Modern Engineering Thermodynamics

16.9 Shock Waves 677
and, for an ideal gas with constant specific heats,
h ox − h x = c p ðT ox − T x Þ = V 2 x /ð2g cÞ
Since c p = kR/ðk−1Þ for constant specific heat ideal gases,
T ox /T x = T osx /T x = 1 + k−1
Vx
2
2 kg c RT x
Similarly, for the downstream region, we can write that
α = − 1 v
∂v
∂p
s
= 1 + k−1
2 M2 x
Since h ox = h oy , then T ox = T oy = T osx = T osy and we can divide the preceding two equations to get
1 + k−1
T x
=
2 M2 y
T y
1 + k−1
2 M2 x
(16.35)
Linear momentum rate balance (LMRB, SS, SF, SI, SO)
F x − F y = ðp x − p y ÞA = _mðV y − V x Þ/g c
or
p x − p y = ð _m/AÞðV y − V x Þ/g c
= ðρ y V 2 y − ρ xV 2 x Þ/g c
Now ρ = p/RT, so
p x − p y = kp y V 2 y /ðkg cRT y Þ − kp x V 2 x /ðkg cRT x Þ
= kðp y M 2 y − p xM 2 x Þ
or
p x
p y
= 1 + kM2 y
1 + kM 2 x
(16.36)
SubstitutingEqs.(16.35)and(16.36)intoEq.(16.34)yieldsanequationforM x ,M y ,andk, which can be
solved for M x ≥ 1 and M y ≤ 1 to give
M 2 y = ðk−1ÞM2 x + 2
2kM 2 x + 1 − k (16.37)
Because Eq. (16.34) is symmetrical in x and y, thex and y subscripts in Eq. (16.37) can be interchanged to
produce an equation for M x in terms of M y and k.
EXAMPLE 16.11
A nuclear explosion produces a normal shock wave that travels through still air with a Mach number of 5.50. The pressure
and temperature of the air in front of the shock wave are 14.7 psia and 70.0°F. Determine the pressure, temperature, and
wind velocity directly behind the shock wave.
Solution
If we attach our reference frame to the moving shock wave, it appears that the air is approaching it with a Mach number of
M x = 5.50. The Mach number behind the shock wave can be determined from Eq. (16.37) with k = 1.40 as
" #
ð1:40 − 1Þð5:50Þ 2 1/2
+ 2
M y =
2ð1:40Þð5:50Þ 2 = 0:409
+ 1 − 1:4
(Continued )