Solutions Manual • Fluid Mechanics, Fifth Edition
682
Assuming an isentropic expansion to Mae ≈ 3.06, we can compute the throat area:
2.38
2(0.38)
ù
A e 36.6
1 é 2 + 0.38(3.06)
=
=
ê
ú
A*
A* 3.06 ë
1.38 + 1
û
2
= 4.65, or A* =
Solve for throat diameter D * ≈ 3.2 ft
36.6
π
= 7.87 ft 2 = D*2
4.65
4
Ans. (b)
9.84 Air flows through a duct as in
Fig. P9.84, where A1 = 24 cm2, A2 = 18 cm2,
and A3 = 32 cm2. A normal shock stands
at section 2. Compute (a) the mass flow,
(b) the Mach number, and (c) the stagnation
pressure at section 3.
Solution: We have enough information
at section 1 to compute the mass flow:
Fig. P9.84
a1 = 1.4(287)(30 + 273) ≈ 349 m/s, V1 = 2.5(349) = 872
m
p
kg
, ρ1 = 1 = 0.46 3
s
RT1
m
& = ρe A e Ve = 0.46(0.0024)(872) ≈ 0.96 kg/s
Then m
Ans. (a)
Now move isentropically from 1 to 2 upstream of the shock and thence across to 3:
Ma1 = 2.5, ∴
A1
24
= 2.64, A*1 =
= 9.1 cm 2 , and
2.64
A*1
A 2 18
=
= 1.98
A*1 9.1
Read Ma 2,upstream ≈ 2.18, po1 = po2 = 40[1 + 0.2(2.5)2 ]3.5 ≈ 683 kPa, across the
shock,
A*3
= 1.57, A*3 = 14.3 cm 2 ,
A*2
A3
= 2.24 |sub , Ma 3 ≈ 0.27
A*3
Ans. (b)
Finally, go back and get the stagnation pressure ratio across the shock:
at Ma 2 ≈ 2.18,
po3
≈ 0.637, ∴ po3 = 0.637(683) ≈ 435 kPa
po2
Ans. (c)
9.85 A large tank at 300 kPa delivers air through a nozzle of 1-cm2 throat area and 2.2-cm2
exit area. A normal shock wave stands in the exit plane. The temperature just downstream