Modern Engineering Thermodynamics

678 CHAPTER 16: Compressible Fluid Flow
EXAMPLE 16.11 (Continued )
Then, the temperature directly behind the shock wave is given by Eq. (16.35) as
2
1 + k−1 3
2
T y = T
2 ðM xÞ 2
1 + 1:40 − 1 3
ð5:50Þ 2
6
7
6
x 4
1 + k−1 5 = ð70:0 + 459:67 RÞ
2
7
4
2 ðM yÞ 2
1 + 1:40 − 1 5 = 3610 R
ð0:409Þ 2
2
and the pressure directly behind the shock wave is given by Eq. (16.34) as
M
p y = p x
x
M y
T y
T x
1/2
= ð14:7 lbf/in 2 Þ
5:50
0:409
1/2
3610 R
= 516 lbf/in 2
70:0 + 459:67 R
Finally, the wind velocity directly behind the shock wave is just the relative velocity V x – V y ,or
pffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffi
V wind = V x − V y = M x c x − M y c y = M x kg c RT x − M y kg c RT y
p
= 5:50 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1:40ð32:174 lbm.ft/lbf × s 2 Þð53:34 ft.lbf/lbm.RÞð70:0 + 459:67 RÞ
p
− 0:409
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1:40ð32:174 lbm.ft/lbf .s 2 Þð53:34 ft.lbf/lbm.RÞð3610 RÞ
= 5:00 × 10 3 ft/s
These extremely high values for p y , T y , and the wind velocity show why large explosions produce so much damage to life and
property.
Exercises
28. Determine the Mach number directly behind the shock wave in Example 16.11 if the Mach number of the shock wave
itself is 4.0 and all the other variables remain unchanged. Answer: M y = 0.434.
29. Use Eq. (16.36) to verify the pressure directly behind the shock wave calculated in Example 16.11. Answer: p y = 516
psia (the same as determined in Example 16.11).
30. If the temperature in front of the shock wave in Example 16.11 is 0.00°F rather than 70.0°F and all the other variables
remain unchanged, determine the wind velocity directly behind the shock wave. Answer: V wind = 4570 ft/s.
31. Determine the constant value assumed by M y as M x → 4. Thus, no matter how high M x is, M y can never be less than this
value. Answer: M y (as M x → 4) = [(k – 1)/2k] 1/2 = 0.378 for air.
Equations (16.35), (16.36), and (16.37) have been tabulated for air ðk = 1:4Þ in Table C.19 in Thermodynamic
Tables to accompany Modern Engineering Thermodynamics. The reader is encouraged to use this table when its direct
entry is convenient. However, rather than interpolating for nondirect entry values, the equations just given can
be used to make accurate direct calculations, since they were used to generate the table.
Finally, an entropy rate balance on the shock wave gives
Entropy rate balance (SRB, SS, SF, SI/SO, A)
_s p = _mðs y − s x Þ ≥ 0
So,
_S p / _m = s y − s x = c p ln T y /T x − R ln py /p x
h i k/ðk−1Þ
= R ln p x /p y Ty /T x ≥ 0
(16.38)
Therefore,
p x /p y ≥ T x /T y
k/ðk−1Þ
and by substituting Eqs. (16.35) and (16.36) into this relation, it can be shown that M x ≥ M y : Consequently,
the second law of thermodynamics stipulates that shock waves can occur only in supersonic flows, from
M x ≥ 1toM y ≤ 1 and can never occur in subsonic flows. Equation (16.38) is shown in Figure 16.23 for air
(k = 1.40).