ME 131B Fluid Mechanics
Solutions to Mid-Term Exam
Winter 1996-97
S. K. Lele
Open book (Fox & McDonald) Only
Need a calculator
February 12, 1997
10:00 a.m. to 10:50 a.m.
Solve all problems
1
1. (total of 30 points)
The sketch below depicts a steady ow. The constant area duct has rough-walls, other
ducts are frictionless.
P0
Station (1)
Station (2)
(1)
(2)
out
In
x
Rough−Walled
Constant Area
Duct
Frictionless
Nozzle
Frictionless
Nozzle
Normal
Shock
True
(a) P0 1 A1 = P0 2 A2?
(b) At the given operating conditions, the mass ow rate will increase if the roughwalls were made smooth.
False
(c) Sketch the pressure distribution along the duct:
;
;
1
P(x)
−−−
P0
P*/P0
decreasin
g
asing
incre
Throat
Nozzle 1
Shock
(1)
Constant Area Duct
2
x
(2)
Nozzle 2
2. (total of 30 points)
The converging-diverging nozzle in the sketch is to be used as the second throat in a
supersonic wind tunnel. The Mach number at this throat (whose area A1 = 100 cm2)
is M1 = 1.4, and the pressure P1 is recorded to be 100 kPa. The exit area A = 250
cm2.
e
(1)
Throat
Station 1
A1 = 100 cm2
Exit Area
Ae = 250 cm2
(a) For isentropic ow throughout the duct, what is the exit Mach number M and
the exit pressure P ?
For M1 = 1:4, we obtain from the isentropic ow table:
e
e
A1 = 1:115;
A1
This gives
P1 = 0:3142
P0 1
;
A1 = 89:69 cm2;
P0 1 = 318:2 kPa
;
A = 250 = 2:787
A 89:69
)
e
e
From the isentropic ow table, we obtain
M = 2:56
e
and
P = 0:05332
P0
e
But P0 = P0 1
;e
;
;e
)
P = 16:97 kPa
e
3
(b) What is the exit pressure P for which a normal shock stands at the exit?
e
e1
For M
1
e;
e2
= 2:56, we obtain from the normal shock table,
P 2 = 7:479
P1
M 2 = 0:5074;
e;
e;
e;
)
P 2 = 126:9 kPa
e;
(c) What is the exit pressure P for which a normal shock stands at the throat?
e
(1)
(2)
For M1 = 1:4, we obtain from the normal shock table,
P0 2 = 0:9582
P0 1
M2 = 0:7397;
;
;
Since P0 1 A1 = P0 2 A2
;
;
)
A2 = 93:60 cm2
A = 2:671
A2
)
e
which gives M = 0:22 and P =P0 = 0:9669. This leads to
e
e
;e
P = (0:9669) (0:9582) (318:2) = 294:8 kPa
e
4
3. (total of 40 points)
A frictionless tube of constant area (sketched below) supplies hydrogen gas in a chemical plant. In the supply tube the gas temperature T is 27 degree Celsius and its
pressure P is 200 kPa but its speed V is not known. A malfunction in the chemical plant triggers the control valve to rapidly shut-o the supply of hydrogen. This
generates a shock wave which propagates back into the tube. A temperature sensor
measuring the gas temperature is located upstream of the valve and records a steady
temperature reading of 58 degree Celsius after the shock wave has gone past the sensor.
g
g
g
H2
Vg
T g , Pg
Temperature Sensor
Control Valve
T = 300 K
P = 200 kPa
c = 1316:1 m/sec
g
g
g
(a) What is the speed of the gas (at the sensor) after the shock wave has gone past?
Since the valve is shut-o, speed of gas = 0.
(b) What is the gas pressure (at the sensor) after the shock wave has gone past?
Transform to the frame moving with the shock (say shock speed is V )
s
(1)
(2)
V1 = Vg + Vs
V2 = Vs
T1 = 300 K
This gives
T2 = 331 K
T2 = 1:10
T1
From the normal shock table, we obtain
M1 = 1:16
M2 = 0:8682
P2 = 1:403
P
1
5
This gives
P2 = 280:6 kPa
(c) What is the speed of the shock wave V ?
For T2 = 331 K, we have c2 = 1382:4 m/sec.
s
M2 = Vc 2 = Vc = 0:8682
2
2
s
)
V = 1200:2 m/sec
s
(d) What is the speed of the gas in the supply tube V ?
g
M1 = V c+ V = 1:16
1
)
V = 326:5 m/sec
g
g
6
s