Resolução Capítulo 11 hibeller

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1134
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Solution
Oil is considered to be incompressible. The flow is steady.
The transition from a laminar boundary layer occurs at a critical Reynolds number 
of (Rex)cr = 5(105).
(Rex)cr =
Uxcr
n
=
Uxcr
mg>g =
gUxcr
mg
5(105) =
(55.1 lb>ft3)(3 ft>s)xcr
31.40(10-3) lb # s>ft4 (32.2 ft>s2)
xcr = 136.36 ft = 136 ft Ans.
11–1. Oil flows with a free-stream velocity of U = 3 ft>s 
over the flat plate. Determine the distance xcr to where 
the  boundary layer begins to transition from laminar 
to  turbulent flow. Take mo = 1.40110-32 lb # s>ft and 
go = 55.1 lb>ft3.
x
U
Ans:
136 ft
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11–2. Water at 15°C flows with a free-stream velocity of 
U = 2 m>s over the flat plate. Determine the shear stress 
on the surface of the plate at point A.
Solution
Water is considered to be incompressible. The flow is steady.
From Appendix A, m = 1.15(10-3) N # s>m2 and n = 1.15(10-6) m2>s. Thus, 
Rex =
Ux
n
=
(2 m>s)(0.2 m)
1.15(10-6) m2>s
= 3.478(105)
Since Rex 6 (Rex)cr = 5(105), the boundary layer is still laminar. Thus,
t0 = 0.332m aU
x
b2Rex
 = 0.33231.15(10-3) N>m24a2 m>s
0.2 m
b23.478(105)
 = 2.25 N>m2 Ans.
A
200 mm
U
Ans:
2.25 Pa
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Solution
Air is considered to be incompressible. The flow is steady.
Here,
u
U
=
y
y + 0.01
u = U a y
y + 0.01
b = a 15y
y + 0.01
b m>s
At y = 0.1 m,
u � y = 0.1 m =
15(0.1)
0.1 + 0.01
= 13.6 m>s Ans.
At y = 0.3 m,
u � y = 0.3 m =
15(0.3)
0.3 + 0.01
= 14.5 m>s Ans.
11–3. The boundary layer for wind blowing over 
rough  terrain can be approximated by the equation 
u>U = 1 y>( y + 0.01)2 , where y is in meters. If the free-
stream velocity of the wind is 15 m>s, determine the velocity 
at an elevation y  = 0.1 m and at y = 0.3 m from the 
ground surface. y
15 m/s
Ans:
u � y = 0.1 m = 13.6 m>s
u � y = 0.3 m = 14.5 m>s
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Solution
The mixture is considered to be incompressible. The flow is steady.
The transition from a laminar boundary layer occurs at a critical Reynolds number 
of (Rex)cr = 5(105).
(Rex)cr =
Uxcr
n
5(105) =
(0.8 m>s)xcr
42(10-6) m2>s
xcr = 26.25 m
Since xcr 7 L = 1.5 m, the boundary layer for the entire length of the plate is 
 laminar. The maximum thickness occurs at the end of the plate where x = L = 1.5 m. 
The Reynolds number at this point is 
Rex =
Ux
n
=
(0.8 m>s)(1.5 m)
42(10-6) m2>s
= 2.857(104)
Then,
dmax = d �x = 1.5 m =
5.0x2Rex
=
5.0(1.5 m)22.857(104)
= 0.04437 m = 44.4 mm Ans.
*11–4. An oil–gas mixture flows over the top surface of 
the plate that is contained in a separator used to process 
these two fluids. If the free-stream velocity is 0.8 m>s, 
determine the maximum boundary layer thickness over the 
plate’s surface. Take n = 42110-62 m2>s. 1.5 m
0.8 m/s
0.75 m
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Solution
The mixture is considered to be incompressible. The flow is steady.
The transition from a laminar boundary layer occurs at a critical Reynolds number 
of (Rex)cr = 5(105).
(Rex)cr =
Uxcr
n
5(105) =
(0.8 m>s)xcr
42(10-6) m2>s
xcr = 26.25 m
Since xcr 7 L = 1.5 m, the boundary layer for the entire length of the plate is 
 laminar. Here, the Reynolds number at x = L = 1.5 m is
ReL =
UL
n
=
(0.8 m>s)(1.5 m)
42(10-6) m2>s
= 2.857(104)
Then,
FD =
0.664brU 2L2ReL
=
0.664(0.75 m)(910 kg>m3)(0.8 m>s)2(1.5 m)22.857(104)
= 2.57 N Ans.
11–5. An oil-gas mixture flows over the top surface of 
the plate that is contained in a separator used to process 
these two fluids. If the free-stream velocity is 0.8 m>s, 
determine the friction drag acting on the surface of the 
plate. Take n = 42110-62 m2>s and r = 910 kg>m3. 1.5 m
0.8 m/s
0.75 m
Ans:
2.57 N
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s i
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Solution
The air is considered to be incompressible. The flow is steady.
From Appendix A, r = 0.00237 slug>ft2 and n = 0.158(10-3) ft2>s. The 
 transition from a laminar boundary layer occurs at a critical Reynolds number of 
(Rex)cr = 5(105).
(Rex)cr =
Uxcr
n
5(105) =
(6 ft>s)x cr
0.158(10-3) ft2>s
x cr = 13.17 ft
Since x cr 7 L = 12 ft, theboundary layer for the entire length of the signboard is 
laminar. Here, the Reynolds number at x = L = 12 ft is
ReL =
UL
n
=
(6 ft>s)(12 ft)
0.158(10-3) ft2>s
= 4.557(105)
Then,
FD =
0.664brU 2L2Re L
=
0.664(6 ft)(0.00237 slug>ft3)(6 ft>s)2(12 ft)24.557(105)
= 0.00604 lb Ans.
11–6. Wind flows along the side of the rectangular sign. 
If the air is at a temperature of 60°F and has a free-stream 
velocity of 6 ft>s, determine the friction drag on the front 
surface of the sign.
12 ft
6 ft6 ft/s
Ans:
0.00604 lb
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Solution
The polymer is considered to be incompressible. The flow is steady.
The Reynolds number at x = 0.5 m is
Rex =
Ux
n
=
U(0.5 m)
4.68(10-6) m2>s
= 1.0684(105)U
Then,
d =
5.0x2Rex
0.01 m =
5.0(0.5 m)21.0684(105)U
U = 0.585 m>s Ans.
11–7. A flat plate is to be coated with a polymer. If the 
thickness of the laminar boundary layer that occurs during 
the coating process at a distance of 0.5 m from the plate’s 
front edge is 10 mm, determine the free-stream velocity of 
this fluid. Take n = 4.68110-62 m2>s.
Ans:
0.585 m>s
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Solution
Both water and air are to be incompressible. The flow is steady. 
From Appendix A, nw = 1.00(10- 6) m2 > s and na = 15.1(10- 6) m2>s. Thus, the 
Reynolds numbers for water and air at the end of the plate x = L = 0.4 m are
(ReL)w =
UL
nw
=
(0.8 m>s)(0.4 m)
1.00(10- 6) m2>s
= 3.2(105)
(ReL)a =
UL
na
=
(0.8 m>s)(0.4 m)
15.1(10- 6) m2>s
= 2.1192(104)
Since (ReL)w 6 (Rex)cr and (ReL)a 6 (Rex)cr, where (Rex)cr = 5(105), the 
 boundary layers for both water and air are laminar. For x = L = 0.4 m,
dw =
5.0x2(Rex)w
=
5.0(0.4 m)23.2(105)
= 0.003536 m = 3.54 mm Ans.
da =
5.0x2(Rex)a
=
5.0(0.4 m)22.1192(104)
= 0.01374 m = 13.7 mm Ans.
*11–8. Compare the thickness of the boundary layer of 
water with air at the end of the 0.4-m-long flat plate. Both 
fluids are at 20°C and have a free-stream velocity of 
U = 0.8 m>s.
0.4 m
0.8 m/s 0.8 m/s
y
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Solution
The liquid is considered to be incompressible. The flow is steady.
The Reynolds number at the x = L is ReL =
rUL
m
 and at the exit d =
a
2
. Thus,
d =
5.0x2Rex
 ; 
a
2
=
5.0xBrUx
m
x =
rUa2
100m
 Ans.
11–9. A liquid having a viscosity m, a density r, and a free-
stream velocity U flows over the plate. Determine the distance 
x where the boundary layer has a disturbance thickness that is 
one-half the depth a of the liquid. Assume laminar flow. a
x
U
Ans:
x =
rUa2
100m
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Solution
The fluid is considered to be incompressible. The flow is steady.
The Reynolds number at x = 0.5 m and 1 m can be determined using
Rex � x = 0.5 m =
Ux
n
=
U(0.5 m)
n
=
0.5U
n
and
Rex � x = 1 m =
Ux
n
=
U(1 m)
n
=
U
n
At x = 0.5 m, d = 0.01 m. Thus,
d =
5.0x2Rex
; 0.01 m =
5.0(0.5 m)A0.5U
n
 
U
n
= 125 000
Thus, at x = 1 m, Rex =
U
n
= 125 000. Then,
d =
5.0x2Rex
=
5.0(1 m)2125 000
= 0.01414 m = 14.1 mm Ans.
11–10. A fluid has laminar flow and passes over the flat 
plate. If the thickness of the boundary layer at a distance of 
0.5 m from the plate’s edge is 10 mm, determine the 
boundary layer thickness at a distance of 1 m.
0.5 m
10 mm
6 m/s
Ans:
14.1 mm
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1144
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Solution
The air is considered to be incompressible. The flow is steady.
From Appendix A, n = 18.9(10- 6) m2>s. Thus, the Reynolds number at x = 4 m is
Rex =
Ux
n
=
(0.5 m>s)(4 m)
18.9(10- 6) m2>s
= 1.0582(105)
Since Rex 6 (Rex)cr = 5(105), the boundary layer is laminar throughout the entire 
length of the duct.
Thus, the displacement thickness is 
d* =
1.721x2Rex
=
1.721(4 m)21.0582(105)
= 0.02116 m = 21.16 mm
The dimension of the square duct at x = 4 m is
a = 200 mm + 2d*
= 200 mm + 2(21.16 mm)
= 242 mm Ans.
11–11. Air at 60°C flows through the very wide duct. 
Determine the required dimension a of the duct at x = 4 m 
so that the central 200-mm core flow velocity maintains the 
constant free-stream velocity of 0.5 m>s.
x � 4 m
200 mm a 0.5 m/s
0.5 m/s
Ans:
242 mm
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s n
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pe
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itte
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1145
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Oil is considered to be incompressible. The flow is steady.
The transition from a laminar boundary layer occurs at a critical Reynolds number 
of (Rex)cr = 5(105).
(Rex)cr =
Uxcr
n
5(105) =
(6 m>s)xcr
40(10- 6) m2>s
xcr = 3.33 m
Since xcr 7 L = 0.05 m, the boundary layer for the entire length of the fin is 
 laminar. Here, the Reynolds number at x = L = 0.05 m is
ReL =
UL
n
=
(6 m>s)(0.05 m)
40(10- 6) m2>s
= 7500
The drag force of both sides of the fin is
FD = 2£ 0.664brU 2L2ReL
§ = 2£ 0.664(0.2 m)(900 kg>m3)(6 m>s)2(0.05 m)27500
§
= 4.97 N Ans.
*11–12. Oil confined in a channel flows past the diverter 
fin at U = 6 m>s. Determine the friction drag acting on both 
sides of the fin. Take no = 40110-62 m2>s and 
ro = 900 kg>m3. Neglect end effects. U
50 mm
200 mm
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 S
tat
es
 co
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law
s 
an
d i
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rov
ide
d s
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ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
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tio
n o
r 
sa
le 
of 
an
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 of
 th
is 
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 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
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nte
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ty 
of 
the
 w
ork
 an
d i
s n
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pe
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itte
d. 
1146
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Air is considered to be incompressible. The flow is steady.
From Appendix A, n = 0.169(10- 3) ft2>s for air at T = 80° F. Thus, the Reynolds 
number at x = 0.2 ft is
Rex =
Ux
n
=
(4 ft>s)(0.2 ft)
0.169(10- 3) ft2>s
= 4733.73
Since Rex 6 (Rex)cr = 5(105), the boundary layer up to x = 0.2 ft is still laminar. 
Thus, its thickness at this point is 
d =
5.0x2Rex
=
5.0(0.2 ft)24733.73
= (0.01453 ft)a12 in.
1 ft
b = 0.174 in. Ans.
Also, the velocity of a particle at the point x = 0.2 ft and y = 3(10- 3) ft can be 
determined using Blasius solution. Here,
y
x
2Rex =
3(10- 3) ft
0.2 ft
 24733.73 = 1.032
Interpolating the values in the table gives
u
U
 ≈ 0.3396
u = 0.3396(4 ft>s) = 1.36 ft>s (approx.) Ans.
11–13. Air at 80°F and atmospheric pressure has a free-
stream velocity of 4 ft>s. If it passes along the surface of a 
smooth glass window of a building, determine the thickness 
of the boundary layer at a distance of 0.2 ft from the leading 
edge of the window. Also, what is the velocity of air 0.003 ft 
away from the window’s surface at this point?
Ans:
d = 0.174 in.
u = 1.36 ft>s (approx.)
This
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cte
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nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1147
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Water is considered to be incompressible. The flow is steady.
From Appendix A, n = 0.664(10- 6) m2>s for water at T = 40° C. Thus, the 
Reynolds number in terms of x is
Rex =
Ux
n
=
(0.3 m>s)x
0.664(10- 6) m2>s
= 4.5181(105)x
At x = 0.2 m and 0.4 m,
Rex � x = 0.2 m = c 4.5181(105) d (0.2 m) = 9.0361(104)
Rex � x = 0.4 m = c 4.5181(105) d (0.4 m) = 1.8072(105)
Since Rex � x = 0.4 m 6 (Rex)cr = 5(105), the boundary layer up to x = 0.4 is still 
 laminar. Thus, its thickness is 
d � x = 0.2 m =
5.0x2Rex
=
5.0(0.2 m)29.036(104)
= 0.003327 m = 3.33 mm Ans.
d � x = 0.4 m =
5.0x2Rex
=
5.0(0.4 m)218.072(104)
= 0.004705 m = 4.70 mm Ans.
11–14. Water at 40°C has a free-stream velocity of 0.3 m>s. 
Determine the boundary layer thickness at x = 0.2 m and at 
x = 0.4 m on the flat plate.
x 
0.4 m
0.3 m/s
Ans:
d � x = 0.2 m = 3.33 mm
d � x = 0.4 m = 4.70 mm
This
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nit
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tat
es
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py
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law
s 
an
d i
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ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
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he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
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ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1148
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Water is considered to be incompressible. The flow is steady.
From Appendix A, m = 0.659(10- 3) N # m>s and n = 0.664(10- 6) m2>s for water 
at T = 40° C. Thus, the Reynolds number in terms of x is 
Rex =
Ux
n
=
(0.3 m>s)x
0.664(10- 6) m2>s
= 4.5181(105)x
At x = 0.2 m and 0.4 m,
Rex � x = 0.2 m = c 4.5181(105) d (0.2 m) = 9.0361(104)
Rex � x = 0.4 m = c 4.5181(105) d (0.4 m) = 1.8072(105)
Since Rex � x = 0.4 m
6 (Rex)cr = 5(105), the boundary layer up to x = 0.4 is still 
 laminar. Thus, the shear stress on the plate’s surface is
t0 � x = 0.2 m
= 0.332m aU
x
b2Rex
= 0.332 c 0.659(10- 3) N # s>m2 d a0.3 m>s
0.2 m
b29.0361(104)
= 0.0987 N>m2 Ans.
t0 � x = 0.4 m = 0.332m aU
x
b2Rex
= 0.332 c 0.659(10- 3) N # s>m2 d a0.3 m>s
0.4 m
b21.8072(105)
= 0.0698 N>m2 Ans.
11–15. Water at 40°C has a free-stream velocity of 0.3 m>s. 
Determine the shear stress on the plate’s surface at x = 0.2 m 
and at x = 0.4 m.
x 
0.4 m
0.3 m/s
Ans:
t0 � x = 0.2 m
= 0.0987 Pa
t0 � x = 0.4 m
= 0.0698 Pa
This
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tat
es
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py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1149
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Water is considered to be incompressible. The relative flow is steady.
From Appendix A, n = 12.2(10- 6) ft2>s for water at T = 60° F. Thus, the Reynolds 
number at the trailing edge of the rudder (x = L = 1.75 ft) is
Rex =
Ux
n
=
(0.7 ft>s)(1.75 ft)
12.2(10- 6) ft2>s
= 1.0041(105)
Since Rex 6 (Rex)cr = 5(105), the boundary layer is laminar for the entire length 
of the rudder. Thus, its thickness and displacement thickness at the trailing edge are
d =
5.0x2Rex
=
5.0(1.75 ft)21.0041(105)
= (0.02761 ft) a12 in.
1 ft
b = 0.331 in. Ans.
and
d* =
1.721x2Rex
=
1.721(1.75 ft)21.0041(105)
= (0.00951 ft) a12 in.
1 ft
b = 0.114 in. Ans.
*11–16.The boat is traveling at 0.7 ft>s through still water 
having a temperature of 60°F. If the rudder can be assumed 
to be a flat plate, determine the boundary layer thickness at 
the trailing edge A. Also, what is the displacement thickness 
of the boundary layer at this point?
2 ft
1.75 ft
A
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law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1150
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Water is considered to be incompressible. The relative flow is steady.
From Appendix A, n = 12.2(10- 6) ft2>s for water at T = 60° F. Thus, the Reynolds 
number at the trailing edge of the rudder (x = L = 1.75 ft) is
Rex =
Ux
n
=
(0.7 ft>s)(1.75 ft)
12.2(10- 6) ft2>s
= 1.0041(105)
Since Rex 6 (Rex)cr = 5(105), the boundary layer is laminar for the entire length 
of the rudder. Therefore, the frictional drag force on both surfaces of the rudder is
F = 2£ 0.664brU 2L2ReL
§
= 2£ 0.664(2 ft)(1.939 slug>ft3)(0.7 ft>s)2(1.75 ft)21.004(105)
§
= 0.0139 lb Ans.
11–17. The boat is traveling at 0.7 ft>s through water having 
a temperature of 60°F. If the rudder can be assumed to be a 
flat plate having a height of 2 ft and a length of 1.75 ft, 
determine the friction drag acting on both sides of the 
rudder.
2 ft
1.75 ft
A
Ans:
0.0139 lb
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tat
es
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py
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1151
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Air is considered to be incompressible. The flow is steady.
We will assume the boundary layer is laminar. From Appendix A, 
n = 0.147(10- 3) ft2>s at T = 40° F. Thus, the Reynolds number in terms of x is
Rex =
Ux
n
=
(0.6 ft>s)x
0.147(10- 3) ft2>s
= 4081.63x
Here, the requirement is d =
1.5
12
 ft. Thus,
d =
5.0x2Rex
; 0.125 ft =
5.0x24081.63x
 x = 2.5510 ft = 2.55 ft Ans.
Using this result,
Rex = 4081.63(2.5510 ft) = 1.041(104)
Since Rex 6 (Rex)cr = 5(103), the boundary layer is laminar as assumed.
11–18. Air at a temperature of 40°F flows at 0.6 ft>s over 
the plate. Determine the distance x where the disturbance 
thickness of the boundary layer becomes 1.5 in.
x
0.6 ft/s
3 in.
Ans:
2.55 ft
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law
s 
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d i
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ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1152
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The point is considered to be incompressible. The relative flow is steady.
The Reynolds number at x = L = 0.5 m is
Rex =
Ux
n
=
(3 m>s)(0.5 m)
42(10- 6) m2>s
= 0.35714(105)
Since Rex 6 (Rex)cr = 5(105), the boundary layer is laminar throughout the 
entire length of the bar. Thus, the total frictional force on the bar is
FD = Σ
0.664brU 2L2ReL
=
0.664(920 kg>m3)(3 m>s)2(0.5 m)20.35714(105)
 [2(0.05 m) + 2(0.002 m)]
= 1.51 N Ans.
11–19. Determine the friction drag on the bar required to 
overcome the resistance of the paint if the force F lifts the 
bar at 3 m>s. Take r = 920 kg>m3 and n = 42(10-6) m2>s.
F
500 mm
2 mm
50 mm
3 m/s
Ans:
1.51 N
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law
s 
an
d i
s p
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ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1153
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
We will assume that steady flow occurs, and the air is incompressible. The Reynolds 
number of the flow at the trailing edge (x = 2 ft) is 
ReL =
raUL
ma
=
(0.00257 slug>ft3)(25 ft>s)(2 ft)
0.351(10-6) lb # s>ft2
= 3.661(105)
Since ReL 6 5(105), laminar flow persist within the boundary layer. Thus, the 
 frictional drag force on the fin surface can be determined using Eq. 11–11.
FDf =
0.664braU
2L2ReL
=
0.664(0.3 ft)(0.00257 slug>ft3)(25 ft>s)2(2 ft)23.661(105)
= 0.001058 lb
Since this force acts on two surfaces, Fig. a, the total drag force is
(FDf)T
= 2 FDf = 2(0.001058 lb) = 0.00212 lb Ans.
*11–20. The diverter fin extends 2 ft within the air duct to 
partition the flow through two separate conduits. Determine 
the friction drag on the fin if it is 0.3 ft wide and the velocity 
of the air is 25 ft>s. Take ra = 0.00257 slug>ft3 and 
ma = 0.351(10-6) lb # s>ft2.
FDf
(a)
FDf
2 ft
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es
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s 
an
d i
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ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1154
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
We will assume that steady flow occurs, and crude oil is incompressible. Appendix A 
gives rco = 880 kg>m3 and mco = 30.2(10-3) N # s>m2. The Reynold number as a 
function of x is 
Rex =
rcoUx
mco
=
(880 kg>m3)(10 m>s) x
30.2(10-3) N # s>m2
= 2.914(105)x
At x = L = 1.5 m, ReL = 2.914(105)(1.5) = 4.371(105) 6 5(105). Thus, laminar 
flow persist within the boundary layer. For the boundary thickness
d =
5.02Rex
 x =
5.022.914(105)x
 x = c 9.2626(10-3)x
1
2 d m
x(m) 0 0.25 0.5 0.75 1.0 1.25 1.5
d(mm) 0 4.63 6.55 8.02 9.26 10.36 11.34
The plot of d vs x is show in Fig. a.
11–21. Crude oil at 20°C flows over the surface of the flat 
plate that has a width of 0.7 m. If the free-stream velocity is 
U = 10 m>s, plot theboundary layer thickness and the shear-
stress distribution along the plate. What is the friction drag 
on the plate?
0
δ (m)
x (mm)
5
10
15
0.25 0.50 0.75 1.0 1.25 1.50 0
50
10
150
0.25 0.50 0.75 1.0 1.25 1.50
(a) (b)
x (mm)
� 0 ( (N m2
10 m/s
1.5 m
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s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1155
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–21. Continued 
For the shear stress,
 t0 = 0.332m aU
x
b2Rex
 = 0.332330.2(10-3) N # s>m24a10 m>s
x
b22.914(105)x
 = a54.12
x
1
2
b N>m2
x(m) 0 0.125 0.25 0.50 0.75 1.0 1.25 1.5
t0(N>m2) ∞ 153.08 108.25 76.54 62.50 54.12 48.41 44.19
The plot of t0 vs x is shown in Fig. b.
For the frictional drag force,
 FDf =
0.664brcoU 2L2ReL
 =
0.664(0.7 m)(880 kg>m3)(10 m>s)2(1.5 m)24.371(105)
 = 92.8 N Ans.
0
δ (m)
x (mm)
5
10
15
0.25 0.50 0.75 1.0 1.25 1.50 0
50
10
150
0.25 0.50 0.75 1.0 1.25 1.50
(a) (b)
x (mm)
� 0 ( (N m2
Ans:
92.8 N
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1156
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
We will assume that steady flow occurs and castor oil is incompressible. The Reynolds 
number as a function of x is 
Rex =
rcoUx
mco
=
(960 kg>m3)(2 m>s) x
985(10-3) N # s>m2
= 1949.24x
At x = L = 2 m, ReL = 1949.24(2) = 3.898(103) 6 5(103). Thus, laminar flow 
persist within the boundary layer. For the boundary thickness, 
d =
5.02Rex
 x =
5.021949.24x
 x = (0.1132x
1
2 ) m
x(m) 0 0.5 1.0 1.5 2.0
d(mm) 0 80.08 113.25 138.70 160.16
The plot of d vs x is show in Fig. a.
11–22. Castor oil flows over the surface of the flat plate 
at a free-stream speed of 2 m>s. The plate is 0.5 m wide and 
1 m long. Plot the boundary layer and the shear stress 
versus x. Give values for every 0.5 m. Also calculate the 
friction drag on the plate. Take rco = 960 kg>m3 and 
mco = 985110-32 N # s>m2.
0
δ (mm)
x (m)
50
150
100
200
0.50 1.0 1.5 2.0
(a)
0
20
10
40
50
30
60
0.5 1.0 1.5 2.0
(b)
x (m)
� 0 ( (N m2
2 m/s
2 m
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1157
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–22. Continued
For the shear stress
 t0 = 0.332m aU
x
b2Rex
 = 0.3323985(10-3) N # s>m24a2 m>s
x
b21949.24x
 = a28.876
x
1
2
b N>m2
x(m) 0 0.5 1.0 1.5 2.0
t0(N>m2) ∞ 40.84 28.88 23.58 20.42
The plot of t0 vs x is shown in Fig. b.
For the frictional drag force,
 FDf =
0.664brcoU 2L2ReL
 =
0.664(0.5 m)(960 kg>m3)(2 m>s)2(2 m)23.898(103)
 = 40.8 N Ans.
0
δ (mm)
x (m)
50
150
100
200
0.50 1.0 1.5 2.0
(a)
0
20
10
40
50
30
60
0.5 1.0 1.5 2.0
(b)
x (m)
� 0 ( (N m2
Ans:
40.8 N
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1158
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The fluid is considered to be incompressible. The flow is steady.
Here, 
u
U
=
y
d
. Substituting this result into the momentum integral equation
 t0 = rU 2 
d
dx L
d
0
u
U
 a1 -
u
U
bdy
 t0 = rU 2 
d
dx L
d
0
y
d
 a1 -
y
d
bdy
 t0 = rU 2 
d
dx L
d
0
ay
d
-
y2
d2 bdy
 t0 = rU 2 
d
dx
 ad
6
b
 t0 =
rU 2
6
 
dd
dx
 (1)
For a laminar boundary layer, Newton’s law of viscosity applies. Thus,
t0 = u 
du
dy
`
y = 0
= u 
d
dy
 cU ay
d
b d =
mU
d
 (2)
Equating Eqs. (1) and (2),
 
mU
d
=
rU 2
6
 
dd
dx
 ddd =
6m
rU
 dx
At the leading edge of the plate, x = 0 and d = 0. Therefore,
L
d
0
ddd =
6m
rU L
x
0
dx
d2
2
`
d
0
=
6m
rU
 x `
x
0
d = B 12m
rU
 x
Substituting this result into Eq. (2),
t0 =
mUB 12m
rU
x
= mUB rU
12mx
= 0.289m aU
x
bBrUx
m
However, Rex = BrUx
m
. Then,
 t0 = 0.289m aU
x
b2Rex Ans.
Note: Compare this result with the one obtained using Blasius’s solution.
11–23. Assume the boundary layer has a velocity profile 
that is linear and defined by u = U1y>d2 . Use the 
momentum integral equation to determine t0 for the fluid 
passing over the flat plate.
U U
y u � U(y/d)d
Ans:
t0 = 0.289m aU
x
b2Rex
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1159
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The flow is steady.
From Appendix A, n = 15.1(10-6) m2>s. Thus, the Reynolds number at x = 6 m is
Rex =
Ux
n
=
(40 m>s)(6 m)
15.1(10-6) m2>s
= 15.894(106)
Since Rex 7 (Rex)cr = 5(105), the boundary layer is turbulent. Thus the displace-
ment thickness is
d* =
0.0463x
(Rex)
1
5
=
0.0463(6 m)
315.894(106) 4 1
5
= 0.01008 m
Thus, the dimension of the square tunnel at exit is 
 a = 1 m + 2d* = 1 m + 2(0.01008 m) = 1.02 m Ans.
*11–24. The wind tunnel operates using air at a temperature 
of 20°C with a free-stream velocity of 40 m>s. If this velocity 
is to be maintained at the central 1-m core throughout 
the tunnel, determine the increased dimensiona at the exit 
in  order to accommodate the growing boundary layer. 
Show  that the boundary layer is turbulent, and use 
d* = 0.0463x>(Rex)
1>5 to calculate the displacement 
thickness.
6 m
1 m
a
a
1 m
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1160
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The fluid is considered to be incompressible. The flow is steady.
Here, 
u
U
= ay
d
b
1
6
. Substituting this result into the momentum integral equation,
 t0 = rU 2 
d
dx L
d
0
u
U
 a1 -
u
U
bdy
 t0 = rU 2 
d
dx L
d
0
ay
d
b
1
6J1 - ay
d
b
1
6 Rdy
 t0 = rU 2 
d
dx L
d
0
Jay
d
b
1
6
- ay
d
b
1
3 Rdy
 t0 =
3rU 2
28
 
dd
dx
 (1)
Using the empirical formula developed by Prandtl and Blasius,
t0 = 0.0225rU 2 a n
Ud
b
1
4
 (2)
0.0225rU 2 a n
Ud
b
1
4
dx =
3rU 2
28
 
dd
dx
d
1
4dd = 0.21a v
U
b
1
4
dx
Assuming that the boundary layer is initially turbulent, then d = 0 at x = 0. Thus,
L
d
0
d
1
4 dd = 0.21a n
U
b
1
4
L
x
0
dx
4
5
 d
5
4 `
d
0
= 0.21a n
U
b
1
4
x `
x
0
d
5
4 = 0.2625 a n
U
b
1
4
x
d = 0.343 a n
U
b
1
5
x
4
5
= 0.343 a n
1
5
U
1
5x
1
5
bx
= 0.343≥
x
aUx
n
b
1
5
¥
However, Rex =
Ux
n
. Then this equation becomes
 d =
0.343x
(Rex)
1
5
 Ans.
11–25. Assume the turbulent boundary layer for a fluid 
has a velocity profile that can be approximated by 
u = U(y>d)1>6. Use the momentum integral equation to 
determine the boundary layer thickness as a function of x. 
Use the empirical formula Eq. 11–19, developed by Prandtl 
and Blasius.
Ans:
d =
0.343x
(Rex)
1
5
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1161
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
We assume steady flow and the air to be incompressible. Appendix A gives 
ra = 1.247 kg>m3, and ma = 17.6(10-6) N # s>m2 at T = 10° C.
The Reynolds number as a function of x is 
Rex =
raUx
ma
=
(1.247 kg>m3)(6 m>s)x
17.6(10-6) N # s>m2
= 4.251(105)x
at x = 1 m, Rex = 4.251(105) 6 5(105), thus, laminar flow within the boundary 
layer.
 d =
5.02Rex
 x =
5.024.251(105)(1)
 (1 m) = 7.67 mm Ans.
 ϴ =
0.6642Rex
 x =
0.66424.251(105)(1)
 (1)
 ϴ = 0.00102 m = 1.02 mm Ans.
11–26. Air enters the square plenum of an air-handling 
system with a velocity of 6 m>s and a temperature of 10°C. 
Determine the thickness of the boundary layer and the 
momentum thickness of the boundary layer at x = 1 m.
300 mm
x
300 mm
Ans:
d = 7.67 mm
ϴ = 1.02 mm
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1162
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
We will assume that steady flow occurs and air is incompressible. Appendix A gives 
ra = 1.247 kg>m3 and ma = 17.6(10-6) N # s>m2 at T = 10° C. The Reynolds number 
as a function of x is 
Rex =
raUx
ma
=
(1.247 kg>m3)(6 m>s)x
17.6(10-6) N # s>m2
= 4.251(105)x
At x = 1 m, Rex = 4.251(105) 6 5(105). Thus, laminar flow persist within the 
boundary layer. Then, Eq. 11–5 can be used to determine the displacement thickness.
d* =
1.7212Rex
 x = £ 1.72124.251(105)x
§ x = 2.6395(10-3)x
1
2
At x = 1 m
d* = 0.0026395 m = 2.64 mm Ans.
Thus, the imaginary cross-sectional area of the duct at x = 1 m is
A′ = 30.3 m - 2(0.0026395 m)42 = 0.08686 m2
The continuity condition requires that the discharge through entrance is the same as 
that through the imaginary cross-section. 
VinAin = V′A′
(6 m>s)(0.3 m)2 = U(0.08686 m2)
 U = 6.22 m>s Ans.
11–27. Air enters the square plenum of an air-handling 
system with a velocity of 6 m>s and a temperature of 10°C. 
Determine the displacement thickness d* of the boundary 
layer at a point x = 1 m downstream. Also, what is the 
uniform speed of the air at this location?
300 mm
x
300 mm
Ans:
6.22 m>s
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1163
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The fluid is considered to be incompressible. The flow is steady.
Here, 
u
U
= ay
d
b
1
6
. Substituting this result into the momentum integral equation,
 t0 = rU 2 
d
dx L
d
0
u
U
 a1 -
u
U
bdy
 t0 = rU 2 
d
dx L
d
0
ay
d
b
1
6J1 - ay
d
b
1
6 Rdy
 t0 = rU 2 
d
dx L
d
0
Jay
d
b
1
6
- ay
d
b
1
3 Rdy
 t0 =
3rU 2
28
 
dd
dx
 (1)
Using the empirical formula developed by Prandtl and Blasius,
t0 = 0.0225rU 2 a n
Ud
b
1
4
 (2)
0.0225rU 2 a n
Ud
b
1
4
dx =
3rU 2
28
 dd
d
1
4dd = 0.21a n
U
b
1
4
dx
Assuming that the boundary layer is initially turbulent, then d = 0 at x = 0. Thus,
L
d
0
d
1
4 dd = 0.21 a n
U
b
1
4
L
x
0
dx
4
5
 d
5
4 `
d
0
= 0.21a n
U
b
1
4
x `
x
0
d
5
4 = 0.2625 a n
U
b
1
4
x
d = 0.343 a n
U
b
1
5
x
4
5
= 0.343 a n
1
5
U
1
5x
1
5
bx
= 0.343≥ x
aUx
n
b
1
5
¥
*11–28. Assume the turbulent boundary layer for a fluid 
has a velocity profile that can be approximated by 
u = U1y>d21>6. Use the momentum integral equation to 
determine the displacement thickness as a function of x 
and Rex. Use the empirical formula, Eq. 11–19, developed 
by Prandtl and Blasius.
y
U
d
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
ei
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1164
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*11–28. Continued
However, Rex =
Ux
n
. Then this equation becomes
d =
0.343x
(Rex)1
5
 (3)
The displacement thickness is
 d* = L
d
0
a1 -
u
U
bdy
 = L
d
0
c 1 - ay
d
b
1
6 ddy
 = ay -
6
7
y
6
7
d
1
6
b `
d
0
 =
d
7
Substituting Eq. 3 into this result,
 d* =
1
7
 £ 0.343x
(Rex)1
5
§ =
0.0490x
(Rex)1
5
 Ans.
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1165
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
We will assume that steady flow occurs, and the fluid is incompressible. Applying the 
boundary conditions at y = 0, u = 0. Then
 0 = C1 + C2 a0
d
b + C3 a0
d
b
2
 C1 = 0 Ans.
And at y = d, u = U. Then
1 = 0 + C2 ad
d
b + C3 ad
d
b
2
C2 + C3 = 1 (1)
Since laminar flow persist within the boundary layer, Newton’s law of viscosity 
t = m 
du
dy
 can be applied. Here
1
U
 
du
dy
=
C2
d
+
2C3
d2 y 
du
dy
= U aC2
d
+
2C3
d2 yb
At y = d, t = 0. Then
 0 = m 
du
dy
`
y =d
= mJU aC2
d
+
2C3
d
b R
 0 =
mU
d
 (C2 + 2C3)
Since 
mU
d
∙ 0, then
C2 + 2C3 = 0 (2)
Solving Eqs. (2) and (3),
 C2 = 2 C3 = -1 Ans.
11–29. The laminar boundary layer for a fluid is assumed to 
be parabolic, such that u>U = C1 + C21y>d2 + C31y>d22. 
If the free-stream velocity U starts at y = d, determine the 
constants C1, C2, and C3.
y
U
d
Ans:
C1 = 0 C2 = 2 C3 = -1
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1166
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
We will assume that steady flow occurs and the fluid is incompressible. Applying the 
boundary condition at y = 0, u = 0. Then
 0 = C1 + C2 a0
d
b + C3 a0
d
b
3
 C1 = 0 Ans.
And at y = d, u = U. Then
1 = 0 + C2 ad
d
b + C3 ad
d
b
3
C2 + C3 = 1 (1)
Since laminar flow persist within the boundary layer, Newton’s Law of viscosity 
t = m 
du
dy
 can be applied. Here
1
U
 
du
dy
=
C2
d
+
3C3
d3 y2 
du
dy
= U aC2
d
+
3C3
d3 y2b
At y = d, t = 0. Then
 0 = m 
du
dy
`
y =d
= mJU aC2
d
+
3C3
d
b R
 0 =
mU
d
 (C2 + 3C3)
Since 
mU
d
∙ 0, then
C2 + 3C3 = 0 (2)
Solving Eqs. (1) and (2),
 C2 =
3
2
 C3 = -
1
2
 Ans.
11–30. The laminar boundary layer for a fluid is assumed 
to be cubic, such that u>U = C1 + C21y>d2 + C31y>d23. 
If the free-stream velocity U starts at y = d, determine the 
constants C1, C2, and C3.
y
U
d
Ans:
C1 = 0 C2 =
3
2
 C3 = -
1
2
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1167
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
 t0 = rU 2 
d
dx L
d
0
u
U
 a1 -
u
U
bdy
 t0 = rU 2 
d
dx L
d
0
ay
d
ba1 -
y
d
bdy
 t0 =
rU 2
6
 
dd
dx
Newton’s law of viscosity,
t0 = m 
U
d
Thus
mU
d
=
rU 2
6
=
dd
dx
L
d
0
d dd = L
x
0
6m
rU
 dx
1
2
 d2 =
6m
ru
 x
d = 3.46A mx
rU
Since Rex = rU * 1m then
 d =
3.46x2Rex
 Ans.
11–31. Assume a laminar boundary layer for a fluid can 
be approximated by u>U = y>d. Determine the thickness 
of the boundary layer as a function of x and Rex.
x
U
Ans:
d =
3.46x2Rex
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1168
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The fluid is considered to be incompressible. The flow is steady.
Substituting 
u
U
= sinap
2
 
y
d
b into the moment integral equation,
t0 = rU 2 
d
dx L
d
0
u
U
 a1 -
u
U
bdy
t0 = rU 2 
d
dx L
d
0
sin ap
2
 
y
d
b c 1 - sin ap
2
 
y
d
b ddy
t0 = rU 2 
d
dx L
d
0
c sin ap
2
 
y
d
b - sin2 ap
2
 
y
d
b ddy
From the trigonometric identity,
sin2 ap
2
 
y
d
b =
1
2
 c 1 - cos apy
d
b d
Then,
t0 = rU 2 
d
dx L
d
0
c sin ap
2
 
y
d
b +
1
2
 cos apy
d
b -
1
2
ddy
t0 = rU 2 
d
dx
 c -
2d
p
 cos ap
2
 
y
d
b +
d
2p
 sin apy
d
b -
1
2
y d `
d
0
t0 = 0.1366rU 2 
dd
dx
 (1)
For a laminar boundary layer, Newton’s law of viscosity applies. Thus,
t0 = m 
du
dy
`
y = 0
= m 
d
dy
 cU sin ap
2
 
y
d
b d `
y = 0
t0 = mU c p
2d
 cos ap
2
 
y
d
b d `
y = 0
t0 =
pmU
2d
 (2)
Equating Eqs. (1) and (2),
pmU
2d
= 0.1366rU 2 
dd
dx
ddd =
11.498m
rU
 dx
*11–32. Assume a laminar boundary layer for a fluid can 
be  approximated by u>U = sin 1py>2d2 . Determine the 
thickness of the boundary layer as a function of x and Rex.
x
U
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1169
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*11–32. Continued
At the leading edge of the plate, x = 0 and d = 0. Thus,
L
d
0
d dd =
11.498m
rU Lx
0
dx
d2
2
`
d
0
=
11.498m
rU
 x `
x
0
d2 =
22.995m
rU
 x
d =
4.7953m
1
2x
1
2
r
1
2U
1
2
d =
4.7953x
r
1
2U
1
2x
1
2
m
1
2
=
4.7953xBrUx
m
Since Rex =
rUx
m
, this equation becomes
 d =
4.80x2Rex
 Ans.
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1170
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The displacement thickness is
 d* = L
d
0
a1 -
u
U
bdy
 = L
d
0
a1 - sin apy
2d
b bdy
 = a1 -
2
p
bd
From solution 11–32,
d =
4.7953x2Rex
So,
d* = a1 -
2
p
b 
4.7953x2Rex
=
1.74x2Rex
11–33. Assume a laminar boundary layer for a fluid can 
be  approximated by u>U = sin 1py>2d2 . Determine the 
displacement thickness d* for the boundary layer as a 
function of x and Rex.
x
U
Ans:
d* =
1.74x2Rex
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1171
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The fluid is considered to be incompressible. The flow is steady.
Substituting 
u
U
= 1.5 ay
d
b - 0.5 ay
d
b
3
 into the moment integral equation,
t0 = rU 2 
d
dx L
d
0
u
U
 a1 -
u
U
bdy
t0 = rU 2 
d
dx L
d
0
J u
U
- a u
U
b
2Rdy
Here,
 
u
U
- a u
U
b
2
= J1.5 ay
d
b - 0.5 ay
d
b
3R - J1.5 ay
d
b - 0.5 ay
d
b
3R 2
= -0.25 ay
d
b
6
+ 1.5 ay
d
b
4
- 0.5 ay
d
b
3
- 2.25 ay
d
b
2
+ 1.5 ay
d
b
Then,
t0 = rU 2 
d
dx L
d
0
J -0.25 ay
d
b
6
+ 1.5 ay
d
b
4
- 0.5 ay
d
b
3
- 2.25 ay
d
b
2
+ 1.5 ay
d
b Rdy
t0 = 0.1393rU 2 
dd
dx
 (1)
For laminar boundary layer, Newton’s law of viscosity applies. Thus,
 t0 = m 
du
dy
`
y = 0
= u 
d
dy
 £UJ1.5ay
d
b - 0.5ay
d
b
3R § †
y = 0
 t0 =
1.5mU
d
 (2)
Equating Eqs. (1) and (2),
1.5mU
d
= 0.1393rU 2 
dd
dx
ddd =
10.769m
rU
 dx
11–34. The, velocity profile for a laminar boundary layer 
of a fluid is represented by u>U = 1.51y>d2 - 0.51y>d23. 
Determine the thickness of the boundary layer as a function 
of x and Rex.
y
U
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1172
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–34. Continued
At the leading edge of the plate, x = 0 and d = 0. Thus,
L
d
0
d dd =
10.769m
rU L
x
0
dx
d2
2
`
d
0
=
10.769m
rU
 x `
x
0
d2 =
21.538m
rU
 x
d =
4.6410m
1
2x
1
2
r
1
2U
1
2
=
4.6410x
r
1
2U
1
2x
1
2
m
1
2
=
4.6410xBrUx
m
Since Rex =
rUx
m
, this equation becomes
 d =
4.64x2Rex
 Ans.
Ans:
d =
4.64x2Rex
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1173
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The fluid is considered to be incompressible. The flow is steady.
Substituting 
u
U
= 1.5 ay
d
b - 0.5 ay
d
b
3
 into the moment integral equation,
t0 = rU 2 
d
dx L
d
0
u
U
 a1 -
u
U
bdy
t0 = rU 2 
d
dx L
d
0
J u
U
- a u
U
b
2Rdy
Here,
 
u
U
- a u
U
b
2
= J1.5 ay
d
b - 0.5 ay
d
b
3R - J1.5 ay
d
b - 0.5 ay
d
b
3R 2
 = -0.25 ay
d
b
6
+ 1.5 ay
d
b
4
- 0.5 ay
d
b
3
- 2.25 ay
d
b
2
+ 1.5 ay
d
b
Then,
t0 = rU 2 
d
dx L
d
0
J -0.25 ay
d
b
6
+ 1.5 ay
d
b
4
- 0.5 ay
d
b
3
- 2.25 ay
d
b
2
+ 1.5 ay
d
b Rdy
t0 = 0.1393rU 2 
dd
dx
 (1)
For a laminar boundary layer, Newton’s law of viscosity applies. Thus,
 t0 = m 
du
dy
`
y = 0
= u 
d
dy
 £U c 1.5ay
d
b - 0.5ay
d
b
3
d § †
y = 0
 t0 =
1.5mU
d
 (2)
Equation Eqs. (1) and (2), 
1.5mU
d
= 0.1393U 2 
dd
dx
ddd =
10.769m
rU
 dx
11–35. The velocity profile for a laminar boundary layer 
of a fluid is represented by u>U = 1.51y>d2 - 0.51y>d23. 
Determine the shear-stress distribution acting on the 
surface as a function of x and Rex.
y
U
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1174
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–35. Continued
At the leading edge of the plate, x = 0 and d = 0. Thus,
L
d
0
d dd =
10.769m
rU L
x
0
dx
d2
2
`
d
0
=
10.769m
rU
 x `
x
0
d2 =
21.538m
rU
 x
d =
4.6410m
1
2x
1
2
r
1
2U
1
2
=
4.6410x
r
1
2U
1
2x
1
2
m
1
2
=
4.6410xBrUx
m
Since Rex =
rUx
m
, this equation becomes
d =
4.6410x2Rex
 (3)
Substituting Eq. (3) into Eq. (2),
 t0 =
1.5mU
4.6410x2Rex
= 0.323m aU
x
b2Rex Ans.
Ans:
t0 = 0.323maU
x
b2Rex
This
 w
ork
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 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1175
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The fluid is considered to be incompressible. The flow is steady.
Here,
u = UJC1 ay
d
b + C2 ay
d
b
2
+ C3 ay
d
b
3R
For the boundary condition u = U at y = d,
U = U3C1(1) + C2(1) + C3(1)4
C1 + C2 + C3 = 1 (1)
Subsequently,
du
dy
= U JC1
d
+
2C2y
d2 +
3C3y
2
d3 R
For the boundary condition 
du
dy
= 0 at y = d,
0 = U c C1
d
+
2C2
d
+
3C3
d
d
C1 + 2C2 + 3C3 = 0 (2)
For the boundary condition 
d2u
dy2 = 0 at y = 0,
 0 =
U
d2 (2C2 + 0) C2 = 0 Ans.
Substituting this result into Eqs. (1) and (2) and solving,
 C1 =
3
2
 C3 = -
1
2
 Ans.
Thus,
u
U
=
3
2
 ay
d
b -
1
2
 ay
d
b
3
Substituting this result into the momentum integral equation,
t0 = rU 2 
d
dx L
d
0
u
U
 a1 -
u
U
bdy
t0 = rU 2 
d
dx L
d
0
J u
U
- a u
U
b
2Rdy
Here,
 
u
U
- a u
U
b
2
= J3
2
 ay
d
b -
1
2
 ay
d
b
3R - J3
2
 ay
d
b -
1
2
 ay
d
b
3R 2
= -
1
4
 ay
d
b
6
+
3
2
 ay
d
b
4
-
1
2
 ay
d
b
3
-
9
4
 ay
d
b
2
+
3
2
 ay
d
b
*11–36. A boundary layer for laminar flow of a fluid over 
the plate is to be approximated by the equation 
u>U = C11y>d2 + C21y>d22 + C31y>d23. Determine the 
constants C1, C2, and C3 using the boundary conditions 
when y = d, u = U; when y = d, du>dy = 0; and when 
y = 0, d2u>dy2 = 0. Find the thickness of the boundary 
layer as a function of x and Rex using the momentum 
integral equation.
U
x
This
 w
ork
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ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1176
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*11–36. Continued
Then,
t0 = rU 2 
d
dx L
d
0
J-
1
4
 ay
d
b
6
+
3
2
 ay
d
b
4
-
1
2
 ay
d
b
3
-
9
4
 ay
d
b
2
+
3
2
 ay
d
bRdy
t0 =
39rU 2
280
 
dd
dx
 (1)
For a laminar boundary layer, Newton’s law of viscosity applies. Thus, 
t0 = m 
du
dy
`
y = 0
= u 
d
dy
£U J3
2
 ay
d
b -
1
2
 ay
d
b
3R § †
y = 0
t0 =
3mU
2d
 (2)
Equating Eqs. (1) and (2),
3mU
2d
=
39rU 2
280
 
dd
dx
ddd =
140m
13rU
 dx
At the leading edge of the plate, x = 0 and d = 0. Thus,
L
d
0
d dd =
140m
13rU L
x
0
dx
d2
2
`
d
0
=
140m
13rU
 x `
x
0
d2 =
280m
13rU
 x
d =
4.6410m
1
2x
1
2
r
1
2U
1
2
=
4.6410x
r
1
2U
1
2x
1
2
m
1
2
=
4.6410xBrUx
m
Since Rex = BrUx
m
, this equation becomes
 d =
4.64x2Rex
 Ans.
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1177
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, n = 15.1(10-6) m2>s for air at T = 20° C. Thus, the Reynolds 
number at x = 18 m is
Rex =
Ux
n
=
(30 m>s)(18 m)
15.1(10-6) m2>s
= 3.576(107)
Since Rex 7 (Rex)cr = 5(105), the boundary layer is turbulent. Assuming that the 
boundary layer is turbulent from x = 0,
 d =
0.371x
(Rex)1
5
=
0.371(18 m)
33.576(107) 4 1
5
= 0.2060 m = 206 mm Ans.
11–37. The train travels at 30 m>s and consists of an engine 
and a series of cars. Determine the approximate thickness 
of the boundary layer at the top of a car, x = 18 m from the 
front of the train. The air is still and has a temperature of 
20°C. Assume the surfaces are smooth and flat, and the 
boundary layer is completely turbulent.
30 m/s
Ans:
206 mm
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1178
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 1.202 kg>m3 and n = 15.1(10-6) m2>s for air at T = 20° C. 
Thus, the Reynolds number at x = 18 m is
Rex =
Ux
n
=
(30 m>s)(18 m)
15.1(10-6) m2>s
= 3.576(107)
Since Rex 7 (Rex)cr = 5(105), the boundary layer is turbulent. Assuming that the 
boundary layer is turbulent from x = 0, the shear stress at any point on the top surface 
of the train is
 t0 =
0.0288rU 2
(Rex)1
5
=
0.0288(1.202 kg>m3)(30 m>s)2
33.576(107) 4 1
5
 = 0.961 N>m2 Ans.
11–38. The train travels at 30 m>s and consists of an 
engine and a series of cars. Determine the approximate 
shear stress acting on the top of a car at x = 18 m from the 
front of the train. The air is still and has a temperature of 
20°C. Assume the surfaces are smooth and flat, and the 
boundary layer is completely turbulent.
30 m/s
Ans:
0.961 Pa
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1179
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Water is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 999.2 kg>m3 and n = 1.15(10-6) m2>s for water at 
T = 15° C. Thus, the Reynolds number at x = L = 100 m is
ReL =
UL
n
=
(10 m>s)(100 m)
1.15(10-6) m2>s
= 8.6957(108)
Here, we assume that the boundary layer is turbulent from x = 0. Since 
107 … ReL 6 109, the frictional drag coefficient is
CD =
0.455
(log10 ReL)2.58
=
0.455
c log103869.57(106) 4 d 2.58 = 0.0015983
Thus, the frictional drag force can be determined from
 F = CD a1
2
 rU 2bbL
 = 0.0015983 c 1
2
 (999.2 kg>m3)(10 m>s)2 d (1 m)(100 m)
 = 7.9849(103) N = 7.98 kN Ans.
11–39. A ship is traveling forward at 10 m>s on a lake. If it 
is 100 m long and the side of the ship can be assumed to be a 
flat plate, determine the drag force on a 1-m-wide strip along 
the entire length of the ship. The water is still and has a 
temperature of 15°C. Assume the boundary layer is 
completely turbulent.Ans:
7.98 kN
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1180
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 1.007 kg>m3 and n = 17.15(10-6) m2>s for air at an altitude 
of 2 km. Here, U = a600 
km
h
ba 1 h
3600 s
ba1000 m
1 km
b = 166.67 m>s. Thus, the Reynolds 
number at x = L = 3 m is
ReL =
UL
n
=
(166.67 m>s)(3 m)
17.15(10-6) m2>s
= 2.915(107)
Here, we assume that the boundary layer is turbulent from x = 0. Since 
107 … ReL 6 109, the frictional drag coefficient is
CD =
0.455
(log10 ReL)2.58
=
0.455
c log1032.915(107) 4 d 2.58 = 0.0025447
Since each of the wings has top and bottom surfaces (4 surfaces altogether), the total 
drag force on the two wings can be determined from
 F = ΣCD a1
2
 rU 2bbL
 = 4 c 0.0025447 c 1
2
 (1.007 kg>m3)(166.67 m>s)2 d (5 m)(3 m) d
 = 2135.44 N = 2.14 kN Ans.
*11–40. An airplane has wings that are, on average, each 
5 m long and 3 m wide. Determine the friction drag on the 
wings when the plane is flying at 600 km>h in still air at an 
altitude of 2 km. Assume the wings are flat plates and the 
boundary layer is completely turbulent.
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1181
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
We will assume that steady flow occurs and sea water is incompressible. The 
Reynolds number at the trailing edge of the hull is
ReL =
rUL
m
=
(1030 kg>m3)(2 m>s)(300 m)
1.14(10-3) N # s>m2
= 5.42(108)
Since 5(105) 6 ReL 6 109, the boundary layer on the hull will be laminar and 
 turbulent along the length. Thus,
CDf =
0.455
(log10 ReL)2.58 -
1700
ReL
=
0.455
e log10 c 5.421(108) d f
2.58 -
1700
5.421(108)
= 0.001694
The frictional drag force acting on each side of the hull can be determined from
FDf = CDf AarU 2
2
b
= 0.00169434.5(103) m24 £ (1030 kg>m3)(2 m>s)2
2
§
= 15.70(103) N = 15.7 kN Ans.
The power required is
P = (FDf)T (V) = 315.70(103) N4 (2 m>s)
= 31.40(103) W
= 31.4 kW Ans.
11–41. The oil tanker has a smooth surface area of 
4.5(103) m2 in contact with the sea. Determine the friction 
drag on its hull and the power required to overcome this 
force if the velocity of the ship is 2 m>s. Take r = 1030 kg>m3 
and m = 1.14110-32 N # s>m2.
2 m/s
300 m
Ans:
FDf = 15.7 kN
W
#
= 31.4 kW
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1182
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
Here, the velocity of the truck relative to still air is 8 m>s + 2 m>s = 10 m>s. Thus, 
U = 10 m>s. From Appendix A, r = 1.202 kg>m3 and n = 15.1(10-6) m2>s. Thus, 
the Reynolds number at x = L = 8 m is
ReL =
UL
n
=
(10 m>s)(8 m)
15.1(10-6) m2>s
= 5.298(106)
Here, we assume that the boundary layer is turbulent from x = 0. Since 
5(105) … ReL 6 107, the frictional drag coefficient is
CD =
0.0740
(ReL)
1
5
=
0.0740
35.298(106)4 1
5
= 0.0033451
Thus, the frictional drag force on surface ABCD can be determined from
F = CDa
1
2
rU 2bbL
= 0.0033451 c 1
2
(1.202 kg>m3)(10 m>s)2 d (4 m)(8 m)
= 6.43 N Ans.
11–42. Wind is blowing at 2 m>s as the truck moves 
forward into the wind at 8 m>s. If the air has a temperature 
of 20°C, determine the friction drag acting on the flat side 
ABCD of the truck. Assume the boundary layer is 
completely turbulent.
8 m
2 m/s
8 m/s
3 m
4 m
A
C
D
E
F
B
Ans:
6.43 N
This
 w
ork
 is
 pr
ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1183
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
Here, the velocity of the truck relative to still air is 8 m>s + 2 m>s = 10 m>s . Thus,
U = 10 m>s. From Appendix A, r = 1.202 kg>m3 and n = 15.1(10-6) m2>s. 
Thus, the Reynolds number at x = L = 8 m is
ReL =
UL
n
=
(10 m>s)(8 m)
15.1(10-6) m2>s
= 5.298(106)
Here, we assume that the boundary layer is turbulent from x = 0. Since 
5(105) … ReL 6 107, the frictional drag coefficient is
CD =
0.0740
(ReL)
1
5
=
0.0740
35.298(106)4 1
5
= 0.0033451
Thus, the frictional drag force on surface BCFE can be determined from
F = CDa1
2
rU 2bbL
= 0.0033451 c 1
2
(1.202 kg>m3)(10 m>s)2 d (3 m)(8 m)
= 4.82 N Ans.
11–43. The wind is blowing at 2 m>s as the truck moves 
forward into the wind at 8 m>s. If the air has a temperature 
of 20°C, determine the friction drag acting on the top 
surface BCFE of the truck. Assume the boundary layer is 
completely turbulent.
8 m
2 m/s
8 m/s
3 m
4 m
A
C
D
E
F
B
Ans:
4.82 N
This
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ork
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y U
nit
ed
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tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1184
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Water is considered to beincompressible. The relative flow is steady.
From Appendix A, r = 999.2 kg>m3 and n = 1.15(10-6) m2>s for water at 
T = 15° C. Thus, the Reynolds number at x = L = 10 m is 
ReL =
UL
n
=
(4 m>s)(10 m)
1.15(10-6) m2>s
= 3.4783(107)
Here, we assume that the boundary layer is turbulent from x = 0. Since 
107 … ReL 6 109, the frictional drag coefficient is
CD =
0.455
(log10 ReL)2.58 =
0.455
3 log1033.4783(107)442.58
= 0.0024785
Thus, the frictional drag force on the bottom surface can be determined from
F = CDa
1
2
rU 2bbL
= 0.0024785 c 1
2
(999.2 kg>m3)(4 m>s)2d (2.5 m)(10 m)
= 495 N Ans.
*11–44. The flat-bottom boat is traveling at 4 m>s on a 
lake for which the water temperature is 15°C. Determine 
the approximate drag force acting on the bottom of the boat 
if the length of the boat is 10 m and its width is 2.5 m. 
Assume the boundary layer is completely turbulent.
10 m
4 m/s
This
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ork
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nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1185
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, n = 0.1779(10- 3) ft2>s for air at an altitude of 5000 ft. Thus, the 
Reynolds number at x = L = 7 ft is
ReL =
UL
n
=
(170 ft>s)(7 ft)
0.1779(10- 3) ft2>s
= 6.689(106)
Here, we assume that the boundary layer is turbulent from x = 0. Thus, the thick-
ness of the boundary layer.
d =
0.371x
(Rex)
1
5
=
0.371(7 ft)
36.689(106) 4 1
5
= 0.1120 ft a12 in.
1 ft
b = 1.34 in. Ans.
11–45. An airplane is flying at 170 ft>s through still air at 
an altitude of 5000 ft. If the wings can be assumed to be flat 
plates, each having a width of 7 ft, determine the boundary 
layer thickness at their trailing or back edge if the boundary 
layer is considered to be fully turbulent.
Ans:
1.34 in.
This
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ork
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cte
d b
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nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1186
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–46. An airplane is flying at 170 ft>s though still air at 
an altitude of 5000 ft. If the wings can be assumed to be flat 
plates, each having a width of 7 ft and a length of 15 ft, 
determine the friction drag on each wing if the boundary 
layer is considered fully turbulent.
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 2.043(10-3) slug>ft3 and n = 0.1779(10-3) ft2>s for air at 
an altitude of 5000 ft. Thus, the Reynolds number at x = L = 7 ft is
ReL =
UL
n
=
(170 ft>s)(7 ft)
0.1779(10-3) ft2>s
= 6.689(106)
Here, we assume that the boundary layer is turbulent from x = 0. Since 
5(105) … ReL 6 107, the frictional drag coefficient is
CD =
0.0740
(ReL)1
5
=
0.0740
36.689(106) 4 1
5
= 0.0031927
Since each wing has two surfaces, top and bottom, the total frictional drag force can 
be determined from
F = ΣCDa1
2
rU 2bbL
= 2 c 0.0031927 c 1
2
 (2.043)(10-3) slug>ft3(170 ft>s)2 d d (15 ft)(7 ft)
 = 19.8 lb Ans.
Ans:
19.8 lb
This
 w
ork
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ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1187
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
Here, the velocity of the airplane is, U = 90 m>s.
From Appendix A, r = 1.112 kg>m3 and n = 15.81(10-6) m2>s for air at an altitude 
of 1 km. Thus, the Reynolds number at x = L = 2.5 m is
ReL =
UL
n
=
(90 m>s)(2.5 m)
15.81(10-6) m2>s
= 1.423(107)
Here, we assume that the boundary layer is turbulent from x = 0. Therefore, the 
thickness of the boundary layer and the shear stress at x = L = 2.5 m are
d =
0.371x
(ReL)1
5
=
0.371(2.5 m)
31.423(107)4 1
5
= 0.03441 m = 34.4 mm Ans.
t0 =
0.0288rU 2
(ReL)1
5
=
0.0288(1.112 kg>m3)(90 m>s)2
31.423(107)4 1
5
= 9.62 N>m2 Ans.
11–47. An airplane is flying at a speed of 90 m>s. If the 
wings are assumed to have a flat surface of width 2.5 m, 
determine the boundary layer thickness d and the shear 
stress at the trailing or back edge. Assume the boundary 
layer is fully turbulent. The airplane flies at an altitude 
of 1 km.
B
2.5 m
90 m/s
A
Ans:
d = 34.4 mm
t0 = 9.62 Pa
This
 w
ork
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ote
cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1188
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
Here, the velocity of the airplane is 90 m>s .
From Appendix A, r = 1.112 kg>m3 and n = 15.81(10-6) m2>s for air at an altitude 
of 1 km. Thus, the Reynolds number at x = L = 2.5 m is
ReL =
UL
n
=
(90 m>s)(2.5 m)
15.81(10-6) m2>s
= 1.423(107)
Here, we assume that the boundary layer is turbulent from x = 0. Since 
107 … ReL 6 109, the frictional drag coefficient can be determined using.
CD =
0.455
(log10ReL)2.58
=
0.455
3 log1031.423(107)442.58
= 0.0028405
Since each wing has two surfaces, top and bottom, the total frictional drag force can 
be determined from
F = ΣCD a1
2
 rU 2bbL
= 2 c0.0028405 c 1
2
(1.112 kg>m3)(90 m>s)2 d (7 m)(2.5 m)d
= 447.73 N = 448 N Ans.
*11–48. An airplane is flying at an altitude of 1 km and a 
speed of 90 m>s. If the wings are assumed to have a flat 
surface of width 2.5 m and length 7 m, determine the 
friction drag on each wing. Assume the boundary layer is 
fully turbulent.
B
2.5 m
90 m/s
A
This
 w
ork
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cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 asse
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1189
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
We will assume that steady flow occurs and the air is incompressible. Appendix A 
gives ra = 2.043(10-3) slug>ft3 and ma = 0.3637(10-6) lb # s>ft2 for air at an altitude 
of 5000 ft. The Reynolds number as a function of x is
Rex =
raUx
ma
=
32.043(10-3)
 
 slug>ft34 (500 ft>s)x
0.3637(10-6) lb # s>ft2
= 2.809(106)x
At the trailing edge where x = L = 1.5 ft, ReL = 2.809(106)(1.5 ft) = 4.213(106). 
Since 5(105) 6 ReL 6 109, the boundary layer will be laminar and turbulent. First, 
we will determine the critical distance xcr where the transition to turbulent flow 
occurs.
(Rex)cr = 5(105); 2.809(106)xcr = 5(105)
xcr = 0.1780 ft
For the laminar boundary layer where x 6 xcr,
d =
5.02Rex
 x =
5.022.809(106)x
 x = c 2.9835(10-3)x
1
2 d ft
x(ft) 0 0.05 0.10 0.15 0.178
d(in.) 0 0.00801 0.0113 0.0139 0.0151
11–49. The tail of the airplane has an approximate width 
of 1.5 ft and a length of 4.5 ft. Assuming the air flow onto 
the tail is uniform, plot the boundary layer thickness d. Give 
values for every increment of 0.05 ft for the laminar 
boundary layer, and every 0.25 ft for the turbulent boundary 
layer. Also, calculate the friction drag on the rudder. The 
plane is flying in still air at an altitude of 5000 ft with a 
speed of 500 ft>s.
500 ft/s
1.5 ft
4.5 ft
This
 w
ork
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d b
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nit
ed
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tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1190
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
For the turbulent boundary layer where x 7 xcr,
d =
0.371
(Rex)1
5
 x =
0.371
32.809(106)x4 1
5
 x = (0.01904x
4
5 ) ft
x(ft) 0.178 0.25 0.50 0.75 1.0 1.25 1.50
d(in.) 0.0574 0.0754 0.1312 0.1815 0.2284 0.2731 0.3160
The plot of the boundary layer is shown in Fig. a.
For the laminar and turbulent boundary layer the frictional drag coefficient can be 
determined from
CDf =
0.455
(log10ReL)2.58
-
1700
ReL
=
0.455
3 log104.213(106)42.58
-
1700
4.213(106)
= 0.003059
Thus, the frictional drag force can be determined by applying
FDf = CDf bLaraU
2
2
b
= 0.003059(4.5 ft)(1.5 ft)• 32.043(10-3) slug>ft34 (500 ft>s)2
2
¶
 = 5.27 lb Ans.
11–49. Continued
0 0.10
δ (in.)
x (ft)
xcr = 0.178
0.10
0.05
0.20
0.15
0.30
0.25
0.35
0.25 0.50 1.00.75 1.25 1.50
(a) Ans:
5.27 lb
This
 w
ork
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nit
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tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1191
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Water is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 999.2 kg>m3 and n = 1.15(10-6) m2>s for water at 
T = 15° C. Thus, the Reynolds number at x = L = 0.25 m is
Re =
UL
n
=
(20 m>s)(0.25 m)
1.15(10-6) m2>s
= 4.348(106)
Here, we assume that the boundary layer is turbulent from x = 0. Thus, the thickness 
of the boundary layer at x = L = 0.25 m is
 d =
0.371x
(Rex)1
5
=
0.371(0.25 m)
3(4.348(106) )4 1
5
= 0.00436 m = 4.36 mm Ans.
Since 5(105) … ReL 6 107, the frictional drag coefficient can be determined using
CD =
0.0740
(ReL)1
5
=
0.0740
34.348(106)4 1
5
= 0.003480
Since each wing has two surfaces, top and bottom, the total frictional drag force can 
be determined from
F = ΣCD a1
2
rU 2bbL
= 2 c0.003480 c 1
2
(999.2 kg>m3)(20 m>s)2 d (4 m)(0.25 m)d
= 1390.88 N = 1.39 kN Ans.
11–50. Two hydrofoils are used on the boat that is traveling 
at 20 m>s. If the water is at 15°C, and if each blade can be 
considered as a flat plate, 4 m long and 0.25 m wide, 
determine the thickness of the boundary layer at the trailing 
or back edge of each blade. What is the drag on each blade? 
Assume the flow is completely turbulent.
20 m/s
Ans:
d = 4.36 mm
F = 1.39 kN
This
 w
ork
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cte
d b
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nit
ed
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tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1192
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Water is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 999.2 kg>m3 and n = 1.15(10-6) m2>s for water at 
T = 15° C. Thus, the Reynolds number at x = L = 0.25 m is
Re =
UL
n
=
(20 m>s)(0.25 m)
1.15(10-6) m2>s
= 4.348(106)
Since 5(105) … ReL 6 109, the frictional drag coefficient considering both laminar 
and turbulent boundary layers is
CD =
0.455
(log10ReL)2.58
-
1700
ReL
=
0.455
3 log1034.348(106)442.58
-
1700
4.348(106)
= 0.0030534
Since each wing has two surfaces, top and bottom, the total frictional drag force can 
be determined from
F = ΣCD a
1
2
rU 2bbL
 = 2 c0.0030533 c 1
2
(999.2 kg>m3)(20 m>s2) d (4 m)(0.25 m)d
 = 1220.37 N = 1.22 kN Ans.
11–51. Two hydrofoils are used on the boat that is traveling 
at 20 m>s. If the water is at 15°C and if each blade can be 
considered as a flat plate, 4 m long and 0.25 m wide, 
determine the drag on each blade. Consider both laminar 
and turbulent flow boundary layers.
20 m/s
Ans:
1.22 kN
This
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ork
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tat
es
 co
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rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
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itte
d. 
1193
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow issteady.
From Appendix A, r = 0.9092 kg>m3 and n = 18.63(10-6) m2>s for air at an 
altitude of 3 km. Here, 
U = a700 
km
h
ba1000 m
1 km
ba 1 h
3600 s
b = 194.44 m>s . Thus, the Reynolds number at 
x = L = 2 m is
ReL =
UL
n
=
(194.44 m>s)(2 m)
18.63(10-6) m2>s
= 2.087(107)
Since 5(105) … ReL 6 109, the frictional drag coefficient considering both laminar 
and turbulent boundary layers is 
CD =
0.455
(log10ReL)2.58
-
1700
ReL
=
0.455
3 log102.087(107)42.58
-
1700
2.087(107)
= 0.002595
Since each wing has two surfaces, top and bottom, the total frictional drag force can 
be determined from
F = ΣCDa1
2
rU 2bbL
= 2 c0.002595 c 1
2
(0.9092 kg>m3)(194.44 m>s)2 d (6 m)(2 m)d
= 1070.65 N = 1.07 kN Ans.
*11–52. An airplane is flying at an altitude of 3 km and 
a speed of 700 km>h. If each wing is assumed to have a 
flat surface of width 2 m and length 6 m, determine the 
friction drag acting on each wing. Consider both laminar 
and turbulent boundary layers.
120 ft
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tat
es
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py
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1194
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Water is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 1.939 slug>ft3 and n = 12.2(10-6) ft2>s for water at 
T = 60° F. Thus, the Reynolds number at x = L = 120 ft is
ReL =
UL
n
=
(15 ft>s)(120 ft)
12.2(10-6) ft2>s
= 1.475(108)
Since 5(105) … ReL 6 109, the frictional drag coefficient considering both laminar 
and turbulent boundary layers is
CD =
0.455
(log10ReL)2.58
-
1700
ReL
=
0.455
3 log1031.475(108) 4 42.58
-
1700
1.475(108)
= 0.0020051
Thus, the frictional drag force on the bottom surface can be determined from
FD = CD a1
2
rU 2bbL
= 0.0020051 c 1
2
(1.939 slug>ft3)(15 ft>s)2 d (25 ft)(120 ft)
= 1312.17 lb
Thus, the power required to overcome FD is
P = FD
# V = (1312.17 lb)(15 ft>s) = a19 683 
ft # lb
s
ba 1 hp
550 ft # lb>s
b
= 35.8 hp Ans.
11–53. The barge is traveling forward at 15 ft>s in still 
water having a temperature of 60°F. If the bottom of the 
barge can be assumed to be a flat plate of length 120 ft and 
width 25 ft, determine the power of the engine required to 
overcome the frictional resistance of the water on the 
bottom of the barge. Consider both laminar and turbulent 
boundary layers.
120 ft
Ans:
35.8 hp
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es
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1195
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The flow is steady.
The resultant forces of the pressure on the top and bottom surfaces of the plate are 
Ft = ptA = c 60(103) 
N
m2 d 32 m(3 m)4 = 360(103) N = 360 kN
Fn = pnA = c 40(103) 
N
m2 d 32 m(3 m)4 = 240(103) N = 240 kN
The pressure drag is equal to the component of the resultant force along the 
direction of the free-stream flow, which in this case is horizontal. Referring to Fig. a,
S+ (FR)x = ΣFx ; FPD = (360 kN) sin 12° - (240 kN) sin 12° = 24.9 kN Ans.
11–54. The plate is 2 m wide and is held at an angle of 12° 
with the wind as shown. If the average pressure under the 
plate is 40 kPa, and on the top it is 60 kPa, determine the 
pressure drag on the plate.
Fpd = (FR)x
Ft = 360 kN
Fb = 240 kN
12°
12°
60 kPa
40 kPa
(FR)y
(a)
3 m
12�
Ans:
24.9 kN
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tat
es
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py
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1196
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The flow is steady.
The resultant force of the pressure on the inclined surface is
F =
1
2
 a80 
N
m2 b 33 m(4 m)4 = 480 N
The pressure drag is equal to the component of the resultant force along the 
direction of the free-stream flow which in this case is horizontal. Referring to Fig. a,
+
S(FR)x = ΣFx; FD = (480 N) sin 30° = 240 N Ans.
11–55. Wind blows over the inclined surface and produces 
the approximate pressure distribution shown. Determine 
the pressure drag acting over the surface if the surface is 
3 m wide.
(FR)y
FPD = (FR)x
F = 480 N
80 Pa
30°
(a)
30�
4 m
10 m/s 80 Pa
Ans:
240 N
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ork
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1197
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*11–56. The sign is subjected to a wind profile that 
produce a pressure distribution that can be approximated 
by r = 1112.5 ey0.62 Pa, where y is in meters. Determine 
the resultant pressure force on the sign due to the wind. The 
air is at a temperature of 20°C, and the sign is 0.5 m wide.
Solution
The air is considered to be incompressible. The flow is steady.
The dynamic pressure can be determined from
p = (112.5ry0.6) Pa
The force of this pressure on a differential area dA = bdy = (0.5 m)dy is 
dF = pdA = (112.5ry0.6)(0.5dy) = 56.25ry0.6dy. Thus, the resultant force on the 
entire surface of the sign is
FR = LA
dF = L
6 m
3 m
56.25ry0.6dy =
56.25ry1.6dy
1.6
`
6 m
3 m
= (414.19r) N
From Appendix A, r = 1.202 kg>m2 for air at T = 20° C. Thus, 
 FR = 414.19(1.202 kg>m2) = 498 N Ans.
3 m
3 m
y
p
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
ofthe
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1198
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–57. The air pressure acting from A to B on the 
surface of a curved body can be approximated as 
p = (5 - 1.5u) kPa, where u is in radians. Determine the 
pressure drag acting on the body from 0 … u … 90°. The 
body has a width of 300 mm.
Solution
The air is considered to be incompressible. The flow is steady.
The force of the pressure on a differential area dA = bds = brdu = (0.3 m)(0.1 m)du
= 0.03du is dF = pdA = (5 - 1.5u)(103)(0.03du) = 30(5 - 1.5u)du. The pressure 
drag is equal to the component of the force along the direction of the free-stream 
flow, which in this case is horizontal. Referring to Fig. a,
+
S(FR)x = ΣFx ; FPD = LA
(dF)x = L
p
2
0
dF cos u = L
p
2
0
30(5 - 1.5u) cos u du
= 3035 sin u - 1.5(cos u + u sin u)4 �p20
 = 124.31 N = 124 N Ans.
(FR)y
FPD (FR)x
dF
d
(a)
=
100 mm
A
B
u
Ans:
124 N
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s 
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d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1199
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–58. The air pressure acting on the inclined surfaces is 
approximated by the linear distributions shown. Determine 
the horizontal force resultant acting on the surface if it is 
3 m wide.
Solution
The air is considered to be incompressible. The flow is steady.
The resultant force of the trapezoidal and triangular pressure prism are
Ftrap =
1
2
c (5 + 3)(103) 
N
m2 d 36 m(3 m)4 = 72(103) N = 72 kN
Ftri =
1
2
c 3(103) 
N
m2 d 36 m(3 m)4 = 27(103) N = 27 kN
Referring to Fig. a,
+
S(FR)x = ΣFx ; FPD = (72 kN) sin 20° + (27 kN) sin 45° = 43.7 kN Ans.
(FR)y
FPD = 
Ftrap = 72 kN
Ftri = 27 kN
5 kPa
3 kPa20°
45°
(a)
(FR)x
5 kPa
3 kPa
3 kPa
6 m
6 m
20�
45�
Ans:
43.7 kN
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ork
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law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1200
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–59. The front of the building is subjected to wind that 
exerts a pressure of p = 10.25y1>22 lb>ft2, where y is in feet, 
measured from the ground. Determine the resultant 
pressure force on the windward face of the building due to 
this loading.
Solution
The air is considered to be . The flow is steady.
The force of the wind pressure on a differential area dA = bdy = (80 ft)dy is 
dF = pdA = (0.25y
1
2 )(80 dy) = 20y
1
2dy.
Thus, the resultant force on the entire windward surface is
FR = LA
dF = L
30 ft
0
20y
1
2dy = 20 a2
3
b y
3
2 `
30 ft
0
= 2190.89 lb = 2.19 kip Ans.
30 ft
80 ft
y
Ans:
2.19 kip
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py
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1201
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*11–60. The building is subjected to a uniform wind 
having a speed of 80 ft>s. If the temperature of the air is 
40°F, determine the resultant pressure force on the front of 
the building if the drag coefficient is 1.43.
Solution
The air is considered to be incompressible. The flow is steady.
From Appendix A, r = 0.00247 slug>ft3 for air at T = 40° F.
FD = CDApr
U 2
2
= 1.43380 ft(30 ft)4 (0.00247 slug>ft3) £ (80 ft>s)2
2
§
 = 27.13(103) lb = 27.1 kip Ans.
30 ft
80 ft
y
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law
s 
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d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1202
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–61. Determine the moment developed at the base A of 
the square sign due to wind drag if the front of the signboard 
is subjected to a 16 m>s wind. The air is at 20°C. Neglect the 
drag on the pole.
Solution
The air is considered to be incompressible. The flow is steady.
From Appendix A, r = 1.202 kg>m3 and n = 15.1(10-6) m2>s for air at T = 20° C. 
Thus, the Reynolds number of the flow is
Re =
UL
n
=
(16 m>s)(2 m)
15.1(10-6) m2>s
= 2.12(106)
Since Re 7 104, the value of CD for the plate in Table 11–3 can be used. For 
b
h
=
2 m
2 m
= 1, CD = 1.10. Here, AP = 2 m(2 m) = 4 m2
FD = CDAP r 
U 2
2
= 1.10(4 m2)(1.202 kg>m3) £ (16 m>s)2
2
§
= 676.97 N
Here, FD will act through the center of the signboard, as shown on its free-body 
diagram, Fig. a. Consider the moment equilibrium about point A.
a+ΣMA = 0; MA - (676.97)(4 m) = 0
MA - 2.707(103) N # m = 2.71 kN # m Ans.
Ax
Ay
MA
FD = 1415.47 N
4 m
(a)
3 m
2 m
2 m
A
Ans:
2.71 kN # m
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law
s 
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d i
s p
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ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1203
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–62. The mast on the boat is held in place by the 
rigging, which consists of rope having a diameter of 0.75 in. 
and a total length of 130 ft. Assuming the rope to be 
cylindrical, determine the drag it exerts on the boat if the 
boat is movingforward at a speed of 30 ft>s. The air has a 
temperature of 60°F.
Solution
We will assume that steady laminar flow occurs, and the air is incompressible. Appendix 
A gives ra = 0.00237 slug>ft3 and ma = 0.374(10-6) lb # s>ft2 at T = 60° F. The 
Reynolds number is
Re =
raUD
ma
=
(0.00237 slug>ft3)(30 ft>s)a0.75
12
 ftb
0.374(10-6) lb # s>ft2
= 1.188(104)
With this Reynolds number, the drag coefficient for the cylinder can be obtained 
using Fig. 11–31, for which CD = 1.3 (approximately). Then the drag force on the 
rope can be determined by applying.
FD = CDAP a
raV
2
2
b
= 1.3 c a0.75
12
 ftb(130 ft) d £ (0.00237 slug>ft3)(30 ft>s)2
2
§
= 11.3 lb Ans.
Ans:
11.3 lb
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ea
ch
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ou
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ing
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en
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arn
ing
. D
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em
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of 
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art
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 th
is 
work
 (in
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 W
orl
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ide
 W
eb
) 
will 
de
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1204
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–63. Wind at 20°C blows against the 100-mm-diameter 
flagpole with a speed of 1.20 m>s. Determine the drag on the 
pole if it has a height of 8 m. Consider the pole to be a smooth 
cylinder. Would you consider this a significant force?
Solution
The air is considered to be incompressible. The flow is steady.
From Appendix A, r = 1.202 kg>m3, n = 15.1(10-6) m2>s for air at T = 20° C. 
Here, it is required that
Re =
UD
n
Re =
1.20 m>s(0.1 m)
15.1(10-6) m2>s
= 7947
With Re = 7947, CD for the cylindrical pole can be determined from the graph.
CD ≈ 1.2 (Approx.)
Thus,
FD = CDAP r 
u2
2
= 1.2(8 m)(0.1 m)(1.202 kg>m3)° (1.20 m>s)2
2
¢
 FD = 0.831 N Ans.
This is a very small force.
Ans:
0.831 N
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for
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e u
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 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
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e i
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of 
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1205
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*11–64. Each of the smooth bridge piers (cylinders) has a 
diameter of 0.75 m. If the river maintains an average speed 
of 0.08 m>s, determine the drag the water exerts on each 
pier. The water temperature is 20°C.
Solution
Water is considered to be incompressible. The flow is steady.
From Appendix A, r = 998.3 kg>m3 and n = 1.00(10-6) m2>s . Thus, the Reynolds 
number of the flow is 
Re =
UD
n
=
(0.08 m>s)(0.75 m)
1.00(10-6) m2>s
= 6(104)
Since the pier is a cylinder (rough), the drag coefficient can be determined by 
entering this Re on the graph which gives CD = 1.4 (approximately). Also, 
AP = 0.75 m(6 m) = 4.5 m2.
FD = CDAP r 
U 2
2
= 1.4(4.5 m2)(998.3 kg>m3)° (0.08 m>s)2
2
¢
= 20.1 N Ans.
6 m
0.75 m
0.08 m/s
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law
s 
an
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rov
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d s
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ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–65. A 60-mi>h wind blows on the side of the truss. If 
the members are each 4 in. wide, determine the drag acting 
on the truss. The air is at 60°F, and CD = 1.2. Note that 
1 mi = 5280 ft.
Solution
The air is considered to be incompressible. The flow is steady.
From Appendix A, r = 0.00237 slug>ft3 for air at T = 60° F. Here,
U = a60 
mi
h
ba5280 ft
1 mi
ba 1 h
3600 s
b = 88 ft>s . The projected area of the truss’s 
members perpendicular to the air stream is
AP = c 6(10 ft) + 32(10 ft)2 + (10 ft)2 d a 4
12
 ftb = 34.14 ft2
FD = CDAp r 
U 2
2
= 1.2(34.14 ft2)(0.00237 slug>ft3) £ (88 ft>s)2
2
§
 = 375.97 lb = 376 lb Ans.
10 ft
B C D
E
F
A
10 ft 10 ft 10 ft
Ans:
376 lb
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law
s 
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for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
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oy
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1207
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–66. A periscope on a submarine has a submerged 
length of 2.5 m and a diameter of 50 mm. If the submarine is 
traveling at 8 m>s, determine the moment developed at the 
base of the periscope. The water is at a temperature of 15°C. 
Consider the periscope to be a smooth cylinder.
Solution
Water is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 999.2 kg>m3 and n = 1.15(10-6) m2>s for water at 
T = 15° C. Thus, the Reynolds number of the flow is
 Re =
UD
n
=
(8 m>s)(0.05 m)
1.15(10-6) m2>s
= 3.48(105)
Since the periscope is a cylinder (smooth), the drag coefficient can be deter-
mined by entering this Re into Fig. 11–31 which gives CD ≅ 0.85 (approx.). Also, 
AP = 0.05 m(2.5 m) = 0.125 m2.
 FD = CDAp r 
U 2
2
= 0.85(0.125 m2)(999.2 kg>m3) £ (8 m>s)2
2
§
 = 3397.28 N
Here, FD acts through the mid-length of the periscope’s submerged length as shown 
in its free-body diagram in Fig. a,
a+ ΣMA = 0; MA - 3397.28 N(1.25 m) = 0
 MA = 4246.6 N # m = 4.25 kN # m Ans.
Ay
W
Ax
MA
FD = 3197.44 N
1.25 m
(a)
Ans:
4.25 kN # m
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law
s 
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rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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1208
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–67. The antenna on the building is 20 ft high and has a 
diameter of 12 in. Determine the restraining moment at its 
base to hold it in equilibrium if it is subjected to a wind 
having an average speed of 80 ft>s. The air is at a 
temperature of 60°F. Consider the antenna to be a smooth 
cylinder.Solution
The air is considered to be incompressible. The flow is steady.
From Appendix A, r = 0.00237 slug>ft3 and n = 0.158(10-3) ft2>s for air at 
T = 60° F. Thus, the Reynolds number for the flow is
 Re =
UD
n
=
(80 ft>s)(1 ft)
0.158(10-3) ft2>s
= 5.06(105)
Since the antenna is a cylinder (smooth), the drag coefficient can be determined 
by entering this Re into the graph, which gives CD ≅ 0.32 (approx.). Also, 
AP = 1 ft(20 ft) = 20 ft2.
 FD = CDAp r 
U 2
2
= 0.32(20 ft2)(0.00237 slug>ft3) c (80 ft>s)2
2
d
 = 48.54 lb
Here, FD acts through the mid-height of the antenna as shown in its free-body 
diagram in Fig. a,
a+ΣMA = 0; MA - (48.54 lb)(10 ft) = 0
 MA = 485 lb # ft Ans.
FD = 48.54 lb
Ax
10 ft
M
(a)
Ans:
485 lb # ft
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law
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d s
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for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
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of 
the
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1209
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*11–68. The truck has a drag coefficient of CD = 1.12 when 
it is moving with a constant velocity of 80 km>h. Determine 
the power needed to drive the truck at this speed if the 
average front projected area of the truck is 10.5 m2. The air 
is at a temperature of 10°C.
Solution
The air is considered to be incompressible. The flow is steady.
From Appendix A, r = 1.247 kg>m3 for air at T = 10° C. Here,
U = a80 
km
h
ba1000 m
1 km
ba 1 h
3600 s
b = 22.22 m>s .
FD = CDAp r 
U 2
2
= 1.12(10.5 m2)(1.247 kg>m3) c (22.2 m>s)2
2
d
= 3620.92 N
Thus, the power needed to overcome the drag is
W
#
= FD
# V = (3620.92 N)(22.2 m>s) = 80.46(103) W
= 80.5 kW Ans.
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s 
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d s
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ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
 an
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s n
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d. 
1210
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–69. The truck has a drag coefficient of CD = 0.86 when 
it is moving with a constant velocity of 60 km>h. Determine 
the power needed to drive the truck at this speed if the 
average front projected area of the truck is 10.5 m2. The air 
is at a temperature of 10°C.
Solution
The air is considered to be incompressible. The flow is steady.
From Appendix A, r = 1.247 kg>m3 for air at T = 10° C. Here, 
U = a60 
km
h
ba1000 m
1 km
ba 1 h
3600 s
b = 16.67 m>s .
FD = CDAP r 
U 2
2
= 0.86(10.5 m2)(1.247 kg>m3) c (16.667 m>s)2
2
d
= 1564 N
Thus, the power needed to overcome the drag is
W
#
= FD
# V = (1564 N)(16.667 m>s) = 26.07(103) W
= 26.1 kW Ans.
Ans:
26.1 kW
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law
s 
an
d i
s p
rov
ide
d s
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ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
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pe
rm
itte
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1211
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–70. Wind at 10°C blows against the 30-m-high chimney 
at 2.5 m>s. If the diameter of the chimney is 2 m, determine 
the moment that must be developed at its base to hold it in 
place. Consider the chimney to be a rough cylinder. 
Solution
The air is considered to be incompressible. The flow is steady.
From Appendix A, r = 1.247 kg>m3 and n = 14.2(10-6) m2>s for air at T = 10° C. 
Thus, the Reynolds number of the flow is
 Re =
UD
n
=
(2.5 m>s)(2 m)
14.2(10-6) m2>s
= 3.52(105)
Since the chimney is a cylinder (rough), the drag coefficient can be determined 
by entering this Re into the graph, which gives CD ≅ 0.5 (approx.). Also,
AP = 2 m(30 m) = 60 m2.
 FD = CDAp r 
U 2
2
= 0.5(60 m2)(1.247 kg>m3) c (2.5 m>s)2
2
d
 = 116.90 N
Here, FD acts through the mid-height of the chimney as shown in its free-body 
diagram in Fig. a,
+ΣMA = 0; MA - (116.90 N)(15 m) = 0
 MA = 1753.59 N # m = 1.75 kN # m Ans.
Ax
Ay
MA
FD = 116.90 N
15 m
W
(a)
2 m
30 m
Ans:
1.75 kN # m
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law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
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d. 
1212
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Oil is considered to be incompressible. The flow is steady.
When AB is the leading edge, 
b
h
=
0.8 m
0.2 m
= 4. From the table, CD = 1.19 for a 
 rectangular plate. Also, AP = (0.8 m)(0.2 m) = 0.16 m2.
(FD)AB = CDAP r 
U 2
2
= 1.19(0.16 m2)(880 kg>m3) c (0.5 m>s)2
2
d
= 20.9 N Ans.
When BC is the leading edge, 
b
h
=
0.8 m
0.4 m
= 2. From the table, CD = 1.15 for a 
 rectangular plate. Also, AP = (0.8 m)(0.4 m) = 0.32 m2.
(FD)BC = CDAP r 
U 2
2h
= 1.15(0.32 m2)(880 kg>m3) c (0.5 m>s)2
2
d
= 40.5 N Ans.
11–71. A rectangular plate is immersed in a stream of oil 
flowing at 0.5 m>s. Compare the drag acting on the plate if it 
is oriented so that AB is the leading edge and then when it 
is rotated 90° counterclockwise so that BC is the leading 
edge. The plate is 0.8 m wide. Take ro = 880 kg>m3.
0.4 m
8 m/s
0.2 m
B
A
C
Ans:
1FD2AB = 20.9 N
1FD2BC = 40.5 N
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law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
 an
d i
s n
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itte
d. 
1213
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
Since the parachutist descends with a constant terminal velocity, the acceleration is 
zero. Referring to the free-body diagram shown in Fig.a,
+ c ΣFy = may; FD - 90(9.81) N = 90(0) FD = 882.9 N
From Appendix A, r = 1.202 kg>m3 for air at T = 20° C. Here, the projected area 
of the parachute perpendicular to the air stream is AP = p(2 m)2 = 4pm2.
FD = CDAP r 
U 2
2
882.9 N = 1.36(4p m2)(1.202 kg>m3)aU 2
2
b
U = 9.27 m>s Ans.
*11–72. The parachute has a drag coefficient of CD = 1.36 
and an open diameter of 4  m. Determine the terminal 
velocity as the man parachutes downward. The air is at 20°C. 
The total mass of the parachute and man is 90 kg. Neglect the 
drag on the man.
FD
a = 0
90(9.81) N
(a)
V
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cte
d b
y U
nit
ed
 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
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pe
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itte
d. 
1214
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
Since the parachutist descends with a constant terminal velocity, the acceleration is 
zero. Referring to the free-body diagram shown in Fig. a,
+ c ΣFy = may; FD - 90(9.81) N = 90(0) FD = 882.9 N
From Appendix A, r = 1.202 kg>m3 for air at T = 20° C. Here, the projected area 
of the parachute perpendicular to the air stream is AP = p ad
2
b
2
=
pd2
4
.
FD = CDAP r 
U 2
2
882.9 N = 1.36apd2
4
b (1.202 kg>m3) c (10 m>s)2
2
d
d = 3.71 m Ans.
11–73. The parachute has a drag coefficient of CD = 1.36. 
Determine the required open diameter of the parachute so 
the man attains a terminal velocity of 10 m>s. The air is at 
20°C. The total mass of the parachute and man is 90 kg. 
Neglect the drag on the man.
FD
a = 0
90(9.81) N
(a)
V
Ans:
3.71 m
This
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tat
es
 co
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1215
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
Since the parachutist descends with a constant terminal velocity, the acceleration is 
zero. Referring to the free-body diagram shown in Fig. a,
+ c ΣFy = may; FD - 90(9.81) N = 90(0) FD = 882.9 N
From Appendix A, r = 1.202 kg>m3 for air at T = 20° C. Here, the projected area 
of the parachute perpendicular to the air stream is AP = p a6 m
2
b
2
= 9p m2.
FD = CDAP r 
U 2
2
882.9 N = CD(9p m2)(1.202 kg>m3) c (5 m>s)2
2
d
CD = 2.08 Ans.
11–74. The man and the parachute have a total mass of 
90 kg. If the parachute has an open diameter of 6 m and the 
man attains a terminal velocity of 5 m>s, determine the drag 
coefficient of the parachute. The air is at 20°C. Neglect the 
drag on the man.
FD
a = 0
90(9.81) N
(a)
V
Ans:
2.08
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tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1216
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 0.00237 slug>ft3 for air at T = 60° F. Here, 
U = a60 
mi
h
ba5280 ft
1 mi
ba 1 h
3600 s
b = 88 ft>s .
FD = CDAP r 
U 2
2
= 0.83(14.5 ft2)(0.00237 slug>ft3) c (88 ft>s)2
2
d
= 110.44 lb
Referring to the free-body diagram shown in Fig. a,
d
+ ΣFx = ma; F - 110.44 lb = 0 F = 110.44 lb
Subsequently, the power needed to produce this drive force is
P = F # U = 110.44 lb(88 ft>s) = a9718.80 
ft # lb
s
ba 1 hp
550 ft
# lb
s
b = 17.7 hp Ans.
11–75. The car has a projected front area of 14.5 ft2. 
Determine the power required to drive at a constant 
velocity of 60 mi>h if the drag coefficient is CD = 0.83 and 
the air is at 60°F. Note that 1 mi = 5280 ft.
FD = 110.44 lb
x W
a = 0
F
N
(a)
60 mi/h
Ans:
17.7 hp
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es
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
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pe
rm
itte
d. 
1217
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 1.007 kg>m3 and n = 17.26(10-6) m2>s for air at an altitude 
of 2 km. Here, U = a12 
km
h
ba1000 m
1 km
ba 1 h
3600 s
b = 3.333 m>s. Thus, the Reynolds 
number is
Re =
UD
n
=
(3.333 m>s)(5 m)
17.26(10-6) m2>s
= 9.656(105)
Entering this Re into the graph for a sphere, CD ≅ 0.16 (approx.). Here, 
AP = p a
5 m
2
b
2
= 6.25p m2.
FD = CDAP r 
U 2
2
 0.16(6.25p m2)(1.007 kg>m3) c (3.333 m>s)2
2
d
= 17.6 N Ans.
*11–76. A 5-m-diameter balloon is at an altitude of 2 km. 
If it is moving with a terminal velocity of 12 km>h, determine 
the drag on the balloon.
This
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ork
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nit
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 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1218
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 1.202 kg>m3 for air at T = 20° C. Here, 
U = a160 
km
h
ba1000 m
1 km
ba 1 h
3600 s
b = 44.44 m>s.
FD = CDAP r 
U 2
2
= 0.28(2.5 m2)(1.202 kg>m3) £ (44.44 m>s)2
2
§
= 831.01 N
Referring to the free-body diagram shown in Fig. a,
d
+ ΣFx = ma; F - 831.01 N = 0 F = 831.01 N
Subsequently, the powerthat must be supplied by the engine to produce this drive 
force is
P = F # U = (831.01 N)(44.44 m>s) = 36.93(103) W
= 36.9 kW Ans.
11–77. The drag coefficient for the car is CD = 0.28, and the 
projected area into the 20°C airstream is 2.5 m2. Determine 
the power the engine must supply to maintain a constant 
speed of 160 km>h.
FD = 831.01 N
x
W
a = 0
F
N
(a)
160 km/h
Ans:
36.9 kW
This
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 S
tat
es
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py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
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pe
rm
itte
d. 
1219
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 1.247 kg>m3 and n = 14.2(10-6) m2>s for air at T = 10° C. 
Thus, the Reynolds number of the air flow is 
Re =
UD
n
=
(60 m>s)(1.25 m)
14.2(10-6) m2>s
= 5.28(106)
Since Re 7 104, the value of CD for the cone in the table can be used. For u = 60°,
CD = 0.8. Here, AP = p a
1.25 m
2
b
2
= 0.390625p m2.
FD = CDAP r 
U 2
2
= 0.8(0.39025p m2)(1.247 kg>m3) £ (60 m>s)2
2
§
= 2.204(103) N = 2.20 kN Ans.
11–78. The rocket has a nose cone that is 60° and a base 
diameter of 1.25 m. Determine the drag of the air on the 
cone when the rocket is traveling at 60 m>s in air having a 
temperature of 10°C. Use Table 11–3 for the cone, but 
explain why this may not be an accurate assumption. 1.25 m
60�
Ans:
2.20 kN
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tat
es
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py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1220
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Water is considered to be incompressible. The relative flow is steady.
Since the flow is along the length of the log and the log has an approximate diameter 
of 0.35 m, 
AP = p a0.35 m
2
b
2
= 0.030625p m2
The drag on the log is
F = CDAP r 
U 2
2
= 0.85a1
2
b (0.030625p m2)(1000 kg>m3) c (2 m>s)2
2
d
= 81.78 N
Referring to the free-body diagram of the log in Fig. a,
d
+ ΣFx = max ; T - 81.78 N = 0
 T = 81.78 N = 81.8 N Ans.
11–79. A boat traveling with a constant velocity of 2 m>s 
tows a half submerged log having an approximate diameter 
of 0.35 m. If  the  drag coefficient is CD = 0.85, determine 
the tension in the tow rope if it is horizontal. The log is 
oriented so that the flow is along the length of the log.
FD = 81.77 N
x W
a = 0
T
N
(a)
Ans:
81.8 N
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es
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py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1221
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 0.00237 slug>ft3 and n = 0.158(10-3) ft2>s for air at 
T = 60° F. Thus, the Reynolds number is
Re =
UD
n
=
(20 ft>s)a 3
12
 ftb
0.158(10-3) ft2>s
= 3.165(104)
Entering this Re into the graph for a sphere, CD ≅ 0.5 (approx.). Here 
AP = p a1.5
12
 ftb
2
= 0.015625p ft2.
FD = CDADr 
U 2
2
= 0.5(0.015625p ft2)(0.00237 slug>ft3) c (20 ft>s)2
2
d
= 0.01163 lb Ans.
Writing the equation of motion along the y axis by referring to the free-body 
 diagram of the ball in Fig. a,
+ T ΣFy = may: 0.25 lb - 0.01163 lb = a 0.25 lb
32.2 ft>s2 ba
a = 30.7 ft>s2 Ans.
*11–80. A 0.25-lb ball has a diameter of 3 in. Determine 
the initial acceleration of the ball when it is thrown vertically 
downward with an initial speed of 20 ft>s. The air is at a 
temperature of 60°F.
FD = 0.01163 lb
y
0.25 lb
a
(a)
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law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1222
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The flow is steady.
From Appendix A, r = 0.00237 slug>ft3 and n = 0.158(10-3) ft2>s for air at 
T = 60° F. When the plate is held normal to the air flow, the drag is contributed 
by pressure drag only. For this case, CD = 1.1 for a square plate where 
b
h
= 1 and 
AP = 1 ft(1 ft) = 1 ft2.
Normal:
FD = CDAP r 
U 2
2
= 1.1(1 ft2)(0.00237 slug>ft3) c (50 ft>s)2
2
d
= 3.26 lb Ans.
When the plate is held parallel to the air flow, the drag is contributed by frictional 
drag only. Here, the Reynolds number for the flow at x = L = 1 ft is
ReL =
UL
n
=
(50 ft>s)(1 ft)
0.158(10-3) ft2>s
= 3.165(105)
Since ReL 6 (Rex)cr = 5(105), the boundary layer throughout the length of the 
plate is laminar. Since there are two surfaces subjected to flow
Parallel:
FD = Σ 
0.664brU 2L2ReL
= 2 c 0.664(1 ft)(0.00237 slug>ft3)(50 ft>s)2(1 ft)23.165(105)
d
= 0.0140 lb Ans.
11–81. A 1 ft by 1 ft square plate is held in air at 60°F, 
which is blowing at 50 ft>s. Compare the drag on the plate 
when it is held normal and then parallel to the air flow.
Ans:
Normal: FD = 3.26 lb
Parallel: FD = 0.0140 lb
This
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es
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py
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1223
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The flow is steady.
Here, FD acts through the mid-height ofthe drum as shown in its free-body diagram 
in Fig. a,
+ c ΣFy = 0; N - 8(9.81)N = 0 N = 78.48 N
S
+ ΣFx = 0; FD - F = 0 (1)
a+ ΣMO = 0; [8(9.81) N]x - FD(0.625 m) = 0 (2)
Assuming sliding occurs first,
F = msN = 0.3(78.48 N) = 23.544 N
Using this result to solve Eqs. (1) and (2),
FD = 23.544 N x = 0.1875 m
Since x 6 0.3 m, the drum will slide before it tips as assumed. From Appendix A, 
r = 1.164 kg>m3 and n = 16.0(10-6) m2>s for air at T = 30° C. Thus, the Reynolds 
number of the flow is
Re =
UD
n
=
U(0.6 m)
16.0(10-6) m2>s
= 37 500U (3)
Also, AP = (0.6 m)(1.25 m) = 0.75 m2.
FD = CDAP r 
U 2
2
23.544 N = CD(0.75 m2)(1.164 kg>m3)aU 2
2
b
U 2 =
53.938
CD
 (4)
The iterations carried out are tabulated as follows:
Iteration Assumed CD U (m>s): Eq. (4) Re: Eq. (3) CD from the graph gives
 1 1.0 7.344 2.75(105) 1.2
 2 1.2 6.704 2.51(105) 1.2
Since the assumed CD is almost the same as that obtained from the graph in iteration 
2, the result of U in the iteration is acceptable. Thus,
U = 6.70 m>s Ans.
11–82. The smooth empty drum has a mass of 8 kg and 
rests on a surface having a coefficient of static friction of 
ms = 0.3. Determine the speed of the wind needed to cause 
it to either tip over or slide. The air temperature is 30°C.
8(9.81) N
O
0.625 m
FD
F
N
(a)
x
1.25 m
0.3 m
Ans:
6.70 m>s
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s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
 an
d i
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itte
d. 
1224
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The flow is steady.
Here, FD acts through the mid-height of the drum as shown in its free-body diagram 
in Fig. a. Considering the equilibrium of the drum,
+ c ΣFy = 0; N - 8(9.81) N = 0 N = 78.48 N
S
+ ΣFx = 0; FD - F = 0 (1)
a+ ΣMO = 0; [8(9.81) N]x - FD(0.625 m) = 0 (2)
Assuming sliding occurs first,
F = msN = 0.6(78.48 N) = 47.088 N
Using this result to solve Eqs. (1) and (2),
FD = 47.088 N x = 0.375 m
Tipping will occur first. Setting x equal to 0.3 m in Eq. (2), FD = 37.6704 N.
From Appendix A, r = 1.164 kg>m3 and n = 16.0(10-6) m2>s for air at T = 30° C. 
Thus, the Reynolds number of the flow is
Re =
UD
n
=
U(0.6 m)
16.0(10-6) m2>s
= 37 500U (3)
Also, AP = (0.6 m)(1.25 m) = 0.75 m2.
FD = CDAP r 
U 2
2
37.67 N = CD(0.75 m2)(1.164 kg>m3)aU 2
2
b
U 2 =
86.301
CD
 (4)
The iterations carried out are tabulated as follows:
Iteration Assumed CD U (m>s): Eq. (4) Re: Eq. (3) CD from the graph
 1 1.0 9.290 3.48(105) 0.8
 2 0.8 10.386 3.89(105) 0.64
 3 0.64 11.61 4.35(105) 0.50
 4 0.50 13.137 4.92(105) 0.33
 5 0.35 15.703 5.89(105) 0.34
Since the assumed CD is almost the same as that obtained from the graph in iteration 
5, the result of U in the iteration is acceptable. Thus,
U = 15.7 m>s Ans.
11–83. The smooth empty drum has a mass of 8 kg and 
rests on a surface having a coefficient of static friction of 
ms = 0.6. Determine the speed of the wind needed to cause 
it to either tip over or slide. The air temperature is 30°C.
8(9.81) N
O
0.625 m
FD
F
N
(a)
x
1.25 m
0.3 m
Ans:
15.7 m>s
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law
s 
an
d i
s p
rov
ide
d s
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ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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 an
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itte
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1225
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The liquid is considered to be incompressible. The relative flow is steady.
The drag on the differential area dA = wdr is shown on the free-body diagram of 
the blade, Fig. a.
dFD = CDAP r 
U 2
2
Here, AP = dA = wdr and U = vr. Thus,
dFD = CD(wdr)r c (vr)2
2
d =
1
2
 CDrwv2r 2dr
Since the blade rotates with a constant angular velocity, moment equilibrium exists 
about the z axis. Thus,
2L
L
0
dFD(r) - T = 0
T = 2L
L
0
dFD(r) = 2L
L
0
1
2
 CDrwv2r 3dr
= CDrwv2L
L
0
r 3dr
= CDrwv2 ar 4
4
b 2 L
0
=
1
4
 CDrwv2L4 Ans.
*11–84. The blades of a mixer are used to stir a liquid 
having a density r and viscosity m. If each blade has a length 
L and width w, determine the torque T needed to rotate the 
blades at a constant angular rate v. Take the drag coefficient 
of the blade’s cross section to be CD.
z
dr
dr
dFD
T
dFDr
r
L
w
L
T
v
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law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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 an
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pe
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itte
d. 
1226
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Oil is considered to be incompressible. The relative flow is steady.
The initial Reynolds number is
Re =
UD
n
=
(0.08 m>s)(0.06 m)
40(10-6) m2>s
= 1200
Using this Re, CD ≅ 0.44. Also, AP = p(0.03 m)2 = 0.9(10-3)p m2.
FD = CDAP r0
U 2
2
= 0.4430.9(10-3)p m24 (880 kg>m3) c (0.8 m)2
2
d
= 0.35033 N
The volume of the ball is V =
4
3
 pr 3 =
4
3
 p(0.03 m)3 = 36(10-6)p m3. Thus, the 
weight of the ball and the bouyant force are
W = mg = rbVg = rb336(10-6)p m34 (9.81 m>s2) = 1.109(10-3)rb
FB = r0Vg = (880 kg>m3) 336(10-6)p m34 (9.81 m>s2) = 0.97635 N
Referring to the free-body diagram in Fig. a,
+ c ΣFy = may; 0.35033 N + 0.97635 N = 1.109(10-3)rb
rb = 1196 kg>m3 = 1.20 mg>m3 Ans.
11–85. A ball has a diameter of 60 mm and falls in oil with 
a terminal velocity of 0.8 m>s. Determine the density of the 
ball. For oil, take ro = 880 kg>m3 and n0 = 40(10-6) m2>s. 
Note: The volume of a sphere is V = 4
3pr 3. 
y
W = 1.109 10–3))
FB = 0.97635 N
FD = 0.36625 N
a = 0
b
(a)
Ans:
1.20 mg>m3
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s 
an
d i
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rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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pe
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itte
d. 
1227
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 0.00237 slug>ft3 and n = 0.158(10-3)ft2>s for air at 
T = 60° F. Thus, the initial Reynolds number is
Re =
UD
n
=
(18 ft>s)a 8
12
 ftb
0.158(10-3) ft2>s
= 7.59(104)
Entering this Re into the graph for a sphere, CD ≅ 0.5 (approx.). Also, 
AP = p a 4
12
 ftb
2
= 0.1111p ft2.
FD = CDAP r 
U 2
2
= 0.5(0.1111p ft2)(0.00237 slug>ft3) £ (18 ft>s)2
2
§
= 0.0670 lb Ans.
The drag force on the ball will not remain constant since the velocity of the ball 
changes. Furthermore, it also depends on the drag coefficient, which is a function of 
velocity.
11–86. A ball has a diameter of 8 in. If it is kicked with a 
speed of 18 ft>s, determine the initial drag acting on the ball. 
Does this force remain constant? The air is at a temperature 
of 60°F.
Ans:
0.0670 lb
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tat
es
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
 an
d i
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pe
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itte
d. 
1228
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
Here, we will assume that Re 6 1. Therefore, the drag is
FD = 3pmUD = 3p318.1(10-6) N # s>m24(U)33(10-6) m4
= 0.1629(10-9)pU
Referring to the free-body diagram in Fig. a,
+ c ΣFy = 0; 0.1629(10-9)pU - £ 42.5(10-12)
1000
§ (9.81) N = 0
U = 0.8147(10-3) m>s
Therefore, the Reynolds number is
Re =
rUD
m
=
(1.202 kg>m3) 30.8147(10- 3) m>s4 33(10-6) m4
18.1(10-6) N # s>m2
= 1.623(10-4) 6 1 (O.K.)
Thus, the time needed to settle
t =
s
U
=
8(103) m
0.8147(10-3) m>s
= 39.820(106) s4a 1 hr
3600 s
ba1 day
24 hr
b
= 113.66 days = 114 days Ans.
11–87. Particulate matter at an altitude of 8 km in the upper 
atmosphere has an average diameter of 3 µm. If a particle 
has a mass of 42.5(10-12) g, determine the time needed for it 
to settle to the earth. Assume gravity is constant, and for air, 
r = 1.202 kg>m3 and m = 18.1(10-6) N # s>m2.
y
42.5 10–12 
(9.81) N
1000
FD = 0.1629 10–9
(a)
U
a = 0
( (
( (
] ]
8 km
Ans:
114 days
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law
s 
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d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
 an
d i
s n
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pe
rm
itte
d. 
1229
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The liquid is considered to be incompressible. The relative flow is steady.
Here, we will assume that Re 6 1. Realizing that m = rLn, then
FD = 3pmUD = 3prLnUD
The volume of the ball is V =
4
3
 pr 3 =
4
3
 p aD
2
b
3
=
p
6
 D3. Thus, the weight of the 
ball and the bouyant force are
W = mg = rbVg = rb c
p
6
D3 d g =
p
6
 rb gD3
FB = rLVg = rLap6 D3bg =
p
6
 rL gD3
Referring to the free-body diagram in Fig. a,
+ c ΣFy = may; 3prLnUD +
p
6
rLgD3 -
p
6
rbgD3 = 0
U =
gD2(rb - rL)
18rLn
Substituting the data,
U =
(9.81 m>s2)(0.02 m)2(3000 kg>m3 - 2300 kg>m3)
18(2300 kg>m3)(0.052 m2>s)
= 0.001276 m>s = 0.00128 m>s Ans.
Thus, the Reynolds number is
Re =
UD
n
=
(0.001276 m>s)(0.02 m)
0.052 m2>s
= 0.4907(10-3) 6 1 (O.K.)
*11–88. A solid ball has a diameter of 20 mm and a density 
of 3.00 Mg>m3. Determine its terminal velocity if it is 
dropped into a liquid having a density of r = 2.30 Mg>m3 
and a viscosity of n = 0.052 m2>s. Note: The volume of a 
sphere is V = 4
3pr 3.
y
W =
FB =
FD = 3
(a)
a = 0
L UD
6 b gD3
6 L gD3
v
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s 
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d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
 an
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pe
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itte
d. 
1230
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, ra = 1.202 kg>m3 and ma = 18.1(10-6) N # s>m2 for air at 
T = 20° C. Thus, the maximum Reynolds number is
(Re)max =
rmaxUmaxD
ma
=
(1.202 kg>m3)(30 m>s) 30.4(10-6) m4
18.1(10-6) N # s>m2
= 0.7969
Since (Re)max 6 1, the drag on the droplet is
FD = 3pmaUD = 3p318.1(10-6) N # s>m24(U)30.4(10-6) m4
= 21.72(10-12)pU
Referring to the free-body diagram of the droplet in Fig. a,
+
SΣFx = max; -21.72(10-12)pU = c 0.4(10-12)
1000
 kg d adU
dt
b
 -54300pL
10(10-6) s
0
dt = L
V
30 m>s
dU
U
 
 -0.543p = ln U �V
30 m>s
 -0.543p = ln 
V
30
 e-0.543p =
V
30
 V = 5.45 m>s Ans.
11–89. Determine the velocity of the aerosol solid particles 
when t = 10 s, if when t = 0 they leave the can with a 
horizontal velocity of 30 m>s. Assume the average diameter 
of the droplets is 0.4 µm and each has a mass of 0.4(10-12) g. 
The air is at 20°C. Neglect the vertical component of 
the velocity. Note: The volume of a sphere is V = 4
3pr 3.
FD = 21.72
(a)
mg
a =
U( (10–12
du
dt
x
30 m/s
Ans:
5.45 m>s
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d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
 an
d i
s n
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pe
rm
itte
d. 
1231
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
Water is considered to be incompressible. The relative flow is steady.
Here, we will assume that Re 6 1. Realizing that m = rLn, then
FD = 3pmUD = 3prLnUD
The volume of the particles is V =
4
3
 pr 3 =
4
3
 p aD
2
b
3
=
p
6
 D3. Thus, the weight of 
the particles and the bouyant force are
W = mg = rVg = rbap6 D3bg =
p
6
 rbgD3
FB = rLVg = rLap6 D3bg =
p
6
 rLgD3
Referring to the free-body diagram in Fig. a,
+ c ΣFy = may; 3prLnUD +
p
6
 rLgD3 -
p
6
 rbgD3 = 0
U =
gD3(rb - rL)
18rLn
From Appendix A, rL = 998.3 kg>m3 and n = 1.00(10-6) m2>s for water at 
T = 20° C. Substituting the data,
U =
(9.81 m>s2) 350(10-6)m42(1600 kg>m3 - 998.3 kg>m3)
18(998.3 kg>m3) 31.00(10-6) m2>s4
= 0.8212(10-3) m>s
Thus, the Reynolds number is
Re =
UD
n
=
(0.8212(10-3) m>s) 350(10-6) m4
1.00(10-6) m2>s
= 0.0411 6 1 (O.K.)
Thus, the time required for the particles to settle is
t =
s
U
=
2 m
0.8212(10-3) m>s
= 2435.42 s a1 min
60 s
b
= 40.6 min Ans.
11–90. Impure waterat 20°C enters the retention tank and 
rises to a level of 2 m when it stops flowing in. Determine 
the shortest time needed for all sediment particles having a 
diameter of 0.05 mm or greater to settle to the bottom. 
Assume the density of the particles is r = 1.6 Mg>m3 or 
greater. Note: The volume of a sphere is V = 4
3pr 3.
y
W =
FB =
FD = 3
(a)
a = 0
L UD
6 b gD3
6 L gD3
v
5 m
2 m
2 m
Ans:
40.6 min
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ork
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s p
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ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
 an
d i
s n
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pe
rm
itte
d. 
1232
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–91. A ball having a diameter of 0.6 m and a mass of 
0.35 kg is falling in the atmosphere at 10°C. Determine its 
terminal velocity. Note: The volume of a sphere is V = 4
3pr 3.
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, ra = 1.247 kg>m3 and na = 14.2(10-6) m2>s for air at T = 10° C. 
Thus, the Reynolds number is
 Re =
UD
na
=
U(0.6 m)
14.2(10-6) m2>s
= 4.225(104)U (1)
The projected area perpendicular to the air stream is AP = p(0.3 m)2 = 0.09p m2.
 FD = CDAP r 
U 2
2
= CD(0.09p m2)(1.247 kg>m3)aU 2
2
b
 = 0.056115pCDU 2
The volume of the ball is V =
4
3
 pr 3 =
4
3
 p(0.3 m)3 = 0.036p m3. Thus, the bouyant 
force is
 FB = raVg = (1.247 kg>m3)(0.36p m3)(9.81 m>s2) = 1.3835 N
Referring to the free-body diagram in Fig. a,
+ c ΣFy = may; 0.056115pCDU 2 + 1.3835 N - 0.35(9.81) N = 0
The iterations carried out are tabulated as follows:
Iteration Asumed CD U (m>s); Eq. (2) Re; Eq. (1) CD from the graph
1 0.5 4.823 2.04(105) 0.46
2 0.46 5.028 2.12(105) 0.45
Since the assumed CD is almost the same as that obtained from the graph in iteration 
2, the result of U in this iteration is acceptable. Thus,
 U = 5.03 m>s Ans.
FB = 1.3835 N
FD = 0.056115 U2
0.35(9.81) N
y
a = 0
Ans:
5.03 m>s
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str
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ea
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the
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ou
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s a
nd
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se
ss
ing
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ud
en
t le
arn
ing
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iss
em
ina
tio
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r 
sa
le 
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eb
) 
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*11–92. A raindrop has a diameter of 1 mm. Determine its 
approximate terminal velocity as it falls. Assume that the air 
has a constant ra = 1.247 kg>m3 and na = 14.2(10-6) m2>s. 
Neglect buoyancy. Note: The volume of a sphere is V = 1
6pD3.
Solution
The air is considered to be incompressible. The relative flow is steady.
The Reynolds number is
 Re =
UD
na
=
U(0.001 m)
14.2(10-6) m2>s
= 70.422U (1)
The volume of the raindrop is V =
p
6
 D3 =
p
6
 (0.001 m)3 = 1.667(10-10)p m3. Thus, 
its weight is 
W = mg = rwVg = (1000 kg>m3) 31.667(10-10)p m34 (9.81 m>s2) = 5.1365(10-6) N
Here, AP = p a0.001 m
2
b
2
= 2.5(10-7)p m2.
 FD = CDAP r 
U 2
2
= CD32.5(10-7)p m24 (1.247 kg>m3)aU 2
2
b
 = 4.897(10-7)CDU 2
Referring to the free-body diagram in Fig. a,
+ c ΣFy = may; 4.897(10-7)CDU 2 - 5.1365(10-6) N = 0
 U 2 =
10.489
CD
 (2)
The iterations carried out are tabulated as follows:
Iteration Assumed CD U (m>s); Eq. (2) Re; Eq. (1) CD from the graph
1 0.5 4.58 323 0.66
2 0.66 3.98 280 0.7
Use CD = 0.7, and
 U = 3.87 m>s Ans.
FD
y
W
a = 0
(a)
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–93. The 2-Mg race car has a projected front area of 
1.35 m2 and a drag coefficient of 1CD2
C
= 0.28. If the car is 
traveling at 60 m>s, determine the diameter of the parachute 
needed to reduce the car’s speed to 20 m>s in 4 s. Take 
1CD2p = 1.15 for the parachute. The air is at 20°C. The 
wheels are free to roll.
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 1.202 kg>m3 and n = 15.1(10-6) m2>s for art at T = 20° C.
(FD)C = (CD)C(AP)Cr 
U 2
2
= 0.28(1.35 m2)(1.202 kg>m3)aU 2
2
b = 0.2272U 2
(FD)P = (CD)P(AP)P r 
U 2
2
= 1.15 ap
4
 d2b (1.202 kg>m3)aU 2
2
b = 0.5428d2U 2
Writing the equation of motion along the x axis by referring to the free-body 
 diagram shown in Fig. a,
+
dΣFx = ma; -0.2272U 2 - 0.5428d2U 2 = 2(103)dU
dt
 - (0.2272 + 0.5428d2)U 2 = 2(103)dU
dt
 J - (0.2272 + 0.5428d2)
2(103)
R L
4 s
0
dt = L
20 m>s
60 m>s
dU
U 2
 -0.002(0.2272 + 0.5428d2) =
1
U
`
20 m>s
60 m>s
 0.002(0.2272 + 0.5428d2) =
1
20
 - 
1
60
 d = 5.50 m Ans.
FD c = 0.2272U2
(a)
x
a =
(( FD P = 0.5428d2U2((
du
dt
Ans:
5.50 m
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s a
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arn
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11–94. A 2-mm-diameter sand particle having a density of 
2.40 Mg>m3 is released from rest at the surface of oil that is 
contained in the tube. As the particle falls downward, 
“creeping flow” will be established around it. Determine 
the velocity of the particle and the time at which Stokes’ law 
becomes invalid, at about Re = 1. The oil has a density of 
ro = 900 kg>m3 and a viscosity of mo = 30.2110-32 N # s>m2. 
Assume the particle is a sphere, where its volume is 
V = 4
3pr 3.
Solution
W = rVg = (2400 kg>m3) c 4
3
 p(0.001 m)3 d (9.81 m>s2) = 9.8621(10-5) N
Fb = roVg = (900 kg>m3) c 4
3
 p(0.001 m)3 d (9.81 m>s2) = 3.6983(10-5) N
FD = 3pm0VD = 3p330.2(10- 3) N # s>m24V(0.002 m)
= 5.6926(10-4)V
We solve Re = 1 to find V:
roVD
m0
= 1
V =
m0
roD
=
30.2(10- 3) N # s>m2
(900 kg>m3)(0.002 m)
= 0.016778 m>s
Now we integrate, starting with Newton’s Second Law: 
+ T ΣFy = ma;
W - Fb - FD = m 
dV
dt
9.8621(10- 5) - 3.6983(10-5) - 5.6926(10-4)V
= (2400) c 4
3
 p(0.001 m3) d dV
dt
6.1638(10-5) - 5.6926(10-4)V = 1.00531(10-5)dV
dt
L
t
0
dt =L
0.016778
0
1.00531(10-5)
6.1638(10-5) - 5.6926(10-4)V
 dV
t = 0.002973 s = 2.97 ms Ans.
FD
Fb
W
(a)
Ans:
V = 16.8 mm>s
t = 2.97 ms
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ea
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ir c
ou
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se
ss
ing
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ud
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–95. Dust particles having an average diameter of 
0.05 mm and an average density of 450 kg>m3 are stirred up 
by an airstream and blown off the edge of the 600-mm-high 
desk into a horizontal steady wind of 0.5 m>s. Determine 
the distance d from the edge of the desk where most of 
them will strike the ground. The air is at a temperature of 
20°C. Note: The volume of a sphere is V = 4
3 pr 3.
Solution
Due to the smallness of a dust particle, the flow can be assumed steady and 
often referred to as creeping flow. Also, the air will be assumed incompressible. 
Appendix  A gives ra = 1.202 kg>m3 and ma = 18.1(10-6) N # s>m2. For creeping 
flow we assume that Re 6 1 so that stokes equation FD = 3pmaVD can be used. 
The bouyant force is Fb = raVg, and the weight of the dust W = mg = rdVg. Since 
the dust is creeping in the vertical direction with its terminal velocity (constant), 
then referring to the FBD in Fig. a, 
+ T ΣFy = 0; rdVg - raVg - 3pmaVD = 0
V =
(rd - ra)Vg
3pmaD
since V =
4
3
 p aD
2
b
3
=
p
6
 D3, the above equation becomes 
V =
(rd - ra)ap
6
D3bg
3pmaD
=
(rd - ra)gD2
18ma
Substitute the numerical data into this equation to find the terminal downward 
velocity, 
V =
(450 kg>m3 - 1.202 kg>m3)(9.81 m>s2) 30.05(10-6) m42
18318.1(10-6) N # s>m24
 = 0.03378 m>s
Then the Reynolds number is 
Re =
raVD
ma
=
(1.202 kg>m3)(0.03378 m>s) 30.05(10-3) m4
18.1(10-6) N # s>m2
= 0.1122 6 1 (O.K.)
FD
Fb
W
(a)
600 mm
0.5 m/s
d
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 in
str
uc
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s i
n t
ea
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ing
 
the
ir c
ou
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s a
nd
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se
ss
ing
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ud
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arn
ing
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iss
em
ina
tio
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r 
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*11–95. Continued
The time for the dust to strike the ground can therefore be determined from
t =
h
v
=
0.6 m
0.03378 m>s
= 17.76 s
Thus, the horizontal distance d is
d = Ut = (0.5 m>s)(17.76 s) = 8.88 m Ans.
Ans:
8.88 m
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e u
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 of
 in
str
uc
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s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
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se
ss
ing
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ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
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art
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is 
work
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orl
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eb
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the
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*11–96. A rock is released from rest at the surface of the 
lake, where the average water temperature is 15°C. If 
CD = 0.5, determine its speed when it reaches a depth of 
600 mm. The rock can be considered a sphere having a 
diameter of 50 mm and a density of rp = 2400 kg>m3. 
Note: The volume of a sphere is V = 4
3pr 3.
Solution
+ T ΣFy =
d(mV)
dt
= m
dV
dt
= m
VdV
ds
Referring to the FBD shown in Fig. a, and realizing that Fb = rw g V, 
FD = CDAp a
rwV2
2
b and W = rsg V,
 rsg V - rwg V - CDAr arwV2
2
b = rsV 
VdV
ds
 2(rs - rw)Vg - CDAprwV2 = 2rsV 
VdV
ds
with the initial condition at s = 0, V = 0, 
 L
s
0
ds = 2rsVL
V
0
VdV
2(rs - rw)Vg - CpAprwV2
Let a = 2(rs - rw)V g and b = CDAprw. Then 
 L
s
0
ds = 2rsVL
V
0
VdV
a - bV2
 s =
2rsV
2( -b)
c ln(a - bV2) d `
V
0
 s =
rsV
b
 ln a a
a - bV2 b
 
bs
rsV
= ln a a
a - bV2 b
 
a
a - bV2 = e 
bs
rsV
 bV2 = a - ae -
bs
rsV
 V = A a
b
(1 - e-bs>rsV) (1)
FD
Fb
W
(a)
600 mm
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str
uc
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s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
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se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
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art
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is 
work
 (in
clu
din
g o
n t
he
 W
orl
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ide
 W
eb
) 
will 
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the
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*11–96. Continued
Substituting the numerical data, rs = 2400 kg>m3, rw = 999.2 kg>m3, s = 0.6 m
V =
4
3
 p(0.025 m)3 = 65.4498(10-6) m3 and Ap = p(0.025 m)2 = 0.625(10-3)p m2,
a = 2(2400 kg>m3 - 999.2 kg>m3) 365.4498(10-6)m34 (9.81 m>s2) = 1.7988
b = CD30.625(10-3)p m24 (999.2 kg>m3) = 1.9619CD
rsV = (2400 kg>m3) 365.4498(10-6) m34 = 0.15708 kg
Then Eq. (1) becomes
 V = c A0.9169
CD
(1 - e-7.4940CD) d m>s (2)
Using CD = 0.5,
at s = 0.6 m V = 1.34 m>s Ans.
The terminal velocity can be obtained by setting s S ∞ . Then Eq. (1) becomes 
 Vt = A0.9169
CD
Again using CD = 0.5,
 Vt = 1.35 m>s Ans.
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e u
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 of
 in
str
uc
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s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
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se
ss
ing
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ud
en
t le
arn
ing
. D
iss
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ina
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n o
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of 
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11–97. The smooth cylinder is suspended from the rail and 
is partially submerged in the water. If the wind blows at 
8 m>s, determine the terminal velocity of the cylinder. The 
water and air are both at 20°C.
Solution
The fluids are considered incompressible. The relative fluid is steady. From Appendix A, 
ra = 1.202 kg>m3 and va = 15.1(10-6) m2>s for air and rw = 998.3 kg>m3 and 
nw = 1.00(10-6) m2>s for water at T = 20° C. If the terminal velocity of the cylinder 
is V0, Ua = 8 m>s - V0 and Uw - V0. Thus, the Reynolds number for air and water 
are 
 (Re)a =
UaD
na
=
(8 - V0)(0.25 m)
15.1(10-6) m2>s
= 1.6556(104)(8 - V0) (1)
 (Re)w =
UwD
nw
=
V0(0.25 m)
1.00(10-6) m2>s
= 2.5(105)V0 (2)
The projected areas perpendicular to the stream for air andwater are 
(AP)a = (0.25 m)(1 m) = 0.25 m2 and (AP)w = (0.25 m)(0.5 m) = 0.125 m2.
 (FD)a = (CD)a(AP)a ra
Ua
2
2
= (CD)a(0.25 m)(1.202 kg>m3) J (8 - V0)2
2
R
 = 0.15025(CD)a(V0
2 - 16n0 + 64)
 (FD)w = (CD)w(AP)wrw
Uw
2
2
= (CD)w(0.125 m)(998.3 kg>m3) c V0
2
2
d
 = 62.394(CD)wV0
2
Writing the equation of motion along the x axis by referring to the free-body 
 diagram in Fig. a, 
S+ ΣFx = max; (FD)a - (FD)w = 0
 (FD)a = (FD)w
 0.15025(CD)a(V0
2 - 16n0 + 64) = 62.394(CD)wV0
2
 (CD)a(V0
2 - 16v0 + 64) - 415.27(CD)wV0
2 = 0 (3)
The iterations carried out are tabulated as follows:
Assumed Value from the graph
Iteration (CD)a (CD)w V0(m>s); Eq. (3) (Re)a; Eq. (1) (Re)w; Eq. (2) (CD)a (CD)w
1 1.3 1.3 0.3742 1.26(105) 9.36(104) 1.4 1.4
Since the assumed CD is almost the same as that obtained from the graph in itera-
tion 1, the result of v0 in the iteration is acceptable. Thus, 
 V0 = 0.374 m>s Ans.
x
W
N1 N2
FD a
(a)
a = 0
( (
FD k( (
1 m
0.5 m
0.25 m
8 m/s
Ans:
0.374 m>s
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 in
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ea
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the
ir c
ou
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s a
nd
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se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
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pe
rm
itte
d. 
1241
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–98. A 5-m-diameter balloon and the gas within it have 
a mass of 80 kg. Determine its terminal velocity of descent. 
Assume the air temperature is at 20°C. Note: The volume of 
a sphere is V = 4
3pr 3.
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, ra = 1.202 kg>m3 and va = 15.1(10-6) m2>s for air at T = 20° C. 
Thus, the Reynolds number is
 Re =
UD
va
=
U(5 m)
15.1(10-6) m2>s
= 3.3113(105)U
The volume of the balloon is V =
4
3
 pr 3 =
4
3
p a5 m
2
b
3
= 65.45 m3. Thus, the bouy-
ant force is
 FB = raVg = (1.202 kg>m3)(65.45 m3)(9.81 m>s2) = 771.76 N
Here, AP = p a5 m
2
b
2
= 6.25 p m2
 FD = CDAP r 
U 2
2
 CD(6.25p m2)(1.202 kg>m3)aU 2
2
b
 = 3.75625 pCDU 2
Referring to the free-body diagram in Fig. a,
+ c ΣFy = may; 3.75625pCDU 2 + 771.76 N - 80(9.81) N = 0
 U 2 =
1.1051
CD
The iterations carried out are tabulated as follows:
Iteration Assumed CD U(m>s); Eq. (2) Re; Eq. (1) CD from the graph
1 0.2 2.351 7.78(105) 0.14
2 0.14 2.809 9.30(105) 0.155
3 0.155 2.670 8.84(105) 0.15
Since the assumed CD is almost the same as that obtained from the graph in itera-
tion 3, the result of U in this iteration is acceptable. Thus, 
 U = 2.67 m>s Ans.
yW = 100 (9.81) N
FD = 3.75625 CDU2
(a)
a = 0
FB = 771.76 N
Ans:
2.67 m>s
This
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tat
es
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s 
an
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ide
d s
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ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1242
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–99. A smooth ball has a diameter of 43 mm and a mass 
of 45 g. When it is thrown vertically upward with a speed of 
20  m>s, determine the initial deceleration of the ball. The 
temperature is 20°C.
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 1.202 kg>m3 and v = 15.1(10-6) m2>s for air at T = 20° C. 
Thus, the initial Reynolds number is
 Re =
UD
v
=
(20 m>s)(0.043 m)
15.1(10-6) m2>s
= 5.695(104)
Entering this Re into the graph for a sphere, CD ≅ 0.5 (aprox.). Here, 
AP = p a0.043 m
2
b
2
= 0.46225(10-3)p m2.
 FD = CDADr 
U 2
2
= 0.530.46225(10-3)p m24 (1.202 kg>m3) J (20 m>s)2
2
R
 = 0.1746 N Ans.
Referring to the free-body diagram of the ball in Fig. a,
+ c ΣFy = may; - 30.045(9.81) N4 - 0.1746 N = 0.045a
 a = 13.7 m>s2 Ans.
y
FD = 0.1746 N
0.045(9.81) N
(a)
a
Ans:
13.7 m>s2
This
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tat
es
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py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1243
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*11–100. The parachutist has a total mass of 90 kg and is 
in free fall at 6 m>s when she opens her 3-m-diameter 
parachute. Determine the time for her speed to be increased 
to 10 m>s. Also, what is her terminal velocity? For the 
calculation, assume the parachute to be similar to a hollow 
hemisphere. The air has a density of ra = 1.25 kg>m3. 
Solution
Relative to the parachutist, the flow is unsteady and uniform since he is decelerating. 
Here, the air is assumed to be incompressible. Applying the momentum equation
+ T ΣFy =
0
0t Lcv
VrdV + Lcs
VrVdA
The control volume considered is the parachute and the parachutist. Since there is 
no opened control surface, Lcs
VrVdA = 0. Also, Vr can be factored out from the 
integral since it is independent of V . Also Lcv
dV = V since the volume of the 
control volume is fixed. Realizing that rV = m, the above equation reduces to 
+ T ΣFy =
d(mv)
dt
= m
dv
dt
Referring to the FBD shown in Fig. a, and realizing that FD = CDAP a
raV
2
2
b ,
 mg - CDAPa
raV
2
2
b = m 
dv
dt
 
2mg - CDAP raV
2
2
= m 
dv
dt
with the initial condition at t = 0, V = Vo,
 L
t
0
dt = 2 mL
V
V0
dV
2 mg - cDAP raV
2
 t =
2 m
2(22mg)2CDAp ra
 ln°22mg + 2CDAP raV22mg - 2CDAP raV
¢ †
V
Vo
t =
m22mgCDAP ra
 £ ln °22mg + 2CDAP raV22mg - 2CDAP raV
¢ - ln°22mg + 2CDAP raVo22mg - 2CDAP raVo
¢ § (1)
Substituting the numerical data, m = 90 kg, CD = 1.4 (table 11–3), AP = p(1.5 m)2
= 2.25 p m2, ra = 1.25 kg>m3, Vo = 6 m>s and V = 12 m>s, we have 
 22mg = 22(90 kg)(9.81 m>s2 ) = 42.02
 2CDAP ra = 21.4(2.25p m2)(1.25 kg>m3 ) = 3.5171
mg
FD 
(a)
V
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d i
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ide
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ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1244
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*11–100. Continued
Then
t =
90
(42.02)(3.5045)
£ ln °42.02 + 3.5171(10)
42.02 - 3.5171(10)
¢ - ln °42.02 + 3.5171(6)
42.02 - 3.5171(6)¢ §
 = 0.805 s Ans.
Terminal velocity occurs when t = ∞ . By inspecting Eq. (1), this condition can be 
satisfied if 
 22mg + 2CDAP raVt = 0
 Vt = A 2mg
CDAP ra
=
42.02
3.5171
= 11.95 m>s = 12.0 m>s Ans.
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es
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law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1245
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
For two wings, A = 2(5 m)(1.75 m) = 17.5 m2. Thus, the lift is
FL = CLAr
U 2
2
= CL(17.5 m2)(1.225 kg>m3) c (70 m>s)2
2
d = 52521.875 CL
The equilibrium along a vertical requires
+ c ΣFy = 0; FL - W = 0
 52521.875 CL - 3000(9.81) N = 0
 CL = 0.560
Entering this value of CL into the graph
a = 5° (approx.) Ans.
11–101. A 3-Mg airplane is flying at a speed of 70 m>s. If 
each wing can be assumed rectangular of length 5 m and 
width 1.75 m, determine the smallest angle of attack a to 
provide lift assuming the wing is a NACA 2409 section. The 
density of air is r = 1.225 kg>m3.
5 m 5 m
Ans:
5° (approx.)
This
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es
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law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1246
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r = 0.9092 kg>m3 for air at an altitude of 3 km. For two wings, 
A = 2(5 m)(1.75 m) = 17.5 m2. Thus, the lift is 
FL = CLAr
U 2
2
= CL(17.5 m2)(0.9092 kg>m3) c (150 m>s)2
2
d = 178998.75 CL
Equilibrium requires
+ c ΣFy = 0; FL - W = 0
 178998.75 CL - 5000(9.81) N = 0
 CL = 0.274 Ans.
11–102. The 5-Mg airplane has wings that are each 5 m 
long and 1.75 m wide. It is flying horizontally at an altitude 
of 3 km with a speed of 150 m>s. Determine the lift 
coefficient. 
5 m 5 m
Ans:
0.274
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tat
es
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1247
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r1 = 0.9092 kg>m3 and r2 = 0.7364 kg>m3 for air at an altitude 
of 3 km and 5 km, respectively. Here, it is required that 
(FL)1 = (FL)2
(CL)1A1r1
U 2
1
2
= (CL)2A2r2
U 2
2
2
U2 = °A (CL)1A1r1
(CL)2A2r2
¢U1
Since A1 = A2 and the angle of attack is the same for both cases, (CL)1 = (CL)2. 
Thus,
U2 = Ar1
r2
U1 = °A0.9092 kg>m3
0.7364 kg>m3
¢(150 m>s) = 167 m>s Ans.
11–103. The 5-Mg airplane has wings that are each 5 m 
long and 1.75 m wide. Determine its speed in order to 
generate the same lift when flying horizontally at an altitude 
of 5 km as it does when flying horizontally at 3 km with a 
speed of 150 m>s.
5 m 5 m
Ans:
167 m>s
This
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ork
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1248
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
For two wings, A = 2(5 m)(1.75 m) = 17.5 m2. Thus, the lift is
FL = CLAr
U 2
2
= CL(17.5 m2)(1.225 kg>m3) c (70 m>s)2
2
d = 52521.875CL
The equilibrium along a vertical requires
+ c ΣFy = 0; FL - W = 0
 52521.875CL - 4000(9.81) N = 0
 CL = 0.747
Entering this value of CL into the graph
a = 8.20° (approx.)
Using this result, the graph gives CD ≅ 0.04 (approx.). For each wing, 
A = 5 m(1.75 m) = 8.75 m2.
FD = CDAr
U 2
2
= 0.04(8.75 m2)(1.225 kg>m3) c (70 m>s)2
2
d
= 1.05 kN Ans.
*11–104. A 4-Mg airplane is flying at a speed of 70 m>s. If 
each wing can be assumed rectangular of length 5 m and 
width 1.75 m, determine the drag on each wing when it is 
flying at the proper angle of attack a. Assume each wing is a 
NACA 2409 section. The density of air is ra = 1.225 kg>m3.
This
 w
ork
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cte
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nit
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 S
tat
es
 co
py
rig
ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1249
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
Solution
The air is considered to be incompressible. The relative flow is steady.
From Appendix A, r1 = 1.007 kg>m3 and r2 = 1.225 kg>m3 for air at an altitude of 
2 km and 0 km, respectively. Here, it is required that
(FL)1 = (FL)2
(CL)1A1r1
U 2
1
2
= (CL)2A2r2
U 2
2
2
U2 = °A (CL)1A1r1
(CL)2A2r2
¢U1
Since A1 = A2 and the angle of attack is the same for both cases, (CL)1 = (CL)2. 
Thus,
U2 = Ar1
r2
U1 = °A1.007 kg>m3
1.225 kg>m3
¢(250 km>h) = 227 km>h Ans.
11–105. The plane can take off at 250 km>h when it is at an 
airport located at an elevation of 2 km. Determine the 
takeoff speed from an airport at sea level.
V
Ans:
227 km>h
This
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tat
es
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py
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
grity 
of 
the
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ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1250
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–106. The glider has a weight of 350 lb. If the drag 
coefficient is CD = 0.456, the lift coefficient is CL = 1.20, 
and the total area of the wings is A = 80 ft2, determine the 
angle u at which it is descending with a constant speed.
Solution
The air is considered to be incompressible. The relative flow is steady.
 ΣFx′ = max′; FD - W sin u = 0 FD = W sin u (1)
 ΣFy′ = may′; FL - W cos u = 0 FL = W cos u (2)
Dividing Eq. (1) by Eq. (2),
 
FD
FL
=
W sin u
W cos u
= tan u (3)
The drag and lift are
 FD = CDAr
U 2
2
 FL = CLAr
U 2
2
Substituting these results into Eq. (3),
 
CDAr
U 2
2
CLAr
U 2
2
= tan u
 tan u =
CD
CL
=
0.456
1.2
 u = 20.8° Ans.
y´
x´
FL
(a)
FD
W
a = 0
u
Ans:
20.8°
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ork
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py
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ht 
law
s 
an
d i
s p
rov
ide
d s
ole
ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
s n
ot 
pe
rm
itte
d. 
1251
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–107. The glider has a weight of 350 lb. If the drag 
coefficient is CD = 0.316, the lift coefficient is CL = 1.20, 
and the total area of the wings is A = 80 ft2, determine if it 
can land on a landing strip that is 1.5 km long and located 
5 km away from where its altitude is 1.5 km. Assume the 
density of the air remains constant.
Solution
The air is considered to be incompressible. The relative flow is steady.
 ΣFx′ = max′; FD - W sin u = 0 FD = W sin u (1)
 ΣFy′ = may′; FL - W cos u = 0 FL = W cos u (2)
Dividing Eq. (1) by Eq. (2),
 
FD
FL
=
W sin u
W cos u
= tan u (3)
The drag and lift are
 FD = CDAr 
U 2
2
 FL = CLAr 
U 2
2
Substituting these results into Eq. (3),
 
CDAr 
U 2
2
CLAr 
U 2
2
= tan u
 tan u =
CD
CL
=
0.316
1.2
 u = 14.75°
Referring to the geometry shown in Fig. a,
 tan 14.75° =
1.5 km
d
 d = 5.7 km
Since 5 km 6 d 6 (5 + 1.5) km = 6.5 km, the glider can land on the landing strip. 
1.5 Km
= 14.75′
(a)
d
u
Ans:
The glider can land.
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str
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ch
ing
 
the
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s a
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 as
se
ss
ing
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ud
en
t le
arn
ing
. D
iss
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r 
sa
le 
of 
an
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art
 of
 th
is 
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 W
orl
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ide
 W
eb
) 
will 
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oy
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of 
the
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1252
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*11–108. Each of the two wings on a 20 000-lb airplane is 
to have a span of 25 ft and an average cord distance of 5 ft. 
When a 1>15 scale model of the wing section (assumed to 
be infinite) is tested in a wind tunnel at 1500 ft>s, using a 
gas for which rg = 7.80110-32 slug>ft3, the total drag is 160 
lb. Determine the total drag on the wing when the plane 
is  flying at a constant altitude with a speed of 400 ft>s,  
where  ra = 1.75110-32 slug>ft3. Assume an elliptical 
lift distribution.
Solution
We will assume that the flow is steady relative to airplane and the air and the gas is 
incompressible. For the model plane, 
 (AP)m = 2 c 1
15
 (25 ft) d c 1
15
 (5 ft) d = 1.1111 ft2
The drag coefficient for the infinite span can be determined using the model plane.
(CD) ∞ =
(FD)m
(AP)ma
rgV
2
m
2
b
=
160 lb
(1.1111 ft2)• 37.80(10-3) slug>ft34 (1500 ft>s)2
2
¶
= 0.01641
Since the plane is flying at a constant altitude, it is in vertical equilibrium. This means 
that the lift is equal to its weight; ie, FL = 20000 lb. Then the lift coefficient is
CL =
FL
APa
raV
2
2
b
=
20000
[2(25)(5 ft)]• 31.75(10-3) slug>ft34 (400 ft>s)2
2
¶
= 0.5714
The total drag coefficient can be determined by applying
CD = (CD) ∞ +
C 2
L
pb2>AP
 = 0.01641 +
0.57142
p(25 ft)2>(25 ft)(5 ft)
 = 0.03720
Thus, the total drag force on the wing is
 FD = CDAP a
raV
2
2
b
 = 0.03720[2(25 ft)(5 ft)]• 31.75(10-3) slug>ft34 (400 ft>s)2
2
¶
 = 1302 lb Ans.
This
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law
s 
an
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d s
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ly 
for
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e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
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s a
nd
 as
se
ss
ing
 st
ud
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arn
ing
. D
iss
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n o
r 
sa
le 
of 
an
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 of
 th
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 (in
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n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
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 an
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d. 
1253
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–109. The glider has a constant speed of 8 m>s through 
still air. Determine the angle of descent u if it has a lift 
coefficient of CL = 0.70 and a wing drag coefficient of 
CD = 0.04. The drag on the fuselage is considered negligible 
compared to that on the wings, since the glider has a very 
long wingspan.
Solution
Since the glider is gliding with a constant velocity, it is in equilibrium. Referring to 
the FBD of the glider in Fig. a,
+
R ΣFx = 0; W sin u - FD = 0
 W sin u - CDAP a
raV
2
2
b = 0
 W sin u = CDAP a
raV
2
2
b (1)
+ c ΣFy = 0; FL - W cos u = 0
 CLAP a
raV
2
2
b - W cos u = 0
 W cos u = CLAP a
raV
2
2
b (2)
Divided Eq. (1) by Eq. (2)
 
W sin u
W cos u
=
CDAP a
raV
2
2
b
CLAP a
raV
2
2
b
 tan u =
CD
CL
 u = tan-1aCD
CL
b = tan-1a0.04
0.7
b = 3.27° Ans.
y
x
FLFD 
W
(a)
8 m/s
u
Ans:
3.27°
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law
s 
an
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rov
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d s
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ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
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n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
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pe
rm
itte
d. 
1254
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–110. The 2000-lb airplane is flying at an altitude of 
5000 ft. Each wing has a span of 16 ft and a cord length of 
3.5 ft. If each wing can be classified as a NACA 2409 section, 
determine the lift coefficient and the angle of attack when 
the plane is flying at 225 ft>s.
Solution
Relative to the airplane, the flow is steady. Also, air is assumed to be incompressible. 
Appendix A gives ra = 2.043(10-3) slug>ft3. Since the air plane is flying at a 
 constant altitude, equilibrium exists along the vertical. Thus 
+ c ΣFy = 0; FL - W = 0
 CLAP a
raV
22
b - W = 0
 CL32(16 ft)(3.5 ft)4 • 32.043(10-3) slug>ft34 (225 ft>s)2
2
¶ - 2000 lb = 0
 CL = 0.345 Ans.
with this value of CL,
 a = 3° (Approx.) Ans.
16 ft
3.5 ft
16 ft
3.5 ft
Ans:
CL = 0.345
a = 3° (Approx.)
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law
s 
an
d i
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rov
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d s
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ly 
for
 th
e u
se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
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pe
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itte
d. 
1255
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–111. The 2000-lb airplane is flying at an altitude of 
5000 ft. Each wing has a span of 16 ft and a cord length 
of 3.5 ft, and it can be classified as a NACA 2409 section. 
If  the plane is flying at 225 ft>s, determine the total drag 
on  the wings. Also, what is the angle of attack and 
the  corresponding velocity at which the condition of 
stall occurs?
Solution
Relative to the airplane, the flow is steady. Also, air is assumed to be incompressible. 
Appendix A gives ra = 2.043(10-3) slug>ft3. Since the air plane is flying at a con-
stant altitude, equilibrium exists along the vertical. Thus,
+ c ΣFy = 0; FL - W = 0
CLApa
raV
2
2
b - W = 0 (1)
CL[2(16 ft)(3.5 ft)]• 32.043(10-3) slug>ft34 (225 ft>s)2
2
¶ - 2000 lb = 0
CL = 0.3453
with this value of CL, 
a = 2.75
with this angle of attack, 
(CD) ∞ = 0.015
The total drag coefficient can be determined using 
CD = (CD) ∞ +
CL
2
pb2>A
= 0.015 +
0.34532
p(16 ft)2>(16 ft)(3.5 ft)
= 0.0233
Thus, the drag force on the airplane caused by the wing is 
FD = CD APa
raV
2
2
b
= 0.023332(16 ft)(3.5 ft)4 • 32.043(10-3) slug>ft34 (225 ft>s)2
2
¶
= 135 lb Ans.
From the text, the condition of stall occurs when the angle of attack is
a = 20° Ans.
And the corresponding lift coefficient is 
CL = 1.50
Again, applying Eq. (1)
CL APa
raV
2
2
b - W = 0
1.5[2(16 ft)(3.5 ft)]• 32.043(10-3) slug>ft34Vs
2
2
¶ - 2000 lb = 0
Vs = 107.95 ft>s = 108 ft>s Ans.
16 ft
3.5 ft
Ans:
FD = 135 lb
a = 20°
Vs = 108 ft>s
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 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
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s a
nd
 as
se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
g o
n t
he
 W
orl
d W
ide
 W
eb
) 
will 
de
str
oy
 th
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nte
gri
ty 
of 
the
 w
ork
 an
d i
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itte
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1256
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
*11–112. If it takes 80 kW of power to fly an airplane at 
20 m>s, how much power does it take to fly the plane at 
25 m>s at the same altitude? Assume CD remains constant.
Solution
We will assume that the flow is steady relative to the airplane and the air is incom-
pressible. When the speed is 20 m>s, 
P = FDV; 80(103)W = (FD)1(20 m>s)
(FD) = 4000 N
Using the drag force equation, 
(FD)1 = CDApa
raV1
2
2
b
4000 N = CD AP c
ra(20 m>s)2
2
d
CD =
20
ra AP
when the speed is 25 m>s,
(FD)2 = CD APa
raV2
2
2
b
= a 20
raAp
b (Ap) c ra(25 m>s)2
2
d
= 6250 N
Thus, the power regained is 
W
#
2 = (FD)2V2 = (6250 N)(25 m>s)
= 156.25(103)W
= 156 kW Ans.This
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se
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str
uc
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n t
ea
ch
ing
 
the
ir c
ou
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s a
nd
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se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
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n t
he
 W
orl
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ide
 W
eb
) 
will 
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str
oy
 th
e i
nte
gri
ty 
of 
the
 w
ork
 an
d i
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pe
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itte
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1257
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–113. The plane weighs 9000 lb and can take off from an 
airport when it attains an airspeed of 125 mi>h. If it carries 
an additional load of 750 lb, what must be its airspeed 
before takeoff at the same angle of attack?
Solution
The air is considered to be incompressible. The relative flow is steady.
Equilibrium along the vertical requires 
+ c ΣFy = 0; FL - W = 0
 FL = W (1)
The lift is FL = CL Ar
U 2
2
. Thus, using Eq. (1),
(FL)1 = (CL)1A1r1
U 1
2
2
= W1 (2)
(FL)2 = (CL)2A2r2
U 2
2
2
= W2 (3)
Dividing Eq. (3) by Eq. (2),
(CL)2A2r2
U 2
2
2
(CL)1A1r1
U 1
2
2
=
W2
W1
U2 = °A (CL)1A1r1W2
(CL)2A2r2W1
¢U1
Here, A1 = A2 and (CL)1 = (CL)2. Thus,
r1 = r2,
U2 = AW2
W1
U1 = °A9750 lb
9000 lb
¢(125 mi>h) = 130 mi>h Ans.
125 mi/h
Ans:
130 mi>h
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s a
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se
ss
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 st
ud
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arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
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 W
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ide
 W
eb
) 
will 
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str
oy
 th
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nte
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of 
the
 w
ork
 an
d i
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1258
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–114. A baseball has a diameter of 73 mm. If it is thrown 
with a speed of 5 m>s and an angular velocity of 60 rad>s, 
determine the lift on the ball. Assume the surface of the ball 
is smooth. Take ra = 1.20 kg>m3 and na = 15.0110-62 m2>s, 
and use Fig. 11–50.
Solution
For the given data,
vD
2V
=
(60 rad>s)(0.073 m)
2(5 m>s)
= 0.438
Re =
VD
v
=
(5 m>s)(0.073 m)
15.0(10-6) m2>s
= 2.43(104)
Since Re is in the range of 104, the figure in the text can be used to determine the lift 
coefficient. Here CL ≈ 0.13. Thus,
FL = CL APa
raV
2
2
b
= 0.133p(0.0365 m)24 £ (1.20 kg>m3)(5 m>s)2
2
§
= 0.00816 N Ans.
5 m/s
60 rad/s
Ans:
0.00816 N
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s 
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se
 of
 in
str
uc
tor
s i
n t
ea
ch
ing
 
the
ir c
ou
rse
s a
nd
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se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
em
ina
tio
n o
r 
sa
le 
of 
an
y p
art
 of
 th
is 
work
 (in
clu
din
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n t
he
 W
orl
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ide
 W
eb
) 
will 
de
str
oy
 th
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nte
gri
ty 
of 
the
 w
ork
 an
d i
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1259
© 2014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
11–115. A 0.5-kg ball having a diameter of 50 mm is thrown 
with a speed of 10 m>s and has an angular velocity of 
400 rad>s. Determine its horizontal deviation d from striking 
a target a distance of 10 m away. Take ra = 1.20 kg>m3 and 
na = 15.0110-62 m2>s, and use Fig. 11–50.
Solution
From the given data, 
vD
2V
=
(400 rad>s)(0.05 m)
2(10 m>s)
= 1.0
Re =
VD
va
=
(10 m>s)(0.05m)
15.0(10-6) m2>s
= 3.33(104)
Since Re is in the range of 104, the figure in the text can be used to determine the lift 
coefficient. Here, CL ≃ 0.270. Thus, 
FL = CLApa
raV
2
2
b
= 0.273p(0.025 m)24 £ (1.20 kg>m3)(10 m>s)2
2
§
= 0.03181 N
The acceleration of the ball in the y-direction is 
+ c ΣFy = may; 0.03181 N = (0.5 kg)ay
 ay = 0.06362 m>s2
The ball travels with a constant velocity V = 10 m>s in the x- direction. Thus, the 
time for the ball to strike the wall is 
t =
Sx
V
=
10 m
10 m>s
= 1s
The displacement d in the y direction for this same time interval is 
+ c sy = (sy)0 + (vy)0t +
1
2
ayt
2;
d = 0 + 0 +
1
2
(0.06362 m>s2)(1 s)2
= 0.03181 m = 31.8 mm Ans.
10 m
v
Top View
d
y
x
Ans:
31.8 mm
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 in
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the
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s a
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se
ss
ing
 st
ud
en
t le
arn
ing
. D
iss
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n o
r 
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le 
of 
an
y p
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 of
 th
is 
work
 (in
clu
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n t
he
 W
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ide
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) 
will 
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oy
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of 
the
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ork
 an
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