Chapter 6: Fluids Engineering
Discussion:
Across a long artery, the cumulative pressure drop may be significant. This is why blood
pressure is measured at a single point on an artery to eliminate any impact of the pressure
175
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2000
sft
slug
1045
3
avg
.
v
s
Convert to gal/min:
gal
7222
s
60
sgal
4817
ft
49620 3
3
...q
Discussion:
Because of the much greater viscosity of oil compared to water, the oil can flow at much
larger velocities and still remain laminar. If the pipe were larger, the maximum flow rates
for both water and oil would decrease.
177
P6.21: Using the volcanic flow model and flow parameters from P6.20, prepare two charts.
(a) On one chart, plot the flow velocity in mph as a function of the slope, varying the
slope from 0o to 90o. (b) On the other chart, plot the flow velocity in mph as a function
178
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Discussion:
(c) The slope angle influences the velocity according to the sine of the angle. This
0
1
2
3
4
5
6
7
0 50 100 150 200 250 300
Thickness (cm)
179
P6.22: A steel storage tank is filled with gasoline. The tank has partially corroded on its
inside, and small particles of rust have contaminated the fuel. The rust particles are
approximately spherical, and they have a mean diameter of 25 μm and density of 5.3
g/cm3. (a) What is the terminal velocity of the particles as they fall through the
gasoline? (b) How long would it take the particles to fall 5 m and settle out of the
liquid?
and its weight is
315
m
kg
Chapter 6: Fluids Engineering
180
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The buoyancy force that develops in gasoline is
315
23
Bm101818
s
m
819
m
kg
680
..F
2
11
s
mkg
104575
.
= 5.457× 10–11 N
The drag force is therefore
FD = (4.254 × 10–10 N) – (5.457× 10–11 N)
= 3.708 × 10–10 N
Following Equation (6.16) for the drag force on a sphere, the terminal velocity becomes
kg
N107083
64
10
.
v
(b) To fall 5 m, it will take time:
m 5
Discussion:
Because this is less than one, we have confirmed that it was acceptable to apply Equation
(6.16). Had we found otherwise, we would have discarded this prediction, and instead
applied Equation (6.14) with the graph of Figure 6.22 for CD.
If the size of the particle increases, then the terminal velocity would increase. This is
181
P6.23: A small water droplet in a mist of air is approximated as being a sphere of diameter
1.5 mil. Calculate the terminal velocity as it falls through still air to the ground. Is it
reasonable to neglect the buoyancy force in this instance?
10–4 ft (Table 3.5) and its volume is:
6
in00150 3
.).(
v
= 1.767× 10–9 in3 =1.023× 10–12 ft3
and its weight is
312
ft
slug
The buoyancy force that develops in air is
312
3
ft
slug
...F
182
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FD = 6.388× 10–11 lb
Following Equation (6.16) for the drag force on a sphere, the terminal velocity becomes
lb103886
11
.
v
Because this is less than one, we have confirmed that it was acceptable to apply Equation
(6.16). Had we found otherwise, we would have discarded this prediction, and instead
applied Equation (6.14) with the graph of Figure 6.22 for CD.
If the size of the droplet increases, then the terminal velocity would increase. This is
Chapter 6: Fluids Engineering
184
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FD = (4.815× 10–10 N) – (7.705× 10–13 N)
= 4.807 × 10–10 N
Following Equation (6.16) for the drag force on a sphere, the terminal velocity is:
N108074
10
.
v
s
(b) To fall 2 m, it will take time
s
m2
Discussion:
As a double–check on the consistency of our solution, we will verify that the terminal
velocity is low enough so that Re < 1 and Equation (6.16) was properly applied. We
calculate the Reynolds number to be
m
kg
62
Because this is less than one, we have confirmed that it was acceptable to apply Equation
(6.16). Had we found otherwise, we would have discarded this prediction, and instead
applied Equation (6.14) with the graph of Figure 6.22 for CD.
If the size of the dust particle increases, then the terminal velocity would increase. This is
s
Chapter 6: Fluids Engineering
186
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FD = (1.357× 10–4 N) – (1.589× 10–5 N)
= 1.198× 10–4 N
Following Equation (6.16) for the drag force on a sphere, the terminal velocity becomes
m00150
sm
kg
102603
N101981
5
4
..
.
v
s
s
(b) With the terminal velocity instead given, and the viscosity unknown:
sm
kg
84740
m 00150
s
m
0103
N101981 4
.
..
.
μ
s
m
Because this is less than one, we have confirmed that it was acceptable to apply Equation
(6.16). Had we found otherwise, we would have discarded this prediction, and instead
applied Equation (6.14) with the graph of Figure 6.22 for CD. In (b), since the velocity is
smaller and the viscosity is larger, Re will be smaller still, indicating that Equation (6.16)
189
P6.28: A low-altitude meteorological research balloon, temperature sensor, and radio
transmitter together weigh 2.5 lb. When inflated with helium, the balloon is spherical
with a diameter of 4 ft. The volume of the transmitter can be neglected when compared
to the balloon’s size. The balloon is released from ground level and quickly reaches its
terminal ascent velocity. Neglecting variations in the atmosphere’s density, how long
does it take the balloon to reach an altitude of 1000 ft?
Assuming laminar flow and Re < 1, calculate the velocity by using Equation (6.16):
s
ft
987
ft4
sft
slug
10833
lb01410
7
.
.
v
190
which is over 600 mph and clearly not laminar. The Reynolds number is:
7
3
s
ft
slug
1083
ft
slug
.
μ
The initial assumption that Equation (6.16) could be applied is not valid since it requires a
Reynolds number less than one. Instead, we use the drag force law (6.14) which is valid for
high speeds, when CD is known.
s
The Reynolds number at this condition is:
3
ft
slug
Discussion:
This demonstrates the important of checking whether or not one’s assumptions are valid.
Without a check of the Reynolds number, one may assume the first calculated velocity of