978-0078027680 Chapter 15 Part 1

subject Type Homework Help
subject Pages 14
subject Words 7139
subject Authors John Cimbala, Robert Turner, Yunus Cengel

Unlock document.

This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
page-pf1
15-1
Solutions Manual
for
Fundamentals of Thermal Fluid Sciences
5th Edition
Yunus A. Çengel, John M. Cimbala, Robert H. Turner
McGraw-Hill, 2017
Chapter 15
EXTERNAL FLOW: DRAG AND LIFT
PROPRIETARY AND CONFIDENTIAL
This Manual is the proprietary property of McGraw-Hill Education and protected by copyright and
other state and federal laws. By opening and using this Manual the user agrees to the following
restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly
returned unopened to McGraw-Hill Education: This Manual is being provided only to authorized
professors and instructors for use in preparing for the classes using the affiliated textbook. No
other use or distribution of this Manual is permitted. This Manual may not be sold and may not
be distributed to or used by any student or other third party. No part of this Manual may be
reproduced, displayed or distributed in any form or by any means, electronic or otherwise,
without the prior written permission of McGraw-Hill Education.
page-pf2
15-2
Drag, Lift, and Drag Coefficients
15-1C
Solution We are to discuss how the local skin friction coefficient changes with position along a flat plate in laminar
15-2C
Solution We are to define the frontal area of a body and discuss its applications.
15-3C
Solution We are to define the planform area of a body and discuss its applications.
15-4C
Solution We are to explain when a flow is 2-D, 3-D, and axisymmetric.
page-pf3
15-3
15-5C
Solution We are to discuss the difference between upstream and free-stream velocity.
15-6C
Solution We are to discuss the difference between streamlined and blunt bodies.
15-7C
Solution We are to discuss applications in which a large drag is desired.
15-8C
Solution We are to define drag and discuss why we usually try to minimize it.
15-9C
Solution We are to define lift, and discuss its cause and the contribution of wall shear to lift.
page-pf4
15-10C
Solution We are to explain how to calculate the drag coefficient, and discuss the appropriate area.
Analysis When the drag force FD, the upstream velocity V, and the fluid density
are measured during flow over a
15-11C
Solution We are to define and discuss terminal velocity.
15-12C
Solution We are to discuss the difference between skin friction drag and pressure drag, and which is more significant
for slender bodies.
15-13C
Solution We are to discuss the effect of surface roughness on drag coefficient.
page-pf5
15-5
15-14C
Solution We are to discuss the effect of streamlining, and its effect on friction drag and pressure drag.
15-15C
Solution We are to define and discuss flow separation.
15-16C
Solution We are to define and discuss drafting.
15-17C
Solution We are to discuss how drag coefficient varies with Reynolds number.
page-pf6
15-18
Solution The drag force acting on a car is measured in a wind tunnel. The drag coefficient of the car at the test
conditions is to be determined.
1.164 kg/m3.
Analysis The drag force acting on a body and the drag
coefficient are given by
2
22
and
2AV
F
C
V
ACF D
DDD
where A is the frontal area. Substituting and noting that 1 m/s = 3.6 km/h, the drag coefficient of the car is determined to be
0.29
N 1
m/skg 1
m/s) 6.3/90)(m 65.125.1)(kg/m 164.1(
N) 220(2 2
223
D
C
Discussion Note that the drag coefficient depends on the design conditions, and its value will be different at different
conditions. Therefore, the published drag coefficients of different vehicles can be compared meaningfully only if they are
determined under identical conditions. This shows the importance of developing standard testing procedures in industry.
15-19
Solution The resultant of the pressure and wall shear forces acting on a body is given. The drag and the lift forces
acting on the body are to be determined.
Analysis The drag and lift forces are determined by decomposing
V
90 km/h
FD
page-pf7
15-7
15-20
Solution The total drag force acting on a spherical body is measured, and the pressure drag acting on the body is
N 3.09.42.5
pressure,friction, DDD FFF
2
2V
V
Air
V
D = 12 cm
page-pf8
15-22E
Solution The frontal area of a car is reduced by redesigning. The amount of fuel and money saved per year as a result
are to be determined.
Assumptions 1 The car is driven 12,000 miles a year at an average speed of 55
lbf 9.40
ft/slbm 32.2
lbf 1
mph 1
ft/s 1.4667
2
mph) 55)(lbm/ft 075.0(
)ft 18(3.0 2
2
23
2
D
F
page-pf9
15-9
15-23E
rho=0.075 "lbm/ft3"
V=55*1.4667 "ft/s"
Eff=0.32
Price=2.20 "$/gal"
efuel=20000 "Btu/lbm"
page-pfa
15-10
15-24
Solution A circular sign is subjected to high winds. The drag force acting on the sign and the bending moment at the
2). The density of air at 100 kPa and 10°C = 283 K is
3
3kg/m 231.1
K) /kg.K)(283mkPa (0.287
kPa 100
RT
P
Analysis The frontal area of a circular plate subjected to normal flow is A
= D2/4. Then the drag force acting on the sign is
N 231
2
23
2
2
m/skg 1
N 1
2
m/s) 6.3/150)(kg/m 231.1(
]4/m) 5.0()[1.1(
2
V
ACF DD
1.5 m
150 km/h
50 cm
SIGN
page-pfb
15-25E
Solution We are to estimate how much money is wasted by driving with a pizza sign on a car roof.
Properties
fuel = 50.2 lbm/ft3, HVfuel = 1.53 107 ftlbf/lbm,
air = 0.07518 lbm/ft3,
air = 1.227 10-5 lbm/fts
Analysis First some conversions: V = 45 mph = 66.0 ft/s and the total distance traveled in one year = L = 10,000
page-pfc
15-12
15-26
Solution An advertisement sign in the form of a rectangular block that has the same frontal area from all four sides is
mounted on top of a taxicab. The increase in the annual fuel cost due to this sign is to be determined.
2.2 (Table 15-1).
Analysis Noting that 1 m/s = 3.6 km/h, the drag force acting on the sign is
N 61.71
m/skg 1
N 1
2
m/s) 6.3/50)(kg/m 25.1(
)m 3.09.0)(2.2(
22
23
2
2
V
ACF DD
Noting that work is force times distance, the amount of work done to overcome this drag force and the required energy input
for a distance of 60,000 km are
kJ/year 1054.1
28.0
kJ/year 1030.4
kJ/year 1030.4km/year) N)(60,000 61.71(
7
6
car
drag
6
drag
W
E
LFW
in
D
Then the amount and cost of the fuel that supplies this much energy are
L/year 509
kJ/kg) 000,42/()kJ/year 10(1.54
/HV
fuel ofAmont
7
in
fuel
E
m
page-pfd
15-27E
Solution A person who normally drives at 55 mph now starts driving at 75 mph. The percentage increase in fuel
consumption of the car is to be determined.
Assumptions 1 The fuel consumption is approximately proportional to the drag force on a level road (as stated). 2 The
2.543
3
3
1
3
2
drag1
drag2
55
75
V
V
W
W
1.86 an increase of 86%. This is large, especially with the high cost of gasoline these days.
15-28
Solution A submarine is treated as an ellipsoid at a specified length and diameter. The powers required for this
submarine to cruise horizontally in seawater and to tow it in air are to be determined.
Assumptions 1 The submarine can be treated as an ellipsoid. 2 The flow is turbulent. 3 The drag of the towing rope is
negligible. 4 The motion of submarine is steady and horizontal.
m/skg 1000
2
22
In air:
kN 1575.0
m/skg 1000
kN 1
2
m/s) 11.11)(kg/m 30.1(
]4/m) 5()[1.0(
22
23
2
2
V
ACF DD
Noting that power is force times velocity, the power needed to overcome this drag force is
kW 1
page-pfe
15-29E
Solution A billboard is subjected to high winds. The drag force acting on the billboard is to be determined.
Assumptions 1 The flow of air is steady and incompressible. 2 The drag force on the supporting poles are negligible. 3
page-pff
15-15
15-30
Solution A semi truck is exposed to winds from its side surface. The wind velocity that will tip the truck over to its
side is to be determined.
Assumptions 1 The flow of air in the wind is steady and incompressible. 2 The edge effects on the semi truck are
page-pf10
15-31
Solution A bicyclist is riding his bicycle downhill on a road with a specified slope without pedaling or breaking. The
0.45 m2 and 1.1 in the upright position, and 0.4 m2 and 0.9 on the racing position.
Analysis The terminal velocity of a free falling object is reached when the drag force equals the component of the
total weight (bicyclist + bicycle) in the flow direction,
sin
total
WFD
sin
2total
2
gm
V
ACD
Solving for V gives
AC
gm
V
D
sin2
total
The terminal velocities for both positions are obtained by substituting the
given values:
Upright position:
km/h 70
km/h 69.7m/s 4.19
)kg/m 25.1)(m 45.0(1.1
kg)sin8 1570)(m/s 81.9(2
32
2
V
Racing position:
km/h 82
km/h 81.8m/s 7.22
)kg/m 25.1)(m 4.0(9.0
kg)sin8 1570)(m/s 81.9(2
32
2
V
Discussion Note that the position of the bicyclist has a significant effect on the drag force, and thus the terminal
velocity. So it is no surprise that the bicyclists maintain the racing position during a race.
15-32
Solution The pivot of a wind turbine with two hollow hemispherical cups is stuck as a result of some malfunction.
For a given wind speed, the maximum torque applied on the pivot is to be determined.
Assumptions 1 The flow of air in the wind is steady and incompressible. 2 The air
mN 0.113 m) N)(0.20 283.0848.0()( 1212max LFFLFLFM DDDD
40 cm
page-pf11
15-17
15-33
Solution The previous problem is reconsidered. The effect of wind speed on the torque applied on the pivot as the
CD2=1.2 "Plain frontal area"
"rho=density(Air, T=T, P=P)" "kg/m^3"
D=0.08 "m"
L=0.25 "m"
A=pi*D^2/4 "m^2"
page-pf12
15-18
15-34E
Solution A spherical tank completely submerged in fresh water is being towed by a ship at a specified velocity. The
required towing power is to be determined.
Assumptions 1 The flow is turbulent so that the tabulated value of the drag coefficient can
be used. 2 The drag of the towing bar is negligible.
15-35
Solution The power delivered to the wheels of a car is used to overcome aerodynamic drag and rolling resistance.
For a given power, the speed at which the rolling resistance is equal to the aerodynamic drag and the maximum speed of the
car are to be determined.
Assumptions 1 The air flow is steady and incompressible. 2 The bearing friction is negligible. 3 The drag and rolling
m/s 8.32
m/skg 1
N 1
2
)kg/m 20.1(
)m 8.1)(32.0( N 372.8
22
23
2
2
V
V
V
ACF DD
(or 118 km/h)
W 1
m/sN 1
W 000,80372.8
m/s kg1
N 1
2
) kg/m20.1(
)m 8.1)(32.0( 2
33
2V
V
or,
000,80372.83456.0 3VV
5 ft
page-pf13
15-36
Solution The previous problem is reconsidered. The effect of car speed on the required power to overcome (a) rolling
rho=1.20 "kg/m3"
g=9.81 "m/s2"
A=1.8 "m2"
C_D=0.32
page-pf14
15-20
15-37
Solution A garbage can is found in the morning tipped over due to high winds the night before. The wind velocity
during the night when the can was tipped over is to be determined.
N 7.1029
m/skg 1
N 1
)m 6998.0)(m/s )(9.81kg/m 150(
m 6998.0m) 1.1](4/m) 90.0([]4/[
2
323
222
V
V
gmgW
HD
N 5.842
m 1.1
m) N)(0.90 7.1029(
0)2/()2/( 0
contact
H
WD
FDWHFM DD
Noting that the frontal area is DH, the wind velocity that will cause this drag force is determined to be
m/s 10.44
m/skg 1
N 1
2
)kg/m 25.1(
]m 90.01.1)[7.0( N 5.842
22
23
2
2
V
VV
ACF DD
which is equivalent to a wind velocity of V = 44.10 3.6 = 159 km/h.
Discussion The analysis above shows that under the stated assumptions, the wind velocity at some moment exceeded
159 km/h. But we cannot tell how high the wind velocity has been. Such analysis and predictions are commonly used in
forensic engineering.
D = 0.9 m

Trusted by Thousands of
Students

Here are what students say about us.

Copyright ©2022 All rights reserved. | CoursePaper is not sponsored or endorsed by any college or university.