solutions to checkpoint problems:...
TRANSCRIPT
Solutions to Checkpoint Problems: EFM10e.
Problem 1.1. Weightwatchers, Inc. has developed “Points”TM, which are used to track food intake.
Points are calculated as a function of calories, grams of fat, and grams of fiber. You’re only supposed
to eat a certain number of PointsTM in a day. Is the PointTM a dimension or a unit?
Solution. The PointTM is a unit. Reasoning.
• Dimensions are what we measure (power, force, etc.). Units are how we quantify (the amountof force is quantified in units of newtons or lbf, etc.).
• Weightwatchers PointsTM are used to quantify food intake. Thus, they are a unit. The
dimension (what is being measured) is food intake.
Problem 1.2 Which conversion ratio is correct?
(a) 1.0 = (3.785 US gallons)/(1.0 L), (b) 1.0 = (1.0 cm)/(2.54 in), (c) 1.0 = (1.0 lbm)/ (2.205 kg),
(d) 1.0 = (3.281 yd)/(1.0 m), (e) 1.0 = (14.7 psi)/(101.3 kPa)
Solution. The way to solve this problem is to look in Table F.1 in the front of the book. This table
shows that Choice e is correct.
Problem 1.3. A force of =10 lbf accelerates a block at a rate of = 5 ft/s2. Using = ,
calculate the mass of the block in units of pounds-mass.
Solution. Apply the grid method:
=
=(10 lbf)
(5 ft s2)
µ322 lbm · ftlbf · s2
¶= 644 lbm
Review. When pounds-mass appear in a calculation, it is common to apply the conversion ratio:
10 =
µ322 lbm · ftlbf · s2
¶
Problem 1.4 If the molar mass of a gas is 35 grams per mole, what is the specific gas constant for
this gas in SI units?
Solution.
• The specific gas constant and the universal gas constant are related by = ,
where is the molar mass.
• Apply the equation.
=
=
µ8314 J
mol · K¶µ
mol
0035 kg
¶= 238
J
kg · K
Problem 2.1An inventor is considering two lubricants: glycerin and SAE 10W-30 oil.
(a) Which lubricant has a higher viscosity at 150◦F?(b) Which lubricant has a higher viscosity at 230◦F?
Solution.
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• Fig. A.2 shows the viscosity of liquids as a function of temperature.• From Fig. A.2, glycerin has a higher viscosity at both temperatures.
Problem 2.2. What is the kinematic viscosity of nitrogen at 7 atmospheres of pressure (absolute)
and a temperature of 15C?
Solution.Kinematic viscosity is defined as the ratio of (absolute viscosity) to (density). Absolute
viscosity of a gas varies with temperature only. Density of a gas varies with temperature and
pressure according to the IGL.Thus
=( = 15 ◦C)
( = 15 ◦C = 7atm)(1)
Step 1. To find viscosity, look up data for nitrogen in Table A.2
= =
µ145× 10−5 m
2
s
¶µ118
kg
m3
¶= 1 711× 10−5 kg
m · s (2)
Step 2. To find density, start with the IGL: = Then, use a ratio:
( = 15 ◦C = 7atm)
( = 15 ◦C = 1atm)=
=7atm
1atm= 7
Now, apply Table A.2 data to find density at 7 atmospheres.
15 ◦C 7 atm = 715 ◦C 1 atm (3)
= 7
µ118
kg
m3
¶= 826 kgm3
Step 3. Plug Eqs. (2) and (3) into Eq. (1):
=( = 15 ◦C)
( = 15 ◦C = 7atm)=1 711× 10−5 kg
m· s826 kgm3
= 2 07× 10−6 m2
s
Problem 2.3 Two capillary tubes are placed in a liquid. The diameter of tube A is twice the
diameter of tube B. Which statement is true?
a. Capillary rise in both tubes is the same
b. Capillary rise in tube A is twice that of tube B
c. Capillary rise in tube B is twice that of tube A
d. None of the above.
Solution. Answer (c) is correct. Reasoning.
• Capillary rise in a round tube is given by
(Capillary Rise) = ∆ =4
• Build a ratio(Capillary Rise: Tube A)
(Capillary Rise: Tube B)=
44
=
=1
2
Thus, the capillary rise in Tube B is twice the capillary rise in Tube A.
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Problem 3.1. What is the mechanical advantage of this hydraulic machine?
(neglect pressure changes due to elevation changes)
= 2 tons, = 0.9
= 3 inch, 2 = 6 inch, 1 = 1 inch
a. 2:1
b. 4:1
c. 6:1
d. 16:1
e. 36:1
Solution. The answer is (e) 36:1. Reasoning.
An equation for mechanical advantage can be derived by applying force equilibrium to each piston
and then solving for the pressure acting on each piston. Since the pressure is the same (ignore the
hydrostatic pressure variation), one can derive the following equation for the mechanical advantage.µMechanical
advantage
¶=(output force)
(input force)=
µ2
1
¶2=
µ6 in
1 in
¶= 36 : 1
Problem 3.2 In the glass of water shown, which location has the highest value of piezometric head?
Which location has the highest value of the piezometric pressure?
a. A
b. B
c. C
d. None of the above
Solution. The correct answer is (d). Reasoning.
Since the water is in hydrostatic equilibrium and the density is constant, the piezometric head (and
piezometric pressure) is the same at all points.
Problem 3.3. Consider a balloon filled with helium (case A) and a balloon filled with air (case B).
Which statement is correct?
a. Buoyant force (case A) Buoyant force (case B)
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b. Buoyant force (case A) Buoyant force (case B)
c. Buoyant force (case A) = Buoyant force (case B)
Solution. The correct answer is (c). Reasoning.
• The buoyant force is equal to the weight of the displaced air.• Both balloons displace the same amount of air.• Therefore, the buoyant That is, the buoyant force depends only on the volume of the balloonand the specific weight of the ambient air.
Problem 4.1. A velocity field is given by V = (+ )i where = = 2 s−1 and ( ) is positionin the field. A particle moving in this field
a. Moves in the x-direction only
b. Moves in the y-direction only
c. Moves in both the x- and y-directions.
Solution. Answer (a) because the velocity field only has a component in the i-direction.
Problem 4.2. As shown, water drains out of a small opening in a container. Which statement is
true? Why?
a. The flow in the container is steady.
b. The flow in the container is unsteady.
Solution. Answer (b) is correct. Reasoning
• Unsteady flow means that the velocity changes with time at one or more points in the flowfield.: 6= 0
• As the level of water in the container drops, the velocity will change with time at most pointsin the flow field.
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Problem 4.3 Restaurants often use large coffee dispensers (see sketch). The sight glass shows the
level of coffee. If the valve is opened, what happens to the level of coffee that is visible in the sight
glass? Will the level go up, go down or stay the same? Why?
Solution. The water level in the sight glass will go down. Reasoning.
1. Select the Bernoulli equation. Assume inviscid and steady flow (these are good assumptions
for this problem). The Bernoulli equation can be written as
(piezometric head) + (velocity head) = + 2
2= a constant
2. Introduce piezometric head. The sight glass (which is a piezometer) measures piezometric head
which is given the symbol
3. Apply the Bernoulli equation (from 1 to 2).
1 + 21
2= 2 +
22
2= a constant
Now, 1 ≈ 0, so Eq. (1) reduces to
1 = 2 + 22
2= a constant
4. Interpret the Bernoulli Equation. Since 22 2 is a positive number, one can conclude that
2 1Thus, the piezometric head at 2 will be less than the piezometric head at 1. Since
piezometric head corresponds to water level, the level in the sight glass will be lower than the
level in the reservoir.
Problem 5.1. Consider flow through two round pipes. Pipe A has twice the diameter of pipe B.
The mean velocity in each pipe is the same. What is ?
a. 1
5
b. 2
c. 4
d. 8
Solution. Answer (c) is correct. Reasoning:
Apply the flow rate equation ( = ) to each pipe. Then use a ratio.
=
=¡2
¢4
(2) 4
=2
2
=(2)
2
2= 4
Problem 5.2 Consider flow through two round pipes. The maximum velocity in each pipe is the
same. The only difference is the velocity distribution. Which pipe has the larger value of mean
velocity? Why?
(a) Pipe A
(b) Pipe B
(c) They both have the same mean velocity
Solution. The correct answer is (b). Reasoning.
Start with the equation for mean velocity and recognize that an integral is defined as a sum
=1
Z
≈ 1
X
∆
For Case B, when ∆ is large (near the outside of the pipe), then is also large. Thus, the sum
for pipe B is larger than the sum for pipe A.
Problem 5.3. Water and alcohol mix in a tank. Can the continuity equation can be used to show
that the outlet flow rate is 2 liters per second?
(a) Yes
(b) No
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Solution. The correct answer is (b). Reasoning.
1. Apply the continuity equation (CV surrounds the tank). Do a term-by-term analysis.
= Σ̇in −Σ̇out (1)
0 = ̇water + ̇alcohol − ̇mix
2. Rewrite Eq. (1), and introduce the flow rate equation
waterwater +alcoholalcohol = mixturemixture (2)
3. Draw conclusions. The continuity equation (2) does not show that the outlet flow rate is 2
liters/second because the general equation does not reduce down to out = Σin
4. Do secondary research. To gain insight, one can do research on the mixing of liquids. According
to Wikipedia (http://en.wikipedia.org/wiki/Ethanol),µOne liter
of water
¶+
µOne liter
of ethanol
¶=
µ1.92 liters
of mixture
¶Review. When applying the continuity equation to liquids, many engineers make the mistake of
leaving density out of the equation and balancing volume flow rates. Because there is no law called
conservation of volume, we recommend starting with the general equation for continuity, and then
doing a term-by-term analysis.
Problem 6.1 A disk in a horizontal plane is rotating in a counterclockwise direction and the speed
of rotation is decreasing. A penny stays in place on the disk due to friction. Which letter (a to h)
best represents the direction of acceleration of the penny? Which letter best represents the direction
of the sum-of-forces vector? Why?
Solution. Answer (h) is the best answer. Reasoning.
1. Idealize the penny as a particle because acceleration is defined for a particle.
2. Analyze the normal and tangential components of acceleration.
• Normal. Because the particle is moving in a circular path there must be a component ofacceleration () inwards towards the center of curvature.
• Tangential. Because the particle is slowing down (rotation speed is decreasing), theremust be a component of acceleration tangent to path ().
• The resultant acceleration (see sketch) acts in the northwest direction (approximately).Thus, answer (h) is the best answer.
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3. Analyze the sum-of-forces vector by applying Newton’s second law. This law asserts that
the sum of forces and the acceleration must be in the same direction (ΣF = a) Thus, the
sum-of-forces vector will also act in the northwest direction (approximately).
Problem 6.2 Pressurized air forces water out of a tank. If the air pressure is increased so that the
exit speed increases from to 2 , what happens to the rate of momentum flow out the bottom of
the tank? The rate
a. Stays the same
b. Increases by 1x
c. Increases by 2x
d. Increases by 3x
e. Increases by 4x
f. Increases by 8x
Solution. Answer (e). Reasoning.
The magnitude of the rate-of-momentum vector is given byµMagnitude of the
rate-of-momentum vector
¶= ̇ = 2
Thus, if is changed to 2 then the rate-of-momentum changes to (2 )2 = 4 × 2, which
represents an increase by a factor of 4.
Problem 6.3 The sketch shows a liquid flowing through a stationary nozzle. Assume steady flow.
Which statements are true? (select all that apply)
a. The momentum accumulation is zero.
b. The momentum accumulation is non-zero.
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c. The sum of forces acting on the CV is zero.
d. The sum of forces acting on the CV is non-zero.
Solution. Answers (a) and (d) are correct. Reasoning:
• Analyze the momentum accumulation. Because the momentum of the matter inside the CV
is constant with time, the amount of momentum is constant with time. Thus, the momentum
accumulation term is zero. Thus, answer (a) is correct.
• Analyze the sum of force. The momentum equation for this problem reduces toµSum of forces acting
on the matter inside the CV
¶= ̇2 − ̇1
Because the right side of this equation is non-zero, the left side must be nonzero. Thus, answer
(d) is correct.
Problem 7.1. A battery is used to power a DC motor which is then used to drive a centrifugal
pump. For the indicated system, which statements are true? Circle all that apply. Assume steady
state operation.
a. ̇ 0
b. ̇ 0
c.̇ 0
d. ̇ 0
e.
0
f.
0
Solution. Correct statements are (a), (c) and (f). Reasoning.
• The work rate is positive because the system (motor + battery) is doing work on the environ-
ment (i.e. the shaft torque is acting through an angular displacement as the shaft rotates the
pump impeller).
• The heat transfer rate is positive because thermal energy (i.e. heat) from the battery and fromthe DC motor is being transferred across the boundary.
• The energy storage term is negative because the system (i.e. the battery) is losing chemical
energy as the battery runs down. (Note that a battery is a device that stores energy in the
form of chemical energy).
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Problem 7.2. As shown, a pump moves water from a lower reservoir to a higher reservoir. The
pipe has a constant diameter. Which statements are true? (circle all that apply)
(a) Pressure head at 1 is zero
(b) Pressure head at 2 is zero
(c) Velocity head at 1 Velocity head at 2
(d) Velocity head at 2 Velocity head at 1
(e) Pump head is negative.
(f) Pump head is positive.
(g) Head loss is positive
(h) Head loss is negative
Solution. Correct statements are (b), (f), and (g). Reasoning.
• Pressure at 1 must be less than atmospheric because— The elevation change from the datum to the section 1 would cause pressure to drop below
atmospheric pressure.
— A centrifugal pump creates a negative suction pressure at the pump inlet.
— Thus, the pressure head at 1 is negative reflecting a negative gage pressure.
• Pressure at 2 is 0 gage; thus, the pressure head at 2 is zero.• Apply the continuity equation to a control volume situated between sections 1 and 2 to showthat ̇1 = ̇2Thus, assuming that the pipe diameter is constant one can conclude that the
velocity heads at 1 and 2 are equal.
• Pumps add energy to a flowing fluid. Thus, pump head is always positive.• Head loss is always positive because head loss represents the loss of energy due to viscouseffects. These viscous effect can only cause energy losses, never energy gains.
Problem 7.3. The shaft on a centrifugal pump is spinning at a rate of 3500 rpm. The pump
discharge is 20 gpm and the pump head is 40 ft. What is the torque in the shaft of the pump in
units of ft-lbf? Assume that the mechanical efficiency of the pump is 100% (this means that all of
the power supplied by shaft goes into increasing the energy of the flowing water)
a. 0.0032
b. 0.30
c. 9.80
d. 27.2
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Solution. The correct answer is (b). Reasoning.
1. Apply the power equation:µPower supplied by
the motor shaft
¶=
µPower delivered to
the flowing fluid
¶(1)
= pump
In Eq. (1), there is one unknown variable ( ) Thus, the problem is cracked. Move on to
calculations.
2. List variables and convert units to consistent units:
= 3500 rpm = 3665 radians/second = 3665 s−1
= pump = 40 ft
= 624 lbf ft3 (water, Table F.5, = 59 ◦F = 1atm)
= 20 gpm = 004456 ft3 s
3. Calculate the torque
=pump
=
¡624 lbf ft3
¢ ¡004456 ft3 s
¢(40 ft)
3665 s−1= 0303 ft-lbf
4. Review.
• The torque carried by the shaft is tiny. Little need to do a stress analysis for this situation!• An easy way to do unit conversions is to use the online calculator at www.onlineconversion.com.
Problem 8.1.When a fluid flows in a round pipe, the shear stress on the walls of the pipe depends
on the viscosity and density of the fluid, the mean velocity, the pipe diameter, and on the roughness
of the pipe wall. The wall roughness is characterized by a variable that has units of meters that is
called the sand roughness height. How many −groups are needed to correlate experimental data?(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
Solution. The correct answer is (b). Reasoning.
1. Write the functional equation (The symbol for wall roughness is )
= ( )
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2. Count the number of variables and the number of primary dimensions.
= # variables = 5
= # primary dimensions =3 (because mass, length and time dimensions appear)
3. Apply the -Buckingham theorem
(number of -groups) = − = 5− 3 = 2
Problem 9.1. The sketch identifies terms that appear in the Navier-Stokes Equation.
a. What are the secondary dimensions of each term? Primary dimensions?
b. What does Term A mean?
c. What does Term B mean?
d. What does this equation mean holistically? That is, what is the physical interpretation?
Solution
• The secondary dimensions are force/volume because the derivation started with forces andthen divides each term by volume.
• To find primary dimensions:[Force]
[Volume]=
µ
2
¶µ1
3
¶=
2 · 2
• Physics (each term).
2
2=
(Resultant force due to viscosity)
(Volume of the CV)
(+ )
=
(Resultant force due to pressure + Weight)
(Volume of the CV)
• Physics (equation: holistic interpretation). Picture an infinitesimal CV located at point
( ). The sum of forces acting on the matter inside this CV is zero. That is, the forces
balance like this:
(Resultant force due to viscosity)
(Volume of the CV)=(Resultant force due to pressure + Weight)
(Volume of the CV)
Review:
• To figure out the physics, one can go through the derivation.• This particular form of the Navier-Stokes equation is a way of expressing Newton’s first law:
ΣF = 0
• The general form of the Navier-Stokes equation is a way of expressing Newton’s second law:
ΣF = a
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Problem 9.2. Suppose the roof of an automobile is idealized as a flat plate. Given the data in the
figure, what is the speed of the car in mph? Assume = 20 ◦C and = 1atm
a. 12.6
b. 14.1
c. 16.9
d. 28.1
e. 34.7
Solution. Correct choice is (d). Reasoning.
• Properties: Air ( = 20 ◦C = 1atmTable A.3): = 151× 10−6m2 s• Use Re = 500 000 as the critical Reynolds number.
Re =
= 500 000
=500 000
=(500 000)
¡151× 10−6m2 s¢(06m)
= 12583m s = 281 mph
Problem 10.1. The figure shows flow through two pipes. Case 1 has half the flow of Case 2. Both
cases involve the same length of pipe, the same friction factor, and the same diameter. What is the
ratio of head loss for Case 1 to head loss for Case 2?
(a) 1:4
(b) 1:2
(c) head loss is the same
(d) 2:1
(e) 4:1
Solution. Correct answer is (a). Reasoning:
1. Since the goal involves , start with the Darcy-Weisbach equation and take a ratio.
(case 1)
(case 2)=
(Case 1)
22
(Case 2)
22
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2. Introduce the flow rate equation ( = ) then, cancel terms:
(case 1)
(case 2)=(Case 1)
2
(Case 2)2=(1)
2
(2)2=1
4
Problem 10.2. Water (15 ◦C) flows in a 100 m length of cast iron pipe. The pipe inside diameter
is 0.15 m and the mean velocity is 0.6 m/s.
(a) What is the value of Reynolds number?
(b) What is the value of ?
(c) What is the value of from the Moody diagram?
(d) What is the value of from the Swamee-Jain correlation?
(e) What is the value of head loss?
Solution:
• Properties:
— Water (15 ◦C = 1atmTable F.5): = 114× 10−6m2 s— Relative roughness: Cast iron pipe (Table 10.4): = 026× 10−3m
• Reynolds number (Re)
Re =
=(06m s) (015m)
(114× 10−6m2 s) = 78947
• Relative roughness ()
=026× 10−3m0015m
= 1733× 10−2
• Friction factor () Both the Moody diagram and the Swamee-Jain correlation will give the
same value of
=025£
log10¡
37
+ 574Re09
¢¤2 = 025hlog10
³1733×10−2
37+ 574
7894709
´i2 = 0047• Head loss ( ) Calculate using the Darcy-Weisbach equation
=
2
2= 0047
µ100m
0015m
¶Ã(06m s)
2
2 (981m s2)
!= 575m
Problem 10.3. The sketch shows an idealization of a garden hose of diameter and length
connected to a pipe bib at a residence. Assume that the supply pressure upstream of the valve is
constant. Assume that the faucet valve has no head loss because it is fully open. Thus, the only
head loss is in the garden hose.
(a) Derive an equation for the mean velocity of the water in terms of the friction factor and other
relevant variables. (b) How much will change if is doubled? Assume remains constant.
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Generate Ideas and Make a Plan.
Because the Darcy-Weisbach equation contains the goal start with this equation.
= =
2
2(1)
Because Eq. (1) involves the unknown apply the energy equation. Locate section 1 upstream of
the valve and section 2 at the outlet. Let = . Let supply represent the pressure upstream of
the valve Then, the energy equation reduces to
supply
= (2)
Now, Eqs. (1) and (2) can be combined to reach the goal.
Take Action (Execute the Plan)
1. Combine Eq. (1) and (2):
supply
=
2
2
2. Solve for
=
s2supply
3. Analyze the change in
(state 2)
(state 1)=
q2su p p ly
(state 2)q2su p p ly
(state 1)
=
s (state 1)
(state 2)=
r
2=
r1
2= 07071
Thus if the length is doubled, the velocity will decrease to 70.7% of its initial value.
Problem 11.1.Consider a car that is traveling in a straight line at constant speed.
Case A: The car speed is 40 km/h. There is no wind.
Case B: The car speed is 80 km/h. There is no wind.
Case C: The car speed is 65 km/h. There is a 15 km/h steady head wind.
The coefficient of drag is the same in all three cases.
Which statement(s) are true? (Select all that apply).
(a) (Drag in Case B) = 2(Drag in Case A).
(b) (Drag in Case B) = 4(Drag in Case A).
(c) (Drag in Case B) = 8(Drag in Case A).
(d) (Drag in Case C) (Drag in Case B).
(e) (Drag in Case C) (Drag in Case B).
(f) (Drag in Case C) = (Drag in Case B).
Solution. Answers (b) and (f) are correct. Reasoning.
• Compare drag without wind.
Drag force (at faster speed)
Drag force (at slower speed)=
ref
³ 2
f a s t
2
´ref
³ 2
s l ow
2
´ = 2fast
2slow
(1)
Thus, drag force varies as speed-squared.
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• Compare Cases A and B. Since speed doubles, the drag force (see Eq. 1) goes up by a factorof 4. Answer (b) is correct.
• Compare drag with headwind. In the drag force equation, the speed is the speed of fluid
referenced to the moving object. In Cases B and C, the speed of the air relative to the car is
80 km/h in both cases. Thus, the drag force is the same. Answer (f) is correct.
Problem 11.2 Suppose you are estimating for an American football oriented so that its long
axis is into the wind. You have available Fig. 11.9. Which choice would you make? I would idealize
the football
(a) As a sphere
(b) As a streamlined body
(c) As one of the other bodies on the figure.
Do you thinking that a spinning football (about its long axis) has a different value of drag than a
non-spinning football?
(d) yes, the spinning football will have lower drag
(e) yes, the spinning football has higher drag
(f) no, the spinning football has the same drag as a non spinning football
Solution.
• Idealizing a football. For drag calculations, idealize the object so that the flow pattern aroundthe real object is similar to the flow pattern around the idealized object. Because a football is
designed to be thrown a long distance, it is shaped for low drag. Thus, idealize a football as
a streamlined body.
• Effect of rotation. In football, the quarterback generally tries to throw the football with spinbecause the football will travel farther and more accurately. Part of the rationale for spin is
to stabilize the flight of the football. But, I also guess that the spin will lower the drag.
• Literature review. The answers to the previous two questions were based on intuition. A quickweb search revealed the following data. Watts and Moore (2003) measured drag and reported
that the drag coefficient for a spinning football is slightly lower than that for a non-spinning
football. Both are in the range of 0.05-0.06. These data are similar to the drag coefficients
reported by Rouse (1946) for the case of an ellipsoid with major diameter/minor diameter
similar to the length/diameter for the football. [Watts, R. G., & Moore, G. (2003). The drag
force on an American football. American Journal Of Physics, 71(8), 791.].
Problem 11.3. Consider a bicycle racer that is traveling in a straight line at constant speed. Case
A: The speed is 20 km/h. There is no wind.
Case B: The speed is 40 km/h. There is no wind.
For both cases, is the same and rolling resistance is negligible.
Which statement is true?
(a) (Power in Case B) = (Power in Case A).
(b) (Power in Case B) = 2(Power in Case A).
(c) (Power in Case B) = 4(Power in Case A).
(d) (Power in Case B) = 9(Power in Case A).
Solution. Answer (d) is correct. Reasoning:
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• To analyze power, select the power equation and develop a ratio.
Power (at faster speed)
Power (at slower speed)=(drag) (fast)
(drag) (slow)=
href
³ 2
f a s t
2
´i(fast)
ref
³ 2
s l ow
2
´ = 2fast
2slow
(1)
• Next, introduce the drag force equation and cancel terms:
Power (at faster speed)
Power (at slower speed)=
href
³ 2
f a s t
2
´i(fast)
ref
³ 2
s low
2
´(fast)
= 3fast
3slow
(2)
Thus, power increases as speed cubed.
• Regarding Cases A and B, since speed doubles, the power (see Eq. 1) goes up by a factor of23 = 8.
Problem 12.1. Consider two airplanes A and B flying at identical speeds. Airplane B is at a higher
altitude where the atmospheric pressure and temperature are lower. Which statement is correct?
(a) (Mach number of plane A) (Mach number of plane B)
(b) (Mach number of plane A) (Mach number of plane B)
(c) (Mach number of plane A) = (Mach number of plane B)
Solution. Choice (b) is correct. Reasoning.
1. Define the Mach number
=(Speed of still air relative to the plane)
(Local speed of sound)=
2. Build a ratio
=(Speed of still air relative to the plane)
(Local speed of sound)=()
()=
3. Introduce the speed of sound, and cancel terms. Assume that = because the specific
heat ratio is nearly constant over the temperature range that corresponds to flight elevations.
=
=
pp
=
√√
4. Using the given information about temperature
=
√√
1
Thus, .
Problem 13.1. Suppose you were interested in measuring the velocity at various points in a small
creek. (a) What are the advantages of using a stagnation tube? Disadvantages? (b) What are the
advantages of using a HWA? Disadvantages? (c) Which instrument would you recommend? Why?
Solution
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• Stagnation tube
— Advantages
∗ Low cost∗ Simplicity∗ Mechanical measurement (not electronic) so one does not need to calibrate
— Disadvantages
∗ Cannot measure time variations in velocity (important for characterizing turbulence)∗ Can plug up∗ Need to know the direction of the velocity vector (stagnation tube needs to be alignedwith the velocity variation)
• Hot-wire Anemometer
— Advantages
∗ Can measure turbulent velocity.∗ Can resolve the direction of the velocity vector.∗ Can measure velocity changes with time.
— Disadvantages
∗ Expensive and complex; can take long time to learn∗ Electronic measurement; usually requires calibration∗ Requires electrical power
• Recommendation. Use a stagnation tube unless there are specific reasons for selecting theHWA (such as gathering turbulence data). In this later case, consider alternatives (such as
the acoustic Doppler velocimeter).
Problem 15.1. What is the hydraulic radius for this channel?
a. (4 + 2)
b.(2 + )
c. 4
d. 2
e.
Solution. Answer (a) is correct. Reasoning.
• Define the Hydraulic radius
(Hydraulic Radius) = =(Section area filled with flow)
(Wetted Perimeter)=
• Apply the definition =
=
22
()=
2
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Problem 15.2. Consider uniform flow of water in two channels. Both have the same slope, the
same wall roughness, and the same section area. (note: section B is a regular hexagon) Which
statement is true?
(a) =
(b)
(c)
The correct answer is (b). Reasoning.
Generate Ideas and Make a Plan
• Both sections are the best section for the given channel shape. Thus, look at an equation forflow rate (the Manning equation for example), and notice that the channel with the largest
value of will have the largest flow rate. Thus, the problem can be solved by comparing the
hydraulic radius of each channel.
• Comparing the hydraulic radius will require relating and This can be done by equating
the section areas.
• The plan is
1. Relate y and L by equating the section areas.
2. Compute the hydraulic radius of each section.
3. Take the ratio of hydraulic radii.
Take Action (Execute the Plan)
1. Relate and by equating the section areas.
(Area of rectangular section) = (Area of hexagonal section)
(2) =3√3
42
Thus,
=
s3√3
8 = 064952
2. Compute the hydraulic radius of each section:
(Rectangle) =
=22
4=
2
(Hexagon) =
=
³3√34
´2
3=
√3
4
.
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3. Determine which section has the highest flow by comparing the hydraulic radius:
(Rectangular section)
(Hexagon section)=
2√34
= 11547
= (11547) (064952) = 0750
Thus, the hexagonal section has the largest hydraulic radius which means it will have the
largest flow rate.
Problem 16.1. Regarding the Navier-Stokes equation (Eq. 16.74 in EFM10e), which statements
are true?
a. The terms on the right side are linear.
b. The equation is invariant.
c. The equation applies to all fluids (all liquids and all gases).
Solution
• True: The nonlinear terms are in the acceleration term.• False: This is a coordinate-specific form of the Navier-Stokes equation; applies to Cartesian
coordinates in this case.
• False: This form of the Navier-Stokes equation only applies to flows with constant properties
(i.e. constant density and constant viscosity). Also, the Navier-Stokes equation (by assump-
tion) only applies to fluids that can be idealized as Newtonian fluids.
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