# 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.

## Reviews

There are no reviews yet.