Solution 9.131
The formula is quite correct and serves as an interesting alternative to Eq. (9.86). Notice that one can
Problem 9.132
Air flows at Ma = 3 and p = 10 lbf/in2 absolute toward a wedge of 16° angle at zero incidence in
Fig. P9.132. If the pointed edge is forward, what will be the pressure at point A? If the blunt edge
is forward, what will be the pressure at point B?
Solution 9.132
For Ma = 3,
= 8°, Eq. 9.86:
Problem 9.133
Air flows supersonically toward the double-wedge system in Fig. P9.133. The (x,y) coordinates
of the tips are given. The shock wave of the forward wedge strikes the tip of the aft wedge. Both
wedges have 15° deflection angles. What is the free-stream Mach number?
Solution 9.133
However tricky the shock reflection might be at the upper (aft) wedge, the problem is solved by
knowing the shock angle at the lower (forward) wedge:
Problem 9.134
When an oblique shock strikes a solid wall, it reflects as a shock of sufficient strength to cause the
exit flow Ma3 to be parallel to the wall, as in Fig. P9.134. For airflow with Ma1 = 2.5 and
p1 = 100 kPa, compute Ma3, p3, and the angle
.
Solution 9.134
With
1 = 40°, we can compute the first shock deflection, which then must turn back the same
amount through the second shock:
Problem 9.135
A bend in the bottom of a supersonic duct flow induces a shock wave which reflects from the
upper wall, as in Fig. P9.135. Compute the Mach number and pressure in region 3.
Solution 9.135
Given
= 10°, find state 2:
11
1
Ma 3.0, 10°,
Eq. 9.86 predicts 27.38°,
==
Problem 9.136
Figure P9.136 is a special application of Prob. 9.135. With careful design, one can orient the bend on
the lower wall so that the reflected wave is exactly canceled by the return bend, as shown. This is a
method of reducing the Mach number in a channel (a supersonic diffuser). If the bend angle is
= 10°, find (a) the downstream width h and (b) the downstream Mach number. Assume a weak
shock wave.
Problem 9.135
A bend in the bottom of a supersonic duct flow induces a shock wave which reflects from the
upper wall, as in Fig. P9.135. Compute the Mach number and pressure in region 3.
Solution 9.136
The important thing is to find the angle
between the second shock and the upper wall, as shown
in the figure. With initial deflection = 10°, proceed forward to “3”:
Problem 9.137
A 6 half–angle wedge creates the reflected shock system in Fig. P9.137. If Ma3 = 2.5, find (a) Ma1;
and (b) the angle
.
Solution 9.137
(a) We have to go backward from region 3 to regions 2 and 1, using Eq. (9.86). In both cases the
Problem 9.138
The supersonic nozzle of Fig. P9.138 is overexpanded (case G of Fig. 9.12b) with Ae/At = 3.0 and
a stagnation pressure of 350 kPa. If the jet edge makes a 4° angle with the nozzle centerline,
what is the back pressure pr in kPa?
Solution 9.138
The nozzle is clearly choked because there are shock waves downstream. Thus
Problem 9.139
Airflow at Ma = 2.2 takes a compression turn of 12° and then another turn of angle
in
Fig. P9.139. What is the maximum value of
for the second shock to be attached? Will the two
shocks inter-sect for any
less than
max?
Solution 9.139
First get the conditions in section (2) and then iterate for
max:
Problem 9.140
The solution to Prob. 9.122 is Ma2 = 2.750 and p2 = 145.5 kPa. Compare these results with an
isentropic compression turn of 5°, using Prandtl-Meyer theory.
Problem 9.122
Supersonic air takes a 5° compression turn, as in Fig. P9.122. Compute the downstream pressure
and Mach number and wave angle, and compare with small-disturbance theory.
Solution 9.140
Find
for Ma = 3 and subtract 5°:
Problem 9.141
Supersonic airflow takes a 5° expansion turn, as in Fig. P9.141. Compute the downstream Mach
number and pressure and compare with small-disturbance theory.
Solution 9.141
Find
for Ma = 3 and add 5°:
Problem 9.142
A supersonic airflow at Ma1 = 3.2 and p1 = 50 kPa undergoes a compression shock followed by
an isentropic expansion turn. The flow deflection is 30° for each turn. Compute Ma2 and p2 if
(a) the shock is followed by the expansion and (b) the expansion is followed by the shock.
Solution 9.142
The solution is given in the form of the two sketches below. A shock wave with a 30° turn is a
Problem 9.143
Airflow at Ma = 3.4 and 300 K encounters a 28º oblique shock turn. What subsequent isentropic
expansion turn will bring the temperature back to 300 K?
Solution 9.143
From Eq. (9.86), for Ma1 = 3.4 and θ = 28º, calculate β:
Problem 9.144
The 10 deflection in Example 9.17 caused the Mach number to drop to 1.64. (a) What turn
angle will create a Prandtl-Meyer fan and bring the Mach number back up to 2.0? (b) What will
be the final pressure?
Solution 9.144
Recall from Ex. 9.17 that the shock wave caused Ma2 = 1.64 and p2 = 17.07 psia. First calculate
the (reduced) stagnation pressure in section 2, plus the Prandtl-Meyer angle
2:
Problem 9.145
Air at Ma1 = 2.0 and p1 = 100 kPa undergoes an isentropic expansion to a down-stream pressure
of 50 kPa. What is the desired turn angle in degrees?
Solution 9.145
This is a real ‘quickie’ compared to what we have been doing for the past few problems.
Isentropic expansion to a new pressure specifies the downstream Mach number:
Problem 9.146
Air flows supersonically over a surface which changes direction twice, as in Fig. P9.146.
Calculate (a) Ma2; and (b) p3.