p 2
p 1
¼
1 þ γM
2
1
1 þ γM
2
2
ð3:30Þ
Similarly, by using the equation of state and the continuity equation it is easy to
show that
ρ 2
ρ 1
¼
M 1
M 2
ffiffiffiffiffi
T 1
T 2
r
and
p 2
p 1
¼
M 1
M 2
ffiffiffiffiffi
T 2
T 1
r
,
while the conservation of energy equation gives
T 2
T 1
¼
1 þ
1
2 γ À 1
ð
ÞM
2
1
1 þ
1
2 γ À 1
ð
ÞM
2
2
:
ð3:31Þ
Let us now establish a relationship between M 1 and M 2 . Using Eqs. (3.25) and
(3.30) we obtain
1 þ γM
2
2
1 þ γM
2
1
¼
γ þ 1
1 À γ
ð
Þþ2γM
2
1
and solving for M 2 we have
M 2 ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 þ γ À 1
ð
ÞM
2
1
2γM
2
1 À γ À 1
ð
Þ
s
,
ð3:32Þ
which give the Mach number M 2 in terms of the Mach number M 1 . It can be seen that
M 2 ¼ 1 when M 1 ¼ 1 and M 2 < 1 when M 1 > 1 and vice versa.
3.9 Entropy Change Across the Shock in Terms of Mach
Number
Equation (3.32) is symmetric in M 1 and M 2 as can be seen by writing it in the form,
2γM
2
1 M
2
2 À γ À 1
ð
Þ M
2
1 þ M
2
2
À
Á À 2 ¼ 0:
3.9 Entropy Change Across the Shock in Terms of Mach Number
105