Mechanics of Fluids 5th Edition By Potter – Test Bank

Solutions for Section 9.1 – 9.4
1. In which one of the following flows must compressibility be considered?
(D) Flow around the wing of a commercial aircraft in its approach to an airport
Speeds must be in excess of 100 m/s (360 kph or 240 mph) before compressibility
must be considered in flow around an object. Commercial aircraft take off at speeds
less than 40 m/s, considerably below M = 0.3. Only (D) qualifies.

3. A traffic flow on a freeway is very dense and it is proposed that a fluid flow could possibly
simulate the flow. If it was decided to study the feasibility of the proposal, which flow would
be a possibility?
(C) A compressible flow with M >1
Consider a primary characteristic of an incompressible flow or a subsonic flow: as
the area decreases, the velocity increases. For a supersonic flow, as the area
decreases the velocity decreases, as demanded by Eq. 9.3.10. We all know that when
a traffic flow encounters a decrease in the number of lanes, the traffic slows down, it
does not speed up. So, it would require a compressible flow model with M > 1.
4. A farmer is using a tank of 20°C nitrogen pressurized to 540 kPa absolute. It exits the tank out
a hose. The hose snaps and the farmer is hit with the expanding nitrogen. Air is primarily
nitrogen so assuming it is air flowing with no losses, the temperature of the exiting nitrogen is
nearest:
(B) −90°C
The pressure ratio between the exit and the tank is 0
/
e
p p = 100/540 = 0.185. From the
isentropic flow Table D.1 we find, at M = 1.76, that
0
0 6175 181 K or 92 C or 134 F
293
.
T T
T
= = = − ° − °
Such a low temperature would require that the farmer be treated for ‘burns.” The
temperature of the nitrogen in the tank is hardly a factor since it would undoubtedly
be in the range of 10°C < T < 35°C.

. Select the incorrect statement for an inviscid flow.
(C) The velocity distribution in the wake of a body can be approximated by neglecting
viscous effects
Viscous effects are responsible for the velocity distribution in the wake region. We
have not attempted to approximate the velocity distribution in the wake of a body.
That is beyond the scope of an introductory or even an intermediate fluids course.

2. If the velocity potential in an inviscid flow is given by φ = Ay , where A is a constant, the
stream function is:
(D) − + Ax B
Equation (8.5.11) relates the velocity potential function and the stream function:
Then which must, by Eq. 8.5.11, equal 0
0 so const and Most often, we let 0
. ( , ) ( )
.
. .
A x y Adx Ax f y
x y
df
y dy x
df f B Ax B B
dy
ψ φ
ψ
ψ φ
ψ
∂ ∂
= − = − ∴ = − = − +
∂ ∂
∂ ∂
= =
∂ ∂
∴ = = = = − + =
∫

3. If an irrotational vortex approximates the air flow in a tornado, the highest possible velocity,
assuming an incompressible flow, is nearest:
(A) 400 m/s
Assume the velocity at the extreme outer edges of the tornado is zero and the
pressure is 100 kPa. The lowest possible pressure is absolute zero, i.e., 2
p = 0 kPa .
Use Bernoulli’s equation and find

2 V1 1
1
2
p
gz
ρ
+ +
2
2 2
2
V p
= + 2
gz
ρ
+
2
2
2
100 000 408 m/s
1 2 2
. .
.
V
= ∴ = V
where standard conditions have been assumed so that ρ = 1.2 kg/m3
. This is
unreasonable since it exceeds the speed of sound, i.e., it would be a compressible
flow, so the assumption of an incompressible flow would not be applicable. But, it
does illustrate why the velocities are so high near the eye of a tornado.

4. A line source of strength 2 m2
/s is superposed with a velocity of 8 m/s, producing a long,
slender body that resembles the one shown. The thickness h of the body is nearest:
(C) 25 cm
Superposition of the uniform flow function and the source function provides
2
8 8
2
y r sin θ
ψ θ θ
π π
= + = +
The streamline at θ π = splits at the stagnation point so its value remains the same at
ψ π = 8r sin + = π π/ 1 along the top of the body (ψ = const along a streamline). But,
using the equation for ψ , we see that ψ = 0 along the positive x-axis where θ = 0. So,
the thickness of the body is
q U h h h = × = ∆ ∴ = ∴ = ( . ( . . /2) 8 /2) 1 0 25 m ψ
using the result of Example 8.9.
Actually, at a large distance from the origin the velocity in the flow would be 8
m/s so, the thickness of the body would be q = hV or 2 = h × 8 so that h = 0.25 m.