fluid_mechanics
628 Chapter 11 ■ Compressible Flow T T y p 0, y T 0 = constant y b Normal shock x p 0, x p* Fanno line T* = constant Normal shock x a Rayleigh line p a T a s s (a) (b) T p 0, x p 0, y T 0 y Normal shock x (c) s F I G U R E 11.26 (a) The normal shock in a Fanno flow. (b) The normal shock in a Rayleigh flow. (c) The normal shock in a frictionless and adiabatic flow. But from Eq. 11.123 for Rayleigh flow we get and p y 1 k p 2 a 1 kMa y p x 1 k p 2 a 1 kMa x Thus, by combining Eqs. 11.137, 11.138, and 11.139 we get p y p x 1 kMa x 2 1 kMa y 2 (11.138) (11.139) (11.140) Equation 11.140 can also be derived starting with Ratios of thermodynamic properties across a normal shock are functions of the Mach numbers. and using the Fanno flow equation 1Eq. 11.1072 As might be expected, Eq. 11.140 can be obtained directly from the linear momentum equation since rV 2 p V 2 RT kV 2 RTk k Ma 2 . For the Fanno flow of Fig. 11.26a, p y p x a p y p* b ap* p x b p p* 1 Ma e 1k 122 12 1 31k 1224Ma f 2 p x r x V x 2 p y r y V y 2 T y T x a T y T* b aT* T x b (11.141)
11.5 Nonisentropic Flow of an Ideal Gas 629 From Eq. 11.101 for Fanno flow we get and T y T* 1k 122 1 31k 1224Ma y 2 T x T* 1k 122 2 1 31k 1224Ma x A consolidation of Eqs. 11.141, 11.142, and 11.143 gives (11.142) (11.143) T y T x 1 31k 12 24Ma x 2 1 31k 1224Ma y 2 (11.144) We seek next to develop an equation that will allow us to determine the Mach number downstream of the normal shock, Ma y , when the Mach number upstream of the normal shock, Ma x , is known. From the ideal gas equation of state 1Eq. 11.12, we can form p y p x a T y T x b a r y r x b (11.145) Using the continuity equation r x V x r y V y with Eq. 11.145 we obtain p y p x a T y T x b a V x V y b (11.146) When combined with the Mach number definition 1Eq. 11.462 and the ideal gas speed-of-sound equation 1Eq. 11.362, Eq. 11.146 becomes 1.0 Thus, Eqs. 11.147 and 11.144 lead to p y a T 12 y b a Ma x b p x T x Ma y (11.147) Ma y 0.0 1.0 Ma x The flow changes from supersonic to subsonic across a normal shock. 40 5.0 which can be merged with Eq. 11.140 to yield p y p x e 1 31k 12 24Ma x 2 1 31k 1224Ma y 2 f 12 Ma x Ma y Ma y 2 Ma x 2 321k 124 32k1k 124Ma x 2 1 (11.148) (11.149) This relationship is graphed in the margin for air. Thus, we can use Eq. 11.149 to calculate values of Mach number downstream of a normal shock from a known Mach number upstream of the shock. As suggested by Fig. 11.26, to have a normal shock we must have Ma x 7 1. From Eq. 11.149 we find that Ma y 6 1. If we combine Eqs. 11.149 and 11.140, we get p y ___ p x 0.0 1.0 Ma x 5.0 p y 2k p x k 1 Ma x 2 k 1 k 1 This relationship is graphed in the margin for air. (11.150)
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628 Chapter 11 ■ Compressible Flow<br />
T<br />
T<br />
y<br />
p 0, y<br />
T 0 =<br />
constant<br />
y<br />
b<br />
Normal shock<br />
x<br />
p 0, x<br />
p*<br />
Fanno line<br />
T* =<br />
constant<br />
Normal shock<br />
x<br />
a<br />
Rayleigh line<br />
p a<br />
T a<br />
s<br />
s<br />
(a)<br />
(b)<br />
T<br />
p 0, x p 0, y<br />
T 0<br />
y<br />
Normal shock<br />
x<br />
(c)<br />
s<br />
F I G U R E 11.26 (a) The normal<br />
shock in a Fanno flow. (b) The normal shock in a<br />
Rayleigh flow. (c) The normal shock in a frictionless<br />
and adiabatic flow.<br />
But from Eq. 11.123 for Rayleigh flow we get<br />
and<br />
p y<br />
<br />
1 k<br />
p<br />
2<br />
a 1 kMa y<br />
p x<br />
<br />
1 k<br />
p<br />
2<br />
a 1 kMa x<br />
Thus, by combining Eqs. 11.137, 11.138, and 11.139 we get<br />
p y<br />
p x<br />
1 kMa x 2<br />
1 kMa y<br />
2<br />
(11.138)<br />
(11.139)<br />
(11.140)<br />
Equation 11.140 can also be derived starting with<br />
Ratios of thermodynamic<br />
properties<br />
across a normal<br />
shock are functions<br />
of the Mach<br />
numbers.<br />
and using the Fanno flow equation 1Eq. 11.1072<br />
As might be expected, Eq. 11.140 can be obtained directly from the linear momentum equation<br />
since rV 2 p V 2 RT kV 2 RTk k Ma 2 .<br />
For the Fanno flow of Fig. 11.26a,<br />
p y<br />
p x<br />
a p y<br />
p* b ap* p x<br />
b<br />
p<br />
p* 1 Ma e 1k 122<br />
12<br />
1 31k 1224Ma f 2<br />
p x r x V x 2 p y r y V y<br />
2<br />
T y<br />
T x<br />
a T y<br />
T* b aT* T x<br />
b<br />
(11.141)