Solution

Chapter 9 • Compressible Flow 643
p/p a = 80.4 = 3001(ρ/1025) 7 − 3000, solve ρ ≈ 1029 kg/m 3
7 7
a = n(B + 1)p ( ρρ / ) / ρ= 7(3001)(101350)(1029/1025) /1029 ≈1457 m/s
a
a
Hardly worth the trouble: One-way distance ≈ a ∆t/2 = 1457(15/2) ≈ 10900 m. Ans.
9.18 Race cars at the Indianapolis Speedway average speeds of 185 mi/h. After
determining the altitude of Indianapolis, find the Mach number of these cars and estimate
whether compressibility might affect their aerodynamics.
Solution: Rush to the Almanac and find that Indianapolis is at 220 m altitude, for
which Table A.6 predicts that the standard speed of sound is 339.4 m/s = 759 mi/h. Thus
the Mach number is
Ma racer = V/a = 185 mph/759 mph = 0.24 Ans.
This is less than 0.3, so the Indianapolis Speedway need not worry about compressibility.
9.19 The Concorde aircraft flies at Ma ≈ 2.3 at 11-km standard altitude. Estimate the
temperature in °C at the front stagnation point. At what Mach number would it have a
front stagnation point temperature of 450°C
Solution: At 11-km standard altitude, T ≈ 216.66 K, a = √(kRT) = 295 m/s. Then
k−1
2
2
Tnose
= To
= T1+ Ma = 216.66[1 + 0.2(2.3) ] = 446 K ≈173°
C Ans.
2
If, instead, T o = 450°C = 723 K = 216.66(1 + 0.2 Ma 2 ), solve Ma ≈ 3.42 Ans.
9.20 A gas flows at V = 200 m/s, p = 125 kPa, and T = 200°C. For (a) air and (b) helium,
compute the maximum pressure and the maximum velocity attainable by expansion or
compression.
Solution: Given (V, p, T), we can compute Ma, T o and p o and then V
max
= (2cpT o):
V 200 200
(a) air: Ma = = = = 0.459
kRT 1.4(287)(200 + 273) 436
k/(k−1)
2
2 3.5
k−1
Then pmax
= po
= p1+ Ma = 125[1 + 0.2(0.459) ] ≈144 kPa Ans. (a)
2
o
2
T = (200 + 273)[1 + 0.2(0.459) ] = 493 K, V = 2(1005)(493) ≈995 m/s Ans. (a)
max