for the height of the chimney using the Bernoulli equation. It turns out that the losses due to friction are negligible.

p , -250

Problem 12

With the substitution of the maximum height h in the Bernoulli equation you will find the vertical velocity component Vy of the nozzle. The force in the y-direction can be calculated from the mass flow and vr Do you find the same answer when you calculate the weight of water the poor fellow has to carry? Problem 16 John and the sewer murder case

The flow conditions in the sewer are turbulent (Re ~ 4 x 106 ) and we can assume f to be practically constant. Because the driving force pg ~h is constant we can write :

79 and we find :

v2

= v1 )

Dh 1/D112

For a half-filled sewer channel Dhl = D, but if the channel is filled to onequarter of its diameter :

4(nR 8 _ R sin (J) 2

Dh 2

=

2

360

2

nR8 180

if we use cos 8/2 = (R - tR)/R =

= 0·59D

f.

Since v 1 = 1 m/s (half-filled channel) we find v 2 = 0·77 m/s and the time necessary for the body to arrive at the fisher village is : t

= ~ = 9360 s = vl

2 h 36 min

(or longer, if the water level was lower than iD~ Thus the corpse must have been dropped into the channel at 9·09 p.m. or earlier, and John came to the right conclusion. ll.J. Flow with negligible energy dissipation John looked from the corpse to the drum, from which wine (a poor quality rose, thought J ohn) still spouted through the two 6 mm holes which two of the bullets had left there. The fat man who had called him had stated that he had been in this isolated shed with only one other person, who had run away twenty minutes ago. John thought: the drum of 0-6 m diameter stands on its fiat bottom, the holes are 1 m from its top and a wet rim indicates that the wine level has decreased by 20 em. He remembered what is outlined in the pages to come and arrested the fat man.

For a great number of flow situations the energy dissipation Err is zero or negligibly small with resp-ect to the amount of mechanical energy which changes from one form into the other (see example 3 of the previous paragraph). In those cases the relation between velocity and driving force is simply given by Bernoulli's law (in the form of equations 11.42 or II.43). This equation can also be written in such a way that the terms acquire the dimension of a length : p

ti2

pg

2g

- + h +-

= constant along a streamline

(11.47)

The three terms are then called pressure head, static head and velocity head respectively ; the sum total of these terms remains constant if there is no energy

80

----

-

------y-

--

- -

11 + .1!...

-pg

-- -vzO

--

1

-

--

--

p

--

r---------f=-=-

1

pg

-T -

hI

-

p

pg

--

v2

+

2g

~

.... --· -

-----

-

--

-

·- ,___~-

·v

h

•

h=O

Figure 11.26 Static head, pressure head and velocity head

change due to friction or mechanical work. These heads also have physical significance, as can be seen from Figure II.26. _. With the left manometer tube only the pressure prevailing in the liquid i~ measured. The lower opening of the right-hand. tube is opposing the stream. Right in front of this opening the velocity is negligible so that all kinetic energy is converted into pressure and causes a thrust equal to !pv 2 • This thrust is used when measuring velocities with a Pitot tube. . We will treat here two well-known examples of flow with negligible ener~n dissipation, the flow from orifices and the flow over a sharp-edged weir. ll.3.1. Flow of a liquid from an orifice

Friction hardly plays a role in this type of problem. For calculation of the outflow velocity use is made of the mass balance and the Bernoulli equation. Using this mass balance it appears that some distance before the orifice the velocity is negligible .compared with the velocity in the orifice. Next, with equation (11.41) it can be derived that for a sharp-edged orifice (Figure II.27) the volume flow ¢v becomes :

Po

( 0 )

(b )

Figure 11.27

(c )

Flow of liquids through holes


A'1 is the surface area of the smallest cross-section of the jet where the pressure should be p 0 , because no radial accelerations occur. The calculations can better be based on the actual surface area of the orifice A 1 : cf>v = CA1j2(p t - Po)/ p

81

C is the flow coefficient and is the product of a coefficient of contraction Cc = A'1/ A 1 and a friction factor C1 . For the sharp-edged orifice, C1 is practically 1, whereas Cc is ca. 0·62 at sufficiently high velocity. For a rounded orifice with smallest cross-section A 1 Figure 11.27b applies: (II.48b) with c, = 0·95-0·99. If a 'diffusor' (angle not greater than 8° to avoid eddying) is attached to the rounded orifice then : (11.49)

where A 0 is the cross-sectional area at the end of the diffusor. (Figure Il.27c). With the diffusor more liquid flows through the smallest cross-section than without the diffusor, because in the former case the pressure in the constriction is lower than p 0 . The diffusor cannot be lengthened infinitely because then the friction losses become too great On the other hand, if the pressure in the throat falls below the vapour pressure of the liquid, vapour generation (cavitation) occurs; the given calculations are then no longer valid. The following table shows the flow coefficients C for various orifices. These coefficients must be used with equation (II.48a). Table ll.2 Flow coefficients for flow through orifices

• d

c

= 0 ·96

c

= 0 · 82

.-

;Iff d--

'1

d/ 2

c ;

0 ·53

82 11.3.2. Flow of gases through orifices

The approach followed in the last section can also be applied to the flow of compressible gases through orifices. Friction losses can be neglected, and we can assume that the change of. state takes place adiabatically. However, due to the adiabatic expansion, the pressure in the jet will be different from the initial pressure p 1 and the final pressure p0 outside the gas jet (see Figure Il.28).

Figure 0.28 Flow of gas through an orifice

In the stationary state the Bernoulli equation (11.41) reads in its differential form (g dh = 0, .,. = 0, E rr = 0) for this situation : 0 =-

f Hdp

+

(II. 50)

vdv}

whereas the energy balance treated in chapter I (equation 1.6) :

0= is simplified to (¢H

0= -

f {ctu =

+ ct(~) + vdv + gdh}m + q,_.-

t/>n

0):

f {ctu + ct(~) + vdv}

= -

J.' {du + pd(~) + ~dp + vdv} (11.51)

From equations (II.50) and (11.51) we thus obtain:

0

= dU +

pd(~)

or, realizing that (if there is no change of state) dU

0 = c. dT

= c" dT :

+ pd(~)

(II.52}

Applying the ideal gas law for a number of moles nina volume V , p V = nRT (R =gas law constant = 8·31 J/mol °K), we find with M = molecular weight of the gas (kg/mol) : p

=

nM V

and

p

d (~) = _L d V = R T d V

. p

nM

MV

83

R = M(cp- c") and

K

= cJcr: we now obtain from equation (II.52):

0

dT

=

T + (K -

dV l)y-

tion of this equation finally yields TVIC ormed by means of the ideal gas law to:

(II.53) 1

= constant, which can be

p VIC = constant and pp -K = constant

last equations describe the adiabatic compression and expansion of ideal Returning to the Bernoulli equation (II.50) for the flow of gases through .nfices under steady-state conditions :

1 0 = -dp p

can now replace p by p -.d p, v we find : _ _1_ l constant - 1

+ v dv

= constant p 11

1(

{ p - 1/K+ 1 _ p - 1/K+ 1} 1

-+1

and by integration between p 1 , v = 0

+ 21v2 = 0 = !v2 +

K (!!_ _ K -

1 p

P1) p1

K

V

2 -

-

2K K -

p 1 {l -(p ) (K- 1}/IC} 1 Pt Pt

(II.54)

the mass flow rate at any place of the variable cross-section A of the gas jet .,...,p,·p_ the pressure is p is given by: ¢m

= pvA = = A

p ) 1/IC p 1vA ( Pt

{2~: ,P.P ·[(:r ·_

(:.r ·>'·Jr

-

\. -

(11.55)

The change of the cross-section of the gas stream with pressure is shown 3Chematically in Figure 11.29, where:

been plotted against pj p 1 _ It can be seen that the cross-section of the jet reaches a minimum at

_!!_ = ( 2 ) K/ (tc- 1) Pt K + 1

84

p/pl - - -

Figure 11.29 Flow of gas through an orifice

At this point the critical conditions Ac, Po vc and velocity vc we find with equation (II.55):

Vc =

P 1)t + 1 Pt

2K

( K

=

(KPc)t Pc

=

Pc are present. For the critical (KRT)t

(11.56)

M

which is the velocity of sound at Pc and Pc (for air at 0°C, vc = 331 m js). For gases consisting of two atoms, K = 1·4 and consequently pcfp 1 = 0·527. The expression developed for the mass flow rate (equation JI.55) can be simplified for a number of conditions. (a) p0 ~ p 1 • Under these conditions the compressibility of gases can be neglected and equation (11.55) simplified to: (II. 57)

where A = cross-section of jet at p = Po, A 1 = cros5-section of orifice and C = contraction factor which is approximately 0·6 < C < 0·7.

(b) Pc < Po < p 1 (for air, Pt by CA 1 with 0·6 < C < 0·7.

=

0·53pt). Equation (11.55) is valid, A being given

(c) p 0 ~ Pc ( =0·53 p 1 for air). When Po is so low that at the cross-section Ac the speed of sound is reached, a further decrease of p0 will not result in a lower

pressure at Ac. The reason for this is that pressure waves travel with the velocity of sound ; hence as soon as the pressure Pcis reached at Ac the pressure difference p 1 - Pc determines the flow rate completely independently of the value of Po·

85 Equation (11.55) becomes:

"'m = p,v.A, = A,{Kp,p,(K!

J.. ,y,.-,,r

= C"A,JP:/);

with: CK

=

HK ! t)"+l)/(x-1)}

(II. 58)

t

and Ac = CcA 1 , where Cc ~ 0·6 for sharp-edged orifices and Cc ~ 0·97 for ;ounded orifices. The factor C K can be calculated for each gas from the known · (air: C" = 0·684 ; steam : C K = 0·668). For flow of gases in pipes or pipe systems the same considerations as those : ast discussed apply. If p0 ~ p 1 the compressibility can be neglected and the :elations discussed in paragraph II.2.2 can be applied. If Po < p 1 the calculation becomes more involved and Figure II.30 can be used for a quick estimation of pressure drop during turbulent flow. This graph is based on air at l7°C. The irawn lines show the calculation of the pressure drop for a mass flow of 0·1 kg/s air at l7°C and 10 atm pressure through a pipe of25 x 10- 3 m diameter and a K w = 4. We find 6p = 0.06 atm (for straight total friction loss factor of pipes Kw = fSLfA ; for circular pipes at Re > 105, Kw ~ 0·02 L/ D). We can further read from the graph that under these conditions v ~ 30 mf s and o ~ 11 kgjm 3 . For gases other than air or for other temperatures, the small inserted graph gives two correction factors, CM and Cr . For these conditions, instead of the gas pressure p 0 a corrected pressure p 0 CMCr has to be used and instead of !he flow velocity v the product vCMCr is found, as well as pf CMCr instead of the specific gravity p. The same applies to the pressure drop determined, which IS then ~pCMCr . So, if we had used C0 2(M = 44) at 100°C, CMCr = 1·3 x 0·9 = 1·17, the example drawn in the graph is valid for an initial pressure ofll·7 atm. The flow velocity fou~d is then 25·6 mjs, the specific gravity is 12·9 kg/m 3 and the pressure drop would, be Ap = 0·052 atm.

I

I 1.3.3. Flow through ~eirs The total liquid flo~ ca~, in this case, be calculated approximately as a function of the height h of the liquid surface above the edge (see Figure II.31), with Bernoulli's law assuming that mainly potential energy is converted into kinetic energy. On the upstream side of the weir at a height h the static pressure is :~0 + pg(h 0 - h) (p 0 = ambient pressure), whereas the velocity may be considered negligibly low. The pressure in the overflowing jet is Po in all places, whereas the velocity of the streamline coming from height h is called vh. Application of the Bernoulli equation yields for the velocity distribution : vh

=

j2g(h 0

-

h)

(II. 59)

86

lnitiaJ tem"'"atwre

roc

CorrectiOn factors for other 90ses than air or ot her temperatures than I7°C

....L:.t....;;~L----Z-:.L..::.::z._:::J____::jr,--?-..:z.....::.L~I:......Z-:.t.-.z..._ Flow veloc ity v (m I s )( cr. eN . v ) 3 ""--......c...--'......c...-""--......L.--'-'--L-.L.-- - " - - Density p ( kg I m ) ( p I C T .cN ) 0

.....c........L.......c...-

Figure 11.30 Pressure drop for

87

Diameter (m) 400 x 10-3

Reference line Mass flow rote (kg/s) :soo Friction loss zoo foetor}:; Kw

300xlo-3 250x lo- 3 200 xl0-3 150xl0-3

I

IOOxi0-3

2

80

X

10- 3

sol( ro- 3

10

X

10-3

8 X 10 - 3

s x ro-3 4xlo- 3 3x lo-3 2·5)( I0- 3

..· .

.

..:bulent flow of gases in circular pipes

88

Figure ll.31

Flow through a narrow-crested weir

So at h = 0 the flow velocity is maximum and v11 = 0 at h = h0 • Hence the volumetric flow rate in the streamline between h and h + dh is :

The total flow rate then becomes theoretically : .·

In reality the result must be multiplied by a flow coefficient C = 0-~2, owing to contraction and some friction loss. Then for L » h0 and h 0 > 0·01 m the practical formula becomes :

4>v = 0-59L~

(11.60)

A weir can be used for measuring liquid streams with a free surface area Besides the rectangular weir there are differently shaped weirs. Figure II.32 gives a survey of the relations applicable to the various weirs. Occasionally a viscous liquid is passed over a narrow-crested weir. In this case, the flow rate is given by:

4>v = 0·285L~{g~~}t if 4>v < 0·2 v

Lv

(11.61)

according to Slocum (Can . J. Chern. Eng., 42, 196, 1964). I !.3.4. Problems 1. The trough of the rotating cell filter from problem 8 in paragraph II.2.4 has a weir edge which is 2·85 m broad. The level of the cellulose slurry is,

as a rule, 2·5 em above the edge. Calculate the amount of slurry flowing over the edge. Answer : 76m 3 jh

2. A free-falling liquid jet contracts owing to acceleration in the field of gravitation. For a given jet a 20 per cent contraction is attained at a distance of 8 em from the outlet opening. Calculate at what distance 20 per cent contraction is found if the outftowing liquid flows 1·5 times as fast. Answer :

18 em

89 Rectangular weir ( L > 2h 0 )



Narrow rectangular notch ( ho

Broad- crested weir ( 0

>

>L

J 9/15

0 · 59 (L-0 ·2 ho ) ~

0 · 6 m /s; ho

> 0 · 1 m)

)

3h 0 )

4> u ( L :

: 0 · 465 (L-0 ·2 h0

)

J911~

width of weir )

Triangular notch

0 · 44

;-;;g-

tong
Figuri ll.32 Flow of water through weirs

•3. In the bottom of a cylindrical vessel (area of horizontal cross-section F = 0·5 m 2 , height h0 = 3m) there is a sharp-edged round orifice (surface area F 2 = 5 x 10 - 4 m 2 ). Calculate the timet necessary for the water-filled vessel to drain thro ugh orifice F2 • The flow coeffici~nt C = 0·632 is assumed to be independent of the water height.

Answer:

1240 s

4. In the house of Professor T there is a rectangular bath tub. If the tap is fully opened and the plug ·is put in the drain the bath-tub is filled in 20

90 minutes. The water height is then 50 em. When the tap is turned off the bath filled up to this height empties in 32 min through the dr.ain in the bottom. The flow resistance in the hole is then determinative of the velocity. One day Professor T wants to take a bath and turns the tap wide open, but in his absent-mindedness he forgets to put the plug in the drain. Calculate the eventual height of the water in the bath-tub. Answer :

32 em

*5. A gas cylinder has a volume of 40 I and an initial pressure of 200 atm. If we open the main valve (2 mm diameter), how long would we have to wait at 20°C before the pressure is reduced to 10 atm, if the cylinder is filled · with : (a) air? (b) hydrogen? Answer : (a) 312 s (b) 82 s 6. A spray column is applied for the extraction of free fatty acids from edible

oils by a solvent, which forms the continuous phase. The oil is supplied under a cylindrical cap (wall thickness 2 mm, height 10 em), the lower circumference of which is equipped with forty V -notches (equilateral, height 15 mm). Furthermore, the top of the cap has thirty holes (sharp-edged) of 4 mm diameter. What oil flow rate can be handled by this distribution device if the oil height in the notches is allowed to be 10 mm? Which percentage of oil flows through the holes? p oil = 800 kg/m 3

Answer:

p solvent = 1000 kgjm 3

1·93 m 3 jh; 40·2 per cent through holes

*7. Show that in the case presented at the beginning of this chapter, John took the right action when he arrested the fat man. Comments on problems Problem 3 For the volumetric flow rate from the drum, with equation (II.48a) : "' = A dH

'+'v

1

dt

= CA J 2(pl 2

-Po)

p

where A 1 = cross-section of drum, A2 = cross-section of hole, H = height of liquid above orifice. Further, the liquid pressure is given by: P1 =Po+ pgH

Substitution and integration between t = 0, H = H 0 , t effluent time.

= te, H = 0 yields the

91

t'1· o., 1-

Problem 5

For air, Pc = 0·53 P1 and the pressure outside the cylinder p 0 is 1 atm, whereas lhe lowest pressure in the cylinder is Pe = 10 atm. Thus Pc > Po , i.e. critical conditions are present and we can use equation (II.58): d(V Pl) - Vdpi - C dt dt- KAcvr:::-;:P1P1

A. 'Pm -

Expressing p 1 in p 1 with the help of the ideal gas law and integrating between 2 5 2 l = Q, p 1 = 200 x 10 N/m and t = te, Pe = 10 x lOs N/m , we can find the efftuent time te. Problem 7 John and the barrel of wine

The flow rate of the wine (irrespective of its quality!) from the barrel at any

height h is given by:

1le er

nbt

:d)

:>n

!f-

ok

¢> IJ

= 2CA

=

1t 2 dh --D 4 dt

here C = flow coefficient ( =0·82 according to Table II.2), A = (n/4) d 2 = cross-section of one hole, D = diameter of barrel and h = height of wine above the holes. Note that we have assumed the barrel to be cylindrical. Integrating between t = 0 and t = t and h = h 0 and h = h, we find t =: 61minutes; in other words, the character running away 20 minutes before John entered could never have fired the fatal shots. For the time being, the behaviour of the fat man is suspicious.

John had been asked to investigate the case of an alcohol manufacturer who claimed that the tax inspector in his factory used a calibration curve based on experiments with water for the alcohol rotameters. In this way, the manufacturer said, he had paid taxes for a production which was 11 per cent higher than his actual production. John asked this man, who presented to be an experienced chemical engineer, whether the calibration curve for water of this ;. .- rotameter was dependent on the temperature of the water. The man answered that chis was nor so and that che tax inspector used the right float. Hence, said John, you are only partially right and you can claim no more than to be overtaxed by 2 per cent. If you are not convinced, I advise you to read the following pages.

To measure the flow or the flow rate (volume flow rate l/>,_ or mass flow e ¢m) of a fluid one can distinguish between: Displacement meters which divide the gas or liquid stream into known volume units and give the number of these units which have flown through during a certain time interval (e.g. wet and dry gas meters, various types of petrol gauge).

92 B. Flow meters with which the mean velocity of the flow can be in various ways, e.g. : (a) dynamic: the indication is based on the flow laws (constriction in pipe-orifice plate~ venturi tube, rotameter and also the weir and Pitot tube (see the previous paragraph)). (b) kinematic : the velocity is measured with a mechanical system mill, blade whee4 water meter). (c) thermic : the increase in temperature of the How as a result of a input of electric energy is determinative of the velocity. We confine ourselves here to treatment of group B(a) and discuss in the venturi tube, the orifice plate and the rotameter. These apparatus designed in such a way that the pressure drop i~ as low as possible (Err ~ I/.4.1. Venturi tube This tube consists of a gradual constriction (angle 25- 30°) and an even gradual widening (angle maximal 8°) ; s~e Figure II.33. With this we can be sure that the stream follows the boundary lines and eddy is excluded. p 1 is measured before (or behind!) the constriction and p 2 in narrowest cross-section (cross-sectional area A2 ) . For the relation tPm and (p 1 - p 2 ) use is made of the mass balance : ¢,. = p (v 1) A 1 = p(v 2 ) A 2

and of the Bernoulli equation: Pt

+ !p(v t ) 2

= P2

+ !p(v2) 2

from which follows :

J ~p Figure 1133 The Venturi meter

T he factor (1 - Ai/Ar)- t, if necessary with a correction factor Ji.w for (slight) energy dissipation, is called the flow coefficient a. of the Venturi tu

93 This factor .Uw is almost 1 (for 0-20 ·< A 2 / A 1 < 0·50 within 10 per cent, provided the Reynolds' number related to the greatest diameter is higher than 2 x 104). A neat finish of the Venturi tube is a prerequisite, so that it is more expensive than the orifice plate to be discussed below. The pressure loss owing to energy dissipation is, however, small (not more than 20 per cent of the pressure difference measured), so that Venturi tubes are used if extra pressure losses are inadmissible or too expensive (very large pipes, water supply). . Il.4.2. Orifice plate The principle of this flow meter is that the fluid is allowed through a constriction in the pipe (see Figure II.34) and the difference in pressure before and after the constriction, p 1 and p 2 , is measured. Essentially the same happens as in the Venturi tube, however : (a) The smallest cross-section does not coincide with the orifice (cross-section A 0 ) because the liquid jet after the onfice still constricts to the cross-section A2 (=,UAo)·

--

---~~~~~~~~~mm~~mm~

Figure I1.34 !

S~arp-edged

orifice meter with corner taps

fb) A considerable energy dissipation is given because the flow separates from the wall and eddying occurs on the side of the orifice.

The relation between the mass flow and the pressure d ifference measured is expressed b y equation (I1.35), in which A 2 has been replaced by ,uA 0 • The factor ,u(l - ,u 2 AUA~) -t times the correction factor for the energy dissipation is called the flow coefficient rJ. for the orifice plate. If necessary, for the transport of gases allowance can be made with the compressibility of the medium by mtroducing, in addition to rJ., another correction factor which lies between 0·9 and 1·0 for the usual orifice plates. The flow coefficient <X depends on Re 1 =

94 ( v) 1Dtfv, on the ratio A 0 / A 1 , on the shape of the orifice and the way in which its

edge has been finished and on the places where p 1 and p 2 are measured. The finish of the orifice has been normalized in many countries. • If tabulated values of a are used it should be noted that the prescribed dimensioning is accurately maintained and that the orifice plate is kept at a distance of at least 50 D 1 behind and 10 D 1 before an irregularity (bend, valve, change in diameter). In the case of careful construction the calibration curve can thus be predicted to within + 1 per cent. Orifice plates are preferably dimensioned in such a way that the flow coefficient in the working range is constant (high Re numbers). (See Table 11.3.) Table 11.3 Flow coefficient o: fo r orifices with sharp edges and corner taps

0·1 A o/A 1 0·05 > 2 X 10 4 > 10 4 (X 0·598 0·603

Re

Kw

1000

240

0·2 >6 X 10 4 0·617

52

0·3 >8 X 104 0·635

19

0·4 0·5 0·6 > 105 > ·105 >2 X 105 0·660 ().693 0·740

8·5

4

2

0·7 >4 X 10 5 0·805 l

A disadvantage of using an orifice plate can be that it causes a considerable pressure loss in the pipe. The kinetic energy of the flow coming out of the orifice is only partly converted into pressure. The resistance value Kw of sharp-edged orifice plates is therefore not zero, as can be seen from Table 11.3.

I I .4.3. Rotameter A rotameter is a tapered vertical tube (usually made of glass) containing a float (Figure Il.35). The greater the upward fluid flow, the higher the level where the float remains. The rotameter can be considered as an orifice plate with . variable free cross-section A 0 (dependent on the position of the float in the tube) and a constant pressure drop which is determined by the dimensions and the weight of the fto~t. The latter can be explained as follows: the pressure on

p,,p I

1

Figure ll.35 Rotameter • Accurate instructions for the design and use of orifice plates can be found in : V Dl-Durchflussmessregeln (DIN, 1952) and R. F. Streams and others, Flow M easurements with Orifice M eters, Kostrand, New York, 1951.

95 the underside of the float is everywhere almost p 1 and on the upper surface so the force on the float which is in equilibrium with the apparent weight of the float G1 = Vg(p1 - p) is (p 1 - p 2 )A1 ~ where Af is the greatest cross-section of the float. Using this in equation (11.35) leads to the pressure drop/flow rate characteristic of the rotameter : p2 ,

4>. = oe(A,-

Ar)J 2p(pf

~ p)gV

(11.63)

If the Re number in the annulus is sufficiently high the coefficient a has a constant value which is still dependent on the shape of the float. It appears from equation (II.63) that{Ab - A1 ) for a certain rotameter fluid combination is proportional to cPm · Since an almost linear dependence of (Ab ·- A1 ) on the . height is ensured, the calibration graph of a rotameter (h as a function of m) is practically linear between 20 and 80 per cent of its full scale. 11.4.4. Problems 1. Calculate the approximate dependence of the loss of mechanical energy

per unit of time on the flow rate in : (a) an orifice plate ; (b) a rotameter.

¢;

Answer : (a) Etrcl>m "' (b) Err4>m "' ¢v

2 (a) Can a rotameter, which has been calibrated for water, be used without· renewed calibration for measuring the mass flow of an alcohol stream? If so, calculate the conversion factor. (b) Can a rotameter destined for measuring a H 2 S04 flow be calibrated with a glycerin/water mixture? What difficulty is encountered? 3 3 (palcohol = 760 kgjm , Pnoat = 7000 kg/ m ) Answer : (a) 0·89 ~

A water supply sy~tem contains a device as sketched in Figure II.36 for measuring the water consumption. In main I an orifice plate Ma with a

I

!

R

Db

Figure 11.36 Water meter

96 sharp-edged circular orifice has been incorporated Parallel pipe II contains an orifice plate M b with a sharp-edged circular opening and a rotameter R. The dimensions have been chosen in such a way that the flow resistance in the parallel pipe is fully determined by the orifice plate. Calculate the maximum capacity of the rotameter needed at a water flow of at most 500m 3 (h in the main.

Da = 0·105 m

ma =

D')2 = (D:

m, =

(~:)

0·345

= 0065

Answer : 2·37 m3/ h *4. Through a pipe (Di = O·lOm) flows an amount of water (T = 20°C) from 12 x 10- 3 to 24 x 10- 3 m 3 fs. The flow must be acc urately m easured with a sharp-edged orifice plat e with ·corner taps'. The following requirements should be met : (i) The flow coefficient of the orifice plate should be constant over the entire range. (ii) The pressure difference to be measured should not exceed 68 em Hg. Determine the limits between which the opening ratio m should be chosen.

Answer : 0·3 ~ m ~ 0·5. *5. Show that John gave the greedy alcohol manufacturer the right answer to the problem described at the beginning of this paragraph (palcohol = 790 kg/m 3 ) .

Comments on problems Problem 4 Calculate the Re number for volume flow : Re = pvl D I '1 Re

=

= P 0·8 Vt D f 0·8 17D 1

12 X 10 - 3 x . 103 = 1·5 0·8 x 10- 3 x 10- 1

X

105

- . v~ P l 0·8 17D 1

(for minimum flow)

(an d for maximum flow 3 x 105 ). F rom Table 11.3 it follows that the upper limit of m is 0·5. In other words, m has to be ~ 0·5. The lower limit of m is controlled by the maximum pressure that

n-

a w

97

is allowed :

Jw


= cxm!nDi~

m

m

~


cxtnDi~

~ 0·33

The combination of the two conditions to be fulfilled leads to: 0·33 ~ m ~ 0·5

John and the alcohol manufacturer The flow rate through a rotameter is given by :

Problem 5

t/Jm = <X(A,- AJ)!p(pf ~ p)gV In this story (Ab - A1) is. fixed. If we assume for a moment that p1 » p we find: cP m alcohol

-

cPmwater

J Pal~hol JP::::r c/>malcobol

= 0·89 mwate r

Here we see the 11 per cent claimed.by the manufacturer! However, the manufacturer admits that the reading of the flow meter is not affected by temperature variations. This information can be used to find a relation between p1 and p: d{p(pf - pJ} , dp

I

= 0

p=pw

He also informs us th.a t the float was selected in such a way that the ex value was independent of the Re number. Now we have:

c/>m alcohol 1·37 X 10 3 c/>malcohol

_ -

mw 1·4 X 103

= 0·98mw

and again. John is right.

- - - - - -- - - - -- - - - -- - - - - -

98

n.s.

Flow arOlmd obstacles Inspector John relaxed for a moment and read a report of an inventor about a gun which fired almost spherical tear gas grenades at a muzzle velocity of 45 mfs. The report claimed that the grenades could travel about 180 m. John observed that the terminal freefall .velocity of these grenades was 25 mjs and wrote the following text in the margin, before he turned co work again: 'I am afraid that when fired at a distance of 200 m from a crowd, they will drop dead just in between the crowd and the police force.' It took his superior one night of studying the following text, before he decided that again John was right.

11.5.1. General approach Flow around obstacles results in a force F of the flowing medium (with relative velocity v,) on the obstacle. Generally there are two different mechanisms, which together account for the total force, viz. the friction resistance (a) and the geometric resistance (b). (a} In the extreme case of flow along a fiat plate (Figure II.37), the force is only caused by the friction resistance. This figure shows that the disturbance of

v,

Figure 11.37

Flow along a fiat plate

the relative flow velocity vr increases in the y-direction if we move from x = 0 to higher values of x. The velocity distribution in this area can be calculated with the aid of the boundary layer theory. This theory shows that the thickness of the hydrodynamic boundary layer is given by: /Jh =

3R:·x

(II.64)

li r

This result is somewhat different from what we found in paragraph II.1.6, when we discussed the penetration of momentum for a stagnant liquid and a moving plate (see page 50). There we had found that:

op

=fiW =

Rvx v

r

99 Apparently, at a given distance~, > br This is due to the fact that in the case of the moving liquid there is also a velocity component in the y-direction which increases the thickness of the boundary layer. This velocity component is zero in the case of the moving plate. The presence of a velocity component vy in the flowing liquid is illustrated by Figure 11.38, where the mean velocities averaged over a constant boundary

x=O

v,

Figure 11.38

Mean velocity of flow along a fiat plate

layer thickness are shown at two values of x. It is evident that (v 1 ) > (v 2 ) and therefore some liquid must move upwards at a velocity vr The shear stress at the wall is now given by : !yxly=O

v,. = 1Jbh =

2 /v 3.Y;; ·2PV, 1

2

(1 1. 65)

and the total force on the plate (both sides, length L , width B) is given by:

1 F = 2B Jo

L

Tw

dx

,L)-tBLtpv;

= 2·66 ( vv

(II.66)

The dimensionless combination (v,L/v ) is here a Reynolds' number related to the relative velocity vr and the length dimension in the direction of flow. (b) If a fiat plate with surface area A is placed perpendicular to the flow

{Figure Il.39) an eJqtrem-e case of geometric resistance is attained. Before the plate a pressure builds up of the order of magnitude of the thrust (!pv;). At the

Figure 11.39 Eddy fonnation behind an obstacle in flow

100 edge of the plate this thrust has been converted into extra kinetic energy which, however, owing to eddying in the ·dead water' behind the plate, is only partly converted into pressure again. As a result, the order of magnitude of the force on the plate is given by:

F = A-!pv; In general, the force on an obstacle ·in the flow is described with a drag coefficient Cw which is defined according to: (II.67) Table 11.4

---- ~

\tP r ~

Ver tical plote L>D

----

~

Cy linder L>D

---

Drag coefficients of various objects (10 3 < Re < 10 5 )

Cw .. 1·4

---

c.... 2

Cw• 1· 2

Sphere

---

c... -

1· 2

Cor

s:;;d~oOd

--

Flat circular plate

Steam enoine

c......

1

-

Bus

Cw.,. 0 ·4

fJ ell( .. 0 ·43

8

Cylinder L

---

- ~ ---

~

Cw• 0 ·62

~D

c0 f

I

~

I o-:J c... -

0 ·4

ffiCJ~j 0

0

Cw• 0·6

101

where A.t is the largest cross-sectional area of the obstacle perpendicular to v,. ( = surface area of the projection perpendicular to the flow) and C w is dependent on Re = vxDfv, with D · characteristic dimension ·of the obst acle. Hence equation (11.67) has been adapted to the mechanism for the geometric resistance (b). At high values of Re this effect predominates the friction resistance so that Cw is constant. In Table II.4 a survey is given of the C 141 values of a number of obstacles in the turbulent range. In the extreme case that the total momentum flux is used for building up the force on the obstacle, this force is given by F = 4>vPV

Comparing this with equation (1!.67) we see that always Cw

~ 2.

I 1.5.2. Spherical particles

In engineering we often come across situations where liquid or solid particles are dispersed in a fluid phase. The flow resistance determines the velocity at which they move under the influence of an external force (gravity, centrifugal accelerating force, magnetic force or electric force) with respect to the continuous phase. Very v iscous f l ow, Re < I

Release of boundary Ioyer

Mainly friction resistance Turbulent flow , 103 < Re <105 Mainly geometric re sistance

Figure 11.40 Flow pattern around a sphere

The flow field round one single sphere depends on the Reynolds ' number, which is here defined as Re = v,Djv, D being the external diameter (see F igure Il.40). At highly viscous flow (Re < 1) no inertia effects at all occur and the streamlines converge again behind the sphere. In this particular case Stokes' law (which can be derives theoretically) applies: F = 3nrrDv,

or, with the definition of C.., in (II.67):

c

= 2411 = 24 w

p v,D

Re

(Re < I)

At higher values ofRe both the friction resistance (a) a nd the geometric resistance b) play a role. The influence of the former becomes relatively smaller with mcreasing Re number. The geometric resistance is largely determined by the place where the boundary layer is released from the surface. For 103 < Re < 10 5 this release

102 takes place a short distance behind the centre of the sphere where Cw is constant and approximately 0 ·43 (Newton's law of resistance). At Re > 2 x 10 5 the friction boundary layer becomes turbulent~ as a result of which the point where the boundary layer is released shifts to the back so that C..., decreases fairly rapidly. Because of the two mechanisms stated above the drag coefficient Cw is B Xt~il'N.&\j'U'S lu'tf\.'"rMa eX R~ CJPIR;Ysi're ro me rhcn'o n foss ractor J~tbr trow in plpeS.

The stationary rate of fall vs of a sphere in a stagnant medium is calculated by assuming its apparent weight to be equal to the resistance force : 3

1td2 1

2

(;dP(pp- p)g = Cw4 p"fPVs 7t

The solution for v5 is found by solving this equation-together with the relation for C..., as a function of Re. This can only take place analytically in the Stokes range (Re < 1) : '

vs

=

(pp- p)gdp2

1817

(Re < 1)

(IL68)

or, in the Newton range (10 3 < Re < 105 ) : v s

=

1·16j(pP - p)gdP

(II.69)

p

In the intermediate range (1 < Re < 103 ) which is often encountered, only a numerical solution is possible. The force balance is then written as follows:

Cw Re 2 =

~ d!p(p~ 2-

p)g

(II.70)

The right-hand term of this equation is a known constant for a given problem. With the help of Figure II.41 the relevant Renumber can now be found, from which the stationary rate of fall can .be calculated according to : 17 Re

v =-s pdp The relations (11.68), (II.69) and (II.70) are also valid for gas bubbles rising in a liquid if they have a rigid surface. Because of the upward direction of the force, instead of (p P - p) of course (p - Pb) has to be used. Gas bubbles with diameters < 0·8 mm are spherical and the drag coefficients for spheres ca n be applied. Bubbles with diameters 0·8 < db < 1·5 mm form oblate spheroids, the drag coefficient of which is approximately two-thirds of the C.., value of a sphere with the same volume. Gas bubbles in liquids have a rigid surface as long as: d, <

~

(11.71)

103

I

N

"'

0:

1

it

u

I

/

/

/

I

I

// t/

/;

-c... =24 Re

to'~--~~~-L~LU---~~-L~~~--~~~~~L~

I

2

4

6 8 tO

2

4

6

8 100

2

3 4

6

8 103

--Re

Figure 11.41

Drag coefficient and related functions for spherical particles

If surface active agents or dirt are present in the liquid. the mobility of the

surface is decreased and equation (II.71) gives too low values of the critical value of db. Thus, air bubbles in water are rigid if db < 2·6 mm. F or bubbles with a mobile surface the flow resistance is lower. The stationar y rate of rise is then given by : (11.72)

Figure II.42 il!ustrates the rising velocity of air bubbles of various diameters 2ja/g ~p

in water. According to equation (II.72) bubbles with diameters of db =

104

5r---------------------------------·-----,

I Equation ( U . 70)

Rigid

Mobile

-

-

d11 (mrn }

Figure II.42 Velocity of rise of air bubbles in water

have the minimum rate of rise. F or very large bubbles (for air in water if d > 15 mm) equation (II. 72) is simplified to: b (Il.73)

(pgdl >>a)

The rising gas bubbles then have the form of spherical caps. Comparison of equations (If.73) and (li.70) written for half a sphere shows that for these conditions C"' = ~at aRe value of > 12,000 (air in water). This C..., value is comparable with that of a circular fiat plate {Ta ble II.4). ·

11.5.3. Free fall of droplets The friction forces acting on a falling liquid droplet will deform the droplet from a purely spherical to an elonga ted form until the droplet breaks up. The surface tension counteracts this deformation and we can assume roughly that the droplets will no longer break up if this ratio is equal to or smaller than unity: 1t 21

2

friction force

C.., 4 dP1 p 8 v

surface tension

nadP

- - - - -- - ---..,...- - =

C d _

w p{JgV

8v -

2

~ 1

105

In the dimensionless group we recognize the Weber number, the critical va)ue of which should then be of the order of magnitude of: pgdpmuvz

W e c r--

8

~- ~

cw

(J

19

(II.74)

(assuming turbulent conditions). Hence free-falling droplets can only be stable for diameters smaller than or equal to d pmax . Experimental results have shown that We ~ 22 for free fall and We ~ 13 if the droplets are sudden] y exposed to drag forces. 11.5.4. Particles in non-stationary flow All relations described in this paragraph so far apply only if the stationary state bas been att ained. If we consider a free-falling sphere, the non-stationary momentum balance reads :

dv 2 M dt = M g - CwAtpv

(II. 75)

if we assume Cw to be constant over the whole range of velocities. In the stationary state (velocity v.J : M g = CwA!pv;

(11.76)

is valid, as we have discussed in the previous section. Thus equation (II.75) can be simplified to :

v2 dv · M. = M g - fttf g- 2 dt vs Integration between v = 0 at

t =

0 and v.1 t yields: v

1 +'

V5

1- ~ vs

= exp (2gt)

(II. 77)

Vs

In Figure II.43 vjv 5 has been plotted against the dimensionless factor 2gtfvs. It appears that the velocity is stationary if2gtjv5 > 5. For 2gtfvs < 1 the velocity can be represented with good approximation by: 1;

gt

- = -

i.e. v = gt ; the relation applies to free fall in vacuum (no friction losses). This means that up to a timet < v5/ 2g friction loss may be neglected, which justifies our as~umption of a constant drag coefficient in equation (JI.76). Velocities of free fall are often measured by measuring the time for free fall over a distance L. In that way, the m1e an velocity ( v) = L/ t is determined. The

106 1·0

o-s

..

I

0 ·6

/~f vs=7 v5 v

I

0

"" ......

!>

0-4

0 ·2 0

0

12

10

0

16

14

18

2gl

v

Figure 0 .43 N on-statjonary fall of spherical particles

mean velocity can be calculated as :

which yields with equation (1177) :

< >=

-

.V

Vs

-

Vs

2gt

1

n

{

2

+ exp (

2

~t)

Vs

+ exp (4

2

gt)'}

Vs

-

(II.78)

This function is also shown in Figure II.43. lt can be used to ·c alculate·(via trial and error, see problem 4) the stationary velocity of free fall from a measured mean velocity.

II 5.5. Rate of sedimentation of a swarm of particles The rate of sedimentation of a swarm of equally large particles in a liquid is lower than that of one separate particle. In a given system of particles and liquid this rate of sedimentation (v.Js appears to depend only on the fraction 4> of the volume which is taken up by the particles. This constant is also connected with the porosity e: this is the volume fraction of the continuous phase so that e = 1 - qJ. There a re various theories about the relation between (vs)s and o or e. A well-usable empirical relation is that of Richardson and Zaki : (vJs = v~e

11

In this equation v~ is the rate of settling of one single particle of the system · question (e = 1) and n a value. dependent on v~4/v = Re~ and on the rei · of the particle diameter and the diameter of the vessel D, in which they

107

For dJD 1 < 0·1 the authors give the following values of n : Re~

<0·1 465

n

•·

1

10

100

>500

4·35

3·53

2·80

2·39

Equation (II. 79) has the advantage that the exponent n is given as a function of a Re number defined for the free settling velocity of a single particle and not, as .h appens in most liteliture, as a function of a Re number based ori the un· known swarm velocity. · ·

11.5.6. Cylinders perpendicular to the direction offlow For an infinite cylinder perpendicular to the direction of flow the picture of Cw = f(Re) is analogous to that for spheres. At Re < 10- 1 , Cw"' 1/Re and for Re > 103 , C.w :::=:: 1·2. If the cylinder is shortened the flow can also pass the ends and Cw becomes lower (Cw = 0·62 for L/D = 1 at Re > 103 ). If the cylinder is not given~ round but a streamlined cross·section, Cw decreases considerably to values between 0·01 and 0 ·1. If, as in another extreme case, it is flattened (perpendicular to the direction of flow), then C.., ~ 2 for LID = co. In practice, banks of tubes are often used as heating-or cooling--elements in a liquid or gas flow. In flow perpendicular to the tubes two different ways of arrangemen t can be distinguished : (a) in line and (b) staggered (see Figure II.44).

\\\\\\\\\\\\\~

- -ooool 0000

v0 -

0 0 0 0 0 0 0

0

-oooo

F .

0.\\'\\m\.\%'

{0 )

-~""6\\ll'b 0 0 \\\\\'

v0 0 0 0 0 0 -0 0 0 . \\\\'V\\\\\\

F

( b )

Figure U.44 Flow perpendicular to a bank of pipes : (a) in line, (b) staggered

In case (a) most pipes are in the 'wake' of thei( .p redecessor, in case (b) they are not. Owing to this shielding the resistance of arrangement (a) is lower than that of (b) and so is the heat transfer. For case (b), as a rule of thumb, we can use the fact that the resistance coefficient of a tube (related to the superficial velocity) in the range 30 < Re < 2000 is about 1·5 times as high as for a single t ube in the flow if the distance between two neighbouring tubes equals the tube diameter. Naturally it also occurs quite often that, for example, in the case of heat exchang~rs the resistance for flow aiong a bank of tubes must be calculated. For turbulent flow use is then made of th~ Fanning equa ti on (II.35) and the hydraulic diameter.

108

ll.5.7. Problems 1. What is the maximum diameter of a rain droplet at 20°C?

Answer : d P = 6·6 mm *2. When driving his car {1·6 m wide, 1·3 m high, 9 50 kg) at a steady 105 kmjh on a highway on a windless day, Klaus notices that, if he pushes the clutch down, it takes 5 seconds for the car to decelerate to 95 kmfh. (a) Estimate the maximum value of the drag coefficient. (b) Driving steadily, which percentage of the petrol used (101/ 100 km) is converted into mechanical energy, if the heat of combustion is 11·000 kcalfkg (p = 700 kgfm 3 )? Answer : (a) Cw < 0·55 (b) l 5·5 per cent 3. Two metal spheres of equal weight but of different diameter fall through air. Calculate the relationship of their stationary rate of fall if the ratio of their diameter is 3 and the flow round the spheres is dynamically similar. Is this supposition correct?

Answer :

1; yes

4. The ' Euromasr tower in Rotterdam bas a height of 104m. It takes a cherry (diameter 1·7 em) 6·5 seconds to fall from the top of the tower to the pleasant park below. (a) What would be the velocity of stationary free fall? (b) What is the drag coefficient? Answer : (a) 20 m/s;

(b) Cw

= 045

5. According to a newspaper report the air 1n Mexico City (situated at 2200 m above sea level) contains per volume unit 25 per cent Jess oxygen than the air at sea level. (a) Ascertain that this statement is correct by calculating from a balance of forces the pressure as a function of the height above sea level.

Given : Air is an ideal gas. Gas constant R = 8310 J/kmol °K Air temperature = 2rc 'Molecular weight' of air = 30 kgfkmol This slighter air density in Mexico City causes the sprinters to meet with lower air resistance. (b) In what time will a sprinter tneoretically be able to cover the 100m in Mexico City because oftbis lower air resistance, if he does the 100m at sea level in 10 s?

Assume:

(i) that both in Mexico City and at sea level he develops the same power.

109 (ii) that he runs at constant speed. (iii) that the time necessary to pick up speed is negligible.

(c) His actual time in Me::dco City is. however. likewise 10 s. What is your conclusion? Answer : (b) 9·2 s (c) Assumptions (ii) and (iii) are wrong.

6. The gas bubbles formed at the bottom of a glass of beer are released as soon as their diameter is ca. 1 mm. (a) Calculate their stationary rate of rise. (b) How many times more rapidly would an equally large water drop faJJ in air?

Answer : (a) 0·105 mj s; (b) 35 times *7. A horizon tal air duct with rectangular cross-section (H x B = 30 x 30 em 2 ; L = lOm;xj Dh = 2 x 10- 3 ) containsanairbeatingelement. This element consists of six rows of fifteen tubes placed transversely to the direction of flow. The rows are installed in a staggered position (Figure fl.27b). The external diameter of the tubes is D = 2 em, the mean temperature in the

duct is 80°Cand the mean pressure is 1·3 x 105 N/m2 . Calculate the pressure

drop in this duct at a mean air velocity of 20 m/s.

Answer: 2860 N j m 2 *8. Show that John indeed gave the right answer to the problem of the gas grenade treated at the beginning of this paragraph. Comments on problems

Problem 2 Klaus' data collected in a quiet hour on the ElO motorway will help us to calculate a drag coefficient that includes not only air resistance effects but also friction losses. The actual C..., value will therefore be smaller than the ·value we calculate from the unsteady-state momentum balance: dv -Mdt Now, from the measured data at f

=

= Cw A-21 pzr' 100 kmfh:

dv = 10 kmfh

dr

5 min

= 0-555 mj s2

and we find : C"' malt

950

=

2·08

X

X

0·555

0-60

X

772

= O·SS

110 If we assume the friction energy loss of the rolling wheels on the road to be of

tbe same order of magnitude as the energy loss due to air friction. we find C\9 ~ 0-30, which compares well with the value stated in Table II.4. The total energy content of 10 litres of petrol is: 10

X

10-

3

.

700 . 11,000. 4-2

X

103 = 327

X

10 5 Joule

and the mechanical energy for moving the car 100 km : dv L =50 x 10 5 Joule dt

- M-

Thus the percentage transformed into mechanical energy is -15·5 per cent. Problem 7

The pressure drop in the channel is caused by friction along the wall and by the flow resistance of the pipes, and we can assume that these two resistances are additive. The pressure drop in the empty channel is then found to be (using Dh = 0·3 m, Reh - 3 x 105 , 4f = 0·024, equation Il.35): tlp 1

= 200 N/ m 2

The drag coefficient for one single pipe would be (with Re II.4) : Cw = 1·2

=2

x 104, Table

Thus, for staggered tubes C~ = 1·5 C..., = 1·8. Thus the force on one pipe is: F = CwAtPV 2 = 2·66 N

The force on all pipes together is then drop:

F.ot = 6·15 F = 239 N

and the pressure

t:.p 2 = 239/0·09 = 2660 N j m 2

John and the tear gas grenades An unsteady-state momentum balance in the x-direction for this case (see the figure) reads: Problem 8

dvx l 2 M dt =_ -C w2 pv x A

l -:... ---- v/ ::0...

II

/

/

-----'

I I

I I I I

'

x.-

( 1)

"'-..

~

111 which yields after integration between 1

t =

0, vxo and t, vx:

l

(2)

Now vx = dx/ dl and after substituting vx from equation (2) and integration between t = 0, x = 0 and te, Xe we find :

x

2M In ( 1 C...,pAv,. 0 c~) + 2M

=

(3)

C wP A

e

Now, for steady-state free fall :

and thus: (4)

2MfCwpA = v;fg

Considering now the velocity component in the y-direction (and neglecting friction losses because g » Cw{pr.lj2M)) we have : d v1

dr

=-

g ; V 11 =

dy

dt = -

gt

and

y

=v

110 t

-

!-gt 2

So, for y = 0 :

te

= 2vy0 /g

Introducing this expression and equation (4) into equation (3)' we obtain~ xe =

v; In ( 1 + 2~~~Vyo) g

vs

Now : 2 .

v"'0 Vy 0 = v0 sm a cos ex

which is maximum for rx· · 45°, i.e. v.xo = v>' 0 = 0·71 v0 , and so the maximum distance is :

If you , as d id the poor inventor, neglect the air friction, you find for the maximum distance: Xe

which yields for

Xemax

= 202m!

=

2

2

.

-V0 SID 0: COS f1.

g

112 11.6. Flow through beds of particles The police doaor rose and stared chac the man had died by town gas poisoning not longer than 10 hours ago. 'Remarkable', John thought. The room where they found the body was locked from Ihe inside and there was no smell of gas at all, although concentrations above one-thousandth of the lethal level can be smelt quite easily. The gas valve was open but due to an explosion at the gas works the supply ceased 8 hours before. A small fan in one of the windows pressed air into che room so thaz lhe pressure was 80 mm water column higher than outside the room. The room had a volume of 80 m 3 and t'J-.ere was about 50m2 of wall (20 em thick'h the porosity of the material being approximately 20 per cent and the particles which constituted most of the material having a diameter of ca. 0-3 mm. Lighting a cigarette, John concluded that the man had died at another place.

The fiow through a bed of particles is of gre~t importance for all branches of technology; in chemical engineering for filtrationycatalytic processes, separation processes in columns with packing materials, etc., and also for petroleum production, soil mechanics and hydraulics. A great deal of theoretical and experimental work has been done by various groups. Outwardly the various formulae we encounter often differ, but on second thoughts they have much in common Two types of particle bed are of great technical imponance : the fixed bed and the fluidized bed. ln the former bed the particles rest on each other and are not moved by the fluid. If in this bed the velocity of the fluid js increased, a situation is created in which the force on the bed becomes equal to the apparent weight of the particles. The flow lifts the bed, as a result of which the particles become suspended. This is called the fluidized state which is characterized by a thorough mixing, especially of the particl~s. The result is that temperature- and/ or concentration differences in the fluidized bed are rapidly eliminated, so that in the bed uniform conditions prevail which are often desired. If the velocity of the fluid is increased still further. the particles are ultimately blown out of the bed and we enter the field of pneumatic transport. We shall now discuss some essential points of the flow in the two types.

II.6.1. Fixed bed This bed may be considered as a collection of particles and also a s a collection of interconnected winding channels. We stick to the former description in order to calculate the energy dissipation per unit of mass in a bed of spherical particles

as a function of the superficial velocity v0 related to the cross-section of the tube which forms the boundary of the bed. The actual velocity (v) in the bed is of course higher than v0 , notably v0 = e ( v) if e again represents the volume fraction voids (the porosity) of the bed (for beds of spheres, in practice, 0·35 < s < 0·45). The train of thought is now

as follows : we calculate the energy which is dissipated around one particle in a

113 ftow of velocity ( v) ; this energy is multiplied b y the total number of particles in the bed in order to obtain the constant ErlPm, ie. the total energy which is dissipated in the bed (see also equation II.41). To this end we base our calculations on the definition equation for Cw 01.67) to determine the force which the flow exerts on one particle :

The drag coefficient in this relation is self-evidently another from that which is found for one sphere in an infinite fluid. From the force F we can now calculate, by multiplication by (v), the energy whic4 is dissipated round one sphere. The total number of spheres in tbe bed is subsequently calcu1ated from the volume the spheres take up in the bed, (1 - s) AL, where Lis the height of the bed So:

ErlPm

= ErrPVoA =

(1 - e)AL

~d3 6 p

1t H

Cw dp,;p(v) 4

3

or: _J.

Err -

.!. 2 2Cw2Vo

1-c;L t3 d p

At a sufficiently high velocity in the bed (high Renumber) the flow is turbulent and, as we know, Cw is constant. Ergun found for this range Cw = 2·3. At low velocities in the bed the flow is no longer turbulent, but shows eddy currents. In this region (which is also called ·laminary') Cw is expected to be inversely proportional to the Re n~mber. The difficulty here is to define the Re number which is characteristic of this flow. The .c haracteristi.c velocity is, of course, (v). However, the characteristic cross-sectional dimension oftheflow between the particles is not so much -d etermined by the particle diameter as by the space remaining between the particles. Now it appears that the hydraulic radius A/ S of the cavities between the particles can be used for correlating Cw as a function of Re11 (= 4p( v) R nf'1S) for all packed beds. This hydraulic radius R11 is connected withe a~d the specific surface area of the particles Sp: A

S=

surface o.f cross-section of cavity volume cavity circumference cavity - wall surface cavity

volume cavitiesfm 3 = wall surface cavitiesj m 3

=

porosity (m 3 jm 3 ) specific wall surface

=

s Su

The specific surface area is, in turn, again a function of e and the particle diameter, namely: _ 6(1 - e)AL nd2 _1_ _ 6(1 - e)

S., -

nd3p

PAL -

dp

114

so that : Re&= ~ PVodp

3 '1(1 - e)

Now it appears that for Re,. < 1, Cw = 152/Re,.. This relation is known as the Carman formula, but is also named after Blake and Kozeny. According to Ergun a good description of the flow resistance in the entire Re range is obtained by adding the relations for laminar and turbulent conditions, with the result : Err v~ 1 - e ( v · ) -=- = 170- (1- e)+ 1-75 3 pL L dP e· v0 d11

· fj,p

(11.80)

It appears that the flow may be considered as tl,lrbulent if Reh > 700. The · constant v 0 dJv is again a Reynolds number and is often used in the case of flow through fixed beds, although somewhat incorrectly. Restrictions in the use of equation (11.80) are that the particles must not deviate too much from the sphere shape and that the diameter of the particles must be considerably smaller than the diameter of the bed ( < 210 ) .

For a number offorms of non·spberical particles Efr has also been determined In such a case, instead of d, in equation (1180) tbe diameter of a sphere having the same volume/ diameter ratio a s the combined particles in the bed should be used, and the pressure drop so found should be multiplied by a correction factor k. In Table 11.5 the correction factors for a number of agricultural and horticultural products are given.

a,

Table D.S Factor k with which equation (11.80) underestimates Err for some agricultural crops (purified pro~ ducts, •random' arrangement)

Product Peas Rapeseed Potat~

k

1·05 beans

Clover seed. wheat

Rye, summer barley

Maize Sugar beets. carrots Oats

1·2 14

1·7

2·7

3·2 3·5 3·8

F or asparagus rather divergent k values are found , just as, for example, for flow through yarn packets and yarn filters. Here it is important to find out whether these cylinder-shaped bodies lie mainly in the direction of flow (k ~ 0·75) or mainly perpendicular to the flow (k :::::: 1·1). In these cases there is hardly ever a question of 'random' arrangement.

115

IJ.62. Filtration through a bed ofparticies

Equation (II.80) can also be used for an idealized description of the filtration process. which is practically always carried out under laminar flow conditions. Rearrangement yields, if Vis the amount of filtrate at timet: s3

dV

vA-A. --0 - '+' II - dt - 170(1 - s) 2

~pAd 2

Yf L

P

(ll.80a)

During filtration, the suspended particles will collect on the filter medium (e.g. filter paper, cloth) and form a cake. If we neglect the flow resistance of the filter medium, the term Lin equation (II.80a) is t he thickness of the filter cake, which is a function of the weight fraction f of solids in the suspension and the amount of filtrate V already produced. A material balance shows: LA(l - e)pp = fp,V

(11.81)

Substituting L into equation (II.80a) and integrating between V = 0 at t = 0 and V, t, we find for the amount of filtrate at any time :

v-- dsr4

j

e3 flppP --t 85(1 - e) rJPd

(11.82)

and for the rate of filtration: A.

'+'u =

dV

dt =

dPA

J

3

e ll.ppP 340(1 - e) Yfp,Jt

(II.83)

The derivation of equations (II.82) and (II.83) was based on the following assumptions: (a) constant e, i.e. the filter cake is incompressible; (b) constant ~p. i.e. the filtration is carried out at constant pressure ; (c) spherical particles; and (d) no How resistance of !tlter medium. In practice, these assumptions are not always permissible. For a gjven system the influence of the flow resistance of the filter medium ca n be allowed fo r by adding, in equation (Il.80a), to the resistance of the filter cake which is represented by: (1!. 84)

a resistance Rm for the filter medium. We o btain dV


~pA

dt = 17(Rc1 + Rm)

(II.85)

116

Substituting Lin equation (11.84) from equation (11.81) we find: R1 = c

v 1A

pr7 170(1 - .s) = Pp e3

fv R A

c

With this expression equation (11.85) becomes:

-~. = dV =

Integration with V

(

dt

'f'v

.1pA V ·

11 f .F.R, ,-J

= 0 at t = 0 yields:

V)

-1 ( 2 A

2

Y) R

fR + ( c

+ Rm

(11.86)

)

A

m

6pr(

(11.87)

=-l

or : (II.87a)

Thus a plot of (A/ V)t versus J1A should yield a straight line with slope -!
Using the Ergun equation (11.80) it follows: t,;20 .d2

' P v2

=

{

v 170- -(l -c)+ 1-75 vOt-d p

}-1

gpd3

_ 2_P(p - p)s3 r~

'I

p

(II.88)

The maximal velocity at which the fluidized bed can just exist occurs in a highly dilute bed (e--+ 1) and is equal to the stationary rate of fan vs of a single particle. The expansion of the bed as a function of v 0 is more difficult to predict because the flow condition in the fluidized bed is in principle unstable. A gas stream, particularly. tends to move rapidly in large bubbles through the bed (aggregative flujdization). A bed fluidized with liquid (provided its length is not

117

much greater than its diameter) is, as a rule, fairly uniform (particulate fluidization). In the latter case the bed expansion can be calculated by using the relation of Richardson and Zaki for the rate ·of settling of a swarm ·of particles . (paragraph 11.5.5). It should be borne in mind that in the fluidized bed the relative velocity between the fluid and the particles is (v) = v0 jc. The relative velocity between the phases in a. settling· test (in a vessel with a. bottom) must be calculated from the velocity of the swarm (vs)s. The settling particles displace liquid upwards, so that on settling the relative velocity vr is equal to the velocity of the swarm plus the velocity of the liquid. Since the volume of particles which settles per unit of time equals the volume of liquid which is displaced per unit of time, the velocity of the liquid on settling is equal to (1 - e)(v.s)Je. So, the relative velocity between fluid and particles is:

v,

= (vs)s

1- e (vs)s . ·- (vs)s = e e

+-

Hence the expansion of a uniformly fluidized bed is given by Richardson and Zaki as follows :

/
e = \ v; }

_

or e- ~Vol -

l /n

v~

I 1.6.4. Problems

*1.

Equation (II.80) can, apart from numerical factors, also be derived from the Fanning equation (II.35) by regarding the bed as a complex of irregular channels with total length Land hydraulic diameter 4Aj S. Check this and calculate that for the bed applies: 4fL'/L = 152/Reh for Reh < 1 and 4fL'jL = 2·3 for Reh > 700.

*2. In a vertical tube (Di = 10 em, L = 25 em) almost spherical anthracite particles are fluidized with air. The particle size distributi~n is :

dp(mm)

Mass( %)

from

to

1·00 0 -84 0 ·71 0·59 0·50

(}84 (}71 (}59 (}50 0·42

20 29

23 25 3

The densit y of the anthracite (measured with a pyknometer with mercury) is 1970 kgjm 3 , the temperature is 32°C and the pressure above the bed is

105 N/ m1 . (a) At which mass flow rate does the fluidization start?

118

(b) At which mass flow rate are the particles with 0.42 mm diameter blown

out of the bed? (c) What is therefore the range of mass flow rates suitable for fluidization? Answer : (a) 0.40 kg,lm 2 s (b) 3·5 kgfm 2 s (c) 1 < c/>';, < 3 kgfm2 s 3. A suspension is filtered at a constant pressure difference 6p over the filter. At t = 0 the filtration begins (the cake thickness is then 0) and after a time t a volume V has been filtered . The following results are fo und :

1·00 1·4l 1·73

!lp

t (s)

(at)

20

0·2

40

Q-2 Q-2

60

Predict the filtered volume after 20 s if the test is repeated under the following

conditions : (a) at a pressure difference of 0-4 at ; (b) with a second material, the particles of which are of the same form as the first but with the linear dimensions smaller by a factor of 2 (at 6p = 0·2at). Answer : (a) 1·41 m 3 {b) 0·5 m 3

4. A mixture of oil and bleaching earth is filtered over a plate and frame press of 20m 2 filtering surface. After 4 minutes 1m3 and after 10 minutes 2m3 of filtrate are collected. How long does it take to filter one charge of 20m 3 ?

Answer :

t

= 460 min.

5. At reguJar intervals the filter press of problem 4 must be cleaned. This cleaning operation takes t 0 minutes, irrespective of the amount of cake to be removed. If the flow resistance of the filter cake is neglected (R m « RJ, how long should the filtration last in order to obtain the maximum filtration capacity over a complete cycle (filtration and cleaning time)? Answer : t

= t0

*6. Show that John reached the right conclusion in the problem stated at the beginning of this paragraph.

l19 Comments on problems Problem I With equation (11.35) we have: L'

!J.p = 4f-!p( v) z

D,

4A

S

Dh =

4edP

Vo

= 6{1 - e)'

and

(v) = -

e

_ 2 pv 0 dP Reh - - ~...;:;........!=--3 '1(1 - c:)

..,-e find :

and :

llp = Err pL L

=

4JL'v~(l - e)3 Ldpe 34

Comparing th is with the expression developed in paragraph 1!.6. 1 : Err-~ 1 zl- e L - 2 w"'iVo d e3 p

and utilizing the information abou t -C.,.. given there we find : L'

4/-

L

152

= -(Reh

Reh

L'

< 1) and 4fL

= 2·3 (Reh >

700}

Problem 2 Since we are dealing here with a mixture of particles of different size we must be careful which diameter to use in equation (II.88). The derivation of this and of the Ergun equation was based on the specific wall surface Sp given as: S

"

=

surface of cavity yolume of particles

= 6(1

- e) dP

For a mixture of particles of different size we must therefore use :

a= P

6

volume of particles = surface of particles

X

L nidt = I

Mi

L nflf L M

1

d;

where M ; = mass fraction of particles with diameter d;. Thus we find :

ap=

1·0 0·23

0·20 0·29 0·25 0·03 0·92 + 0·78 + 0·65 + 0·55 + 0-46

= 0- 69 mm

120

Estimating c ~ 0-4, we now find from equation (11.88) (17 p = 1·3

kgfm 3}:

rf>':..o;

=

17 x 10- 6 Ns/m2,

= PV 0 ; = 0·40 kgjm 2 s

From equation (II. 70) we find for the steady-state velocity of free fall of the smallest particles v = 2·65 m/s ; thus ':na11 = 3·5 kgfm 2 s. Problem 6

John and the body in the locked room

From equation (11.80) we find the flow rate of air leaking through the wall to be (dp = 800 N /m 2 , laminar flow) : {j>,.

= Vo A =

7·9

X

lO - 3 m3 fs

Assuming the gas in the room to be perfectly mixed the material balance : de

vdt=- c¢ 11

leads, after integration between t

= 0, c 0 and t, c to : c

ln-

c0

=

(j)v - -t

V

Thus, after 8 hours, the gas concentration should still be one-seventeenth of the lethal level and John should be abJe to smell the gas.

H. 7. Stining and mixing John saw chat the only man on duty was very busy crying to speed up the circulation of tile people past the crashed car. He saw a parallel to a mechanical .stirrer and remembered that increasing the speed by only 25 per cent would increase che frictional power by a factor cwo. Therefore, he cold him not to pay too much attention to the crowd's speed, but, now it Juu:l started moving, to concentrate on keeping it flowing smoothly .

Mixing and stirring are two conceptions which partly overlap. Mixing is the uniform distribution to a certain scale of inhomogeneities over a certain space. M ixing can be attained by stirring. The object of stirring can also be to promote heat or mass transfers or to make a suspension or emulsion. Three transport mechanisms generally occur during mixing : (a) convection, i.e. the material present is moved ; (b) dispersion, i.e. different parts of the material move at different speeds; and (c) diffusion.

11.7.1. Types of stirrer and flow patterns Figure II.45 shows a few types of stirrer which are applied in practice. T ypes (a) and (b) will induce radial flow, type (c) axial flow, whereas type (d) will induce

121

I

I

{0 )

( b)

lc)

(d)

Flat blade turbtne

Flat blade paddle

Propeller

Inclined blade turbine

Figure 11.45 Various types of impellers

flow in both directions. Figure II.46 shows the flow patterns obtained for two type s or impellers. I

I I

------ I ------

{I

I

Axial flow type impeller

Radial flow type impeller

Figure II.46 Flow patterns of impellers

In the circulation mainstream of radial tlow type impellers fo_UI different zones can be distinguished :

(i) the radial liquid flow leaving the stirrer blades ; (ii) the vertical flow along the wall ; {iii) the horizontal flow back to the stirrer shaft ; and (iv) the axial flow along the shaft to the stirrer. Measurements have shown that in the radial liquid flow leaving the stirrer blades : rfr

= constant

being the distance to the Stirrer Shaft and Vr the radial Velocity Of the liquid flow at position r , averaged over the time and stirrer height. The flow pattern of axial flow type impellers is less distinct. The downward directed liquid flow (these stirrers are usual1y made to pump downwards) changes direct ion at the bottom and flows upwards along the wall of the vessel. T

122

Here again a circulation pattern can clearly be distinguished in the flow pattern. A propeller stirrer. however. causes not only axial but also rather strong radial and tangential ftow in the vessel. The flow pattern of a propeller stirrer is therefore less systematic compared with that of a radially pumping stirrer. When a draft tube is placed around a propeller stirrer the flow pattern in the vessel becomes more systematic. When the vessel has not been fitted with baffles, vortexing occurs at high velocities so that gas can be sucked into the liquid. Baffles counteract this effect Sufficient activity of the baffles is attained when the product of number times width of the baffles divided by the vessel diameter is ~0·4. This situation (mostly reached by applying 4 baffles of 0·1 D width) is called 'fully baffled'.

11.7.2. Power consumption With dimensional analysis it can be derived that for geometrically similar situations the power number P 0 must be a function of the Reynolds' number Re related to the stirrer and the Froude number Fr: 2

2

P0 = P3 d 5 = f(Re Fr) = f (pnd - ; -n d) pn '7 g The Froude number only plays a role when vortexing occurs at the liquid surface. Figure 11.47 shows a typical power curve for a stirrer. The lower power consumption in the tank without baffles is caused because air is sucked in, as a result of which the density of the mixture decreases. The shape of the power curve ofbaffied systems may be physically interpreted as follows. If we consider one of the impeller blades moving through a liquid. then the force exerted on an object in a stream of fluid moving at mean velocity v is given by :

where A is the cross·sectional area of the object and Cw the drag coefficient Since A - d 2 , and v corresponds to the tip speed of the blade (xnd) multiplied by a slip factor, the force on the blade is proportional to pn2 d4.. The powe: needed to overcome this force is equal to force x arm x number of revolutions per unit time, or -Fnd. Therefore, the power number P0 = P/ pn3 tP is proportional to the drag coefficient Cw. The shape of the power curve is similar to that of the drag coefficient versus the R eynolds number. F or Re > 104 the power number P0 becomes a constant. In this turbulent region P0 is only a function of the geometry of the system. At values of Re < 10, P 0 is proportional to 1/ Re, thus the power consumption is given b y :

P = P0 pn 3 d 5

(Re > 104) (Re < 10)

Table 11.6 gives a survey of some approximate values for P0 and impellers.

P'o for variolA

123 Lomtnor--------Tronsition region - - - - - --Turbulent

150 With baffles

2

Wtthout baffles 10 2

10

103

- - - pnd2./"f/

Power consumption curve for turbine stirrer

Figure ll.47

The energy consumption for suspending particles in a vessel by means of a stirrer depends on the minimal stirring speed n* at which the particles no longer settle. For this minimal rpm applies :

{g(p.;dp)dr = c(~r~·-13

(11.90)

where dis the stirrer diameter,ftbe weight percentage of the particles and C a constant of the order 1 (for turbines approximately 1·5) which depends on the geometry. The power consumption then is: p = Popn*3ds

where pis the mean specific gravity of the suspension. Table II.6 Rough estimations of P0 and K Impeller

Re < lO

Re > 1~

K,

Po

Flat blade turbine six blades, width/ diameter-1 / 5 four blades. djtto

70 70

6 4·5

0·6

Flat blade paddle six blades, width/diameter- l/ 6-1/8

70

3

1·3

pitch = diameter pitch = two times diameter

42 42

01.0·3}

0·4-D·S

Inclined blade turbine six blades, width/ diameter - 1/8

70

1·5

0·8

Propeller, three blades

1·3

.·

123 Lominar--...o-----Tronsitioo region - - - - - -Turbulent

Wi th baffles

W1thout baffles

10 2

10 .

- --

10 4

10~ pnd

10 5

2

!')

Power consumption curve for turbine stirrer

Figure U.47

The energy consumption for suspending particles in a vessel by means of a stirrer depends on the minimal stirring speed n* at which the particles no longer settle. For this minimal rpm applies:

{g(p:;dp)dr

=

c(~)*j0•!3

(1190)

where dis the stirrer diameter,/ the weight percentage of the particles and C a constant of the order I (for turbines approximately 1·5) which depends on the geometry. The power consumption then is:· p = Popn*3ds

where pis the mean specific gravity of th e suspension. Table II.6

Rough estimations of P0 and K

Impeller

Re< 10 Po

Re > 104 Po

Re > JQ4

Flat blade turbine six blades. width/diameter-l/5 four blades, c:litto

70 70

6

4-5

1·3 0·6

Flat blade paddle six blades, widtb/diameter- 1}6- l/S

70

3

1·3

Propeller, three blades pitch = diameter pitch = two times diameter

42 42

0·3} 1.()

0·4-0·5

Inclined blade turbine six blades, width/diameter - l/8

70

1·5

0·8

K,

.·

124 I1.7.3. Pumping capacity and mixing time If we consider only the biggest velocity component of the liquid flowing from the impeller, the pumping capacity is given by the product of the mean velocity of the liquid leaving the impeller and the area described by the blade tips. For the propeller this leads to (d - diameter of impeller) :

whereas for a turbine we find (W = width of blade): v = iindW

If we assume ii to be proportional to the tip speed of the blades (1tnd) we find for a given geometry (for turbines W ,_ d) : · (11.91)

In the turbulent region. K v is a constant for a given impeller geometry. Some values of K Dare listed in Table TI.6. In many chemical engineering operations, e.g. the blending of gasoline; the mixing of reactants in a chemical reactor, etc., knowledge of the time necessary to reach homogeneity may be very important. Usually the mixing time is defined as the time needed for homogenization to the molecular scale. Since measurements on this scale are beyond experimental capabilities, an investigator is only able to measure the terminal mixing time, being the time required to attain homogeneity on the scale of observation. In the turbulent region, for many types of impellers the relation (tm = mixing time) : ntm = constant holds, for geometrically similar situations. The mixing time is about four times the circulation time of the vessel contents. Thus, if no dead spaces are present:

v

tm ~ 4 tPc

is a good estimation of the mixing time. 1!.7.4. Problems 1. A catalyst precipitation process has been developed on a pilot plant scale (diameter reactor 0·25 m). It appeared tha t using a standard flat blade turbine (six blades, fully baffled tank), a minimum stirring speed of 700 rpm was required to obtain a satisfactory product (p liquid ~ 1300 kglm 3 ) . A geometrically similar factory scale reactor (0·6 m diameter) has. been designed. Calculate which stirring speed should be applied and what the power consumption would be if we keep constant: (a) the power input per unit volume; (b) the pumping capacity per unit volume;

125 (c) the mixing time; (d) the size of the partic1es that are only just not suspended in the liquid

Answer: (a) (b) (c) (d)

391 rpm, 685 W 700 rpm, 4000 W 700 rpm. 4000 W 391 rpm, 685 w

2. Show that John reached the right conclusion in the problem described at the beginning of this paragraph.

ll.8. Residence time distribution The sergeant on dury in front of the supermarket informed John that the m.an they we,·e looking for had entered the

supermarket together with his wife 15 minutes ago. He said tluu they seemed to be doing their shopping for the next week and the man looked relaxed. John, knowing from his own experience that shopping for the week in this supermarket cook on the average about 30 niinutes, mode a swift calculation. He decided that he had a probability of 60 per cent to catch the man if he searched the supermarket directly with his platoon. He therefore ordered them accordingly, but they failed to find the man. At night, he read the following chapter and came to the conclusion that the probability ofan event and the actual course ofan evenl are, ash£ should know, two different things.

Flow through an apparatus practically always results in a certain residence time distribution. Fluid elements entering the apparatus simultaneously will generally leave it at different times. This distribution in residence time can be described quantitatively with two distribution functions which are closely related, the F and the E functions. Although residence time distribution is of great importance to the design of process equipment, we will restrict our discussion to the principle. A more comprehensive treatment of non-ideal flow, including its influence on the design for chemical reactions, is given by Levenspiel (Chemical Reaction Engineering). Il.8.1. The F function I< 0 , C

=0

I ~ 0, C = Co

Figure D .48 Measurement of the F curve

Somphng point

tracer cone. C

Consider the continuous flow system shown in Figure Il.48. From a time t = .o on, we replace the feed stream by a fluid containing a tracer material (e.g. colouring agent) at a concentration C 0 . If we measure th e mean concentra-

126

T irne 1 -

Ftgme U.49 The F curve

tion of the tracer material in the effluent stream·we will find that the tracer concentration (and also
F(t} = 0

for

t

=0

(i.e. no fluid remains infinitely in the apparatus) or, formulated more precisely, F(t) = 1

for

= oc

l

(i.e. no fluid remains infinitely in the apparatus) or, formulated more precisely. mathematically : lim (1 - F(t))

=0

(i.e. all fluid originally present in the apparatus will have left it ultimately). Funhermore. from a mass balance we find:

foco f/J

0

(Co - ( C) ) dt

=

VC 0

and using F = C/ C 0 we obtain : ::o

f

0

(1 - f) dt ·

v =-

cPv

= 't'

(Il.92 t

127

t--

Figure 11.50

Determinat ion o f T

The mean residence time thus can be determined from experimentally found F(t) curves, as illustrated by Figure II. 50. A line at t = t 1 = constant creates the areas A 1 , A 2 and A 3 such that :

rl = A t + A3

Because: .T

=

Lro (1 -

F) dt

=

A2

+ A3

this line represents the mean residence time r if A 1 equals A 2 • As a check on the measurements, r can be calculated as -r = V flf>v from measured values of V and f/J,. Often, instead of the time r. a dimensionless time parameter 8 = t/-r is used and we will apply this quantity in all following considerations. It will be evident that for the F(8) function equation (II.92} becomes: f oCX) (1 - F) d8 = 1

I 1.8.2. The E function The residence time distribution function of the fluid in the apparatus is given by the E function, which therefore represents the age distribution of the fluid leaving the vessel. Figure II.S 1 shows a typical E curve. The fraction of material in the exit stream with a residence time between (J and (J + d(J is given by E d8, from which fol1ows that :

f oO() E d8

=1

The fraction of effiuent material younger than 8 1 ) which is also represented by the F function, is then given by:

(11.93)

127

t--Figure 11.50 Determination of -r

The mean residence time thus can be determined from experimentally found EU) curves, as illustrated by Figure II.50. A line at t = t 1 = constant creates the areas A 1 , A 2 and A 3 such that:

Because :

this line represents the mean residence time r if A 1 equals A 2 • As a check on the measurements, t can be calculated as r = V /4>., from measured values of ;·and tPv· Often, instead of the time c, a dimensionless time parameter 8 = t/-c is used and we will apply this quantity in all following considerations. It will be evident that for the F(8) function equation (II.92) becomes:

f

00

(1- F )d8 = 1

0

11.8.2. The E function The residence time distripution function of the fluid in the apparatus is given by the E function, which therefore represents the age distribution of the fluid leaving the vessel. Figure II.51 shows a typical E curve. The fraction of material in the exit stream with a residence time between e and 8 + d8 is given byE d8, from which follows that:

Jo:o Ed6 = 1 The fraction of efiluent material younger than 8 1 , which is also represented by the F function, is then given by:

(11.93)

128

G

Figure 0.51 The E curve

and the fraction older than 8 1 is :

1 - F(8 1 )

ra)

= J~

8t

Ed(}

=l

r8,

- J ~ E dO 0

The E curve can be measured by injecting a tracer material into a flow system for a very short time and by measuring its concentration as a function of time at a downstream point. A typical plot of these concentrations versus time is shown in Figure II .52. The area under the curve :

LCX) ( C ) dt

~ L c L\t

(11.94)

Area ; 0 : total amount of lf'ljected

tracer materlol

t -

Figure D.52 signal

Response curve to a tracer injection

represents the total amount o f uacer substance injected, Q. In order to find E(t), this area must be unity ; thus all concentration readings must be divided by the total amount of tracer injected:

( C)

( C)

E(r) =

s~ Cdt

=

( C)

Q ~ L ( C ) lit

(11.95}

129

The mean residence time can then be calculated via : t

=

co

J 0

C)Llt L tE(t} At = L~L t(( C) At 00

rE(t} dt ~

0

(11.96)

Since 9 = t/ t and E{8) = ~E(t) we are now able to construct the actual E curve in dimensionless parameters. As we will see later, it is often desirable to know the variance of the distribution curve, which can be calculated according to : U2

=

Jco (8o

1)2 E d8

~

L 8 E il8- 1 = L t 2

2

r2 L

( C> L\t- 1

L\t

(II.97)

11.8.3. Simple applications ofF and E functions l/.8.3.1. Perfect mixer. A perfect mixer is a vessel in which stirring is so effective that the composition of its contents is identical at all places. Naturally the effluent from this vessel has the same composition as its contents. Keeping in mind that Cout = C a mass balance over this vessel reads : C o/~ 0 A.

-

dC C = Vo/v dt A.

(Il98)

For the determination of the F curve from timet= 0 on, the tracer concentration of the feed stream is changed from Cin = 0 to Cin = C 0 • Integration of equation (II.98) therefore yields, via:

Jr

c - dC l/J dt ----! o Co - C- o V

J

F =

5:._ = Co

1 - e -e

(ll.99)

For the determination of the E curve at time c = 0, a tracer is injected into the vessel for a short time. The amount of tracer added would, when well mixed, give a concentration C 0 . Integration of equation (13) for this case yields . (C,n = Ol) :

fc dC = Jc C

-

0

E

=

(' lf>a: dt

Jo V

C = e -8

Co

(11.100)

As can be seen, the results obtained agree with equation (Il.93), which can be written as E = dF/d8. I /.8.3.2. Plug .flow. During plug flow all parts of the fluid move at the same speed and there is no axial dispersion and no residence time distribution.

130

Therefore, the F curve is described by : F = 0 fort < -r F = 1 fort > -r

and for the E curve : E = 0 for t #

but E

= tracer input signal at t =

T

r!

I 1.8.3.3. Laminar flow in circular tube. Here, the residence time distribution is caused by the differences in velocity and by diffusion. If we neglect diffusion, we can write for any streamline with velocity v and residence time t :

<:> =2(1- :,) =i= 9

(11. 101)

Because the maximum velocity at the centre is twice the mean velocity, both the F and the E functions are zero for: v

(v) < 2

and

tf 'C <

.

! respecuvely

For times > r/ 2 we find for the F curve :

F

= s~:~:R~r = 1 _

(1 _

~',r = 1 -

;t: = , - Uol'

(11.102)

and for the E curve:

dF E = d8

4 = (28)3

(II.103)

The F and E functions derived for the three cases treated here are shown in

Figure 11.53.

Plug flow oreo

Lommor flow

oreo = 1

112

-e

Figure ll.S3

oo~--~--~------8 ~ -~---J

112

F and E curves for three simple cases

131 I I .8.4. Continuous flow models

The analysis applied in the last chapter to three idealized cases of flow is not applicable to most practical flow situations. It is therefore desirable to describe actual fiow conditions by models. Two of these model systems, the 'dispersion model ' and the ' tanks in series model ', are widely applied, and we will analyse both models in the followi ng paragraphs. 11.8.4.1. Dispersion model. The dispersion model is based on the plug ftow of a fluid with a certain amount of intermixing in axial direction. This intermixing can be due to diffusion by Brownian motion or by turbulence (Figure 11.54). dx I I X

~I fdeol plug f low

I I

::

Dis~rsed

plug

f low

F~

D.S4 Dispersed plug flow

If we assume the mass flow rate for dispersion in the x-direction to be proportional to t he concentration gradient a mass balance over a length dx shows:

Adx dC = dt

(vxCA- DAdC) - (vx CA- DAdC) dx dx JC

x+dx:

from which follows: dC -d t

+

d2C V.x-d = D-d 2 .X X dC

(Il.l04)

Here, D is the dispersion coefficient. For the case of an instantaneous tracer injection (at timet = 0, place x = 0) this differential equation .can be solved to yield : E

=

~o =

g

2 tr.8vxL

exp{- (:8--_:_l}

(II. lOS)*

1

L:,r-

(Levenspiel, Chern. Eng. Sci. , 6, 227, 1957). Equation (II.105) describes the distribution of the tracer concentration as a function of time 8 and distance from the point of injection x = vxt. Figure 11.55 shows E curves for various *This model description holds only when the convective veiocity is higher than the velocity of diffusion (i.e. vx > 20/L). The reason for this lim itation is that for lower velocities the diffusion against the direction of flow presents a problem which can only be solved by specifying the flow and the dispersion mechanism in detail for the area x < 0.

132

o/v}( L =

l

e

Figure ll.55 Dispersion modeL E curves

values of the parameter Dfv xL according to this relation. For DfvxL « 1 equation (II.105) can be simplified. For small values of Dfvx L theE curve will be zero but at 8 ~ 1. Now with 8 ~ 1 we obtain:

-

D

VxL

« 0·01 (II.l06)

which is the normal Gauss distribution. For low values of D fvxL the maximum of E occurs at(}= 1. For higher values ofDfvxL (>0·01) the maximum occurs at

fJ

< 1.

In order to fit an experimentally determined E curve to one of the possible theoretical curves the variances of the curves can be compared. For a flow system with negligible entrance effects {ie. for small values of D/v'J:L), the variance can be calculated from equation (II.105) via: 2

a =

J~ (x - x)f(x) dx

J~f(x)dx

to be : (11.107)

133

which is simplified for low Dfv;rL to : D v:r;L

.,-2 = 2 -

(II.107a)

The F curve can be obtained from theE curve by integration according to: F

=

s:

=I

E(O) dO

(II.108)

E AO

Integration of equation (Ill OS) is not possible, but the F curve can be constructed from theE curve by graphical integration of the latter. Equation (11.106), which is valid in the case of very little dispersion [(DfvxL) « 0.01], can be integrated directly. The corresponding F curve is then : F =

J 6

E(O) d8 =

0

1£ f

8

1

2n: _!!_ v~

exp

{

-

(1 -

0

4-

8)2} d9

D

(II.109)

v%L

which can be solved to yield:

(II.llO) where erf y is the normaJ error function: erf y

=]; J:

e-tl de; erf(m) = 1; erf(O) = 0

which is tabulated in many -books (see also paragraph m .2).

I 1'.8.4.2. Tanks in series model. For a number n of perfect mixers in series with w tal residence time -r the E function can be shown to be : (C)

E = -C0 For large n (n

> 5) the .expression

(11. 111) becomes :

=

'lffY" -1 (n - 1)!

n!

e- "'

(Illll)

~ -,r: e-"fon is valid and equation (n > 5)

(Il.llla)

Figure II.56 shows E curves according to equations (11.111) and (II.llla) for various numbers of stirred tank reactors in series. The variance of the E curve for n stirred vessels in series is given by : (II.112)

134 H~r------------------------;---,

l

8 Figure ll.56 Tanks in series model, E curves

and analogous to the.development of equation (11.110) we find for the F curve for· n tanks in series : F -

-

1-e!'e {1 +n(] +2! (nB)2 (n0)3 (n9Y' + 3J + ... +(n-1)!

1 }

(II.113)

F curves for different values of n are shown in Figure II.57. The analysis of experimentally determined F curves is more complicated than that of E curves. Two cases can be distinguished : F curves for n < 50 or Df(vxL > 0.01, and F curves with very little dispersion, i.e n > 50 or D/v;cL < 0.0 1. Both cases wil1 be treated separately. (a) High dispersion : n < 50 or Df vxL > 0·01 There are two possible methods of analysis : (i) Via the E curve Plot the measured F values versus time and determine graphically at different times the slope dF(t)/ dt E(t). The values determined for E(t} at the approximate time can then be analysed as described in paragraph II.8.2.

=

135

1·0

e--

Figure ll.57 Tanks in series model, F curves

(ii) Via the slopes of the F(8) curve at 8 = 1 The slope of an F curve at 8 = 1 can be shown to be given by: d8 (dF)

8= 1

~ (E)e..,l

n"

=

(n- 1)! e-n

The relation between this slope and the number of stirred vessels in

series is shown in Figure ll.58.

3·0.------- - - - - -- -- - -- ----,

-..

n

Figure U.58 Tanks in series model, dFfdO at 8 = 1

136

To analyse a measured F curve, determine the mean residence time as shown in Figure II. 50. Determine graphically dFfd8 at 8 = l and find the number of stirred vessels in series from Figure II. 58. (b) Low dispersion: n > 50 or D(v:r;L < 0·01 By plotting the F curve on probability paper a straight line is obtained. The mean residence time is found from this plot on the t-axis at the 50percentile point, whereas the time difference between the 16- and the 84percentile points equals two standard deviations. Thus: 2(],

= tlr=0·84 -

(II.l14)

t1F=O·I6

· With <1:} = l ju;, the variance and hence the number of ideal mixers can be • determined using equation II.ll2. · 11.8.4.3. Comparison of dispersion/ tanks in series models. At high dispersion (small n, large Djv%L) the two models y1eld E or F curves of different shapes. At very low dispersion (large n, small D /vxL) the two curves become more similar and are identical at n--+- oo or Df vxL--+- 0. The number of vessels n of the tank in series model and Dfv::~:L of the dispersion model are related via the variance :

(J

2

= -1= 2D- n ~L

2 (- D )

2 {

v~

Tonks in series rnodel

n"' 5

2

xL)}

1 - exp ( -v-

8--

Figure U.59 Comparison of models

D

(11.115)

137

bich becomes for small dispersion numbers :

~2=~=2.~

(.~dt)

(II.ll5a)

Figure 11.59 shows an Ecurve calculated for both models on the basis of identical ~ariance (equation Il.115). As can be seen the E curve of the tanks in series model yields a higher proportion of material staying only for a short time in the system, whereas the E curve of the dispersion model (which shows also a smaller maximum E value at higher 0) shows a greater p ortion of liquid staying for a longer time in the system. It is evident from the above that on analysing and presenting the residence time distribution of a continuous reactor system the model must be selected carefully in order to present the reactor system as well as possible. If, for example, the reactor system consist s of a series of vessels or compartments in which mixing occurs, the tanks in series model should be chosen, whereas in the case of a tubular reactor the dispersion model will generally represent the system more satisfactorily.

11.8.5. Dispersion inflow systems The dispersion of fluids flowing through tubes is satisfactorily described by the dispersion model. For laminar flow in tubes the dispersion coefficient is found to be (d, = diameter of tube):

D

=5x

2

l0_,(v~) + 0

( if Re < 2000and ReSc


1 Lommor f low

r

Re Sc

r

«

3 0L/d1

r,.. r o 2 L-~-L~~u__J~~~~--L-~~~

104 -

105 Re Sc

10

Turbulent flow

- - • Re

6

~ d,lv 11/D ~ ~~d, IO

Figure ll.60 Dispersion number for flow in tubes

138 whereas for turbulent flow : D

~

Q. 2v ;xtl1

(if Re > 10 5)

(IJ.117)

Figure 11.60 shows more detailed information about the dispersion coefficient for flow in tubes. In the literature the degree of fluid dispersion is often not described by the dispersion coefficient but by stating the height of a mixing urut, HMU. For the tanks in series model this length is given by: L HMU= -

(11.118)

n

and states the length of pipe that is equivalent to one perfect mixer. For the . dispersed plug Bow model, applying equation (II.llS), we find:

For fully developed flow through beds of particles (e.g. fixed beds, packed columns) the height of a mixing unit is found to be equal to one particle diameter: HMU

~

dP

in the range 0·1 < Re < 2000. This· relation also applies for flow around cylinders (e.g. in heat exchangers), as shown in Figure II.27, if instead of dP the cylinder diameter ~ is used. Since for packed columns and fixed beds mostly dP « L , the number of mixers in series is so big that plug flow is approximated. Knowledge of the dispersion coefficients in pipe flow enables us, for instance. to calculate

the degree of mixing in pipelines, if one fluid is followed by another Jiquid miscible with the first. The situation is schematically drawn in Figure ll.61. At t = 0 (entrance of tube) thert~ is no mixing at all, whereas at t ·= 't some intermixing has occurred. The problem is now to determine the volume of liquid leaving the pipe, in which the concentration of A and Bare lower than the preset values C~ and Cl. This volume often has to be considered as a loss. ·

This problem is most easily solved by using the definition of the F curve: C 8 == F(O) and CA = 1 - F({}). Hence, the considered volume, which leaves the pipe between 81 and 92 ,

•c*\

I

: I

I :

0

0=0

Figure ll.61

02

I

=

T

0 =I

Mixing in pipelines

G,

e

Equotion{ll·IIO)

139 can be calculated from :

c; =

F(8 1)

and

C~

= 1-

F(8 2 )

If we concur to the model of pipe flow with axial dispersion and if CA and q are nor too low, we might use a linearization of the F curve around the mean residence time :

F~ !2 + (dF) (9 dO s=t

1) "" _!_

2

+ E(8 =

1)(8 - 1)

Replacing E (8 = l) with t he help of equation (11.106} we obtain :

F

1

~-

1

+

2~

(8 -

(11.119)

1)

Equation (II.ll9) could have been derived directly jf we h~dy known that the penetration depth of both fluids into each other is given by 2.j1tDLfv:r (see paragraph IV.2). Introducing equation (11.119) into the analysis of the contaminated volume, which occupies a length of pipe L as it reaches the end of a pipe of total length Lo. gives the following result: (II.l20) Hence, the contaminated volume: increases with the square root of the dispersion coefficient ; increases with the square root of the total pipe length ; decreases with the square root of the velocity of the fluids; and decreases with increasing and

c:

c:.

For turbulent flow in pipes (Re > I 0 5 ) we flnd, with equation (II.ll 7): L ~ -

Lo

2(1- C*A

,.....)lff2trd '--B -1

-

L0

0

~ -.1

1 -o ro~

·e Q

c:

c0 0

20-80

v

30-~40-60

tl)

=

'ii 0. 5.

\

5-95

0

Pe«ent contommoHon

10-9 0

c:

g

uc

ct i0 - 4 10-6

10-3 10-2 Dlvx L = ( Dlvx d1 Hd1 /L)

Figure H.62 Mixing in pipelines

140 or, roughly: L ~ (1 -

q - q")./"Q,

indicating that the length of the contaminated zone depends on the square root of the product of the total length of the pipeline and its diameter. Because of the fact that we have approximated the concentration profiJe by a straight line, equation (11.120) will give results that can be a factor 2- 3 too low at very low values of C~ and c;_ Levenspiel has given an exact solution of this problem and has prepared a chart for an easy calculation of L/L0 , as shown in Figure 11.62 (from Petr. Refiner, March 1958).

11.8.6. Problems 1. De.velop an expression for the E and F functions of two perfect mixers in series (each with a residence time r:/2) and show that the results are in agreement with equations (II.lll) and (11.115). •2. The feed stream to a continuous deodorizer (a vacuum steam stripper applied for rendering edible oils tasteless) is at time t = 0 changed from bean oil to coconut oil and the refractive index of the effluent stream is recorded. From the refractive index the percentage of coconut oil is determined: Time (min)

Percentage of coconut oil

Time (min)

Percentage of coconut oil

30

0

40 45

16·5

5

60 65

34·5

80

70·5 83 92

50 55

52

70

99

(a) What is the mean residence time of the oil? (b) How many perfect mixers in series would yield a corresponding residence time distribution? (c) What would be the dispersion factor if we wanted to descnbe the residence time distribution by the dispersion model? Answer : (a) 55·1 min (b) 25 vessels (c) 0·02 3. Calculate the height of a mixing unit for air and water flowing through a circular pipe at Re numbers of 1, 100 and 10,000.

Air

Water

2·5 x lO- s '2 X 10- 9

1·7 x 10-s 10- 3

141 Answer :

1

Air Water

100

10,000

0-Sd,

d, d,

SOOd,

4. A relatively small waste water fiow from a factory is bled into a big effiuent stream of 0.1 m 3 /s and discharged in to the sea. If the waste water flow is stopped, the concentration of impurities in the effluent stream at the end of the drain appears to fall alniost exponentially to nil. The concentration at the outlet is then halved every 100 minutes. If t he plant is running and 1 kg of very harmful material, uniformly distributed over 10 minutes, enters the waste water flow, calculate the maximal concentration of this material at the outlet. Answer :

•s.

1·14 gjm 3

In order to determine the mixing efficiency in a cascade of six stirred vessels the following experiment is carried out. A continuous stream of water is passed through and at time r = 0 an amount of acid is injected into the first reactor. At regular intervals samples from the effiuent of the last reactor are taken and analysed : Time (min)

Concentration (g/1)

Time (min)

Concentration (gfl)

0 5 10 15 20 25

0 0-10 1·63 3·23 3·96 3·71

30

3·00 2·12 1·39 0·51 0·10 0

35

40

50

60

70

(a) What is the mean residence time in the cascade?

(b) How many perfect mixers in series would yield the same residence time

distribution? (c) What would be the dispersion factor if we wanted to describe the residence time distribution by the dispersion model? (d) How many grams of acid were injected if the fluid flow rate was 100 ljmin? Answer:

(a) 25·5 min (b) 5·5 vessels (c) 0·10 (d) 10·2 kg

142

6. Show that equation (II.ll5a), which relates the model of a cascade of mixers to the model of flow with axial dispersion and which is based on the criterion of equal variance, also predicts that when both models are matched in this way : the E values at (} = 1 are the same for both models and, hence, the slopes dF/d(} at(} = I are equal.

*7. Show that in the problem stated at the beginning of this paragraph John calculated the correct probability of catching the man in the supermarket.

Comments 011 problems Problem 2 A plot of the F values given versus time (see the figure) yields, after graphical integration, -r = 55·1 min. At r , we find graphically for the slope of the line:

IdF) \ dt

0-0357 min- 1

= t= "t

or :

(dF) de

8= 1

From Figure 1158 this leads to n number is then found via ()'2

(dF) dt

= ~

=

I

0'

r =

1·96

25 perfect mixers in series. The dispersion

D

2v;cL

1·0-~

0

z=t

1 n

=-

S:

v

~

~

~

1. ~~,~

~~

ll. ~

~

~

04 -1

- - - Slope =0·0357 mrn

0-2~

i~ 00

~""""'~

20

~ 40

J

I

60 r ( min)

I

I

eo

I

I

100

143 Problem 5 With equations (11.96) and (11.97) we find: < = 2:tcAt

I

cat

=

2600 = 25-Smin 101-8

cr = rL:l:cllt ~c ~t- 1 = o-t8t 2

Using equation (II.107) we find by trial and error :

_Q_ v;xL

=

0·10

The amount of tracer injected is found as:

L c /j,t = 101~8 =

M ; M = 10·2 kg acid

v

Problem 7

John and ~he supermarket problem John regarded the supermarket as a perfect mixer with a mean residence time of 30 minutes. Since the man had entered the supermarket 15 minutes ago, the chance that he was still in the shop is given by ~

1 - F = e- 9

....:.

e - o-s = 0-606

or 60 per cent. However, John .neglected to realize that be was looking for one man only, whereas the concept of residence time distribution is applicable only to a very great number of elements!

ep( ifl hob/em Sclv1n9 SOlve ror Re "' pvQ . '7

,

Solt<..

.,. - L ~

,

ror

Prtx~CH)

4.

ft: my\lie

I!· Id ~

L "'Jn ll ~~ !•f,Cit!l!. r;, • D~ • D1 / In !~

ftlf

)I ? I .

'

I

• J~r.r~,

ft

h , gm l tr

I{

J)l

u't ·~11· wl otral!r

('r II. 11

r.:

1-•':'·•~V~' VI l:;u( • hJ

.

c/:

'· VJ!:f\J" r.::t ftt f

T.,..c

CHAPTER III

t

t.J, {l:
rrt , ~ T - LIT; ( 1ern_,)

f~ r

Jchft

9

ft~•\.

t11t -

U, , l'tldoh

-llv..n 1..,. fqn 11

1~~

In this chapter the transport of energy is studied on the understanding that the changes in heat content are much greater than the changes in mechaniGa] energy. In this case the law of conservation of energy is written as :

d:, =

lPrninOill -

¢ m out0out -

lPH+ ¢ A

where E, = pcPT V and 0 = cPT* (as long as n o phase transition occurs~ ¢ H = heat that is exchanged with the surroundings and f/>.,. = mechanical energy added. With this law of conservation all problems of heat transpon in a system can be solved ; the following t herefore does nothing but ascertain the consequences of this balance. The transport of heat..plays an important role in many processes in engineering. It may be that we wish to promote the transport from one medium to other (generation of steam, heating or cooling of liquids, removal or supply heat of reaction, cooling of transmitter tubes, removal of heat generated nuclear reactors, etc.) or to suppress the transport to restrict loss of beat cold (e.g. insulatiol) of pipes, heated vessels and cold stores). Three different mechanisms according to which heat can be tr (invariably from higher to lower temperature) may be distinguished, namely: (a) H eat conduction

In non-moving media in the presence of a temperature gradient heat ·

transported by the molecular movemen t (from high t o low tenmerat1ure1 According to F ourier, the heat flux tP:/t (Jjm 2 s) occurring a t this .........n.er·... +~,·­ gradient is given by:
~

. dT

- , 1, -

dx

We have taJcen here for the heat capacity the vaJue which is correct if the process proceeds constant pressure. F undamentally, this is not q uite accurate, because the presence of a flow that there are presf.u re differences.. In mcst cases this is not a great problem because either cp c. are almost equal to each other (e.g. when solid substances and in compressible liquitts involved) or the process is almost isobaric If we want to proceed in a p recise manner, we read, instead of c,, .c. + T(6p/f>n9 !JVj ()T, this constant o nly being calculated for every"'"".;_ meot if it is known which change in the volume per un it of mass ()()takes place within the

at which the temperature changes witb ~T.

145

Since cp''g is positive in the direction of decreasing temperature, the minus sign has been added. ln many correlations for heat transport the beat conductivity A. is found in the combination AjpcP which, because of its analogy with the diffusion coefficient, is called thermal diffusivity a (m 2 /s). In Table 111.1 approximate values for the heat conductivity and for the thermal diffusivity are collected. Table W.l

Approximate values of land a for various materials a (m 2 / s)

Gases

Liquids Solids (non-metallic) Metals

0·02

0-2 (water 0·6) 2

20-200

20 X J0- 6 0-1 x to- 6 1 x to- 6

s-so x to- 6

(b) Flow (convection)

In this case heat is entrained with the moving conducting medium. The flow may be the result of external causes (forced flow or convection} or spontaneous, e.g. as a result of density differences caused by temperature differences (thermal, free flow or convection), or of bubble formation (boiling).

(c) Radiation This is the transfer of beat by electromagnetic waves between surfaces of various temperatures which are separated by a me
perature.

These three mechanisms often occur in combination. In addition, for the heat transport by (b) the heat conductivity). is a very important property, for a ftowing medium can only absorb beat from a hot wall if it is conducting. Temperature-levelling in moving media takes place by convection (mixing of cold and hot) as well as by conduction. Just as in the theory of'fiow phenomena, for cases of stationary heat conduction knowledge of the relationship between the heat flow an d the driving force is sufficient to rapidly SC?lve one-dimensional problems. In the theory o f flow Newton's law was the st arting point ; in this case it will be Fourier's law. ID.l. Stationary beat conduction John realized char the man found in eire storage room for deep. frozen foods was dead. Although he wore protectir;e clothing his body was quite cold and had scarted to freeze. The doctor informed John that one hour ago the temperature measured at the arm had still been ffC . John concluded,

realizing that the storage temperature was - J(f C, that the man must have died approximately hours ago and started questiqning his colleagues.

Jt

146 lll.l .I . Heat conduction through a wall

In the case of stationary heat conduction through a wall (thickness d~ other dimensions infinite, temperature difference oveT the wall {j. T = T1 - T0 ) the beat flow through any plane parallel to the end faces will be of the same magnitude (i.e. no accumulation of heat in the wall takes place because of its stationary state). Now that the heat flux 'H appears to be constant (independent of x), A. dT/dx will have to be constant (Fourier); so if A. is constant in the temperature range from T1 to T0 , the temperature distribution over the wall will be linear. If the heat conductivity A. of the material is known, the beat flux at a given temperature difference can be calculated :

" - ). Tt - To H- • d

(IIL2a)

Methods for measuring the heat conduction of solid substances are bas¢ on this equation. If the wall consists of more ta yers (Figure II I.l ), for the sanie reasons as in the foregoing~ rf>'H must be constant in every cross-section in every layer and so in every layer dT/d.x must be constant (and inversely proportional to the conductivity of that layer!). If for every layer we write equation (JII.2a) as follows: A,.H H

'f'

~d;) A.

= flT:I = T:T:IT. l I

I

and add up all these relations for all layers we get : AT.

tot

=

T _ 7: = 1

4

.J.n (X2- X1 '¥ B 1

+

ll. t

X3- X2 •

+

1.2

X4 ,

x 3)

(111.3)

1\.3

--K Figure ill.l Stationary heat conduction through walls

The terms (x2 - x .}/ }..., etc., can be considered as heat resistances, and the above equation says that the total temperature difference (T1 - TJ equals the heat flux multiplied by the sum total of the resistances placed between .x 1 and x 4 • (Compare this with Ohm's law in electricity.) The sum total of all partial resistances. which are also called the partial heat resistances, is described as the

147

total heat resistance ·for which usually (and also here) the symbol 1/ U is used. 'U' is then the total heat transfer coefficient. . ll1.1.2. Heat conducti9.n)hr.ough, cylindrical walls .·· The next example is· the stationary heat conduction in the thick wall of a cylindrical tube (see Figure 111.2). In this case the heat flows ¢~ (per unit of . tubt:: length) through the surfaces r- = R 1 and r = R 2 and through all surfaces in between must be equal: A-' ,dT 2 'PH :::; '"7 1.

dr 1_tr

=

constant

(III.4)

· {ntegraHon yields: . .

T = constant ln r + c;, and with tlte boundary conditions indicated in the figure we find for the. tern. perature distribution in the wa:Il:

a 0;:kd

.J /

dr

(IIL5)

A

' .

Figure m.2 Stationary ·heat conduction through a cylindrical wall

Subsequently, the heat flow·J,er unit oflength can be calculated a s a function of the temperature difference a pplied: · , , dT 2nJ.(T1 - T2 ) ¢ H = -2tul. = In R2/ R 1

qr

(III.6)

which is indeed independent of r. Equation (IJl.6) can be used to ~alculate the optimum th ickness of thermal insulations of a pipe. There are two op~i.mization criteria. The first js purely technological. With increasing diameter of the insulation the heat transport res}i;tance increases and so does the outer surface of the insulation. Due to this

148

a situation may arise where, with increasing insulation thickness, the radial heat loss per meter length of pipe ¢'n goes through a maximum. If the beat transfer coefficient (by free convection and radiation) at the outer surface is called a. 0 , we find from equation (111.6) after addition of the transport resistances in the insulation and at the outer surface (T3 = air temperature): (Tt - T3)

¢' _ R-

ln R d Rt 1 -+ 2nA. 2nR 2 cx0

(IIL 7}

From this equation it can be calculated that the maximum heat loss is obtained for a.0 R .j). = 1. For values of rxoR) J. higher than 1, ¢'s decreases monotonously with increasing outer radius R 2 of the insulation. for most insulation materiais l = 0.05 W/ m oc and cx0 is of the order of magnitude of 10 W /m 2 oc (see also paragraph IIL6). In those cases the critical value of R 2 a t which the above phenomenon occurs is R 2 = 5·x 10- 3 m. With such insulation materials and under such conditions there is therefore no risk that fitting an insulation jacket increases the loss of beat. A different situation arises when, for example, for the insulation of laboratory apparatus asbestos is used (A.= 0·15 W/ m 0 C} ; in that case the critical value of R 2 = 1·5 x 10- 2 m! :The second optimization criterion is connected with the consideration that for technical installations the choice of the insulation thickness is determined for economic reasons. At greater thickness ·the loss of heat decreases but the costs of insulation rise. By writing down the equation for the total capital loss as a function of the insulation thickness (loss of beat, interest and depreciation) and by differentiating this equation to the insulation thickness, the optimal insulation thickness can be calculated_ IIJ.J.3. Heat conduction around a sphere

As the last example of stationary conduction we sha11 treat the heat transfer from a sphere to a stationary medjurn (sphere diameter 2R, sphere t~mperature T, ambient temperature T00 , constant A. of the medium around the sphere). Here also applies that the heat flow through each surfacer = constant must have the same value : lf>H = -AdT 47tT 2 =constant (independent ofr)

dr

from which follows the temperature distribution in the medium around the sphere: T- Tcc R

---= -

(IlLS)

This result is again sufficient for calculating the heat flow which is transferred between sphere and medium at a given temperature : (III.9)

149

Newton described the cooling of bodies of any form with a pragmatic cooling law in wh.ich he put the heat flow proportional to the outer surface of the body and to the prevailing temperature difference (T1 - Tee,). The proportionality constant, which is a purely phenomenological quantity and contains all that we do not know of the transfer process, be called the heat transfer coefficient a. For the sphere from the previous example this cooling Jaw is therefore formulated as follows : (III.lO)

It follows that in this case !X = J..R. In our j argon this is summarized to Nu = 2. where the Nusselt number Nu is defined as a2R/ A.. The Nu number represents the relation between the heat resistance which is estimated on account of a characteristic dimension of the body {in the above example 2Rj ).) and the actual heat resistance (1/ a). We may also consider this quantity as the relatio n between the dimension of the body (2R) and the thickness of the Jayer over which the drop in temperature takes place (/.fa). Analogously for a fiat plate and for a cylinder we find Nu = rxdjl = 1 and Nu = a2Rj ). = 2/ln R 2 j R 1 respectively. IJu • hD ) rt•h• tr -H.11 l,caJ -lr:ntrrttr!d

~~ tnltt f. ~Whlenl lronrr rt ~ ~ h~n~cd ~i ~ill ,,. lu· 1 (lllJ 11 I I I .].4. General approach for the calculation of temperature distributions It· ' 1..,. . (i (Q ·tc "'' rill -

K

The general approach for the calculation of temperature distributions is a microbalance which here will be given for a Cartesian volume element (A and cP constant). The approach is completely analogous to that which we followed for calculation of velocity distributions during laminar flow. Her~ again is the restriction that assoon as the eddies take 9ver the heat transport from the molecular transport, exact calculation is no longer possible, b~cause we can descr~be the molecular transport of heat mathematically (Fourier's law). but not the . eddy ·transport. This means that pure conduction problems and t~~nsport to laminar flows can be dealt with e~actly, but he~t transfer t_o turbulen't flows cannot.

z

Figure ID.3 The Cartesian volume element

The microbalance is as follows. In the x-direction an amount of heat equal to ( - J.(d T/dx) + pcpvxT)dydz flows in at position x. The net transport ;in the volume element through the walls x and x + dx is therefore:

.,

150

Analogous expressions can be written for the y- and z-directions. The beat content of the volume is pcPT dx dy dz and the production of beat per unit of time is q dx dy dz (hence q is production per unit of time and per unit of volume). The balance, reading: accumulation _ fi . fl production . f . - ow m - ow out + . f . umt o t1me un1t o t1me

now becomes : dpT c, dr

+

dpvxT cP dx

+

dpvyT dpvzT _ , d 2 T c" dy + cP dz - 1\. dx 2

,d2T

+A

dy 2

+

A.d 2 T dz 2 + q

If we use the continuity equation once to modify this beat balance Uust as

the micromomentum balance) the result is:

dT peP ( dt

+

dT dT vx dx + v:v dy

..-

2

+

2

~th

2

dT) (d T d T d T) v: dz =A. dx 2 + dy 2 + dz2

+q

(HI.ll)

In cylindrical coordinates this equation reads : dT dT v8 dT pep ( dt + vr dr + dO

r

dT)

+ v, dz

(' 1 = )..

T

d( r~n 1 d'T d'Tl) dr + r d8 + dz + q 2

2

2

(Ill.l2)

The reader may check that the Simple cases treated in the first three sections of this paragraph can be derived from these equations. !n the case of conduction . through a cylindrical wall treated in paragraph .JII.l.2, for example, q = 0, vx = v>' = vz = 0 , dT/dt = 0 (stationary state), dTjdfJ = 0 (constant temperature in all radial directions) and dT/d2 = 0 (constant temperature in axial direction). So we find from equation (III.12) :

~ +~l r

dr

= 0· r '

ddTr = constant

which is identical with equation flll.4).

III.l5. Temperature distribution in a cylinder with uniform heat production An electric current flowing through a wire will, because of the electric resistance of the material, produce an amount of heat q per unit of volume (Wjml)_ By beat conduction the heat produced is transported to the surface of the wire which has a constant temperature T0 . From the surface ·of the wire there is a heat flow to the surro'undings which, in steady-state conditions, equals the amount of heat produced Thus :

c/>nlr=R = qnR 2 L

(fiLl

151

L

Figure lll.4

Temperature distribution in a .cylinder

In order to find the temperature distribution in the wire we set up a microenergy balance over a cylindrical shell ofthickness dr (Figure III.4}. For steady-

state conditions, equation (111.1) can be written as:

dE,

dt =

0

= 2nr L4>8L. -

21t1'L
'·

Dividing by 27tL dr we obtain:

. O

=

rcP'HI, -

r~'Hlr+dr

dr

d(rcp'H) dr

+ rq = -

+ rq

Integrating this equation we find : ,~.,." 'PH=

q . cl lT ;-7

The integration constant C 1 must be zero because, at r = 0, if>'H cannot be infinite. Thus, applying Fourier's law (equation III.2):

¢'H = ~r = -).~T 2

dr

A second integration (assuming). to be constant) yields: . T

= _!i.,.:z + c2 4).

and, using th e boundary condition T = T0 for r

= R , we finally obtain: (111.14)

Thus, the temperature distribution is parabolic, the maximum temperature being at the centre (r = 0): · (III.15)

152 The average temperature is given b y:

foR 2nrT dr ( T) =

fR Jo 2w dr

-

s:

Tor dr +

J.R £
For the heat flow at the surface we find:
=

dT I _ = nR 2 Lq -2rcRLJ.-d T r"'R

which is in agreement with the overall balance equation (III.l3) set up for steadystate conditions. It is interesting to compare the results just obtained with the equations developed for the shear force and flow velocity distribution for laminar flow in a circular pipe (paragraphs IIJ.2 and 111.3). The microenergy balance applied here is analogous to the micromomentum balance of paragraph Il.l.3 and the correlations for Jarrunar flow found in paragraph 111.2 can be translated into the temperature distributions just developed if we use the conversions given in Table 111.2. Apparently, the processes of laminar fiow through pipes and of uniform heat production in a cylinder are analogous. Table

m.l

Analogy between laminar How and uniform heat production

Laminar flow in cylindrical pipe 2

~/to ]

Trz

[In/s] [Ns/m2 ]

~

'1 P 1-

PlfL =

- ~[N/m 3]

Cylinder with uniform heat production

4>11 TA.

q

To

(W/ m2]

rCJ

[W/m 0 C] [W/m 3]

Ill.l.6. Problems 1. An oven wall consists in succession of a layer of firebricks (Ar = 1·21 J/ m oc s), a layer of lagging bricks {111 = 0.080 1/m oc s) and a layer of bricks (),b = 0·69 J/ m oc s). Each layer is 10 em thick. The temperature on the inside of the wall is 87rC, on the outside 32°C. (a) If the surface area of the wall is 42m 2 , calculate how much heat is lost by conduction every 24 hours. (b) Calculate the temperature T'" in the middle of the layer oflagging bricks. Answer : (a) 2·1 x 109 J ;

153 2. The Chemisch Weekblad of 30 September 1966 gave the following news item:· In Elsa, a mining settlement in the Yukon district, situated only 290 km south of the polar circle where the temperature sometimes falls to - 70°C. the water supply was seriousJy hampered by freezing. This pro blem has been solved by putting a floating pumping engine in the nearby lake. The construction is such that the water inlet tube is well below the level of the ice. The 12·5 an thick pipe connecting the pumping engine with the storage tank in town-a distance of ca. 5 Jan-was insulated with a 5 em thick layer of urethane foam. The flow rate of the water in the pipe is on an average 6751/min Even at extremely low temperatures water of 2SC can be pumped from the lake to tbe town without heating being necessary.

What follows from these data for the heat conductivity of urethane foam? Answer :

0·03 W/m °C

3. A management considers installation of double windows in a large office building. According to ·the bookkeeper the fixed costs (depreciation, maintenance, insurance) are Dfl. lOOOper year more than for single windows. Natural gas costs 2 cents per decamega joule (107 J ), a year has 2 x 10 7 heating seconds and the average temperature difference inside-outside is l0°C. The heat transfer coefficients inside and outside are both 20 w/ m 2 °C, a pane is 3 mm thick (l glass = 1 w/ m °C) and betweeii the double windows there is a 5 mm air space (). air = 0·025 w/ m °C). (a} Calculate the loss of energy per year for the single and the double windows. (b) What would the total window surface area be if the additional expenses of double windows could be paid for from the saving in heating costs?

Answer :

(a) single

windows : 65 x 10 J/m 2 (b) at least 388m 2 7

194 x 107 J/rr/;

double

windows:

4. A length of steam pipe (outer diameter 0·05 m) bas to be insulated. Depending on the thickness of the glass wool insulation a Dutch firm in November 1970 quoted the following prices (including glass wool insulation, outer layer of aluminium plate 1 mm thick, labour, transport and tax) : Thickness of insulation (mm)

Price Dft.j m

10 20

13·50

30 40

50

15-00

17·00 19·50 22·50

Which thickness of insulation represents the econ omic optimum?

154 Data :

capital costs per year= 20 per cent (10 per cent depreciation, 3 per cent maintenance, 7 per cent loss of interest) of investment steam temperature = 180°C; mean air temperature = ooc J. glass wool = 0-05 W/m oc steam price = Dft. 1().00 per ton heat of vaporization AH = 480 kcal/kg The steam pipe is used for 5550 hours per year. 30 mm ; total costs per m

Answer :

= Dfi. 5·70 per year

5. The rate of heat production of a Newtonian liquid per unit of volume by viscous dissipation in a cylindrical pipe is given by q = '1(dvf dr) 2 [Wf m 3 ]. (a) If the pipe has a constant wall temperature Tw, develop an expressio~ for the steady-state radial temperature. ·distribution during laminar

flow.

(b) Show that for most practical cases the radial temperature gradient

caused by viscous dissipation can be neglected. 2

Answer : (a) T- T. =

~~> (1-

;:)

1/Jj Mft!l..J'6. The hydrogenation of edible oils is carried out with a suspended catalyst. ~.~~~:¥1 · What is the maximum temperature difference between the catalyst surface 01 • ' 13 .-u' and the oil? '1,{~-t.tt.- 411 rl ~ ; . ,3) '·li m1 'h ",v,r~tr&

1! :

J ;.r

1

, ('.,'"'-' 1':1

r ~

'

·•

~f

~ · 1· ..• t·.

spherical particles, dp = 2.5 X 10 - 6 m ~ '-;4' n ) ~· (I J 'i · 1 apparent specific gravity= 1750 kg/m 3 it ·rr ~ .. ~r 0·2 per cent by weight of catalyst applied
.

"

•

Datil :

I ·I'!·' 1l

/

,_ ...

f..,} '

J

·-•

4

0

J

-

'1 I

- l

,J , ,:

off f

1

l.'l

-~

:

Answer :

)

;

3 x 10- 4 oc

''1'"

-J

"Ol0Cd

_:rvl i; {.

I '' ~~ - ·*7 , . A copper pipe (length 0·5 m , outside diameter 2-5 em, wall thickness 2 mm)

bas a thick heat insulation and the ends of the pipe are being kept at ooc (i.copper = 400 w / m °C). Calculate the temperature distribution in the pipe wall if an electric current flows through the pipe giving 20 W in the stationary staLe.

= 360( ~

2

Answer : T

-

x2 )

·c

*8. Show that John d rew the right conclusion in the case of the deep frozen body, treated at the beginning of tllis paragraph.

· 155

Comments on problems Problem 6

A heat balance over one catalyst particle reads: rAH

H=--=aA~T

n

where n =number of catalyst particles per m 3 oil = 1·25 x 1014• Now for these small particles Nu ~ 2; thus a ~ 2ld and &T can be calculated. Problem 7 A beat balance over a length of pipe dx reads :

dT dx

q

+ -dx

r/>8 =-A).-

L

x

dT

= - A).-

dx

x+dx

dJI

L

or :

Integration yields:

q

. dT dx

= -

J..LA x + C 1

but since, at x = 0, d T fdx must be zero, C 1 = 0. A second integration leads to :

T and, since T = 0 at is given by :

X

= +fL, c2 T

=

-q

= 2A.LA x 2 + C 2 = qLj &}.A. Thus the temperature distribution

q 2J.LA

{L4

2

-

x

2}

156 John and the frozen body If we regard the body as a cylinder (L » R) we can set up the following heat balance per unit length :

Problem 8

A.'

'+'H =

=

T - T,. ln R / R 1 2

2nA.

1

+ - --

_McpdT L dt

2nR 2 cx 0

or, collecting aU constant values into one constant C :

(T d T

JroT - 1:

=-

C

f' dt; Jo

In T -

Ya - -

To - 1:

Ct

The information that T0 = 8°C at t = 0 and T = ooc at t = 1 hat T'g =:= - 30°C allows us to calculate the value of the constant C as C = 0·236 h - I . So, assuming the poor fellow had no fever, his temperature was 37°C when still alive but ooc after t hours and we find, using the same equation, that death must have occurred 3·4 h ago.

ill.2. NoD-stationary beat conduction When John entered the empty room, he found a hot iron on an asbestos plate of 1 em thickness. He lifted the plate which proved still to be cold on the underside. He decided that somebody had been in the room less than 5 minutes ago and ordered his men to search the apartment.

If we want to describe the variation of the mean temperature of a body with time caused by heat conduction we have to start by setting up a microbalance. From this balance we can calculate the temperature as a function of time and place. This temperature distribution then enables us to calculate the mean temperature of the body. This complicated procedure is necessary because during beat conduction large temperature gradients occur in the medium that necessitate a careful averaging procedure to find the mean temperature. We have already come across the microheat balance in the previous paragraph (equations III.ll and III.12). For pure conduction without heat production (vx = v:v = v:: = 0, q = 0) equation (111.11) is simplified to:

In order to solve this equation we generally need to know : the geometric boundaries of the conducting medium ; the temperature distribution at a certain time, e.g. at t = 0 (initial condition): the conditions for the temperature (or temperature gradient) at the boundaries of the medium (boundary conditions).

157

The analytical solutions of equation (lll1 7) are known for a great number of problems and can be found in the literature {e.g. in Carslaw and Jaeger). For non-stationary heat conduction the duration of the heat penetration process is of great importance. If this time is so short that the heat penetrating into one side of a body has not yet reached the other side we speak of heat penetration into a semi-infinite medium. Accordingly, if the penetrating heat has reached the geometric boundaries of the medium we speak of penetration into a finite medium.

lll.2.1. Heat penetration into a semi-infinite medium ~t us consider a solid body with a flat boundary at x = 0 and an infinite length in the positive x-direction, as shown in Figure III.5. At t = 0 , the temperature of the medium is T0 in all places (for all values of x; initial condition). At t ~ 0, the wall at x = 0 is brought to a constant temperature T1 (boundary conclition). A second boundary condition is given by the assumption that the medium .is semi-infinite, which means that at t > 0 and x = co , T = T0 (no penetration to other wall). For this problem equation (111.17) is simplified to: d2 T

dT

(111.18)

dt =a dx 2 with :

initial condition : T = To fort=

ot X ~ 0

boundary condition 1 : T = T1 for t boundary condition 2 : T =

;'i:?;

To for t >

0, x

=0

0, x = co

1 r, ~-------t_=ex:._ _ _ _ __

0

Figure

m..s

medium

X Penetration of heat into a semi-infinite

Equation (Ill.l 8) can now be solved by anticipating that the temperature distribution can be described b y two dimensionless variables: () =

T - To T1

Via :

-

and

T0

de dB d~ dt = d~ dt =

~

=

x

..j4cU

1 x de -2 4ar3 cte

158 and :

d B= d B(de) 2 2

dx 2

2

d~

2

de d 2 ~ + d~ dx2

dx

_ -

d e2 _t_ . 1

d~ 4at

equation (111.18) is simplified to:

d2 B d( 2

+ 2~ d~

oo}

8 = 0 for ~ = with B = 1 for = 0

dB

e

= 0

(111.19)

Integration yield s:

and:

e= c,J: .-<' d~ + c, With the boundary conditions we find C 2

= 1 (since f~ e- ~, d~ =

0) and:

B = 0 = C1 fooo e- ~ de+ 1 = ! C1.j;t + 1

2

Thus C 1 = - 2/~ and we find as the final solution :

(111.21} ..1.

or:

2

l _ fJ

=

l _ T- T0

T1 - T = ~ f" ./4.ii e -~l T1- To j; Jo 1

7;.- To

=

cl .. /c ,/IC.p

d~

(III.22)

The right-hand term of this. equation is called the error function, which is often represented by the symbol erf, defined by:

~-T

TJ - Ja

erf Y

2

=

.fi.

JY

o e - ~2

cte

In Table III.3 a few values of this integral are tabulated ; complete tables can be found in the handbooks. Equation (III.22) enables us to calculate the temperature distribution in the medium at any time t and in any place x. The type of curve found is illustrated in Figure III.6, which shows the temperature penetration into a semi-infinite slab of stainless steel (a= 7-5 x 10- 6 mz/s) at t 1 = 40, t 2 = 1 and t 3 = 0·1 s for T1 - T0 l0°C. It appears that, for small penetration depths x, the temperature curves can be represented by straight lines. The slope of these tina

=

159 Table 111.3 The error function

y

erfy

0

0 0·11

0.1

f/J:

0·22

0·2

0·4

1·0 1·2

1·5

t

If~

h.. lli'Jn

~rat){fet eel llltf'f lCit.~Tf

fS~*

.::.

1·00'

co

IJ

~ - u.. lrr-rq)/I-

0-60 0·74 0-84 0·91 ,0·97

0·8

(1r T.)J--~~L,

I J." 1 '.l.IV~Jata

().43

0·6

=

I ~~---L----------~------------~~

0

lrrct1

/rrat,

)rrat3

x-

Figure Ifi.6 Temperature distribution in a semi-infinite medium

is found from equation (III.22) to be :

dT dx

dO

x=O

= (:(1 - ToF. · dx

x=O

-

(T1

-

T0 ) d8

!A::

--; d">

y 4at

( =0

Using equation (III.20) we find: dT dx

2 T1

x=O -

-

-

T1

T0

,fi .J4Qi

= -

-

T0

Fa£

(III.23)

showing that the tangent to the temperature curve at x = 0 goes through the point T = T0 , x = The distance x = ~ is called the penetration depth, which represents the distance x over which the temperature difference T1 ~ T0 at x = 0 has dropped to 20 per cent of its original value. The last statement can be checked by substituting x =~into equation (III.22).

Fat.

160 The heat flux through the wall. of the medium (x = 0) is found, using equation (III.23), as : (III.24)

Equation (III.24) is one of the most important results of physical technology; it is known as the result of the penetration theory. The number of successful applications of equation (Ill24} is enormous, because it is often possible to define a characteristic time for the heat penetration process. Take, for instance, the heating or cooling of particles in a rotating drum. The contact time of the particles with the wall follows from the volume of particles and the rotational speed of the drum In most cases the penetration depth (~f smaller than half the smallest dimension (D/ 2) of the particles, so that the medium can indeed be regarded as-semi-infinite. The condition for semi-infiniteness of the medium :

is

D

for «2 is often written as :

at Fo =D2 - « 0·1

(III.25)

( < 0.05)

where the dimensionless number Fo is the Fourier number. The physical meaning of this number is :

*

Fo

~=

=

D2

( penetration depth )

2

thickness of medium

The heat fl ux through the plane x with a heat transfer coefficient tX:

= 0 (equation 111.24) can also be described (III.26)

Thus: (IIl26a) If the penetration process lasts a time te , the mean beat transfer coefficie:tit is given by : ( a ) = -1

te

f'•a dt = !ff·PC --f i'• 0

Jtle

0

1r. dt = 2 V L

!f:·PCp 1tt,

(Ill.27)

161

Applications of this fonriula are numerous ; see, for example, problems 2, 4, 6 and 7 at the end of this paragraph.

II 1.2.2. Heat penetration into a finite medium If the penetration depth is greater than the dimensions of the body (Fo > 0·1), the penetration theory proves imidequate, so another description must be found. The situation then is as·sketched in Figure III.?. After a certain period of time tin the figure for t > ·t 3 , which corresponds with . Fo > 0·1) the consecutive :em perature distributions remain·geometrically similar, which means:

T - T0

= (T1. -

T0 ){1 - A(t)f(x( R)}

The function A(t) only depends on the time; it decreases with progtessing time (A(oo) 4 ' 0). The temperature distribution is further given as a function of the place by the time-independent funCtion f(x/R) which must satisfy the following conditions:

f(+ 1) d" Jl'll

~

= 0 and df fdxlx= o = 0

-l_qr dx

J'

Y -

f :co

·

,- "

To

F"-.gu:re 01.7 Heat penetration into a finite medium (for 0 < t. < t 2 penetration theory ; for t 3 < t < a:> constant a:}

t--~---"""'t""''----........---1

-R

0

+R

-x

The non·stationary temperature distributions in a body of any form·.should therefore satisfy this general relation for larger times (Fo > 0·1). We shall now ascertain .what the consequences will be by further analysing the case of the flat plate. If we apply the general relation for the temperature distribution of the microbalance (equation III.l8) to the flat plate, the result after a slight , adaptation will be: 1 d 2j f.(x/ R) d(x/ R) 2

R 2 1 dA

=a A(t) dt

(IIL29)

The left-hand term is only a function of the place, the right-hand term only a iunction of the time. T herefore the two terms should be constant, e.g. equal to - {3 2 (a real, negative number). Only then can the two terms for all x and all t be equal to each other and only then will A(t) for c -4 XJ asymptotically approach zero. It follows that : A ..... exp {- {1 2

at

I R2)

and

f(x/ R) "" cos (f3x/r)

162 Thus the differential equation for the flat plate and the boundary conditions are satisfied, but not, however, the starting condition, which after all could be expected of a solution which applies only to large times. Now the value of /J 2 has still to be determined. It follows from the consideration that only cosine functions of certain wavelengths neatly fit the range - R < x < R (just as when vibrating a string the permissible wavelengths are determined by the length of the string). The highest possible wavelength is here apparently equal to 4R, further possibilities in this problem being (considering the boundary conditions f( + 1) = 0) 4R/ 3, 4R/ 5, etc. The relevant fJ values are : rr/ 2, 3x/2, 5rr/ 2, etc. (as the reader may establish for himself). These p values, which indicate which of the cosine functions fit, are called the ~igenwerte' of the problem. The lowest P value (here 1t/ 2) is the most important for long times, for they need an A function whic~ decreases less rapidly with time than the A fllDctions belonging to the other f3 values. So, of all the possible solutions (which, for example, have to be summed up with certain weighing factors if also the initial condition has to be satisfied ) for long times, only that with p = 7t/2 remains : ·

~ ~ ;.- exp (-:::)cos(;;)

(Hat plate)

(III.30}

The proportionality constant (which is of the order of 1 and here equals 4/n) cannot bC found in this way. A complete analysis is necessary (it is this weighing factor which reminds us of the previous rustory oft = 0). It is, however, sufficient for practical purposes to know that it is almost 1 ; the cumbersome full analysis is seldom required. For times in the range oft = 0 (F o > 0·1) it yields a poorly converging series of cosine terms as solution. A better usable solution for short times is .already available (see the previous paragraph). For longer times we are also often interested not so much in the actual temperature distribution as in the heat transfer coefficient ex, which we define on the basis of the mean temperature difference :* cx(T1

-

( T )) = Q>'fi

=

-A.dT dx

(1Il.31) x=R

From equation (III.30) we find for the temperature distribution : dT d x = - C t(Tt -

(

2 1t

at)

tt

.

1tX

To} exp - 4R2 2R sm 2R

(111.32)

and at x = R : (IIL32a)

163 The mean temperature ( T ) is given by:

(T) =

~

s: {

T, - C,(T, -To)

exp (-:::)cos;;}

dx

(IIL33)

which yields for the mean temperature difference : ( T)- 7; = -C,

~(T, -

T0 )exp (-

;R~)

(III.33a)

Thus we find with equation (III.31): A._::_ 2R n 2 ). et.=--= - 2 4 R

(IIL34)

or:

cx2R 1t2 Nu = = - = 4-93 l 2

(ill.34a)

(Fo > 0·1)

Due to the fact that for Fo > 0·1 the shapes of the temperature distributions do not change with time, the Nu values become constant. The value of the Nusselt number depends only on the form of the body. For a flat plate we found Nu = 4·93 and for a sphere Nu = 6·6. For bodies which can be enclosed by a sphere we can estimate the Nu numbers by applying NUsphcre = a.D tfA. = 6·6 and taking for D 1 the diameter of the sphere having the same volume as the body. For a cube (with side D) we find: 1t aD 6·6 = 6·6 ~7t V = D 3 = 6- D 31•· Nucube =.. = -D ). Dl 6- = 5·3

(exact solution 4·9). For a cylinder with L = D we find analogously:

I

Nu

1 = cy

6-6

3

{f

5·8

\}3

(exact solution 5.6). Since for Fo > 0·1 the -Nu ana·a values .are constant a heat balance over the total body becomes: d (1) pc P V dt = et.A(T1

-

(IIL3S)

( T ))

Integrating, with ( T ) = T0 at t = 0, we find : In T1 - ( T ) T1 - T0

=_

~At pcPV

=

_aD ~ _i_t A. pep VD

= _

Nu a~ D

(III.36)

This solution is identical to the relation found with equation (JII.33a). Both relations show that a plot of the relative temperature equalization

164

1-0

..._I

---..._

....... -.. ...... I

-Fo=at/0 2

Figwe JD.8 Mean temperature during non-stationary heat conduction d..

(~

- ( T))/(T1 - T0 ) against Fo = ! tjD2 should yield a straight line for Fo > 0·1. This is indeed the case, as shown by Figure lll.8. An analogous graph can be produced for the relative temperature equalization in the centre of the body, as shown in Figure 111.9. Since ¢'H = a(T1 - ( T )) they-axis of the first graph also represents: -1-

)J

T1

-

T1

-

- --4>~ =-T0

cr(T1

-

T0 )

(III.38)

Figure 111.8 can also be used to find the heat flux into the medium at any time. Il/.2.3. Influence of an outside heat transfer coefficient In paragraphs Ul.2.1 and III2 ? we h ave considered cases where the beat transport outside the medium is very big (au- oo) compar·e d with the beat conduction in the medium. This situation is, for example, present if a cold piece

165

._o

...I

Flat plote

Square cyl inder, L :oo \---\--Cylinder, L "00 \----1~~,--

Cu be

\--\-___.:lr----1\--- C yll nde r, L = 0

\---+---+---

-+----+--Sphere

10- 3 ~~~~~~~~~~~~~-L-A~-L~~

0

0 ·1

0 ·2

0 ·3

Fo-=at (D 2

0 ·4

0 ·5

Figure Ill.9 Centre temperature during non-stationary heat

conduction

of wood (poor heat conductivity) is -placed in a steam atmosphere (good heat transfer) with a saturation -temperature T1 • The opposite situation is realized if a piece of metal (good heat conduction) of temperature T0 is placed in air (poor heat transfer) with a temperatU:re T1 . In this case the heat transport outside the medium is so slow that temperature equalization in the metal is practically complete ; the temperature T inside the m edium is the same in all places and the rate of heating or cooling is independent of the heat conductivity of the material. A macroscopic heat balance for this case leads to an exponentially decreasing temperature equalization like equation (111.36), but with the outside beat transfer coefficient a:" instead of ex: cxuAt --pcpV

(III.38)

166 If both heat transport resistances, the outside (1/cx11) and the inside (1/<Xi "' Df J.), are important, it is possible to construct graphs like Figures IIL8 and III.9 for a series of values ofcc.,. These graphs can be found in a number of handbooks (e.g. VDI Wlirmeatlas). We can, however, approximate the exact solution by assuming that also during non-stationary conduction the heat resistances are additive (as is the case during stationary beat conduction). For the region with constant Nu values (Fo > 0·1; i.e. long penetration times) we can then write instead of equation (11!.36): ·

-ln Tl - ( T ) T1

-

T0

= ~{_!_ + ~}-t pcPV

~i

au

or, if we call/the function, shown in Figure £11.8:

ln Tt - ( T) T 1 - T0

=

f{ a~(l + ~ ) D~

~{1 + « t

= Nu

Ciw

a

D

-t}

}-1

(III.39)

11

(III.40)

Apparently, we can also use Figure II18 to solve problems with an outside heat resistance if we use instead of at./02 the corrected parameter at ( Dz 1 + cr:.j !Xu)- 1

i.e. the times necessary to reach the same relative temperatu re equalization with and without an external heat resistance form a ratio of (1 + a) aj : 1. In equations (IIL39) and (IIL40) T 1 is the outside temperature and not the surface temperature T! of the medium. Because of the resistance 1/t:~.u, T1 is lower than T1 • The surface temperature can calculated from a heat balance:

be

(III.41) Equation (III.41) appears to be in good agreement with the known exact solutions. The heating or cooling times· calculated with the above procedure arc! within 10 per cent of the exact solutions. In the literature other cases are also known where an exact analysis showed that place- or time-dependent resistances in series are not exactly additive, but that only a small error is made by assuming additiveness. We can therefore extend the approach applied to cases where the internal heat resistance is time-dependent, i.e. to short penetration times, F o < 0·1. The times needed to reach a certain temperature equalization with and without external resistance form a ratio of(l + ( a.;)/'XJ: L In this equation ( !X) is the mean internal beat transfer coefficient averaged over the time of the heat transfer process te (eq uation III.27). We can, of course, also use the relations :

1

_!_ = - U

( a i)

+ _!_

ctu

and

¢'li =

U(T1

-

T0 )

(III.42)

where ( o:i) is the mean inside heat transfer coefficient, in order to calculate the heat flux into the medium.

167

111.2.4. Problems 1. A packet of butter of 250 g has a temperature of 20°C. It is put in a refrigerator of 4°C After 120 min the centre of the packet appears to have a temperature of 6°C.

(a) After bow many minutes (from the beginning) is this temperature 5°C? (b) How long does it take for the interior of a 500 g packet (equal in geometry to the former) to reach a temperature of 6°C?

Answer : (a) 160 min; (b) 190 min 2. Check that if two conducting media a and b with uniform temperatures T: and 7b are brought into contact with each other, immediately a constant temperature 7~ (contact temperature) prevails at the surface, which is given by:

T, - 7;

T:: :-

~

~ J<~pce), (Apcp)a

*3. A cooling drum (external diameter D 11 = 1m) has a surface temperature of 20°C and rotates a t an angular velocity of2n min - t. A bituminous product with a temperature of 80°C (at this temperature the bitumen is just liquid) is brought as a layer onto the drum and after a half-revolution is scraped off again. The requirement is that the temperature of the cooled bitumen 3 does not exceed 25°C. Properties of bitumen: c, = 920 J/kg oc ; p == 10 kgjm 3 ; J.. == 0·17 W/ m (a) What is the maximum thickness of the layer which still fulfils this requirement under the con
oc.

Answer : (a) 2·2 mm ; (b) 1·73 4. Objects are submerged in a hot fluidiz~d bed to warm them up. The particles (diameter 1 mm, ). = 1 W/m oc, cp = 4 X 10 6 J jm 3 0 C} remain on an average for 0·3 son the surface of these objects and ca. 60 per cent of the surface is occupied by particles. Calculate the heat transfer coefficient between the fluidized bed and the objects. Assume that every particle remains on the surface equally long, that the gas with which the particles are fluidized does not contribute to the heat transfer and that the objects are submerged in the bed much longer than 0·3 s.

Answer : 2400 W jm 2 oc. 5. Elsevier's Culinary Encyclopedia recommends the following boiling times for obtaining hard-boiled, non-crumbly eggs from various birds:

168 Bird

Egg weight (g)

Turkey

Goose Duck

Chicken (category 2) Guinea fowl Partridge, pewit

Boiling time (s)

95 85 70

850 780 620

60 40

420

500

36

Pheasant Pigeon

390 350

30

300

24

How can these data be summarized scientifically? 6. A solid glass sphere with a diameter of0·2 m is s'uspended in a room which at the beginning has the same temperature as the sphere. The ambient temperature suddenly increases by l0°C. (a) If the heat transfer coefficient is a = 8 Wf m 2 °C, how long does it take for the centre of the sphere t o adopt the temperature change for i? And if a = oc? (b) Answer the same questions for the mean temperature of the sphere. Explain the difference between these answers and those to question (a). In which case is the difference the greatest and why? A.cJus

Answer :

= ().8 Wfm oc; aglass

= 44

X

10- 8 m2/s

(a) 4·5 and 1·26 h; (b) 4·0 and 0·63 h

7. A pan with custard must be cooled from IOOOC to 20°C. To this end the · CJ>u(i..t "' ~~- pan is put in a sink filled with water of l0°C to the same height as the custard· 1 in the pan. Water is added to keep the temperature around the pan at nr.:.r ~
,,((i

dt

\

•AC. c

_ ;:{1)

Data :

},A

' - · -Fip;l t • l
]

i. 0

Thermal diffusivity oftae custard is 10- 7 m 2 j s and .l = 0·5 Wf m cc . over the entire temperature range. The custard is so viscous that free convection is fully suppressed. Heat transport through the bottom is neglig1ble. The pan diameter is 20 em.

Answer : (a) 146 min ; (b) 12 min

8. 1f water droplets (temperature 20°C) fall from a given height onto a hot

'I

!•

plate, the droplets will no longer moisten the plate above a certain plate temperature (Tp) (Leidenfrost phenom~non~ In th~ foJJowing table this _ ) f A \ ! . - ,,.., :.4 !., r~ J~ Zr> :::c - ( s ,J r t. 1 m (' 1 1 r ( 7[' J 01. ~- 'ol --;--· VI

-

r~Qi\lll• •)i"

1

.1 \.

169

temperature is given for four materials. Determine whether. this phenomenon occurs at a contact temperature characteristic for the droplet. Material of plate

). [Wj m oq

1 2 3

382 140

4

14

Answer :

r, r cJ

pcp(106(Jfm 3 °C)]

3·51

199

244

206

1·45 2·05

72

219 244

r;, = 190 + 2°C

*9. Tins filled with spinach (cliameter and height 12 em) are put in a room filled with steam of l20°C to sterilize the spinach. (a) The heat transfer coefficient of the steam to the tin is 10,000 W /m 2 °C. After how many seconds does the heat resistance in the tin determine the rate of h~at-transfer? (b) Make a (schematic) sketch of the temperature distribution in and around the tin at the moment the temperature of the centre of the tin begins to rise. (c) If the temperature of the tins is originally 20°C, for how long should the tins be put in the room with steam; so as to ensure a spinach temperature of at least ll0°C everywhere in the tin? Properties of spinach: A. = 0·7 Jf m oc; p = 1200 kgjm 3 ; cP = 400 Jfkg

oc

Answer : (a) 1 s; (c) 2·5 h *10. Show that John took the right steps in the case of the empty apartment discussed at the beginning ofthis paragraph (a~sbenos = 10- 7 m 2 /s). (.){::,

Comments on problems

l l\'IU-1-

HJ"2:/r

X "" {trdi

x ~ o-107 rn

Problem 3

The situation is as drawn in the left picture; there is non-stationary heat conduction. The figure on the right shows the temperature distribution in the film. xP "' 1 /. T- 1q = 0. 01

X .,

'J

--·

p t ~O -

r.0 =SO"C - _L. - ---

25"C

SO'"C

170 (a) The temperature equalization after a half-revolution (at time c = nf 2n = t min == 30 s) is :

T1

-

= 20- 25 = 0 _0833

T,.

20-80

T 1 - T0

and Figure 11!.9 yields a corresponding value (flat plate, since heating only on one side, 2d must be used instead of D) for the Fourier number of:

Fo

at at 0-27 = -2 = -2 D 4d

=

from which we find d = 2·2 x 10- 3 m. (b) Fo stays con stant, but c2 = !t 1 ; thus:

and the capacity :

Problem 9 The figure shows the temperature distribution in the spinach tins during non-stationary heating. (a) · The heat transfer resistance in the tin after a short time is oj A-spinach, with We can assume that the inside resistance determines the·overall

· o= JMt.

heat transfer resistance as soon as :

which is the case after: t

= 1s

f1j =1 200C 1-

\ T.

0

\

,,

=zooc ~~'--"""-!roc-~"--~ I:

0

~---L-------1

R

0

- x (b) See the figure at t 2 .

R

·... 171 (c) Temperature equalization : _T,.=,. l_-_T...:;..: m = Q.l

Tl ~To

and Figure Ill.9 yields : Fo = 0·09

at

= D2

and we find: t =

2-Sh.

John in the empty ap~ent The penetration depth of heat into the asbestos plate is {) = fo, < 10- 2 m, and therefore t < 5 min, meaning that somebody must have been in the apartment less than 5 minutes before. Problem 10

ill.3. Heat transfer by forced convection In pipes It was near midnight when John entered his fiat. An open poe of water was boiling in the kicchen. John rhought: 'The pot has a diameter of 20 em and could have contained at most 6 I of waw. The boccom of the pot is covered with a 3 TTUn layer of scale and the temperature controller of che healing plate is pur at 15fr C.' He remembered che pages w come and looked for other signs of whether somebody was at home or not.

In this paragraph we will treat, subsequently, heat transfer during laminar pipe flow a nd during turbulent flow and heat transport from one fluid to an•~ther through a solid wall.

1113.1. H eat transfer during laminar flow in pipes If a fluid moves through a pipe as a plug the penetration theory discussed in paragraph III.2 can be applied for calculating heat flows if one substitutes t = xj(v). H owever, if .radial velOcity distributions occur, the penetration

- - · -- --

X=O

Figure ID.lO

X=L

Heat transport during laminar flow

172

theory cannot be applied as such, because the flowing fluid transports heat as well. Consider the situation shown in Figure IIl.lO. A viscous liquid passes as a laminar flow through a circular tube. At x < 0 the liquid and the pipe wall have uniform temperatures T0 . At x ~ 0, the temperature of the pipe wa11 is Tw (Tw > T0 ). Because of this temperature gradient there will be a heat flux q from the wall into the liquid by conduction in the radial direction. Since there is also a temperature gradient in the x -direction, there will be conductive heat transport in the x-direction as well and, furthermore, heat is produced by viscous dissipation and heat is transported in the x-direction by the flowing liquid An energy balance over a ring-shaped control volume 2rrr dx dr consists of the following expressions :

+ qrlr21tr ~ - q~lr+dr21t(r + dr) dx + qxlx 2nr dx - qxlx +d:r 2nr dx

conduction in radial direction:

conduction in axial direction:

2

heat produced by viscous dissipation : -21t dr dxrrx ddxvx = 11 d vx2tu dr dx

dx2

energy transported by fluid : pcpvx(T- T0 )lx21tr dr - pcPvx(T- T0 )1x+d.r2nr dr

Equating heat input and beat output over the control volume, neglecting heat conduction in the x-direction and beat production b y viscous dissipation (which is allowed in the majority of cases), we find:

or: d(rq,.) . dT --T-rpc v - = 0 dr P x dx

(111.43)

Applying Fourier's law (equation 111.2):

qr = - ).

dT; dr

-..\~(r dT)

d(rqr) = dr

dr

dr

wre find :rom equation (lll.43) :

dT)

~ d ( r - + rpc

-.A -

dr

dr

dT = 0 dx

lix-

P

(III.44)

DR wlocity distribution for laminar flow in a circular pipe is given by (equation

v, =2( v) ( I - ; : )

)

n

.173

Introducing this expression into equation (Ill.44) we .find:

~ d ( d T) ·( r ) - J. dr r dr . + 2rpcP 1 .... . R2 2

d

T

dx = 0·.

(III.45)

This differential equation (the so-called Graetz equation) has been solved f9r a number of bound,ary ~onditions. For the conditions st'lt~ in Figure 111.9 the solution is, for great distances from the pipe entrance :. Nu =

aD

T

ax D 2 > 0·1

= 3·66 for

(III.46a)

and for the region near the pipe entrance:

r~.D

Nu =

T

( ax ) - t

for

= 1·08 ( v)D 2

ax . (v)D 2 < 0·05

(IIL46b)

The mean N u number over the entrance region is accordingly given by :

· 1

(Nu) = ~ U:z "' p< )j

r:.

Xe 1

ixe Nu dx = 1·62 (
( ,

V

1 = _:. )

'1

-t

)!

for

~2 < 0·05

(III.46c)

The term axf(v) D 2 is called the Graetz number (Gz); it has here the same funcfion as the Founer number for non-stat10nary conduction In the entrance region of the pipe (short contact time) the thermal boundary layer is built up and therefore Nu "'x--t. At great distances from the entrance the temperature distributions will not change with x and the Nu number becomes constant The above Nu values and beat transfer coefficients are related to a mean fluid temperature .(T) which is defined as being proportional to the heat flux transported by the liquid. Thus: 1t

rD/2

D 2 (v) = Jo 4

2nrv,T,.dr . .

(III.47)

This is the temperature we would measure if the fiow was collected in a vessel and mixed (mean cup temperature). · Equations (Ill.46a) and (lll.46c) can also be written as:

( Nu) =· 1·62 Ret Prt (;;

r

t;

( Nu ) m;• = 3·66

as the reader m ay check. The reader may also check tha1 for laminar flow the Nu numbers for pipes of varying diameter and the same length are equal if the volumetric flow rate through the pipes is equal. The ratioS' of the heat transfer coefficients are then ff.Ifoc2 = D 2jD 1 , whereas the total beat fiows in both pipes are equal. The Prandtl number is defined by Pr = vja = Cp7f/A and represents the ratio of the rate of momentum and beat transport. For other types of velocity distributions and other pipe geometries temperature distributions and heat transfer coefficients have also been calculated

174 theoretically. The results can alwa)'S be presented by an equation like equation (Ill.46b), but the numeric constants change with the geometry of the flow and the rheology of the fluid Generally, we can state that the greater the velocity gradient near the wall the greater the heat transfer coefficient will be. For that reason, under laminar conditions non-Newtonian (pseudoplastic) liquids often show higher heat transfer coefficients than Newtonian fluids. Measurements about temperature distributions and temperature equalization in laminar flow are scarce. Practical experience indicates that, especially in laminar flow, the temperature dependence of the physical properties of the fluid (especially viscosity) has great influence on the rate of heat transport. During heating up of a liquid the warmer liquid near the wall will have a lower viscosity than the liqutid near the centre of the tube. Since the shear stress distribution is fixed. (see paragraph II.1.2), this means that due to the heat transfer process the temperature distribution will be flattened (more uniform). This increases the heat transport. Differences in specific gravity caused by temperature differences may cause extra flow (free convection) and also tend to increase the heat transfer. An experimentally checked relation for the heat transfer during laminar flow of Newtonian liquids is given by Siedt;r and Tate as :

= 1·86 Re•

Pr•(;r ·(~"J t

( Nu)min = 3·66 where 11 is the viscosity of the fluid at mean fluid temperature YJ,., the fluid Viscosity at wall temperature Tw.

(111.48]

(7: - T )/2 and 0

ll/.3.2. Heat transfer during turbulent flow nNt ftY' The best·known empirical correlation for heat transfer during turbulent flow ~he.n u.re in circular pipes is: nept

7

h [r ,~ l:z) T-..

11

gi 'IU'l , eh·l f)~leci

f

Nu 'I

1

= rt~ = 0.027 Re0 ·8 Pr-t (!!.._)~ 'lw

1-

{IJI.49)

which is valid in the region 2 X 103 < Re < 10 5 and Pr ~ 0-7. We can gain some insight into this type of relation by using the boundary layer concept (Figure 111.11). For turbulent flow in pipes the to tal velocity gradient can be thought to be located in a layer of thickness lJh at the wall (see paragraph 11.2.1)- The thickness of this hydrodynamic boundary layer is defined by (equation 11.24): !w

( v)

j

= Yf (>b =

2p(v) 2

which gives for the ratio of pipe diameter to hydrodynamic boundary layer (equation 11.25): ·

D f - =-Re bh

2

175

V=O



{T)

- - -v

T=Tw

---T

Fiaure lli.ll Boundary layers in turbulent pipe flow An analogous model can be set up for the temperature distribudon by assuming t hat the total temperature gradient is located in a layer of thickness {>T· So we can write :

or: (III.SO) We have already found (paragraph Il21) that the ratio of hydrodynamic boundary layer to thermal boundary layer is given by (equation II.29): (Jh =

br

Prt

Combined with eq uation (III.SO) we obtain :

Nu

=f

2

Re P rt

(III.Sl)

and using the Blasius equation (11.37) to substitute for the friction factor we finally obtain for turbulent flow in pipes with smooth walls: Nu = 0·04 Re0 . 75 Pr+ which is quite similar to equation (III.49) if we add again the Sieder and Tate · correction for the influence of the radial viscosity gradients. During turbulent ftow there is also an entrance region where higher heat transfer coefficients occur. We can c9rrect for this effect by m ultiplying equation (111.49) by (1 + 0·7 D/L). Since for most practical heat exchangers L » D, this correction term is rarely necessary, however. Equation (IJ T.49) will yield approximate values for heat transfer in pipes of other than circular crosssection if instead of the pipe diameter the hydraulic diameter of the channel is used

176

If a circular pipe is wound to a spiral (diameter 2R) the heat transfer coefficient compared to that in a straight tube is increased by a factor of: (1 +

3·5~)

(2~~yt < 200

if 10 <

(III.52)

due to the occurrence of systematic eddies. III.33. Partial and total heat transfer coefficients Heat must often be transferred from one flowing medium to another, which are separated from each other by a wall. We then have (e.g. as shown in Figure III.l2) cx 1 , a wall with thickness t4 and conductivity A,.,, moreover possibly ~ deposit or dirt layer with dd and .A.d and finally cx 2 • The overall heat transf~r­ coefficient U can now be formally defined according to : ~c
.,

l/J':I = U{( T1 )

iTDl

Q,, .

( T2 )}

-

,-~.

Ul

-UJ1r

1 ~

~

h

._ "" dw

A\

-

'l'-!1

r,.J

tr

:=

~

-

7-

Fluid I

Wa ll

d~t

-----

h

F lu!d 2

)

1~II ~ () -

-1

l<'vrA~

dd

t('dA, ~

_! _

al

~ A5

x,..

)

",; P.lil

'Ad

-- ~ az

A\ -\ Ao

Tor ..

To/l- ·

1--.

:2

~

R

j. T

I

t

..{III.53) --1

-Tor

1il

~ L oyer of d i r t

~ ( ~ ~\~ ~ 1~ (Ti -, ~~~ I:J ha )j

h4 A~t

1

~

~T _ T

,

AL- Av

- - - - --- -


~Ai

1 - - - 1- -

Ao

~

Lli

K

1T

=

Di stance

Llr

--

Figure ID.12 Heat resistances in series

1

x

..

1JCJ

Ll1~1

<

<

where T1) and T2 ) are the mean temperatures of the flowing liquids. 1ust as _, with heat conduction through various solid layers {Figure III .1 ), here in the onedimensional case the total resistance 1/ U equals the sum total of the partial

resistances. Thus :

(III. 54)

Table 111.4 Survey of heat transfer coefficients (W/m2 C) 0

Heat flow - to:

1

from :

Gas (free convection) ~ = 5- 15

Gas (free convection) ~ = 5- 15

Gas (jl owing) IX= 10-JQo

Liquid (free convection) C( = 50-1000

Liquid (flowing) water (x = 3000-10,000

other liquids (J. = 500-2000

Room/outside air through glass ·

Boiling liquid water C(

= 3500-~,000

other liquids C( · = HJ00-20,000

- - - - - ·----·-··· - - - · - - - - - -

Superheaters

u=

3-10 .

u = t- 2 Oven · - - - - - - - - - - - - - - ' - - -···--·--······ ..•. .. ·--·-··- . - - - - - - - - u = 10-40 + Gas (flowing) Heat exchangers Gas boiler radiation {X= 10-100 for gases u = t0-50 u =: 10-30 --··------- --·- ··-·'·--., - - ~-----~--------- ---· Oil bath for heating Liquid (jree convection) Cooling coil (): = 50-10000 u = 25-500 u = 500-1500 if stirred Radiator central heating Liquid (flowing) Gas coolers Heating coil in vessel · Heat exchanger u = 5---15 . waterfwater ·· u = 10-50 water waterfwater 0: = 3000-10,000 withO\!-t stirring -u = 9oo--.zsoo other liquids u = 5o-250 water/other 0: = 500-3000 with stirring liquids u = 200-1000 u 500-2000

Boiler u·= 10::-40

radiatioll

+

Evaporators of cooling units, hrine coolers u = 300-1000

=

- - - - - - - - - - - - - - - - - - - . . , . . . . . - - - - - , . . . - - - - - - - - - - - - -··------

Condensing vapour water 0: = 500(}-30,000 other liquids 0: = 1000-4000

Steam radiators u = 5-20

Air heaters

u=

1()-50

Steam jackets around vessels with stirrers. water · u = 300- 1000 otlier liquids u = 150--500

· Condensers

-· .steamjwater

u = 1000-4000

other v,apour/water iJ = 300-1000 :

Evaporators ·

steamfwater

u=

l5()(H)()QO

steam/other liquids· u = 300-200()'·

178 When the various surface areas are not of the same size, e.g. in the case of thickwalled cylinders, the varying size of inner and outer surface areas should be taken into account. The value of U is mainly determined by the greatest resistance in series-by the lowest value of oc or ).j d. In metal apparatus mostly ).wfdw » oc (A. steel ~ 45 Wfm ac). However, if a dirt layer is present, A.Jd4 can have a considerable influence on the overall U (e.g. in water-cooled heat exchangers A.4 jd4 ~ 0·3/ 5 x 10- 4 = 600). The partial heat transfer coefficients increase in the order gas (free convection), gas (forced convection), liquid (free convection), liquid (forced convection), condensing vapour and boiling liquid. Table III.4 gives a rough survey of values for the partial heat transfer coefficient occurring ~n practice and the consequences for the overall coefficient if two partial coefficie~ts are combined. · In the case of laminar flow equation (III.54) yields only approximate results. For an exact solution ofthis problem equation (Ill.45) should be solved applying the correct boundary conditions.

III.3.4. Problems L A very long horizontal heat exchanger for warming up liquids consists of internally smooth pipes which are kept at constant temperat ure ~ the flow through the pipes is turbulent (a) If the pressure drop is doubled per unit of length of the pipes, by which factor does the heat transfer coefficient increase? (b) By which factor should the pipe length be increased in order to attain the same heating up as in the former case? Now answer the same questions for a laminar flow if the entrance region is negligibly short with respect to the total pipe length.

Answer:

turbulentflow (a) (b} Laminar flo w (a) (b)

oc 1-39 times as large L 1-08 times as large a constant L 1·2 times as large

*2. On the outside of a steel tube (outside diameter 1 in, wall thickness 312 in) steam condenses, whereas there is a turbulent fiow of cooling water at a rate v through the tube. From meas urements it is found that on variation of v the total heat transfer coefficient U can be represented by : 1 - = 0·001

u

0·004

+ -vo·s-

where U is expressed in BTUj ft 2 h oy and based on the external pipe diameter ; v is expressed in ft/s. Using the same tube some time later the same experiments are carried out and it is then found that :

~ = U

0·0025

004 + 0vo·s "

179 The difference should be ascribed to a deposit layer on the cooling water side. Determine the heat resistance caused by the condensate layer, tube wall and dirt layer and a relation for the a on the cooling water side. Find out whether the latter relation agrees approximately with the existing empirical correlations for this case. The ..l. of steel is 26 BTU/ft 2 h °F . Answer : dirt layer 0·0015 ft 2 b °FJBTU tube wall 0·0001 ft 2 h °FJBTU condensate layer 0·0009 ft 2 h °FjBTU ex = 250 v 0 · 8 ; agrees within 10 per cent

3. For a completely turbulent gas flow (Re = 10 6 ) in a rough pipe (D = 0.10 m, relative roughness 10- 2 ) the following relation is valid between the N u number and the friction factor: Nu

= f Re 2

Calculate a., if the heat conductivity of the gas is 0.02 W/m oc and of the pipe material 200 W/m oc. This rough pipe (wall thickness 2 mm) is surrounded by a steam jacket which keeps the temperature of the outside at 120°. The inlet temperature of the gas is 20°C, the outlet temperature 30°C. What will be the outlet temperature if the gas throughput is doubled, and b y which factor is the total amount of heat, which the pipe transfers per unit of time, increased? Discuss the (very simple) answer. Answer : ex = 950 W jm 2 oc;

4>w = twice as large;

~u' = 30°C

4. On the outside of a tube a hydrocarbon is evaporated. The heat necessary for the evaporation is supplied by condensing steam in the tube. During this process a thin layer deposits on the tube wall. Measurements have shown that the amount of deposit is proportional to the total amount of heat transferred. Prove that the heat flux at time t is given by: A-" ( )

'fJH

t

~

{(

· d1 -1 +~s .A.l

+

-1 ) <X~t

2

t}-

2c ~ T + -.,-------

t

AT

u

).z

where :

a. - the partial heat transfer coefficient on the steam side, d 1 = thickness of the tube wall, i. 1 = heat conductivity of the tube wall, cxk = the partial heat transfer coefficient on the hydrocarbon side, c = volume of deposit per unit of heat transferred, ). 2 = heat conductivity of the deposit, ll T = the difference between the mean temperatures of the steam and of the hydrocarbon.

180 5. A heat exchanger must be designed for heating oil from 40 to 60°C. The beating medium is condensing steam of l00°C. For this situation we can assume that the heat resistance is entirely on the oi1 side. The oil flow through the pipe is laminar (Gz > 0-1). For an oil flow of 10m3jh.100 pipes with an internal diameter of 2-2 em and a length of 1m are found to be necessary. (a) If 100 pipes with an internal diameter of 3-3 em are chosen, what should be the pipe length? (b) If 15 m 3/h of oil must be heated in the original heat exchanger from 40 to 60°C, what should be the pipe length? Answer:

(a) 1 m ;

(b) 1·5 m

*6. For a .flow through a circular pipe (steady-state condition) the velocity distribution and the temperature distribution are measured to be parabolic functions of the radius. Calculate the Nusselt number for this flow. Answer :

Nu

=5

*7. Show that John took the right steps in the case of the boiling water discussed at the beginning of this paragraph. Comments on problems Problem 2 The overall heat transfer coefficient is given by :

1

t

1

ds

u

rxt'

rx..,

~~s

d4 A..,

- =-+ -+ - + For the steel wall we find d3 j). 3 (dd = 0) yielded:

= 0·0001 ft 1 h oF/BTU. .J"l:te first - 0-004

1

- = 0·001 + -0·8 -= U

V

1 !Xc

1

d

!Xw

As

+ - + -' 5

Now only ~w is dep endent on the fluid velocity. Thus!Xw Further:

i.e. :

..:_ = 0·0009 ft 2 h °F/BTU. o:c The second measurement analogously yields : dd = ·0·0015 ft 2 h °F!BTU ;(d

measurement

I

= 250 v0 ' 8 BTU/ ft 2 b

0

F.

181

We can calculate ocw with equation (111.49) as :

a.-

~0-021(p~D) •·• Pr• =

41BOv'·•wtm' ·c

= 283v0 ' 8

BTU/ ft 2 h oF

Problem 6

The velocity distribution of the flowing fluid is given by

~~)

1-

Vm., (

V -

We now know, furthermore, that the temperature distribution is parabolic, i.e. T- r 2 j R 2 • Defining that T = Tw at r = Rand T = T.. at r = 0, the temperature distribution is therefore given by:

T-_T;.;:_ = 2 _

T.,.,-1; ..____

R

..

The mean temperature of the flow in' the pipe can be calculated from:

to be:

The heat tlux through the pipe wall is now given by :

lPHir=R

= A~~ R =

i.(Tw-

~)~ =

x(Tw - ( T ) )

from which we find :

= C(2R = 4(Tw -:- 7;.)

N u

A.

Tw-

=5

182 Problem 7

John and the boiling water John assumed that aD heat transfer resistance is caused by the layer of scale. Thus :

4>H = ~ ~ T

=

157 w

- 4>n - J.A l!:.T -- \.e.g -r x l o- s kgjs 4>Ill !1H d!l.H In 4 hours. therefore, only about 1 kg of water will evaporate.

m.4.

Heat exchangers John had just poured out a cup of coffee and filled his pipe when his boss asked to see him for a j ew minutes. 'HeW, John thought. 'the coffee will be cold before I am back.' Remembering the pages to come, John put milk and sugar in his coffee and went to see his boss.

Heat exchangers play an important role in technology, not only for conditioning liquid ftows but also for attaining a favourable heat economy. On the basis of tbe demands made on the heat exchange (the amount of heat rPHwhich must be t ransferred per unit of time and the available temperature difference) and of data on the beat resistances, the required exchanger area A of a heat exchanger can be calculated from the formula:

l/>8 =

UA~T

(111.55)

Usually both U and ..1 Tare dependent on the place in the heat exchanger, so that the formula for the entire heat exchanger should read :

(111.56) We have already discussed the determination of the overaU heat transfer coefficient U for simple pipe flow (paragraph Ill3) and we will return to this subject in later paragraphs (see also Table lll.S). Here, we will find out how the mean temperature difference can be determined in a number of simple cases and we will consider the problems which have to be solved when d esigning heat exchangers. 111.4.1. Determinacion of mean temperarure differe nce

Let us consider a simple model of a heat exchanger- two concentric tubes (Figure III.13). Here, for example, liquid' gives up beat to liquid", either cocurrently or countercurrently. In the former case TIL should invariably be ~ T{ . In the latter case the outlet temperature T~ can also be higher than T~­ ln the first instance a heat balance can be set up over the entire heat exchanger, which shows that in the staionary state just as much heat is supplied as removed. If tne heat exchange with the surroundings can be neglected and no heat is

183 T il L

Liquid

.P 'm' c~,

1

c:- ---_..:...__----Jil! ~ -

_

To'--.Noc---!-:;I::::::::::::::::::::::::X::::_.-7,7"!-·I I

1

x=o

Liquid

L

I I

JJ

"'..~,11 m ' c" p

t

0

r. '

x =L

r,lf o

0

-x L Figure Ill.l3 Heat exchanger and temperature gradients for co- and countercurrent flow -x

L

produced in the apparatus, the heat balance will read:

+ countercurrent {-

cocurrent

(lll.57)

This relation is valid for any heat exchanger which satisfies the above conditions* and is indispensable for solving a heat exchanger problem. In order to calcula.t e the mean value~ T ovei'the entire apparatus in equation (II 1.56), we should first know T' and T" as a function of the place. For the type of heat exchanger in Figure III.13 we .make a heat balance over a small part dx:

-(
= +(cJ>,.,cP)" dT" = dl/>H

(III. 58)

Then, according to equation. (III .55), the following applies: (III.59)

d¢n = (T' - T")US dx

where S = circumference of the exchange area. After introduction of T ' - T" we find from equation (II1.57):

~T =

1

l\TL - l\1;, =

-
* If heat

exchangers are considered in which phase transitions also take place (evaporation or condensation) the heat balance should be given a more general form: ¢~(H~ - H~) =

where H = the enthalpy per unit of mass.

±

¢;{H~ - H'L)

= ¢> 8

184 and from equation (111.58): d T ' - d T " = d!l T =

de/> H [

l + l (cf>mcp)' (cJ>mcp'f'

J

Elimination of the term in brackets and of dc/>H using equation (IIL59) finally gives the following differential equation for~ T:

d6TT

=

1).

~or a few

u

(A 1i. ¢H- .1 T

0)

S

(II 1.60)

dx

special cases the integration can be carried out easily.

(a) If U is independent of x or T (often the case with gases). Then integration ~f

equation (II 1.60) leads to: y

=

!!.TL- ~To USLinf!..TJll.To = USL(!!.T'hog

(IIL61)

10

2

9

e 7 6

5 4

3 2 1· 5

r

~ ()

0

1-..J

10 9 <J

8 7

6

5 4

3 2

1·5

ld!

8

6

5

4

3

2

15

10

8

6

5

4

3

2

1· 5

l!. To[oc] - - -

Figure DI.J4 Logarithmic mean temperature difference as a function of and LlT0

~ T,

185

We see that for this case ~ T in equation (111.56) equals the logarithmic mean of ~To and~ TL, both cocurrently and countercurrently. Figure II I .14 gives a simple graph which can be used for a rough estimation of the logarithmic mean temperature difference. From this figure we find, for example, for ATL = 40oC and AT0 = 100C that (A 11toc = 22°C. {b) If U is dependent on the temperature (hence also on the place). In firs t approximation U can be assumed linearly in aT : U = Uo + (U L

-

U 0 )(AT- AT0 )/(11TL - AT0 }

Substitu tion in equation (III.60) and integration between 0 to L gives : H

= SL U L AT0 - U 0 !:iTL ln (U L !:iT0 / U 0 AT~..)

(III.62)

So here the product U!:iT is used instead of !:iT in equation (111.56). If the value of U depends on ~Tin a more complicated way, equation (111.61} can be applied to a number of sections of the heat exchanger or the integration can be carried out graphically. Cases where this is necessary are, for example, those where combined condensation and after-cooling or prebeating and evaporation occur. During the transition between the two mechanisms the value of the transfer coefficient U may change rapidly. Finally, we have to add that most heat exchangers applied in practice do not belong to the simple 'single pass' type discussed here but to the 'multiple pass' principle illustrated in Figure III.l5. Also, for these apparatus a mean temperature difference can be calculated. In practice, however, we prefer to use the ~

r" 0

I

r.=================~ ---~

~========~-

ro'

r" 0

r" L.

-L

Figure (11.15 Principle and temperature gradients of a multiple p ass h eat exchanger (one shell, two tube passes)

186 logarithmic mean temperature difference of a countercurrent heat exchanger extended by a correction term lj;: ATL - llT0

I

ll1'c.

n llTo

Values for the correction fa-ctor 1/1 can be obtained from engineering handbooks where graphs like Figure lll.l6 can be found for various situations. These graphs show t/1 as a function of the ratio X of the temperature drop of one fluid and the maximum temperature difference, (T£. - T 0}/(T 0 - T 0), and of the ratio R of the temperature changes of both streams, (T 0 - T{)/(TL - T~J- . In order to ensure a high efficiency of a multiple pass heat exchanger, we should try to keep t/1 > ().7 5.

Figure 111.16 Correction factor t/J for multiple pass heat exchangers (one shell, even number of tube passes) as a function of R and X

Jl/.4.2. Height of a transf er unit If we consider a heat exchanger with constant wall temperature Tw we can set up the following heat balance :

= 4>rnc9 (T0

n = UA To - ( T )

To -T.

In ( T ) -

-

( T ))

~w

or :

T0 - Tw ln ( T ) - T..,

u·A

=

USL

-¢ -mc-P = -¢ -mc-P

i.e. the mean temperature difference b etween the fl uid and wall decreases by a factor 1/ e ( =0-37) over a length Le given by:

L = 4>mce e

US

(I Il.63)

187 This length, which is characteristic of the heat transfer process, is called one HTU, ·height of a transfer unit', and is used in many books on chemical engineermg. We will use the concept of HTUs here to show how we can allow for fluid dispersion when designing a heat exchanger. If fluid dispersion occurs in a heat exchanger, extra heat transport in the direction of flow will be the consequence, and the actual temperature gradients over the length of the exchanger may differ from those shown in the previous section (where dispersion was neglected). This is illustrated by Figure U1.17 which shows the temperature gradients over the length of a heat exchanger (like Figure III.13) for cocurrent plug flow and for dispersed flow of liquid. It will be clear that in the latter case the mean temperature difference between the two fluids is smaller and, consequently, a larger surface area (or greater length) of beat exchanger is required for the transfer of the same amount of heat. We can easily correct for this effect of dispersion by using instead of the height of a transfer unit HTU the HETU, the height equivalent of a transfer unit which is given by the sum of the HTU and the height of a mixing unit HMU:

HETU = HTU r.0 '

(III.64)

+ HMU

No di spersion

, ......I ........

1

.......

With d ispersiOfl

__\\

-

....

TL '

L

Figure ID.17 Temperature gradients with and without dispersiori

The height of a mixing·unit can be determined by the relations given in paragraph 11.8.5. From what is stated there we· can conclude that for turbulent flow and for flow around obstac1es HTU » HMU (i.e. dispersion can be neglected) but tha t under laminar flow conditions a correction for the effect of dispersion can be necessary. 1I /.4.3 Design of heat exchangers

With the knowledge now available it is possible to determine the main dimensions of a heat exchanger. lf the overall heat transfer coefficient, aB temperatures and the fluid flows are known. we can easily determine the total surface of the exchanger required. The task remains to determit}e the number N, diameter D and length L of the pipes as well as the diameter of the shell Ds and the geometric pattern of the

00

00

Tube material

~~------- Hastelloy C

-

Pressure index (0/o )

Shell 6 16

Tube

20 atm 30 aim

.2 ...

40 atm 60 atm

29 44

14 23

r--"'

N

E

'

D

1..-J d)

0.

0

2

7

0>

304SS

Length lndel'. (%)

1'-

Admlrolity

Shell material Index (%)

304 ss 316 ss C r ~ Mo

5C r ~ Mo

I· 2m 4 i 1· 6m 27 2·4rn !2 3·0m 9 3·6m 5

9 12

4

Kettle or bellow

4 ·2 m

+ 16% 10

5Cr ~ Mo

cs

2

0 -3

2.0

50

100

200

Total oreo [m 2]

Figure m.t8 Costs in 1970 of standard carbon steel shell and tube heat exchanger (3/4 in diameter, 15/ 16 in pitch, L0/10 atm) in Dfl/m2 for all materials; cost index CS 6·5 per cent/year, all others 4·5 per centjyear-

189

pipes in the shell. F or the solution of this problem two momentum balances are available which restrict the applicable velocities, diameters and lengths due to limits put on pressure drops in the pipes and in the shell. Furthermore, we have an economic balance which is used to find the cheapest design for a given problem. Once we have -chosen the type of heat exchanger these three balances determine N , L and D, and geometric considerations then further determine shell diameter, etc. If we realize that, besjdes the solutions of the above problems, questions on construction material and a great number of constructional details must be solved, we can understand why big engineering firms often have departments specialized in designing heat exchange equipment. For a rough estimation of the costs of a heat exchanger Figure II1.18 can be used This figure shows, for instance, that a heat exchanger of 100m2 with stainless steel 316 pipes and aSS 304 shell, suitable for 300 psi in the-shell and 600 psi in the tubes, and with a pipe lengtb of 10 feet will cost per rn 2 surface in 1973: 445

+ (9 + 6 +

14

+9+

15)% =Oft. 683/m 2

Finally, in Table III.5 some heat transfer correlations are summarized, which are useful when designing heat exchangers. A few of these relations have already been discussed in paragraph 111.3 ; others will be treated in following paragraphs.

III.4.4. Problems 1. Through a horizontal smooth pipe (internal diameter 18 mm) water of 90°C is forced up at a rate of 0·8 m/s. Determine the distance from the entrance at which the water begins to boil if the pressure drop is negligible and the pipe wall over its entire length is kept at l10°C.

Answer : L = 1·5 m 2 A closed vessel filled with 86 1 of water which is stirred thoroughly bas a temperature of l00°C whereas the air temperature is 20°C. If the total outer surface area of _the vessel is I m 2 and the mean heat transfer coefficient U over this area is 25 Wfm2 oc, how long does it take for the temperature to fall to 50°C ? Answer :

4h

3. In a double-tube beat exchanger benzene is heated from 20 to 60oC countercurrently with water, which during this process cools down from 88 to 48°C. During an overhaul the direction of the water flow is accidentally reversed. Determine the final ~emperature of the benzene. (The beat transfer coefficient is independent of 6. T.)

Answer: 52°C

190 Table IU.S Heat transfer correlations for heaL exchangers Flow siluotioll

Heat transfer correlation

Flow throu9fl S1r•119ht

Re" I'<~>O < 2300. 2 5< Pr <4000

circular pipes

I

:~D·I·S& Re'j

. ~~ r

0

Rtmorks

-

Flow o!Qng circular

• dp 3 < Re • ., -•-

pipes witt. bofflts

fm

2

•~0 •O·Oa7

'

2

OptniiWJo mi'm

L p

Flow a long flot plate Cone 5idt)

I

!

(t )'

I

Tovt Physico• properties 01 -r,.+ - 2Mton orto ovolloble for flow

{S·d

}! I

2 2 lo • -;s!O· d l4r P( O·nd

:

s•• equotton em 65)

I

Tout Physical properties at 7in+ --2-

• 0·33Re2 Pr3

s~ tO$
~

~ttcot

/ :: ::::f

won

f

I 3

.,

•

c

In prae11ce 4< a< 6 mm

~<">21

----

011<1

<0 ~2 8 •

I

0

-.

foetor 2 ) toke c • O·S ond

-1 ·'

{

\ om

-;;z} ! ~z'A:;

l

{~,2,~3} s I

0 4 '

t

! 3 Pr

{).pt., nN}~

•

su

r'w,..zg t+..Ji

•

a,.,,b • 0 ·2 a tom

I

• I 13

• • 0 ·57

a

Re • .FCII < 1600

<•> • 0·0087 Re

~.~

3

c 'le Pr ( ~.. )"

Re>3200

surface htot

for rough es11male (within

I

a
'

.,.

8oollf19

R •

•2·0 5 Rt

:::: ~ -.!

.

I

Pr 3

n • number of pipes

uc~ongtrs

FCIII1119 film on

Sc:~

< 500

= •~L < 3 x 10:1

=0 037 Reo.e Pr Plate heat

"·

2 lliO ,· 0 ·5 < Pr

<";fl • Q-22 Reo.n

=~ ~

I

L

(.!. )7

0 ·25

"

Re

;;::::>

-<'

f)"'

z

I

I

3 0 Re " Pr 4

<

Physical properties ot thnds at > T,~-+ To~t f'or cor·~~oon for spiralled p1pe see equottcn II 52

,.,n•3· 66

2300 07

uiT -o.... -'s j

I

Pr! (hr { ~.) y

Ftc-., along c.rculor popes without bofflts

s

I

I

Re

0· 2

·r.tr.

Phvsl col properties at ~ 2 number at rows al s.c raper blodes N • numbe r d revolut1ons CofrtciiCn foetor i' • I for wa ter 0·2 < oy< Q-6 for viscous ll q11ids If •

Figure m.24

*4. A turbulent liquid stream of 25°C flows through a straight p ipe, the wall temperature of which is lOOoC. At 3m distance from the entrance the mean liquid temperature is 50°C. (a) Determine at what distance from the entrance the liquid has a mean temperature of 90°C. (b) To what degree does this distance change if the flow rate is doubled? Answer :

(a) 15m ; (b) increases by 15 per cent

191 Table IlLS (cont) Heat transfer correlations for heat exchangers Flow

Slf\XIliOf'l

Heo: tronsfer correlation

Notvr41 conveerio,. Otoun4

1000 < Gr Pr

k

...

&' 9 t:>p < tf?

honzontal pi pes


<~O ·<>-SS { Gr

Physocal pn:per1res crt

Pr}4 I

Pr}4 I

=<•~L.•o-55 {Gr Gt Pr > i0 8

•<~L·o

Ccndensotion

2

l nflu ~~nCe o f c ombined roatural / fa tc.ed corovectrOt\ on < a > su

poroqroph

m.s

lnfl~nce

of lnett CiQS oe

I

t3 { Gr Pt} )

I • o-n{t;/'Dt}AT 13AH}4

en

nor tzonto l prpe

rw11+ r b.. lk

l 1700
""

venrcol swfoces

Remarks

Ccnd.nsotton o~ ver11col ptpe

condensotron see

~110'1

m79

PllySICOI prOperties of fh.ildS

ot T• ~

-1

AT

For qu••k est ornatton of < a > su Fogure m.23

Ccndei'SOIIOII 11'1 honzonlal

Index c "' condensot e

PIPe

Condepsation in vertical pipe

tndex

~

c

vopOUT

Usi! correlation for eoPuenspt ioro on vertical p1pe If considerable vop04.1r veloc ity,

*5. A gas stream of 10,000 m 3 jh must be cooled from 200°C to 50°C. Use is made of a bank of steel tubes installed in the open air. The gas flows through the pipes. Water is sprayed onto the bank of tubes and wind blows between the pipes. In this way the outer wall of each tube is kept at a constant temperature of 20°C. Design the most economical heat exchanger for the job for the case where the pressure drop over the tubes must b e less than 0·1 atm. State the length, number and inner diameters of the tubes and the approximate 1970 price of the unit. Data : aair = 10- 5 m 2/s 5 v. m2/2 a1r = 10material of construction = stainless steel 316 Answer :

L -. 4-8 m ; N

=

87; D = 0·05 m ; Dfi. 33,000·00

192 *6. Show that John drew the right conclusion regarding his cup of coffee discussed at the beginning of this paragraph. Comments on problems Problem 4 A heat balance in the stationary state reads :

'" 1

L2

,._------~----~----------

n ""' et.rr.D L A T;08

= t/>

11

pc P tl T

(a) Writing the above equation for two lengths of pipe 1 and 2 we obtain (cx 1 = et2 , ¢u. = ¢ r:2 ) after division of tbe equations:

L

1

= L tlll: ll1;og

I

A'Ji = 14·9

AT.1 log2

m

(b) Now the fluid velocity v varies and· with it 4>v "' v and a "' v 0 · 8 . Using the same procedure we find :

L2 = Ll AT.og lllT2alt!Ju2 = 14·9

11 'lioa2 f1 Tl CX2 tP v 1

X

20· 2

=

17·1 m

Problem 5

We can set up the following heat balance: . H ·=

UA llTjog = ifJ.uPCP ll T

Assuming a1 » J••.Jdw and applying equation (IIL49) we arrive at :

ND0·8L

= ¢vpcPtJ.Tvo·s = 0.027v0 · 8 An L\ 7;01

38·5(ml·8)

Figure ITI.l7 shows that costs are minimal for L = 16 feet = 4-8m (length of standard pipe). lf we choose now three practical pipe dimensions, we can, with the help of the Fanning equation and Figure II1.17, calculate the following table : D (m) N v(m/s) 4f

6.p (N/m 2) A (m 2 } Cost (Dfl./m 2) Total cost (Dfi.)

(}025

158 40-5 0.019 3900 59·5 . 520 31,200

(}()5

87 16·2 0·019 310 66 500 33,000

0·10

51

6-95 0·019 28·6

77

470 36,200

193

We see that any diameter above approximately 0·04 m will satisfy the pressure drop criterion. The smaller the pipe diameter, the lower the total cost. Thus we would choose the smallest practical diameter above 0.()4 m, probably D = 2 in = 0.05 m. Problem 6 John and the cup of coffee John wants to keep his coffee as bot as possible. If he added milk and sugar directly, the heat balance:

Mcp(To- T) = Mscps(T- Ys)

+ Mmcpm(T- Tm)

shows thit the temperature of the mixture is:·

T

=

+ MscpsT;. + M'"cP'"T'" McP + Mscps + Mmcpm

McpTo

The further cooling of the coffee is then described by :

which can be integrated to yield :

T - 1'air

Yair

To -

~

''

I

/ '

,

.........

UAt) = exp (-

.M cP

Sugar addition after visit to bos-s .........

..... .......

.......

J....

7a l r

0

-

•

_,

The decrease in temperature is shown in the neighbouring graph. If John had left the coffee without milk and sugar, the temperature difference, T - T,1p would be bigger. Thus the total system would lose ·more heat and the final temperature after addition of milk and sugar would be lower.

194

m.s. Heat t!'amfer by fOI'ced convection around obstacles The body was lying aLa draughty corner. The police doctor stated that the·man could only have been dead for 20 minutes and that death was due to a stab in the heart with a pQinted weapon of approximately 3 em diameter. The murder weapon had not yet been found. John noticed that the jacket and shirt of the victim were wet. He estimated the air temperature tc be 20° C and the wind velocity at the comer tc be 8 mfs. He made a quick calculation and concluded that the man could have been stabbed with an 'icicle.

l/1.5.1. Flow along ajlat plate This situation is drawn in Figure 1!.37. A thermal boundary layer is present, the thickness f>r of which increases in the x-direction. This causes the local heat transfer coefficients to decrease in the positive x-direction. For the calculation of the heat transfer coefficients we can apply the same technique as used in paragraph 111.3 for the calculation of heat transfer coefficients for turbulent flow in pipes. We first calculate the thickness of the hydrodynamic boundary layer from equation (11.64):

Using our knowledge about the ratio of the hydrodynamic and the thermal boundary layer thickness (equation 11.29):

~ = · (~) t =

oT

a

Prt

we can now calculate the Nusselt number as:

-v~ --; Prt

Nu = rtx = _:: = .:._ Prt = 0·332 ).

{)T

bh

(111.65)

This equation is valid for Re = vxf v < 3 . 10s.. At higher Revalues the boundary layer is of constant thickness. 111.52. Heat transfer to falling films The derivation of a relation for heat transfer to falling films follows the lines set out in paragraph III.3 when discussing laminar and turbulent pipe flows. During laminar film flow the Graetz number is practically always:

Gz =

ax

., < 0·05 -


.

i.e. the Nusselt number is constant. Under this condition we find for flow between two parallel plates from an analysis identical to the one applied in

195

paragraph 111.3.1 to circular pipes : (X 2<5 Nu=-= 3·76

l

Consequently. for a laminar film on a fiat plate : Nu = rxDh = a4<5 = 2 x 3·76 = 7·52

l

.A.

Substituting fJ by means of (see problem 2 paragraph ll.l):

0

= '{31jf. = J3~(v)~

vi>iW

pg

we find, after some rearrangement : « -

7~2~f<~~r {g~:A'f

Thus the heat transfer coefficient to a falling film is :

a = 2·05 {

p(v)4b} - t{gp2 17

tT

J..3}t

2

(III.66a)

for : Re

= p (v)40 = 4
1600

.

w,.,

Equation (ll1.66a) is in agreement with experimental results. Above Re = 3200 the film flow is turbulent and the heat transfer coefficient can be described by the empirical correlation : (J.

= 0·0087

{

p(v)~}o-4{gp2 17

;._3} t{v }t -

rf

a

(IIL66b)

. f.I1.5.3. Flow around spheres and cylinders · During flow around cylinders, spheres and similar bodies a boundary layer is similarly formed which is thicker as x (the distance measured along the surface from the edge) is larger. As h as been discussed when treating the fiow resistance of bocties (paragraph 11.5.1), the boundary layer becomes detached somewhere behind the greatest thickness of the body ; behind it there is the 'dead water, with eddies. The mean heat transfer coefficient (a ) over the entire surface is mainly determined by what happens at the flow side of the body. The relations for ( a:) for such bodies therefore have a form which very much resembles equa tion (II1.65). Of th e great number of empirical rela tions for cylinders and spheres the following are mentio~ed as a good ap proximation fo r gases and liquids (Pr > 0·7).

196 (a) Long cylinders with circular cross-section perpendicular to the Bow :

(a)D11 ).

(

y)0·20 , . 0·=>7 _(v,.Du) 0·50(")~ 0·33(1 - v-

= 042 ~

< Re < 10 4 )

(ITI.67)

< Re < 104 )

(111.68)

(b) Flow around spheres:

( )D11 a ).

=

2·0

+ 1·3 ( ~)O·ls + 0·66(VrvD lo·so( ~)0·33 (1 11

In both equations the physical properties of the fluid should be taken at the average film temperature unless the correction term (tt/ t7w)0 . 14 is added. The extra term containing the Pr number is connected with the curved shape of these bodies. At Re >ca. 104 the exponent of Re is slightly higher than (}5. More detailed information can be found in various handbooks. At Re = 0 we find·for a sphere Nu = 2, the value for pure conductive transport. At high flow velocities the convective transport is much greater than the conductive transport. For this situation we can estimate the thickness of the hydrodynamic boundary layer on the basis of the penetration of momentum {paragraph 11.1.6) as :

O~r =

foi

Further, reafuing that lJJ !Jr = Prt and that the timet available for creating the boundary layer is r ~ Djv we find:

Nu = -rxD -= -D ~ 0·56~v - Prt

A

lJT

v

which is in complete agreement with equation (III.68). The same argumentation applied to cylinders perpendicular to the dircx:tion of flow will yield equation (III.67). For cylinders parallel to the direction of fiow the boundary layer will have a constant thickness at any place and we can apply the relations for heat transfer in laminar and turbulent flow in pipes discussed in paragraph III.3. In the latter case we can apply equation {111.43) ~we use the hydraulic diameter D11 instead o f the pipe diameter D. Tbe two given relations (equations III.67 and III.68) are applicable in situatiom v.·here the heat transport by convection (in the x-direction) is much more imponant than the conductivity in that direction, which means:

dT .d 2 T pcpvr dx » /, dxl 'I.e .._

197

The left-hand side of this inequality is a dimensionless constant which is called the Peclet number (Pe = v)Jja~ The right-hand term is a value which follows from the temperature distribution_ This value differs for spheres and cylinders but is of the order of 1. So it could be said that the above relations hold if

Pe : .: . 1.

Both equations must be applied with care at low flow velocities. In these cases we have to check whether heat transfer by free convection participates significantly (see paragraph II1.6}.

lll.5.4. Heat transfer in packed beds The heat transfer between the fluid and the particles in a packed be~ can also be described with the relations for a boundary layer flow (equation 111.65). provided the hydraulic diameter Dh is taken as the characteristic dimension of the successive short channels of which the bed is built up (see paragraph Il.6.1). The result for a bed of spheres is :

( a ) Dh _ ~ ( u.) eDu _ ( 2v0 Du ). - 3 (l _ e)). - constant x 3 v(l _ e)

)t

t Pr

{111.69)

For the constant for which approximately the value 0·66 is expected, a value of (0·75 + 0·15) is found in the range 10 < Reh = 2v0 D.,/3v(l -e) < 10 3 . Many authors state that for them~ Nu number of a packed bed (e = 0·40):

(ct~D. =

(1-8 +

0-3)(v~f(~r

(III.69a)

applies, pro,ided 30 < v0 DJv < 3 x 10 3 and Pr ~ 1. The reader may ascertain that this corresponds with the above more fundamental result So it is again advisable when reading the literature to ascertain which definition o f the Reynolds number is used (just as in paragraph 11.6.1 for the pressure drop relations in fixed beds the choice is: Rehor v 0 Djv, of which the first is fundamen-

tally correct)The heat transfer between the tube wall and the fixed bed (i.e. particles+ fluid~ which, for example, is important for catalytic reactions in this type of reactor, can still only poorly be described. The problem is that of the total -temperature difference between the wall and the centre of the bed an appreciable part is found in the outer layer of particles (thus at the wall) but that considerable temperature differences a lso occur over the diameter of the bed_ It is therefore obvious to try a model description in which two heat resistances occur : one in the boundary layer at the wall and one in the bed itself. The resistance at the wall {1/(a.)) can, in principle, be described with the same relation as that between particles and fluid (equations III.69 or III.69a) because the-wall can be considered as a (somewhat large and strange) particle. In practice this appears to be the case, only the constant is ca. 30 per cent lower than in equations (III.69) or (111.69a). The reason is that the mean dimension of the channels at the wall is slightly larger than in the heart of the bed, where the particles lie

198

closer to each other. The resistance in the bed is usually described with the aid of an effective heat conductivity. Regarding the heat transfer the bed (particles + fluid) is taken as a solid body with a heat capacity equal to (1 - e)(pc~rticle.s

+ t(pcJnuid

and an effective ).en· This heat conductivity is then, of oourse, dependent on the porosity of the bed, the ~ of the particles, the ~ of the ftuid and of the flow situation in the bed. A well~known rule of thumb for the effe.c tive heat conductivity is : (III.70)

where ;.:r is the effect ive conductivity of the bed if no flow occurs. It can be: .calculated with the help of Figure llt19 in which J.:cr/A.r has been plotted against J...,JA.r for various values of r..

-

* ';

-<

10

20 50 100 200 500 1000

Figure lll.19 Effective heat conductivity A:Hjn a packed bed without flow

At high temperatures.the heat transport from particle to particle is controlled by radiation transfer. This m echanism works parallel to the heat conduction through the fluid (gas~ Even then, Figure Ill.19 can be used provided the conductivity ;., of the fluid is increased by 4 eaT 3 D11 (the la tter term can be understood after reading the paragraph on radiation, paragraph III.9).

!11.5.5. Heat transfer in fluidized beds The problem here is that the fluid usually moves through the bed in the form of bubbles (see paragraph 11.6.3~ The contact between the fluid and the particles is therefore poorer than expected for a bed in which bubbles are a bsent (particulate fluidization). Usually the transfer resistance between the bubbles and

199 their surroundings is greater than the transfer resistance between the particles and the fluid that flows round the particles. In the case of 'particulate fluidization' the situation is analogous to that in the fixed bed. For not too high porosity (e < 0.75), when the channel model can still be used the transfer between the particles and the fluid is described with equation (III.69), whereas a t higher porosity the particles 'notice' so little from each other that the transfer is described with the relation of the single sphere, provided the actual speed in the bed is used as the characteristic speed. Also, the heat transfer between a wall and the bed can be described by the considerations discussed for fixed beds, but owing to the mobility ofthe particles the effective conductivity in the bed is so great that we may assume the temperature in the bed and outside the 'boundary layer' at the wall as being uniform. The heat transfer coefficient at the wall is again given by equation (III.69) with a constant factor which lies ca. 30 per cent lower than indicated there. Because of the presence of the solid wall the channel model remains valid up to very high porosities ( e < ().9).

111.5.6. Problems * 1. A stirrer (power consumption 4 kW) stirs a fluid in a horizontal cylindrical vessel (D = 1 m, L = 3 m) with an insulated bottom Air is blown at a rate of 9 m/s perpendicular to tb~ axis·of the vessel. If the vessel is completely filled with water, how will the temperature of the water change with the temperature if at t = 0, T ·= T0 = 'Fair?

Answer : T - T0 = 41·5 {1 - exp ( -9·8 x 10- 6 t)} *2. Lead shot is made by allowing drops of melted lead (p = 11,340 kgjm 3 ) to fall from a height L through air. The temperature of the air is 20°C. What should L be to ensure that drops of 2 mm diameter reach the bottom completely solidified? The drops reach the stationary rate of fall very soon after being released The heat conductivity of lead is so high that in the drops a uniform temperature can be assumed equal to the melting temperature (Ys = 327°C) oflead (melting heat of lead 23-5 x 103 Jjkg).

Answer : 14·8 m 3. T wo equally heavy metal spheres of good heat cond uctivity fall stationary through air. The ratio of the diameters of the spheres is 2: 1. With the help of dimension analysis it can be proved that the product of the stationary ra te of fall and the diameter of the one sphere is equal to that of the other sphere (vD = constant). lithe initial temperature of the spheres is T1 and that of the air Teo, calculate the ratio between the distances the spheres s hould travel to be cooled to temperature T2 • The product of the density and the specific heat is the same for both spheres.

Answer :

Distance large sphere Distance small sphere

=

2

200 4. If a thermometer with temperature T0 is suddenly put in a flowing fluid with a temperature 1f, the mean temperature T of the thermometer will, in course of time, approach the value T,. The time constant 1: of a thermometer is defined as the time necessary to reach a value of (1f - T), which is equal to (1f - T0 )j e, where e = the base of the natural logarithms. If the thermometer conducts the heat infinitely well, has an external surface area A and a total heat capacity c (Jr C), and if the heat transfer coeffi~ent between the flow and the thermometer is o:, tben derive that : 7:

c =A~

We now consider a thermometer consisting of a mercury-filled iron cylinder · (D,. = 20 mm. length L = 200 mm, wall thickness d = 1·5 mm) under the following conditions: (a) in air of 20°C at a flow rate of 10 m /s, perpendicular to the cylinder; · (b) in water of 20°C at a flow rate of 2 m/s~ perpendicular to the cylinder; (c) in condensing steam with a mean beat transfer coefficient (X = 15 x 103 W/m 2 °C. Calculate t in all cases, both for a clean thermometer and for a thermometer covered with a 0·5 mm thick dirt .layer (A.dirt = 0·6 W / m oc). Neglect the effect of the ends. The specific heat of Fe is:

that of Hg is : cP =

1·3 x 10 2 Jikg°C

Answer : (a) clean: 1 = 150 s; dirty : -r:::;: 159 s (b) clean:-r = 5·9s;dirty :-r~ 17s (c) clean: r = 0·65 s; dirty: t = 8·8 s '

*5. Show that John drew the right conclusion in the case of the wet body discussed at the beginning of this paragraph. Comments on problems

Problem 1 The outside heat transfer coefficient can be calculated with equation (III.67) to be a = 10·26 W jm °C. A heat balance over the entire system reads:

H = 4 x 103

dT

= vpcP dt +

e>:A(T - ·T0 )

and after integration between t = 0, T0 and t, T we obtain :

T - T0 = 41-5{1 - exp ( -9·8 x 10 - 6 t)}

201 Problem 2

The velocity of free fall is found to be (equation II.69) v = 24 m/s and with equation (III.68) we find for the outside heat transfer coefficient a = 469 2 W /m oc. The heat balance (steady-state conditions) reads for this situation: 1t

l/>Hl

from which we find t

= (;d!p Ml&=and; !l Tt

= 0·616 sand L

= 14·8 m.

Problem 5 John and the body at the comer The heat transfer coefficient to a cylindrical icicle is given by equation (IIJ.67) and, using A.= 0 ·025 W f m oc, v = 14·2 x w- 6 m 2 js, a = 20 x 10- 6 m 2 /s, we find : (X=

9·5 v-t

(Wjm 2 °C)

The heat balance per unit length of icicle reads : ~~

'I'H

= - rr.D

dD

dt p llHm = rxnD

A

u

T

Substituting for rx the value found above and integrating between t = 0, D = D0 and t, D = 0, we finally obtain (with Mlm = 334 x 103 Jfkg) t = 610 s, i.e. an icicle would melt away in approximately 10 minutes.

ID.6. Heat transfer during natural convection · Yes,' John thought, 'the alibi of the suspected farmer seems to be waterright.' The only way by which he could have returned home in time for the police from the house of the murdered banker on the other side of the lake .would have been to cross the lake. The air temperature had been (f C for weeks and had dropped quite suddenly w - N C 15 hours ago. An ice layer of 3 em thickness was necessary on this lake to carry a man. John remembered an old lesson, made a swift calculation and arrested the farmer. II 1.6 .1. Heat transfer during natural convection

Heat transport by natural convection is of importance in cases where there is no forced flow, e.g. during cooling of solids suspended in quiet air, or heating or cooling of liquid s through walls v.rithout forced flow caused by pressure differences, or stirring. Consider, for example, a vertical surface with wall temperat ure Tw in a stagnant fluid which has a temperature Ta) at a great distance from the surface (Figure III.20). Near the wall the fluid is heated so that the specific gravity there is (usually) lower than at a greater distance from the wall. In the stationary state a velocity distribution adjusts itself (in Figure III.20 drawn for a height x), which is partly determ ined by the heat transport from the wall. The beat

202 Width of velodly and

z

ltmperaue field

X

f I / / / /

-Y

Figure m.20

wall

Free convection along a vertical

transport in tum is influenced by the velocity distribution. In this case. therefore. the equation of motion (Navier-Stokes) and the energy equation are coupled. For laminar natural convection, which in practice occurs most frequently, the velocity and temperature distributions can be calculated for simple geometric situations. As a measure for the heat transfer, once again the local partial beat transfer coefficient ex can be introduced. It appears that in the case of natural convection along surfaces in general :

applies, where A., a and v should be taken at the mean boundary layer temperature and the coefficient of expansion pat the ambient temperature T':X>· This relationship shows that ex decreases with increasing x because the boundary layer thickness increases with greater x. For a vertical surface with height L , the mean heat transfer coefficient ( ex) can be represented by :

The quantity .BITw -

Tool can

also be expressed in densities:

The dimensionless group which determines the heat transfer can then be written as:

G, .:

-z- ti(Hot)te 1rr•n ;nk:j .J.•l•tJ Y'fl .9 X iJT

p

)J.$ - --

203

Gr is the dimensionless Grashof number* which is often encountered when describing transport phenome~during- fiow caused by d ensity differences (although in heat transfer to liquids (Pr ~ 1) the product Gr Pr almost always occurs). For vertical surfaces applies (theoretically and experimentally):

( a)L

-

1

- =
=

(103 < Gr Pr < 10 8 )

().55(Gr Pr)-l

(111.71)

By approximation this relation can also be used for horizontal pipes, if L is

replaced by D. Equation (III.71) can be understood by realizing that the force per surface area (gc5h J1p) in the positive x-direction (Figure fii.20) must be in equilibrium with the shear force at the wall, which can be estimated to be:

Thus: -

gbh &p

= rJ

(I1I.72)

Vm.ax

!>,

The thickness bh of the hydrodynamic boundary layer can now be estimated with the help of the penetration theory as:

bh = y c::; /LV£ =

/2vx

(lll.7~)

-

Vma:tt

Substituting vmax using equation (III.72), we find for tbe thickness of the hydrodynamic boundary layer~" :

{7tV2Xp}*

[) =

llp g

h

Remembering that the ratio of the thermal and hydrodynamic boundary layer thickness (see paragraph !1.2.1) is given by {equation II.29):

c5r = (~) - t

o

a

11

we find for the Nu number:

Nu

= - = - Prt X

X

br

b11

= {X

= Pr--t

3t::.p g}f 1tV

2

p

P rt

= 0·75 Gr+ Prt

(III.73)

Hence we see that with the simple model chosen we obtain a relation quite similar to the exact solution given by equation (III.71). • Gr can be regarded as a combination of Re = vLJv and the Froude number v2 p/Lg Ap where the velocity no longer occurs:

Gr

------- - - - - - -

= Re 2/Fr

204

If Gr Pr becomes > 108 • the flow along the surface is turbulent and ex no longer depends on x and by approximation (ex) no longer depends on L. It is clear that in this case in a relation of the type of equation (111.71) the exponent of Gr Pr must be equal to ! . For this case has been found :

( Nu)

= 0·13 {Gr Pr)t

(III.74)

A similar situation occurs on cooling a horizontal plate. Above the plate thermally generated eddies occur. If the dimension of the plate Lis great with respect to the size of the eddies (condition: Gr Pr. > 107) (ex) is no longer dependent on L . In that case: ( Nu) = 0-17 (Gr Pr)t

(111.75):

During heat transfer by forced convection, in principle free convection is always involved, because there are temperature differences in the medium. The combination of both effects is very complicated and as yet little is known about it. The effect of free convection on heat transfer is negligible compared to that of forced convection if the liquid flow velocities are : Vrrce coovcc:tion

<

Vcorced convection

From equations (III.72) and {III.72a) the maximum ftuid velocity can be calculated and we find: Vcreeconvc:ction

=

VfiitilP -p <

Vrorccd conwclion

and, by multiplying both sides of this equation by

~<

yp;r-

or:

x/v, we finally obtain :

XVrorced convection

.jitGr <

y

Re

(III.76)

for

So we see that free convection heat transfer effects can be neglected if Re > but that convective heat transport is most important if Re « In the region Re ~for we can add the fluid velocities due to free and forced convection and use the sum total to calculate N u as a function of the Re number.

for.

111.6.2. Problems 1. Water of 50°C is passed through a non-insulated horizontal pipe in air of 20°C. Through the wall an am.oun t of heat of 100 W is lost. The heat loss is completely controlled by free convection. Later on water of 80°C is passed through the pipe. Calculate the heat loss. Answer : 0·24 kW

205 2. Liquid condenses in an air-cooled pipe of 10 em diameter and 2m length. Calculate the ratio of the condensation capacities if the pipe '1s put in the horizontal and vertical positiQns respectively and if (a) g ~P

vpwa

=3

x 106

and (b) g Ap = 3

vpwa

X

108

Answer : (a) horizontal : vertical = 2·12 : 1 (b) horizontal : vertical= 2·7:1 *3. A metal thermometer well (see F igure III.21) (outside diameter D , wall thickness b) is inserted over a length L into a gas . pipe t~rough which a

Figure UI.21 Thermometer well

warm gas with temperature Yg flows. The tempe_rature T~ of the well ~ear the wall is lower than T,, so that by conduction through the metal (A.nJ heat flows through the well to the wall. In the stationary state this beat is supplied by the gas (heat transfer coefficient ex). Show that for the temperature of the

206 sleeve applies : d 2 6.T dx 2

-

d !iT= 0 dx

X= 0, X

a Ab AT = 0, !iT = T - T,

= L,

/i Tw

=

Tw -

~

The solution of this equation is:

"T _ u",..,..cosh px 1. w h e re p = "'cosh pL •

u

jfg -

A.<5

There is no heat flow through the plane x = 0. Calculate T0 , if~ = l00°C , Tw = 60°C, D = 20 mm, lJ = 1 mm, L = 0·15 m , .A. = 100 W/m oc and ex= 25 W j m 2

oc.

Answer :

T0 = 92·6°C

*4. Show that John took the right steps during the case of the d ead banker described at the beginning of this paragraph.

Comments on problems Problem 3 A heat balance over a short length dx of the thermometer well reads in the stationary state : 0

=

- 'HnDJix+ dx

lx

-

etnD dx( T -

7;)

or:

0=

-

d¢" nDb_____!!_ - anD(T - T.) 1 dx

Now th e heat ftux in the x -direction is also given by ¢'H = - A. dT/ dx. Thus:

We simplify this equation by writing it in the following form : 2

d !:iT __ a fl.T __ dx 2

...t<S

2

p

"T· u

,

where p

fa

= "V M

207 This differential equation must now be integrated between the boundary

conditions :

X X

= T..., ' 6 T

L, T

=

=

o.

c/J'H

=

T..., - T,

11

= 0, d T dx

= 0

Multiplying through with d 11 T /dx yields :

---ax-Tl

2

d 11 T d 0. T _ d (d ~ 2 dX dx2 - dx

2

d llT 2 (IXP 2

=

ctaT -= dx

d

aT = dx(p 6T)

2

+p6.T -

and thus :

LlT = Aepx + Be - P"

Using the boundary conditions stated, the values of the constants A and B can be found to be : A = B

and

T.,., - T.8

= A ePL + B e - pL

So we find, finally, for the temperature distribution along the thermometer well : T - T, ----::.. = T., - T,

epx e PL

+ e- px cosh px = + e- pL cosh pL

The temperature measured at x

=

.

With p =

~ .At5

0 is thus given by :

and with the data given we finally find for the temperature measured T0 = 92·65°C instead of the actual gas temperature T, = l00°C This example shows that considerable mistakes can be made in temperature measurement if thermometer wells are applied The deviations increase with decreasing p and thus with decreasing beat transfer coefficients (gases, free convection) and increasing heat conductivity and wall thickness of the well material. Problem 4 J ohn on the ice The figure shows the situation during freezing up of t~e lake. Assuming Yc ~ T..., (we will check later!) we can calculate the heat transfer coefficient due to free convection by equation (III.75) to be ( ex) = 8·75 W/ m 2 oc (using A = 0 ·025 W im oc, v = JO - 5 m 2 is, Pc = 1·29 k gjm 3, Pa = 1·385 k g/rn 3 , a = 17·3 x 10 - () m 2 j s).

208

Air

lee

Woter

For setting up a heat balance over a thin layer dd we can neglect the specific heat of the ice formed because at most: ~'IIceCpice :::::; 4·2 X 10

Thus our heat balance becomes: 'H = a.('I; -

4

«

61fice

= 3·35

1rw - ~)

~) =

X

Ml

== -

10 5 Jjkg

Pice~~

Solving for the ice surface temperature 7; we find:

(for conditions here 4 = -2·l°C; thus, indeed, in the heat balance leads to:

r;:::::;

Tw). Substituting

r;

.

rJ.d)dd ( 1 + -;. -dt

Tw-

Ta

= -ct---

tili p

and, after integration between t = 0, d = 0 and t, d, we find that it takes roughly 13·5 h to form an ice layer of 3 em thickness. Thus, the farmer could have been back in time, as John rightly concluded.

ill.7. Heat transfer during condensation and boiling Condensation and boiling heat transfer can be regarded as special cases of convective heat transfer. The heat transfer is attended by a change of state (vapour - liquid or vice versa} and the great difference in density between the two phases causes flows which greatly promote the heat transfer. Therefore much higher heat transfer coefficients can be obtained than in the case of free

209 convection without phase transition and even in forced convection. The beat transfer coefficient further depends in principle on the temperature difference, just as was the case in free convection, in one phase. II/.7.1. Film condensarion We shall use the classic example of Nusselt for the calculation of the heat transfer coefficient <~ > averaged over Lin film condensation of a pure vapour on a vertical plane with height L (see Figure III.22). We assume that the condensate film covers the surface entirely and flows down laminarly and that the difference ll T between the condensation temperature 7; and wall temperature T.., is constant over the entire height. When {J is the thickness of the liquid film the local heat transfer coefficient is oc · )./b (J. = conductivity of condensate). In order to calculate ( oc) over the height L, we must know~ as a function of x . Wolf

x=O

Conde ... sate

L

-f

-

+d~.,

tl>v

l! X

=L

Vapour

df.ll

=WS+d (< v> WS )

.,7~

- - --~---'

liL

Figure IU.22 Fihn condensation

When on a surface element of height dx ·and width W an amount d~o condenses, the condensate flow 4>v increases over d:X by an amount d~ 11 , which is equal to d((v) Wb) (mass balance). Furthermore, on the conde·nsate surface an amount of heat tv! oP d¢" is released which in the case of laminar flow should be transported to the wall by conduction only; so the energy balance will be : ).

l:lH,.p dv = -g~T W dx

Now it is assumed that for each value of x the -relation between given by:

0

4J., and

~

is

. J 3q( v)lJ pg

(see problem 11.1.2) although strictly speaking this equation applies only to laminar vertical films of constant thickness. So : cf>v = pgb 3 W/371,

and

2

dcf>v = pgb W db 17

- -- - - -- - -

- -

-

210 Substitution into the energy balance gives·: !lT l dx

= llH

11

2 03 p g db

71

After integration over x between 0 and Land over b between 0 and OL we get:

J.llT L = AHoplgo1_

.

4'7

The average heat transfer coefficient (a) is defined by: ( ex) AT WL = MlvP4>r:lx =L =

Miup 2 gWo1. 371

From the last two equations we find:

<(1.>=

4 A. -3 c5L

but in this form the solution is hardly usable because oL is not an easily measurable constant. Elimination of bL from the three previous equations gives :

( a ) = (}94[AHuPl

).3gJ *

(111.77)

LYf6T

So ( (1.) is inversely proportional to .jL;for this reason horizontal condensers often applied and an increase in the capacity is sought in Wrather than in L . U sually horizontal internally cooled t ubes are used. For on e horizontal tube a formula analogous to equation (ll1.77) but with a coefficient of 0·72 can be derived if instead of L the external pipe diameter Du is used. This relation can be extended to the case where n horizontal pipes are placed under each other. The general equation for this is then :

~ are

3g]t

llH &1 p2). ( ex) = 0·72 [ nD,J7 6 T

(III.78)

The values of ( a:) found in practice for laminar condensate films a re on an average 20 per cent higher than follows from theory. Figure III.23 gives a nomograph from which partial heat transfer coefficients for c;:ondensation can be rapidly found. The validity is limited to those cases where the condensate film still flows laminarly (4(v)~J:v < 103 ) . The presence of a non-condensable gas in the vapour greatly reduces the rate of condensation. The vapour must now diffuse through the gas which has accumulated at the condenser surface. The partial vapour pressure at the condc!nser surface is then co,:15iderably lower than the total pressure. The presence of relatively small amounts of gas can therefore decrease the temperature at '*The constants p, equals 1f = Tc -

rr and ~ should be taken at a mean temperature T1 o f the condensate film which

f ~T.

211 14 { l!.H,:'i3}'

1

·(·Ov)·

' (4H) .

r

r.,:r.-~t::.T c 4

Lt:.T

1

-30 0 .30

ntXlT

. ~0

go··

120 ISO 180 210

10

liquid ·

( I ) HzO

( 2)

CH30H

( 3)

C2 H 50 H

'(4) (5)

. ,.

c 6 H6 NH

. :

3

.

•'

~100

·.·

Figure ffi.23 Nomograph for the calculation of heat transfer coefficients of condensing liquids if 4(v) oLfv = 4cpnJW1] < 1000

which the vapour condenses .considerably., so that.a sm~ller t~mperature difference AT is avaUable for the heat transport through ~he condens~te 'film. If 0·01 < PtnerJPtotal < 0·4 we can correct for this ,effe,c t by *e following empirical

rule:

·

·

CXmixture

··

· :;::::;;

C!pure vapour .

O·l ·

rP::;;;_

Vp;:::

.

.· ..

(III.79)

-Equation (III. 79) shows that the presence of, for ex~mp'le, 20 per ·cent of air by vol~m~ in a condensing vapour will d~crease the heat trailsf~r coefficien: by a factQr 4. If P inerJPtotal >. 0·4 a.heat transfer coefficient should be calculated · assuming no condensation hutjust cooling of the gas mixture. '<

.

•

.

I I 1.7.2. Drop wise condensation

[f the cooled surfac.e is poorly wettable by .the condensate, so-called dropwise condensation occurs. Parts of the surface are then almost dry and have a very .h igh local ~ea.t. transfer coefficient..As s,oon as so~~ condensate has formed a drop collects which after having grown sufficiently rolls down and sweeps part of the surface clean. The average ·heat transfer coefficients can be up to ten times higher than on film condensation.

212

I 11. 7.3. Boiling The boiling phenomenon is complex. There are eight principal situations, as shown by the following table of possibilities (a), (b) and (c): (a) J.Y all horizontal vertical

(b) Liquid

a t boiling temperature undercooled

(c) M edium flowing stagnant

Most is known of the heat transfer from horizontal walls to a stagnant medium at boiling temperature (water kettle on the fire), the so-called ·rree boiling'. The greater part of the literature deals with this situation. Dependent ~n the degree of superheating (6T with respect to the 1ocal boiling point) four boiling regimens can be distinguished : , ( I ) Low superheating (6 T < ca. 2°C), free convection (with or without forced convection), no bubble formation yet. A slight superheating (and consequent-

ly overpressure) is necessary to form bubbles because of the following effects: tlp due to surface tension (=Zaf R and so tlp is rugh for small bubbles), ll.p so that the growing bubble can displace the liquid, tlT to transport the heat of evaporation through the liquid to the bubble andif no allowance has been made on selecting the boiling temperature (reference for £\T)-a !lp due to the hydrostatic pressure on the immersed wall.

(2) Nucleate boiling (for water : 2 < l'iT < 25 - 75°C) : in certain 'active' places (small indentations in the wall)·vapour bubbles (nuclei) are formed. which in an undercooled liquid 'collapse' because of condensation-as soon as the tops of the bubbles reach into the cold liquid as a result of growing (very high <XS). In a liquid at boiling temperature they grow until, because of their buoyancy and momentum, they become detached from the indentation. Every active place yields bubbles of a size characteristic thereof. At low /'iTs those places first become active which yield relatively large bubbles. If superheating is increased the frequ~ncy at which nuclei are formed in these places becomes higher until the maximum rate of discharge of the bubbles formed has been attained for the relative active place. Next, smaller indentations which produce smaller bubbles are produced at a higher frequency an d bring about more 'agitation' in the interface, so that a increases rather strongly with ll T (approximately proportional to ~ T 3). It appears from the foregoing that in this regime the s urface conditions of the wall (nature and size of the irregularities, wettability) play an important role. It is therefore difficult to carry out reproducible exp eriments, particularly because when a wall has been exposed to boiling once its surface changes (ageing). ·

213

-

(3) Leidenfrost regime: the bubbles of neighbouring active places coalesce and form a closed vapour layer. As soon as this takes place the heat transfer coefficient decreases considerably with respect to the values which were obtained in reginie (2), because the vapour layer has a fair heat resistance. If instead of a given wall temperature a given thermal load is applied (electric heating element, nuclear reactor), the temperature of the wall rises considerably on transition from regime (2) to (3), as a result of which the heat transfer coefficient decreases further until finally the heating element melts. In these cases one should therefore stay at a safe distance from the critical ~ T where the Leidenfrost phenomenon occurs. (4) A stable vapour layer has formed from which bubbles with a dimension characteristic of the liquid are released (for stagnant, clean water approximately of 15 mm) ; in the vapour layer _the heat transfer takes place by con duction and particularly by radiation (tiT for water > 500°C).

III. 7.4. Heat transfer in evaporators Evaporators are used for concentrating solutions, if necessary combined with crystallization of the dissolved substances. There are a great many versions, from 'boiling pans' to pipe evaporators, in which the solution flows through banks o~ for example, vertical pipes . which are usu~y heated by condensing steam. During the evaporation in vertical pipes a combination of heat transfer mechanisms occurs. In the lower part of the pipe the incoming liquid is heated to the boiling point (heat transfer by forced fiow). In the upper part the evaporation begins. After a short boiling height so much vapour has been produced that in the upper part of the pipe mainly a vapour flow occurs which entrains the liquid which has not yet evaporated. In this range the heat transfer is determined by the flow conditions of the two-phase mixture rather than by the boiling phenomenon. In certain types of pipe evaporator the Hquid is pumped into the pipes (forced circulation). How~ver, there are also many evaporators with natural circulation (thermosiphon action), the rate of circulation irt them depending ·Qn the vapour production and on the heat transfer. The latter is influenced ·by the rate of circulation. Pipe evaporator design is largely based o n experience supplemented by experimental data which have been obtained with model set-ups. For dilute aqueous solutions the as usually lie between 2000 and 5000 W/m2 oc and for hydrocarbons in the range from 500 to 1000 Wj m 2 Figure III.24 gives a survey of experimentally determined heat transfer coefficients during boiling as a function of heat flux. This figure can be used for quick estimation of boiling beat transfer coefficients. In order to avoid the 'Leidenfrost' regime j ust discussed it is advisable to limit the vapour velodty at the evaporator surface to roughly Q,2 to 0-3 m 3 j m 2 s.

oc.

214

1 u 0

N

....E

3 ......... 0

Free convection

Figure ID.l4

Nudeott boiling

Heat transfer coefficients during boiling of liquids

II 1.7.5. Problems *1. A chemical reaction is carried out. in ethanol solution under r~flux. The reflux condenser consists of a straight inner glass tube (A.,ta.ss = 1 W/m °C) of 12 mm diameter arid 2 mm wall thickness and an outer pipe of 30 inner diameter and 40 em length. (a) If 15 1/min cooling water of l5°C are available, calculate the amount of heating energy that can be supplied to the reaction mixture so that just no vapour leaves the condenser. (b) What would be the amount if chemists could be persuaded to use stainJess steel equipment?

mm

Answer : (a) 315 W ; (b) 700 W 2 A condenser consists of a number of copper pipes (D j = 22·8 mm, wall thickness d = 1·25 mm). On the outside of the pipes 90 80 kg/h Freon 12 condenses at 32·2°C. The heat of condensation (137·5 kJfkg) is removed by cooling water in the pipes. The flow rate of the water is 1·22 m/ s, the

inlet and outlet temperatures are 15·5 and 17·8°C respectively. and the total heat transfer coefficient Freon-water related to the external pipe

215 surface is 1010 W j m 2 '0C. Calculate the·number of pipes n the condenser consists of and the length L of these pipes. Answer : n

= 72 pipes ;

L = 3·9 m

3. A packet of 6 x 6 horiZontal tubes of 30 mm outer diameter and 1·5 m length is.surrounded by saturated steam of 180°C. (a) If the surface of the tubes is kept at 120°C, calculate the heat fiux through the wall. (b) What would it be if the pipes were in a vertical position? Answer:

(a) 300,000 Wj m 2 ; (b) 236,000 Wjm 2

Comments on problems Problem 1

The maximum amount of heat that can be removed is given by cPH = U_A llT. Now AT and A are known. Thus, after determining the overall heat transfer coefficient U , ¢ 8 can be calculated. For the various partial heat transfer coefficients we find
ill.8. Heat ttansfer in stirred vessels Heat transfer in stirred vessels is strongly dependent on the flow pattern produced by the stirrer. As we have already seen in paragraph II.7:1 (Figure II.46) ~ propeller-type stirrer creates two systematic eddies which take up nearly half the vessel, the flow in which is 'parallel to the wall and to the shaft of the stirrer. A paddle-type stirrer mainly produces an eddy perpendicular to the stirrer shaft. whereas a turbine-type stirrer produces two eddies. one above and one below the stirrer. For the flow along the wall of the vessel thls means that a propeller or turbine stirrer produces a flow in the axial direction, whereas a paddle-type stirrer .induces flow in the tangential direction. The axial flow along the wall is caused by a jet of liquid which comes from the stirrer and hits the wall at a well-defined point. F rom this point a boundary layer is built up along the wall. The thickness of this boundary layer increases with distance from the stirrer. Also during tangential flow a boundary layer along the wall of the vessel is formed, but this layer has a constant thickness. It is understandable that it is not possible to produce a simple heat transfer model on the basis of this complex hydrodynamic picture, especially if we realize that the rate at which the boundary layer during axial flow along the wall is formed is dependent on the energy of the eddy and thus on the Renumber of the stirrer nd2/v. For relatively low Renumbers tbe momentum transport from the eddy to the wall boundary layer is small and we can make a rough guess of the

216 heat transfer coefficient at the wall on the basis of momentum penetration: a. ,....., -

l

<>r

A D~t A. .~.. =- = -Pr~ ~ Dh Dr

b"

A Prt ~TCvxjv..,

where xis the distance along the wall from the point of contact with the liquid jet and vw the mean velocity of the flow along the wall (which is proportional to nd). This model predicts that :

or :

.~. ~fv)tfv) + Nu '""Re~ Pr>\d \x where the diameter ofthe stirrer d and the diameter of the vessel Dare introduced because ex is independent of v·. For relatively high Renumbers-the momentum transport between the core of the eddy and the boundary layer determines the boundary layer thickness which is practically uniform (analogous to the boundary layer thickness during turbulent pipe flow) if the flow situation in the vessel is completely turbulent. This results in (see paragraph 111.3.2): Nu - Re

0 8 '

/D]O·S

Prt \d

For turbine-type stirrers wbich.induce mainly axial flow, the experimentally found exponent of theRe number varie~ between 0·5 (low Re, Re < 104 ) and 0·7 to 0·8 (very high Renumbers, Re > 5 .x 10 5 ). A rough description of the dependency of the N~ value on theRe number in the region of practical interest (103 < Re < 106 ) leads toRe exponents of 0·65 to 0·70. · The dependency of Nu on Pr1 as predicted by both q1odels (for high and low Re numbers) has been experimentally substantiated by a number of authors. The influence of the parameters Dfd and Dfx has been studied by only a few scientists and no agreement has been reached. Both theories predjct too strong a dependency of Nu on Df d or on Df x. This is due to the fact that the ratio v...,jnd (which we a ssumed to be constant) is a function of d/ D and of xfD. For turbine stirrers, for example, we might prove that v,..,/rui ,_ df D , because the width of the jet of liquid leaving the stirrer remains constant until it reaches the wall (see paragraph II.7.1); it is easily proven that in this case :

Only a few results of measurements of local beat transfer coefficients (see Figure 111.26) are available because measurement of these data is complicated

217

and, anyway, for design purposes a knowledge ·of heat transfer coefficients averaged over the height of the vessel is sufficienf . For flow creating a systematic eddy perpendicular to the axis, as is the case with paddle-type stirrers, we practically always find :

Nu...., Rel Prt which is another reason why for turbine- and propeller-type stirrers (which always produce tangential flow) the mean Re exponent is found to be ca. 0·65 for 103 < Re < 106 . In the foregoing we have considered heat transfer to the wall of a stirred vessel. The presence of a coil makes no principal difference but complicates the picture because the fiow of liquid is mare complicated. The presence of a coil will decrease wall beat transfer coefficients (by up·to 25 per cent for normal configurations) because of decreased flow velocity along the wall. Coil heat

transfer data are correlated with a coil-Nusselt number and are generally found to be: rt.dc

Nu = -

l

-. Re

t

.J..

Pr~

Because the combination of vessel and stirrer is not completely standardizedwhich is partly due to the fact that a stirrer is often not selected for heat transfer

performance only but also for other tasks (e.g. suspending, mixing, etc.)-th~ chemical engineer designing a stirred vessel has to search literature in order to find heat transfer data whlch can be applied for solving his problem. In doing so, the chemical engineer will have to answer the following questions: Which type of liquid flow is required? Which Re range is required? Are the literature data found applicable to the specific geometric configuration of the design? Are local or average heat transfer data required?

---o 0 0

0 0

H

"

l

0'

0

Figure ID.25

de

~

0

•

!

1.

:

H ~

.

3

D

I

I

"i"

:

l

I

~

1

0

- m ( 4 baffles)

5

( 6 blades )

4 d 6 - I

Standard configuration of stirred vessel

218 In the following, a few .experimental correlations for heat transfer to a 'standard turbine stirred vessel' are given (see Figure III.25). The average beat transfer c~fficient to the wall Qf a standard vessel according to Figure III.25 is:

J
(Nu )

14

(111.80)

for 10 3 < Re < 10 5 . For slightly different configurations the relation:

(nd2lt Prt ('-7w)0·14(d)0·13(L)O·l2 -

(rx)D - = 1·01 - ·

( Nu) = -

A.

v

D

t1

D

(III.80a)

can be used (0·17 < d/D < 0·75; 0·1 < L/D < 07). For the standard coli;. figuration equation (III.80a) yields equation (111.80) with a constant of 0·77. Local heat transfer data to the wall can be calculated from the relation -(standard configuration with h·= vertical distance from stirrer): Nu =

~ = 0·17Re+Pr•('1:)"'..(~)"'"

if

~ < 15

or: Nu

=

1·5 Ret Pr~ ( : 17

0·14 )

D if - > 15 h

if 103 < Re < 106 • Figure Ill.26 illustrates the dependence of the ratio of rx1ocaJ(a.) on the height of the vessel. The average heat transfer .values calculated from the local values by integration over the vessel height are in agreement with the values found from

0

,---

Height of stirrer

a/(a) -

- --

Figure 111.26 Local heat transfer coefficients in standard vessel

219

equation (111.80), i.e. the surface area under the line a./ ( rx) = f(D fh) in Figure III.26 is unity. · The average heat transfer coefficient to coils in a vessel with the configuration of Figure III.25 has been found to be:

( Nu )

=
.

= 0·17 Re 0 . 67

Pr 0 ·~ 7 ( Dd) 0 ·1

(d; )

0· 5

(III.81 }

for d d ' . 5 0·25 < -D < 0·58 ,· 0·03 <...!. < D 0·15 and 400 < Re < 2 x 10

A much-applied scaling-up rule for geometrieally similar vessels is to keep the stirrer energy input per unit of volume of the vessel (which is "' n 3 D 2 ) constant. In this case, for double-walled vess~ls, the amount of heat transferred per unit of time and of volume (which is -r:zD 2 j D 3 ) is proportional to D-4 (foi low Revalues) to v - 1019 (for average Revalues). For geometrically similar vessels. therefore, the heat transport decreases with increasing size of the vessel if the scaling-up rule of constant power per unit of volume is applied Dy.ring scaling-up steps have therefore to be taken to increase the heat flow (extra surface, larger temperature difference) or higher power consumption of the stirrer has to be accept~d. If vessels with heating or cooling coils are used and the heat transfer surface per unit of volume is kept constant, the heat transferred per unit of time and of volume is proportional to D- t to D-t if the above scaling-up rule is applied. Under these circumstances only a slight correction of stirrer speed (and thus of power consumption) is required to keep the heat transport during scaUng-up constant. , lll.8. 1. Problem 1. Two water! ike liquids which react instantaneously with each other are pumped

through a continuously stirred tank reactor. Pilot plant experiments with a double-walled tank (diameter of vessel = height of double wall = D = 04 m ; stirring speed n = 360 rpm) according to Figure 111.25 showed that the production capacity of the reactor is determined by the rate of removal of the heat of reaction. It appeared that the heat transfer coefficient was independent of the cooling water flow rate. (a) By wha t factor would the production capacity per unit volume of the pilot plant reactor be increased if additional cooling by means of a coil (O.J m diameter. 0·01 m tube diameter. 15 turns) was applied? (b) T he reaction is to be carried out on a factory scale in a double-walled reactor, according to Figure. III.25, of 3 m diameter. By what factor is the production capacity per unit reactor volume increased if the scaling·up rule of constant stirrer power consumption per unit reactor volume is applied?

220 (c) What stirring speed would have to be applied in situation (b) if the same production capacity per unit volume as in the pilot plant reactor is desired? (d) The chief engineer of the factory wants to use a vessel with a cooling coil instead of a double wall A coil of 2·5 m diameter, made from pipe of 5 em diameter, is available. How many loops of coil are needed to assure the same production capacity per unit volume as obtained with the reactor of situation (b) if again the scaling·up rule of constant power consumption per unit reactor volume is applied? (e) What is the ratio of the cooling surfaces of cases (b) and (d)? Answer : (a) 2·13 x ; (b) 0·11 x ; (c) 2700 rpm ; (d) ca. 15 turns ; (e) AwarJAcoil = 1·6.

ID.9 Heat tramport by radiation 'Heat rays' are electromagnetic waves (0·5 < ). < 10 pm)* which are radiated by a body owing to thermal agitation of the composing molecules. Between two surfaces radiation exchange may take place if the interspace is more or less diathermanous for the entire or part of the wavelength range. There is, th~ heat transport in both directions and, if the temperatures of the surfaces differ., the difference between the opposite heat flows does not equal nil By means of thermodynamics it was derived that the heat flow which is emitted per unit of surface cJ>; by a so-called black surface with temperature T is given by the Stefan-Boltzmann law: (III.82)

where the radiation constant (J = _5·75 x 10- 8 W(m 2 °K 4 and T the absolute . temperature of the surface. For a black body the intensity distribution over the wavelength range is given by the Planck radiation law. The wavelength at which this intensity 1s maximal is inversely proportional to T (Wien's displacement law):

Amax T = 2880 (micron °K)

(III.83)

At room temperature )·max = 10 ).lm (fa{ infrared) and at 6000°K ~ax = 0·5 J1rt1 (yellow). ·A completely black surface emits the maximal amount of radiation relevant to its temperature; it has the emission coefficient e = 1. According to Kirchhoff's law (see equation III.84) the absorption coefficient fo r the radiation on the surface is then also a = 1. A body which absorbs all radiation and does not reflect an y is indeed black. In general, technical surfaces have an emission and an absorption coefficient smaller than 1 ~ the value depends on the material and the roughness of the surface, on the temperature and on tbe wavelength of the radiation. If in a *Other exampl~ of electromagnetic waves are radio waves (.4 > o-8 Jllll), X-rays (J. ;:::: 10- 4 .um) and 1-rays {i. R: w-6 pm).

J
< ). <

221 certain wavelength range preferential absorption occurs, then just ·as in the case of visible light the surface is 'coloured'; In engineering, mean absorption coefficients over the entire wavelength range will often suffice. In other words, such surfaces are considered as ~grey• between the extremes 'black' -and 'white'. Of most non-metallic surfaces

a > 0·8, so they are almost black for the infrared radiation. Metals reflect the heat rays only satisfactorily if polished and clean; the absorpt~on ·coefficient a· lies then between 0·05 and 0·20. Oxidation and contamination· increase a. Figure 111.27 shows absorption coefficients fo_r a. number of differ~nt ma~erial~. .. . .. '

Pol ished metals

I

.

,

I Oxidized metals

. I.

·I

f-

: I{ rr1-----;---~---..,.-----+-~ l 1-lI I

1

.

..

Non-.me!als ~

~~~~~linin~ Asbestos

I'

Gypsum I Soot 0·8 r 1---:----.:._-41- - - - - - . . . J .I _ Refractor y bricks

r-- -- - ---+----[ron _ ___.__Cu ., 0 ·6 Bross

I

r-------+1-AI paint r

I

r--------r:- Ni iron

!""--Cu

Ni

0·2-

.1

I

I I I I

1 !=====~~---1---A t

F-==~A~l- Pt,Zn t0 -Ag

-,

I

04 r

~Cost

I I I

I

1 1

I

Figure Ill.27 The absorption. coeffiCient for heat radiation

'

The amount of heat which is transported between two or' m
'

'the spatial configuration (how fat can the surfaces 'see' each other?); the absorption and emission capacity of these surfaces ; the possible· absorption -by the intermediate medium (water v.apour and ·carbon dioxide, for example, have strong absorption bands in the infrare¢) ; . the emission of the intermediat'e medium-(luminous flames)'. Only the first two will be further treated here_ For calculating the net transfer by radiation a goo d ·balance of the radiation transport has to be drawn up. This can best be done by using PoJjak,s c~culations scheme. Poljak ·assumed that for every wall the total energy'flux 4>" of the radiation which is emitted by the wall equals the sum total of the energy flux of its own emission' euT 4 ( = e¢;) and the energy flux of that part o f the foreign radiation (")on the wall which is refiec;ted it ((1 - a)¢i'): . · .

by

(HL84a)

222 (total radiation = own emission + reflection foreign radiation). Moreover, the energy flux of the net transport which is emitted by the wall(¢;;) equals the difference between its own emission and that part of the foreign radiation which is absorbed by the wall:

4>:_= e~ - iuM .

(III.84b)

(net radiation = own emission - absorption foreign radiation). Much can be concluded frpm these relations:

(a) If the wall is completely enclosed by black bodies~ then in the case of thermodynamic equilibrium ¢~ = 0 and ; = <1T4 . The Poljak system can be formulated even more practically by converting equation (111.84) algebraically as follows : (I II.85a} (IU.85b)

using Kircbhotrs law. The practical course of calculation is then that with equation (lli.8Sb) for each wall the total radiation is expressed as a function of the temperature of that wall (l/>~ = .aT 4 ) and of the still unknown net transport 4>~- With the relations for the total radiation of each wall thus obtained the foreign radiation f>'( on each waiJ can be calculated. These ¢;' values used in equation (UI.85a) then give the net energy fluxes t/>~ for each wall as functions of the wall temperatures. In order to calculate the foreign radiation '(" for a wall with number k, we must know the fractions Ejk which indicate which part of the total radiation of wall j directly reaches wall k. These view factors Fare calculated by cumbersome integrations (summation o~·er small surface elements) and we have to take into account that the radiation intensity depends on the angle cjJ between the direction of the radiation and the norm~ to the surface (approximately proportional to cos
223

A 1/ A 2 of the radiation which is emitted by wall 2 reaches wall 1 ;·the fraction (1 - AdA 2 ) again reaches walll). With these data we can now ·calculate the net radiation transport between two bodies of which the one (1) is completely enclosed by the other (2). The foreign radiation over wall 1 ( =A 1 f'1 ) is then equal to F21 times the total radiation of wall 2 ( = F21 A 2 cf>;). It follows that the energy flux of the radiation Qn the smallest wall ( = rfJi'1 ) equals the total energy flux of the radiation emitted by the larger surface ( = cf>;). So, using equation (III.85): .A._,

'¥111

=

A., II -

o/1

A.,~l

'+'12

=

A., II

o/1

-

~II

o/2

=

.A._II

'f'z l

-

l -

a

a ·l

1


.A._ JI

o/:2

+ 1 -a Q2.A._II '¥1'12 2

Between the net radiation transport of body 1 (¢; 1A 1) and that of body 2 (
cp;1A1 +

~2A2 = 0

because the energy lost by wall 1 most be taken up by wall 2 (law of energy conservation). From the two latter relations we find for the net radiation transport from body 1 :

~

l

+ 1 -a

a1 1

+ A1 A

2

1-

a

a2) cplld = cp"zt. _
2

1'!) 2

(III.S6)

This is an important relation which is often used for calculating radiation

transport If the temperature difference {T1 - T2 ) is not too great, (Ji - T~) can be expressed approximately as 4 P(T1 - T2) . In these cases a heat transfer coefficient for radiation Cl.r can be defined, which according to equation (Ill.86) equals:

a2 ) a,.="W_"7'1 l 3 ~ 1 + 1 - a 1 + -A1 -1 ---= .A

a1

A2

a2

1

(III.86a)

This means that at room temperature a,. is about 6 Wjm 2 oc and can therefore not be neglected with respect to free convection. At higher temperatures a,. ~ increases considerably and hence so does the radiation contribution to the total heat transport.

II 1.9. I. Problems 1. A long rod (diameter l ·l em, emission coefficient 0·8) with a temperature of 327°C is suspended in quiet air of 2rc. Calculate the contribution of free convection and radiation to the total heat transfer coefficient. Assume the air to be an ideal gas with a density of l·Okgjm3 at 27°C.

Answer : horizontal rod: 46 % free convection 54 % radiation vertical rod :

29 % free convection 71% radiation

224 *2. A small ceramic sphere (diameter 1 em), e.g. of a thermocouple holder, is situated in the middle of a long tube with a diameter of 20 em. Air of l50°C is passed through the tube at a mean velocity of 20 m/s. The temperature of the tube wall is l00°C. Calculate the temperature of the sphere.

Answer :

l47°C

3. Check that if between two large flat walls with equal emission coefficients a third equal waU is placed the net heat transport by radiation is halved. An example of this type of radiation shield is the layers of aluminium foil in the walls of refrigerators and cold store rooms. 4. The maximum intensity of sunlight occurs at a wavelength of approximately 0·5 x 10- 6 m. Calculate the surface temperature of the sun. Comments on problems Problem 2 In steady-state conditions, heat transport by radiation from the sphere must equal convective transport to the sphere. Thus : ( ex)

(T, - TJ = a.,(7;

- Tw)

thus ( a ) = 183 Wfm 2 oc (equation 111.68). Assuming 7; ~ T, , 'f = (T, + TJ/2 = 125°C ~ 400°K. Thus,employingequation(lll.86a) we find (a 1 = 0·9, A1/A2 « 1): Now Re

= 7000,

CHAPTER IV

Mass Transport

In this chapter the transport of matter will be treated. The starting point of our analysis is the law of conservation of matter, which reads for a substance A (see paragraph 1.1): (IV.l)

The following pages are no more than an elaboration of this law.

Apart from agreement in the description of the transport of matter and heat transfer, there are distinct differences as, for example :

(a) The 'production' of mass (by chemical reaction) is usually a compJjcated function of the molar concentrations of the reactants (and of the temperature), whereas the production of impulse and heat is seldom a strong function of concentration or temperature. (b) The interface between two media between which matter is transferred (e.g. between two liquids or a gas and a liquid) is usually mobile, contrary to the interface between two media between which heat is transferred (usually a solid wall). (c) The order of magnitude of the diff~sion coefficients ID of liquid and solid substances is much smaller than, for example, the thermal diffusivity or the kinematic viscosity (for gases at 20° C and 1 atm, D ~ 10- 5 m 2/ s ; for liquids at 20°C , D ~ 10- 9 m 2 j s ; and for solid substances D varies widely between 10- 10 and 10- 1 5 m 2f s). This means forliquids and solid substances that in the first place the penetration of matter in a given time takes place over (much) smaller distances than the penetration of heat or momentum ; that in the second place already with convective currents of low velocities, wruch frequently occur in the case of mobile interfaces, the convective transport exceeds the diffusion transport in the direction of flow. From the foregoing there appears to be a dilemma in the description of the mass transport. We formulated a law of conservation of mass because mass is the most characteristic property of matter. Accordingly we defined the diffusion coefficient in paragraph 1.2, equation (I.13), as the proportionality constant

226

between the mass flux and the gradient in the mass concentration: ,~.,

= -ll}

'+'mA, :r.

A

dpA

(IV.2a)

(Fick)

d;c

In addition, production rates in chemical reactions are functions of the molar concentrations, which we. shall indicate by cA, etc. (kmolfm 3). So we can imagine that for many calculations it offers advantages to define the diffusion coefficient as the ratio between the mole ftux (kmoljm 2 s) and the gradient in

the molar concentration :

,.~,.,

'+'mol A,x

=-

dcA

n LY A

(IV.2b}

(Fick)

dx

Both thus defined diffusion coefficients are in principle different (although in practice they are often almost equal) and in applying them we should realize on which concentration the description is based. The diffusion coefficient has been defined in such a way that the net transport through a fixed plane (in Figure 1.6 the plane n = constant) is zero. If we restrict ourselves to binar'y systems (with components A and B) for practical purposes two cases can be distinguished in which the net transport is zero:

A For every unit of mass A which diffuses through a fixed plane a unit of mass B moves in the opposite direction by diffusion ('barocentric' system). B. For every mole A a mole B moves in the other direction through the plane.

sub A. Net mass flow

= 0. p = constant

For the diffusion transport of substance A we can write : ¢':rrA,:r

=-

[)AB :A

whereas for the transport of substance B applies: A.."

'f'w.B,x

= - 0

BA

dps

dx

where ¢':nA.x is the mass flux of A (kgjm 2 s) in the x-direction for a net mass flow = 0 ; PA is the mass concentration of A (kgfm )). Because the conditions have been such that: ,/,."

"f'rnA.x

+

A-,•1

'f'mB .x

=

0

whereas :

PA

+ p8

= p =constant

or dpA

+ dp8

= 0

it follows from these relations that DAB . DnA. In other words, the definitions have been determined in such. a way that the diffusion coefficients of A in B and of B in A have the same numerical value. These definitions are used for describing the dlffusion transport in solid substances and liquids ; here, in the

227

case of not too high concentrations of, for example, A in B the density may be assumed to be constant. · sub B . Net mole ftux =

o. the total concentr~tion c =constant (kmoljm 3 )

Now we can write for the mole flux. (kmol/m 2 s) in the x-direction: n

dcA dx

A." 'f'mol A..x

= -

A." 'YmolB.x

= -:- 0)BA dc dx

11..11 AB

and : 8

and cB are molar concentrations (lcmoljm 3); for ideal gases and constant pressure and temperature their sum total is constant and equals c. These definitions are very suitable for describing the diffusion transport in gases. In the case of equimolar diffusion :

cA

A." 'f'mol A.x

A." 0 + '#'mol B,;r =

Also, here, the diffusivities have been defined in such a way that : _[]) AB

= . O)BA

According to the two definitions the dimension of II) is (length)2/ time, so that Din the MKS system is expressed in (m2 /s). This was likewise the case for the two other transport coefficients v (kinematic viscosity) and a (thermal diffusivity). However, in the case of diffusion it is .advisable to distinguish between two cases: p = constant and c = constant For these cases []) has to be defined using respectively the gradient of the mass or of the molar concentration. . IV.l. Stationary diffosion and mass transfer John noticed that, before m had left his office, the lawyer who had disappeared had prepared a pot ofcoffee. A cup on his desk was still half .fuU and from the scale on the sides. caused by evaporacton ofrhe wacer, he could see that the cup had been .filled to 1 em below the rim. He estimated rhe cup co be 8 em high. remembered the pages to come and concluded that the lawyer muse have left h is office approximately 17

days before.

IV.J.l . Stationary diffusion During most practical cases of diffusion neither the net mass flux nor the ~et mole flux discussed in the foregoing paragraph will be zero. A practical example in which the net mole flux is almost zero is the distillation in which, for every mole of the light component which passes into the gas phase, approximately one mole of the heavy component enters the liquid. However, on absorption and extraction of one component from a mixture the net transfer is invariably

~

:_ t;A·'

I

' ;~ ij'

. I

,,

228

, I

It

~

~·

.

unequal to zero. AI~() guring a chemical reaction e.g. at a catalyst ·s urface, the mole fluxes of-the re~9.~~t$ ~o be f~d. ~n(:i the mole fluxes of the product ·to be discharged are usu~UY y~~gll~ ·<~~g, ·in ~ 9.tm~ri~tign). When describing q}~ m~ ifi:iil~f~r til th~§~ ~@.~s ~!lgw~Qc~ ll1l!st b~ m~qe for the entrain~ent of &pbstilP.f~1i fP ~h~ ~PW ~~~q b.Y th~ ma-s~ tnm&fer as SIJCh. Fqr ~ biJl~ry mi~tl.q-e J1lO.Vliig wit~ fl velocity v relative to .a plane X = constant, t}le steady-state diffusion flux of component A relative to this velocity is·:

(vA

v)cA

-

de A

= - 0 dx

Tht~ rate of molar transport relative to ·t he plar~

':not A t

VC

=

'

" ! :·


.


I.~

I

mol A,;.;

\,

. .

=-

((jJ"

[DdCA

dx +

mol A,x

+

x = constant is then, since . . ..

if>" . m ol

B,.x)

CA C

{1V.3a)

:' . , ·,.. . , ··;. ,.... whereas'!ihe mass ·tra:nsport is given by : \,

first right-hand teqn in these equations repre~nts the mass flux resulting from diffusio'n, whereas the secopd is a. m~a~!ll'~ the mass flux result~ng fro~ ~o~~~ mola~ flow, Equations (11I~3a) and (IIIJb) can only be solved if the rel.atio~ b.~twe~n' Hi~'fripl~at ff.~x~s of Aand B;s kl}gwn. We cari di~tinguish two exJreJ11~ s~wat1ons: . .. . i •

Th~

term

of

I

(a) 4>';.. +
•

) . :, •

::-: • : \ . . . :;-;, . .

.

·.~ ~..

.. '

..'i

:

.

:~· :·~-~ · ~"! ~

..

••

i

• !.

:\

l

~

:

,

1 l ~..!~ _1 :·. .. ~~ ··:·. . ~~~{L:.: '· i.' ·.~ 1

I :.:\ 1 \,

•

·\

•

\

:l.

~. ~. ' . .

••

• • •

In;·practice,. sit.'"l;lti.on.s; be,tw.een . these .extr,em~s,:"w.ill ofteq ·o9cur, but sine¥ ~he qiff~rence~ P~tw~ep ~s~s (a).an~(~) ~t;~.o.ftF9.&JA~U, it is ~earl:y 1atways p~.ssible to>ptoduce :a good es:timate·of the:molar flu,x by making a ahoice between these ca~es~ :.l n)the·following, we~.will.anatyse 'the t:Wo extreme:.cases. st~~ed ·in.greater ,_ ' •

•

•

de~u.

•

•

:·

•-

· ·.. .

:1 ~ '':...,(

•

- ·~/· ··. :-

·• •

-.

••

-·

-. • "

•••

' ,.. •

•

~

•

•

'

. ,,

(aj ~Etl~iiiJD.lar· d.iif:J!..siPn, ;·= 0.

••

•

·

.

~quation {IV.,3~ can now· be simpljfied t~ :

~f :we

consiq¢.r. one-dimension.a l

station~ry , diffusion

in the x-direction, the

mass balance (IV.l) yields (there is no production of A) :

. or ¢':nolA,;= coqstant for every plane

.e quation · yields, ; with

! .

'

··

..

':

=constant. Solution' of this the boundary conditions · cA= cA·; at x ' x 1 and X

.'i

229

c: .~

~c:

&I

u c: 0

u

l

- - Distance

Figure IV.l Concentration profile during equimolar diffusion (drawn line) and during diffusion through a stagnant fluid (dotted line)

cA

= cA 2 at

x = x 2, the concentration -distribution : CA- CAl

X -

X1

(IV5t

We see that the concentration distribution (Figure IV.l) is linear. · The mole flux of A is found ·after substitution of dc.Jdx in equation (IV.4) by the value calculated'from equation (IV.5) to be : (IV.6)

So, during equimolar diffusion the mole flux (and analogously also 'the mass flux) is directly proportional to the concentration gradient llcjAx. We notice that mass transport by equimolar diffusion is analogous to the transport of heat by conduction. For one--dimensional beat conduction we had found in paragraph 111.1 .1 (equation II1.2a): A-"

'+-' H

= - /,• T2 -

T1

x 2 - Xt

=-

peP~ - pcPT2 Q--"' - - - - - . _ _ _

x2-

xl

This equation is identical to equation (IV.6) if we replace the molar diffusivity 10 by the thermal diffusivity a and the molar concentrations c (moljm 3 ) by the 'heat co ncentratio ns, pc PT ('W jm 3 ). We can therefore also ad apt the relations found for stationary heat conduction around cylinders and spheres to diffusion. Using equations (lll.6) and (Ill. 8) we find for equimolar diffusion

230 between coaxial cylinders with diameters D 1 and D 2 and concentrations CAl and C Al : (IV.7) and for equimolar diffusion between concentric spheres (diameters D 1 and D 2 , concentrations cA 1 and cA2) : mol A

2-nD D

= R) D2 -1 D21 (cAl

-

CA z)

(IV.8)

(b) Diffusion through a stagnant body. cb':notB = 0 This process can be observed, for example, during selective absorption oi extraction of material A from a mixture of A and B. For this situation equation (JV.3a) can be simplified to: _

A."

'PmolA,x- -

c

0 C-

dcA

dx

CA

(IV.9)

Application of the mass balance (equation IV.l) yields, again, that under stationary conditions and no production of A the molar flux of A through aU planes x =constant must be the same. So :

de A - = constant

dx

The above equation can be integrated and, applying the boundary conditions x 1 , cA 1 and x 2, cA2, we find for ~he concentration distribution: C -

CA

C-

CAl

=,1

CA>) (x-.xl)/(xl -.x!)

C-

\c- CAl

(IV.lO)

This concentration distribution is also shown in Figure IV.l for materials A (e.g. solvent) and B (e.g. solute). With the aid of equation (IV.lO) we find from equation (IV.9) for the molar flux for diffusion through a stagnant body : ,~,.,.

_

'f'moiA,x -

In c - cA 2

Oc X2 -

C -

X1

CAl

(IV.ll)

If cA is at all places much smaller than the total concentration c, the logarithmic factor in this relation can be approximated by : }n C C -

CA2 ~ _

CA2 -

CAL

CA l

C

and we find :

(IV.12)

the same result as for equimolar diffusion.

231

Mass transport through a stagnant fluid is a factor of I' _ JD-

C CAl -

} CA2

n

C -

CA2 _

C -

CAl

-

1

+

1

2

CAt

+

CA2

2

+ ...

bigger than during equimolar diffusion. The correction factor fD is named after Stefan; if trus factor is applied to equations (IV. 7) and (IV .8) these equations will also correctly describe, for the geometries considered, diffusion through a stagnant fluid. ·

I V.J.2. Mass transfer coefficients Analogous to the introduction o f the heat transfer coefficient ex in paragraph 111.1.3 whlch was defined by: dT


ex =

T... - ( T ) = -

A dx

x=O

Tw - ( T )

we can now introduce a mass transfer coefficient k which is defined as the ratio of the mass flux and the concentration gradient (which constitutes the driving force) in o n e phase :

kA

=

A.." '+'mol,A

cAo - (cA)

dx

_ _ rn.

-

!LJIA

x•O

cA, - (cA)

(IV.l3)

Here, 4>':no1A is the mass flux o f A into phase Band cA, - ( cA) is the difference between the concentration at the interface and the average concentration of A in phase B (mean cup concentration). For diffusion from a sphere into an infu1ite stagnant medium we ·find, combining equations (IV.13) and (IV.8) (the latter for the case D 2 ~ ex:>):

The dimensionless number kAD/ 0 is called the Sherwood number Sh, which plays the same role in mass transfer as the Nusselt number in heat transfer. For diffusion from a sphere into a stagnant infinite fluid Sh = 2, analogous to Nu = 2 for heat conduction. The dimension of the mass transfer coefficient is the same as that of velocity, distance{unit of time.* If a certain component is exchanged between two mobile phases (e.g. gas absorption, extraction, distillation, etc.) we encounter an overall mass transfer coefficient K which is composed of the partial transfer coefficients in the different

* In mass transfer with gases the partial pressure

instead of the molar concentration is often used for calculations. A mass transfer coefficient is theo defined as: 4J:C<>1A = kr{pA.w - PA.r)

since for ideal gases c = p/ RT, k, = kJ R T.

232

phases. Contrary to heat transfer (paragraph·111.2.2), however, in the case of mass transfer it is not possible to calculate the overall coefficient by summing up the partial resistances (1/ k) to the total resistance (1/ K). This is due to the fact that the concentrations in both pbases are generally not equal if the two phases are in equilibrium. Figure IV.2 shows an interface between two fluid phases through which a stationary mass flux ~ol A

=

k'(c~.w - C~.J)

= k"(c;._.J

- c~.w)

(IV.l4) :

It may now be assumed that there is equilibrium between the two phases at the interface, so that c:...wand cA,ware related according to the equilibrium relation for this system. i'-lnterfaee

~

0...

c •uc

I Phase

u 0

I

1

c'A,w

10 . P nose

I

~

I

cA,f

A,w

Distance

Fig..e IV.2 Concentration profiles near the interface of two liquid phases

In principle, with this equilibrium relation cA,w and c;.,w can be eliminated from the previous relations, so that a relationship between
1 (2_ + _ _)- (c" - cA,JJ k" mk' m 1

A."

If' mol A

=

A ,f

=

K"('c"

A,f

- ci.JJ m

(IV.l5a)

t Henry's law for gas-liquid and Nernst's distribution law for liquid-liquid systems: the relation between the distribution coefficient m and the Henry coefficient He is m = HefRT.

233 or also : II

-

¢moi A -

l

m

k•

1) -t

+ k'

!I

!

t

If

(meA,/ - cA./)= K (meA./ -

I

c,..,1)

(IV.15b)

It appears that the total mass transfer resistance (1/ K", if related to the phase and 1/ K' is related to the phase ') is composed additively of the partial resistances 1/lt and 1/ kH, but that at the same time the relation of the equilibrium J'

concentrations is involved. The reader may check that the partial resistance of the phase in which the equilibrium concentration is lowest is relatively of the most importance. Further. it appears from these relations forK that the concentration difference, or 'the driving force' with the help of which k 11 and k' have been defined, represents the deviation from the equilibrium between the two phases. If we write instead of cA..rfm., c~~f (i.e. the concentration phase " would have if it was in equilibrium with cA..r ), and instead of meA./ , cA:1 , we get the more general definition for k' and k" : A." 'f'mol A

= K '(c'* A./ - c'A,/ )

= K "(c"A,/

- c"A./ *)

When in a certain apparatus (mass exchanger) a mass flow mol A of a certain component must be transferred from one medium to another, the required exchange area A follows from an equation of the form :

where K and c are related to one of the two phases. If K does not vary too much over the apparatus, in the case of cocurrent or countercurrent flow of both phases, for ( cA- c~) the logarithmic mean between the driving forces at the inlet and outlet of the apparatus can be taken (see also paragraph IV.4~ IV.l.3. General approach for the calculation of concentration distributions The partial differential equation which forms the basis for calculating concentration distributions is obtained if the law of conservation of matter is applied to a small (e.g. Cartesian) volume element of the medium. If the fiow which may come from the diffusion itself is neglected, so that Fick's law (equation IV.2) can be used for a fixed surface element in the medium, the derivation is completely analogous to that of the general differential equation for the heat transport (see paragraph III.1.4~ The resulting equations which we can now derive are for a component A :

(IV.16a)

234 where r A = production of A in kg/m3 s (barocentric system) and :

dcA dt

+

dcA

Vx d x

+

dcA Vy

dy

dcA

+ V: dz

= 2

ff)Al dx; (d c

2

2 d c d c } + dy; + dz2A

+

RA

(IV.l6b)

whereRA = production of Ainkmoljm 3 s. Strictlyspeakingt these two equations only apply at constant IJ) A and if PA « p orcA« c. In that case they can also be used if more than one component is present in a low concentration in the medium. TI1e diffusion equation for low concentrations with constant ID and without . chemical reactidn (also called Fick's second law) corresponds entirely with · Fourier's differential equation (equation I11.2~ The solution of a great many diffusion problems can therefore be found in books on heat conductivity (see, for example, Carslaw and Jaeger). The distribution of the neutron concentration in a nuclear reactor is in principle also described with an equation such as (IV. l6), where rA, the neutron production by nuclear fission, is proportional to the neutron concentration. The production terms rA and RA usually depend on the concentration in the mixture in a way which is determined by the chemical kinetics of the reaction. For simple kinetics and for simple geometric situations solutions of equation (IV.16) are known, especially for the stationary statet We shall return to this subject in paragraph IV.S. IV./.4. Film theory

Analogous io the boundary layer theory in heat transfer (see paragraph 111.3.2) Lewis and Whitman introduced the concept of a boundary layer to mass transfer. They assumed the mass transfer resistance to b e located in a stagnant boundary layer of thickness be (see .F igure IV.3) through which mass transport has to take place. The mass balance (equation IV.16b) reads for this case (stationary state): (IV.l7) At the interface (x = 0) the equilibrium concentration cAo is present, whereas on the other side of the film (.x = <5,) the concentration equals the bulk con· centration cAt· Solving equation (IV.17) with these boundary conditions yields the concentration distribution: CA -CAl = CAo-CAI

1-

X

8,

(IV.l8)

235

I ---x

Figure IV.3 Film theory eoncentration distribution

With the aid of this concentration distribution we can now calculate the mass transfer coefficient for which we find: k = - DA

deA l dX

CAO -

x=O

( CA) =

[)A

tJ~

(IV.l9)

Thus, the Sherwood number is given by : kD

D

[)A

~c

Sh= - = -

(IV.20)

and we see the formal analogy between the Shand the Nu number (Nu = D/ br) and the dimensionless expression jJRe ( = D/ll,J, which .all represent the ratio between geometrical scale and the thickness of the boundary Layer for the transport process concerned We may not conclude from equation (TV 20) that the mass transfer coefficient is linearly dependent on the diffusivity because !Jc is also dependent on [)1, as will be discussed later. The thickness of~' of the mass transfer boundary layer cannot be determ.i ned directly but can be calculated if the mass transfer coefficient is known. So, the film theory does not actually help us to predict k values. The advantage of this theory is the simplicity of the model and the possibility to estimate the order of magnitude of the film thickness over which the mass transfer process occurs. A further advantage of this· theory is the fact that the influence of a chemical reaction on the mass transfer process can easily be studied: We will therefore use the film model extensively in paragraph IV.5 when treating mass transfer with chemical reaction.

236

I V.l .5. Problems *1. The vapour pressure of naphthalene at room temperature is approximately 0·05 mm Hg. If a mothball (diameter 1 em) consisting of this material is suspended in still air, calculate the initial rate of vaporization. How long does it take before the diameter of the mothball is reduced to half its initial value([)) Of naphthalene in air = 0-7 X 1o- S m 2/ S, p - 1150 kg/m3 )? Attswer :

4>m

= 1·15

x 10- 10 kgfs; t

= 23 years

2. If in a gas phase a reactant A diffuses to a catalyst surface for dimerization, the mole flux of A is given by: ..1,.11

'f'molA,.;>~'

=

_[])A

2c

de A

2c- cA dx

Prove this. 3. If in a desiccator a wet plate is at a distance of L = 0.1 m fTom a layer of a drying agent, calculate the drying time if: total concentration c = 4-5 x 10- 2 kmoljm 3 ; equilibrium vapour concentration of water under the prevailing conditions C = 1·1 X 10- J k:Jnolfm 3 ; diffusion coefficient of water vapour 0 = 2·5 x 10- 5 m 2 /s;

surface area of the plate

= 12

x 10- 4 m 2 ;

amount of water to be evaporated = 1·2 x 10- 3 kg.

The desiccator can be considered to be isothermal and convective :flow can be neglected.

Answer: 56 h 4. In a closed gas burette two equal volumes of a pure gas and a pure liquid are brought into contact with each other under atmospheric pressure and at room temperature. After shaking vigorously for some time the pressure appears to have become two-thirds atmospheric, whereas the temperature has not change
Answer :

* 5.

m = (concentration in liquid/concentration in gas phase) = 0.5

Small bubbles (original diameter d 0 ) of a pure gas are brought into a liquid in which they ascend very 'Slowly. During ascending the bubbles disappear by absorption (Sh = 2). It appears that the time of solution t and the

original bubble diameter satisfy the relation d~ = ~nstant x t where the constant has the value.2S x 10- 9 m 2 / s. Give a theoretical explanation for this -relation and calculate from the constant the diffusion coefficient of the gas in theliquid,ifthesolubility of the gas is given by m = 0·5.(concentration in the gas phase divided by concentration in the liquid phase). Assume that I

237 the pressure and consequently the concentration in the bubbles remains constant. Answer:

fi)

= 25/ 16 x 10 - 9 m 2 f s

• 6_ A bottle filled with dietbyl ether is connected with the surrounding air (T = l5°C) by a long vertical tube (L = 50 em, D = 2 em). The diffusion in the tube is determinative of the rate of evaporation of the ether. Per second, 16·5 x 10- 9 kg ether evaporates. (a) How much ether t/Jm would evaporate per second if the tube were twice as long and twice as wide? (b) What is the diffusion coefficient of ether in air? (c) Make a rough estimate of the time one should wait after filling the bottle before the diffusion coefficient can be measured with this set-up. The vapour pressure of. ethyl ether at l5°C is 4·85 x 10 4 N jm 2 .

Answer:

(a) 3·3 x 10- 8 kgjs (b) 1·30 X w - s m 2 js (c) ca 120 min

7. The air pressure (1·5 atm) in the tyres of Klaus' car decreased by 0·05 atm in 6 months. What is the diffusivity of air in rubber? Data: volume of tyre 251 surface of tyre 0·7 m 2 wall thickness 1 em solubility of air in rubber m = 0·07 m 3 j m 3

Answer : 2·6 x 10 - 1 1 m 2 js 8. Somebody measures the diffusion coefficient of water in oil by pouring a layer of oil onto a layer of water in a glass vessel (cf> = 6·8 em). The vessel is put in a thermostat (21°C) and the space above the oil is kept dry with P 20 5 • The weight decrease of the vessel is measured as a function oftime. The results are as follows : Layer thickness (em) 0·00052 0·13

0-37 0.63

t-35

2·26

Weight loss glass vessel (gfday) 3-14 0.0116 ().0042 0.0032

().0046

00053

Calculate the diffusion coefficient of water in this oil and explain the results. The solubility of water in the oil at 20°C is 0·004 i>er cent by weight ; p oil = 880 kgfm 3 •

Answer :

0 = 1·4 x 10- 9 m2/s

238

*9. Show that John drew the right conclusion in t he case of the missing lawyer discussed at the beginning of this paragraph. Comments on problems Problem I Since the air is stagnant. Sh = 2 is valid (paragraphs IV.l.2 and IV. l.4-)_ We can assume the partial pressure of naphthalene to be equal to zero at a great distance from the mothball and be equal to the saturation pressure p* at the surface of the ball (no resistance at the interface). So, the molar flow rate is given by : ..l.. 'l'mo1

= kAc* = 2

°nD _.!!_ RT 2

D

which leads to ¢mol = l- 18 x 10- 12 kmoljs and, with the molar weight of naphthalene being M = 106 kg/kmol :

4>m = MQ>mol = 1·25 X 10- 10 kgls A

mass balance of the mothball shows: dV

cPm = - P dt

=

-4npD

2 dD

dE= Mc/Jmol

Substituting for ¢mol the expression found above and integrating between t = 0, D 0 ; t, D we obtain : p(D6- D 2 )RT t = 8M[)p*

which leads tot = 7·12 x to - ss ~ 23 years. In reality, the evaporation of the mothball does not take so long. This is because, in reality, the air is not st agnant. Problem 5 The concentration of the pure gas A in the bubbles is cA.t = 1 m3/m 3 and we can assume the concentration in the bulk of the liquid to be cA,L = 0. We further assume the concentration at the in terface cA..i to equal the equilibrium concentration c! = cA.Jm (no resistance of interface, pure gas, therefore no gas phase resistance)_ The volumetric flow rate of A from the bubble is then : ,~.

'l' v

dV dt

= Akc* = - -

A

and, using Sh = 2 to sub stitute fork, we find after integration that : [0=

constant m 8[}cA,g

=25 - x l o- 9 m 21s 16

239

Problem 6

We can assume diffusion of ether vapour through stagnant air and equation

(IV.ll) can be applied. The partial vapour pressure of ether at the liquid surface

is the equilibrium pressure p* (no transport resistance at the interface); at the end of the tube the partial pressur~ is zero. In order to roughly estimate the time before the vapour starts to leave the tube we assume a sharp interface ether vapour/air which moves at a velocity dL/dt. A material balance then shows :

r~.

o/mol

=

* A dL dt

C

..

and after combining this with equation (IV.ll) and integrating between t :::; 0, L = 0 and t, L we find t = 121 minutes. Problem 9 John and the case of the disappeared lawyer

The situation is illustrated by the sketch. At the top of the cup, the water vapour pressure is zero {assuming dry air in the office); at the coffee surface the equilibrium water vapour pressure p* is present (,.., 20 mm Hg at room temperature). Water vapour diffuses through stagnant air and equation (IV.ll) yields the molar flow as a function of the distance L from the top of the cup (total concentration c = pj RT = 10 5 /8·2 x 10 3 x 298 = 0·042 kmol/m 3 ). ·

A mass balance, on the other hand, gives for the mass flow of water vapour: _

dV

m = PCft

=

dL pAdc

Combining both equations we find (M = molar weight of water): mot

AOp p - 0 = LRTln p- p*

m pA dL M = M dt

Integrating with the bounda ry conditions t = 0, L = L 1 = 1 em and t, L 2 = 4cm, we find, with IJ) = 2·5 x 10- s m 2 /s, t = 533 h ~ 17 days and, again, J ohn is right.

240

IV.2. Non--stationary diffusion The skin diver, busily filling rhe cylinders first with oxygen and then with nitrogen at the desired p ressure, cold John that the search for the drowned man had been delayed by the fact that no bottles with pressurized air were available.

He had found, however, cylinders with pure oxygen and nitrogen and he would start the search in a few minutes. John thought: the diver's cylinders are standing upright and each hasadiameterof20anand a length of60 em. In order to avoid another casualty, he suggested rhat the diver fill the cylinders when they were in the horizontal positicm and wait half an hour before using them.

between the micro balances for mass transfer (equation • IV.16) and that for the heat transport (equation III.4) the non-stationary penetration of matter into a medium can be treated easily and shortly. If in a rigid medium the concentration is cAo and at time t = 0 the interfacial concentration is brought to cAh the mass transfer is given as a function of the time with the solutions of paragraph II12 if the following substitutions are made : Bc~cause of the analogy

a = ).Jpc P (pcpT)-+

--+ I!) A CA

(pcpTo)-+ cAo (pcPTl)-+ ,/,.11

cAi

.~,. ,

'Y H -+ o/mol A

For shorter times (Fo = 'D AtfD 2 < 0·05) the penetration theory applies and we find for the mole flux: (IV.21)

(analogous to equation Ill.24). For longer times (Fo = 'OAtfD2 > 0·1) Sh = constant applies and Figures 1Il8 and 111.9-after the correct substitutionscan be used. (It is customary to indicate the characteristic number for nonstationary heat (and for mass penetration) as the Fourier number, Fo.) We see that in the case of short times (equation IV.21) the mass transfer coefficient kat any timet is g]ven by :

(lV.22) If the total mass transfer operation lasts a time te • the average mass transfer coefficient is found as:

(IV.23)

241

Since the penetration depth for mass transfer ( = jrl)i) is usually very small, because of the low values for the diffusion coefficient and the short contact times characteristic of mass transfer (usually « 1 s), the penetration theory is even more frequently used for the description of mass transfer than for heat transfer. Especially in cases where the interface (because it flows) is refreshed, the penetration theory is usable. For ascending bubbles in a liquid the maximal contact time of the surface elements is given b yte = d/v (d = bubble diameter, v ~ rate of ascension, the 'jacket' of the bubble is renewed every time it has moved a diameter). For a C0 2 bubble.which ascends in water at a rate of0·5 m/s and which has a diameter of 10- 2 m, te = 2, x 10- 2 s. Further. fflco 2 = 2 · x 10- 9 m 2 /s (20°C, 1 atm). so in this case the m~an mass transfer coefficient in the liquid phase (k) = 2~ = 3·6 X 10- 4 mjs (mass transfer coefficients for aqueous solutions under normal conditions are invariably in this order of magnitude). Based on the idea that in many apparatus for liquid- liquid or gas-liquid contact the interface is refreshed because there is flow, it has been proposed to introduce an age distribution for surface elements in the apparatus, and on this basis to calculate mean mass transfer coefficients ( k) with the penetration theory. If 1/f(t) dt is the fraction of the surface with ages between t and t + de, according to this theory : ( k) =

f <Xlt/J(t)

Jo

(0 dt y;t

(IV.24)

Usually the age distribution t/l(t) is unknown. In the literature two theories about f/l(t) can be found. Higbie assumed that the probability to nnd a surface element of age t for all times between t = 0 and t = te (the maximum possible age) is equally great, so:

..

dt t/l(t) dt = te

for

t/l(t) = 0

for

0

~

t ~ t,

and : t > te

This is a supposition which is correct for, for example, the absorption of a gas in jets of liquid, in laminar liquid films and in swarms of falling, rigid liquid droplets. In these cases the mean mass transfer coefficient is :

as we had already found with equation (IV.23 ). Danckwerts supposed that every surface element, independent of its previous history, has at any moment a

242

chance s dt to disappear in the subsequent period of time dt. In this case:

This can be realized by writing this relation as follows: -dlfr(t) = t/t(t)s dt

which in words means : 'The chance that the surface element disappears in the time between t and t + dt equals the chance that it is still there at timer multiplied by s dt. ' Using this supposition we find with equation (IV.24):

= foS

(IV.2S)

The constants is called the surface renewal frequency (unit : s- 1 ) . The practical use of Danckwerts • theory is restricted to those cases where a priori something can be said about the renewal frequency; those cases are rare, unfortunately. This theory can be applied, for example, for calculating the mass transfer . coefficients for liquids flowing over a packing material. If the liquid has at any contact place between two packings a chance p to be completely mixe~ the frequency s is given by the ratio of p and the residence time on one piece of packing. The same is valid for a dispersion of small, separate droplets which have a certain chance (independent of their previous history) to leave the dispersion. The droplets must stay in the continuous phase for a short time only, so that they are not completely extracted. For small droplets staying in the continuous phase for a long time, k = 2rJ/dP describes the mass transfer coefficient in the continuous phase; ·

IV.2.1. Problems 1. A jet of water (diameter 2 mm, temperature 20°C) with uniform velocity distribution (v = 5 mjs) falls vertically through practically pure C0 2 gas under atmospheric pressure. How much 00 2 is absorbed per unit of time by the first 10 em of the jet? The solubility of C02 in water at 20°C and 1 atm C0 2 pressure is 1·73 kg C0 2 / m 3 . The diffusion coefficient D = 1·7 x 10- 9 m2 fs.

Answer : 3·6 x 10- 1 kgjs 2 A waJl is coated on one side with a 0·2 mm thick layer of paint consisting of a very volatile component and a heavier, non-volatile component. The paint is aUowed to dry, i.e. the volatile component is caused to evaporate. (a) Calculate with a simple physical model how long it takes before the drying front has reached the interface paint wall. (b) How long does it take before 99 per cent of the volatile component has evaporated?

243

The diffusion coefficient of the volatile component in the paint is. D = 2·2 x 10- 11 m2 /s. Answer :

(a) 10 min

(b) 5·7 h

*3. A layer of water and a layer of toluene are brought together at time t

= 0.

Both layers contain 10.kg/m3 of an iodine compound. The ratio of the equilibrium concentrations of this iodine GOmpound in toluene and in water is 10 : 1 ; the ratio of the diffusion coefficients of the iodine compound in water and in toluene is 4. In what direction is the iodine compound transported? Calculate for relatively short times the concentration at the interface for both phases, assuming that the transport takes place by diffusion only. Sketch the concentration distribution in both phases. . Answer:

ciw = 2·4 kgfm 3 ; cit= 25 kgjm 3

*4. A short laminar jet of pure water (temperature 20°C) falls through pure S02 . The gas has a temperature of20°C and a pressure of 1 atm. With what theory can the rate of absorption of the gas be described? (a) Calculate the surface temperature of the Vfater jet if under the prevailing conditions the solubility of so2 in water is 1·54kmolfm 3, the heat of solution Is 6·7 kcal/ mole and the Lewis number (Le = aj D) is 90. In this calculation neglect the heat transpon to the gas phase. (b) How much higher wi11 the rate of absorption be if the jet falls twice as fast? (c) What is the influence of the jet diameter on the rate of absorption? (In practice it appears that the jet contracts because it is accelerated by gravity. What will be the influence of the contraction on the rate of absorption considering the answer to the ·p revious,question?)

Answer : (a) T = 21·09°C (b) higher (c) no influence if flow rate is constant

fi

5. The discontinuous phase of a continuously operated emulsion reactor shows the residence time distribution of an ideal mixer (63 per cent leaves the reactor within 5 s). Give an estimate of the lower limit of the partial mass transfer coefficient in the conti.n uous phase. Can the partial mass transfer coefficient in the discontinous phase also be estimated ([} > 0·5 x 10 - 9 m 2/ s)? Answer :

k ~ 10- 5 m fs; yes, ~ 10- 5 mj s if dp > 2

X

10- 4 m

6. For large bubbles with a free-flowing interface (model: a half-sphere with diameter D) the rate of ascension is :

v,.

=

iJgf)

244 Prove that the mass transfer coefficient for the free-ftowing interface of these bubbles is given by : ( k) =

!·6,f!!f

7. A water droplet with a diameter of 2 mm falls 4 m through surrounding air (20°C). Originally the droplet contains no oxygen. What is its mean oxygen concentration after 4 m? If it contains originally 5 mg 0 2 / l, what would have been its mean oxygen ·concentration after 4 m? The distribution coefficient of 0 2 in H 2 0 is 0·033 (20°C). Answer:

0·85 mg 0

2

/1;

5·34 mg O J.!]

*8. The liquid hold-up H, in a packed column at very low gas velocities (below loading point) can be estimated from the relations for a free-falling liquid film to be :

H, =

af;~T

(mfm

13

3

3

)

where v0 = the superficial liquid velocity (m 3fm 2 s). Prove this. So the mean residence time of the liquid is given by : t1

L

= H1 -

vo

Estimate the mass transfer coefficient in a column packed with 1 inch Raschig rings (a= 200m 3fm 3) at a liquid load of v0 = 10- 3 rp. 3 jm 2 s of water.

Answer: k = 4·2 x lo -s m/s *9. Show that John gave the right answer to the diver in the case of the drowned man discussed at the beginning of this paragraph. Comments on problems

Problem 3 Since ctfcw at equilibrium = 10, the iodine compound is transported into the toluene phase. For short times we can apply the penetration theory and we find :

¢;,

= (c,., - c;..,)

= (c;, -

C1)

,ff; -

l

1tl

Assuming equilibrium a t the interface, thus c;Jc;. . . = 10, we fi nd with the given ratio of diffusion coefficients : cil = 25 k gjm 3 ;

c,.,.,

= 2·5 kgfm 3

The penetration depth is given by :

b= JnOi

245

and since Dr= !Dw , we find : A-U 2Vw

Vr -

which enables us to draw the concentration distribution curves in both phases. Problem 4

Applying the penetration theory the mass flux of S0 2 into the jet at any time is given by:

If we assume that : aU heat o f solution is produced at the surface of the jet ; there is no heat loss into the gas phase ;

the temperature in the centre of the jet: remains constant

we can describe the transport of heat into the jet by the penetration theory :

l/J'H = pc p l!T

/a y;i

Further. the heat to be transported is related to the mass flux of S0 2 into the jet by:

and we find :

4T

=

ci ~Hs pc,

va{[£ .

1·09°C

The total mass flow into the jet is given by (using t = L/v) :

( 50,) =

2c1ttdL~ = 4c1~

So, if the jet velocity is doubled, ~o is doubled and c/Jso 2 jncreases by a factor .o f We further see tha~ a t constant cPv• the diameter has no influence on the mass flow rate.

.fi.

Problem 8 For the thickness of a falling laminar film we found in paragraph II.l :

d

= f 3y(v)d}t ..

g

where :


- - -- -- --

- -

--

cPv

=

-

vL

vL

Vo

width= -A- = -aV - =a-

246

Since H 1 = ad we find :

{3v v}'i

0 _ H ~-a-ga

If we assume the probability t hat the liquid .is completely mixed after each packing particle to be p = 1 we find for the frequency of surface renewal:

1 s= -r:P

=

L

dpTL

v0

= Hrdp

1 {v~ga} !

= adP ·~ .

= ·1 ·77s- l

.

and with Dankwaert's theory we find:

i[) .'

k =

.jii)s ~

JI0~ 9 x 1-77

=

4·2 x to- s m/s

Problem 9 John and the case of the drowned man The contents of the bottles will be practically homogeneous if: ~t

Fo- - 2 4L

~

.

, !.

0·2

(see Figure III.8)~ Estimating the diffusion coefficient·of 0 2 in N 2 at 2 x. lo-s m 2 j s, John thus finds t = 4 hours for the bottles filled while standing upright.

If the bottles are filled when they are in the horizontal position the distance L over which concentration equalization has to occur is 1 and so t!'te time needed is ~ of 4 hours. IV.3. Mass transfer with forced convection John was busy doing work he hated, namely cleaning his pipes. He realized that the rate of growth of rhe tar layer in the stem was completely determined by mass transfer. Contemplating that ~he air flow in the pipe was laminar, he concluded that he ·could decrease the growth rate of the layer by a factor of 0-79 if he halved the rate of air flow.

Wondering whether this measure would influence his smo~ing pleasure, he started to fill the pipe from the humidor on his desk. IV.3.1. Analogy with heat transfer

For mass transfer between a wall and a convective flow the· same analysis can be followed as for heat transfer between a wall and a forced flow (paragraph III.3) provided mass transfer is considered at low concentrations and without ·

chemical reaction. This can be understood from the fact that with the following

'transpositions' the entire description of the mass transfer (without chemical reaction) is also valid for the description of the heat transfer. Heat transfer pcPT

i./peP = a ¢~

a/peP

Mass transfer (component A)

247

Thus, if for a certain geometric case of heat transfer with forced convection it is concluded that, for example: Nu =

f 1(Re,Pr, Gr, Fo, Gz, ... )

it follows from the above that for the same case with the sam e type of b oundary conditions applies: Sh = / 2 (Re, Sc, G r, F o, G z, ... ) We have already come across the Sherwood number, Sh = kD/ 0, in paragraph IV. l. The Schmidt number, Sc = vj [[), performs in mass transfer the role which is played by the Pr number in heat transfer; it gives the ratio between the thickness of the hydrodynanric and of the concentration boundary layer. For gas mixtures Sc is about 1, for liquids Sc is considerably higher (102- 103 for normal liquids), because [b is so much lower than v. Whereas for gases Sc depends hardly on the temperature, Sc d ecreases strongly for liquids ~if the temperature increases. Before it can be concluded that the results for beat transfer (i.e. the function f 1 ) can be transposed to a reliable relation for mass transfer (function f 1 = f 2 ), it should be ascertained whether the mass transfer process does indeed proceed entirely analogously to fhe (assumed to be known) heat transfer process. This means that the functions / 1 and f 2 are only equal to each other if the Re range used for mass and heat transfers is equal and if the Sc range in which the relation has to be applied equals the Pr range in which the heat transfer relation applies. For gases where Sc ~ Pr this 'transposition' can be carried out frequently ; the relation for heat transfer for liquids cannot always be transposed to usable relations for mass transfer because for liquids Sc » Pr. On the basis of the analysis for heat transfer between a wall and a turbulent flow (paragraph.III.3.2) the reader can now establish for himself that in the range 2 x 103 < Re < 105 , Sc > Q. 7 a usable relation for mass transfer between a wall and a turbulent flow is given by : Sh

= 0·027 Re

0

·~

Sc0 ' 33

(IV.26)

With these analogy relations mass transfer values, for example, can be :·~IcuJated from t heoretical or experimental data about heat transfer and vice versa. Now, particularly if Re is not too low, the local or the mean transfer coefficients over a surface can be calculated with good approximation from fo rmulae such as: Nu = CRem Pr11

Sh = C Re'" Sen The exponent m varies from ca. j (for the laminar entrance region in a pipe) via 0·5 (in the case of flow around spheres) to ca. 0·8 (for turbulent flow in pipes), but in all these cases (excep t with liquid metals) the exponent n of P r (and he~ce also o f Sc) is about equal to t. Chilton and Colburn have used this for representing the analogy between heat and mass transfers in the following

- -- -- - - - -

248 form. They defined a heat transfer

v~l'\lt; :

j 8 = N~ ~~- 1

Pr-t

and a mass transfer value:

....

j D = Sh Re- 1 sc-i

Using these definitions the analogy for geometrically similar cases and not too low values of Re becomes : (IV.27)

For turbulent flow through pipes and along fia t plates this analogy can be extended to: . . 1j (IV.28) )g = }D = 2 where f is the so-called Fanning friction factor (see paragraph 11.2, equation 11.22). This analogy with momentum transfer, however, applies only in those cases where f is caused by wall friction. The flow resistance of a body caused by eddies which can be an important part of the overall flow resistance is not a ttended by analogous effects in t:he. beat or mass transfers. So fo r these cases equation (IV.28) does not apply, but equation (lV.27) does. The analogy between heat and mass transfers also applies to free convection. The density difference ap which occurs in the Grashof number (see paragraph Ill.6) might originate from concentration d ifferences as well as from temperature differences. If there are differences both in concentration and in temperature (combined heat and mass transfers) the influence of both on Gr can be ~aken into account.

IV.3.2. Ma ss transfer during laminar flow The mass transfer process can be calculated accurately for a number of simple flow situations, the results then being collected in a catalogue. Although a certain problem is hardly ever completely equal to a case from this catalogue, it is possible by proper estimation to approach the actual problem with a catalogue case. We can tell the 'chem~cal engineer' by this 'estimation '. It should be borne in mind that three data on the velocity field near the boundary layer are of decisive importance for the mass transfer. These data are, in order of importance : (1) the velocity of the boundary layer and that of the fluid (li; and vx); (2) the velocity gradient perpendicular to the boundary layer and, (3) the velocity perpendicular to the interface near the surface of the mass transfer boundary layer (v1) . We shall discuss here five cases which are represented in Figure IVA. These examples contain the characteristics for the flow field just mentioned. The transfer process is described with the following mass balance (see equation

249 Case t

JC

Case 2.

Case 3.

v..,

=wy+v;

Vy::

Case 4.

0

vQ;)} Ou1side

vx

=

vy

=

v.r

* vQ;)l vy *OJ Inside

0

boundary layer

vi = 0

boundary Ioyer

case 5.

Figure IV.4 Five cases of laminar ftow situations

IV.I6b) and the boundary conditions: 0=

b2 c by

1[}2 -

be x5x

oc

v -- v 1 by

(IV.29)

with c = c 0 for y = 0 and for all x (at the interface the equilibrium concentration prevails), c = 0 if x = 0 for all y (the infiowing liquid has the concentration zero) and c = 0 if y = co for all x (no accumulation of substance very far from the interface). Case I. Uniform velocity parallel to the interface Here vx = vi and v, = 0, and the boundary layer also moves at a velocity vi. Tbis case bas b een discussed by Higbie. By moving with the ftow, it becomes a

250 case of non-stationary diffusion (penetration theory wi th t = xjv1) . The solution is:

k =

v(0-:m

ro;;; -.J-;;

=

or : kx = Sb = [j)

X

1 -( v;x)t(~)t Jnv [) =

0-565 Ret Set

(IV.30)

X

The mean mass transfer coefficient is given by :

.!._

(k) =

t ·k cbc =

LJo

2kL

The absorption of gases in liquid jets and in laminar falling films proceeds according to this description. Case 2. Flow with a constant velocity gradient

Here vx = wy and vY = 0, and ·the boundary layer is stationary. Leveque* has solved this problem : kx -0 = Sh X

)t = 0·539(wx 1

[)

(IV.31)

and : ( k ) = ~kL

For a laminar, Newtonian pipe flow (diameter D) :

dv;

W=-

dy

y= O

-

s· D

The reader can compare this result with the analogous heat transfer problem of Graetz (equation III.46b). For mass transfer this case is not so important but it is a borderline case of the following case in which both an interface velocity (vi) and a velocity gradient (w = dvJd yly=o) p lay a role. Case 3. lnteiface velocity and velocity gradient

Here Vx = vi + wy and vy = 0. The complete solution of this case is known. In pracrice, approximations of chis solution are sufficient. These approximations apply tow sufficiently small and vi s ufficientl y small. The characteristic dimensionless value is w 2 '0xfvt. If this value is lower than 1, the penetration theory (case 1) gives to within 15 per cent the correct mass transfer coefficient. If this value is higher than 1, the Leveque theory (case 2) gives the correct result to within 15 per cent "' Ann. des Mines, 13, 201, 305, 381 (1928).

251 In general, w is of the otder of magnitude of vifL.. The dimensionless group mentioned can then be written as fD/v 1L (Peclet number). If we consider bubbles and droplets•with a diameter at least 1 em and a velocity of at least 0·1 m js, both for gases (D ~ 10- 5 m 2 /s) and fot liquids (Ul ~ 10- 9 m 2 fs), this value is lower than I. So in these cases the penetration theory can be used. For gas bubbles with small diameters and velocities the penetration theory is no longer applicable.

of

Case 4. Stationary iitterface and a velocity vx intelface

~ Vex;

sufficiently far from the

. For the velocity and shear stress distribution~ along the interface the following appllies, according to SGhlichting's boundary layer theory :

Also applies :

where ~his the thickness of the hydrodynamic boundary layer. Further applies: Sh · = kx = x :r

[[}

(J

t:

Jf the mass transfer boundary layer {Jc is much thinner than the hydrodynamic boundary layer ~h for the relation of the two boundary layer thicknesses applies:

from which follows: Sh X = 0-332 RetX Sc~

(IV.32)

and: -Now ~c « ~h only applies when Sc » 1. But for Sc = 1 equation (IV.32) also yields reasonably accurate results. For separate solid particles, e.g. in a fluidized bed or in a n ot too dense swarm of rigid liquid droplets, equati9n {IV.32) describes the mass transfer satisfactorily if ·sc » 1 has been satisfied (L is in this case the particle diameter dp). In th~ case of particles in a suspension or rigid droplets in an emulsion eq uat]on (IV.32) can also be used. The problem in these cases is that the relative velocity vcc of the particle with respect to the fluid is difficult to determine. This velocity is mainly determined by the type of stirrer, the number of revolutions, the diameter of the stirrer D and the diameter of the particle dP.

252 Case 5. M oving interface and a velocity far from the interf ace This problem lies between case 4 (v; = 0) and case 1 (v 1 = v 00 ). The complete solution is known.* In this case no simple expression is found for the mass transfer coefficient The result of thls analysis can be seen in Figure IV.5. This figure shows that for Sc > 200 the penetration theory can be applied if the interface velocity v; is higher than 10 per cent of the velocity of the bulk u::o (case 1 at velocity vJ From this figu re it also follows that fo r Sc = 1 (e.g. with a gas fiow through a column with packing over which liquid drips) the boundary layer theory (case 4) can be applied to predict the mass transfer in the gas phase, if we u se as the characteristic velocity the sum total of the effective gas rate and the sudace _ rate of the film (v 00 + v;). This result has been ~onfumed experimentally. 0·6 0 0 . - - - - - - - - - - - - - - - -- -----, Cow 1. wi1h

v,-="co

Cose 1. wi th cttOI'octeristic velocity v;

I

Figure IV .5 The mass transfer coefficient in the flow sit uatio n for case 5 • Appf_ Sci. Res. lOA, 241 {1961).

253 ;

IV.3.3. Mass transfer during turbulent .flow

.

If the laminar boundary layer exceeds a certain thickness, the boundary layer flow becomes turbulent (approximately for VxlJJv > 5 X 10 4 ). The eddies then penetrate into the area where the mass transfer takes place. The {statistical) velocity components in the y-direction contribute, together with the diffusion, to the transport perpendicular to the interface. Quite near the interface where the eddies hardly penetrate (laminar sublayer) transport takes place exclusively as a result of diffusion. At larger distances from the interface the {statistical) transport prevails by entrainment In between lies an area where both processes are equally imponant. Since the transport by turbulent motion is statistical, just as the diffusion, this transport is described by means of an eddy mass diffusivity E which depends on y. Equation (IV.30) then becomes:

':nol.y = -{0 +

E)d:;

{IV.33)

Since the transport of matter and substance proceeds analogously because of the eddies, the momentum flux can be described with the same diffusivity E: "t

=

-

dv.x

{IV.34)

+ E) dy

p(v

The next thing to do is to measure vx(Y); to assume t in all layers near to the interface to be constant and equal to the shear stress on the wall ; to determine E(y) using equation (1V.34) and then to use this result for calculating k using equation (1V.33). To this end equation (IV.33} is written as follows: Sh X

=

kx = [J)

x¢;~ol,y

ro [) J

o de A

= .

x{~ IX)

dy

~l

o \1

}-

1

(IV.35)

+ tO} .

A drawback to this method, of course, is that particularly close to the wall where the resistance to the mass transfer is found v:x:{y) is difficult to measure. As a consequence the E(y) ·values derived from the velocity distribution are inaccurate, which causes errors in the estimation·of Shx by this route. Most data are known about steady-state turbulent flow along a solid flat wall. The number of suggestions for E(y) is large. All these suggestions can be summarized a s follows: _k = St = _ _f_/_ 2 -== v:n

1 + g(ScVJ72

where : St (Stanton number)= Sh/Re Sc f = Fanning friction factor for the solid wall g(Sc) = function of Sc ·

= 2tw/ pv~

(IV.36)

254 In the case wbere Sc = l , g == 0 applies and 2k = fvr:t> (Reynolds' analogy). Further, g is the following function of Sc :

Sc

1

5

lO

100

500

1000

g

0

27

52

330 ( ±10 %)

1000 (± 15 %)

(±20%)

1800

2000 2800

(±30%)

The greater error in the function g for higher values of Sc is mainly caused by the various approximations of E(y) near the wall DeissJer, for example, finds g = 9·0 Sc~ (Sc > 200); Reichardt finds in the same Sc range g = 8·8 Sc-i and Ling = 18 Sc~. These results can be used in the description of mass transfer between a pipe wall and a turbulent flow (e.g. the free space in a packed bed can be considered as a system of irregular channels). The friction factor fin equation (IV.36) should then exclusively describe the friction resistance and not the form resistance. This factor should also be related to a Reynold s n umber in which the hydraulic diameter of the channels occurs as the characteristic length.

I V.3.4. Problems 1. An air freshener is suspended vertically in still air. Calculate the thickness which evaporates each day. Data:

molecular weight : 200 v = 1·5 x lo- s m 2j s

10

A nswer:

= 0.5

x

to- 5 m2/s

vapour pressure: 10- 3 atm L = 25cm p = 1500 kgjm 3

0·1 mmjday

2. The inner wall of a tube has, in the course of time, been caked with a solid layer of brine of 2 m.m. The heat transfer coefficient ~ at the wall is 1000 Wj m 2 0 C. Calculate the time necessary to dissolve the salt if the tube is fi usbed under the same conditions with dis tilled water. (PsaJt = 2500 kglm3 , the Lewis number under these conditions is Le = a/ fD = 100 and the solubility of the sa1t is 300 kgjm 3 .) Answer:

25 min

3. A metal foil has t o be etched. To this end it is drawn through an etching liquid which flows at the samespeed (1 m/s)with the foil. The etching process is fully determined b y mass transfer. By what factor does the velocity of the etching process change if the foH is stopped and the liquid continues to flow (Sc = 10 3 )? Answer :

0·2 times

*4. The Obersalzberg, a mountain in the G erman/ Austrian Alps near Salzburg, consists of rela tively pure sea salt.. The salt is m ined by drilling and digging

255 out chambers of about 60 x 60 m and 1 m height some hundreds of metres deep into the mountain. These chambers are filled with water. Fresh water is added and a 90 per cent saturated salt solution is withdrawn · continuously. Part of the mines can be visited and the guide will tell you that the height ofthese chambers increases at a rate of approximately 1 em/day. Wha t speed of solution of the. salt would you predict1 (p5111 = 2165 kg/m 3 , solubility = 300 kg/m 3 Psaturated solution = 1200 kg/m 3, [}Nact = 1·5 X lO - 9 m2/ s and v = 10- 6 m2 /s). I

Answer :

1·48 em/day

*5. In .a diluted emulsion the rate of circulation of the surface of the droplets is one-quarter of the relative velocity between the two phases. After a smal1 amount of surface-active substance bas been added, the droplets have become rigid. By what factor has the external mass transfer coefficient changed after the addition (Sc = 103 )?

0·37 times

Answer:

6. In a cylindrical vessel {diameter 2m) a 5 mm thick crust ofKMn0 4 crystals has deposited on the side wall in the course of a crystallization process.

The crystals are removed by filling the vessel with a diluted K:Mn0 4 solution (c0 = 3 kg/m 3 ) of 200C and by stirring vigorously. It is known that under the same conditions the heat transfer coefficient liquid wall is 1100 Wjm 2 °C. After what time has the entire crust dissolved? The solubility of KMn0 4 at 20°C is 63·2 kgjm3 ; the density of crystallized KMnO.._ is 2700 kgjm 3 and the diffusion coefficient of KMn0 4 in water is 10 - 9 m2fs. Let all other substance properties be equal to those of water at 20°C. Answer :

517 min

7. A liquid flows turbulently along a rough wall. The temperature of the wall rises by 5°C, thus changing the Sc number from 103 to 500. How does the mass transfer coefficient change? Answer :

1·8 times

*8. A small cooking salt crystal is dissolved at 20°C in a large amount of clean water which is stirred thoroughly. At what value does the difference between the water and crystal temperatures adjust itself? Data :

1-5 x 10- 9 m 2/s AH5 = 9 ·3 X 104 Jjkg solubility = 300 kg/m 3 D NaCI =

Answer : 0·3°C

*9. Show that John drew the right conclusion in the case of his pipe discussed at th e beginning of thls paragraph.

--- -- - - - - -

256

Comments on problems Prob~m. 4

We assume the bulk of the salt solution to have a uniform concentration of 90 per cent of saturation, so c<Xl = 270 kgfm 3 . At the ceiling, a thin film of liquid is completely saturated, ci = 300 kg,lm3 • Because of the difference in specific gravity free convection occurs, which enhances mass transfer. By analogy with equation (HI. 75) we can write : 3

( Sb) = (k)L = 0·17{GrSc}1· = 0·17{L g fJ.p ~}i' 10 v 2p l 0 Estimating p co : p<XJ

= 1000 + ().9 x 200 = 1180 kg/m 3

we find: ( k)

= 1·23

X

10- 5 m js

A material balance shows that :

and we find :

dH

-dt - = 1·48 em/day The salt contains small amounts ofgypsum and iron oxide which do not dissolve but settle at the bottom of the chamber. Apparently, these insoluble impurities slightly lower the speed of dissolution. Problem 5

The drawings show the flow situations around the stagnant and the mobile droplet. These flow situations were discussed in paragraph IV.3.2 (Figure IV.4, cases 4 and 5). So for the stagnant droplet equation (IV.32) is valid:

sb =

vCD

constonl

0

Stognont droplet

().332("~"d·r Set

/

VQ)

:

cons toni

v,

I

0

I

/~

MobWe dr ople t

257 whereas for the mobile droplet, because Sc > 200, the penetration theory, -equation (IV.30) can be applied:

Consequently :

0·332(v"") t Sc-t = 0·37

k stagnant --'--= -- -

kmobile

because

V;

=

0·565

Vj

!vco·

Problem 8

We assume that all heat of solution produced at the interface crystal-liquid is transported .into the liquid with uniform bulk temperature T00 • A simple · ·· heat balance shows : ¢'H = ct(J; - . T~) ,.._4>'/n b.Hs

=

kci AHs ·

Now, because of the analogue transport mechanism of heat and mass :

~: = (~;r

or

~ = ~(~r

so:

Problem 9 John and the case of the pipe In the stem of John's pipe there is laminar flow with flow velocity zero at the boundary (case 1 of paragraph IV.3.2). For this case the mass tra~sfer coefficient is proportional to the third root of the velocity gradient, w·!- (Lefeque, equation IV.31). Since for laminar pipe flow w = 8(vx)/D, the mass transfer coefficient is proportional to ( vx> t and the rate of accumulation oftar is also proportional to ( Vx) 1-. If, therefore, the volumetric flow rate is halved, the rate of accurnula tion : is decreased to (t)t = 0·79 of its initial value. N .4. Mass exchangers John put down his cup of tea with distaste and thought about the detrimental effect of water pollution on his drinking habits. He knew that his secretary used a coal filter to decrease the concentration of impurities in the water by a factor of10. Remembering chat because ofthe long dry period the level of impurities had increase_d by afactor 3· he advised his secretary to i_ncrease the length of the coal column by

48 per cent.

258 The calculation of mass exchangers in which no chemical reactions occur proceeds analogously to the calculation of beat ex<:hangers. Mass transfer usually takes place in apparatus in which the two exchanging phases are brought into close contact with each other. To this end, the one phase is finely dispersed in the other so that a Jarge exchange area is formed For selecting the correct type of mass transfer apparatus and for determining the main dimensions, the following data are of importance : (a) the thermodynamic equilibrium between the phases in question under the prevailing condition s (temperature and pressure) ; (b) the rate of mass transport in the chosen apparatus as a functio n of the operational variables (e.g. flow rate, stirring intensity, etc.). IVA.J. Thermodynamic equilibrium

During heat transport under equilibrium conditions the temperatures in all phases are equal, i.e. T., = ~ . AT= 0. During mass transport under equilibrium conditions the activities of a given substance in both phases are equal. These activities are seldom known and therefore the more practical concept of a distribution coefficient (see also paragarph 1V.l.2) was introduced. We will use here exclusively the simplest correlation possible, cA = mc8 . This simplifies the calculations considerably but it is not essential for the approach chosen. For a. large number of cases them value and consequently the slope (gradient) of the equilibrium line is constant at constant temperature. In such cases there are sufficient thermodynamic data for calculating apparatus for physical absorption if the value of m, which is sometimes also called the Bunsen absorption coefficient, is known as a function of the temperature. Table IV.l gives a survey of distribution coefficients of some gases in water. Table IV.l

Distribution coefficient m for various gases in water at 20°C. In 1m 3 water at 20°C m m3 gas is dissolved (reduced t<> ooc and 1 bar)

Gas Air Ammonia

Carbon dioxide Chlorine Hydrogen Hydrogen chlonde Nitrogen

Oxygen

Sulphur d ioxide

m 0 ·018 685 0·86

2·25 0.01 8

430

0·015

0·031 38·6

In older handbooks the thermodynamic distribution law over these media is described as follows : cA = He P 8 , where P 8 is the par tial pressure of the gas to be absorbed at the interface. Using the ideal gas law it foJJows that m = H e RT

259 (T in oK), He being the Henry coefficient. Table I V.l shows that solubilities may

differ by a factor of 104 and that a small difference in chemical structure of the component to be absor:bed may have a great influence on the solubility. As soon as the thermodynamics are known. some considerations regarding the attainable equilibrium should be made prior to selecting the type of mass exchanger. The thermodynamic equilibrium between the phases under the conditions prevailing in the apparatus indicates what could be achieved in the case of very intense contact between the phases and at 'very long contact times. To this end, we take as an example the case where a wel1-m~ed gas ·flow ijJ with inlet concentration c0 comes to equilibrium by absorption, with a likewis~ well-mixed liquid flow 4J1 with inlet concentration zero. From a mass balance·(mass flow in = mass tlo~ . ou~):

cpg(c0

c)

-

= mc¢L

·

it appears that the concentration in the emerging liquid is given by:· co c=---

l

(.JV.37)

+ mg

To ensure the highest possible outlet concentration· in the liquid, the ratio m¢JJ 4>8 , which is called the extraction factor, should be as small as possible. . e.g. ¢>, = IOmL· The desired ratio between the flows r/>8/t/>L is therefore determined by the distribution coefficient. For poorly soluble gases such as oxygen it is sufficient to choose r/>g/c/>L ~ 0·3. For readily soluble gases such as C02 , g/
260 Since when choosing physical absorption apparatus it bas first to be decided whether the liquid or the gas has to be distributed, it is clear on the· basis of the above considerations that readily soluble gases can only be transferred efficiently at high values of. 4>,/t/JL, so that the liquid will be distributed For some types of gas-liquid contact apparatus approximate values are given in Table IV.2, with which the semi-quantitative considerations, on choosing the apparatus, can be supported.

Table IV.2 Some approximate values for gas-liquid contact apparatus. The flow rate is v0 L= 2 x J0- 2 m3/m2 s Type of contactor

Maximum gas Bow rate lio,

(m 3/m2 s)

Bubble column

0·20

Distillation stage

().3

(with bubble caps) Stirred tanlc

0.15

reactor

Maximum interfacial area

a(m 2/m3 )

Liquid fraction (m l/ m3)

Power consumpti<;m per kg liquid (W/kg)

300

>0·6

2

75

>0·6

2·5

1000

>0.5

6

Sieve plate

1

150

> 0·15

l

Falling film

2

100

<0·05

1

Packed column (two phases)

l

250

< 0·05

1·5

If on the basis of thermodynamic considerations the preference of some types of apparatus has been expressed, some other consideration may play a part, such as the pressure drop over the apparatus, the heat transfer, corrosion and the question of whether a countercurrent way of contacting the phases is possible. IV.4.3. Size determination of the mass exchanger If on the basis of the complex of qualitative and quantitative considerations discussed in paragraph IV.4.2 the type of apparatus bas been determined, we have to know the mass transfer coefficient and the specific interface as a function of the operational conditions of the apparatus, so that this apparatus can be designed.

In the case of physical gas absorption the resistance to mass transfer is entirely on the liquid side and the total transfer coefficient related to the Hquid phase equals the partial transfer coefficient in this phase. The values of the mass transfer coefficients can be calculated from the data of paragraph IV.4.2 and appear to be in the order of 10- 4 m/ s for all practical cases of gas absorption.

The same relations can also be used for calculating mass transfer coefficients between two liquid phases if their flow behaviour is known.

261 The area of the interface between ~he two phases is more difficult to estimate and Table IV.2 only gives rough indications. For gas-liquid contacting in a packed column, the surface of the packing material can be taken as the .actual interface _(accurate within + 20 per cent)~ The size of the gas- liquid interlace in a bubble column or in a . stirred v~ssel is strongly dependent ·on power inpu~ gas flow rate, gas hold-up and properties of the liquid. Reliable correlations for the estimation of the size of the interface over a wide range of operating conditions are stilllackjng. If the volume fraction of gas (I - s) and the average diameter of the gas bubbles is known, the specitic interface can be calculated from:

a=

6(1 - s) --· -

(IV.38)

dp

This relation can also be used for calculating the size of the specific interface for dispersions of Jiquids. Having determined the-size of the mass transfer interface we can find the necessary dimensions of the mass exchanger from a mass balanc·e (equation IV.l). As a first example we take the case in which two phases 1 and 2 with flow rates 4>v 1 and 4>v 2 are contacted iii·countercurrent plug flow in a column. For this case we can set up the following mass balance (see also Figure IV.6) for phase 1 (K 1 = total mass transfer coefficient based on phase 1) over a differential height dx:

I j

l4>v1

- -- ~-r-----0

___ _

. Q

---0 _ 1 ........

I

Figure IV.6 Contact between two phases in a column

j

..

dx

0

f 4> v2 .

Integrating over the total apparatus we obtain :

J

c l (.x - L)

-

c.<.x=o> c 1

de 1 -

mc 2

=

JL --Adx Kla JL = 0

<j>v 1

0

dx

HTU

-

L

HTU

(IV.39)

The terms of equation (IV.39) are dimensionless. The left-hand term is called the number of transfer units (n). The right-hand term is a quotient of two lengths : the length of the column (L) and the height of a transfer unit (HTU). The HTU is a measure of _the efficiency of -the column (small HTU -i> good ·transfer). The HTU can-be defined as that length of the column for which the integral of the left-hand term of equation (IV.39) has the value of 1.

262 The HTU as used in equation (IV.39) is called an ·overall' HTU because the overall d riving force (c 1 - mc 2 ) has been used (here related to the '2' phase). If we had only considered one pliase (e.g. the '1' phase) with driving force = c 1 - C1 1 (where C1 ; is the concentration at the interface of phases 1 and 2) we would have o btained a partial height of a transfer unit (htu) 1 • It can easily be derived from equations (IV.15) a nd (IV.38) that : (btu}l

(HTU) 1 =

(IV.40)

+ E(htuh

where E = m tf¢ 2 is the extraction factor. So the HTU is dependent on the extraction factor. In fact, equation (IV.39) is the same as the design equation for a plug flow heat exchanger obtained in paragraph III.4. This will now be shown. lntegratio11 of equation (IV.39) gives : · elo - ell., In t1co = L 6 c0 - L\cL LlcL HTU

(IV.41)

where llc 0

= c10

11eL

=

-

mc20

elL - mczL

If we set up a material balance over the entire a pparatus we obtain :

0=

r/Jr;lCi o -

4>vle1L

+ l/Jm

where m is the mass flow from phase 1 to phase2. Substituting the concentration difference c 1 0 - e 1L with the help of equation (IV.41) we find for the mass flow rate, if both phases are in countercurrent plug flow : ·

4>m

L

=
~c0 -

~

llcL

ln~

= K 1aAL

L\c 0 -

~

llcL

ln~

= K 1a AL(6 c)108

(JV.42)

llcL

A cL

which can be compared with equation (111.61). In the foregoing we have neglected the influence of liquid dispersion . If dispersion or longitudinal mixing occurs in the flows, the overall driving force in the apparatus will decrease (as we have a lready d iscussed in paragraph 111.4.2). Equation (IV.39) d oes not adequately describe such a case. However. for practical purposes, cases in which dispersion has to be taken into account can still be written in the form of this equation: t(X.,L) [

cl(x=o)

d Cl

l

ct - mc2 - HETU

(IV.43)

which is the definitio n of the ·height equivalent of a transfer unit' (HETU). This HETU is, in general, the sum of the HTU a nd of fo rms which account for the mixing in both phases.

263

As a second example we treat again the situation of Figure IV.6, but now for the case that the two flows can be treated as plug flows in combination with a dispersion process with dispersion coefficients 0 1 and []) 2 respectively. For this case the following result is obtained : (IV.44)

The determination of these dispersion coefficients has already been discussed in paragraph II.8.5. There we saw (equation Il.118) that the height of a mixing unit is given by:

L

HMU=-

n

where n is the number of perfect mixers in series yielding a corresponding residence time distribution. For small dispersion we found (equation Il.115a): 1 D _......,2-

n

vL

and consequently in this case we find (analogous to equation II1.64) : HETU = HTU

+ «HMU)1 + !(HMU) 2

(IV.45)

For the system water- air in a packed column (HMU)watcr ~ dP and dP < (HMU)air < 5dP.Sincetheheightofatransferunit is8dP < HTU < 40dP. in practically all cases plug flow of the liquid phase can be assumed Of course, for the exceptional case that one of the phases is perfectly mix~ the foregoing analysis (which is restricted to small dispersion coefficients) is not of much help. In such a case equation (IV.43) and the HETU can no longer be used. However, the approach here is again very simple. Considering that the bulk concentration of the well-mixed phase is uniform (and equal to its exit concentration) we can write down the mass balance for the other phase over a differential height dx. This equation can then be integrated in a straightforward way, because it contains only one concentration which is dependent on the location. If, as our third example, the other phase flows in plug flow, :equation (IV.39) results again, the index 2 now indicating the perfectly mixed phase and c2 being constant, e.g. c2(x = L). I V.4.4. Th e concepc of theoretical plates

Another frequently used expression in the calculation of continuous mass exchangers is the 'height equivalent of a theoretical plate' (HETP). One HETP is that column height for which for the outgoing flows applies that their concentrations come up to the equilibrium condition. This has been elucidated in F igure IV.7, which shows a material balance over a segment of a mass exchanger with the length of 1 HETP. So the equilibrium condition is: c 1(x)

= mc 2 (x +

HETP)

264 c1 ( K+ HETP);cp

.. t

x+ HETP

4>v 1 ;c(x} 1

Figure IV.7 Material balance of a mass exchanger over a height of one HETP

In order to determine the correlation berween HETP and HTU use is made of the definition equation (1V.40) of the HTU: -

HETP --

HTU

Jr+

de 1

HETP

c1

x

-

(IV.46}

mc 2

The integral on the right-hand side of equation (IV.46) can be calculated if it is borne 'in mind that a relation between c 1 and c2 can be found from a mass balance. For the case of plug flow this mass balance reads: - tPv 1C1

+

v2C2

= COnstant

Equation (IV.46) can then be elaborated to :

H ETP HTU

1 1- E

In

/(1 - E)c1(x + HETP)- mc (x + HETP) + Ec (x + HETP)) 2

\

(1 - E)c 1(x)- mc 2 (x

1

+ HETP) + Ec 1(x + HETP)

and using the definition of the HETP, c 1(x) = mc2(X + HETP), that can be reduced to:

{.!..

HETP _ 1 Jn c 1(x + HETP) - mc2 (x + HETP)} _ ln E HTU - 1 - E E c1 (x + HETP) - mcz(x + HETP) - E- 1

(IV.47)

So only if the extraction factor approaches the value of 1, which will be chosen in many practical cases, will HETP ~ HTU. For packing materials such as Raschig rings and Berl saddles the HTU is of the order of magnitude of 6-60 em, 15 em being a good average. The reader may check for himself what this means for the HETP if E is respectively 0·25, 0·50, 0·75 and 1-5.

265

Because the HETP is a stronger function of the extraction factor E than the HTU, the Ia tter should be considered as a funclC!J!Ientall y more correct expression for describing the operation of an extraction column. IV.4.5. Problems 1. A column is filled with a layer of silica gel to a height of 30 em. Air with a moisture content of 1000 ppm is passed through. When the air flows very

slowly the moisture content after passing the column is 0·1 ppm, while under working conditions the moisture content of the air after passing the column is 10 ppm. Calculate the layer of silica gel necessary to obtain air with a moisture content of 2 ppm. 4~

Answer:

em

- 2. A H 2S-containing water stream (concentration of H 2 S is 50gjm 3 , tem-

perature 25°C) is treated countercurrently with air in order to desorb the H 2 S. At 25°C the distribution coefficient of H 2 S (=concentration in air: concentration in water) is 0-44. (a) Calculate the m inimum amount of air necessary for fully degassing 1m 3 water. (b) What is the lowest concentration of H 2 S which can be achieved in the discharged water if 1m 3 air is used per m 3 water?

Answer: (a) 2·3 m 3/ m 3 (b) 28 gfm 3

3. Ajr flows th rough a column filled with benzoic acid spheres. The outgoing flow appears to be for 90 per cent saturated with benzoic acid vapour. Now, instead of air, water is passed through the column in such a way that the Reynolds number is the same .for both flows. What is the .degree of saturation of the effluent water with benzoic acid? Data: Kinematic viscosity Water Air

Answer :

1 x 10- 6 m 2 { s 2 x 10- 6 m 2 j s

Diffusion coefficient benzoic acid 10- 9 m 2f s 2 x 10- 6 m 2fs 1 x

2-3 %

*4. A water stream of 1000 1/h must be freed from Ca2 + ions. The initial concentration is 500 mgjl while the required final concentration is 10 mgfl. This process is carried out in a column filled with ion exchanger consisting of almost spherical particles. The Ca2 + ion concentration which is in equilibrium with the exchanger is negligibly small. The rate~determining step during ion exchange is the mass transfer from liquid to the particles.

266 Determine the product of length and diameter of the column required for this process.

Data:

dp = 2 X DeaH l ollS

Answer : DH

=

w- 3 m, volume fraction voids e = =

10- 9 m 2 is, v = 10- 6 m 2/s

0-4

0·12mz

5. An extraction is carried out by causing droplets of water to descend through an oil layer. The resistance to mass transfer is entirely in the oil phase. The mass transfer can be described with the penetration theory by introducing as contact time the time a droplet needs to cover a distance equal to twice its own diameter. The flow around the droplets can be considered as a laminar flow round a rigid sphere (Stokes' law applica"Qle). (a) By what factor dQes the mass transfer coefficient increase or decrease when the droplets are made twice as small? (b) How much does the product kS (i.e, the product of transfer coefficient and total surface area of all droplets in 1 m3 of the extraction column) increase or decrease if at this droplet size reduction the total water flow (m 3/ s) remains unchanged?

Answer : (a) 0·71 times as great (b) 5·6 times as great

*6. An organic liquid has to be saturated to 90 per cent with a certain substance A However, only a saturated solution of A in water ·is available. The organic liquid which is insoluble in water is now caused to ascend in small

rigid droplets (diameter : d11 = 2 mm) through the saturated solution of A in water (rate of rise : v = 10"" 2 m/s). What should be the height of the water column to attain a 90 per cent saturation of the organic liquid with A? The diffusion of A both in water and in the organic liquid is 10- 9 m 2/s and the solubility of A in the organic liquid is 200 t imes as low as in water. The continuous water phase is present in excess a nd is not exhausted. The initial concentration of A in the droplets is zero.

Answer : 2m 7. In order to saturate an air stream with water vapour, the air is passed through a wash bottle filled to a height of 10 em with pure water of 2SOC. The air is divided through a sieve plate into bubbles with a diameter of 3 mm. The bubbles behave as rigid spheres. The volume fraction of water in the dispersion is 0·8. (a) Calculate the rate of rise of the swarm of bubbles using the relation of Richardson and Zaki (paragraph II.S.5). (b) What liquid height is necessary to ensure a relative humidity of 99·9 per cent in the outflowing air'? (c) If the water is originally air-free how long does it take to saturate the \iqui.d for 99 \)et: cent of air.

267 Data :

[)) H10 inair

= 2·5 X

DairinH1 o

=

w- 5 m 2/ s

5 x 10- 9 m1 /s

Answer : (a} 0 -175 m/ s (b) 1·1 em (c) 121 s

8. A dry air stream must be saturated with 99 per cent water vapour. This bas to be attained by passing the air flow through a wetted wall column (a pipe in which a liquid film fiows downwards along the inside wall). The thickness of the liquid film is very small compared with the column diameter. Data :

oair

= 6·9 X 10- ID3 / S, D internalc:otumn 3

Pair = 1 kgjml, 'lair = 17 X 10 Ns/m 5 2 D HzOioair = 2·5 X 10 - m /s 6

2

=5

X

I o -z m

For the mass transfer coefficient in wetted wall columns applies, according to Gilliand and Sherwood (Ind. Eng. Chem, 26, 516, 1934) :

Sh

=

0·023 Reo·s3 sco·44(2 x 103 < Re < 3-5 x 0·60 < Sc < 2·5

104)

(a) What is the required length of the column? (b) A physicist maintains that the column becomes undoubtedly much shorter if a recirculation of the gas flow over the column is applied. Show quantitatively whether his statement is cor rect for a recirculation ratio 1 : 1 (i.e. just as much gas is recirculated as leaves the apparatus). (c) What column length would be calculated as the answer to question (a) if instead of the given correlation for Sh use was made of the a~alogy between mass and heat transfers?

Answer: (a) 9·6 m ; (b) 9·2 m ; (c) 10·5 m *9. An S0 3- air mixture containing 7 mole por cent S0 3 is introduced at the bottom of an absorption column which is filled with 1! inch Raschig rings (a = 140 m 2/ m 3 , e = 0.92). The gas rate calculated on the empty column is v0 = 2 m/s. Concentrated H 2 SO~ is passed over the Raschig rings downwards. The S0 3 equilibrium pressure of strong H 2 S0 4 is almost zero. (a) What length of column is required to absorb 98 per cen t of the S0 3 passed through if plug flow of the gas phase is assumed? (b) What length of column would be required if measurements on a 2m long column had shown that the residence time d istribution resembled that of ten perfect mixers connected in series?

Answer: (a) 3·65 m (b) 4·05 m

268 *10. Show that John reached the right conclusion during his deliberations about his cup oft~ discussed at the beginning of this paragraph. Comments on problems

Problem 4 A mass balance over a short height dx of the column yields after integration (c = c0 at H = 0) : 2

With cjc0

=

kanD H} -c =exp { - -~Co 44>"

0.02 and a

= 6(1 -

s)/dp = 1800 m 2/ m 3 we find :

kD 2 H = 7·65 x 10- 7 m 4 /s

(1)

The mass transfer process is analogous to bea.t transfer during flow through

a packed bed, so we can adapt equation (III.69a) for the estimation of the mass transfer coefficient :

and we find :

kD = 6-3 x 10- 6 m 2/s

(2)

Combining this with equation (1) we obtain for the desired product :

DH

= 0·12m2

So a small column of, for example, D = 0·2 m and H for this purpose. ·

= 0·6 m would be suitable

Problem 6 The only difficulty of this problem is to decide which mass transfer coefficient is important. This problem has already been discussed in paragraph 1V.1.2, so we can be very brief here. The figure shows the concentrations that play a

c* 2 Water

O r-gonic liquid

269 role and, since there is no accumulation of matter at the interface, Y'fC can Write :


k1(mc~ - c l } = k 2 (c 2

-

cf>

Eliminating lhe unknown c~ , we find for the mass flux :

cf>" nc

= mc 2 - c 1 m 1 -+kl

k2

Since the diffusion coefficient of the transported compound is the same in both phases k 1 = k2 , but because m = cf!c; = 1/200 the mass transport in the droplet is rate-determining and equation (1) becomes :

cf>':n

= k 1(mc 2

-

c 1)

The concentration equalization in the droplets is :

c1 - ( c)= c~- 0·9 c! = O·l c 1 - c0 c~

and we find from Fi~e III.8 that Fo

= Otfd; =

0·05 and L

= vt =2m.

Problem 9 The rate of absorption is determined by the gas phase mass transfer resistance because the S0 3 vapour pressure of concentrated H 2 S04 is practically zero. The gas phase mass transfer coefficient can be calculated with the help of the boundary layer theory (equation IV.32; see . .paragraph IV.3.2, case 5) by using instead of vCX) the sum V; + vCX). Here we can assume v, « V 00 and we find (vCX) = vj£ = 2·18 rnjs) :

k = 0·765 x 10- 2 .m2/s and . ( k) = 1·53 x 10- 2 m/s

A mass balance over the differential height dx of the column yields after iii· tegration (c = c0 at H = 0):

_c = exp (- kanDz c0 4
H)

= exp ( __ kaH) = exp . v

(-~·H-) HTU

Now cfc0 = 0·02 and HTU = vfka = 0·935 m , so we find :

H

=

-HTUln0·02 = 3·65m

If we want to account for longitudinal dispersion, we first have to calculate the height of a mixing unit : L n

2 = 0·2m 10

HMU = - = -

So we find for the height equivalent of a transfer unit:

HETU

= HTU + !HMU =

1·035 m

270 and, writing more correctly :

we find H = 4·05 m. Problem 10 John and his cup of tea We assume the mass transfer resistance to be situated outside the coal particles. A material balance over a differential height element dx of the coal column then leads (with c = c0 at H = 0) to the expression :

~ = c 0

exp (-

kmtDz) 4¢

= exp (-

0

_!!_) HTU

The height of a transfer unit stays constant and from the information about the normal operation of the column (c = 0·01 c0 ) we find that H 1 = 2·3 HTU. With the impure water, for the same final concentration (now c = 00033 c0 ) a column height of H 2 = 34 HTU is required~ i.e. 48 per cent more. IV .5. Mass transfer witb chemical reaction John looked

at

the fiercely burning haystack. The police

agent informed him that at 5 o'clock in the nwrning, when he discovered the fire, the stack appeartd to be 4 m high. Now, 3 hours later che stack was still 2·5 m high. John thought: the mitial height of the stack was 6 m and therefore the fire must have started at I o'clock or later. Lighting his pipe and carefully extinguishing the nuuch he inquired for what sum the hay was insured.

Mass transport can be increased if the substance transported into another phase reacts there chemically, but a chemical reaction can be slowed dow.n by a ·m~ transfer resistance (the transport rate is then lower than the conversion rate). We distinguish between homogeneous and heterogeneous reactions. During homogeneous reactions the chemical conversion occurs at aU places where the reactants are present. The reaction rate term has to be included in the mass balance, equation (IV. l ), which for the situations we shall discuss can be simplified to : 2 d cA) · (IV.48) DA( dxl + RA = 0 (no concentration gradient in the x-and z-directions, no tlow velocity,stationary state). Since the transported substance is converted at all places, the concentration gradient is increased and a sufficiently fast reaction will therefore increase mass transfer. During heterogeneous reactions diffusion and reaction are separated. The reaction occurs at a reaction interface only (e.g. at the catalyst surface) and the mass transfer coefficient is not influenced by th e chemical reaction.

271

The reaction rate term in equation (lV.48) is, for a first-order chemical reaction, given by : _(IV.49) RA = - k,cA For a nth-order reaction this expression becomes: (IV.50)

RA ~ -. k,c~

whereas the reaction rate for a second-order bimolecular reaction .between reactants A and B is given by: (IV.51)

We will restrict ourselves here to irreversible first-, second- and nth-order

homogeneous reactions and to first- and second-or,der heterogeneous reactions. For analysing these problems we will make use of the film theory (paragraph IV.l.4), i.e. we will assume that the bulk of the liquid is well mixed and that all mass transfer resistance is localized. in a thin boundary layer o_f thickness ~c · Thus, the mass transfer coefficient is given by k = 0 /bc. . . For homogeneous reactions we can distinguish two extreme cases : the chemical react_ion is so slow that ~t ~curs mainly in the bulk of the liquid and the reaction is so fast that it occurs partly or completely in the boundary layer. IV.5.1. Slow

homogefl2ous first-ord~!"

r:eactions

If the chemical reaction is slow, we ca~ neglect the reaction occurring in the thin boundary layer and assum-e tha.t the e~tire reaction takes place in the bulk of the liquid: In that case we will find a concentr~tion gradient over the laminar boundary layer as shown in Figure IV.8. At the interface the concentration is cA; and on the other side of the film·-the ~oncentration equals the bulk concentration. The assumption. of negligible reaction in the film is justified, if the ... /CAi

I I I I I I.

1

1,)

I I

CA

-x

Figure IV.8 Concentration distribution during mass transfer with slow homogeneous first-order chemical reaction

272 amount transported through the film (""' AkAcAi) is much bigger than the amount of A converted in the film by chemical reaction (,.., Aok,.cAJ} Thus, if:

or, making u se of the relation~ .:-. Of kA, if :

This is genera!Jy written as:

[fi;.k, = Ha <

Vk.f"

1

(in practice < 0·3)

Jo

The dimensionless group Ak.Jki is called the Hatta number H~ which will be discussed in greater detail in the next section. The reader should furthermore realize that in most practical situations the film volume is very small as compared with the bulk volume_ If we take. for exampie, a stirred gas-liquid reactor with a= 1000m 2/ m 3, we find l> = OjkA ~ lo- s and the total film volume is then 10- 2 m 2 jm 3 , i.e. just 1 per cent of the total volume. · Returning to the situation shown in Figure IV.8 we can now set up a macromass balance, realizing that in the stationary state all reactant A that diffuses through the boundary layer has to react in the bulk volume. Thus: (IV.52)

where A is the size of the interface and V the bulk volume. Eliminating the unknown bulk concentration we find for the mass flux through the interface:

(IV.53) If the reaction is very slow, i.e. Vk, « AkA or AkAfY k,. >> 1, the bulk concentration will be almost equal to the concentration at the interface, cA ~ cAI. From equation (IV.53) we find, then, that the mass flux is given by: n

V

cf>A = Ak,cAi

(JV.54)

This means that under these conditions the rate of the transport process is completely determined by the rate of the {slow) chemical reaction. If, on the other hand, the chemical reaction is very fast, Vk, » AkA or A kAfli k, « l, we find:

(IV.55)

273

because the bulk concentration cA is now very small (due to the fast conversion of A~ The rate of the transport process is in this case determined by the ma.Ss transport through the boundary layer.

IV.5.2. Fast homogeneous first-order reactions If the reaction is so f~st that all substance A is converted in the boundary layer, we can easily solve the problem o f diffusion with chemical reaction by using the mass balance (equation IV.48):

where RA = - k,c A. The solution of this differential equation with the boundary conditions cA = cA; at x = 0 and cA = 0 for x = ro is: cA

= cA, exp (

-xJft )

(IV.56)

is

This concentration distribution shown in Figure IV.9. From equation (IV. 56) we find for the mass flux through the boundary layer :

(IV.57)

.I I

CAi

I

I

I. I

t

I

I

l I

~

~A. =exp:{x~} I AI

l\

l I -----X

Figure IV.9 Concentration distribution during mass transfer with fast homogeneous first-order chemical reaction

In this case the mass transfer coefficient is given by~ and is solely depen-

dent on the diffusion coefficient and the reaction rate constant k, . The ratio jD Ak,fk! which represents the ratio between the mass transfer coefficients with and without chemical reaction is called the Hatta (Ha) number, which plays an importan t part in mass transfer with chemical reaction.

----==~~====~--------------~--

274 From the concentration distribution, equation (IV.56}, we see that the penetration depth ~r of the concentration distribution with chemical reaction is given approximately by : (IV.58)

Consequently, the ratio of the thickness b = 'JJfkA of the boundary layer and the penetration depth b, is given by :

!__= 8,.

~=Ha

~~

which shows another physical meaning of the Ha number. ff the reaction occurs partly in the boundary layer and partly in the bulk of the liquid (Figure IV.lOd) (where. because of the fast reaction, cA very Jow), we can show that the mass flux is given by:

is

(1V.59)

If the reaction is very fast, i.e. Ha = j k,.IO Afk~ » 1 (for all practical purposes Ha > 3), equation (IV.59) can be simplified to equation (IV.57), which we have already derived In this case the total reaction occurs in the boundary layer. If the reaction is relatively slow, Ha = J krDAik! « 1 (Ha < 0·3), we find from equation (IV.59) the relation for simple physical adsorption :

lj>A

=

kACA i

and only a negligible part of the reaction occurs in the boundary layer. The Ha number is therefore the criterion for whether the reaction occurs completely in the bulk of the liquid (Ha < 0·3) or completely in the boundary layer (Ha > 3). If 0·3 < Ha < 3, reaction in both the bulk and the film is important. Figure IV.lO shows the five possible cases of mass transport with homogeneous first-order chemical reaction which we have discussed. If we want to determine to which class a certain problem belongs we start by calculating the Ha number. If Ha < 0·3, as a second criterion the expression Ak.JVk, has to be determined (compare equation IV.53). Equation (IV.59} and Figure IV.IO also show that the rate of physical absorption is increased by a factor : (IV.60)

by the occurrence of a homogeneous first-order chemical reaction. This increase

J

in adsorption rate by a factor 1 + Ha 2 in the case of a chemical reaction is valid under all flow conditions. The thickness b, of the boundary layer where the chemical reaction occurs is so small that it is not influenced by the flow conditions. Also. during turbulent flow, the eddies cannot penetra te this thin layer. The enhancement factor Fe= j 1 + Ha2 is therefore universally applicable.

275 _,...-cAi

~

(0 )

t

u

: I I

I I

J

CA-.::.CAi

-

Ak Ha < 0.3 ; ~ >> l Vk,

No reaction in boundary layer

Ak Ha < 0.3 · ~ ~ I ' Vk, (b)

t/J~ : : : : k;.CAi{

l

A.kA Vk,

}

1+ -

No reaction in boundary Jayer

Ak Ha < 0·3; _ A» J Vk, ( c)

No reaction in boundary layer

Ak 0.3 < Ha < 3 ; ~ « 1 Vk, (d)

¢'A = kAcA;j i

+ Ha2

Reaction partly in boundary layer

AkA

Ha < 3 ; - k « 1

v ,

(e)

¢;. = kAcAIHa Reaction c:Omp1etely in boundary layer

Figme IV.JO Five cases of mass transfer witb homogeneous irreversible first-order

reaction

IV.5.3. Homogeneous nth-order reactions The analysis developed in paragraphs IV.5.1 and IV.5.2 can also he applied t o nth-order chemical reactions, where the production term of equation (IV.48) is given by (equation IV. 50):

276 The results of this analysis are completely analogous to that of first-order reactions and, again, the five situations summarized in Figure IV.l 0 can be distinguished. For this type of reaction, the thickness of the reaction boundary layer (equation IV.58) becomes : (IV.61)

and the definition of the Hatta number is now:

1

Ha = -

kA

2 krCAt n- l [)A. n + 1

(IV.62)

With the above Ha number, the relations developed in paragraphs IV.5.1 and IV.5.2 for first-order reactions also describe nth-order reactions. There is one important practical difference: for first-order reactions b, and Ha are only (through k,) dependent on temperature, but for nth-order reactions these parameters are also dependent on the concentration of A at the interface.

IV.5.4. Homogeneous second-order reactions A component A is absorbed and reacts with a component B which is present in the bulk and which is not desorbed The production term in equation (IV..48) is given by (equation lV.Sl): RA = -krcAcB If the concentration of component B is uniform and equal to c 8 CXJ, the bulk concentration, the solution to this problem is simple : the reaction can be regarded as being of a pseudo first-order type with a first-order reaction rate constant k~ = k,c8 00 . Thus :

(JV_63) The solution to this problem has already been treated in paragraphs IV.S.l and IV.5.2. There are two conditions under which the assumption of a pseudo first-order reaction is justified : (a) If rhe reaction takes place in the bulk so £hat :

-~ _ -~OA H ak2- A

(b) If the reacta nt B can be transported quickly enough into the film where the reaction occurs so that the concentration of B in the film can be assumed to equal the bulk concentration (see Figure IV.ll), i.e. if: (IV.64)

277

- -x

~, 8

Fieure IV.11 Concentratjon distribution during mass transfer and second-order reaction if Ha > 3, kBcsaJfkAcAi > 10 Ha

In practice this condition is fulfilled if: kBcBa:; > IOjk,.c 8 CX)IDl A CAo or : (IV.65)

If equation (IV.65) is fulfilled, the correction factor Fe for the acceleration of mass transport by chemical reaction is given by : Fe = j l + Ha2

=

1+

k,.c~ifi)A

. (IV.66)

If the reaction is so fast that insufficient B is transported into the boundary layer, the transport of B will influence the reaction rate in the film. If:

the reaction is so fast that A and B cannot exist without reacting instantly. A reaction front is formed at x = xR (see Figure IV.l2) where cA = c8 = 0. The film theory yields for the transport rates of both components to the reaction front : (IV.67)

Hence, the chemical acceleration factor is fo und to be, bearing in mind the

definition of the H a number given in paragraph IV.5.2 and realizing tha t for this

278

0

-x

Figure IV.12 Concentration distribution during mass transfer with second·order reaction if Ha > 3, kaescrlkAcAi < 0·1 Ha

case always Ha > 3 :

(IV.68) where f>A and <58 are the thicknesses of the boundary layers during physical absorption only. The penetration theory predicts that :

bA-

SbB f>a - ShA -

(OA)t . lOa

(see case 1 of paragraph IV.3.2) whereas the boundary theory (cases 2 and 4) predict a relation of the form :

{)A= (0 A) -t <>a

De

(IV.69) l ay~r

and the Leveque (IV.70)

If D A and 0 8 are of the same order of magnitude (which is often the case) the enhancement factor Fe is only slightly dependent on the physical model chosen and we can use equation (IV.68), together with equation (1V.69) or (IV. 70) to produce a reasonably accurate estimate of FeIn the in termediate area between a fast reaction (k 8 c8 d kAcA1 > 10 Ha) and an instantaneous reaction (k8 c8 a>f kAcAi < 0·1 Ha), so in the area where (see Figure IV.l3) : kaeaa> kAcA

~ Ha

the F, value is a function of the Ha number and of

279

l

-x

Figure IV.13 Concentration distribution during mass transfer with second-order reaction ifHa > 0·3 k8 c8 .,jkAcAi ~ Ha

where the exponent m again is a function of the physical model chosen (m = -! for penetration theory, m = t for boundary layer theory)~ The dependency ofF, on these two expressions is shown in Figure IV.l4 which can be used for estimating the ch.emical acceleration factor. Table IV.3 gives a summary of the various solutions for mass transfer with second-order reactions discussed 1000~--------------------------------------------------------~

10

10

100

1000

Ha

Figure IV.14 Values for the enhancement factor Ha > 0.3, k8 c8 «J/kAcA1 ~ H a

Fe for second-order reactions if

280 Table IV.3

Summary of mass transfer situations with irreversible second-order reactions

kacaao

Ha

Solution

kACAc

<03

Pseudo first -order reaction with

solutions of Figure IV.lO apply

> l ORa >0·3

~

= k,cBrr;

Situation given in Figure IV.13

~ Ha

~

=

FckAGAi

with Fe from Figure IV.l4 < 0.1 Ha

>0.3

Instantaneous reaction, Fjgure 1V.12

~~ = {( ~:r + ( ~;

r-l

C;:}kACAO

where!< m
the material ba,lance for any of the five flow situations discussed in paragraph IV.3.2 becomes:

(IV.71) The boundary condition s are : X = x

0, y

> O,y

~

0, CA

= CA«>,

Cs

=

CBCD , CR

=

Cs

= 0

= oo , CA = CAoo • CB = Ca.:c • c 1 = c 5 = 0

dcA

x > O,y = 0, O A dy =

n

u..v 8

dc 8 dy

=

-

[) R

dcR rn. des dy = -ILls dy

= R A(c,.., c8 , ca,cs )

The last boundary condition implies that there is no accumulation of reactants at the interface. The reaction rate RA is related to the area of the interface (in the case of porous catalysts to the outer catalyst surface). Because the reaction kinetic term is found only in the boundary condition and not in the mass balance, the chemical enhancement factor F, for heterogeneous reactions is unity. Under steady-state conditions a mass balance at the interface shows (no accumulation of reactants at interface) : kA(CAq; - CAi) =

kB(CB CD -

CsJ =

kRCR i

= KsCs; =

- RA(CA; , CB; ,

CR£,

CsJ (IV.72)

281

These equations are, in general, easily solved. The .rate of conversion is in all cases given by: (IV.73)

We will now discuss the special cases of first- and second-order reactions. (a) First-order reaction. Equation (IV.73)·for tbis case reads : ~ = kA(cAoo - cAi) = - RA

= T<.cAi

(IV.74)

The reaction rate constant k~ is here related to the outer size of the interface (ma:croscopic reaction rate constant) and has the dimension (m/s). It is related to the well-known chemical reaction rate constant k, (s -l) via the specific interface area a (m 2 jm 3 ) as follows: k,

= ak', *

From equation (IV.74) the ratio cAJ cAoo can be calculated to be: 1 ---KA 1 k;

kA

CAi

-=

k~

+

+-

kA

and the overall conversion rate is found from equation (IV.74) as:

(IV.75)

Apparently, here the quotient kAfk~ fulfils the role the expression Ak.JVk, played during homogeneous reactions (paragraph IV.S.i). We can distinguish three principal situations :

A. The reaction is so slow compared. with the mass transfer process that mass transfer does not decrease the reaction rate and the concentration of A is uniform and equals cA-:c· B.

k~

»

kA, CAi ~

0, tP~

=

kACAoo

The reaction is so fast that the concentra~ion of A at the in terface is practically zero and physical mass transport determines the overall reaction rate. * In the case of heterogeneous catalysis often a microkinetic reaction rate constant the total pore surface of the catalyst is used; 'K,. is related to k;. and k, by :

k;

related to

k r = ap k"r = ak'r 2 3 where a = outer specific surface of catalyst particles and aP = specific pore area of catalyst (m /m ).

282

c. Both chemical reaction and mass transfer determine the overall conversion rate. Figure IV.lS gives a summary of these three situations and shows the concentration distributions which occur. The reader should compare this figure with the first three cases of Figure IV.lO, which refer to ho mogeneous first· order reactions.

A.

-)(

B.

c.

~ =

kAcAco

Ftpre IV.l5 Three cases of irreversible heterogeneous first-order reactions

(b) Second-order heterogeneous reactions. Equation (IV.73) for this case reads :
Elimination of cs; leads to the following expression for cAJcA«J:

x¢(CAi)2 + {.x(l -
CA«J

CA~

(IV.76)

283 Calculation of c,.JcAoo from equation (lV.76) enables 'us to find the rate o_f

conversion as :

·

= kACAo ( 1 -

cf>~ ·

CA·) '

CAo

(IV.77)

The reader may check that for rjJ-+ 0 second-order heterogeneous reactions can be regarded as pseudo first-order reactions. At this stage, the authors had Gonsidered inserting -a section on the design of simple chemical reactors. This would have been logical in the lay-out of the book The main features of the .design of a heat exchanger (paragraph ITI.4) fo-Ilowed the fundamental considerations on heat transfer in the preceding sections and the design of a mass exchanger (paragraph IV.4) concluded the theory of mass transfer without chemical reaction 1n the beginning of this chapter. However, simple and isothermal chemical reactors seldom occur in practice and we prefer to refer the reader to a more elaborated text on the subject of chemical reactors, e.g. Levenspiel 's Chemical Reaction Engineering. Simple calculations can already be made on the basis of what is offered here, as the following problems will show-it is only and again applying the law of mass conservation.

IV ..5.6. Problems 1. On the bottom of a cylindrical vessel (diameter D = 1 m, height H = l m) with stirrer is a 1 em layer of solid substance. To remove this layer the vessel js filled with a liquid in which the solid dissolves. In solution the solid decomposes according to a first-order chemical reaction. Data :

mass transfer coefficient (without chemical reaction) k = 10- 4 mfs solubility of the solid substance in the liquid = 50 kgjm 3 density of the solid substance p = 2000 kgjm 3 diffusivity [} = 10- 9 m 2 js reaction rate constant kr = 10- 2 s-

1

•

(a) Show that the mean concentration of the dissolved solid in the liquid is much lower than the solubility of the solid substance. (b) Show that the conversion by chemical reaction in the .boundary layer

is negligible. (c) How long does it take for the layer of solid substance to be ~issolved completely? (d) A chemist states that the process can no doubt be accelerated by adding a catalyst which makes the reaction rate constant 1000 times as great. Demonstrate quantitatively whether his statement is correct. Answer : (a) c = 0·01 c, (b) be = lo-s m ~ <0-3 per cent of the total conversion in· film (c) 4000 s (d) j2 times as fast

284 2. If under otherwise equal conditions pure oxygen or oxygen -from the surrounding air is absorbed in a strong Na2 S0 3 solution to which a large amount of cobalt salts have been added, the difference in rate of adsorption appears to be a factor of 10. Determine the order of the reaction between 0 2 and Na2 S0 3 with respect to the oxygen (the reaction is very fast, Ha > 3).

Answer: 2 3. In a laminar liquid film (c~ntact time O·l s) C0 2 (m = l) is absorbe~ from the pure gas (20°C. I atm). The liquid film contains 0·04 kmolfm 3 NaOH. Under the prevailing conditions the second-order reaction rate constant between C0 2 and NaOH is 6 x 103 m 3 jkmol s. Calcula te the chemical enhancement factor and the mole flu x in this situation and draw the c~ntration distribution in the film. Assume []lc02 = DoH- ~ 2 x 10 - 9 rn 2 /s; because 1 mole C02 uses two moles NaOH, for CsQ) in this "case tcNaOH must be used.

oon-

Answer: F, = 1·5,

q,;or =

9·85 x 10- 6 kmoljm 2 s

4. What will be the chemical enhancement facto r and the mole flux if in the situation of problem 3 a temperature of 70°C is taken (k.- 2 x 105 m 3 fkmol s) and the concentration ofNaOH is chosen equal to 04 kmol/m 3 (m = 042)?

Answer: Fe = 13·5; ¢mo1 = 5·6 x 10- " kmo1/m 2 s. *5. In a packed column of catalyst particles a component from the liquid stream is converted by a chemical reaction. The reaction is an irreversible heterogeneous reaction of the first-order and proceeds at the surface of the particles. How high.should be the catalyst bed to ensure a degree of conversion of0·63 ( = 1 - e - 1 )? Data :

ro-s

reaction rate constant k, = 4 X m{s (macrokinetic) 9 2 diffusion coefficient II) = 10 - m js superficial fiow rate in the bed v0 . = 4 cmjs kinematic viscosity of the liquid v = 10- 6 m 2 /s bed porosity e = ().40 particle diameter d P = 1·0 mm

Answer : 0·376 m *6. During fat hydrogenation pure H 2 is absorbed in liquid fat containing catalyst particles at which the reaction between H 2 and unsaturated fatty acid takes place. The rate of conversion per unit liquid volume related to H 2 is independent of the degree of conversion of the oil and appears to be linearly proportional to the solubility of H 2 • In a certain experiment 1 the ratio between these two constants appears to be 0.03 min- •

285 If the interface between gas and liquid is increased by a factor of 2, the specific rate of conversion will be 20 per cent higher. If the catalyst concentration is increased b y a factor of :2. the specific rate of conversion will be 50 per cent higher. Calculate for this experiment the gas- liquid interfacia] area per unit liquid volume and the {macro)reaction rate constant

o f the reaction between H 2 and unsaturated fat. Data :

the partial mass transfer coefficients in the liquid at the interface with gas kl = 1o- 4 m/s and at the interface with the catalyst k2 =

w-s mfs

the interface of the catalyst per unit volume of liquid Answer:

= 150 m 2 fm3

15 m 2 /m 3 ; lo -s m/s

7. One of the components (A) of a gas·mixture is absorbed in a solution in

which it reacts very fast with a component B which is present in excess according to a second·order reaction of the type R = kcAcs (i.e. the reaction takes place entirely in the boundary layer). At a certain concentration of B the overall mass transfer coefficient is found to be K = 10- 4 mjs. In order to find for this case the partial mass transfer coefficient in the gas phase, the overall mass transfer coefficient is measured again at a 4 times higher concentration of B. The result is K = 1·5 x 10- 4 m/s. Calculate the partial mass transfer coefficient in the gas phase if the given mass transfer coefficients are related to the gas phase. Answer:

kg= 3 x 10 - 4 m/s

*8. Reith carried out the following tests:

Two equal streams of 0 2 and N 2 were mixed and passed through a stirred gas- liquid reactor containing a solution with an excess of sulphite and sufficient cobalt ions to realize chemical absorption in the range of the fast reaction (Ha > 3). Subsequently, he passed under otherwise the same conditions the two streams 0 2 and N 2 separately into the gas-liquid reactor. In the second experiment he found absorption rates which were 40 per cent higher than in the first test. How do you expiain this? Answer :

There is complete segregation in the bubble phase under the conditions of this experiment.

9. The chemical absorption of C0 2 (A) in aqueous NaOH solutions can in firs t approximation be described by a second-order reaction :

286 ln a laminar jet (di~eter 0·66 mm, t>i = 5 m js, T = 20°C) the following absorption rates, are measured at various jet lengths L : L (m) .

2.() X 10- 2 45 x w- 2

tPm

(kg/s)

2.()4

X

4-6{) X

10- 8 10- 8

where 2 c 8 ~ = (OH --)ro = 1·05 kmoljm 3 , cAo = C0 2 = 0·238 kg/m 3 , 1 U)A = 10- 9 m2/sand lOB= 1·6 x 10- 9 m /s. Calculate the reaction rate constant k, (at 20°C) from these data. Answer :

k,

=

5-75 x 103 m 3fkmol s (k,.. = 4-48 x 10 - 4 mfs Ha = 4·6)

*10. C02 has to be washed out of a continuous stream of hydrogen at 30°C and 7 atm absolute pressure in a packed column by means of a countercurrently flowing monoethano]amine (MEA= HO-C2 H 4 -NH2 ; one mole C0 2 uses two moles MEA} solutio~ which is reactivated and recycled. If the C0 2 concentration in the hydrogen has to be decreased from 20 to 0-01 per cent by volume and if the initial concentration of free MEA in the wash solution is 3 kmoljm 3, how high should the packing in the column be?

Data : column diameter D = 1m

packing : Raschig rings, dP = 7·5 em·(a = 60 m 2 /m 3 , s = 0·95) gas flow rate c/>g = 4000 Nm 3fh liquid flow rate 4>L = 30m3fh solubility of C0 2 in MEA solution m = 1 second-order reaction rate constant between C0 2 and MEA, k, = 7500 m3 jkmol s p, = 0-56 kg/m~ (at 7 atm), PL = 1000 kgjm 2 rJ1 = 9·3 x 10- 6 Nsjm 2 2 Dco2 inH:z = 5·5 x 10 - 5 m f s . - 9 2 Dco2 inH:zO = 2 x 10 m js IOMEA lnHl o = 10- 9 m2/s vL = 10 - 6 m 2/s

Answer :

15·4 m

*1 1. Show that John drew the right conclusion in the case o f the burning haysta ck related at the beginning of this paragraph. Comments on problems Problem 5

A material balance over a short height dh of the packed column reads (A = cross-sectional area of column):

- cf>L del..= ka(cL - ci)A dh = krac;A dh

287 and isolating and substituting the unknown interface concentration ci we obtain :

or, after integration between h CL

cLo

~

0, cLo and h, cL : } !) k+ k

{

- ah

= exp . (~ Vo

r

Now the degree of conversion was given as (cL 0 - cJ JcLo cJcLo = e - 1 and therefore :

=

1 - e - 1, so

We find the mass transfer coefficient with the aid of equation (III.69a) (making use of the analogy between heat and mass transfer) to be k = 1·14 x 10- 4 mjs and with a = 6(1 - e)jdP = 3600 m 2 fm 3 we finally obtain h = 0·376m.

Problem 6 The reaction rate between hydrogen and (cL = H 2 concentration in liquid phase):

fat~

given by a relation of the form

Now it is stated that the reaction rate is linearly dependent on the hydrogen concentration, so ex = 1, and independent of the degree of conversion of the fat, hence f3 = 0, i.e. the reaction is of the pseudo first-C?rder type. We can write :

(kmoljm 3 s)

c,.

Hydrogen has to be transported from the gas phase (con~ntration mass transfer coefficient k,. specific interface a, ) to the liquid phase (concentration cL) and from there to the surface of the catalyst particles (concentration c;, mass transfer coefficient kL , specific interface ai) where it reacts chemically. Under steady-state conditions all these mass transfer steps proceed at the same rate (no accumulation of H 1 in any phase) and we can write (k; = a!<,, where k, = mascroscopic rate constant in m/s and k; = chemical rate constant in s - 1 ) that: (1)

288

From this equation we isolate and eliminate the unknown concentrations and c; and obtain : _c/J_m_ol = _ _ _ m_c.!!..~---

1 kgag

V

cL

(2)

1 1 +--+ kLai k,.a1

This equation completely describes the overall hydrogenation process. With the aid of the information given we can evaluate the various terms of this equation as follows: 1 given : 4>mol =0·03 min - 1 = 5 x 10- 4 s- 1 = (3) ~~ _1_ + _1_+_1_ k 8 a8

kLa 1

k,a,

given: if a~ is doubled, 4>mor increases by 20 per cent~ so:

1

1·2

=

1

1

--+-+-

( 4)

k,ag kLai k,.a1 1 1 1 -- + -+ 2kgag kLai k,.a;

given : if~ is doubled (via catalyst concentration), ¢aw1 increases by 50 per cent,

so:

1

1

1

k Lat

k,.ai

- + -- + - -

1·5 = k,ag

(5)

1 1 1 - + --+-kgag 2kLai 2k,a1

When this equation is worked out it appears to be identical with equation (4), so the information given is consistent. From equa~ions (3) and (4) k 8 a, can be calculated and is found to be k 8a, = 1·5 x 10- 3 s- 1, so that with the given k8 value we find a8 = 15 m 2 /m 3 (which is very low for a gas liquid reactor). Using the values given forkLand ai we find 5 from equation (4) that the macroscopic rate constant kr = m/s, so the 1 3 chemical rate constant is given by k; = 1·5 x 10- s- •

w-

Problem 8 Sodium sulfite is oxidized in solution to sodium sulfate by oxygen if cobalt ions which act as catalysts are present:

0

2

+ 2S0 3 2 - ~ 2S04 2 -

At a pH of 7-8 and 15-35°C the reaction is :

10- 6 < CeoH < 10 - 3 km.olj m 3 , second-order in 0 2 (see problem 2),

first-Order in C0 2 + if 3

X

first-order in S0 3 2 - if c503 ~- < 0·4 kmol/m 3 , zero-order in SQ3 2 - if Cso3 2- > 0·4 kmo}jm 3 •

For the above range of Co2 - concentrations and c503 2 ~ > 0·4 kmolfm 3 the pseudo second-order reaction rate constant is :

2 x 10$ < k, < 108 m 3 fkmol s The only explanation of the phenomenon Reith measured is segregation of the gas bubbles which influences the driving force of the mass transfer process. H we neglect the concentration decrease in the gas bubbles we can write for Ha > 3 : ¢>mol

=

Ac;Jik,c;ID = Am~ c't

In Reith's first experiment c, = 50 per cent by volume in all bubbles, so all A. In the second experiment c, = 100 per cent by volume in half of the bubbles. so 50 per cent of A if complete segregation occurs. The ratio of the two mass flux rates would be :

4>

mol sepuate streams

4>mol mi~lure

=

I; 1

{1)

23

1 = X

J2 = 1·41

1

Theoretically for complete segregation of the bubbles we would expect a 41 per cent higher absorption rate during the second experiment, which indeed was measured. This strong segregation (no coalescence of bubbles) is caused by the high ion concentration of the sulfite system. In distilled water and in most organic solvents considerable coalescence and redispersion of bubbles would occur. Problem 10

The sketch of the absorption column shows the gas and liquid flows and the corresponding concentrations in all streaii?-s. To begin with we assume that the reaction between C0 2 and MEA proceeds in the boundary layer, i.e. that Ha > 3. We will check this assumption later on, when we calculate the Ha number. P =0· 56 kg / m3 9

3

r#JL =3 0 m /h 3 cL =3 ·0 kmol/m 1

.P9 =507 ~ 3/h (7 ot m l 3 Cg =0·0 I vol % = 2·8 x 10- 5 kmol 1m

t

'

109 =7 37 x 10 - 4

3

log =153 kmol /m

t

P 9

3

kmol/m

= 2· 3 1 kg / m3

1/>9 =635 m3 /h ( 7 otm ) c9 = 20 vol % =0 ·056 xmol/m3

290 Since the .Qeight of the col~ will be greater than 10 packing particles, we can assume countercurrent plug tlow of both fluids and the mass flow rate of C0 2 into the MEA solution is given by equation (IV .53) as:

4> = K aAL !leo - Jlcl '"

llc

L

ln-0 llc 1

where K.L. is the overall mass transfer coefficient related to the Liquid phase and Ac0 and Ac 1 the concentration differences at the column entrance and exit. Thus:

· .lle0 = me,0

-

c~. 0

llc 1 = mcg 1

-

eL 1 =

0·056 - 0 = 0·056 kmo]/m3

.

2·8 x 10- 5

-

0 = 2-8

x

10..,. 5 kmoljm 3

since the C02 concentration in the bulk liquid is zero. Consequently, the logarithmic mean driving force is given by : (lleh01 =

b.c0 - t:.c 1 Ac ln-0

=

7·37 x 10- 3 kmolj m 3

Ac 1

We shall now calculate the overall mass transfer coefficient KL which

given by (see equation IV.15b) :

KL = {

m

kg +

1 }-

IS

1

k F

L c

since the mass transport is enhanced by chemical reaction. The partial gas pha.se mass trans.fer coeffici,ent kg is fou~d with the aid of the boundary layer: theory (case 5 of paragraph IV.3.2; see also problem 9 of paragraph IV.4) to be (with v0 = 0 ·20 mjs, v = 1·66 x lo-s m 2 js): ( kg) = 10- 2 rnjs The partial liquid phase mass transfer coefficient KL is found analogously to problem 8 of paragraph IV.2 (see comments on problem 8, page 246) to be (v 0 = 1·05 x 10- 2 m/ s, s = 6·2 s- 1 ) kL = 1·1 x 10- 4 m/ s

We shall now determine the enhancement factor F,. For the Ha number we find using the mean MEA concentration :

1 Ha = K

L

.jk Cs a:> [)A= r

l ·l

1 X

10-

4 V /7500

x 1·53 x 2 x 10- 9

= 43·5

This confirms our initial assumption that Ha > 3 was correct and the reaction proceeds in the boundary layer.

291 At the column entrance (~ottom) .(see 'p,aragraph tV.5.4): k BcBa:> = [JlB cBco kAc Ai

[Jl A CAi

= ~ 0·62 = 5-53 ~ Ha 2 ().056 ..

and we have to use Figure IV.l4 in order to find . Fc. With :

. /DA) t +·Csoo([JJA) - t = 14 + cA i Da

\lOs

. 0·62 0-0.56 x .1·4

=

9·3

we find from this figure Fe ~ 9·0. At the column exit (top) : k BCBct:l kACAt

1

= 22·8

3 X

= ·53 '500

10- S

which is·much bigger than 10 Ha, so here we have a pseudo first--:o rder .reaction with Fe = Ha = 43·5. For further calculation we will use as a rough approximation the logarithmic mean F, value of tog ·= 20·8. With the foregoing results we find for the overall mass transfer coefficient:

m

KL

1 } kLFc

= { k8 +

-J

{

=

1 10-

2

+ 1·1

. X

10

1

20

4 X

X

8 =

}-

t

1·86 x

to- 3 mfs

Since the mass flow of C02 from the gas to the liquid·phase is l/Jm = fb8 llc, = 635 x 0·056/3600 = 9·9 x 10- 3 kmql/~ the necessary height of. the column is found as:

L

.

~·

l/>m .

9·9 ~ 10~3

-

K LaA(Ac)1og

·

·

1-86 x

1~- 3

·

·x

60 7-37 x 1t

4

',

· 3 to-

=154m ....

·. '

'

~

··'

' J'

Problem .11 ., Let us assume that the'haystack has the form of a half-sphere of radius r t·that all burning occurs at the surface. of the·stack and that the rate of .burning is completely controlled by transfer of oxygen froin the·air. With these assumptions the rate of volume d ecrease of.the haystack is dependent on the oxygen transport from the air : dV dt

-

=

dr 4r 2 - ,_ k 2nr 2(c - c.)

dt

g

g

I

or, assuming cg - ci = constant : dr

dt = constant,

r.- c1t + c2

Evaluating the constants with the information given, viz. r = 6 m at t = 0, r = 4 m, t = x and r = 2·5 m, t = x + 3 we find x = 4 h, so the fire must have started at 1 o 'clock . .

I,

292 If the haystack has a form other than that of a half-sphere, it will have a higher surfacejvolume ratio and will therefore bum down more rapidly.

IV.6. Combined heat ud mass tramport The transport ofmatter from one phase to another is, in principle, attended by a heat effect. When, for example, on absorption a certain component from the gas phase dissolves in the absorbing liquid, heat of solution is released; in addition, an extra heat effect may occur if the dissolved component reacts with the liquid or with the reagents in it The temperature of the interface adjusts itself at a value which depends on the possibility of heat removal (or supply in the case of a negative effect) through the two phases. The absorption rate, however, is influenced by the interfacial temperature (e.g. via the solubility)~ so that in such a case the mass transfer and the heat transport are coupled. Examples in which the influence of the beat transport on the mass transfer rate could be neglected were given in paragraph IV.2, problem 4, and in paragraph IV.3, problem 8. IV.6.1. Drying A special case of combined heat and mass transfer is obtained if, for example, a drop of liquid (or an inert object moistened with liquid) is put in a gas flow which has not been saturated with the vapour of the liqwd Assume that the liquid originally has the same temperature as the gas. Because the gas is not saturated at that temperature, evaporation occurs from the liquid surface (mass transfer). As a result~ heat of evaporation is withdrawn from the liquid so that its temperature begins to fall The lower the temperature of the liquid) the greater the temperature difference between gas and liquid and the more heat will be transported) from the gas to the liquid phase. Aft~r some time a stationary state is reached in which the temperature and hence also the heat content of the liquid no longer changes with time; all heat required for the evaporation is then supplied. by tile passing gas. In this situation of dynamic equilibrium the following energy balance applies (see Figu.re IV.16) :

¢'JJ where

tllfe,A

=

t/J':nol AflHe,A

=the molar heat of evaporation of A at Tw . 4> 1r H

.

II

~ c;mol A

\ \

\

\

\

'__,.~"--

Go$ flow

Figure IV.16 surface

Evaporation from a wet

(IV.78)

For the local heat fiux 'H to

the surface can be written : t/>i

293

=ex(~ - Tw)

For the local rate of evaporation per unit of area applies, if PA .w is sufficiently · small with respect to the total pressure: A,. ll

'PmolA

k (p = RT A,w

-

PA,g)

(IV.79)

where (or Ttbe mean between T; and T., can be taken. By substitution of these equations for 4>~ and ¢~oJA in the energy balance {equation IV.78) we find : PA,w- PA.t = ~ - Tw

RT ~ tllfe,A

(IV.80)

k

In this relation PA,w and Tw are coupled via the vapour pressure curve of liquid A. If the evaporating drop is so small that the values of PA.r and T, do not change appreciably, rxj k in this relation can be replaced by the quotient of the mean transfer coefficients over the entire area ( rx)/(k). The value of this quotient can now be calculated with the help of the analogy between heat aDd mass transports. Both ( ex} and ( k) refer to the same geometric situation and to the same flow condition (the same value of Re). If Re is not too small, the analogy relation (paragraph IV.3.1) can be applied, which leads to: ( ex) = c o (Le)!

(k}

(l\".81t

pr

where the product Cp!J must be taken at the mean temperature T and Le = aj D = ScjPr is the dimensionless Lewis number. For gas mixtures the value of Le lies between 0·5 and ca. 2 and for liquids, appreciably higher than 1. With this relation is found as ultimate result: PA,w - P~g - RTcpP ~ - J:, ~ flli e ,A

·

(IV.82)

The temperature Tw is called the wet-bulb temperature because this temperature is reached by a moist sphere if it is placed in a gas flow.* This temperature is an important constant in the description and calculation of drying of materials in an air stream. Another application is found in spraydrying towers in which hot s,olutions in the form of droplets or liquid layers are brought countercurrently irito contact with the outside air; the droplets cool down because of evaporation but their outlet temperature cannot be lower than the wet-bulb temperature, belonging to the temperature and the partial water vapour pressure of the ambient air. The wet-bulb tempera ture takes its name from psychrometry, according to which T, and Tw are measured with a 'dry' and a 'wet' thermometer respectively *In principl~ the same effect occurs in Jt}ass transfer in a liquid flow, e.g. on dissolving crystals. Ascertain that jn this case usually very 'small" temperature differences occur.

- -- --

-

294 in a gas stream. Using equation (IV.82) PA,g can then be calculated. The psychrometer is usually used for the system air/ water, but application to other liquids is in principle also possible. The method is particularly useful because T.., is independent of the gas rate. However, in such a measurement we should make sure that the heat flow to the 'wet' sphere by conduction or radiation is small with respect to the heat flow from the gas. The wet-bulb temperature must not be confused with the so-called adiabatic saturation temperature ~- This is the temperature a gas (e.g. air with a temperature T, and water vapour pressure PA.J a ttains if it is saturated with vapour, the gas itself supplying the required heat of evaporation by cooling from Yg to J;. The calculation of 7:; is based on an energy balance over a unit of mass of the gas, in which. tb.e transport of heat and mass does not play a role. Derive that for 7; (and the relevant equilibrium -vapour pressure PA,s) applies : PA,s- PA,g _

~ - 1;

-

RTcPp filie,A

(IV.83)

Except for the factor (Le)i, (69) corresponds with (68). For all gas mixtu~es Le lies near 1 and for the system air/water vapour (Le)i = 0·95. Therefore, sometimes (wrongly) no distinction is made between Tw and 7,, although with regard to mechanism, they are of completely different origin.

IV.6.2. Problems *1 . Air of20°C and atmospheric pressure with a relative humidity of79 per cent

is heated to 50°C. What will be the partial pressure of the water vapour and the relative humidity? 2 This warm air flo~s along a completely moist ma~erial, r:t being 35 W /m °C. How much water ';,.' evaporates per unit of a;rea of the . material if the heat of evaporati.o n is entirely supplied .bY ~he a1r? Temperature 7; \C)

Vapour pressure Ps (10 3 N /m 2 )

20-0 22-0 25.0 27.0 '30.0 34·0

2·38 2·69 3·23 3·64 4·33

37.()

40·0 50-0

Answer :

5·13

642

7·50 12-33

1-88 x 103 N/m2 ; 15 per cent~= 0·34 x 10- 3 kglm2 s

2 A wet object of good heat conductivity is put vertically in dry stagnant air to dry. After a short time a stationary state has set jn and the object has reached the temperature T..,..,. The heat transfer air-o bject is described

295 by ( Nu)

= constant

Answer:

p - p RT w g = - - peP(Le)t T,- Tw Mle

(Gr Pr)~. p, and ~ change very little during the

drying process, Pw is small with respect to the total pressure and there occur no dry spots on the object Find a relation forT..,.

Comments on problems

Problem 1 The partial water vapour pressure in air at 20°C and 79 per cent saturation is 2 3 3 PH2 o = 0-79 x 2·38 x 10 = 1·88 x 10 N /m . If this air is heated to 50°C, PJi 1 o stays constant, but the saturation becomes 1·88 x 10 3 / 12·33 x 103 = 15 per cent The wall temperature is found via: Pw- Pg

T, -

= Pw-

1·86 x 103 = RTcPp Let= 69 50 - Tw !lHe

Tw

by trial and error to be Tw = 26°C and Pw = 3·37 x 103 N/m 2 • The mass flux of the evaporating water is given by equation (IV.79)and with the aid of equation (IV.81) we find :

A

~hd

wj

(K. /J

H .t

.p::,,, = :T(pw - p,) = c:::ci:..~+~~ = 18·6 x 10-• kmol/m2 s - -.,cdktJ tt4lt.

'r

J-ktnlw

rirtJ

1>a~rn

a 3·35 x 10-4 kgfm2 s 11JQ-mal

"7

undscAnn-1~

Cf

1

~-&1

lA.

•r

ID o -62f4-ma)t1J ID o -6~~2 u1 ,:_,. cc-,,,·d t,~f a /al'Cf 'f a Jruw ti11u!aj;~ 1 O· 2133 .!!.... Ir1rr 't-Ie I"~Cif/ T !OJ «) . K q= P'ft If .rn k "'"cl a-Jfrlch T r illf\A'a·ltr

c·

r,-r

311), J . ·h.JflrJ . @ -lilt r

d . ]OJ

r1

}eJ'4-/tJ 'i

fnlrn~are, !nt ,

·m·e}a)

-1

C/

)' ~ d, , ,'

,,

t:iftUW.

k = 6 . 'l
h

\ ()_

(I

/11 .rbr

"

.

I

I

Jrf/J(. lucri /oJr an(l tnfl.4 cuw·

=

-

t -JGI6 3" ·i K.

oii'(J

nlt "n~·rr

...-i)eJl

J'ic/f


?l c

Gj+l•'cl

tnttrf

..!7:rl ""~

iw~tr

• rJ

l"oi tii ·; OL

e ~'c) oc . r -:--- ..

I

l·,l ,v

•.,. r f·

\\;h ~

iiJ~

,, q ' ' l

1

II'U:tl{)

1. ,_. ( ft - To . )) { T ,, - Ia, '

~

~

I

.....!..

- -

\l

_(

{jh )

I

.

.3 II

;::.-

2-

'3

a

I

-

7 11 op

-"()

r...

'I

lJ -qqoC

\

(

o.q ,nqq'c)

{J •

- --;~

~11t- '' = q,l- •c {tH·-~g)] /I~ K70S-qo)/(44 - 7~] =-

C•rt•-dtd ~ )/J!JTrll :.

.:=

'I

LJ 1, - L1h)/ In ( LJY, f A17 )

=-- 1 (?O~ -qo) -

Ti. ..

""-- T .' 0

Index DT1 =

>

==-

1~.qq Condensation effect of peonanent gas,.i11 heat transfer coefficients, 210

Absorption coefficient for heat radiation,

221 '

Absorption of gases, 258, 269, 285

Analogy between heat and mass transfer, 240 between heat and momentum transfer, 152 Annulus · laminar flow in, 42 with moving inner cylinder, 46 with rotating inner cylinder, 47 Average temperature difference, 182 Average velocity in annulus, 43 in falling fitms, 40 in tube ftow, 41 Bernoulli equation, 65 Binary distillation, 32 Bingham fluid, 38 Blake Kozeny equation, 114 Blasius formula, 59 Bohzmann's constant, 220 Boundary layer thickness concentration, 234

hydraulic, 50, 58

thermal, 58 with chemical reaction, 274 Bubbles flow of, 104 mass transfer to, 241,261

Catalytic reactions, 280, 286 Chemical reactions heterogeneous, 280

homogeneous,270, 273,276 Chemical reactors, 283 Concentration boundary layer, 59, 234 Concentrations, definition, 226

Conservation laws. 3 ·.

Conversion factors. 22

Diffusion coefficients, 12, 15, ~~5, 227 equimolar, 228 through stagnant fluid, 230 Dimensional analysis, 21

Dispersion, 131, 137 Distillation, binary, 32 Distribution coefficients, 232, 258

Drag coefficient, 100 Droplets, 104

Eddy diffusivity, 58 Eddy viscosity, 56 Effective thermal conductivity in granular materials, 198

Emission of radiant energy, 220 Energy balance macroscopic system, 7, 144 microscopic sysf~· 150 Ergun equation, 114 Evaporators, 213 Extraction factor, 259 FalJing film in condensing systems, 209

laminar flow, 52 Fanning friction factor, 57 Fick's law, 12 Film theory, 234,272

Filtration, 115 Fixed bed heat transfer, 197 pressure d rop, 112 Flow around obstacles, 98

;

297 Flow meters, 94 Flow through orifices, 80 Flow through weirs, 85 Fluidized bed

heat transfer, 198 pressure drop, 116 Fourier number, 160 Fourier's law, 12

Free convection heat transfer, 201 mass transfer, 248, 254 Free faU of droplets, 104

Friction Joss factors, 57, 62, 68

Gas absorption

from bubbles, 241, 261 in_ packed columns, 242, 261, 265, 286

wttb chemical reaction, 270 Gas bubbles mass transfer, 241, 261 movement o( 104

Graetz number, 173

Grashof number, 203 Hagen- Poiseuille law, 41 Hatta number, 272

Heat conduction non stationary, 15.6 penetration, 159 statio~ary, 146 Heat exchangers, 182

Heat transfer . analogy with momentum transport, 152 by natural convection, 201 · by radiation, 220

coefficients, 177, 190

combined with mass transfer, 152 during boiling, 212 ' d uring condensation, 208 in fluidized beds, 197 in stirred vessels. 215 on a flat plate, 194 to falling films. 194 to laminar ftow, l71 to spheres and cylinders, 195 to turbulent flow, 174 Height of a transfer unit, 186, 261 Henry coefficient, 259 Hydraulic diameter, 61 , 63 Hydrodynamic boundary layer, 58

Kinematic viscosity, 12, 15

Laminar flow

between fiat plates, 37 by moving surfaces, 46 in circular tubes. 40 in non-circular tubes, 48 non station~. 49 througn an annulus. 42

Laplace number, 27 Lewis number, 293

Macroscopic baJances, 3 M ass balance, 6 Mass exchangers driving force, 262 extraction factor, 259 gas- liquid, 260 liquid dispersion, 262 packed oolumns, 242, 263 size determination, 260 unpacked columns, 266 Mass transport analogy with heat transport, 246 combined with heat transport, 292

during laminar flow, 248

during turbulent flow, 253 in packed columns, 242, 261, 265, 286

overall coefficient, 232

stationary diffusion, 227 to bubbles, 241, 261 with heterogeneous chemical reaction~ 280 with homogeneous chemical r~action 7 270 Microbalances, 16 ·

Mixing in pipes, 137 in vessels, 120

Momentum balance, 8, 17 Newtonian fluid, 38 Newton's law, 12 Non stationary flow, 49 free faU, 105 mass transport, 240 momentum transport, 49 Nusselt number, 149 Orifice ftow, meter, 93 Ostwald fluid , 38 Overall heat transfer coefficient, 176 OveralJ mass transfer coefficient, 232

298 Packed columns mass transfer. 242. 261. 265, 286 pressure drop, 112 Peclet number, 25 Penetmtion theory derivation, 157 diagrams, 164 Physical properties of materials, 24 Plank's radiation law, 220 Power consumption of stirrers, 123

Spheres--continued mass transfer, 246 Stoke's law, 101 terminaJ velocity, 102 Statistical transport, 12 Stefan Boltzmann law, 220 Stirred vessels, 120, 215 Stoke's law, 101

Tanks in series, 133 Temperature equalization diagrams, 164 Thermal boundary layer, 58 Thermal conductivity, 145 Thermal diffusivity, 12, 15, 145

Prandtl number, 58, 173

Pressure drop, 60, 64

of heat, 220 Reaction force on pipe bend, 10 Residence time distribution, 125 Reynolds number, 26, 57

Radi~tion

TurbulC?nt fiow, SS

Units and conversion factors, 22

SchnUdtnUJDber,59,247 Sedimentation of particles, 106 Sherwood number, 231 SI-Units, 21 Simultaneous heat and masS transport, 292 Spheres flow around, 101 free fall of, 104 heat transfer. 196

r l•'"r r ..:· I

J

!LlL'•lGl

~t- kt ~lu-

then

Mil

!

d/1 1

a~+

(

(

Xp

;::

) . 6'f

d..

==.

L

w

P4 ==

--o· IT3

(>-rt) (!o«)

• li1.(.L

I L!

bvtt- ~J tJ>+.

M 1r

uJe4

't

1-

em

tktW.9~'*- 1h~

Weber number, 104 Weirs. 85 Wien's displacement law, 220 I rhrCJc

.rerw

af

a pd

j_

~

'- -

{~W~ j

r~m!q~ ~ 14 J 411 • 11~ ~m ~·

.Jik. ~wt- pet~ttra+td through fh.t; -1~ kellle~ bcrlbrr1 vva! hf~' aJIJ·hmt e 9··~ r ! .r•~ -~-~(.! 1J tJ. ll·J 'fll/ '~ ~ r (J-= str kJ f'I J C?t1 a Sp. ~a+ I~ :!._

2:r•c . M :"- ~ o.r prfl

tnal- a.rrum;nj -1-httl-l 't- 4J ~tl~r ~c,~h,

"r

Velocity distribution in laminar flow, 41 in turbuJent flow, 60. Venturi meter, 92 Void fraction, 112

1-)wv dtt.~

~o4

(91<