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Mechanics of Fluids SI Edition 5th Edition by Potter - Test Bank
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,Mechanics of Fluids Chapter 1: Basic ConsiderationsSolutions for Sections 1.1 – 1.3
1. If force was selected as a fundamental dimension rather than mass, the units on density
would be:
(B) N·s2/m4
[ρ ] = [m]/[ V ] = (F ⋅ T 2 /L)/L3 = F ⋅ T 2 /L4 . Using SI units, this is N ⋅ s 2 /m 4
2. Which of the following is not acceptable in the SI system of units?
(A) N·cm/s
cm is only allowed when used as cm, cm2, or cm3
3. The energy usage in the United States in 2010 was about 100 × 1018 J. This can be
written as:
(D) 100 trillion MJ
A trillion is 1012. So, 100 × 1012 × 106 J = 100 × 1018 J
4. The fluids that are considered in this text:
(D) Move when subjected to any shear stress
The fluids of interest in Fluid Mechanics always move when subjected to a shearing
stress no matter how small that shearing stress may be.
5. The mean free path where commercial aircraft fly (about 30,000 ft) is nearest:
(C) 0.00076 mm
m
λ = 0.225
ρ d2
4.8 ×10−26 kg
= 0.225 −10 2
= 1.72 ×10−7 m
(0.00089 × 515.4) kg/m × (3.7 ×10 ) m
3 2
The factor 515.4 came from Table A.1 in the Appendix.
2
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
,Mechanics of Fluids Chapter 1: Basic Considerations
Solutions for Sections 1.4 – 1.7
1. The specific weight of air at 5 km is nearest:
(C) 7.26 N/m3
The density is found in Table B.3 in the Appendix. Using a straight-line interpolation
0.8194 − 0.6602
ρ = 0.8194 − = 0.7398 or 0.740 kg/m3
2
The interpolation will result in less accuracy so 4 significant digits is not appropriate.
The specific weight is then, using Eq.1.5.1,
γ = ρ g = 0.740 × 9.81 = 7.26 N/m3
2. The specific gravity of an unknown liquid is known to be 1.26. The liquid is most likely:
(A) Glycerin
γ x = S γ water = 1.26 × 9810 = 12 360 N/m3
Reviewing the specific weights of the various liquids in Table B.5 of the Appendix
suggests the unknown liquid is glycerin.
3. Estimate the torque needed to rotate the inner cylinder at 1000 rpm if SAE-30 oil at 40°C fills
the gap. Assume a linear velocity distribution in the gap. Neglect end effects.
(D) 20.2 N·m
The velocity of the outer cylinder is 0 and that of the inner cylinder is
1000 × 2π
V = rω = 0.08 × = 8.38 m/s
60
Using Eq.1.5.6, the shear stress on the inner cylinder is
ΔV 8.38 − 0
τ =μ = 0.1× = 838 N/m 2
Δr 0.001
The viscosity is found in the Appendix Fig. B.1. The torque is related to the shear stress
as follows:
T = τ Ar = τ × (2π r × L) × r = 2π r 2τ L
= (2π × 0.082 ) m 2 × 838 N/m 2 × 0.6 m = 20.2 N ⋅ m
4
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Mechanics of Fluids, 4th Edition Chapter 1: Basic Considerations
4. Two clean, glass plates spaced 0.2 mm apart are placed in 20°C water. How far up the plates
will the water climb if the contact angle is 0°?
(A) 75 mm
A free-body of the water between the plates would appear as in Fig. 1.10 with β = 0°.
Summing forces provides
σ × 2L = γ × V = γ tLh or 0.0736 × 2L = 9810 × 0.0002Lh
∴ h = 0.0750 m or 75 mm
5. A farmer is applying compressed nitrogen to a field from a tank originally at 800 kPa and
10°C. The temperature of the nitrogen as it leaves the tank is approximately:
(B) −120°C
Assume the nitrogen is an ideal gas which undergoes an isentropic process so that
(k −1) / k
⎛p ⎞
0..4
⎛ 100 ⎞
T2 = T1 ⎜ 2 ⎟ = (273 + 10) ⎜ ⎟ = 156 K or −117°C
⎝ p1 ⎠ ⎝ 800 ⎠
The assumption of a constant entropy process as the nitrogen expands is quite
acceptable since there is no heat transfer and the process is relatively frictionless.
6. Thunder is heard 2.2 seconds after Lightning is observed in Arizona when the temperature is
50°C. The best estimate of how far away the lightning struck is nearest:
(B) 790 m
We assume the lightning is observed instantly. The speed with which a sound wave
travels is given by
c = kRT = 1.4 × 287 J/(kg ⋅ K) × (273 + 50) K = 360 m/s
where we have used
J N⋅m m2
= =
kg ⋅ K (N ⋅ s 2 /m) ⋅ K s 2 ⋅ K
The distance is then estimated to be the velocity multiplied by the time:
d = Vt = 360 × 2.2 = 792 m
5
© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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