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Modern Engineering Thermodynamics

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16.9 Shock Waves 677<br />

and, for an ideal gas with constant specific heats,<br />

h ox − h x = c p ðT ox − T x Þ = V 2 x /ð2g cÞ<br />

Since c p = kR/ðk−1Þ for constant specific heat ideal gases,<br />

T ox /T x = T osx /T x = 1 + k−1<br />

<br />

Vx<br />

2 <br />

2 kg c RT x<br />

Similarly, for the downstream region, we can write that<br />

α = − 1 v<br />

<br />

∂v<br />

∂p<br />

s<br />

= 1 + k−1<br />

2 M2 x<br />

Since h ox = h oy , then T ox = T oy = T osx = T osy and we can divide the preceding two equations to get<br />

1 + k−1<br />

T x<br />

=<br />

2 M2 y<br />

T y<br />

1 + k−1<br />

2 M2 x<br />

(16.35)<br />

Linear momentum rate balance (LMRB, SS, SF, SI, SO)<br />

F x − F y = ðp x − p y ÞA = _mðV y − V x Þ/g c<br />

or<br />

p x − p y = ð _m/AÞðV y − V x Þ/g c<br />

= ðρ y V 2 y − ρ xV 2 x Þ/g c<br />

Now ρ = p/RT, so<br />

p x − p y = kp y V 2 y /ðkg cRT y Þ − kp x V 2 x /ðkg cRT x Þ<br />

= kðp y M 2 y − p xM 2 x Þ<br />

or<br />

p x<br />

p y<br />

= 1 + kM2 y<br />

1 + kM 2 x<br />

(16.36)<br />

SubstitutingEqs.(16.35)and(16.36)intoEq.(16.34)yieldsanequationforM x ,M y ,andk, which can be<br />

solved for M x ≥ 1 and M y ≤ 1 to give<br />

M 2 y = ðk−1ÞM2 x + 2<br />

2kM 2 x + 1 − k (16.37)<br />

Because Eq. (16.34) is symmetrical in x and y, thex and y subscripts in Eq. (16.37) can be interchanged to<br />

produce an equation for M x in terms of M y and k.<br />

EXAMPLE 16.11<br />

A nuclear explosion produces a normal shock wave that travels through still air with a Mach number of 5.50. The pressure<br />

and temperature of the air in front of the shock wave are 14.7 psia and 70.0°F. Determine the pressure, temperature, and<br />

wind velocity directly behind the shock wave.<br />

Solution<br />

If we attach our reference frame to the moving shock wave, it appears that the air is approaching it with a Mach number of<br />

M x = 5.50. The Mach number behind the shock wave can be determined from Eq. (16.37) with k = 1.40 as<br />

" #<br />

ð1:40 − 1Þð5:50Þ 2 1/2<br />

+ 2<br />

M y =<br />

2ð1:40Þð5:50Þ 2 = 0:409<br />

+ 1 − 1:4<br />

(Continued )

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