Mecánica de Fluidos_Merle C. Potter, David C. Wiggert_3ed Solucionario

Mecánica de los fluidos

1. http://www.elsolucionario.net

2. 1
CHAPTER 1

Basic Considerations


1.1 Conservation of mass — Mass — density
Newton’s second law — Momentum — velocity
The first law of thermodynamics — internal energy — temperature

1.2 a) density = mass/volume = ML/
3

b) pressure = force/area = FLMLTLMLT/ / /
2 22 2
= =
c) power = force´velocity = FLTMLTLTMLT´ = ´ =/ / / /
2 2 3

d) energy = force´distance = MLTLMLT/ /
2 2 2
´=
e) mass flux = rAV = M/L
3
´L
2
´L/T = M/T
f) flow rate = AV = L
2
´L/T = L
3
/T

1.3 a) density =
M
L
FTL
L
FTL
3
2
3
2 4/
/=
b) pressure = F/L
2

c) power = F´velocity = F ´L/T = FL/T
d) energy = F´L = FL
e) mass flux =
M
T
FTL
T
FTL= =
2
/
/
f) flow rate = AV = L
2
´L/T = L
3
/T

1.4 (C) m = F/a or kg = N/m/s
2
= N
.
s
2
/m.

1.5 (B) [m] = [t/du/dy] = (F/L
2
)/(L/T)/L = F
.
T/L
2
.

1.6 a) L = [C] T
2
. \[C] = L/T
2

b) F = [C]M. \[C] = F/M = ML/T
2
M = L/T
2

c) L
3
/T = [C] L
2
L
2/3
. \[C] = LTLL LT
3 2 23 13
/
/ /
××=
Note: the slope S0 has no dimensions.

1.7 a) m = [C] s
2
. \[C] = m/s
2

b) N = [C] kg. \[C] = N/kg = kg × m/s
2
× kg = m/s
2

c) m
3
/s = [C] m
2
m
2/3
. \[C] = m
3
/s×m
2
× m
2/3
= m
1/3
/s

1.8 a) pressure: N/m
2
= kg × m/s
2
/m
2
= kg/m× s
2

b) energy: N× m = kg × m/s
2
´m = kg× m
2
/s
2
c) power: N× m/s = kg × m
2
/s
3

d) viscosity: N× s/m
2
=
kgm
s
s
1
m
kg/ms
2 2
×
× = ×

3. 2
e) heat flux: J/s =
Nm
s
kgm
s
m
s
kgm/s
2
2 3×
=
×
×=×
f) specific heat:
J
kgK
Nm
kgK
kgm
s
m
kgK
m/Ks
2
2 2
×
=
×
×
=
×
×
×
= ×

1.9 kg
m
s
m
s
m
2
+ +=c k f. Since all terms must have the same dimensions (units) we
require:
[c] = kg/s, [k] = kg/s
2
= Ns/msN/m,
2 2
× ×= [f] =kgm/sN.
2
× =
Note: we could express the units on c as [c] = kg/sNs/msNs/m
2
=× ×=×

1.10 a) 250 kN b) 572 GPa c) 42 nPa d) 17.6 cm
3

e) 1.2 cm
2
f) 76 mm
3


1.11 a) 1.25´10
8
N b) 3.21´10
-5
s c) 6.7´10
8
Pa
d) 5.6´10
-12
m
3
e) 5.2 ´10
-2
m
2
f) 7.8´10
9
m
3


1.12 (A)
89
2.361023.61023.6 nPa.
--
´=´=

1.13
222
0.06854
0.2250.738
0.001943.281
mm
dd
l
rr
==
´

where m is in slugs, r in slug/ft
3
and d in feet. We used the conversions in the
front cover.

1.14 a) 20 cm/hr =
520
/36005.55510m/s
100
-
=´
520
/36005.55510m/s
100
-
=´
b) 2000 rev/min = 20002´p/60 = 209.4 rad/s
c) 50 Hp = 50´745.7 = 37 285 W
d) 100 ft
3
/min = 100´0.02832/60 = 0.0472 m
3
/s
e) 2000 kN/cm
2
= 2´10
6
N/cm
2
´100
2
cm
2
/m
2
= 2´10
10
N/m
2

f) 4 slug/min = 4 ´14.59/60 = 0.9727 kg/s
g) 500 g/L = 500´10
-3
kg/10
-3
m
3
= 500 kg/m
3
h) 500 kWh = 500´1000´3600 = 1.8´10
9
J

1.15 a) F = ma = 10 ´40 = 400 N.
b) F - W = ma. \ F = 10 ´40 + 10 ´9.81 = 498.1 N.
c) F - W sin 30° = ma. \ F = 10 ´40 + 9.81 ´0.5 = 449 N.

1.16 (C) The mass is the same on earth and the moon: [4(8)]32.
du
rr
dr
tmmm===

1.17 The mass is the same on the earth and the moon:

4. 3
m =
60
322
1863
.
..= \ Wmoon = 1.863´5.4 = 10.06 lb

1.18 (C)
shear
sin4200sin302100 N.FF q===
o


shear
4
2100
=84 kPa
25010
F
A
t
-
==
´

1.19 a) l
r
= =
´
´´
=´
-
-
-
. .
.
. (. )
.43225 225
4810
1843710
10
2
26
102
6m
d
m or0.00043 mm
b) l
r
= =
´
´´
=´
-
-
-
. .
.
. (. )
.225 225
4810
001033710
7710
2
26
102
5m
d
m or 0.077 mm
c)
26
2102
4.810
.225.225.0039m
.00002(3.710)
m
d
l
r
-
-
´
===
´´
or 3.9 mm

1.20 Use the values from Table B.3 in the Appendix.
a) 52.3 + 101.3 = 153.6 kPa.
b) 52.3 + 89.85 = 142.2 kPa.
c) 52.3 + 54.4 = 106.7 kPa (use a straight-line interpolation).
d) 52.3 + 26.49 = 78.8 kPa.
e) 52.3 + 1.196 = 53.5 kPa.

1.21 a) 101 - 31 = 70 kPa abs. b) 760 -
31
101
´ 760 = 527 mm of Hg abs.
c) 14.7 -
31
101
´ 14.7 = 10.2 psia. d) 34 -
31
101
´ 34 = 23.6 ft of H2O abs.
e) 30 -
31
101
´ 30 = 20.8 in. of Hg abs.

1.22 p = po e
-gz/RT
= 101 e
-9.81 ´ 4000/287 ´ (15 + 273)
= 62.8 kPa
From Table B.3, at 4000 m: p = 61.6 kPa. The percent error is
% error =
62.8 61.6
61.6
-
´ 100 = 1.95 %.

1.23 a) p = 973 +
22,560 20,000
25,000 20,000
-
-
(785 - 973) = 877 psf
T = -12.3 +
22,560 20,000
25,000 20,000
-
-
(-30.1 + 12.3) = -21.4°F
b) p = 973 + .512 (785 - 973) +
.512
2
(-.488) (628 - 2 ´ 785 + 973) = 873 psf
T = -12.3 + .512 (-30.1 + 12.3) +
.512
2
(-.488) (-48 + 2 ´ 30.1 - 12.3) = -21.4°F
Note: The results in (b) are more accurate than the results in (a). When we use a
linear interpolation, we lose significant digits in the result.

5. 4

1.24 T = -48 +
33,000 30,000
35,000 30,000
-
-
(-65.8 + 48) = -59°F or (-59 - 32)
5
9
= -50.6°C

1.25 (B)
1.26 p =
n
F
A
=
4
26.5 cos 42
15210
-
´
o
= 1296 MN/m
2
= 1296 MPa.

1.27
4
n
4
t
(120000).2102.4N
20.210.0004N
F
F
-
-
ü=´´= ï
ý
=´´= ïþ
F =
22
nt
FF+ = 2.400 N.
q = tan
-1

.0004
2.4
=.0095°

1.28 r =
m
V-
=
02
1801728
.
/
= 1.92 slug/ft
3
. t = rg = 1.92 ´ 32.2 = 61.8 lb/ft
3
.

1.29 r = 1000 - (T - 4)
2
/180 = 1000 - (70 - 4)
2
/180 = 976 kg/m
3

g = 9800 - (T - 4)
2
/18 = 9800 - (70 - 4)
2
/180 = 9560 N/m
3

% error for r =
976978
978
-
´ 100 = -.20%
% error for g =
95609789.81
9789.81
-´
´
´ 100 = -.36%

1.30 S = 13.6 - .0024T = 13.6 - .0024 ´ 50 = 13.48.
% error =
13.4813.6
13.6
-
´ 100 = -.88%

1.31 a) m =
WV
g
g
=
6
12 40050010
9.81g
-
´´
= = 0.632 kg
b) m =
6
12 40050010
9.77
-
´´
= 0.635 kg
c) m =
6
12 40050010
9.83
-
´´
= 0.631 kg

1.32 S =
/
water
mVr
r
=
10/
.1.2
water
V
r
= .
1.94
\V = 4.30 ft
3


1.33 (D)
22
3
water
(4)(804)
10001000968 kg/m
180180
T
r
--
=-=-=

6. 5
1.34 t = m
du
dr
= 1.92 ´ 10
-5

2
30(21/12)
(1/12)
éù´
êú
ëû
= 0.014 lb/ft
2


1.35 T = force ´ moment arm = t 2pRL ´ R = m
du
dr
2pR
2
L = m
2
0.4
1000
R
æö
+
ç÷
èø
2pR
2
L.
\m =
22
2
0.0026
0.40.4
1000210002.010.2
12
T
RL
R
pp
=
æöæö
++´´
ç÷ç÷
èøèø
= 0.414 N
.
s/m
2
.

1.36 Use Eq.1.5.8: T =
3
2RL
h
pwm
=
( )2.5/12p
p
´ ´
´
´´
320002
60
4006
0112
.
./
= 2.74 ft-lb.
Hp =
Tw
550
274209
550
=
´. .4
= 1.04 Hp

1.37 Fbelt = m
du
dy
A= ´
-
13110
10
002
3
.
.
(.6 ´ 4) = 15.7 N.
Hp =
FV´
=
´
746
15710
746
.
= 0.210 Hp

1.38 Assume a linear velocity so .
dur
dyh
w
= Due to the area
element shown, dT = dF ´ r = tdA ´ r = m
du
dy
2pr dr ´ r.

dr
r
t


T =
3
0
2
R
rdr
h
mwp
ò =
2
4
23610
4002
60
312
20812
4
5 4
pmw
p
p
h
R
=
´ ´ ´
´
´
´
-
. (/)
./
= 91 ´ 10
-5
ft-lb.

1.39
2
30(21/12)
(1/12)
éù´
êú
ëû
22
00
[32/]32/.
du
rrrr
dr
tmmm=== \tr = 0 = 0,

tr=0.25 = 32 ´ 1 ´ 10
-3
´
2
.25/100
(.5/100)
= 3.2 Pa, tr=0.5 = 32 ´ 1 ´ 10
-3
´
2
.5/100
(.5/100)
= 6.4 Pa

1.40 (A)
3
[105000]101050000.021 Pa.
du
r
dr
tmm
-
==´=´´´=
1.41 The velocity at a radius r is rw. The shear stress is tm=
D
D
u
y
.
The torque is dT = trdA on a differential element. We have

7. 6

0.08
0
= =2
0.0002
r
TrdArdx
w
tmpòò ,
20002
209.4 rad/s
60
p
w
´
==
where x is measured along the rotating surface. From the geometry 2x=r, so that

0.080.08
23
00
209.4/2329000
=0.12329000(0.08)
0.00023 2
xx
Tdxxdx p
´
==òò = 56.1 N
.
m

1.42 If tm=
du
dy
= cons’t and m = Ae
B/T
= Ae
By/K
= Ae
Cy
, then
Ae
Cy

du
dy
= cons’t. \
du
dy
= De
-Cy
.
Finally,
00
yu
Cy
duDedy
-
=òò or u(y) =
0
y
CyD
e
C
-
- = E (e
-Cy
- 1)
where A, B, C, D, E, and K are constants.

1.43
m= =
=
ü
ý
þ
Ae Ae
Ae
BT B
B
/ /
/
.
.
001
000357
293
353
\A = 2.334 ´ 10
-6
, B = 1776.



m40 = 2.334 ´ 10
-6
e
1776/313
= 6.80 ´ 10
-4
N
.
s/m
2


1.44 m = rV. Then dm = rdV + Vdr. Assume mass to be constant in a volume
subjected to a pressure increase; then dm = 0. \rdV = -Vdr, or
dV
V
.
dr
r
=-

1.45 B =
V
-
p
V
D
D
2200 MPa.= V\D
V-
=
210
2200
p
B
D-´
= = -0.00909 m
3
or -9090 cm
3

1.46 Use c = 1450 m/s. L = cDt = 1450 ´ 0.62 = 899 m

1.47 =
BV
p
D
D-
V
= -2100
-13
20
.
= 136.5 MPa

1.48 a) 327,000144/1.93c=´ = 4670 fps b) 327,000144/1.93c=´ = 4940 fps
c) 308,000144/1.87c=´ = 4870 fps

1.49 VD=3.8 ´ 10
-4
´ -20 ´ 1 = .0076 m
3
.
Dp = -B
VD
V
.0076
2270
1
-
=- = 17.25 MPa

1.50 p =
220741
510
6
s
R
=
´
´
-
.
= 2.96 ´ 10
4
Pa or 29.6 kPa. Bubbles: p = 4s/R = 59.3 kPa

8. 7
1.51 Use Table B.1: s = 0.00504 lb/ft. \p =
44.00504
1/3212R
s´
=
´
= 7.74 psf or 0.0538 psi

1.52 See Example 1.4: h =
4cos40.07360.866
0.130 m.
10009.810.0002gD
sb
r
´´
==
´´


1.53 (D)
6
4cos40.07361
3 m or 300 cm.
10009.811010
h
gD
sb
r
-
´´
===
´´´


1.54 See Example 1.4: h =
4cos40.032cos130
1.9413.632.20.8/12gD
sb
r
´
=
´´´
o

= -0.00145 ft or -0.0174 in

1.55 force up = s ´ L ´ 2 cosb = force down = rghtL. \h =
2sb
r
cos
.
gt

1.56 Draw a free-body diagram:
The force must balance:
W = 2sL or
p
rs
d
Lg L
2
4
2
æ
è
ç
ö
ø
÷=.
\=d
g
8s
pr




W
sL sL
needle


1.57 From the free-body diagram in No. 1.47, a force balance yields:
Is
p
r
d
g
2
4
< 2s?
p(.)
. .
004
4
785098120741
2
´ <´
0.968 < 0.1482 \No

1.58 Each surface tension force = s ´ p D. There is a force on the outside
and one on the inside of the ring.
\F = 2spD neglecting the weight of the ring.



F
D



1.59

h(x)
h
dW
sdl



From the infinitesimal free-body shown:
cos.dghxdxsqra =l cosq =
dx
dl
.

/ddxd
h
gxdxgx
ss
rara
\==
ll

We assumed small a so that the element
thickness is ax.

9. 8


1.60 The absolute pressure is p = -80 + 92 = 12 kPa. At 50°C water has a vapor
pressure of 12.2 kPa; so T = 50°C is a maximum temperature. The water would
“boil” above this temperature.

1.61 The engineer knew that water boils near the vapor pressure. At 82°C the vapor
pressure from Table B.1 is 50.8 (by interpolation). From Table B.3, the elevation
that has a pressure of 50.8 kPa is interpolated to be 5500 m.

1.62 At 40°C the vapor pressure from Table B.1 is 7.4 kPa. This would be the
minimum pressure that could be obtained since the water would vaporize below
this pressure.

1.63 The absolute pressure is 14.5 - 11.5 = 3.0 psia. If bubbles were observed to form
at 3.0 psia (this is boiling), the temperature from Table B.1 is interpolated, using
vapor pressure, to be 141°F.

1.64 The inlet pressure to a pump cannot be less than 0 kPa absolute. Assuming
atmospheric pressure to be 100 kPa, we have
10 000 + 100 = 600 x. \x = 16.83 km.

1.65 (C)

1.66 r= =
´ +
=
p
RT
1013
028727315
.
. ( )
1.226 kg/m
3
. g = 1.226 ´ 9.81 = 12.03 N/m
3


1.67
3
in
101.3
1.226 kg/m.
0.287(15273)
p
RT
r===
´+

3
out
85
1.19 kg/m.
0.287248
r==
´

Yes. The heavier air outside enters at the bottom and the lighter air inside exits at
the top. A circulation is set up and the air moves from the outside in and the inside
out: infiltration. This is the “chimney” effect.

1.68
375044
0.1339 slug/ft.
1716470
p
RT
r
´
===
´
mVr= 0.1339152.01 slug.=´=

1.69 (C)
pV
m=
8004
59.95 kg
0.1886(10273)RT
´
==
´+


1.70
p
WV
RT
=
100
(10204)9.819333 N.
0.287293
g=´´´´=
´

10. 9
1.71 Assume that the steel belts and tire rigidity result in a constant volume so that m1
= m2:
V1V=
1122
2
12
2
21
1
or .
150460
(3514.7)67.4 psia or 52.7 psi gage.
10460
mRTmRT
pp
T
pp
T
=
+
\==+=
-+


1.72 The pressure holding up the mass is 100 kPa. Hence, using pA = W, we have

10000019.81.10200 kg.mm´=´\=
Hence,

pV
m=
3
1004/3
10200.12.6 m or 25.2 m.
0.287288
r
rd
RT
p´
==\==
´


1.73
221
0(10).2032.2.25.4 fps.
2
KEPEmVmgVV=D+D=+-\=´\=
221
0(20).4032.2.35.9 fps.
2
mVmgVV=+-\=´\=

1.74
22
1-2
1
.a) 20005(10).19.15 m/s.
2
ff
WKEVV=D´=´-\=
b)
10
22
0
1
2015(10).
2
f
sdsV=´-ò

2
22101
2015(10).15.27 m/s.
22
ff
VV´=´-\=
c)
10
22
0
1
200cos15(10).
202
f
s
dsV
p
=´-ò
22201
200sin15(10).16.42 m/s.
22
ff
VV
p
p
´=´-\=

1.75
2
121221
1
.10400.20.40000.
2
EEuuuu=´´+=+\-= %%%%

40000
.55.8C where comes from Table B.4.
717
vv
ucTTcD=D\D==
o
%
The following shows that the units check:

2 2222
car
222
air
kgm/smkgCmkgC
C
kgJ/(kgC)Nms(kgm/s)ms
mV
mc
éù´ ×××××
====êú
×××××××êúëû
oo
o
o

where we used N = kg
.
m/s
2
from Newton’s 2
nd
law.

11. 10
1.76
2
2
21HO
1
..
2
EEmVmcT==D

2
611001000
150010002000104180.69.2C.
23600
TT
-´æö
´´=´´´D\D=
ç÷
èø
o

We used c = 4180 J/kg
.
C
o
from Table B.5. (See Problem 1.75 for a units check.)

1.77
water
.0.2400001004.18.19.1C.
ff
mhmcTTT=D´=´D\D=
o

The specific heat c was found in Table B.5. Note: We used kJ on the left and kJ
on the right.

1.78. (B)
icewatericewaterwater
.320.EEmmcTD=D´=´D

63
5(4010)1000320(210)10004.18.7.66C. TT
--
´´´´=´´´D\D=
o

We assumed the density of ice to be equal to that of water, namely 1000
kg/m
3
. Ice is actually slightly lighter than water, but it is not necessary for
such accuracy in this problem.

1.79. WpdV=
mRT
V
=ò dV
dV
mRT=ò
V
ln
V
mRT=ò
2
V
2
1
1
ln
p
mRT
p
=
since, for the T = const process,
1
pV1 2
pV= 2. Finally,

1-2
41
1716530ln78,310 ft-lb.
32.22
W=´´=-
The 1
st
law states that
0.78,310 ft-lb or 101 Btu.
v
QWumcTQW-=D=D=\==-- %

1.80 If the volume is fixed the reversible work is zero since the boundary does not
move. Also, since V
12
12
,
mRTTT
ppp
== so the temperature doubles if the
pressure doubles. Hence, using Table B.4 and Eq. 1.7.17,
2002
a) (1.0040.287)(2293293)999 kJ
0.287293
v
QmcT
´
=D=-´-=
´

b)
2002
(1.0040.287)(2373373)999 kJ
0.287373
v
QmcT
´
=D=-´-=
´

c)
2002
(1.0040.287)(2473473)999 kJ
0.287473
v
QmcT
´
=D=-´-=
´


1.81 WpdV= (pV= 2V-
1
1). If = const,
T
p
V
2
1
T
V
=
21
2
so if 2,TT=ò
then V22V= 1 and (2WpV= 1V-1)pV= 1 1
.mRT=
a) 20.287333191 kJW=´´=
b) 20.287423243 kJW=´´=

12. 11
c) 20.287473272 kJW=´´=

1.82 =1.4287318357 m/s. 3578.322970 m.ckRTLct =´´==D=´=

1/ 0.4/1.4
2
21
1
500
(20273)151.8 K or 121.2C
5000
kk
p
TT
p
-
æö æö
==+=-ç÷ ç÷
èøèø
o


1.83 We assume an isentropic process for the maximum pressure:
/1 1.4/0.4
2
21
1
423
(150100)904 kPa abs or 804 kPa gage.
293
kk
T
pp
T
-
æö æö
==+=ç÷ ç÷
èøèø

Note: We assumed patm = 100 kPa since it was not given. Also, a measured
pressure is a gage pressure.

1.84
/1 1.4/0.4
2
21
1
473
100534 kPa abs.
293
kk
T
pp
T
-
æö æö
===ç÷ ç÷
èøèø


21
()(1.0040.287)(473293)129 kJ/kg.
v
wucTT=-D=--=---=-
We used Eq. 1.7.17 for cv.

1.85 a) 1.4287293343.1 m/sckRT==´´=
b) 1.4188.9293266.9 m/sckRT==´´=
c) 1.4296.8293348.9 m/sckRT==´´=
d) 1.441242931301 m/sckRT==´´=
e) 1.4461.5293424.1 m/sckRT==´´=
Note: We must use the units on R to be J/kg
.
K in the above equations.

1.86 (D) For this high-frequency wave, 287323304 m/s.cRT==´=

1.87 At 10 000 m the speed of sound 1.4287223299 m/s.ckRT==´´=
At sea level, 1.4287288340 m/s.ckRT==´´=

340299
% decrease 10012.06 %.
340
-
=´=

1.88 a) =1.4287253319 m/s. 3198.322654 m.ckRTLct =´´==D=´=
b) =1.4287293343 m/s. 3438.322854 m.ckRTLct =´´==D=´=
c) =1.4287318357 m/s. 3578.322970 m.ckRTLct =´´==D=´=

13. 12
C HAPTER 2

Fluid Statics

2.1 S D D
DD
Fmapzps
yz
a
y y y y
= - =: sinar
2

S D D
DD DD
Fmapyps
yz
a g
yz
z z z z
= - = +: cosar r
2 2

Since D Ds ycosa= and D Ds z sin a=, we have

pp
y
a
y y
-=r
D
2
and ( )pp
z
ag
z z
-= +r
D
2

Let Dy®0 and Dz®0:

pp
pp
y
z
-=
-=
ü
ý
þ
0
0

\==ppp
y z
.

2.2 p = gh. a) 9810 ´ 10 = 98 100 Pa or 98.1 kPa
b) (0.8 ´ 9810) ´ 10 = 78 480 Pa or 78.5 kPa
c) (13.6 ´ 9810) ´ 10 = 1 334 000 Pa or 1334 kPa
d) (1.59 ´ 9810) ´ 10 = 155 980 Pa or 156.0 kPa
e) (0.68 ´ 9810) ´ 10 = 66 710 Pa or 66.7 kPa

2.3 h = p/g. a) h = 250 000/9810 = 25.5 m
b) h = 250 000/(0.8 ´ 9810) = 31.9 m
c) h = 250 000/(13.6 ´ 9810) = 1.874 m
d) h = 250 000/(1.59 ´ 9810) = 16.0 m
e) h = 250 000/(0.68 ´ 9810) = 37.5 m

2.4 (C) (13.69810)(28.50.0254)96600 Pa
Hg
phg==´´´=
2.5 S =
20144
62.420
p
hg
´
=
´
= 2.31. r = 1.94 ´ 2.31 = 4.48 slug/ft
3
.

2.6 a) p = gh = 0.76 ´ (13.6 ´ 9810) = 9810 h. \h = 10.34 m.
b) (13.6 ´ 9810) ´ 0.75 = 9810 h. \h = 10.2 m.
c) (13.6 ´ 9810) ´ 0.01 = 9810 h. \h = 0.136 m or 13.6 cm.

2.7 a) p = g1h1 + g2h2 = 9810 ´ 0.2 + (13.6 ´ 9810) ´ 0.02 = 4630 Pa or 4.63 kPa.
b) 9810 ´ 0.052 + 15 630 ´ 0.026 = 916 Pa or 0.916 kPa.
c) 9016 ´ 3 + 9810 ´ 2 + (13.6 ´ 9810) ´ 0.1 = 60 010 Pa or 60.0 kPa.

y
z
p
z
Dy
p
y
Dz
pDs
Ds
a
Dz
Dy
rgDV

14. 13
2.8 Dp = rgh = 0.0024 ´ 32.2 (–10,000) = –773 psf or –5.37 psi.


2.9 (D)
0
840001.009.81400044760 Pappgh r=-=-´´=


2.10

inside
1009.81
313.51 Pa
.287253
1009.81
311.67 Pa
.287293
outsideo
o
base
i
i
pg
pghh
RT
p
pg
pghh
RT
r
r
´ ü
D=D=D=´=
ï
´ ï
\Dý
´
ï
D=D=D=´=
ï´
þ
= 1.84 Pa
If no wind is present this Dp
base
would produce a small infiltration since the higher
pressure outside would force outside air into the bottom region (through cracks).


2.11 p = rgdh where h = –z. From the given information S = 1.0 + h/100 since S(0) = 1 and
S(10) = 1.1. By definition r = 1000 S, where rwater = 1000 kg/m
3
.
Then dp = 1000 (1 + h/100) gdh. Integrate:
dp h gdh
p
= +òò
10001 100
0
10
0
(/)
p= ´ +
´
100098110
10
2100
2
.( ) = 103 000 Pa or 103 kPa
Note: we could have used an average S: Savg = 1.05, so that r
avg
= 1050 kg/m
3
.

2.12
v
Ñ= + +p
p
x
i
p
y
j
p
z
k
¶
¶
¶
¶
¶
¶
$ $ $

= –rai
x
$
–ra
y
j
$
–ra
z
k
$
–rgk
$
= –( )raiajak
x y z
$$$
++ –rgk
$

= –r
v
a–r
v
g
\Ñ=-+
v
vv
p agr()


2.13
/
00
[()/]
gR
atm
ppTzT
a
a=-
= 100 [(288 - 0.0065 ´ 300)/288]
9.81/.0065 ´ 287
= 96.49 kPa
pp gh
atm
= - = -
´
´ ´r 100
100
287288
9813001000
.
. / = 96.44 kPa
% error =
96 96
96
100
.44.49
.49
-
´ = -0.052%
The density variation can be ignored over heights of 300 m or less.

15. 14
2.14
/
0
0
0
gR
atmatm
Tz
ppppp
T
a
aæö-
D=-=- ç÷
èø

= 100
288006520
288
1
9810065287
- ´æ
è
ç
ö
ø
÷ -
é
ë
ê
ù
û
ú
´
.
./.
= -0.237 Pa or -0.000237 kPa
This change is very small and can most often be ignored.

2.15 Eq. 1.5.11 gives 310,000144.
dp
d
r
r
´= But, dp = rgdh. Therefore,

7
4.46410
gdhdrr
r
´
= or
27
32.2
4.46410
d
dh
r
r
=
´

Integrate, using
0
r = 2.00 slug/ft
3
:

27
20
32.2
4.46410
h
d
dh
r
r
r
=òò
´
. \
11
2r
æö
--
ç÷
èø
= 7.21 ´ 10
-7
h or
7
2
114.4210 h
r
-
=
-´

Now,

7
77
00
22
ln(114.4210)
114.421014.4210
hh
gg
pgdhdhh
h
r
-
--
===-´òò
-´-´

Assume r = const:
2.032.264.4pghhhr==´´=

a) For h = 1500 ft: paccurate = 96,700 psf and pestimate = 96,600 psf.
96,60096,700
% error 1000.103 %
96,700
-
=´=-
b) For h = 5000 ft: paccurate = 323,200 psf and pestimate = 322,000 psf.
322,000323,200
% error 1000.371 %
323,200
-
=´=-
c) For h = 15,000 ft: paccurate = 976,600 psf and pestimate = 966,000 psf.
966,000976,600
% error 1001.085 %
976,600
-
=´=-

2.16 Use the result of Example 2.2: p = 101 e
-gz/RT
.

a) p = 101 e
-9.81 ´10 000/287 ´273
= 28.9 kPa.
b) p = 101 e
-9.81 ´10 000/287 ´288
= 30.8 kPa.
c) p = 101 e
-9.81 ´10 000/287 ´258
= 26.9 kPa.

2.17 Use Eq. 2.4.8: p =
9.81
0.0065287
101(10.0065/288). z
´
-

a) z = 3000. \p = 69.9 kPa. b) z = 6000. \p = 47.0 kPa.
c) z = 9000. \p = 30.6 kPa. d) z = 11 000. \p = 22.5 kPa.

16. 15
2.18 Use the result of Example 2.2:
/
0
=.
gzRTp
e
p
-


0
ln.
pgz
pRT
=-
0.00132.2
ln.
14.71716455
z
=-
´
\z = 232,700 ft.

2.19 p = gh = (13.6 ´ 9810) ´ 0.25 = 33 350 Pa or 33.35 kPa.

2.20 a) p = gh. 450 000 = (13.6 ´ 9810) h. \h = 3.373 m
b) p + 11.78 ´ 1.5 = (13.6 ´ 9810) h. Use p = 450 000, then h = 3.373 m
% error is 0.000 %.

2.21 Referring to Fig. 2.6a, the pressure in the pipe is p = rgh. If p = 2400 Pa, then
2400 = rgh = r ´ 9.81 h.
a) r=
´
2400
98136..
= 680 kg/m
3
. \gasoline
b) r=
´
2400
981272..
= 899 kg/m
3
. \benzene
c) r=
´
2400
981245..
= 999 kg/m
3
. \water
d) r=
´
2400
981154..
= 1589 kg/m
3
. \carbon tetrachloride

2.22 Referring to Fig. 2.6a, the pressure is p = rwgh =
21
.
2
a
Vr Then V
gh
w
a
22
=
r
r
.
a) V
22100098106
123
=
´ ´´..
.
= 957. \V = 30.9 m/s
b) V
22194322312
00238
=
´ ´ ´. ./
.
= 13,124. \V = 115 ft/sec
c) V
2210009811
123
=
´ ´´..
.
= 1595. \V = 39.9 m/s
d) V
22194322512
00238
=
´ ´ ´. ./
.
= 21,870. \V = 148 ft/sec

2.23 (C) 0300000.398100.18020 Pa
watmxxwaterw
pphh gg=+-=+´-´=

2.24 See Fig. 2.6b: p
1
= –g1h + g2H.
p
1
= –0.86 ´ 62.4 ´
5
12
+ 13.6 ´ 62.4 ´
95
12
.
= 649.5 psf or 4.51 psi.

2.25
011223344
ppghghghgh rrrr=++++
= 3200 + 917´9.81´0.2 + 1000´9.81´0.1 + 1258´9.81´0.15 + 1593´9.81´0.18
= 10 640 Pa or 10.64 kPa

17. 16

2.26 ( )( )( )pppp pp pp
1 4 1 2 2 3 3 4-=-+-+- (Use D Dpgh=r)
40 000 – 16 000 = 1000´9.81(–.2) + 13 600´9.81´H + 920´9.81´.3.
\H = .1743 m or 17.43 cm

2.27 ( )( )( )pppp pp pp
1 4 1 2 2 3 3 4-=-+-+- (Use D Dpgh=r)
po – pw = 900´9.81(–.2) + 13 600´9.81(–.1) + 1000´9.81´.15
= –12 300Pa or –12.3 kPa

2.28 ( )( )( )( )pppp pp pp pp
1 5 1 2 2 3 3 4 4 5-=-+-+-+-
p
1
= 9810(–.02) + 13 600´9.81(.–04) + 9810(–.02)+13 600´9.81´.16
= 15 620 Pa or 15.62 kPa

2.29 pw + 9810 ´ .15 – 13.6 ´ 9810 ´ .1 – .68 ´ 9810 ´ .2 + .86 ´ 9810 ´ .15 = po.
\pw – po = 11 940 Pa or 11.94 kPa.

2.30 pw – 9810 ´ .12 – .68 ´ 9810 ´ .1 + .86 ´ 9810 ´ .1 = po.
With pw = 15 000, po = 14 000 Pa or 14.0 kPa.

2.31 a) p + 9810 ´ 2 = 13.6 ´ 9810 ´ .1. \p = –6278 Pa or –6.28 kPa.
b) p + 9810 ´ .8 = 13.6 ´ 9810 ´ .2. \p = 18 835 Pa or 18.84 kPa.
c) p + 62.4 ´ 6 = 13.6 ´ 62.4 ´ 4/12. \p = –91.5 psf or –0.635 psi.
d) p + 62.4 ´ 2 = 13.6 ´ 62.4 ´ 8/12. \p = 441 psf or 3.06 psi.

2.32 p – 9810 ´ 4 + 13.6 ´ 9810 ´ .16 = 0. \p = 17 890 Pa or 17.89 kPa.

2.33 (A) (13.69810)0.1621350 Pa.
a
pHg=-=-´´=-

,
21350100001135013.69810.0.0851 m
aafterafterafter
pHH =-+=-=´\=

2.34 8200 + 9810 ´ .25 = 1.59 ´ 9810 ´ H. \H = 0.683 m
Hnew = .683 + .273 = .956 m. DH =
.273
2
= .1365.
p + 9810 (.25 + .1365) = 1.59 ´ 9810 ´ .956.
\p = 11 120 Pa or 11.12 kPa.


2.35 p + 9810 ´ .05 + 1.59 ´ 9810 ´ .07 – .8 ´ 9810 ´ .1 = 13.6 ´ 9810 ´ .05.
\p = 5873 Pa or 5.87 kPa.

Note: In our solutions we usually retain 3 significant digits in the answers (if a number
starts with “1” then 4 digits are retained. In most problems a material property is used,
i.e., S = 1.59. This is only 3 sig. digits! \ only 3 are usually retained in the answer!

H
DH
DH

18. 17
2.36 Before pressure is applied the air column on the right is 48" high. After pressure is
applied, it is (4 – H/2) ft high. For an isothermal process
1pV1 2pV= 2 using
absolute pressures. Thus,
14.7 ´ 144 ´ 4A = p2(4 – H/2)A or p2 =
8467
4 2-H/
.
From a pressure balance on the manometer (pressures in psf):
30 ´ 144 + 14.7 ´ 144 = 13.6 ´ 62.4 H +
8467
4 2-H/
,
or H
2
– 15.59 H + 40.73 = 0. \H = 12.27 or 3.32 ft.

2.37 a) ( )( )( )( )pppp pp pp pp
1 5 1 2 2 3 3 4 4 5-=-+-+-+-
4000 = 9800(0.16–0.22) + 15 600(0.10–0.16) + 133 400H + 15 600(0.07–H).
\H = .0376 m or 3.76 cm
b) 0.6´144 = 62.4(–2/12) + 99.5(–2/12) + 849H + 99.5(2.5/12 – H).
\H = .1236 ft or 1.483 in.

2.38 a)
22
22
1 1232
2/
22()/
HDd
p Ddgggg
D
=
D -++-

2
2
2(.1/.005)
9800215 6002(133 40015 600)(.1/.005)
=
-+´+-
= ´
-
8 10
6
.487H
\ = ´ ´
-
DH8 10400
6
.487 = 0.0034 m or 3.4 mm
b) DH=
- +´ + -
´
242
6229952 99542
06144
2
2
(/.)
.4 .(849.)(/.)
. = 0.01153 ft or 0.138 in.

2.39 ( )( )( )pppp pp pp
1 4 1 2 2 3 3 4-=-+-+- (poil = 14.0 kPa from No. 2.30)
15 500 – 14 000 = 9800(0.12 + Dz) + 680(0.1 – 2Dz) + 860(–0.1 – Dz).
\Dz = 0.0451 m or 4.51 cm

2.40 a) pair = –6250 + 625 = –5620 Pa.
–5620 + 9800(2 + Dz) – 13 600 ´ 9.81(0.1 + 2Dz) = 0. \Dz = 0.0025.
\h = 0.1 + 2Dz = .15 m or 15 cm
b) pair = 18 800 + 1880 = 20 680 Pa.
20 680 + 9800(0.8 + Dz) – 13 600 ´ 9.81(0.2 + 2Dz) = 0. \Dz = 0.00715 m
\h = .2+ 2Dz = .214 or 21.4 cm
c) pair = –91.5 + 9.15 = –82.4 psf.
–82.4 + 62.4(6 + Dz) – 13.6 ´ 62.4(4/12 + 2Dz) = 0. \Dz = 0.00558 ft.
\h = 4/12 + 2 (0.00558) = 0.3445 ft or 4.13 in.
d) pair = 441 + 44.1 = 485 psf
485 + 62.4(2 + Dz) – 13.6 ´ 62.4(8/12 + 2Dz) = 0. \Dz = 0.0267 ft.
\h = 8/12 + 2 (0.0267) = 0.7205 ft or 8.65 in.

19. 18
2.41FhA=g= 9810 ´ 10 ´ p ´ .3
2
/4 = 6934 N.

2.42
5155
(2)(2)[980013(2)].32670 N
3333
PP´´=´´´´´´´\= a) F = pc A = 9800 ´ 2
´ 4
2
= 313 600 N or 313.6 kN
b)
22
98001(24)980029800198000 N or 98.0 kN
33
c
FpA==´´´+´´+´´=
c) F = pc A = 9800 ´ 1 ´ 2 ´ 4 ´ 2 = 110 900 N or 110.9 kN
d) F = pc A = 9800 ´ 1 ´ 2 ´ 4/.866 = 90 500 N or 90.5 kN

2.43 For saturated ground, the force on the bottom tending to lift the vault is
F = pc A = 9800 ´ 1.5 ´ (2 ´ 1) = 29 400 N.
The weight of the vault is approximately
WgVr= 24009.81walls=´ [2(2´1.5´.1) + 2(2´1´.1) + 2(.8´1.3´.1)] = 28 400 N.
The vault will tend to rise out of the ground.

2.44 F = pc A = 6660 ´ 2 ´ p ´ 2
2
= 167 400 N or 167.4 kN
Find g in Table B.5 in the Appendix.

2.45 a) F = pc A = 9800 (10 - 2.828/3) (2.828 ´ 2/2) = 251 000 N or 251 kN
where the height of the triangle is (3
2
- 1
2
)
1/2
= 2.828 m.
b) F = pc A = 9800 ´ 10 (2.828 ´ 2/2) = 277 100 N or 277.1 kN
c) F = pc A = 9800 (10 - 2.828 ´ .866/3) (2.828 ´ 2/2) = 254 500 N or 254.5 kN

2.46 a) FhA= = ´ ´=g 62273324.4. 40,930 lb.
y
p= +
´
´
2733
6836
273324
3
.
/
.
= 27.46'. \y = 30 – 27.46 = 2.54'.
8/5.46 = 3/x. \x = 2.05’. (2.05, 2.54) ft.


b) F = 62.4 ´ 30 ´ 24 = 44,930 lb. The centroid is the center of pressure.
y = 2.667'. 8/5.333 = 3/x. \x = 2.000' (2.000, 2.667) ft.

c) F = 62.4 (30 – 2.667 ´ .707) ´ 24 = 42,100 lb.
y
p= +
´
´
3977
6836
397724
3
.
/
.
= 39.86'. y = 42.43 – 39.86 = 2.57'
8/5.43 = 3/x. \x = 2.04'. (2.04, 2.57) ft.

2.47 (B) The force acts 1/3 the distance from the hinge to the water line:
5155
(2)(2)[980013(2)].32670 N
3333
PP´´=´´´´´´´\=


(x, y)
y
x

20. 19
2.48 a)FhA= = ´´g p981062
2
= 739 700 N or 739.7 kN.
yy
I
Ay
p=+=+
´
´
6
24
46
4
p
p
/
= 6.167 m. \(x, y)p = (0, –0.167) m

b)FhA= = ´´g p981062 = 369 800 N or 369.8 kN.
y
p=+
´
´
6
28
26
4
p
p
/
= 6.167 m. x
2
+ y
2
= 4
xF
x
pdA xyxdy y ydy
p
= = - = - -ò òò
--
2 2
6
2
4 6
2
2
2
2
2
g g
() ( )().
\ ´= -- + =
-
ò
x yyydy
p
g p
g
g62
2
2446 32
2
2
2
3
( ) .\xp = 0.8488 m


\(x, y)p = (0.8488, –0.167) m

c) F = 9810 ´ (4 + 4/3) ´ 6 = 313 900 N or 313.9 kN.
y
p= +
´
´
5333
3436
53336
3
.
/
.
= 5.500 m. \y = –1.5
4/2.5 = 1.5/x. \x = 0.9375. \(x,y)p = (0.9375, –1.5) m


d) F= ´+´98104
2
3
4( sin36.9°) ´ 6 = 330 000 N
y
p=+
´
´
56
5236
656
3
.
.4/
.
= 5.657 m. \y = 0.343 m


3 cos 53.13° = 1.8, 2.5 – 1.8 = 0.7, 2.4/2.057 = ./7
1
x. \x
1
= 0.6.
x = 1.8 + 0.6 = 2.4. \(x,y)p = (2.4, 0.343) m.

2.49 FhA= = ´´´g 6211610.4 ( ) = 41,180 lb.
yy
I
yA
p=+=+
´
´
11
61012
1160
3
/
= 11.758'.
(16 – 11.758) 41,180 = 10P. \P = 17,470 lb.


2.50 FhA= = ´´g 9810620 = 1.777 ´ 10
6
N, or 1177 kN.
yy
I
Ay
p=+=+
´
´
75
4512
7520
3
.
/
.
= 7.778 m.
(10 – 7.778) 1177 = 5 P. \P = 523 kN.


y
x
(x, y)
dA
dy
x
y
3 4
53.13
o
y
p
P
F

21. 20
2.51 FhA= = ´´g 98101220 = 2.354 ´ 10
6
N, or 2354 kN.
yy
I
Ay
p=+=+
´
´
15
4512
1520
3
/
= 15.139 m.
(17.5 – 15.139) 2354 = 5 P. \P = 1112 kN.

2.52 yy
I
Ay
HbH
bHH
HH
H
p=+=+
´
=+=
2
12
226
2
3
3
/
/
. y
p
is measured from the surface.
\From the bottom, HyH H H
p
-=- =
2
3
1
3
.
Note: this result is independent of the angle a, so it is true for a vertical area or a sloped
area.

2.53
31
sin403.(2)sin40.2(2).
23
l
FllFlPllPgg=´´=+\=+
oo

a) 9810´2
3
= 2(2 + 2)P. \P = 9810 N
b) 9810´4
3
= 2(4 + 2)P. \P = 52 300 N
c) 9810´5
3
= 2(5 + 2)P. \P = 87 600 N

2.54 h= -12
2 2
..4 = 1.1314 m. A = 1.2 ´ 1.1314 + .4 ´ 1.1314 = 1.8102 m
2

Use 2 forces: FhA
c1 1980056571211314= = ´ ´´g . (..) = 7527 N
FhA
c2 2
9800
11314
3
11314= = ´ ´´g
.
(.4.) = 1673 N
y
p1
2
3
11314=(.). yy
I
Ay
p2
2
2
3
11314
3
1131436
113142113143
=+ = +
´
´ ´
. .4. /
.4././
= 0.5657 m
SM
hinge
=0: 7527
11314
3
1673113140565711314´ + ´ - -
.
(. .).P = 0. \P = 3346 N.

2.55 To open, the resultant force must be just above the hinge, i.e., yp must be just less than h.
Let yp = h, the condition when the gate is about to open:
yhH AhHI hHhH=+ =+ = + +( )/,( ),[( )]()/3 2 36
2 3

\=
+
+
+
+ +
=
+
+
+
=
+
y
hH hH
hHhH
hHhHhH
p
3
2 36
3 3 6 2
4
2
( )/
( )( )/

a) h
hH
=
+
2
. \h = H = 0.9 m
b) h = H = 1.2 m
c) h = H = 1.5 m

2.56 The gate is about to open when the center of pressure is at the hinge.
a) y H H
b
Hb
p=+= ++
´
+
12 182
1812
9 18
3
. (./ )
./
(.).
. \H = 0.

22. 21
b) y H H
b
Hb
p=+= ++
´
+
12 202
212
1 2
3
. (./ )
/
( )
. \H = 0.6667 m.
c) y H H
b
Hb
p=+= ++
´
+
12 222
2212
11 22
3
. (./ )
./
(.).
. \H = 2.933 m.

2.57 (A) The gate opens when the center of pressure in at the hinge:

3
1.211.2(1.2)/12
5.51.2.
22(1.2)(11.2)/2
p
hIhbh
yyy
Ayhbh
+++
=+=+=+=+
++

This can be solved by trial-and –error, or we can simply substitute one of the
answers into the equation and check to see if it is correct. This yields h = 1.08 m.

2.58 F
H
bH bH
1
2
2
1
2
= ´=g g
FHbbH
2
=´=g gl l

1
2 3 2
2
g gbH
H
bH´= ´l
l
. \=H3l
a) H=´32 = 3.464 m b) H = 1.732 m c) H = 10.39' d) H = 5.196'


Assume 1 m deep
2.59 The dam will topple if the moment about “O” of F1 and F3 exceeds
the restoring moment of W and F2.
a) W= ´ ´+´(.4 )( /)2981065024502 = 21.19 ´ 10
6
N
d
w
=
´+´
+
3002760016
300600
= 19.67 m.
F2 = 9810 ´ 5 ´ 11.09 = 0.544 ´ 10
6
N. d
2
1109
3
=
.
= 3.697 m.

F
1
9810
45
2
45= ´´ = 9.933 ´ 10
6
N. d1 = 15 m.
F
3
9810
4510
2
30= ´
+
´ = 8.093 ´ 10
6
N. d
3
294315515020
29435150
=
´+ ´
+
. .
. .
= 18.18 m.

WdFd
FdFd
w+ = ´ ×
+ = ´ ×
ü
ý
þ
22
6
11 33
6
418810
296110
.
.
Nm
Nm
\will not topple.

b) W = (2.4 ´ 9810) (6 ´ 65 + 65 ´ 12) = 27.55 ´ 10
6
N.
dw =
3902778016
390780
´+´
+
= 19.67 m.
F
2
6
05410@ ´. N. d
2
[email protected].
F1 = 9810 ´ 30 ´ 60 = 17.66 ´ 10
6
N. d1 = 20 m.
F
3
9810
6010
2
30= ´
+
´ = 10.3 ´ 10
6
N. d
3
294315735820
29437358
=
´+ ´
+
. .
. .
= 18.57 m.
F
2
F
1
H/3
l/2
F
1
F
2
F
3
W
O

23. 22

WdFd
FdFd
w+ = ´ ×
+ = ´ ×
ü
ý
þ
22
6
11 33
6
543910
544510
.
.
Nm
Nm
\it will topple.

c) Since it will topple for H = 60, it certainly will topple if H = 75 m.
assume 1 m deep
2.60 The dam will topple if there is a net clockwise moment about “O.”
a) WWWW=+ =´´´ ´
1 2 1
6431622. ( ).4.4 = 38,640 lb.
W
2
24432622=´ ´ ´( /).4.4 = 77,280 lb.
W
3402233262=´ ´( ./).4 = 27,870 lb @ 20.89 ft.
F
1
6220401= ´´´.4 ( ) = 49,920 lb @ 40/3 ft.
F
2
625101= ´´´.4 ( ) = 3120 lb @ 3.33 ft

1
3
2
= 18,720 lb @ 15 ft

= 28,080 lb @ 20 ft
p
p
F
F
F
ì
ï
=í
ïî




O
MS: (49,920)(40/3) + (18,720)(15) + (28,080)(20) - (38,640)(3)
- (77,280)(14) - (27,870)(20.89) - (3120)(3.33) < 0. \won’t tip.

b) W1 = 6 ´ 63 ´ 62.4 ´ 2.4 = 56,610 lb. W2 = (24 ´ 63/2) ´ 62.4 ´ 2.4 = 113,220 lb.
F
1
623060= ´´.4 = 112,300 lb.
3
(6022.86/2)62.4W=´´ = 42,790 lb.
F
2
62510= ´´.4 = 3120 lb
F
p1
621030= ´´.4 = 18,720 lb. F
p2
6250302= ´´.4 / = 46,800 lb.

O
MS: (112,300)(20) + (18,720)(15) + (46,800)(20) - (56,610)(3)
- (113,220)(14) - 42,790(21.24) = 799,000 > 0. \will tip.

c) Since it will topple for H = 60 ft., it will also topple for H = 80 ft.

2.61 SM
hinge
= 0. 2.5P – dw ´ W – d1 ´ F1 = 0.
\P= ´ ´´+
´
´ ´
´
´
é
ë
ê
ù
û
ú
1
25
2
3
980018
42
3
9800
2
4
4
2
. p
p
= 62 700 N
Note: This calculation is simpler than that of Example 2.7. Actually, We could
have moved the horizontal force FH and a vertical force FV (equal to W)
simultaneously to the center of the circle and then 2.5P = 2FH.=2F1. This
was outlined at the end of Example 2.7.


2.62 Since all infinitesimal pressure forces pass thru the center, we can place the resultant
forces at the center. Since the vertical components pass thru the bottom point, they
produce no moment about that point. Hence, consider only horizontal forces:

()98102(410)784 800N
()0.869810120168 700N
waterH
Hoil
F
F
=´´´=
=´´´=

SMP: . .. 27848216872= ´- ´ \P = 616.1 kN.

F
1
F
2
F
3
W
O
W
3
F
1
d
w
d
1
PW

24. 23
2.63 Place the resultant force
vv
FF
H V
+ at the center of the circular arc.
v
F
H
passes thru the
hinge showing that PF
V=.
a) PF
V
== ´´+´=9810624 4594( )p 200 N or594.2 kN.
b) P = FV = 62.4 (20 ´ 6 ´ 12 + 9p ´ 12) = 111,000 lb.

2.64 (D) Place the force
vv
FF
H V
+ at the center of the circular arc. FH passes through the
hinge:

2
41.29800(1.2/4)9800300000.5.16 m.
V
PFwww p\==´´+´´=\=

2.65 Place the resultant
vv
FF
H V
+ at the circular arc center.
v
F
H
passes thru the hinge so that
PF
V=. Use the water that could be contained above the gate; it produces the same
pressure distribution and hence the same F
V.
PF
V= = 9810 (6 ´ 3 ´ 4 + 9p) = 983 700 N or 983.7 kN.

2.66 Place the resultant
vv
FF
H V
+ at the center.
v
F
V
passes thru the hinge
2 ´ (9810 ´ 1 ´ 10) = 2.8 P. \P = 70 070 N or 70.07 kN.

2.67 The incremental pressure forces on the circular quarter arc pass through the hinge so that
no moment is produced by such forces. Moments about the hinge gives:
3 P = 0.9 W = 0.9 ´ 400. \P = 120 N.

2.68 The resultant
vv
FF
H V
+ of the unknown liquid acts thru the center of the circular arc.
v
F
V

passes thru the hinge. Thus we use only ().F
Hoil
Assume 1 m wide.
a) SM
R R
R
R
S
R
R
R
R
x
: .
3
9810
2
4
3
9800
4 2
2
æ
è
ç
ö
ø
÷+
æ
è
ç
ö
ø
÷=
æ
è
ç
ö
ø
÷
p
p
g \g
x=4580 N/m
3

b) SM
R R
R
R
S
R
R
R
R
x
: . . .
3
624
2
4
3
624
4 2
2
æ
è
ç
ö
ø
÷+
æ
è
ç
ö
ø
÷=
æ
è
ç
ö
ø
÷
p
p
g \g
x=29.1 lb/ft
3


2.69 The force of the water is only vertical (FV)w, acting thru the center. The force of the oil
can also be positioned at the center:
a) PF
Ho= =´ ´´()(. )..0898100336 = 8476 N.
SF WF F
y Vo Vw
==+ -0 ()()
0 = S ´ 9810 p ´ .6
2
´ 6 + .
.
36
36
4
-
æ
è
ç
ö
ø
÷
p
´ 6 ´ (.8 ´ 9810) – 9810 ´ p ´ .18 ´ 6
- ´´´-98108266
2
. . . \=S0955..

b) gVr .W= = 1996 lb.
SF WF F
y Vo Vw
==+ -0 ()()

25. 24
0 = S ´ 62.4 ´ p ´ 2
2
´ 20 + 4
4
4
-
æ
è
ç
ö
ø
÷
p
´ 20 ´ .8 ´ 62.4 – 62.4 ´ p ´ 2 ´ 20
-´´´´62482220
2
.. . \=S0955..

2.70 The pressure in the dome is
a) p = 60 000 – 9810 ´ 3 – 0.8 ´ 9810 ´ 2 = 14 870 Pa or 14.87 kPa.
The force is F = pAprojected = (p ´ 3
2
)

´ 14.87 = 420.4 kN.

b) From a free-body diagram of the dome filled with oil:
Fweld + W = pA
Using the pressure from part (a):
Fweld = 14 870 ´ p ´ 3
2
– (.8 ´ 9810) ´
1
2
4
3
3
3
p´
æ
è
ç
ö
ø
÷= –23 400 N
or –23.4 kN


2.71 A free-body diagram of the gate and water is shown.

H
FdWHP
w
3
+ =´.
a) H = 2 m. F = 9810 ´ 1 ´ 4 = 39 240 N.
W xdy
y
dy= = =
´
òò
98102 98102
2
29810
2
2
32
12
0
2
0
2 32/ /
/
= 26 160 N.

dx
x
xdy
xdy
xdx
xdx
w
== = =
æ
è
ç
ö
ø
÷
ò
ò
ò
ò
2
1
2
4
4
1
2
14
13
3
0
1
2
0
1
/
/
= 0.375 m.
\=´ + ´P
1
3
39
0375
2
26 160 240
.
= 17 980 N or 17.98 kN.

b) H = 8'. F = 62.4 ´ 4 ´ 32 = 7,987 lb.
W xdy xdx= = ´ = ´´òò
624 6244 621623
2
0
2
3
.4 .4 .4 / = 2,662 lb.
dx
xdx
xdx
w
== =
æ
è
ç
ö
ø
÷
ò
ò
1
2
4
4
1
2
164
83
3
0
2
2
0
2
/
/
= 0.75'.
18
7,9870.752,6622910 lb
83
P
æö
=´+´=
ç÷
èø


2.72 (A) WVg=
9009.8198100.0115.6 m ww´=´´\=
W
pA
Fweld
dA=xdy
y
x
F
h/3

26. 25
2.73 W = weight of displaced water.
a) 20 000 + 250 000 = 9810 ´ 3
2
(6/2).dd+ \d
2
+ 12d – 18.35 = 0.

\d = 1.372 m.
b) 270 000 = 1.03 ´ 9810 ´ 3
2
(6/2).dd+ d
2
+ 12d – 17.81 = 0. \d = 1.336 m.

2.74 25 + FB = 100. \FB = 75 = 9810 -V. \ -V= 7.645 ´ 10
-3
m
3

g ´ 7.645 ´ 10
-3
= 100. or 7645 cm
3

\g = 13 080 N/m
3
.

2.75 3000 ´ 60 = 25 ´ 300 Dd ´ 62.4. \Dd = 0.3846' or 4.62".

2.76 100 000 ´ 9.81 + 6 000 000 = (12 ´ 30 + 8h ´ 30) 9810
\h = 1.465 m. \distance from top = 2 – 1.465 = 0.535 m

2.77 T + FB = W. (See Fig. 2.11 c.)
T = 40 000 – 1.59 ´ 9810 ´ 2 = 8804 N or 8.804 kN.

2.78 The forces acting on the balloon are its weight W, the buoyant force FB, and the weight of
the air in the balloon Fa. Sum forces:

FB = W + Fa or
4
3
1000
4
3
3 3
pr prRg Rg
a
= +

4
3
5
100981
287293
1000
4
3
5
100981
287
3 3
p p´
´
´
= +´
´.
.
.
.
.
T
a
\Ta = 350.4 K or 77.4°C

2.79 The forces acting on the blimp are the payload Fp, the weight of the blimp W, the buoyant
force FB, and the weight of the helium Fh:
FB = Fp + W + Fh
1500150
100981
287288
2
p´ ´
´
´
.
.
= Fp + 0.1 Fp + 1500 p ´ 150
2
´
100981
2077288
´
´
.
.


4
o
/64.Idp= . Npeople =
98610
800
8
.´
= 1.23 ´ 10
6

Of course equipment and other niceties such as gyms, pools, restaurants, etc., would add
significant weight.

2.80 Neglect the bouyant force of air. A force balance yields
FB = W + F
= 50 + 10 = 60 = 9800 -V. \-V= .006122 m
3

Density: gVr .W=
r´´981006122.. = 50. \r = 832.5 kg/m
3

Specific wt: g = rg = 832.5 ´ 9.81 = 8167 N/m
3

Specific gravity: S = r/rwater = 832.5/1000 = 0.8325

27. 26
2.81 From a force balance FB = W + pA.
a) The buoyant force is found as follows (h > 16'):
cos ,q=
--h R
R
15
Area = qR
2
– (h – 15 – R) R sinq
\FB = 10 ´ 62.4[pR
2
- qR
2
+ (h – 15 – R) R sinq].
FB = 1500 + ghA.
The h that makes the above 2 FB’s equal is found by trial-and-
error:
h = 16.5: 1859
?
1577 h = 16.8: 1866
?
1858
h = 17.0: 1870
?
1960 \h = 16.82 ft.

b) Assume h > 16
1
3
ft. and use the above equations with R = 1.333':
h = 16.4: 1857
?
1853 \h = 16.4 ft.
c) Assume h < 16
2
3
ft. With R = 1.667',
FB = 10 ´ 62.4[qR
2
- (R – h + 15) R sinq].
FB = 1500 + ghA. cos q=
-+Rh
R
15

Trial-and-error for h:

h = 16: 1849
?
1374 h = 16.2: 1853
?
1765
h = 16.4: 1857
?
2170 \h = 16.25 ft.



2.82 a)
WF
B
=. [ ]0011361000 01549819810
2
. . . / . .+ ´ ´´ ´ = -h Vp
-=
´
´+
´
´= ´
-
V
p p.
.
.
. . .
015
4
15
005
4
06276910
2 2
53
m \h = 7.361 ´ 10
-3
m
\ =m
Hg
2
13.61000.015/4 hp´´´ = 0.01769 kg
b) (.01 + .01769) 9.81 = 9810
p p´
´+
´
´
é
ë
ê
ù
û
ú
.
.
.
. .
015
4
15
005
4
12
2 2
S
x
\Sx = 0.959.
c) (.01 + .01769) 9.81 = 9810
p´
´
.
.
015
4
15
2
Sx. \Sx = 1.045.

2.83 (. ).
.
.
.
..01 9819810
015
4
15
005
4
12
2 2
+ =
´
´+
´
´
é
ë
ê
ù
û
ú
m
Hg
p p
\mHg = 0.01886.
a) (.01 + .01886) 9.81 = 9810
p´
´
.
.
015
4
15
2
Sx. \Sx = 1.089.
b) mHg = 0.01886 kg.

h - 15
qR
pA
FB
W
h - 15
qR

28. 27
2.84 a)
44
o
(10/12)
6464
d
I
pp ´
== = 0.02367 ft
4
.
-= =
´ ´´ ´
V
W
r
HO
2
862 5121212
62
2
. .4 (/) /
.4
p
= 0.4363. depth =
.4363
(/)p512
2
= 0.8'
\ = --GM. /.4363(..4)02367 5 = –0.0457'. \It will not float with ends horizontal.

b) Io = 0.02367 ft
4
, -V= 0.3636 ft
3
, depth = 0.6667'
GM= --. /. ()/0236736365412 = –0.01823'. \It will not float as given.

c) -V = 0.2909, depth = 6.4", GM =
.
.
.02367
2909
432
12
-
-
= 0.0147. \It will float.

2.85 With ends horizontal
4
o
/64.Idp= The displaced volume is
-= ´ = ´
-
V dh d
x x
gp g
2 5 3
49800801410/ . since h = d. The depth the cylinder will
sink is
depth =
-
= ´ = ´
- -V
A
dd d
x x
801410 4102010
5 3 2 5
. // .g p g
The distance CG is CG
h
d
x
=- ´
-
2
10210 2
5
. /g . Then
GM
d
d
d
d
x
x=
´
-+ ´ >
-
-p
g
g
4
5 3
564
801410 2
10210 20
/
.
. / .
This gives (divide by d and multiply by gx):
612.5 – .5 gx + 5.1 ´ 10
-5
g
x
2
> 0.
Consequently,
gx > 8369 N/m
3
or gx < 1435 N/m
3


2.86
3
3
.
water
waterwater
SdW
VSd
g
gg
-===
3
3
.
water
waterwater
SdW
VSd
g
gg
-=== \h = Sd.
GM
d
Sd
d Sd d
S
S
= - - = -+
4
3
12
2 2
1
12
1
22
/
(/ /)( ).
If GM = 0 the cube is neutral and 6S
2
– 6S + 1 = 0.
\=
± -
S
63624
12
= 0.7887, 0.2113.

The cube is unstable if 0.2113 < S < 0.7887.
Note: Try S = 0.8 and S = 0.1 to see if GM>0. This indicates stability.

2.87 As shown, y=
´+´
+
169164
1616
= 6.5 cm above the bottom edge.
G
S
S
A
A
=
´+´+ ´
´+´+ ´
495168516 4
5828 16
g g g
g g g
. .
.
= 6.5 cm.
C
G
h

29. 28
\130 + 104 SA = 174 + 64 SA. \ SA = 1.1.

2.88 a) y=
´+´+´
++
1648187
1688
= 4. x=
´+´+´
++
1618484
1688
= 2.5.
For G: y=
´´+´´+´´
´+´+´
121645811587
121658158
. . .
. . .
= 4.682.

x=
´+´´+´´
´+´+´
12165841584
121658158
. . .
. . .
= 2.364.

G must be directly under C.
tan
.
.
.q=
136
682
\q =11.3°.


b) y=
´+´+´
++
422
1
2
235
422
.
= 2. x=
´+´+´
++
4
1
2
2222
422
= 1.25
For G:y=
´´+´+´
´+´+´
124251157
12452152
. . .
. . .
= 2.34.x=
´+´+´
´+´+´
12254154
12452152
. . .
. . .
= 1.182
Dy = 0.34, Dx = 0.068. tan
.
.
.q=
068
34
\q = 11.3°.

2.89 The centroid C is 1.5 m below the water surface. \CG = 1.5 m.
Using Eq. 2.4.47: GM=
´
´´
-= -= >
l
l
812
83
1517771502770
3
/
.. .. .
\The barge is stable.


2.90 y=
´ + ´
+
8 3 16971
8 1697
.485.414.
.485.
= 1.8 m. \CG=-1815.. = 0.3 m.
Using Eq. 2.4.47: GM=
´
-=-=
l
l
8 12
3497
313116
3
.485/
.
..46... \Stable.

2.91 (A)
2
5
20000200006660(1.2)24070 Pa
9.81
240700.0230.25 N
plug
plugplug
ph
FpA
g
p
=+=+´´=
==´´=
.

2.92 a) tan
.
.a= =
20
9814
H
\H = 8.155 m. pmax = 9810 (8.155 + 2) = 99 620 Pa
b) pmax = r(g + az) h = 1000 (9.81 + 20) ´ 2 = 59 620 Pa
c) pmax = 1.94 ´ 60 (–12) – 1.94 (32.2 + 60) (–6) = 2470 psf or 17.15 psi
d) pmax = 1.94 (32.2 + 60) (–6) = 1073 psf or 7.45 psi

0.682
C
0.136
G

30. 29
2.93 The air volume is the same before and after.
\ 0.5 ´ 8 = hb/2. tan
.
.a= =
10
981
h
b

4
981
10
=
h
h
2
.
.
\h = 2.856. \Use dotted line.
25
1
2
252 4. ..452.w+´´ = \w = 0.374 m.

a) pA = –1000 ´ 10 (0 – 7.626) – 1000 ´ 9.81 ´ 2.5 = 51 740 Pa or 51.74 kPa
b) pB = –1000 ´ 10 (0 – 7.626) = 76 260 Pa or 76.26 kPa
c) pC = 0. Air fills the space to the dotted line.

2.94 Use Eq. 2.5.2: Assume an air-water surface as shown in the above figure.
a) 60 000 = –1000 ax (0–8) – 1000 ´ 9.81 025
8
981
- -
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
.
.
a
x

4 =
h
a
x
2
981
2
´.
60 = 8 ax + 24.52 – 9.81
8
981
a
x
.
. ax – 4.435 = 1.1074 a
x.
a
x
2
– 10.1 ax + 19.67 = 0 \ax = 2.64, 7.46 m/s
2

b) 60 000 = –1000 ax (–8) – 1000 (9.81 + 10)-+
æ
è
ç
ö
ø
÷25
8
981
.
.
.
a
x

60 = 8 ax + 49.52 – 19.81
8
1981
a
x
.
. ax – 1.31 = 1.574 a
x.
a
x
2
– 5.1 ax + 1.44 = 0 \ax = 0.25, 4.8 m/s
2

c) 60 000 = –1000 ax (–8) – 1000 (9.81 + 5) (–2.5 +
8
1481
a
x
.
).
60 = 8 ax + 37.0 – 14.81
8
1481
a
x
.
. ax – 2.875 = 1.361 a
x.
a
x
2
– 7.6 ax + 8.266 = 0 \ax = 1.32, 6.28 m/s
2


2.95 a) ax = 20 ´ .866 = 17.32 m/s
2
, az = 10 m/s
2
. Use Eq. 2.5.2 with the peep hole as
position 1. The x-axis is horizontal passing thru A. We have
pA = –1000 ´ 17.32 (0 – 1.232) – 1000 (9.81 + 10) (0 – 1.866) = 58 290 Pa
b) pA = –1000 ´ 8.66 (0 – 1.848) – 1000 (9.81 + 5) (0 – 2.799) = 57 460 Pa
c) The peep hole is located at (3.696, 5.598). Use Eq. 2.5.2:
pA = –1.94 ´ 51.96 (0 – 3.696) – 1.94 (32.2 + 30) (0 – 5.598) = 1048 psf
d) The peep hole is located at (4.928, 7.464). Use Eq. 2.5.2:
pA = –1.94 ´ 25.98 (–4.928) – 1.94 (32.2 + 15) (–7.464) = 932 psf

2.96 a) The pressure on the end AB (z is zero at B) is, using Eq. 2.5.2,
p(z) = –1000 ´ 10 (–7.626) – 1000 ´ 9.81(z) = 76 260 – 9810 z
a
b
h
A
B
z
1
xw
C

31. 30
\ = -ò
F zdz
AB
( )
.
7626098104
0
25
= 640 000 N or 640 kN
b) The pressure on the bottom BC is
p(x) = –1000 ´ 10 (x – 7.626) = 76 260 – 10 000 x.
\ = -ò
F xdx
BC
( )
.
76260100004
0
7626
= 1.163 ´ 10
6
N or 1163 kN
c) On the top p(x) = –1000 ´ 10 (x – 5.174) where position 1 is on the top surface:
\ = -ò
F xdx
top
( )
.
51740100004
0
5174
= 5.35 ´ 10
5
N or 535 kN
2.97 a) The pressure at A is 58.29 kPa. At B it is
pB = –1000 ´ 17.32 (1.732–1.232)
– 1000 (19.81) (1–1.866) = 8495 Pa.
Since the pressure varies linearly over AB, we
can use an average pressure times the area:
F
AB
=
+
´´
582908495
2
152

. = 100 200 N or 100.2 kN


b) pD = 0. pC = –1000 ´ 17.32 (–.5–1.232) - 1000 ´ 19.81(.866–1.866) = 49 810 Pa.
F
CD
=´ ´´
1
2
49810152 . = 74 720 N or 74.72 kN.
c) pA = 58 290 Pa. pC = 49 810 Pa. \ =
+
´F
AC
58294981
2
15
. .
. = 81.08 kN.


2.98 Use Eq. 2.5.2 with position 1 at the open end:
a) pA = 0 since z2 = z1.
pB = 1000 ´ 19.81 ´ 0.6 = 11 890 Pa.
pC = 11 890 Pa.

b) pA = –1000 ´ 10 (.9–0) = –9000 Pa.
pB = –000 ´ 10 (.9)–1000 ´ 9.81(-.6) = –3114 Pa
pC = –1000 ´ 9.81 ´ (–.6) = 5886 Pa.

c) pA = –1000´20 (0.9) = –18 000 Pa.
pB = –1000 ´ 20 ´ 0.9–1000´19.81(-0.6) = –6110 Pa. pC = 11 890 Pa
d) pA = 0. pB = 1.94 ´ (32.2-60)
25
12
æ
è
ç
ö
ø
÷ = -112 psf. pC = –112 psf.
e) pA = 1.94 ´ 60 -
æ
è
ç
ö
ø
÷
375
12
.
= -364 psf.
pB = 1.94 ´ 60 -
æ
è
ç
ö
ø
÷
375
12
.
– 1.94 ´ 32.2 -
æ
è
ç
ö
ø
÷
25
12
= –234 psf.
x
z
A
BC
x
z
1

32. 31
pC = –1.94 ´ 32.2 -
æ
è
ç
ö
ø
÷
25
12
= 130 psf.
f) pA = 1.94 ´ 30
375
12
.æ
è
ç
ö
ø
÷ = 182 psf.
pB = –1.94(–30)
375
12
.æ
è
ç
ö
ø
÷– 1.94 ´ 62.2 -
æ
è
ç
ö
ø
÷
25
12
= 433 psf.
pC = –1.94 ´ 62.2 ´ -
æ
è
ç
ö
ø
÷
25
12
= 251 psf.

2.99 Use Eq. 2.6.4 with position 1 at the open end:
w
p
=
´502
60
= 5.236 rad/s.
a) p
A=
´
´´
10005236
2
615
2
2.
(..) = 11 100 Pa.
p
B
=
1
2
´ 1000 ´ 5.236
2
´ .9
2
+ 9810 ´ .6 = 16 990 Pa.
pC = 9810 ´ .6 = 5886 Pa.


b) p
A
=
1
2
´ 1000 ´ 5.236
2
´ 0.6
2
= 4935 Pa.
p
B
=
1
2
´ 1000 ´ 5.236
2
´ 0.6
2
+ 9810 ´ 0.4 = 8859 Pa.
pC = 9810 ´ 0.4 = 3924 Pa.
c) p
A
=
1
2
´ 1.94 ´ 5.236
2
´
375
12
2
.æ
è
ç
ö
ø
÷ = 259.7 psf.
p
B
=
1
2
´ 1.94 ´ 5.236
2
´
375
12
62
25
12
2
.
.4
æ
è
ç
ö
ø
÷+ ´ = 389.7 psf.
pC =62
25
12
.4´ = 130 psf.
d) p
A
=
1
2
´ 1.94 ´ 5.236
2
´
225
12
2
.æ
è
ç
ö
ø
÷ = 93.5 psf.
p
B
=
1
2
´ 1.94 ´ 5.236
2
´
225
12
2
.æ
è
ç
ö
ø
÷+ 62
15
12
.4´ = 171.5 psf.
pC = 62
15
12
.4´ = 78 psf.

A
BC
z
1
r
w

33. 32
2.100 Use Eq. 2.6.4 with position 1 at the open end.
a) p
A
=
1
2
´ 1000 ´ 10
2
(0 – 0.9
2
) = –40 500 Pa.
pB = –40 500 + 9810 ´ 0.6 = –34 600 Pa.
pC = 9810 ´ 0.6 = 5886 Pa.
b) p
A
=
1
2
´ 1000 ´ 10
2
(0 – 0.6
2
) = –18 000 Pa.
pB = –18 000 + 9810 ´ 0.4 = –14 080 Pa.

pC = 9810 ´ 0.4 = 3924 Pa.
c) p
A
=
1
2
´ 1.94 ´ 10
2
0
375
144
2
-
æ
è
ç
ö
ø
÷
.
= –947 psf.
pB = -947 + 62.4 ´
25
12
= –817 psf. pC = 62.4 ´
25
12
= 130 psf.
d) p
A
=
1
2
´ 1.94 ´ 10
2
-
æ
è
ç
ö
ø
÷
225
12
2
2
.
= –341 psf.
pB = –341 + 62.4 ´
15
12
= –263 psf. pC = 62.4 ´
15
12
= 78 psf.

2.101.1Use Eq. 2.6.4 with position 1 at the open end and position 2 at the origin. Given: p2 = 0.
a) 0 =
1
2
´ 1000 w
2
(0 – 0.45
2
) – 9810 (0 – 0.6). \w = 7.62 rad/s.
b) 0 =
1
2
´ 1000 w
2
(0 – 0.3
2
) – 9810 (0 – 0.4). \w = 9.34 rad/s.
c) 0 =
1
2
´ 1.94 w
2
0
1875
12
2
2
-
æ
è
ç
ö
ø
÷
.
– 62.4-
æ
è
ç
ö
ø
÷
25
12
. \w = 7.41 rad/s.
d) 0 =
1
2
´ 1.94 w
2
-
æ
è
ç
ö
ø
÷
1125
12
2
2
.
– 62.4-
æ
è
ç
ö
ø
÷
15
12
. \w = 9.57 rad/s.



2.102 The air volume before and after is equal.
\ =´´
1
2
62
0
2 2
p prh ... \rh
0
2
= 0.144.
a) Using Eq. 2.6.5: r
0
2 2
52´/= 9.81 h
\h = 0.428 m
\pA =
1
2
´ 1000 ´ 5
2
´ 0.6
2
– 9810 (–0.372)
= 8149 Pa.


b) r
0
2 2
72´/= 9.81 h. \h = 0.6 m.
\pA =
1000
2
´ 7
2
´ 0.6
2
+ 9810 ´ 0.2 = 10 780 Pa.
A
BC
z
1
r
w
z
1
r
w
2
1
2
h
z
rA
r
0

34. 33
c) For w = 10, part of the bottom is bared.
p p p´´= -.. .62
1
2
1
2
2
0
2
1
2
1
rh rh
Using Eq. 2.6.5:

w
2
0
2
2
r
g
h=,
w
2
1
2
2
r
g
h=
1.
\ = -0144
2 2
2
2
21
2
.
g
h
g
h
w w
or
hh
2
1
2
2
014410
2981
-=
´
´
.
.
.


Also, h – h1 = 0.8. 1.6h – 0.64 = .7339. \h = 0.859 m, r1 = 0.108 m.
\pA =
1
2
´ 1000 ´ 10
2
(0.6
2
– 0.108
2
) = 17 400 Pa.
d) Following part (c): hh
2
1
2
2
014420
2981
-=
´
´
.
.
. 1.6h – .64 = 2.936.\ h = 2.235 m.
\pA =
1
2
´ 1000 ´ 20
2
(0.6
2
– 0.265
2
) = 57 900 Pa r1 = 0.265 m

2.103 The answers to Problem 2.102 are increased by 25 000 Pa.
a) 33 150 Pa b) 35 780 Pa c) 42 400 Pa d) 82 900 Pa

2.104 pr rg h() [(.)].= - --
1
2
08
22
rw r
pr r h() (.)= + -500 98108
22
w if h < .8.
pr rr() ( )= -500
22
1
2
w if h > .8.
a) Fprdr r rdr= = +òò
2 2125003650
3
0
6
p p( )
.
= 6670 N.
(We used h = .428 m)


b) Fprdr r rdr= = +òò
2 2245001962
3
0
6
p p( )
.
= 7210 N. (We used h = 0.6 m)
c) Fprdr r rdr= = -
-
òò
2 2 50000 108
3
108
6
2
p p( (. )
.
.
= 9520 N. (We used r1 = 0.108 m)
d) Fprdr r rdr= = -òò
2 2200000 265
3
265
6
2
p p( (. )
.
.
= 26 400 N. (r1 = 0.265 m)




1
h
z
rA
r
0
h
1
dr
dA = 2prdr

35. 34
CHAPTER 3

Introduction to Fluids in Motion

3.1

3.2 Pathlines: Release several at an instant in time and take a time exposure of the
subsequent motions of the bulbs.

Sreakline: Continue to release the devises at a given location and after the last
one is released, take a snapshot of the “line” of bulbs. Repeat this
for several different release locations for additional streaklines.

3.3

3.4

streakline
pathline
streamline
streakline
pathline hose
boytime t
t = 0
streamlines
t = 2 hr
pathline
t = 2 hr
streakline at t = 3 hry
x

36. 35
3.5 a)u
dx
dt
t v
dy
dt
t==+ ==22 2
xttc ytc=++ =+
2
1
2
2
2
=+y y2
\- +=x xyy y
2 2
2 4 \parabola.

b)xttc c c=++\=- =-
2
1 1 2
2 8 4. , . and
=++±+-y y42 48( )
\- ++- =x xyy x y
2 2
2 8120. \parabola.

3.6
ˆˆˆ ()
ˆ ˆˆˆˆ ˆˆˆˆ using , .
z
VdrudyvdxVuivjwk
drdxidyjdzk ijkjik
ü ´=-=++ ï
ý
=++ ´=´=-ïþ
vv v
v


3.7 Lagrangian: Several college students would be hired to ride bikes around the
various roads, making notes of quantities of interest.

Eulerian: Several college students would be positioned at each intersection
and quantities would be recorded as a function of time.

3.8 a) At t V= = =2 000 22
2
and m/s(,,) .
At t V= - = +=2 120 323606
2 2
and m/s(,,) . .
b) At t V= =2 000 0 and (,,) .
At t V= - =-+-=2 120 2 8 8246
2 2
and m/s(,,) ()(). .
c) At t V= =-=2 000 4 4
2
and m/s(,,) () .
At t V= - = +-+-=2 120 2 4 4 6
2 2 2
and m/s(,,) ()() .

3.9 (D)
5ˆ(51.4 10) j
-
-´
A simultaneous solution yields 4/5 and 3/5.
xy
nn== (They must
both have the same sign.

3.10 a) cos
$
/()/ .. .a a=× =+ += \=
v
o
ViV12320832 3369
2 2


v
Vn ijninj
n n
nn
n n
n n
x y
x y
x y
y x
x x
×= +×+ =
+ =
+=
ü
ý
þ
\
=-
+ =
$.(
$$
)(
$$
).032 0
3 2 0
1
3
2
9
4
1
2 2
2 2

\= =- = -n n n ij
x y
2
13
3
13
1
13
23, $ (
$$
). or
39.8
o
y
x
streamlines
t = 5 s
(27, 21)
(35, 25)

37. 36
b) cos
$
/ /()() ..a a=× =- -+-=- \=
v
o
ViV2 2 8 02425 104
2 2


v
Vn ijninj
nn
nn
n n
nn
x y
x y
x y
x y
y y
×= --×+ =
-- =
+=
ü
ý
þ
\
=-
+=
$.(
$$
)(
$$
).0 28 0
28 0
1
4
16 1
2 2 2 2
\= =- = -+n n n ij
y x
1
17
4
17
1
17
4, $ (
$$
). or

c) cos
$
/ / ().. .a a=× = +-= \=-
v
o
ViV55 806202 5167
2 2


v
Vn ijninj
n n
nn
n n
nn
x y
x y
x y
x y
y y
×= -×+ =
- =
+=
ü
ý
þ
\
=
+=
$.(
$$
)(
$$
).0 58 0
5 8 0
1
8
5
64
25
1
2 2
2 2

\= = = +n n n ij
y x
5
89
8
89
1
89
85, $ (
$$
). or

3.11 a) [ ]
vv
Vdr xixtjdxidyj´= ++´ + =0 2 0.()
$$
(
$$
).
\+ - =
+
=() .xdyxtdx t
xdx
x
dy2 0
2
or
Integrate: [ ]t
xdx
x
dytxnx yC
+
= - +=+
òò
2
2 2. . l
21232( ) .- =-+ \ln C C =0.8028.
[ ]txnx y- +=+2 2 08028l .

b) [ ]
vv
Vdr xyiyjdxidyj´= - ´ + =0 2 0
2
.
$ $
(
$$
).
\ + = =-xydyydx
dx
x
dy
y
2 0
2
2
or .
Integrate: 2 21 2l l l lnx nyC n n C=- =--(/). () (/).
\=- =- - \ =-C nx ny xy2 2 2
2 2
. (/). . l l

c) [ ]
vv
Vdr x iytjdxidyj´= +- ´ + =0 4 0
2 2
.( )
$ $
(
$$
).
( ) .x dyytdx
tdx
x
dy
y
2 2
2 2
4 0
4
+ + =
+
=- or
Integrate:
t x
C
y
C
2 2
1 2
2
1
2
1
2
1 1
tan . tan .
- -
+
æ
è
ç
ö
ø
÷= +
æ
è
ç
ö
ø
÷=-
\=- -
æ
è
ç
ö
ø
÷=
-
C yt
x
09636
2
096362
1
.. tan .

38. 37
3.12 (C)
2ˆˆˆˆˆˆ2(2)(22)16816.
VVVV
auvwxyyiyxiyjiij
txyz
¶¶¶¶
=+++=--=--+
¶¶¶¶
vvvv
v


22
81617.89 m/sa\=+=

3.13 a)
DV
Dt
u
V
x
v
V
y
w
V
z
V
t
v v v vv
= + + +
¶
¶
¶
¶
¶
¶
¶
¶
=0.
b) u
V
x
v
V
y
w
V
z
V
t
xiyjxiyj
¶
¶
¶
¶
¶
¶
¶
¶
v v vv
+ + += + = +222244(
$
)(
$
)
$$
= 84
$$
ij-
c) u
V
x
v
V
y
w
V
z
V
t
xtxtiytjxytxtjztkxixyj
¶
¶
¶
¶
¶
¶
¶
¶
v v vv
+ + += + + + ++
2 2
2 2 22 2 2(
$ $
) (
$ $
)
$

+ =- -2 6810054yzki jk
$ $$

d) u
V
x
v
V
y
w
V
z
V
t
xiyzjxyzxzjtzxyjtkzk
¶
¶
¶
¶
¶
¶
¶
¶
v v vv
+ + +=- - - +- ++(
$ $
) (
$
)(
$$
)
$
2 2 2 2
=xiyzxyzxyztjztzk
$
( )
$
( )
$
- - + + +24 2
22 2

=211415
$ $$
i jk- +
3.14
v
W= -
æ
è
ç
ö
ø
÷+ -
æ
è
ç
ö
ø
÷+ -
æ
è
ç
ö
ø
÷
1
2
1
2
1
2
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
w
y
v
z
i
u
z
w
x
j
v
x
u
y
k
$ $ $
.
a)
v
W=- =
1
2
20
¶
¶
u
y
k yk
$ $
= -20
$
k
b)
v
W= -+ -+ -
1
2
00
1
2
00
1
2
00()
$
()
$
()
$
i j k = 0
c)
v
W= -+ -+ -
1
2
20
1
2
00
1
2
20( )
$
()
$
( )
$
zti j ytk = 62
$$
ik-
d)
v
W= + + -+--
1
2
02
1
2
00
1
2
2 0( )
$
()
$
( )
$
xyi j yzk = -+23
$$
ik

3.15 The vorticity
vv
w=2W. Using the results of Problem 3.7:
a)
v
w=-40
$
i b)
v
w= 0 c)
v
w = 124
$$
ik- d)
v
w = -+46
$$
ik

3.16 a) e
¶
¶
e
¶
¶
e
¶
¶
xx yy zz
u
x
v
y
w
z
== == ==0 0 0, , .
e
¶
¶
¶
¶
e
¶
¶
¶
¶
xy xz
u
y
v
x
y
u
z
w
x
= +
æ
è
ç
ö
ø
÷=-= = +
æ
è
ç
ö
ø
÷=
1
2
2020
1
2
0, ,
e
¶
¶
¶
¶
yz
v
z
w
y
= +
æ
è
ç
ö
ø
÷= \ =
é
ë
ê
ê
ê
ù
û
ú
ú
ú
1
2
0
0200
2000
000
. rate-of strain

39. 38
b)
e e e
e e e
xx yy zz
xy xz yz
= = =
= = =
2 2 0
0 0 0
, , .
, , .


rate-of strain =
200
020
000
é
ë
ê
ê
ê
ù
û
ú
ú
ú


c)
e e e
xx yy zz
xt xt yt== = = = =-28 2 8 2 4, , .
e e e
xy xz yz
yt zt= =- = = = =
1
2
2 2
1
2
00
1
2
2 6(), (), ().
rate-of strain =
820
286
064
-
-
-
é
ë
ê
ê
ê
ù
û
ú
ú
ú

d)
e e e
xx yy zz
xz t= =-=- ==1 2 12 2, , .
e e e
xy xz yz
yz xy=- = = = =- =
1
2
2 3
1
2
00
1
2
2 2(), (), ( ).
rate-of strain =
130
3122
022
-
é
ë
ê
ê
ê
ù
û
ú
ú
ú


3.17 a)
a
r r r r r
r=-
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷ -+
æ
è
ç
ö
ø
÷ -
æ
è
ç
ö
ø
÷-10
40 80
10
40
1
40
2 3 2 2
cos cos
sin
(sin)q q
q
q
- +
æ
è
ç
ö
ø
÷ =- - -
1
10
40
10211251
2
2
2
r r
sin ( .5)().()q = 9.375 m/s
2
.
a
r r r r r
q q q
q
q=-
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷ ++
æ
è
ç
ö
ø
÷ +
æ
è
ç
ö
ø
÷10
40 80
10
40
10
40
2 3 2 2
cos sin
sin
cos
- -
æ
è
ç
ö
ø
÷
1
100
1600
4
r r
sincosqq = 0 since sin 180° = 0.
a
f
= 0.
b) w w w q q
qr z
r r r r
= = =-+
æ
è
ç
ö
ø
÷ - -
æ
è
ç
ö
ø
÷-0 0
1
10
40 1
10
40
2 2
, , sin (sin) = 0.
At (4, 180°)
v
w = 0 since
v
w = 0 everywhere.

3.18 a) a
r r r r r
r= -
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷ -+
æ
è
ç
ö
ø
÷ - -
æ
è
ç
ö
ø
÷10
80 240
10
80
10
80
3 4 3 3
cos cos
sin
(sin)q q
q
q
-+
æ
è
ç
ö
ø
÷ = - -10
80
875193751
3
2 2
r r
sin
.()(.)()
q
= 8.203 m/s
2

a
q
= 0 since sin 180° = 0. a
f
= 0 since v
f
=0.
b) w
r
= 0, w
q
= 0, w
f
= 0, since sin 180° = 0.

40. 39
3.19
VV
au
tx
¶¶
¶¶
=+
vv
v
v+
V
w
y
¶
+
¶
v
ˆ.
Vu
i
zt
¶
¶
¶
=
¶
v
For steady flow ¶¶ut a/ .= =0 0 so that
v


3.20 Assume u(r,x) and v(r,x) are not zero. Then, replacing z with x in the appropriate
equations of Table 3.1 and recognizing that v
q ¶¶q= =0 0 and / :

rx
vvuu
avuavu
rxrx
¶¶¶¶
¶¶¶¶
=+=+

3.21 a) u e t
t
=- - = =¥
-
2101
10
()( ) .
/
2 m/s at
()a
u
t
e t
x
t
==-
æ
è
ç
ö
ø
÷ = =
-¶
¶
210
1
10
02 0
10
() . .
/
m/s at
2

b) u e t
t
=- - = =¥
-
2101
2 10
(.5)( ) .
/
1.875 m/s at
a e t
x
t
=-
æ
è
ç
ö
ø
÷= =
-
210 2
1
10
00125 0
2 2 10
(.5/) . .
/
m/s at
2

c) u e t
t
=- - =
-
21221
2 2 10
( /)( ) .
/
0 for all
a e t
x
t
=-
æ
è
ç
ö
ø
÷=
-
2122
1
10
0
2 2 10
( /) .
/
for all

3.22
DTT
u
Dtx
¶
¶
= v+
T
w
y
¶
¶
+
2
20(1)sin0.5878
1001005
TTt
y
zt
¶¶ppp
¶¶
æö
+=--=-´
ç÷
èø

= -0.3693 °C/s.

3.23
D
Dt
u
x
v
y
w
zt
e
r ¶r
¶
¶r
¶
¶r
¶
¶r
¶
= + + +=-´
-- ´
-
1012310
4300010
4
(. ) = -´
×
-
91110
4
. .
kg
ms
3


3.24
D
Dt
u
x
v
y
w
zt
r¶r
¶
¶r
¶
¶r
¶
¶r
¶
= + + +=-
æ
è
ç
ö
ø
÷10
1000
4
= -
×
2500
kg
ms
3
.

3.25
D
Dt
u
x
r ¶r
¶
= =´401(.) = 0.04 kg/m
3
×s

3.26 (D)
2
2
10
[10(4)]
(4)
x
uuuuu
auvwux
txyzxx x
-¶¶¶¶¶¶
=+++==-
¶¶¶¶¶¶ -


32
2
10101
10(2)(1)(4)206.25 m/s.
48(4)
x
x
-
=---=´´=
-

3.27
D
Dt
V
t
=×Ñ+
vv¶
¶
observing that the dot product of two vectors
v
AAiAjAk
x y z
= + +
$ $ $

and
v vv
BBiBjBkABABABAB
x y z xx yy zz
=++ ×= + +
$$$
. is

41. 40

3.28
a
u
t
Vu
a
v
t
Vv
a
w
t
Vw
a
V
t
VV
x
y
z
=+×Ñ
=+×Ñ
=+×Ñ
ü
ý
ï
ï
ï
þ
ï
ï
ï
\=+×Ñ
¶
¶
¶
¶
¶
¶
¶
¶
vv
vv
vv
v
v
vvv
()

3.29 Using Eq. 3.2.12:
a)
vv
v
vvvvv
v
v
Aa
ds
dt
V r
d
dt
r=+ +´+´´+´
2
2
2W WW
W
( )
= 220420201(
$$
)
$
(
$
.5
$
)ki k k i´+ ´ ´ = 160600
$ $
j i- m/s
2

b)
vvvvvv
o
A V r k j k ki=´+´´= ´- +´ ´2 2202030 20203W WW( )(
$
cos
$
)
$
(
$$
) = -507
$
i
3.30
v
W=
´´
= ´
-2
246060
727210
5p
$
.
$
k k rad/s.

v
v i k i k=- - =- -5707707 3 3(.
$
.
$
).535
$
.535
$
m/s.

vvvvvv
A V r=´+´´2W WW( )
= 2727210 3 3 727210
5 5
´ ´ ´- - + ´ ´
- -
.
$
(.535
$
.535
$
).
$
k i k k
[.
$
(.
$
.
$
)]727210 610707707
5 6
´ ´´ - +
-
k i k = -´ +
-
51410 00224
5
.
$
.
$
j i m/s
2
.
Note: We have neglected the acceleration of the earth relative to the sun since it is quite small
(it is dsdt
2 2v
/). The component
5ˆ(51.4 10) j
-
-´ is the Coriolis acceleration and causes air
motions to move c.w. or c.c.w. in the two hemispheres.

3.31 a) two-dimensional (r, z) b) two-dimensional (x, y)
c) two-dimensional (r, z) d) two-dimensional (r, z)
e) three-dimensional (x, y, z) f) three-dimensional (x, y, z)
g) two-dimensional (r, z) h) one-dimensional (r)

3.32 Steady: a, c, e, f, h Unsteady: b, d, g

3.33 b. It is an unsteady plane flow.

3.34 a) d) e)



3.35 f, h

42. 41

3.36 a) inviscid. b) inviscid. c) inviscid.
d) viscous inside the boundary layer.
e) viscous inside the boundary layers and separated regions.
f) viscous. g) viscous. h) viscous.

3.37 d and e. Each flow possesses a stagnation point.

3.38


3.39 (C) The only velocity component is u(x). We have neglected v(x) since it is
quite small. If v(x) in not negligible, the flow would be two-dimensional.

3.40 Re=VL/n= 2 ´ .015/.77 ´ 10
-6
= 39 000. \Turbulent.

3.41 Re= =
VL
n
.2 ´ .8/1.4 ´ 10
-5
= 11 400. \Turbulent.

3.42 Re
.
.
= =
´
´
-
VL
n
406
1710
5
= 14 100. \Turbulent.
Note: We used the smallest dimension to be safe!

3.43 a) Re
..
.51
= =
´
´
=
-
VD
n
12001
1 10
795.
5
Always laminar.

b) Re
.
.51
= =
´
´
=
-
VD
n
121
1 10
79
5
500. May not be laminar.

3.44 Re = 3 ´ 10
5
=
TVx
n
. nmr=/ where mm=().T
a) T = 223 K or -50°C. \=´ ×
-
m110
5
.5 Ns/m.
2

\=
´
´
=´
-
-
n
1510
3376123
2510
5
5.
. .
. . m/s
2

310
9001000
3600210
5
5
´ =
´
´´
-
x
T
.5
. \xT = 0.03 m or 3 cm
b) T = -48°F. \m = 3.3 ´ 10
-7
lb-sec/ft
2
. n=
´
=´
-
-3310
00089
3710
7
4.
.
. ft
2
/sec.
310
6005280
36003710
5
4
´ =
´
´´
-
x
T
.
. \xT = 0.13' or 1.5"

43. 42
3.45 Assume the flow is parallel to the leaf. Then 3 ´ 10
5
=Vx
T/.n
\=´ =´´´ =
-
x V
T
310 31014106817
5 5 4
n/ .5 . / . m.
The flow is expected to be laminar.

3.46 a) M
V
c
==
´ ´
=
100
14287236
0325
.
.. For accurate calculations the flow is
compressible. Assume incompressible flow if an error of 4%, or so, is
acceptable.
b) M
V
c
==
´ ´
=
80
14287288
0235
.
.. \Assume incompressible.
c) M
V
c
==
´ ´
=
100
14287373
0258
.
.. \Assume incompressible.
3.47
D
Dt
u
x
v
y
w
zt
r ¶r
¶
¶r
¶
¶r
¶
¶r
¶
= + + +=0. For a steady, plane flow
¶r¶/t=0 and w=0. Then
u
x
v
y
¶r
¶
¶r
¶
+ =0.

3.48
D
u
Dtx
r¶r
¶
= v
y
¶r
¶
+ w+
zt
¶r¶r
¶¶
+ 0.= \incompressible.

3.49 (B)
2
98100.800
.113 m/s.
21.23
water
air
hVp
V
g
rr
´
===\=

3.50
V p
2
2
=
r
. Use r = 0.0021 slug/ft
3
.
a) v p= =´´2 231440021/ . /.r = 203 ft/sec.
b) v p= =´´2 291440021/ . /.r = 351 ft/sec.
c) v p= =´´2 2091440021/ . /.r = 111 ft/sec.

3.51 p
V
= =
´æ
è
ç
ö
ø
÷r
2 2
2
123
1201000
3600
2. / = 683 Pa.
\F = pA = 683 p ´ 0.075
2
= 12.1 N.

3.52
V p
2
2
0+=
r
. \=
-
=
´
V
p2 22000
123r .
= 57.0 m/s

44. 43
3.53 (C)
222
121
.0.2000.600.29.810.4002.80 m/s.
222
VVVp
V
gggg
+=+=\=´´=

3.54 (B) The manometer reading h implies:
22
21122
22
2
or (6010.2). 9.39 m/s
221.13
VpVp
VV
rr
+=+=-\= The
temperature (the viscosity of the water) and the diameter of the pipe are
not needed.

3.55 a)
22
0
22
VVp
r
+=
2
2(10)
. . 50
2
oo
o
pp xp
ppx r
rrr
-
++=\=-
b)
22
0
22
VVp
r
+=
2
2(10)
. . 50
2
oo
o
pp yp
ppy r
rrr
++=\=-
3.56
22
22
UpVp
rr
¥¥
+=+ .
a) v vU rr
r cq
q= = = - -
¥
0 180 1 1
2 2
and
o
: ( /)().
( )\= -= -
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
¥ ¥
p Uv U
r
r
r
r
r
c c
r r
2 2
2
2 2 2
2
2
4
.
b) Let rrp U
c T
= =
¥
:
r
2
2

c) ( ) [ ]v rrv U p v U
r c
= =- \= -= -
¥ ¥ ¥
0 2
2 2
14
2 2 2
and = U
2
: sin. sin
q q
q
r r
q
d) Let q r= =-
¥
90
3
2
90
2o
: p U

3.57
22
22
UpVp
rr
¥¥
+=+ .
a) v
q
q= =0 180 and
o
: ( )p Uv U
r
r
r
r
r
c c
= -=
æ
è
ç
ö
ø
÷-
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
¥ ¥
r r
2 2
2
2 2 2
3 6
.
b) Let rrp U
c T
= =
¥
:
1
2
2
r.
c) ( ) [ ]v rrp v U
r c
= = -= -
¥ ¥
0
2 2
14
2 2 2
and = U
2
: sin
r r
q
q

d) Let q r= =-
¥
90
3
2
90
2o
: p U

45. 44
3.58
22
22
UpVp
rr
¥¥
+=+ .
a) ( )p Uu
x x
= -= -+
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú= -+
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú¥
r r p
p
r
2 2
1010
20
2
5011
1
2 2 2
2 2

=- +
æ
è
ç
ö
ø
÷50
21
2
r
xx

b)
1
0 when 1. 50(21)50uxp rr
-
==-=--+=
c) ( )
22
222 601
303045011
222
pUu
xx
rrp
r
p
¥
éùéù
æöæö
=-=-+=-+ êúêúç÷ç÷
èøèøêúêúëûëû
=- +
æ
è
ç
ö
ø
÷450
21
2
r
xx

d)
1
0 when 1. 450(21)450uxp rr
-
==-=--+=

3.59
VpV p
V pp
1
2
1 2
2
2
1 1 2
2 2
0 20+= + = -=
r r
. and kPa.
( )V pp V
2
2
1 2 2
2 2
1000
20 632= -= \=
r
( . 000)=40. m/s

3.60 Assume the velocity in the plenum is zero. Then

22
21122
22
2
or (6010.2). 9.39 m/s
221.13
VpVp
VV
rr
+=+=-\=
We found r=113. kg/m
3
in Table B.2.

3.61 Bernoulli from the stream to the pitot probe: p
V
p
T
= +r
2
2
.
Manometer: .
THg
pHHhphgggg+--=-
Then,
2
2
Hg
V
pHHprgg++-= .
2
(2)
Hg
VH
gg
r
-
\=

a) V V
213619800
1000
2004 314=
-
´ \=
(.)
(.). . m/s
b) V V
213619800
1000
201 497=
-
´ \=
(.)
(.). . m/s
c) V V
21361624
194
2212 1162=
-
´ \=
(.).
.
( /). . fps
d) V V
21361624
194
2412 1644=
-
´ \=
(.).
.
( /). . fps

46. 45
3.62 The pressure at 90° from Problem 3.56 is
2
90
3/2.pU r
¥
=- The pressure at the
stagnation point is
2
/2.
TpUr
¥= The manometer provides: pHp
T-=g
90


2213
1.20498000.041.204. 12.76 m/s
22
UUU
¥¥¥
´-´=-´\=
3.63 The pressure at 90° from Problem 3.57 is
2
90
3/2.pU r
¥
=- The pressure at the
stagnation point is
2
/2.
TpUr
¥= The manometer provides: pHp
T-=g
90


2213
1.20498000.041.204. 12.76 m/s
22
UUU
¥¥¥
´-´=-´\=

3.64 Assume an incompressible flow with point 1 outside in the room where p
10=
andv
10=. The Bernoulli’s equation gives, with p h
w2 2=g,

2
1
2
V
1
p
r
+
2
22
.
2
Vp
r
=+
a) 0
2
9800002
1204
1804
2
2
2
= +
- ´
\=
V
V
.
.
. . m/s
b) 0
2
9800008
1204
361
2
2
2
= +
- ´
\=
V
V
.
.
. . m/s
c) 0
2
624112
000233
668
2
2
2
= +
- ´
\=
V
V
./
.
. . fps
d) 0
2
624412
000233
1336
2
2
2
=+
- ´
\=
V
V
./
.
. . fps

3.65 Assume incompressible flow (V < 100 m/s) with point 1 outside the wind tunnel
where p V
1 10 0= = and . Bernoulli’s equation gives
0
2
1
2
2
2
2
2 2
2
=+ \=-
V p
p V
r
r.
a)r= =
´
= \=-´ ´ =-
p
RT
p
90
0287253
1239
1
2
1239100 6195
2
2
.
. . . kg/m Pa
3

b)r= =
´
= \=-´ ´ =-
p
RT
p
95
0287273
1212
1
2
1212100 6060
2
2
.
. . . kg/m Pa
3

c)r= =
´
= \=-´ ´ =-
p
RT
p
92
0287293
1094
1
2
1094100 5470
2
2
.
. . . kg/m Pa
3

d) r= =
´
= \=-´ ´ =-
p
RT
p
100
0287313
1113
1
2
1113100 5566
2
2
.
. . . kg/m Pa
3


3.66 (A)
2
1
2
V
g
2
122
2
pVp
ggg
+=+
2
2
2
800000
..40 m/s.
981029.81
V
V=\=
´

47. 46

3.67 a) p h V hh
A A A== ´= = =g9800439 0
2 200 Pa, Using . ,

2
2
A
V
g
A
A
p
h
g
++
2
22
2
2
Vp
h
gg
=++
2
2
2
.
2
A
V
pp
g
g=-
= -
´
´ =-39
14
2981
980058
2
200 700 Pa
.


b) 0 and 0.
BBpV== Bernoulli’s eq. gives, with the datum through the pipe,

V
g
p
h
V
g
p
h p
B B
B
2
2
2
2
2 2
2
2 2
4
14
2981
980058++=++ =-
´
æ
è
ç
ö
ø
÷ =-
g g
.
.
700 Pa

3.68 Bernoulli:
2
22
2
Vp
gg
+
2
11
2
Vp
gg
=+
Manometer:
2
2
12
2
Hg
V
pzHHzp
g
ggggg++--=+
Substitute Bernoulli’s into the manometer equation:
( )
2
1
11
.
2
Hg
V
pHp
g
ggg+-=+
a) Use H = 0.01 m:
V
V
1
2
1
9800
2981
13619800001 1
´
´
= - ´ \=
.
(.) . .572 m/s
Substitute into Bernoulli:
p
VV
g
1
2
2
1
2 2 2
2
201
2981
9800198=
-
=
-
´
´ =g
.572
.
600 Pa
b) Use H = 0.05 m:
V
V
1
2
1
9800
2981
13619800005 3
´
´
= - ´ \=
.
(.) . .516 m/s
Substitute into Bernoulli:
p
VV
g
1
2
2
1
2 2 2
2
203
2981
9800193=
-
=
-
´
´ =g
.516
.
600 Pa
c) Use H = 0.1 m:
V
V
1
2
1
9800
2981
1361980001 4972
´
´
= - ´ \=
.
(.) . . m/s
Substitute into Bernoulli:
p
VV
g
1
2
2
1
2 2 2
2
204972
2981
9800187=
-
=
-
´
´ =g
.
.
400 Pa

48. 47
3.69 Bernoulli across nozzle:
2
1
2
V
2
122
2
pVp
rr
+=+
21
. 2/Vp r\=
Bernoulli to max. height:
2
1
2
V
g
1
1
p
h
g
++
2
2
2
V
g
=
2
p
g
+
221
. /.hhp g+\=

a) V p
2 1
2 2700 10003742= =´ =/ / .r 000 m/s
hp
2 1 700= =/g 000/9800=71.4 m

b) V p
2 1
2 21 10005292= =´ =/ / .r 400 000 m/s
hp
2 1= =/g1 400 000/9800=142.9 m

c) V p
2 1
2 2100 1941218= =´´ =/ /. .r 144 fps
hp
2 1= =´/g100144/62.4=231 ft

d) V p
2 1
2 2200 1941723= =´ ´ =/ /. .r 144 fps
hp
2 1 200= = ´/g 144/62.4=462 ft

3.70 a) Apply Bernoulli’s eq. from the surface to a point on top of the downstream
flow:

2
1
2
V
g
1
p
g
+
2
22
1
2
Vp
h
gg
+=+
22
. 2()hVgHh+\=-

b) Apply Bernoulli’s eq. from a point near the bottom upstream to a point on the
bottom of the downstream flow:

2
1
2
V
g
2
122
12
.
2
pVp
hh
ggg
++=++
Using pHph hh V gHh
1 2 1 2 22= = = = -g g, , ( ) and

3.71
2
1
2
V
2
122
.
2
pVp
rr
+=+ p2 = -100 000 Pa, the lowest possible pressure.
a)
600
2
2
2
000
1000
100 000
1000
=-
V
. \V
2
= 37.4 m/s.

b)
300
2
2
2
000
1000
100 000
1000
=-
V
. \V
2
= 28.3 m/s.

49. 48
c)
80144
1.94
14.7144
1.94
´
=-
´V
2
2
2
. \V
2
= 118.6 ft/sec.

d)
40144
1.94
14.7144
1.94
´
=-
´V
2
2
2
. \V
2
= 90.1 ft/sec.

3.72 A water system must never have a negative pressure, since a leak could ingest
impurities. \ The least pressure is zero gage.

Vp
gz
Vp
gz
1
2
1
1
2
2
2
2
2 2
++= ++
r r
. VV
1 2=. Let z
10=,and p
20=.

500 000
1000
=981
2
..z \z
2
= 51.0 m.

3.73 a) ( ) ( )p VV
1 2
2
1
2 2 2
2
1000
2
210= - = -
r
= -48 000 Pa
b) ( ) ( )
2222
121
902
21043300 Pa
22
pVV
r
=-=-=-
c) ( ) ( )
2222
121
680
21032600 Pa
22
pVV
r
=-=-=-
d) ( ) ( )
2222
121
1.23
21059.0 Pa
22
pVV
r
=-=-=-

3.74
VpVp
1
2
1 2
2
2
2 2
+= +
r r
. ( ) ( )
2222
121
1.23
28
22
pVV
r
=-=- = -36.9 Pa

3.75 (D) ( ) ( )
2222
121
902
3015304400 Pa
22
pVV
r
=-=-=

3.76 Apply Bernoulli’s equation between the exit (point 2) where the radius is R and
a point 1 in between the exit and the center of the tube at a radius r less than R:

VpVp
p
VV
1
2
1 2
2
2
1
2
2
1
2
2 2 2
+= + \=
-
r r
r. .
Since VV
2 1<, we see that p
1
is negative (a vacuum) so that the envelope would
tend to rise due to the negative pressure over most of its area (except for a small
area near the end of the tube).

3.77 Re .=
VD
n
For air n@´
-
110
5
.5 . Use reasonable dimensions from your
experience!

50. 49
a) Re
.
.5
.=
´
´
=´
-
20003
110
410
5
4
\Separate
b) Re
.
.5
.=
´
´
=
-
200005
110
6700
5
\Separate
c) Re
.5
. .=
´
´
=´
-
202
110
2710
5
6
\Separate
d) Re
.
.5
.=
´
´
=
-
50002
110
670
5
\Separate
e) Re
.5
. .=
´
´
=´
-
202
110
2710
5
6
\Separate
f) Re
.5
.=
´
´
=´
-
1003
110
210
5
7

\It will tend to separate, except streamlining the components
eliminates separation.

3.78 A burr downstream of the opening will create a region that
acts similar to a stagnation region thereby creating a high
pressure since the velocity will be relatively low in that region.


3.79 D Dp
V
R
n= = ´ =r
2 2
1000
10
005
002
.
.40 000 Pa Along AB, we
expect V V
A B
> <10 10 m/s and m/s.


3.80 The higher pressure at B will force the fluid toward the lower
pressure at A, especially in the wall region of slow moving
fluid, thereby causing a secondary flow normal to the pipe’s
axis. This results in a relatively high loss for an elbow.


3.81 Refer to Bernoulli’s equation:
VpVp
1
2
1 2
2
2
2 2
+=+
r r

pp
A B> since VV
A B<

pp
C D< since VV
C D>

pp
B D> since VV
D B>
stagnation
region
A
B
VA
VB

51. 50
CHAPTER 4

The Integral Forms of the
Fundamental Laws

4.1 a) No net force may act on the system: S
v
F=0.
b) The energy transferred to or from the system must be zero: Q - W = 0.
c) If
332
ˆˆˆ10() 0
nVVnij=×=×-=
v
is the same for all volume elements then
S
v v
F
D
Dt
Vdm=
ò
, or S
v v
F
D
Dt
mV=(). Since mass is constant for a system
S
v
v
Fm
DV
Dt
= . Since
DV
Dt
a Fma
v
v vv
= =, . S

4.2 Extensive properties: Mass, m; Momentum, mV
v
; kinetic energy,
1
2
mV
2
;
potential energy, mgh; enthalpy, H.
Associated intensive properties (divide by the mass): unity, 1; velocity,
v
V;V
2
/2;
gh; H/m = h (specific enthalpy).
Intensive properties: Temperature, T; time, t; pressure, p; density, r; viscosity, m.

4.3 (B)

4.4

System () tV=1
c.v.()tV=1
System () ttV+D= 1V+2
c.v.()ttV+D= 1




4.5

System () tV
=1V+2
c.v.()tV=1V+2
System () ttV+D= 2V+3
c.v.()ttV+D= 1V+2


1 2
1
2
3
pump

52. 51
4.6 a) The energy equation (the 1
st
law of Thermo).
b) The conservation of mass.
c) Newton’s 2
nd
law.
d) The energy equation.
e) The energy equation.

4.7





4.8


4.9

4.10 $ $ $
.(
$$
)n i j ij
1
1
2
1
2
0707=- - =- +. $.
$
.5
$
n i j
208660= - . $ $
n j
3=-.

111
ˆˆˆˆ10[0.707()]7.07 fps
n
VVniij=×=×-+=-
v

VVn i i j
n2 2 210086605866=×=× - =
v
$ $(.$.$). fps

332
ˆˆˆ10() 0
nVVnij=×=×-=
v




4.11 flux = hr$nVA×
v

flux1 = hr hr[.(
$$
)]
$
/.- +× =-0707 10 070710ij iA A
flux2 = hr hr(.
$
.5
$
)
$
/.08660 10 086610i jiA A- × =
flux3 = hr(
$
)
$
-× =jiA10 0
3



ˆn
v ˆn
v
w
ˆn
v
ˆn
v
ˆn
v
ˆn
v ˆn
ˆn
ˆn
v
v
v
v
ˆnˆn
ˆn
ˆn
v
v
v
v
ˆn ˆn
v
v

53. 52
4.12 ($) (.5
$
.
$
)
$
( )
v
BnA i jj× = + ×´ 150 0866 1012
=´ ´ =1508661201559. cm
3

Volume = 156010121559 sin cm
3o
´´=

4.13 The control volume must be independent of time. Since all space coordinates are
integrated out on the left, only time remains; thus, we use an ordinary derivative
to differentiate a function of time. But, on the right, we note that r and h may be
functions of (x, y, z, t); hence, the partial derivative is used.

4.14

4.15


4.16

4.17 If fluid crosses the control surface only on areas A1 and A2,
r r r$ $ $
..
nVdA nVdA nVdA
AAcs
× = ×+ × =òòò
v v v
0
21

For uniform flow all quantities are constant over each area:
r r
11 1 22 2 0
21
$ $nVdA nVdA
AA
× + × =òò
v v

Let A
1
be the inlet so $nV V A
1 1 1 2
×=-
v
and be the outlet so $ .nVV
2 2 2
×=
v
Then
- + =r r
111 2220VA VA
or
r r
222 111AV AV=
1
2
1
system (Dt) is in
volumes 1 and 2
c.v. (0) = c.v. (Dt)
= volume 1
1
2
3
system (Dt) = V
1
+ V
2
+ V
3
c.v. (Dt) = V
1
+ V
2
system boundary
at (t + Dt)

54. 53

4.18 Use Eq. 4.4.2 with m
V
representing the mass in the volume:
0= + ×ò
dm
dt
nVdA
V
cs
r$
..
v
= + -
dm
dt
AV AV
V
r r
22 11

= +-
dm
dt
Qm
V
r &.
Finally,

dm
dt
mQ
V
=-& .r

4.19 Use Eq. 4.4.2 with m
S
representing the mass in the sponge:
0= + ×
ò
dm
dt
nVdA
S
r$
v
= + + -
dm
dt
AV AV AV
S
r r r
22 33 11

= ++ -
dm
dt
m AV Q
S
& .
2 33 1
r r
Finally,

dm
dt
Qm AV
S
= --r r
1 2 33
& .

4.20 (D)
2200
0.04700.837 kg/s
0.287293
p
mAVAV
RT
rp===´´=
´
& .

4.21 A1V1 = A2V2. p ´
125
144
2
.
´ 60 = p ´
25
144
2
.
V2. \V2 = 15 ft/sec.
& .
.
m AV= = ´r p194
125
144
60
2
= 3.968 slug/sec. Q = AV = p
125
144
2
.
´ 60 =2045. /. ftsec
3


4.22 A1V1 = A2V2. p ´ .025
2
´ 10 = (2p ´ .6 ´ .003)V2. \V2 = 1.736 m/s.
& .mAV= = ´ ´r p100002510
2
= 19.63 kg/s. Q = AV=p ´ .025
2
´ 10 =001963. . m/s
3


4.23 &m
in
= rA1V1 + rA2V2. 200 = 1000 p ´ .025
2
´ 25 + 1000 Q2. \Q2 =01509. . m/s
3


4.24 r
1
1
1
40144
1716520
= =
´
´
p
RT
= .006455 slug/ft
3
. r
2
7144
1716610
=
´
´
= .000963 slug/ft
3
.
& .
& .
( /).
.mAV V
m
A
= \= =
´
r
r p

1
11
2
2
2144006455
\V1 = 355 fps.
& .. ( /).m V
2 2
0200096323144== ´´ \V2 = 4984 fps.

55. 54
4.25 r r r r
111 222 1
1
2
500
287393
4
1246
287522
8317AV AV
p
RT
= = =
´
= =
´
=.
.
.433.
.
.
kg
m

kg
m
3 3

4.433 p ´ .05
2
´ 600 = 8.317 p ´ .05
2
V2. \V2 = 319.8 m/s.
&m AV=r
111
= 20.89 kg/s. QAV
1 11
= =4712. . m/s
3
Q2 =2.512 . m/s
3


4.26 r r
111 222AV AV=

p
RT
AV
p
RT
AV
1
1
11
2
2
22
=

200
293
00540
120
003120
2
2
2
p p´ ´= ´ ´. . .
T

\= -T
2
1899 83. . K or C
o


4.27 a) AVAV
11 22= . (2 ´ 1.5 + 1.5 ´ 1.5) 3 =p
d
2
2
4
2´. \d2 = 3.167 m
b) (2 ´ 1.5 + 1.5 ´ 1.5) 3 =p
d
2
2
4
2
2
´. \d2 = 4.478 m
c) (2 ´ 1.5 + 1.5 ´ 1.5) 3 =
1
3 2
866 2
2
pR
R
R-´
æ
è
ç
ö
ø
÷´. .
\R = 3.581 m. \d2 = 7.162 m



4.28 (A) Refer to the circle of Problem 4.27:

2375.72
(0.40.100.40sin75.5)30.516 m/s.
360
QAV p
´
==´´-´´´=
o


4.29 a) v
r
r
rVvdA
r
r
rdr r
r
r
dr
rr r
= -
æ
è
ç
ö
ø
÷ = = -
æ
è
ç
ö
ø
÷ = -
æ
è
ç
ö
ø
÷òò ò
101 101 2 20
0
0
2
000
2
00
00 0
. . p p p
\= -
æ
è
ç
ö
ø
÷=V
r
rr20
23
10
3
0
2
0
2
0
2
= 3.333 m/s.
& . .mAV= = ´´ ´r p1000 04333
2
= 1675. .kg/s Q = AV =001675. . m/s
3

b) v
r
r
rV
r
r
rdr
rr
r
= -
æ
è
ç
ö
ø
÷ = -
æ
è
ç
ö
ø
÷ = -
æ
è
ç
ö
ø
÷ò
101 101 2 20
24
2
0
2 0
2
2
0
2
0
0
2
0
20
. . p p p \V = 5 m/s
& .mAV= = ´´ ´r p1000 045
2
= 2513. kg/s. Q = AV =002513. m/s.
3

c) v
r
r
rV
r
r
rdr r
r
r
= -
æ
è
ç
ö
ø
÷ = -
æ
è
ç
ö
ø
÷ +
ò
201 201 2 10 4
0
0
2
02
0
2
0
0
. /.
/
p p p \V = 5.833 m/s
& . .mAV= = ´´ ´r p1000 045833
2
= 2932. kg/s. Q =002932. . m/s
3


qR
cosq = 1/2
q = 60
o

56. 55
4.30 a) Since the area is rectangular, V = 5 m/s.
& . .mAV= = ´ ´´r 10000885 = 320 kg/s. Q =
&m
r
= 032. . m/s
3

b) v
y
h
y
h
= -
æ
è
ç
ö
ø
÷40
2
2
with y = 0 at the lower wall.
\ = -
æ
è
ç
ö
ø
÷=´ò
Vhw
y
h
y
h
wdy
h
w
h
40 40
6
2
2
0
. \V = 6.667 m/s.
& . . .mAV= = ´ ´´r 10000886667 =4267. kg/s. Q = 04267. . m/s
3

c) V ´ .08 = 10 ´ .04 + 5 ´ .02 + 5 ´ .02. \V = 7.5 m/s.
& . . .5mAV= = ´ ´´r 1000088 7 = 480 kg/s.
&
&
Q
m
=
r
= 048. . m/s
3


4.31 a) AV vdA
11 2
=
ò
. p p p´
æ
è
ç
ö
ø
÷´= -
æ
è
ç
ö
ø
÷ =ò
1
24
6 1 2 2
4
2
0
2
0
2
0
20
v
r
r
rdrv
r
r
max max.
With r
0
1
24
=, v
max
= 12 fps. \ =vr()121576
2
( ).-r fps
b) AV vdA
11 2
=
ò
.
1
12
6 1
4
3
2
2
´´= -
æ
è
ç
ö
ø
÷=
-
ò
w v
y
h
wdyvw
h
h
h
max max .
With h =
1
24
, v
max
= 9 fps. \ =vy()91576
2
( ).-y fps
c)
112 .AVvdA=ò
0
22
2 0
maxmax
2
0
0
0.012122.
4
r
rr
vrdrv
r
ppp
æö
´´=-= ç÷ò
ç÷
èø

With r0 = 0.01 m,
max
v= 4 m/s. ()vr\=
2
4(110000) m/s.r-
d) ˆn
2
maxmax
2
4
0.0221.
3
h
h
yh
wvwdyvw
h-
æö
´´=-= ç÷ò
ç÷
èø

With h = 0.01 m,
max
v = 3 m/s. ()vy\=
2
3(110000) m/s.y-

4.32 If dmdt/ ,=0 then r r r
111 222 333AV AV AV= + . In terms of &m Q
2 3 and this
becomes, letting rrr
1 2 3==,
1000 00212 1000001
2
2
´´ ´=+ ´p. & ..m \ =& . .m
2508 kg/s

4.33 vdAAV
r
1
0
22
1
ò
= . v
r
r
rdr
r
max . .
0
2
1
2
2
1
1 2 00252ò
-
æ
è
ç
ö
ø
÷ =´ ´p p
\ =´ ´2
005
4
00252
2
2
p pv
max
.
. . \v
max
= 1 m/s. \ =vr()1
005
2
2
-
æ
è
ç
ö
ø
÷
r
.
.m/s

57. 56

4.34 & & &. . ( ) . &.
.
m m m y ydy m
in out= + ´´´= - +´´´
é
ë
ê
ù
û
ú+ò
r r r2210 10201002 1210
2
0
1

Note: We see that at y = 0.1 m the velocity u(.1) = 10 m/s. Thus we integrate to
y = 0.1, and between y = 0.1 and 0.2 the velocity u = 10.
4
4
3
2r rr= +
é
ë
ê
ù
û
ú
+&.m \=&.m06667r = 0.82 kg/s.

4.35 Vh uydy
h
11
0
=ò
(). 10051020100
2
0
´= -ò
. ( )y ydy
h

= -
æ
è
ç
ö
ø
÷1010
100
3
2 3
h h.
\666.7 h
3
- 200 h
2
= -1. This can be solved by trial-and-error:
h = .06: -.576
?
-1. h = .07: -.751
?
-1.
h = .08: -.939
?
-1. h = .083: -.997
?
-1.
h = .084: -1.016
?
-1. \h = 0.0832: or 8.32 cm.
Note: Fluid does not cross a streamline so all the flow that enters on the left leaves on
the right. The streamline simply moves further from the wall.

4.36 ( )& .. ( )
/
m VdA yyy dy= = - - ´òò
r 22135456925
0
13
2

( )= - - +ò
2262127 9 319
2 2 3
0
13
y y y ydy. .
/
= 4.528 slug/sec.
V u= =´=
2
3
2
3
2
4
3
max
.fps (See Prob. 4.31b).
r=
+22194
2
..
= 2.07 slug/ft
3
. \ = ´´´
æ
è
ç
ö
ø
÷rVA207
4
3
5
1
3
. = 4.6 slug/sec.
Thus, rVAm¹& since r = r(y) and V = V(y) so that VVrr¹.

4.37 AVAV
11 22= . p p´´=´´. (..) .0182204
2
2
Vcos 30
o
\V2 = 0.05774 m/s.

4.38 2000
4
3
0015 90005
3
3
3
3
´´ ´ ´p.
m of HO
m of air
m of air
s
2
= 1.5 ´ (1.5h). \h = 0.565 m.

4.39 Use Eq. 4.3.3: 0
11 11
= -+ ×
ò
¶r
¶
r
t
dV VnA
v
$.
v
Vn V
1 1 1
×=-$ .
\ =-r
¶r
¶
111
AV
t
V
tire
.
( .)
.
37147144
1716520
1
96
180 17
2
+
´
´´
æ
è
ç
ö
ø
÷´ =´p
¶r
¶t

58. 57
\ =
¶r
¶t
30110
5
. .´
-
-
slug
ftsec
3


4.40 & &&.m mm
in
=+
2 3
V
1
= 20 m/s (see Prob. 4.31c).
2010000210100002
2 2
3
´ ´ =+ ´p p. . .V \V
3
= 12.04 m/s.

4.41 0
2 3 1
= + = ++-
d
dt
m m
d
dt
m mmm
cv net cv.. ..
& &&&
\ =--= ´´´-- ´ ´
d
dt
m mmm
cv..
&&& . .
1 2 3
2 2
1000 02201010000210p p
= 2.57 kg/s.

4.42 The control surface is close to the
interface at the instant shown.
\Vi = interface velocity.
r r
eee iiiAV AV= .

1 15300
8000
287673
1
2 2
.5.
.
.´´´ =
´
´p p V
i

\Vi = 0.244 m/s.

4.43 Assume an incompressible flow:
4
1 22QAV= . 415006024
2´ =´/ ().V \=V
212.5. fps

4.44 For an incompressible flow (low speed air flow)
udAAV
A
=ò 22
1
. 20 08 015
15 2
2
0
02
y dy V
/
.
. . .´ =´ò
p
2008
5
6
02 015
65 2
2
´ =´.. . .
/
p V \=V
2273. . m/s

4.45 AV vdAAV
ee11 2
+ =
ò

p p p(. .)
.
.
.
0100254 2001
0025
2 01
2 2
2
2
2
0
0025
- ´+ -
æ
è
ç
ö
ø
÷ =´ò
r
rdr V
e

011780196300314. . . .+ = V
e
\=V
e100. . m/s

4.46 Draw a control volume around the entire set-up:
0
22 11
= + -
dm
dt
VA VA
tissue
r r
= +
-æ
è
ç
ö
ø
÷-& &
(tan)
&
m
dd
h h h
tissuerp rp f
2
2 2
2 1
2
1
4

V
e
ˆn
ˆn
V
i

59. 58
or
&
& &
tan.m
dd
hhh
tissue
=
-
+
é
ë
ê
ù
û
ú
rp f
2
2
2
2 1
2
1
2
4


4.47 The width w of the channel is constant throughout the flow. Then
0
22 11
= + -
dm
dt
AV AVr r . 0
22 11
= + -
d
dt
whL AV AV( )r r r
0 100028402= ´+ ´- ´r r r
dh
dt
w w w. .. \=
&
. .h0008 m/s

4.48 0
22 11
= + -
dm
dt
AV AVr r
=+ ´ ´ -´
-
& (. . /).m10000003002101060
2 6
p \= ´
-
&. .m39910
4
kg/s

4.49 r r
111 222AV AV= . &m AV
1 222=r .
400 1090002 005
10100 6 2
e V
e
- -
´ ´ =´´
/
. . .p \=V
e207 m/s.

4.50 0
33 111 2
=+ - -
dm
dt
Q AVmr r & where m Ah=r.
a) 0100006 1000066010000021010
2 2
= ´ + ´ - ´ ´-p p.
&
./ . .h
\=
&
.h00111 m/s or 11.1 mm/s
b) 01000061000001020
2
= ´ + ´ --p.
&
. .h
\=
&
. .h000884 m/s or 8.84 mm/s
c)
22
010000.610001.0/6010000.02510. hpp=´+´-´´-
&

0.000339 m/s or 0.339 mm/s.h\=
&


4.51 AVAV
11 22= where A
2
is an area just under the top surface.
a) p p´ ´ =´
-
00210 60
2 10 2
. (tan)
/
e h
dh
dt
t o

\ =
-
hdh edt
t2 10
0001333. .
/
\=- +
-
h e
t3 10
004 004. ..
/

Finally,
ht e
t
().( ).
/ /
= -
-
03421
1013

b) 0041010 6010
10
. (tan)
&/
´´ = ´
-
e h h
t o

\ =
-
hdh edt
t
02309
10
. .
/
\=- +
-
h e
t2 10
462 462. ..
/

Finally,

/101/2
()2.15(1).
t
hte
-
=-

60. 59
4.52
&
WTpAV
du
dy
A
belt
=+ +w m
=´´ +´´´+´ ´´´
-
205002604000401018110100008
5
p/ ..5 . .5.
= ++ =104780000007241847. W

4.53 If the temperature is essentially constant, the internal energy of the c.v. does not
change and the flux of internal energy into the pipe is the same as that leaving
the pipe. Hence, the two integral terms are zero. The losses are equal to the heat
transfer exiting the pipe.



4.54 80% of the power is used to increase the pressure while 20% increases the
internal energy (
&
Q=0 because of the insulation). Hence,
&
~
.
&
mu WD=02
1000002418 02500´ ´ =´. . . .DT \ =DT0836. . C
o


4.55 (D)
22
21
2
P
WVV
Qgg
-
=
&
21
1200200
..
0.040
P
ppW
ggg
- -
+=
´
&


40
40 kW and energy req'd = 47.1 kW.
0.85
P
W\==
&


4.56
&
.W
QH
P
P
p
=
g
h
5746
980020
087
´ =
´ ´Q
.
. \=Q001656. . m/s
3

4.57 - =-´
&
&
..
W
mg
T
40089
a)
&
. .W
T=´ ´´ =4008920098169 850 W
b)
&
.( /).W
T=´ ´ ´ =400899000060981523 900 W
c)
&
.( / ).W
T
=´ ´´ ´ =400898103600981776
6
100 W

4.58
10000000
.0.8950.1.273 m/s
1003609.8
T
T
W
zV
AVgV
h
r
-=D=´\=
´´´´
&

4.59
V
g
p
z
V
g
p
z
1
2
1
1
2
2
2
2
2 2
++=++
g g
.

12
2322
6
36
644
2 2
2
2 2
´
+= +
. .
.
h
h
8236
201
2
2 2
.
.
.= +
h
h Continuity: 3 ´ 12 = h2 V2.
3 ft
V
1
h
2
V
2

61. 60
This can be solved by trial-and-error.
2
8': 8.24 ? 8.31h=
2
7.9': 8.24 ? 8.22h= Qh
2=793..'

2
1.8': 8.24 ? 8.00h=
2
1.75': 8.24 ? 8.31h= Qh
2=176.'.

4.60
V
g
z
V
g
zh
L
1
2
1
2
2
2
2 2
+=++. \
´
+=
´
++
4
2981
2
16
2981
02
2
2
2 2
. .
..
h
h
\26150815
2
2
2
. ./ .= +hh Trial-and-error provides the following:
h
222615263=
=
.5:.
?
. h
22456152=
=
.:.
?
.59. 2 \=h
2
247. m
h
206526152=
=
.:.
?
.58 h
2064615263=
=
.:.
?
.. 2 \=h
2
0646. m

4.61 Manometer: Position the datum at the top of the right mercury level.
981049810
2
10009810136498102
2 2
2
2
1
´+ ++´ = ´ ´+ ´+. ( .).zp
V
p
Divide by g=9810: . .. .4
2
13642
2
2 2
2
1
+++= ´++z
pV
g
p
g g
(1)
Energy:
V
g
p
z
V
g
p
z
1
2
1
1
2
2
2
2
2 2
++=++
g g
. (2)
Subtract (1) from (2): With z
1
= 2 m,
V
g
1
2
2
1264= ´... \V1 = 9.94 m/s

4.62 The manometer equation (see Prob. 4.61) is
04
2
13642
2
2 2
2
1
. .. .+++ = ´++z
pV
g
p
g g
(1)
Energy:
V
g
p
z
V
g
p
z
V
g
1
2
1
1
2
2
2
2
2
2
2 2
005
2
++=+++
g g
. . (2)
Subtract (1) from (2): With z
1
= 2 m, and with V2 = 4V1 (continuity)

18
2
12604
1
2
.
...
V
g
= ´ \V1 = 7.41 m/s.

4.63 (A)
2
2
0
V
=
2
121
2
Vpp
g
--
+
2
2
2
120
.0.7200000 Pa.
29.89810
p
p
g
-
=+\=
´


4.64 Q = 120 ´ 0.002228 = p ´
1
12
2
1
æ
è
ç
ö
ø
÷V. \V1 = 12.25 fps.

62. 61
Continuity: p p´
æ
è
ç
ö
ø
÷=´
æ
è
ç
ö
ø
÷
1
12
1
12
2
1
2
2
V V
.5
. \V2 = 5.44 fps.
Energy:
V
g
pV
g
p V
g
1
2
1 2
2
2 1
2
2 2
037
2
+= ++
g g
. .
\=´+ -
é
ë
ê
ù
û
ú
p
2
2 2
60144624063
1225
644
544
644
..
.
.
.
.
= 8702.9 psf or 60.44 psi

4.65 Q = 600 ´ 10
-3
/60 = p ´ .02
2
V1. \V1 = 7.958 m/s.
3
20.02
33
332
0
11
101
0.026.670.02
y
VdAwdy
AVw
a
æö
==- ç÷òò
ç÷
´´ èø

V
AV
A
2
11
2
2
2
047958
06
= =
´. .
.
= 3.537 m/s.
Energy:
V
g
pV
g
p
h
L
1
2
1 2
2
2
2 2
+=++
g g
.
\=
-
´
+
-
h
L
79583
2981
690000700000
9810
2 2
. .537
.

= 1.571 m

4.66 VQA
1 1 2
008
03
= =
´
/
.
.p
= 28.29 m/s. \V2 = 9V1 = 254.6 m/s.
Energy:
V
g
pV
g
p V
g
1
2
1 2
2
2 1
2
2 2
2
2
+=++
g g
..
\=
´
-
´
é
ë
ê
ù
û
ú
p
1
2 2
9810
2546
2981
08
2829
2981
.
.
.
.
.
= 32110
6
. .´ Pa

4.67 a) Across the nozzle:
22
12.07.025.VVpp´=´ \V2 = 7.84 V1.
Energy:
V
g
pV
g
p
1
2
1 2
2
2
2 2
+=+
g g
.
2
2
11
7.841
9810.
29.81
pV
-
\=
´

For the contraction: p p´ =´. . .07 05
2
1
2
3
V V \V3 = 1.96 V1.
Energy:
V
g
pV
g
p
1
2
1 3
2
3
2 2
+=+
g g
.
Manometer: g g´+= ´+. .. .15 13615
1 3p p \= ´+
p p
1 3
12615
g g
.. .
Subtract the above 2 eqns:
V
g
V
g
V
g
1
2
3
2
21
2
2
12615
2
196
2
+´==.. . .
\ - = ´´(. ) .. .1961 126152
2
1
2
V g \V1 = 3.612 m/s. \p1 = 394 400 Pa.

63. 62
From the reservoir surface to section 1:

V
g
p
z
V
g
p
z
0
2
0
0
1
2
1
1
2 2
++=++
g g

H= +
3612
1962
2
.
.
394 400
9810
= 40.0 m.
b) Manometer: g g´+= ´+. .. .2 1362
1 3p p \= ´+
p p
1 3
1262
g g
.. .
Energy:
V
g
pV
g
p
1
2
1 3
2
3
2 2
+=+
g g
. Also, V3 = 1.96 V1.
\ +´=
V
g
V
g
1
2 2
1
2
2
1262
196
2
..
.
. \V1 = 4.171 m/s.
The nozzle is the same as in part (a): \p1 = 534 700 Pa.

From the reservoir surface to the nozzle exit:

V
g
p
z
V
g
p
z
0
2
0
0
2
2
2
2
2 2
++=++
g g
. \==
´
H
V
g
2
2 2
2
327
2981
.
.
= 54.5 m.

4.68 a) Energy:
V
g
p
z
V
g
p
z
0
2
0
0
2
2
2
2
2 2
++=++
g g
. \= =´ ´V gz
2 0
2 298124. .= 6.862 m/s.
Q = AV = .8 ´ 1 ´ 6.862 = 549. . m/s
3

For the second geometry the pressure on the surface is zero but it increases
with depth. The elevation of the surface is 0.8 m.
\=+z
V
g
h
0
2
2
2
. \= -=´ ´V gzh
2 02 29812( ) . = 6.264 m/s.
\Q = .8 ´ 6.264 = 501. . m/s
3

Note: z0 is measured from the channel bottom in the 2nd geometry.
\z0 = H + h.

b)
V
g
p
z
V
g
p
z
0
2
0
0
2
2
2
2
2 2
++=++
g g
. \= =´ ´+
æ
è
ç
ö
ø
÷=V gz
2 0
2 23226
2
2
2123. . fps.
\Q = AV = (2 ´ 1) ´ 21.23 = 42.5 cfs.
For the second geometry, the bottom is used as the datum:
\=++z
V
g
h
0
2
2
2
0. \ =+-
V
g
Hhh
2
2
2
( ).
\= =´ ´V gH
2
2 23226. = 19.66 fps. \Q = 39.3 cfs.

64. 63
4.69 From the reservoir surface to the exit: Continuity:

V
g
p
z
V
g
p
zK
V
g
0
2
0
0
2
2
2
2
1
2
2 2 2
++=+++
g g
. VV
1 2
2
2
03
08
=
.
.
= .1406 V2.
10
2
5
1406
2
2
2 2
2
2
=+´
V
g
V
g
.

\V2 = 13.36 m/s. \Q = 13.36 ´ p ´ .015
2
=000944. . m/s
3

The velocity in the pipe is V1 = 1.878 m/s.
Energy 0®A: 10
1878
29819810
8
1878
2981
3
2 2
=
´
+ +
´
+
.
.
.
.
.
.
p
A
\pA = 65 500 Pa.
Energy 0®B: 10
1878
29819810
20
1878
2981
10
2 2
=
´
+ +
´
+
.
.
.
.
.
.
p
B
\pB = -5290 Pa.
Energy 0®C: 10
1878
29819810
1228
1878
2981
2 2
=
´
+ ++
´
.
.
.
.
.
.
p
C
\pC = -26 300 Pa.
Energy 0®D: 10
1878
29819810
05
1878
2981
2 2
=
´
+ ++
´
.
.
.
.
.
p
D
\pD = 87 500 Pa.

4.70
V
g
p
z
V
g
p
z
0
2
0
0
2
2
2
2
2 2
++=++
g g
.
80 000
9810
+=
´
4
2981
2
2
V
.
. \V2 = 19.04 m/s.
a) QAV= =´ ´
22
2
0251904p. . = 00374. . m/s
3

b) QAV= =´ ´
22
2
091904p. . = 0485. . m/s
3

c) QAV= =´ ´
22
2
051904p. . = 01495. . m/s
3


4.71 a)
p
z
V
g
V
g
0
0
2
2
1
2
2
1
2g
+=+.54.
80 000
9810
+= +4
16
2
1
2
1
2
1
2
V
g
V
g
.54. \V1 = 3.687 m/s.
QAV= =´ ´
11
2
053687p. . = 00290. . m/s
3

b) AVAV
11 22= . V V V
1
2
22 2
09
05
324= =
.
.
..

80 000
9810
+=+4
2
23
324
2
2
2 2
2
2
V
g
V
g
.
.
. \V2 = 3.08 m/s. \QAV=
22
= 00784. . m/s
3

c)
80 000
9810
+=+4
2
1
2
2
2
2
2
V
g
V
g
.5. \V2 = 9.77 m/s. \QAV=
22
= 00767. . m/s
3


4.72 (C) Manometer:
2
2
12
2
V
Hpgp
g
gr+=+ or
2
2
1
98100.02.
2
V
pg
g
r´+=
Energy:
2
1000007.96
.3.15.
29.819810
KK =\=
´

65. 64
Combine the equations:
2
1
198100.021.2.18.1 m/s.
2
V
V´=´\=

4.73 Manometer: .6.13
21 pzHpzH ++=++ gggg \= +
p
H
p
1 2
126
g g
. .
Energy:
pV
g
pV
g
1 1
2
2 2
2
2 2g g
+=+.
Combine energy and manometer: 126
2
2
2
1
2
. .H
VV
g
=
-

Continuity: V
d
d
V
2
1
2
2
2
1
= . \= ´ -
æ
è
ç
ö
ø
÷V Hg
d
d
1
2 1
4
2
4
1262 1. / .
\= =
´
-
æ
è
ç
ö
ø
÷QV
d Hg
dd
d
1
1
2
1
4
2
4
12
1
2
44
1262
1
p
p.
/
/
= 1235
1
2
2
2
1
4
2
4
12
.
/
dd
H
dd-
æ
è
ç
ö
ø
÷

4.74 Use the result of Problem 4.73:
a) Q = 12.35 ´..
.
..
/
1608
2
1608
2 2
4 4
12
´
-
æ
è
ç
ö
ø
÷ = 00365. . m/s
3

b) Q = 12.35 ´..
.
..
/
2408
4
2408
2 2
4 4
12
´
-
æ
è
ç
ö
ø
÷ = 00503. . m/s
3

c) Using English units with g = 32.2: Q dd
H
dd
=
-
æ
è
ç
ö
ø
÷2237
1
2
2
2
1
4
2
4
12
. .
/

Q= ´
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
-
æ
è
ç
ö
ø
÷2237
1
2
1
4
1012
25
2 2
4 4
12
.
/
.5.
/
= 1.318 cfs.
d) Q = 22.37 ´1
1
3
1512
13333
2
2
4 4
12
´
æ
è
ç
ö
ø
÷
-
æ
è
ç
ö
ø
÷
/
.
/
= 2.796 cfs.

4.75 (B)
2
2
0.040
.7.96 m/s.
2 0.04
L
VpQ
hKV
gAg p
D
=====
´


2
1000007.96
.3.15.
29.819810
KK =\=
´


4.76 a) Energy from surface to outlet:
V
g
H
2
2
2
=. \ =V gH
2
2
2.
Energy from constriction to outlet:
pV
g
pV
g
1 1
2
2 2
2
2 2g g
+=+.

66. 65
Continuity: V V
1 24=. With p1 = pv = 2450 Pa and p2 = 100 000 Pa,

2450
9810
16
2981
2
100000
9810
1
2981
2+
´
´ = +
´
´
. .
.gH gH

\H = 0.663 m.

b) With p1 = 0.34 psia, p2 = 14.7 psia,

.
.
.
.
.
34144
624
16
2
2
147144
624
1
2
2
´
+ =
´
+
g
gH
g
gH \H = 2.21 ft.

4.77 Continuity: V V
1 24=. Energy ¾ surface to exit:
V
g
H
2
2
2
=.
Energy¾ constriction to exit:
pV
g
pV
g
v
g g
+=+
1
2
2 2
2
2 2
.
\=+
-
=- = -´´pp
V V
g
p H
v 2
2
2
2
2
2
16
2
15 10000015659810g g . = 4350 Pa.
From Table B.1, T = 30°C.

4.78 Energy ¾ surface to surface: zzh
L0 2=+. \=+30202
2
2
2
V
g
.
Continuity: V1 = 4V2. \V
1
2
= 160 g. \V
2
2
= 10 g.
Energy ¾ surface to constriction: 30
160
2
94000
9810
1
= +
-
+
g
g
z
( )

\z1 = -40.4 m. \H = 40.4 + 20 = 60.4 m.

4.79 Continuity: V V
2
2
2 1
10
6
= = 2.778 V1.
Energy:
V
g
pV
g
p
1
2
1 2
2
2
2 2
+= +
g g
.
V
g
V
g
1
2 2
1
2
2
200 2778
2
2450
9810
+ = +
000
9810
.
.

\V
1
= 7.67 m/s. \Q = p ´ .05
2
´ 7.67 = 00602. . m/s
3


4.80 Velocity at exit = V
e. Velocity in constriction = V
1. Velocity in pipe = V
2.
Energy — surface to exit:
V
g
H
e
2
2
=. \ =V gH
e
2
2.
Continuity across nozzle: V
D
d
V
e2
2
2
= . Also, V V
1 24=.
Energy — surface to constriction: H
V
g
p
v
=+
1
2
2g
.

67. 66
a) 5
1
2
16
2
25
97
4
4
= ´´´
æ
è
ç
ö
ø
÷+
-
g
D
g
.
.
550
9810
\=D0131. m

b) 15
1
2
16
812
215
34147144
624
4
4
= ´´
æ
è
ç
ö
ø
÷+
-
g
D
g
(/)
(. .)
.
. \= ¢ ¢¢D0446. or 5.35

4.81 Energy — surface to exit: 3
2
4
2
1177
2
2
2
2
2
2
=+ \ =
V
g
V
g
V. ..
Energy — surface to “A”: 3
1177
2981
1176100
31
1177
2981
2
=
´
+
-
+++
´
.
.
( ).5
.
.
.
000
9810
H
\=H8.57. m

4.82 & . .mAV= = ´´
æ
è
ç
ö
ø
÷´ =r p194
1
12
1205079
2
slug/sec.

&
. .
. .
/. , .W
P
= ´
-
´
+
´é
ë
ê
ù
û
ú
=5079322
30120
2322
120144
624
08512950
2 2
ft-lb
sec
or 23.5 Hp

4.83 & . .mAV= = ´´ ´=r p1000 02405027
2
kg/s.
20
40
29819810
082
2
000=50.279.81
10
2
´
-
´
+
é
ë
ê
ù
û
ú
.
/..
Dp
\= ´Dp108810
6
. . Pa

4.84 (C)
22
21
2
P
WVV
Qgg
-
=
&
.
p
g
D
+

16
0.04040016 kW.18.0 kW.
0.89
P
P
W
WQp
h
=D=´===
&
&


4.85 -=´ ´
-
´
+
-é
ë
ê
ù
û
ú
´
&
.
.
.
..W
T
21000981
0102
2981
600
087
2
000
9810
\ = ´
&
. .W
T130410
6
W
We used VQA
2 2 2
2
25
102= =
´
=/
.
.
p
m/s.

4.86 V V
1 2 2 2
450
3
159
450
375
1019=
´
= =
´
=
p p
..
.
. . fps fps
-
æ
è
ç
ö
ø
÷´=´´
-
´
+
-é
ë
ê
ù
û
ú10000
1
746
550450194322
1019159
2322
18140144
624
2 2
,
.
. .
. .
.
( )
.
.h
T

\=h
T
0924.

68. 67
4.87 a) &&& ( ).QWmg
VV
g
pp
zz
c
g
TT
S
v
-=
-
+-+-+ -
é
ë
ê
ù
û
ú
2
2
1
2
2
2
1
1
2 1 2 1
2 gg

The above is Eq. 4.5.17 with Eq. 4.5.18 and Eq. 1.7.13.
g g
1
1
1
2
2 2
85981
287293
992
600981
287
20
= =
´
´
= =
´
=
pg
RT T T
.
.
. .
.
.
. N/m

500
3

\-- ´
´
+ - + -
é
ë
ê
ù
û
ú
(
.
.5
.
( ).1
2981
600
20
85 716
981
293
2
2
500 000)=59.81
200 000
500
000
9.92
2
T
T
\=T
2
572 K or 299C .
o

Be careful of units! p c
v2
600 716= =
×
000 Pa,
J
Kkg
.5

b) -60 000 + 1 500 000 = same as above. \=T
2
560 K or 287C.
o


4.88 g g
1
1
1
2
147144322
1716520
00764
60144322
1716760
0213= =
´´
´
= =
´´
´
=
pg
RT
. .
. .
.
. .
lb
ft

lb
ft
3 3

c mgAVgAV
v
= = = = ´´
æ
è
ç
ö
ø
÷´ =4296 213
1
24
600697
2
ft-lb
slug-R
lb/sec.
o
.& . .r g p
Use Eq. 4.5.17 with Eqs. 4.5.18 and 1.7.13:

&&& ( ) .QWmg
VV
g
ppc
g
TTzz
c
v
+=
-
+-+ -+-
é
ë
ê
ù
û
ú
2
2
1
2
2
2
1
1
2 1 2 1
2 gg

-´´ +=
´
+
´
-
´
+ -
é
ë
ê
ù
û
ú
10778697 697
600
2322
60144
213
147144
0764
4296
322
30060
2
.
&
.
. .
.
. .
( )W
c

\=& .W
c
40 600
ft-lb
sec
or 73.8 Hp

4.89 Energy — surface to exit: - = -+
é
ë
ê
ù
û
ú
& & . .W mg
V
g
V
g
TTh
2
2
2
2
2
2045
2

V mgQ
2 2
15
6
1326 159810147=
´
= ==´ =
p
g
.
. & m/s. 150 N/s.
-´=
´
-+
´
é
ë
ê
ù
û
ú \=&.
.
.
.
.
. & .W W
T T08147
2981
2045
1326
2981
2
150
13.26
5390 kW
2


4.90 (D)
22
4.587.16
36.0153.2. 416 000 Pa
29.81981029.81
B
B
p
p+=++\=
´´

In the above energy equation we used

2
2
0.2
with 4.42 m/s.
2 0.2
L
VQ
hKV
gA p
====
´

69. 68

4.91 Energy — surface to “C”:

&
.&
.
.
.
..W mg
P
´+´=
´
+ +
´
é
ë
ê
ù
û
ú
8 10
10
2981
200
77
10
2981
7705
2 2
000
9810

(& . . .5
&
.mgAVg W
P= = ´´´´ = \ =r p1000 0510981770 52
2
N/s.) 700 W

Energy — surface to “A”: 30
10
29819810
1
10
2981
169
2 2
=
´
+ +
´
\=
.
.5
.
. .
p
p
A
A
300 Pa
Energy — surface to “B”:
22 2
22
BOBO B
PPBO
VVpp V
WmgzzK
gg
h
g
éù--
=++-+êú
êúëû
& &
52
29819810
3015
10
2981
706
2
700.8=770.5
10
100 Pa
2
´
´
+ -+
´
é
ë
ê
ù
û
ú\=
.
.
.
. .
p
p
B
B


4.92 Manometer: g g g g r´++= ´+++
20
12
136
20
12 2
1 1 2 2
2
2
zp zp
V
. .
\++= ´+++
20
12
136
20
12 2
1
1
2
2 2
2
z
p
z
pV
gg g
. .
Energy:
V
g
z
p
Hz
pV
g
T
1
2
1
1
2
2 2
2
2 2
++=+++
g g
.
\= ´+- =
´
æ
è
ç
ö
ø
÷
=
20
12
136
20
122
18
1
3
516
1
2
1 2
. . .
V
g
H V
T fps.
p

\ = ´+
´
= = = ´´´H W QH
T T TT
126
20
12
516
2322
623 624189623
2
.
.
.
.'.
&
. . . gh
=62 115, .980
ft-lb
sec
or Hp

4.93 Energy—across the nozzle:
22 2
1122
211
2
5
. 6.25.
22 2
pVpV
VVV
gggg
+=+==

222
11
1
6.25400 000
. 4.58 m/s
981029.8129.81
VV
V\+=\=
´´
, 7.16 m/s
AV= ,
228.6 m/s.V=

Energy—surface to exit:

222
28.64.587.16
151.53.2.
29.8129.8129.81
P
H+=++
´´´
36.8 m.
PH\=

2
/9810(.01)28.636.8/.853820 W.
PPPWQHghp\==´´´´=&

70. 69
Energy —surface to “A”:

22
7.167.16
153.2. 39 400 Pa
29.81981029.81
A
A
p
p=++\=
´´

Energy —surface to “B”:

22
4.587.16
36.0153.2. 416 000 Pa
29.81981029.81
B
B
p
p+=++\=
´´


4.94 (A) V
Q
A
==
´
=
01
04
1989
2
.
.
.
p
m/s.
Energy —surface to entrance: H
V
g
p
zK
V
g
P
=+++
2
2
2
2
2
2
2 2g
.
\ =
´
+ ++
´
=H
P
1989
2981
180
5056
1989
2981
2014
2 2
.
.
.
.
.
.
000
9810
m.
\ = = ´´ =
&
/ . ./. .W QH
P P Pg h9810012014075263 000 W

4.95 Energy —surface to exit: 10
2
22
2
2
2
2
2
2
2
=+++
V
g
p
z
V
gg
..
\= = = ´ \=V Q d d
2 2
2
2
783 002783 4 00570. . . /. . m/s. m.p

4.96 Depth on raised section = y
2. Continuity: 33
22´=Vy.
Energy (see Eq. 4.5.21):
3
2
3
2
04
2
2
2
2
g
V
g
y+=+ +(. ).
\ = + - + =3059
9
2
3059 41280
2
2
2 2 2
3
2
2
. , . .
gy
y y y

or
Trial-and-error:
2
2
2
2.0: .11 ? 0.
1.85 m.
1.8: .05 ? 0.
y
y
y
=- ü
\=ý
=+
þ


2
2
2
2.1: .1 ? 0.
2.22 m.
2.3: .1 ? 0.
y
y
y
=- ü
\=ý
=+
þ

The depth that actually occurs depends on the downstream conditions.
We cannot select a “correct” answer between the two.




4.97 Mass flux occurs as shown. The velocity
of all fluid elements leaving the top and
bottom is approximately 32 m/s. The
distance where
u y= =±32 m/s is 2 m.

m
1
m
3
m
2
m
3
.
.
..

71. 70
To find &m
3
use continuity:
&& &. ( ) &.mm m ydym
1 2 3
2
3
0
2
2 41032228 102=+ ´´= + +ò
r r
\ = - ´+
é
ë
ê
ù
û
ú
=& ..m
364010282
8
3
533r r r
Rate of K.E. loss = & &m
V
m
V u
dy
1
1
2
3
1
2 3
0
2
2
2
2
2
2
10- -ò
r
= - - +ò
1280
32
2
533321028
2
2 23
0
2
r r r. ( )ydy
[65536054579507320]115000 . Wr=--=

4.98 The average velocity at section 2 is also 8 m/s. The kinetic-energy-
correction factor for a parabola is 2 (see Example 4.9). The energy
equation is:

V
g
p V
g
p
h
L
1
2
1
2
2
2
2
2 2
+= ++
g
a
g
.

8
2981
150
2
8
2981
110
2 2
´
+ =
´
+ +
. .
.
000
9810
000
9810
h
L


\=h
L0815. . m

4.99 V
A
VdA ydy= = + = ´+
é
ë
ê
ù
û
ú=òò
1 1
2
28
1
2
282
2
3
2933
2
3
0
2
( ) . m/s
a= =
´
+ò ò
1 1
22933
28
3
3
3
23
0
2
AV
VdA ydy
.
( )
[ ]=
´
´+´ ´ +´´ + =
1
22933
2823282332825271005
3
3 2 3 5 7
.
/ / / .

4.100 a)
224
0.01
2222
0
11200.010.01
10125 m/s
20.010.010.0140.01
r
VVdArdr
A
p
p
æöéù
==-=-= ç÷ êúòò
ç÷
´´ êúèøëû

a
p
p= =
´ ´
-
æ
è
ç
ö
ø
÷ò ò
1 1
0015
101
001
2
3
3
2 3
3
2
2
3
0
001
AV
VdA
r
rdr
. .
.

=
´
-
´
´
+
´
´
-
´
æ
è
ç
ö
ø
÷=
2000
0015
001
2
3001
4001
3001
6001
001
8001
200
2 3
2 4
2
6
4
8
6
.
. .
.
.
.
.
.
.

72. 71
b)
23
0.02
22
0
11100.02
1010.026.67 m/s
0.020.02 0.0230.02
y
VVdAwdy
Aw
æöæö
==-=-= ç÷ç÷òò
ç÷ç÷
´
èøèø


3
20.02
33
332
0
11
101
0.026.670.02
y
VdAwdy
AVw
a
æö
==- ç÷òò
ç÷
´´ èø

=
´
-
´
´
+
´
´
-
´
æ
è
ç
ö
ø
÷=
1000
002667
002
3002
3002
3002
5002
002
7002
1
3
3
2
5
4
7
6
. .
.
.
.
.
.
.
.
.541

4.101 V
A
VdA
R
u
r
R
rdru
n
n
n
n
nR
= = -
æ
è
ç
ö
ø
÷ =-
+
-
+
æ
è
ç
ö
ø
÷òò
1 1
1 2 2
21 1
2
1
0
p
p
max
/
max

KE V
V
dA u
r
R
rdru R
n
n
n
n
nR
..
max
/
max=
æ
è
ç
ö
ø
÷= -
æ
è
ç
ö
ø
÷ = -
+
-
+
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
úòò
r
r
p rp
2
3
3
3 2
0
2 2
1 2
323


a) V u u=- -
æ
è
ç
ö
ø
÷=2
5
11
5
6
0758
max max.
KE Ru Ru.. .
max max= -
æ
è
ç
ö
ø
÷=rp rp
23 235
8
5
13
024
a
r
rp
rp
= =
´
=
KE
AV
Ru
R u
.. .
.
.
max
max
1
2
024
1
2
0758
1102
3
23
2 33


b) V u u=- -
æ
è
ç
ö
ø
÷=2
7
15
7
8
0817
max max.
KE uR Ru.. .
max max= -
æ
è
ç
ö
ø
÷=rp rp
3 2 237
10
7
17
0288
a
r
rp
rp
=
æ
è
ç
ö
ø
÷
=
´
æ
è
ç
ö
ø
÷
=
KE
AV
V
Ru
R u
u
.. .
.
.
.
max
max
max
2
23
2
22
2
0288
0817
0817
2
1056

c) V u u=- -
æ
è
ç
ö
ø
÷=2
9
19
9
10
0853
max max.
KE Ru Ru.. .
max max= -
æ
è
ç
ö
ø
÷=rp rp
23 239
12
9
21
0321
a
r
rp
rp
= =
´
=
KE
AV
Ru
R u
.. .
.
.
max
max
1
2
0321
1
2
0853
1034
3
23
2 33

73. 72
4.102 Engine power = FVm
VV
uu
D´+
-
+-
æ
è
ç
ö
ø
÷
¥
&
~~2
2
1
2
2 1
2

& & ( )mgFVm
VV
cTT
ff D v= +
-
+ -
é
ë
ê
ù
û
ú¥
2
2
1
2
2 1
2


4.103
&
W FV
Dh=´

10
5
930
100
3600
015
1340
1000
100
3-
æ
è
ç
ö
ø
÷´
æ
è
ç
ö
ø
÷´
æ
è
ç
ö
ø
÷´
æ
è
ç
ö
ø
÷´ = ´
m
km
kg
m
kJ
kg
km
s
000
3600
3
3
q
f
.
\=q
f
48 030 kJ/kg

4.104 0
2 2
32
2
2
2
2
2
1
2
1
1 2
= ++---+a
g g
nV
g
p
z
V
g
p
z
LV
gD

02
2981
03532
10180
981002
2 6
2
=
´
- +´
´
´
-
V V
.
.
. .
.
V V V Q
2 5
14434340 0235 73710+ - = \= = ´
-
. . . . . m/s and m/s
3


4.105 Energy from surface to surface: H
V
g
p
z
V
g
p
zK
V
g
P
=++---+
2
2
2
2
1
2
1
1
2
2 2 2g g
.
a) H
Q
Q
P
=+
´ ´´
=+405
0042981
40507
2
2
2
p. .
.
Try Q H H
P P= = =025 432 58.: . (energy). (curve)
Try Q H H
P P= = =030 446 48.: . (energy). (curve)
Solution: Q=032. . m/s
3


b) H
Q
Q
P
=+
´ ´´
=+40
20
0042981
40203
2
2
2

p. .

Try Q H H
P P= = =025 527 58.: . (energy). (curve)
Solution: Q=027. m/s
3

Note: The curve does not allow for significant accuracy.


4.106 Continuity: AVAVAV
11 22 33
= +
p p p´ ´=´ ´+´ \=0065 00220 003 1111
2 2 2
3 3
. . . . .V V m/s
Energy: energy in + pump energy = energy out
& & & &m
Vp
W m
Vp
m
Vp
P P1
1
2
1
2
2
2
2
3
3
2
3
2 2 2
+
æ
è
ç
ö
ø
÷+´= +
æ
è
ç
ö
ø
÷+ +
æ
è
ç
ö
ø
÷
r
h
r r

74. 73
10000065
5
2
120
085 100000220
20
2
300
2
2
2
2
p p´ ´ +
æ
è
ç
ö
ø
÷+ = ´ ´ +
æ
è
ç
ö
ø
÷. .
&
.
000
1000
000
1000
W
P

+ ´ ´ +
æ
è
ç
ö
ø
÷10000031111
1111
2
500
2
2
p. .
. 000
1000

\ =
&
W
P26 700 W

4.107 (A) After the pressure is found, that pressure is multiplied by the area of the
window. The pressure is relatively constant over the area.

4.108
V
g
pV
g
p
1
2
1 2
2
2
2 2
+= +
g g
.
2
211 2
4 .
(/2)
d
VVV
d
==
a)
V V
1
2
1
2
2981
200 16
2981´
+ =
´. .
.
000
9810

\=V
15164. m/s.
( )pAFmVV
11 2 1-= -& .
20000003 1000035164451645164
2 2
p p´-= ´´ ´ -. . .(. .).F \=F339 N.

b)
V V
1
2
1
2
2981
400 16
2981´
+ =
´. .
.
000
9810

\=V
17303. m/s.
40000003 1000037303473037303
2 2
p p´-= ´´ ´ -. . .(. .).F \=F679 N.
c)
22
11 16 200 000
.
29.81981029.81
VV
+=
´´

15.164 m/s.V\=

22
200 000.061000.065.164(45.1645.164). Fpp´-=´´´- 1356 N.F\=
d)
22
11
1
16 30144
.17.24 fps.
232.262.4232.2
VV
V
´
+=\=
´´


222
301.51.94(1.5/12)17.24(41).127 lb. FFpp´´-=´´´-\=
e)
22
11
1
16 60144
.24.38 fps.
232.262.4232.2
VV
V
´
+=\=
´´


222
601.51.94(1.5/12)24.38(41).254 lb. FFpp´´-=´´´-\=
f)
22
11
1
16 30144
.17.24 fps.
232.262.4232.2
VV
V
´
+=\=
´´


222
3031.94(3/12)17.24(41).509 lb.FFpp´´-=´´´-\=

4.109
V
g
pV
g
p
1
2
1 2
2
2
2 2
+= +
g g
. V V V
2
2
21 1
9
3
9= =.

V
1
2
2981
2 81
2981´
+ =
´. .
.
000 000
9810
V
1
2
\=V
1
2
50.

75. 74
pAFmVVmV
11 2 1 18-= -=&( )&
2 045 1000045850
2 2
000 000p p´ -= ´ ´´. .F
\=F10 180 N.

4.110
22
1122
.
22
VpVp
gggg
+=+
0
2
.01.006.15. 11.1 m/s.
ee
VVVp´=´´\=
SFmVV
x x x= -&( ).
2 1

a) V V V
2
2
2 1 1
10
8
1= =.562.
V V
1
2
1
2
2981
400 2441
2981´
+ =
´.
.
.
.
000
9810

\=V
123.56 . m/s
\ -= -pAFmVV
11 2 1
&( ).
400 05 10000523 23
2 2
000p p´-= ´´ ´. . .56(.562.56).F \=F692 N.
b) V V V
2
2
2 1 1
10
6
2778= =. .
V
g
V
g
1
2
1
2
2
400 7716
2
+ =
000
9810
.
. \=V
11091. . m/s
400 05 100005109117781091
2 2
000p p´-= ´´ ´. . .(. .).F \=F1479 N.
c) V V V
2
2
2 1 1
10
4
625= =. .
V
g
V
g
1
2
1
2
2
400 3906
2
+ =
000
9810
.
. \=V
14.585 . m/s
400 05 1000054 5254
2 2
000p p´-= ´´ ´. . .585(. .585).F \=F2275 N.
d) V V V
2
2
2 1 1
10
2
25= = .
V
g
V
g
1
2
1
2
2
400 625
2
+ =
000
9810

. \=V
11132. . m/s
400 05 1000051132241132
2 2
000p p´-= ´´ ´. . .( .).F \=F2900 N.

4.111 (C)
22
1122
22
VpVp
gggg
+=+
22
1
(6.251)12.73
.98103085000 Pa.
29.81
p
-´
=´=
´


2
1121 ().30850000.0510000.112.73(6.251)pAFQVVF rp-=-´´-=´´-
17500 N.F\=

4.112 V V p
VV
g
2 1 1
2
2
1
2 2 2
4 120
2
624
12030
2322
13= = =
-é
ë
ê
ù
û
ú=
-
´
é
ë
ê
ù
û
ú= fps. 080 psf.g .
.
,

22
1121
1.51.5
()13,080 1.9430(12030)1072 lb.
1212
xxFpAmVV pp
æöæö
=--=-´´--=
ç÷ç÷
èøèø
&

4.113 V V
V
g
pV
g
p V
g
p
2 1
1
2
1 2
2
2 1
2
1
4
2 2
15
2
= +=+ \ =

. . .
g g g

a) V
1
22981
159810
2000002667=
´
´
´ =
.
.. \= =V V
1 2516 207. . m/s, m/s.
pAFmVV
x x x11 2 1-= -&( ). \= ´ + ´ ´ =F
x
200 000041000045161139
2 2 2
p p. . . . N

76. 75
FmVV
y y y
= -&( ).
2 1
\= ´´ =F
y100004516207537
2
p. .(.) . N

b) V
1
22981
159810
4000005333=
´
´
´ =
.
.. \= =V V
1 2730 292. . m/s, m/s.
pAFmVV
x x x11 2 1-= -&( ). \= ´ + ´ ´ =F
x
400 00004100004732280
2 2 2
p p. . . . N
FmVV
y y y
= -&( )
2 1
=100004732921071
2
p´ ´´ =. .(.) . N

c) V
1
22981
159810
8000001067=
´
´
´ =
.
.. \= =V V
1 21033 413. . m/s, m/s.

2222
1111
800 000.041000.0410.334560 N.
x
FpAAV rpp=+=´+´´=
FmV
y y
=&()
2
=10000410334132140
2
p´ ´ =. .(.) . N

4.114 V V
2
2
21
40
10
80= = m/s.
V
g
pV
g
p
1
2
1 2
2
2
2 2
+=+
g g

\=
´
-
´
é
ë
ê
ù
û
ú
= ´p
1
2 2
6
9810
80
2981
5
2981
31910
. .
. Pa.

pAFmVV
x x11 2 1-= -&( ). \= ´ ´- ´´ -=F319102100025805353
6 2 2
. . . ( )p p 000 N.

4.115 AVAV
11 22= . p p´ ´= -. (. .).0254 02502
2 2 2
2
V
\=V
21111. m/s.
pV
g
pV
g
1 1
2
2 2
2
2 2g g
+=+.

22
1
11.114
981053700 Pa.
29.81
p
æö -
==ç÷
ç÷
´
èø


1121 (). pAFmVV-=-&

22
53 700.0251000.0254(11.114)49.6 N.F pp\=´-´´-=



4.116 Continuity: . .. .7 1 7
1 2 2 1 V V V V= \=
Energy:
V
g
p
z
V
g
p
z
1
2
1
1
2
2
2
2
2 2
++=++
g g




V V
V V
1
2
1
2
1 2
2981
7
49
2981
1 0495 3467
´
+=
´
+ \= =
.
.
.
.. ., . m/s.
Momentum: FFRmVV
x1 2 2 1--= -&( )
98103571981005011 10001134673467495´ ´- ´ ´-= ´´´ -.(..5) .(..5) (..5).(..)R
x

\=R
x1986 N.
\R
x
acts to the left on the water, and to the right on the obstruction.
F
V
2
p
1
A
1
F
p
1
A
1
F
2
F
1
R
x

77. 76

4.117 Continuity: 6 2 30
1 2 2 1 V V V V= \=.. .
Energy (along bottom streamline):

V
g
p
z
V
g
p
z
1
2
1
1
2
2
2
2
2 2
++=++
g g


22
22/900
60.2.
29.8129.81
VV
+=+
´´


21 10.67, .36 m/s.VV\==

Momentum: FFFmVV
1 2 2 1--= -&( )
98103649810124 1000241067106736´´- ´´-= ´´´ -() .(.) (.).(..)F
\=F618 000 N. (F acts to the right on the gate.)

4.118 a) 86
22´=. .Vy FFmVV
1 2 2 1-= -&( ).
g g r´´- = ´
´
-
æ
è
ç
ö
ø
÷.. .
.
.36
2
68
86
8
2
2
2
w
y
yw w
y


g
r
2
36 488
6
6
4882
981
2
2 2
2
22
(. ).
.
. (. )
.
.
.-= ´
-
\+ =
´´
y
y
y
yy
yy y
2
2
2 2
6 78290 2+ - = \=. . . .51. m (See Example 4.12.)
b) y y y
g
yV
2 1 1
2
11
2 2 21
2
8 1
2
44
8
981
412 323=-+ +
é
ë
ê
ê
ù
û
ú
ú
=-+ + ´´
é
ë
ê
ù
û
ú=..
.
. .. m
c) y y y
g
yV
2 1 1
2
11
2 2 21
2
8 1
2
22
8
322
220 612=-+ +
é
ë
ê
ê
ù
û
ú
ú
=-+ + ´´
é
ë
ê
ù
û
ú=
.
.. ft
d) y y y
g
yV
2 1 1
2
11
2 2 21
2
8 1
2
33
8
322
330 11=-+ +
é
ë
ê
ê
ù
û
ú
ú
=-+ + ´´
é
ë
ê
ù
û
ú=
.
.54. ft

4.119 Continuity: VyVyVy y y
22 11 21 2 14 4= = \=. .
Use the result of Example 4.12: y yy
g
yV
2 1 1
2
11
2
12
1
2
8
=-++
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
/

a) y
24832=´=... m

1/2
22
1
18
3.2.8.8.8.
29.81
V
éù
æö
=-++´´êúç÷
èøêúëû
\=V
1886. m/s.
b) y
2428=´= ft.
8
1
2
22
8
322
2
2
1
2
12
=-++ ´´
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
.
.
/
V \=V
1254. fps.

F
2
F
1
F

78. 77
4.120 V=
´
=
9
33
1 m/s.
1
2981
3
2981
2
1
2
1
´
+=
´
+
. .
.
V
y Vy
1113=´.
\ = +305
1962
3
1
2
1
.
.
.
V
V
Trial-and-error:
V
V
1
1
7305293
72305306
=
=
=
=
ü
ý
ï
ï
þ
ï
ï
:.
?
.
.:.
?
.


V
1719
417
=
=
.
.
m/s.
y m.
1


1/2
22
2
18
.417.417.4177.191.90 m.
29.81
y
éù
æö
=-++´´=êú ç÷
èøêúëû

V
219719417´=´.... V
21=.58 . m/s

4.121 Refer to Example 4.12: g g r
y
yw w w
y
Vy
1
1
1
11
2
36 61010
60
610-´´=´´ -
æ
è
ç
ö
ø
÷ =×.( ).
\ -=
-æ
è
ç
ö
ø
÷
g
r
2
36600
6
1
2 1
1
( ) .y
y
y
\ + = = \= =( )
.
.. ., ..y y y V
1 1 1 1
6
1200
322
3727 38 158 ft fps

4.122 Continuity: 20 015 03
2
2
2
´´ = ´p p. ..V
\=V
25 m/s.
Momentum: pApAmVV
11 22 2 1- = -&( ).


60 03 03100001520520
2
2
2 2
000p p p´-´ = ´ ´ -. . . ( ).p \=p
2135 kPa.

4.123
2
11222
2
.05
2. 1530 m/s.
2.025
VAVAV
p
p
´
===
´


pV
g
pV
g
p
1 1
2
2 2
2
1
2 2
2 2
9810
3015
2981
337
g g
+=+ \=
-
´
=.
.
500 Pa.
( )21111 . ().
xxxFmVVpAFmVS=--=-&&
\= + = ´+ ´´ =FpAmV
11 1
2 2 2
337 05100005154420& . . . 500 Np p

4.124 & . .m
1
2
100003123393= ´ ´=p kg/s.
& . .m
3
2
10000281005= ´ ´=p kg/s.
\ =-= = ´ \=&&& . . . .mmm V V
2 1 3
2
2 2
2388100003 8446p m/s.

Energy from 1 ® 2:
V
g
pV
g
p
p
1
2
1 2
2
2
2
2
2 2
500
8446
2981
9810+= + \=
-
´
´
g g
.
.
.
000+
12
2

= 536 300 Pa.
p
1
A
1
p
2
A
2
p
1
A
1 p
2
A
2
p
3
A
3
R
yRx

79. 78
Energy from 1 ® 3:
V
g
pV
g
p
1
2
1 3
2
3
2 2
+=+
g g
.
\= +
-
´
=p
3
2 2
500
128
2981
9810540 000 000 Pa.
.

pApARmVmVmV
x x x x11 22 22 33 11- -= + -& & &.
\= ´ - ´ + ´- ´ =R
x
500 03536 0333931223888446103
2 2
000 300 Np p. . . . . .
pARmVmVmV
y y y y33 33 22 11
-= + -& & &.
\= ´- ´-=R
y540 10058759 000.02 N
2
p .() .

4.125 a) ( )SFmVV FmV V
m
A
x x x
= - -=- =& . &.
&
2 1 1 1
1

r

=
´
300
100005
2
p.
= 38.2 m/s
\= ´ =F30038211. . 460 N

b) -= - -FmVV
r B
&( )(cos).
1 1a
\= ´ -=F300
282
382
382106250
.
.
(. ) . N


c)-= - -FmVV
r B
&( )(cos).
1 1a
\= ´ --=F300
482
382
3821018
.
.
(.()) . 250 N
4.126 a) -= -FmVV
x x
&( ).
2 1
200194
125
12
2
1
2
=
æ
è
ç
ö
ø
÷´.
.
.p V \=V
155 fps.
b)-= - -FmVV
r B
&( )(cos).
1 1a 200194
125
12
30
2
1
2
=
æ
è
ç
ö
ø
÷ -.
.
( ).p V \=V
185 fps.
c)-= - -FmVV
r B
&( )(cos).
1 1a 200194
125
12
30
2
1
2
=
æ
è
ç
ö
ø
÷ +.
.
( ).p V \=V
125 fps.

4.127 a) -= -FmVV
x x
&( ).
2 1
-= ´ ´ -700100004 30
2
11 1
p. (cos ).VV V
o
\=V
13224. m/s.
\= = ´ ´ =& . . . .mAVr p
11
2
10000432241621 kg/s
b)-= - -FmVV
r B
&( )(cos).
1 1a -= ´ - -700100004 88661
2
1
2
p.( )(. ).V \=V
14024. m/s.
\= = ´ ´ =& . . .mAVr p
11
2
1000044024202 kg/s
c)-= - -FmVV
r B
&( )(cos).
1 1a -= ´ + -700100004 88661
2
1
2
p.( )(. ).V \=V
12424. m/s.
\= = ´ ´ =& . . . .mAVr p
11
2
10000424241218 kg/s

4.128 (D)
21
()10000.010.250(50cos6050)2500 N.
xxx
FmVV-=-=´´´-=-
o
&
F
V
1
V
2

80. 79
4.129 a)-= - = ´
æ
è
ç
ö
ø
÷´ -RmVV
x x x
&( ). (cos ).
2 1
2
194
1
12
12012060120p
o
\=R
x305 lb.
RmVV
y y y
= - =
æ
è
ç
ö
ø
÷´´ ´&( ). (.).
2 1
2
194
1
12
120120866p \=R
y
528 lb.
b) -= - -= ´
æ
è
ç
ö
ø
÷´´ -RmVV
x r B
&( )(cos). (.5).
1
2
1194
1
12
6060 1a p \=R
x762.. lb
RmVV
y r B
= - =
æ
è
ç
ö
ø
÷´´´&( )sin . (.).
1
2
194
1
12
6060866a p \=R
y
132 lb.
c) -= - -= ´
æ
è
ç
ö
ø
÷´´ -RmVV
x r B
&( )(cos). (.5).
1
2
1194
1
12
180180 1a p \=R
x686 lb.
RmVV
y r B
= - =
æ
è
ç
ö
ø
÷´´ ´&( )sin . (.).
1
2
194
1
12
180180866a p \=R
y
1188 lb.
4.130 VR
B= =´=w03015.5 m/s.
-= - -= ´ ´´ -RmVV
x B
&( )(cos) . (.5).
1
2
1100002540251a p \=R
x982 N.
\= =´´=
&
.W RV
xB10 1098215147 300 W
4.131 a)-= - = ´ - -RmVV
x x x
&( ) . ( cos ).
2 1
2
40240040060400p
o
\=R
x1206 N.
RmVV
y y y= - =´´&( ) . (sin).
2 1
2
40240040060p
o
\=R
y
696 N.
b)-= - -= ´ --RmVV
x r B
&( )(cos ) . (.).
1
2 2
120140230051
o
p \=R
x679 N.
RmVV
y r B= - =´´ ´&( )sin . ..
1
2 2
402300866ap \=R
y
392 N.
c) -= - -= ´ --RmVV
x r B
&( )(cos ) . (.).
1
2 2
120140250051
o
p \=R
x1885 N.
RmVV
y r B= - =´´ ´&( )sin . ..
1
2 2
402500866ap \=R
y
1088 N.

4.132 -= - -=´´ - --FmVV
x B
&( )(cos ) . ( )(.5).
1
2 2
1201402400180 1
o
p \=R
x365 N.
V
B=´=12150180. m/s.
&
.W=´´ =15365180986 000 W
The y-component force does no work.

4.133 (A)
2
21
()10000.0260(40cos4540)884 N.
xrr xx
FmVV p-=-=´´´´-=
o
&
Power8842017700 W.
xB
FV=´=´=

4.134 a) Refer to Fig. 4.16:
111
2
11
750sinsin45507 fps.

750cos300cos45
rr
r
r
VV
VV
b
b
ü== ï
\ý
=-= ïþ
o
o

Note: VV V VV V V
x x r B r B r2 1 2 2 1 1 1 2 1-=- +- -=- +cos cos (cos cos).a a a a
\= + =
æ
è
ç
ö
ø
÷´´ + =RmV
x r
&(cos cos).
.5
(cos cos).
1 2 1
2
015
12
75050730 45489a a p
o o
lb.

81. 80
\= =´ ´ =
&
. ,W RV
xB
15 15489300220000
ft-lb
sec
or 400 Hp.
b)
750 60
750 300 60
554
1 1
1 1
1 2
sin sin
cos cos
.
b
b
=
- =
ü
ý
þ
=
V
V
V V
r
r
r r
o
o
fps =
\= + = ´
æ
è
ç
ö
ø
÷´´ + =RmV
x r
&(cos cos).
.5
(cos cos).
1 2 1
2
015
12
75055430 60464a a p
o o
lb.
\= =´ ´ =
&
. ,W RV
xB
15 15464300209000
ft-lb
sec
or 380 Hp.
c)
750 90
750 300 90
687
1 1
1 1
1 2
sin sin
cos cos
.
b
b
=
- =
ü
ý
þ
=
V
V
V V
r
r
r r
o
o
fps =
\= + = ´
æ
è
ç
ö
ø
÷´´ +=RmV
x r
&(cos cos).
.5
(cos ).5
1 2 1
2
015
12
75068730036a a p
o
lb.
\= =´ ´ =
&
. ,W RV
xB
15 15365300164300
ft-lb
sec
or 299 Hp.

4.135 a) Refer to Fig. 4.16:
10030
1003020
369 833
1 1
1 1
1 1
sin sin
cos cos
., .
o
o
o=
-=
ü
ý
þ
\= =
V
V
V
r
r
r
a
a
a m/s.

22
22
22
sin6083.3sin
71.5, 48.
cos6083.3cos20
V
V
V
a
a
a
ü= ï
==ý
=- ïþ
o
o
o

-= - = ´ ´ - - \=RmVV R
x x x x
&( ) . (.5cos cos).
2 1
2
100001510071 6010030 8650p
o o
N.
\= =´´ = ´
&
. .W VR
Bx12 1220865020810
6
W

b)
10030
1003040
47 6835
1 1
1 1
1 1 2
sin sin
cos cos
, .
o
o
o=
-=
ü
ý
þ
\= ==
V
V
VV
r
r
r r
a
a
a m/s.

22
22
22
sin6068.35sin
38.9 m/s, 29.5.
cos6068.35cos40
V
V
V
a
a
a
ü= ï
==ý
=- ïþ
o
o
o

-= - = ´ ´ - - \=RmVV R
x x x x
&( ) . (.cos cos).
2 1
2
10000151003896010030 7500p
o o
N.
\= =´´ = ´
&
. .W VR
Bx12 1240750036010
6
W

c)
10030
1003050
538 6196
1 1
1 1
1 1 2
sin sin
cos cos
., .
o
o
o=
-=
ü
ý
þ
\= ==
V
V
VV
r
r
r r
a
a
a m/s.

22
22
22
sin6061.76sin
19.32 m/s, 15.66.
cos6061.96cos50
V
V
V
a
a
a
ü= ï
==ý
=- ïþ
o
o
o

82. 81
-= - = ´ ´ - - \=RmVV R
x x x x
&( ) . (.cos cos).
2 1
2
100001510019326010030 6800p
o o
N.
\= =´ ´= ´
&
. .W RV
xB12 1268005040810
6
W

4.136 a) Refer to Fig. 4.16:
5030
5030
2500866
1 1
1 1
1
2 2sin sin
cos cos
.
o
o
=
-=
ü
ý
þ
\= - +
V
VV
V VV
r
B r
r B B
a
a


22 222
21
22
30sin60sin
90030.
30cos60cos
r
rrBB
rB
V
VVVV
VV
a
a
ü= ï
\==++ý
-= ïþ
o
o

Combine the above: V
B=1372. m/s. Then, a a
1 2
594 421= =., ..
o o

-= - = ´´- - \=RmVV R
x x x x
&( ) . (cos cos).
2 1
2
1000015030605030 916p
o o
N.
\= =´ ´=
&
. .W VR
Bx15 151372916188 500 W

b)
5030
5030
2500866
1 1
1 1
1
2 2sin sin
cos cos
.
o
o
=
-=
ü
ý
þ
\= - +
V
VV
V VV
r
B r
r B B
a
a
\=V
B1494. m/s.

22 22
2
22
30sin70sin
90020.52.
30cos70cos
r
rBB
rB
V
VVV
VV
a
a
ü= ï
\=++ý
-= ïþ
o
o
a
1
414=.
o
,a
2
482=.
o


2
21
()1000.0150(30cos7050cos30). 841 N.
xxxx
RmVVR p-=-=´´--\=
oo
&
\= =´ ´=
&
. .W VR
Bx15 151494841188 500 W

c)
5030
5030
2500866
1 1
1 1
1
2 2sin sin
cos cos
.
o
o
=
-=
ü
ý
þ
\ = - +
V
VV
V VV
r
B r
r B B
a
a
\=V
B1649. m/s

22 22
2
22
30sin80sin
90010.42.
30cos80cos
r
rBB
rB
V
VVV
VV
a
a
ü= ï
\=++ý
-= ïþ
o
o
a
1
43=
o
, a
2
537=.
o

-= - = ´´- - \=RmVV R
x x x x
&( ) . (cos cos).
2 1
2
1000015030805030 762p
o o
N.
\= =´ ´=
&
. .W VR
Bx15 151649762188 500 W

4.137 To find F, sum forces normal to the plate: ()nout
n
FmVS=&
1
.
n
V
éù
-
ëû

a) [ ]\= ´´´-- =F1000024404060 11.. (sin) .
o
080 N (We have neglected friction)

22331
0()40sin30.
t
FmVmVmS==+--´
o
&&& Bernoulli: VVV
1 2 3==.

\ =--
=+
ü
ý
þ
\= =´=
=

Continuity:

kg/s.
kg/s.
0 75 75320240
80
2 3 1
1 2 3
2 1
3
&&.5&
&&&
&.&.
&
mm m
mmm
m m
m

b)
120
1.94120(120sin60)3360 lb.
1212
F\=-´´´-=
o
(We have neglected friction)
SF mVmVm
t
== +--´0 12030
22 3 3 1
& &()& sin.
o
Bernoulli: VVV
1 2 3==.

83. 82

21231
123
3
20
.75.751.94120 00.5
144
Continuity:
22.6 slug/sec. and 9.7 slug/sec.
mmmmm
mmm
m
\==´´´\=-- ü
ý
=+
þ
==
&&&&&
&&&
&


4.138 FmV
r rn
= = ´´´+ =&() ..( )sin
1
2
10000244020 6024
o
940 N.
F W
x= = \= ´=24 3021 21 20432 940 600 N. 600 000 Wcos
&
.
o


4.139 FmV V F V
r rn B x B
= = ´´ - = -&() ..( )sin. ( )sin.
1
2 2 2
100002440 60 840 60
o o


&
( ). ( ).WVFV V V VV
Bx B B B B B= = - ´= - [phone]
2 2 3


dW
dV
V V V
B
B B B
&
( ). .= - + = \=61600160 3 0 1333
2
m/s.

4.140 (A) Let the vehicle move to the right. The scoop then diverts the water to the
right. Then
21
()10000.05260[60(60)]720000 N.
xx
FmVV=-=´´´´--=&

4.141
1(
rFmV=&
2
)(cos1)1000.1.6()(2)120.
BBBBVVVV a--=´´--=
At
2
1201000
0: 120133 300 N.
3600
tF
´æö
==´=
ç÷
èø


2133 300
1.33 m/s
100 000
o
a==


-
= =
-
\- =òò
F
m
dV
dt
V dV
V
dt
B B B
B
t
120
100
0012
2
2
03333
1667
000
. . .
.
.

\ -
é
ë
ê
ù
û
ú
= \=
1
1667
1
3333
0012 266
. .
. .. t. sect

4.142
1()(cos1)90.82.513.89(13.89)(1)34700 N.
rBFmVV a=--=´´´´--=&

501000
13.89 m/s 3470013.89482 000 W or 647 Hp.
3600
BVW
´æö
==\=´=
ç÷
èø
&


4.143 See the figure in Problem 4.141.
FmVV VV
r B B B
= - -= ´´´--=&( )(cos) .. ()()
1
11000062 224a V.
B
2

-= \- =FmV
dV
dx
V V
dV
dx
B
B
B B
B
. . 24 5000
2

- = - = - \=òò
24
5000
24
5000
2778250 458
250
2778
0

m
dx dV
V
xn n x
B
B
x
. . . .
.
l l
V
2
V
1
= 0
F

84. 83


4.144 -= - -= ´
æ
è
ç
ö
ø
÷- -FmVV VV
r B B
&( )(cos).
.
)().
1 1
2
1194
125
12
2a p
2
(
\= - =F VV
dV
dt
B
B
01323 20
1
2
.( ) .
At t V
dV
dt
V
B
B
= = =0 0 01323
1
2
, . . . Then 20
With
dV
dt
V
B
= =6 301
1
, .. fps
For t
dV
V
dt
V
V
B
B B
B
V
B
>
-
= =
-
- \=òò
0
301
0006615 001323
1
301
1
301
857
2
0
2
0
,
(. )
. ..
. .
. . . fps

4.145 For this steady-state flow, we fix the boat and move the upstream air. This
provides us with the steady-state flow of Fig. 4.17. This is the same as
observing the flow while standing on the boat.

&
. . .( .WFV F F V=
´
\= =
1 1
501000
3600
1440 1389 20 000= N m/s)
FmVV
V
V V= - = ´
+
- \=&( ). .
.
( .). .
2 1
22
2 2
14401231
1389
2
1389 306 m/s.p
\= =´
+
=QAV
33
2
1
3061389
2
699p
. .
. m/s.
3

h
p
V
V
== =
1
3
1389
2224
0625
.
.
. . or 62.5%

4.146 Fix the reference frame to the aircraft so that V
1
2001000
3600
55=
´
=.56 m/s.
V m
2
23201000
3600
8889 12 11
55 8889
2
329=
´
= \=´´
+
=. &. .
.56.
.5 m/s. kg/s.p
F= - =329888955 10.5(. .56) 980 N.
= ´ \=D Dp pp11 2890
2
.. . Pa

&
.WFV=´= ´
1
10 98055.56=610 000 W or 818 Hp

4.147 Fix the reference frame to the boat so that V
1
20
88
60
2933=´=. fps.
V FmVV
2 2 1
2
40
88
60
5867 194
10
12
29335867
2
58672933=´= \= -= ´
æ
è
ç
ö
ø
÷
+
-. &( ).
. .
(. .) fps. p
=5460 lb.

&
. , .WFV=´= ´ =
1
54602933160000
ft-lb
sec
or 291 Hp
V
2
F
V
B

85. 84
&.
. .
. .m= ´´
æ
è
ç
ö
ø
÷
+
=194
10
12
29335867
2
1862
2
p slug/sec

4.148 Fix the reference frame to the boat: V V
1 210 20= = m/s, m/s.
\Thrust = &( ) .( ) .mVV
2 110000220102000-= ´ -= N

&
.WFV=´= ´=
1
20001020 000 W or 26.8 Hp

4.149 02 210 1 2 2001
11 1 1 1 1
. ... ()(.).= =´´ \= \ = \ = -VAV V V Vy y m/s. m/s.
max

flux in = 2 21000201 800
1
3
267
2 2 2
3
0
1
0
1
rVdy ydy= ´ - = =òò
(.)
.
..
000 N.
The slope at section 1 is -20. \ =-+Vy yA
2 20() .
Continuity: AVAV V V
11 22 2 1
2 2= \= =. m/s.
V A
V A
VA
2
2
2
0
05 1
12
()
(.)
/.
=
=-
ü
ý
þ
\=-
2 12 2 220
2=- \= \ =-A A Vy y/. .5. ().5 .
flux out = 21000220 800
125
3
800
000153
2
3
0
05
0
05
(.5 )
(.)
[. ]
.
.
- =
-é
ë
ê
ù
û
ú=ò
ydy
y
000
000
3

=4083. N. \change = 408 - 267 = 141 N.

4.150 a) b= =
-
´´
= =
ò
ò
VdA
VA
ydy2
2
2 2
0
1
2
3
2201
1210
4000
1
3
4
3
(.)
..
.
.
.

b) See Problem 4.149: Vy y y V
2 2
200125 05 0 2()(. ),. .= - ³³ = m/s.
b= =
-
´´
=
-
=
ò
òVdA
VA
y dy
y
2
2
2 2
0
05
2
3
0
05
220 125
2110
2000
125
3
1021
(.)
..
(.)
..
.
.


4.151 From the c.v. shown: ( ) .ppr rL
w o1 2 0
2
2- =p tp
\= = \ =
´´
´´ ´
-
t m
w
o
w w
pr
L
du
dr
du
dr
D

2
0031447512
23023610
5
.
. ./
.

=191
ft/sec
ft
.




4.152 Write the equation of the parabola: VrV
r
r
() .
max= -
æ
è
ç
ö
ø
÷1
2
0
2

p
1
A
1
p
2
A
2
t
w
2pr
o
L

86. 85
Continuity: p p´ ´= -
æ
è
ç
ö
ø
÷ \ =ò
.
.
.
max max
.
0068 1
006
2 16
2
2
2
0
006
V
r
rdr V m/s.
Momentum: pApAF VdAmV
11 22
2
1
- - = -
òDrag
r &.
40 006 1000161
006
2 1000 00688
2 2
2
2
2
2
0
006
000
Drag
p p p´ - = ´ -
æ
è
ç
ö
ø
÷ - ´´ ´´ò
.
.
.
.
F
r
rdr
4 96517238.524 . ..- = -F
Drag
\ =F
Drag
N211..

4.153 & () . ( ) .m AV VydA ydy
top= - = ´´- +
é
ë
ê
ù
û
ú=òò
r r
11 2
2
0
2
1232103228 10 656 kg/s.
-= + - = + + ´- ´´òò
F
VdAmVmV y dy
top
2
12328 10656321232032
2
1 11
22 2
0
2
r & & .( ) . . .
\=F3780 N.

4.154 a) & && () .. ( )
.
m mm AV uydA y y dy
top=-= - = ´´- - ´
é
ë
ê
ù
û
úòò1 2 11
2
0
1
1231282010082r r
= = =0656 01 8. . ()). kg/s. (Note: for y uy
Momentum: - = - +´-´´´ò
F y ydy
Dragr r642010026568 128
22 2
0
1
( ) . .
.

= ´ + - ´ \ =123683525123128 21. . . . .. . N
Drag
F
b) To find h: 8 820100
2
0
1
h y ydy= -ò
( ).
.

\=
´
-
´
=h
201
2
100001
3
00667
2
. .
. m.
Momentum: - = - -´ ´´ò
F y ydy
Drag12364201002123066728
22 2
0
1
. ( ) .. .
.

= ´ -12368310. . .50. \ =F
Drag
N21..

4.155 a) Energy:
V
g
z
V
g
zh
L
1
2
1
2
2
2
2 2
+=++. See Problem 4.118(a).

8
2981
06
1912
2981
2
2 2
´
+=
´
+ +
.
.
.
.
.51.h
L
\=h
L1166. m.
\ = ´´´´ =losses = 900 W/m of width.gAVh
L11 9810618116654(.) .

b) See Problem 4.120:
V
g
z
V
g
zh
L
1
2
1
2
2
2
2 2
+=++.

87. 86

719
2981
417
1
2981
19
2 2
.
.
.
.58
.
. .
´
+ =
´
++h
L
\=h
L1025. m.
\ = ´ ´´ ´ =losses = 300 WgAVh
L11 98104173719102590. . .

c) See Problem 4.121:
517
2981
116
3
2981
2
2 2
.
.
.
.
.
´
+ =
´
++h
L
\=h
L00636. m.
\ = ´ ´ ´ =losses = /m of width.gAVh
L11 9810116517006363740 W. . .

4.156 See Problem 4.122: V V p p
1 2 1 220 5 60 135= = = = m/s, m/s, kPa, kPa.
Then,
V
g
pV
g
p
h h
L L
1
2
1 2
2
2
2 2
2 2
20
2981
60 5
2981
135
+= ++
´
+ =
´
+ +
g g
.
. .
.
000
9810
000
9810

\= =
´
\=h K
g
K K
L
1147
2
20
2981
0562
2
.
.
. .. m=
V

1
2


4.157 Continuity: VDVd V
d
D
V
1
2 2
1
2
2
= \=. .
Energy:
V
g
Ht
V
g
V gHt
1
2 2
2 2
2+ = \=() . ().
Momentum: SFF
d
dt
VdVmVV
ds
dt
a
x Ix x
cv
x x
x
x
- = -+ -
æ
è
ç
ö
ø
÷=ò
() &( ). .
..
r
2 1
2
2

v

\- = =-òamt
d
VVmtm
d
Vtdt
x o
t
() ().() ().r
p
r
p
2 2
0
4 4

But, V
dH
dt
dH
dt
d
D
gH
dH
H
d
D
gdtH
gd
D
tH
o1
2
2 12
2
2
12
2
2
2 2
2
2
=- \-= \- = \ = +. . . .
/
/

\= +
æ
è
ç
ç
ö
ø
÷
÷
+
æ
è
ç
ç
ö
ø
÷
÷
-
é
ë
ê
ê
ù
û
ú
ú
ò
a
d
g
gd
D
tH
d
g
gd
D
tHdtm
x o
t
o o
rp
r
p
2 2
2
2
2
0
2
2
4
2
2
2 4
2
2
2


4.158 This is a very difficult design problem. There is an optimum initial mass of water
for a maximum height attained by the rocket. It will take a team of students
many hours to work this problem. It involves continuity, energy, and
momentum.

4.159 V
m
A
V
e
e
= =
´´´
=
&
.
. .
r p
4
10004 004
1989
2
m/s. Velocity in arm =
v
v
v
M r VdV ri kViAdr
I
cv
=´ ´ -= ´- ´ò ò
( )
$
(
$$
)
..
.
2 4 2
0
3
W Wr r

88. 87
=- =-ò
8 036
0
3
r rAVkrdr AVkW W
$
.
$
.
.

S
v
v
v
v
vv
M
d
dt
rVdV rVVndAi Vj VkVA
cv
e e e e
cs
= ´ -= ´ × =´ +ò ò
0 0 3707 707 and ( ) . ($) .
$
(.
$
.
$
) .
.. ..
r r r
The z-component of
v
vv
rVVndA VA
ee
cs
´ × =´ò
($) .. .
..
r r3707
2

Finally, - = =´´ =(). .. ,M AV VA AVAV
Iz ee ee
036 43707
2
r rW . Using
036437071989. .. ..W=´´ ´ \=W469. . rad/s



4.160 A moment
v
Mresists the motion thereby
producing power. One of the arms is shown.


v
M ri kViAdr AVkrdr AVk
I= ´-´ =- =-òò
4 2 8 2778
0
1012
0
25
$
(
$$
)
$
.
$
.
/.
W W Wr r r
S
v
v
v
v
vv
MMk
d
dt
rVdV rVVndA VA k
cv
e e
cs
= ´ -= ´ × =´ ´ò ò
$
, ( ) , ($)
$
.
.. ..
and 0
10
12
4
2
r r
Thus, M+ ´
æ
è
ç
ö
ø
÷´ ´= ´´
æ
è
ç
ö
ø
÷´2778194
75
12
200
9
30200
10
12
194
14
12
4
2
2
2
. .
.
.
/
.p p
\=M309 ft-lb.

&
.WM= = ´=W309309270 ft-lb/sec

4.161 & . . .m AV V V== = ´ \=10 100001 318
2
0 0
r p m/s.
Continuity: V V V r
e0
2 2
01 01 00605p p´=´+´ -. . .(.).

0
2
.01.006.15. 11.1 m/s.
ee
VVVp´=´´\=
\=- - = -VV rV r
e0
19105 424212.(.) . .

v
M ri kViAdr ri k riAdr
I= ´+´ + ´+ ´ -òò
2 2 2 2 424212
0
05
2
0
05
$
(
$$
)
$
[
$
(. )
$
]
.
..
W Wr r
= + -òò
4 4 424212
0
2
05
2
0
05
W WVAkrdr Ak r rdrr r
$ $
(. )
.
..

=´ ´ ´´ +´ ´4318100001
05
2
4100001
2
2
2
W W. .
.
$
.p pk

424
2
205
212
3
205
2 2 3 3.
(..) (..)
$
- - -
é
ë
ê
ù
û
ú
k
x
y
r
V
e
V
W

89. 88
= + =(. .)
$
.
$
.00503 035W W Wk k
riVjV dr rdrk k
e e
$
(
$
). . .
$
.
$
.
.
.
.
.
´- ´ =- ´ ´ =-òò
r006 1111000006 1386
2
05
2
05
2

\- =-035 1386. ..W \=W396. . rad/s

4.162 1000
1000
500
2= \= = ×M MW. Nm.

v
M ri kVri rdr
I r r
= ´- ´ ´
ò
$
(
$
()
$
) .2 202W rp
= ò
008
2
0
. ()
$
.pWrVrdrk
R

Continuity: Vrr V R Vr RVr
r r
(). cos .. (). /.202 302 02 0866p p´= ´ \ =
o


v
vv
o o
rVVndARRV V R k V Vk
r r r r
cs
´ × =- + ´ =- +ò
($) ( sin)cos .
$
. (.)
$
.
..
r rpW 30 302 02 00301355
\-- =- + \- - =ò
21632 0030135 52113330
0
15
2
. . (.5). . .
.
Vrdr V V V V
r r r r r
\= ± +´ =V
r
1
2
52152141333709
2
(. . ). m/s.
The flow rate is QAV
er
= =´´´ ´ =cos .. .. . .3021502709866116
o
p m/s
3


4.163 See Problem 4.159. V V
e
= = ´ =1989
008
02
1989318
2
2
.
.
.
. . m/s. m/s.

v
M ri kVi
d
dt
kriAdrA A
I e= ´-´+-
æ
è
ç
ö
ø
÷´
é
ë
ê
ù
û
ú
=´ =´ò
4 2 01 004
0
3
2 2$
(
$$
)
$$
. ., ..
.
W
W
r p p
=- - =- -ò ò
8 4 360 36
0
3
2
0
3
r rAVkrdrA
d
dt
krdr AVk A
d
dt
kW
W
W
W
$ $ $ $
.
. .

( )($)
$
.
..
v
vv
rVVndA VAk
z ee
cs
´ × =ò
r 212
2

Thus, 360 36 212 318373
2
AV A
d
dt
VA
d
dt
ee
W
W W
W+ = + = or . .
The solution is W= +
-
Ce
t318
1173
.
..
The initial condition is W().00= \=-C1173..
Finally, W= -
-
11731
318
.( ) .
.
e
t
rad/s

4.164 This design problem would be good for a team of students to do as a project. How large a
horsepower blower could be handled by an average person?

90. 89
CHAPTER 5

The Differential Forms
of the
Fundamental Laws


5.1 0= -+ ×ò ò
¶r
¶
r
t
dV VndA
cv cs.. ..
$.
v
Using Gauss’ theorem:
0= -+Ñ× -= +Ñ×
é
ë
ê
ù
û
ú
-ò òò
¶r
¶
r
¶r
¶
r
t
dV VdV
t
VdV
cv cvcv.. ....
() ().
vv vv

Since this is true for all arbitrary control volumes (i.e., for all limits of
integration), the integrand must be zero:

¶r
¶
r
t
V+Ñ× =
vv
().0
This can be written in rectangular coordinates as
-= + +
¶r
¶
¶
¶
r
¶
¶
r
¶
¶
r
tx
u
y
v
z
w() ()().
This is Eq. 5.2.2. The other forms of the continuity equation follow.

5.2 && .mm
m
t
in out
element
- =
¶
¶

()r q r
¶
¶
r qvrddz v
r
vdrrdrddz
r r r( ) ( )- +
é
ë
ê
ù
û
ú
+
+ - +
é
ë
ê
ù
û
ú
r r
¶
¶q
rq
q q qvdrdzv vddrdz()
+ +
æ
è
ç
ö
ø
÷ - +
é
ë
ê
ù
û
ú
+
æ
è
ç
ö
ø
÷ = +
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
r q r
¶
¶
r q
¶
¶
r qvr
dr
ddrv
z
vdzr
dr
ddr
t
r
dr
ddrdz
z z z
2 2 2
() .
Subtract terms and divide by rddrdzq:
- -
+
- -
+
=
+r ¶
¶
r
¶
¶q
r
¶
¶
r
¶
¶
r
q
v
r r
v
rdr
r
v
rz
v
rdr
r t
rdr
r
r
r z
() () ()
/ /
.
1 2 2

Since dr is an infinitesimal, ( )/ ( /)/ .rdrr rdr r+ = + =1 2 1 and Hence,

¶r
¶
¶
¶
r
¶
¶q
r
¶
¶
r r
q
tr
v
r
v
z
v
r
v
r z r
+ + + + =() ()() .
1 1
0 This can be put in various forms.

91. 90
5.3 && .mm
m
t
in out
element
- =
¶
¶

r q qfr
¶
¶
r q qfvrdr d v
r
vdrrdrdrdr d
r r r()sin ()( )( )sin- +
é
ë
ê
ù
û
ú
+ +
+ +
æ
è
ç
ö
ø
÷ - +
é
ë
ê
ù
û
ú
+
æ
è
ç
ö
ø
÷r qfr
¶
¶q
rq qf
q q qvdrr
dr
d v vddrr
dr
d
2 2
sin () sin
+ +
æ
è
ç
ö
ø
÷- +
é
ë
ê
ù
û
ú
+
æ
è
ç
ö
ø
÷r qr
¶
¶f
rf q
f f f
vdrr
dr
d v vddrr
dr
d
2 2
()
= +
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
¶
¶
r qqf
t
r
dr
drd d
2
2
sin
Because some areas are not rectangular, we used an average length ( /).rdr+ 2
Now, subtract some terms and divide by rdddrqf:
- - -
+
-
+
r qr q
¶
¶
r q
¶
¶q
r q
q
v v
r
v
rdr
r
v
r
dr
r
r r r
sin sin ()sin
( )
() sin
2
2

-
+
=
+
æ
è
ç
ö
ø
÷
¶
¶f
r
¶r
¶
q
f
() sinv
r
dr
r t
r
dr
r
2 2
2

Since dr is infinitesimal ( )/ ( /)/ .rdrrr rdr r+ = + =
2
2 1 and Divide by rsinq
and there results

¶r
¶
¶
¶
r
¶
¶q
r
q
¶
¶f
r r
q f
tr
v
r
v
r
v
r
v
r r
+ + + + =() ()
sin
()
1 1 2
0

5.4 For a steady flow
¶r
¶t
=0. Then, with vw==0 Eq. 5.2.2 yields

¶
¶
r r
r
x
u
du
dx
u
d
dx
() .= + =0 0 or
Partial derivatives are not used since there is only one independent variable.

5.5 Since the flow is incompressible
D
Dt
r
=0. This gives

323
12001200
ˆˆˆˆ cos2sin2
rr
pp
piiii
rr rrr
qq
¶¶rr
qq
¶¶q
éù
\Ñ=+=--
êú
ëû
v
or
u
x
w
z
¶r
¶
¶r
¶
+ =0.
Also,
vv
Ñ×=V0, or
¶
¶
¶
¶
u
x
w
z
+=0.

92. 91
5.6 Given:
¶
¶
¶r
¶t z
= ¹0 0, . Since water can be considered to be incompressible, we
demand that
D
Dt
r
=0. \u
x
w
z
¶r
¶
¶r
¶
+ =0, assuming the x-direction to be in the
direction of flow. Also, we demand that
vv
Ñ×=V0, or
¶
¶
¶
¶
u
x
w
z
+=0.

5.7 We can use the ideal gas law, r=
p
RT
. Then, the continuity equation

D
Dt
V
r
r=-Ñ×
vv
becomes, assuming RT to be constant,
1
RT
Dp
Dt
p
RT
V=-Ñ×
vv
or

1
p
Dp
Dt
V=-Ñ×
vv
.

5.8 a) Use cylindrical coordinates with vv
zq==0:

1
0
rr
rv
r
¶
¶
()=
Integrate:
rvC
r=. \=v
C
r
r
.

b) Use spherical coordinates with vv
q f
==0:

1
0
2
2
rr
rv
r
¶
¶
()=
Integrate:
rvC
r
2
=. \=v
C
r
r 2
.

5.9
D
Dt
V
u
x
v
y
r
r r
¶
¶
¶
¶
=-Ñ×=- +
æ
è
ç
ö
ø
÷=- ´+´=-
×
vv
23200140011380.( ) .
kg
ms
3


5.10 In a plane flow, uuxy vvxy= =(,) (,). and Continuity demands that
¶
¶
¶
¶
u
x
v
y
+=0.
If u
u
x
= =const, then
¶
¶
0 and hence
¶
¶
v
y
=0. Thus, v=const also.

93. 92
5.11 If uC vC= =
1 2 and , the continuity equation provides, for an incompressible
flow,

¶
¶
¶
¶
¶
¶
¶
¶
u
x
v
y
w
z
w
z
wC++= \ = =0 0
3
. . and
The z-component of velocity w is also constant.
We also have

D
Dt t
u
x
v
y
w
z
r ¶r
¶
¶r
¶
¶r
¶
¶r
¶
==+ + +0
The density may vary with x, y, z and t. It is not, necessarily, constant.

5.12
¶
¶
¶
¶
u
x
v
y
+=0. \+= A
v
y
¶
¶
0. \ =-+ vxy Ayfx(,) ().
But, vxo fx(,) ().==0 \=- vAy.

5.13
¶
¶
¶
¶
u
x
v
y
+=0. \=-=-
+ -
+
=-
-
+
¶
¶
¶
¶
v
y
u
x
xy xx
xy
x y
xy
( )5()
( ) ( )
2 2
2 22
2 2
2 22
52 5 5

\ =
-
+
+ =
+
+ = \=
+
ò
vxy
y x
xy
dyfx
y
xy
fxfx v
y
xy
(,)
( )
() ().(). .
5 5 5
0
5
2 2
2 22 2 2 2 2


5.14 From Table 5.1:
1 1 1
10
4
2
rr
rv
r
v
r r
r
¶
¶
¶
¶q
q
q
()
.
sin.=- =- +
æ
è
ç
ö
ø
÷
\ = +
æ
è
ç
ö
ø
÷ + = -
æ
è
ç
ö
ø
÷ +ò
rv
r
drf r
r
f
r 10
4
10
4
2
.
sin ()
.
sin().q q qq
.(.,) .
.
.
sin(). ().22 102
4
2
0 0v f f
rq qq q=´-
æ
è
ç
ö
ø
÷ + = \ =
\= -
æ
è
ç
ö
ø
÷v
r
r
10
04
2
.
sin.q

5.15 From Table 5.1:
1 1 20
1
1
2
rr
rv
r
v
r r
r
¶
¶
¶
¶q
q
q
() cos.=- =
-
+
æ
è
ç
ö
ø
÷
\ =- +
æ
è
ç
ö
ø
÷ + =- -
æ
è
ç
ö
ø
÷ +ò
rv
r
drf r
r
f
r 201
1
20
1
2
cos () cos ().q q qq
v f f
r(,) ()cos(). ().1 2011 0 0q qq q=-- + = \ =
\=- -
æ
è
ç
ö
ø
÷v
r
r
201
1
2
cos.q

94. 93
5.16 From Table 5.1, spherical coordinates:
1 1
2
2
rr
rv
r
v
r
¶
¶ q
¶
¶q
q
q
()
sin
(sin).=-
\ = +
æ
è
ç
ö
ø
÷
1 1
10
40
2
2
2
3
rr
rv
r r
r
¶
¶ q
qq()
sin
sincos.
\ = +
æ
è
ç
ö
ø
÷ + = -
æ
è
ç
ö
ø
÷ +ò
rv r
r
drf r
r
f
r
2
3
2
10
40
2 10
80
cos () cos ()q q qq
42 102
80
2
0 0
2
v f f
r(,) cos (). ().q qq q= ´-
æ
è
ç
ö
ø
÷ + = \ =
\= -
æ
è
ç
ö
ø
÷v
r
r
10
80
3
cos.q

5.17 Continuity:
¶
¶
r r
r
x
u
du
dx
u
d
dx
(). .= \ + =0 0
r= =
´
´
= =
-
´
=
p
RT
du
dx
18144
1716500
000302
526453
2212
219
3
. .
/

slug
ft
fps/ft.
\ =- =- ´ =-
d
dx u
du
dx
r r .
. .
00302
486
219000136 slug/ft
4


5.18 [ ]
¶
¶
¶
¶
¶
¶
u
x
v
y x
e e
x x
+= - - =-
- -
0 201 20. ( )
Hence, in the vicinity of the x-axis:

¶
¶
v
y
e v yeC
x x
= = +
- -
20 20 and .
But v y C= = \=0 0 0 if . .
v ye e
x
= = =
- -
20 2002 0
2
(.) .541 m/s

5.19 [ ]
1
0 201 20
rr
rv
v
z z
e e
r
z z z¶
¶
¶
¶
¶
¶
() . ( )+ = - - =-
- -

Hence, in the vicinity of the z-axis:

1
20
2
20
2
rr
rv e rv
r
eC
r
z
r
z¶
¶
() .= = +
- -
and
But v r C
r= = \=0 0 0 if . .
v re e
r
z
= = =
- -
10 1002 0271
2
(.) . m/s

5.20 The velocity is zero at the stagnation point. Hence,
010
40
2
2
=- \=
R
R. m
The continuity equation for this plane flow is
¶
¶
¶
¶
u
x
v
y
+=0. Using
¶
¶
u
x
x=
-
80
3
,

95. 94
we see that
¶
¶
v
y
x=-
-
80
3
near the x-axis. Consequently, for small Dy,
D Dv xy=-
-
80
3
so that v=-- =
-
803010296
3
()(.). . m/s

5.21 The velocity is zero at the stagnation point. Hence
0
40
10 2
2
=- \=
R
R. m
()
1 1
4010
20
2
2
2
2
rr
rv
rr
r
r
r
¶
¶
¶
¶
= - =-( ) .
Near the negative x-axis continuity provides us with
( )
1 20
r
v
rsin
sin .
q
¶
¶q
q
q
=
Integrate, letting q=0 from the y-axis:
v C
qq qsin cos=- +20
Since v
q=0 when q= =90 0
o
, . C Then, with a= =
-
tan
.
.,
101
3
1909
o

v
q
q
q
=- =- =- =20 20
88091
88091
20
00333
0999
0667
cos
sin
cos.
sin.
.
.
. m/s

5.22 Continuity:
¶
¶
¶
¶
u
x
v
y
v
y
u
x
+= \ =-=-
-
´
=-0
135113
2005
220.
. .
.
.
m/s
m
D
D
D
D

\=-=- \=-´=-D Dvv y v0220 220004088. . . . m/s
b) au
u
x
x
= = ´+ =
¶
¶
1262202772.() . m/s
2


5.23 SFma
y y
= . For the fluid particle occupying the volume of Fig. 5.3:
t
¶t
¶
t
¶t
¶
t
¶t
¶
yy
yy
zy
zy
xy
xy
y
dy
dxdz
z
dz
dxdy
x
dx
dydz+
æ
è
ç
ö
ø
÷ + +
æ
è
ç
ö
ø
÷ + +
æ
è
ç
ö
ø
÷
2 2 2

- -
æ
è
ç
ö
ø
÷ - -
æ
è
ç
ö
ø
÷ - -
æ
è
ç
ö
ø
÷t
¶t
¶
t
¶t
¶
t
¶t
¶
yy
yy
zy
zy
xy
xy
y
dy
dxdz
z
dz
dxdy
x
dx
dydz
2 2 2

+ =r rgdxdydzdxdydz
Dv
Dt
y

Dividing by dxdydz ,and adding and subtracting terms:

¶t
¶
¶t
¶
¶t
¶
r r
xy yy zy
y
x y z
g
Dv
Dt
+ + + = .

5.24 Check continuity:

¶
¶
¶
¶
¶
¶
u
x
v
y
w
z
xy xx
xy
xy yy
xy
++=
+ -
+
+
+ -
+
=
( )10 ()
( )
( )10 ()
( )
.
2 2
2 22
2 2
2 22
102 102
0

96. 95
Thus, it is a possible flow. For a frictionless flow, Euler’s Eqs. 5.3.7 give, with
gg
x y
==0:
.
uup
uv
xyx
¶¶¶
rr
¶¶¶
+=-

2222
2222222222223
1010101020100()
()()()
pxyxyxyxyy
x xyxyxyxyxy
¶
rrr
¶
--+
\=--=
+++++
.
vvp
uv
xyy
¶¶¶
rr
¶¶¶
+=-

2222
2222222222223
1020101010100()
()()()
pxxyyxyxyy
y xyxyxyxyxy
¶
rrr
¶
--+
\=--=
+++++


222222222
100100100
ˆˆˆˆˆˆ ().
()()()
ppxy
pijijxiyj
xy xyxyxy
¶¶rrr
¶¶
\Ñ=+=+=+
+++
v


5.25 Check continuity (cylindrical coord from Table 5.1):

1 1 10
1
1 10
1
1
0
2 2
rr
rv
r
v
r r r r
r
¶
¶
¶
¶q
q q
q
() cos cos .+ = +
æ
è
ç
ö
ø
÷ +
-
+
æ
è
ç
ö
ø
÷ = \ It is a possible
flow. For Euler’s Eqs. (let n=0 in the momentum eqns of Table 5.1) in
cylindrical coord:

22
22
223
1001120
1sin101cos
rr
r
vv vvp
v
rrrrr rrr
qq ¶¶¶r
rrrqrq
¶¶¶q
æöæöæö
=--=+--
ç÷ç÷ç÷
èøèøèø

- +
æ
è
ç
ö
ø
÷ -
æ
è
ç
ö
ø
÷
10
1
1
10
10
2
2
2
r
q
r r r
sin .

4
11001
1sincos
r
r
vvvvvp
v
rrrrr r
qqqq
¶¶¶r
rrrqq
¶q¶¶q
æö
=---=-
ç÷
èø

- -
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷- +
æ
è
ç
ö
ø
÷101
1 20100
1
1
2 3 2
2
r qq
r
qq
r r r r
cossin sincos.

323
12001200
ˆˆˆˆ cos2sin2
rr
pp
piiii
rr rrr
qq
¶¶rr
qq
¶¶q
éù
\Ñ=+=--
êú
ëû
v



5.26 This is an involved problem. Follow the steps of Problem 5.25. Good luck!

( )¶
¶
r r
¶
¶
r
¶
¶q
q f
q
p
r
vv
r
v
v
r
v
r
v
r
r r
=
+
- -
2 2


1
r
p vv
r
v
v
r
v
r
v
r
r
¶
¶q
r r
¶
¶
r
¶
¶q
q q q q
=- - -
()

97. 96
5.27
22
. .
33
ppVppV
mm
ll
æöæö
\=-+Ñ×\-=-+Ñ×
ç÷ç÷
èøèø
vvvv


¶
¶
$$ $ $
.
s
s
s
s
n
R
n
R
@=- =-
D
D
Da
Da


¶
¶
¶q
¶
$$$
$.
s
t
s
t
n
t
n
t
@= =
D
D
Dq
D




2
ˆˆ .
DVVVV
VsVn
DttstR
¶¶¶q
¶¶¶
æö
æö
\=++- ç÷ç÷
ç÷
èø
èø
v

For steady flow, the normal acc. is -
æ
è
ç
ö
ø
÷
V
R
2
, the tangential acc. is V
V
s
¶
¶
.

5.28 For a rotating reference frame (see Eq. 3.2.15), we must add the terms due to
v
W.
Thus, Euler’s equation becomes
2().
DVd
Vrrpg
Dtdt
rr
æö W
+W´+W´W´+´=-Ñ-
ç÷
èø
vv
vvvvv
vvv


5.29 2
xx
u
p
x
¶
tm
¶
=-+ Vl+Ñ×
vv
30 psi.=-
tt
yy zz
p==-=-30 psi.

xy
uv
yx
¶¶
tm
¶¶
=+
55 .1
103014401810 psf.
12
--æö éù
=-´=´ç÷ êú
ëûèø

tt
t
t
xz yz
xx
== =
´
´
= ´
-
-
0
1810
30144
41710
5
8
. . .
xy

5.30
223
9/5213/59/5213/5
1616816
. (,)().
3
vuyyyy
vxyfx
yx CxCxCxCx
¶¶
¶¶
=-=-\=-+
vxo fx C C(,). (). . ..
/
= \ = = \=0 08 1000 00318
45

\ = -
- -
uxy yx yx(,) .
/
629 9890
4/5 285

vxy yx yx(,) .
/ /
= -
- -
252 5270
295 3135

t m
¶
¶
xx
p
u
x
=-+ =-+=-2 1000100 kPa.
tt
yy zz
p==-=-100 kPa.

54/55
21062910005.0110 Pa.
xy
uv
yx
¶¶
tm
¶¶
---æö
éù=+=´´=´
ç÷
ëû
èø

tt
xz yz
==0.

98. 97
5.31
Duu
Dtt
¶
¶
= ().uvwuVu
xyz
¶¶¶
¶¶¶
æö
+++=×Ñ
ç÷
èø
vv


Dvv
Dtt
¶
¶
= ().uvwvVv
xyz
¶¶¶
¶¶¶
æö
+++=×Ñ
ç÷
èø
vv


Dww
Dtt
¶
¶
= ()uvwwVw
xyz
¶¶¶
¶¶¶
æö
+++=×Ñ
ç÷
èø
vv


ˆˆˆˆˆˆ ()().
DVDuDvDw
ijkVuivjwkVV
DtDtDtDt
\=++=×Ñ++=×Ñ
v
vvvvv


5.32 Follow the steps that lead to Eq. 5.3.17 and add the term due to compressible
effects:

2 ˆˆˆ
333
DV
pgVViVjVk
Dtxyz
m¶m¶m¶
rrm
¶¶¶
=-Ñ++Ñ+Ñ×+Ñ×+Ñ×
v
vvvvvvvvv

=-Ñ++ + + +
æ
è
ç
ö
ø
÷Ñ×
v v v vv
pg V
x
i
y
j
z
kVrmÑ
m¶
¶
¶
¶
¶
¶
2
3
$ $ $

\ =-Ñ++Ñ+ÑÑ×r rm
mDV
Dt
pg V V
v
v v v vvv
2
3
().

5.33 If u=u(y), then continuity demands that
¶
¶
v
y
vC= \=0. .
But, at y=0 (the lower plate) v=0. \= =C vxy0 0, (,). and

222
222
.
x
uuuDuuuuup
uvwg
Dttxyzx xyz
¶¶¶¶¶¶¶¶
rrrm
¶¶¶¶¶ ¶¶¶
æöæö
\=+++=-++++ ç÷ç÷
ç÷
èø èø

\=-+0
2
2
¶
¶
m
¶p
x
u
ay
.
0.
Dvp
Dty
¶
r
¶
==-
r
¶
¶
r
¶
¶
r
Dw
Dt
p
z
g
p
z
g==-+- \=--0 0(). .

5.34 Continuity: ()0. .
rr
rvrvC
r
¶
¶
=\= At 0, . 0.
rrvC=¹¥\=

1
0.
r
Dv p
Dtr
¶
r¶
==-

1
0.
Dv p
Dtr
q ¶
r¶q
==-

99. 98

zz
Dvv
Dtt
¶
rr
¶
=
rv+
z
vv
r
q¶
¶
+
zz
z
vv
v
rz
¶¶
¶q¶
+
22
222
11
zzz
vvvp
zrr rr
¶¶¶¶
m
¶¶ ¶¶q
æö
ç÷ =-+++
ç÷
èø
2
2
z
v
z
¶
¶
+
æö
ç÷
ç÷
èø

\=-+ +
æ
è
ç
ö
ø
÷0
1
2
2
¶
¶
m
¶
¶
¶
¶
p
z
v
rr
v
r
z z
.

5.35 Continuity:
1
0 0 0
2
2 2
1
rr
rv rvC rrv C
r r r
¶
¶
(). . , . .= \ = = = \= At

2
2
2
cot.
vv p
rr r
qq
¶
rmq
¶
æö
-=-+-
ç÷
èø


2
222
11
0
sin
vvp
r
rrr rr
qq
¶¶¶
m
¶q¶¶ q
éù æö
=-+-
êú ç÷
èøëû

0
1
=-
r
p
sin
.
q
¶
¶f


5.36 For an incompressible flow
vv
Ñ×=V0. Substitute Eqs. 5.3.10 into Eq. 5.3.2 and
5.3.3:
r
¶
¶
m
¶
¶
¶
¶
m
¶
¶
¶
¶
¶
¶
m
¶
¶
¶
¶
r
Du
Dtx
p
u
x y
u
y
v
x z
u
z
w
x
g
x= -+
æ
è
ç
ö
ø
÷+ +
æ
è
ç
ö
ø
÷+ +
æ
è
ç
ö
ø
÷+2 .
=-+ + + + ++
æ
è
ç
ö
ø
÷+
¶
¶
m
¶
¶
m
¶
¶
m
¶
¶
m
¶
¶
¶
¶
¶
¶
¶
¶
r
p
x
u
x
u
y
u
z x
u
x
v
y
w
z
g
x
2
2
2
2
2
2

\ =-+ + +
æ
è
ç
ö
ø
÷+r
¶
¶
m
¶
¶
¶
¶
¶
¶
r
Du
Dt
p
x
u
x
u
y
u
z
g
x
2
2
2
2
2
2
.

2.
y
Dvuvvvw
pg
Dtxyxyyzzy
¶¶¶¶¶¶¶¶
rmmmr
¶¶¶¶¶¶¶¶
æöæöæö
=++-++++
ç÷ç÷ç÷
èøèøèø

=-+ + + + ++
æ
è
ç
ö
ø
÷+
¶
¶
m
¶
¶
m
¶
¶
m
¶
¶
m
¶
¶
¶
¶
¶
¶
¶
¶
r
p
y
v
x
v
y
v
z y
u
x
v
y
w
z
g
y
2
2
2
2
2
2

\ =-+ + +
æ
è
ç
ö
ø
÷+r
¶
¶
m
¶
¶
¶
¶
¶
¶
r
Dv
Dt
p
y
v
x
v
y
v
z
g
y
2
2
2
2
2
2
.

2
z
Dwuwvww
pg
Dtxzxyzyzz
¶¶¶¶¶¶¶¶
rmmmr
¶¶¶¶¶¶¶¶
æöæöæö
=++++-++
ç÷ç÷ç÷
èøèø èø

100. 99
=-+ + + + ++
æ
è
ç
ö
ø
÷+
¶
¶
m
¶
¶
m
¶
¶
m
¶
¶
m
¶
¶
¶
¶
¶
¶
¶
¶
r
p
z
w
x
w
y
w
z z
u
x
v
y
w
z
g
z
2
2
2
2
2
2

\ =-+ + +
æ
è
ç
ö
ø
÷+r
¶
¶
m
¶
¶
¶
¶
¶
¶
r
Dw
Dt
p
z
w
x
w
y
w
z
g
z
2
2
2
2
2
2
.

5.37 If we substitute the constitutive equations (5.3.10) into Eqs. 5.3.2 and 5.3.3., with
mm=(,,)xyz we arrive at
222
222
2
x
Dupuuuuuvuw
g
Dtxxxyyxzzx xyz
¶¶¶¶¶m¶¶m¶¶¶m¶¶
rrm
¶¶¶¶¶¶¶¶¶ ¶¶¶
æö æö æö
=-+++++++++ ç÷ ç÷ ç÷
ç÷
èøèøèø


5.38 If plane flow is only parallel to the plate, vw==0. Continuity then demands
that ¶¶ux/ .=0 The first equation of (5.3.14) simplifies to

uu
u
tx
¶¶
r
¶¶
+ v+
u
w
y
¶
¶
+
up
zx
¶¶
¶¶
æö
=-ç÷
èø
xgr+
2
2
u
x
¶
m
¶
+
22
22
uu
yz
¶¶
¶¶
++
æö
ç÷
ç÷
èø


2
2
uu
t y
¶¶
rm
¶ ¶
=
We assumed g to be in the y-direction, and since no forcing occurs other than
due to the motion of the plate, we let ¶¶px/ .=0

5.39 From Eqs. 5.3.10, -
++
=- ++
æ
è
ç
ö
ø
÷-Ñ×
ttt m¶
¶
¶
¶
¶
¶
l
xx yy zz
p
u
x
v
y
w
z
V
3
2
3

vv
.

22
. .
33
ppVppV
mm
ll
æöæö
\=-+Ñ×\-=-+Ñ×
ç÷ç÷
èøèø
vvvv


5.40
ˆˆˆ()()VVuvwuivjwk
xyz
æö¶¶¶
×Ñ=++++
ç÷
¶¶¶èø
vv

ˆ()
wwwvvv
VVuvwuvwi
yxyzzxyz
éùæöæö¶¶¶¶¶¶¶¶
Ñ×Ñ=++-++ êúç÷ç÷
¶¶¶¶¶¶¶¶èøèøëû
×
vv

ˆ
uuuwww
uvwuvwj
zxyzxxyz
éùæöæö¶¶¶¶¶¶¶¶
+++-++êúç÷ç÷
¶¶¶¶¶¶¶¶èøèøëû

ˆ
vvvuuu
uvwuvwk
xxyzyxyz
éùæöæö¶¶¶¶¶¶¶¶
+++-++êúç÷ç÷
¶¶¶¶¶¶¶¶èøèøëû

Use the definition of vorticity:
ˆˆˆ()()()
wvuwvu
ijk
yzzxxy
w
¶¶¶¶¶¶
=-+-+-
¶¶¶¶¶¶
v

101. 100
ˆˆˆ()()()()()
wvuwvx
Vuivjwk
yzxzxyxyz
w
éù¶¶¶¶¶¶¶¶¶
×Ñ=-+-+-++
êú
¶¶¶¶¶¶¶¶¶ëû
v
v

ˆˆˆ()()()()
wvuwvu
Vuvwijk
xyzyzzxxy
w
éùéù¶¶¶¶¶¶¶¶¶
×Ñ=++-+-+-
êúêú
¶¶¶¶¶¶¶¶¶ëûëû
v
v

Expand the above, collect like terms, and compare coefficients of ˆ,i ˆ,j and
ˆ
.k

5.41 Studying the vorticity components of Eq. 3.2.21, we see that /
z uyw¶¶=- is the
only vorticity component of interest. The third equation of Eq. 5.3.24 then
simplifies to

D
Dt
z
z
w
nw=Ñ
2


2
2
z
y
¶w
n
¶
=
since changes normal to the plate are much larger than changes along the plate,
i.e., .
zz
yx
¶w¶w
¶¶
>>

5.42 If viscous effects are negligible, as they are in a short section, Eq. 5.3.25 reduces
to

D
Dt
z
w
=0
that is, there is no change in vorticity (along a streamline) between sections 1 and
2. Since (see Eq. 3.2.21), at section 1,
w
¶
¶
¶
¶
z
v
x
u
y
=-=-10
we conclude that, for the lower half of the flow at section 2,
.10=
y
u
¶
¶

This means the velocity profile at section 2 is a straight line with the same slope
of the profile at section 1. Since we are neglecting viscosity, the flow can slip at
the wall with a slip velocity u
0; hence, the velocity distribution at section 2 is
uyu y
2 010() .=+ Continuity then allows us to calculate the profile:
VAVA
11 22=

1
2
10004004 100022002 03
0 0
( .)(.)( ./)(.). .´ =+´ \=wu w u m/s.
Finally,
uy y
2 0310().=+

102. 101
5.43 No. The first of Eqs. 5.3.24 shows that, neglecting viscous effects,

x
xyz
D uuu
Dtxyz
w ¶¶¶
www
¶¶¶
=++
so that w
y
, which is nonzero near the snow surface, creates w
x
through the term
/,
y
uyw¶¶ since there would be a nonzero ¶¶uy/ near the tree.

5.44 kTndA
t
V
gzudV
V
gzu
p
VndA
cvcs cs
v v
Ñ×= ++
æ
è
ç
ö
ø
÷-+ +++
æ
è
ç
ö
ø
÷×òò ò
$ ~
)
~ $
.... ..
¶
¶
r
r
r
2 2
2 2


vv vv
Ñ×Ñ-= ++
æ
è
ç
ö
ø
÷-+Ñ× +++
æ
è
ç
ö
ø
÷-òò ò
()
~ ~
.... ..
kTdV
t
V
gzudV V
V
gzu
p
dV
cvcv cv
¶
¶
r r
r
2 2
2 2

\-Ñ+ ++
æ
è
ç
ö
ø
÷+Ñ× +++
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú-=ò
kT
t
V
gzu V
V
gzu
p
dV
cv
2
2 2
2 2
0
¶
¶
r rr r
r
~ ~
.
..
vv


222
222
VVpV
VgzV
tt
¶¶r
rrr
¶r¶
æö
+Ñ×++=+Ñ× ç÷
ç÷
èø
vvvv Vp
VVVgz
t
¶
r
¶r
æö éù Ñ
+×+×Ñ++Ñç÷ êú
ëûèø
vv
vvvv
0.=
continuity momentum
\-Ñ+ +×Ñ= \ =ÑkT
t
uVu
Du
Dt
kT
2 2
0
¶
¶
rr r
~ ~
.
~
.
vv


5.45 Divide each side by dxdydz and observe that

222
222
,,
ydyyxdxxzdzz
TTTTTT
yyxxzz TTT
dxdydz xxz
+++
¶¶¶¶¶¶
-
--
¶¶¶¶¶¶ ¶¶¶
===
¶¶¶

Eq. 5.4.5 follows.

5.46
(/)DuDhpDhDppDDhDpp
V
DtDtDtDtDtDtDt
rr
rrrrr
rr
-
éù==-+=-+-Ñ×
ëû
v%

where we used the continuity equation: /.DDtVrr=-Ñ×
v
Then Eq. 5.4. 9
becomes
2DhDpp
VKTpV
DtDt
rr
r
éù-+-Ñ×=Ñ-Ñ×
ëû
vv

which is simplified to

2DhDp
KT
DtDt
r=Ñ+

103. 102
5.47 See Eq. 5.4.9:
~
. .ucT c
T
t
u
T
x
v
T
y
w
T
z
kT= \ + + +
æ
è
ç
ö
ø
÷=Ñ r
¶
¶
¶
¶
¶
¶
¶
¶
2

Neglect terms with velocity: r
¶
¶
c
T
t
kT=Ñ
2
.


5.48 The dissipation function
Finvolves viscous effects. For flows with extremely
large velocity gradients, it becomes quite large. Then
rc
DT
Dt
p
=F
and
DT
Dt
is large. This leads to very high temperatures on reentry vehicles.

5.49 u r
u
r
r r y= - \=-´10110000 210
5
( ). . ) ( takes the place of
2 ¶
¶

From Eq. 5.4.17, F=
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
= ´2
1
2
410
2
2 10
m
¶
¶
m
u
y
r .
At the wall where r= =´ ´´´ = ×
-
001 18104011072
5 2 10
. . . . m, N/ms
2
F
At the centerline
¶
¶
u
r
= =0 0 so F.
At a point half-way: F=´ ´´ ´ = ×
-
181040051018
5 2 10
. . . N/ms
2


5.50 (a) Momentum:
¶
¶
n
¶
¶
u
t
u
y
=
2
2

Energy: r
¶
¶
¶
¶
m
¶
¶
c
T
t
K
T
y
u
y
= +
æ
è
ç
ö
ø
÷
2
2
2
.
(b) Momentum: r
¶
¶
m
¶
¶
¶m
¶
¶
¶
u
t
u
y y
u
y
= +
2
2

Energy: r
¶
¶
¶
¶
m
¶
¶
c
T
t
K
T
y
u
y
= +
æ
è
ç
ö
ø
÷
2
2
2
.

104. 103
CHAPTER 6

Dimensional Analysis
and Similitude

6.1
g
V
V
g
p
g
z
g
V
V
g
p
g
z
1
2
1
2
1
1
1
2
2
2
2
2
2 2
++
é
ë
ê
ù
û
ú= ++
é
ë
ê
ù
û
ú
r r
.

1
2
1
2
1
1
2
1
1
2
2
2
1
2
2
1
2
2
1
2
+ + = + +
p
V
gz
V
V
V
p
V
gz
Vr r
.
or
1
2
1
2
1
1
2
1
1
2
2
2
2
2
2
2
2
2
1
2
+ +=+ +
æ
è
ç
ç
ö
ø
÷
÷
p
V
gz
V
p
V
gz
V
V
Vr r


6.2 a) []& . .m
FT
L
==
×
×
=
×
\
kg
s
Ns
ms
Ns
m

2

b) []p
F
L
= \
N
m

2
.
2

c) []r= =
×
×
=
×
\
kg
m
Ns
mm
Ns
m

3
2
3
2
4
. .
FT
L
2
4

d) []m=
×
\
Ns
m

2
.
FT
L
2

e) []W FL=× \Nm.
f) []
&
.W
FL
T
=
×
\
Nm
s

g) []s= \N/m .
F
L


6.3 (A) The dimensions on the variables are as follows:

22
2322
/
[][],[],[],[]
LLMLMLTML
WFVMdLpV
TTTTLLT
=´=´==D===
&

First, eliminate T by dividing W
&
by Dp. That leaves T in the denominator so
divide by V leaving L
2
in the numerator. Then divide by d
2
. That provides

2
W
pVd
p=
D
&


6.4
1
252
, , , .
Ter
f
RRRRR
m
rwrw
æö
\= ç÷
ç÷
èø
l

105. 104

6.5 (A) (, , , , ).The units on the variables on the rhs are as follows:Vfdlg wm=

1
2
[], [], [], [], []
LML
dLlLgT
TT
wm
-
=====
Because mass M occurs in only one term, it cannot enter the relationship.

6.6 [] [] [] []Vf V
L
T
L
M
L
M
LT
= = = = =(,,). , , , .l lrm r m
3

\There is one p-term: p
r
m
1=
Vl
.
\= = \ = =p p r
m
1 12
0
f
V
C() , ReConst. or Const.
l


6.7 [] [] [] []Vf d V
L
T
M
T
M
L
dL= = = = =(,,). , , , .sr s r
2 3

\= \= = \ =p
s
r
p p
s
r
1 2 1 12
0
2
Vd
f C
Vd
C. () , onst. or We = Const.

6.8 [] [] [] []VfHgm V
L
T
g
L
T
mMHL= = = = =(,,). , , , .
2

\= \= \=p p
1
0
2 1
gHm
V
C VgHC. . /.

6.9 [] [] [] [] [] []VfHgm V
L
T
HLg
L
T
mM
M
L
M
LT
= = = = = = =(,,,,). , , , , , .rm r m
2 3

Choose repeating variables Hg,,r (select ones with simple dimensions-we couldn’t
select V, H, and g since M is not contained in any of those terms):
p rp r pm r
1 2 3
1 1 1 2 2 2 3 3 3
= = =VHg mHg Hg
abc abc abc
, , .
\= = = = =p
r
p
r
p
m
r
m
r
1
0
2 3 3 32
3
V
gH
V
gH
m
H gH gH
. . .
/

\ =
æ
è
ç
ç
ö
ø
÷
÷
V
gH
f
m
H gH
1 3
3
r
m
r
, .
Note: The above dimensionless groups are formed by observation: simply combine the
dimensions so that the p-term is dimensionless. We could have set up equations similar
to those of Eq. 6.2.11 and solved for
111222333
, , and , , c and , , .abcababc But the
method of observation is usually successful.

6.10 [] [] [] [] []FfdV F
ML
T
dLV
L
T
M
LT
M
L
D D
= = = = = =(,,,,). , , , , .lmr m r
2 3

106. 105
p rp r pm r
1 2 3
1 1 1 2 2 2 3 3 3
= = =FV dV V
D
abc bca abc
l l l, , .
\= = =p
r
p p
m
r
1 2 2 2 3
F
V
d
V
D
l l l
, , .
\ =
æ
è
ç
ö
ø
÷
F
V
f
d
V
D
r
m
rl ll
22 1
, .
We could write
p
p p
p
p
1
2
2
2
2
3
2
1
=
æ
è
ç
ö
ø
÷f , or
F
dV
f
ddV
D
r
m
r
22 2=
æ
è
ç
ö
ø
÷
l
, . This is equivalent
to the above. Either functional form must be determined by experimentation.

6.11 [] [] [] [] []FfdV F
ML
T
dLV
L
T
M
LT
M
L
D D
= = = = = =(,,,,). , , , , .lmr m r
2 3

p m p m prm
1 2 3
1 1 1 2 2 2 3 3 3
= = =FdV d V d V
D
abc abc abc
, , . l
By observation we have
123 , , .
D
F Vd
Vdd
r
ppp
mm
===
l

\ =
æ
è
ç
ö
ø
÷
F
Vd
f
d
Vd
D
m
r
m
1
l
, .
Rather than p pp
1 12 3=f(,), we could write

p
p
p
p
1
3
2 2
3
1
=
æ
è
ç
ö
ø
÷f,, an acceptable form:
F
Vd
f
dVd
D
r
m
r
22 2=
æ
è
ç
ö
ø
÷
l
, .

6.12 [] [] [] [] [] []hfd ghL
M
T
dL
M
LT
g
L
T
= = = = = = =(,,,,). , , , , , .sgb s g b
2 22 2
1
Select dg,,g as repeating variables.
p g psg pb
1 2
1 1 1 2 2 2
= = =hdg d g
abc abc
, , .
3

\= = =p p
s
g
pb
1 2 2 3
h
d d
, , .
\=
æ
è
ç
ö
ø
÷
h
d
f
d
1 2
s
g
b,. Note: gravity does not enter the answer.

6.13 [] [] [] []FfmRF
ML
T
mM
T
RL
C C
= = = = =(,,). , , , .w w
2
1

\= = \ = \=p w
w w
w
1 2 2
2
FmR
F
mR
F
mR
C FCmR
C
abc C C
C
. .


6.14 [] [] [] []s s= = = = =fMyI
M
LT
M
ML
T
yLIL(,,). , , , .
2
2
2
4
\=ps
1
MyI
abc
.

107. 106
Given that 1b=-,
1
Const. C.
IMy
yMI
s
ps==\=

6.15 [] [] []Vfd
dp
dx
V
L
T
M
LT
dL
dp
dx
M
LT
=
æ
è
ç
ö
ø
÷ = = =
é
ë
ê
ù
û
ú
=(,,. , , , .m m
22

\=
æ
è
ç
ö
ø
÷p m
1
Vd
dp
dx
ab
c
. Let’s start with the ratio
m
dpdx/
so that “M” is accounted for.
Then the p
1
-term is
mV
dpdxd/
.

2
Hence,
p
m
m
1 2
2
= = \=
V
dpdxd
V
ddpdx
/
/
.

Const. Const

6.16 [] [] [] []VfHg V
L
T
HLg
L
T
M
L
= = = = =(,,). , , , .r r
2 3

\= = = \=p r
r
1
0
VHg V
gH
V gH
abc
Const. Const. .
Density does not enter the expression.

6.17 [] [][] [] [] []
32
(,,,,). ,,,,,.
LMML
VfHgdVHLgdL
TLT LT
mrmr=======
p r pmr p r
1 2
1 1 1 2 2 2 3 3 3
= = =VH g H g dH g
abc abc abc
, , .
3

Repeating
variables
ü
ý
þ
Hg,,.r
p p
m
r
p
1 2 32 3
= = =
V
gH gH
d
H
, , .
/

\= =
æ
è
ç
ç
ö
ø
÷
÷
p pp
m
r
1 12 3 1
3
f
V
gH
f
gH
d
H
(,), ,. or

6.18 DpfVdL=(,,,,,).ner
[] [] [][] [][][]
2
23
,,,,,,.
MLLM
pVdLLLL
TTLTL
nerD=======
Repeating variables: Vd,,.r
p rpn rp rpe r
1 2 3 4
1 1 1 2 2 2 3 3 3 4 4 4
= = = =DpVd Vd LVd Vd
abc abc abc abc
, , , .
\= = = =p
r
p
n
p p
e
1 2 3
Dp
V Vd
L
d d
, , , .
2 4

p ppp
r
n e
1 12 3 4 2 1= \ =
æ
è
ç
ö
ø
÷f
p
V
f
Vd
L
dd
(,,). ,,.
D

108. 107
6.19 FfV chrw
D
=(,,,,,,,,)rm fawhere c is the chord length, h is the maximum thickness, r
is the nose radius, f is the trailing edge angle, and a is the angle of attack. Repeating
variables: Vc,,.r Thep-terms are
p
r
p
r
m
p p pfp pa
1 22 2 3 4 5 7
= = = = = = =
F
Vc
Vc c
h
c
r
c
w
D
, , , , , , .
6

Then,

F
Vc
f
Vcc
h
c
r
c
w
D
r
r
m
fa
22 1
=
æ
è
ç
ö
ø
÷,,,,,

6.20 [] [][] [][][]
3
2
2
(,,,,). ,,,,1,.
LL
QfRAeSgQRLALeLsg
T T
=======
There are only two basic dimensions. Choose two repeating variables, R and g. Then,
p p p p
1 2 3 4
1 1 2 2 3 3 4 4
= = = =QRg ARg eRg sRg
ab ab ab ab
, , , .
\= = = =p p p p
1 52 2 2 3 4
Q
gR
A
R
e
R
s
/
, , , .
\= \ =
æ
è
ç
ö
ø
÷p ppp
1 12 34
5
1 2
f
Q
gR
f
A
R
e
R
s(,,). ,,.

6.21 [][] [] []
223
(,,,). ,,,,.
pp
LLMM
VfhgVhLg
T TTL
srsr éù======
ëû

Repeating variables: hg Vhg h g
p
abc abc
,,. , .r p r psr \= =
1 2
1 1 1 2 2 2

\= = \ =
æ
è
ç
ö
ø
÷p p
s
r
s
r
1 2 2 1 2
V
hg gh
V
gh
f
gh
p p
, . .

6.22 FfV eId
D
=(,,,,,).mr Repeating variables: Vd,,.r
[] [] [] [] [][][]
23
,,,,,1,.
D
MLLMM
FVeLIdL
TLTTL
mr=======
p r pmr p r p r
1 2 3 4
1 1 1 2 2 2 3 3 3 4 4 4
= = = =FV d V d eV d IV d
D
abc abc abc abc
, , , .
\= = = =p
r
p
m
r
p p
1 22 2 3 4
F
Vd Vd
e
d
I
D
, , , .
\ =
æ
è
ç
ö
ø
÷
F
Vd
f
Vd
e
d
I
D
r
m
r
22 1
,,.


6.23 FfV Dg
D s
=(,,,,,).rrm Repeating variables: VD,,.r
[] [] [] [] [] [][]
2332
,,,,,,.
Ds
MLLMMML
FVDLg
TLTTLLT
rrm=======

109. 108
p r prr pmr p r
1 2 3 4
1 1 1 2 2 2 3 3 3 4 4 4
= = = =FV D V D V D gV D
D
abc
s
ab c abc ab c
, , , .
\= = = =p
r
p
r
r
p
m
r
p
1 22 2 3 4 2
F
VD VD
gD
V
D s
, , , .
\ =
æ
è
ç
ö
ø
÷
F
VD
f
VD
gD
V
D s
r
r
r
m
r
22 1 2
, ,.

6.24 FfV derc
D
=(,,,,,,).mr Repeating variables: Vd,,.r
[] ,[],[] ,[] ,[],[],[],[] .F
ML
T
V
L
T
M
LT
M
L
dLeLrLc
L
D
= = = = = = = =
2 3 2
1
m r
p r pmr p r p r p r
1 2 3 4 5
1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
= = = = =FV d V d eV d rV d cV d
D
abc abc abc abc abc
, , , , .
\= = = = =p
r
p
m
r
p p p
1 22 2 3 4 5
2F
Vd Vd
e
d
r
d
cd
D
, , , , .
\ =
æ
è
ç
ö
ø
÷
F
Vd
f
Vd
e
d
r
d
cd
D
r
m
r
22 1
2
,,,.

6.25 fg Vd f
T
M
LT
M
L
V
L
T
dL= = = = = =(,,,).[],[] ,[] ,[],[].mr m r
1
3

Repeating variables, Vd fVd Vd
abc abc
,,. , rp rpm r
1 2
1 1 1 2 2 2
= =
\= = \=
æ
è
ç
ö
ø
÷p p
m
r
m
r
1 2 1
fd
V Vd
fd
V
g
Vd
, . .

6.26 FfVc t
L c=(,,,,). , r al Repeating variables: V
c,,. rl
[] ,[],[],[] ,[],[],[].F
ML
T
V
L
T
c
L
T
M
L
LtL
L c
= = = = = = =
2 3
1 r al
p r p r p r par
1 2 3 4
1 1 1 2 2 2 3 3 3 4 4 4
= = = =FV cV tV V
L
ab
c
c ab
c
c ab
c
c ab
c
c
l l l l, , , .
\= = = =p
r
p p pa
1 22 2 3 4
F
V
c
V
t
L
c cl l
, , , .
\ =
æ
è
ç
ö
ø
÷
F
V
f
c
V
t
L
c cr
a
22 1
l l
,,.

6.27
2
23
1
(, , , , ). [], []=,[]=,[],[],[].
MLMM
TfdtTdLtL
TLTTL
wrmwrm=====
Repeating variables: d Td d td
abc abc abc
,,. , , .
3
wrp wrpmwrp wr
1 2
1 1 1 2 2 2 3 3 3
= = =

110. 109
123 252
, , .
Tt
ddd
m
ppp
rwrw
\===

35
11
2522
,. ,.
Ttt
fWdf
ddddd
mm
rw
rwrwrw
æöæö
\== ç÷ç÷
ç÷ç÷
èøèø
&


6.28 FfV dL
D c=(,,,,,,) rm rwwheredis the cable diameter, Lthe cable length, r
c

the cable density, andw the vibration frequency.
Repeating variables: Vd,,. r The p-terms are
p
r
p
r
m
p p
r
r
p
w
1 22 2 3 4 5
= = = = =
F
Vd
Vd d
L
V
d
D
c
, , , ,
We then have

F
Vd
f
Vdd
L
V
d
D
c
r
r
m
r
rw
22 1
=
æ
è
ç
ö
ø
÷,,,

6.29 DpfDh dd=(,,,,,). wr
1 0
Repeating variables: D,,. wr
[] ,[],[],[],[] ,[],[]Dp
M
LT
DLhL
T
M
L
dLdL= = = = = = =
2 3 1 0
1
w r
p
rw
p p p
1 22 2 3
1
4
0
= = = =
Dp
D
h
D
d
D
d
D
, , , .
\ =
æ
è
ç
ö
ø
÷ =
Dp
D
f
h
D
d
D
d
D
W
rw
22 1
1 0
,,.& force ´ velocity = DpD D
2
´w.
\=
æ
è
ç
ö
ø
÷
&
,,.W Df
h
D
d
D
d
D
rw
35
1
1 0


6.30 TgfdHNh=(,,,,,,). , w rl Repeating variables: w r,,. d
[] ,[],[],[],[],[],[],[],[] .T
ML
T
f
T T
dLHL LN hL
M
L
= = = = = = = = =
2
2 3
1 1
1 w rl
p
rw
p
w
p p p p
1 25 2 3 4 5 6
= = = = = =
T
d
f H
d d
N
h
d
, , , , , .
l

\ =
æ
è
ç
ö
ø
÷
T
d
g
fH
d
N
h
drw w
25 1
,,,,.
d

l


6.31 QfHwg=(,,,,,). mrs Repeating variables: Hg,,. r
[] ,[],[],[] ,[] ,[] ,[] .Q
L
T
HLwLg
L
T
M
LT
M
L
M
T
= = = = = = =
3
2 3 2
m r s
\= = = =p p p
m
r
p
s
r
1
5
2 3
3
4 2
Q
gH
w
H gH gH
, , , .

111. 110
\ =
æ
è
ç
ç
ö
ø
÷
÷
Q
gH
f
w
H gH gH
5
1
3
2
, , .
m
r
s
r


6.32 dfVVD
j a
=(,,,,,,). srmr Repeating variables: VD
j
,,. r
[],[],[],[],[] ,[] ,[] ,[] .dLV
L
T
V
L
T
DL
M
T
M
L
M
LT
M
L
j a
= = = = = = = = s r m r
2 3 3

p p p
s
r
p
m
r
p
r
r
1 2 3 2 4 5= = = = =
d
D
V
V VD VD
j j j
a
, , , , .
\=
æ
è
ç
ç
ö
ø
÷
÷
d
D
f
V
V VDVD
j j j
a
1
2
, , , .
s
r
m
r
r
r


6.33 TfHhRt=(,,,, ).w mr , , Repeating variables: w r,,. h
[] ,[],[],[],[],[],[] ,[] .T
ML
T T
HLhLRLtL
M
LT
M
L
= = = = = = = =
2
2 3
1
w m r
p
rw
p p p p
m
rw
1 25 2 3 4 5 2
= = = = =
T
d
H
h
R
h
t
h h
, , , ,
\ =
æ
è
ç
ö
ø
÷
T
d
f
H
h
R
h
t
h hrw
m
rw
25 1 2
,,, .

6.34 m r=fDHgV(,,,,,) l . D= tube dia., H= head above outlet, l= tube length.
Repeating variables: DV
VD
H
D D
gD
V
,,. , , , rp
m
r
p p p
1 2 3 4 2
= = = =
l

\ =
æ
è
ç
ö
ø
÷
m
rVD
f
H
DD
gD
V
1 2
,, .
l


6.35 TfR er=(,,,, ). , , wr ml Repeating variables: R,,.wr
[] ,[],[],[] ,[],[],[] ,[].T
ML
T
RL
T
M
L
eLrL
M
LT
L= = = = = = = =
2
2 3
1
w r m l
p
rw
p p p p
m
rw
1 25 2 3 4 5 2
= = = = =
T
R
e
R
r
R R R
, , , ,
l


1
252
, , , .
Ter
f
RRRRR
m
rwrw
æö
\= ç÷
ç÷
èø
l


6.36 yfVy g
2 11
=(,,,). r Neglect viscous wall shear.
[],[],[],[] ,[] .yLV
L
T
yL
M
L
g
L
T
2 1 1 3 2
= = = = = r Repeating variables: Vy
1 1
,,. r

112. 111
p p
1
2
1
2
1
1
2
= =
y
y
gy
V
, . (r does not enter the problem).
\=
æ
è
ç
ö
ø
÷
y
y
f
gy
V
2
1
1
1
2
.

6.37 fgd V f
T
dL L
M
L
M
LT
V
L
T
= = = = = = =(,,,,).[],[],],[] ,[] ,[]. [ l lrm r m
1
3

Repeating variables: dV,,. r (l= length of cylinder).
p p p
m
r
m
r
1 2 3 1.
, , . , .= = = \=
æ
è
ç
ö
ø
÷
fd
V d Vd
fd
V
f
dVd

l l


6.38
Q
Q
V
V
p
p
V
V
F
F
V
V
m
p
mm
pp
m
p
mm
pp
pm
pp
mmm
ppp
= = =
l
l
l
l
2
2
2
2
22
22
, ,
()
()

D
D
r
r
r
r


t
t
r
r
r
r
r
r
m
p
mm
pp
m
p
mmm
ppp
m
p
mmm
ppp
V
V
T
T
V
V
Q
Q
V
V
= = =
2
2
23
23
32
32
, ,
&
&

l
l
l
l

(
&
Q has same dimensions as .)W
&


6.39 (A) ReRe.
mm
mp
m
VL
n
=
pp
p
VL
n
= . 129108 m/s.
p
mp
m
L
VV
L
\==´=

6.40 A)
5
6
1.5110
ReRe. . 410461 m/s.
1.3110
ppp
mmm
mpmp
mpmp
VLLVL
VV
L
n
nnn
-
-
´
==\==´=
´


6.41 a) ReRe. . .
m p
mm
m
pp
p
m
p
p
m
VdVd V
V
d
d
= = \ ==
n n
7

Q
Q
V
V
QQ
V
V
m
p
mm
pp
m p
m
p
m
p
= \ = =´´=
l
l
l
l
2
2
2
2 2
157
1
7
0214. . . . m/s
3


&
&
.
&
.
W
W
V
V
W
m
p
mmm
ppp
m
= =´= \ =´ =
r
r
32
32
3
2
7
1
7
7 72001400
l
l
kW
b) ReRe.
.
.
..
m p
m
p
p
m
m
p
V
V
d
d
= \ = =´=
n
n
7
9
13
485

3
2
3
2
1
1.54.850.148 m/s.
7
1
4.85200466 kW
7
m
m
Q
W
=´´=
=´´=&

113. 112
6.42 a) ReRe. . .
m p
mm
m
pp
p
m
p
p
m
VdVd V
V
d
d
= = \ ==
n n
5

2
22
11
5. 800/5160 kg/s.
55
mmmm
mp
p
ppp
mV
mm
m V
r
r
==´\===
& l
&&
&
l


D
D
D D
p
p
V
V
p p
m
p
mm
pp
m p
= = \ = =´ =
r
r
2
2
2
5 25 2560015. . 000 kPa
b) ReRe.
.
.
..
m p
m
p
p
m
m
p
V
V
d
d
= \ = =´ =
n
n
5
8
114
351

2
2
1
8003.51112 kg/s. 6003.517390 kPa .
5
mm
mp=´´=D=´=&

6.43 a) ReRe. . .
m p
mm
m
pp
p
m
p
p
m
VdVd V
V
d
d
= = \ ==
n n
10

F
F
V
V
FF
m
p
mmm
ppp
m p
= =´ \==
r
r
22
22
2
10 10
l
l
1
10
=1. lb
2
.
b) ReRe.
.
.41
..
m p
m
p
p
m
m
p
V
V
d
d
= \ = =´ =
n
n
10
106
1
752

p
pmFF
r
=
mr
22
2
222
1
101017.68 lb.
7.52
pp
mm
VL
VL
=´´=

6.44 ReRe. . .
m p
mm
m
pp
p
m
p
p
m
m
p
m
p
V V V
V
= = \ = = = assuming
l l l
ln n
n
n
n
n
10 1
\= =V V
m p
10 1000 km/hr.
This velocity is much too high for a model test; it is in the compressibility
region. Thus, small-scale models of autos are not used. Full-scale wind
tunnels are common.

6.45 ReRe. . .
m p
mm
m
pp
p
m
p
p
m
m
p
V V V
V
= \ = \ =
l l l
ln n
n
n

Water:
V
V
V V
m
p
p
m
m p m p
== = \= =
l
l
10 10 900 assuming km/hr.nn.
Air: VV
m p
p
m
m
p
= =´
´
´
=
-
-
l
l
n
n
9010
1510
110
13
5
6
.
500 km/hr.
Neither a water channel or a wind tunnel is recommended. Full-scale
testing in a water channel is suggested.

114. 113
6.46 ReRe. . / / .
m p
mm
m
pp
p
m p p m m p
V V
VV= = \ = = = if
l l
ll
n n
nn10
\=´=V
m1050500 m/s.
This is in the compressibility range so is not recommended. Try a water channel for the
model study. Then

6
5
110
100.662.
1.510
p
mm
pmp
V
V
n
n
-
-
´
==´=
´
l
l
33.1 m/s.
m
V\=
This is a possibility, although 33.1 m/s is still quite large.

22
2
222
() 10001
0.6623.56.
()1.23 10
Dmmmm
Dp
ppp
FV
F V
r
r
==´´=
l
l


6.47
5
3
1.0610
ReRe. . 2.51
5.510
ppp
mmm
mpmp
mpmp
VdVVd
dd
V
n
nnn
-
-
´
==\==´´
´
= 0.0048 ft.
Find
oil
n using Fig. B.2. Then

2
2
2
1.94
11.11.
1.940.9
mmm
p
pp
pV
p V
r
r
D
==´=
D´


6.48 ReRe. . .. .
m p
mm
m
pp
p
m p
p
m
m
p
p
m
V V
VV= = \= =´´ ´
-

l l l
l
l
ln n
n
n
0102510
3

If l
l
l
p
p
m
m
V@ = = =5
5
0025
2000 0005 cm, then and m/s.
.
.
We could try l
p m
V@ =50 005 cm, but m/s.. Each of these V
m'sis quite
small — too small for easy measurements. Let’s try a wind tunnel. Then,
VV
m p
p
m
m
p
p
m
p
= =´´ ´
´
´
= =
-
-
-
l
l
l
l
l
n
n
0102510
110
1810
028 5
3
3
5
..
.
. m/s if cm. Or,
if l
p m
V= =50 28 cm, m/s.. This is a much better velocity to work with
in the lab. Thus, choose a wind tunnel.

6.49 ReRe. . . . .
m p
mm
m
pp
p
m p
m
mm
p
pp
m
p
V V V
g
V
g
V
V
= \ = = \ = \ = FrFr
l l
l ln n
2 2
1
30


V
V
m
p
p
m
m
p
m
p
m
p
m
= = = \ = \=´
-
l
l
n
n
n
n
n
n
n30
1
30
1
164
6110
9
. . . . m/s Impossible!
2


6.50 (C)
2
FrFr.
m
mp
mm
V
lg
=
2
p
pp
V
lg
=
1
.20.5 m/s.
4
m
mp
p
l
VV
l
\==´=

115. 114
6.51 (A) From Froude’s number
m
mp
p
l
VV
l
= . From the dimensionless force we have:

22
**2
222222
or .102525156000 N.
pppm
mppm
mmmpppmm
FVlF
FFFF
VlVlVlrr
==\==´´=

6.52 FrFr m/s
m p
m
mm
p
pp
m p
m
p
V
g
V
g
VV= \ = \= = =. . . .
2 2
10
1
60
129
l l
l
l


()
()
.() () . .
F
F
V
V
F
V
V
F
Dm
Dp
mmm
ppp
Dp
p
m
p
m
Dm= \ =´ =´´= ´
r
r
22
22
2
2
2
2
2 6
60601021610
l
l
l
l
N

6.53 FrFr
m p
m
mm
p
pp
m
p
m
p
V
g
V
g
V
V
= = \ =. .
2 2
l l
l
l
.
a)
Q
Q
V
V
QQ
V
V
m
p
mm
pp
m p
m
p
m
p
= \ = =´ ´ =
l
l
l
l
2
2
2
2 2
2
1
10
1
10
000632. . . m/s
3

b)
F
F
V
V
FF
V
V
m
p
mmm
ppp
p m
p
m
p
m
= \= =´´=
r
r
22
22
2
2
2
2
2
121010
l
l
l
l
. . 12 000 N


6.54 Neglect viscous effects. FrFr fps
m p
m
p
m
p
p
V
V
V= \ = = \=. . ..
l
l
1
10
632

F
F
V
V
FF
V
V
m
p
mmm
ppp
p m
p
m
p
m
= \= =´´=
r
r
22
22
2
2
2
2
2
081010800
l
l
l
l
. . . lb

6.55 Neglect viscous effects, and account for wave (gravity) effects.
FrFr
m p
m
mm
p
pp
m
p
m
p
m
p
m m
p p
V
g
V
g
V
V
V
V
= \ = \ = =. . .
/
/
.
2 2
l l
l
l
l
l
w
w

\ = = ´ ´=ww
m p
m
p
p
m
V
V
l
l
600
1
10
101897 rpm.

T
T
V
V
TT
V
V
m
p
mmm
ppp
p m
p
m
p
m
= \= =´´= ×
r
r
23
23
2
2
3
3
3
121010120
l
l
l
l
. . . 000 Nm

6.56 FrFr
6
100

m p
m
mm
p
pp
m
p
m
p
m
p
p
m
V
g
V
g
V
V
= \ = \ = = \=. . . . .
2 2
278
l l
l
l
l
l
l
l

116. 115
6.57 Check the Reynolds number:
Re .
p
pp
p
Vd
= =
´
=´
-
n
152
10
3010
6
6

This is a high-Reynolds-number flow.
Re
/
. .
m
=
´
= ´
-
2230
10
13310
6
5

This may be sufficiently large for similarity. If so,

&
&
. .
W
W
V
V
m
p
mmm
ppp
= = ´ = ´
-r
r
32
32
3
3 2
62
15
1
30
26310
l
l

\ =´ ´ =
-&
(.)/. .W
p2215263101633
6
kW

6.58 This is due to the separated flow downwind of the stacks, a viscous effect.
\Re is the significant parameter. Re
.
. .
p
=
´
´
= ´
-
104
1510
26710
5
5
This is a
high-Reynolds-number flow. Let’s assume the flow to be Reynolds
number independent above Re=´510
5
(see Fig. 6.4). Then

5
5
4/20
Re510. 37.5 m/s.
1.510
m
mm
V
V
-
´
=´=\³
´


6.59 Re
.
. .
p
=
´
´
= ´
-
2010
1510
13310
5
6
This is a high-Reynolds-number flow.
Let Re
.4
.
. .
m
m
m
V
V= =
´
´
\³
-
10
1510
375
5
5
m/s for the wind tunnel.
Re
.
. .
m
m
m
V
V= =
´
´
\³
-
10
1
110
10
5
6
m/s for the water channel.
Either could be selected. The better facility would be chosen.

F
F
V
V F
F
m
m
mmm
mmm m
m
1
2
1 1 1
2 2 2 2
2
22
22
2
2
2
2
32
32
1000
123
2
15
1
416= = \ = ´=
r
r
l
l
.
. .
.
.4.
.4
.. N

&
&
.
.
&
( .)
.
.
W
W
V
V
W
m
p
mmm
ppp
p
= =
´
´
\ =´ ´ =
r
r
32
32
3 2
3 2
3
3
2
2
154
2010
1532
20
15
10
4
71
l
l
100 W

6.60 Re is the significant parameter. This is undoubtedly a high-Reynolds-
number flow. If the model is 4' high then
l
l
p
m
=250,and the model’s diameter is
45/250 = 0.18'. For Re ,
m
=´310
5
we have
Re
.
.
.
m
m
m
V
V=´=
´
´
\³
-
310
18
1510
250
5
4
fps,and a study is possible.

117. 116
6.61 Mach No. is the significant parameter. M M
m p
=.
a) MM. . 200 m/s.
p
m
mpmp
mp
VV
VV
cc
=\=\==

F
F
V
V
F
m
p
mmm
ppp
p
= \=´´ =
r
r
22
22
2 2
101204000
l
l
. . N
b)
255.7
200186 m/s.
296
pp
pmm
mm
cT
VVV
cT
====
FF
V
V
p m
mpp
mmm
= =´´ ´=
r
r
22
22
2
2
2
10601
186
200
202080
l
l
. . N
c)
223.3
200174 m/s.
296
pp
pmm
mm
cT
VVV
cT
====
FF
V
V
p m
p
m
p
m
p
m
= =´´ ´ =
r
r
2
2
2
2
2
2
2
10338
174
200
201023
l
l
. . N

6.62
273
MM. . 250276 m/s.
223.3
p
m
mpm
mp
VV
V
cc
=\=\==

223.3
. 290262 m/s.
273
mmm
p
ppp
VcT
V
VcT
==\==

p
p
V
V
pp
V
V
m
p
mm
pp
p m
p
m
p
m
o
o
= \= = =
r
r
r
r
r
r
2
2
2
2
2
2
80
338
8
262
290
346.
.
.
. . kPa, abs
a
p=5
o
for similarity. (Note: we use r
m
at 2700 m where T = 0°C.)

6.63 a) FrFr
m p
m
mm
p
pp
m
p
m
p
V
g
V
g
V
V
= = \ =. . .
2 2
l l
l
l


w
w
w
m
p
m
p
p
m
m
V
V
= = ´ \ = ´ =
l
l
1
10
10 2000
10
10
6320. . rpm
b) ReRe. . .
m p
mm
m
pp
p
m
p
p
m
V V V
V
= = \ ==
l l l
ln n
10

w
w
w
m
p
m
p
p
m
m
V
V
= =´= \ =
l
l
10
1
10
1 2000. . rpm

6.64 There are no gravity effects nor compressibility effects. It is a high-Re
flow.
T
T
V
V
TT
V
V
m
p
mmm
ppm
p m
p
m
p
m
= \= =´ ´= ×
r
r
23
23
2
2
3
3
2
2
3
12
15
60
10750
l
l
l
l
. . Nm

118. 117

w
w
ww
m
p
m
p
p
m
p m
p
m
m
p
V
V
V
V
= \ = = ´´=
l
l
l
l
. . . rpm500
15
60
1
10
125

6.65 ReRe. .
m p
mm
m
pp
p
m p
p
m
V V
VV= \ = \= =´= m/s.
l l l
ln n
1010100
This is too large for a water channel. Undoubtedly this is a high-Re
flow. Select a speed of 5 m/s. For this speed,
Re
.
,
m
=
´
´
=´
-
501
110
510
6
5
where we used 0.1 (1 m,
mp
==ll i.e., the
dia. of the porpoise). w w
m p
m
p
p
m
V
V
= =´´=
l
l
1
5
10
105 motions/second.

6.66 r
r
r
* * * * * *
, , , , , .= = = = = =
o
ttfu
u
V
v
v
V
x
x
y
y

l l
Substitute in:

*****
o***
()()
0.
oo
VuVv
f
txy
¶r¶r¶r
rrr
¶¶¶
++=
ll

Divide by r
oV/:l
\ + + =
f
Vt x
u
y
v
f
V
l l¶r
¶
¶
¶
r
¶
¶
r
*
* *
**
*
**
() (). .0 parameter =
6.67
*********
2
, , , , , , , , .
Vuvwxyzp
Vuvwxyzpttf
UUUU Ur
=========
v
v
lll

Substitute into Euler’s equation and obtain:
Uf
V
t
U
u
V
x
U
v
V
y
U
w
V
z
U p¶
¶
¶
¶
¶
¶
¶
¶
r
r
v
l
v
l
v
l
v
l
v
*
*
*
*
*
*
*
*
*
*
*
**
.+ + + =-
Ñ
2 2 2 2

Divide by U
2
/:l

f
U
V
t
u
V
x
v
V
y
w
V
z
p
l
v v v v
¶
¶
¶
¶
¶
¶
¶
¶
*
*
*
*
*
*
*
*
*
*
*
**
.+ + + =-Ñ Parameter =
f
U
l


6.68
v
v
l
v
l
v
l
V
V
U
t
tU
p
p
U
h
h
* * * * *
, , , , .= = Ñ=Ñ = =
r
2
Euler’s equation is then
r r r
UDV
Dt
U
p g h
2 2
l
v
l
v
l
l
v
*
*
** **
.=- Ñ- Ñ
Divide by rU
2
/:l

DV
Dt
p
g
U
h
v
v
l
v
*
*
** **
.=-Ñ-Ñ
2
Parameter =
g
U
l
2
.

6.69 There is no y- or z-component velocity so continuity requires that¶¶ux/ .=0There
is no initial pressure distribution tending to cause motion so¶¶px/ .=0The

119. 118
x-component Navier-Stokes equation is then

uu
u
tx
¶¶
¶¶
+ v+
u
w
y
¶
¶
+
1up
zx
¶¶
¶r¶
=-
2
2x
u
g
x
¶
n
¶
++
22
22
uu
yz
¶¶
¶¶
++
æö
ç÷
ç÷
èø

(wide plates)
This simplifies to

¶
¶
n
¶
¶
u
t
u
y
=
2
2
.
a) Let uuUyyhttUh
* * *
/, / /.= = = and Then

U
h
u
t
U
h
u
y
2
2
2
¶
¶
n¶
¶
*
*
*
*2
=
The normalized equation is

¶
¶
¶
¶
u
t
u
y
*
*
*
*2
Re
=
1
2
where Re=
Uh
n

b) Let uuUyyhtth
* * *
/, / /.= = = and Thenn
2


n¶
¶
n
¶
¶
U
h
u
t
U
h
u
y
2 2
2*
*
*
*2
=
The normalized equation is

¶
¶
¶
¶
u
t
u
y
*
*
*
*2
=
2

6.70 The only velocity component is u. Continuity then requires that ¶¶ux/=0
(replace z with x and vz with u in the equations written using cylindrical
coordinates). The x-component Navier-Stokes equation is

r
u
v
t
¶
¶
+
vu
r
q¶
¶
+
uu
u
rx
¶¶
¶q¶
+
1
x
p
g
x
¶
r¶
=-+
22
222
11uuu
rrrr
¶¶¶
n
¶¶¶q
+++
2
2
u
x
¶
¶
+
æö
ç÷
ç÷
èø

This simplifies to

¶
¶ r
¶
¶
n
¶
¶
¶
¶
u
t
p
x
u
rr
u
r
=- + +
æ
è
ç
ö
ø
÷
1 1
2
2

a) Let
****2*
/, /, /, =/ and /:uuVxxdttVdppVrrd r====

V
d
u
t
V
d
p
x
V
d
u
r r
u
r
2 2
2
2
1¶
¶
r
r
¶
¶
n¶
¶
¶
¶
*
*
*
*
*
*2 *
*
*
=- + +
æ
è
ç
ö
ø
÷
The normalized equation is

¶
¶
¶
¶
¶
¶
¶
¶
u
t
p
x
u
r r
u
r
*
*
*
*
*
*2 *
*
*
Re
=-+ +
æ
è
ç
ö
ø
÷
1 1
2
where Re=
Vd
n

b) Let
***2*2*
/, /, /, =/ and /:uuVxxdttdppVrrd nr====

n¶
¶
r
r
¶
¶
n¶
¶
¶
¶
V
d
u
t
V
d
p
x
V
d
u
r r
u
r
2
2
2
2
1
*
*
*
*
*
*2 *
*
*
=- + +
æ
è
ç
ö
ø
÷
The normalized equation is

120. 119

¶
¶
¶
¶
¶
¶
¶
¶
u
t
p
x
u
rr
u
r
*
*
*
*
*
*2 *
*
*
Re=- + +
2
1
where Re=
Vd
n


6.71 Assume w
z
= =0 0 and
¶
¶
. The x-component Navier-Stokes equation is then

u
t
¶
¶
u
uv
x
¶
¶
++
uu
w
yz
¶¶
¶¶
+
1p
x
¶
r¶
=-
222
222x
uuu
g
xyz
¶¶¶
n
¶¶¶
++++
æö
ç÷
ç÷
èø

With gg
x= the simplified equation is
u
u
x
g
u
x
u
y
¶
¶
n
¶
¶
¶
¶
=+ +
æ
è
ç
ö
ø
÷
2
2
2
2

Let uuVxxhyyh
* * *
/, / /.= = = and Then

V
h
u
u
x
g
V
h
u
x
u
y
2
2
2 2
*
*
*
*
*2
*
*2
¶
¶
n
¶
¶
¶
¶
=+ +
æ
è
ç
ö
ø
÷
The normalized equation is
u
u
x
u
x
u
y
*
*
*
*
*
*
*
Re
¶
¶
¶
¶
¶
¶
= + +
æ
è
ç
ö
ø
÷
11
2
2
2
2
2
Fr
where Fr and = =
V
hg
Vh
Re
n

6.72
******222
, , , , , .
o
uvTxy
uvTxy
UUT
=====Ñ=Ñ l
ll

r
¶
¶
¶
¶
c
UTT
x
UTT
y
K
TT
p
o o
o
l l l
*
*
*
*
**
.+
é
ë
ê
ù
û
ú= Ñ
2
2

Divide by rcUT
p o
/:l

¶
¶
¶
¶ r
T
x
T
y
K
cU
T
p
*
*
*
*
**
.+ = Ñ
l
2
Parameter =
K
cU
p
m
m
rl
=
11
PrRe
.

6.73 r
r
r
* * * * * * *
, , , , , , .= = = Ñ=ÑÑ=Ñ = =
o o o
V
V
U
t
tU
p
p
p
T
T
T

v
v
l
v
l
v
l
1 1
2
2
2

momentum: rr
m m
o
oUDV
Dt
p
p
U
V
U
V
*
*
*
** ** * * *
( ).
2
2
2
2
3l
v
l
v
l
v
l
vvv
=-Ñ+ Ñ + ÑÑ×
Divide by r
o
U
2
/:l
[ ]r
r
m
r
*
*
*
** ** * * *
( ).
DV
Dt
p
U
p
U
V V
o
o o
v
v
l
vvvv
=- Ñ+ Ñ +ÑÑ×
2
2

energy: rr
*
*
*
** ** *
.cT
UDT
Dt
K
TTp
U
pV
voo o o
l l l
vv
= Ñ - Ñ×
2
2

Divide by r
ovo
cTU/:l
r
r r
*
*
*
** ** *
.
DT
Dt
K
cU
T
p
cT
pV
ov
o
ovo
= Ñ - Ñ×
l
vv
2

121. 120
The parameters are:
p
U
RT
U
kRT
kU
c
kU k
o
o
o o
r
2 2 2
2
2 2
1
= = = =
M
.
m
r
o
Ul
=
1
Re
.
K
cU
K
c
c
cU
K
ov p
p
vor m
m
rl l
= =
PrRe
.



p
cT
RT
cT
cc
c
K
o
ovo
o
vo
p v
v
r
= =
-
=-1.
The significant parameters are K,,Re,Pr. M

122. 162
CHAPTER 8

External Flows



8.1


8.2
Re .
.
. .== \=
´ ´
= ´
-
-
5
515110
20
37810
5
5VD
D
n
m


8.3

8.4


8.5 (C)

8.6 (C)

A
B
C
A-B: favorable
B-C: unfavorable
A-C: favorable
separated
region
inviscid
flow
boundary layer
near surface
inviscid
flow
viscous flow
near sphere
no separation
separation
wake
separation
separated
region
building
inviscid
flow
boundary
layer
wake
A-B: favorable
B-C: unfavorable
A-D: favorable
C-D: undefined
A
BC
D
separated
flow

123. 163
8.7 ( B)
6
0.80.008
Re4880.
1.3110
VD
n
-
´
===
´


8.8 5 5 512210
8
12
000915
5
= \= =´ ´ =
-VD
V D V
n
n a) fps./. . /
.
.
b) V=´ ´ =
-
538810
8
12
000291
5
. /
.
. fps. c) V=´´ =
-
51610
8
12
0012
4
. /
.
. fps.

8.9 Re
.
. .= =
´
´
= ´
-
VD D
D
n
20
15110
132510
5
5

a) Re . . .= ´ ´=´13251067910
5 6
\Separated flow.
b) Re . . . .= ´´=´132510067910
5 4
\Separated flow.
c) Re . . .= ´´ =1325100067950
5
\Separated flow.

8.10 F pdApA p rrdrp p
D
A
= - = - = -
æ
è
ç
ö
ø
÷=
òò backback
front
0
2
0 0
0
1
1 2 2
1
2
1
4
1
2
( )p p p
Bernoulli: p Vp p
¥ ¥
+ = \=´ ´ =
1
2
1
2
12120242
2
0 0
2
r . . Pa.
\= =F
D
1
2
242380p() N
C
F
VA
D
D
= =
´
´´´
=
1
2
2
2 2
2380
12120 1
05
r
p.
.

8.11 F F F
total bottom top
000.3.3+10 000 N.= + = ´´ ´´=20 332700..
F
lift
cos 10 N= =2700 2659
o

F
drag 10 N= =2700 469sin
o

C
F
VA
L
L
= =
´
´´´
=
1
2
2
2
22659
1000533
236
r
..
.
C
F
VA
D
D
= =
´
´´´
=
1
2
2
2
2469
1000533
0417
r
..
.
8.12 FpA LwFpA
Lw
Lw
u uul ll o
= = ´ = = ´ =26 8000
25
4015 000 .
cos

FF F Lw
L u
= - =
l
o o
cos cos5 1021 950
FF F Lw
D u
= - =
l
o o
sin sin5 101569

124. 164
C
F
VA
Lw
Lw
L
L
= =
´
´
=
1
2
2
2
221
3119750
025
r
950
.
.
C
F
VA
Lw
Lw
D
D
= =
´
´
=
1
2
2
2
21569
3119750
00179
r
.
.

8.13 If C
D
=10. for a sphere, Re = 100 (see Fig. 8.8). \
´
=
V.
,
1
100
n
n V=1000.
a) V F
D
= ´ ´ = \=´ ´ ´ ´
-
1000146100146
1
2
1220146 0510
5 2 2
. . .. . . m/s. p
= ´
-
32510
7
. . N
b) V F
D
= ´
´
´
= \=´ ´ ´ ´ ´
-
1000
14610
015122
0798
1
2
015122798 0510
5
2 2.
. .
. (. .). . . m/s. p
= ´
-
45810
5
. . N
c) V F
D
= ´ ´ = \=´ ´ ´ ´
-
10001311000131
1
2
1000001310510
6 2 2
. . . . . m/s. p
= ´
-
67410
6
. . N

8.14 a) Re
.
.
. .= =
´
´
=´ \ =
-
VD
C
D
n
65
1510
210 045
5
5
from Fig. 8.8.
\= =´ ´´´´=F VAC
D D
1
2
1
2
1226 2545194
2 2 2
r p N.. .. .
b) Re
.
.
. .=
´
´
=´ \ =
-
155
1510
510 02
5
5
C
D
from Fig. 8.8.
\= =´ ´´´´=F VAC
D D
1
2
1
2
12215 25254
2 2 2
r p N.. ...

8.15 (B) Assume a large Reynolds number so that CD = 0.2. Then

2
2211801000
1.2350.24770 N.
223600
DFVACrp
´æö
==´´´´´=
ç÷
èø


8.16 (D) Assume a Reynolds number of 10
5
. Then CD = 1.2.

2211
.601.234041.2.0.0041 m.
22
D
FVACDDr=\=´´´´´\=

5
6
400.0041
Re1.6410.1.2. The assumption was OK.
10
D
VD
C
n
-
´
===´\=

125. 165
8.17 The velocities associated with the two Re's are
V
D
1
1
5 5
3101510
0445
101= =
´ ´´
=
-
Re .
.
n
m/s,
V
D
2
2
4 5
6101510
0445
20= =
´ ´´
=
-
Re .
.
n
m/s.
The drag, between these two velocities, is reduced by a factor of 2.5
() ()[ ]C C
D Dhigh low
and = =05 02. .. Thus, between 20 m/s and 100 m/s the drag is
reduced by a factor of 2.5. This would significantly lengthen the flight of the
ball.

8.18 a) F VAC V C VC
D D D D
= \=´ ´
æ
è
ç
ö
ø
÷ \ =
1
2
05
1
2
00238
2
12
4810
2 2
2
2
r p . . . . .
Re
/
.
. .: .= =
´
´
= = = ´
-
VDV
V C V
D
n
412
1610
2080 5 98
4
Try fps, Re=210
5

Try fps, Re=2.310
5
C V
D
= = ´.: .4 110
b) C V V
D
= =´ ´
æ
è
ç
ö
ø
÷´ \=02
1
2
00238
2
12
2 155
2
2
.: . .. 0.5 fps.p

8.19 42
1
2
1000 1 0267
2
10
210
2 2 2
6
5
. . . ..Re
.
.=´ ´ \ = =
´
=´
-
V C VC
V
V
D D
p
Try C V
D
= \= = ´ \05 073 14610
5
.: . Re. . m/s. OK.

8.20 Re
.
. . .= =
´
´
=´ \ =
-
VD
C
D
n
402
1510
5310 07
5
6
. (This is extrapolated from
Fig. 8.8.) \=´ ´´´´=F
D
1
2
12240260781
2
. ( ). . 900 N
M = 81 900 ´ 30 = 24610
6
. .´ × Nm

8.21 a) Re
.
.
. .Re . .Re . .
1 5
5
2
5
3
52505
10810
1210 1810 2410=
´
´
=´ =´ =´
-
Assume a
rough cylinder (the air is highly turbulent).
() () ()\ = = =C C C
D D D1 2 3
07 08 09., ., ..
\=´ ´ ´´+ ´´+´´=F
D
1
2
145250510707515812091380
2
. (. .. .. .) . N
M=´ ´ ´´´+ ´´´ +´´´= ×
1
2
14525051074007515827512091025700
2
. (. . . . .. . ) . Nm
b) Re
.
.
. .Re . ,Re . .
1 5
4
2
5
3
52505
16510
7610 11410 1510=
´
´
=´ = ´ =´
-

126. 166
() () ()\ = = = =
´
=C C C
D D D1 2 3
8 7 8
101
287308
117., ., ..
.
. . kg/m
3
r
\=´ ´ ´´+ ´´+´´=F
D
1
2
117250510807515712081020
2
. (. .. .. .) . N
M=´ ´ ´´´+ ´´´ +´´´= ×
1
2
11725051084007515727512081019
2
. (. . . . .. . ) . 600 Nm

8.22 Atmospheric air is turbulent. \Use the "rough" curve. \ =C
D
07..
F V D VD
V V
D
==´ ´´ \ =
´
´
-
10
1
2
00238 67 2000 10
2000
1610
2 5
2
4
. .. .
/
.
. =
2


2222
mino
0.0024
= 3010411.8 psf.
22
pUv
r
¥
éùéù\=--=-
ëûëû

\ = \= =VD V D
2
2370 148 0108. . .'. fps

8.23 Since the air cannot flow around the bottom, we
imagine the structure to be mirrored as shown. Then
LD C C
D D
/ / . . .= = \ =
¥
4058 066
Re
.
. ....
min
min
= =
´
´
=´ \=´=
-
VD
C
D
n
302
1510
410 106666
5
6

\=´ ´ ´
+
´
æ
è
ç
ö
ø
÷´=F
D
1
2
12230
28
2
206636
2
. . . 000 N



8.24 .
BDW
FFF+=

3223414
9810100098107.82.
323
D
rVrCrppp´+´=´´
Re .=
´
=´
-
Vr
Vr
2
10
210
6
6
\ =VC r
D
2
178


a) r VVC
D
= \ =. , ..05 89
2
m. Re=10
5
Assume a smooth sphere.
Try C V
D
=\= ´.: . .5 422 m/s. Re=4.2210
5
This is too large for Re.
Try C V
D
=\= ´.: . .2 667 m/s. Re=6.6710
5
OK.
b) r VVC
D
= ´ =. , ..025 445
2
m. Re=510
4

Try C V
D
= = ´.: . .2 472 m/s. Re=2.410
5
OK.
c) r VVC
D
= =. , ..005 089
2
m. Re=10
4

Try C V
D
= = ´.: . .5 133 m/s. Re=1.3310
4
OK.
d) r VVC
D
= ´ =. , ..001 0178
2
m. Re=210
3

Try C V
D
= = ´.: . .4 067 m/s. Re=1.3310
3
OK.

W
F
B
F
D

127. 167
8.25
323
2410110410
. .077.0023862.4.
312212312
BDWDFFFVCS ppp
æöæöæö
+=´+´=
ç÷ç÷ç÷
èøèøèø


32
4
10/12
Re5.210. 1.0139810
1.610
D
V
VVCS
-
´
==´\+=
´

a) S VC
D
= =.. .005 219
2
Assume atmospheric turbulence, i.e., rough.
Try C V C V
D D
= = =´ \ = =.: ..Re. . .. .4 234 1210 3 27
5
fps fps
b) S VC C V
D D
= = = = =´ \.. . .: .Re. .02 1090 4 52 2710
2 5
Try fps OK.
c) S VC C V
D D
= = = =10 58 4 381
2
.. . .: . 200 Try fps

8.26
Assume a 180 lb, 6' sky diver, with components
as shown. If V is quite large, then Re > 2 ´ 10
5
.
FF
D W
=.


1
2
0023823
1
2
107 25
1
2
107
18
12
2510
4
12
4180
2
´ ´´´´ ´´´´+´´+´
æ
è
ç
ö
ø
÷´
é
ë
ê
ù
û
ú
=. .. . .. .. . .V +2 p
We used data from Table 8.1. \=V140 fps.

8.27 From Table 8.2 C F V V
D D
= =´ ´´ =035
1
2
122 32035683
2 2
.. . ... .
a) F W
D
= ´
´æ
è
ç
ö
ø
÷= \=
´
=.
&
.683
801000
3600
337 337
801000
3600
7500
2
N. W or 10 Hp
b) V F W
D
= = ´ = \= ´=25 68325427 4272510
2
m/s. N. 700 W or 14.3 Hp.
&
.
c)V F W
D
= = ´ = \= ´ =278 683278527 52727814
2
. . .
&
. . m/s. N. 700 W or 19.6 Hp

8.28 12 11400
1
2
11
2
. . . . .F F VAC C
D D D D
=´ = = r
12
1
2
122 231111400
2
. . ().. .´´ ´´´=´V
\=V95. . m/s


8.29 Re
(
. . .= =
´
= ´ \ =
VD
C
D
n
40
44210 035
5 000/3600)0.6
1.5110

-5
from Fig. 8.8.
a) F VAC
D D
= =´ ´ ´´´ =
1
2
1
2
120440 066035936
2
r . ( . . . 000/3600) N
2

b) F LD
D
= ´ = = =936068637 60610.. . / /. . N where
c) F LD
D
= ´ = =936076711 20.. . / N where we can use since only one end is
free. The ground acts like the mid-section of a 12-m-long cylinder.
3 ft
6 in
8 in. dia.
2.5 ft
6 in
2.5 ft
18 in
1.1 m
1.2 m
F
W
F
D
F
y
F
x

128. 168

8.30 a) Curled up, she makes an approximate sphere of about 1.2 m in diameter (just
a guess!). Assume a rough sphere at large Re. From Fig. 8.8, C
D
=04.:
F VAC
D D
=
1
2
2
r
8098
1
2
121 0604 537
2 2
´=´ ´ ´ ´ \=. . . .. . .V Vp m/s
Check Re: Re
..
.
. .=
´
´
= ´ \
-
53712
15110
42710
5
6
OK.
b) F VAC
D D
=
1
2
2
r . From Table 8.2, CD = 1.4:
8098
1
2
121 414 429
2 2
´=´´ ´´ \=. . .. . .V Vp m/s
Check Re: Re
.
.
. .=
´
´
= ´
-
4298
15110
22710
5
6
Should be larger but the velocity
should be close.
c)
21
2
DD
FVAC r=

221
809.81.2111.4. 17.2 m/s.
2
VVp´=´´´´\=
Check Re: Re
.
.
. .=
´
´
=´
-
1721
15110
11410
5
6
This should be greater than 10
7
for
C
D
to be acceptable. Hence, the velocity is approximate.

8.31 With the deflector the drag coefficient is 0.76 rather than 0.96. The required
power (directly related to fuel consumed) is reduced by the ratio of 0.76/0.96.
The cost per year without the deflector is
Cost = (200 000/1.2) ´ 0.25 = $41,667.
With the deflector it is
Cost = 41,667 ´ 0.76/0.96 = $32,986.
The savings is $41.667 - 32,986 = $8,800.

8.32
2211
.0023888(62)1.1122 lb.
22
DD
FVAC r==´´´´´=

&
, .WFV
D
=´= ´=1228810700
ft-lb
sec
or 19.5 Hp

8.33 F VAC
D D
= =´ ´ ´ ´´´=
1
2
1
2
12227816 05111043
2 2 2
r p.(..) . . . N.
& .(..) .WFV
D=´´= ´ ´´=21043278162226 W or 1.24 Hp

129. 169
8.34 The projected area is
(.)
..
203
2
446
+
´= m
2

F VAC
D D
= =´ ´ ´´=
1
2
1
2
118204604434
2 2
r . .. N.
Since there are two free ends, we use Table 8.1 with LD/ /. .,= =4115347 and
approximate the force as
F
D
= ´ =434062269. . N

8.35 The net force acting up is (use absolute pressure)

33
up
44120
0.41.219.80.50.49.82.16 N
332.077293
F pp=´´´--´´=
´

From a force triangle (2.16 N up and F
D
to the right), we see that
tan /.a=FF
Dup

a) F
D
= =216 800381./tan ..
o

0381
1
2
121 0402 250
2 2
. . . .. .=´ ´ ´ \=V Vp m/s.
Check Re: Re
..
.
. .=
´
´
= ´
-
2508
15110
13310
5
5
Too low. Use C
D
=05.:
0381
1
2
121 0405 158
2 2
. . . .. .=´ ´ ´ \=V Vp m/s
b) F
D
= =216 700786./tan ..
o

0786
1
2
121 0402 360
2 2
. . . .. .=´ ´ ´ \=V Vp m/s.
Check Re: Re
..
.
. .=
´
´
=´
-
3608
15110
1910
5
5
Too low. Use C
D
=05.:
0786
1
2
121 0405 227
2 2
. . . .. .=´ ´ ´ \=V Vp m/s
c) F
D
= =216 60125./tan ..
o

125
1
2
121 0405
2 2
. . . .. .=´ ´ ´ \=V Vp 2.86 m/s
Check Re: Re
. .
.
. .=
´
´
=´
-
28608
15110
1510
5
5
\OK.
d) F
D
= =216 50181./tan ..
o

181
1
2
121 0405
2 2
. . . .. .=´ ´ ´ \=V Vp 3.45 m/s
Check Re:
5
5
3.450.8
Re1.810.
1.5110
-
´
==´
´
Close, but OK.

8.36 Assume each section of the tree is a cylinder. The average diameter of the tree is
1 m. The top doesn't have a blunt end around which the air flows, however,

130. 170
the bottom does; so assume LD/(/) .= ´=5225 So, use a factor of 0.62 from
Table 8.1 to multiply the drag coefficient. The force acts near the centroid of the
triangular area, one-third the way up. Finally,
Fd´=5000

1
2
121504062
5
3
065000
2
´ ´´
é
ë
ê
ù
û
ú
´+
æ
è
ç
ö
ø
÷= =.().. . . .V V 54.2 m/s

8.37 Power to move the sign:
FV VACV
D D
= ´
1
2
2
r
=´ ´ ´ ´´ =
1
2
1211111072111111657
2
. . . . . J/s.
This power comes from the engine:
65712 03 182510
4
= ´ ´ \= ´
-
( &.. &. 0001000) kg/s.m m
Assuming the density of gas to be 900 kg/m
3
,
182510103600652
1000
900
030
4
. .$683´ ´´ ´´´ ´ =
-


8.38 The power expended is FVV
D
´ =´ =. ( /)/. . m/s25886032811118

1
2
1211118056
1
2
121 04 08
3 3
´´ ´ ´=´´´´´. . . . . .C V C
D D

\=V1347. . m/s or 30.1 mph

8.39 & .W FV VACV ACV
D D D
=´ =´= ´=40746
1
2
1
2
2 3
h r r
\´´=´ ´´ \=407469
1
2
1223035 347
3
. . .. . .V V m/s or 125 km/hr

8.40 (C)
5
40.020.02
Re5000.St0.21.
41.610
VDfDf
Vn
-
´´
===\===
´


4 m/s
42 Hz (cycles/second).distance = 0.095 m /cycle.
42 cycles/s
V
f
f
\===

8.41
5
.003
40Re10 000. 40<10 000. 0.2<50 m/s.
1.510
V
V
-
´
<<<\<
´


low
.003
St = 0.12 = . 8 Hz.
.2
f
f
´
\=
St=.21=
.003
50
Hz.
lhigh
f
f
´
\ =. 3500
The vortices could be heard over most of the range.

131. 171
8.42
5
5
6
40. 8.1310 ft.
1.2210
VDD
D
n
-
-
>=\<´
´

10
6
12210
0020
5
000< ft or 0.24"
VD D
D
n
=
´
\>
-
.
. . .

8.43 From Fig. 8.9, Re is related to St. St=
fD
V V
´
=
´021..
.
Re
.
.
. .= =
´
´
= \
-
VD V
V
n
1
1510
0095
5
Try St=.21: m/s. Re=630.
This is acceptable. \=V0095. . m/s

8.44 St= Re= Use Fig. 8.9.
fD
V V
VDV
=
´
=
´
-
.
. .
0022 2
10
6
n

Try St=0.21: m/s Re=3810 OK.
3
V= ´ \. . .0191

8.45 Let S
t
= 0.21 for the wind imposed vorticies. When this frequency equals the
natural frequency, or one of its odd harmonics, resonance occurs:
fTLd=/rp
22


220.2110
30 000/78500.016. 0.525 m
0.016
LL p
´
=´´\=
Consider the third and fifth harmonics:
f TLd=3
22
/ .rp \=L1.56 m.
f TLd=5
22
/ .rp \=L2.62 m.

8.46 (C) By reducing the separated flow area, the pressure in that area increases
thereby reducing that part of the drag due to pressure.

Fig. 8.8 Table 8.1
8.47 Re
/
.
. . . ..=
´
´
=´ =´ ´ ´´´´
æ
è
ç
ö
ø
÷=
-
88612
1610
2810
1
2
00238881086
6
12
4
5 2
22 lb.F
D

The coefficient 1.0 comes from Fig. 8.8 and 0.8 from Table 8.1.
&
.WFV
D
=´=´=22881946 ft-lb/sec or 3.5 Hp
()C F W
D Dstreamlined
lb.
ft-lb
sec
or 0.12 Hp= \= =0035 077 678.. .
&
. .

8.48 Re
.
.
.= =
´
´
=
-
VD
n
308
1510
16
5
000 \=´ ´´ ´´´=F
D
1
2
1223008212780822
2
. (. ).. . N
The coefficient 1.2 comes from Fig. 8.8 and 0.78 from Table 8.1.
()C F
D D
streamlined
N. % reduction = = \= \
-
´=.. .
. .
.
.35 024
0822024
0822
100708%

132. 172

8.49 Re
.
. .= =
´
=´ \ =
-
VD
C
D
n
208
10
1610 045
6
6
. from Fig. 8.8.

L
D
C
D
= = \ = ´ =
4
08
5 062045028
.
. . . ..
Because only one end is free, we double the length.
F VAC
D D
= =´ ´´´´ =
1
2
1
2
10002082028
2 2
r . . .900 N
If streamlined, C
D
= ´ =00306200186. . ..
\=´ ´´´´ =F
D
1
2
1000208200186
2
. . .60 N

8.50 V=´ =5010003600139/ . m/s.
Assume the ends to not be free. \Use C
D
from Fig. 8.8.
()Re
..
.
. . .. .=
´
´
= ´ \= =
-
139002
1510
18510 12 03
5
4

streamlined
C C
D D


&
. . . . .WFV VAC
D D
=´= =´´ ´ ´´=
1
2
1
2
121390022012773
3 3
r W or 1.04 Hp

&
. . . .W
streamlined
W or 0.26 Hp=´´ ´ ´´=
1
2
121390022003193
3


8.51 V=´ =5010003600139/ . m/s. Re
..
.
. . .=
´
´
=´ \ =
-
13903
1510
2810 04
5
5
C
D

We assumed a head diameter of 0.3 m and used the rough sphere curve.
F
D
=´´ ´ ´=
1
2
12139 03404
2 2
. .(./). .p 3.3 N
F
D
=´´ ´ ´ =
1
2
12139 0340035
2 2
. .(./). .p 0.29 N

8.52 s
r
g=
-
=
-
´
=+ =
¥
¥
pp
V V
p hp
v
1
2
07
150
1000
150
2 2
..
0001670
1
2
where 000 Pa.
atm

\=V20.6 m/s.

8.53 C
F
VA
L
L
= =
´ ´´´
= \@
1
2
200
100012410
069 3
2 2
r
a
000
1
2

.
.. .
o

C
F
F
D
D
D
= =
´ ´´´
\= N.
.
. .0165
1
2
100012410
4800
2

133. 173
s
crit

000)1670
1
2
=
>
´+ -
´ ´
=.
?( .
..75
98104101
100012
143
2
\No cavitation.

8.54 C
F
VA
L
L
= =
´ ´´´
= \=
000
1
2

1
2
50
19435
16
12
30
105 73
2 2
r
a
.
.. ..
o

C
F
F
D
D
D
= =
´´´´
\= lb.
.
. .027
1
2
19435
16
12
30
1280
2

s
crit

1
2
=
>
´ + -´
´ ´
=16
6241612211725144
19435
182
2
.
?. / .
.
.. \No cavitation.

8.55 p p
v¥
= ´+ =
´
=´98105101 1670 1610
6
000=150 000 Pa. Pa. Re=
20.8
10
-6
.
s s=
-
´ ´
= \ = +=+=
150
100020
074 01 317452
2
0001670
1
2
.. ()().(.).CC
D D

\= =´ ´ ´´´=F VAC
D D
1
2
1
2
100020 452
2 2 2
r p.. .52 000 N
Note: We retain 2 sig. figures since C
D
is known to only 2 sig. nos.

8.56 For a 6° angle of attack we find from Table 8.4 C
L
=095..
F VAC L
L L
= =´ ´´´ ´= ´
1
2
1
2
1000154049512 98
2 2
r .. .. 000
\=L069.. m

8.57 SFma a a= - ´´= \=. .
.
. . a) 1.75 m/s
2
4009810
4
3
2
400
981
3
p
b) 4009810
4
3
2
400
981
1
2
1000
4
3
2
3 3
- ´´= +´ ´´
æ
è
ç
ö
ø
÷\=p p.
.
.. .a a 1.24 m/s
2


8.58 Fma Va a
F
V
m V
a
= = ´´- \=
-
=´ -
1 1 1
100012
1200
021000. . . . .
Fmma a
F
V V
F
V
a
a
=+ \=
-+ -
=
-
( ). .
2 2 2
1200 200 1400
is true acceleration.
\
-
´=
-
-
-
-
´ =% .. error =
aa
a
F
V
F
V
F
V
2 1
2
100
1400 1200
1400
100167%

134. 174

8.59 (B)
1
2
2
From Fig. 8.12a 1.1..
L
LL
F
CC
VcLr
==

2 2212009.81
1088.33.0 m/s.
1.23161.1
L
W
VV
cLCr
´´
\===\=
´´

8.60 C
F
VA
C
L
L
D
= =
´
´ ´ ´
= \= =
1
2
1000981
4128015
0496 32 0065
2 2
r
a
.
.
.. .. ..
1
2

o


&
. . .WFV
D= =´ ´ ´´
æ
è
ç
ö
ø
÷´=
1
2
412801500658010
2
300 W or 13.8 Hp

8.61 a) C
V
V
L
= =
´ +
´ ´´
\=122
15009813000
1
2
122 20
2
.
.
.
. . 34.5 m/s
b) ()C
V
V
Lmax
50 m/s= =
´ +
´ ´´
\=172
15009813000
1
2
412 20
2
.
.
.
. . (at 10 000 m)
c)
&
. .WFV
D= =´ ´ ´´
æ
è
ç
ö
ø
÷´=
1
2
412802000658013
2
700 W or 18.4 Hp
where we found C
D
as follows:
()C C
L Dcruise
=
´ +
´ ´ ´
= \ =
15009813000
1
2
4128020
67 0065
2
.
.
.. ., from Fig. 8.12.
\Power =
184
045
409
.
.
..= Hp

8.62 C
V
V
L
= =
´ +
´ ´´
\=122
15009813000
1
2
1007 20
2
.
.
.
. . 38.0 m/s

8.63 ()C C
L Dcruise
=
´ +
´ ´ ´
= \ = =
15009813000
1
2
10078020
0275
0275
48
00057
2
.
.
..
.
..
\= =´ ´ ´´ =
&
. .WFV
D
1
2
100780200005729
3
400 W or 39.4 Hp
% change =
394184
184
100114%
. .
.
-
´ = increase
The increased power is due to the increase in air density.

135. 175
8.64 C
V
V
L
= =
´ +
´ ´´
\=122
15009819000
1
2
122 20
2
.
.
.
. . 39.9 m/s

8.65 C
V
V
L
= =
´
´ ´´´
\=172
250 981
1
2
122 608
2
.
.
.
. .
000
69.8 m/s

8.66 a) C
V
V
L
= =
´
´´´´
\=172
250 981
1
2
105 608
752
2
.
.
.
. . .
000
m/s
% change =
752698
698
100777%
. .
.
.
-
´ = increase
b) C
V
V
L= =
´
´ ´´
\= =
´
=
æ
è
ç
ö
ø
÷172
250 981
1
2
1515 608
626
1013
287233
1515
2
.
.
.
. .
.
.
.
000
m/s kg/m
3
r
% change =
626698
698
100103%
. .
.
.
-
´ =-
c) C
V
V
L= =
´
´ ´´
\= =
´
=
æ
è
ç
ö
ø
÷172
250 981
1
2
1093 608
737
1013
287323
1093
2
.
.
.
. .
.
.
.
000
m/s kg/m
3
r
% change =
737698
698
100563%
. .
.
.
-
´ = increase

8.67 For a conventional airfoil assume CC C
L D L
/ . ..= =476 03 at
03
981
1
2
052622220030
23810
2
6
.
.
.
. .=
´
´ ´ ´ ´
\= ´
m
m kg

&
.
.
.
WFV
D
= =´ ´ ´ ´´ =
1
2
052622220030
03
476
490
3
000 W or 657 Hp

8.68
v
v
vvv
v
v
Ñ´ +×Ñ+
Ñ
-Ñ
é
ë
ê
ù
û
ú=
¶
¶ r
n
V
t
VV
p
V() .
2
0

v
v
vv
v
v
v
vv
Ñ´=Ñ´= Ñ´
Ñ
=Ñ´Ñ=
¶
¶
¶
¶
¶w
¶ rr
V
tt
V
t
p
p( ) . .
1
0

v v vv v
Ñ´Ñ=ÑÑ´=Ñ() ( )
2 2 2
V V w (we have interchanged derivatives)

2211
()()()
22
VVVVVV
éù
éùÑ´×Ñ=Ñ´Ñ-´Ñ´=Ñ´Ñ
ëû êú
ëû
vvvvvvvvvvv
()Vw-Ñ´´
vv v

()Vw=Ñ×
vvv
()Vw-Ñ×
vvv
()()VVww+×Ñ-×Ñ
vvvvvv

136. 176

, . .xLuUU
y
¶y
¶
==\= =×Ñ-×Ñ Ñ×=Ñ×Ñ´=()() ( ).
vvvvvv vvvvv
V V Vww w where 0
There results:
¶w
¶
ww nw
v
vvvvvv v
t
V V+×Ñ-×Ñ-Ñ=()() .
2
0
This is written as
D
Dt
V
v
vvv vw
w nw=×Ñ+Ñ() .
2


8.69 x-comp:
¶w
¶
¶w
¶
¶w
¶
¶w
¶
w
¶
¶
w
¶
¶
w
¶
¶
nw
x x x x
x y z x
t
u
x
v
y
w
z
u
x
u
y
u
z
+ + + = + + +Ñ
2

y-comp:

¶w
¶
¶w
¶
¶w
¶
¶w
¶
w
¶
¶
w
¶
¶
w
¶
¶
n
¶w
¶
¶w
¶
¶w
¶
y y y y
x y z
y y y
t
u
x
v
y
w
z
v
x
v
y
v
z x y z
+ + + = + + + + +
æ
è
ç
ç
ö
ø
÷
÷
2
2
2
2
2
2

z-comp:

¶w
¶
¶w
¶
¶w
¶
¶w
¶
w
¶
¶
w
¶
¶
w
¶
¶
n
¶w
¶
¶w
¶
¶w
¶
z z z z
x y z
z z z
t
u
x
v
y
w
z
w
x
w
y
w
z x y z
+ + + = + + + + +
æ
è
ç
ö
ø
÷
2
2
2
2
2
2


8.70
x
w
y
¶
w
¶
=
v
z
¶
¶
- 0.
y
u
z
¶
w
¶
==
w
x
¶
¶
- 0. 0.
z
vu
xy
¶¶
w
¶¶
==-¹

22 z
() ; .
z
zz
DD
w
DTDt
ww
wnwnw=×Ñ+Ñ\=Ñ
vv


If viscous effects are negligible, then
D
Dt
z
w
=0.

Thus, for a planer flow, w
z=const if viscous effects are negligible.

8.71 a)
vv
Ñ´= -
æ
è
ç
ö
ø
÷+ -
æ
è
ç
ö
ø
÷+ -
æ
è
ç
ö
ø
÷=V
w
y
v
i
u
z
w
x
j
v
x
u
y
k
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶z
$ $ $
.0 \irrotational

¶f
¶
f
x
x xfy= \= +10 5
2
. ()

¶f
¶
¶
¶y
f
y
y fyC C== \= + =20 10 0
2
. . . Let \= +f5 10
2 2
x y
b)
vv
Ñ´=++-=Vij k00880
$$
()
$
. \irrotational

¶f
¶
f
x
y xyfyz= \= +8 8. (,) .
¶f
¶
¶
¶
¶
¶y
x
f
y
x
f
y
ffz=+= \= =8 8 0. (). and

¶f
¶z
df
dz
z f zC C==- \=-+ =6 3 0
2
. . . Let

137. 177
\= -f8 3
2
xyz
c)
vv
Ñ´=++
- +
+
-
- +
+
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
=
- -
Vij
yxy x
xy
xxy y
xy
k00
1
2
2
1
2
2
0
2 212
2 2
2 212
2 2
$$
( ) ( )
$
.
/ /
\irrotational

¶f
¶
f
x
x
xy
xyfy=
+
\= ++
2 2
2 2
. ()

¶f
¶
¶
¶
¶
¶y
xy y
f
y
y
xy
f
y
fC C= + +=
+
\=\= =
-1
2
2 0 0
2 212
2 2
( ) . . . .
/
Let
\= +fxy
2 2

d)
vv
Ñ´=++
-
+
-
-
+
é
ë
ê
ù
û
ú=Vij
yx
xy
xy
xy
k00
2 2
0
2 22 2 22
$$
()
( )
()
( )
$
. \irrotational

¶f
¶
f
x
x
xy
nxy fy=
+
\= ++
2 2
2 21
2
. ( )() l

¶f
¶
¶
¶
¶
¶y
y
xy
y
xy
f
y
f
y
fC C=
+
=
+
+ \=\= =
2 2 2 2
1
2
2
0 0. . . . Let
\= +flnxy
2 2


8.72
¶y
¶
¶y
¶
2
2
2
2
0
x y
+ =. This requires two conditions on x and two on y.
At , . .xLuUU
y
¶y
¶
=-=\=
At , . .xLuUU
y
¶y
¶
==\=
At yh=-, . =0 y

At , =h. yhUy=´ (See Example 8.9).
The boundary conditions are stated as:

¶y
¶
¶y
¶
y y
y
LyU
y
LyU xh xhUh(,),(,),(,),(,) .- = = -= = 0 2

8.73 u
y
yfxv
x
df
dx
f xC== \= + =-=-= \=-+
¶y
¶
y
¶y
¶
100 100 50 50. (). . .
\ = -y(,) .xy y x10050 (We usually let C = 0.)
u
x
xfyv
y
df
dy
f yC== \= + === \= +
¶f
¶
f
¶f
¶
100 100 50 50. (). . .
\ = +f(,) .xy x y10050
y = h
y = 0
x = -L
y
x
U

138. 178

8.74 a) y q=40.
b)
1 1 1
40
1
00
2
rr r rrr r
¶
¶
¶y
¶q
¶y
¶q¶
¶
¶
¶
¶q
æ
è
ç
ö
ø
÷+-
æ
è
ç
ö
ø
÷= +-
æ
è
ç
ö
ø
÷=() ().
\It is incompressible since the above continuity equation is satisfied.
Note: The continuity equation is found in Table 5.1.
c)
¶f
¶
¶y
¶q
f q
rr r
nrf= = \= +
1 40
40. () l

¶f
¶q
¶
¶q
¶y
¶
==- = \= =
f
r
r
fC C0 0. . . Let
\=f40lnr
d) v
r
v av
v
rrr
r r r
r
= = = = -
æ
è
ç
ö
ø
÷=-
40
0
4040
10
2
, . .
q
¶
¶

\=r543. m

8.75 u
y
y
xy x
y
x
fy==
+
= \=- +
-¶y
¶
¶f
¶
f20
2
40
2 2
1
. tan ().
v
y
x
yx
f
y
x
xy
f
y
x
xy
fC C==-
+
+=-
+
+=-
+
\= =
¶f
¶
¶
¶
¶
¶
40
1
40
20
2
0
2 2 2 2 2 2
/
/
. . . Let
f=-
-
40
1
tan.
y
x


8.76 a)
¶y
¶
¶y
¶
¶y
¶
2
2
2
2
2 22
0 10 2
x y x
yxy x+ = = +
-
. ( )().

¶y
¶
2
2
2 22
20
x
yxy= -
-
( )- +
-
80
2 2 23
xyxy( )

¶y
¶y
xy yxy y=- + + +
- -
1010 10 2
2 21 2 22
( ) ( )().

¶y
¶
2
2
2 22
20
y
yxy= +
-
( )+ + - +
- -
40 80
2 22 32 23
yxy yxy( ) ( ).
\ + =
+
-
+
+
+
-
+
¶y
¶
¶y
¶
2
2
2
2 2 22
2
2 23 2 22
3
2 23
20 80 60 80
x y
y
xy
xy
xy
y
xy
y
xy( )( )( )( )

=
+
+
-
+
-
+
=
+ - -
+
=
80 80 80 80 80 80 80
0
2 2
2 23
2
2 23
3
2 23
2 3 2 3
2 23
yxy
xy
xy
xy
y
xy
xy y xy y
xy
( )
( ) ( )( ) ( )
.
b) In polar coord: yq q
q
q q(,) sin
sin
sin sin.r r
r
r
r
r
= - = -10
10
10
10
2


1
10
10
10
10
2
r r r r
f
¶y
¶q
q
¶f
¶
f q q=-
æ
è
ç
ö
ø
÷ = \=+
æ
è
ç
ö
ø
÷ +cos . cos ().

139. 179

22
111010
10sin10sinsin. 0.
dfdf
fC
rrdrd rr
¶f¶y
qqq
¶qq¶q
æö
=-+=-=--==
ç÷
èø
.
\= +
æ
è
ç
ö
ø
÷ = +
+
f q f10
1
10
10
2 2
r
r
xy x
x
xy
cos (,) , or
where we let r x rxycos .q= =+ and
2 2 2

c) Along the x-axis, v
x
=-=
¶y
¶
0 where we let y = 0 in part (a) and
u
y xy
y
xy x
y= =-
+
+
+
=- =
¶y
¶
10
10 20
10
10
0
2 2
2
2 22 2
( )
. with
Euler’s Eq: r
¶
¶
¶
¶
r
¶
¶
u
u
x
p
x xx
p
x
=- \ -
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷=-. . 10
1020
2 3

\= -
æ
è
ç
ö
ø
÷=-+
é
ë
ê
ù
û
ú
+ =ò
p
x x
dx
xx
CCr r
200200 50100
50
5 3 4 2
. 000.
= -
é
ë
ê
ù
û
ú
+1000
10050
50
2 4
xx
000 Pa. (Could have used Bernoulli!)
d) Let u
x
x= =- \=±\0010
10
1
2
: . . Stag pts: (1, 0), (-1, 0)

8.77 a)
¶f
¶
¶f
¶
¶
¶
¶
¶
2
2
2
2 2 2 2 2
2 2
2 22
10
10 10 102
x y x
x
xy y
y
xy
xy xx
xy
+ = +
+
é
ë
ê
ù
û
ú+
+
é
ë
ê
ù
û
ú=
+ -
+
( )10 (
( )

+
+ -
+
=
+ - + + -
+
=
( )10 ()
( ) ( )
.
xy yy
xy
x y x x y y
xy
2 2
2 22
2 2 2 2 2 2
2 22
102 10 10 20 10 10 20
0
b) Polar coord: f q= +10 5
2
r nrcos .l (See Eq. 8.5.14.)

¶f
¶
q
¶y
¶q
y q q
r
r
rr
r fr= + = \= + +10
101
10 10
2
cos . sin ()

1
10sin10sin.
df
fC
rrdr
¶f¶y
qq
¶q¶
=-=-=--\= . 10sin10. ryqq\=+
\ = +
-
y(,) tan.xy y
y
x
1010
1

c) v
y
y
xy
==
+
¶f
¶
10
2 2
. Along x-axis (y = 0) v = 0.
u
x
x
xy
==+
+
¶f
¶
10
10
2 2
. Along x-axis u
x
=+10
10
.
Bernoulli:
2
2
Vp
gz
r
++
2
2
Vp
gz
r
¥¥
¥
=++ (assume ) zz
¥
=

( /)
. .
1010
2
10
2
10050
21
2 2
2
+
+= + \= - +
æ
è
ç
ö
ø
÷
x p
p
xxr r
100 000
kPa

140. 180
d) u
x
x= =+ \=-\0010
10
1: . . Stag pt: (-1, 0)
e)
y
av=
vv
u
yx
¶¶
¶¶
+ 0 on axis.
x
u
xauv
x
¶
¶
=-=+
2
1010
10.
u
yx x
¶
¶
æöæö
=+-
ç÷ ç÷
èø èø


210
(2,0)(105)12.5 m/s.
4
xa
æö
\-=--=-
ç÷
èø


8.78
23
2 5
(,)5.
23
yy
uxyyyC
y
¶y
y
¶
=-=\=-+
231
. (310).
6
yyy\=-
qudy yydy= =- = - = ´
-
òò
( )
. .
. .
..
5
02
2
5
02
3
666710
2
2 3
3
0
2
0
2
m/s
2

yy
2 1
2 3 31
6
30210020666710-= ´ -´ -= ´
-
(. .) . . m/s
2

w
¶
¶
=-=-+¹
u
y
y1100. \f doesn’t exist.

8.79 y
p
p
q qq= + = +30
5
2
30
5
2
y rsin .
a) v
r r
r
= = +=
1
30
5
2
0
¶y
¶q
qcos .
At qp= = \=, . .' .
5
2
30 00833
r
r
s
s

Stag. pt: (",).-10


b) At qp y
p p p
= == +, sin . =.0833, r r
s
5
2
30
2
5
22

\= =ry
inter.0119 ft.
c) qUH H H=´= \ = \Dy
p p
. . . =
5
60
30
5
2
Thickness = 2
5
30
H=
p
ft or 1.257".
d) v u
r
(,) cos . .. ..1 30
5
2
3025275 275p p= +=-+=- \= fps

8.80 [ ] [ ] [ ]f
p
p
p
p
= ++ - -+ + = ++
2
1
2
1 10
1
4
1
2
12
2 2
12
2 2
l l lnx y nx y x nx y() () ()
/ /

[ ]- -++
1
4
1 10
2 2
lnx y x() .
u
x
x
x
x
x x x
v y
y
= =
+
+
-
-
-
+=
+
-
-
+ = =
=
¶f
¶
0
2 2
1
4
21
1
1
4
21
1
10
1
21
1
21
10 0 0
[()]
()
[()]
() ()()
. . if
30 fps
y
x= 0
= 5p/2

141. 181
At the stagnation point, u
x x x
=\
+
-
-
+=\
-
=0
1
21
1
21
100
2
1
20
2
.
()()
. .
\= \=± \ ´x x
2
11 1049.. . . m. oval length = 21.049=2.098 m
All the flow from the source goes to the sink, i.e., p
p
m/s, or m/s for
2 2
2
0y>.
uy
x y y y
x
()
() ()
.= =
+
-
-
+
+=
+
+
=
¶f
¶
0
2 2 2
1
4
2
1
1
4
2
1
10
1
1
10

1
2
0
1
10. tan10.
221
h
qdyhh
y
pp
-
æö
=+=\+=ç÷ò
ç÷
+
èø


h = 0.143 m so that thickness = 2h = 0.286 m.
The minimum pressure occurs on the oval surface at (0,h).
There u=
+
+=
1
1143
101098
2
.
. m/s.
Bernoulli:
V pVp p
2 2 2 2
2 2
1098
2 1000
10
2
+=+ + = +
¥ ¥
r r
.
.
.
10 000
1000

Pa
min\ =-p 280.

8.81 [ ] [ ] [ ]f
p
p
p
p
= ++ +
-
-+ += ++
2
2
1
2
2
1 2
1
2
1
2 2
12
2 2
12
2 2
l l lnx y nx y x nx y() () ()
/ /

[ ]- -++
1
2
1 2
2 2
lnx y x() .
u
x
x
x y
x
x y
v
y
x y
y
x y
==
+
++
-
-
-+
+ =
-+
-
-+
¶f
¶
1
2
21
1
1
2
21
1
2
1 1
2 2 2 3 2 2 2 2
()
()
()
()
.
() ()

Along the x-axis (y = 0), v = 0 and u
x x
=
+
-
-
+
1
1
1
1
2.
Set u
x x
x x=
-
-
+
= =\=±0
1
1
1
1
2 2 2
2
: , . . or
Stag. pts.: (,),(,).20 20 -
u v(,) .(,).40
1
41
1
41
2 400=
-+
-
--
+= - =1.867 m/s
u v(,) . (,) .04
1
14
1
14
2 04
4
14
4
14
0
2 2 2 2
=
+
-
-
+
+= =
+
-
+
=2.118 m/s

8.82 f
p
p
p
p
= +- + ++
2
2
1
2
2
1
2 212 2 212
l lnxy nxy[()] [()]
/ /

= +-+ ++
1
2
1
1
2
1
2 2 2 2
l lnxy nxy[()] [()].
(0, h)
y
x

142. 182
u
x
x
xy
x
xy
==
+-
+
++
¶f
¶
2 2 2 2
1 1() ()
.
v
y
y
xy
y
xy
==
-
+-
+
+
+-
¶f
¶
1
1
1
1
2 2 2 2
() ()
.
At (0, 0) u = 0 and v = 0. At (1, 1)

22222
2211
00.4 m/s. 1.2 m/s.
521121
vu=+===+=
++


\=+
v
V i j1204.$.$. m/s


8.83 f
p
p
p
p
= -+ + ++ +
¥
2
2
1
2
2
1
2 212 2 212
l lny x ny x Ux[() ] [() ] .
/ /

= -++ +++
¥
1
2
1
1
2
1
2 2 2 2
l lny x ny xUx[() ] [() ] .
a) Stag. pts. May occur on x-axis, y =0.
u
x
x
x
x
x
y
= =
+
+
+
+
=
¶f
¶
0
2 2
1 1
10.

2
0.210. xx\++= \no stagnation points exist on the x-axis.

(They do exist away from the x-axis.)
Along the y-axis: uy qudy
h
(). ()= = = =ò10
1
2
2
0
m/s.
2
pp
\= = \=ò
p1010 0314
0
dy h h
h
. m. .
b) u
x
x
xx x=
+
+\++= \=-
2
1
1 210 1
2
2
. . m.
Stag. pt.: (-1, 0)
Along the y-axis: u h h= \=´ \=10 1 314.. .. . mp

c) u
x
x
x x x=
+
+ \++= \=- -
2
1
02 1010 990
2
2
.. . ., 0.10 m.
Stag. pts.: (-9.9, 0) , (-0.1, 0).
Along the y-axis: u h h= \= \=02 02 1571.. . .. . mp

8.84 f q q= +
60
8
r
rcos cos.
a) v
r r r
r==- + =-
æ
è
ç
ö
ø
÷
¶f
¶
q q q
60
8 8
60
2 2
cos cos cos.
At the cylinder surface v
r
=0 for all q. Hence,
x
y
x
y
x
y

143. 183

60
8 2739
2
r
r
c
c
= \=. . m
b) Bernoulli: Dp
U
= = =
¥
r
2 2
2
1000
8
2
32 000 Pa or 32 kPa
c) v
r r
q
¶f
¶q
q q= =- -
1 60
8
2
sin sin.
At rrv
c= =- - =-, sin sin sin
q q q q8 8 16
d) Dp
v
= = =r
90
2 2
2
1000
16
2
128
o
000 Pa or 128 kPa

8.85 y
p
p
q
p
p
q= + =+
4
2
20
2
210l lnr nr
At (,)(,),(,)(,/).xy r= =01 12 q p
v
r
r
(,/) ().1 2
1 1
1
22p
¶y
¶q
= = =
v
r
q
p
¶y
¶
(,/) .1 2
10
1
10=-=-=-
v
r
(.,/)
.
.17 4
2
17
118p == , v
q
p(.,/)
.
.17 4
10
17
588=
-
=-
v
r
(.,)
.
.320
2
32
0625= = , v
q
(.,)
.
.320
10
32
3125=
-
=
v
r
(,/) .6 4
2
6
0333- ==p , v
q
p(,/) .6 4
10
6
167- =
-
=-, etc.
Note: We scaled the radius at each 45° increment to find r.

b) v
r
v
r
r
= =-
2 10
and
q
. From Table 5.1 (use the l.h.s. of
momentum)
a
Dv
Dt
v
r
v
v
r
v
r
r
r
r
r
= += -
q q
¶
¶
2 2
=-
æ
è
ç
ö
ø
÷- =-
22100 104
1
2 3
rr r

=-104 m/s
2

a
Dv
Dt
vv
r
v
v
r
vv
r
r
r
r
q
q q q q
¶
¶
= + = + =
æ
è
ç
ö
ø
÷+
-
=
210210
0
2 3
rr r
()
.
\ =-
v
a(,)( ,)01 1040 m/s
2








x
y

144. 184
c) v v
r
(.,/)/. .,(.,/)
.
.1414 42141401414 1414 4
10
1414
0707p p
q
= = =- =-
v v
r(.,/)/. ,(.,/) /. .01220120 0121001100p p
q= = =- =-
Bernoulli:
20 000
1.2
13 760 Pa+
+
=+
+
\=
014140707
2 12
20100
2
2 2 2 2
. .
.
.
p
p
We used r
air
3
kg/m at standard conditions.=12.

8.86 Along the y-axis v v
r
r
= =--0 10
40
2
and
q
.
We have set q
p
=
2
in Eq. 8.5.27. r
c
= =
40
10
2.
b) v
r
r
= - - Þ10
40
43 51269
2
cos cos.(,) (,.).q q
o

v
r
v v
rq q
q q=- - \=- =-10
40
696 928
2
sin sin. . , . . m/s m/s

c) Use Eq. 8.5.28: pp U=-
¥0
2 2
2r asin
Drag = p rdLp rLpp U
c c
cos . .
/
/
aa r
p
p
-´ =-
¥
-
ò 90 90 0
2
2
2
2 2
= - -´
¥ò
2 2 2
0
2 2
90
0
2
( sin)cos
/
p U rLdp rL
c cr a a a
p

[ ]= -
é
ë
ê
ù
û
ú
-- =
¥ ¥ ¥
2 2
3
2 2
8
3
0
2
3
0
2
0
2 2
rLp U p UrL rLU
c c c
sin
sin
.
/
ar
a
r r
p


C
UA
rLU
UrL
D
c
c
= = ==
¥
¥
¥
Drag
1
2
83
1
2
2
8
3
2667
2
2
2
r
r
r
(/)
..

8.87 vU
r
U rU
r c
= - = = =´=
¥ ¥ ¥
cos cos. , .q
m
q m
2
2 2
4 144 Let
For qp= =-+, . v
r
r
4
4
2

b) v
r
U
r
c
q
¶y
¶
q
mq
q q=-=- - =--
æ
è
ç
ö
ø
÷ =-
¥sin
sin
sin sin.
2 2
4
4
1
8

c) pp
V v
c
=+ - = + ´-
¥
¥
r r
q
q
2 2 2 2 2
2 2
1000
4
2
1000
8
2
50 000
sin
.
\=-p
c
5832
2
sinq kPa.

10 m/s
20 m/s
da
p(a)
a
p
90
x = -1
-x
-v
r

145. 185
d) Drag = 25832 11 2621
2
0
2
( sin)cos
/
- ´´ -´´ò
a a a
p
d
= -
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
-=25832
1
3
5242.7 kN. (See the figure in Problem 8.86c.)

8.88 On the cylinder
1000
2sin60sin ,
223.651
c
vU
r
q
qq
pp
¥
G
=--=--
´
where we have
Used
400
3.651 ft.
30
cr
U
m
¥
===
If
22222222
6666
(,).0318
(6)(2)(6)(2)(6)(2)(6)(2)
xxxx
uxy
xyxyxyxy
éù --++
=-+++ êú
-+--++++-+++êúëû
227, 313.q\=
oo

Stag. pts.: (3.651 ft, 227°) , (3.651 ft, 313°).

Max. pressure occurs on the cylinder at a stagnation pt.:

2222
maxo
0.0024
= 3001.08 psf.
22
pUv
r
¥
éùéù\=--=
ëûëû

Min. pressure occurs at the top of the cylinder where q=90
o
and the velocity is:

90
o
1000
2sin230104 fps
223.651
vU
r
q
pp
¥
G
=--=-´-=
´


2222
mino
0.0024
= 3010411.8 psf.
22
pUv
r
¥
éùéù\=--=-
ëûëû


8.89 v
q
q
p
=-´ -
´
220
24
sin
.
.
G
For one stag. pt.: v
q
q= =0 270 at
o
:
0220270
24
220241005=-´ -
´
\=´´´=sin
.
. . .
o G
G
p
p m/s.
2

G
G
= \= =
´
=2
2
1005
24
2
2 2
pw w
p p
r
r
c
c
.
.
.
. 100 rad/s (See Example 8.12.)
Min. pressure occurs where v
q
is max, i.e., qp=/.2 There
v
q
p
=-´´-
´
=2201
1005
24
80
.
.
m/s.
\ =+ - =+ ´ - ´ =-
¥
¥
p p
V v
min
Pa
2 2 2 2
2 2
0
20
2
122
80
2
1223660r r
q
. . .

146. 186
8.90 G= =´´ ´ = = =´=2 261202602842 63108
2 2 2 2
pwp p mr rU
c c
. / . . . m/s. m/s.
2 3

\=-´ -
´
\ =-v
q
q
p
q23
2842
26
1256sin
.
.
sin ... Impossible. \Stag. pt. is off the
cylinder at q= >270
o
, . but rr
c
From Eq. 8.5.29,
v
r
U
r r r r
q
¶y
¶
q
m
q
p p
=-=- - - =--- -- =
¥
sin sin ()
.
()
.
.
2 2
2
31
108
1
2842
2
0
G

\+ = \- + = \=3
1084523
15080360 121
2
2. .
. . . . .
r r
r r r m.
Stag. pt.: (1.21, 270°). ()
.
.
.v
q
p
90
23
2842
26
1354
o=-´-
´
=- m/s.
Min. pressure occurs at q= = = -
æ
è
ç
ö
ø
÷=-90
3
2
1354
2
122106
2 2
o
, :
.
. . at Pa
min
rrp
c

Max. pressure occurs at q= = = -
æ
è
ç
ö
ø
÷=-270
3
2
154
2
122404
2 2
o
, :
.
. . . at Pa
max
rrp
c


8.91 At 15,000 ft, r=. .0015 slug/ft
3

Lift = rUL
¥= ´ ´ ´=G. , .00153501500060472,000 lb

8.92 Place four sources as shown. Then, with q=2p for each:
uxy
x
x y
x
x y
x
x y
(,)
()()()()()()
=
-
-+-
+
+
++-
+
-
-++
2
2 2
2
2 2
2
2 2
2 2 2 2 2 2
+
+
+++
x
x y
2
2 2
2 2
()()


vxy
y
x y
y
x y
y
x y
(,)
()()()()()()
=
-
-+-
+
+
-++
+
-
++-
2
2 2
2
2 2
2
2 2
2 2 2 2 2 2
+
+
+++
y
x y
2
2 2
2 2
()()



8.93 Place four sources with q=02. m/s, as shown.
2



uxy
x
x y
x
x y
x
x y
x
x y
(,).
()()()()()()()()
=-
-
-+-
+
-
-++
+
+
++-
+
+
+++
é
ë
ê
ù
û
ú0318
6
6 2
6
6 2
6
6 2
6
6 2
2 2 2 2 2 2 2 2
vxy
y
x y
y
x y
y
x y
y
x y
(,).
()()()()()()()()
=-
-
-+-
+
-
++-
+
+
-++
+
+
+++
é
ë
ê
ù
û
ú0318
2
6 2
2
6 2
2
6 2
2
6 2
2 2 2 2 2 2 2 2
where
q
2
2
2
0318
pp
=
-
=-
.
..
x
y
x
y
(6, 2)

147. 187
At (4,3) u(,). . .43 0318
2
41
2
425
10
1001
10
10025
000922=-
-
+
+
-
+
+
+
+
+
é
ë
ê
ù
û
ú
= m/s
v(,). . .43 0318
1
41
1
1001
5
425
5
10025
001343=-
+
+
+
+
+
+
+
é
ë
ê
ù
û
ú
=- m/s

8.94 Re . / .
crit
= \=´ =
¥
Ux
x
T
T
n
n n610 3002000
5

a) n= ´ \= ´ ´ =
- -
15610 200015610 0312
4 4
. . .' ft/sec. or 3.74"
2
x
T

b) n
m
r
==´ \= ´´ =
- -
2110 20002110 042
4 4
. . .' ft/sec. or 5.04"
2
x
T

c) n= ´ \= ´ ´ =
- -
34710 2000347100694
4 4
. . .' ft/sec. or 8.33"
2
x
T


8.95 a) Use Re /. .
crit
=´ = ´
-
31010 15110
5 5
x
T
\=x
T
0453. . m
b) Use Re /. .
crit
= = ´
-
1010 15110
6 5
x
T
\=x
T
151.. m
c) Use Re /. .
crit
=´ = ´
-
31010 15110
5 5
x
T
\=x
T
0453. . m
d) Use Re /. .
crit
=´ = ´
-
31010 15110
5 5
x
T
\=x
T
0453. . m
e)
45
growthgrowth
Re61010/1.5110.0.091 m or 9.1 cmxx
-
=´=´\= .
Note: A rough plate, high free-stream disturbances, or a vibrated smooth plate
all experience transition at the lower Re.
crit


8.96 a) Use Re / .
crit
=´ =
-
31010 10
5 6
x
T
\=x
T
003 3. . m or cm
b) Use Re / .
crit
= =
-
1010 10
6 6
x
T
\=x
T
01 10. . m or cm
c) Use Re / .
crit
=´ =
-
31010 10
5 6
x
T
\=x
T
003 3. . m or cm
d) Use Re / .
crit
=´ =
-
31010 10
5 6
x
T
\=x
T
003 3. . m or cm
e)
22
()20 0002 1000 10sin(/2)pxx=-´´

8.97 Re .
.
.
crit
For a wind tunnel: =´ =
´
´ =
´
´
¥ ¥
-
610
2
610
2
1510
5 5
5
U U
n

m/s.\ =
¥U 45.
For a water channel: 610
2
10
03
5
6
´ =
´
\ =
¥
- ¥
U
U. . m/s .

8.98 The x-coordinate is measured along the cylinder surface as shown in Fig. 8.19.
The pressure distribution (see solution 8.86) on the surface is
pp U rx
c
=- =
¥0
2 2
2r a a asin ( where is zero at the stagnation point). Then

22
()20 0002 1000 10sin(/2)pxx=-´´
=-20200 2
2
sin(/)x kPa
The velocity U(x) at the edge of the b.l. is U(x) on the cylinder wall:

148. 188
vr
q q q pa a() sin sin sin() sin==- - =- -=210 10 20 20
\ =Ux x() sin(/)20 2

8.99 Uxvr v Ux x xr
c c() . sin. ()sin .= = = \ = =
q q a a at since 1 8 8
px x() sin sin=- =-5832 5832
2 2
a kPa

8.100 The height h above the plate is hxmx m m() ... . .=+ =´+\=-41 24 15
\ =- ´= \ =
-
hx x Uxh Ux
x
()... . (). ()
.
..
0415 64
24
0415
Continuity: or
Ux
x
()
.
.=
-
16
267

Euler’s Eqn: r
¶
¶
¶
¶
ru
u
x
p
x
dp
dx x x
=- \=
- -
.
. (. )

16
267
16
267
2

=
-
256
267
3
(. )
.
x


8.101 a)
topoutin
0000
mmmudyudydxudyudydx
xx
dddd
rrrr
¶¶
=-=+-= òòòò
¶¶
&&&
b)
0
()()()
2
x
dp
FpdxpdpdpddtdddS=-++-++
0
higher order termsdxdptd=--+
222
outintop
0000
2
00
()
()
mommommomudyudydxudyUxudydx
xx
udydxUxudydx
xx
dddd
dd
rrrr
rr
æö¶¶
--=+-- òòòò ç÷
¶¶ èø
æö¶¶
=-òò ç÷
¶¶
èø
&&&

23
25452
3434
4.
224.651.51034.65(1.5103)
yd
y
dx
d
--
éù ´
=-+êú
´´´´´´´êúëû


8.102 t d r r
dd
0
2
00
=- + -òò
dp
dx
Ux
d
dx
udy
d
dx
udy()
=- + -
æ
è
ç
ö
ø
÷-ò òò
d r r r
d dd
dp
dx
d
dx
uUdy
dU
dx
udy
d
dx
udy
0
2
00

where we have used g
df
dx
dfg
dx
f
dg
dx
Ugf udy=- = =
æ
è
ç
ö
ø
÷ò
. , . Here r
d
0

\=- + - - =òò
t d r r r
dd
0
00
dp
dx
d
dx
uUudy
dU
dx
udy( ) .( const.)

149. 189

8.103
dp
dx
d
dx
U U
dU
dx
dU
dx
Udy U Udy=- =- =-
æ
è
ç
ö
ø
÷ =ò ò
r
r r
d d
d d
2
1 1
2
0 0
where .
\=--
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
+ -
ò ò
t dr
d
rq r
d d
0
0
2
0
1dU
dx
Udy
d
dx
U
dU
dx
udy()
= + - = +ò
rq r rq r d
d
d
dx
U
dU
dx
Uudy
d
dx
U
dU
dx
U
d() ( ) () .
2 2
0


8.104 If dpdx
dU
dx
d
dx
uUudy/ ( ).= = = -
¥ò
0 0
0
0
then and tr
d

tr
p
d
p
d
r
d
p
p
d
r
d
p
d
d d
0
2
0
2
0
2
2
1
2
2
22
2
2
= -
æ
è
ç
ö
ø
÷= - -
é
ë
ê
ù
û
ú
= -
æ
è
ç
ö
ø
÷
¥ ¥ ¥ò
d
dx
U
y y
dyU
d
dx
yy
U
d
dx
sin sin cos
tm
¶
¶
m
p
d
0
0
2
0= =
=
¥
u
y
U
y
cos.
\ = \ = \=
¥ ¥
¥ ¥
m
p
d
r
d
dd
n
d
n
U U
d
dx
d
U
dx
x
U2
137 115 479
2
. . . . . .
b) tm
p
n
m
n
0
2
1
479
0328= =
¥
¥
¥
¥
U
U
x
U
U
x.
. .
c)
¶
¶
¶
¶
p
n
¶
¶
¶
¶
u
x
U
x
y U
x
U
x
a
x
U
ax a
x
v
y
=
´
æ
è
ç
ö
ø
÷=
æ
è
ç
ö
ø
÷= -
æ
è
ç
ö
ø
÷ =-
¥
¥
¥ ¥
-
sin
.
sin cos .
/
2479 2
32

\=
æ
è
ç
ö
ø
÷=
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
¥
¥ ¥
¥
¥ ¥
òò
vU
y
x
U
y
U
x
dy U
U U
ydy
.
cos. . cos. .
/
164
328 0316 189
32
00
n n n n
dd


8.105 uU
y d
dx
U
y y
dy= = -
æ
è
ç
ö
ø
÷¥ ¥ò
d
tr
d d
d
.
0
2
0
1
= -
æ
è
ç
ö
ø
÷=
¥ ¥r
dd
r
dd
dx
U U
d
dx
2 2
23
1
6
.


tm
¶
¶
m
d
m
d
r
d
dd
n
0
21
6
6= = \ = \ =
¥ ¥
¥
¥
u
y
U U
U
d
dx
d
U
dx. .
\= = =
¥ ¥
¥
¥
d
n
d
n
t m
n
2
012 346 0289
U
x x
x
U
U
U
x
.(). . . .
%error in d()
.
.x=
-
´ =
5346
5
100308% low.
U
d u = Uy/d
y

150. 190
%error in lowt
0
332289
332
10013%()
..
.
.x=
-
´ =

8.106 t r
d d
r
d d
d
dd
0
2 2
6
2
0
6
3 13
1
3
1
1
3
= -
æ
è
ç
ö
ø
÷+ +
æ
è
ç
ö
ø
÷--
æ
è
ç
ö
ø
÷
ì
í
ï
î
ï
¥ ¥òò
d
dx
U
y y
dy U
y y
dy
/
//

+ +
æ
è
ç
ö
ø
÷--
æ
è
ç
ö
ø
÷
ü
ý
ï
þ
ï
¥ò
r
d d
d
d
U
y y
dy
2
2
3
2
3
1
3
2
3
/

= = \ =
¥
¥
¥
d
dx
U
U d
dx
Ur dm
d
d
d
mr
2
01358
3
2208(. ) . ./ .
Thus, d t r r(). /,().
.
. Re.
/
x vxU x U
v
Ux
U
x
= =
æ
è
ç
ö
ø
÷=
¥ ¥
¥
¥
-
665 01358
665
2
0451
0
2 2 12

%error for d=
-
´ =
6655
5
10033%
.
. %error for t
0
04510332
0332
10036%=
-
´ =
. .
.


8.107 Continuity from entrance to x: UH uydyUxH
0
0
2 2= + -ò
() ()( ).d
d

Write UxUxdyUxdy() () ().d
dd
= =òò
00
Then, continuity provides
UH uUdyUH
0
0
2= - +ò
( )
d
= - -ò
UH Uudy2
0
( )
d

= - \ =
-
UH U Ux
UH
H
d
d
2
2
0
d
d
. () .
If we were to move the walls out a distance d
d
x(), then Ux() would be constant
since ( )[ ]H
d d
- +2 2d d would be constant; then UxU() .=
0
For a square wind
tunnel, displace one wall outward 4 0d
ddpdx for / .=

8.108 The given velocity profile is that used in Example 8.13. There we found
d n= = = = =
¥
-
548 54810 10000173 00017330003
6
. / . / . . .xU x x m.
Assume the streamline is outside the b.l. Continuity is then
10002 10
2
003003
00310
2
2
0
003
´ = -
æ
è
ç
ö
ø
÷+-ò
.
. .
(.)
.
y y
dyh
= + - \=00210003 0021. .. .h h m or 2.1 cm
[ ]d
d
y y
dy= - +
æ
è
ç
ö
ø
÷= -+=ò
1
10
10
20
003
10
003
1
10
0303010001
2
2
0
003
. .
... .
.
m
h-=-=221201. . cm or 0.001 m.
The streamline moves away from the wall a distance d
d
.

151. 191

8.109 From Prob. 8.107 we found that we should displace the one wall outward 4d
d
.
From the definition of d
d
:
hx
y y
dy
d()= = - +
æ
è
ç
ö
ø
÷= -+
æ
è
ç
ö
ø
÷=ò
4
4
10
10
2010
4
3
4
3
2
2
0
d
d d
dd
d
d
d

=
´
´
æ
è
ç
ç
ö
ø
÷
÷
=
-
4
3
548
18610 10
160287303
000735
5
.
. /
/(. )
.
x
x m
We used d()x found in Example 8.13, r nmr= =pRT/, /. and

8.110 a) uU
y y
U
U
y y
dy
d= -
æ
è
ç
ö
ø
÷ = - +
æ
è
ç
ö
ø
÷=-+=
¥
¥
¥ò
3
2
1
2
1
1
3
2
1
2
3
4
1
8
375
3
3
3
3
0
d d
d
d d
ddd d
d
. ..
From Eq. 8.6.16, d
n n
d
x
U
x
U
= ´ =
¥ ¥
. . . .375465 174 %error = 1.2%.
q
d d d d
d
d
= -
æ
è
ç
ö
ø
÷- +
æ
è
ç
ö
ø
÷=
¥
¥ò
1 3
2
1
2
1
3
2
1
2
0139
2
2
3
3
3
3
0
U
U
y y y y
dy..
\= ´ =
-
´ =
¥ ¥
q
n n
. . . .
..
.
.139465 0648
648644
644
100062%
x
U
x
U
%error =
b) uU
yy yy
dy
d= -
æ
è
ç
ö
ø
÷ = -+
æ
è
ç
ö
ø
÷=-+=
¥ ò
2 1
2
3
3
2
2
2
2
0
dd
d
dd
dd
d
d
d
. /. See Example 8.13.
\= =
-
´ =
¥ ¥
d
n n
d
x
U
x
U
548
3
183
183172
172
10064%
.
. .
. .
.
. %error = .
q
dd dd
dddddd d
d
= -
æ
è
ç
ö
ø
÷-+
æ
è
ç
ö
ø
÷=--++-=ò
2 12
1
3
4
3
2
4
2
4
1
5
1333
2
2
2
2
0
yy yy
dy . .
\= ´ =
-
´ =
¥ ¥
q
n n
. . . .
..
.
.1333548 0731
731644
644
100135%
x
U
x
U
%error = .
c)
0
2
1sin0.363. See Problem 8.104. 4.79.
2
d
yx
dy
U
d
pdn
dddd
dp
¥
æö
=-=-==òç÷
èø

\= ´ =
-
´ =
¥ ¥
d
n n
d
x
U
x
U
0363479 174
174172
172
10012%. . . .
. .
.
. %error =
q
p
d
p
d
d
p
p
d
dd
p
d
dd
= -
æ
è
ç
ö
ø
÷=- -+-
é
ë
ê
ù
û
ú
=-+=ò
sin sin cos sin ..
y y
dy
yy
2
1
2
2
22 2
2
0137
00
term
\= ´ =
-
´ =
¥ ¥
q
n n
. . . .
..
.
..137479 0654
654644
644
10016%
x
U
x
U
%error =

152. 192
8.111 a) d
n
= =
´ ´æ
è
ç
ö
ø
÷=
¥
-
465 465
161020
12
00759
4
12
. .
.
. .
/
x
U
ft
b) t r
n
0
2 2
4
12
5
323 323002412
1610
2012
91110= =´ ´
´
´
æ
è
ç
ö
ø
÷= ´
¥
¥
-
-
. ..
.
. .
/
U
xU
psf
c) Drag =
21
20151.29
2
U
LU
n
r
¥
¥
´´´

1/2
4
211.610
.0024123001.290.0546 lb.
22012
-æö´
=´´´´= ç÷
ç÷´
èø

d) d
¶
¶ d d
d
x
u
x
U
y yd
dx
=
-
¥
=
´ ´
= = - +
é
ë
ê
ù
û
ú10
4
2
3
4
465
161010
12
00416
3
2
3
2
.
.
. . ft.
\=-
´
+
´
é
ë
ê
ù
û
ú
´
´
=- +
-
¶
¶
u
x
y y
y y12
3
20416
3
20416
465
2
1610
1012
27916140
2
3
4
4
3
.. ..
. .
. .
\=- = ´ - ´ =ò
v
u
x
dy
¶
¶
d
279
2
0416
16140
4
041600121
2 4
0
.
. . . . fps

8.112 a) d
n
= =
´ ´æ
è
ç
ö
ø
÷=
¥
-
465 465
15106
4
00221
5
12
. .
.
. .
/
x
U
m
b) t r
n
0
2 2
5
12
0323 3231224
1510
64
000498= =´´
´
´
æ
è
ç
ö
ø
÷=
¥
¥
-
. . .
.
. .
/
U
xU
Pa
c) Drag =
1
2
129
1
2
122465129
1510
64
0299
2 2
5
12
rULw
v
LU
¥
¥
-
´ =´´´´´
´
´
æ
è
ç
ö
ø
÷=. . .
.
. .
/
N
d)
3
24
33
22
uyyd
U
xdx
¶d
¶ dd
¥
éù
=-+êú
êúëû


23
25452
3434
4.
224.651.51034.65(1.5103)
yd
y
dx
d
--
éù ´
=-+êú
´´´´´´´êúëû

[ ]\=- +´
´
=- + ´
-
¶
¶
u
x
y y y y4616625310
465
2
1510
4
1
3
64126310
73
5
53
.
..
. . .
\=- = ´ -
´
´ =òv
u
x
dy
¶
¶
d
641
2
0156
26310
4
0156000391
2
5
4
0
.
.
.
. . , m/s
where d
x=
-
=
´ ´
=
3
5
465
15103
4
01560.
.
. m.

153. 193
8.113 a) d
n
= =
´ ´æ
è
ç
ö
ø
÷=
¥
-
5 5
15102
10
000866
5
12
x
U
.
. .
/
m Use t r
n
0
2
332=
¥
¥
. .U
xU

Drag = t
0
2
5
0
33212210
1510
10
2
12
40561wdx
L
= ´ ´
´
´=
-
ò
. .
.
/
. . N
b) d=´
´
´
æ
è
ç
ö
ø
÷=
-
.
.
. .
.
382
1510
102
00453
5
2
m
Drag =
1
2
074
1
2
1221024074
1510
102
215
2
2
2
5
2
r
n
ULw
UL
¥
¥
-
´
æ
è
ç
ö
ø
÷=´ ´ ´´´
´
´
æ
è
ç
ö
ø
÷=. . .
.
..
. .
N

8.114 a) d t=´
´
´
æ
è
ç
ö
ø
÷= =´ ´´
´
´
æ
è
ç
ö
ø
÷
- -
.
.
. . . .
.
. .
386
1510
206
00949
1
2
12220059
1510
206
5
2
0
2
5
2
m
=..6 Pa
b) d t=´
´
æ
è
ç
ö
ø
÷= =´ ´´
´
æ
è
ç
ö
ø
÷
- -
. . . .
. .
386
10
206
00552
1
2
100020059
10
206
6
2
0
2
6
2
m
=286 Pa.

8.115 uy U
u
y
Uy
u
y
U
y
() . . /.
/ /
== = =
¥ ¥
- -
=
¥d
¶
¶
d
¶
¶
d
d

1
7
1
7
6717


¶
¶
d
u
y
y=
should be zero. Thus, this condition is not satisfied.
tm
¶
¶
m d
0
0
171
7
1
0
= = =¥
=
¥
-u
y
U
y
/
. Thus, this is unacceptable and


¶
¶
u
y
at, and near, the wall is not valid.
uU
y y
= -
æ
è
ç
ö
ø
÷
¥
3
2
1
2
3
3
d d
.
uU
y
=
æ
è
ç
ö
ø
÷¥
d
17/
.


8.116 a) Drag =

1
2
0024201215074
15810
2012
1060
15810
2012
031
2
4
2
4
´ ´ ´´
´
´
æ
è
ç
ö
ø
÷-
´
´
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
=
- -
. ( ).
. .
..
.
lb


cubic
turb (power-law)
U
y
u
u

154. 194
b) Drag =

1
2
0024201215074
15810
2012
1700
15810
2012
027
2
4
2
4
´ ´ ´´
´
´
æ
è
ç
ö
ø
÷-
´
´
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
=
- -
. ( ).
. .
..
.
lb
c) Drag =

1
2
0024201215074
15810
2012
2080
15810
2012
025
2
4
2
4
´ ´ ´´
´
´
æ
è
ç
ö
ø
÷-
´
´
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
=
- -
. ( ).
. .
..
.
lb

8.117 a) Drag =
1
2
10001212074
10
121
1060
10
121
521
2
6
2
6
´ ´ ´´
´
æ
è
ç
ö
ø
÷-
´
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
=
- -
.().
. .
..
.
N
b) Drag =
1
2
10001212074
10
121
1700
10
121
444
2
6
2
6
´ ´ ´´
´
æ
è
ç
ö
ø
÷-
´
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
=
- -
.().
. .
..
.
N
c) Drag =
1
2
10001212074
10
121
2080
10
121
399
2
6
2
6
´ ´ ´´
´
æ
è
ç
ö
ø
÷-
´
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
=
- -
.().
. .
..
.
N

8.118 U
¥
= = =´
´
´
æ
è
ç
ö
ø
÷=60
1000
3600
1667 38100
166710
235
5
2
. .
.
.
.
m/s. 000
1.510
m
-5
d
t r
0
2 2
5
5
2
1
2
1
2
1221667 059
1510
166710
00618= =´ ´ ´
´
´
æ
è
ç
ö
ø
÷
é
ë
ê
ê
ù
û
ú
ú
=
¥
-
Uc
f
. . .
.
.
. .
.
Pa
b) t r
0
2 2
5
5
2
1
2
1
2
1221667
455
06
166710
1510
0151= =´ ´
´
´
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
=
¥
-
Uc
n
f
. .
.
.
.
.
. .
l
Pa
\= = \ =
´
+ \=
-
u n
t
d
d
.
.
.
.
.
.
.
.
.. .
151
122
351
1667
351
244
351
1510
74 585
5
m/s. ml
Both (a) and (b) are in error, however, (b) is more accurate. 0.
p
x
¶
¶
<

8.119 a) 5=
u
tnd
n
(See Fig. 8.24 b). \=
´´
= ´
-
-
d
n
51510
351
21410
5
4.
.
. . m
b) d
m
d d
t t
d
d
d
dd
n
d
U
U udy
U
n
y
dy
u
U
n
y
dy= - = -
æ
è
ç
ö
ø
÷+ -
¥
¥
¥ ¥
òòò
1
25244 374
15
15
0
( ) .. .
.
.
l l
( )= -- -
æ
è
ç
ö
ø
÷ - -
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
¥
= =
u
U
yn
y
y yn
y
y
v
t
n
d
d d
dd
d d
2515 244 374
15878
878
585
.. . .
. .
.
l l
= +- + - =
.
.
[ . ]..
351
1667
2196200082188951437 m
Note: We cannot use zero as a lower limit since the ln-profile does not go to the

155. 195
wall. Hence, we use d
n; the lower limit provides a negligible
contribution to the integral.

8.120 a) Use Eq. 8.6.40: c
n
f
=
´
´
æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
=
-
.
.
.
. .
455
06
30020
15810
000212
4
2
l

b) t r
0
2 21
2
1
2
0024300002120229= =´ ´ ´ =
¥
Uc
f
. . . . psf u
t
= =
.
.
.
229
0024
977 fps.
c) d
n
n
t
==´ ´ = ´
- -5
51581097780910
4 5
u
. /. . . ft
d)
300
977
244
977
15810
74 0228
4
.
.
.
.
.. . .=
´
+ \=
-
ln
d
d ft

8.121 a) t r
0
2 2
6
2
1
2
1
2
100010
455
06
103
10
110= =´ ´
´æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
=
¥
-
Uc
n
f
.
.
.
l
Pa
\= = \==
´
= ´
-
-
u
u
t n
t
d
n110
1000
332
5510
332
15110
6
5
.
.
. . m/s. m
b) uu= =´ =5 5332166
t
. . . m/s
c) y=..15d — Do part (d) first! \=´ =y.. . .1503330005 m
d)
10
332
244
332
10
74 00333
6
.
.
.
.. . .= + \=
-
ln
d
d m

8.122 Assume flat plates with dpdx
n
f/ .
.
.
. .= =
´æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
=
-
0
523
06
10100
10
00163
6
2 C
l

\Drag = 2
1
2
1000101010000163
2
´´ ´ ´´´ =. .163 000 N
To find d
tmax we need u.
t
t0
2
6
2
1
2
100010
455
06
10100
10
709
709
1000
0266=´ ´
´æ
è
ç
ö
ø
÷
é
ë
ê
ù
û
ú
= \= =
-
.
.
.
.
.
ln
u Pa. m/s.

10
266
244
266
10
74 089
6
.
.
.
.. ..= + \ =
-
ln
d
d m
max


8.123 a) Assume a flat plate of width pD.
8
5
15600
Re610.
1.510
UL
n
-
´
===´
´

156. 196

281/5211
drag0.073(610)1.21560010032600 N
22
f
CULDrpp
-
==´´´´´´´=
power3260015489000 W or 655 hp or 1 64 hp/engine
D
FU=´=´= .

b)
3
helium
100
0.167 kg/m.
2.077288
p
RT
r ===
´


airheliumB
FWWV r=-=D´
27
(1.20.167)9.850600/22.3810 p=-´´´´=´
payload =
666
23.8109.81.2101210 N
BFW-=´-´´=´

8.124 u
y
u
xxy
v
x
u
yy
u
y y
= = =- = =
¶y
¶
¶
¶
¶y
¶¶
¶y
¶
¶
¶
¶y
¶
¶
¶
¶y
¶
, , , , .
2 2
2
2
2
3
3

Substitute into Eq. 8.6.45 (with dpdx/ ):=0

¶y
¶
¶y
¶¶
¶y
¶
¶y
¶
n
¶y
¶yxyxy y
2 2
2
3
3
- = .

8.125 We also have

¶y
¶
¶y
¶f
¶f
¶
¶y
¶h
¶h
¶
¶y
¶¶
¶¶y¶
¶f
¶f
¶
¶¶y¶
¶h
¶h
¶x x xxy
y
x
y
x
= + = +,
(/) (/)

2

Recognizing that ¶f¶ ¶f¶ ¶h¶ n/ ,/ ,/ /,x y x
y
U x= = =-
¥1 0
2
3
and
¶h¶ n/ /,yU x=
¥

¶y
¶ nf
¶y
¶h
¶y
¶
¶y
¶f n
¶y
¶hy
U
x
yU
x
= = -
¥ ¥
,
2
3


¶y
¶¶ nf
¶y
¶f¶h nf
¶y
¶h nf
¶y
¶h nf
2 2
3
2
2 3
2xy
U U U yU
= + -
æ
è
ç
ö
ø
÷
¥ ¥ ¥ ¥
-
1
2


¶y
¶ nf
¶y
¶h nf
¶y
¶ nf
¶y
¶hnf
2
2
2
2
3
3
3
3
y
U U
y
U U
=
æ
è
ç
ö
ø
÷ =
¥ ¥ ¥ ¥
,
Equation 8.6.47 then becomes, using U y
¥ =/ /,nfh

h¶y
¶h
h¶y
¶f¶h
h¶y
¶h
h¶y
¶h
¶y
¶f
h¶y
¶h
h¶y
¶hy y yx yx x y
2 2 2
2
2
2
2
2
2 2 2
- -
æ
è
ç
ö
ø
÷- -
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
=
¥
n
n
h¶y
¶h
U
xy
3
3

Multiply by y
2 2
/h and Eq. 8.6.49 results:
-
æ
è
ç
ö
ø
÷+ - =
¥
1
2
2
2 2
2
3
3
f
¶y
¶h
¶y
¶f¶h
¶y
¶h
¶y
¶f
¶y
¶h
n
¶y
¶hnf
U

157. 197
8.126 u
y
Ux
dF
dy
UxF
U
x
UF== = =
¥ ¥
¥
¥
¶y
¶
n
h
¶h
¶
n h
n
h '() '().
We used Eq. 8.6.50 and Eqs. 8.6.48.
( )v
x x
UxF
U
x
FUx
F
x
=-=- =- -
¥
¥
¥
¶y
¶
¶
¶
n
n
n
¶
¶h
¶h
¶

1
2

=- - -
æ
è
ç
ö
ø
÷
¥
¥
¥ -1
2
1
2
32U
x
FUxFy
U
x
n
n
n
'
/

=- + = -
¥ ¥ ¥ ¥1
2 2
1
2
U
x
F
yU
x
U
x
F
U
x
FF
n
n
n n
h ' (').

8.127 The results are shown in Table 8.5.

8.128 a) t
0
2
5
03321225
1510
25
00124= ´ ´
´
´
=
-
. .
.
. . Pa
b) d=
´ ´
=
-
5
15102
5
00122
5
.
. . m
c) v
U
x
FF
max
max
m/s= -
é
ë
ê
ù
û
ú
=
´ ´
´ =
¥
-
n
h
1
2
15105
2
8605000527
5
(')
.
. . .
d) QudyU
dF
d
dyU
dF
d
d
vx
U
= = =
¥ ¥
¥
ò òò
0 00
d dd
h h
h
= - =
´ ´
´ =
¥
¥
-
U
x
U
F F
n
d[()()]
.
. . /05
15102
5
328004
5
ms/m
2


8.129 a) t
0
2
4
4
332002415
1610
615
23910=´ ´
´
´
= ´
-
-
..
.
. . psf
b) d=
´ ´
=
-
5
16106
15
004
4
.
. ft.
c) v
U
x
FF
max
max
fps= -
é
ë
ê
ù
û
ú
=
´ ´
´ =
¥
-
n
h
1
2
161015
6
860500172
4
(')
.
. . .
d) QudyU
x
U
F= = =
´ ´
´ =
¥
¥
-
ò
n
d
d
()
.
. . / .15
16106
15
3280394
4
0
ftsec/ft
2


8.130 At x = 2m, Re = 5 ´ 2/10
-6
= 10
7
. \Assume turbulent from the leading edge.
a) t r
0
2
2
1
2
0455
006
=
¥
U
n
x
.
(.Re)l

158. 198
=´ ´
´
=
1
2
10005
0455
00610
321
2
72
.
(. )
.
ln
Pa
b) u
ttr= = =
0 321100001792/ ./ . m/s

5
01792
244
01792
10
74 00248
6
.
.
.
.. .= + \=
-
ln m or 24.8 mm
d
d
c) Use the 1/7 the power-law equation:
Q y dy= =ò
500248 0109
17
0
00248
(/.) .
/
.
m/s/m
3


8.131 From Table 8.5 we would select h=6:
a)
5
1.5102
660.0147 m
5
x
U
n
d
-
¥
´´
===
b)
5
15.8106
660.047 ft or 0.57 in.
15
x
U
n
d
-
¥
´´
===

8.132 From Table 8.5 we interpolate for F'.=05 to be
h=
-
-
-+=
0503298
0629803298
211157
..
. .
() .
=
´ ´
\=
-
y y
5
15102
000385
5
.
. . m or 3.85 mm
( )v
U
x
FF=
æ
è
ç
ö
ø
÷-
¥
n
h
1
2
' =
´ ´
=
-
15105
2
0207000127
5
.
(.). m/s

uv
yx
¶¶
tm
¶¶
=+
2
"FU
xU
n
r
¥
¥
æö
=ç÷
èø
5
21.510
0.291(1.2)50.011 Pa
25
-
´
==
´


8.133

If v y v y vy= = > = <0 10 0 0 at and at then d d ¶¶, / and continuity demands that
/0. The component, for must then be g reater than ,uxuyU¶¶d >> as shown
in (b); there should be a slight “overshoot”. Also, consider the control volume of
(c) where the lower boundary is just above y v y= =d. , If at large say 0
y=10d, then continuity demands that u out the right area be greater than
:U an “overshoot”. It is not reasonable to assume that v = const as in (a);
y
y = d
v
v
y
y = d
v = 0
v
(a) (b) (c)
U
v
v = 0
u > U

159. 199
reality would demand a profile such as that sketched in (b). The overshoot
would be quite small and is neglected in boundary layer theory.

8.134 uU
y y
= -
æ
è
ç
ö
ø
÷
¥
3
2
1
2
3
3
d d

For the Blasius profile: see Table 8.5.
(This is only a sketch. The student is encouraged
to draw the profiles to scale.)




8.135


8.136 A: 0. (favorable)
p
x
¶
¶
<
B:
¶
¶
p
x
@0.
C: 0. (unfavorable)
p
x
¶
¶
>
D: 0.
p
x
¶
¶
>
E: 0.
p
x
¶
¶
<




y
U
cubic
Blasius
A B C D
y y y y
2U¥
zero velocity
gradient
separation
streamline
backflow
inviscid
profile
low velocity
outside b.l.
y
A
B
C
D
E
d
D
d
B
d
C
d
A
d
E

160. 200
CHAPTER 9

Compressible Flow


9.1
Btuft-lblbmft-lb
0.24 778 32.26012
Btusluglbm-Rslug-R
p
c==
oo

ccR
v p=-= - = =601217164296 4296
1
778
1
322

ft-lb
slug-R

ft-lb
slug-R
Btu
ft-lb
slug
lbm
o o
.


Btu
0.171
lbm-R
=
o


9.2 ccR ckc c
c
k
Rc
k
R
p v p v p
p
p=+ = \=+ -
æ
è
ç
ö
ø
÷=. . . or 1
1

\= -cRkk
p
/().1

9.3 If Ds=0, Eq. 9.1.9 can be written as
cn
T
T
Rn
p
p
n
T
T
n
p
p
p
c R
p
l l l l
2
1
2
1
2
1
2
1
=
æ
è
ç
ö
ø
÷=
æ
è
ç
ö
ø
÷ or
It follows that, using ccR cck
p v p v
=+ = and / ,

T
T
p
p
p
p
Rc
k
p
2
1
2
1
2
1
1
1
=
æ
è
ç
ö
ø
÷=
æ
è
ç
ö
ø
÷
-/
.
Using Eq. 9.1.7,

T
T
p
p
p
p
p
p
k
k
2
1
21
21
2
1
1
1
1
2
2
1
1
= =
æ
è
ç
ö
ø
÷ =
æ
è
ç
ö
ø
÷
- -
r
r
r
r
or
/
.
Finally, this can be written as
p
p
k
2
1
2
1
=
æ
è
ç
ö
ø
÷
r
r
.

9.4 Substitute Eq. 4.5.18 into Eq. 4.5.17 and neglect potential energy change:

&&
&
~~
.
QW
m
VV pp
uu
S
-
=
-
+-+-
2
2
1
2
2
2
1
1
2 1
2 rr

161. 201
Enthalpy is defined in Thermodynamics as hupvup=+=+
~ ~
/.r Therefore,

&&
&
.
QW
m
VV
hh
S
-
=
-
+-
2
2
1
2
2 1
2

Assume the fluid is an ideal gas with constant specific heat so that D DhcT
p
= .
Then
( )
&&
&
.
QW
m
VV
cTT
S
p
-
=
-
+ -
2
2
1
2
2 1
2

Next, let ccR kcc cRkk
p v p v p
=+ = =- and so that / / ().1 Then, with the ideal gas
law TpR=/,r the first law takes the form

&&
&
.
QW
m
VV k
k
pp
S-
=
-
+
-
-
æ
è
ç
ö
ø
÷
2
2
1
2
2
2
1
12 1rr


9.5 Differentiate p c dxyydxxdy
k
r
-
= = + using () :
r r r
- --
- =
k k
dppk d
1
0.
Rewrite:

dp
d
k
p
rr
=.

9.6 The speed of sound is given by
cdpd= /.r
For an isothermal process TRp K K= =/ ,r where is a constant. This can be
differentiated:
dpKdRTd= =r r.
Hence, the speed of sound is
cRT= .

9.7 Eq. 9.1.4 with
&&
QW
V
cT
S p
== + =0
2
2
is: cons't.

V
cT
VV
cTT
V VV V
cTcT
p p p p
2 2 2 2
2 2
2
2
+ =
+
+ + =
+ +
+ +
( )
( )
()
.
D
D
D D
D

2
2()
0
22
VVVDD
\=+ . .
ppcTVVcTh+D\-D=D=D
We neglected (DV)
2
. The velocity of a small wave is Vc hcV= \=-. . D D

9.8 For water
r
r
dp
d
= ´211010
6
Pa
Since r=1000 kg/m we see that
3
,

162. 202
cdpd= /r
= ´ =21101010001453
6
/ m/s

9.9 For water c
pdp
d
= @ =
´
=
D
Dr r
211010
1000
1453
6
m/s.
L= ´ ´velocity time = 1453 0.6 = 872 m.

9.10 Since c = 1450 m/s for the small wave, the time increment is
Dt
d
c
== =
10
1450
00069. seconds

9.11 a) M==
´ ´
=
V
c
200
14287288
0588
.
..
b) M600/1.417164660.567.=´´=
c) M= ´ ´ =200142872230668/. ..
d) M600/1.417163920.618.=´´=
e) M= ´ ´ =200142872380647/. ..

9.12 ckRT dct= = ´ ´ = \== ´ =14287263256 256121309. . . m/s. m

9.13 a) Assume T = 20°C:
ckRT= = ´ ´ =14287293343. m/s.
dct== ´=D3432686 m

b) Assume T = 70°F:
ckRT= = ´ ´ =1417165301130. fps.
dct== ´=D113022260 ft.
For every second that passes, the lightning flashed about 1000 ft away. Count 5
seconds and it is approximately one mile away.

9.14 c
M
c
V
= ´ ´ = ==14287263256
1
. sin . m/s. a

1000
sin0.256.tan0.2648.3776 m L
L
aa=\==\=
Dt= =
3776
1000
3776.. s




1000 m
V
L

163. 203
9.15 Use Eq. 9.2.13:
a)
c
V
V= =
´ ´
=sin
.
sin
a or m/s
14287288
22
908
o


b)
c
V
V= =
´ ´
=sin
.
sin
a or fps
141716519
22
2980
o


9.16 Eq. 9.2.4: D
D D
V
p
c
p
kRT
=-=- =-
´ ´
=-
r r
03
00237141716519
0113
.
. .
. . fps
Energy Eq:
V
cT
VV
cTT VVcT
p p p
2 2
2 2
0+ =
=
+ + \= +
( )
( ). .
D
D D D
\ =- =- ´ ´ - ´ ´ =D DTcVc
p
/ . (.)/(. .). .1417165191130247783220021
o
F
Note: c
p
= ´ ´ =´ ´. . . . .24 778 322 24778322
Btu
lbm-F
ft-lb
Btu
lbm
slug
ft-lb
slug-F
o o

Then
ft
ft-lb/(slug-F)
ftlb-secF
secft-lb-ft
F.
2 2 2
2
/sec
2
o
o
o
=
- -
-
= (units can be a pain!)

9.17 a) r r r r r r r r rAV AV AdVAVd AddVVdA dAdVVddAddAdV= + + + + + + +
Keep only the first order terms (the higher order terms—those with more
than one differential quantity—will be negligible):
0= + +r rrAdVAVd VdA
Divide by rAV:

dV
V
ddA
A
++=
r
r
0
b) Expand the r.h.s. of Eq. 9.3.5 (keep only first order terms):

V k
k
pV VdV k
k
pdp
d
2 2
2 1
2
2 1
+
-
=
+
+
-
+
+r rr
.
Hence,
0
2
2 1
= +
-
+
+
-
æ
è
ç
ö
ø
÷
VdV k
k
pdp
d
p
rrr

= +
-
+ --
+
æ
è
ç
ö
ø
÷VdV
k
k
pdpppd
d1
2
rr r r
rrr

= +
-
-æ
è
ç
ö
ø
÷VdV
k
k
dppd
1
2
r r
r

where we neglected rr rd compared to
2
. For an isentropic process Eq. 9.2.8
gives r rdpkpd= , so the above becomes

164. 204
0
1
2
= +
-
-
VdV
k
k
kpdpdr r
r

= +
-
-
VdV
k
k
kpd
1
1
2
()r
r
= +VdVk
p
d
r
r
2

But d dVVdAArr/ / /=- - so that the above equation is
0= + --
æ
è
ç
ö
ø
÷VdVk
pdV
V
dA
Ar

which can be written as

dA
A
V
kp
dV
V
= -
æ
è
ç
ö
ø
÷
2
1
r
.
Since ckp
2
=/,r and M = V/c, this is put in the form

dA
A
V
c
dV
V
= -
æ
è
ç
ö
ø
÷
2
2
1 or ( )
dA
A
dV
V
= -M
2
1
c) Substituting in V cckRT Rck k
p= = =-M and we find, , /()/,
2
1

T
T
V
cT
c
cT
kRT
cT
p p p
0
2 22 2
2
1
2
1
2
1= += += +
M M

=
-
+=+
-M
M
2
21
2
11
1
2
kk
k
k()
.
d) &
/()
/
mp
k
TR
A
p
k
T
k
k
R
A
k k
= =
+
-æ
è
ç
ö
ø
÷
+
-æ
è
ç
ö
ø
÷
-
-
M
M
M
M
0
2
1
0
2
12
1
1
2
1
1
2

= +
-æ
è
ç
ö
ø
÷
+
-
p
k
RT
A
k
k
k
0
0
2
1
21
1
1
2
M M
()

At the critical area A
* *
, . M=1 Hence,
& .
*
()
mp
k
RT
A
k
k
k
=
+æ
è
ç
ö
ø
÷
+
-
0
0
1
211
2

e) Since &m is constant throughout the nozzle, we can equate Eq. 9.3.17 to
Eq. 9.3.18:
p
k
RT
A
k
p
k
RT
A
k
k
k
k
k
0
0
2
1
21
0
0
1
21
1
1
2
1
2
M M+
-æ
è
ç
ö
ø
÷ =
+æ
è
ç
ö
ø
÷
+
-
+
-()
*
()

or

A
A
k
k
k
k
*
()()
=
+-
+
é
ë
ê
ù
û
ú
+
-12 1
1
2
1
21
M
M

165. 205
9.18 a)
atm
1069.91079.9 kPa abs.
s
pp=+=+=

1 69.9 kPa abs.p=

From 1 ® s :
V pp p
p
s
s
s
s
k
1
2
1
1
1
1
1
114
2
906
799
699
0997+= =
æ
è
ç
ö
ø
÷=
æ
è
ç
ö
ø
÷=
rr
rr. .
.
.
. .
/
/.
kg/m
3

\ + = \=
V
V
1
2
1
2
69 79
773
900
.906
900
.997
m/s. . .
b) p p
s= += =26410364 264
1. . . kPa abs. kPa abs.
From 1 ® s :
V pp p
p
s
s
s
s
k
1
2
1
1
1
1
1
114
2
0412
364
264
0518+= =
æ
è
ç
ö
ø
÷=
æ
è
ç
ö
ø
÷=
rr
rr. .
.
.
. .
/
/.
kg/m
3


V
V
1
2
1
2
26 36
111+ = \=
400
.412
400
.518
m/s. .

9.19 a)
1/ 1/1.42
311
1
11
105
. 1.221.254 kg/m.
2101
k
ss
s
s
ppVp
p
rr
rr
æö æö
+==== ç÷ ç÷
èøèø


2
1
1
1.4101 000105 0001.4
. 81.3 m/s.
2.41.221.254.4
V
V+=\=
b)
2
1
1
400081.3-81
. 81.0 m/s. % error = 1000.42%.
21.2281.3
V
V=\=´=

9.20 Is p p
r< ´= kPa.. ? . .5283 052832001057
0

a) p p VkRT p
r e e e e
< \ \ = \= = choked flow. M kPa.. . . . .5283 1 1057
0
2

1000298
14287
2
1000 2481 3158´ =
´
+ \= =
.
. . .
T
T T V
e
e e e
K, m/s.
r p
e
m=
´
= \= ´´ ´ =
1057
2872481
1484 1484 01315801473
2.
. .
. .&. . .. . kg/m kg/s
3

b) p p
V
r e
e
> \ < ´ + =
æ
è
ç
ö
ø
÷ M 1000298=
000

e
2
e
. . .
.
.
.
.
.
5283 1
2
14
4
130 130
2002338
0
14
r
r

r r
0
200
287298
2338 17187 2579=
´
= \= \=
.
.. . . . kg/m m/s.
3
e e
V
\= ´´ ´ =&. . .. .m17187 01257901393
2
p kg/s

9.21 Is p p
r< ´= psia.. ? . .5283 05283301585
0

a) p p VkRT
r e e e
< \ = = = 15 choked flow and M psia. .. , . .85 1 1585
2

024530
141716
2778322
024 4417 1030.
.
( .)
. . .´ =
´ ´
´
+ \= =
T
T T V
e
e e e
R, fps.
o

V
s1
V
s
=0

166. 206
315.85144
0.003011 slug/ft.
1716441.7
e
r
´
==
´

2
.5
.00301110300.01692 slug/sec.
12
m p
æö
\=´´=
ç÷
èø
&
b) 15.85. M1, and 20 psia.
ree
pp>\<=
3
0
30144
.00475 slug/ft.
1716530
r
´
==
´

1/1.4
320
.00475.003556 slug/ft.
30
er
æö
\==
ç÷
èø

2
1.420144
0.24530(77832.2).
2.4.003556
eV ´
´´=+
\= \= ´
æ
è
ç
ö
ø
÷´ =V m
e
8389 003556
5
12
8389001627
2
. &.
.
.. . fps. slug/secp
(Note:
ft-lbft-lb
0.24 Btu/lbm-R=0.247780.2477832.2.)
lbm-Rslug-R
c
p
=´=´´
o
oo


9.22 a) p p p T
r e e e< \ =\= ´= = ´= M kPa. K.. . . . . . .5283 1 52832001057 83332982483
0

r
e e
V=
´
= = ´ ´ =
1057
2872483
1483 1428724833159
.
. .
. . . . . kg/m m/s.
3

\= ´´ ´ =&. . .. .m1483 01315901472
2
p kg/s
b) p p p
p
p
T T
r e
e
e e> \= = \ = = kPa, M . . .. ., .5283 130 065 81 884
0
0
0

r
e e
V=
´
= = ´ ´ =
130
2872634
1719 811428726342635
. .
. , .. . . kg/m m/s.
3

\= ´´ ´ =&. . .. .m1719 01263501423
2
p kg/s

9.23 a) p p p
r e e< \ =\= ´= M psia. . . . . .5283 1 5283301585
0

T
e
= ´ =. .83335304416
o
R.
\=
´
´
= = ´ ´ =r
e eV
1585144
17164416
003012 14171644161030
.
.
. . . .
slug
ft
fps.
3

&.
.
. .m= ´
æ
è
ç
ö
ø
÷´ =003012
5
12
1030001692
2
p slug/sec
b) p p p
p
p
T T
r e
e
e e> \= == \ = = psia. M . . .. .. . .5283 20
20
30
6667 785 0890
0
0
0

\=
´
´
= = ´ ´ =r
0
20144
1716472
00356 785141716472836. . . . fps.V
e

\=
æ
è
ç
ö
ø
÷´ =&.
.
. .m00356
5
12
836001664
2
p slug/sec

167. 207

9.24 p T
e e= ´ = = ´ =. . . . .52834002113 83333032525 kPa abs K.
V m
e
= ´ ´ = \=
´
´ ´ =1428725253185
2113
2872525
053185729
2
. . . &
.
. .
. .. . m/s. kg/sp

9.25 p p p T
e e= = \= = ´=. . . . . .5283 101 1912 83332832358
0 0 kPa kPa abs K.
V m
e
= ´ ´ = \=
´
´ ´ =1428723583078
101
2872358
033078130
2
. . . &
. .
. .. . m/s. kg/sp
p p p T
e e0 0219123824 5283 2020 2358=´ = = = =. . . . . kPa abs. kPa abs. K.
V m
e e
= = \=
´
´ ´ =3078 1
202
2872358
033078260
2
. .&
. .
. .. . m/s since M kg/sp

9.26 p p p T
e e
= = \= = ´ =. . . . . . .5283 147 2783 83335004166
0 0
psia psia R.
o

1.41716416.61000 fps.
e
V=´´=
3
0.3203 kg/m and 199.4 kPa abs.
ee
pr\==

00
227.83. 0.5283 29.4 psia, 416. 6R, 1000 fps.
eee
pppTV=´====
o

\=&. .m0202 slug/sec

9.27 Treat the pipeline as a reservoir. Then, p p
e= =. .5283 2645
0 kPa abs.
M and m/s.
e eV= = ´ ´ =1 1428783332833078. (. ) .
&
.
. (. )
..m=
´ ´
´´ ´ =
-2645
2878333283
30103078361
4
kg/s.
D
D
-= =
´´
´ ´
=V
mt& .
./(.. )
.
r
361660
26452878333283
333 m
3


9.28 5193300
16672077
2
5193 225 200
225
300
1667
667
´ =
´
+ \ = \ =
æ
è
ç
ö
ø
÷
.
.
.
.
K.
T
T T p
e
e e e

=97.45 kPa.
Next, T p V
t t t= = \= ´ ´ =225 9745 166720772258826 K, kPa; m/s.. . .

322
97.45
= 0.2085 kg/m. 0.2085 × × .03 × 882.6 = 0.075
2.077225
tee
Vrprp=´
´

5193300
2
1667
667
200
2002077300
1330
2
1667
1667
´ =+ =
´
æ
è
ç
ö
ø
÷=
V p
p
e e
e
e
e
e
.
.
.
/.
.
.
r
r
r kPa.
=+ ´ ´
-
V
V
e
e
2
3 667
2
332410954. .
.

or 311610 6342010 918
6 2 3 667
. . .
.
´ =+ ´ =
-
V V V
e e e
Trial-and-error: m/s.

3
0.3203 kg/m and 199.4 kPa abs.
ee
pr\==

168. 208
9.29 r r
1
1
1
2
114
300100
287293
4757 4757
340
400
4236= =
+
´
= =
æ
è
ç
ö
ø
÷=
p
RT.
. . . . .
/.
kg/m kg/m
3 3

V V V V
1
2
2
2
2 1
475710 42365 4492´ ´ =´ ´ \=. . . . .

V k
k
pV k
k
pV V
1
2
1
1
2
2
2
2
1
2 2
1
2
2 1 2 1 2
14
4
400 4492
2
14
4
340
+
-
=+
-
+ = +
r r
.
.
.
. .
.
.
000
4.757
000
4.236

\=V
1
3735. .
m
s

\= = ´´ ´ =& . . . . .m AVr p
111
2
4757 0537351395 kg/s

9.30 r
1
1
1
45147144
1716520
0009634= =
+
´
=
p
RT
( .)
. .
slug
ft

3

slug/ft
3
r
2
114
009634
507
597
008573=
æ
è
ç
ö
ø
÷=.
.
.
. .
/.

V V V V
1
2
2
2
2 1
0096344 0085732 4495´ ´=´ ´ \=. . . . .

V V
V
1
2 2
1
2
1
2
14
4
5971444495
2
14
4
507144
1219+
´
= +
´
\=
.
.
. . .
.
.
. .
.009634 .008573
fps.
\= ´ ´ =&. (/) .. .m009634 212121901025
2
p slug/sec

9.31 Energy 0 ® 2: 1000303
2
1000 3
2
2
2 2 2
´ =+ =
V
TV kRT .

2
1.6272032.5 kPa.p\=´=
l



1.4
.4 3
22
107.95.39
2005.390 kPa. 0.1740 kg/m.
303.287107.9
p r
æö
\====
ç÷
´èø

Energy 0 ® 1: 1000303
2
1000
14287
3184 2523
1
2
1 1
´ =+
´
\= =
V
V T
V
m/s, K.
1
2
.
. . .
p
1
14
4
1
200
2523
303
1054
1054
2872523
1455=
æ
è
ç
ö
ø
÷= =
´
=
.
.
.
. .
. .
.
.
kPa. kg/m
3
r
Continuity: 1455053184174
4
3142871079 02065
2 2
2
2
. . .. . .. . .p p´ ´ = ´ ´ ´ \=
d
d m

9.32 VkRT
T
T T V
t t
t
t t t
2
1000293
14287
2
1000 2440 3131= ´ =
´
+ \= =.
.
. . .

K. m/s.
\=
æ
è
ç
ö
ø
÷= \=
´
=p
t t
500
244
293
2635
2635
287244
3763
14
4
.
.
.
.
.
. . kPa abs. kg/m
3
r
0 1 2

169. 209

2
22
1.41.4
1.4263 500
1000293. 3.763.025313.1.075.
2.4 3.763
eee
ee
e e
Vpp
Vprp
r r
´=+´´´=´=

2
6.4
293 000=1.01410. Trial-and-error: 22.2 m/s, 659 m/s.
2
e
ee
V
VV
-
\+´=
\= \=r
e e
p589701987 4942 429.,. . . ,. . kg/m kPa kPa abs
3


9.33
*
0
9. 0.997 from Table D.1. 500.997498.5 kPa.
ee
e
Ap
p
pA
=\=\=´=
and
0
0.00855 from Table D.1. 4.28 kPa abs.
e
e
p
p
p
=\=

9.34 M psia, R.
t
=\= ´ = = ´ =1 5283120634 83335204333. . . . .p T
t t
o

\=r
t. .01228
slug
ft
3
& . . .. . .m
d
d
t
t
== ´ ´ \=101228
4
1417164333 0319
2
p
ft

p
p
T V
e e
0
15
120
125 2014 552520287 2014141716287= = \ = =´= = ´ ´.. .,. . . M R,
e
o

= 684 fps.

A
A
d
d
e
e*
.. .
.
. . .= \ =
´
\=1708
4
1708
319
4
0417
2 2
ft
p p


9.35
*
M4. 10.72, .006586200013.17 kPa, .23812 9369.76 K.
eee
A
pT
A
===´==´=
For
A
A
p p
e*
., .. . . . .= = \= = ´ =1072 0584 9976 9976200019952
0
M kPa abs
e


9.36 Let
1
150
M1. Neglect viscous effects. M0.430.
1.4287303
t
===
´´


22
1
*
.05
1.5007. . 0.0816 m or 8.16 cm.
1.50071.50074
t
tt
dAA
Ad
A
pp´
\=\===\=

9.37 p T
e es= ´= = ´=. . . ..52834002113 83333032525 kPa abs.

303
.96. 254.5 K. 1.4287254.5319.8 m/s.
303252.5
e
ee
T
TV
-
=\=\=´´=
-


2211.3
.05319.87.27 kg/s.
.287254.5
m p\=´´=
´
&

170. 210
9.38 Isentropic flow. Since k = 1.4
for nitrogen, the isentropic table
may be used.
M = =3 4235: ..
*
A
A


3100
31.42973731181 m/s. .9027 kg/m.
.297373
ii
V r=´´===
´

\= =
´
= \= =A
m
V
A
i
ii
t
&
.
. .
.
.
. .
r
10
90271181
000938
00938
4235
000221 m m
2 2

At M = = =3 3571 02722
0 0,. ,. .T Tp p
\== = = ==TT p p
e e0 0
373
3571
1044
100
02722
3670
.
.
.
. K or 772C kPa
o


9.39 Isentropic flow. Since k = 1.4
for nitrogen, the isentropic table
may be used.
M = =3 4235: ..
*
A
A

V
i i= ´ ´ = =
´
´
=31417766603840
15144
1776660
001843. . . fps.
slug
ft
3
r
A A
i t
=
´
= \= =
.
.
. .
.
.
. .
2
0018433840
0283
0283
4235
000667 ft ft
2 2

At M = = =3 3571 02722
0 0,. ,. .T Tp p
\== = == =TT pp
e e0 0
660
3571
1848
15
02722
551
.
.
.
.
o o
R or 1388F psia

9.40 Assume p
e e
= =
´
=101
101
1891273
4198 kPa. Then kg/m
3
r
.
. .
Momentum:
22280 0009.81
. .4198.25.
6
FmVAVV rp
´
===´&
1260 m/s.V\=

9.41 FmVAV= = =
´
=& .
.
. .r r
2 101
287873
403 kg/m (Assume gases are air.)
3


42
1009.81.40320010. 349 m/s. VV
-
´=´´\=

9.42 M M
t e
= =\ = =1 4 294 02980
0
. ; .,. .
*
A
A
p p
e
e


0
.3665 .3665300109.95 K,
e
TT==´=

00
100.0298 . 3356 kPa abs.
e
ppp==\=

i
t
e
V
e
= 0
M
t = 1M > 1
M < 1
~
i
t
e
V
e
= 0
M
t = 1M > 1
M < 1
~
V
e
F
B
p
0
A
0

171. 211
\= ´ ´ =V
e
2941428710995618.. . m/s.
\=
-
´
´ ´ + ´=F
B
100
28710995
056183 2
2 2 2
. .
. . .p p 356 000 412 000 N

9.43 Assume an isentropic flow; Eq. 9.3.13 provides

103
1
1
2
2
1
1.
.
p
p
k k
=+
-æ
è
ç
ö
ø
÷
-
M
Using k = 1.4 this gives M or M
2
00424 0206= =. ..
For standard conditions V c== ´´ =M m/s02061428728870. . .

9.44 a)
22220.98501000. 80 0000.9851000(1000) VpVr´=-=´-
V p
2
2 2
2
2
1
1000
2
14
4
2872830
80
287283
9850
-
+ - ´
æ
è
ç
ö
ø
÷= =
´
=
æ
è
ç
ö
ø
÷
.
.
.
.
. .
r
r kg/m
3

2 2
22
2
10001.4
(9851 065 000)284 300 = 0
22.4985
VV
V-+-+-
\ - + \= =3 3784 784 261 3774
2
2
2 2 2
V V V 300=0. m/s. kg/m
3
r . .
Substitute in and find p
2808= kPa.
M K or 473C
1 2
1000
14287283
2966
808
2873774
746=
´ ´
= =
´
=
.
..
. .
.T
o

M
2
261
14287746
0477=
´ ´
=
.
..
b) M M0.477. kPa
1 2 2 1100014287283297 1012 8096= ´ ´ = \ = = =/. .. . . .p p
T
2 2
2644283748
8096
287748
3771= ´ = \=
´
=. .
.
.
. . K or 475C kg/m
3o
r

9.45 a) r r
1 22
12144
1716500
002014 0020143000=
´
´
= ´ =. .. .
slug
ft

3
V
Momentum:
2212144.0020143000(3000).pV´-=´-
22
22
2
30001.4
17165000.
2.4
Vp
r
æö-
+-´=ç÷
èø

V
V
V
2
2 2 2
2
6
30007
6042
198546042 600610- +
æ
è
ç
ö
ø
÷ - - ´
.
(, . ). = 0.
\ - +´ \= =6 23 1510 833 000725
2
2
2
6
2 2V V V, . .000 =0. fps.
slug
ft

3
r
p
21029=. . psia
M R or 731F
1 2
3000
141716500
274
1029144
171600725
1191=
´ ´
= =
´
´
=
.
..
.
.
.T
o o

172. 212
M
2
833
1417161191
0492=
´ ´
=
.
..
b)
122
M3000/1.417165002.74. M0.493. 8.592121 03.1 psia.p=´´=\==´=
T
2 2
23865001193
1031144
17161193
000725= ´ = \=
´
´
=. .
.
. .
o o
R or 733F slug/ft
3
r

9.46
[ ]
r
r
2
1
2
1
1
2
1
2 2
1
2
1
2
1
2
1
2
1
2
2 1
1
1
1
1
2
4 22
1
2 1
= =
-+
+
+
+
-é
ë
ê
ù
û
ú
-+
=
+
+-
p
p
T
T
k k
k
k
k
k k
k
k
M M
M M
M
M
() ()
()
.
M
1
2 2
1
1
2
1
2
=
+
+
-k
k
p
p
k
k
. (This is Eq. 9.4.12). Substitute into above:

r
r
2
1
2
1
2
1
2
1
2 2
1
1 1 1
4 1 1 1
1 1 1
1 1 1
=
+ + +-
é
ë
ê
ù
û
ú
+- + +-
ì
í
î
ü
ý
þ
=
+ + +-
é
ë
ê
ù
û
ú
++- +
()() ()
()() ()
()()
()()()
.
k k
p
p
k
kk k
p
p
k
k k
p
p
k
k k k
p
p

=
-++
++-
k kpp
k kpp
1 1
1 1
2 1
2 1
()/
()/
.
For a strong schock in which
p
p
k
k
2
1
2
1
1
1
1
>> =
+
-
, .
r
r


9.47 Assume standard conditions: T
1 1
15 101= =
o
C, kPa.r
\= ´ ´ =V
1
214287288680. m/s.

122M2. M.5774. 1.688288486 K. T=\==´=

2 4.5101454 kPa.p=´=

2
.57741.4287486255 m/s.V\=´´=
induced12
680255425 m/s.VVV\=-=-=


The high pressure and high induced velocity cause extreme damage.

9.48 If M then M m/s
2 1 1
5 2645 264514287293908= = \= ´ ´ =., .. . . .V
p
2 2
8002001600
1600
2872285293
833= ´ = =
´ ´
=. .
. (. )
. . kPa abs kg/m
3
r

9.49 If M then M fps
2 1 1
5 2645 26451417165201118= = \= ´ ´ =., .. . . .V
p
2 2
80030240
240144
17162285520
001695= ´= =
´
´ ´
=. .
(. )
. . psia slug/ft
3
r

V
1
V
2
stationary
shock

173. 213
9.50 p T
1 1 12615101264 2233 1000142872233334= ´= = = ´´ =. . . /. ... kPa. K. M
\ = = ´ = = ´ =M kPa. K.
2 2 24578 1285264339 310122336925.. . . . . .p T
For isentropic flow from ‚ ® €: For M = .458, p = .866p
0 and
T T p T= \= = = =. . /. . ./. .960 339866391 6925960721
0 0 0
kPa abs K or 448C
o


9.51 After the shock M kPa abs.
2 24752 10338008264= = ´ =., .p
For isentropic flow from ‚ ® €: For M = .475, p = .857p
0.
\= =p
082648579640/. . kPa abs

9.52
A
A
p p p
e*
. .. . /. . .= \ = = \= =4 147 985 1019851025
0 0
M kPa abs
e

M kPa. K.
t= = ´ = = ´=1 528310255415 83332982483. . . . . .p T
t t
\=
´
= = ´´ =r
t t
V
5415
2872483
7599 1428724833159
.
. .
. . . . . kg/m m/s.
3

\= ´´ ´ =&. . .. .m7599 02531590471
2
p kg/s If throat area is reduced, M
t

remains at 1, r p
t
m= = ´´ ´ =. &. . .. .7599 7599 0231590302
2
kg/m and kg/s
3


9.53 p p
A
A
pp
e
= =\ = =101 4 294 9918
2 1 2 1
kPa = M and . . ., / ..
*

\= = = =p pp
1 1 010199181018 294 0298/. . .,/ .. kPa. At M
\= =p
010180298342./. . kPa abs
M kPa abs K.
t= = ´ = = ´=1 5283342181 83332932441,. . . .p T
t t

\= ´ ´ =V
t
142872441313. . . m/s
M kPa abs K.
1 1 1294 1018 36652931074= = = ´=., . . . .p T
\= ´ ´ =V
1
294142871074611.. . . m/s
M kPa K.
2 24788 101 260910742802= = == ´ =., . . . .p TT
e e

\= ´ ´ =V
2
4788142872802161. . . . m/s

9.54 p p
A
A
pp
e
= =\ = =147 4 294 9918
2 1 2 1
. . . ., / ..
*
psia = M and
\= = = =p pp
1 1 014799181482 294 0298./. . .,/ .. psia. At M
\= =p
014820298497./. . . psia
M psia R.
t
= = ´ = = ´ =1 5283497263 83335204333,. . . . . .p T
t t
o

\= ´ ´ =V
t
14171643331020. . . fps
M psia R.
1 1 1
294 1482 36655201906= = = ´ =., . . . .p T
o

\= ´ ´ =V
1
29414171619061989.. . . fps

174. 214
M psia R.
2 2
4788 147 260919064973= = == ´ =., . . . . .p TT
e e
o

\= ´ ´ =V
2
47881417164973523. . . . fps

9.55 M kPa K.
t t tp T= = ´= = ´=1 5283500264 83332982483. . . . .

A
A
p
1
2
2 1 1
8
5
256 247 06135003065
*
.. .,. ..== \ = = ´ = M
T V
1 1
4512981344 247142871344574= ´ = \= ´ ´ =. . .. . . K. m/s
M kPa K.
2 2 2516 6953065213 210813442833= = ´ = = ´ =., . . . . . .p T
After the shock it’s isentropic flow. At M = =., ..
*
516 1314
A
A
p A
02
2
5115002555
04
1314
003825= ´ = =
´
=. .
.
.
. .
*
kPa. m
2p


A
A
p p
e
e r*
.
.
.. . . . . ..=
´
= \= ´ = =
p05
003825
205 9402555240 298
2
kPa abs = M
e

T V
e e
=
æ
è
ç
ö
ø
÷= \= ´ ´ =2833
213
240
2738 29814287273899
2857
. . . . . .
.
K. m/s

9.56 p p T
t t
= = ´ = =
æ
è
ç
ö
ø
÷ =. .
./.
546 5461200655 673
655
1200
585
0
313
kPa. K.
\=
´
= = ´ ´ = =r
t t
V
655
462585
242 13462585593 1
.
. . . ( .) kg/m m/s. M
3
t

& . . . . .m AV
d
d
ttt
t
t
= \= ´
´
´ \=r
p
m or 6 cm4242
4
593 0060
2

T
e e
=
æ
è
ç
ö
ø
÷ = \=
´
=673
101
1200
3802
101
4623802
575
313./.
.
. .
. . K kg/m
3
r

V
e
2
2
187238021872673+ ´ = ´. . (Energy from € ® e .) (c
p
= ×1872 J/kgK)
\= \= ´ \=V
d
d
e
e
e
1050 4575
4
1050 0092
2
m/s. m or 9.2 cm. . . .
p


9.57 M kPa.
e= = =´ =1 546 5461000546
0. . .p p
e

T
e e
=
æ
è
ç
ö
ø
÷= \=
´
=623
546
1000
542
546
462542
218
3
13
.
.
.
. . K.
kg
m
3
r
V
d
d
e
e
e
= ´ ´ = = ´ \=13462542571 15218
4
571 0124
2
. . . . . m/s. m or 12.4 cm
p

175. 215
9.58 M psia. R.
e
= = ´ = =
æ
è
ç
ö
ø
÷=1 546150819 1160
819
150
1009
3
13
. . .
.
.
.
p T
e e
o

\=
´
´
= = ´ ´ =r
e eV
819144
27621009
00423 13276010091903
.
. . .
slug
ft
fps.
3

.. . . .2500423
4
1903 0199
2
= ´ \=
pd
d
e
e
ft. or 2.39"

9.59 M kPa. K.
t
= = ´ = =
æ
è
ç
ö
ø
÷ =1 5461200655 673
655
1200
585
313
. .
./.
p T
t t

\= ´ ´ = =
´
=V
t t
13462585593
655
462585
242.
.
. . m/s. kg/m
3
r
\= ´´ ´ =&. . .m242 00755930254
2
p kg/s per nozzle
T
e
=
æ
è
ç
ö
ø
÷ =673
120
1200
396
313./.
. K

9.60 M
1
800
14287303
229=
´ ´
=
.
..
From Fig. 9.15, b=4679
o o
,.
a)
1n46. M2.29sin461.65. b=\==
oo


2n2 M.654Msin(4620). \==-
oo
\ =M
2149..





p T
2 2301401204 1423303431= ´= = ´=. . . . kPa abs K.
V
2
14287431149620= ´ ´ ´ =. . . m/s
b) b= \ = = \ = = -79 22979225 541 7920
2
o o o o
. .sin .. . sin( ). M M M
1n 2n

\ =M
20631..
p T
2 257440230 190303576= ´= = ´=. . . kPa abs K.
V
2
14287576631303= ´ ´´ =. . . m/s


c)


V
1
V
2
= 20
o
V
1
= 35
o
a detached
shock

176. 216
9.61 b q
1
40 10= \=
o o
. .
M M M M=1.58.
1n 2n
= = \ = = - \240129 791 4010
2 2
sin .. . sin( ).
o o o

If q b
2 2
10 158 5115851 123= = = = \ =
o o o
then, with M M M
2n 2n
., ..sin . ..
\ = = - \ = =-=-=M M M
3n
. sin( ). .. .824 5110 126 10511041
3 3 2
o o o
bb

9.62 M M K.
1n 2n
= = \ = = ´ =3535201 576 1696303514
2
.sin .. .. .
o
T
M
2 1 2 2
576
3520
226 20 47=
-
= = = \=
.
sin( )
.. . .
o o
o o
q q b
M M M M
2n 3n 3
= = \ = = - \ =22647165 654 4720 144
3
.sin .. . sin( ). ..
o o o

T V kRT
3 3 3 3
1423514731 14414287731780= ´ = = = ´ ´ =. .. . K. M m/s

9.63 M M R.
1n 2n
= = \ = = ´ =3535201 576 1696490831
2
.sin .. .. .
o o
T
M
2 1 2 2
576
3520
226 20 47=
-
= = = \=
.
sin( )
.. . .
o o
o o
q q b
M M M M
2n 3n
= = \ = = - \ =22647165 654 4720 144
3 3
.sin .. . sin( ). ..
o o o

T V kRT
3 3 3 3
14238311180 14414171611802420= ´= = = ´ ´ =. .. .
o
R. M fps
9.64 M M M
1n 2n1 1
3 10 28 328141 736= = \= = = \ =, . . sin .. ..q b
o o o

\= ´=p
2215340861. . kPa.
M kPa
2 3
736
2810
238 6442861555=
-
= \= ´ =
.
sin( )
.. . . .
o o
p
() . .p
3 103340413
normal kPa= ´=

9.65 At M
1 1 1
3 498 1947= = =, ., ..q m
o o
(See Fig. 9.18.)
qq
1 2 2
49825748 478+= += \ =. .. ..
o
M
From isentropic flow table: pp
p
p
p
p
2 1
0
1
2
0
20
1
02722
002452180= =´ ´ =
.
. . . kPa

02
212
10
1
253.1795127K or 146C. 12.08.
.3571
TT
TT
TT
m==´´=-=
oo

V
2
478142871271080 902570531208324= ´ ´ = =+- - =.. . . . .. m/s a
o


9.66 q q
1
264 4 658= = =.. , ..
o o
For M (See Fig. 9.18.)
\= - =q658264394. . ..
o

TT
T
T
T
T
V
2 1
0
1
2
0
2
273
1
5556
2381117 414287117867= = ´ = \= ´ ´ =
.
. . . K. m/s
T
2
156=-
o
C.

177. 217
9.67 q q= = =264 4 658.. , ..
o o
For M \= - =q658264394. . ..
o


02
21
10
1
490.2381210R or 250F.
.5556
TT
TT
TT
==´=-
oo

V
2
41417162102840= ´ ´ =. . fps

9.68 a) q q
1 2 2
391 3915441 272 20
1
0585
04165= = += \ = = ´.. . .. ..
.
.
o o
M
u u
p
= 14.24 kPa.
For q b= = = = = \ =5 25 27 2527113 889
o o o
and M M M
1n 2n
., . .sin .. ..
\= ´=p
213220264
l. . . kPa
M M
l o o
= =
-
=
2
889
275
237
.
sin( )
..
b) M M M
1n 2n
= = \= = = =272 5 25 27225115 875., . . .sin ., ..q b
o o o

M
2u
=
-
=
.
sin( )
..
875
255
256
o o

For M =36.0 For M
2
= =+= =237 36541 258., . , ..q q
o o
l

c) Force on plate = (. .) .26414241000- ´ ´=AF
C
F
VA
A
A
L
= =
´ ´
´´ ´
=
cos ..
..
..
5
1
2
1229961000
1
2
142520000
0139
11
2 2
o
r

d) C
F
VA
A
A
D
= =
´ ´
´´ ´
=
sin . .
..
..
5
1
2
12210000872
1
2
142520000
00122
11
2 2
o
r




9.69 b= = = \= ´= =19 419130 180520361 786
2
o o
. sin .. . . .. M kPa. M
1n 2n
p
M M
2 1 2 3
786
195
325 5436 5936 355=
-
= = = \ =
.
sin( )
.. .. .. ..
o o
q q
pp
p
p
p
p
3 2
0
2
3
0
361
1
0188
0122234= = ´ ´ =.
.
. . kPa.



C
A A
VA
D
=
-
æ
è
ç
ö
ø
÷
=
´
´´´
=
361
2
234
2
5
1
2
6350872
1
2
14420
00025
1
2 2
. .sin
..
.
..
o
r


Lift
Drag
F
Airfoil
surface
M
1
M
2
M
3
p
2
p
3
shock

178. 218
9.70 If q b= = ®5 4
1
o o
with M then Fig. 9.15 =18, .

1n2nM4sin181.24. M.818. ==\=
o

2
1.6272032.5 kPa.p\=´=
l

\ =
-
=M
2l o o
.
sin( )
..
818
185
364


At M At M
2u u1 1 2 1
02
0
4 658 758 488 20
002177
006586
= = = = =, .. . ..
.
.
q
o o
pp
p
p
p
p

= 6.61 kPa.
C
VA
A A A
A
L= =
-´ -´ ´
´´´
=
Lift
1
2
325 520 2661 2 10
1
2
14420
00854
1
2 2
r
.cos /. /cos
.
..
o o

C
VA
A A
A
D
= =
- ´ ´
´´´
=
Drag
1
2
325 5661 2 10
1
2
14420
0010
1
2 2
r
.sin . /sin
.
..
o o







M
1
M
2u
M
2l
shock
shock