Mechanics and Thermodynamics of Propulsion (1st Edition)

80 f---+-+-- j--+--',

E

2-3

50 ~--+---+---+--fr-f--+/~

s - s, ~

~ 40

R In 1'_ , /h

(2- 22)

in which the subscript I again denotes a reference value. The function

"~ 30~--+---+-f7f/~<.c

..L

=

e{4J-tP l> / H.

p,

10 I------j~~,p-.'+-+-_t_____j

For any given datum state I this isentropic pressure ratio p, may be expressed by 2000

4000

Temperature, OK

FIG. 2-5.

Enthalpy-temperature relationships for common gases [8].

For the particular case of constant specific heat and for an isentropic process, Eq. (2-17) may be integrated, with the results that

(~t"·

(2-18)

Defining the specific heat ratio 'Y,

and using Eq. (2-16), Eq. (2-18) may be expressed in the form

l!... ~

(I.-)YI<>-".

(2-19)

T,

p,

Using the perfect gas law, this can be transformed to

which may also be tabulated as a function of lemperature. Then if the temperature changes from any value TA to another value Till the isentropic pressure change may be found from

PE

p,(Til)

P.·'

",(T A )

,

in which p,(Tn) and p,(TA ) are obtained from tabulated functions. The state of a gas having only two independent properties may be designated by a si ngle point on a graph whose x- and y-axes correspond to any two properties. A process between states may be indicated by T a line. The temperature-entropy diagram P, shown in Fig. 2-6 is an example of such a graph. . The se ries of lines marked with various pressures correspond to possible constant-pressure processes. The shape of each line may be obtained from Eq. (2-17), L-~

!'-~

(2-20)

!'-

(2-21 )

p,

or

p,

in which p

I/ u is the fluid density.

_________________

s

FIG. 2- 6. Constant-pressure lines on a T-s diagram.

From this it is clear that the lines have pos itive slopes, increasing with temperature. Further, if c1) is a function of T only, the slope of the curve is a function of T only. Hence the curves are of the same shape displaced horizontally.

20

MECHAN ICS AND THERMODYNAM ICS OF FLU ID FLOW

2-4

From Eq. (2-17), it is also clear that

nt higher values of Since this derivati ve is always positive , the upper curves represe pressure. 2-4

EQUILIBRIUM COMBUSTION THERMODYNAMICS; CHEMICAL REACTIONS

ents change Up to this point thermo dynami c processes in which system compon is of vi ta l tion combus Since ussed. disc been not their chemical identity have how chemic al importa nce to many propuls ion method s, it is necessa ry to show is possibl e to sim reaction s are go verned by th ermody nam ic laws. Fortun ately it Fi rst, chemic al plify the discuss ion a great dea l by adoptin g certa in restrict ions. between gases reaction s of interest in propuls ion systems nearly always occur Second , with gas. perfect -a by r behavio in imated which can be closely approx nt chemic al importa the es, process ion expans rapid very th e excepti on of a few in equ ilibrium . processcs usually take place between states which are approx imately gases capable of T he basic problem may be stated as follows: G iven a mixture of will occur, a chemic al reaction and a set of conditi ons u nder which the reaction be asked might one e, exampl For ion? react al chemic that of lt what will be the resu t produc ts of a to determ ine the compos ition and temper ature of the exhaus te method s for constan t-pressu re adiabat ic combus tion process . To formula separat e topics: answering this kind of quest ion, it is convenient to consider three (I) Mixtur es of gases, (2) Chemical transfo rmation , (3) Compo si tion of pioduct s.

Mixtures of gases ly exist as a The reactan ts and produc ts of a chemic al process each general mixture may be gaseous mixtu re of chemic a] compou nds. The propert ies of the Gibbs-Dalton Law, determ ined from the propert ies of the constitu ents by using the wh ich is express ed by Keenan [I ) as follows : pressur es of I. The pressur e of a mixture of gases is equal to the sum of the at the temmixture the of each constitu ent when each occupie s alone the vol ume peratur e of the mixture . respectively, to 2. The interna l energy and the entropy of a mixture are equal, when each ents constitu its of es entropi the and the sums of the internal e nergies m ixture. the of ture a temper the at ure xt mi the of occupie s alone the volume

EQUILIBRIUM COMBUSTION THERMODYNAMICS

2-4

21

Thus for a mixture of 11 constitu ents:

= Tz

...

Tempe ratu re

T m = Tl

Pressure

Pm ='PI +]12 + 1'3

Volume

'U m

Energy

Em

Entropy

Sm

Enthalp y

H"

=

T/!,

(2-23a)

+ ... + PIl,

(2-23 b)

= 9))I Vl = 9R em = 9)1,e, + 9R msm = 9)I , s 1 + IDLJ1m = 9)1 1"1 + 9)}mvm lll

9)l2 V, 9)/,e, 9)I,s,

=

...

=

mCnvl/ ,

+ ... + IDlne, + + BRlIs

"2 +

9)1 2

!,

ll ,

... + 9)l"",,,

(2-23c) (2-23d) (2- 23e) (2-23f)

subscri pts 1,2, in which ID] signifies mass, subscri pt In refers to the mixture and uents. constit of ••• , 11 refer to a series Since de = c, dT, then

+

9)I,c,.1 + 9.1I,c"

(2-24)

Simi larly, C plII

9J"I.o.'C:'/if''iT-'+-·-· _·_+~:,m':.I,!:"C::.t/ll! '. c"I'ccl _+-,--'. 9)-1,,1 = ,, 9)1",

t alone in The pressur es P j, Pz, Pa. Pn that would be exerted by each co nstituen partial the called are mixture the of ature the volume of the mixture at the temper of number the to lated re is e pressur partial pressure s of eac h constit uent. The i, ent constitu any For . present moles of a consti tuent

or, in te rms of the unive rsal c0as consta nt

R, 11.

PiVi=

M~Tm'

where M i is th e mo lecular weigh t of constitu ent i. Since Vi

'U m = ~R i '

Pi =

~n whi~h

9.1] ,.J?'TI/j

-=..--

M /U IIl

Since is the !lumber of moles of constitu ent i present in the mixture . tiai par he t of m su aCCo rding to the Gibbs- Dalton Law, the mixture pressu re is t he pressures, fl i

1~

Pili

IIi

(2- 25)

22

M ECHANICS AND THERMO DYNAMICS OF FLUID FLO W

2- 4

T hus the ratio of the partial pressure to the mi xture press ure is just the ratio of the num ber of moles of the constitue nt to the total number of moles. This ratio is called the mole fraction, X:

(2- 26)

co nst ituents , eac h behav ing as tho ugh the other consti tue nts were not there. It is interes ting to no te that during a process wh ich is isent ropic fo r th e mixture , the co nsti tuents do not necessa ri ly unde rgo isentro pic changes . There is ge nerally a redistribution of e ntro py amo ng the co nstituents.

In determ ining part ial pressures, it is convenient to use moles rather than unit masses. For this purpose one can define molal specific heats by the equations

To avoid confusio n the va lues per mole of a property are usua lly denoted wit h a bar, as

23

In this example, as in all exoth ermic reactions , the hea t o f formatio n is nega tive,

since the system must give off heat to maintain (or return to) the reference temperature. As anot her exa mple, the formation of monatom ic o xygen from nat urall y occurring oxygen is ~02 ~ O. In this endothermic reac ti on the heat o f formation

(+59.162 T he entropy cha nge of the mixture is the sum of the entropy changes of all the

EQ UILIBRIUM COM BUSTION THERMODYNAMICS

2-4

~cal/gm-mole

at 3000 K) is positive, since the heat transfer to the system

must be p()sitive.

These heat quantities may be determ ined experimentally with a simple consta ntpressure contain er. Practically , it may be difficult to produce a reacti on at constant temperatu re and pressure. It is not necessary , in any eve nt, since fr om the stand-

poin t of thermodyna mics only the end states of the reacti o n are of interest. The intermediate conditions do not ma tter (altho ugh they may be important to the preserva ti on of laboratory equipment).

A sleadr jlolV process is another ex peri mental method for determi ni ng the heat of reaction Q/ . Equation (2-8) states that, for a stead y-fl ow process with no shaft work, the heat transfer of the fluid flowing through a control volume is equal to the change in enthal py (neglecting changes in veloci ty and elevation). T hus

dh = Mdh = 1'" dT.

Q = MI or, per moM o f product formed,

Chemical tran sfo rmation

Q A general c hemical reac tion, in w hic h a moles of reac tan t A a nd {3 moles o f reac ta nt B, etc., combi ne to form J.L mo les o f prod uct M plus v mo les o f product N , etc ., may be rep resented by

a A + {3B + . . .

-->

jJ.M + vN

+

(2- 27)

The coefficients a , {3, etc., are kn own fr om the mixture ratio of the reactants, a nd it may be assumed for the prese nt that the prod uct co mposit io n, that is, /.L, v, etc. , is also know n. Accompanyin g a chemical reacti on there is genera lly an excha nge of energy with the surro undin gs . Those reac tions w hich tend to give o ff hea t are ca ll ed exothermic and o nce star ted they will , under the proper cond iti ons, proceed unaided. The other kind , ca lled endothermic, req uire an energy input to proceed. Com bust ion processes, for insta nce, are exotherm ic. T he magn itud e o f these energy inte rac tio ns ca n be determi ned from the compositi o n a nd the heat s of formatio n o f the reac tan ts and products. The co nsta ntpress ure heat of formation of a substance is the heat in teracti on which occ urs whe n one mol e o f the substa nce is form ed at constan t press ure and temperature from its elements as they occur in nature . For examp le, the hea t of for mation of car bon diox ide ( - 94. 052 kca l/ gm -m ole) is the hea t interact io n accompanyin g o the follo wing reactio n from reac tant s to products at 300 K :

=

Q,

=

Hproduct -

H rcactant;;.

(2- 2 8 )

For example, the formation of carbon dioxide in steady flow is represented by Fig. 2-7. Reactant s at T1

Steady-flow reaction chamber

O2

Product at Tf CO 2

t

Qleo, FIG . 2-7 . Schematic of stea dy-fl ow process for formation of CO 2 , with Tj being the reference teJ1lperatu re.

Table 2-1 gives the heats of format ion for some commo n products of combustio n, T/ being the reference temperature at which Q/ is measured. The heats of format ion of natura ll y occurrin g elements are, by defi ni tion, ze ro. ] n co ntr;lSt, most cbmm o n reactio ns are com plex in the sense that the reacta nts are not na tu ra ll y occurring elements and the prod ucts incl ude more tha n o ne substa nc~ . Also, they do not ge nerally occur at co nsta nt temperature. Comple x reac tIOns cj n ~e treated as a co mb inatio n o f simpler reac tions by the followin g meth ?d . (o.nslder the steady-ft o.w r~ac tion ind icated in F ig. 2-8, where Q is the heat 1I1terac lJo n necessa ry to ma1l1ta1l1 constant temperature. It is conv enie nt to consider tris reacti on to be the Slim of dist inct decomp os it ion and form ation reactions.

2-4

MECHANICS AND THERMODYNAMICS OF FLUID FLOW

24

Table 2-1

Table 2- 1

Heals of Formation

Formula

Compound

formation,

Tf , oK

(I)

5 5 5 9

(g) +124.5 (g) + 97.2 + 23.050 (g)

298 298 300

5 5 6

(g)

298 300 300 300

5

Boron Boron atom Bromine atom

B, B

Carbon Carbon (graphite) Carbon dioxide Carbon monoxide

C C CO, CO (CH,,),NNH,

Ammonium nitrate Ammonium perchlorate

Aniline

D imethylhydrazine (unsymmetrical) (UDMH) Fluorine

Br

Fz F

Hydrazine Hydrogen Hydrogen atom

NzH·l H, H HBr HF H202

Hydrogen peroxide

Hydrazine hydrate Hydroxyl

N zH4HzO

OH

Isobutane (2-methylpropane)

C1HIO

JP-3 JP- 4

C7.:10r, H ll.442

Melhyl akohol

CH"OH

Monomethylhydrazine

CH"NHNH2

Nitric acid

* kcal/gm in this casc o

(s)

(g) (g)

Formula

formation,

7 7 10

+ 12.734 (I)

H / C 1.93

HNO"

-

3.0 (I) (g) 0 + 18.906 (g) + 12.000 (I) 1.92 (I) + 52.092 (g) 8.66 (g) - 64.20 (g) - 33.74 (g) - 44.84 (I) - 47.36 (s) - 10.3 (I) + 10.063 (g) - 31.489 (g) - 36. 169 (I) - 26.66 (g) - 0.423* (I) - 48.08 (g) - 57 .04 (I) + 13.109 (I) - 41.35 (I) - 31.92 (g)

300 300 298 300 300 298 298 300

14% N0 2 1% H,O NO N, N NO, N2 0 5 NZ04 N2 0 ::I N,O CSHl S

N itric acid, red fuming N itric oxide Nitrogen N itrogen atom Nitrogen dioxide Nitrogen pentoxide Nitrogen tetroxide Nitrogen trioxide Nitrous oxide N-octane Oxygen

02

Oxygen atom Ozo ne

0 03

-

45.3 21.6 0 +112.5 + 8.091 + 3.6 + 2.30 + 20.0 + 19.55 - 49.88 - 59.795

9 7 5 5 8

300

7

300 300

13

298

5 14

298

10

300

8

8

(g)

(g) (g) (g)

(g) (g) (g)

(g)

SO, H2O

-

57.802 (g)

Complex reaction chamber

300 298 300 300 298 298 300 298 298 300

Reference number

8

5 15 5 5 8 5 5 7

(I)

Water

at T = Tj

12 7

(g)

Sulfur diox ide

Reactants

300 300

(I)

0 (g) + 3.08 (I) + 59.162 (g) + 34.00 (g) (I) + 30.3 - 70.96 (g)

6 10

Tr. OK

kcal/ gm-mole

298

Fluorine atom

Hydrogen bromide Hydrogen fluoride

+171.698 0 - 94.052 - 26.413

(g) (s) (s)

Compound

Reference

298 298 298 300

-399.09 - 87.27 - 69.42 + 6.11

Heats of Formation (Confinued)

number

kca ljgm-mole AL20;; NH,NO" NH,CLO, C(jH5N Hz

25

Heal of

Heat of

Aluminum oxide

EQUILIBRIUM COMBUSTION THERMODYNAMICS

2- 4

300 300 300

7 8

298

5

300

7

~llcts at T = 1j

t

Q Compou nd reactants

at~

Decompos ition reacti on chambers

T

Elemental reactants at T= T..r

Formation reaction chambers

1

Products

Mixing

Multiple products

at T=Tj

chamber

at T = T..r

1

1 F IG. 2- 8. Sc?ematic of complex reaction at constant temperature, showing equivalent e ementary reactions.

26

2-4

MECHANICS AND THERMODYNAMICS OF flUID FLOW

EQUILIBRIUM COMBUSTION THERMODYNAMICS

2-4

The decom position reaction is just the reverse of the formation reacti on ; hence it is a sim ple one. Howeve r hypothetical t he reverse process may be, the first law tells us t hat the decomp osition heat in te raction will be equ al and opposite to the format ion heat interaction. Thus, summing ove r all the decomposition reactions,

Reactants

at T I

(2-29)

Hea l exchanger

Q=HHf - i-i HI

Reactants

Complex

at Tj

reactio n chambe r

t

Q ~ H ltl'!

Products at Tj

Heat exchanger

27

Produc ts at

T2

Q = H p2 - HFf

FIG. 2- 9. Reaction at variable temperatures.

where lli is the number of moles of reactant i and Q/i is its heat of formation per

mole. Once the reactants arc in elemental form , the remainder of the reaction may be treated as the su m of separate form ations, so that (2-30) where Ilj is the number of moles of product j . Since the mi xing process in the t hird chambe r is ad ia batic, t he net heat interaction fo r th e complex reac tion at constant tem pera ture Tj is

L: (n j Q /i)produc ts - L: ( l1 i Q /i)rcactants'

the reference state (Tj ), and may be calcu lated from t ables of the mixture properties or by the methods .of Section 2-3 . The reference tempe rature Tj is simply that temperature for which the heat of react ion, H RP / . is ava ilable. Let us take a typical problem: we wish t o determine the adiabat ic flam e tem perature for a give n reactant mixtu re. Once we know the product composition and det~rm i ne the reference temperature, we can calculate H R b H fl,f, and H RP/. Then, sll1ce Q ~ 0, we can determi ne the product en th alpy H p " a nd fro m this the prod uct tem perature. II

(2- 31)

j

Also, us ing Eq. (2- 28),

Qu =

}fproducls -

I1HI'(

Hrcactnnts =

:L: (njQIi)producta - l:::

I

(n i Q /i) rcnctanb '

Reactants

j

-t--~~~--_t--~T~I---1~----~~~~--------~lf2---+--~T

This difference in ent halpy is given the symbol H ill'! and may be interpreted as the cha nge in enthalpy from products to reactants at th e co nstant reference tem-

/B - - - - - - - - - - - - - - - - -

perature TI . Hence H nPI = H pJ -

H ili =

L

(nj Qfj)prodllct~

-

L

(lI i QI'i )rcuctant~ .

(2- 32)

F,G. 2- 10.

Enthalpy~temperature

diagram for ad ia batic combustion.

j

The va lue of HIl !'! is tab ula ted for many common reactions, but if such a tabulati o n is not available for a particular reaction it can be calculated from Eq. (2-32) and a table suc h as Ta ble 2-\' Note that this calculat ion req ui res knowledge of t he composition of the product. If t he products or reacta nts arc not at the reference temperat ure Tj, the fo regoing procedure must be modi fi ed so mewha t. Hypot hetica ll y, it is possi ble to b rin g the reactants to t he reference temperat ure by heat t ransfer, allow the reaction to occur at consta nt temperature, and then brin g the products to th e final tem perature by anot her heat transfer. The heat interactions required are indicated in Fig. 2-9. The net heat interaction from reactants at T 1 to products at T 2 may be written as the sum of t he three heat-transfer rates indicated in Fig. 2-9. Thus

(2-33)

This ca n a ll be presented graphica ll y on an en t halpy-temperatu re diagram whi ch shows Ihe r elationship between entha lpy a nd tempera ture fo r bo th product and reaClant mIXtu res. Figure 2-10 desc ribes a hypothelical path of the process. The reactants at sta te (Dare brought to T/ a t sta te A, after which conversion to products occ urs a t cons tant te mperature along line A B. Then the prod ucts are heated to t he ~et~t ~et st~te at~ and T 2 is the outlet temperature. Since there is no net heat transinsern.thls partlcul.a: example, ~Rl must t herefore eq ual H p 2 [as may be seen by Ihi tlng Ihe definItion of H ltP! 111 Eq. (2-33) and sell ing Q eq ual to ze ro]. From . s dl3gram It can be see n tha t ir the specific heats* of reactant and product lTIlxtures are unequal , then th e heat of reaction H RI'/ must be a functi on of temperatu re.

---•

Each of the pare nthesized terms is a change in enth alpy from the actual state to

In

>I<

".fhe specific heat of the mixture is give n by the slope of its enlhalpy lem

Fig. 2- 9.

~

l peru ure cur ve

28

2-4

MECHANICS AND THERMODYNAMICS OF FLUID FLOW

2-4

EQUILIBRIUM COMBUSTION THERMODYNAMICS

Composition of products

......

/

]0 all the aforementioned rela tionships, OUf calculations have depended on knowledge of the composition of the products of reaction. Once this composition is known, calculating the state of the product is a rather simple matter. Howeve~, determining the product composition is not always easy. The product composition depends on the temperature, which in turn has just been shown to dep.end on the composi ti on. This suggests that a trial-aod-error procedure may be req UIred for the solution. For certain reactions, it is relatively simple to determine product composition. For example, at sufficiently low temperature, the combustion of a particu lar 110nstoich iometric mixture of octane and oxygen yields the following products:

/

+

150 2 --> SCO z

P.l,/

I Pure A

C,H I s

+

+ bHzO + cO, + dH, + eC + ICO + iO + jOH + kC sH 18 + other hydrocarbons,

150 , --> aCO,

+ gH

(2- 34)

where a, b, c, d, etc., are the unknown numbers of moles of each substance. A reasonable first approximation may be made by recognizing that for all practical purposes Cs l-I I R and the "other hydrocarbon" molecules are present in negligi ?le quantities. However, th ere are still unknowns-a, h, c, d, e, /, g, i, j - for which we can obtain only three equations in terms of numbe rs of atoms. Atom balances yield: For carbon, a e 1= 8;

+ + + 2d + g + j = 18; 2a + b + 2c + I + j + i 2b

For hydrogen, For oxygen,

=

30.

Six more independe nt equations are needed in order to determine th e p roduct composition. They may be obtained from the requirements for chemical equilibrium.

Chemical equilibrium. Consider a reaction in which a stoichiomet ric com bination of A and B reacts to fo rm products M and N with no A and B left: "A

+ f3B --> I"M + vN.

(2- 35)

At the same time thi s reaction is taking place, there is ge nerally some tendency for the reverse reaction, I"M vN --> "A f3B

+

+

A,B,M,N',

I

/

"PJI

"

\

\ Pure M \

P = P.I+Pfj+P.lI+PS

\

I

\

l

T = constant

"

\\

I

II Pure B

Pure N /

jJ/j ~,

I

\

+ 9H zO + 2.50 z.

In this case, the composition is easily determined by requ iring that the number of atoms of each element taking part in th e reaction be constant. Actuatly, at high temperatures this expression may be a conside rable oversi mplification, because the compound s CO 2 , H 2 0 , and O 2 dissociate to some extent into free atoms and radica ls. A m ore general sta tement is

"-

/

/

/

\

C S HI 8

/-- ---- -- - "

29

/

"',........

/

......

FIG. 2- 11.

_------

....-//

Reaction chamber for derivation of equilibrium constants.

to occur, especially at high temperatures. Equilib rium is that state at which bot h forward a nd reverse reactions are occurring at equal ra les. When eq uilibrium obtains, the concentrations of A, B, M , and N are constant. Using the seco nd law of thermodynamics, we can express the conditions for equili brium in algebraic form. Suppose that an equil ibrium mixt ure resulting from the combi nation of A and B is contained in a chamber which communicates through semipermeable membranes with each individual cons tituent substa nce in a pure form, as in Fig. 2- 11. The entire apparatus is surrounded by a constant-temperature bath. A semipermeable membrane is a hypothetical surface through which onc substa nce may pass freely, though all others are preve nted from passi ng. Such membranes are only approxi mated by certain real substances, but they are nevertheless useful ana lytical concepts. For example, in Fig. 2-11, substance A may pass freely into or out of the mixture at cylinder A, and the pressure of th c pure A wilt be equa l to the partial pressure of A in the mixture, whereas B, M, and N cannot pass through this membrane. Similarly, the membranes of cy linders B, M, and N adm it only substances B, M, and N respective ly. The amount of any particular substance within the mixture chamber can be cont rolled by simply pushing or pull in g on the piston in the cylinder containing the pure substa nce. This control process is reversible if done slowly. Starti ng with an equilibrium mixture resulti ng from the combination of A and B in any propo rtion , consider the introduction of additiona l small amounts of A and B through the semipe rmeable membranes into the mixture chamber, in the stoichiometric ratio kaA + k(3B, where k is somc small number. Once in co ntact Within the main cham ber, the A and B tend to react, forming M and N. There are two ways to restore the contcnts of the ma in chamber to their origina l condition. One is to wi thdraw from the chamber kaA + kf3B. The othe r is to with-

30

MECHANICS AND THERMODYNAMICS OF FLUID FLOW

d raw kp. M

2-4

+

- a R Inp., - (3 R Inp B

reached equilibrium, its composition will be restored to its former value. As a result of the reaction

+

k{3 8

->

kp.M

+

~

(d

q) ~

T

k vN ,

":p;''. P PAP H

T

K,- p(CO)p(H ,O) p(CO,)p(H ,)

-

dSnli~tu re = O. ch alllbc r

dS.\(

+kI'S.lI,

dS e

~ k{3SB ,

dS ,y

+ kVSN.

p(NO)p( H 2) [peN))' 'p(H,o)

K _ [p(H,)I'p(O,j r Ip(H 20 )),

In add ition, the partial pressure of each constituent does not change and hence the pre~s ure in each cylinder does not change. Since the temperature is also constant in each cyl inder. the entropy pel' mole of each pure substance must be constant and the change in entropy within the cy linders may be written -kaSA.

K- - ~ )

KG_ l p( H,) )' ' p(O':!) p(H,o)

K _ p(C0 2) ,- p(CO)[p(O, )) ' ,

{3SB - I'Su - vS.y ~ -

~p ~ fi() --r T.

(2- 36)

The entropy of a perfect gas per mole can be obtained by integrating Eq. (2-17) with the result: 7'

-

-

S = -Rlnp+

r

ciT )T\cPT +

p(NO) [piN ,)!' 'lp(O, ))' ,

K - - peN)' I

p eNt)

10 32 10

10'_8

10 - 1

10·2.11 - - -

I

§ 10- Cl

IG

o

.§ 10- '

.~ =s

.t ~

.~

1\

,

10 1

£

1O- !1

"'

~

lOS

\

10.1

10- 13 10

l ;j

:;--"LLL-,-,LL--1_

-L---L---.l

2000 4000 Temperature, oK (a)

-I R npl.

where the subscript 1 refers to an arbitrary datum state. Thus. for a given datum

\

0

10- 11

+

p(CO)

s - p(C)[p(O,)I ' ,

~

or " -SA

K _

[p( H, )J ' 2

o

~ 10- 3

where S is the entropy per mole of pure substance. The total entropy change is equal to thc sum of the changes in the mixture chamber and in the cylinders: k H il I' dS ~ - kaS., - k {3S n + k p.S.l1 + kvSx T

(2- 37)

p( H,o)

.

K,-

dSA

~ K ,,(T).

K, - p( H, )p(O)

For perfect gases it may be shown that H il l' is a funct ion of tempe rature only. Thus for a given , value of k the change in entropy of the system is a function of temperature only : dS ~ kf(T). Since the temperature, volume, mass, and composition of the mixture do not change, there will be no change of state of the system in the mi xture chamber. Thus

~ g(T),

The term Kp is called the equilibrium constant for th is mixture in terms of parLial pressures. Figure 2- 12 gives typical values of the equilibriulll constant Kp. Several general statements should be made regarding equilibrium constants before their use is demonstrated in a detailed example.

kH 1l 1' .

re v

+ p.71lnp.l1 + vRlnp x

whe re g(T) is another!unction of temperature only. Adding the logarithmic terms and wri ting g(T) ~ R In Kp(T) leads to the foll owing relationship between the partial pressures :

there will be a heat interaction of magnitude kH rt p with the constant-temperature bath surrounding the apparatus, since the reaction occurs at the constant temperature T of the bath. In the limit th is heat interaction will be reversible. Then, for a system composed of all the gases in the mixture and in thc cy li nders, dS

31

state, Eq. (2- 36) can be written

k vN. In both cases the total number of atoms of each element in the chamber will have returned to its original value. After the mi xture has again

k~A

EQUILIBRIUM COMBUSTION THERMODYNAMICS

2-4

F,G. 2- 12.

atmospheres.

I?

lO- .,

6000

Equilibrium constants [8] in terms

t)

~K

l, ~

~I

---

I

1000 2000 3000 4000 5000 6000 Temperature, oK (b)

of partial pressures.

Pressure

in

32

MECHANICS AND THERMODYNAMICS OF FLU ID FLOW

2- 4

First, it may be noted that the partial pressures of the prod uct appear in the numerator and those of the reactants in the denominator, each with an exponent taken from Eq. (2-35). Qualitatively, then, one can expect a large Kp for subreactions which approach completion a nd small Kp for those th at do not. The number of constituents considered in this deriva tion is four, but the result can easily be extended to include any number of reactants a nd products. Second, the exponents are the stoichiometric coefficients. For example, consider the relationship between ca rbon d ioxide, carbon monoxide, and oxygen which would be of interes t in our octane~combustion example:

co, -->

CO

+

EQUILIBRIUM COMBUSTION THERMODYNAMICS

2-4

to form x moles of Hand x / 2 moles of O 2 , accord ing to the reaction

xH,O

-->

The relative concentrations of the constituents of the mi xture may be tabulated as shown below. Moles presen t at Mole fraction Xi eq uilibrium Constituents

~02'

p(CO)[P(02)] 1/2 = K p (C0 2 ) p'

H2

x

O2

2:

K II

= XII, ( xo, ) X

where Kp is the equilibrium constant based on partial pressures. Note that Kn, whi le mo re convenient in calculati ng mole fractions, is a function of two variables, pressure and temperature, and is thus inconvenient to tab ulate. In Eq. (2-39), Pm must be in the units for which Kp is tab ula ted. The actual use of these concepts ca n best be demonstrated by an example which contai ns a considerably simplified description of an actua l chemical reaction. ~ 1?,.sec

One mole of H 0 is raised in temperature to 6000°F, and it may be assumed 2 that it dissociates to a n equili brium mixture of H 2 0, H 2• and O 2 at a total pressure of 1 atm according to H 20 ~ H2 + !O 2· The equilibrium constant for this react ion a t 60000 P is Kp = 0.23 atml / 2. The concentration of t he mixture may be determined as follows. During the d issociation, x moles o f H 20 break down

- _ K -

-1/2

pPm

.

Substituting the equilibrium concentrat io ns, we obtain

x

) (

+ x/ 2

x/ 2

I I

1+ (2-39)

1/ '

I[:P

(2- 38)

cr+f3-P.-"K K n - Pm P'

Example 1

1+ x / 2

The relativc conce ntrations may be determined as follows:

(I

Since, according to the G ibbs-Dalton Law, Xi = p;/p,", it may be shown that

1+ x / 2 x 1+ x / 2 X 2( 1 + x / 2)

x

Total

than partial press ures : =

I -x

x

In this case the equilibrium equation is

Kn

X

+ 2: 0,.

xH,

H 2O

It may be noted that these e"ponents a re not directly related to n, c, or f of Eq. (2-34) for the overall rcaction . TI1is equilibrium equation provides one of the six relationships necessary to the determin at ion of product composition after t he reaction (2-34). In a similar way we can state five other equilibrium requirements to govern t he relat ive pa rtia l pressures o f the products. T hird , since the object of this calculation is to determine composition, it is convenient to define an equilibri um constant K n , based on mole fractions rather

33

+ x/2

)

II'

0.23.

x x/2

T;e quant ity x may then be determined by trial and e rror and the mole fracti ons H ,0,0" and H , at equilibrium obtained from the above tabulation TI he foregoi ng calculation oversimplifies t he dissociati on. Strictly spe~king the foI owmg three dlssociat'IOn reactIOns . ' are also of importa nce at 6000°F :

°

O2

-->

20,

H,

-->

2H.

Determination of the equilibrium concentrations of H 0 H 0 0 d H en req Ulre th . I 2, 2, 2, ,an th kno I s e Slmu taneous solution of three eq uilibrium co nditions wit h wn va ues of the three equilibr ium constants for the a bove react ions. Exam ple 2 (from Reference [3])

Consider a complete set of six products for a hydrogen-oxygen reactio n :

AH2

+ B0 2 --> ilH,oH,O + 110,_,0, + I1 H, H ,+ 1100 + I1 HH + /loaOH. (2-40)

34

MECHANICS AN D THERMODYNAMICS OF FLUID FLOW

2-4

Since two equations can be obtained from the conservat ion of hydrogen and

oxygen atoms, four equilibrium relationships are required. The following method is due to Penner [3], who has shown that it is conven ient

srn,

to define yet another equi librium constant based on number of moles rather than mole fract ions. T he use of this equili brium constant results in simplified calculations, and it is introd uced here because the technique has value beyond this example. Using again the typical reaction (2- 35),

EQUILIBRIUM COMBUSTION THERMODYNAM ICS

2-4

35

and the corresponding relationships in terms of moles may be writ ten from Eq. (2-41):

-no - , 1/ 2

st n3

110I-f =

1/ 2

1/ 2

n0 2 1111 2

n0 2

A procedure for obtaini ng a solution for the six un kn owns is now out lined.

STEP I: From the equilibrium equations write the number of moles of each nelV species appearing in the product in te rms of the reactants and the proper equilibrium constant. In terms of sr n :

(2- 41 )

no (2-45)

and usi ng Eq. (2-25),

(2- 42)

nOlI

STEP 2: Use the equilibrium relationships (2- 45) to elimi nate the number of moles

Strictly speaking, the use of this equilibrium constant introduces another unknown, 11, along with an additional equation which simply states that 11 is the total number of moles in the actual product mixture (not the subreacti o n). However, the simplification attained by th is approach lies in the fact that 11 can be rather easi ly approximated, and in the fact that the results of the calculation are not strongly dependent on II. Thus, as illustrated in this example, II is at first assumed known and later checked. An iterative procedure can be used if greater accuracy is desired . Returning to the example in Eq. (2-40), the two equations resulting from conservation of atoms are :

+ 2nH, + nII + nOH ~ 2A, nll ,o + 2no, + no + nOlI ~ 2B,

For hydrogen,

NH ~ 2nH,0

(2-43)

For oxygen,

No ~

(2-44)

where NH and No are the known num bers of hydrogen and oxygen atoms, respectively. The equilibrium information available is of the following form:

of new species in one of the atom-conservat ion stituting Eq. (2-45) in (2-43), we have

-->

STEP 3. Use this expression to eli minate no, from Eqs. (2-45).

no

=

(2-47)

nOH

dln411H2

0, 112

PlI z

Si'''211 ~1,2) . + Jr1l3nl/22

(NH - 211H, 2hl'n411H2

PH ,

Thus, sub-

(2- 46)

"

t02

~quat ions .

STEP 4: Use these expressions along with Eq. (2-46) to eliminate all but the rema ini ng atom conservation equation (2- 44).

1111 ,

from

-

POT-[ 1/2 1/2

jJo zPn 2 pl-r 20

112

PII zPO z

No·

(2- 48)

36

MECHANICS AND THERMODYNAMICS OF FLUID FLOW

Equation (2- 48) can now be solved for

"H. if the various ,\1'" are known.

2- 4

2-4

The

Table 2-2

EQUILIBRIUM COMBUSTION THERMODYNAMICS

37

Enthalpies and Heats of Formation for a Given Product Composition

,fil n's may be eva lu ated from Eg. (2-42) for k~own values of K p , Pm, and n. The

total number of moles in the mixture is actually unknown, but it can be rather easily approximated , and the results are not very sensitive to II anyway. The act ual product compos ition is not Car from that resulLing from "complete" react ion (unless the temperature is very high) so that Il is approximately equal to the number of moles resulting from "complete" reaction. For thi s example, it is

AH2

+

BO, -> A H 2 0

+

2BH,O if there is excess H ,. Hence

Il

+ (A

-

n b.7i

Qj,

kca l/gm-mole

nQj

"

kca ljgm-mole

2 I

0 0

0 0 0

0 0

0 0 0

0.317 0.1008 0.05399 0.1090 0.2333 1.5119

25.693 28.263 16.028 15.897 25.907 36.920

8.145 2.849 0.865 1.733 6.044 55.819 75.455

0 0 59.162 52.092 10.063 -57.802

0 0 3.194 5.678 2.348 -75. 11 8 -63.898

Hz 0, Products

(2B - A)O,

H2

if there is excess O 2 , or

/J.7i,

Reactants

2B)H,

can be readily approximated in terms of the know n

reactant quantities. For a numerica l example. consid er the equi lib rium compos iti on of2H 2 + O 2 at 35000 K (centigrade absolute) and 300 psia. Then NH ~ 4 and No ~ 2. For

0, 0 H OH H, O

complete reaction, the number of moles of products would be 2, and we write 2H2

+

Adding these gives us the tota l number of moles (which was assumed to be 2.2):

O 2 -> 2H,O.

However, since the temperature is rather high, II is assumed to be 2.2. Assumptions of this nature come from exper ience in this type of calcu latio n ; but as we shall see, iterations are easily made and the ass umption is not cr itica L From the

known Kp (pressure in atmosphe res in this case) and pressure (20.42 at m), and the assumed 11, the new equilibrium terms can be calculated . Using Eq. (2-42), ,\1',,1 ~

Kpl C;;,t I12~ 0.17003 , ,\1',,3 ~

K p3 =

1.3046,

,\1',,2 ~ ,\1'" , ~

K P4

Kp, (P~n) -I/2 ~ 0.19362, (~;n)' /2 ~ 15.020.

These values are then used wit h various assumed values of

IlH2

to ca lculate the

left-hand side of Eq. (2- 48). The correct value is of course No ~ 2. T rial results are tabulated below. Calculated No from Eg. (2-48) 2.2561 2.9549 2.0224 1.9917

0.2500 0.3400 0.3000 0.3200

By linear in terpolation between the last two figures,

flH2

was found to be

0.3170. Then Eqs. (2-46) and (2-47) give nH z

Jln

0.3170, 0.1090,

0.1008, 0.2333,

"0

0.05399, 1.5119.

11H z

Il ~

2.326.

The next step is to recalculate "If .. elc., assuming It

~

2.326. When we do this

we find no apprec iable change in -composition. Hence eve n an inexperienced person cou ld probably complete the calculat ion in two, or at mos t three, iterations.

This solution method is not limited to reactions of only two reactants. The generalization of the method for additional reactants is discussed in Reference [3]. In order to take 2 moles of H, and I mole of O 2 from 3000 K and transform them at constant pressure to the above composition at 3500o K, it would be necessary to accompa ny the reacti on by a heat transfer Q which can be determi ned by Eg. (2- 33). If we use Table 2-1 to com pute the heat of reaction we must let TJ ~ 300°K. Since the inlet temperature is also 3000 K in this exa:nple H H - 0 ' III RJ . Table 2-2 shows how the terms (Hp , - HpJ) and H RPJ are com puted from enthalpy-temperature information (e.g. Fig. 2-5), heats of for ma tion (Table 2-2) , and tlelgiven ' .. compositions. 111 the tab le, 11 is th e number of moles and b.h the ent~alpy per mole of eacl~ const ituent above the datum temperature Tj . As before, QJ s the heat of for matIO n of each constituent, per mole. From Table 2-2 the heat of rea c t'Jon H RPj IS . - 63 .8 98 kcaljgm-m ole of oxygen reactant A lso' the enthalpy d'~ (H ' . Th I erence 1' 2 - H PJ ) IS 75.455 kcal/gm-m ole of oxygen reacta nt erefore, from Eq. (2-33), the heat transfer necessary to take two moles of H' and One mole of O at 300 0 K and transform them to the product composit ion o} 2 Table 2- 2 at 3500 K is: 0

Q

(H p2

-

H/2) -

(Hill -

75.455 - 0 - 63.898

~

H HJ)

+

H UPJ

11.6 kcal/gm-mole·02.

38

MECHANICS AND THERMODYNAMICS OF FLUID FLOW

2-4

EQUILIBRIUM COMBUSTION THERMODYNAMICS

2- 4

P .. oblerns

References 1. KEENAN, J. H. , Thermodynamics. New Yo rk: John Wiley & So ns, 1941 H" The Dynamics and Thermodynamics of Compressible Fluid Flow, Volume l. New Yo rk: The Ronald Press Company, 1953; C hapters 1- 8 and 16 3. PENNER , S. 5., Chemistry Problems ill Jet Propulsion. New York: Pergamon Press, 2. S HAPI RO, ASCHER

195 1 4. KEENAN, J. H. and J. KAYE, Gas Tables. New York: John Wi ley & Sons, 1957 5. ROSSINI, F. D., et al. , "Selected Values of Properties of Hydrocarbons." Circula r No. 500, National Bureau of Standards, Washington, D. c., Feb. 1, 1952 6. "Selected Values of Chemical Thermodynam ic Properties." Series III, National Bureau of Standards, Washington, D. c., April I, 1954 7. ROSSINI, F. D., el al. , "Selected Values of Properties of Hydroca rbo ns." Circular No. 461, National Bureau of Standards, Washington, D. c., Nov. 1947 8. MARTINEZ, J. S. and G. W. E L VERUM, JR " Memo randum No. 20- 121, "A Method of Calculating the Performance of Liquid·Pro pellant Systems Containing the Species C, H, 0, N, F, and One Other Halogen, with Tables Required; Thermochemical Proper· tics to 6000o K. " Jet Propu lsion Laboratory, California Institute of Technology, Pasa· dena, Californ ia, Dec. 6, 1955 9. RICE, H. E., "Performance Calculations of RFNA·Aniline and Three·Component Systems Using Hyd rogen." Memorandum No. 9- 2, Jet Propulsion Laboratory, Pasa· dena , California , Ju ne 30, 1947 10. " Heat of Fo rmation of Hydrazine, Un symmetrical Dimethylhyd razi ne, and Monomethylhydrazine." Inorganic Research and Development Dept., FMC Corporation, Princeto n, N. J. , May 14, 1959 11. JOHNSTON, S. A., "A Short·Cut Method for Calculating the Performance of F uels Containing C, H, 0, and N with HN0 3, Oz, or NH-l-N03 ." Progress Report No. 20-202 . Jet Propu lsion Laboratory, Pasadena, California, November 9, 1953 12. CLARK, C. E., "Hydrazine," first ed itio n. Technical Bulletin No. 123.1, Olin Mathieson C hemical Corporation, Balt imore, Maryland, 1953 13. CARTER, J. M., and M. PROTTEAU, "Thermochemical Calculations on the Whi te Fumi ng N itric Acid and JP·3 Propellant Co mbi nation." Research Technical Memora n· dum No. 58. Aerojet.Genera l Engineering Corporation, Azusa, Califo rnia, April S, 1950 14. PERRY, J. H., Chemical Engilleer's Handbook, third edition. New York: McGraw-

39

1. A ?e rfe~t gas of molecular weight M = 20 and specific heat ratio Y = 1.2 expands adia batIcally from a pressure of 1000 psia, temperature 5000 o R, and velocity 1000 fps to a pressure of 14.7 psia. (a) If the fina l temperature is 26000 R a nd the fina l velocity negligible, how much wo rk has been done by the gas during expansion? (b) If.no work i~ done by or on this gas during expa nsion to 14.7 psia, what is the ma ximum pOSSible velocity at the end of the process? (c) If any ad iabatic expansion process to the same final pressu re is permissi ble is it possible for the exit temperat ure to be 300 0 R? State your reasoning. ' 2. An inventor suggests ~ propulsion device shown sc hematically in the figure. Would s~c h a scheme be. feaSible for steady operat ion over lo ng periods of time? D isc uss rigoro usly and bn efly.

Eleclric

I

generalOr

Q

Fluid circuit s

Propel lant

storagc

Nuclea r

tank

~-"--'===--- Cot11ro l ~--_ _ _---.,("""'-~

PROBLEM 2

Rockct Il oule

Hi ll, 1950 15. GAY DON, A. G., Dissociation Energies and Spectra of Diatomic Molecules, second ed ition. London: Chapman and Hall, 1953 PROBLEM 3 3. A strut of cho rd . I d· . except ~ tl '11 CIS P ace In a n Incompressible flow wh ich is everywhere uniform far dOw~\ le I usftrlated effe~t ~f the strut. Ve locity measurements at a certa in plane . . .s ream 0 t lC strut Indica te the presence of a wake as shown T I I·t , . le ve OCI y d IstnbutlOn ac ross the wa k" e IS app rox imately a cos ine function; that is, Il

U

1 -

(J

(1 +

cos 27bry)

for

b -2h < y<2'

40

2- 4

M ECHANICS AN D THERM O DYNAMICS OF FLU ID FLOW

where U is the veloc ity of the un ifo rm flow field fa r fro m the strut and its wake. If the measured values of a and bl e are 0 .10 and 2.0, respectively. determine t he drag coefficient C1) of the strut. The drag coefficient is defined by C1) = TJ/ !pU2 c , where is the total d rag per un it length on the strut d ue to pressure and viscous forces, and p is the fl uid density .

m

Pro pella nt

\

fl ow

,

/h

EQ UILIBRIUM CO M BUSTION THERMODYNAM ICS

2- 4

7, Calculate the combustion temperatu re of an oxygen-hydrogen mixture whose mass ratio is 3: 1 (0 2: H 2). Since this mea ns a co nsiderable excess of hyd rogen ove r the stoic hiometric proportion, assume that the product temperature wi ll be low enough to prevent dissociation. T he temperature of the reactants entering t he cha mber is 20°C, and the heat transfer to the coolant fro m the cha mber walls is 2 kcal per mole of prod uct. As a fi rst approximation, the mean specific heats may be taken as in the table below, an d the heat of reaction as -57.802 kcal/gm-mole at 3000 K (see Table 2-1). Average specific heat, cal/gm·mo le· oC

p"

~

Reactants Products T U RBOJET

R OCKET

P ROBL EM

4

4 . Ca lculate the static t hrusts of the rocket and turbojet engines described in the figure. The th rust T is the fo rce necessary to prevent horizonta l movement of the engine. Both engines ex ha ust a mass flow of 120 Ib/ sec. The ratio of air and fuel mass flowing into the turbojet is 50: 1 and in its exhaust plane the veloci ty is 1500 ft/ sec and the pressure is the same as the ambient pressure . The rocket contains a ll its own propel2 lant and ex hausts it at a velocity of 9000 it/ sec thro ugh an area of 2 ft . The press ure in the exha ust plane of the rocket is 22.5 psia and ambient pressure is 14.7 psia. S. An idea lized supersonic ramjet diffuser consists of an axisymmetric cente r body loca ted in a cy lindrical duct as shown in the fig ure. The flow at stat ionsCD andQ) may be ass umed unifo rm at the values indicated a nd the strea m tube which enters the inlet has a d ia meter D fa r upstream. Show that the aerodynamic drag on the center body and struts is ~ = -

4 D - [Pl tl1 (U2

"

?

-

/1})

+ P2

-

P IJ·

Constituent

6. A mixtu re of gases con taining 10lb of n itrogen, 10 lb o f hydrogen, and 1S lb of helium is contained at a pressure of 100 psia and tempera ture SOOoR . If the constituents are taken to be perfect gases a nd the G ibbs-Da lton law hol ds, what is the molecul ar we ight and specific heat ratio of th e mixtu re?

02

H2O

6.5 7.4

7.5 8.3

11.0

Q" Btu/ lb-mole at 14.7 psia and 7JOF

CHu A

- 130,000

C0 2 (g) O 2 (g)

-1 69,300 0 0 - 104,000

(g) H 2O

5

H,

8. A mixture of air and a hydrocarbon fuel whose average compositio n is indicated by CH5A undergoes complete combustion. Twice as much ai r is present as needed to burn all the fu el. How high is the fi nal temperatu re ? T he heats of fo rmation a re given in t he table be low. It may be assumed that air is 79% nitrogen and 21 % oxygen by volume. T he reactants have a temperature of 77°F befo re combustion . T he adiabatic flame temperature may be fo und eit her by using average values of the specific heats, or enthalpy tables fo r each constituent. A very a pproximate estimate may be made by si mpl y assuming that the product mixture has the same specific heat as air at the same temperature.

N2

PROBLEM

41

9. Find the flame temperature of the products of combustion of hydrogen a nd oxygen whose composition and average specific heats are given in the table below. T he reactants enter the adiabatic combustion cham ber at OaF . In th is ta ble, Cp p is the ~vera~ ~ola r specific heat o f the compo nent between 77°F and t he fla me temperaure, C pr IS the average mola r specific heat of the component between OaF and 77°F ' and Q/ is the heat of forma tio n at 77°F . Component

Gas

Mol wt, M

Speci fic heat ratio, 'Y

02 H2 0

Nitrogen H ydrogen Helium

28 2 4

1.4 1.4 1.67

OR

H

H 2O

Moles 0 .1008 0 .3 170 0 .054 0.109 0 .233 1.512

Cp p,

Btu/ lb-m ole· oR 8.8 8.0 5.0 5.0 8.0 11.2

Cpr Btu/ Q" lb-mole 7.0 7.0

0 0 +1 05,400 + 92,910 + 18,000 - 104,000

MECHANICS AND THERMODYNAMICS OF FLUID FLOW

42

2-4

10. One mole of C02 is heated at atmospheric pressure to 50aaoF. What is the equilibrium mixture composition if only CO, 02. and C02 are present? 11. An inventor claims that a secret device shown figuratively below can take airstreams anctQ) and convert them into output, streamsG) and.c±> without energy exchange with its environment. [nside the device are mechanisms for mass, momentum, and energy exchange between the streams. Is it possible that such a device can be made to operate in the steady state? It may be assumed that ve locities at all stations are negligib le.

CD

3 Steady One-Dimensional Flo"" of a Perfect Gas

PR013LE1vl

11

12. Show that if any cons tant pressure line p J is given on a temperature-entropy plane, a line corresponding to any other constant pressure p can be obtained by horizo nta l shift of magn itude (for a perfect gas):

I1S ~ -R In 1'.. . PI

13. Two st reams of ai r mix in a constant-area mixing tube. The primary st rea m enters with a veloc ity of 1000 [LIsee and a temperature of the mixing tube at station ISOooR. The secondary stream enters with velocity 100 ft/ sec and temperature 50QoR. The flow at stationsQ) andQ) may be assumed one-d imensional. The p~es. sure at s(ationCDis 15 psia and the rat io or prima ry to secondary flow areas at station

CD

(Dis 1:3.

;;'l~::===t-

Primary

""

Sccond--~ar~Y~:..-j-I----;=,II.'_ _ _ __ _i7::'\

CD

PROBLEM

CD

INTRODUCTION

3-1

In most of the propulsion methods discussed in this volume, the working fluid can be quite satisfactori ly app roximated by a perfect gas. In this chapter the basic laws stated in the first chapter are applied to perfect gases flowing in channels . For algebra ic cOllvenien~e the specific heat of the gas is usually assumed constant herein.

Useful analyses of actual channel flows are re latively easy if it is assumed that ~he ~uid co ndition~ vary in t he streamline direction only. In this case, the flow IS saId to be one-dImensional. The flow in t he immediate v ici nity of a waH can never be one-d im ensional, s ince the velocity at the wa ll surface is zero and there are significan t property va ri ations across the s tream lines . However , with the ex~eption of a thin layer of fluid next to the wall , fluid condit ions are often fairly umform over a large part of a flow. Even if the bulk of the flow is not uniform on any s urface norma l to the stream lines , the one·ciime ns ional approx imation may still lead to useful solutions for the st reamwise va ri ation of averacre fluid . 0 propertIes. One-dimensional so lut ions exhibit the major c haracter istics of many Impor tant flows, including t hose in nozzles, diffusers, and combustors.

13

(a) Using co ntinuity, momentum, and energy equations, a l ~l~g with the perfect-gas law, show how the flow at Q) may be determined from condltl~ns atG). (b) Determine the velocity, temperature, and pressure at slalIon(~).

3-2

GENERAL ONE-DIMENSIONAL FLOW OF A PERFECT GAS

Fluid flows arc governed by the equations of conti n u ity, momentum and energy alan . h . ' , WIt · an appropriate eq uation of state. In this sec ti on these equations are applied to a one-dimensional diITeren tial cont r ol volume within a duct , as indi-

?

43

3- 2

STEADY ONE· DIMENSIONAL FLOW OF A PERFECT GAS

44

cated in Fig. 3- 1. The ra tes of heat and work tran sfe r are dQ and dO'" respectively. A body force of magnit ude X per unit volume of fluid may act in the stream direction. For steady flow t he continuity requirement [from Eq. (2- 2)] is Lpu, n dA = O.

Since the flow of Fig. 3- 1 is ass um ed one-dime nsi onal, t his expression may be reduced to (d/ dx)(pllA) = 0, o r

+

+

dp dA dll = 0 pAll

(3- 1)

for the cont rol vol um e of infinitesimal le ngth ind ica ted in Fig. 3-1 .

A

" -

p

r

du II+-:r:..dx

I I

I p+ dp dx(, x

dp

~ ~/.1

TO \

x----

3- 1.

dQ

=

j' Q.

dA"

=

l'iiI

L . ,

(h

u,

?Q;;;;;;:~

Control

dfr

+

d6'"

u du = 0,

F = Lpu(u' n)dA.

+ roCdX) +

XA dx

_(0: + TI)~) + X dx A To

+

Which may be integrated to

Writing 'LF as the sum of ..pressure. wall shear, and body forces, and aga in sidering th e fl ow of F ig. 3-1, this equation can be written

where

d1)

The stagnatioll state is defined as that state which would be reached by a fluid if it were brought to rest reve rsibly, adiabatically, and with out work. The energy equation for such a decelcration is obtained from Eq . (3-3 b),

volume

Iro '" fr

or

r X· u

){'\.

where A{,!l indicates the ent ire cont rol surface area (as opposed to the area A of the duct cross section). The gravitational potential energy term has been neglected, as is usually permissible. Further, except for specia l cases to be co nsidered in Chapters 5 and 16, the body force X will be ze ro. In this case the e nergy equatio n reduces to Ih(dh + II dll) = dQ - dO', (3- 3a) or dfr + II dll = dq - dll', (3-3 b)

The momcntum eq ua ti on for steady fl ow [from Eq. (2- 4)] is

(A dp

pu . n dA _

Stagnation state

>-

One-dimensional flow throu gh a control volume.

L

r (Ir + 1~2)-

)es

]f the fl ow a rea, heat tra nsfer, work, shea r stress, and body force are known, changes in the sta te of lh c fluid stream may be obtained from th ese four equat ions. ]n genera l, solutions cannot be expressed in simple algebraic form. There are a numbe r of important special cases, however, which can be in tegrated in closed fo rm . Before we proceed to these special cases, let us int rod uce two concepts useful fo r all cases: the stag nat ion state and the Mach num ber.

X

I -:-_---~

"wc:::e:r FIG.

For steady fl ow the ene rgy req ui reme nt [from Eq. (2-8)] is

dA

A+Tdx (X

I

1

p

45

where liz is the mass flow rat e puA (a constant) and q and 1\1 a re the heat and work tran sfer per unit mass. If the fluid ca n be assumed a perfect gas (as it will be in this chapter), the equa· ti on of state is p = pRT. (3-4)

d (J's

%

GENERA L ONE· DIMENSIONAL FLOW OF A PERFECT GAS

3-2

=

=

is the wall shear stress and c is th e duct circ um ference.

2

(3- 5)

COIl -

plIA dll

plI!!'!. ' dx

/1

+ "2 .

(3 - 2)

The constant of in tegratio n, fro, is cal led the stagnat ion e nlhalpy . For any ad iabat ic fl ow viewed from a reference frame in which no wor k occ urs the stagna tion enthalpy is co nstant. Note parti cularly that this co nclusion h old~ for both reve rsible and irreversible flows. U nlik e ent ha lpy, the press ure reached When there is ad iabatic zero-wor k deceleration of a fluid is decrea sed by irre~ersible ~rocesses . This is ill~s trated in Fig. 3-2, an enthalpy-en tr o~ diagram Or a typIcal gas. The stagnation state corresponding to state is ® a nd the stagnation press ure is POI. which is the pressure atta ined by the flui d afte r it has

CD

46

STEADY ONE-DIM ENSIONAL FLO W OF A PERFECT GAS

3-2

3-3

ISENTROPIC FLOW

47

where a is the local speed of sound in the fluid. The speed of sound is the speed of propagatIOn of very small pressure disturbances. For a perfect gas it is given by [ I]

IIi

flu]

(3- 8)

2

3-3 pz

p,

M.any a~ tual processes, such as flows in nozzles and diffusers, are ideally isentrop'c. It 15, therefore, worthwhile to study the isentropic flow of a perfect gas in the absence of work and body forces . The simple results obtained with the constan t specific heat ~pproximation are useful even for large temperature changes, so long as approprIate average va lues of cp and)' are used . For constant specific heat, Eq. (3-5) may be written

Entropy

FIG. 3-2.

D efinition of stagna tion state.

been stop ped isentropically. If the flow is brought to rcst irreversibly, as at @ (l he increased ent ropy is a measure of the irreversibility), the stagnation enthalpy is the same as before but the pressure attained is less than Po l' For any process, slIch as to..0), the stagnat ion state gene rall y undergoes a change li ke the one from @ to (Q;). The decrease in pressure from Po 1 to P02 is, in the absence of work, an indication of the irreversibility of the process. If the flu id is a perfect gas, the stagnat ion tel11peralUre is related to the stagnation enthalpy through the

CD

re lation

'I'

ho - h =

ISENTROPIC FLOW

f OcI' dT.

!.-~ = l+!/ T

(3- 9)

2cI'T

which, with the definition of Mach number, becomes To = I

T

+ ~M' 2'

(3- 10)

Equation (3-6) may then be wri tten

7'

The stagnation temperature To of the fluid is that temperature which would be reached upon adiabatic, zero-work deceleration. For a perfect gas with constant specific heats the stagnation temperature and pressure are related [using Eq . (2-19)] by

l Po _ ( 'Y - 1 2)I' /n - ) - I +-- M .

P

2

From the perfect gas la w and Eqs. (3- 10) and (3- 11),

(3- 6)

Po

-=

P

since the deceleration process is, by definition, isentropic. Other properties of the fluid at the stagnation state may be determined from Po, To . and the equation of state . The stagnation conditions may be regarded as local fluid properties. Aside frolll analytical convenience, the definition of the stagnation state is useful experimentally since To and Po are relat ively easily measured. It is usually much more convenien t to measure the stagnation tempera ture To than the temperature T. Mach number

A very conven ient variable in compressible-flow problems is the Mach number M, defined as (3- 7) M = !!.. a

(3- 11 )

(I + )' -- 1 M-")

2

1/("(- 1)

.

(3- 12)

For allY flow of a perfect gas these can be used to relate the local conditions (T,p~ p) t? the local stagna tion conditions To, Po , Po, since this relationship is by definff /oll IsentropIC whatever the actual flow considered may be. The mass flow per unit area is liz A = pli.

Using Eqs. (3-7), (3-8), and (3-10), we may express the velocity as !I

= M

J--

0fFo- -

I

+

-y ;

I M' .

48

3-3

STEADY ONE-DIMENSIONAL FLOW OF A PERFECT GAS

10

f II

f~

-

,,-

~-

I2

"-

"-

/

""

I" 'I, I~~ I i

I

A

A*

I I

\

1\ ~ \ .1 \

0.02

I

0.1

0.2

2

0.4

l Po

\

\

4

Mach number, M

FIG. 3-3.

2)J'H ')/2(7-1) .

(3- 15)

1"0 3-4

,

One-dimensional isentropic flow of a perfect gas.

FRICTIONLESS CONSTANT-AREA FLOW WITH STAGNATION TEMPERATURE CHANGE

Anot her special case for which the basic equations may be solved in closed form

I

I

! I

0.0 I

r'

-

a

is frictionless flow in a constant -area duct in which stagnation enthalpy change occurs. Four processes of interest in which the stagnation temperature of a moving stream changes are:

\ I p\:-r" - I

Po

+

perfect gas of y ~ 1.4. Tabulations for other Y are given by Shapiro [I], Keenan and Kaye [2], and others.

_\

.1

,

(3- 14)

appropr iate 'Y. F igure 3-3 gives the one-d imensiona l isentropic function s for a

,

0.04

1' +1

For a given isentropic flow (given 'Y, R, Po, To , 111), it is clear that A* is a constant so tha t it may be used as in Eq . (3- 15) to normalize the actual flow area A. Eq uations (3-10), (3-1 1), and (3- 12) relate the fl uid properties to the Mach number M and Eq. (3- 15) shows how the Mach number depends on the flow area. Expressing fl ow variables in terms of Mach number is a matter of great convenience, since the various flow functions can be plotted or tabulated as a funct ion of M for the

I

I

.

) (1' + 1) / 2(1' - 1)

I [ -2- ( I+ y -I M M y I 2

- ~ -

\\

v RTo

(2 - -

Combining Eqs. (3-13) and (3- 14), we have

"

\\

I

0.1

±

'-

I

A*

I

49

1 with an asterisk, the maximum flow per unit area is, from th Po - ~ -- ...n

V

I\"

I

I

0.2

'

I

,

/

~

I "

I

I

0.4

--

,

,/

/

CONSTANT-AREA FLOW, STAGNATION TEMPERATURE CHANGE

of the flow at M Eq. (3-13),

.f--L-

fA I A'

I-

4

3-4

,

I. Combustion, 2. Evaporation or condensation of liquid drops traveling with stream,

I

3. Flux of electric current through a fluid of finite conductivity (Joule heating), and 4. Wall heat Iransfer.

\0 l' = 1.4

(After Shapiro [11 ·)

Then, using this expression with Eq. (3- 12), we may write the mass flow rate as (3-13)

For a given fluid (y, R) and inlet state (Po , T o), it may readily be shown that the mass fl ow per Unt't area ,'s maximum at M = I . If we indicate those properties

In any real flow, frictional effects are always present, especially near solid boundaries. As we shall discuss in Chapter 4, wall heat transfer and wall friction are in fact so closely related that it is not rea list ic to discuss the forme r with out at the same time talk ing about the latter. Nevertheless, the study of frictionless flow in a consta nt-area duct in which a To c hange occurs illustra tes some important features of the real flows in which the fi rst three effects above are present. Combustion and evaporat ion or condensation processes may result in variations of mo lecu lar we ight an d gas consta nts, but for th e sake of simplicity these may be considered negl igib le. The stagnation-entha lpy cha nge in the processes can be determined fr om I:1h o = q - w,

/

L

3-4

STEADY ONE-DIMENSIONAL FLOW OF A PERFECT GAS

50

. For th is case d 1\I are net heat and work transfers . per unit fmass Eof fluid where q an (3 I) (3 2) and . ·t momentum and energy equations are rom qs. the contlOUI y, ' (3-3), respectively: 1

+

dp

Momentum:

dp

Energy:

dll o = dh

-pu dll,

+

II

and, using the above relation between static pressure rat io and Mach number, T r--;

To

dll.

T Ol

The first two of these can be readily integrated to

P _ p,

in which the subscript also be expressed as

pll = PIll!! = - (pII)(U - Ill) = -(P,II,)(II - II,),

sign ifies an initial condition. The second of these may ,

P - PI = PI III - pll

2

. duc·mg the Mach number , we can transform this to By mtro

p _ I + 1'M ; . I + 1' M'

p; Further, by use 0 f Eq.

(3 -11) we can derive the stagnation pressure ratio from ,

this equation, with the result that

(

l' _

I + -

.fJll.. POI

Th

I + 1'Mi) ( I + "'1 M 2

I

,)''''-1)

- M(3- 16)

2 I +

l' ;

1 M;

the change in stagnation pressure is directly related to the Mach number.

;~e dependence of Mach number on stagnation enthalpy (or stagna l10n tem-

perature) can in turn be obtained as follows : Using the perfect-gas law, T Tl

I' P,

=

P-; p '

or invo king the continuity condition, pll = PIll " it may be seen that T

r-; Now if th e Mach num ber reI at shown that

·lon M = II/V"'IRT is introduced , it can e;1sily be T

51

=

[I ++ 'Y M' (M)J2 'Y M ;

I

M~

.

From Eq. (3- 10) and this expression, we can show that the stagnati on temperature ra tio is governed by

II

=

CONSTANT-AREA FLO W, STAGNATION TEMPERATURE CHANGE

,

du = 0,

Continuity:

P

-

3-4

1 +1'M;

[I

=

M - 2 1+ 1'- 2- 1 M-' ) + 1'M' (M)] ( I + M; .

~~

(3- 17)

Thus the Mach number depends uniquely on the rali o of stagnati on temperatures and the ini tial Mach number. It is desirable in this case, as in the isentropic case, to simpl ify these relationships by the choice of a convenient reference state. Since stagnation conditions are not. co nsta nt, the stagnation state is not suitable for this purpose. However, the state corresponding to unity Mac h number is suita ble si nce, as Eqs. (3-16) and (3-17) show, condit ions the re are constant fo r a given flow (i.e. , for a given Po" To" M I). Again using an asterisk to signify properties a t M = I , it can be shown from the equations above that

T T* To T~ ..P....

(1 +1')' , I + 1'M " M- , 2(1' + I)M 2 (I +

l' ;

1 M2)

(3- 21 )

( I + 1'M" F

1+ 1'

P*

I + 1'M2

Po

(_ 2 l' + I

Po*

(3-20)

rn-I) (

(3- 22) I + l' ) (

I + 1'M"

'Y _ I ?)'Y !('Y - 1 ) M. + 2I

(3-23)

In this way the fluid properties may be presented as a fu nction of a single argument, the local Mach number. Equations (3- 20) through (3-23) are shown grap hically in Fig. 3-4 for the case l' = 1.4. G iven particular entrance conditions, To 1, Po 1, M 1, the exit conditions after a given change in stagnation temperature may be obtained as follows: The value M, fixes the val ue of T orlT I; and th~s the va lue TJ, since T O l is know n. The exit state is then fixed by T02 /TJ, detenmned by or Then M "p z/p* ,Poz/p~ are all fixed by the value of T oz/TI; (for a given value of 1').

• 52

3-4

STEADY ONE-DIMENSIONAL FLOW OF A PERFECT GAS

3-5 4.00

/

/

I

, ,::::,

l-

0.60

t-

~,

-

\

Ij 0.40

Momentum:

T

\ I '

~

+ u du o

~ dh

For flow in long pipes it will be shown (Fig. 4-14) that the wall shear stresses can be correlated as follows:

p'

//

°

Energy:

\

l/

0. 10 0.08

Continuity:

To

kTZ

/

0.20

CONSTANT-AREA FLOW WITH FRICTION

respectively,

p~ /

- - - - - - 1- - -l-

f1.00 0.80

/

p"

1--

53

Another solution of Eqs. (3- 1) through (3- 4) can be obtained for constant-area adiabatic flow with friction but no body force or work. In this case, from Eqs. (3-1 ), (3- 2), and (3-3) the continuity, momen tum , and energy relations are,

/

I

2.00

CONSTANT-AREA FLOW WITH FRICTION

3-5

TO

or

0.0 6 0.04

0.0 2

0.0

,

diameter.

-

For Mach numbers greater than unity, the Mach number is also a

significant variable. The elTects of roughness (or at least of variations in roughness) are usually fairly un important for surfaces which have a reasonably smooth finish.

II); 0.04

FIG. 3-4.

,

in which C, is the skin friction coefficient To / (pu z/ 2), Re is the Reynolds number puD/ !" ~ is the average height of surface roughness elements, and D is the duct

, 0.1

0.2 0.4 0.6 0.8 Mach number, M

2

3

"( = 1.4

This point is discussed in more detail in Chapter 4. T he way in which friction alTects the Mach number may be shown as follows. First the conservation equations are transformed (by ihtroducing the perfect gas law and the definition of Mach number) to read:

Frictionless flow of a perfect gas in a constant-area duct with stagna tion

It may be seen from Fig. 3-4 that inc reasing the stagnat ion tempera ture drives the Mach number toward un ity whether the flow is supersonic o r subsonic. After the M ac h numbe r has approached unity in a give n duct, further increase in stagnation enthalpy is possible only if the initi al conditions cha nge. The flow may be

said to be thermally choked . It is int eres tin g to note the variations in sta gna ti o n press ure as the stream is subjected to energy transfer. The stagnation pressure alwa ys drops when energy is added to the stream and rises when energy is transferred from the stream. Thus, whether the flow is subsonic o r supersonic. there may be s ignificant JOSS of stagnation press ure due to combustion in a moving s tream. Conversely, cooling tends to increa se the stagnat ion pressure.


Continuity:

temperature change. (Afte r Shapiro [1].)

T

2 u-z.

2

0,

= " M 2 (4 C/ dX) _ 2 D

Momentum:

Energy :

~ dll

o ~ dT

T

+

,,- l lIdll RT'

- ,,-

Since

M

the n dM 2

M2

II

V"Y RT '

dll , RT

II

54

STEADY ONE-DIMENSIONAL FLOW OF A PERFECT GAS

3-5

3-5

CONSTANT- AREA FLOW W ITH FRICTION

55

/00

And now, if we use the con tinuity eq uation to eliminate du 2 / U 2 , it follows that

60 40 20

Also, combin ing the momentum and energy eq uations, we can easily show that

\

\

Po

~, I

dp = - 1'M' (4 C dX) P 2 D j

+

0

_1'_ dT 'Y- I T

6 4

Combining these last two expressions yields

dM' = M'

-M ' Y

(4C dX) D

2

,

,, ~

1/

f

I ; /

1

T T'

which, for constant stagnatio n temperat u re [us ing Eq. (3- 10)], becomest 0.6

4C U

0.4

/ DII

(3- 24) 0.2

Equation (3- 24) shows that dM' > 0 for M < I and dM' < 0 for M > I. That is, frict ion a/ways changes the Mach number toward unity. For constant Cj, Eq . (3- 24) ca n be in tegrated between the limi ts M = M and M I to yield

V\ 1\

/ \

I

0.1

~\

P'

0.06 U 4Cf D

0.04

(3- 25) 0.02

in which L* is the le ngth of d uct necessary to cha nge t he Mach n umbcr of the flow from M to un ity. Co nsider a duct of given cross-sectional area and variable length. If the inle t stale, mass flow rate, and average skin fr iction coefficient are fixed, there is a maximum length of duct which can transm it the flow. Since the Mach number is unity at the ex haust plane in that case, the length is des ignated L* and the flow may be said to be fric ti on-choked. F rom Eq . (3-25), we can see that at any poin t I in the duct, the variable Cf L* / D de pends o nl y on M 1 and 1'. Since D is constant and C, is assumed cons tant , then at some other poin t a distance x (x < L*) downstrea m from point I ,

0.01 O.t

0.2

0.4 2 Mach number, M

4

/0 1 = 1.4

FIG. 3-5. Ad iabatic flow of a perfect gas in a consta nt-area duct with friction. (After Shapiro [\].) by lIsing the definition of Mach number and the continu ity and energy re lations are : T

1' + 1

(3- 26)

T*

p

•

(3- 27)

From this we can determine M z . Other relationships which may be derived [I] Po

t Fo r noncirc ular ducts, the hydraulic diameter mation in place of the diameter D .

DII

=

4A / c can be used as an approxi-

Po*

(3-28)

56

3-5

STEADY ONE-DIMENSIONAL FLOW OF A PERFECT GAS

1.7

1.5

I-

0.9 1'1

1)::,..

!,.?

_}-I~

1-p<, l:V'l' /~ 0' Pj ,"-.Vi

'" I

o. 7 L-~~~~~~~~~~~~~~~~~ o 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.t8 s

FIG. 3- 6.

Typical temperature-entropy diagram for choking process (subsonic).

SHOCKS

3-6

57

Equations (3- 26), (3-27), and (3-28) are plotted in Fig. 3-5 for l' = 1.4. Again the asterisk denotes properties at that point where M = 1. In Sections 3-3, 3-4, and 3-5, three special problems have been treated to show the separate effects of area change, energy transfer, and fr iction. In each case, reference was made to a state where M = 1 and properties at that point were disti nguished by an asterisk. Note that properties at that point are used as convenient normalizing quantities only, and the state need not actually exist in the real flow studied. For a given inlet condition, the M = 1 conditions represent three distinctly different states. This is clearly indicated in Figs. 3- 6 and 3- 7. For two initial states, one subsonic and the other supersonic, three processes to reach a Mach number of one are shown on a typical T-s diagram. The energy-transfer path is called a Rayleigh line and th e friction path is called a Fanno line. On both these lines the * condition occurs on the nose of the curve. The notation *s, *J, *11 indicates the M = 1 states reached by isentropic, friction, and heating processes, respec tively. Subsonic states are above the nose of the appropria te curve, while supersonic states are below it. Note also that travel in eit her direction is possible along the isentropic and energy-transfer curves, while only one direction is possible on the frict ion curve. If it is important to study combined effects of these processes, then simple solutions are not available. However, the complete equations may be integrated numerically according to procedures described by Shapiro [I].

3-6

SHOCKS

A shock is a discontinuity in a (partly) supersonic flow fluid. Fluid crossing a stati onary shock front rises suddenly and irreversibly in pressure and decreases in veloci ty. It also changes its direction except when passing through a shock which is perpendicular (normal) to the approaching flow direction. Such plane normal shocks are the easiest to analyze.

'L T01

1.0

Norma l shocks

F or flow through a normal shock, with no direction change, area change, or work done, the continuity, momentum, and energy equations are

,

I

0.10

0.12

0.t4

0.16

s

F IG. 3-7. Typica l temperature-entropy diagram for choking process (supersonic).

Continuity:

PIli}

Momentum:

PI - pz =

Energy :

To 1 = T 02 ,

=

P2 U 2 , PIUl(U2

-

lit),

where the subscripts 1 and 2 indicate initial and final states, respectively. Two solutions to these equations are possible, one of which states simply that no change

58

STEADY ONE-DIM ENSIONA L FLOW O F A PERFECT GAS

3-6

3-6

SHOCKS

59

occ urs. In term s of Mach number, the interes tin g solu tion may be stated [I ] th at Co nt rol volume

M' + _ 2_ 'Y- I

I

I

M~

., 2'Y - - - M; 'Y-I

,

! 'Y 'Y

+

I L _ _ _ _

I I '

)]Y!(Y-l) + - -- M; 'Y -

.

III

Streamlines

s

1

2

P02

s

PO I

FIG. 3- 9.

T?

Oblique shocks

T; = In F ig. 3-8, M , and PO,/POI are plo tted for the case 'Y = 1.4. Note especially th a t rather large losses of stagn ation press ure occ ur across Mach numbers greater than abo ut 1.5. Tables of all the functions for var io us 'Y are provided by Keenan and Kaye [2]. I. I'\.

o

o. 8

o. 6

"

""

I

""-

P02 =

'\.

1\01

l iT

I

SlalgnatIO[~ pre~sur~ I 1,1

1.1;10 I-

"'- "- ~ M,l---'\

o.2

o. t F IG. 3-8.

If a plane shock is inclined at an a ngle to t he fl o w, t he fluid passing through suffers n ot only a sudden rise in pressure a nd dec rease in speed but a lso a s udde n change of d irec ti on. The situat io n is illustrated in F ig. 3- 9 fo r an obliq ue shock s-s' in one-dimens ional flow. I n passing through the shoc k t he fluid is deflected through a n an gle O. The basic equati ons ap plied to the indicated control volume are : Continuity:

Pllfl l l

Energy:

TOl

1

P2 1l 2n,

T 02 .

r

Downslrca: 1

\

I

jJ I -

(b) Momen tum parallel to the shock,

o=

IIU

'Y =

1.4

1\ 4 6 2 Upstream Mach number, M I

8

.,

(a) M ome ntum normal to the shoc k,

jJ2

=

?

P 21f21l -

Plill ll (1I21 -

111

P lili n.

1111),

in wh ich the subscripts f1 and ( indicate d irect ions normal to a nd parallel to t he shock, res pect ively. F rom (b),

1\

\

=

=

Two momentum equations may be written for this flow, o ne for changes mo mentum perpendicu lar to the shock, the olhe r for cha nges pa rallel to it :

Mach n umber

o. 4

Oblique shock geometry.

10

Normal shock fun ct ions. (After Shapiro [1].)

=

Ill!'

Since the veloc ity co m ponent pa ralle l to t he shock is the same o n both sides of it. it may be see n that an obl iq ue shock becomes a normal shock relative to a Coo rd inate systcm moving with ve locity "11 = " 2 /. This fact may permit us to Use the normal shoc k equations to calculate obl ique shocks. For cxample, o-ive n the. upstream (or initial) Mach number M J a nd th e overall stagnation pre~sure ratIO, the compone nts M In and M 21< could be determined from Fig. 3- 8. The presSU r e a,nd t cmp~ra ture ratios p 1 a nd T 2/ T I co uld th en be obtained by rep lacing M 1 With M III In the appro priate normal shock eq uations. T he shock angle 0' could

ez/

D efl ectLOn ang 1e 0 - 0° ( n ormal shock)

90

\ 80 \

\

\

II "

7

/'

---

1- -

V

I

/

'l_

o \

I

-t--

l.--~V:::--l-/ ./

/

-+- --{-=- M''''''[

\

SHOCKS 3.>-

---

1

f

--

35~

(min)\

Of

o

3-6

I

2

V IL

3.0

~ t'--- ---l\~ '"\ ~'" ~ "~ r----- '?i--- -

~

-

i

MaChD7IV !/

t-

O

~~~ = ~ t-t-w~~~ ""---

t---

0

20 0

20

1.0

0

2. 0

"

"LL:=

I4

j

3.5

3.0

/15V'

I. 5

/

Initial Mach numbe r, M1

FIG. 3- 10. Shock angle ve rsus inlet Mach number and tu rning angle [1]. Curve~ above dashed li ne hold for M , < I and curves below ho ld for M z > 1. (After ShapIro[!].)

... ~

1.0

0.9

~:-("~

c 0.8

~

-"0 8

~ 0.7

~~

~

------ Gr

t------

~ ~" ~ ""::::: I ~L\, , I'---- 30~

,!2

ee 0.6 o

~

~ O. 5

Normal 4

s~:ck , ~ ~> ~

3

~-...-:: "-35"--

~

O. 2 y~

o.1

1.0

1.5

2.0 2. 5 Initial Mach' nurnber, M\

3.0

I I

V

1.,4

3.5

FIG. 3- 11. Stagnation pressure ratio ver sus inlet Mac h number , with turn ing angle as pa ramete r [I]. Curves above dashed line hold for M z > 1 a nd cu rves below hold for M , < 1. (After Shapiro [11.)

shoc~

1.0

1.5

/rV

5mlx or

-~--,

o.5

/

7V 1 7 ~ ~~~H'\~,:= __

Normal

-tw--

I-- f-..-

V

// V/ 2~ / / V/ f/ / ;;;/ / V

~\ t~1\

\

50 10 0

~~,

I.

V

/

V 1/

V

1/

11VI 0/1

Mach wa ve

Deflection angle 0= 00

5~~V V

I/ V lr

u

~

2.5

2.0

1.5

Dd1ection angle

0

.c

,~

10

1/ / 1/ VV VV

2. 5

E

----r:::: ~ ~

30

~,

61

I I

:

2.0 2.5 Initial Mnch !lumber, M\

: 3.0

3.5 )' = 1.4

FIG. 3- 12 . Exit Mach number versus inlet Mach number, with turning ang le as parame ter [1]. Curves above dash ed line cor respond to small (J and curves below corn. spond to large~. (After Shapiro [I].)

be determ ined from M

In

T 2 / T, and the condi tion

and M 1 and the d ownstream Mach num ber from M 211, lilt

= "2 /-

Alternatively, the above four eq ua ti ons, alo ng with the equa tion of sta te, can be reduced to a set of four independ e nt equati o ns relati ng the va riables M I , M 2, ",II'" p ,;p 1, fT , and 0 [I]. If any two variables are give n, for exam ple M 1 and the turni ng angle 0, then all others are de term ined and the downstream (or fina l) conditions can be founel in terms of th e upstream conditi o ns. The solution of these equations from Shapi ro [IJ is given in Figs. 3- 10, 3- 11 ancl 3- 12 fo r th e case f = l .4 and fo r various v:.1 1ucs of o. Axisym metric oblique shocks arc di scussed by Shapiro [3], who presents a comp lete solution for the case of th e conical shoc k as we ll as methods of treatment of the general axisymmet ri c shock problem .

62

STEADY ONE-DIMENSIONAL FLOW OF A PERFECT GAS

3- 6

References 1. SHAPI RO, ASCHER H., Tire Dynamics and Thermodynamics 0/ Compressible Flllid Flow Volume I. New York : The Rona ld Press Company, 1953; Chapters 1-8, 16 2. 'KEENAN , J. H., and J. KAYE, Gas Tables. New York: John Wiley & Sons, 1957 3. SHA PIRO, ASCHER H., op. cit., Volume H, Chapter 17

3-6

SHOCKS

63

5. A Mach 2 flow passes th rough an oblique shock as show n, a nd deflects 10°. A second obliq ue shock reflects from the solid wall. Wha t is the pressure ratio across the twoshock system? It may be assu med that there is no boundary layer nea r the wall; i.e., the flow is uniform in each of the regions bou nded by the shocks.

M =2

10° • ~

Refl ec ted

oblique shock ~ PROBLEM

Problems 1. Show that for a pure substance the stagnatio n pressure cannot increase for an adiabatic zero-wo rk process. During what physica l processes might it increase for a zero-wor k process? 2. Show that, for a perfect gas, the Mach number at the nose of the Fanno and Rayleigh lines is indeed unity. The nose is defined by ds/(rr = O. 3. A perfect fluid expa nds in a frictionless nozzle fr om stagnation conditions Po = 600 psia, T o = SOOooR , to ambien t pressure of 15 psia . If the expansion is isentropic, determ ine the followin g conditions a t the fin al pressure : (l) velocity, (2) Mach number, (3) temperatu re, a nd (4) area pe r un it ma ss flo w. How does the fi nal flow a rea compare with the th roat area for a give n mass flo w? The specific hea t ra tio 'Y is 1.4 and the molecul ar wei ght M is 30.

MI~ ~ f-:o,.--:=----- ----,=_ T

P:~

L----

2

M, T, PI

PROBLEM 4

4. Air fl ows through a cy lindrica l combustion c hamber of diameter 1 ft a nd lengt h 10 ft. StationsQ) andQ) are the inlet and ou tlet, res~cti velY .. The inlet stagna ti on temperatu re is SOooR and thel in let stagn at ion pressure IS 200 P SIa. (a) H the skin friction coefficient is C[ = 0.0040 (a pproximately uniform) and no comb ustion ta kes pla ce, wha t will the inlet Mach number be if it is subsonic and the exit Mac h number is un ity? What are t,he values of stat ic and stagnat ion pressures a t the exit? (b) If combustion takes place and the heat of reaction is 300 Btu per pound of mixture, neglecting fri ctional effects, and if the inlet Mach numbe r is 0.25, what is the exit Mach number? (c) If, in a simi lar problem, the effects of combust ion an d skin fri ction on ~he flow are o f the sa me order of magn itude, how might thei r simu lta neo us effects be estimated ? The flui d may be cons idered a perfect gas with 'Y = 1.4 a nd M = 29.

5

6. Air ente rs a constant-a rea duct at Mach 3 a nd stagnation cond itions 13000 R a nd 185 psia. In the duct it undergoes a frictionless energy-tra nsfer process such that the exit Mach number is unity. Consider two cases : (a) normal shock at in let to the duct, a nd (b) shockfree supersonic heating. Determine the stagnat ion temperat ure and press ure at the exit in each case. Is there any reaso n why the total energy transfer should d iffer (or be equal) in the two cases? 7. A uniform mixture of very sma l1 solid particles and a perfect gas expands adiabatica lly fro m a give n cha mber tempera ture through a given pressure ratio in a nozzle. The particles a re so small th at they may be considered to trave l with the loca l gas speed. Since the density of the solid material is much higher tha n the gas phase, the total volume of the pa rticles is negligible . The rat io of solid to gas flow ra tes is Ji., where o < J1. < 1. Consider two cases : (a) no heat tra nsfer between solid a nd gaseous phases, and (b) the solid particles have the loca l gas temperature at all points in the flow. Qualitatively, how does the presence of the so lid pa rticles affect the ve loc ity at the end of the expansion? Show that fo r both cases the mi xture is equivalent to the adia ba tic expansion of a homogeneous gas of different molec ular weight M and speci fic heat rat io 'Y . 8. A 30-ft3 rocket- propellan t chamber is filled with combustion gases at 60000 R and 1000 psia at the instant combustion ceases. If the throat area of the nozzle is 1 ft 2 esti mate the time it takes for the chamber pressure to drop to 100 ps ia. Assume tha~ at the insta nt combust i ~ ceases the chamber propella nt supply stops completely, a nd also that l' ~ 1.2 and M ~ 20. 9.

(a) Compare the work of compression pcr unit mass of ai r for both reversible ~~i~batic a nd reversible isothermal compress ion through a press ure ratio of 10 with InIttal conditio ns SOQoR and 14.7 psia. Is the importa nce of specific hea t variation large? Why should cool ing d ur ing compression red uce the wo rk ? . (b) In a n effort to provide a continuous cooling process during compression wate r IS sprayed int? a n airstrea m ente ri ng the axia l compressor of a gas turb ine ~ngine. The compressIOn may be taken to be reversible an d the evaporatio n rate is ap proximately.dw/ dT = k, where HI is the wa ter-a ir ma ss ratio , T is the mixture temperatu re, and k IS a constant. If w« I , the process may be simplified by consider ing the sale effect of the evapo ration to be e nergy trans ferred from the gas phase . Show how the fi na l temperat ure and the work of compression depend o n k.

4-1

THE VISCOUS BOUNDARY LAYER 2

4

"\

supercritica;/

1,,\~,SUb:itical

Boundary Layer

-2

Mechanics and

-3

Heal Transfer

4-1

65

THE VISCOUS BOUNDARY LAYER

Historically the development of boundary layer theory fo~lowed the development of mathematical solutions for flows assumed to be nonvlscou~. In th~ latter half of the nineteenth century-after Euler had formulated the ba~lc equatlOns of motion-Helmholtz, Kelvin, Lamb, Rayleigh, and other~ ap~hed them t.o a number of physical situations. By neglecting the effects of VISCOSIty they obtaIned elegant mathematical descriptions of various flow field~. .I~ was argued that a fluid like air, for example, which is so rarefied as to be InvlSlble, must ~ave negligible viscosity. However, with some exceptions, the so-called perfect-flUId theory disagreed strongly with the results of experiment.

(a) Perfect fluid

(b) Actual fluid, very low velocity

(c) Actual fluid, low velocity

FIG. 4-1. Flow patterns about a cylinder. One particular example, which came to. be kn~wn as d'Alembert's .paradox, attracted a great deal of attention. As an illustratIOn.' th~ flow of a flUId over a cylinder as predicted by nonviscous theory is shown In Fig. 4-:-1 (a). The theory predicts that the velocity and pressure fields are both s~mmetflcal about a plane norma I t 0 th e ups t re am velo city vectors ' Thus . the flUid , can . .exert no net forces (lift or drag) on the cylinder. Moreover , .th~s c?ncluslOn IS Independent o.f the shape of the body; if the fluid is truly invlscld, It can be shown mathematically 64

\ o

)(

I

I /

r---

r! - --

~.../ '/ 2

- - - Perfect-fluidtheo ry prediction -

-

- Measured results

FIG. 4-2. Pressure distribution on a cylinder. (After Flachsbart [11].) that for any body shape the net integrated pressure force upon the body will be zero. Of course, this conclusion directly contradicts experience. Everyone knows that it takes a finite force to hold a body immersed in a moving stream or to propel it through a stationary mass of fluid; hence the paradox. Observations of the actual streamline pattern around the cylinder reveal something like Fig. 4-1 (b), which suggests that the streamline symmetry is grossly disturbed. On the rear side of the cylinder the fluid seems almost stagnant when the stream velocity is quite low. For higher speeds, fairly violent oscillations take place, which affect the entire streamline pattern near the cylinder. Under certain conditions it is possible to observe a periodic shedding of vortices alternately from the top and bottom sides, as suggested by Fig. 4-1 (c). This downstream flow pattern has been called a vortex street. The drag of the actual fluid on the cylinder is largely due to the asymmetry of the pressure distribution on its surface. Figure 4-2 shows measured pressure distributions on a cylinder and , for comparison, the prediction of perfect-fluid theory. The symbol p stands for fluid density and the pressure and velocity far upstream are denoted by poo and uoo . It can be seen that two experimental pressure distributions may be measured (the difference between the two will be discussed later) and that both differ greatly from the theoretical result. The fact that the average pressure on the cylinder is much higher on the upstream hemi-surface explains most of the drag. In addition to this, the viscous force of the fluid on the cylinder imposes a drag force which, in this particular case, is considerably smaller than the pressure drag. For " streamlined" bodies, which have much less pressure drag, the viscous force is relatively more important. For quite a period the discrepancy between existing theory and physical results was unexplained, and a marked rift developed between the theoreticians, who continued to expound the perfect-fluid theory, and the engineering-oriented workers like Bernoulli and Hagen , who actively developed the empirical body of knowledge generally called hydraulics . The equations of motion which do take viscosity into account-the Navier-Stokes equations- were formulated but were

66

BOUNDARY LAYER MECHANICS AND HEAT TRANSFER

4-1

so very complicated that they could not, in general , be solved. It appeared exceedingly difficult to improve mathematically on the perfect-fluid theory except in a few very special cases. Then Prandtl (in 1904) made a major contribution to the solution of the problem by introducing the concept of the viscous boundary layer. . He showed that even for fluids of vanishingly small viscosity there is a thin regIOn near the wall in which viscous effects cannot be neglected. Since the velocity of the fluid particles on the wall is zero, there may be, under certain conditions, a region of large velocity gradient next to it. Then, even though the viscositY.M may be s.malI,. the shear stress on a fluid layer, T = McaU/ ay), may be large In that regIOn since au/ ay is large. This region is called the viscous boundary layer. ~ bound.ary layer of thickness 0 is indicated in Fig. 4-3. Outside t~e laye.r, the velocity gradient au/ay will be so small that viscous shear may be qUIte unlmportant, so that the fluid behaves as though it actually had zero viscosity.

FIG. 4-3. Boundary layer velocity distribution. Thus Prandtl showed that the flow field may be broken into two parts; a thin viscous zone near the waIl, and an outer zone where the nonviscous theory is adequate. This suggested the major simplification o~ the ~avier-Stokes e~uati~ns which is the basis of boundary layer theory. For fairly hIgh-speed flow, In WhICh the boundary layer is so thin that the pressure gradient normal to the wall is negligible, Prandtl and others showed how to treat the boundary lay~r mathematically for special cases, many of which have been solved in the past SIX~y y~ars. Given that a boundary layer exists on a body, it is easy to see qualitatively why the case of vanishing viscosity is fundamentally different from the case of zero viscosity . Figure 4- 2 shows that for zero viscosity a particle on ~h~ su:face streamline of the cylinder rises in pressure to the stagnation value when It Impinges upon the cylinder (8 = 0). Then the pressure falls to a minimu~ (~nd the velocity rises to a maximum) at the position 8 = 7r/2. From that pOint It travel~ to the rearward stagnation point (e = 7r), having just enough momentum to chmb the "pressure hill" and arrive at the stagnation point. . However small the viscosity, the viscous case is fundamentally different. The fluid near the wall has been slowed by viscous action. When a particle in the boundary layer reaches e = 7r/ 2 it will have reduced momentu~ so th~t only a small pressure rise will stop its forward motion, or actually send It movmg back-

THE VISCOUS BOUNDARY LAYER

4-1

--

/ / /

67

Compressor cascade

FIG. 4-4.

Subsonic diffuser

Rocket nozzle

Some practical devices in which separation may occur.

ward. For this reason the particle can no longer follow the contour of the wall. The accumulation of stagnant fluid on the back part of the cylinder deflects the outer streamlines, as suggested in Fig. 4-1 Cb). This in turn greatly modifies the wall pressure distribution, as shown in Fig. 4-2. When the streamlines in the vicinity of the wall cease to follow it, they are said to separate. This accumulation of stagnant or near-stagnant fluid and the resultant gross distortion of the streamlines means that the boundary layer behavior can have an important influence on overall flow behavior even though the quantity of fluid directly affected by viscosity is a small fraction of the total flow. Three separated flows of practical significance are shown in Fig. 4-4. Both the compressor cascade and the subsonic diffuser are designed so that the mainstream velocity decreases in the flow direction; hence the flow is against an adverse pressure gradient. If the pressure rise is too great the boundary layer will separate, creating regions of near-stagnant fluid as shown. Within rocket nozzles the pressure gradient is usually favorable, so that separation need not be a problem. However, under certain off-design conditions shocks can occur within rocket nozzles and the sudden pressure rise across the shock can cause separation. Separation and its consequences in these devices will be discussed in the following chapters where appropriate. Our purpose in this chapter is to establish the nature and cause of separation and to indicate how one might predict its occurrence (although this is seldom possible with purely theoretical methods). Separation is, of course, only one phase of boundary layer behavior. We shall be interested also in the behavior of non-separated boundary layers, since they are of importance to heat transfer and other phenomena. The classic picture of separation is illustrated in Fig. 4-5, which shows a series of velocity profiles in a boundary layer as it approaches and passes the separation point. As the pressure rises the free-stream velocity U falls, as the Bernoulli equation would predict. However, the effect of a given change in pressure is greater on the slow-moving fluid near the wall than on the free-stream fluid.

68

4-1

BOUNDARY LAYER MECHANICS AND HEAT TRANSFER

4- 1

THE VISCOUS BOUNDARY LAYER

69

cussed SUbsequently.) With these assumptions , the momentum equation requires that

1)1-______ u y

j

J

./

7

!

J

~

/

~/ ~/~~~ ~/

]

/ Streamlin e -Z-Which Ie ft ~___ the wall Recirculating flow ~~

-7

f--- x

=

-u du (boundary layer),

where, in keeping with boundary layer assumptions, the free stream and the boundary layer experience the same pressure change. Equating the two expressions for dp and defining the ratio a = u/ U, there results

J

/

p

p

---1

J

dp

dp = _ U dU (free stream),

da = _

d~ (~ -

pU2

(

2 )

.

a

Hence if the pressure rises (dp > 0), a decreases and the boun.dary layer fluid is decelerated more than the free stream. Noting that -dp/ pU 2 is simply dU/ U, this expression can be integrated (for constant density), yielding

Separation point

FIG. 4-5.

A simplified picture of the development of separation.

I -

The fluid near the wall can be slowed to a stagnant state and then be made to flow backward if the adverse pressure gradient continues. The separation point is generally defined as that point where a streamline very near the wall leaves the wall. In two-dimensional flow this corresponds to the point at which the velocity profile has zero slope at the wall. That is ,

(au) ay

=

y=O

°

ai

where the subscript 1 refers to an initial condition. The thickness 0 of the boundary layer can be found from the equation for continuity which requires that

or, again for constant density, OIl_Ul •

(two-dimensional separation).

OIU

Figure 4-5 suggests that, past the separation point, there is a region of recirculating flow. In practice this region is highly unstable. It may be quite difficult to determine analytically, since once separation occurs there is a strong interaction between the body of stagnant fluid and the free stream. Often there results an adjustment of the free-stream geometry such that the adverse pressure gradient is reduced to practically zero, in which case there is no force to drive the recirculating fluid. The effects of pressure gradient on the shape of the boundary layer, which are shown in Fig. 4-5, U may be explained in a simplified manner as follows . Let the boundary layer be replaced by a hypothetical discontinuous flow shown in Fig. 4- 6, in u which the continuous variation in velocity normal J to the wall is replaced by two step changes. Fur~ ther in order to focus on the effect of pressure gradient, let us for the moment neglect shear FIG. 4-6. Discontinuous stresses. (The effects of shear stresses will be dis- boundary layer. ~

~

~

~

T

The two quantities a and 0 are plotted against U/ U l in Fig. 4-7 for the particular case al = l It may be seen that this boundary layer separates (a -70, 0 -7 U) ) when the free-stream velocity has decreased by less than 14%. With this picture of the relatively large effect of a given pressure rise on the more slowly moving fluid, it would appear that any pressure rise applied to a real boundary layer would cause separation since, if the velocity goes continuously to zero at the wall, there will always be some fluid with insufficient momentum to negotiate the pressure rise. However, this tendency toward separation is resisted ~p to a point, by visc?us stress. That is, ~he slowly moving fluid can be "dragged'; hrough a pressure flSe by the faster flUid farther from the wall. An important ~etho~ of reducing. the separation tendency is to enhance this "drag" force by InCreasmg the effective shear stress through turbulent motion . The mention of turbulent motion introduces another important aspect of boundary layer behavior. In 1883 Reynolds showed that under certain conditions dye particles injected into a pipe flow moved smoothly along streamlines without late~al fluctuations . In the same pipe at higher velocities, this smooth laminar motLOn broke down into turbulent motion consisting of rapid random fluctua-

a 4-1

BOUNDARY LAYER MECHANICS AND HEAT TRANSFER

70

1.0,----- --

t -/

5 4

\ \

\_1

I

o0.7

( 0.8

0.3

....

0.2

-

=0

pDD. j..L

0.1

1.0

V

00

u

V

VI

FIG. 4- 7. "Growth" of step boundary layer near separation for a 1 = t·

FIG. 4-8. Comparison of laminar and turbulent velocity profiles.

tions superposed on the mean motion. After the flow had become t~rbulent the dye lines rapidly mixed with the surrounding fluid, the macroscoplC tu.rbul~nt mixing process being at least an order of magnitude faster than molecular diffusIOn (the only agency for mixing in laminar flow). . . The same transition behavior takes place in the boundary layer, resulting In a significant reduction in its tendency to sep~rate. The reaso~ for .this is that the turbulent fluctuations constitute a mechanism for transportmg high momentum from the outer part of the layer to the region near the wall.. The effect is ~s if there were an increased shear stress or, as will be shown, an Increased coefficlent of viscosity. This tends to raise the wall shear stress, but it also means that the turbulent boundary fluid can climb higher up the " pressure hill" than the fluid in the laminar layer. A comparison of typical turbulent and laminar boundary layer velocity profiles is indicated in Fig. 4-8. The fact that the turbulent layer tends to have much higher momentum (and shear stress) near the wall gives it an increased resistance to sep.arati.o~. As a method of specifying the resistance of a boundary layer to separatIOn, 1t 1S useful to define a pressure coefficient C p as f::.p

Cp

71

geometry is completely specified (far from the entrance) by pipe diameter D and wall roughness. The pertinent fluid properties are density p , viscosity j..L,and the average velocity D. These variables can be arranged in a dimensionless form known as the Reynolds number, Re, so that Re

0 0.9

THE BOUNDARY LAYER EQUATIONS

y Ii

ex

ii

',ii i

--------,

0.4

fa -

-

0.5

4-2

=

lxPU2'

where U is the free-stream velocity at the point where the pressure begins to rise . and f::.p is the total increase in pressure up to the point in question. For an approximate comparison of cases in which the adverse pressu~e gradIent is not abrupt (i.e., where the shear stress has time to act) it may be saId that the laminar boundary layer will support a pressure rise given by 0.15 < C p < 0.2, whereas the turbulent boundary layer can tolerate without separation a pressure rise corresponding to 0.4 < C p < 0.8, depending on the Reynolds number and the upstream pressure distribution of the flow. . .' The onset of turbulence, or transition as it is called, IS dependent In a given device on fluid properties and velocity, the general or upstream turbulence level, and the geometry of the particular device. In the case of pipe flow, for example,

Reynolds found that for "smooth" pipes (commercial pipes and tubing are usually "smooth") and very quiet inlet conditions, the transition process could in fact be correlated with a critical Reynolds number below which the flow was laminar and above which it was turbulent. More generally it is found that transition in any series of geometrically similar devices can be correlated with the Reynolds number, where D is replaced by some characteristic length Land D by some characteristic velocity U. Thus Re

=

pUL. j..L

Obviously, a critical value for Re can have meaning only within the geometry for which it was defined . With these facts in mind, the explanation of the two pressure distributions observed on the cylinder of Fig. 4-2 becomes straightforward. For reasonably low velocities (below a "critical" Reynolds number) the boundary layer remains laminar and separates readily on the back of the cylinder' before much pressure rise has taken place. At the critical Reynolds number, the boundary layer becomes turbulent and is able to flow considerably further around the back of the cylinder before separation . As it does so it rises in pressure, thus reducing the discrepancy between inviscid and actual pressure distributions. Since the drag on a blunt body such as a cylinder is primarily a pressure force (as opposed to skin friction), the onset of turbulent flow in this case is actually accompanied by decreased drag. In most practical fluid machines to be discussed in this book the bound ary layers a.re turbulent, and so long as this is true the influence of Reynolds-number variations on overall performance is usually minor. However, under extreme conditions, as for exampl e at very high altitudes, the Reynolds number can become low enough that portions of the flow revert to laminar flow. There usually follows a rather substantial reduction in efficiency of performance, due to the increased tendency toward separation.

4-2

THE BOUNDARY LAYER EQUATIONS

T~e general equations for the flow of Newtonian viscous fluids are called the

Nav~er-Stokes e~uations [1,2].

Because of their mathematical complexity, it is pOSSIble to obtam exact solutions of them for only a few physical situations. I many cases, .howeve:, it is legitimate to make simplifying approximations an~ thereby obtaIn solutions whIch have a useful range of validity. The boundary

72

4-2

BOUNDARY LAYER MECHANICS AND HEAT TRANSFER

4-2

THE BOUNDARY LAYER EQUATIONS

73

y

Table 4-1

u

_ _ _+-__TX_+ ..

aTx

Surface

ay dy

Mass flux

(}),---- -----,Cll I I

I I I ~ I

I~ I

I I p

pV

I I

4-9. Small control volume for development of boundary layer equations. FIG.

~ x

+-__

_ _+-1

Control volume

+ aTy d

p( u + ~ dX)

+ fJ! dx ax

-----J(D

layer equations are much more tractable than the Navier-Stokes equations. They may be derived by neglecting certain terms in the Navier-Stokes equations which can be shown to be quite unimportant in boundary layer flows. The method is developed by Schlichting [I J in careful detail. In the following discussion , equations will be developed less rigorously, but more directly, for the special case of the steady, two-dimensional, incompressible boundary layer. Shapiro [3J uses this approach to develop equations for compressible flow, and includes a good discussion of the range of validity of the simplifications. The object in view is to lay a foundation for a quantitative description of both laminar and turbulent boundary layers, with particular emphasis on skin friction, separation, and heat transfer. Although the solutions of the boundary layer equations which are available are seldom directly applicable to the complex flows in real machines, they at least illustrate the forces at work and the behavior typical of the boundary layer. Consider a very small control volume within a boundary layer, as shown in Fig. 4-9. Velocities in the x- and y-directions are u and v, respectively, while the shear stresses in these directions are Tx and Ty • To apply the continuity and momentum equations to this control volume, we shall need to know the mass and momentum flux across each surface and the force (shear and pressure) acting on each surface. The outward fluxes of mass and x-momentum are shown in Table 4-1. Since the flow is taken to be two-dimensional, no fluid crosses the surfaces parallel to the xy-plane. The continuity equation for steady, two-dimensional, incompressible flow requires that p (u

or

+ :~ dX) dy

-

+ p (v+ ~~ dY) ~':!. + av = o.

pu dy ax

ay

dx -

+ ~: dX) dy

CW)

p (u

@@

-pudy

COO)

p (v

@)(D

-pvdx

+ :; dY)

(u + ~: dX)

2

dy

dx

p (v

+ :;dY)

(u

+ ~;dY) dx

-pvu dx

The momentum equation in the x-direction, using Table 4-1 to obtain the net outward momentum flux, is (applying Eq. (2-4»

L Fx =

p (u

+ ~~ dX) 2 dy -

2

pu dy

+ p (v + :; dY )

(u

+ ~; dY ) dx -

puv dx.

Expanding and disregarding second-order terms, we have

L Fx = p (2U axau + u ayav + v ayau) dx dy. Using the continuity equation, we may simplify this to

L Fx

= p (u au

ax

+ v au) dx dy. ay

~ince only pressure and shear forces act on the control surface, the force summatiOn may be written, in accordance with Fig. 4-9, as

p pdy - (p + aax dX) dy + (Tx + aTX) ay dy dx (-

Tx dx

~~ + ~?) dxdy.

Thus the momentum equation may be written

X

(4-1)

p

_pu 2 dy

p (u aau

pu dx = 0

Flux of x-momentum

+

~p. + aTayx .

u v aa ) = _ y ax

Th~ shear ~tress T developed in a Newtonian fluid is proportional to the rate of sheanng straIn. For the particular case of the boundary layer, in which there is a

74

4-2

BOUNDARY LAYER MECHANICS AND HEAT TRANSFER

large velocity gradient close to the wall, the shear stress is very nearly equal to

4-2

THE BOUNDARY LAYER EQUATIONS

Thus this equation may be written

f

au ay

f..t-'

h

o

where f..t is the coefficient of viscosity. Using this definition, the momentum equation for the x-direction can be written

75

(u

au ax

h

au + v oy

_

dV) U dx dy

f

=

0 /I

2 a u ay2 dy.

Integrating the right-hand term and noting that the wall shear stress is given by TO

=

f..t

(au) , ay y=o

while au/ oy is zero outside the boundary layer, we can write (for constant As a consequence of the thinness of the boundary layer, pressure changes normal to the wall may be neglected inside the layer, so that it may be assumed that p is a function of x only. Then the momentum equation may be written

h

u ou / o ( ax

+ v au

v dV) d

_

oy

dx

= _

y

/I)

TO.

P

I ntegrating the second term in the integral by parts, we have (4--2)

where /I =::0 f..t/ p is the kinematic viscosity. The momentum equation as derived here assumes a fluid of constant viscosity and constant density. Although these conditions may be satisfied within many low-speed laminar boundary layers, we shall see that the "effective" viscosity in turbulent flow may vary strongly through the boundary layer. Further, p and f..t could vary as the result of temperature variation within the boundary layer, due either to high heat-transfer rates or, in compressible flow, to the variation of static temperature in response to velocity variation. In such cases a complete set of equations must include at least an energy equation and an equation of state. However, even in cases not requiring the simultaneous solution of an energy equation, Eqs. (4-1) and (4-2) can present considerable difficulties in the solution for u and v (where p is assumed a known function of x). It is desirable, indeed necessary in most cases, to utilize approximate methods for solving the boundary layer equations. An extremely useful approximation, the momentum integral method, has been developed and widely applied. The object of the method is to find a solution which will satisfy the integral of the momentutn equation across the boundary layer even though it may fail to satisfy the equation at particular points. The integration of the momentum equation (4-2) may proceed as follows:

foh(u auax + v ayau) dy = fh 0

1 dp

-

P dx dy +

fh 0

lJ

a2u ay2 dy,

where h is an undefined distance from the wall outside the boundary layer. the free stream, I dp UdU. dx dx

P

In

h hVaudY f o oy

vul

=

j'h u~1!.dY.

_ 0

From the continuity equation, v at y

= h is given by

h

v

/

ay

0

av -dy o ay

h

au

= - /0 -dy. ax

Thus

{ h au v ay dy = -

} 0

v

fl< au 0 ax dy

+

fh 0

au u ax dy.

The momentum equation may then be written

l< fo

(2U ax au

-

v au

ax

-

v dV) dx

dy

=

TO

p

or, rearranging,

fo h

a~ [u(V -

u)] dy

+ ~~ fo h (U

- u) dy =

+ ~o.

(4--3)

Both the integrands are zero outside the boundary layer, so h can be indefinitely large. . Equation (4-3) is arranged in this particular form because of the physical sig?Ificance (and the common usage) of the individual integrals. In the second Integral, (U - u) is the mass flux defect (actually I/p times the defect) that occurs as the result of the deceleration of the boundary layer fluid. In the case of flow within a duct, for example, if there is a mass flux defect near the walls there must be an increased mass flux within the free stream. This effect on the free stream would be the same if there were no boundary layer but the duct walls were dis-

76

BOUNDARY LAYER MECHANICS AND HEAT TRANSFER

4-3

4-3

lAMINAR BOUNDARY lAYER SOLUTIONS I. 0

placed inward an amount 0* so that Vo*

=

10'"

(V -

0.8

u

U 0.4

0*

=

faoo

(I -

~)

e = faoo ~ (I - i ) dy.

(4-4a) (4-4b)

I

R

1.0

=

Ux v

o 1.08 X 10 5 o 3.64 XIO·; /:,. 7.28x I05

2.0

3.0

4.0

5.0

5.646.0

7.0

q=y!!Z.

~~ Velocity distribution in the laminar boundary layer on a flat plate at zero InCidence, as measured by Nikuradse [IJ . ~IG. 4-10.

The simplest flow is that over a flat plate where dpl dx is zero. In this case, Eqs. (4-1) and (4-2) reduce to

With these definitions, the momentum integral equation, Eq. (4-3), may be written

u au

ax

+ v au ay

=

with the boundary conditions

or

(4-5) As it stands this equation applies to both laminar and turbulent boundary layers. The solution, say for a given Vex), requires knowledge of the relationship between 0* and e, which may be found if an approximate velocity profile shape is assumed (0* 1 e is often called the shape factor H), and the skin friction TO is known. Note that shear stress is evaluated only at the wall where, even in turbulent flow, it is given by M(aul ay), since turbulent fluctuations must go to zero at the wall. 4-3

i

I

o o

.

dy,

/

V

( 00

ve=}o u(V-u)dy. Here e is called the momentum thickness of the bounda ry layer. The two terms are then defined as follows (for incompressibldow):

/1'"

0.6

0.2

~

/ 11

u) dy.

With respect to mass flux then, the boundary layer has the effect of making the thickness 0* unavailable for free-stream flow ; 0* is called the displacement thickness of the boundary layer. The first integral can be treated in a similar ma nner. Note first that since the limits of integration are not functions of x, the partial differentiation al ax can be carried outside the integral where it becomes d(dx. In this case we interpret u as l i p times the actual mass flu x, while (V - II) is the momentum defect per unit mass of the boundary layer fluid . Hence the integral is the momentum defect in the boundary layer which, as above, could be accounted for by the loss of free-stream thickness e such that 2

77

y = 0:

u

y= 00:

u = V = constant.

v

=

=

0,

A stream function t/; may be defined such that v

_ at/;.

=

ax

The convenience of this form is that it automatically satisfies the continuity equation so that the above two equations reduce to

LAMINAR BOUNDARY LAYER SOLUT:ONS

A variety of exactt solutions of Eqs. (4-1) and (4-2) are available, two of which will be mentioned here .

(4-6) This equation may be considerably simplified by substituting

t Exact in the sense that they do not neglect anI of the terms in the boundary layer equations. Most solutions are, however, expressed in terms of series expansions.

1/

= y ".,! V /vx

and

if;

= vvxU /(1/) ,