Chapter 6: Fluids Engineering
151
Chapter 6
Solutions
152
P6.1: Convert the viscosity of mercury (1.5×10–3 kg/(m s)) to the dimensions of
slug/(ft s) and centipoise.
P
smkg
sm
= 1.5 cP
m
slug
kg
153
P6.2: Michael Phelps won a record-setting 8 gold medals at the 2008 Beijing Olympics.
Now imagine if Phelps had competed in a pool filled with pancake syrup. Would you
expect his race times to increase, decrease, or stay the same? Research the issue and
prepare an approximately 250-word report supporting your answer. Cite at least two
references for your information.
The report may outline a number of different concepts, including the impact of viscous
forces, inertial forces, or body size on Phelps’ performance. Some of the basic ideas that
the students should be able to discuss and explore in the report are summarized in the
following:
157
P6.6: A 6 m high, 4 m wide rectangular gate is located at the end of an open freshwater
tank. The gate is hinged at the top and held in place by a force F. From Equation (6.4),
the pressure is proportional to the depth of the water, and the average pressure pavg
exerted on the gate by the water is
2
Δ
avg
p
p
where ∆p is the difference in pressure between the bottom of the gate (p1) and the
surface (p0). The resulting force of the water on the gate is
ApF avgwater
where A is the area of the gate the water acts upon. The resulting force acts 2 m from the
bottom of the gate because the pressure increases with depth. Determine the force
required to hold the gate in place.
Approach:
Using the density of water from Table 6.1, calculate the pressure at the bottom of the tank
using Equation (6.4). Calculate the pressure difference between the bottom of the tank and
the surface to eliminate the surface pressure in the calculations. The average pressure is
then half of the pressure difference. The water acts on the surface of the wall, therefore
158
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2
2
avg ,
Calculate the area the water is acting upon:
m
m
3
160
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m 24)m 4)(m 6( A
Calculate the force exerted by the water (Equation (6.1)):
N
2
161
P6.9: An ancient king’s supposedly golden crown had a mass of 3 kg, but it was actually
made by a dishonest metal smith from an equal mix of gold (1.93×104 kg/m3) and silver
(1.06×104 kg/ m3). (a) Suppose that Archimedes suspended the crown from a string and
lowered it into water until it was fully submerged. If the string was then connected to a
balance scale, what tension would Archimedes have measured in the string? (b) If the
test was repeated, but this time with the crown replaced by a 3-kg block of pure gold,
what tension would be measured?
6.1.
Solution:
(a) Average crown density:
3
3
3
3
m
kg
109514
m
kg
1010.6)(19.3
2
1 .ρ
Volume:
Chapter 6: Fluids Engineering
(b) Volume:
Discussion:
163
P6.10: Scuba divers carry ballast weights to have neutral buoyancy. At that condition, the
buoyancy force on the diver exactly balances weight, and there is no tendency either to
float toward the surface or to sink. In freshwater, a certain diver carries 10 lb of lead
alloy ballast of density 1.17×104 kg/m3. During an excursion in seawater, the diver
must carry 50% more ballast to remain neutrally buoyant. How much does this diver
weigh?
3
3
2
m
kg
10711
s
m
819
..
Ballast buoyancy force =
m
kg
Sea water:
Chapter 6: Fluids Engineering
164
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736 N or 165 lb
Discussion:
A weight of 165 lb is reasonable from an order of magnitude perspective. The diver must
carry more ballast in sea water because it is denser than fresh water which makes the diver
more buoyant. Alternatively, equations (1) and (2) could be used to solve for V which then
could be used to solve for w.
168
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170
P6.16: (a) Determine the volumetric flow rate of blood in the artery of P6.15. (b) Calculate
the maximum velocity of blood across the artery’s cross section. (c) Determine the
amount by which the blood pressure decreases along each 10 cm length of the artery.
Approach:
The volumetric flow rate is given by Equation (6.10), where the cross–sectional area and
average velocity are known. For laminar flow (the result of P6.14), the maximum velocity
is twice the average value (Equation 6.11). Likewise, the pressure drop Δp can be obtained
from Equation (6.12).
Solution:
(a) Volumetric flow rate
Cross–sectional area:
25
2m102571m0040
4
Volumetric flow rate:
m
m
3
625
(b) Maximum velocity (laminar flow):
s
(c) Pressure drop (Equation 6.12):
4
128
Lq
p
3
6
m
kg