Mechanics of Fluids

whence
p2
p1
= 1 + γ M2 1
1 + γ M 2 2
(11.24)
If there is no net heat transfer to or from the stream-tube considered, the
adiabatic energy equation (11.10) may be used. Hence
cpT1 + 1
2 u2 1 = cpT2 + 1
2 u2 2
(11.25)
Thus, if cp remains constant, the stagnation temperature (defined by
eqn 11.13) does not change across the shock. Putting u = aM = M � (γ RT)
for a perfect gas we obtain
whence
cpT1 + 1
2 M2 1γ RT1 = cpT2 + 1
2 M2 2γ RT2
T2
T1
= cp + 1
2M2 1γ R
cp + 1
2 M2 2
1
1 + 2
=
γ R (γ − 1)M2 1
1 + 1
2 (γ − 1)M2 2
From the equation of state and from eqn 11.22
T2
T1
= p2ρ1
p1ρ2
= p2u2
p1u1
= p2M2
�
�γ �
RT2
p1M1 γ RT1
whence
T2
T1
=
� �2 p2M2
p1M1
Substitution from eqns 11.24 and 11.26 now gives
1 + 1
2 (γ − 1)M2 1
1 + 1
2 (γ − 1)M2 =
2
� 1 + γ M 2 1
1 + γ M 2 2
� 2 M 2 2
M 2 1
(11.26)
(11.27)
(11.28)
An obvious and trivial solution of this equation is M2 1 = M2 2 ; that is,
conditions upstream and downstream are identical and no shock exists.
Simplification in which M2 1 − M2 2 is factored out, however, gives
M 2 2 = 1 +[(γ − 1)/2]M2 1
γ M2 1 − (γ − 1)/2
(11.29)
From this result the downstream Mach number may be calculated and the
ratios of pressure, temperature, density and velocity then obtained from
eqns 11.24, 11.26 and 11.27. Equation 11.29 shows that if M1 = 1 then
M2 = 1, and that if M1 > 1, M2 < 1.
Shock waves 501