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11.5 Nonisentropic Flow of an Ideal Gas 625<br />

This relationship is graphed in the margin on the previous page for air.<br />

From Eqs. 11.125, 11.126, and 11.128 we see that<br />

2.0<br />

r a<br />

r V 11 k2Ma<br />

Ma c<br />

V a 1 kMa d 2<br />

(11.129)<br />

V___<br />

Va<br />

r<br />

___ a ,<br />

ρ<br />

1.0<br />

0.0<br />

0.1<br />

1.0<br />

Ma<br />

10<br />

This relationship is graphed in the margin for air.<br />

The energy equation 1Eq. 5.692 tells us that because of the heat transfer involved in Rayleigh<br />

flows, the stagnation temperature varies. We note that<br />

T 0<br />

T 0,a<br />

a T 0<br />

T b a T T a<br />

b a T a<br />

T 0,a<br />

b<br />

(11.130)<br />

Unlike Fanno flow,<br />

the stagnation temperature<br />

in Rayleigh<br />

flow varies.<br />

We can use Eq. 11.56 1developed earlier for steady, isentropic, ideal gas flow2 to evaluate T 0T and<br />

T aT 0a because these two temperature ratios, by definition of the stagnation state, involve isentropic<br />

processes. Equation 11.128 can be used for TT a . Thus, consolidating Eqs. 11.130, 11.56, and<br />

11.128 we obtain<br />

2.0<br />

21k 12Ma 2 a1 k 1 Ma 2 b<br />

T 0<br />

2<br />

<br />

T 0,a 11 kMa 2 2 2<br />

(11.131)<br />

___ T 0<br />

T 0,a<br />

1.0<br />

0.0<br />

0.1<br />

1.0<br />

Ma<br />

10<br />

This relationship is graphed in the margin for air.<br />

Finally, we observe that<br />

p 0<br />

p 0,a<br />

a p 0<br />

p b a p p a<br />

b a p a<br />

p 0,a<br />

b<br />

(11.132)<br />

We can use Eq. 11.59 developed earlier for steady, isentropic, ideal gas flow to evaluate p 0p and<br />

p ap 0,a because these two pressure ratios, by definition, involve isentropic processes. Equation<br />

11.123 can be used for pp a . Together, Eqs. 11.59, 11.123, and 11.132 give<br />

p 0 ___<br />

2.0<br />

1.0<br />

p 0<br />

p 0,a<br />

<br />

11 k2<br />

11 kMa 2 2<br />

2<br />

ca<br />

k 1 b a1 k 1<br />

2<br />

k1k12<br />

Ma 2 bd<br />

(11.133)<br />

p 0,a<br />

GIVEN<br />

0.0<br />

0.1<br />

1.0<br />

Ma<br />

10<br />

This relationship is graphed in the margin for air.<br />

Values of pp a , TT a , rr a or VV a , T 0T 0,a , and p 0p 0,a are graphed in Fig. D.3 of Appendix D<br />

as a function of Mach number for Rayleigh flow of air 1k 1.42. The values in Fig. D.3 were calculated<br />

from Eqs. 11.123, 11.128, 11.129, 11.131, and 11.133. The usefulness of Fig. D.3 is illustrated<br />

in Example 11.16.<br />

See Ref. 7 for a more advanced treatment of internal flows with heat transfer.<br />

E XAMPLE 11.16<br />

Effect of Mach Number and Heating/Cooling<br />

for Rayleigh Flow<br />

The information in Table 11.2 shows us that subsonic<br />

Rayleigh flow accelerates when heated and decelerates when<br />

cooled. Supersonic Rayleigh flow behaves just opposite to subsonic<br />

Rayleigh flow; it decelerates when heated and accelerates<br />

when cooled.<br />

FIND Using Fig. D.3 for air 1k 1.42, state whether velocity,<br />

Mach number, static temperature, stagnation temperature, static<br />

pressure, and stagnation pressure increase or decrease as subsonic<br />

and supersonic Rayleigh flow is 1a2 heated, 1b2 cooled.

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