1. This document summarizes key concepts from a chapter on basic considerations in thermodynamics and fluid mechanics. It defines concepts such as density, pressure, power, energy, mass flux, and flow rate using mathematical equations involving mass, length, time, and other fundamental units. 2. The document also discusses dimensional analysis and demonstrates how to determine the dimensions of physical quantities like force, viscosity, heat flux, and specific heat. Dimensional equations are derived and simplified to fundamental units. 3. Examples are provided for performing unit conversions between the English and SI systems. Problems are also presented and solved for determining values like density, pressure, and flow rates using the defined equations and given data.

1. 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 = M L
/
3
b) pressure = force/area = F L ML T L M LT
/ / /
2 2 2 2
= =
c) power = force × velocity = F L T ML T L T ML T
× = × =
/ / / /
2 2 3
d) energy = force × distance = ML T L ML T
/ /
2 2 2
× =
e) mass flux = ρAV = M/L3
× L2
× L/T = M/T
f) flow rate = AV = L2
× L/T = L3
/T
1.3 a) density =
M
L
FT L
L
FT L
3
2
3
2 4
/
/
=
b) pressure = F/L2
c) power = F × velocity = F × L/T = FL/T
d) energy = F × L = FL
e) mass flux =
M
T
FT L
T
FT L
= =
2
/
/
f) flow rate = AV = L2
× L/T = L3
/T
1.4 (C) m = F/a or kg = N/m/s2
= N.
s2
/m.
1.5 (B) [µ] = [τ/du/dy] = (F/L2
)/(L/T)/L = F.
T/L2
.
1.6 a) L = [C] T2
. ∴[C] = L/T2
b) F = [C]M. ∴[C] = F/M = ML/T2
M = L/T2
c) L3
/T = [C] L2
L2/3
. ∴[C] = L T L L L T
3 2 2 3 1 3
/
/ /
⋅ ⋅ =
Note: the slope S0 has no dimensions.
1.7 a) m = [C] s2
. ∴[C] = m/s2
b) N = [C] kg. ∴[C] = N/kg = kg ⋅ m/s2
⋅ kg = m/s2
c) m3
/s = [C] m2
m2/3
. ∴[C] = m3
/s⋅m2
⋅ m2/3
= m1/3
/s
1.8 a) pressure: N/m2
= kg ⋅ m/s2
/m2
= kg/m⋅ s2
b) energy: N⋅ m = kg ⋅ m/s2
× m = kg⋅ m2
/s2
c) power: N⋅ m/s = kg ⋅ m2
/s3
d) viscosity: N⋅ s/m2
=
kg m
s
s
1
m
kg / m s
2 2
⋅
⋅ = ⋅

2. 2
e) heat flux: J/s =
N m
s
kg m
s
m
s
kg m / s
2
2 3
⋅
=
⋅
⋅ = ⋅
f) specific heat:
J
kg K
N m
kg K
kg m
s
m
kg K
m / K s
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/s2
= N s / m s N / m,
2 2
⋅ ⋅ = [f] = kg m / s N.
2
⋅ =
Note: we could express the units on c as [c] = kg / s N s / m s N s / m
2
= ⋅ ⋅ = ⋅
1.10 a) 250 kN b) 572 GPa c) 42 nPa d) 17.6 cm3
e) 1.2 cm2
f) 76 mm3
1.11 a) 1.25× 108
N b) 3.21× 10−5
s c) 6.7× 108
Pa
d) 5.6× 10−12
m3
e) 5.2 × 10−2
m2
f) 7.8 × 109
m3
1.12 (A) 8 9
2.36 10 23.6 10 23.6 nPa.
− −
× = × =
1.13 2 2 2
0.06854
0.225 0.738
0.00194 3.281
m m
d d
λ
ρ ρ
= =
×
where m is in slugs, ρ in slug/ft3
and d in feet. We used the conversions in the
front cover.
1.14 a) 20 cm/hr = 5
20
/3600 5.555 10 m/s
100
−
= × 5
20
/3600 5.555 10 m/s
100
−
= ×
b) 2000 rev/min = 2000 2
× π/60 = 209.4 rad/s
c) 50 Hp = 50 × 745.7 = 37 285 W
d) 100 ft3
/min = 100 × 0.02832/60 = 0.0472 m3
/s
e) 2000 kN/cm2
= 2× 106
N/cm2
× 1002
cm2
/m2
= 2× 1010
N/m2
f) 4 slug/min = 4 × 14.59/60 = 0.9727 kg/s
g) 500 g/L = 500 × 10−3
kg/10−3
m3
= 500 kg/m3
h) 500 kWh = 500 × 1000× 3600 = 1.8× 109
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
r r
dr
τ µ µ µ
= = =
1.17 The mass is the same on the earth and the moon:

3. 3
m =
60
32 2
1863
.
. .
= ∴ Wmoon = 1.863× 5.4 = 10.06 lb
1.18 (C) shear sin 4200sin30 2100 N.
F F θ
= = =
o
shear
4
2100
= 84 kPa
250 10
F
A
τ −
= =
×
1.19 a) λ
ρ
= =
×
× ×
= ×
−
−
−
. .
.
. ( . )
.43
225 225
4 8 10
184 3 7 10
10
2
26
10 2
6
m
d
m or 0.00043 mm
b) λ
ρ
= =
×
× ×
= ×
−
−
−
. .
.
. ( . )
.
225 225
4 8 10
00103 3 7 10
7 7 10
2
26
10 2
5
m
d
m or 0.077 mm
c)
26
2 10 2
4.8 10
.225 .225 .0039m
.00002 (3.7 10 )
m
d
λ
ρ
−
−
×
= = =
× ×
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.

4. 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
152 10−
×
o
= 1296 MN/m2
= 1296 MPa.
1.27
4
n
4
t
(120000) .2 10 2.4N
20 .2 10 .0004N
F
F
−
−

= × × = 

= × × = 

F = 2 2
n t
F F
+ = 2.400 N.
θ= tan−1 .0004
2.4
=.0095°
1.28 ρ=
m
V
−
=
0 2
180 1728
.
/
= 1.92 slug/ft3
. τ = ρg = 1.92 × 32.2 = 61.8 lb/ft3
.
1.29 ρ= 1000 − (T − 4)2
/180 = 1000 − (70 − 4)2
/180 = 976 kg/m3
γ = 9800 − (T − 4)2
/18 = 9800 − (70 − 4)2
/180 = 9560 N/m3
% error for ρ=
976 978
978
−
× 100 = −.20%
% error for γ =
9560 978 9.81
978 9.81
− ×
×
× 100 = −.36%
1.30 S = 13.6 − .0024T = 13.6 − .0024 × 50 = 13.48.
% error =
13.48 13.6
13.6
−
× 100 = −.88%
1.31 a) m =
W V
g
γ
=
6
12 400 500 10
9.81
g
−
× ×
= = 0.632 kg
b) m =
6
12 400 500 10
9.77
−
× ×
= 0.635 kg
c) m =
6
12 400 500 10
9.83
−
× ×
= 0.631 kg
1.32 S =
/
water
m V
ρ
ρ
=
10/
. 1.2
water
V
ρ
= .
1.94
∴ V = 4.30 ft3
1.33 (D)
2 2
3
water
( 4) (80 4)
1000 1000 968 kg/m
180 180
T
ρ
− −
= − = − =

5. 5
1.34 τ= µ
du
dr
= 1.92 × 10−5
2
30(2 1/12)
(1/12)
 
×
 
 
= 0.014 lb/ft2
1.35 T = force × moment arm = τ2πRL × R = µ
du
dr
2πR2
L = µ
2
0.4
1000
R
 
+
 
 
2πR2
L.
∴µ=
2 2
2
0.0026
0.4 0.4
1000 2 1000 2 .01 0.2
12
T
R L
R
π π
=
   
+ + × ×
   
   
= 0.414 N.
s/m2
.
1.36 Use Eq.1.5.8: T =
3
2 R L
h
π ω µ
=
( )
2 .5 / 12
π
π
× ×
×
× ×
3 2000 2
60
4 006
01 12
.
. /
= 2.74 ft-lb.
Hp =
Tω
550
2 74 209
550
=
×
. .4
= 1.04 Hp
1.37 Fbelt = µ
du
dy
A = × −
131 10
10
002
3
.
.
(.6 × 4) = 15.7 N.
Hp =
F V
×
=
×
746
15 7 10
746
.
= 0.210 Hp
1.38 Assume a linear velocity so .
du r
dy h
ω
= Due to the area
element shown, dT = dF × r = τdA × r = µ
du
dy
2πr dr × r.
dr
r
τ
T = 3
0
2
R
r dr
h
µω π
∫ =
2
4
2 36 10
400 2
60
3 12
2 08 12
4
5 4
πµω
π
π
h
R
=
× × ×
×
×
×
−
. ( / )
. /
= 91 × 10−5
ft-lb.
1.39
2
30(2 1/12)
(1/12)
 
×
 
 
2 2
0 0
[32 / ] 32 / .
du
r r r r
dr
τ µ µ µ
= = = ∴τr = 0 = 0,
τr=0.25 = 32 × 1 × 10−3
× 2
.25/100
(.5/100)
= 3.2 Pa, τr=0.5 = 32 × 1 × 10−3
× 2
.5/100
(.5/100)
= 6.4 Pa
1.40 (A) 3
[10 5000 ] 10 10 5000 0.02 1 Pa.
du
r
dr
τ µ µ −
= = × = × × × =
1.41 The velocity at a radius r is rω. The shear stress is τ µ
=
∆
∆
u
y
.
The torque is dT = τrdA on a differential element. We have

6. 6
0.08
0
= = 2
0.0002
r
T rdA rdx
ω
τ µ π
∫ ∫ ,
2000 2
209.4 rad/s
60
π
ω
×
= =
where x is measured along the rotating surface. From the geometry 2
x = r, so that
0.08 0.08
2 3
0 0
209.4 / 2 329000
= 0.1 2 329000 (0.08 )
0.0002 3
2
x x
T dx x dx
π
×
= =
∫ ∫ = 56.1 N .
m
1.42 If τ µ
=
du
dy
= cons’t and µ= AeB/T
= AeBy/K
= AeCy
, then
AeCy du
dy
= cons’t. ∴
du
dy
= De−Cy
.
Finally,
0 0
y
u
Cy
du De dy
−
=
∫ ∫ or u(y) = 0
y
Cy
D
e
C
−
− = E (e−Cy
− 1)
where A, B, C, D, E, and K are constants.
1.43
µ = =
=



Ae Ae
Ae
B T B
B
/ /
/
.
.
001
000357
293
353
∴A = 2.334 × 10−6
, B = 1776.
µ40 = 2.334 × 10−6
e1776/313
= 6.80 × 10−4
N.
s/m2
1.44 m = ρV . Then dm = ρd V + V dρ. Assume mass to be constant in a volume
subjected to a pressure increase; then dm = 0. ∴ρd V = − V dρ, or
d V
V
.
d ρ
ρ
= −
1.45 B =
V
−
p
V
∆
∆
2200 MPa.
= V
∴∆
V
−
=
2 10
2200
p
B
∆ − ×
= = −0.00909 m3
or −9090 cm3
1.46 Use c = 1450 m/s. L = c∆t = 1450 × 0.62 = 899 m
1.47 =
B V
p
∆
∆ −
V
= −2100
−13
20
.
= 136.5 MPa
1.48 a) 327,000 144/1.93
c = × = 4670 fps b) 327,000 144/1.93
c = × = 4940 fps
c) 308,000 144/1.87
c = × = 4870 fps
1.49 V
∆ =3.8 × 10−4
× −20 × 1 = .0076 m3
.
∆p = −B
V
∆
V
.0076
2270
1
−
= − = 17.25 MPa
1.50 p =
2 2 0741
5 10 6
σ
R
=
×
× −
.
= 2.96 × 104
Pa or 29.6 kPa. Bubbles: p = 4σ/R = 59.3 kPa

7. 7
1.51 Use Table B.1: σ= 0.00504 lb/ft. ∴p =
4 4 .00504
1/32 12
R
σ ×
=
×
= 7.74 psf or 0.0538 psi
1.52 See Example 1.4: h =
4 cos 4 0.0736 0.866
0.130 m.
1000 9.81 0.0002
gD
σ β
ρ
× ×
= =
× ×
1.53 (D)
6
4 cos 4 0.0736 1
3 m or 300 cm.
1000 9.81 10 10
h
gD
σ β
ρ −
× ×
= = =
× × ×
1.54 See Example 1.4: h =
4 cos 4 0.032cos130
1.94 13.6 32.2 0.8/12
gD
σ β
ρ
×
=
× × ×
o
= −0.00145 ft or −0.0174 in
1.55 force up = σ× L × 2 cosβ= force down = ρghtL. ∴h =
2σ β
ρ
cos
.
gt
1.56 Draw a free-body diagram:
The force must balance:
W = 2σL or
π
ρ σ
d
L g L
2
4
2





 = .
∴ =
d
g
8σ
πρ
W
σL σL
needle
1.57 From the free-body diagram in No. 1.47, a force balance yields:
Is
π
ρ
d
g
2
4
< 2σ?
π(. )
. .
004
4
7850 9 81 2 0741
2
× < ×
0.968 < 0.1482 ∴No
1.58 Each surface tension force = σ× πD. There is a force on the outside
and one on the inside of the ring.
∴F = 2σπD neglecting the weight of the ring.
F
D
1.59
h(x)
h
dW
σdl
From the infinitesimal free-body shown:
cos .
d gh xdx
σ θ ρ α
=
l cosθ=
dx
dl
.
/
d d x d
h
g xdx g x
σ σ
ρ α ρ α
∴ = =
l l
We assumed small αso that the element
thickness is αx.

8. 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 ρ= =
× +
=
p
RT
1013
0 287 273 15
.
. ( )
1.226 kg/m3
. γ = 1.226 × 9.81 = 12.03 N/m3
1.67 3
in
101.3
1.226 kg/m .
0.287 (15 273)
p
RT
ρ = = =
× +
3
out
85
1.19 kg/m .
0.287 248
ρ = =
×
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 3
750 44
0.1339 slug/ft .
1716 470
p
RT
ρ
×
= = =
×
m V
ρ
= 0.1339 15 2.01 slug.
= × =
1.69 (C)
pV
m =
800 4
59.95 kg
0.1886 (10 273)
RT
×
= =
× +
1.70
p
W V
RT
=
100
(10 20 4) 9.81 9333 N.
0.287 293
g = × × × × =
×

9. 9
1.71 Assume that the steel belts and tire rigidity result in a constant volume so that m1
= m2:
V 1 V
= 1 1 2 2
2
1 2
2
2 1
1
or .
150 460
(35 14.7) 67.4 psia or 52.7 psi gage.
10 460
m RT m RT
p p
T
p p
T
=
+
∴ = = + =
− +
1.72 The pressure holding up the mass is 100 kPa. Hence, using pA = W, we have
100000 1 9.81. 10200 kg.
m m
× = × ∴ =
Hence,
pV
m =
3
100 4 / 3
10200. 12.6 m or 25.2 m.
0.287 288
r
r d
RT
π
×
= = ∴ = =
×
1.73 2 2
1
0 ( 10). 20 32.2. 25.4 fps.
2
KE PE mV mg V V
= ∆ + ∆ = + − ∴ = × ∴ =
2 2
1
0 ( 20). 40 32.2. 35.9 fps.
2
mV mg V V
= + − ∴ = × ∴ =
1.74 2 2
1-2
1
. a) 200 0 5( 10 ). 19.15 m/s.
2
f f
W KE V V
= ∆ × = × − ∴ =
b)
10
2 2
0
1
20 15( 10 ).
2
f
sds V
= × −
∫
2
2 2
10 1
20 15( 10 ). 15.27 m/s.
2 2
f f
V V
× = × − ∴ =
c)
10
2 2
0
1
200cos 15( 10 ).
20 2
f
s
ds V
π
= × −
∫
2 2
20 1
200sin 15( 10 ). 16.42 m/s.
2 2
f f
V V
π
π
× = × − ∴ =
1.75 2
1 2 1 2 2 1
1
. 10 40 0.2 0 . 40000.
2
E E u u u u
= × × + = + ∴ − =
% % % %
40000
. 55.8 C where comes from Table B.4.
717
v v
u c T T c
∆ = ∆ ∴∆ = = o
%
The following shows that the units check:
2 2 2 2 2
car
2 2 2
air
kg m / s m kg C m kg C
C
kg J/(kg C) N m s (kg m/s ) m s
m V
m c
 
× ⋅ ⋅ ⋅ ⋅ ⋅
= = = =
 
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
 
 
o o
o
o
where we used N = kg.
m/s2
from Newton’s 2nd
law.

10. 10
1.76 2
2
2 1 H O
1
. .
2
E E mV m c T
= = ∆
2
6
1 100 1000
1500 1000 2000 10 4180 . 69.2 C.
2 3600
T T
−
×
 
× × = × × × ∆ ∴∆ =
 
 
o
We used c = 4180 J/kg.
C
o
from Table B.5. (See Problem 1.75 for a units check.)
1.77 water . 0.2 40000 100 4.18 . 19.1 C.
f f
m h m c T T T
= ∆ × = × ∆ ∴∆ = 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) ice water ice water water
. 320 .
E E m m c T
∆ = ∆ × = × ∆
6 3
5 (40 10 ) 1000 320 (2 10 ) 1000 4.18 . 7.66 C.
T T
− −
× × × × = × × × ∆ ∴∆ = o
We assumed the density of ice to be equal to that of water, namely 1000
kg/m3
. Ice is actually slightly lighter than water, but it is not necessary for
such accuracy in this problem.
1.79. W pdV
=
mRT
V
=
∫ d V
d V
mRT
=
∫
V
ln
V
mRT
=
∫
2
V
2
1 1
ln
p
mRT
p
=
since, for the T = const process, 1
p V 1 2
p V
= 2. Finally,
1-2
4 1
1716 530ln 78,310 ft-lb.
32.2 2
W = × × = −
The 1st
law states that
0. 78,310 ft-lb or 101 Btu.
v
Q W u mc T Q W
− = ∆ = ∆ = ∴ = = − −
%
1.80 If the volume is fixed the reversible work is zero since the boundary does not
move. Also, since V 1 2
1 2
,
mRT T T
p p p
= = so the temperature doubles if the
pressure doubles. Hence, using Table B.4 and Eq. 1.7.17,
200 2
a) (1.004 0.287)(2 293 293) 999 kJ
0.287 293
v
Q mc T
×
= ∆ = − × − =
×
b)
200 2
(1.004 0.287)(2 373 373) 999 kJ
0.287 373
v
Q mc T
×
= ∆ = − × − =
×
c)
200 2
(1.004 0.287)(2 473 473) 999 kJ
0.287 473
v
Q mc T
×
= ∆ = − × − =
×
1.81 W pdV
= (
p V
= 2 V
− 1
1). If = const,
T
p
V
2
1
T
V
= 2 1
2
so if 2 ,
T T
=
∫
then V 2 2V
= 1 and (2
W p V
= 1 V
− 1) pV
= 1 1.
mRT
=
a) 2 0.287 333 191 kJ
W = × × =
b) 2 0.287 423 243 kJ
W = × × =

11. 11
c) 2 0.287 473 272 kJ
W = × × =
1.82 = 1.4 287 318 357 m/s. 357 8.32 2970 m.
c kRT L c t
= × × = = ∆ = × =
1/ 0.4/1.4
2
2 1
1
500
(20 273) 151.8 K or 121.2 C
5000
k k
p
T T
p
−
   
= = + = −
   
 
 
o
1.83 We assume an isentropic process for the maximum pressure:
/ 1 1.4/0.4
2
2 1
1
423
(150 100) 904 kPa abs or 804 kPa gage.
293
k k
T
p p
T
−
   
= = + =
   
 
 
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
2 1
1
473
100 534 kPa abs.
293
k k
T
p p
T
−
   
= = =
   
 
 
2 1
( ) (1.004 0.287)(473 293) 129 kJ/kg.
v
w u c T T
=−∆ = − − = − − − = −
We used Eq. 1.7.17 for cv.
1.85 a) 1.4 287 293 343.1 m/s
c kRT
= = × × =
b) 1.4 188.9 293 266.9 m/s
c kRT
= = × × =
c) 1.4 296.8 293 348.9 m/s
c kRT
= = × × =
d) 1.4 4124 293 1301 m/s
c kRT
= = × × =
e) 1.4 461.5 293 424.1 m/s
c kRT
= = × × =
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, 287 323 304 m/s.
c RT
= = × =
1.87 At 10 000 m the speed of sound 1.4 287 223 299 m/s.
c kRT
= = × × =
At sea level, 1.4 287 288 340 m/s.
c kRT
= = × × =
340 299
% decrease 100 12.06 %.
340
−
= × =
1.88 a) = 1.4 287 253 319 m/s. 319 8.32 2654 m.
c kRT L c t
= × × = = ∆ = × =
b) = 1.4 287 293 343 m/s. 343 8.32 2854 m.
c kRT L c t
= × × = = ∆ = × =
c) = 1.4 287 318 357 m/s. 357 8.32 2970 m.
c kRT L c t
= × × = = ∆ = × =

12. 12
C HAPTER 2
Fluid Statics
2.1 Σ ∆ ∆
∆ ∆
F ma p z p s
y z
a
y y y y
= − =
: sinα ρ
2
Σ ∆ ∆
∆ ∆ ∆ ∆
F ma p y p s
y z
a g
y z
z z z z
= − = +
: cosα ρ ρ
2 2
Since ∆ ∆
s y
cosα = and ∆ ∆
s z
sin α = , we have
p p
y
a
y y
− = ρ
∆
2
and ( )
p p
z
a g
z z
− = +
ρ
∆
2
Let ∆y → 0 and ∆z → 0:
p p
p p
y
z
− =
− =



0
0
∴ = =
p p p
y z .
2.2 p = γh. 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/γ. 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) (13.6 9810) (28.5 0.0254) 96600 Pa
Hg
p h
γ
= = × × × =
2.5 S =
20 144
62.4 20
p
h
γ
×
=
×
= 2.31. ρ = 1.94 × 2.31 = 4.48 slug/ft3
.
2.6 a) p = γh = 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 = γ1h1 + γ2h2 = 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
pz∆y
py∆z
p∆s
∆s
α
∆z
∆y
ρg∆V

13. 13
2.8 ∆p = ρgh = 0.0024 × 32.2 (–10,000) = –773 psf or –5.37 psi.
2.9 (D) 0 84000 1.00 9.81 4000 44760 Pa
p p gh
ρ
= − = − × × =
2.10
inside
100 9.81
3 13.51 Pa
.287 253
100 9.81
3 11.67 Pa
.287 293
outside o
o
base
i
i
pg
p g h h
RT
p
pg
p g h h
RT
ρ
ρ
× 
∆ = ∆ = ∆ = × = 
× 
∴∆

× 
∆ = ∆ = ∆ = × =

× 
= 1.84 Pa
If no wind is present this ∆pbase would produce a small infiltration since the higher
pressure outside would force outside air into the bottom region (through cracks).
2.11 p = ρgdh where h = –z. From the given information S = 1.0 + h/100 since S(0) = 1 and
S(10) = 1.1. By definition ρ = 1000 S, where ρwater = 1000 kg/m3
.
Then dp = 1000 (1 + h/100) gdh. Integrate:
dp h gdh
p
= +
∫
∫ 1000 1 100
0
10
0
( / )
p = × +
×
1000 9 81 10
10
2 100
2
. ( ) = 103 000 Pa or 103 kPa
Note: we could have used an average S: Savg = 1.05, so that ρavg = 1050 kg/m3
.
2.12
v
∇ = + +
p
p
x
i
p
y
j
p
z
k
∂
∂
∂
∂
∂
∂
$ $ $
= – ρa i
x
$ – ρα y j
$ – ρα z k
$ – ρgk
$ = – ( )
ρ a i a j a k
x y z
$ $ $
+ + – ρgk
$
= – ρ
v
a – ρ
v
g
∴ ∇ = − +
v v v
p a g
ρ( )
2.13 /
0 0
[( ) / ]g R
atm
p p T z T α
α
= −
= 100 [(288 − 0.0065 × 300)/288]9.81/.0065 × 287
= 96.49 kPa
p p gh
atm
= − = −
×
× ×
ρ 100
100
287 288
9 81 300 1000
.
. / = 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.

14. 14
2.14
/
0
0
0
g R
atm atm
T z
p p p p p
T
α
α
 
−
∆ = − = −
 
 
= 100
288 0065 20
288
1
9 81 0065 287
− ×


 

 −






×
. . /.
= −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 310,000 144 .
dp
d
ρ
ρ
× = But, dp = ρgdh. Therefore,
7
4.464 10
gdh d
ρ ρ
ρ
×
= or 2 7
32.2
4.464 10
d
dh
ρ
ρ
=
×
Integrate, using 0
ρ = 2.00 slug/ft3
:
2 7
2 0
32.2
4.464 10
h
d
dh
ρ
ρ
ρ
=
∫ ∫
×
. ∴
1 1
2
ρ
 
− −
 
 
= 7.21 × 10-7
h or
7
2
1 14.42 10 h
ρ
−
=
− ×
Now,
7
7 7
0 0
2 2
ln(1 14.42 10 )
1 14.42 10 14.42 10
h h g g
p gdh dh h
h
ρ −
− −
= = = − ×
∫ ∫
− × − ×
Assume ρ = const:
2.0 32.2 64.4
p gh h h
ρ
= = × × =
a) For h = 1500 ft: paccurate = 96,700 psf and pestimate = 96,600 psf.
96,600 96,700
% error 100 0.103 %
96,700
−
= × = −
b) For h = 5000 ft: paccurate = 323,200 psf and pestimate = 322,000 psf.
322,000 323,200
% error 100 0.371 %
323,200
−
= × = −
c) For h = 15,000 ft: paccurate = 976,600 psf and pestimate = 966,000 psf.
966,000 976,600
% error 100 1.085 %
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
0.0065 287
101(1 0.0065 /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.

15. 15
2.18 Use the result of Example 2.2: /
0
= .
gz RT
p
e
p
−
0
ln .
p gz
p RT
= −
0.001 32.2
ln .
14.7 1716 455
z
= −
×
∴z = 232,700 ft.
2.19 p = γh = (13.6 × 9810) × 0.25 = 33 350 Pa or 33.35 kPa.
2.20 a) p = γh. 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 = ρgh. If p = 2400 Pa, then
2400 = ρgh = ρ × 9.81 h.
a) ρ =
×
2400
981 36
. .
= 680 kg/m3
. ∴gasoline
b) ρ =
×
2400
981 272
. .
= 899 kg/m3
. ∴benzene
c) ρ =
×
2400
981 245
. .
= 999 kg/m3
. ∴water
d) ρ =
×
2400
9 81 154
. .
= 1589 kg/m3
. ∴carbon tetrachloride
2.22 Referring to Fig. 2.6a, the pressure is p = ρwgh = 2
1
.
2
aV
ρ Then V
gh
w
a
2 2
=
ρ
ρ
.
a) V2 2 1000 9 81 06
123
=
× × ×
. .
.
= 957. ∴V = 30.9 m/s
b) V2 2 194 32 2 3 12
00238
=
× × ×
. . /
.
= 13,124. ∴V = 115 ft/sec
c) V2 2 1000 9 81 1
123
=
× × ×
. .
.
= 1595. ∴V = 39.9 m/s
d) V2 2 194 32 2 5 12
00238
=
× × ×
. . /
.
= 21,870. ∴V = 148 ft/sec
2.23 (C) 0 30000 0.3 9810 0.1 8020 Pa
w atm x x water w
p p h h
γ γ
= + − = + × − × =
2.24 See Fig. 2.6b: p1 = –γ1h + γ2H.
p1 = –0.86 × 62.4 ×
5
12
+ 13.6 × 62.4 ×
9 5
12
.
= 649.5 psf or 4.51 psi.
2.25 0 1 1 2 2 3 3 4 4
p p gh gh gh gh
ρ ρ ρ ρ
= + + + +
= 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

16. 16
2.26 ( ) ( ) ( )
p p p p p p p p
1 4 1 2 2 3 3 4
− = − + − + − (Use ∆ ∆
p g h
= ρ )
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 ( ) ( ) ( )
p p p p p p p p
1 4 1 2 2 3 3 4
− = − + − + − (Use ∆ ∆
p g h
= ρ )
po – pw = 900×9.81(–.2) + 13 600×9.81(–.1) + 1000×9.81×.15
= –12 300Pa or –12.3 kPa
2.28 ( ) ( ) ( ) ( )
p p p p p p p p p p
1 5 1 2 2 3 3 4 4 5
− = − + − + − + −
p1 = 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) (13.6 9810) 0.16 21350 Pa.
a
p H
γ
= − = − × × = −
, 21350 10000 11350 13.6 9810 . 0.0851 m
aafter after after
p H H
= − + = − = × ∴ =
2.34 8200 + 9810 × .25 = 1.59 × 9810 × H. ∴H = 0.683 m
Hnew = .683 + .273 = .956 m. ∆H =
.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
∆H
∆H

17. 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 1
p V 1 2
p V
= 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 H2
– 15.59 H + 40.73 = 0. ∴H = 12.27 or 3.32 ft.
2.37 a) ( ) ( ) ( ) ( )
p p p p p p p p p p
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)
2 2
2 2
1 1 2 3 2
2 /
2 2( ) /
H D d
p D d
γ γ γ γ
∆
=
∆ − + + −
2
2
2(.1/.005)
9800 2 15 600 2(133 400 15 600)(.1/.005)
=
− + × + −
= × −
8 10 6
.487 H
∴ = × ×
−
∆H 8 10 400
6
.487 = 0.0034 m or 3.4 mm
b) ∆H =
− + × + −
×
2 4 2
62 2 99 5 2 99 5 4 2
06 144
2
2
( /. )
.4 . (849 . )( /. )
. = 0.01153 ft or 0.138 in.
2.39 ( ) ( ) ( )
p p p p p p p p
1 4 1 2 2 3 3 4
− = − + − + − (poil = 14.0 kPa from No. 2.30)
15 500 – 14 000 = 9800(0.12 + ∆z) + 680(0.1 – 2∆z) + 860(–0.1 – ∆z).
∴∆z = 0.0451 m or 4.51 cm
2.40 a) pair = –6250 + 625 = –5620 Pa.
–5620 + 9800(2 + ∆z) – 13 600 × 9.81(0.1 + 2∆z) = 0. ∴∆z = 0.0025.
∴h = 0.1 + 2∆z = .15 m or 15 cm
b) pair = 18 800 + 1880 = 20 680 Pa.
20 680 + 9800(0.8 + ∆z) – 13 600 × 9.81(0.2 + 2∆z) = 0. ∴∆z = 0.00715 m
∴h = .2+ 2∆z = .214 or 21.4 cm
c) pair = –91.5 + 9.15 = –82.4 psf.
–82.4 + 62.4(6 + ∆z) – 13.6 × 62.4(4/12 + 2∆z) = 0. ∴∆z = 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 + ∆z) – 13.6 × 62.4(8/12 + 2∆z) = 0. ∴∆z = 0.0267 ft.
∴h = 8/12 + 2 (0.0267) = 0.7205 ft or 8.65 in.

18. 18
2.41 F hA
= γ = 9810 × 10 × π × .32
/4 = 6934 N.
2.42
5 1 5 5
(2 ) (2 ) [9800 1 3 (2 )]. 32670 N
3 3 3 3
P P
× × = × × × × × × × ∴ = a) F = pc A = 9800 × 2
× 42
= 313 600 N or 313.6 kN
b)
2 2
9800 1 (2 4) 9800 2 9800 1 98000 N or 98.0 kN
3 3
c
F p A
= = × × × + × × + × × =
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
W gV
ρ
= 2400 9.81
walls = × [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 × π × 22
= 167 400 N or 167.4 kN
Find γ 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 (32
− 12
)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) F hA
= = × × =
γ 62 27 33 24
.4 . 40,930 lb.
yp = +
×
×
27 33
6 8 36
27 33 24
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.
yp = +
×
×
39 77
6 8 36
39 77 24
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:
5 1 5 5
(2 ) (2 ) [9800 1 3 (2 )]. 32670 N
3 3 3 3
P P
× × = × × × × × × × ∴ =
(x, y)
y
x

19. 19
2.48 a) F hA
= = × ×
γ π
9810 6 22
= 739 700 N or 739.7 kN.
y y
I
Ay
p = + = +
×
×
6
2 4
4 6
4
π
π
/
= 6.167 m. ∴(x, y)p = (0, –0.167) m
b) F hA
= = × ×
γ π
9810 6 2 = 369 800 N or 369.8 kN.
yp = +
×
×
6
2 8
2 6
4
π
π
/
= 6.167 m. x2
+ y2
= 4
x F
x
pdA x y xdy y y dy
p = = − = − −
∫ ∫
∫
−
−
2 2
6
2
4 6
2
2
2
2
2
γ γ
( ) ( )( ) .
∴ × = − − + =
−
∫
x y y y dy
pγ π
γ
γ
6 2
2
24 4 6 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.
yp = +
×
×
5333
3 4 36
5 333 6
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 = × + ×
9810 4
2
3
4
( sin 36.9°) × 6 = 330 000 N
yp = +
×
×
5 6
5 2 36
6 5 6
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 . ∴ x1 = 0.6.
x = 1.8 + 0.6 = 2.4. ∴(x,y)p = (2.4, 0.343) m.
2.49 F hA
= = × × ×
γ 62 11 6 10
.4 ( ) = 41,180 lb.
y y
I
yA
p = + = +
×
×
11
6 10 12
11 60
3
/
= 11.758'.
(16 – 11.758) 41,180 = 10P. ∴P = 17,470 lb.
2.50 F hA
= = × ×
γ 9810 6 20 = 1.777 × 106
N, or 1177 kN.
y y
I
Ay
p = + = +
×
×
7 5
4 5 12
7 5 20
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
yp
P
F

20. 20
2.51 F hA
= = × ×
γ 9810 12 20 = 2.354 × 106
N, or 2354 kN.
y y
I
Ay
p = + = +
×
×
15
4 5 12
15 20
3
/
= 15.139 m.
(17.5 – 15.139) 2354 = 5 P. ∴P = 1112 kN.
2.52 y y
I
Ay
H bH
bH H
H H
H
p = + = +
×
= + =
2
12
2 2 6
2
3
3
/
/
. yp is measured from the surface.
∴From the bottom, H y H H H
p
− = − =
2
3
1
3
.
Note: this result is independent of the angle α, so it is true for a vertical area or a sloped
area.
2.53 3
1
sin40 3 . ( 2) sin40 . 2( 2) .
2 3
l
F l l F l P l l P
γ γ
= × × = + ∴ = +
o o
a) 9810× 23
= 2(2 + 2)P. ∴P = 9810 N
b) 9810× 43
= 2(4 + 2)P. ∴P = 52 300 N
c) 9810× 53
= 2(5 + 2)P. ∴P = 87 600 N
2.54 h = −
122 2
. .4 = 1.1314 m. A = 1.2 × 1.1314 + .4 × 1.1314 = 1.8102 m2
Use 2 forces: F h A
c
1 1 9800 5657 1 2 11314
= = × × ×
γ . ( . . ) = 7527 N
F h A
c
2 2 9800
11314
3
11314
= = × × ×
γ
.
(.4 . ) = 1673 N
yp1
2
3
11314
= ( . ). y y
I
A y
p2
2
2
3
11314
3
11314 36
11314 2 11314 3
= + = +
×
× ×
. .4 . /
.4 . / . /
= 0.5657 m
ΣMhinge = 0: 7527
11314
3
1673 11314 0 5657 11314
× + × − −
.
( . . ) . 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:
y h H A h H I h H h H
= + = + = + +
( ) / , ( ) , [ ( )]( ) /
3 2 36
2 3
∴ =
+
+
+
+ +
=
+
+
+
=
+
y
h H h H
h H h H
h H h H h H
p
3
2 36
3 3 6 2
4
2
( ) /
( ) ( ) /
a) h
h H
=
+
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
H b
p = + = + +
×
+
12 18 2
18 12
9 18
3
. ( . / )
. /
(. ) .
. ∴H = 0.

21. 21
b) y H H
b
H b
p = + = + +
×
+
12 2 0 2
2 12
1 2
3
. ( . / )
/
( )
. ∴H = 0.6667 m.
c) y H H
b
H b
p = + = + +
×
+
12 2 2 2
2 2 12
11 2 2
3
. ( . / )
. /
( . ) .
. ∴H = 2.933 m.
2.57 (A) The gate opens when the center of pressure in at the hinge:
3
1.2 11.2 (1.2 ) /12
5. 5 1.2.
2 2 (1.2 ) (11.2 ) / 2
p
h I h b h
y y y
Ay h b h
+ + +
= + = + = + = +
+ +
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
= × =
γ γ
F H b b H
2 = × =
γ γ
l l
1
2 3 2
2
γ γ
bH
H
b H
× = ×
l
l
. ∴ =
H 3l
a) H = ×
3 2 = 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 )( / )
2 9810 6 50 24 50 2 = 21.19 × 106
N
dw =
× + ×
+
300 27 600 16
300 600
= 19.67 m.
F2 = 9810 × 5 × 11.09 = 0.544 × 106
N. d2
11 09
3
=
.
= 3.697 m.
F1 9810
45
2
45
= × × = 9.933 × 106
N. d1 = 15 m.
F3 9810
45 10
2
30
= ×
+
× = 8.093 × 106
N. d3
2 943 15 5150 20
2 943 5150
=
× + ×
+
. .
. .
= 18.18 m.
Wd F d
Fd F d
w + = × ⋅
+ = × ⋅



2 2
6
1 1 3 3
6
418 8 10
2961 10
.
.
N m
N m
∴will not topple.
b) W = (2.4 × 9810) (6 × 65 + 65 × 12) = 27.55 × 106
N.
dw =
390 27 780 16
390 780
× + ×
+
= 19.67 m.
F2
6
0 54 10
≅ ×
. N. d2 3 70
≅ . m.
F1 = 9810 × 30 × 60 = 17.66 × 106
N. d1 = 20 m.
F3 9810
60 10
2
30
= ×
+
× = 10.3 × 106
N. d3
2 943 15 7358 20
2 943 7 358
=
× + ×
+
. .
. .
= 18.57 m.
F2
F1
H/3
l/2
F1
F2
F3
W
O

22. 22
Wd F d
Fd F d
w + = × ⋅
+ = × ⋅



2 2
6
1 1 3 3
6
543 9 10
544 5 10
.
.
N m
N m
∴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) W W W W
= + = × × × ×
1 2 1 6 43 1 62 2
. ( ) .4 .4 = 38,640 lb.
W2 24 43 2 62 2
= × × ×
( / ) .4 .4 = 77,280 lb.
W3 40 22 33 2 62
= × ×
( . / ) .4 = 27,870 lb @ 20.89 ft.
F1 62 20 40 1
= × × ×
.4 ( ) = 49,920 lb @ 40/3 ft.
F2 62 5 10 1
= × × ×
.4 ( ) = 3120 lb @ 3.33 ft
1
3
2
= 18,720 lb @ 15 ft
= 28,080 lb @ 20 ft
p
p
F
F
F


= 


O
M
Σ : (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.
F1 62 30 60
= × ×
.4 = 112,300 lb. 3 (60 22.86/2) 62.4
W = × × = 42,790 lb.
F2 62 5 10
= × ×
.4 = 3120 lb
Fp1 62 10 30
= × ×
.4 = 18,720 lb. Fp2 62 50 30 2
= × ×
.4 / = 46,800 lb.
O
M
Σ : (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 ΣMhinge = 0. 2.5P – dw × W – d1 × F1 = 0.
∴ P = × × × +
×
× ×
×
×






1
2 5
2
3
9800 1 8
4 2
3
9800
2
4
4
2
. π
π
= 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:
( ) 9810 2 (4 10) 784 800N
( ) 0.86 9810 1 20 168 700N
water
H
H oil
F
F
= × × × =
= × × × =
ΣM P
: . . .
2 784 8 2 168 7 2
= × − × ∴P = 616.1 kN.
F1
F2
F3
W
O
W3
F1
dw
d1
P
W

23. 23
2.63 Place the resultant force
v v
F F
H V
+ at the center of the circular arc.
v
FH passes thru the
hinge showing that P FV
= .
a) P FV
= = × × + × =
9810 6 2 4 4 594
( )
π 200 N or 594.2 kN.
b) P = FV = 62.4 (20 × 6 × 12 + 9π × 12) = 111,000 lb.
2.64 (D) Place the force
v v
F F
H V
+ at the center of the circular arc. FH passes through the
hinge:
2
4 1.2 9800 ( 1.2 /4) 9800 300000. 5.16 m.
V
P F w w w
π
∴ = = × × + × × = ∴ =
2.65 Place the resultant
v v
F F
H V
+ at the circular arc center.
v
FH passes thru the hinge so that
P FV
= . Use the water that could be contained above the gate; it produces the same
pressure distribution and hence the same FV .
P FV
= = 9810 (6 × 3 × 4 + 9π) = 983 700 N or 983.7 kN.
2.66 Place the resultant
v v
F F
H V
+ at the center.
v
FV 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
v v
F F
H V
+ of the unknown liquid acts thru the center of the circular arc.
v
FV
passes thru the hinge. Thus we use only ( ) .
FH oil Assume 1 m wide.
a) ΣM
R R
R
R
S
R
R
R
R
x
: .
3
9810
2
4
3
9800
4 2
2





 +





 =






π
π
γ ∴γ x = 4580 N/m3
b) ΣM
R R
R
R
S
R
R
R
R
x
: . . .
3
62 4
2
4
3
62 4
4 2
2





 +





 =






π
π
γ ∴γ x = 29.1 lb/ft3
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) P FH o
= = × × ×
( ) ( . ) . .
0 8 9810 0 3 3 6 = 8476 N.
ΣF W F F
y V o V w
= = + −
0 ( ) ( )
0 = S × 9810 π × .62
× 6 + .
.
36
36
4
−


 


π
× 6 × (.8 × 9810) – 9810 × π × .18 × 6
− × × × −
9810 8 2 6 6
2
. . . ∴ =
S 0 955
. .
b) g V
ρ .
W
= = 1996 lb.
ΣF W F F
y V o V w
= = + −
0 ( ) ( )

24. 24
0 = S × 62.4 × π × 22
× 20 + 4
4
4
−


 


π
× 20 × .8 × 62.4 – 62.4 × π × 2 × 20
− × × × ×
62 4 8 2 2 20
2
. . . ∴ =
S 0 955
. .
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 = (π × 32
) × 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 × π × 32
– (.8 × 9810) ×
1
2
4
3
33
π ×


 

 = –23 400 N
or –23.4 kN
2.71 A free-body diagram of the gate and water is shown.
H
F d W H P
w
3
+ = × .
a) H = 2 m. F = 9810 × 1 × 4 = 39 240 N.
W xdy
y
dy
= = =
×
∫
∫
9810 2 9810 2
2
2 9810
2
2
3 2
1 2
0
2
0
2 3 2
/ /
/
= 26 160 N.
d x
x
xdy
xdy
x dx
x dx
w = = = =






∫
∫
∫
∫
2
1
2
4
4
1
2
1 4
1 3
3
0
1
2
0
1
/
/
= 0.375 m.
∴ = × + ×
P
1
3
39
0 375
2
26 160
240
.
= 17 980 N or 17.98 kN.
b) H = 8'. F = 62.4 × 4 × 32 = 7,987 lb.
W xdy x dx
= = × = × ×
∫
∫
62 4 62 4 4 62 16 2 3
2
0
2
3
.4 .4 .4 / = 2,662 lb.
d x
x dx
x dx
w = = =






∫
∫
1
2
4
4
1
2
16 4
8 3
3
0
2
2
0
2
/
/
= 0.75'.
1 8
7,987 0.75 2,662 2910 lb
8 3
P
 
= × + × =
 
 
2.72 (A) W V
γ
=
900 9.81 9810 0.01 15 . 6 m
w w
× = × × ∴ =
W
pA
Fweld
dA=xdy
y
x
F
h/3

25. 25
2.73 W = weight of displaced water.
a) 20 000 + 250 000 = 9810 × 3 2
(6 /2).
d d
+ ∴d2
+ 12d – 18.35 = 0. ∴d = 1.372 m.
b) 270 000 = 1.03 × 9810 × 3 2
(6 /2).
d d
+ d2
+ 12d – 17.81 = 0. ∴d = 1.336 m.
2.74 25 + FB = 100. ∴FB = 75 = 9810 −
V . ∴ −
V = 7.645 × 10−3
m3
γ × 7.645 × 10−3
= 100. or 7645 cm3
∴γ = 13 080 N/m3
.
2.75 3000 × 60 = 25 × 300 ∆d × 62.4. ∴∆d = 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
π ρ π ρ
R g R g
a
= +
4
3
5
100 9 81
287 293
1000
4
3
5
100 981
287
3 3
π π
×
×
×
= + ×
×
.
.
.
.
.
Ta
∴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
1500 150
100 9 81
287 288
2
π × ×
×
×
.
.
= Fp + 0.1 Fp + 1500 π × 1502
×
100 981
2 077 288
×
×
.
.
4
o /64.
I d
π
= . Npeople =
986 10
800
8
. ×
= 1.23 × 106
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 m3
Density: g V
ρ .
W
=
ρ × ×
9 81 006122
. . = 50. ∴ρ = 832.5 kg/m3
Specific wt: γ = ρg = 832.5 × 9.81 = 8167 N/m3
Specific gravity: S = ρ/ρwater = 832.5/1000 = 0.8325

26. 26
2.81 From a force balance FB = W + pA.
a) The buoyant force is found as follows (h > 16'):
cos ,
θ =
− −
h R
R
15
Area = θR2
– (h – 15 – R) R sinθ
∴FB = 10 × 62.4[πR2
− θR2
+ (h – 15 – R) R sinθ].
FB = 1500 + γhA.
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[θR2
− (R – h + 15) R sinθ].
FB = 1500 + γhA. cos θ =
− +
R h
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) W FB
= . [ ]
0 01 136 1000 015 4 9 81 9810
2
. . . / . .
+ × × × × = −
h V
π
− =
×
× +
×
× = × −
V
π π
.
.
.
. . .
015
4
15
005
4
06 2 769 10
2 2
5 3
m ∴h = 7.361 × 10−3
m
∴ =
mHg
2
13.6 1000 .015 / 4
hπ
× × × = 0.01769 kg
b) (.01 + .01769) 9.81 = 9810
π π
×
× +
×
×






.
.
.
. .
015
4
15
005
4
12
2 2
Sx ∴Sx = 0.959.
c) (.01 + .01769) 9.81 = 9810
π×
×
.
.
015
4
15
2
Sx. ∴Sx = 1.045.
2.83 (. ) .
.
.
.
. .
01 9 81 9810
015
4
15
005
4
12
2 2
+ =
×
× +
×
×






mHg
π π
∴mHg = 0.01886.
a) (.01 + .01886) 9.81 = 9810
π×
×
.
.
015
4
15
2
Sx. ∴Sx = 1.089.
b) mHg = 0.01886 kg.
h − 15
θ R
pA
FB
W
h − 15
θ R

27. 27
2.84 a)
4 4
o
(10/12)
64 64
d
I
π π ×
= = = 0.02367 ft4
.
− = =
× × × ×
V
W
rH O
2
8 62 5 12 12 12
62
2
. .4 ( / ) /
.4
π
= 0.4363. depth =
.4363
( / )
π 5 12 2
= 0.8'
∴ = − −
GM . /.4363 (. .4)
02367 5 = –0.0457'. ∴It will not float with ends horizontal.
b) Io = 0.02367 ft4
, −
V = 0.3636 ft3
, depth = 0.6667'
GM = − −
. /. ( ) /
02367 3636 5 4 12 = –0.01823'. ∴It will not float as given.
c) −
V = 0.2909, depth = 6.4", GM =
.
.
.
02367
2909
4 3 2
12
−
−
= 0.0147. ∴It will float.
2.85 With ends horizontal 4
o /64.
I d
π
= The displaced volume is
− = × = × −
V d h d
x x
γ π γ
2 5 3
4 9800 8 014 10
/ . since h = d. The depth the cylinder will
sink is
depth =
−
= × = ×
− −
V
A
d d d
x x
8 014 10 4 10 20 10
5 3 2 5
. / / .
γ π γ
The distance CG is CG
h
d
x
= − × −
2
10 2 10 2
5
. /
γ . Then
GM
d
d
d
d
x
x
=
×
− + × >
−
−
π
γ
γ
4
5 3
5
64
8 014 10 2
10 2 10 2 0
/
.
. / .
This gives (divide by d and multiply by γx):
612.5 – .5 γx + 5.1 × 10-5
γ x
2
> 0.
Consequently,
γx > 8369 N/m3
or γx < 1435 N/m3
2.86
3
3
.
water
water water
S d
W
V S d
γ
γ γ
− = = =
3
3
.
water
water water
S d
W
V S d
γ
γ γ
− = = = ∴h = Sd.
GM
d
Sd
d Sd d
S
S
= − − = − +
4
3
12
2 2
1
12
1
2 2
/
( / / ) ( ).
If GM = 0 the cube is neutral and 6S2
– 6S + 1 = 0.
∴ =
± −
S
6 36 24
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 =
× + ×
+
16 9 16 4
16 16
= 6.5 cm above the bottom edge.
G
S
S
A
A
=
× + × + ×
× + × + ×
4 9 5 16 8 5 16 4
5 8 2 8 16
γ γ γ
γ γ γ
. .
.
= 6.5 cm.
C
G
h

28. 28
∴130 + 104 SA = 174 + 64 SA. ∴ SA = 1.1.
2.88 a) y =
× + × + ×
+ +
16 4 8 1 8 7
16 8 8
= 4. x =
× + × + ×
+ +
16 1 8 4 8 4
16 8 8
= 2.5.
For G: y =
× × + × × + × ×
× + × + ×
12 16 4 5 8 1 1 5 8 7
1 2 16 5 8 15 8
. . .
. . .
= 4.682.
x =
× + × × + × ×
× + × + ×
12 16 5 8 4 15 8 4
12 16 5 8 15 8
. . .
. . .
= 2.364.
G must be directly under C.
tan
.
.
.
θ =
136
682
∴θ =11.3°.
b) y =
× + × + ×
+ +
4 2 2
1
2
2 3 5
4 2 2
.
= 2. x =
× + × + ×
+ +
4
1
2
2 2 2 2
4 2 2
= 1.25
For G:y =
× × + × + ×
× + × + ×
12 4 2 5 1 15 7
12 4 5 2 15 2
. . .
. . .
= 2.34. x =
× + × + ×
× + × + ×
12 2 5 4 15 4
1 2 4 5 2 15 2
. . .
. . .
= 1.182
∆y = 0.34, ∆x = 0.068. tan
.
.
.
θ =
068
34
∴θ = 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
8 12
8 3
15 1777 15 0 277 0
3
/
. . . . .
∴The barge is stable.
2.90 y =
× + ×
+
8 3 16 97 1
8 16 97
.485 .414 .
.485 .
= 1.8 m. ∴CG = −
18 15
. . = 0.3 m.
Using Eq. 2.4.47: GM =
×
− = − =
l
l
8 12
34 97
3 1 3 116
3
.485 /
.
. .46 . . . ∴Stable.
2.91 (A)
2
5
20000 20000 6660 (1.2 ) 24070 Pa
9.81
24070 0.02 30.25 N
plug
plug plug
p h
F p A
γ
π
= + = + × × =
= = × × =
.
2.92 a) tan
.
.
α = =
20
9 81 4
H
∴H = 8.155 m. pmax = 9810 (8.155 + 2) = 99 620 Pa
b) pmax = ρ(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

29. 29
2.93 The air volume is the same before and after.
∴ 0.5 × 8 = hb/2. tan
.
.
α = =
10
9 81
h
b
4
9 81
10
=
h
h
2
.
.
∴h = 2.856. ∴Use dotted line.
2 5
1
2
2 5 2 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 0 2 5
8
981
− −














.
.
ax
4 =
h
ax
2
9 81
2
× .
60 = 8 ax + 24.52 – 9.81
8
9 81
ax
.
. ax – 4.435 = 1.1074 ax .
ax
2
– 10.1 ax + 19.67 = 0 ∴ax = 2.64, 7.46 m/s2
b) 60 000 = –1000 ax (–8) – 1000 (9.81 + 10) − +






2 5
8
9 81
.
.
.
ax
60 = 8 ax + 49.52 – 19.81
8
19 81
ax
.
. ax – 1.31 = 1.574 ax .
ax
2
– 5.1 ax + 1.44 = 0 ∴ax = 0.25, 4.8 m/s2
c) 60 000 = –1000 ax (–8) – 1000 (9.81 + 5) (–2.5 +
8
14 81
ax
.
).
60 = 8 ax + 37.0 – 14.81
8
14 81
ax
.
. ax – 2.875 = 1.361 ax .
ax
2
– 7.6 ax + 8.266 = 0 ∴ax = 1.32, 6.28 m/s2
2.95 a) ax = 20 × .866 = 17.32 m/s2
, az = 10 m/s2
. 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
α
b
h
A
B
z
1
x
w
C

30. 30
∴ = −
∫
F z dz
AB ( )
.
76 260 9810 4
0
2 5
= 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.
∴ = −
∫
F x dx
BC ( )
.
76 260 10 000 4
0
7 626
= 1.163 × 106
N or 1163 kN
c) On the top p(x) = –1000 × 10 (x – 5.174) where position 1 is on the top surface:
∴ = −
∫
F x dx
top ( )
.
51 740 10 000 4
0
5 174
= 5.35 × 105
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:
FAB =
+
× ×
58 290 8495
2
15 2
. = 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.
FCD = × × ×
1
2
49 810 15 2
. = 74 720 N or 74.72 kN.
c) pA = 58 290 Pa. pC = 49 810 Pa. ∴ =
+
×
FAC
58 29 49 81
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


 

 = −112 psf. pC = –112 psf.
e) pA = 1.94 × 60 −


 


37 5
12
.
= −364 psf.
pB = 1.94 × 60 −


 


37 5
12
.
– 1.94 × 32.2 −


 


25
12
= –234 psf.
x
z
A
B
C
x
z
1

31. 31
pC = –1.94 × 32.2 −


 


25
12
= 130 psf.
f) pA = 1.94 × 30
37 5
12
.


 

 = 182 psf.
pB = –1.94(–30)
37 5
12
.


 

 – 1.94 × 62.2 −


 


25
12
= 433 psf.
pC = –1.94 × 62.2 × −


 


25
12
= 251 psf.
2.99 Use Eq. 2.6.4 with position 1 at the open end:
ω
π
=
×
50 2
60
= 5.236 rad/s.
a) pA =
×
× ×
1000 5 236
2
6 15
2
2
.
(. . ) = 11 100 Pa.
pB =
1
2
× 1000 × 5.2362
× .92
+ 9810 × .6 = 16 990 Pa.
pC = 9810 × .6 = 5886 Pa.
b) pA =
1
2
× 1000 × 5.2362
× 0.62
= 4935 Pa.
pB =
1
2
× 1000 × 5.2362
× 0.62
+ 9810 × 0.4 = 8859 Pa.
pC = 9810 × 0.4 = 3924 Pa.
c) pA =
1
2
× 1.94 × 5.2362
×
37 5
12
2
.





 = 259.7 psf.
pB =
1
2
× 1.94 × 5.2362
×
37 5
12
62
25
12
2
.
.4





 + × = 389.7 psf.
pC =62
25
12
.4 × = 130 psf.
d) pA =
1
2
× 1.94 × 5.2362
×
22 5
12
2
.





 = 93.5 psf.
pB =
1
2
× 1.94 × 5.2362
×
22 5
12
2
.





 + 62
15
12
.4 × = 171.5 psf.
pC = 62
15
12
.4 × = 78 psf.
A
B
C
z
1
r
ω

32. 32
2.100 Use Eq. 2.6.4 with position 1 at the open end.
a) pA =
1
2
× 1000 × 102
(0 – 0.92
) = –40 500 Pa.
pB = –40 500 + 9810 × 0.6 = –34 600 Pa.
pC = 9810 × 0.6 = 5886 Pa.
b) pA =
1
2
× 1000 × 102
(0 – 0.62
) = –18 000 Pa.
pB = –18 000 + 9810 × 0.4 = –14 080 Pa.
pC = 9810 × 0.4 = 3924 Pa.
c) pA =
1
2
× 1.94 × 102
0
37 5
144
2
−






.
= –947 psf.
pB = -947 + 62.4 ×
25
12
= –817 psf. pC = 62.4 ×
25
12
= 130 psf.
d) pA =
1
2
× 1.94 × 102
−






22 5
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 ω2
(0 – 0.452
) – 9810 (0 – 0.6). ∴ω = 7.62 rad/s.
b) 0 =
1
2
× 1000 ω2
(0 – 0.32
) – 9810 (0 – 0.4). ∴ω = 9.34 rad/s.
c) 0 =
1
2
× 1.94 ω2
0
18 75
12
2
2
−






.
– 62.4 −


 


25
12
. ∴ω = 7.41 rad/s.
d) 0 =
1
2
× 1.94 ω2
−






11 25
12
2
2
.
– 62.4 −


 


15
12
. ∴ω = 9.57 rad/s.
2.102 The air volume before and after is equal.
∴ = × ×
1
2
6 2
0
2 2
π π
r h . . . ∴ r h
0
2
= 0.144.
a) Using Eq. 2.6.5: r0
2 2
5 2
× / = 9.81 h
∴h = 0.428 m
∴pA =
1
2
× 1000 × 52
× 0.62
– 9810 (–0.372)
= 8149 Pa.
b) r0
2 2
7 2
× / = 9.81 h. ∴h = 0.6 m.
∴pA =
1000
2
× 72
× 0.62
+ 9810 × 0.2 = 10 780 Pa.
A
B
C
z
1
r
ω
z
1
r
ω
2
1
2
h
z
r
A
r0

33. 33
c) For ω = 10, part of the bottom is bared.
π π π
× × = −
. . .
6 2
1
2
1
2
2
0
2
1
2
1
r h r h
Using Eq. 2.6.5:
ω2
0
2
2
r
g
h
= ,
ω2
1
2
2
r
g
h
= 1.
∴ = −
0144
2 2
2
2
2 1
2
.
g
h
g
h
ω ω
or
h h
2
1
2
2
0 144 10
2 981
− =
×
×
.
.
.
Also, h – h1 = 0.8. 1.6h – 0.64 = .7339. ∴h = 0.859 m, r1 = 0.108 m.
∴pA =
1
2
× 1000 × 102
(0.62
– 0.1082
) = 17 400 Pa.
d) Following part (c): h h
2
1
2
2
0144 20
2 9 81
− =
×
×
.
.
. 1.6h – .64 = 2.936.∴ h = 2.235 m.
∴pA =
1
2
× 1000 × 202
(0.62
– 0.2652
) = 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 p r r g h
( ) [ (. )].
= − − −
1
2
0 8
2 2
ρω ρ
p r r h
( ) (. )
= + −
500 9810 8
2 2
ω if h < .8.
p r r r
( ) ( )
= −
500 2 2
1
2
ω if h > .8.
a) F p rdr r r dr
= = +
∫
∫ 2 2 12 500 3650
3
0
6
π π ( )
.
= 6670 N.
(We used h = .428 m)
b) F p rdr r r dr
= = +
∫
∫ 2 2 24 500 1962
3
0
6
π π ( )
.
= 7210 N. (We used h = 0.6 m)
c) F p rdr r r dr
= = −
−
∫
∫ 2 2 50 000 108
3
108
6
2
π π ( ( . )
.
.
= 9520 N. (We used r1 = 0.108 m)
d) F p rdr r r dr
= = −
∫
∫ 2 2 200 000 265
3
265
6
2
π π ( ( . )
.
.
= 26 400 N. (r1 = 0.265 m)
1
h
z
r
A
r0
h1
dr
dA = 2πrdr

34. 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
boy
time t
t = 0
streamlines
t = 2 hr
pathline
t = 2 hr
streakline at t = 3 hr
y
x

35. 35
3.5 a)u
dx
dt
t v
dy
dt
t
= = + = =
2 2 2
x t t c y t c
= + + = +
2
1
2
2
2
= +
y y
2
∴ − + =
x xy y y
2 2
2 4 ∴parabola.
b) x t t c c c
= + + ∴ = − = −
2
1 1 2
2 8 4
. , .
and
= + + ± + −
y y
4 2 4 8
( )
∴ − + + − =
x xy y x y
2 2
2 8 12 0. ∴parabola.
3.6
ˆ
ˆ ˆ ( )
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
using , .
z
V dr udy vdx
V ui vj wk
dr dxi dy j dzk i j k j i k
 × = −
= + + 

= + + × = × = −


v
v 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 0 0 0 2 2
2
and m / s
( , , ) .
At t V
= − = + =
2 1 2 0 3 2 3 606
2 2
and m / s
( , , ) . .
b) At t V
= =
2 0 0 0 0
and ( , , ) .
At t V
= − = − + − =
2 1 2 0 2 8 8 246
2 2
and m/ s
( , , ) ( ) ( ) . .
c) At t V
= = − =
2 0 0 0 4 4
2
and m / s
( , , ) ( ) .
At t V
= − = + − + − =
2 1 2 0 2 4 4 6
2 2 2
and m / s
( , , ) ( ) ( ) .
3.9 (D) 5 ˆ
( 51.4 10 )
j
−
− ×
A simultaneous solution yields 4/5 and 3/5.
x y
n n
= = (They must
both have the same sign.
3.10 a) cos $ / ( )/ . . .
α α
= ⋅ = + + = ∴ =
v o
V i V 1 2 3 2 0 832 33 69
2 2
v
V n i j n i n j
n n
n n
n n
n n
x y
x y
x y
y x
x x
⋅ = + ⋅ + =
+ =
+ =



∴
= −
+ =
$ . ( $ $) ( $ $) .
0 3 2 0
3 2 0
1
3
2
9
4
1
2 2
2 2
∴ = = − = −
n n n i j
x y
2
13
3
13
1
13
2 3
, $ ( $ $).
or
39.8o
y
x
streamlines
t = 5 s
(27, 21)
(35, 25)

36. 36
b) cos $ / / ( ) ( ) . .
α α
= ⋅ = − − + − = − ∴ =
v o
V i V 2 2 8 0 2425 104
2 2
v
V n i j n i n j
n n
n n
n n
n n
x y
x y
x y
x y
y y
⋅ = − − ⋅ + =
− − =
+ =



∴
= −
+ =
$ . ( $ $) ( $ $) .
0 2 8 0
2 8 0
1
4
16 1
2 2 2 2
∴ = = − = − +
n n n i j
y x
1
17
4
17
1
17
4
, $ ( $ $).
or
c) cos $ / / ( ) . . .
α α
= ⋅ = + − = ∴ = −
v o
V i V 5 5 8 0 6202 51 67
2 2
v
V n i j n i n j
n n
n n
n n
n n
x y
x y
x y
x y
y y
⋅ = − ⋅ + =
− =
+ =



∴
=
+ =
$ . ( $ $) ( $ $) .
0 5 8 0
5 8 0
1
8
5
64
25
1
2 2
2 2
∴ = = = +
n n n i j
y x
5
89
8
89
1
89
8 5
, $ ( $ $).
or
3.11 a) [ ]
v v
V dr x i xtj dxi dyj
× = + + × + =
0 2 0
. ( )$ $ ( $ $) .
∴ + − =
+
=
( ) .
x dy xtdx t
xdx
x
dy
2 0
2
or
Integrate: [ ]
t
xdx
x
dy t x n x y C
+
= − + = +
∫
∫ 2
2 2
. .
l
2 1 2 3 2
( ) .
− = − + ∴
ln C C = 0.8028.
[ ]
t x n x y
− + = +
2 2 0 8028
l .
b) [ ]
v v
V dr xyi y j dxi dyj
× = − × + =
0 2 0
2
. $ $ ( $ $) .
∴ + = = −
xydy y dx
dx
x
dy
y
2 0
2
2
or .
Integrate: 2 2 1 2
l l l l
nx n y C n n C
= − = − −
( / ). ( ) ( / ).
∴ = − = − − ∴ = −
C nx n y x y
2 2 2
2 2
. ( / ). .
l l
c) [ ]
v v
V dr x i y tj dxi dyj
× = + − × + =
0 4 0
2 2
. ( )$ $ ( $ $) .
( ) .
x dy y tdx
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 .
− −
+





 = +





 = −
∴ = − −





 =
−
C yt
x
0 9636
2
0 9636 2
1
. . tan .

37. 37
3.12 (C) 2
ˆ ˆ ˆ ˆ ˆ ˆ
2 (2 ) (2 2 ) 16 8 16 .
V V V V
a u v w xy yi y xi yj i i j
t x y z
∂ ∂ ∂ ∂
= + + + = − − = − − +
∂ ∂ ∂ ∂
v v v v
v
2 2
8 16 17.89 m/s
a
∴ = + =
3.13 a)
DV
Dt
u
V
x
v
V
y
w
V
z
V
t
v v v v v
= + + +
∂
∂
∂
∂
∂
∂
∂
∂
=0.
b) u
V
x
v
V
y
w
V
z
V
t
x i y j xi yj
∂
∂
∂
∂
∂
∂
∂
∂
v v v v
+ + + = + = +
2 2 2 2 4 4
( $) ( $) $ $ = 8 4
$ $
i j
−
c) u
V
x
v
V
y
w
V
z
V
t
x t xti ytj xyt xtj ztk x i xyj
∂
∂
∂
∂
∂
∂
∂
∂
v v v v
+ + + = + + + + +
2 2
2 2 2 2 2 2
( $ $) ( $ $) $
+ = − −
2 68 100 54
yzk i j k
$ $ $
d) u
V
x
v
V
y
w
V
z
V
t
x i yzj xyz xzj tz xyj tk zk
∂
∂
∂
∂
∂
∂
∂
∂
v v v v
+ + + = − − − + − + +
($ $) ( $) ( $ $) $
2 2 2 2
= xi yz x yz xyzt j zt z k
$ ( )$ ( ) $
− − + + +
2 4 2
2 2 2
=2 114 15
$ $ $
i j k
− +
3.14
v
Ω = −





 + −





 + −






1
2
1
2
1
2
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
w
y
v
z
i
u
z
w
x
j
v
x
u
y
k
$ $ $.
a)
v
Ω = − =
1
2
20
∂
∂
u
y
k yk
$ $ = −20 $
k
b)
v
Ω = − + − + −
1
2
0 0
1
2
0 0
1
2
0 0
( )$ ( )$ ( ) $
i j k = 0
c)
v
Ω = − + − + −
1
2
2 0
1
2
0 0
1
2
2 0
( )$ ( )$ ( ) $
zt i j yt k = 6 2
$ $
i k
−
d)
v
Ω = + + − + − −
1
2
0 2
1
2
0 0
1
2
2 0
( )$ ( )$ ( )$
xy i j yz k = − +
2 3
$ $
i k
3.15 The vorticity
v v
ω = 2Ω. Using the results of Problem 3.7:
a)
v
ω = −40$
i b)
v
ω = 0 c)
v
ω = 12 4
$ $
i k
− d)
v
ω = − +
4 6
$ $
i k
3.16 a) ε
∂
∂
ε
∂
∂
ε
∂
∂
xx yy zz
u
x
v
y
w
z
= = = = = =
0 0 0
, , .
ε
∂
∂
∂
∂
ε
∂
∂
∂
∂
xy xz
u
y
v
x
y
u
z
w
x
= +





 = − = = +





 =
1
2
20 20
1
2
0
, ,
ε
∂
∂
∂
∂
yz
v
z
w
y
= +





 = ∴ =










1
2
0
0 20 0
20 0 0
0 0 0
. rate - of strain

38. 38
b)
ε ε ε
ε ε ε
xx yy zz
xy xz yz
= = =
= = =
2 2 0
0 0 0
, , .
, , .
rate-of strain =
2 0 0
0 2 0
0 0 0










c) ε ε ε
xx yy zz
xt xt yt
= = = = = = −
2 8 2 8 2 4
, , .
ε ε ε
xy xz yz
yt zt
= = − = = = =
1
2
2 2
1
2
0 0
1
2
2 6
( ) , ( ) , ( ) .
rate-of strain =
8 2 0
2 8 6
0 6 4
−
−
−










d) ε ε ε
xx yy zz
xz t
= = − = − = =
1 2 12 2
, , .
ε ε ε
xy xz yz
yz xy
= − = = = = − =
1
2
2 3
1
2
0 0
1
2
2 2
( ) , ( ) , ( ) .
rate-of strain =
1 3 0
3 12 2
0 2 2
−










3.17 a) a
r r r r r
r = −











 − +





 −





 −
10
40 80
10
40
1
40
2 3 2 2
cos cos
sin
( sin )
θ θ
θ
θ
− +





 = − − −
1
10
40
10 2 1 125 1
2
2
2
r r
sin ( .5)( ) . ( )
θ = 9.375 m/s2.
a
r r r r r
θ θ θ
θ
θ
= −











 + +





 +






10
40 80
10
40
10
40
2 3 2 2
cos sin
sin
cos
− −






1
100
1600
4
r r
sin cos
θ θ = 0 since sin 180° = 0.
aφ = 0.
b) ω ω ω θ θ
θ
r z
r r r r
= = = − +





 − −





 −
0 0
1
10
40 1
10
40
2 2
, , sin ( sin ) = 0.
At (4, 180°)
v
ω = 0 since
v
ω = 0 everywhere.
3.18 a) a
r r r r r
r = −











 − +





 − −






10
80 240
10
80
10
80
3 4 3 3
cos cos
sin
( sin )
θ θ
θ
θ
− +





 = − −
10
80
8 75 1 9375 1
3
2 2
r r
sin
. ( )(. )( )
θ
= 8.203 m/s2
aθ = 0 since sin 180° = 0. aφ = 0 since vφ = 0.
b) ω r = 0, ω θ = 0, ω φ = 0, since sin 180° = 0.

39. 39
3.19
V V
a u
t x
∂ ∂
∂ ∂
= +
v v
v
v
+
V
w
y
∂
+
∂
v
ˆ.
V u
i
z t
∂
∂
∂
=
∂
v
For steady flow ∂ ∂
u t 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 vθ ∂ ∂θ
= =
0 0
and / :
r x
v v u u
a v u a v u
r x r x
∂ ∂ ∂ ∂
∂ ∂ ∂ ∂
= + = +
3.21 a) u e t
t
= − − = = ∞
−
2 1 0 1 10
( )( ) .
/
2 m / s at
( )
a
u
t
e t
x
t
= = −





 = =
−
∂
∂
2 1 0
1
10
0 2 0
10
( ) . .
/
m / s at
2
b) u e t
t
= − − = = ∞
−
2 1 0 1
2 10
( .5 )( ) .
/
1.875 m / s at
a e t
x
t
= −





 = =
−
2 1 0 2
1
10
0 0125 0
2 2 10
( .5 / ) . .
/
m / s at
2
c) u e t
t
= − − =
−
2 1 2 2 1
2 2 10
( / )( ) .
/
0 for all
a e t
x
t
= −





 =
−
2 1 2 2
1
10
0
2 2 10
( / ) .
/
for all
3.22
DT T
u
Dt x
∂
∂
= v
+
T
w
y
∂
∂
+ 2
20(1 ) sin 0.5878
100 100 5
T T t
y
z t
∂ ∂ π π π
∂ ∂
 
+ = − − = − ×
 
 
= −0.3693 °C/s.
3.23
D
Dt
u
x
v
y
w
z t
e
ρ ∂ρ
∂
∂ρ
∂
∂ρ
∂
∂ρ
∂
= + + + = − × − − × −
10 1 23 10 4 3000 10 4
( . ) = − ×
⋅
−
9 11 10 4
. .
kg
m s
3
3.24
D
Dt
u
x
v
y
w
z t
ρ ∂ρ
∂
∂ρ
∂
∂ρ
∂
∂ρ
∂
= + + + = −






10
1000
4
= −
⋅
2500
kg
m s
3
.
3.25
D
Dt
u
x
ρ ∂ρ
∂
= = ×
4 01
(. ) = 0.04 kg/m3⋅s
3.26 (D)
2
2
10
[10(4 ) ]
(4 )
x
u u u u u
a u v w u x
t x y z x x
x
−
∂ ∂ ∂ ∂ ∂ ∂
= + + + = = −
∂ ∂ ∂ ∂ ∂ ∂
−
3 2
2
10 10 1
10( 2)( 1)(4 ) 20 6.25 m/s .
4 8
(4 )
x
x
−
= − − − = × × =
−
3.27
D
Dt
V
t
= ⋅∇ +
v v ∂
∂
observing that the dot product of two vectors
v
A A i A j A k
x y z
= + +
$ $ $
and
v v v
B B i B j B k A B A B A B A B
x y z x x y y z z
= + + ⋅ = + +
$ $ $ .
is

40. 40
3.28
a
u
t
V u
a
v
t
V v
a
w
t
V w
a
V
t
V V
x
y
z
= + ⋅∇
= + ⋅∇
= + ⋅∇









∴ = + ⋅∇
∂
∂
∂
∂
∂
∂
∂
∂
v v
v v
v v
v
v
v v v
( )
3.29 Using Eq. 3.2.12:
a)
v v
v v v v v v
v
v
A a
d s
dt
V r
d
dt
r
= + + × + × × + ×
2
2
2Ω Ω Ω
Ω
( )
= 2 20 4 20 20 1
( $ $) $ ( $ .5$)
k i k k i
× + × × = 160 600
$ $
j i
− m / s
2
b)
v v v v v v o
A V r k j k k i
= × + × × = × − + × ×
2 2 20 20 30 20 20 3
Ω Ω Ω
( ) ( $ cos $) $ ( $ $) = −507$
i
3.30
v
Ω =
× ×
= × −
2
24 60 60
7 272 10 5
π $ . $
k k rad/s.
v
v i k i k
= − − = − −
5 707 707 3 3
( . $ . $) .535$ .535$ m/s.
v v v v v v
A V r
= × + × ×
2Ω Ω Ω
( )
= 2 7 272 10 3 3 7 272 10
5 5
× × × − − + × ×
− −
. $ ( .535$ .535 $) . $
k i k k
[ . $ ( . $ . $)]
7 272 10 6 10 707 707
5 6
× × × − +
−
k i k = − × +
−
51 4 10 0 0224
5
. $ . $
j i m / s2
.
Note: We have neglected the acceleration of the earth relative to the sun since it is quite small
(it is d s dt
2 2
v
/ ). The component 5 ˆ
( 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

41. 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/ ν = 2 × .015/.77 × 10-6 = 39 000. ∴Turbulent.
3.41 Re = =
VL
ν
.2 × .8/1.4 × 10-5 = 11 400. ∴Turbulent.
3.42 Re
.
.
= =
×
× −
VL
ν
4 06
17 10 5
= 14 100. ∴Turbulent.
Note: We used the smallest dimension to be safe!
3.43 a) Re
. .
.51
= =
×
×
=
−
V D
ν
1 2 0 01
1 10
795.
5
Always laminar.
b) Re
.
.51
= =
×
×
=
−
VD
ν
12 1
1 10
79
5
500. May not be laminar.
3.44 Re = 3 × 105 = T
Vx
ν
. ν µ ρ
= / where µ µ
= ( ).
T
a) T = 223 K or −50°C. ∴ = × ⋅
−
µ 1 10 5
.5 N s / m .
2
∴ =
×
×
= ×
−
−
ν
1 5 10
3376 1 23
2 5 10
5
5
.
. .
. .
m /s
2
3 10
900 1000
3600 2 10
5
5
× =
×
× × −
xT
.5
. ∴xT = 0.03 m or 3 cm
b) T = −48°F. ∴µ = 3.3 × 10−7
lb-sec/ft2. ν =
×
= ×
−
−
3 3 10
00089
3 7 10
7
4
.
.
. ft2/sec.
3 10
600 5280
3600 3 7 10
5
4
× =
×
× × −
xT
.
. ∴xT = 0.13' or 1.5"

42. 42
3.45 Assume the flow is parallel to the leaf. Then 3 × 105 =VxT / .
ν
∴ = × = × × × =
−
x V
T 3 10 3 10 1 4 10 6 8 17
5 5 4
ν / .5 . / . m.
The flow is expected to be laminar.
3.46 a) M
V
c
= =
× ×
=
100
14 287 236
0 325
.
. . For accurate calculations the flow is
compressible. Assume incompressible flow if an error of 4%, or so, is
acceptable.
b) M
V
c
= =
× ×
=
80
14 287 288
0 235
.
. . ∴Assume incompressible.
c) M
V
c
= =
× ×
=
100
14 287 373
0 258
.
. . ∴Assume incompressible.
3.47
D
Dt
u
x
v
y
w
z t
ρ ∂ρ
∂
∂ρ
∂
∂ρ
∂
∂ρ
∂
= + + + = 0. For a steady, plane flow
∂ρ ∂
/ t = 0 and w = 0. Then
u
x
v
y
∂ρ
∂
∂ρ
∂
+ = 0.
3.48
D
u
Dt x
ρ ∂ρ
∂
= v
y
∂ρ
∂
+ w
+
z t
∂ρ ∂ρ
∂ ∂
+ 0.
= ∴incompressible.
3.49 (B)
2
9810 0.800
. 113 m/s.
2 1.23
water
air
h
V p
V
γ
ρ ρ
×
= = = ∴ =
3.50
V p
2
2
=
ρ
. Use ρ = 0.0021 slug/ft3.
a) v p
= = × ×
2 2 3 144 0021
/ . / .
ρ = 203 ft/sec.
b) v p
= = × ×
2 2 9 144 0021
/ . / .
ρ = 351 ft/sec.
c) v p
= = × ×
2 2 09 144 0021
/ . / .
ρ = 111 ft/sec.
3.51 p
V
= =
×






ρ
2 2
2
123
120 1000
3600
2
. / = 683 Pa.
∴F = pA = 683 π × 0.0752 = 12.1 N.
3.52
V p
2
2
0
+ =
ρ
. ∴ =
−
=
×
V
p
2 2 2000
123
ρ .
= 57.0 m/s

43. 43
3.53 (C)
2 2 2
1 2 1
. 0.200 0.600. 2 9.81 0.400 2.80 m/s.
2 2 2
V V V
p
V
g g g
γ
+ = + = ∴ = × × =
3.54 (B) The manometer reading h implies:
2 2
2
1 1 2 2
2 2
2
or (60 10.2). 9.39 m/s
2 2 1.13
V p V p
V V
ρ ρ
+ = + = − ∴ = The
temperature (the viscosity of the water) and the diameter of the pipe are
not needed.
3.55 a)
2
2
0
2 2
V
V p
ρ
+ =
2
2
( 10 )
. . 50
2
o o
o
p p
x p
p p x ρ
ρ ρ ρ
−
+ + = ∴ = −
b)
2
2
0
2 2
V
V p
ρ
+ =
2
2
(10 )
. . 50
2
o o
o
p p
y p
p p y ρ
ρ ρ ρ
+ + = ∴ = −
3.56
2
2
2 2
U p
V p
ρ ρ
∞ ∞
+ = + .
a) v v U r r
r c
θ θ
= = = − −
∞
0 180 1 1
2 2
and o
: ( / )( ).
( )
∴ = − = −














∞ ∞
p U v U
r
r
r
r
r
c c
ρ ρ
2 2
2
2 2 2
2
2
4
.
b) Let r r p U
c T
= = ∞
:
ρ
2
2
c) ( ) [ ]
v r r v U p v U
r c
= = − ∴ = − = −
∞ ∞ ∞
0 2
2 2
1 4
2 2 2
and = U 2
: sin . sin
θ θ
θ
ρ ρ
θ
d) Let θ ρ
= = − ∞
90
3
2
90
2
o
: p U
3.57
2
2
2 2
U p
V p
ρ ρ
∞ ∞
+ = + .
a) vθ θ
= =
0 180
and o
: ( )
p U v U
r
r
r
r
r
c c
= − =





 −














∞ ∞
ρ ρ
2 2
2
2 2 2
3 6
.
b) Let r r p U
c T
= = ∞
:
1
2
2
ρ .
c) ( ) [ ]
v r r p v U
r c
= = − = −
∞ ∞
0
2 2
1 4
2 2 2
and = U 2
: sin
ρ ρ
θ
θ
d) Let θ ρ
= = − ∞
90
3
2
90
2
o
: p U

44. 44
3.58
2
2
2 2
U p
V p
ρ ρ
∞ ∞
+ = + .
a) ( )
p U u
x x
= − = − +











 = − +












∞
ρ ρ π
π
ρ
2 2
10 10
20
2
50 1 1
1
2 2 2
2 2
= − +






50
2 1
2
ρ
x x
b) 1
0 when 1. 50 ( 2 1) 50
u x p ρ ρ
−
= = − = − − + =
c) ( )
2 2
2 2 2 60 1
30 30 450 1 1
2 2 2
p U u
x x
ρ ρ π
ρ
π
∞
   
   
= − = − + = − +
   
   
   
   
   
= − +






450
2 1
2
ρ
x x
d) 1
0 when 1. 450 ( 2 1) 450
u x p ρ ρ
−
= = − = − − + =
3.59
V p V p
V p p
1
2
1 2
2
2
1 1 2
2 2
0 20
+ = + = − =
ρ ρ
. and kPa.
( )
V p p V
2
2
1 2 2
2 2
1000
20 6 32
= − = ∴ =
ρ
( .
000) = 40. m / s
3.60 Assume the velocity in the plenum is zero. Then
2 2
2
1 1 2 2
2 2
2
or (60 10.2). 9.39 m/s
2 2 1.13
V p V p
V V
ρ ρ
+ = + = − ∴ =
We found ρ = 113
. kg / m3
in Table B.2.
3.61 Bernoulli from the stream to the pitot probe: p
V
p
T = +
ρ
2
2
.
Manometer: .
T Hg
p H H h p h
γ γ γ γ
+ − − = −
Then,
2
2
Hg
V
p H H p
ρ γ γ
+ + − = . 2
(2 )
Hg
V H
γ γ
ρ
−
∴ =
a) V V
2 13 6 1 9800
1000
2 0 04 3 14
=
−
× ∴ =
( . )
( . ). . m / s
b) V V
2 13 6 1 9800
1000
2 0 1 4 97
=
−
× ∴ =
( . )
( . ). . m / s
c) V V
2 13 6 1 62 4
194
2 2 12 1162
=
−
× ∴ =
( . ) .
.
( / ). . fps
d) V V
2 13 6 1 62 4
194
2 4 12 16 44
=
−
× ∴ =
( . ) .
.
( / ). . fps

45. 45
3.62 The pressure at 90° from Problem 3.56 is 2
90 3 /2.
p U
ρ ∞
= − The pressure at the
stagnation point is 2
/2.
T
p U
ρ ∞
= The manometer provides: p H p
T − =
γ 90
2 2
1 3
1.204 9800 0.04 1.204 . 12.76 m/s
2 2
U U U
∞ ∞ ∞
× − × = − × ∴ =
3.63 The pressure at 90° from Problem 3.57 is 2
90 3 /2.
p U
ρ ∞
= − The pressure at the
stagnation point is 2
/2.
T
p U
ρ ∞
= The manometer provides: p H p
T − =
γ 90
2 2
1 3
1.204 9800 0.04 1.204 . 12.76 m/s
2 2
U U U
∞ ∞ ∞
× − × = − × ∴ =
3.64 Assume an incompressible flow with point 1 outside in the room where p1 0
=
and v1 0
= . The Bernoulli’s equation gives, with p h
w
2 2
= γ ,
2
1
2
V 1
p
ρ
+
2
2 2
.
2
V p
ρ
= +
a) 0
2
9800 0 02
1 204
18 04
2
2
2
= +
− ×
∴ =
V
V
.
.
. . m / s
b) 0
2
9800 0 08
1 204
36 1
2
2
2
= +
− ×
∴ =
V
V
.
.
. . m / s
c) 0
2
62 4 1 12
0 00233
66 8
2
2
2
= +
− ×
∴ =
V
V
. /
.
. . fps
d) 0
2
62 4 4 12
0 00233
133 6
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 1
0 0
= =
and . Bernoulli’s equation gives
0
2
1
2
2
2
2
2 2
2
= + ∴ = −
V p
p V
ρ
ρ
.
a) ρ = =
×
= ∴ = − × × = −
p
RT
p
90
0 287 253
1 239
1
2
1239 100 6195
2
2
.
. . .
kg / m Pa
3
b) ρ = =
×
= ∴ = − × × = −
p
RT
p
95
0 287 273
1 212
1
2
1212 100 6060
2
2
.
. . .
kg / m Pa
3
c) ρ = =
×
= ∴ = − × × = −
p
RT
p
92
0 287 293
1 094
1
2
1094 100 5470
2
2
.
. . .
kg / m Pa
3
d) ρ = =
×
= ∴ = − × × = −
p
RT
p
100
0 287 313
1113
1
2
1113 100 5566
2
2
.
. . .
kg / m Pa
3
3.66 (A)
2
1
2
V
g
2
1 2 2
2
p V p
g
γ γ
+ = +
2
2
2
800000
. . 40 m/s.
9810 2 9.81
V
V
= ∴ =
×

46. 46
3.67 a) p h V h h
A A A
= = × = = =
γ 9800 4 39 0 2
200 Pa, Using
. ,
2
2
A
V
g
A
A
p
h
γ
+ +
2
2 2
2
2
V p
h
g γ
= + +
2
2
2
.
2
A
V
p p
g
γ
= −
= −
×
× = −
39
14
2 9 81
9800 58
2
200 700 Pa
.
b) 0 and 0.
B B
p V
= = 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
2 9 81
9800 58
+ + = + + = −
×





 = −
γ γ
.
.
700 Pa
3.68 Bernoulli:
2
2 2
2
V p
g γ
+
2
1 1
2
V p
g γ
= +
Manometer:
2
2
1 2
2
Hg
V
p z H H z p
g
γ γ γ γ γ
+ + − − = +
Substitute Bernoulli’s into the manometer equation:
( )
2
1
1 1.
2
Hg
V
p H p
g
γ γ γ
+ − = +
a) Use H = 0.01 m:
V
V
1
2
1
9800
2 9 81
13 6 1 9800 0 01 1
×
×
= − × ∴ =
.
( . ) . .572 m / s
Substitute into Bernoulli:
p
V V
g
1
2
2
1
2 2 2
2
20 1
2 9 81
9800 198
=
−
=
−
×
× =
γ
.572
.
600 Pa
b) Use H = 0.05 m:
V
V
1
2
1
9800
2 9 81
13 6 1 9800 0 05 3
×
×
= − × ∴ =
.
( . ) . .516 m / s
Substitute into Bernoulli:
p
V V
g
1
2
2
1
2 2 2
2
20 3
2 9 81
9800 193
=
−
=
−
×
× =
γ
.516
.
600 Pa
c) Use H = 0.1 m:
V
V
1
2
1
9800
2 9 81
13 6 1 9800 0 1 4 972
×
×
= − × ∴ =
.
( . ) . . m / s
Substitute into Bernoulli:
p
V V
g
1
2
2
1
2 2 2
2
20 4 972
2 9 81
9800 187
=
−
=
−
×
× =
γ
.
.
400 Pa

47. 47
3.69 Bernoulli across nozzle:
2
1
2
V 2
1 2 2
2
p V p
ρ ρ
+ = + 2 1
. 2 /
V p ρ
∴ =
Bernoulli to max. height:
2
1
2
V
g
1
1
p
h
γ
+ +
2
2
2
V
g
= 2
p
γ
+ 2 2 1
. / .
h h p γ
+ ∴ =
a) V p
2 1
2 2 700 1000 37 42
= = × =
/ / .
ρ 000 m / s
h p
2 1 700
= =
/ γ 000 / 9800 = 71.4 m
b) V p
2 1
2 2 1 1000 52 92
= = × =
/ / .
ρ 400 000 m / s
h p
2 1
= =
/ γ 1 400 000/ 9800 = 142.9 m
c) V p
2 1
2 2 100 194 1218
= = × × =
/ / . .
ρ 144 fps
h p
2 1
= = ×
/ γ 100 144 / 62.4 = 231 ft
d) V p
2 1
2 2 200 194 172 3
= = × × =
/ / . .
ρ 144 fps
h p
2 1 200
= = ×
/ γ 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
γ
+
2
2 2
1
2
V p
h
g γ
+ = + 2 2
. 2 ( )
h V g H h
+ ∴ = −
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
1 2 2
1 2.
2
p V p
h h
g
γ γ
+ + = + +
Using p H p h h h V g H h
1 2 1 2 2 2
= = = = −
γ γ
, , ( )
and
3.71
2
1
2
V 2
1 2 2
.
2
p V p
ρ ρ
+ = + p2 = −100 000 Pa, the lowest possible pressure.
a)
600
2
2
2
000
1000
100 000
1000
= −
V
. ∴V2 = 37.4 m/s.
b)
300
2
2
2
000
1000
100 000
1000
= −
V
. ∴V2 = 28.3 m/s.

48. 48
c)
80 144
1.94
14.7 144
1.94
×
= −
×
V2
2
2
. ∴V2 = 118.6 ft/sec.
d)
40 144
1.94
14.7 144
1.94
×
= −
×
V2
2
2
. ∴V2 = 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.
V p
gz
V p
gz
1
2
1
1
2
2
2
2
2 2
+ + = + +
ρ ρ
. V V
1 2
= . Let z1 0
= , and p2 0
= .
500 000
1000
= 9 81 2
. .
z ∴ z2 = 51.0 m.
3.73 a) ( ) ( )
p V V
1 2
2
1
2 2 2
2
1000
2
2 10
= − = −
ρ
= −48 000 Pa
b) ( ) ( )
2 2 2 2
1 2 1
902
2 10 43300 Pa
2 2
p V V
ρ
= − = − = −
c) ( ) ( )
2 2 2 2
1 2 1
680
2 10 32600 Pa
2 2
p V V
ρ
= − = − = −
d) ( ) ( )
2 2 2 2
1 2 1
1.23
2 10 59.0 Pa
2 2
p V V
ρ
= − = − = −
3.74
V p V p
1
2
1 2
2
2
2 2
+ = +
ρ ρ
. ( ) ( )
2 2 2 2
1 2 1
1.23
2 8
2 2
p V V
ρ
= − = − = −36.9 Pa
3.75 (D) ( ) ( )
2 2 2 2
1 2 1
902
30 15 304400 Pa
2 2
p V V
ρ
= − = − =
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:
V p V p
p
V V
1
2
1 2
2
2
1
2
2
1
2
2 2 2
+ = + ∴ =
−
ρ ρ
ρ
. .
Since V V
2 1
< , we see that p1 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 .
=
V D
ν
For air ν ≅ × −
1 10 5
.5 . Use reasonable dimensions from your
experience!

49. 49
a) Re
.
.5
.
=
×
×
= ×
−
20 0 03
1 10
4 10
5
4
∴Separate
b) Re
.
.5
.
=
×
×
=
−
20 0 005
1 10
6700
5
∴Separate
c) Re
.5
. .
=
×
×
= ×
−
20 2
1 10
2 7 10
5
6
∴Separate
d) Re
.
.5
.
=
×
×
=
−
5 0 002
1 10
670
5
∴Separate
e) Re
.5
. .
=
×
×
= ×
−
20 2
1 10
2 7 10
5
6
∴Separate
f) Re
.5
.
=
×
×
= ×
−
100 3
1 10
2 10
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 ∆ ∆
p
V
R
n
= = × =
ρ
2 2
1000
10
0 05
0 02
.
. 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:
V p V p
1
2
1 2
2
2
2 2
+ = +
ρ ρ
p p
A B
> since V V
A B
<
p p
C D
< since V V
C D
>
p p
B D
> since V V
D B
>
stagnation
region
A
B
VA
VB

50. 50
CHAPTER 4
The Integral Forms of the
Fundamental Laws
4.1 a) No net force may act on the system: Σ
v
F = 0.
b) The energy transferred to or from the system must be zero: Q - W = 0.
c) If 3 3 2
ˆ ˆ
ˆ 10 ( ) 0
n
V V n i j
= ⋅ = ⋅ − =
v
is the same for all volume elements then
Σ
v v
F
D
Dt
V dm
= ∫ , or Σ
v v
F
D
Dt
mV
= ( ). Since mass is constant for a system
Σ
v
v
F m
DV
Dt
= . Since
DV
Dt
a F ma
v
v v v
= =
, .
Σ
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;V2/2;
gh; H/m = h (specific enthalpy).
Intensive properties: Temperature, T; time, t; pressure, p; density, ρ; viscosity, µ.
4.3 (B)
4.4
System ( )
t V
= 1
c.v.( )
t V
= 1
System ( )
t t V
+ ∆ = 1 V
+ 2
c.v.( )
t t V
+ ∆ = 1
4.5
System ( )
t V
= 1 V
+ 2
c.v.( )
t V
= 1 V
+ 2
System ( )
t t V
+ ∆ = 2 V
+ 3
c.v.( )
t t V
+ ∆ = 1 V
+ 2
1 2
1
2
3
pump

51. 51
4.6 a) The energy equation (the 1st law of Thermo).
b) The conservation of mass.
c) Newton’s 2nd law.
d) The energy equation.
e) The energy equation.
4.7
4.8
4.9
4.10 $ $ $ . ($ $)
n i j i j
1
1
2
1
2
0 707
= − − = − + . $ . $ .5$
n i j
2 0 866 0
= − . $ $
n j
3 = − .
1 1 1
ˆ ˆ ˆ
ˆ 10 [ 0.707( )] 7.07 fps
n
V V n i i j
= ⋅ = ⋅ − + = −
v
V V n i i j
n
2 2 2 10 0866 05 866
= ⋅ = ⋅ − =
v
$ $ ( . $ . $) . fps
3 3 2
ˆ ˆ
ˆ 10 ( ) 0
n
V V n i j
= ⋅ = ⋅ − =
v
4.11 flux = ηρ$
n VA
⋅
v
flux1 = ηρ ηρ
[ . ($ $)] $ / .
− + ⋅ = −
0 707 10 0 707 10
i j iA A
flux2 = ηρ ηρ
( . $ .5$) $ / .
0 866 0 10 0 866 10
i j iA A
− ⋅ =
flux3 = ηρ( $) $
− ⋅ =
j iA
10 0
3
n̂
v n̂
v
ω
n̂ v n̂
v
n̂
v
n̂
v n̂
n̂
n̂
v v
v
v
n̂
n̂
n̂
n̂
v
v
v
v
n̂ n̂ v
v

52. 52
4.12 ( $) ( .5$ . $) $( )
v
B n A i j j
⋅ = + ⋅ ×
15 0 0 866 10 12
= × × =
15 0 866 120 1559
. cm 3
Volume = 15 60 10 12 1559
sin cm 3
o
× × =
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 ρ and η 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,
ρ ρ ρ
$ $ $
. .
n VdA n VdA n VdA
A
A
c s
⋅ = ⋅ + ⋅ =
∫
∫
∫
v v v
0
2
1
For uniform flow all quantities are constant over each area:
ρ ρ
1 1 1 2 2 2 0
2
1
$ $
n V dA n V dA
A
A
⋅ + ⋅ =
∫
∫
v v
Let A1 be the inlet so $
n V V A
1 1 1 2
⋅ = −
v
and be the outlet so $ .
n V V
2 2 2
⋅ =
v
Then
− + =
ρ ρ
1 1 1 2 2 2 0
V A V A
or
ρ ρ
2 2 2 1 1 1
A V A V
=
1
2
1
system (∆t) is in
volumes 1 and 2
c.v. (0) = c.v. (∆t)
= volume 1
1
2
3
system (∆t) = V1 + V2 + V3
c.v. (∆t) = V1 + V2
system boundary
at (t + ∆t)

53. 53
4.18 Use Eq. 4.4.2 with mV representing the mass in the volume:
0 = + ⋅
∫
dm
dt
n VdA
V
c s
ρ$
. .
v
= + −
dm
dt
A V A V
V
ρ ρ
2 2 1 1
= + −
dm
dt
Q m
V
ρ &.
Finally,
dm
dt
m Q
V
= −
& .
ρ
4.19 Use Eq. 4.4.2 with mS representing the mass in the sponge:
0 = + ⋅
∫
dm
dt
n VdA
S
ρ$
v
= + + −
dm
dt
A V A V A V
S
ρ ρ ρ
2 2 3 3 1 1
= + + −
dm
dt
m A V Q
S
& .
2 3 3 1
ρ ρ
Finally,
dm
dt
Q m A V
S
= − −
ρ ρ
1 2 3 3
& .
4.20 (D) 2
200
0.04 70 0.837 kg/s
0.287 293
p
m AV AV
RT
ρ π
= = = × × =
×
& .
4.21 A1V1 = A2V2. π ×
125
144
2
.
× 60 = π ×
2 5
144
2
.
V2. ∴V2 = 15 ft/sec.
& .
.
m AV
= = ×
ρ π
194
1 25
144
60
2
= 3.968 slug/sec. Q = AV = π
125
144
2
.
× 60 = 2 045
. / .
ft sec
3
4.22 A1V1 = A2V2.π × .0252 × 10 = (2π × .6 × .003)V2. ∴V2 = 1.736 m/s.
& .
m AV
= = × ×
ρ π
1000 025 10
2
= 19.63 kg/s. Q = AV=π × .0252 × 10 =0 01963
. .
m / s
3
4.23 &
min = ρA1V1 + ρA2V2. 200 = 1000 π × .0252 × 25 + 1000 Q2. ∴Q2 =0 1509
. .
m / s
3
4.24 ρ1
1
1
40 144
1716 520
= =
×
×
p
RT
= .006455 slug/ft3. ρ2
7 144
1716 610
=
×
×
= .000963 slug/ft3.
& .
& .
( / ).
.
m AV V
m
A
= ∴ = =
×
ρ
ρ π
1
1 1
2
2
2 144 006455
∴V1 = 355 fps.
& . . ( / ) .
m V
2 2
0 2 000963 2 3 144
= = × × ∴V2 = 4984 fps.

54. 54
4.25 ρ ρ ρ ρ
1 1 1 2 2 2 1
1
2
500
287 393
4
1246
287 522
8 317
A V A V
p
RT
= = =
×
= =
×
=
.
.
.433 .
.
.
kg
m
kg
m
3 3
4.433 π × .052 × 600 = 8.317 π × .052 V2. ∴V2 = 319.8 m/s.
&
m A V
= ρ1 1 1 = 20.89 kg/s. Q A V
1 1 1
= = 4 712
. .
m / s
3
Q2 =2.512 .
m / s
3
4.26 ρ ρ
1 1 1 2 2 2
A V A V
=
p
RT
A V
p
RT
A V
1
1
1 1
2
2
2 2
=
200
293
0 05 40
120
0 03 120
2
2
2
π π
× × = × ×
. . .
T
∴ = −
T2 189 9 83
. .
K or C
o
4.27 a) A V A V
1 1 2 2
= . (2 × 1.5 + 1.5 × 1.5) 3 =π
d2
2
4
2
× . ∴d2 = 3.167 m
b) (2 × 1.5 + 1.5 × 1.5) 3 =π
d2
2
4
2
2
× . ∴d2 = 4.478 m
c) (2 × 1.5 + 1.5 × 1.5) 3 =
1
3 2
866 2
2
πR
R
R
− ×


 

 ×
. .
∴R = 3.581 m. ∴d2 = 7.162 m
4.28 (A) Refer to the circle of Problem 4.27:
2 3
75.7 2
( 0.4 0.10 0.40 sin75.5 ) 3 0.516 m /s.
360
Q AV π
×
= = × × − × × × =
o
4.29 a) v
r
r
r V vdA
r
r
rdr r
r
r
dr
r
r r
= −





 = = −





 = −






∫
∫ ∫
10 1 10 1 2 20
0
0
2
0
0
0
2
0
0
0
0 0
. .
π π π
∴ = −





 =
V
r
r r
20
2 3
10
3
0
2
0
2
0
2
= 3.333 m/s.
& . .
m AV
= = × × ×
ρ π
1000 04 3 33
2
= 16 75
. .
kg / s Q = AV =0 01675
. .
m / s
3
b) v
r
r
r V
r
r
rdr
r r
r
= −





 = −





 = −






∫
10 1 10 1 2 20
2 4
2
0
2 0
2
2
0
2
0
0
2
0
2
0
. .
π π π ∴V = 5 m/s
& .
m AV
= = × × ×
ρ π
1000 04 5
2
= 25 13
. kg / s. Q = AV =0 02513
. m / s.
3
c) v
r
r
r V
r
r
rdr r
r
r
= −





 = −





 +
∫
20 1 20 1 2 10 4
0
0
2
0
2
0
2
0
0
. / .
/
π π π ∴V = 5.833 m/s
& . .
m AV
= = × × ×
ρ π
1000 04 5833
2
= 29 32
. kg / s. Q =0 02932
. .
m / s
3
θ R
cosθ = 1/2
θ = 60o

55. 55
4.30 a) Since the area is rectangular, V = 5 m/s.
& . .
m A V
= = × × ×
ρ 1000 08 8 5 = 320 kg / s. Q =
&
m
ρ
= 0 32
. .
m / s
3
b) v
y
h
y
h
= −






40
2
2
with y = 0 at the lower wall.
∴ = −





 = ×
∫
Vhw
y
h
y
h
wdy
h
w
h
40 40
6
2
2
0
. ∴V = 6.667 m/s.
& . . .
m A V
= = × × ×
ρ 1000 08 8 6 667 = 426 7
. kg / s. Q = 0 4267
. .
m / s
3
c) V × .08 = 10 × .04 + 5 × .02 + 5 × .02. ∴V = 7.5 m/s.
& . . .5
m A V
= = × × ×
ρ 1000 08 8 7 = 480 kg / s. & &
Q
m
=
ρ
= 0 48
. .
m / s
3
4.31 a) A V v dA
1 1 2
= ∫ . π π π
×





 × = −





 =
∫
1
24
6 1 2 2
4
2
0
2
0
2
0
2
0
v
r
r
rdr v
r
r
max max .
With r0
1
24
= , vmax = 12 fps. ∴ =
v r
( ) 12 1 576 2
( ) .
− r fps
b) A V v dA
1 1 2
= ∫ .
1
12
6 1
4
3
2
2
× × = −





 =
−
∫
w v
y
h
wdy v w
h
h
h
max max .
With h =
1
24
, vmax = 9 fps. ∴ =
v y
( ) 9 1 576 2
( ) .
− y fps
c) 1 1 2 .
AV v dA
= ∫
0
2
2
2 0
max max
2
0 0
0.01 2 1 2 2 .
4
r r
r
v rdr v
r
π π π
 
× × = − =
 
∫  
 
With r0 = 0.01 m, max
v = 4 m/s. ( )
v r
∴ = 2
4(1 10000 ) m/s.
r
−
d) n̂
2
max max
2
4
0.02 2 1 .
3
h
h
y h
w v wdy v w
h
−
 
× × = − =
 
∫  
 
With h = 0.01 m, max
v = 3 m/s. ( )
v y
∴ = 2
3(1 10000 ) m/s.
y
−
4.32 If dm dt
/ ,
= 0 then ρ ρ ρ
1 1 1 2 2 2 3 3 3
A V A V A V
= + . In terms of &
m Q
2 3
and this
becomes, letting ρ ρ ρ
1 2 3
= = ,
1000 0 02 12 1000 0 01
2
2
× × × = + ×
π . & . .
m ∴ =
& . .
m2 5 08 kg / s
4.33 v dA A V
r
1
0
2 2
1
∫ = . v
r
r
rdr
r
max . .
0
2
1
2
2
1
1 2 0025 2
∫ −





 = × ×
π π
∴ = × ×
2
005
4
0025 2
2
2
π π
vmax
.
. . ∴vmax = 1 m/s. ∴ =
v r
( ) 1
005
2
2
−






r
.
.
m / s

56. 56
4.34 & & &. . ( ) . &.
.
m m m y y dy m
in out
= + × × × = − + × × ×





 +
∫
ρ ρ ρ
2 2 10 10 20 100 2 1 2 10
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
2
ρ ρ ρ
= +





 + &.
m ∴ =
& .
m 0 6667ρ = 0.82 kg/s.
4.35 V h u y dy
h
1 1
0
= ∫ ( ) . 10 05 10 20 100 2
0
× = −
∫
. ( )
y y dy
h
= −






10 10
100
3
2 3
h h .
∴666.7 h3 − 200 h2 = −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 y y y dy
= = − − ×
∫
∫ρ 2 2 1 3545 6 9 2 5
0
1 3
2
( )
= − − +
∫
22 6 2 127 9 3 19
2 2 3
0
1 3
y y y y dy
. .
/
= 4.528 slug/sec.
V u
= = × =
2
3
2
3
2
4
3
max .
fps (See Prob. 4.31b).
ρ =
+
2 2 194
2
. .
= 2.07 slug/ft3. ∴ = × × ×






ρV A 2 07
4
3
5
1
3
. = 4.6 slug/sec.
Thus, ρVA m
≠ & since ρ = ρ(y) and V = V(y) so that V V
ρ ρ
≠ .
4.37 A V A V
1 1 2 2
= . π π
× × = × ×
. ( . . ) .
01 8 2 2 04
2
2
V cos 30o
∴V2 = 0.05774 m/s.
4.38 2000
4
3
0015 9000 5
3
3
3
3
× × × ×
π .
m of H O
m of air
m of air
s
2
= 1.5 × (1.5h). ∴h = 0.565 m.
4.39 Use Eq. 4.3.3: 0 1 1 1 1
= − + ⋅
∫
∂ρ
∂
ρ
t
d V V n A
v
$ .
v
V n V
1 1 1
⋅ = −
$ .
∴ = −
ρ
∂ρ
∂
1 1 1
A V
t
Vtire .
( . )
.
37 14 7 144
1716 520
1
96
180 17
2
+
×
× ×





 × = ×
π
∂ρ
∂t

57. 57
∴ =
∂ρ
∂t
3 01 10 5
. .
×
−
− slug
ft sec
3
4.40 & & & .
m m m
in = +
2 3 V1 = 20 m/s (see Prob. 4.31c).
20 1000 02 10 1000 02
2 2
3
× × = + ×
π π
. . .
V ∴ V3 = 12.04 m/s.
4.41 0 2 3 1
= + = + + −
d
dt
m m
d
dt
m m m m
c v net c v
. . . .
& & & &
∴ = − − = × × × − − × ×
d
dt
m m m m
c v
. .
& & & . .
1 2 3
2 2
1000 02 20 10 1000 02 10
π π
= 2.57 kg/s.
4.42 The control surface is close to the
interface at the instant shown.
∴Vi = interface velocity.
ρ ρ
e e e i i i
A V A V
= .
1 15 300
8000
287 673
1
2 2
.5 .
.
.
× × × =
×
×
π π Vi
∴Vi = 0.244 m/s.
4.43 Assume an incompressible flow:
4 1 2 2
Q A V
= . 4 1500 60 2 4 2
× = ×
/ ( ) .
V ∴ =
V2 12.5 .
fps
4.44 For an incompressible flow (low speed air flow)
udA A V
A
=
∫ 2 2
1
. 20 0 8 0 15
1 5 2
2
0
0 2
y dy V
/
.
. . .
× = ×
∫ π
20 0 8
5
6
0 2 015
6 5 2
2
× = ×
. . . .
/
π V ∴ =
V2 27 3
. .
m / s
4.45 A V v dA A V
e e
1 1 2
+ =
∫
π π π
( . . )
.
.
.
01 0 025 4 200 1
0 025
2 0 1
2 2
2
2
2
0
0 025
− × + −





 = ×
∫
r
rdr Ve
0 1178 01963 0 0314
. . . .
+ = Ve ∴ =
Ve 10 0
. .
m / s
4.46 Draw a control volume around the entire set-up:
0 2 2 1 1
= + −
dm
dt
V A V A
tissue
ρ ρ
= +
−





 −
& & ( tan ) &
m
d d
h h h
tissue ρπ ρπ φ
2
2 2
2 1
2
1
4
Ve
n̂
n̂
Vi

58. 58
or
& & & tan .
m
d d
h h h
tissue =
−
+






ρπ φ
2
2
2
2 1
2
1
2
4
4.47 The width w of the channel is constant throughout the flow. Then
0 2 2 1 1
= + −
dm
dt
A V A V
ρ ρ . 0 2 2 1 1
= + −
d
dt
whL A V A V
( )
ρ ρ ρ
0 100 0 2 8 4 0 2
= × + × − ×
ρ ρ ρ
dh
dt
w w w
. . . ∴ =
& . .
h 0 008 m / s
4.48 0 2 2 1 1
= + −
dm
dt
A V A V
ρ ρ
= + × × − × −
& ( . . / ).
m 1000 0 003 0 02 10 10 60
2 6
π ∴ = × −
& . .
m 3 99 10 4
kg / s
4.49 ρ ρ
1 1 1 2 2 2
A V A V
= . &
m A V
1 2 2 2
= ρ .
400 10 900 0 2 0 05
10 100 6 2
e Ve
− −
× × = × ×
/
. . .
π ∴ =
Ve 207 m / s.
4.50 0 3 3 1 1 1 2
= + − −
dm
dt
Q A V m
ρ ρ & where m Ah
= ρ .
a) 0 1000 0 6 1000 0 6 60 1000 0 02 10 10
2 2
= × + × − × × −
π π
. & . / . .
h
∴ =
& .
h 0 0111 m / s or 11.1 mm / s
b) 0 1000 0 6 1000 0 01 0 20
2
= × + × − −
π . & . .
h
∴ =
& . .
h 0 00884 m / s or 8.84 mm/ s
c) 2 2
0 1000 0.6 1000 1.0/60 1000 0.02 5 10.
h
π π
= × + × − × × −
&
0.000339 m/s or 0.339 mm/s.
h
∴ =
&
4.51 A V A V
1 1 2 2
= where A2 is an area just under the top surface.
a) π π
× × = ×
−
0 02 10 60
2 10 2
. ( tan )
/
e h
dh
dt
t o
∴ = −
h dh e dt
t
2 10
0 001333
. .
/
∴ = − +
−
h e t
3 10
0 04 0 04
. . .
/
Finally,
h t e t
( ) . ( ) .
/ /
= − −
0 342 1 10 1 3
b) 0 04 10 10 60 10
10
. ( tan ) &
/
× × = ×
−
e h h
t o
∴ = −
hdh e dt
t
0 2309 10
. .
/
∴ = − +
−
h e t
2 10
4 62 4 62
. . .
/
Finally,
/10 1/2
( ) 2.15(1 ) .
t
h t e−
= −

59. 59
4.52 &
W T pAV
du
dy
Abelt
= + +
ω µ
= × × + × × × + × × × ×
−
20 500 2 60 400 0 4 0 10 181 10 100 0 0 8
5
π / . .5 . .5 .
= + + =
1047 800 0 000724 1847
. 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,
& ~ . &
m u W
∆ = 0 2
1000 0 02 4 18 0 2 500
× × = ×
. . . .
∆T ∴ =
∆T 0 836
. .
C
o
4.55 (D)
2 2
2 1
2
P
W V V
Q g
γ
−
=
&
2 1 1200 200
. .
0.040
P
p p W
γ γ γ
− −
+ =
×
&
40
40 kW and energy req'd = 47.1 kW.
0.85
P
W
∴ = =
&
4.56 & .
W
Q H
P
P
p
=
γ
η
5 746
9800 20
0 87
× =
× ×
Q
.
. ∴ =
Q 0 01656
. .
m / s
3
4.57 − = − ×
&
&
. .
W
mg
T
40 0 89
a) & . .
WT = × × × =
40 0 89 200 9 81 69 850 W
b) & . ( / ) .
WT = × × × =
40 0 89 90 000 60 9 81 523 900 W
c) & . ( / ) .
WT = × × × × =
40 0 89 8 10 3600 9 81 776
6
100 W
4.58
10000000
. 0.89 50. 1.273 m/s
100 3 60 9.8
T
T
W
z V
AVg V
η
ρ
− = ∆ = × ∴ =
× × × ×
&
4.59
V
g
p
z
V
g
p
z
1
2
1
1
2
2
2
2
2 2
+ + = + +
γ γ
.
12
2 32 2
6
36
64 4
2 2
2
2 2
×
+ = +
. .
.
h
h
8 236
20 1
2
2 2
.
.
.
= +
h
h Continuity: 3 × 12 = h2 V2.
3 ft
V1
h2
V2

60. 60
This can be solved by trial-and-error.
2 8': 8.24 ? 8.31
h = 2 7.9': 8.24 ? 8.22
h = Qh2 = 7 93
. .
'
2 1.8': 8.24 ? 8.00
h = 2 1.75': 8.24 ? 8.31
h = Qh2 = 176
. '.
4.60
V
g
z
V
g
z hL
1
2
1
2
2
2
2 2
+ = + + . ∴
×
+ =
×
+ +
4
2 9 81
2
16
2 9 81
0 2
2
2
2 2
. .
. .
h
h
∴2 615 0 815 2
2
2
. . / .
= +
h h Trial-and-error provides the following:
h2 2 2 615 2 63
= =
.5: . ? . h2 2 45 615 2
= =
. : . ? .59.
2 ∴ =
h2 2 47
. m
h2 0 65 2 615 2
= =
. : . ? .58 h2 0 64 615 2 63
= =
. : . ? . .
2 ∴ =
h2 0 646
. m
4.61 Manometer: Position the datum at the top of the right mercury level.
9810 4 9810
2
1000 9810 13 6 4 9810 2
2 2
2
2
1
× + + + × = × × + × +
. ( . ) .
z p
V
p
Divide by γ = 9810: . . . .
4
2
13 6 4 2
2
2 2
2
1
+ + + = × + +
z
p V
g
p
γ γ
(1)
Energy:
V
g
p
z
V
g
p
z
1
2
1
1
2
2
2
2
2 2
+ + = + +
γ γ
. (2)
Subtract (1) from (2): With z1 = 2 m,
V
g
1
2
2
12 6 4
= ×
. . . ∴V1 = 9.94 m/s
4.62 The manometer equation (see Prob. 4.61) is
0 4
2
13 6 4 2
2
2 2
2
1
. . . .
+ + + = × + +
z
p V
g
p
γ γ
(1)
Energy:
V
g
p
z
V
g
p
z
V
g
1
2
1
1
2
2
2
2
2
2
2 2
0 05
2
+ + = + + +
γ γ
. . (2)
Subtract (1) from (2): With z1 = 2 m, and with V2 = 4V1 (continuity)
18
2
12 6 0 4
1
2
.
. . .
V
g
= × ∴V1 = 7.41 m/s.
4.63 (A)
2
2
0
V
=
2
1 2 1
2
V p p
g
− −
+
2
2
2
120
. 0 . 7 200000 Pa.
2 9.8 9810
p
p
γ
−
= + ∴ =
×
4.64 Q = 120 × 0.002228 = π ×
1
12
2
1





 V . ∴V1 = 12.25 fps.

61. 61
Continuity: π π
×





 = ×






1
12
1
12
2
1
2
2
V V
.5
. ∴V2 = 5.44 fps.
Energy:
V
g
p V
g
p V
g
1
2
1 2
2
2 1
2
2 2
0 37
2
+ = + +
γ γ
. .
∴ = × + −






p2
2 2
60 144 62 4 0 63
12 25
64 4
5 44
64 4
. .
.
.
.
.
= 8702.9 psf or 60.44 psi
4.65 Q = 600 × 10-3/60 = π × .022 V1. ∴V1 = 7.958 m/s.
3
2
0.02
3 3
3 3 2
0
1 1
10 1
0.02 6.67 0.02
y
V dA wdy
AV w
α
 
= = −
 
∫ ∫  
× ×  
V
A V
A
2
1 1
2
2
2
04 7 958
06
= =
×
. .
.
= 3.537 m/s.
Energy:
V
g
p V
g
p
hL
1
2
1 2
2
2
2 2
+ = + +
γ γ
.
∴ =
−
×
+
−
hL
7 958 3
2 9 81
690 000 700 000
9810
2 2
. .537
.
= 1.571 m
4.66 V Q A
1 1 2
0 08
03
= =
×
/
.
.
π
= 28.29 m/s. ∴V2 = 9V1 = 254.6 m/s.
Energy:
V
g
p V
g
p V
g
1
2
1 2
2
2 1
2
2 2
2
2
+ = + +
γ γ
. .
∴ =
×
−
×






p1
2 2
9810
254 6
2 9 81
0 8
28 29
2 9 81
.
.
.
.
.
= 32 1 106
. .
× Pa
4.67 a) Across the nozzle: 2 2
1 2
.07 .025 .
V V
π π
× = × ∴V2 = 7.84 V1.
Energy:
V
g
p V
g
p
1
2
1 2
2
2
2 2
+ = +
γ γ
.
2
2
1 1
7.84 1
9810 .
2 9.81
p V
−
∴ =
×
For the contraction: π π
× = ×
. . .
07 05
2
1
2
3
V V ∴V3 = 1.96 V1.
Energy:
V
g
p V
g
p
1
2
1 3
2
3
2 2
+ = +
γ γ
.
Manometer: γ γ
× + = × +
. . . .
15 136 15
1 3
p p ∴ = × +
p p
1 3
12 6 15
γ γ
. . .
Subtract the above 2 eqns:
V
g
V
g
V
g
1
2
3
2
2 1
2
2
12 6 15
2
196
2
+ × = =
. . . .
∴ − = × ×
( . ) . . .
196 1 12 6 15 2
2
1
2
V g ∴V1 = 3.612 m/s. ∴p1 = 394 400 Pa.

62. 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
+ + = + +
γ γ
H = +
3 612
19 62
2
.
.
394 400
9810
= 40.0 m.
b) Manometer: γ γ
× + = × +
. . . .
2 136 2
1 3
p p ∴ = × +
p p
1 3
12 6 2
γ γ
. . .
Energy:
V
g
p V
g
p
1
2
1 3
2
3
2 2
+ = +
γ γ
. Also, V3 = 1.96 V1.
∴ + × =
V
g
V
g
1
2 2
1
2
2
12 6 2
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
+ + = + +
γ γ
. ∴ = =
×
H
V
g
2
2 2
2
32 7
2 9 81
.
.
= 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
+ + = + +
γ γ
. ∴ = = × ×
V gz
2 0
2 2 9 81 2 4
. . = 6.862 m/s.
Q = AV = .8 × 1 × 6.862 = 5 49
. .
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 g z h
2 0
2 2 9 81 2
( ) . = 6.264 m/s.
∴Q = .8 × 6.264 = 5 01
. .
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
+ + = + +
γ γ
. ∴ = = × × +





 =
V gz
2 0
2 2 32 2 6
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
H h h
2
2
2
( ) .
∴ = = × ×
V gH
2 2 2 32 2 6
. = 19.66 fps. ∴Q = 39.3 cfs.

63. 63
4.69 From the reservoir surface to the exit: Continuity:
V
g
p
z
V
g
p
z K
V
g
0
2
0
0
2
2
2
2
1
2
2 2 2
+ + = + + +
γ γ
. V V
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 × π × .0152 =0 00944
. .
m / s
3
The velocity in the pipe is V1 = 1.878 m/s.
Energy 0 → A: 10
1878
2 9 81 9810
8
1878
2 9 81
3
2 2
=
×
+ +
×
+
.
.
.
.
.
.
pA
∴pA = 65 500 Pa.
Energy 0 → B: 10
1878
2 9 81 9810
2 0
1878
2 9 81
10
2 2
=
×
+ +
×
+
.
.
.
.
.
.
pB
∴pB = −5290 Pa.
Energy 0 → C: 10
1878
2 9 81 9810
12 2 8
1878
2 9 81
2 2
=
×
+ + +
×
.
.
.
.
.
.
pC
∴pC = −26 300 Pa.
Energy 0 → D: 10
1878
2 9 81 9810
0 5
1878
2 9 81
2 2
=
×
+ + +
×
.
.
.
.
.
pD
∴pD = 87 500 Pa.
4.70
V
g
p
z
V
g
p
z
0
2
0
0
2
2
2
2
2 2
+ + = + +
γ γ
.
80 000
9810
+ =
×
4
2 9 81
2
2
V
.
. ∴V2 = 19.04 m/s.
a) Q A V
= = × ×
2 2
2
025 19 04
π . . = 0 0374
. .
m / s
3
b) Q A V
= = × ×
2 2
2
09 19 04
π . . = 0 485
. .
m / s
3
c) Q A V
= = × ×
2 2
2
05 19 04
π . . = 0 1495
. .
m / s
3
4.71 a)
p
z
V
g
V
g
0
0
2
2
1
2
2
1
2
γ
+ = + .54 .
80 000
9810
+ = +
4
16
2
1
2
1
2
1
2
V
g
V
g
.54 . ∴V1 = 3.687 m/s.
Q A V
= = × ×
1 1
2
05 3 687
π . . = 0 0290
. .
m / s
3
b) A V A V
1 1 2 2
= . V V V
1
2
2 2 2
09
05
3 24
= =
.
.
. .
80 000
9810
+ = +
4
2
2 3
3 24
2
2
2 2
2
2
V
g
V
g
.
.
. ∴V2 = 3.08 m/s. ∴Q A V
= 2 2 = 0 0784
. .
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. ∴Q A V
= 2 2 = 0 0767
. .
m / s
3
4.72 (C) Manometer:
2
2
1 2
2
V
H p g p
g
γ ρ
+ = + or
2
2
1
9810 0.02 .
2
V
p g
g
ρ
× + =
Energy:
2
100000
7.96
. 3.15.
2 9.81 9810
K K
= ∴ =
×

64. 64
Combine the equations:
2
1
1
9810 0.02 1.2 . 18.1 m/s.
2
V
V
× = × ∴ =
4.73 Manometer: .
6
.
13 2
1 p
z
H
p
z
H +
+
=
+
+ γ
γ
γ
γ ∴ = +
p
H
p
1 2
12 6
γ γ
. .
Energy:
p V
g
p V
g
1 1
2
2 2
2
2 2
γ γ
+ = + .
Combine energy and manometer: 12 6
2
2
2
1
2
. .
H
V V
g
=
−
Continuity: V
d
d
V
2
1
2
2
2 1
= . ∴ = × −






V H g
d
d
1
2 1
4
2
4
12 6 2 1
. / .
∴ = =
×
−






Q V
d H g
d d
d
1
1
2
1
4
2
4
1 2
1
2
4 4
12 6 2
1
π
π .
/
/
= 12 35 1
2
2
2
1
4
2
4
1 2
.
/
d d
H
d d
−






4.74 Use the result of Problem 4.73:
a) Q = 12.35 ×. .
.
. .
/
16 08
2
16 08
2 2
4 4
1 2
×
−





 = 0 0365
. .
m / s
3
b) Q = 12.35 ×. .
.
. .
/
24 08
4
24 08
2 2
4 4
1 2
×
−





 = 0 0503
. .
m / s
3
c) Using English units with g = 32.2: Q d d
H
d d
=
−






22 37 1
2
2
2
1
4
2
4
1 2
. .
/
Q = ×












−






22 37
1
2
1
4
10 12
25
2 2
4 4
1 2
.
/
.5 .
/
= 1.318 cfs.
d) Q = 22.37 ×1
1
3
15 12
1 3333
2
2
4 4
1 2
×






−






/
.
/
= 2.796 cfs.
4.75 (B)
2
2
0.040
. 7.96 m/s.
2 0.04
L
V p Q
h K V
g A
γ π
∆
= = = = =
×
2
100000
7.96
. 3.15.
2 9.81 9810
K K
= ∴ =
×
4.76 a) Energy from surface to outlet:
V
g
H
2
2
2
= . ∴ =
V gH
2
2
2 .
Energy from constriction to outlet:
p V
g
p V
g
1 1
2
2 2
2
2 2
γ γ
+ = + .

65. 65
Continuity: V V
1 2
4
= . With p1 = pv = 2450 Pa and p2 = 100 000 Pa,
2450
9810
16
2 9 81
2
100 000
9810
1
2 9 81
2
+
×
× = +
×
×
. .
.
gH gH ∴H = 0.663 m.
b) With p1 = 0.34 psia, p2 = 14.7 psia,
.
.
.
.
.
34 144
62 4
16
2
2
14 7 144
62 4
1
2
2
×
+ =
×
+
g
gH
g
gH ∴H = 2.21 ft.
4.77 Continuity: V V
1 2
4
= . Energy  surface to exit:
V
g
H
2
2
2
= .
Energy constriction to exit:
p V
g
p V
g
v
γ γ
+ = +
1
2
2 2
2
2 2
.
∴ = +
−
= − = − × ×
p p
V V
g
p H
v 2
2
2
2
2
2
16
2
15 100 000 15 65 9810
γ γ . = 4350 Pa.
From Table B.1, T = 30°C.
4.78 Energy  surface to surface: z z hL
0 2
= + . ∴ = +
30 20 2
2
2
2
V
g
.
Continuity: V1 = 4V2. ∴V1
2
= 160 g. ∴V2
2
= 10 g.
Energy  surface to constriction: 30
160
2
94 000
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
p V
g
p
1
2
1 2
2
2
2 2
+ = +
γ γ
.
V
g
V
g
1
2 2
1
2
2
200 2 778
2
2450
9810
+ = +
000
9810
.
.
∴V1 = 7.67 m/s. ∴Q = π × .052 × 7.67 = 0 0602
. .
m / s
3
4.80 Velocity at exit = Ve . Velocity in constriction = V1. Velocity in pipe = V2 .
Energy — surface to exit:
V
g
H
e
2
2
= . ∴ =
V gH
e
2
2 .
Continuity across nozzle: V
D
d
Ve
2
2
2
= . Also, V V
1 2
4
= .
Energy — surface to constriction: H
V
g
pv
= +
1
2
2 γ
.

66. 66
a) 5
1
2
16
2
2 5
97
4
4
= × × ×





 +
−
g
D
g
.
.
550
9810
∴ =
D 0131
. m
b) 15
1
2
16
8 12
2 15
34 14 7 144
62 4
4
4
= × ×





 +
−
g
D
g
( / )
(. . )
.
. ∴ = ′ ′′
D 0 446
. 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
2 9 81
1176 100
3 1
1177
2 9 81
2
=
×
+
−
+ + +
×
.
.
( ) .5
.
.
.
000
9810
H
∴ =
H 8.57 .
m
4.82 & . .
m AV
= = × ×





 × =
ρ π
1 94
1
12
120 5 079
2
slug / sec.
& . .
. .
/ . , .
WP = ×
−
×
+
×





 =
5 079 32 2
30 120
2 32 2
120 144
62 4
0 85 12 950
2 2
ft - lb
sec
or 23.5 Hp
4.83 & . .
m AV
= = × × × =
ρ π
1000 02 40 50 27
2
kg / s.
20
40
2 9 81 9810
0 82
2
000 = 50.27 9.81
102
×
−
×
+






.
/ . .
∆p
∴ = ×
∆p 1088 106
. .
Pa
4.84 (C)
2 2
2 1
2
P
W V V
Q g
γ
−
=
&
.
p
γ
∆
+
16
0.040 400 16 kW. 18.0 kW.
0.89
P
P
W
W Q p
η
= ∆ = × = = =
&
&
4.85 − = × ×
−
×
+
−





 ×
& .
.
.
. .
WT 2 1000 9 81
0 10 2
2 9 81
600
0 87
2
000
9810
∴ = ×
& . .
WT 1304 106
W
We used V Q A
2 2 2
2
25
10 2
= =
×
=
/
.
.
π
m / s.
4.86 V V
1 2 2 2
450
3
15 9
450
3 75
10 19
=
×
= =
×
=
π π
. .
.
. .
fps fps
−





 × = × ×
−
×
+
−






10 000
1
746
550 450 194 32 2
1019 159
2 32 2
18 140 144
62 4
2 2
,
.
. .
. .
.
( )
.
.
ηT
∴ =
ηT 0924
.

67. 67
4.87 a) & & & ( ) .
Q W mg
V V
g
p p
z z
c
g
T T
S
v
− =
−
+ − + − + −






2
2
1
2
2
2
1
1
2 1 2 1
2 γ γ
The above is Eq. 4.5.17 with Eq. 4.5.18 and Eq. 1.7.13.
γ γ
1
1
1
2
2 2
85 9 81
287 293
9 92
600 9 81
287
20
= =
×
×
= =
×
=
p g
RT T T
.
.
. .
.
.
.
N / m
500
3
∴− − ×
×
+ − + −






(
.
.5
.
( ) .
1
2 9 81
600
20
85 716
9 81
293
2
2
500 000) = 5 9.81
200 000
500
000
9.92
2
T
T
∴ =
T2 572 K or 299 C .
o
Be careful of units! p cv
2 600 716
= =
⋅
000 Pa,
J
K kg
.5
b) −60 000 + 1 500 000 = same as above. ∴ =
T2 560 K or 287 C.
o
4.88 γ γ
1
1
1
2
14 7 144 32 2
1716 520
0 0764
60 144 32 2
1716 760
0 213
= =
× ×
×
= =
× ×
×
=
p g
RT
. .
. .
.
. .
lb
ft
lb
ft
3 3
c mg AVg AV
v = = = = × ×





 × =
4296 213
1
24
600 697
2
ft - lb
slug - R
lb /sec.
o
. & . .
ρ γ π
Use Eq. 4.5.17 with Eqs. 4.5.18 and 1.7.13:
& & & ( ) .
Q W mg
V V
g
p p c
g
T T z z
c
v
+ =
−
+ − + − + −






2
2
1
2
2
2
1
1
2 1 2 1
2 γ γ
− × × + =
×
+
×
−
×
+ −






10 778 697 697
600
2 32 2
60 144
213
14 7 144
0764
4296
32 2
300 60
2
. & .
. .
.
. .
( )
Wc
∴ =
& .
Wc 40 600
ft - lb
sec
or 73.8 Hp
4.89 Energy — surface to exit: − = − +






& & . .
W mg
V
g
V
g
T T
η 2
2
2
2
2
20 4 5
2
V mg Q
2 2
15
6
1326 15 9810 147
=
×
= = = × =
π
γ
.
. &
m / s. 150 N / s.
− × =
×
− +
×





 ∴ =
& .
.
.
.
.
. & .
W W
T T
08 147
2 981
20 4 5
1326
2 9 81
2
150
13.26
5390 kW
2
4.90 (D)
2 2
4.58 7.16
36.0 15 3.2 . 416 000 Pa
2 9.81 9810 2 9.81
B
B
p
p
+ = + + ∴ =
× ×
In the above energy equation we used
2
2
0.2
with 4.42 m/s.
2 0.2
L
V Q
h K V
g A π
= = = =
×

68. 68
4.91 Energy — surface to “C”:
& . &
.
.
.
. .
W mg
P × + × =
×
+ +
×






8 10
10
2 9 81
200
7 7
10
2 9 81
7705
2 2
000
9810
( & . . .5 & .
mg AVg WP
= = × × × × = ∴ =
ρ π
1000 05 10 9 81 770 52
2
N / s.) 700 W
Energy — surface to “A”: 30
10
2 9 81 9810
1
10
2 9 81
169
2 2
=
×
+ +
×
∴ =
.
.5
.
. .
p
p
A
A 300 Pa
Energy — surface to “B”:
2 2 2
2 2
B O B O B
P P B O
V V p p V
W mg z z K
g g
η
γ
 
− −
= + + − +
 
 
 
& &
52
2 9 81 9810
30 15
10
2 981
706
2
700 .8 = 770.5
10
100 Pa
2
×
×
+ − +
×





 ∴ =
.
.
.
. .
p
p
B
B
4.92 Manometer: γ γ γ γ ρ
× + + = × + + +
20
12
13 6
20
12 2
1 1 2 2
2
2
z p z p
V
. .
∴ + + = × + + +
20
12
13 6
20
12 2
1
1
2
2 2
2
z
p
z
p V
g
γ γ
. .
Energy:
V
g
z
p
H z
p V
g
T
1
2
1
1
2
2 2
2
2 2
+ + = + + +
γ γ
.
∴ = × + − =
× 





=
20
12
13 6
20
12 2
18
1
3
516
1
2
1 2
. . .
V
g
H V
T fps.
π
∴ = × +
×
= = = × × ×
H W Q H
T T T T
12 6
20
12
516
2 32 2
62 3 62 4 18 9 62 3
2
.
.
.
. ' . & . . .
γ η
= 62 115
, .
980
ft - lb
sec
or Hp
4.93 Energy—across the nozzle:
2 2 2
1 1 2 2
2 1 1
2
5
. 6.25 .
2 2 2
p V p V
V V V
g g
γ γ
+ = + = =
2 2 2
1 1
1
6.25
400 000
. 4.58 m/s
9810 2 9.81 2 9.81
V V
V
∴ + = ∴ =
× ×
, 7.16 m/s
A
V = , 2 28.6 m/s.
V =
Energy—surface to exit:
2 2 2
28.6 4.58 7.16
15 1.5 3.2 .
2 9.81 2 9.81 2 9.81
P
H + = + +
× × ×
36.8 m.
P
H
∴ =
2
/ 9810 ( .01 ) 28.6 36.8/.85 3820 W.
P P P
W QH
γ η π
∴ = = × × × × =
&

69. 69
Energy —surface to “A”:
2 2
7.16 7.16
15 3.2 . 39 400 Pa
2 9.81 9810 2 9.81
A
A
p
p
= + + ∴ =
× ×
Energy —surface to “B”:
2 2
4.58 7.16
36.0 15 3.2 . 416 000 Pa
2 9.81 9810 2 9.81
B
B
p
p
+ = + + ∴ =
× ×
4.94 (A) V
Q
A
= =
×
=
0 1
04
19 89
2
.
.
.
π
m / s.
Energy —surface to entrance: H
V
g
p
z K
V
g
P = + + +
2
2
2
2
2
2
2 2
γ
.
∴ =
×
+ + +
×
=
H P
19 89
2 9 81
180
50 5 6
19 89
2 9 81
201 4
2 2
.
.
.
.
.
.
000
9810
m.
∴ = = × × =
& / . . / . .
W QH
P P P
γ η 9810 0 1 2014 0 75 263 000 W
4.95 Energy —surface to exit: 10
2
2 2
2
2
2
2
2
2
2
= + + +
V
g
p
z
V
g
γ
. .
∴ = = = × ∴ =
V Q d d
2 2
2
2
7 83 0 02 7 83 4 0 0570
. . . / . .
m / s. m.
π
4.96 Depth on raised section = y2. Continuity: 3 3 2 2
× = V y .
Energy (see Eq. 4.5.21):
3
2
3
2
0 4
2
2
2
2
g
V
g
y
+ = + +
( . ).
∴ = + − + =
3 059
9
2
3 059 4 128 0
2
2
2 2 2
3
2
2
. , . .
g y
y y y
or
Trial-and-error:
2
2
2
2.0: .11 ? 0.
1.85 m.
1.8: .05 ? 0.
y
y
y
= − 
∴ =

= + 
2
2
2
2.1: .1 ? 0.
2.22 m.
2.3: .1 ? 0.
y
y
y
= − 
∴ =

= + 
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.
m1
m3
m2
m3
.
.
.
.

70. 70
To find &
m3 use continuity:
& & & . ( ) & .
m m m y dy m
1 2 3
2
3
0
2
2 4 10 32 2 28 10 2
= + × × = + +
∫
ρ ρ
∴ = − × +





 =
& . .
m3 640 10 28 2
8
3
53 3
ρ ρ ρ
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
− − ∫
ρ
= − − +
∫
1280
32
2
53 3 32 10 28
2
2 2 3
0
2
ρ ρ ρ
. ( )
y dy
[655360 54579 507320] 115000 .
W
ρ
= − − =
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
hL
1
2
1
2
2
2
2
2 2
+ = + +
γ
α
γ
.
8
2 9 81
150
2
8
2 9 81
110
2 2
×
+ =
×
+ +
. .
.
000
9810
000
9810
hL
∴ =
hL 0 815
. .
m
4.99 V
A
VdA y dy
= = + = × +





 =
∫
∫
1 1
2
28
1
2
28 2
2
3
29 33
2
3
0
2
( ) . m / s
α = =
×
+
∫ ∫
1 1
2 29 33
28
3
3
3
2 3
0
2
AV
V dA y dy
.
( )
[ ]
=
×
× + × × + × × + =
1
2 29 33
28 2 3 28 2 3 3 28 2 5 2 7 1005
3
3 2 3 5 7
.
/ / / .
4.100 a)
2 2 4
0.01
2 2 2 2
0
1 1 20 0.01 0.01
10 1 2 5 m/s
2
0.01 0.01 0.01 4 0.01
r
V VdA rdr
A
π
π
   
= = − = − =
   
∫ ∫  
× ×
 
   
α
π
π
= =
× ×
−






∫ ∫
1 1
0 01 5
10 1
0 01
2
3
3
2 3
3
2
2
3
0
0 01
AV
V dA
r
rdr
. .
.
=
×
−
×
×
+
×
×
−
×





 =
2000
0 01 5
0 01
2
3 0 01
4 0 01
3 0 01
6 0 01
0 01
8 0 01
2 00
2 3
2 4
2
6
4
8
6
.
. .
.
.
.
.
.
.

71. 71
b)
2 3
0.02
2 2
0
1 1 10 0.02
10 1 0.02 6.67 m/s
0.02 0.02
0.02 3 0.02
y
V VdA wdy
A w
   
= = − = − =
   
∫ ∫    
×
   
3
2
0.02
3 3
3 3 2
0
1 1
10 1
0.02 6.67 0.02
y
V dA wdy
AV w
α
 
= = −
 
∫ ∫  
× ×  
=
×
−
×
×
+
×
×
−
×





 =
1000
0 02 6 67
0 02
3 0 02
3 0 02
3 0 02
5 0 02
0 02
7 0 02
1
3
3
2
5
4
7
6
. .
.
.
.
.
.
.
.
.541
4.101 V
A
VdA
R
u
r
R
rdr u
n
n
n
n
n
R
= = −





 = −
+
−
+






∫
∫
1 1
1 2 2
2 1 1
2
1
0
π
π
max
/
max
K E V
V
dA u
r
R
rdr u R
n
n
n
n
n
R
. . max
/
max
=





 = −





 = −
+
−
+












∫
∫
ρ
ρ
π ρπ
2
3
3
3 2
0
2 2
1 2
3 2 3
a) V u u
= − −





 =
2
5
11
5
6
0 758
max max
.
K E R u R u
. . .
max max
= −





 =
ρπ ρπ
2 3 2 3
5
8
5
13
0 24
α
ρ
ρπ
ρπ
= =
×
=
K E
AV
R u
R u
. . .
.
.
max
max
1
2
0 24
1
2
0 758
1102
3
2 3
2 3 3
b) V u u
= − −





 =
2
7
15
7
8
0 817
max max
.
K E u R R u
. . .
max max
= −





 =
ρπ ρπ
3 2 2 3
7
10
7
17
0 288
α
ρ
ρπ
ρπ
=






=
×






=
K E
AV
V
R u
R u
u
. . .
.
.
.
max
max
max
2
2 3
2
2 2
2
0 288
0 817
0 817
2
1056
c) V u u
= − −





 =
2
9
19
9
10
0 853
max max
.
K E R u R u
. . .
max max
= −





 =
ρπ ρπ
2 3 2 3
9
12
9
21
0 321
α
ρ
ρπ
ρπ
= =
×
=
K E
AV
R u
R u
. . .
.
.
max
max
1
2
0 321
1
2
0 853
1034
3
2 3
2 3 3

72. 72
4.102 Engine power = F V m
V V
u u
D × +
−
+ −






∞
& ~ ~
2
2
1
2
2 1
2
& & ( )
m g F V m
V V
c T T
f f D v
= +
−
+ −






∞
2
2
1
2
2 1
2
4.103 &
W F V
D
η = ×
10
5
930
100
3600
015
1340
1000
100
3
−





 ×





 ×





 ×





 × = ×
m
km
kg
m
kJ
kg
km
s
000
3600
3
3
qf .
∴ =
qf 48 030 kJ / kg
4.104 0
2 2
32
2
2
2
2
2
1
2
1
1 2
= + + − − − +
α
γ γ
ν
V
g
p
z
V
g
p
z
LV
gD
0 2
2 9 81
0 35 32
10 180
9 81 0 02
2 6
2
=
×
− + ×
×
×
−
V V
.
.
. .
.
V V V Q
2 5
14 4 3 434 0 0 235 7 37 10
+ − = ∴ = = × −
. . . . .
m / s and m / s
3
4.105 Energy from surface to surface: H
V
g
p
z
V
g
p
z K
V
g
P = + + − − − +
2
2
2
2
1
2
1
1
2
2 2 2
γ γ
.
a) H
Q
Q
P = +
× × ×
= +
40 5
0 04 2 9 81
40 50 7
2
2
2
π . .
.
Try Q H H
P P
= = =
0 25 43 2 58
. : . (energy). (curve)
Try Q H H
P P
= = =
0 30 44 6 48
. : . (energy). (curve)
Solution: Q = 0 32
. .
m / s
3
b) H
Q
Q
P = +
× × ×
= +
40
20
0 04 2 9 81
40 203
2
2
2
π . .
Try Q H H
P P
= = =
0 25 52 7 58
. : . (energy). (curve)
Solution: Q = 0 27
. m / s
3
Note: The curve does not allow for significant accuracy.
4.106 Continuity: A V A V A V
1 1 2 2 3 3
= +
π π π
× × = × × + × ∴ =
0 06 5 0 02 20 0 03 1111
2 2 2
3 3
. . . . .
V V m / s
Energy: energy in + pump energy = energy out
& & & &
m
V p
W m
V p
m
V p
P P
1
1
2
1
2
2
2
2
3
3
2
3
2 2 2
+





 + × = +





 + +






ρ
η
ρ ρ

73. 73
1000 0 06 5
5
2
120
0 85 1000 0 02 20
20
2
300
2
2
2
2
π π
× × +





 + = × × +






. . & .
000
1000
000
1000
W P
+ × × +






1000 0 03 1111
1111
2
500
2
2
π . .
. 000
1000
∴ =
&
WP 26 700 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
p V
g
p
1
2
1 2
2
2
2 2
+ = +
γ γ
.
2
2 1 1
2
4 .
( /2)
d
V V V
d
= =
a)
V V
1
2
1
2
2 9 81
200 16
2 9 81
×
+ =
×
. .
.
000
9810
∴ =
V1 5164
. m / s.
( )
p A F m V V
1 1 2 1
− = −
& .
200 000 03 1000 03 5 164 4 5 164 5164
2 2
π π
× − = × × × −
. . . ( . . ).
F ∴ =
F 339 N.
b)
V V
1
2
1
2
2 9 81
400 16
2 9 81
×
+ =
×
. .
.
000
9810
∴ =
V1 7 303
. m / s.
400 000 03 1000 03 7 303 4 7 303 7 303
2 2
π π
× − = × × × −
. . . ( . . ).
F ∴ =
F 679 N.
c)
2 2
1 1
16
200 000
.
2 9.81 9810 2 9.81
V V
+ =
× ×
1 5.164 m/s.
V
∴ =
2 2
200 000 .06 1000 .06 5.164(4 5.164 5.164).
F
π π
× − = × × × − 1356 N.
F
∴ =
d)
2 2
1 1
1
16
30 144
. 17.24 fps.
2 32.2 62.4 2 32.2
V V
V
×
+ = ∴ =
× ×
2 2 2
30 1.5 1.94 (1.5/12) 17.24 (4 1). 127 lb.
F F
π π
× × − = × × × − ∴ =
e)
2 2
1 1
1
16
60 144
. 24.38 fps.
2 32.2 62.4 2 32.2
V V
V
×
+ = ∴ =
× ×
2 2 2
60 1.5 1.94 (1.5/12) 24.38 (4 1). 254 lb.
F F
π π
× × − = × × × − ∴ =
f)
2 2
1 1
1
16
30 144
. 17.24 fps.
2 32.2 62.4 2 32.2
V V
V
×
+ = ∴ =
× ×
2 2 2
30 3 1.94 (3/12) 17.24 (4 1). 509 lb.
F F
π π
× × − = × × × − ∴ =
4.109
V
g
p V
g
p
1
2
1 2
2
2
2 2
+ = +
γ γ
. V V V
2
2
2 1 1
9
3
9
= = .
V1
2
2 9 81
2 81
2 9 81
×
+ =
×
. .
.
000 000
9810
V1
2
∴ =
V1
2
50.

74. 74
p A F m V V m V
1 1 2 1 1
8
− = − =
&( ) &
2 045 1000 045 8 50
2 2
000 000π π
× − = × × ×
. .
F
∴ =
F 10 180 N.
4.110
2 2
1 1 2 2
.
2 2
V p V p
g g
γ γ
+ = + 0
2
.01 .006 .15. 11.1 m/s.
e e
V V V
π × = × × ∴ =
ΣF m V V
x x x
= −
&( ).
2 1
a) V V V
2
2
2 1 1
10
8
1
= = .562 .
V V
1
2
1
2
2 9 81
400 2 441
2 9 81
×
+ =
×
.
.
.
.
000
9810
∴ =
V1 23.56 .
m / s
∴ − = −
p A F m V V
1 1 2 1
&( ).
400 05 1000 05 23 23
2 2
000π π
× − = × × ×
. . .56(.562 .56).
F ∴ =
F 692 N.
b) V V V
2
2
2 1 1
10
6
2 778
= = . .
V
g
V
g
1
2
1
2
2
400 7 716
2
+ =
000
9810
.
. ∴ =
V1 10 91
. .
m / s
400 05 1000 05 10 91 1778 10 91
2 2
000π π
× − = × × ×
. . . ( . . ).
F ∴ =
F 1479 N.
c) V V V
2
2
2 1 1
10
4
6 25
= = . .
V
g
V
g
1
2
1
2
2
400 39 06
2
+ =
000
9810
.
. ∴ =
V1 4.585 .
m / s
400 05 1000 05 4 5 25 4
2 2
000π π
× − = × × ×
. . .585( . .585).
F ∴ =
F 2275 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
. ∴ =
V1 1132
. .
m / s
400 05 1000 05 1132 24 1132
2 2
000π π
× − = × × ×
. . . ( . ).
F ∴ =
F 2900 N.
4.111 (C)
2 2
1 1 2 2
2 2
V p V p
g g
γ γ
+ = +
2 2
1
(6.25 1) 12.73
. 9810 3085000 Pa.
2 9.81
p
− ×
= × =
×
2
1 1 2 1
( ). 3085000 0.05 1000 0.1 12.73(6.25 1)
p A F Q V V F
ρ π
− = − × × − = × × −
17500 N.
F
∴ =
4.112 V V p
V V
g
2 1 1
2
2
1
2 2 2
4 120
2
62 4
120 30
2 32 2
13
= = =
−





 =
−
×





 =
fps. 080 psf.
γ .
.
,
2 2
1 1 2 1
1.5 1.5
( ) 13,080 1.94 30( 120 30) 1072 lb.
12 12
x x
F p A m V V π π
   
= − − = − × × − − =
   
   
&
4.113 V V
V
g
p V
g
p V
g
p
2 1
1
2
1 2
2
2 1
2
1
4
2 2
15
2
= + = + ∴ =
. . .
γ γ γ
a) V1
2 2 9 81
15 9810
200 000 26 67
=
×
×
× =
.
. . ∴ = =
V V
1 2
516 20 7
. .
m / s, m / s.
p A F m V V
x x x
1 1 2 1
− = −
&( ). ∴ = × + × × =
Fx 200 000 04 1000 04 5 16 1139
2 2 2
π π
. . . .
N

75. 75
F m V V
y y y
= −
&( ).
2 1 ∴ = × × =
Fy 1000 04 5 16 20 7 537
2
π . . ( . ) .
N
b) V1
2 2 9 81
15 9810
400 000 53 33
=
×
×
× =
.
. . ∴ = =
V V
1 2
7 30 29 2
. .
m / s, m / s.
p A F m V V
x x x
1 1 2 1
− = −
&( ). ∴ = × + × × =
Fx 400 000 04 1000 04 7 3 2280
2 2 2
π π
. . . .
N
F m V V
y y y
= −
&( )
2 1 =1000 04 7 3 29 2 1071
2
π× × × =
. . ( . ) .
N
c) V1
2 2 9 81
15 9810
800 000 106 7
=
×
×
× =
.
. . ∴ = =
V V
1 2
10 33 41 3
. .
m / s, m / s.
2 2 2 2
1 1 1 1 800 000 .04 1000 .04 10.33 4560 N.
x
F p A AV
ρ π π
= + = × + × × =
F m V
y y
= &( )
2 =1000 04 10 33 41 3 2140
2
π× × =
. . ( . ) .
N
4.114 V V
2
2
2 1
40
10
80
= = m / s.
V
g
p V
g
p
1
2
1 2
2
2
2 2
+ = +
γ γ
∴ =
×
−
×





 = ×
p1
2 2
6
9810
80
2 9 81
5
2 9 81
3 19 10
. .
. Pa.
p A F m V V
x x
1 1 2 1
− = −
& ( ). ∴ = × × − × × − =
F 3 19 10 2 1000 2 5 80 5 353
6 2 2
. . . ( )
π π 000 N.
4.115 A V A V
1 1 2 2
= . π π
× × = −
. (. . ) .
025 4 025 02
2 2 2
2
V
∴ =
V2 1111
. m / s.
p V
g
p V
g
1 1
2
2 2
2
2 2
γ γ
+ = + .
2 2
1
11.11 4
9810 53700 Pa.
2 9.81
p
 
−
= =
 
 
×
 
1 1 2 1
( ).
p A F m V V
− = −
&
2 2
53 700 .025 1000 .025 4(11.11 4) 49.6 N.
F π π
∴ = × − × × − =
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
+ + = + +
γ γ
V V
V V
1
2
1
2
1 2
2 9 81
7
49
2 9 81
1 0 495 3 467
×
+ =
×
+ ∴ = =
.
.
.
. . . , . m / s.
Momentum: F F R m V V
x
1 2 2 1
− − = −
&( )
9810 35 7 1 9810 05 0 1 1 1000 1 1 3 467 3 467 495
× × − × × − = × × × −
. (. .5) . ( . .5) (. .5) . ( . . )
Rx
∴ =
Rx 1986 N.
∴Rx acts to the left on the water, and to the right on the obstruction.
F
V2
p1
A1
F
p1A1
F2
F1
Rx

76. 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
+ + = + +
γ γ
2 2
2 2
/900
6 0.2.
2 9.81 2 9.81
V V
+ = +
× ×
2 1
10.67, .36 m/s.
V V
∴ = =
Momentum: F F F m V V
1 2 2 1
− − = −
&( )
9810 3 6 4 9810 1 2 4 1000 2 4 10 67 10 67 36
× × − × × − = × × × −
( ) . (. ) (. ) . ( . . )
F
∴ =
F 618 000 N. (F acts to the right on the gate.)
4.118 a) 8 6 2 2
× =
. .
V y F F m V V
1 2 2 1
− = −
& ( ).
γ γ ρ
× × − = ×
×
−






. . .
.
.
3 6
2
6 8
8 6
8
2
2
2
w
y
y w w
y
γ
ρ
2
36 4 8 8
6
6
4 8 8 2
9 81
2
2 2
2
2 2
(. ) .
.
. (. )
.
.
.
− = ×
−
∴ + =
× ×
y
y
y
y y
y y y
2
2
2 2
6 7 829 0 2
+ − = ∴ =
. . . .51 .
m (See Example 4.12.)
b) y y y
g
y V
2 1 1
2
1 1
2 2 2
1
2
8 1
2
4 4
8
9 81
4 12 3 23
= − + +








= − + + × ×





 =
. .
.
. . .
m
c) y y y
g
y V
2 1 1
2
1 1
2 2 2
1
2
8 1
2
2 2
8
32 2
2 20 6 12
= − + +








= − + + × ×





 =
.
. .
ft
d) y y y
g
y V
2 1 1
2
1 1
2 2 2
1
2
8 1
2
3 3
8
32 2
3 30 11
= − + +








= − + + × ×





 =
.
.54 .
ft
4.119 Continuity: V y V y V y y y
2 2 1 1 2 1 2 1
4 4
= = ∴ =
. .
Use the result of Example 4.12: y y y
g
y V
2 1 1
2
1 1
2
1 2
1
2
8
= − + +














/
a) y2 4 8 3 2
= × =
. . .
m
1/2
2 2
1
1 8
3.2 .8 .8 .8 .
2 9.81
V
 
 
= − + + × ×
 
 
 
 
 
∴ =
V1 8 86
. m / s.
b) y2 4 2 8
= × = ft.
8
1
2
2 2
8
32 2
2
2
1
2
1 2
= − + + × ×












.
.
/
V ∴ =
V1 25 4
. fps.
F2
F1
F

77. 77
4.120 V =
×
=
9
3 3
1 m / s.
1
2 9 81
3
2 9 81
2
1
2
1
×
+ =
×
+
. .
.
V
y V y
1 1 1 3
= × .
∴ = +
3 05
19 62
3
1
2
1
.
.
.
V
V
Trial-and-error:
V
V
1
1
7 3 05 2 93
7 2 3 05 3 06
=
=
= =







: . ? .
. : .
?
.
V1 7 19
417
=
=
.
.
m / s.
y m.
1
1/2
2 2
2
1 8
.417 .417 .417 7.19 1.90 m.
2 9.81
y
 
 
= − + + × × =
 
 
 
 
 
V2 19 7 19 417
× = ×
. . . . V2 1
= .58 .
m / s
4.121 Refer to Example 4.12: γ γ ρ
y
y w w w
y
V y
1
1
1
1 1
2
3 6 6 10 10
60
6 10
− × × = × × −





 = ⋅
. ( ).
∴ − =
−






γ
ρ
2
36 600
6
1
2 1
1
( ) .
y
y
y
∴ + = = ∴ = =
( )
.
. . . , . .
y y y V
1 1 1 1
6
1200
32 2
37 27 3 8 15 8
ft fps
4.122 Continuity: 20 015 03
2
2
2
× × = ×
π π
. . .
V
∴ =
V2 5 m / s.
Momentum: p A p A m V V
1 1 2 2 2 1
− = −
&( ).
60 03 03 1000 015 20 5 20
2
2
2 2
000π π π
× − × = × × −
. . . ( ).
p ∴ =
p2 135 kPa.
4.123
2
1 1 2 2 2 2
.05
2 . 15 30 m/s.
2 .025
V A V A V
π
π
×
= = =
×
p V
g
p V
g
p
1 1
2
2 2
2
1
2 2
2 2
9810
30 15
2 9 81
337
γ γ
+ = + ∴ =
−
×
=
.
.
500 Pa.
( )
2 1 1 1 1
. ( ).
x x x
F m V V p A F m V
Σ = − − = −
& &
∴ = + = × + × × =
F p A mV
1 1 1
2 2 2
337 05 1000 05 15 4420
& . . .
500 N
π π
4.124 & . .
m1
2
1000 03 12 33 93
= × × =
π kg / s.
& . .
m3
2
1000 02 8 10 05
= × × =
π kg / s.
∴ = − = = × ∴ =
& & & . . . .
m m m V V
2 1 3
2
2 2
23 88 1000 03 8 446
π m / s.
Energy from 1 → 2:
V
g
p V
g
p
p
1
2
1 2
2
2
2
2
2 2
500
8 446
2 9 81
9810
+ = + ∴ =
−
×
×
γ γ
.
.
.
000+
122
= 536 300 Pa.
p1A1
p2A2
p1A1 p2A2
p3A3
Ry
Rx

78. 78
Energy from 1 → 3:
V
g
p V
g
p
1
2
1 3
2
3
2 2
+ = +
γ γ
.
∴ = +
−
×
=
p3
2 2
500
12 8
2 9 81
9810 540
000 000 Pa.
.
p A p A R m V m V m V
x x x x
1 1 2 2 2 2 3 3 1 1
− − = + −
& & & .
∴ = × − × + × − × =
Rx 500 03 536 03 33 93 12 23 88 8 446 103
2 2
000 300 N
π π
. . . . . .
p A R m V m V m V
y y y y
3 3 3 3 2 2 1 1
− = + −
& & & .
∴ = × − × − =
Ry 540 10 05 8 759
000 .02 N
2
π . ( ) .
4.125 a) ( )
ΣF m V V F mV V
m
A
x x x
= − − = − =
& . & .
&
2 1 1 1
1
ρ
=
×
300
1000 052
π .
= 38.2 m/s
∴ = × =
F 300 38 2 11
. .
460 N
b) − = − −
F m V V
r B
& ( )(cos ).
1 1
α
∴ = × − =
F 300
28 2
38 2
38 2 10 6250
.
.
( . ) .
N
c) − = − −
F m V V
r B
& ( )(cos ).
1 1
α
∴ = × − − =
F 300
48 2
38 2
38 2 10 18
.
.
( . ( )) .
250 N
4.126 a) − = −
F m V V
x x
&( ).
2 1 200 1 94
1 25
12
2
1
2
=





 ×
.
.
.
π V ∴ =
V1 55 fps.
b) − = − −
F m V V
r B
& ( )(cos ).
1 1
α 200 1 94
125
12
30
2
1
2
=





 −
.
.
( ) .
π V ∴ =
V1 85 fps.
c) − = − −
F m V V
r B
& ( )(cos ).
1 1
α 200 1 94
125
12
30
2
1
2
=





 +
.
.
( ) .
π V ∴ =
V1 25 fps.
4.127 a) − = −
F m V V
x x
&( ).
2 1 − = × × −
700 1000 04 30
2
1 1 1
π . ( cos ).
V V V
o
∴ =
V1 32 24
. m / s.
∴ = = × × =
& . . . .
m A V
ρ π
1 1
2
1000 04 32 24 162 1 kg / s
b) − = − −
F m V V
r B
& ( )(cos ).
1 1
α − = × − −
700 1000 04 8 866 1
2
1
2
π . ( ) (. ).
V ∴ =
V1 40 24
. m / s.
∴ = = × × =
& . . .
m A V
ρ π
1 1
2
1000 04 40 24 202 kg / s
c) − = − −
F m V V
r B
& ( )(cos ).
1 1
α − = × + −
700 1000 04 8 866 1
2
1
2
π . ( ) (. ).
V ∴ =
V1 24 24
. m / s.
∴ = = × × =
& . . . .
m A V
ρ π
1 1
2
1000 04 24 24 1218 kg / s
4.128 (D) 2 1
( ) 1000 0.01 0.2 50(50cos60 50) 2500 N.
x x x
F m V V
− = − = × × × − = −
o
&
F
V1
V2

79. 79
4.129 a) − = − = ×





 × −
R m V V
x x x
&( ) . ( cos ).
2 1
2
1 94
1
12
120 120 60 120
π o
∴ =
Rx 305 lb.
R m V V
y y y
= − =





 × × ×
& ( ) . ( . ).
2 1
2
1 94
1
12
120 120 866
π ∴ =
Ry 528 lb.
b) − = − − = ×





 × × −
R m V V
x r B
& ( )(cos ) . (.5 ).
1
2
1 194
1
12
60 60 1
α π ∴ =
Rx 76 2
. .
lb
R m V V
y r B
= − =





 × × ×
& ( )sin . ( . ).
1
2
194
1
12
60 60 866
α π ∴ =
Ry 132 lb.
c) − = − − = ×





 × × −
R m V V
x r B
& ( )(cos ) . (.5 ).
1
2
1 1 94
1
12
180 180 1
α π ∴ =
Rx 686 lb.
R m V V
y r B
= − =





 × × ×
& ( )sin . ( . ).
1
2
194
1
12
180 180 866
α π ∴ =
Ry 1188 lb.
4.130 V R
B = = × =
ω 0 30 15
.5 m / s.
− = − − = × × × −
R m V V
x B
&( )(cos ) . (.5 ).
1
2
1 1000 025 40 25 1
α π ∴ =
Rx 982 N.
∴ = = × × =
& .
W R V
x B
10 10 982 15 147 300 W
4.131 a) − = − = × − −
R m V V
x x x
&( ) . ( cos ).
2 1
2
4 02 400 400 60 400
π o
∴ =
Rx 1206 N.
R m V V
y y y
= − = × ×
& ( ) . ( sin ).
2 1
2
4 02 400 400 60
π o
∴ =
Ry 696 N.
b) − = − − = × − −
R m V V
x r B
& ( )(cos ) . ( . ).
1
2 2
120 1 4 02 300 5 1
o
π ∴ =
Rx 679 N.
R m V V
y r B
= − = × × ×
& ( )sin . . .
1
2 2
4 02 300 866
α π ∴ =
Ry 392 N.
c) − = − − = × − −
R m V V
x r B
& ( )(cos ) . ( . ).
1
2 2
120 1 4 02 500 5 1
o
π ∴ =
Rx 1885 N.
R m V V
y r B
= − = × × ×
& ( )sin . . .
1
2 2
4 02 500 866
α π ∴ =
Ry 1088 N.
4.132 − = − − = × × − − −
F m V V
x B
& ( )(cos ) . ( ) ( .5 ).
1
2 2
120 1 4 02 400 180 1
o
π ∴ =
Rx 365 N.
VB = × =
12 150 180
. m / s. & .
W = × × =
15 365 180 986 000 W
The y-component force does no work.
4.133 (A) 2
2 1
( ) 1000 0.02 60 (40cos45 40) 884 N.
x r r
x x
F m V V π
− = − = × × × × − =
o
&
Power 884 20 17700 W.
x B
F V
= × = × =
4.134 a) Refer to Fig. 4.16:
1 1 1
2
1 1
750sin sin45 507 fps.
750cos 300 cos45
r r
r
r
V V
V
V
β
β

= =

∴

=
− = 

o
o
Note: V V V V V V V
x x r B r B r
2 1 2 2 1 1 1 2 1
− = − + − − = − +
cos cos (cos cos ).
α α α α
∴ = + =





 × × + =
R mV
x r
& (cos cos ) .
.5
(cos cos ) .
1 2 1
2
015
12
750 507 30 45 48 9
α α π o o
lb.

80. 80
∴ = = × × =
& . ,
W R V
x B
15 15 48 9 300 220 000
ft - lb
sec
or 400 Hp.
b)
750 60
750 300 60
554
1 1
1 1
1 2
sin sin
cos cos
.
β
β
=
− =



=
V
V
V V
r
r
r r
o
o
fps =
∴ = + = ×





 × × + =
R mV
x r
& (cos cos ) .
.5
(cos cos ) .
1 2 1
2
015
12
750 554 30 60 46 4
α α π o o
lb.
∴ = = × × =
& . ,
W R V
x B
15 15 46 4 300 209 000
ft - lb
sec
or 380 Hp.
c)
750 90
750 300 90
687
1 1
1 1
1 2
sin sin
cos cos
.
β
β
=
− =



=
V
V
V V
r
r
r r
o
o
fps =
∴ = + = ×





 × × + =
R mV
x r
& (cos cos ) .
.5
(cos ) .5
1 2 1
2
015
12
750 687 30 0 36
α α π o
lb.
∴ = = × × =
& . ,
W R V
x B
15 15 36 5 300 164 300
ft - lb
sec
or 299 Hp.
4.135 a) Refer to Fig. 4.16:
100 30
100 30 20
36 9 83 3
1 1
1 1
1 1
sin sin
cos cos
. , .
o
o
o
=
− =



∴ = =
V
V
V
r
r
r
α
α
α m / s.
2 2
2 2
2 2
sin60 83.3sin
71.5, 48 .
cos60 83.3cos 20
V
V
V
α
α
α

= 
= =

= − 

o
o
o
− = − = × × − − ∴ =
R m V V R
x x x x
&( ) . ( .5cos cos ).
2 1
2
1000 015 100 71 60 100 30 8650
π o o
N.
∴ = = × × = ×
& . .
W V R
B x
12 12 20 8650 2 08 106
W
b)
100 30
100 30 40
47 68 35
1 1
1 1
1 1 2
sin sin
cos cos
, .
o
o
o
=
− =



∴ = = =
V
V
V V
r
r
r r
α
α
α m / s.
2 2
2 2
2 2
sin60 68.35sin
38.9 m/s, 29.5 .
cos60 68.35cos 40
V
V
V
α
α
α

= 
= =

= − 

o
o
o
− = − = × × − − ∴ =
R m V V R
x x x x
&( ) . ( . cos cos ).
2 1
2
1000 015 100 38 9 60 100 30 7500
π o o
N.
∴ = = × × = ×
& . .
W V R
B x
12 12 40 7500 3 60 106
W
c)
100 30
100 30 50
53 8 6196
1 1
1 1
1 1 2
sin sin
cos cos
. , .
o
o
o
=
− =



∴ = = =
V
V
V V
r
r
r r
α
α
α m / s.
2 2
2 2
2 2
sin60 61.76sin
19.32 m/s, 15.66 .
cos60 61.96cos 50
V
V
V
α
α
α

= 
= =

= − 

o
o
o

81. 81
− = − = × × − − ∴ =
R m V V R
x x x x
&( ) . ( . cos cos ).
2 1
2
1000 015 100 19 32 60 100 30 6800
π o o
N .
∴ = = × × = ×
& . .
W R V
x B
12 12 6800 50 4 08 106
W
4.136 a) Refer to Fig. 4.16:
50 30
50 30
2500 86 6
1 1
1 1
1
2 2
sin sin
cos cos
.
o
o
=
− =



∴ = − +
V
V V
V V V
r
B r
r B B
α
α
2 2 2 2 2
2 1
2 2
30sin60 sin
900 30 .
30cos60 cos
r
r r B B
r B
V
V V V V
V V
α
α

= 
∴ = = + +

− = 

o
o
Combine the above: VB = 13 72
. m / s. Then, α α
1 2
59 4 42 1
= =
. , . .
o o
− = − = × × − − ∴ =
R m V V R
x x x x
&( ) . ( cos cos ).
2 1
2
1000 01 50 30 60 50 30 916
π o o
N.
∴ = = × × =
& . .
W V R
B x
15 15 13 72 916 188 500 W
b)
50 30
50 30
2500 86 6
1 1
1 1
1
2 2
sin sin
cos cos
.
o
o
=
− =



∴ = − +
V
V V
V V V
r
B r
r B B
α
α
∴ =
VB 14 94
. m / s.
2 2 2 2
2
2 2
30sin70 sin
900 20.52 .
30cos70 cos
r
r B B
r B
V
V V V
V V
α
α

= 
∴ = + +

− = 

o
o
α 1 41 4
= . o
,α 2 48 2
= . o
2
2 1
( ) 1000 .01 50( 30cos70 50cos30 ). 841 N.
x x x x
R m V V R
π
− = − = × × − − ∴ =
o o
&
∴ = = × × =
& . .
W V R
B x
15 15 14 94 841 188 500 W
c)
50 30
50 30
2500 86 6
1 1
1 1
1
2 2
sin sin
cos cos
.
o
o
=
− =



∴ = − +
V
V V
V V V
r
B r
r B B
α
α
∴ =
VB 16 49
. m / s
2 2 2 2
2
2 2
30sin80 sin
900 10.42 .
30cos80 cos
r
r B B
r B
V
V V V
V V
α
α

= 
∴ = + +

− = 

o
o
α 1 43
= o
, α 2 53 7
= . o
− = − = × × − − ∴ =
R m V V R
x x x x
&( ) . ( cos cos ).
2 1
2
1000 01 50 30 80 50 30 762
π o o
N.
∴ = = × × =
& . .
W V R
B x
15 15 16 49 762 188 500 W
4.137 To find F, sum forces normal to the plate: ( )
n out n
F m V
Σ = & 1 .
n
V
 
−
 
a) [ ]
∴ = × × × − − =
F 1000 02 4 40 40 60 11
. . ( sin ) .
o
080 N (We have neglected friction)
2 2 3 3 1
0 ( ) 40sin30 .
t
F m V m V m
Σ = = + − − × o
& & & Bernoulli: V V V
1 2 3
= = .
∴ = − −
= +



∴ = = × =
=
Continuity:
kg / s.
kg / s.
0 75 75 320 240
80
2 3 1
1 2 3
2 1
3
& & .5 &
& & &
& . & .
&
m m m
m m m
m m
m
b)
1 20
1.94 120( 120sin60 ) 3360 lb.
12 12
F
∴ = − × × × − =
o
(We have neglected friction)
ΣF m V m V m
t = = + − − ×
0 120 30
2 2 3 3 1
& & ( ) & sin .
o
Bernoulli: V V V
1 2 3
= = .

82. 82
2 1
2 3 1
1 2 3
3
20
.75 .75 1.94 120
0 0.5
144
Continuity:
22.6 slug/sec. and 9.7 slug/sec.
m m
m m m
m m m
m
∴ = = × × ×
∴ = − − 

= +  = =
& &
& & &
& & &
&
4.138 F m V
r r n
= = × × × + =
& ( ) . . ( ) sin
1
2
1000 02 4 40 20 60 24
o
940 N.
F W
x = = ∴ = × =
24 30 21 21 20 432
940 600 N. 600 000 W
cos & .
o
4.139 F m V V F V
r r n B x B
= = × × − = −
& ( ) . . ( ) sin . ( )sin .
1
2 2 2
1000 02 4 40 60 8 40 60
o o
& ( ) . ( ).
W V F V V V V V
B x B B B B B
= = − × = − +
8 40 75 6 1600 80
2 2 3
dW
dV
V V V
B
B B B
&
( ) . .
= − + = ∴ =
6 1600 160 3 0 13 33
2
m / s.
4.140 (A) Let the vehicle move to the right. The scoop then diverts the water to the
right. Then
2 1
( ) 1000 0.05 2 60 [60 ( 60)] 720000 N.
x x
F m V V
= − = × × × × − − =
&
4.141 1
(
r
F m V
= & 2
)(cos 1) 1000 .1 .6 ( )( 2) 120 .
B B B B
V V V V
α
− − = × × − − =
At
2
120 1000
0 : 120 133 300 N.
3600
t F
×
 
= = × =
 
 
2
133 300
1.33 m/s
100 000
o
a = =
−
= =
−
∴ − = ∫
∫
F
m
dV
dt
V dV
V
dt
B B B
B
t
120
100
0012
2
2
0
3333
16 67
000
. . .
.
.
∴ −





 = ∴ =
1
16 67
1
33 33
0012 26 6
. .
. . .
t. sec
t
4.142 1
( )(cos 1) 90 .8 2.5 13.89 ( 13.89)( 1) 34700 N.
r B
F m V V α
= − − = × × × × − − =
&
50 1000
13.89 m/s 34700 13.89 482 000 W or 647 Hp.
3600
B
V W
×
 
= = ∴ = × =
 
 
&
4.143 See the figure in Problem 4.141.
F m V V V V
r B B B
= − − = × × × − − =
& ( )(cos ) . . ( )( )
1 1 1000 06 2 2 24
α V .
B
2
− = ∴− =
F mV
dV
dx
V V
dV
dx
B
B
B B
B
. .
24 5000
2
− = − = − ∴ =
∫
∫
24
5000
24
5000
27 78 250 458
250
27 78
0
m
dx dV
V
x n n x
B
B
x
. . . .
.
l l
V2
V1
= 0
F

83. 83
4.144 − = − − = ×





 − −
F m V V V V
r B B
& ( )(cos ) .
.
) ( ).
1 1
2
1 194
125
12
2
α π
2
(
∴ = − =
F V V
dV
dt
B
B
0 1323 20
1
2
. ( ) .
At t V
dV
dt
V
B
B
= = =
0 0 0 1323 1
2
, . . .
Then 20
With
dV
dt
V
B
= =
6 30 1
1
, . .
fps
For t
dV
V
dt
V
V
B
B B
B
VB
>
−
= =
−
− ∴ =
∫
∫
0
30 1
0 006615 0 01323
1
30 1
1
30 1
8 57
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.
& . . . ( .
W FV F F V
=
×
∴ = =
1 1
50 1000
3600
1440 13 89
20 000 = N m / s)
F m V V
V
V V
= − = ×
+
− ∴ =
& ( ). .
.
( . ). .
2 1
2 2
2 2
1440 123 1
13 89
2
13 89 30 6 m / s.
π
∴ = = ×
+
=
Q A V
3 3
2
1
30 6 13 89
2
69 9
π
. .
. m / s.
3
η p
V
V
= = =
1
3
13 89
22 24
0 625
.
.
. .
or 62.5%
4.146 Fix the reference frame to the aircraft so that V1
200 1000
3600
55
=
×
= .56 m / s.
V m
2
2
320 1000
3600
88 89 1 2 11
55 88 89
2
329
=
×
= ∴ = × ×
+
=
. & . .
.56 .
.5
m / s. kg / s.
π
F = − =
329 88 89 55 10
.5( . .56) 980 N.
= × ∴ =
∆ ∆
p p
π 11 2890
2
. . .
Pa
& .
W F V
= × = ×
1 10 980 55.56 = 610 000 W or 818 Hp
4.147 Fix the reference frame to the boat so that V1 20
88
60
29 33
= × = . fps.
V F m V V
2 2 1
2
40
88
60
58 67 1 94
10
12
29 33 58 67
2
58 67 29 33
= × = ∴ = − = ×






+
−
. &( ) .
. .
( . . )
fps. π
= 5460 lb.
& . , .
W F V
= × = × =
1 5460 29 33 160 000
ft - lb
sec
or 291 Hp
V2
F
VB

84. 84
& .
. .
. .
m = × ×






+
=
194
10
12
29 33 58 67
2
186 2
2
π slug/ sec
4.148 Fix the reference frame to the boat: V V
1 2
10 20
= =
m / s, m / s.
∴Thrust = &( ) . ( ) .
m V V
2 1 1000 0 2 20 10 2000
− = × − = N
& .
W F V
= × = × =
1 2000 10 20 000 W or 26.8 Hp
4.149 0 2 2 10 1 2 20 0 1
1 1 1 1 1 1
. . . . ( ) ( . ).
= = × × ∴ = ∴ = ∴ = −
V A V V V V y y
m / s. m / s.
max
flux in = 2 2 1000 20 1 800
1
3
267
2 2 2
3
0
1
0
1
ρV dy y dy
= × − = =
∫
∫ (. )
.
.
.
000 N.
The slope at section 1 is −20. ∴ = − +
V y y A
2 20
( ) .
Continuity: A V A V V V
1 1 2 2 2 1
2 2
= ∴ = =
. m / s.
V A
V A
V A
2
2
2
0
05 1
1 2
( )
(. )
/ .
=
= −



∴ = −
2 1 2 2 2 20
2
= − ∴ = ∴ = −
A A V y y
/ . .5. ( ) .5 .
flux out = 2 1000 2 20 800
125
3
800
0 00153
2
3
0
05
0
05
( .5 )
( . )
[ . ]
.
.
− =
−





 =
∫ y dy
y
000
000
3
= 408 3
. N. ∴change = 408 − 267 = 141 N.
4.150 a) β = =
−
× ×
= =
∫ ∫
V dA
V A
y dy
2
2
2 2
0
1
2
3
2 20 1
1 2 1 0
4000
1
3
4
3
(. )
. .
.
.
.
b) See Problem 4.149: V y y y V
2 2
20 0125 05 0 2
( ) ( . ), . .
= − ≥ ≥ = m / s.
β = =
−
× ×
=
−
=
∫ ∫
V dA
V A
y dy
y
2
2
2 2
0
05
2
3
0
05
2 20 125
2 1 10
2000
125
3
1021
( . )
. .
( . )
. .
.
.
4.151 From the c.v. shown: ( ) .
p p r r L
w o
1 2 0
2
2
− =
π τ π
∴ = = ∴ =
× ×
× × × −
τ µ
w
o
w w
p r
L
du
dr
du
dr
∆
2
0 03 144 75 12
2 30 2 36 10 5
.
. . /
.
= 191
ft / sec
ft
.
4.152 Write the equation of the parabola: V r V
r
r
( ) .
max
= −






1
2
0
2
p1
A1
p2
A2
τw2πro
L

85. 85
Continuity: π π
× × = −





 ∴ =
∫
.
.
.
max max
.
006 8 1
006
2 16
2
2
2
0
006
V
r
rdr V m / s.
Momentum: p A p A F V dA mV
1 1 2 2
2
1
− − = −
∫
Drag ρ & .
40 006 1000 16 1
006
2 1000 006 8 8
2 2
2
2
2
2
0
006
000 Drag
π π π
× − = × −





 − × × × ×
∫
.
.
.
.
F
r
rdr
4 9 651 7 238
.524 . . .
− = −
FDrag ∴ =
FDrag N
2 11
. .
4.153 & ( ) . ( ) .
m A V V y dA y dy
top = − = × × − +





 =
∫
∫
ρ ρ
1 1 2
2
0
2
123 2 10 32 28 10 65 6 kg / s.
− = + − = + + × − × ×
∫
∫
F
V dA m V m V y dy
top
2
1 23 28 10 65 6 32 1 23 20 32
2
1 1 1
2 2 2
0
2
ρ & & . ( ) . . .
∴ =
F 3780 N.
4.154 a) & & & ( ) . . ( )
.
m m m A V u y dA y y dy
top = − = − = × × − − ×






∫
∫
1 2 1 1
2
0
1
123 1 2 8 20 100 8 2
ρ ρ
= = =
0 656 0 1 8
. . ( ) ).
kg / s. (Note: for
y u y
Momentum: − = − + × − × × ×
∫
F y y dy
Drag ρ ρ
64 20 100 2 656 8 1 2 8
2 2 2
0
1
( ) . .
.
= × + − × ∴ =
1 23 6 83 5 25 123 12 8 2 1
. . . . . . . N
Drag
F
b) To find h: 8 8 20 100 2
0
1
h y y dy
= −
∫ ( ) .
.
∴ =
×
−
×
=
h
20 1
2
100 001
3
0 0667
2
. .
. m.
Momentum: − = − − × × ×
∫
F y y dy
Drag 1 23 64 20 100 2 123 0667 2 8
2 2 2
0
1
. ( ) . . .
.
= × −
1 23 6 83 10
. . .50. ∴ =
FDrag N
2 1
. .
4.155 a) Energy:
V
g
z
V
g
z hL
1
2
1
2
2
2
2 2
+ = + + . See Problem 4.118(a).
8
2 9 81
0 6
1912
2 9 81
2
2 2
×
+ =
×
+ +
.
.
.
.
.51 .
hL ∴ =
hL 1166
. m.
∴ = × × × × =
losses = 900 W / m of width.
γA V hL
1 1 9810 6 1 8 1166 54
(. ) .
b) See Problem 4.120:
V
g
z
V
g
z hL
1
2
1
2
2
2
2 2
+ = + + .

86. 86
7 19
2 9 81
417
1
2 9 81
1 9
2 2
.
.
.
.58
.
. .
×
+ =
×
+ + hL ∴ =
hL 1025
. m.
∴ = × × × × =
losses = 300 W
γA V hL
1 1 9810 417 3 7 19 1025 90
. . .
c) See Problem 4.121:
5 17
2 9 81
116
3
2 9 81
2
2 2
.
.
.
.
.
×
+ =
×
+ + hL ∴ =
hL 0 0636
. m.
∴ = × × × =
losses = / m of width.
γA V hL
1 1 9810 116 5 17 0 0636 3740 W
. . .
4.156 See Problem 4.122: V V p p
1 2 1 2
20 5 60 135
= = = =
m / s, m / s, kPa, kPa.
Then,
V
g
p V
g
p
h h
L L
1
2
1 2
2
2
2 2
2 2
20
2 9 81
60 5
2 9 81
135
+ = + +
×
+ =
×
+ +
γ γ
.
. .
.
000
9810
000
9810
∴ = =
×
∴ =
h K
g
K K
L 11 47
2
20
2 9 81
0 562
2
.
.
. . .
m =
V1
2
4.157 Continuity: V D Vd V
d
D
V
1
2 2
1
2
2
= ∴ =
. .
Energy:
V
g
H t
V
g
V gH t
1
2 2
2 2
2
+ = ∴ =
( ) . ( ).
Momentum: ΣF F
d
dt
V d V m V V
d s
dt
a
x I x x
c v
x x
x
x
− = − + −





 =
∫
( ) & ( ). .
. .
ρ 2 1
2
2
v
∴− = = − ∫
a m t
d
V V m t m
d
V t dt
x o
t
( ) ( ). ( ) ( ) .
ρ
π
ρ
π
2 2
0
4 4
But, V
dH
dt
dH
dt
d
D
gH
dH
H
d
D
gdt H
gd
D
t Ho
1
2
2 1 2
2
2
1 2
2
2
2 2
2
2
= − ∴− = ∴− = ∴ = +
. . . .
/
/
∴ = +







 +







 −








∫
a
d
g
gd
D
t H
d
g
gd
D
t H dt m
x o
t
o o
ρπ
ρ
π
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
= =
× × ×
=
&
.
. .
ρ π
4
1000 4 004
19 89
2
m / s. Velocity in arm =
v v v
M r V d V ri k Vi Adr
I
c v
= × × − = × − ×
∫ ∫
( ) $ ( $ $)
. .
.
2 4 2
0
3
Ω Ω
ρ ρ

87. 87
= − = −
∫
8 0 36
0
3
ρ ρ
AV k rdr AV k
Ω Ω
$ . $.
.
Σ
v v v v v v
M
d
dt
r V d V r V V n dA i V j V k V A
c v
e e e e
c s
= × − = × ⋅ = × +
∫ ∫
0 0 3 707 707
and ( ) . ( $) . $ (. $ . $) .
. . . .
ρ ρ ρ
The z-component of
v v v
r V V n dA V A
e e
c s
× ⋅ = ×
∫ ( $) . . .
. .
ρ ρ
3 707 2
Finally, − = = × × =
( ) . . . ,
M AV V A AV A V
I z e e e e
0 36 4 3 707 2
ρ ρ
Ω . Using
0 36 4 3 707 19 89
. . . . .
Ω = × × × ∴ =
Ω 46 9
. .
rad / s
4.160 A moment
v
M resists the motion thereby
producing power. One of the arms is shown.
v
M ri k Vi Adr AV k rdr AV k
I = × − × = − = −
∫
∫4 2 8 2 778
0
10 12
0
25
$ ( $ $) $ . $.
/
.
Ω Ω Ω
ρ ρ ρ
Σ
v v v v v v
M Mk
d
dt
r V d V r V V n dA V A k
c v
e e
c s
= × − = × ⋅ = × ×
∫ ∫
$, ( ) , ( $) $.
. . . .
and
0
10
12
4
2
ρ ρ
Thus, M + ×





 × × = × ×





 ×
2 778 194
75
12
200
9
30 200
10
12
194
1 4
12
4
2
2
2
. .
.
.
/
.
π π
∴ =
M 309 ft - lb.
& .
W M
= = × =
Ω 309 30 9270 ft - lb /sec
4.161 & . . .
m AV V V
= = = × ∴ =
10 1000 01 31 8
2
0 0
ρ π m / s.
Continuity: V V V r
e
0
2 2
01 01 006 05
π π
× = × + × −
. . . ( . ).
0
2
.01 .006 .15. 11.1 m/s.
e e
V V V
π × = × × ∴ =
∴ = − − = −
V V r V r
e
0 191 05 42 4 212
. ( . ) . .
v
M ri k V i Adr ri k r i Adr
I = × + × + × + × −
∫
∫2 2 2 2 42 4 212
0
05
2
0
05
$ ( $ $) $ [ $ ( . )$]
.
.
.
Ω Ω
ρ ρ
= + −
∫
∫
4 4 42 4 212
0
2
05
2
0
05
Ω Ω
V Ak rdr Ak r r dr
ρ ρ
$ $ ( . )
.
.
.
= × × × × + × ×
4 31 8 1000 01
05
2
4 1000 01
2
2
2
Ω Ω
. .
. $ .
π π
k
42 4
2
2 05
212
3
2 05
2 2 3 3
.
(. . ) (. . ) $
− − −





k
x
y r
Ve
V
Ω

88. 88
= + =
( . . )$ . $.
0 05 0 3 0 35
Ω Ω Ω
k k
ri V j V dr rdr k k
e e
$ ( $) . . . $ . $.
.
.
.
.
× − × = − × × = −
∫
∫ ρ 006 111 1000 006 13 86
2
05
2
05
2
∴ − = −
0 35 13 86
. . .
Ω ∴ =
Ω 39 6
. .
rad /s
4.162 1000
1000
500
2
= ∴ = = ⋅
M M
Ω. N m.
v
M ri k V r i r dr
I r r
= × − × ×
∫ $ ( $ ( )$ ) .
2 2 02
Ω ρ π
= ∫
0 08 2
0
. ( ) $.
πΩ r V r drk
R
Continuity: V r r V R V r RV r
r r
( ) . cos . . ( ) . / .
2 02 30 2 02 0 866
π π
× = × ∴ =
o
v v v o o
r V V n dA R R V V R k V V k
r r r r
c s
× ⋅ = − + × = − +
∫ ( $) ( sin ) cos . $ . ( . )$.
. .
ρ ρ π
Ω 30 30 2 02 00301 35 5
∴− − = − + ∴ − − =
∫
2 16 32 00301 35 52 1 1333 0
0
15
2
. . ( .5 ). . .
.
V r dr V V V V
r r r r r
∴ = ± + × =
Vr
1
2
52 1 52 1 4 1333 70 9
2
( . . ) . m / s.
The flow rate is Q A V
e r
= = × × × × =
cos . . . . . .
30 2 15 02 70 9 866 116
o
π m / s
3
4.163 See Problem 4.159. V V
e = = × =
19 89
008
02
19 89 318
2
2
.
.
.
. .
m / s. m / s.
v
M ri k Vi
d
dt
k ri Adr A A
I e
= × − × + −





 ×





 = × = ×
∫
4 2 01 004
0
3
2 2
$ ( $ $) $ $ . . , . .
.
Ω
Ω
ρ π π
= − − = − −
∫ ∫
8 4 360 36
0
3
2
0
3
ρ ρ
AV k rdr A
d
dt
k r dr AV k A
d
dt
k
Ω
Ω
Ω
Ω
$ $ $ $.
. .
( ) ( $) $.
. .
v v v
r V V n dA V A k
z e e
c s
× ⋅ =
∫ ρ 212 2
Thus, 360 36 212 31 8 373
2
AV A
d
dt
V A
d
dt
e e
Ω
Ω Ω
Ω
+ = + =
or . .
The solution is Ω = +
−
Ce t
31 8
11 73
.
. .
The initial condition is Ω( ) .
0 0
= ∴ = −
C 1173
. .
Finally, Ω = − −
11 73 1 31 8
. ( ) .
.
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?

89. 89
CHAPTER 5
The Differential Forms
of the
Fundamental Laws
5.1 0 = − + ⋅
∫ ∫
∂ρ
∂
ρ
t
d V V ndA
c v c s
. . . .
$ .
v
Using Gauss’ theorem:
0 = − + ∇ ⋅ − = + ∇ ⋅





 −
∫ ∫
∫
∂ρ
∂
ρ
∂ρ
∂
ρ
t
d V V d V
t
V d V
c v c v
c v
. . . .
. .
( ) ( ) .
v v v v
Since this is true for all arbitrary control volumes (i.e., for all limits of
integration), the integrand must be zero:
∂ρ
∂
ρ
t
V
+ ∇ ⋅ =
v v
( ) .
0
This can be written in rectangular coordinates as
− = + +
∂ρ
∂
∂
∂
ρ
∂
∂
ρ
∂
∂
ρ
t x
u
y
v
z
w
( ) ( ) ( ).
This is Eq. 5.2.2. The other forms of the continuity equation follow.
5.2 & & .
m m
m
t
in out
element
− =
∂
∂
( )
ρ θ ρ
∂
∂
ρ θ
v rd dz v
r
v dr r dr d dz
r r r
( ) ( )
− +





 +
+ − +






ρ ρ
∂
∂θ
ρ θ
θ θ θ
v drdz v v d drdz
( )
+ +





 − +





 +





 = +












ρ θ ρ
∂
∂
ρ θ
∂
∂
ρ θ
v r
dr
d dr v
z
v dz r
dr
d dr
t
r
dr
d drdz
z z z
2 2 2
( ) .
Subtract terms and divide by rd drdz
θ :
− −
+
− −
+
=
+
ρ ∂
∂
ρ
∂
∂θ
ρ
∂
∂
ρ
∂
∂
ρ
θ
v
r r
v
r dr
r
v
r z
v
r dr
r t
r dr
r
r
r z
( ) ( ) ( )
/ /
.
1 2 2
Since dr is an infinitesimal, ( )/ ( / )/ .
r dr r r dr r
+ = + =
1 2 1
and Hence,
∂ρ
∂
∂
∂
ρ
∂
∂θ
ρ
∂
∂
ρ ρ
θ
t r
v
r
v
z
v
r
v
r z r
+ + + + =
( ) ( ) ( ) .
1 1
0 This can be put in various forms.

90. 90
5.3 & & .
m m
m
t
in out
element
− =
∂
∂
ρ θ θ φ ρ
∂
∂
ρ θ θ φ
v rd r d v
r
v dr r dr d r dr d
r r r
( ) sin ( ) ( ) ( )sin
− +





 + +
+ +





 − +





 +






ρ θ φ ρ
∂
∂θ
ρ θ θ φ
θ θ θ
v dr r
dr
d v v d dr r
dr
d
2 2
sin ( ) sin
+ +





 − +





 +






ρ θ ρ
∂
∂φ
ρ φ θ
φ φ φ
v dr r
dr
d v v d dr r
dr
d
2 2
( )
= +












∂
∂
ρ θ θ φ
t
r
dr
drd d
2
2
sin
Because some areas are not rectangular, we used an average length ( / ).
r dr
+ 2
Now, subtract some terms and divide by rd d dr
θ φ :
− − −
+
−
+
ρ θ ρ θ
∂
∂
ρ θ
∂
∂θ
ρ θ
θ
v v
r
v
r dr
r
v
r
dr
r
r r r
sin sin ( )sin
( )
( ) sin
2
2
−
+
=
+






∂
∂φ
ρ
∂ρ
∂
θ
φ
( ) sin
v
r
dr
r t
r
dr
r
2 2
2
Since dr is infinitesimal ( ) / ( / ) / .
r dr r r r dr r
+ = + =
2
2 1
and Divide by r sinθ
and there results
∂ρ
∂
∂
∂
ρ
∂
∂θ
ρ
θ
∂
∂φ
ρ ρ
θ φ
t r
v
r
v
r
v
r
v
r r
+ + + + =
( ) ( )
sin
( )
1 1 2
0
5.4 For a steady flow
∂ρ
∂t
= 0. Then, with v w
= = 0 Eq. 5.2.2 yields
∂
∂
ρ ρ
ρ
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
ρ
= 0. This gives
3 2 3
1 200 1 200
ˆ ˆ ˆ ˆ
cos2 sin2
r r
p p
p i i i i
r r r r r
θ θ
∂ ∂ ρ ρ
θ θ
∂ ∂θ
 
∴∇ = + = − −
 
 
v
or
u
x
w
z
∂ρ
∂
∂ρ
∂
+ = 0.
Also,
v v
∇ ⋅ =
V 0, or
∂
∂
∂
∂
u
x
w
z
+ = 0.

91. 91
5.6 Given:
∂
∂
∂ρ
∂
t z
= ≠
0 0
, . Since water can be considered to be incompressible, we
demand that
D
Dt
ρ
= 0. ∴u
x
w
z
∂ρ
∂
∂ρ
∂
+ = 0, assuming the x-direction to be in the
direction of flow. Also, we demand that
v v
∇ ⋅ =
V 0, or
∂
∂
∂
∂
u
x
w
z
+ = 0.
5.7 We can use the ideal gas law, ρ =
p
RT
. Then, the continuity equation
D
Dt
V
ρ
ρ
= − ∇ ⋅
v v
becomes, assuming RT to be constant,
1
RT
Dp
Dt
p
RT
V
= − ∇ ⋅
v v
or
1
p
Dp
Dt
V
= −∇ ⋅
v v
.
5.8 a) Use cylindrical coordinates with v vz
θ = = 0:
1
0
r r
rvr
∂
∂
( ) =
Integrate:
rv C
r = . ∴ =
v
C
r
r .
b) Use spherical coordinates with v v
θ φ
= = 0:
1
0
2
2
r r
r vr
∂
∂
( ) =
Integrate:
r v C
r
2
= . ∴ =
v
C
r
r 2
.
5.9
D
Dt
V
u
x
v
y
ρ
ρ ρ
∂
∂
∂
∂
= − ∇ ⋅ = − +





 = − × + × = −
⋅
v v
2 3 200 1 400 1 1380
. ( ) .
kg
m s
3
5.10 In a plane flow, u u x y v v x y
= =
( , ) ( , ).
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.

92. 92
5.11 If u C v C
= =
1 2
and , the continuity equation provides, for an incompressible
flow,
∂
∂
∂
∂
∂
∂
∂
∂
u
x
v
y
w
z
w
z
w C
+ + = ∴ = =
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
ρ ∂ρ
∂
∂ρ
∂
∂ρ
∂
∂ρ
∂
= = + + +
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. ∴ = − +
v x y Ay f x
( , ) ( ).
But, v x o f x
( , ) ( ).
= =
0 ∴ = −
v Ay.
5.13
∂
∂
∂
∂
u
x
v
y
+ = 0. ∴ = − = −
+ −
+
= −
−
+
∂
∂
∂
∂
v
y
u
x
x y x x
x y
x y
x y
( )5 ( )
( ) ( )
2 2
2 2 2
2 2
2 2 2
5 2 5 5
∴ =
−
+
+ =
+
+ = ∴ =
+
∫
v x y
y x
x y
dy f x
y
x y
f x f x v
y
x y
( , )
( )
( ) ( ). ( ) . .
5 5 5
0
5
2 2
2 2 2 2 2 2 2
5.14 From Table 5.1:
1 1 1
10
4
2
r r
rv
r
v
r r
r
∂
∂
∂
∂θ
θ
θ
( )
.
sin .
= − = − +






∴ = +





 + = −





 +
∫
rv
r
dr f r
r
f
r 10
4
10
4
2
.
sin ( )
.
sin ( ).
θ θ θ θ
. (. , ) .
.
.
sin ( ) . ( ) .
2 2 10 2
4
2
0 0
v f f
r θ θ θ θ
= × −





 + = ∴ =
∴ = −






v
r
r 10
0 4
2
.
sin .
θ
5.15 From Table 5.1:
1 1 20
1
1
2
r r
rv
r
v
r r
r
∂
∂
∂
∂θ
θ
θ
( ) cos .
= − =
−
+






∴ = − +





 + = − −





 +
∫
rv
r
dr f r
r
f
r 20 1
1
20
1
2
cos ( ) cos ( ).
θ θ θ θ
v f f
r ( , ) ( )cos ( ) . ( ) .
1 20 1 1 0 0
θ θ θ θ
= − − + = ∴ =
∴ = − −






v
r
r 20 1
1
2
cos .
θ

93. 93
5.16 From Table 5.1, spherical coordinates:
1 1
2
2
r r
r v
r
v
r
∂
∂ θ
∂
∂θ
θ
θ
( )
sin
( sin ).
= −
∴ = +






1 1
10
40
2
2
2
3
r r
r v
r r
r
∂
∂ θ
θ θ
( )
sin
sin cos .
∴ = +





 + = −





 +
∫
r v r
r
dr f r
r
f
r
2
3
2
10
40
2 10
80
cos ( ) cos ( )
θ θ θ θ
4 2 10 2
80
2
0 0
2
v f f
r ( , ) cos ( ) . ( ) .
θ θ θ θ
= × −





 + = ∴ =
∴ = −






v
r
r 10
80
3
cos .
θ
5.17 Continuity:
∂
∂
ρ ρ
ρ
x
u
du
dx
u
d
dx
( ) . .
= ∴ + =
0 0
ρ = =
×
×
= =
−
×
=
p
RT
du
dx
18 144
1716 500
0 00302
526 453
2 2 12
219
3
. .
/
slug
ft
fps / ft.
∴ = − = − × = −
d
dx u
du
dx
ρ ρ .
. .
00302
486
219 0 00136 slug/ ft 4
5.18 [ ]
∂
∂
∂
∂
∂
∂
u
x
v
y x
e e
x x
+ = − − = −
− −
0 20 1 20
. ( )
Hence, in the vicinity of the x-axis:
∂
∂
v
y
e v ye C
x x
= = +
− −
20 20
and .
But v y C
= = ∴ =
0 0 0
if . .
v ye e
x
= = =
− −
20 20 0 2 0
2
( . ) .541 m / s
5.19 [ ]
1
0 20 1 20
r r
rv
v
z z
e e
r
z z z
∂
∂
∂
∂
∂
∂
( ) . ( )
+ = − − = −
− −
Hence, in the vicinity of the z-axis:
1
20
2
20
2
r r
rv e rv
r
e C
r
z
r
z
∂
∂
( ) .
= = +
− −
and
But v r C
r = = ∴ =
0 0 0
if . .
v re e
r
z
= = =
− −
10 10 0 2 0 271
2
( . ) . m / s
5.20 The velocity is zero at the stagnation point. Hence,
0 10
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
,

94. 94
we see that
∂
∂
v
y
x
= − −
80 3
near the x-axis. Consequently, for small ∆y,
∆ ∆
v x y
= − −
80 3
so that v = − − =
−
80 3 0 1 0 296
3
( ) ( . ) . .
m / s
5.21 The velocity is zero at the stagnation point. Hence
0
40
10 2
2
= − ∴ =
R
R
. m
( )
1 1
40 10
20
2
2
2
2
r r
r v
r r
r
r
r
∂
∂
∂
∂
= − = −
( ) .
Near the negative x-axis continuity provides us with
( )
1 20
r
v
r
sin
sin .
θ
∂
∂θ
θ
θ =
Integrate, letting θ = 0 from the y-axis:
v C
θ θ θ
sin cos
= − +
20
Since vθ = 0 when θ = =
90 0
o
, .
C Then, with α = =
−
tan
.
. ,
1 0 1
3
1909o
vθ
θ
θ
= − = − = − =
20 20
88 091
88 091
20
0 0333
0 999
0 667
cos
sin
cos .
sin .
.
.
. m / s
5.22 Continuity:
∂
∂
∂
∂
u
x
v
y
v
y
u
x
+ = ∴ = − = −
−
×
= −
0
13 5 11 3
2 005
220
.
. .
.
.
m /s
m
∆
∆
∆
∆
∴ = − = − ∴ = − × = −
∆ ∆
v v y v
0 220 220 004 0 88
. . . .
m / s
b) a u
u
x
x = = × + =
∂
∂
12 6 220 2772
. ( ) .
m /s2
5.23 ΣF ma
y y
= . For the fluid particle occupying the volume of Fig. 5.3:
τ
∂τ
∂
τ
∂τ
∂
τ
∂τ
∂
yy
yy
zy
zy
xy
xy
y
dy
dxdz
z
dz
dxdy
x
dx
dydz
+





 + +





 + +






2 2 2
− −





 − −





 − −






τ
∂τ
∂
τ
∂τ
∂
τ
∂τ
∂
yy
yy
zy
zy
xy
xy
y
dy
dxdz
z
dz
dxdy
x
dx
dydz
2 2 2
+ =
ρ ρ
g dx dy dz dx dy dz
Dv
Dt
y
Dividing by dx dy dz, and adding and subtracting terms:
∂τ
∂
∂τ
∂
∂τ
∂
ρ ρ
xy yy zy
y
x y z
g
Dv
Dt
+ + + = .
5.24 Check continuity:
∂
∂
∂
∂
∂
∂
u
x
v
y
w
z
x y x x
x y
x y y y
x y
+ + =
+ −
+
+
+ −
+
=
( )10 ( )
( )
( )10 ( )
( )
.
2 2
2 2 2
2 2
2 2 2
10 2 10 2
0

95. 95
Thus, it is a possible flow. For a frictionless flow, Euler’s Eqs. 5.3.7 give, with
g g
x y
= = 0:
.
u u p
u v
x y x
∂ ∂ ∂
ρ ρ
∂ ∂ ∂
+ = −
2 2 2 2
2 2 2 2 2 2 2 2 2 2 2 2 3
10 10 10 10 20 100( )
( ) ( ) ( )
p x y x y xy x y y
x x y x y x y x y x y
∂
ρ ρ ρ
∂
− − +
∴ = − − =
+ + + + +
.
v v p
u v
x y y
∂ ∂ ∂
ρ ρ
∂ ∂ ∂
+ = −
2 2 2 2
2 2 2 2 2 2 2 2 2 2 2 2 3
10 20 10 10 10 100( )
( ) ( ) ( )
p x xy y x y x y y
y x y x y x y x y x y
∂
ρ ρ ρ
∂
− − +
∴ = − − =
+ + + + +
2 2 2 2 2 2 2 2 2
100 100 100
ˆ ˆ ˆ ˆ ˆ ˆ
( ).
( ) ( ) ( )
p p x y
p i j i j xi yj
x y x y x y x y
∂ ∂ ρ ρ ρ
∂ ∂
∴∇ = + = + = +
+ + +
v
5.25 Check continuity (cylindrical coord from Table 5.1):
1 1 10
1
1 10
1
1
0
2 2
r r
rv
r
v
r r r r
r
∂
∂
∂
∂θ
θ θ
θ
( ) cos cos .
+ = +





 +
−
+





 = ∴It is a possible
flow. For Euler’s Eqs. (let ν = 0 in the momentum eqns of Table 5.1) in
cylindrical coord:
2
2
2 2
2 2 3
100 1 1 20
1 sin 10 1 cos
r r
r
v v
v v
p
v
r r r r r r r r
θ θ
∂ ∂
∂ ρ
ρ ρ ρ θ ρ θ
∂ ∂ ∂θ
     
= − − = + − −
     
     
− +





 −






10
1
1
10
10
2
2
2
ρ
θ
r r r
sin .
4
1 100 1
1 sin cos
r
r
v v v v v
p
v
r r r r r r
θ θ θ θ
∂ ∂
∂ ρ
ρ ρ ρ θ θ
∂θ ∂ ∂θ
 
= − − − = −
 
 
− −











 − +






10 1
1 20 100
1
1
2 3 2
2
ρ θ θ
ρ
θ θ
r r r r
cos sin sin cos .
3 2 3
1 200 1 200
ˆ ˆ ˆ ˆ
cos2 sin2
r r
p p
p i i i i
r r r r r
θ θ
∂ ∂ ρ ρ
θ θ
∂ ∂θ
 
∴∇ = + = − −
 
 
v
5.26 This is an involved problem. Follow the steps of Problem 5.25. Good luck!
( )
∂
∂
ρ ρ
∂
∂
ρ
∂
∂θ
θ φ θ
p
r
v v
r
v
v
r
v
r
v
r
r r
=
+
− −
2 2
1
r
p v v
r
v
v
r
v
r
v
r
r
∂
∂θ
ρ ρ
∂
∂
ρ
∂
∂θ
θ θ θ θ
= − − −
( )

96. 96
5.27
2 2
. .
3 3
p p V p p V
µ µ
λ λ
   
∴ = − + ∇ ⋅ ∴ − = − + ∇ ⋅
   
   
v v v v
∂
∂
$ $ $ $
.
s
s
s
s
n
R
n
R
≅ = − = −
∆
∆
∆α
∆α
∂
∂
∂θ
∂
$ $ $
$ .
s
t
s
t
n
t
n
t
≅ = =
∆
∆
∆θ
∆
2
ˆ ˆ.
DV V V V
V s V n
Dt t s t R
∂ ∂ ∂θ
∂ ∂ ∂
 
 
∴ = + + −
 
   
   
v
For steady flow, the normal acc. is −






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
Ω.
Thus, Euler’s equation becomes
2 ( ) .
DV d
V r r p g
Dt dt
ρ ρ
 
Ω
+ Ω × + Ω × Ω × + × =−∇ −
 
 
v v
v v v v v
v v v
5.29 2
xx
u
p
x
∂
τ µ
∂
= − + V
λ
+ ∇ ⋅
v v
30 psi.
= −
τ τ
yy zz p
= = − = −30 psi.
xy
u v
y x
∂ ∂
τ µ
∂ ∂
= + 5 5
.1
10 30 1440 18 10 psf.
12
− −
   
= − × = ×
   
 
 
τ τ
τ
τ
xz yz
xx
= = =
×
×
= ×
−
−
0
18 10
30 144
4 17 10
5
8
. . .
xy
5.30
2 2 3
9/5 2 13/5 9/5 2 13/5
16 16 8 16
. ( , ) ( ).
3
v u y y y y
v x y f x
y x C x C x Cx C x
∂ ∂
∂ ∂
= − = − ∴ = − +
v x o f x C C
( , ) . ( ) . . . .
/
= ∴ = = ∴ =
0 0 8 1000 0 0318
4 5
∴ = −
− −
u x y yx y x
( , ) .
/
629 9890
4/ 5 2 8 5
v x y y x y x
( , ) .
/ /
= −
− −
252 5270
2 9 5 3 13 5
τ µ
∂
∂
xx p
u
x
= − + = − + = −
2 100 0 100 kPa.
τ τ
yy zz p
= = − = −100 kPa.
5 4/5 5
2 10 629 1000 5.01 10 Pa.
xy
u v
y x
∂ ∂
τ µ
∂ ∂
− − −
   
= + = × × = ×
   
 
τ τ
xz yz
= = 0.

97. 97
5.31
Du u
Dt t
∂
∂
= ( ) .
u v w u V u
x y z
∂ ∂ ∂
∂ ∂ ∂
 
+ + + = ⋅∇
 
 
v v
Dv v
Dt t
∂
∂
= ( ) .
u v w v V v
x y z
∂ ∂ ∂
∂ ∂ ∂
 
+ + + = ⋅∇
 
 
v v
Dw w
Dt t
∂
∂
= ( )
u v w w V w
x y z
∂ ∂ ∂
∂ ∂ ∂
 
+ + + = ⋅∇
 
 
v v
ˆ ˆ
ˆ ˆ ˆ ˆ
( ) ( ) .
DV Du Dv Dw
i j k V ui vj wk V V
Dt Dt Dt Dt
∴ = + + = ⋅∇ + + = ⋅∇
v
v v v v v
5.32 Follow the steps that lead to Eq. 5.3.17 and add the term due to compressible
effects:
2 ˆ
ˆ ˆ
3 3 3
DV
p g V Vi Vj Vk
Dt x y z
µ ∂ µ ∂ µ ∂
ρ ρ µ
∂ ∂ ∂
=−∇ + + ∇ + ∇ ⋅ + ∇ ⋅ + ∇ ⋅
v
v v v v v v v v
v
= −∇ + + + + +





∇ ⋅
v v v v v
p g V
x
i
y
j
z
k V
ρ µ∇
µ ∂
∂
∂
∂
∂
∂
2
3
$ $ $
∴ = −∇ + + ∇ + ∇ ∇ ⋅
ρ ρ µ
µ
DV
Dt
p g V V
v
v v v v v v
2
3
( ).
5.33 If u=u(y), then continuity demands that
∂
∂
v
y
v C
= ∴ =
0. .
But, at y=0 (the lower plate) v=0. ∴ = =
C v x y
0 0
, ( , ) .
and
2 2 2
2 2 2
.
x
u u u
Du u u u u p
u v w g
Dt t x y z x x y z
∂ ∂ ∂
∂ ∂ ∂ ∂ ∂
ρ ρ ρ µ
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
 
 
∴ = + + + = − + + + +
 
   
   
∴ = − +
0
2
2
∂
∂
µ
∂
p
x
u
ay
.
0 .
Dv p
Dt y
∂
ρ
∂
= = −
ρ
∂
∂
ρ
∂
∂
ρ
Dw
Dt
p
z
g
p
z
g
= = − + − ∴ = − −
0 0
( ). .
5.34 Continuity: ( ) 0. .
r r
rv rv C
r
∂
∂
= ∴ = At 0, . 0.
r
r v C
= ≠ ∞ ∴ =
1
0 .
r
Dv p
Dt r
∂
ρ ∂
= = −
1
0 .
Dv p
Dt r
θ ∂
ρ ∂θ
= = −

98. 98
z z
Dv v
Dt t
∂
ρ ρ
∂
= r
v
+ z
v
v
r
θ
∂
∂
+ z z
z
v v
v
r z
∂ ∂
∂θ ∂
+
2 2
2 2 2
1 1
z z z
v v v
p
z r r
r r
∂ ∂ ∂
∂
µ
∂ ∂
∂ ∂θ
 
  = − + + +
 
 
2
2
z
v
z
∂
∂
+
 
 
 
 
∴ = − + +






0
1
2
2
∂
∂
µ
∂
∂
∂
∂
p
z
v
r r
v
r
z z
.
5.35 Continuity:
1
0 0 0
2
2 2
1
r r
r v r v C r r v C
r r r
∂
∂
( ) . . , . .
= ∴ = = = ∴ =
At
2
2
2
cot .
v v
p
r r r
θ θ
∂
ρ µ θ
∂
 
− = − + −
 
 
2
2 2 2
1 1
0
sin
v v
p
r
r r r
r r
θ θ
∂
∂ ∂
µ
∂θ ∂ ∂ θ
 
 
= − + −
 
 
 
 
0
1
= −
r
p
sin
.
θ
∂
∂φ
5.36 For an incompressible flow
v v
∇⋅ =
V 0. Substitute Eqs. 5.3.10 into Eq. 5.3.2 and
5.3.3:
ρ
∂
∂
µ
∂
∂
∂
∂
µ
∂
∂
∂
∂
∂
∂
µ
∂
∂
∂
∂
ρ
Du
Dt x
p
u
x y
u
y
v
x z
u
z
w
x
gx
= − +





 + +





 + +





 +
2 .
= − + + + + + +





 +
∂
∂
µ
∂
∂
µ
∂
∂
µ
∂
∂
µ
∂
∂
∂
∂
∂
∂
∂
∂
ρ
p
x
u
x
u
y
u
z x
u
x
v
y
w
z
gx
2
2
2
2
2
2
∴ = − + + +





 +
ρ
∂
∂
µ
∂
∂
∂
∂
∂
∂
ρ
Du
Dt
p
x
u
x
u
y
u
z
gx
2
2
2
2
2
2
.
2 .
y
Dv u v v v w
p g
Dt x y x y y z z y
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
ρ µ µ µ ρ
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
     
= + + − + + + +
     
     
= − + + + + + +





 +
∂
∂
µ
∂
∂
µ
∂
∂
µ
∂
∂
µ
∂
∂
∂
∂
∂
∂
∂
∂
ρ
p
y
v
x
v
y
v
z y
u
x
v
y
w
z
gy
2
2
2
2
2
2
∴ = − + + +





 +
ρ
∂
∂
µ
∂
∂
∂
∂
∂
∂
ρ
Dv
Dt
p
y
v
x
v
y
v
z
gy
2
2
2
2
2
2
.
2 z
Dw u w v w w
p g
Dt x z x y z y z z
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
ρ µ µ µ ρ
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
 
   
= + + + + − + +
 
   
   
 

99. 99
= − + + + + + +





 +
∂
∂
µ
∂
∂
µ
∂
∂
µ
∂
∂
µ
∂
∂
∂
∂
∂
∂
∂
∂
ρ
p
z
w
x
w
y
w
z z
u
x
v
y
w
z
gz
2
2
2
2
2
2
∴ = − + + +





 +
ρ
∂
∂
µ
∂
∂
∂
∂
∂
∂
ρ
Dw
Dt
p
z
w
x
w
y
w
z
gz
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
µ µ
= ( , , )
x y z we arrive at
2 2 2
2 2 2
2
x
Du p u u u u u v u w
g
Dt x x x y y x z z x
x y z
∂ ∂ ∂ ∂ ∂µ ∂ ∂µ ∂ ∂ ∂µ ∂ ∂
ρ ρ µ
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂
     
= − + + + + + + + + +
     
   
 
 
5.38 If plane flow is only parallel to the plate, v w
= = 0. Continuity then demands
that ∂ ∂
u x
/ .
= 0 The first equation of (5.3.14) simplifies to
u u
u
t x
∂ ∂
ρ
∂ ∂
+ v
+
u
w
y
∂
∂
+
u p
z x
∂ ∂
∂ ∂
 
= −
 
 
x
g
ρ
+
2
2
u
x
∂
µ
∂
+
2 2
2 2
u u
y z
∂ ∂
∂ ∂
+ +
 
 
 
 
2
2
u u
t y
∂ ∂
ρ µ
∂ ∂
=
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 ∂ ∂
p x
/ .
= 0
5.39 From Eqs. 5.3.10, −
+ +
= − + +





 − ∇ ⋅
τ τ τ µ ∂
∂
∂
∂
∂
∂
λ
xx yy zz
p
u
x
v
y
w
z
V
3
2
3
v v
.
2 2
. .
3 3
p p V p p V
µ µ
λ λ
   
∴ = − + ∇ ⋅ ∴ − = − + ∇ ⋅
   
   
v v v v
5.40 ˆ
ˆ ˆ
( ) ( )
V V u v w ui vj wk
x y z
 
∂ ∂ ∂
⋅∇ = + + + +
 
∂ ∂ ∂
 
v v
ˆ
( )
w w w v v v
V V u v w u v w i
y x y z z x y z
 
   
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
∇ ⋅∇ = + + − + +
 
   
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
   
 
×
v v
ˆ
u u u w w w
u v w u v w j
z x y z x x y z
 
   
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
+ + + − + +
 
   
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
   
 
ˆ
v v v u u u
u v w u v w k
x x y z y x y z
 
   
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
+ + + − + +
 
   
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
   
 
Use the definition of vorticity: ˆ
ˆ ˆ
( ) ( ) ( )
w v u w v u
i j k
y z z x x y
ω
∂ ∂ ∂ ∂ ∂ ∂
= − + − + −
∂ ∂ ∂ ∂ ∂ ∂
v

100. 100
ˆ
ˆ ˆ
( ) ( ) ( ) ( ) ( )
w v u w v x
V ui vj wk
y z x z x y x y z
ω
 
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
⋅∇ = − + − + − + +
 
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
 
v
v
ˆ
ˆ ˆ
( ) ( ) ( ) ( )
w v u w v u
V u v w i j k
x y z y z z x x y
ω
   
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
⋅∇ = + + − + − + −
   
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
   
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 u y
ω ∂ ∂
= − is the
only vorticity component of interest. The third equation of Eq. 5.3.24 then
simplifies to
D
Dt
z
z
ω
ν ω
= ∇2
2
2
z
y
∂ ω
ν
∂
=
since changes normal to the plate are much larger than changes along the plate,
i.e., .
z z
y x
∂ω ∂ω
∂ ∂
>>
5.42 If viscous effects are negligible, as they are in a short section, Eq. 5.3.25 reduces
to
D
Dt
z
ω
= 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,
ω
∂
∂
∂
∂
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 u0 ; hence, the velocity distribution at section 2 is
u y u y
2 0 10
( ) .
= + Continuity then allows us to calculate the profile:
V A V A
1 1 2 2
=
1
2
10 0 04 0 04 10 0 02 2 0 02 0 3
0 0
( . )( . ) ( . / )( . ). .
× = + × ∴ =
w u w u m / s.
Finally,
u y y
2 0 3 10
( ) .
= +

101. 101
5.43 No. The first of Eqs. 5.3.24 shows that, neglecting viscous effects,
x
x y z
D u u u
Dt x y z
ω ∂ ∂ ∂
ω ω ω
∂ ∂ ∂
= + +
so that ω y , which is nonzero near the snow surface, creates ω x through the term
/ ,
y u y
ω ∂ ∂ since there would be a nonzero ∂ ∂
u y
/ near the tree.
5.44 k T ndA
t
V
gz u d V
V
gz u
p
V ndA
c v
c s c s
v v
∇ ⋅ = + +





 − + + + +





 ⋅
∫
∫ ∫
$ ~) ~ $
. .
. . . .
∂
∂
ρ
ρ
ρ
2 2
2 2
v v v v
∇⋅ ∇ − = + +





 − + ∇ ⋅ + + +





 −
∫
∫ ∫
( ) ~ ~
. .
. . . .
k T d V
t
V
gz u d V V
V
gz u
p
d V
c v
c v c v
∂
∂
ρ ρ
ρ
2 2
2 2
∴ − ∇ + + +





 + ∇ ⋅ + + +











 − =
∫ k T
t
V
gz u V
V
gz u
p
d V
c v
2
2 2
2 2
0
∂
∂
ρ ρ ρ ρ
ρ
~ ~ .
. .
v v
2 2 2
2 2 2
V V p V
V gz V
t t
∂ ∂ρ
ρ ρ ρ
∂ ρ ∂
 
+ ∇ ⋅ + + = + ∇ ⋅
 
 
 
v v v v V p
V V V g z
t
∂
ρ
∂ ρ
   
∇
+ ⋅ + ⋅∇ + + ∇
   
 
 
v v
v v v v
0.
=
continuity momentum
∴− ∇ + + ⋅∇ = ∴ = ∇
k T
t
u V u
Du
Dt
k T
2 2
0
∂
∂
ρ ρ ρ
~ ~ .
~
.
v v
5.45 Divide each side by dxdydz and observe that
2 2 2
2 2 2
, ,
y dy y
x dx x z dz z
T T
T T T T
y y
x x z z
T T T
dx dy dz
x x z
+
+ +
∂ ∂
∂ ∂ ∂ ∂
−
− −
∂ ∂
∂ ∂ ∂ ∂
∂ ∂ ∂
= = =
∂ ∂ ∂
Eq. 5.4.5 follows.
5.46
( / )
Du D h p Dh Dp p D Dh Dp p
V
Dt Dt Dt Dt Dt Dt Dt
ρ ρ
ρ ρ ρ ρ ρ
ρ ρ
−
 
= = − + = − + − ∇ ⋅
 
v
%
where we used the continuity equation: / .
D Dt V
ρ ρ
= − ∇ ⋅
v
Then Eq. 5.4. 9
becomes
2
Dh Dp p
V K T p V
Dt Dt
ρ ρ
ρ
 
− + − ∇ ⋅ = ∇ − ∇ ⋅
 
v v
which is simplified to
2
Dh Dp
K T
Dt Dt
ρ = ∇ +

102. 102
5.47 See Eq. 5.4.9: ~ . .
u cT c
T
t
u
T
x
v
T
y
w
T
z
k T
= ∴ + + +





 = ∇
ρ
∂
∂
∂
∂
∂
∂
∂
∂
2
Neglect terms with velocity: ρ
∂
∂
c
T
t
k T
= ∇2
.
5.48 The dissipation function Φ involves viscous effects. For flows with extremely
large velocity gradients, it becomes quite large. Then
ρc
DT
Dt
p = Φ
and
DT
Dt
is large. This leads to very high temperatures on reentry vehicles.
5.49 u r
u
r
r r y
= − ∴ = − ×
10 1 10 000 2 105
( ). . )
( takes the place of
2 ∂
∂
From Eq. 5.4.17, Φ =














= ×
2
1
2
4 10
2
2 10
µ
∂
∂
µ
u
y
r .
At the wall where r = = × × × × = ⋅
−
0 01 18 10 4 01 10 72
5 2 10
. . . .
m, N / m s
2
Φ
At the centerline
∂
∂
u
r
= =
0 0
so Φ .
At a point half-way: Φ = × × × × = ⋅
−
18 10 4 005 10 18
5 2 10
. . .
N / m s
2
5.50 (a) Momentum:
∂
∂
ν
∂
∂
u
t
u
y
=
2
2
Energy: ρ
∂
∂
∂
∂
µ
∂
∂
c
T
t
K
T
y
u
y
= +






2
2
2
.
(b) Momentum: ρ
∂
∂
µ
∂
∂
∂µ
∂
∂
∂
u
t
u
y y
u
y
= +
2
2
Energy: ρ
∂
∂
∂
∂
µ
∂
∂
c
T
t
K
T
y
u
y
= +






2
2
2
.

103. 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
+ +





 = + +






ρ ρ
.
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
V
ρ ρ
.
or
1
2
1
2
1
1
2
1
1
2
2
2
2
2
2
2
2
2
1
2
+ + = + +








p
V
gz
V
p
V
gz
V
V
V
ρ ρ
6.2 a) [ ]
& . .
m
FT
L
= =
⋅
⋅
=
⋅
∴
kg
s
N s
m s
N s
m
2
b) [ ]
p
F
L
= ∴
N
m2
. 2
c) [ ]
ρ = =
⋅
⋅
=
⋅
∴
kg
m
N s
m m
N s
m
3
2
3
2
4
. .
FT
L
2
4
d) [ ]
µ =
⋅
∴
N s
m2
.
FT
L2
e) [ ]
W FL
= ⋅ ∴
N m.
f) [ ]
& .
W
FL
T
=
⋅
∴
N m
s
g) [ ]
σ = ∴
N / m.
F
L
6.3 (A) The dimensions on the variables are as follows:
2 2
2 3 2 2
/
[ ] [ ] , [ ] , [ ] , [ ]
L L ML ML T M L
W F V M d L p V
T T
T T L LT
= × = × = = ∆ = = =
&
First, eliminate T by dividing W
& by ∆p. That leaves T in the denominator so
divide by V leaving L2
in the numerator. Then divide by d2
. That provides
2
W
pVd
π =
∆
&
6.4 1
2 5 2
, , , .
T e r
f
R R R
R R
µ
ρω ρω
 
∴ =  
 
 
l

104. 104
6.5 (A) ( , , , , ). The units on the variables on the rhs are as follows:
V f d l g ω µ
=
1
2
[ ] , [ ] , [ ] , [ ] , [ ]
L ML
d L l L g T
T
T
ω µ
−
= = = = =
Because mass M occurs in only one term, it cannot enter the relationship.
6.6 [ ] [ ] [ ] [ ]
V f V
L
T
L
M
L
M
LT
= = = = =
( , , ). , , , .
l l
ρ µ ρ µ
3
∴There is one π − term: π
ρ
µ
1 =
Vl
.
∴ = = ∴ = =
π π ρ
µ
1 1 2
0
f
V
C
( ) , Re
Const. or Const.
l
6.7 [ ] [ ] [ ] [ ]
V f d V
L
T
M
T
M
L
d L
= = = = =
( , , ). , , , .
σ ρ σ ρ
2 3
∴ = ∴ = = ∴ =
π
σ
ρ
π π
σ
ρ
1 2 1 1 2
0
2
V d
f C
V d
C
. ( ) ,
onst. or We = Const.
6.8 [ ] [ ] [ ] [ ]
V f H g m V
L
T
g
L
T
m M H L
= = = = =
( , , ). , , , .
2
∴ = ∴ = ∴ =
π π
1
0
2 1
gHm
V
C V gH C
. . / .
6.9 [ ] [ ] [ ] [ ] [ ] [ ]
V f H g m V
L
T
H L g
L
T
m M
M
L
M
LT
= = = = = = =
( , , , , ). , , , , , .
ρ µ ρ µ
2 3
Choose repeating variables H g
, ,ρ (select ones with simple dimensions-we couldn’t
select V, H, and g since M is not contained in any of those terms):
π ρ π ρ π µ ρ
1 2 3
1 1 1 2 2 2 3 3 3
= = =
VH g mH g H g
a b c a b c a b c
, , .
∴ = = = = =
π
ρ
π
ρ
π
µ
ρ
µ
ρ
1
0
2 3 3 3 2 3
V
g H
V
gH
m
H gH gH
. . .
/
∴ =








V
gH
f
m
H gH
1 3 3
ρ
µ
ρ
, .
Note: The above dimensionless groups are formed by observation: simply combine the
dimensions so that the π − term is dimensionless. We could have set up equations similar
to those of Eq. 6.2.11 and solved for 1 1 1 2 2 2 3 3 3
, , and , , c and , , .
a b c a b a b c But the
method of observation is usually successful.
6.10 [ ] [ ] [ ] [ ] [ ]
F f d V F
ML
T
d L V
L
T
M
LT
M
L
D D
= = = = = =
( , , , , ). , , , , .
l µ ρ µ ρ
2 3

105. 105
π ρ π ρ π µ ρ
1 2 3
1 1 1 2 2 2 3 3 3
= = =
F V dV V
D
a b c b c a a b c
l l l
, , .
∴ = = =
π
ρ
π π
µ
ρ
1 2 2 2 3
F
V
d
V
D
l l l
, , .
∴ =






F
V
f
d
V
D
ρ
µ
ρ
l l l
2 2 1 , .
We could write
π
π π
π
π
1
2
2 2
2
3
2
1
=






f , or
F
d V
f
d dV
D
ρ
µ
ρ
2 2 2
=






l
, . This is equivalent
to the above. Either functional form must be determined by experimentation.
6.11 [ ] [ ] [ ] [ ] [ ]
F f d V F
ML
T
d L V
L
T
M
LT
M
L
D D
= = = = = =
( , , , , ). , , , , .
l µ ρ µ ρ
2 3
π µ π µ π ρ µ
1 2 3
1 1 1 2 2 2 3 3 3
= = =
F d V d V d V
D
a b c a b c a b c
, , .
l
By observation we have 1 2 3
, , .
D
F Vd
Vd d
ρ
π π π
µ µ
= = =
l
∴ =






F
Vd
f
d
Vd
D
µ
ρ
µ
1
l
, .
Rather than π π π
1 1 2 3
= f ( , ), we could write
π
π
π
π
1
3
2 2
3
1
=






f , , an acceptable form:
F
V d
f
d Vd
D
ρ
µ
ρ
2 2 2
=






l
, .
6.12 [ ] [ ] [ ] [ ] [ ] [ ]
h f d g h L
M
T
d L
M
L T
g
L
T
= = = = = = =
( , , , , ). , , , , , .
σ γ β σ γ β
2 2 2 2
1
Select d g
, ,
γ as repeating variables.
π γ π σ γ π β
1 2
1 1 1 2 2 2
= = =
hd g d g
a b c a b c
, , .
3
∴ = = =
π π
σ
γ
π β
1 2 2 3
h
d d
, , .
∴ =






h
d
f
d
1 2
σ
γ
β
, . Note: gravity does not enter the answer.
6.13 [ ] [ ] [ ] [ ]
F f m R F
ML
T
m M
T
R L
C C
= = = = =
( , , ). , , , .
ω ω
2
1
∴ = = ∴ = ∴ =
π ω
ω ω
ω
1 2 2
2
F m R
F
m R
F
m R
C F Cm R
C
a b c C C
C
. .
6.14 [ ] [ ] [ ] [ ]
σ σ
= = = = =
f M y I
M
LT
M
ML
T
y L I L
( , , ). , , , .
2
2
2
4
∴ =
π σ
1 M y I
a b c
.

106. 106
Given that 1
b = − , 1 Const. C .
I My
yM I
σ
π σ
= = ∴ =
6.15 [ ] [ ] [ ]
V f d
dp
dx
V
L
T
M
LT
d L
dp
dx
M
L T
=





 = = =





 =
( , , . , , , .
µ µ 2 2
∴ =






π µ
1 V d
dp
dx
a b
c
. Let’s start with the ratio
µ
dp dx
/
so that “M” is accounted for.
Then the π1 − term is
µV
dp dx d
/
.
2
Hence,
π
µ
µ
1 2
2
= = ∴ =
V
dp dx d
V
d dp dx
/
/
.
Const. Const
6.16 [ ] [ ] [ ] [ ]
V f H g V
L
T
H L g
L
T
M
L
= = = = =
( , , ). , , , .
ρ ρ
2 3
∴ = = = ∴ =
π ρ
ρ
1
0
VH g V
g H
V gH
a b c
Const. Const. .
Density does not enter the expression.
6.17 [ ] [ ] [ ] [ ] [ ] [ ]
3 2
( , , , , ). , , , , , .
L M M L
V f H g d V H L g d L
T LT L T
µ ρ µ ρ
= = = = = = =
π ρ π µ ρ π ρ
1 2
1 1 1 2 2 2 3 3 3
= = =
VH g H g dH g
a b c a b c a b c
, , .
3
Repeating
variables



H g
, , .
ρ
π π
µ
ρ
π
1 2 3 2 3
= = =
V
gH gH
d
H
, , .
/
∴ = =








π π π
µ
ρ
1 1 2 3 1 3
f
V
gH
f
gH
d
H
( , ), , .
or
6.18 ∆p f V d L
= ( , , , , , ).
ν ε ρ
[ ] [ ] [ ] [ ] [ ] [ ] [ ]
2
2 3
, , , , , , .
M L L M
p V d L L L L
T T
LT L
ν ε ρ
∆ = = = = = = =
Repeating variables: V d
, , .
ρ
π ρ π ν ρ π ρ π ε ρ
1 2 3 4
1 1 1 2 2 2 3 3 3 4 4 4
= = = =
∆pV d V d L V d V d
a b c a b c a b c a b c
, , , .
∴ = = = =
π
ρ
π
ν
π π
ε
1 2 3
∆p
V Vd
L
d d
, , , .
2 4
π π π π
ρ
ν ε
1 1 2 3 4 2 1
= ∴ =






f
p
V
f
Vd
L
d d
( , , ). , , .
∆

107. 107
6.19 F f V c h r w
D = ( , , , , , , , , )
ρ µ φ α where c is the chord length, h is the maximum thickness, r
is the nose radius, φ is the trailing edge angle, and α is the angle of attack. Repeating
variables: V c
, , .
ρ Theπ − terms are
π
ρ
π
ρ
µ
π π π φ π π α
1 2 2 2 3 4 5 7
= = = = = = =
F
V c
V c c
h
c
r
c
w
D
, , , , , , .
6
Then,
F
V c
f
V c c
h
c
r
c
w
D
ρ
ρ
µ
φ α
2 2 1
=






, , , , ,
6.20 [ ] [ ] [ ] [ ] [ ] [ ]
3
2
2
( , , , , ). , , , , 1, .
L L
Q f R A e S g Q R L A L e L s g
T T
= = = = = = =
There are only two basic dimensions. Choose two repeating variables, R and g. Then,
π π π π
1 2 3 4
1 1 2 2 3 3 4 4
= = = =
QR g AR g eR g sR g
a b a b a b a b
, , , .
∴ = = = =
π π π π
1 5 2 2 2 3 4
Q
gR
A
R
e
R
s
/
, , , .
∴ = ∴ = 





π π π π
1 1 2 3 4 5 1 2
f
Q
gR
f
A
R
e
R
s
( , , ). , , .
6.21 [ ] [ ] [ ] [ ]
2 2 3
( , , , ). , , , , .
p p
L L M M
V f h g V h L g
T T T L
σ ρ σ ρ
 
= = = = = =
 
Repeating variables: h g V h g h g
p
a b c a b c
, , . , .
ρ π ρ π σ ρ
∴ = =
1 2
1 1 1 2 2 2
∴ = = ∴ =






π π
σ
ρ
σ
ρ
1 2 2 1 2
V
hg gh
V
gh
f
gh
p p
, . .
6.22 F f V e I d
D = ( , , , , , ).
µ ρ Repeating variables: V d
, , .
ρ
[ ] [ ] [ ] [ ] [ ] [ ] [ ]
2 3
, , , , , 1, .
D
ML L M M
F V e L I d L
T LT
T L
µ ρ
= = = = = = =
π ρ π µ ρ π ρ π ρ
1 2 3 4
1 1 1 2 2 2 3 3 3 4 4 4
= = = =
F V d V d e V d I V d
D
a b c a b c a b c a b c
, , , .
∴ = = = =
π
ρ
π
µ
ρ
π π
1 2 2 2 3 4
F
V d V d
e
d
I
D
, , , .
∴ =






F
V d
f
V d
e
d
I
D
ρ
µ
ρ
2 2 1 , , .
6.23 F f V D g
D s
= ( , , , , , ).
ρ ρ µ Repeating variables: V D
, , .
ρ
[ ] [ ] [ ] [ ] [ ] [ ] [ ]
2 3 3 2
, , , , , , .
D s
ML L M M M L
F V D L g
T LT
T L L T
ρ ρ µ
= = = = = = =

108. 108
π ρ π ρ ρ π µ ρ π ρ
1 2 3 4
1 1 1 2 2 2 3 3 3 4 4 4
= = = =
F V D V D V D gV D
D
a b c
s
a b c a b c a b c
, , , .
∴ = = = =
π
ρ
π
ρ
ρ
π
µ
ρ
π
1 2 2 2 3 4 2
F
V D VD
gD
V
D s
, , , .
∴ =






F
V D
f
VD
gD
V
D s
ρ
ρ
ρ
µ
ρ
2 2 1 2
, , .
6.24 F f V d e r c
D = ( , , , , , , ).
µ ρ Repeating variables: V d
, , .
ρ
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
F
ML
T
V
L
T
M
LT
M
L
d L e L r L c
L
D = = = = = = = =
2 3 2
1
µ ρ
π ρ π µ ρ π ρ π ρ π ρ
1 2 3 4 5
1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
= = = = =
F V d V d eV d rV d cV d
D
a b c a b c a b c a b c a b c
, , , , .
∴ = = = = =
π
ρ
π
µ
ρ
π π π
1 2 2 2 3 4 5
2
F
V d Vd
e
d
r
d
cd
D
, , , , .
∴ =






F
V d
f
Vd
e
d
r
d
cd
D
ρ
µ
ρ
2 2 1
2
, , , .
6.25 f g V d f
T
M
LT
M
L
V
L
T
d L
= = = = = =
( , , , ). [ ] , [ ] , [ ] , [ ] , [ ] .
µ ρ µ ρ
1
3
Repeating variables, V d f V d V d
a b c a b c
, , . ,
ρ π ρ π µ ρ
1 2
1 1 1 2 2 2
= =
∴ = = ∴ =






π π
µ
ρ
µ
ρ
1 2 1
fd
V Vd
fd
V
g
Vd
, . .
6.26 F f V c t
L c
= ( , , , , ).
, ρ α
l Repeating variables: V c
, , .
ρ l
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
F
ML
T
V
L
T
c
L
T
M
L
L t L
L c
= = = = = = =
2 3
1
ρ α
l
π ρ π ρ π ρ π α ρ
1 2 3 4
1 1 1 2 2 2 3 3 3 4 4 4
= = = =
F V cV tV V
L
a b
c
c a b
c
c a b
c
c a b
c
c
l l l l
, , , .
∴ = = = =
π
ρ
π π π α
1 2 2 2 3 4
F
V
c
V
t
L
c c
l l
, , , .
∴ =






F
V
f
c
V
t
L
c c
ρ
α
2 2 1
l l
, , .
6.27
2
2 3
1
( , , , , ). [ ] , [ ]= ,[ ]= , [ ] , [ ] , [ ] .
ML M M
T f d t T d L t L
T LT
T L
ω ρ µ ω ρ µ
= = = = =
Repeating variables: d Td d t d
a b c a b c a b c
, , . , , .
3
ω ρ π ω ρ π µ ω ρ π ω ρ
1 2
1 1 1 2 2 2 3 3 3
= = =

109. 109
1 2 3
2 5 2
, , .
T t
d
d d
µ
π π π
ρω ρω
∴ = = =
3 5
1 1
2 5 2 2
, . , .
T t t
f W d f
d d
d d d
µ µ
ρω
ρω ρω ρω
   
∴ = =
   
   
   
&
6.28 F f V d L
D c
= ( , , , , , , )
ρ µ ρ ω where d is the cable diameter, L the cable length, ρc
the cable density, andω the vibration frequency.
Repeating variables: V d
, , .
ρ The π − terms are
π
ρ
π
ρ
µ
π π
ρ
ρ
π
ω
1 2 2 2 3 4 5
= = = = =
F
V d
Vd d
L
V
d
D
c
, , , ,
We then have
F
V d
f
Vd d
L
V
d
D
c
ρ
ρ
µ
ρ
ρ ω
2 2 1
=






, , ,
6.29 ∆p f D h d d
= ( , , , , , ).
ω ρ 1 0 Repeating variables: D, , .
ω ρ
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ]
∆p
M
LT
D L h L
T
M
L
d L d L
= = = = = = =
2 3 1 0
1
ω ρ
π
ρω
π π π
1 2 2 2 3
1
4
0
= = = =
∆p
D
h
D
d
D
d
D
, , , .
∴ = 




 =
∆p
D
f
h
D
d
D
d
D
W
ρω 2 2 1
1 0
, , . & force × velocity = ∆pD D
2
×ω .
∴ =






& , , .
W D f
h
D
d
D
d
D
ρω3 5
1
1 0
6.30 T g f d H N h
= ( , , , , , , ).
,
ω ρ
l Repeating variables: ω ρ
, , .
d
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
T
ML
T
f
T T
d L H L L N h L
M
L
= = = = = = = = =
2
2 3
1 1
1
ω ρ
l
π
ρω
π
ω
π π π π
1 2 5 2 3 4 5 6
= = = = = =
T
d
f H
d d
N
h
d
, , , , , .
l
∴ = 





T
d
g
f H
d
N
h
d
ρω ω
2 5 1 , , , , .
d
l
6.31 Q f H w g
= ( , , , , , ).
µ ρ σ Repeating variables: H g
, , .
ρ
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
Q
L
T
H L w L g
L
T
M
LT
M
L
M
T
= = = = = = =
3
2 3 2
µ ρ σ
∴ = = = =
π π π
µ
ρ
π
σ
ρ
1 5 2 3 3 4 2
Q
gH
w
H gH gH
, , , .

110. 110
∴ =








Q
gH
f
w
H gH gH
5 1 3 2
, , .
µ
ρ
σ
ρ
6.32 d f V V D
j a
= ( , , , , , , ).
σ ρ µ ρ Repeating variables: V D
j , , .
ρ
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
d L V
L
T
V
L
T
D L
M
T
M
L
M
LT
M
L
j a
= = = = = = = =
σ ρ µ ρ
2 3 3
π π π
σ
ρ
π
µ
ρ
π
ρ
ρ
1 2 3 2 4 5
= = = = =
d
D
V
V V D V D
j j j
a
, , , , .
∴ =








d
D
f
V
V V D V D
j j j
a
1 2
, , , .
σ
ρ
µ
ρ
ρ
ρ
6.33 T f H h R t
= ( , , , , ).
ω µ ρ
, , Repeating variables: ω ρ
, , .
h
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
T
ML
T T
H L h L R L t L
M
LT
M
L
= = = = = = = =
2
2 3
1
ω µ ρ
π
ρω
π π π π
µ
ρω
1 2 5 2 3 4 5 2
= = = = =
T
d
H
h
R
h
t
h h
, , , ,
∴ =






T
d
f
H
h
R
h
t
h h
ρω
µ
ρω
2 5 1 2
, , , .
6.34 µ ρ
= f D H g V
( , , , , , )
l . D = tube dia., H = head above outlet, l = tube length.
Repeating variables: D V
VD
H
D D
gD
V
, , . , , ,
ρ π
µ
ρ
π π π
1 2 3 4 2
= = = =
l
∴ = 





µ
ρVD
f
H
D D
gD
V
1 2
, , .
l
6.35 T f R e r
= ( , , , , ).
, ,
ω ρ µ l Repeating variables: R, , .
ω ρ
[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .
T
ML
T
R L
T
M
L
e L r L
M
LT
L
= = = = = = = =
2
2 3
1
ω ρ µ l
π
ρω
π π π π
µ
ρω
1 2 5 2 3 4 5 2
= = = = =
T
R
e
R
r
R R R
, , , ,
l
1
2 5 2
, , , .
T e r
f
R R R
R R
µ
ρω ρω
 
∴ =  
 
 
l
6.36 y f V y g
2 1 1
= ( , , , ).
ρ Neglect viscous wall shear.
[ ] , [ ] , [ ] , [ ] , [ ] .
y L V
L
T
y L
M
L
g
L
T
2 1 1 3 2
= = = = =
ρ Repeating variables: V y
1 1
, , .
ρ

111. 111
π π
1
2
1
2
1
1
2
= =
y
y
gy
V
, . ( ρ does not enter the problem).
∴ =






y
y
f
gy
V
2
1
1
1
2
.
6.37 f g d V f
T
d L L
M
L
M
LT
V
L
T
= = = = = = =
( , , , , ). [ ] , [ ] , ] , [ ] , [ ] , [ ] .
[
l l
ρ µ ρ µ
1
3
Repeating variables: d V
, , .
ρ (l = length of cylinder).
π π π
µ
ρ
µ
ρ
1 2 3 1
. , , . , .
= = = ∴ =






fd
V d Vd
fd
V
f
d Vd
l l
6.38
Q
Q
V
V
p
p
V
V
F
F
V
V
m
p
m m
p p
m
p
m m
p p
p m
p p
m m m
p p p
= = =
l
l
l
l
2
2
2
2
2 2
2 2
, ,
( )
( )
∆
∆
ρ
ρ
ρ
ρ
τ
τ
ρ
ρ
ρ
ρ
ρ
ρ
m
p
m m
p p
m
p
m m m
p p p
m
p
m m m
p p p
V
V
T
T
V
V
Q
Q
V
V
= = =
2
2
2 3
2 3
3 2
3 2
, ,
&
&
l
l
l
l
( &
Q has same dimensions as .)
W
&
6.39 (A) Re Re . m m
m p
m
V L
ν
=
p p
p
V L
ν
= . 12 9 108 m/s.
p
m p
m
L
V V
L
∴ = = × =
6.40 A)
5
6
1.51 10
Re Re . . 4 10 461 m/s.
1.31 10
p p p
m m m
m p m p
m p m p
V L L
V L
V V
L
ν
ν ν ν
−
−
×
= = ∴ = = × =
×
6.41 a) Re Re . . .
m p
m m
m
p p
p
m
p
p
m
V d V d V
V
d
d
= = ∴ = =
ν ν
7
Q
Q
V
V
Q Q
V
V
m
p
m m
p p
m p
m
p
m
p
= ∴ = = × × =
l
l
l
l
2
2
2
2 2
15 7
1
7
0 214
. . . .
m / s
3
&
&
. & .
W
W
V
V
W
m
p
m m m
p p p
m
= = × = ∴ = × =
ρ
ρ
3 2
3 2
3
2
7
1
7
7 7 200 1400
l
l
kW
b) Re Re .
.
.
. .
m p
m
p
p
m
m
p
V
V
d
d
= ∴ = = × =
ν
ν
7
9
13
4 85
3
2
3
2
1
1.5 4.85 0.148 m /s.
7
1
4.85 200 466 kW
7
m
m
Q
W
= × × =
= × × =
&

112. 112
6.42 a) Re Re . . .
m p
m m
m
p p
p
m
p
p
m
V d V d V
V
d
d
= = ∴ = =
ν ν
5
2
2 2
1 1
5. 800/5 160 kg/s.
5
5
m m m m
m p
p p p p
m V
m m
m V
ρ
ρ
= = × ∴ = = =
& l
& &
& l
∆
∆
∆ ∆
p
p
V
V
p p
m
p
m m
p p
m p
= = ∴ = = × =
ρ
ρ
2
2
2
5 25 25 600 15
. .
000 kPa
b) Re Re .
.
.
. .
m p
m
p
p
m
m
p
V
V
d
d
= ∴ = = × =
ν
ν
5
8
114
3 51
2
2
1
800 3.51 112 kg/s. 600 3.51 7390 kPa.
5
m m
m p
= × × = ∆ = × =
&
6.43 a) Re Re . . .
m p
m m
m
p p
p
m
p
p
m
V d V d V
V
d
d
= = ∴ = =
ν ν
10
F
F
V
V
F F
m
p
m m m
p p p
m p
= = × ∴ = =
ρ
ρ
2 2
2 2
2
10 10
l
l
1
10
= 1. lb
2
.
b) Re Re .
.
.41
. .
m p
m
p
p
m
m
p
V
V
d
d
= ∴ = = × =
ν
ν
10
1 06
1
7 52
p
p m
F F
ρ
=
m
ρ
2 2
2
2 2 2
1
10 10 17.68 lb.
7.52
p p
m m
V L
V L
= × × =
6.44 Re Re . . .
m p
m m
m
p p
p
m
p
p
m
m
p
m
p
V V V
V
= = ∴ = = =
assuming
l l l
l
ν ν
ν
ν
ν
ν
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 Re Re . . .
m p
m m
m
p p
p
m
p
p
m
m
p
V V V
V
= ∴ = ∴ =
l l l
l
ν ν
ν
ν
Water:
V
V
V V
m
p
p
m
m p m p
= = = ∴ = =
l
l
10 10 900
assuming km / hr.
ν ν .
Air: V V
m p
p
m
m
p
= = ×
×
×
=
−
−
l
l
ν
ν
90 10
15 10
1 10
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.

113. 113
6.46 Re Re . . / / .
m p
m m
m
p p
p
m p p m m p
V V
V V
= = ∴ = = =
if
l l
l l
ν ν
ν ν
10
∴ = × =
Vm 10 50 500 m / s.
This is in the compressibility range so is not recommended. Try a water channel for the
model study. Then
6
5
1 10
10 0.662.
1.5 10
p
m m
p m p
V
V
ν
ν
−
−
×
= = × =
×
l
l
33.1 m/s.
m
V
∴ =
This is a possibility, although 33.1 m/s is still quite large.
2 2
2
2 2 2
( ) 1000 1
0.662 3.56.
( ) 1.23 10
D m m m m
D p p p p
F V
F V
ρ
ρ
= = × × =
l
l
6.47
5
3
1.06 10
Re Re . . 2.5 1
5.5 10
p p p
m m m
m p m p
m p m p
V d V
V d
d d
V
ν
ν ν ν
−
−
×
= = ∴ = = × ×
×
= 0.0048 ft.
Find oil
ν using Fig. B.2. Then
2
2
2
1.94
1 1.11.
1.94 0.9
m m m
p p p
p V
p V
ρ
ρ
∆
= = × =
∆ ×
6.48 Re Re . . . . .
m p
m m
m
p p
p
m p
p
m
m
p
p
m
V V
V V
= = ∴ = = × × ×
−
l l l
l
l
l
ν ν
ν
ν
0 1 025 10 3
If l
l
l
p
p
m
m
V
≅ = = =
5
5
0025
2000 0 005
cm, then and m /s.
.
.
We could try lp m
V
≅ =
50 0 05
cm, but m /s.
. Each of these Vm' s is quite
small — too small for easy measurements. Let’s try a wind tunnel. Then,
V V
m p
p
m
m
p
p
m
p
= = × × ×
×
×
= =
−
−
−
l
l
l
l
l
ν
ν
01 025 10
1 10
18 10
0 28 5
3
3
5
. .
.
. m / s if cm. Or,
if lp m
V
= =
50 2 8
cm, m / s.
. This is a much better velocity to work with
in the lab. Thus, choose a wind tunnel.
6.49 Re Re . . . . .
m p
m m
m
p p
p
m p
m
m m
p
p p
m
p
V V V
g
V
g
V
V
= ∴ = = ∴ = ∴ =
Fr Fr
l l
l l
ν ν
2 2
1
30
V
V
m
p
p
m
m
p
m
p
m
p
m
= = = ∴ = ∴ = × −
l
l
ν
ν
ν
ν
ν
ν
ν
30
1
30
1
164
6 1 10 9
. . . .
m / s Impossible!
2
6.50 (C)
2
Fr Fr . m
m p
m m
V
l g
=
2
p
p p
V
l g
=
1
. 2 0.5 m/s.
4
m
m p
p
l
V V
l
∴ = = × =

114. 114
6.51 (A) From Froude’s number m
m p
p
l
V V
l
= . From the dimensionless force we have:
2 2
* * 2
2 2 2 2 2 2
or . 10 25 25 156000 N.
p p p
m
m p p m
m m m p p p m m
F V l
F
F F F F
V l V l V l
ρ ρ
= = ∴ = = × × =
6.52 Fr Fr m /s
m p
m
m m
p
p p
m p
m
p
V
g
V
g
V V
= ∴ = ∴ = = =
. . . .
2 2
10
1
60
1 29
l l
l
l
( )
( )
. ( ) ( ) . .
F
F
V
V
F
V
V
F
D m
D p
m m m
p p p
D p
p
m
p
m
D m
= ∴ = × = × × = ×
ρ
ρ
2 2
2 2
2
2
2
2
2 6
60 60 10 216 10
l
l
l
l
N
6.53 Fr Fr
m p
m
m m
p
p p
m
p
m
p
V
g
V
g
V
V
= = ∴ =
. .
2 2
l l
l
l
.
a)
Q
Q
V
V
Q Q
V
V
m
p
m m
p p
m p
m
p
m
p
= ∴ = = × × =
l
l
l
l
2
2
2
2 2
2
1
10
1
10
0 00632
. . .
m / s
3
b)
F
F
V
V
F F
V
V
m
p
m m m
p p p
p m
p
m
p
m
= ∴ = = × × =
ρ
ρ
2 2
2 2
2
2
2
2
2
12 10 10
l
l
l
l
. .
12 000 N
6.54 Neglect viscous effects. Fr Fr fps
m p
m
p
m
p
p
V
V
V
= ∴ = = ∴ =
. . . .
l
l
1
10
63 2
F
F
V
V
F F
V
V
m
p
m m m
p p p
p m
p
m
p
m
= ∴ = = × × =
ρ
ρ
2 2
2 2
2
2
2
2
2
0 8 10 10 800
l
l
l
l
. . .
lb
6.55 Neglect viscous effects, and account for wave (gravity) effects.
Fr Fr
m p
m
m m
p
p p
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
ω
ω
∴ = = × × =
ω ω
m p
m
p
p
m
V
V
l
l
600
1
10
10 1897 rpm.
T
T
V
V
T T
V
V
m
p
m m m
p p p
p m
p
m
p
m
= ∴ = = × × = ⋅
ρ
ρ
2 3
2 3
2
2
3
3
3
1 2 10 10 120
l
l
l
l
. . .
000 N m
6.56 Fr Fr
6
100
m p
m
m m
p
p p
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

115. 115
6.57 Check the Reynolds number:
Re .
p
p p
p
V d
= =
×
= ×
−
ν
15 2
10
30 10
6
6
This is a high-Reynolds-number flow.
Re
/
. .
m =
×
= ×
−
2 2 30
10
1 33 10
6
5
This may be sufficiently large for similarity. If so,
&
&
. .
W
W
V
V
m
p
m m m
p p p
= = × = × −
ρ
ρ
3 2
3 2
3
3 2
6
2
15
1
30
2 63 10
l
l
∴ = × × =
−
& ( . ) / . .
Wp 2 2 15 2 63 10 1633
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 =
×
×
= ×
−
10 4
15 10
26 7 10
5
5
This is a
high-Reynolds-number flow. Let’s assume the flow to be Reynolds
number independent above Re = ×
5 105
(see Fig. 6.4). Then
5
5
4/20
Re 5 10 . 37.5 m/s.
1.5 10
m
m m
V
V
−
×
= × = ∴ ≥
×
6.59 Re
.
. .
p =
×
×
= ×
−
20 10
15 10
13 3 10
5
6
This is a high-Reynolds-number flow.
Let Re
.4
.
. .
m
m
m
V
V
= =
×
×
∴ ≥
−
10
15 10
3 75
5
5
m / s for the wind tunnel.
Re
.
. .
m
m
m
V
V
= =
×
×
∴ ≥
−
10
1
1 10
1 0
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
m m m
m m m m
m
1
2
1 1 1
2 2 2 2
2
2 2
2 2
2
2
2
2
3 2
3 2
1000
1 23
2
15
1
416
= = ∴ = × =
ρ
ρ
l
l
.
. .
.
.4 .
.4
. .
N
&
&
.
. & ( . )
.
.
W
W
V
V
W
m
p
m m m
p p p
p
= =
×
×
∴ = × × =
ρ
ρ
3 2
3 2
3 2
3 2
3
3
2
2
15 4
20 10
15 3 2
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 = ×
3 105
we have
Re
.
.
.
m
m
m
V
V
= × =
×
×
∴ ≥
−
3 10
18
15 10
250
5
4
fps,and a study is possible.

116. 116
6.61 Mach No. is the significant parameter. M M
m p
= .
a) M M . . 200 m/s.
p
m
m p m p
m p
V
V
V V
c c
= ∴ = ∴ = =
F
F
V
V
F
m
p
m m m
p p p
p
= ∴ = × × =
ρ
ρ
2 2
2 2
2 2
10 1 20 4000
l
l
. .
N
b)
255.7
200 186 m/s.
296
p p
p m m
m m
c T
V V V
c T
= = = =
F F
V
V
p m
m p p
m m m
= = × × × =
ρ
ρ
2 2
2 2
2
2
2
10 601
186
200
20 2080
l
l
. .
N
c)
223.3
200 174 m/s.
296
p p
p m m
m m
c T
V V V
c T
= = = =
F F
V
V
p m
p
m
p
m
p
m
= = × × × =
ρ
ρ
2
2
2
2
2
2
2
10 338
174
200
20 1023
l
l
. .
N
6.62
273
M M . . 250 276 m/s.
223.3
p
m
m p m
m p
V
V
V
c c
= ∴ = ∴ = =
223.3
. 290 262 m/s.
273
m m m
p
p p p
V c T
V
V c T
= = ∴ = =
p
p
V
V
p p
V
V
m
p
m m
p p
p m
p
m
p
m
o
o
= ∴ = = =
ρ
ρ
ρ
ρ
ρ
ρ
2
2
2
2
2
2
80
338
8
262
290
34 6
.
.
.
. .
kPa, abs
α p = 5o
for similarity. (Note: we use ρm at 2700 m where T = 0°C.)
6.63 a) Fr Fr
m p
m
m m
p
p p
m
p
m
p
V
g
V
g
V
V
= = ∴ =
. . .
2 2
l l
l
l
ω
ω
ω
m
p
m
p
p
m
m
V
V
= = × ∴ = × =
l
l
1
10
10 2000
10
10
6320
. .
rpm
b) Re Re . . .
m p
m m
m
p p
p
m
p
p
m
V V V
V
= = ∴ = =
l l l
l
ν ν
10
ω
ω
ω
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
T T
V
V
m
p
m m m
p p m
p m
p
m
p
m
= ∴ = = × × = ⋅
ρ
ρ
2 3
2 3
2
2
3
3
2
2
3
12
15
60
10 750
l
l
l
l
. .
N m

117. 117
ω
ω
ω ω
m
p
m
p
p
m
p m
p
m
m
p
V
V
V
V
= ∴ = = × × =
l
l
l
l
. . .
rpm
500
15
60
1
10
12 5
6.65 Re Re . .
m p
m m
m
p p
p
m p
p
m
V V
V V
= ∴ = ∴ = = × = m / s.
l l l
l
ν ν
10 10 100
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 =
×
×
= ×
−
5 01
1 10
5 10
6
5
where we used 0.1 ( 1 m,
m p
= =
l l i.e., the
dia. of the porpoise). ω ω
m p
m
p
p
m
V
V
= = × × =
l
l
1
5
10
10 5 motions / second.
6.66 ρ
ρ
ρ
* * * * * *
, , , , , .
= = = = = =
o
t tf u
u
V
v
v
V
x
x
y
y
l l
Substitute in:
* * * * *
o * * *
( ) ( )
0.
o o
V u V v
f
t x y
∂ρ ∂ ρ ∂ ρ
ρ ρ ρ
∂ ∂ ∂
+ + =
l l
Divide by ρoV / :
l
∴ + + =
f
V t x
u
y
v
f
V
l l
∂ρ
∂
∂
∂
ρ
∂
∂
ρ
*
* *
* *
*
* *
( ) ( ) . .
0 parameter =
6.67 * * * * * * * * *
2
, , , , , , , , .
V u v w x y z p
V u v w x y z p t tf
U U U U U
ρ
= = = = = = = = =
v
v
l l l
Substitute into Euler’s equation and obtain:
Uf
V
t
U
u
V
x
U
v
V
y
U
w
V
z
U p
∂
∂
∂
∂
∂
∂
∂
∂
ρ
ρ
v
l
v
l
v
l
v
l
v
*
*
*
*
*
*
*
*
*
*
*
* *
.
+ + + = −
∇
2 2 2 2
Divide by U2
/ :
l
f
U
V
t
u
V
x
v
V
y
w
V
z
p
l
v v v v
∂
∂
∂
∂
∂
∂
∂
∂
*
*
*
*
*
*
*
*
*
*
*
* *
.
+ + + = −∇ Parameter =
f
U
l
6.68
v
v
l
v
l
v
l
V
V
U
t
tU
p
p
U
h
h
* * * * *
, , , , .
= = ∇ = ∇ = =
ρ 2
Euler’s equation is then
ρ ρ ρ
U DV
Dt
U
p g h
2 2
l
v
l
v l
l
v
*
*
* * * *
.
= − ∇ − ∇
Divide by ρU 2
/ :
l
DV
Dt
p
g
U
h
v
v l v
*
*
* * * *
.
= −∇ − ∇
2
Parameter =
g
U
l
2
.
6.69 There is no y- or z-component velocity so continuity requires that∂ ∂
u x
/ .
= 0 There
is no initial pressure distribution tending to cause motion so∂ ∂
p x
/ .
= 0 The

118. 118
x-component Navier-Stokes equation is then
u u
u
t x
∂ ∂
∂ ∂
+ v
+
u
w
y
∂
∂
+
1
u p
z x
∂ ∂
∂ ρ ∂
= −
2
2
x
u
g
x
∂
ν
∂
+ +
2 2
2 2
u u
y z
∂ ∂
∂ ∂
+ +
 
 
 
 
(wide plates)
This simplifies to
∂
∂
ν
∂
∂
u
t
u
y
=
2
2
.
a) Let u u U y y h t tU h
* * *
/ , / / .
= = =
and Then
U
h
u
t
U
h
u
y
2
2
2
∂
∂
ν ∂
∂
*
*
*
*2
=
The normalized equation is
∂
∂
∂
∂
u
t
u
y
*
*
*
*2
Re
=
1 2
where Re =
Uh
ν
b) Let u u U y y h t t h
* * *
/ , / / .
= = =
and Then
ν 2
ν ∂
∂
ν
∂
∂
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 ∂ ∂
u x
/ = 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
∂
∂
+
v
u
r
θ
∂
∂
+
u u
u
r x
∂ ∂
∂θ ∂
+
1
x
p
g
x
∂
ρ ∂
= − +
2 2
2 2 2
1 1
u u u
r r
r r
∂ ∂ ∂
ν
∂
∂ ∂θ
+ + +
2
2
u
x
∂
∂
+
 
 
 
 
This simplifies to
∂
∂ ρ
∂
∂
ν
∂
∂
∂
∂
u
t
p
x
u
r r
u
r
= − + +






1 1
2
2
a) Let * * * * 2 *
/ , / , / , = / and / :
u u V x x d t tV d p p V r r d
ρ
= = = =
V
d
u
t
V
d
p
x
V
d
u
r r
u
r
2 2
2
2
1
∂
∂
ρ
ρ
∂
∂
ν ∂
∂
∂
∂
*
*
*
*
*
*2 *
*
*
= − + +






The normalized equation is
∂
∂
∂
∂
∂
∂
∂
∂
u
t
p
x
u
r r
u
r
*
*
*
*
*
*2 *
*
*
Re
= − + +






1 1
2
where Re =
Vd
ν
b) Let * * * 2 * 2 *
/ , / , / , = / and / :
u u V x x d t t d p p V r r d
ν ρ
= = = =
ν ∂
∂
ρ
ρ
∂
∂
ν ∂
∂
∂
∂
V
d
u
t
V
d
p
x
V
d
u
r r
u
r
2
2
2
2
1
*
*
*
*
*
*2 *
*
*
= − + +






The normalized equation is

119. 119
∂
∂
∂
∂
∂
∂
∂
∂
u
t
p
x
u
r r
u
r
*
*
*
*
*
*2 *
*
*
Re
= − + +
2
1
where Re =
Vd
ν
6.71 Assume w
z
= =
0 0
and
∂
∂
. The x-component Navier-Stokes equation is then
u
t
∂
∂
u
u v
x
∂
∂
+ +
u u
w
y z
∂ ∂
∂ ∂
+
1 p
x
∂
ρ ∂
= −
2 2 2
2 2 2
x
u u u
g
x y z
∂ ∂ ∂
ν
∂ ∂ ∂
+ + + +
 
 
 
 
With g g
x = the simplified equation is
u
u
x
g
u
x
u
y
∂
∂
ν
∂
∂
∂
∂
= + +






2
2
2
2
Let u u V x x h y y h
* * *
/ , / / .
= = =
and Then
V
h
u
u
x
g
V
h
u
x
u
y
2
2
2 2
*
*
*
*
*2
*
*2
∂
∂
ν
∂
∂
∂
∂
= + +






The normalized equation is
u
u
x
u
x
u
y
*
*
*
*
*
*
*
Re
∂
∂
∂
∂
∂
∂
= + +






1 1
2
2
2
2
2
Fr
where Fr and
= =
V
hg
Vh
Re
ν
6.72 * * * * * *2 2 2
, , , , , .
o
u v T x y
u v T x y
U U T
= = = = = ∇ = ∇
l
l l
ρ
∂
∂
∂
∂
c
UT T
x
UT T
y
K
T T
p
o o
o
l l l
*
*
*
*
* *
.
+





 = ∇
2
2
Divide by ρc UT
p o / :
l
∂
∂
∂
∂ ρ
T
x
T
y
K
c U
T
p
*
*
*
*
* *
.
+ = ∇
l
2
Parameter =
K
c U
p
µ
µ
ρ l
=
1 1
Pr Re
.
6.73 ρ
ρ
ρ
* * * * * * *
, , , , , , .
= = = ∇ = ∇ ∇ = ∇ = =
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: ρ ρ
µ µ
o
o
U DV
Dt
p
p
U
V
U
V
*
*
*
* * * * * * *
( ).
2
2
2
2
3
l
v
l
v
l
v
l
v v v
= − ∇ + ∇ + ∇ ∇ ⋅
Divide by ρoU 2
/ :
l
[ ]
ρ
ρ
µ
ρ
*
*
*
* * * * * * *
( ) .
DV
Dt
p
U
p
U
V V
o
o o
v
v
l
v v v v
= − ∇ + ∇ + ∇ ∇ ⋅
2
2
energy: ρ ρ
*
*
*
* * * * *
.
c T
U DT
Dt
K
T T p
U
p V
v o o o o
l l l
v v
= ∇ − ∇ ⋅
2
2
Divide by ρo v o
c T U / :
l
ρ
ρ ρ
*
*
*
* * * * *
.
DT
Dt
K
c U
T
p
c T
p V
o v
o
o v o
= ∇ − ∇ ⋅
l
v v
2

120. 120
The parameters are:
p
U
RT
U
kRT
kU
c
kU k
o
o
o o
ρ 2 2 2
2
2 2
1
= = = =
M
.
µ
ρoUl
=
1
Re
.
K
c U
K
c
c
c U
K
o v p
p
v o
ρ µ
µ
ρ
l l
= =
Pr Re
.
p
c T
RT
c T
c c
c
K
o
o v o
o
v o
p v
v
ρ
= =
−
= −1.
The significant parameters are K, , Re, Pr.
M

121. 162
CHAPTER 8
External Flows
8.1
8.2 Re .
.
. .
= = ∴ =
× ×
= ×
−
−
5
5 1 51 10
20
3 78 10
5
5
VD
D
ν
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
B C
D
separated
flow

122. 163
8.7 ( B)
6
0.8 0.008
Re 4880.
1.31 10
VD
ν −
×
= = =
×
8.8 5 5 5 1 22 10
8
12
000915
5
= ∴ = = × × =
−
VD
V D V
ν
ν a) fps.
/ . . /
.
.
b) V = × × =
−
5 388 10
8
12
000291
5
. /
.
. fps. c) V = × × =
−
5 1 6 10
8
12
0 012
4
. /
.
. fps.
8.9 Re
.
. .
= =
×
×
= ×
−
VD D
D
ν
20
1 51 10
13 25 10
5
5
a) Re . . .
= × × = ×
13 25 10 6 7 9 10
5 6
∴Separated flow.
b) Re . . . .
= × × = ×
13 25 10 06 7 9 10
5 4
∴Separated flow.
c) Re . . .
= × × =
13 25 10 006 7950
5
∴Separated flow.
8.10 F pdA p A p r rdr p p
D
A
= − = − = −





 =
∫
∫ back back
front
0
2
0 0
0
1
1 2 2
1
2
1
4
1
2
( ) π π π
Bernoulli: p V p p
∞ ∞
+ = ∴ = × × =
1
2
1
2
1 21 20 242
2
0 0
2
ρ . . Pa.
∴ = =
FD
1
2
242 380
π( ) N
C
F
V A
D
D
= =
×
× × ×
=
1
2
2
2 2
2 380
1 21 20 1
0 5
ρ π
.
.
8.11 F F F
total bottom top 000 .3 .3 +10 000 N.
= + = × × × × =
20 3 3 2700
. .
Flift cos 10 N
= =
2700 2659
o
Fdrag 10 N
= =
2700 469
sin o
C
F
V A
L
L
= =
×
× × ×
=
1
2
2
2
2 2659
1000 5 3 3
2 36
ρ . .
.
C
F
V A
D
D
= =
×
× × ×
=
1
2
2
2
2 469
1000 5 3 3
0 417
ρ . .
.
8.12 F p A Lw F p A
Lw
Lw
u u u
l l l o
= = × = = × =
26 8000
2 5
4015
000 .
cos
F F F Lw
L u
= − =
l
o o
cos cos
5 10 21 950
F F F Lw
D u
= − =
l
o o
sin sin
5 10 1569

123. 164
C
F
V A
Lw
Lw
L
L
= =
×
×
=
1
2
2
2
2 21
3119 750
0 25
ρ
950
.
.
C
F
V A
Lw
Lw
D
D
= =
×
×
=
1
2
2
2
2 1569
3119 750
0 0179
ρ .
.
8.13 If CD = 1 0
. for a sphere, Re = 100 (see Fig. 8.8). ∴
×
=
V .
,
1
100
ν
ν
V = 1000 .
a) V FD
= × × = ∴ = × × × ×
−
1000 1 46 10 0146
1
2
1 22 0146 05 1 0
5 2 2
. . . . . .
m / s. π
= × −
3 25 10 7
. .
N
b) V FD
= ×
×
×
= ∴ = × × × × ×
−
1000
1 46 10
015 1 22
0 798
1
2
015 1 22 798 05 1 0
5
2 2
.
. .
. (. . ) . . .
m /s. π
= × −
4 58 10 5
. .
N
c) V FD
= × × = ∴ = × × × ×
−
1000 1 31 10 00131
1
2
1000 00131 05 1 0
6 2 2
. . . . .
m /s. π
= × −
6 74 10 6
. .
N
8.14 a) Re
.
.
. .
= =
×
×
= × ∴ =
−
VD
CD
ν
6 5
1 5 10
2 10 0 45
5
5
from Fig. 8.8.
∴ = = × × × × × =
F V AC
D D
1
2
1
2
1 22 6 25 45 1 94
2 2 2
ρ π N.
. . . .
b) Re
.
.
. .
=
×
×
= × ∴ =
−
15 5
1 5 10
5 10 0 2
5
5
CD from Fig. 8.8.
∴ = = × × × × × =
F V AC
D D
1
2
1
2
1 22 15 25 2 5 4
2 2 2
ρ π N.
. . . .
8.15 (B) Assume a large Reynolds number so that CD = 0.2. Then
2
2 2
1 1 80 1000
1.23 5 0.2 4770 N.
2 2 3600
D
F V AC
ρ π
×
 
= = × × × × × =
 
 
8.16 (D) Assume a Reynolds number of 105. Then CD = 1.2.
2 2
1 1
. 60 1.23 40 4 1.2. 0.0041 m.
2 2
D
F V AC D D
ρ
= ∴ = × × × × × ∴ =
5
6
40 0.0041
Re 1.64 10 . 1.2. The assumption was OK.
10
D
VD
C
ν −
×
= = = × ∴ =

124. 165
8.17 The velocities associated with the two Re's are
V
D
1
1
5 5
3 10 1 5 10
0445
101
= =
× × ×
=
−
Re .
.
ν
m /s,
V
D
2
2
4 5
6 10 1 5 10
0445
20
= =
× × ×
=
−
Re .
.
ν
m /s.
The drag, between these two velocities, is reduced by a factor of 2.5
( ) ( )
[ ]
C C
D D
high low
and
= =
0 5 0 2
. . . 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 V AC V C V C
D D D D
= ∴ = × ×





 ∴ =
1
2
0 5
1
2
00238
2
12
4810
2 2
2
2
ρ π .
. . . .
Re
/
.
. . : .
= =
×
×
= = = ×
−
VD V
V C V
D
ν
4 12
1 6 10
2080 5 98
4
Try fps, Re = 2 105
Try fps, Re= 2.3 105
C V
D = = ×
. : .
4 110
b) C V V
D = = × ×





 × ∴ =
0 2
1
2
00238
2
12
2 155
2
2
. : . . .
0.5 fps.
π
8.19 4 2
1
2
1000 1 0 267
2
10
2 10
2 2 2
6
5
. . . . . Re
.
.
= × × ∴ = =
×
= ×
−
V C V C
V
V
D D
π
Try C V
D = ∴ = = × ∴
0 5 0 73 1 46 105
. : . Re . .
m /s. OK.
8.20 Re
.
. . .
= =
×
×
= × ∴ =
−
VD
CD
ν
40 2
15 10
53 10 07
5
6
. (This is extrapolated from
Fig. 8.8.) ∴ = × × × × × =
FD
1
2
122 40 2 60 7 81
2
. ( ) . .
900 N
M = 81 900 × 30 = 2 46 106
. .
× ⋅
N m
8.21 a) Re
.
.
. . Re . . Re . .
1 5
5
2
5
3
5
25 05
1 08 10
1 2 10 1 8 10 2 4 10
=
×
×
= × = × = ×
−
Assume a
rough cylinder (the air is highly turbulent).
( ) ( ) ( )
∴ = = =
C C C
D D D
1 2 3
0 7 0 8 0 9
. , . , . .
∴ = × × × × + × × + × × =
FD
1
2
1 45 25 05 10 7 075 15 8 1 20 9 1380
2
. (. . . . . . ) .
N
M = × × × × × + × × × + × × × = ⋅
1
2
1 45 25 05 10 7 40 075 15 8 27 5 1 20 9 10 25 700
2
. (. . . . . . . ) .
N m
b) Re
.
.
. . Re . , Re . .
1 5
4
2
5
3
5
25 05
1 65 10
7 6 10 1 14 10 1 5 10
=
×
×
= × = × = ×
−

125. 166
( ) ( ) ( )
∴ = = = =
×
=
C C C
D D D
1 2 3
8 7 8
101
287 308
1 17
. , . , . .
.
. .
kg / m3
ρ
∴ = × × × × + × × + × × =
FD
1
2
1 17 25 05 10 8 075 15 7 1 20 8 1020
2
. (. . . . . . ) .
N
M = × × × × × + × × × + × × × = ⋅
1
2
1 17 25 05 10 8 40 075 15 7 27 5 1 20 8 10 19
2
. (. . . . . . . ) .
600 N m
8.22 Atmospheric air is turbulent. ∴Use the "rough" curve. ∴ =
CD 0 7
. .
F V D V D
V V
D = = × × × ∴ =
×
× −
10
1
2
00238 6 7 2000 10
2000
1 6 10
2 5
2
4
. . . .
/
.
.
= 2
2 2 2 2
min o
0.0024
= 30 104 11.8 psf.
2 2
p U v
ρ
∞
   
∴ = − − = −
   
∴ = ∴ = =
V D V D
2
2370 148 0 108
. . . '.
fps
8.23 Since the air cannot flow around the bottom, we
imagine the structure to be mirrored as shown. Then
L D C C
D D
/ / . . .
= = ∴ = ∞
40 5 8 0 66
Re
.
. . . . .
min
min
= =
×
×
= × ∴ = × =
−
VD
CD
ν
30 2
1 5 10
4 10 1 0 66 66
5
6
∴ = × × ×
+
×





× =
FD
1
2
1 22 30
2 8
2
20 66 36
2
. . .
000 N
8.24 .
B D W
F F F
+ =
3 2 2 3
4 1 4
9810 1000 9810 7.82 .
3 2 3
D
r V r C r
π π π
× + × = × ×
Re .
=
×
= ×
−
V r
Vr
2
10
2 10
6
6
∴ =
V C r
D
2
178
a) r V V CD
= ∴ =
. , . .
05 89
2
m. Re = 105
Assume a smooth sphere.
Try C V
D = ∴ = ×
. : . .
5 4 22 m / s. Re = 4.22 105
This is too large for Re.
Try C V
D = ∴ = ×
. : . .
2 667 m / s. Re = 6.67 105
OK.
b) r V V CD
= × =
. , . .
025 4 45
2
m. Re = 5 104
Try C V
D = = ×
. : . .
2 4 72 m / s. Re = 2.4 105
OK.
c) r V V CD
= =
. , . .
005 089
2
m. Re = 104
Try C V
D = = ×
. : . .
5 133 m / s. Re = 1.33 104
OK.
d) r V V CD
= × =
. , . .
001 0178
2
m. Re = 2 103
Try C V
D = = ×
. : . .
4 067 m / s. Re = 1.33 103
OK.
W
FB
FD

126. 167
8.25
3 2 3
2
4 10 1 10 4 10
. .077 .00238 62.4 .
3 12 2 12 3 12
B D W D
F F F V C S
π π π
     
+ = × + × =
     
     
3 2
4
10/12
Re 5.2 10 . 1 .0139 810
1.6 10
D
V
V V C S
−
×
= = × ∴ + =
×
a) S V CD
= =
. . .
005 219
2
Assume atmospheric turbulence, i.e., rough.
Try C V C V
D D
= = = × ∴ = =
. : . . Re . . . . .
4 23 4 1 2 10 3 27
5
fps fps
b) S V C C V
D D
= = = = = × ∴
. . . . : . Re . .
02 1090 4 52 2 7 10
2 5
Try fps OK.
c) S V C C V
D D
= = = =
1 0 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 × 105.
F F
D W
= .
1
2
00238 2 3
1
2
1 0 7 2 5
1
2
1 0 7
18
12
2 5 1 0
4
12
4 180
2
× × × × × × × × × + × × + ×





×





 =
. . . . . . . . . .
V +2 π
We used data from Table 8.1. ∴ =
V 140 fps.
8.27 From Table 8.2 C F V V
D D
= = × × × =
0 35
1
2
1 22 3 2 0 35 683
2 2
. . . . . . .
a) F W
D = ×
×





 = ∴ =
×
=
. & .
683
80 1000
3600
337 337
80 1000
3600
7500
2
N. W or 10 Hp
b) V F W
D
= = × = ∴ = × =
25 683 25 427 427 25 10
2
m /s. N. 700 W or 14.3 Hp
. & .
c)V F W
D
= = × = ∴ = × =
27 8 683 27 8 527 527 27 8 14
2
. . . & . .
m /s. N. 700 W or 19.6 Hp
8.28 1 2 1 1 400
1
2
1 1
2
. . . . .
F F V AC C
D D D D
= × = =
ρ
1 2
1
2
1 22 2 3 1 1 1 1 400
2
. . ( ) . . .
× × × × × = ×
V
∴ =
V 95
. .
m / s
8.29 Re
(
. . .
= =
×
= × ∴ =
VD
CD
ν
40
4 42 10 0 35
5
000 / 3600)0.6
1.51 10-5
from Fig. 8.8.
a) F V AC
D D
= = × × × × × =
1
2
1
2
1 204 40 0 6 6 0 35 93 6
2
ρ . ( . . .
000 / 3600) N
2
b) F L D
D = × = = =
93 6 0 68 63 7 6 0 6 10
. . . / / . .
N where
c) F L D
D = × = =
93 6 0 76 71 1 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
FW
FD
Fy
Fx

127. 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, CD = 0 4
. :
F V AC
D D
=
1
2
2
ρ
80 9 8
1
2
1 21 0 6 0 4 53 7
2 2
× = × × × × ∴ =
. . . . . . .
V V
π m /s
Check Re: Re
. .
.
. .
=
×
×
= × ∴
−
53 7 1 2
1 51 10
4 27 10
5
6
OK.
b) F V AC
D D
=
1
2
2
ρ . From Table 8.2, CD = 1.4:
80 98
1
2
121 4 14 4 29
2 2
× = × × × × ∴ =
. . . . . .
V V
π m / s
Check Re: Re
.
.
. .
=
×
×
= ×
−
4 29 8
151 10
2 27 10
5
6
Should be larger but the velocity
should be close.
c) 2
1
2
D D
F V AC
ρ
=
2 2
1
80 9.8 1.21 1 1.4. 17.2 m/s.
2
V V
π
× = × × × × ∴ =
Check Re: Re
.
.
. .
=
×
×
= ×
−
172 1
151 10
114 10
5
6
This should be greater than 107 for
CD 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 2 2
1 1
.00238 88 (6 2) 1.1 122 lb.
2 2
D D
F V AC
ρ
= = × × × × × =
& , .
W F V
D
= × = × =
122 88 10 700
ft - lb
sec
or 19.5 Hp
8.33 F V AC
D D
= = × × × × × × =
1
2
1
2
122 27 8 16 05 11 1043
2 2 2
ρ π
. ( . . ) . . . N.
& . ( . . ) .
W F V
D
= × × = × × × =
2 1043 27 8 16 2 226 W or 1.24 Hp

128. 169
8.34 The projected area is
( . )
. .
2 0 3
2
4 4 6
+
× = m2
F V AC
D D
= = × × × × =
1
2
1
2
1 18 20 4 6 0 4 434
2 2
ρ . . . N.
Since there are two free ends, we use Table 8.1 with L D
/ / . . ,
= =
4 1 15 3 47 and
approximate the force as
FD = × =
434 0 62 269
. .
N
8.35 The net force acting up is (use absolute pressure)
3 3
up
4 4 120
0.4 1.21 9.8 0.5 0.4 9.8 2.16 N
3 3 2.077 293
F π π
= × × × − − × × =
×
From a force triangle (2.16 N up and FD to the right), we see that
tan / .
α = F FD
up
a) FD = =
2 16 80 0 381
. /tan . .
o
0 381
1
2
1 21 0 4 0 2 2 50
2 2
. . . . . .
= × × × ∴ =
V V
π m / s.
Check Re: Re
. .
.
. .
=
×
×
= ×
−
2 5 0 8
1 51 10
1 33 10
5
5
Too low. Use CD = 0 5
. :
0 381
1
2
1 21 0 4 0 5 1 58
2 2
. . . . . .
= × × × ∴ =
V V
π m / s
b) FD = =
2 16 70 0 786
. /tan . .
o
0 786
1
2
1 21 0 4 0 2 3 60
2 2
. . . . . .
= × × × ∴ =
V V
π m / s.
Check Re: Re
. .
.
. .
=
×
×
= ×
−
3 6 0 8
1 51 10
1 9 10
5
5
Too low. Use CD = 0 5
. :
0 786
1
2
1 21 0 4 0 5 2 27
2 2
. . . . . .
= × × × ∴ =
V V
π m / s
c) FD = =
2 16 60 1 25
. /tan . .
o
1 25
1
2
1 21 0 4 0 5
2 2
. . . . . .
= × × × ∴ =
V V
π 2.86 m / s
Check Re: Re
. .
.
. .
=
×
×
= ×
−
2 86 0 8
1 51 10
1 5 10
5
5
∴OK.
d) FD = =
2 16 50 1 81
. /tan . .
o
1 81
1
2
1 21 0 4 0 5
2 2
. . . . . .
= × × × ∴ =
V V
π 3.45 m / s
Check Re: 5
5
3.45 0.8
Re 1.8 10 .
1.51 10−
×
= = ×
×
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,

129. 170
the bottom does; so assume L D
/ ( / ) .
= × =
5 2 2 5 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,
F d
× = 5000
1
2
1 21 5 0 4 0 62
5
3
0 6 5000
2
× × ×





 × +





 = =
. ( ) . . . . .
V V 54.2 m /s
8.37 Power to move the sign:
F V V AC V
D D
= ×
1
2
2
ρ
= × × × × × =
1
2
1 21 11 11 0 72 1 1 11 11 657
2
. . . . . J /s.
This power comes from the engine:
657 12 0 3 1 825 10 4
= × × ∴ = × −
( & . . & .
000 1000) kg /s.
m m
Assuming the density of gas to be 900 kg/m3,
1825 10 10 3600 6 52
1000
900
030
4
. . $683
× × × × × × × =
−
8.38 The power expended is F V V
D × = × =
. ( / ) / . . m / s
25 88 60 3 281 11 18
1
2
121 1118 0 56
1
2
121 0 4 08
3 3
× × × × = × × × × ×
. . . . . .
C V C
D D
∴ =
V 1347
. .
m / s or 30.1 mph
8.39 & .
W F V V AC V AC V
D D D
= × = × = × =
40 746
1
2
1
2
2 3
η ρ ρ
∴ × × = × × × ∴ =
40 746 9
1
2
122 3 035 34 7
3
. . . . . .
V V m / s or 125 km / hr
8.40 (C)
5
4 0.02 0.02
Re 5000. St 0.21 .
4
1.6 10
VD fD f
V
ν −
× ×
= = = ∴ = = =
×
4 m/s
42 Hz (cycles/second). distance = 0.095 m/cycle.
42 cycles/s
V
f
f
∴ = = =
8.41
5
.003
40 Re 10 000. 40< 10 000. 0.2< 50 m/s.
1.5 10
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.

130. 171
8.42 5
5
6
40 . 8.13 10 ft.
1.22 10
VD D
D
ν
−
−
> = ∴ < ×
×
10
6
122 10
0020
5
000 < ft or 0.24"
VD D
D
ν
=
×
∴ >
−
.
. . .
8.43 From Fig. 8.9, Re is related to St. St =
f D
V V
×
=
×
02 1
. .
.
Re
.
.
. .
= =
×
×
= ∴
−
VD V
V
ν
1
15 10
0 095
5 Try St =.21: m / s. Re = 630.
This is acceptable. ∴ =
V 0095
. .
m / s
8.44 St = Re = Use Fig. 8.9.
fD
V V
VD V
=
×
=
×
−
.
. .
002 2 2
10
6
ν
Try St = 0.21: m / s Re = 38 10 OK.
3
V = × ∴
. . .
0191
8.45 Let St = 0.21 for the wind imposed vorticies. When this frequency equals the
natural frequency, or one of its odd harmonics, resonance occurs:
f T L d
= / ρ π
2 2
2 2
0.21 10
30 000/7850 0.016 . 0.525 m
0.016
L L
π
×
= × × ∴ =
Consider the third and fifth harmonics:
f T L d
= 3 2 2
/ .
ρ π ∴ =
L 1.56 m.
f T L d
= 5 2 2
/ .
ρ π ∴ =
L 2.62 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
/
.
. . . . .
=
×
×
= × = × × × × × ×





 =
−
88 6 12
1 6 10
2 8 10
1
2
00238 88 1 0 8 6
6
12
4
5 2
22 lb.
FD
The coefficient 1.0 comes from Fig. 8.8 and 0.8 from Table 8.1.
& .
W F V
D
= × = × =
22 88 1946 ft - lb /sec or 3.5 Hp
( )
C F W
D D
streamlined
lb.
ft - lb
sec
or 0.12 Hp
= ∴ = =
0 035 0 77 67 8
. . . & . .
8.48 Re
.
.
.
= =
×
×
=
−
VD
ν
3 08
1 5 10
16
5
000 ∴ = × × × × × × =
FD
1
2
122 3 008 2 12 78 0822
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 0 24
0822 0 24
0822
100 708%

131. 172
8.49 Re
.
. .
= =
×
= × ∴ =
−
VD
CD
ν
2 0 8
10
1 6 10 0 45
6
6
. from Fig. 8.8.
L
D
CD
= = ∴ = × =
4
0 8
5 0 62 0 45 0 28
.
. . . . .
Because only one end is free, we double the length.
F V AC
D D
= = × × × × × =
1
2
1
2
1000 2 08 2 028
2 2
ρ . . .
900 N
If streamlined, CD = × =
0 03 0 62 0 0186
. . . .
∴ = × × × × × =
FD
1
2
1000 2 0 8 2 0 0186
2
. . .
60 N
8.50 V = × =
50 1000 3600 13 9
/ . m /s.
Assume the ends to not be free. ∴Use CD from Fig. 8.8.
( )
Re
. .
.
. . . . .
=
×
×
= × ∴ = =
−
13 9 0 02
1 5 10
1 85 10 1 2 0 3
5
4
streamlined
C C
D D
& . . . . .
W F V V AC
D D
= × = = × × × × × =
1
2
1
2
1 2 13 9 0 02 20 1 2 773
3 3
ρ W or 1.04 Hp
& . . . .
Wstreamlined W or 0.26 Hp
= × × × × × =
1
2
1 2 13 9 0 02 20 0 3 193
3
8.51 V = × =
50 1000 3600 139
/ . m / s. Re
. .
.
. . .
=
×
×
= × ∴ =
−
13 9 0 3
1 5 10
2 8 10 0 4
5
5
CD
We assumed a head diameter of 0.3 m and used the rough sphere curve.
FD = × × × × =
1
2
12 139 03 4 04
2 2
. . ( . / ) . .
π 3.3 N
FD = × × × × =
1
2
12 139 03 4 0035
2 2
. . ( . / ) . .
π 0.29 N
8.52 σ
ρ
γ
=
−
=
−
×
= + =
∞
∞
p p
V V
p h p
v
1
2
0 7
150
1000
150
2 2
. .
000 1670
1
2
where 000 Pa.
atm
∴ =
V 20.6 m /s.
8.53 C
F
V A
L
L
= =
× × × ×
= ∴ ≅
1
2
200
1000 12 4 10
0 69 3
2 2
ρ
α
000
1
2
.
. . .
o
C
F
F
D
D
D
= =
× × × ×
∴ = N
.
.
. .
0165
1
2
1000 12 4 10
4800
2

132. 173
σcrit
000) 1670
1
2
= >
× + −
× ×
=
.
? ( .
. .
75
9810 4 101
1000 12
143
2
∴No cavitation.
8.54 C
F
V A
L
L
= =
× × × ×
= ∴ =
000
1
2
1
2
50
194 35
16
12
30
105 7 3
2 2
ρ
α
.
. . . .
o
C
F
F
D
D
D
= =
× × × ×
∴ = lb
.
.
. .
027
1
2
194 35
16
12
30
1280
2
σcrit
1
2
= >
× + − ×
× ×
=
16
62 4 16 12 2117 25 144
194 35
182
2
.
? . / .
.
. . ∴No cavitation.
8.55 p pv
∞ = × + =
×
= ×
9810 5 101 1670 16 106
000 = 150 000 Pa. Pa. Re =
20 .8
10-6
.
σ σ
=
−
× ×
= ∴ = + = + =
150
1000 20
0 74 0 1 3 1 74 52
2
000 1670
1
2
. . ( )( ) . ( . ) .
C C
D D
∴ = = × × × × × =
F V AC
D D
1
2
1
2
1000 20 4 52
2 2 2
ρ π . . .
52 000 N
Note: We retain 2 sig. figures since CD is known to only 2 sig. nos.
8.56 For a 6° angle of attack we find from Table 8.4 CL = 0 95
. .
F V AC L
L L
= = × × × × × = ×
1
2
1
2
1000 15 4 0 4 95 12 98
2 2
ρ . . . .
000
∴ =
L 069
. .
m
8.57 ΣF ma a a
= − × × = ∴ =
. .
.
. .
a) 1.75 m /s2
400 9810
4
3
2
400
9 81
3
π
b) 400 9810
4
3
2
400
9 81
1
2
1000
4
3
2
3 3
− × × = + × × ×





 ∴ =
π π
.
.
. . .
a a 1.24 m /s
2
8.58 F ma V a a
F
V
m V
a
= = × × − ∴ =
−
= × −
1 1 1
1000 1 2
1200
0 2 1000
. . . . .
F m m a a
F
V V
F
V
a
a
= + ∴ =
− + −
=
−
( ) . .
2 2 2
1200 200 1400
is true acceleration.
∴
−
× = −
−
−
−
× =
% . .
error =
a a
a
F
V
F
V
F
V
2 1
2
100 1400 1200
1400
100 16 7%

133. 174
8.59 (B) 1
2
2
From Fig. 8.12a 1.1. .
L
L L
F
C C
V cL
ρ
= =
2 2 2 1200 9.81
1088. 33.0 m/s.
1.23 16 1.1
L
W
V V
cLC
ρ
× ×
∴ = = = ∴ =
× ×
8.60 C
F
V A
C
L
L
D
= =
×
× × ×
= ∴ = =
1
2
1000 9 81
412 80 15
0 496 3 2 0065
2 2
ρ
α
.
.
. . . . . .
1
2
o
& . . .
W F V
D
= = × × × ×





 × =
1
2
412 80 15 0065 80 10
2
300 W or 13.8 Hp
8.61 a) C
V
V
L = =
× +
× × ×
∴ =
122
1500 981 3000
1
2
122 20
2
.
.
.
. .
34.5 m / s
b) ( )
C
V
V
L max
50 m /s
= =
× +
× × ×
∴ =
1 72
1500 9 81 3000
1
2
412 20
2
.
.
.
. . (at 10 000 m)
c) & . .
W F V
D
= = × × × ×





 × =
1
2
412 80 20 0065 80 13
2
700 W or 18.4 Hp
where we found CD as follows:
( )
C C
L D
cruise
=
× +
× × ×
= ∴ =
1500 9 81 3000
1
2
412 80 20
67 0065
2
.
.
. . . , from Fig. 8.12.
∴Power =
184
045
409
.
.
. .
= Hp
8.62 C
V
V
L = =
× +
× × ×
∴ =
122
1500 9 81 3000
1
2
1007 20
2
.
.
.
. .
38.0 m / s
8.63 ( )
C C
L D
cruise
=
× +
× × ×
= ∴ = =
1500 9 81 3000
1
2
1 007 80 20
0 275
0 275
48
0 0057
2
.
.
. .
.
. .
∴ = = × × × × =
& . .
W F V
D
1
2
1 007 80 20 0 0057 29
3
400 W or 39.4 Hp
% change =
39 4 18 4
18 4
100 114%
. .
.
−
× = increase
The increased power is due to the increase in air density.

134. 175
8.64 C
V
V
L = =
× +
× × ×
∴ =
122
1500 981 9000
1
2
122 20
2
.
.
.
. .
39.9 m / s
8.65 C
V
V
L = =
×
× × × ×
∴ =
172
250 981
1
2
122 60 8
2
.
.
.
. .
000
69.8 m / s
8.66 a) C
V
V
L = =
×
× × × ×
∴ =
172
250 981
1
2
105 60 8
752
2
.
.
.
. . .
000
m / s
% change =
75 2 69 8
69 8
100 7 77%
. .
.
.
−
× = increase
b) C
V
V
L = =
×
× × ×
∴ = =
×
=






172
250 9 81
1
2
1515 60 8
62 6
1013
287 233
1515
2
.
.
.
. .
.
.
.
000
m / s kg / m3
ρ
% change =
62 6 69 8
69 8
100 10 3%
. .
.
.
−
× = −
c) C
V
V
L = =
×
× × ×
∴ = =
×
=






172
250 9 81
1
2
1093 60 8
737
1013
287 323
1093
2
.
.
.
. .
.
.
.
000
m / s kg / m3
ρ
% change =
73 7 69 8
69 8
100 5 63%
. .
.
.
−
× = increase
8.67 For a conventional airfoil assume C C C
L D L
/ . . .
= =
47 6 0 3
at
0 3
9 81
1
2
0 526 222 200 30
2 38 10
2
6
.
.
.
. .
=
×
× × × ×
∴ = ×
m
m kg
& .
.
.
W F V
D
= = × × × × × =
1
2
0 526 222 200 30
0 3
47 6
490
3
000 W or 657 Hp
8.68
v
v
v v v
v
v
∇ × + ⋅∇ +
∇
− ∇





 =
∂
∂ ρ
ν
V
t
V V
p
V
( ) .
2
0
v
v
v v v v
v
v v
∇ × = ∇ × = ∇ ×
∇
= ∇ × ∇ =
∂
∂
∂
∂
∂ω
∂ ρ ρ
V
t t
V
t
p
p
( ) . .
1
0
v v v v v
∇ × ∇ = ∇ ∇ × = ∇
( ) ( )
2 2 2
V V ω (we have interchanged derivatives)
2 2
1 1
( ) ( ) ( )
2 2
V V V V V V
 
 
∇ × ⋅∇ = ∇ × ∇ − × ∇× = ∇ × ∇
   
 
v v v v v v v v v v v
( )
V ω
− ∇ × ×
v v v
( )
V ω
= ∇ ⋅
v v v
( )
V
ω
− ∇ ⋅
v v
v
( ) ( )
V V
ω ω
+ ⋅∇ − ⋅∇
v v v v
v v

135. 176
, . .
x L u U U
y
∂ψ
∂
= = ∴ = = ⋅ ∇ − ⋅∇ ∇ ⋅ = ∇ ⋅ ∇ × =
( ) ( ) ( ) .
v v v v v v v v v v v
V V V
ω ω ω
where 0
There results:
∂ω
∂
ω ω ν ω
v v v v v v v v
t
V V
+ ⋅∇ − ⋅∇ − ∇ =
( ) ( ) .
2
0
This is written as
D
Dt
V
v
v v v v
ω
ω ν ω
= ⋅∇ + ∇
( ) .
2
8.69 x-comp:
∂ω
∂
∂ω
∂
∂ω
∂
∂ω
∂
ω
∂
∂
ω
∂
∂
ω
∂
∂
ν ω
x x x x
x y z x
t
u
x
v
y
w
z
u
x
u
y
u
z
+ + + = + + + ∇2
y-comp:
∂ω
∂
∂ω
∂
∂ω
∂
∂ω
∂
ω
∂
∂
ω
∂
∂
ω
∂
∂
ν
∂ ω
∂
∂ ω
∂
∂ ω
∂
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
+ + + = + + + + +








2
2
2
2
2
2
z-comp:
∂ω
∂
∂ω
∂
∂ω
∂
∂ω
∂
ω
∂
∂
ω
∂
∂
ω
∂
∂
ν
∂ ω
∂
∂ ω
∂
∂ ω
∂
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
+ + + = + + + + +






2
2
2
2
2
2
8.70 x
w
y
∂
ω
∂
=
v
z
∂
∂
− 0. y
u
z
∂
ω
∂
= =
w
x
∂
∂
− 0. 0.
z
v u
x y
∂ ∂
ω
∂ ∂
= = − ≠
2 2
z
( ) ; .
z
z z
D D
w
DT Dt
ω ω
ω ν ω ν ω
= ⋅∇ + ∇ ∴ = ∇
v
v
If viscous effects are negligible, then
D
Dt
z
ω
= 0.
Thus, for a planer flow, ω z = const if viscous effects are negligible.
8.71 a)
v v
∇ × = −





 + −





 + −





 =
V
w
y
v
i
u
z
w
x
j
v
x
u
y
k
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
z
$ $ $ .
0 ∴irrotational
∂φ
∂
φ
x
x x f y
= ∴ = +
10 5 2
. ( )
∂φ
∂
∂
∂
y
f
y
y f y C C
= = ∴ = + =
20 10 0
2
. . .
Let ∴ = +
φ 5 10
2 2
x y
b)
v v
∇ × = + + − =
V i j k
0 0 8 8 0
$ $ ( )$ . ∴irrotational
∂φ
∂
φ
x
y xy f y z
= ∴ = +
8 8
. ( , ).
∂φ
∂
∂
∂
∂
∂
y
x
f
y
x
f
y
f f z
= + = ∴ = =
8 8 0
. ( ).
and
∂φ
∂z
df
dz
z f z C C
= = − ∴ = − + =
6 3 0
2
. . .
Let

136. 177
∴ = −
φ 8 3 2
xy z
c)
v v
∇ × = + +
− +
+
−
− +
+










=
− −
V i j
y x y x
x y
x x y y
x y
k
0 0
1
2
2
1
2
2
0
2 2 1 2
2 2
2 2 1 2
2 2
$ $
( ) ( )
$ .
/ /
∴irrotational
∂φ
∂
φ
x
x
x y
x y f y
=
+
∴ = + +
2 2
2 2
. ( )
∂φ
∂
∂
∂
∂
∂
y
x y y
f
y
y
x y
f
y
f C C
= + + =
+
∴ = ∴ = =
−
1
2
2 0 0
2 2 1 2
2 2
( ) . . . .
/
Let
∴ = +
φ x y
2 2
d)
v v
∇ × = + +
−
+
−
−
+





 =
V i j
y x
x y
x y
x y
k
0 0
2 2
0
2 2 2 2 2 2
$ $ ( )
( )
( )
( )
$ . ∴irrotational
∂φ
∂
φ
x
x
x y
n x y f y
=
+
∴ = + +
2 2
2 2
1
2
. ( ) ( )
l
∂φ
∂
∂
∂
∂
∂
y
y
x y
y
x y
f
y
f
y
f C C
=
+
=
+
+ ∴ = ∴ = =
2 2 2 2
1
2
2
0 0
. . . .
Let
∴ = +
φ ln x y
2 2
8.72
∂ ψ
∂
∂ ψ
∂
2
2
2
2
0
x y
+ = . This requires two conditions on x and two on y.
At , . .
x L u U U
y
∂ψ
∂
= − = ∴ =
At , . .
x L u U U
y
∂ψ
∂
= = ∴ =
At y h
= − , .
= 0
ψ
At , = h.
y h U
ψ
= × (See Example 8.9).
The boundary conditions are stated as:
∂ψ
∂
∂ψ
∂
ψ ψ
y
L y U
y
L y U x h x h Uh
( , ) , ( , ) , ( , ) , ( , ) .
− = = − = =
0 2
8.73 u
y
y f x v
x
df
dx
f x C
= = ∴ = + = − = − = ∴ = − +
∂ψ
∂
ψ
∂ψ
∂
100 100 50 50
. ( ). . .
∴ = −
ψ( , ) .
x y y x
100 50 (We usually let C = 0.)
u
x
x f y v
y
df
dy
f y C
= = ∴ = + = = = ∴ = +
∂φ
∂
φ
∂φ
∂
100 100 50 50
. ( ). . .
∴ = +
φ( , ) .
x y x y
100 50
y = h
y = 0
x = −L
y
x
U

137. 178
8.74 a) ψ θ
= 40 .
b)
1 1 1
40
1
0 0
2
r r r r r r r
∂
∂
∂ψ
∂θ
∂ ψ
∂θ∂
∂
∂
∂
∂θ





 + −





 = + −





 =
( ) ( ) .
∴It is incompressible since the above continuity equation is satisfied.
Note: The continuity equation is found in Table 5.1.
c)
∂φ
∂
∂ψ
∂θ
φ θ
r r r
nr f
= = ∴ = +
1 40
40
. ( )
l
∂φ
∂θ
∂
∂θ
∂ψ
∂
= = − = ∴ = =
f
r
r
f C C
0 0
. . .
Let
∴ =
φ 40ln r
d) v
r
v a v
v
r r r
r r r
r
= = = = −





 = −
40
0
40 40
10
2
, . .
θ
∂
∂
∴ =
r 5 43
. m
8.75 u
y
y
x y x
y
x
f y
= =
+
= ∴ = − +
−
∂ψ
∂
∂φ
∂
φ
20
2
40
2 2
1
. tan ( ).
v
y
x
y x
f
y
x
x y
f
y
x
x y
f C C
= = −
+
+ = −
+
+ = −
+
∴ = =
∂φ
∂
∂
∂
∂
∂
40
1
40
20
2
0
2 2 2 2 2 2
/
/
. . .
Let
φ = −
−
40
1
tan .
y
x
8.76 a)
∂ ψ
∂
∂ ψ
∂
∂ψ
∂
2
2
2
2
2 2 2
0 10 2
x y x
y x y x
+ = = + −
. ( ) ( ).
∂ ψ
∂
2
2
2 2 2
20
x
y x y
= − −
( ) − + −
80 2 2 2 3
x y x y
( )
∂ψ
∂y
x y y x y y
= − + + +
− −
10 10 10 2
2 2 1 2 2 2
( ) ( ) ( ).
∂ ψ
∂
2
2
2 2 2
20
y
y x y
= + −
( ) + + − +
− −
40 80
2 2 2 3 2 2 3
y x y y x y
( ) ( ) .
∴ + =
+
−
+
+
+
−
+
∂ ψ
∂
∂ ψ
∂
2
2
2
2 2 2 2
2
2 2 3 2 2 2
3
2 2 3
20 80 60 80
x y
y
x y
x y
x y
y
x y
y
x y
( ) ( ) ( ) ( )
=
+
+
−
+
−
+
=
+ − −
+
=
80 80 80 80 80 80 80
0
2 2
2 2 3
2
2 2 3
3
2 2 3
2 3 2 3
2 2 3
y x y
x y
x y
x y
y
x y
x y y x y y
x y
( )
( ) ( ) ( ) ( )
.
b) In polar coord: ψ θ θ
θ
θ θ
( , ) 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
∂ψ
∂θ
θ
∂φ
∂
φ θ θ
= −





 = ∴ = +





 +
cos . cos ( ).

138. 179
2 2
1 1 10 10
10 sin 10sin sin . 0.
df df
f C
r r d r d
r r
∂φ ∂ψ
θ θ θ
∂θ θ ∂ θ
 
= − + = − = − − = =
 
 
.
∴ = +





 = +
+
φ θ φ
10
1
10
10
2 2
r
r
x y x
x
x y
cos ( , ) ,
or
where we let r x r x y
cos .
θ = = +
and 2 2 2
c) Along the x-axis, v
x
= − =
∂ψ
∂
0 where we let y = 0 in part (a) and
u
y x y
y
x y x
y
= = −
+
+
+
= − =
∂ψ
∂
10
10 20
10
10
0
2 2
2
2 2 2 2
( )
.
with
Euler’s Eq: ρ
∂
∂
∂
∂
ρ
∂
∂
u
u
x
p
x x x
p
x
= − ∴ −











 = −
. .
10
10 20
2 3
∴ = −





 = − +





 + =
∫
p
x x
dx
x x
C C
ρ ρ
200 200 50 100
50
5 3 4 2
. 000.
= −





 +
1000
100 50
50
2 4
x x
000 Pa. (Could have used Bernoulli!)
d) Let u
x
x
= = − ∴ = ± ∴
0 0 10
10
1
2
: . . Stag pts: (1, 0), (−1, 0)
8.77 a)
∂ φ
∂
∂ φ
∂
∂
∂
∂
∂
2
2
2
2 2 2 2 2
2 2
2 2 2
10
10 10 10 2
x y x
x
x y y
y
x y
x y x x
x y
+ = +
+





 +
+





 =
+ −
+
( )10 (
( )
+
+ −
+
=
+ − + + −
+
=
( )10 ( )
( ) ( )
.
x y y y
x y
x y x x y y
x y
2 2
2 2 2
2 2 2 2 2 2
2 2 2
10 2 10 10 20 10 10 20
0
b) Polar coord: φ θ
= +
10 5 2
r n r
cos .
l (See Eq. 8.5.14.)
∂φ
∂
θ
∂ψ
∂θ
ψ θ θ
r
r
r r
r f r
= + = ∴ = + +
10
10 1
10 10
2
cos . sin ( )
1
10sin 10sin .
df
f C
r r dr
∂φ ∂ψ
θ θ
∂θ ∂
= − = − = − − ∴ = . 10 sin 10 .
r
ψ θ θ
∴ = +
∴ = +
−
ψ( , ) tan .
x y y
y
x
10 10
1
c) v
y
y
x y
= =
+
∂φ
∂
10
2 2
. Along x-axis (y = 0) v = 0.
u
x
x
x y
= = +
+
∂φ
∂
10
10
2 2
. Along x-axis u
x
= +
10
10
.
Bernoulli:
2
2
V p
gz
ρ
+ +
2
2
V p
gz
ρ
∞ ∞
∞
= + + (assume )
z z∞
=
( / )
. .
10 10
2
10
2
100 50
2 1
2 2
2
+
+ = + ∴ = − +






x p
p
x x
ρ ρ
100 000
kPa

139. 180
d) u
x
x
= = + ∴ = − ∴
0 0 10
10
1
: . . Stag pt: (−1, 0)
e) y
a v
=
v v
u
y x
∂ ∂
∂ ∂
+ 0 on axis. x
u
x a u v
x
∂
∂
= − = +
2
10 10
10 .
u
y x x
∂
∂
 
 
= + −
  
  
2
10
( 2,0) (10 5) 12.5 m/s .
4
x
a
 
∴ − = − − = −
 
 
8.78
2 3
2 5
( , ) 5 .
2 3
y y
u x y y y C
y
∂ψ
ψ
∂
= − = ∴ = − + 2 3
1
. (3 10 ).
6
y y
ψ
∴ = −
q udy y y dy
= = − = − = × −
∫
∫ ( )
. .
. .
.
.
5
02
2
5
0 2
3
6 667 10
2
2 3
3
0
2
0
2
m / s
2
ψ ψ
2 1
2 3 3
1
6
3 0 2 10 02 0 6667 10
− = × − × − = × −
( . . ) . .
m / s
2
ω
∂
∂
= − = − + ≠
u
y
y
1 10 0. ∴φ doesn’t exist.
8.79 ψ
π
π
θ θ θ
= + = +
30
5
2
30
5
2
y r sin .
a) v
r r
r = = + =
1
30
5
2
0
∂ψ
∂θ
θ
cos .
At θ π
= = ∴ =
, . . '.
5
2
30 0 0833
r
r
s
s
Stag. pt: ( " , ).
−1 0
b) At θ π ψ
π π π
= = = +
, sin .
=.0833,
r r
s
5
2
30
2
5
2 2
∴ = =
r yinter .0119 ft.
c) q U H H H
= × = ∴ = ∴
∆ψ
π π
. . .
=
5
60
30
5
2
Thickness = 2
5
30
H =
π
ft or 1.257".
d) v u
r ( , ) cos . . . . .
1 30
5
2
30 2 5 27 5 27 5
π π
= + = − + = − ∴ = fps
8.80 [ ] [ ] [ ]
φ
π
π
π
π
= + + − − + + = + +
2
1
2
1 10
1
4
1
2 1 2 2 2 1 2 2 2
l l l
n x y n x y x n x y
( ) ( ) ( )
/ /
[ ]
− − + +
1
4
1 10
2 2
ln x y x
( ) .
u
x
x
x
x
x x x
v y
y
= =
+
+
−
−
−
+ =
+
−
−
+ = =
=
∂φ
∂ 0
2 2
1
4
2 1
1
1
4
2 1
1
10
1
2 1
1
2 1
10 0 0
[ ( )]
( )
[ ( )]
( ) ( ) ( )
. .
if
30 fps
y
x
= 0
= 5π/2

140. 181
At the stagnation point, u
x x x
= ∴
+
−
−
+ = ∴
−
=
0
1
2 1
1
2 1
10 0
2
1
20
2
.
( ) ( )
. .
∴ = ∴ = ± ∴ ×
x x
2
1 1 1 049
. . . .
m. oval length = 2 1.049 = 2.098 m
All the flow from the source goes to the sink, i.e., π
π
m /s, or m / s for
2 2
2
0
y > .
u y
x y y y
x
( )
( ) ( )
.
= =
+
−
−
+
+ =
+
+
=
∂φ
∂ 0
2 2 2
1
4
2
1
1
4
2
1
10
1
1
10
1
2
0
1
10 . tan 10 .
2 2
1
h
q dy h h
y
π π
−
 
= + = ∴ + =
 
∫  
+
 
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
1 143
10 10 98
2
.
. m /s.
Bernoulli:
V p V p p
2 2 2 2
2 2
10 98
2 1000
10
2
+ = + + = +
∞ ∞
ρ ρ
.
.
.
10 000
1000
Pa
min
∴ = −
p 280 .
8.81 [ ] [ ] [ ]
φ
π
π
π
π
= + + +
−
− + + = + +
2
2
1
2
2
1 2
1
2
1
2 2 1 2 2 2 1 2 2 2
l l l
n x y n x y x n x y
( ) ( ) ( )
/ /
[ ]
− − + +
1
2
1 2
2 2
ln x y x
( ) .
u
x
x
x y
x
x y
v
y
x y
y
x y
= =
+
+ +
−
−
− +
+ =
− +
−
− +
∂φ
∂
1
2
2 1
1
1
2
2 1
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.: ( , ), ( , ).
2 0 2 0
−
u v
( , ) . ( , ) .
4 0
1
4 1
1
4 1
2 4 0 0
=
− +
−
− −
+ = − =
1.867 m / s
u v
( , ) . ( , ) .
0 4
1
1 4
1
1 4
2 0 4
4
1 4
4
1 4
0
2 2 2 2
=
+
−
−
+
+ = =
+
−
+
=
2.118 m / s
8.82 φ
π
π
π
π
= + − + + +
2
2
1
2
2
1
2 2 1 2 2 2 1 2
l l
n x y n x y
[ ( ) ] [ ( ) ]
/ /
= + − + + +
1
2
1
1
2
1
2 2 2 2
l l
n x y n x y
[ ( ) ] [ ( ) ].
(0, h)
y
x

141. 182
u
x
x
x y
x
x y
= =
+ −
+
+ +
∂φ
∂ 2 2 2 2
1 1
( ) ( )
.
v
y
y
x y
y
x y
= =
−
+ −
+
+
+ −
∂φ
∂
1
1
1
1
2 2 2 2
( ) ( )
.
At (0, 0) u = 0 and v = 0. At (1, 1)
2 2 2 2 2
2 2 1 1
0 0.4 m/s. 1.2 m/s.
5
2 1 1 2 1
v u
= + = = = + =
+ +
∴ = +
v
V i j
12 04
. $ . $ .
m/ s
8.83 φ
π
π
π
π
= − + + + + + ∞
2
2
1
2
2
1
2 2 1 2 2 2 1 2
l l
n y x n y x U x
[( ) ] [( ) ] .
/ /
= − + + + + + ∞
1
2
1
1
2
1
2 2 2 2
l l
n y x n y x U x
[( ) ] [( ) ] .
a) Stag. pts. May occur on x-axis, y =0.
u
x
x
x
x
x
y
= =
+
+
+
+
=
∂φ
∂ 0
2 2
1 1
10.
2
0.2 1 0.
x x
∴ + + = ∴no stagnation points exist on the x-axis.
(They do exist away from the x-axis.)
Along the y-axis: u y q udy
h
( ) . ( )
= = = =
∫
10
1
2
2
0
m / s.
2
π π
∴ = = ∴ =
∫
π 10 10 0314
0
dy h h
h
. m
. .
b) u
x
x
x x x
=
+
+ ∴ + + = ∴ = −
2
1
1 2 1 0 1
2
2
. . m.
Stag. pt.: (−1, 0)
Along the y-axis: u h h
= ∴ = × ∴ =
1 0 1 3 14
. . . .
. m
π
c) u
x
x
x x x
=
+
+ ∴ + + = ∴ = − −
2
1
02 10 1 0 9 90
2
2
. . . . , 0.10 m.
Stag. pts.: (−9.9, 0) , (−0.1, 0).
Along the y-axis: u h h
= ∴ = ∴ =
0 2 02 1571
. . . . .
. m
π
8.84 φ θ θ
= +
60
8
r
r
cos cos .
a) v
r r r
r = = − + = −






∂φ
∂
θ θ θ
60
8 8
60
2 2
cos cos cos .
At the cylinder surface vr = 0 for all θ. Hence,
x
y
x
y
x
y

142. 183
60
8 2 739
2
r
r
c
c
= ∴ =
. . m
b) Bernoulli: ∆p
U
= = =
∞
ρ
2 2
2
1000
8
2
32 000 Pa or 32 kPa
c) v
r r
θ
∂φ
∂θ
θ θ
= = − −
1 60
8
2
sin sin .
At r r v
c
= = − − = −
, sin sin sin
θ θ θ θ
8 8 16
d) ∆p
v
= = =
ρ 90
2 2
2
1000
16
2
128
o
000 Pa or 128 kPa
8.85 ψ
π
π
θ
π
π
θ
= + = +
4
2
20
2
2 10
l l
n r n r
At ( , ) ( , ), ( , ) ( , / ).
x y r
= =
0 1 1 2
θ π
v
r
r ( , / ) ( ) .
1 2
1 1
1
2 2
π
∂ψ
∂θ
= = =
v
r
θ π
∂ψ
∂
( , / ) .
1 2
10
1
10
= − = − = −
vr ( . , / )
.
.
1 7 4
2
1 7
1 18
π = = , vθ π
( . , / )
.
.
1 7 4
10
1 7
5 88
=
−
= −
vr ( . , )
.
.
3 2 0
2
3 2
0 625
= = , vθ ( . , )
.
.
3 2 0
10
3 2
3 125
=
−
=
vr ( , / ) .
6 4
2
6
0 333
− = =
π , vθ π
( , / ) .
6 4
10
6
1 67
− =
−
= − , etc.
Note: We scaled the radius at each 45° increment to find r.
b) v
r
v
r
r = = −
2 10
and θ . 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
= + = −
θ θ
∂
∂
2 2
= −





 − = −
2 2 100 104
1
2 3
r r r
= −104 m /s2
a
Dv
Dt
v v
r
v
v
r
v v
r
r
r
r
θ
θ θ θ θ
∂
∂
= + = + =





 +
−
=
2 10 2 10
0
2 3
r r r
( )
.
∴ = −
v
a( , ) ( , )
0 1 104 0 m /s2
x
y

143. 184
c) v v
r ( . , / ) / . . , ( . , / )
.
.
14 14 4 2 14 14 0 1414 14 14 4
10
14 14
0 707
π π
θ
= = = − = −
v v
r ( . , / ) / . , ( . , / ) / . .
0 1 2 2 0 1 20 0 1 2 10 0 1 100
π π
θ
= = = − = −
Bernoulli:
20 000
1.2
13 760 Pa
+
+
= +
+
∴ =
0 1414 0 707
2 1 2
20 100
2
2 2 2 2
. .
.
.
p
p
We used ρair
3
kg / m at standard conditions.
= 1 2
.
8.86 Along the y-axis v v
r
r = = − −
0 10
40
2
and θ .
We have set θ
π
=
2
in Eq. 8.5.27. rc = =
40
10
2.
b) v
r
r = − − ⇒
10
40
4 3 5 126 9
2
cos cos . ( , ) ( , . ).
θ θ o
v
r
v v
r
θ θ
θ θ
= − − ∴ = − = −
10
40
6 96 9 28
2
sin sin . . , . .
m /s m / s
c) Use Eq. 8.5.28: p p U
= − ∞
0
2 2
2ρ α
sin
Drag = p r d L p r L p p U
c c
cos . .
/
/
α α ρ
π
π
− × = − ∞
−
∫ 90 90 0
2
2
2
2 2
= − − ×
∞
∫
2 2 2
0
2 2
90
0
2
( sin )cos
/
p U r Ld p r L
c c
ρ α α α
π
[ ]
= −





 − − =
∞ ∞ ∞
2 2
3
2 2
8
3
0
2
3
0
2
0
2 2
r L p U p U r L r L U
c c c
sin
sin
.
/
α ρ
α
ρ ρ
π
C
U A
r L U
U r L
D
c
c
= = = =
∞
∞
∞
Drag
1
2
8 3
1
2
2
8
3
2 667
2
2
2
ρ
ρ
ρ
( / )
. .
8.87 v U
r
U r U
r c
= − = = = × =
∞ ∞ ∞
cos cos . , .
θ
µ
θ µ
2
2 2
4 1 4 4
Let
For θ π
= = − +
, .
v
r
r 4
4
2
b) v
r
U
rc
θ
∂ψ
∂
θ
µ θ
θ θ
= − = − − = − −





 = −
∞ sin
sin
sin sin .
2 2
4
4
1
8
c) p p
V v
c = + − = + × −
∞
∞
ρ ρ
θ
θ
2 2 2 2 2
2 2
1000
4
2
1000
8
2
50 000
sin
.
∴ = −
pc 58 32 2
sin θ kPa.
10 m/s
20 m/s
dα
p(α)
α
p90
x = −1
−x
−vr

144. 185
d) Drag = 2 58 32 1 1 26 2 1
2
0
2
( sin ) cos
/
− × × − × ×
∫ α α α
π
d
= −











 − =
2 58 32
1
3
52 42.7 kN. (See the figure in Problem 8.86c.)
8.88 On the cylinder
1000
2 sin 60sin ,
2 2 3.651
c
v U
r
θ θ θ
π π
∞
Γ
= − − = − −
×
where we have
Used
400
3.651 ft.
30
c
r
U
µ
∞
= = =
If
2 2 2 2 2 2 2 2
6 6 6 6
( , ) .0318
( 6) ( 2) ( 6) ( 2) ( 6) ( 2) ( 6) ( 2)
x x x x
u x y
x y x y x y x y
 
− − + +
= − + + +
 
− + − − + + + + − + + +
 
 
227 , 313 .
θ
∴ = o o
Stag. pts.: (3.651 ft, 227°) , (3.651 ft, 313°).
Max. pressure occurs on the cylinder at a stagnation pt.:
2 2 2 2
max o
0.0024
= 30 0 1.08 psf.
2 2
p U v
ρ
∞
   
∴ = − − =
   
Min. pressure occurs at the top of the cylinder where θ = 90o
and the velocity is:
90
o
1000
2 sin 2 30 104 fps
2 2 3.651
v U
r
θ
π π
∞
Γ
= − − = − × − =
×
2 2 2 2
min o
0.0024
= 30 104 11.8 psf.
2 2
p U v
ρ
∞
   
∴ = − − = −
   
8.89 vθ θ
π
= − × −
×
2 20
2 4
sin
.
.
Γ
For one stag. pt.: vθ θ
= =
0 270
at o
:
0 2 20 270
2 4
2 20 2 4 100 5
= − × −
×
∴ = × × × =
sin
.
. . .
o Γ
Γ
π
π m / s.
2
Γ
Γ
= ∴ = =
×
=
2
2
100 5
2 4
2
2 2
π ω ω
π π
r
r
c
c
.
.
.
.
100 rad /s (See Example 8.12.)
Min. pressure occurs where vθ is max, i.e., θ π
= / .
2 There
vθ
π
= − × × −
×
=
2 20 1
100 5
2 4
80
.
.
m /s.
∴ = + − = + × − × = −
∞
∞
p p
V v
min Pa
2 2 2 2
2 2
0
20
2
1 22
80
2
1 22 3660
ρ ρ
θ
. . .

145. 186
8.90 Γ = = × × × = = = × =
2 2 6 120 2 60 28 42 6 3 1 08
2 2 2 2
π ω π π µ
r r U
c c
. / . . .
m /s. m /s.
2 3
∴ = − × −
×
∴ = −
vθ θ
π
θ
2 3
28 42
2 6
1256
sin
.
.
sin . .
. Impossible. ∴Stag. pt. is off the
cylinder at θ = >
270o
, .
but r rc From Eq. 8.5.29,
v
r
U
r r r r
θ
∂ψ
∂
θ
µ
θ
π π
= − = − − − = − − − − − =
∞ sin sin ( )
.
( )
.
.
2 2
2
3 1
1 08
1
28 42
2
0
Γ
∴ + = ∴ − + = ∴ =
3
1 08 4 523
1 508 0 36 0 1 21
2
2
. .
. . . . .
r r
r r r m.
Stag. pt.: (1.21, 270°). ( )
.
.
.
vθ
π
90
2 3
2842
2 6
1354
o = − × −
×
= − m / s.
Min. pressure occurs at θ = = = −





 = −
90
3
2
1354
2
122 106
2 2
o
, :
.
. .
at Pa
min
r r p
c
Max. pressure occurs at θ = = = −





 = −
270
3
2
154
2
122 4 04
2 2
o
, :
.
. . .
at Pa
max
r r p
c
8.91 At 15,000 ft, ρ =. .
0015 slug / ft3
Lift = ρU L
∞ = × × × =
Γ . , .
0015 350 15 000 60 472,000 lb
8.92 Place four sources as shown. Then, with q = 2π for each:
u x y
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
( ) ( )
v x y
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 = 0 2
. m / s, as shown.
2
u x y
x
x y
x
x y
x
x y
x
x y
( , ) .
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
= −
−
− + −
+
−
− + +
+
+
+ + −
+
+
+ + +






0318
6
6 2
6
6 2
6
6 2
6
6 2
2 2 2 2 2 2 2 2
v x y
y
x y
y
x y
y
x y
y
x y
( , ) .
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
= −
−
− + −
+
−
+ + −
+
+
− + +
+
+
+ + +






0318
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
π π
=
−
= −
.
. .
x
y
x
y
(6, 2)

146. 187
At (4,3) u( , ) . . .
4 3 0318
2
4 1
2
4 25
10
100 1
10
100 25
0 00922
= −
−
+
+
−
+
+
+
+
+





 = m /s
v( , ) . . .
4 3 0318
1
4 1
1
100 1
5
4 25
5
100 25
0 01343
= −
+
+
+
+
+
+
+





 = − m /s
8.94 Re . / .
crit = ∴ = × =
∞
U x
x
T
T
ν
ν ν
6 10 300 2000
5
a) ν = × ∴ = × × =
− −
1 56 10 2000 1 56 10 0 312
4 4
. . . '
ft /sec. or 3.74"
2
xT
b) ν
µ
ρ
= = × ∴ = × × =
− −
2 1 10 2000 2 1 10 0 42
4 4
. . . '
ft /sec. or 5.04"
2
xT
c) ν = × ∴ = × × =
− −
3 47 10 2000 3 47 10 0 694
4 4
. . . '
ft /sec. or 8.33"
2
xT
8.95 a) Use Re / . .
crit = × = × −
3 10 10 1 51 10
5 5
xT ∴ =
xT 0 453
. .
m
b) Use Re / . .
crit = = × −
10 10 1 51 10
6 5
xT ∴ =
xT 1 51
. .
m
c) Use Re / . .
crit = × = × −
3 10 10 1 51 10
5 5
xT ∴ =
xT 0 453
. .
m
d) Use Re / . .
crit = × = × −
3 10 10 1 51 10
5 5
xT ∴ =
xT 0 453
. .
m
e) 4 5
growth growth
Re 6 10 10 /1.51 10 . 0.091 m or 9.1 cm
x x
−
= × = × ∴ = .
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 = × = −
3 10 10 10
5 6
xT ∴ =
xT 0 03 3
. .
m or cm
b) Use Re / .
crit = = −
10 10 10
6 6
xT ∴ =
xT 0 1 10
. .
m or cm
c) Use Re / .
crit = × = −
3 10 10 10
5 6
xT ∴ =
xT 0 03 3
. .
m or cm
d) Use Re / .
crit = × = −
3 10 10 10
5 6
xT ∴ =
xT 0 03 3
. .
m or cm
e) 2 2
( ) 20 000 2 1000 10 sin ( /2)
p x x
= − × ×
8.97 Re .
.
.
crit For a wind tunnel:
= × =
×
× =
×
×
∞ ∞
−
6 10
2
6 10
2
1 5 10
5 5
5
U U
ν
m / s.
∴ =
∞
U 4 5
.
For a water channel: 6 10
2
10
0 3
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
p p U r x
c
= − =
∞
0
2 2
2ρ α α α
sin (
where is zero at the stagnation point). Then
2 2
( ) 20 000 2 1000 10 sin ( /2)
p x x
= − × ×
= −
20 200 2
2
sin ( / )
x kPa
The velocity U(x) at the edge of the b.l. is U(x) on the cylinder wall:

147. 188
v r
θ θ θ π α α
( ) sin sin sin( ) sin
= = − − = − − =
2 10 10 20 20
∴ =
U x x
( ) sin( / )
20 2
8.99 U x v r v U x x x r
c c
( ) . sin . ( ) sin .
= = = ∴ = =
θ θ α α
at since
1 8 8
p x x
( ) sin sin
= − = −
58 32 58 32
2 2
α kPa
8.100 The height h above the plate is h x mx m m
( ) . . . . .
= + = × + ∴ = −
4 1 2 4 15
∴ = − × = ∴ =
−
h x x U x h U x
x
( ) . . . . ( ) . ( )
.
. .
0 4 15 6 4
2 4
0 4 15
Continuity: or
U x
x
( )
.
.
=
−
16
2 67
Euler’s Eqn: ρ
∂
∂
∂
∂
ρ
u
u
x
p
x
dp
dx x x
= − ∴ =
− −
.
. ( . )
16
2 67
16
2 67 2
=
−
256
2 67 3
( . )
.
x
8.101 a) top out in
0 0 0 0
m m m udy udydx udy udydx
x x
δ δ δ δ
ρ ρ ρ ρ
∂ ∂
= − = + − =
∫ ∫ ∫ ∫
∂ ∂
& & &
b) 0 ( ) ( )( )
2
x
dp
F p dx p d p dp d
δ τ δ δ δ
Σ = − + + − + +
0 higher order terms
dx dp
τ δ
= − − +
2 2 2
out in top
0 0 0 0
2
0 0
( )
( )
mom mom mom u dy u dydx u dy U x udydx
x x
u dydx U x udydx
x x
δ δ δ δ
δ δ
ρ ρ ρ ρ
ρ ρ
 
∂ ∂
− − = + − −
∫ ∫ ∫ ∫
 
∂ ∂
 
 
∂ ∂
= −
∫ ∫
 
∂ ∂
 
& & &
2 3
2 5 4 5 2
3 4 3 4
4 .
2
2 4.65 1.5 10 3 4.65 (1.5 10 3)
y d
y
dx
δ
− −
 
×
= − +
 
× × × × × × ×
 
 
8.102 τ δ ρ ρ
δ
δ
0
2
0
0
= − + − ∫
∫
dp
dx
U x
d
dx
udy
d
dx
u dy
( )
= − + −





 −
∫ ∫
∫
δ ρ ρ ρ
δ δ
δ
dp
dx
d
dx
uUdy
dU
dx
udy
d
dx
u dy
0
2
0
0
where we have used g
df
dx
dfg
dx
f
dg
dx
U g f udy
= − = =






∫
. , .
Here ρ
δ
0
∴ = − + − − =
∫
∫
τ δ ρ ρ ρ
δ
δ
0
0
0
dp
dx
d
dx
u U u dy
dU
dx
udy
( ) . ( const.)

148. 189
8.103
dp
dx
d
dx
U U
dU
dx
dU
dx
Udy U Udy
= − = − = −





 =
∫ ∫
ρ
ρ ρ
δ δ
δ δ
2
1 1
2
0 0
where .
∴ = − −














+ −
∫ ∫
τ δ ρ
δ
ρ θ ρ
δ δ
0
0
2
0
1
dU
dx
Udy
d
dx
U
dU
dx
udy
( )
= + − = +
∫
ρ θ ρ ρ θ ρ δ
δ
d
dx
U
dU
dx
U u dy
d
dx
U
dU
dx
U d
( ) ( ) ( ) .
2 2
0
8.104 If dp dx
dU
dx
d
dx
u U u dy
/ ( ) .
= = = −
∞
∫
0 0 0
0
then and τ ρ
δ
τ ρ
π
δ
π
δ
ρ
δ
π
π
δ
ρ
δ
π
δ
δ δ
0
2
0
2
0
2
2
1
2
2
2 2
2
2
= −





 = − −





 = −






∞ ∞ ∞
∫
d
dx
U
y y
dy U
d
dx
y y
U
d
dx
sin sin cos
τ µ
∂
∂
µ
π
δ
0
0
2
0
= =
=
∞
u
y
U
y
cos .
∴ = ∴ = ∴ =
∞ ∞
∞ ∞
µ
π
δ
ρ
δ
δ δ
ν
δ
ν
U U
d
dx
d
U
dx
x
U
2
137 11 5 4 79
2
. . . . . .
b) τ µ
π
ν
µ
ν
0
2
1
4 79
0 328
= =
∞
∞
∞
∞
U
U
x
U
U
x
.
. .
c)
∂
∂
∂
∂
π
ν
∂
∂
∂
∂
u
x
U
x
y U
x
U
x
a
x
U
ax a
x
v
y
=
×





 =





 = −





 = −
∞
∞
∞ ∞
−
sin
.
sin cos .
/
2 4 79 2
3 2
∴ =





 =














∞
∞ ∞
∞
∞ ∞
∫
∫
v U
y
x
U
y
U
x
dy U
U U
y dy
.
cos . . cos . .
/
164
328 0316 189
3 2
0
0
ν ν ν ν
δ
δ
8.105 u U
y d
dx
U
y y
dy
= = −






∞ ∞
∫
δ
τ ρ
δ δ
δ
. 0
2
0
1
= −





 =
∞ ∞
ρ
δ δ
ρ
δ
d
dx
U U
d
dx
2 2
2 3
1
6
.
τ µ
∂
∂
µ
δ
µ
δ
ρ
δ
δ δ
ν
0
2
1
6
6
= = ∴ = ∴ =
∞ ∞
∞
∞
u
y
U U
U
d
dx
d
U
dx
. .
∴ = = =
∞ ∞
∞
∞
δ
ν
δ
ν
τ µ
ν
2
0
12 3 46 0 289
U
x x
x
U
U
U
x
. ( ) . . . .
%error in δ( )
.
.
x =
−
× =
5 3 46
5
100 30 8% low.
U
δ u = Uy/δ
y

149. 190
%error in low
τ 0
332 289
332
100 13%
( )
. .
.
.
x =
−
× =
8.106 τ ρ
δ δ
ρ
δ δ
δ
δ
δ
0
2 2
6
2
0
6
3 1 3
1
3
1
1
3
= −





 + +





 − −











∞ ∞
∫
∫
d
dx
U
y y
dy U
y y
dy
/
/
/
+ +





 − −











∞
∫ρ
δ δ
δ
δ
U
y y
dy
2
2
3
2
3
1
3
2
3
/
= = ∴ =
∞
∞
∞
d
dx
U
U d
dx
U
ρ δ µ
δ
δ
δ
µ ρ
2
0 1358
3
22 08
( . ) . . / .
Thus, δ τ ρ ρ
( ) . / , ( ) .
.
. Re .
/
x vx U x U
v
U x
U x
= =





 =
∞ ∞
∞
∞
−
665 01358
665
2
0451
0
2 2 1 2
%error for δ =
−
× =
6 65 5
5
100 33%
.
. %error for τ 0
0 451 0 332
0 332
100 36%
=
−
× =
. .
.
8.107 Continuity from entrance to x: U H u y dy U x H
0
0
2 2
= + −
∫ ( ) ( )( ).
δ
δ
Write U x U x dy U x dy
( ) ( ) ( ) .
δ
δ
δ
= = ∫
∫ 0
0
Then, continuity provides
U H u U dy UH
0
0
2
= − +
∫( )
δ
= − −
∫
UH U u dy
2
0
( )
δ
= − ∴ =
−
UH U U x
U H
H
d
d
2
2
0
δ
δ
. ( ) .
If we were to move the walls out a distance δd x
( ), then U x
( ) would be constant
since ( )
[ ]
H d d
− +
2 2
δ δ would be constant; then U x U
( ) .
= 0 For a square wind
tunnel, displace one wall outward 4 0
δ d dp dx
for / .
=
8.108 The given velocity profile is that used in Example 8.13. There we found
δ ν
= = = = =
∞
−
5 48 5 48 10 10 0 00173 0 00173 3 0 003
6
. / . / . . .
x U x x m.
Assume the streamline is outside the b.l. Continuity is then
10 0 02 10
2
003 003
003 10
2
2
0
003
× = −





 + −
∫
.
. .
( . )
.
y y
dy h
= + − ∴ =
0 02 10 0 03 0 021
. . . .
h h m or 2.1 cm
[ ]
δ d
y y
dy
= − +





 = − + =
∫
1
10
10
20
003
10
003
1
10
03 03 01 0 001
2
2
0
003
. .
. . . .
.
m
h − = − =
2 2 1 2 0 1
. . cm or 0.001 m.
The streamline moves away from the wall a distance δ d .

150. 191
8.109 From Prob. 8.107 we found that we should displace the one wall outward 4δ d .
From the definition of δ d :
h x
y y
dy
d
( ) = = − +





 = − +





 =
∫
4
4
10
10
20 10
4
3
4
3
2
2
0
δ
δ δ
δ δ
δ
δ
δ
=
×
×







 =
−
4
3
5 48
1 86 10 10
160 287 303
0 00735
5
.
. /
/(. )
.
x
x m
We used δ( )
x found in Example 8.13, ρ ν µ ρ
= =
p RT
/ , / .
and
8.110 a) u U
y y
U
U
y y
dy
d
= −





 = − +





 = − + =
∞
∞
∞
∫
3
2
1
2
1
1
3
2
1
2
3
4
1
8
375
3
3
3
3
0
δ δ
δ
δ δ
δ δ δ δ
δ
. . .
From Eq. 8.6.16, δ
ν ν
d
x
U
x
U
= × =
∞ ∞
. . . .
375 4 65 1 74 %error = 1.2%.
θ
δ δ δ δ
δ
δ
= −





 − +





 =
∞
∞
∫
1 3
2
1
2
1
3
2
1
2
0 139
2
2
3
3
3
3
0
U
U
y y y y
dy . .
∴ = × =
−
× =
∞ ∞
θ
ν ν
. . . .
. .
.
.
139 4 65 0 648
648 644
644
100 0 62%
x
U
x
U
%error =
b) u U
y y y y
dy
d
= −





 = − +





 = − + =
∞ ∫
2 1
2
3
3
2
2
2
2
0
δ δ
δ
δ δ
δ δ
δ
δ
δ
. / .
See Example 8.13.
∴ = =
−
× =
∞ ∞
δ
ν ν
d
x
U
x
U
5 48
3
1 83
1 83 1 72
1 72
100 6 4%
.
. .
. .
.
.
%error = .
θ
δ δ δ δ
δ δ δ δ δ δ δ
δ
= −





 − +





 = − − + + − =
∫ 2 1 2
1
3
4
3
2
4
2
4
1
5
1333
2
2
2
2
0
y y y y
dy . .
∴ = × =
−
× =
∞ ∞
θ
ν ν
. . . .
. .
.
.
1333 5 48 0 731
731 644
644
100 13 5%
x
U
x
U
%error = .
c)
0
2
1 sin 0.363 . See Problem 8.104. 4.79 .
2
d
y x
dy
U
δ π δ ν
δ δ δ δ
δ π ∞
 
= − = − = =
∫  
 
∴ = × =
−
× =
∞ ∞
δ
ν ν
d
x
U
x
U
0 363 4 79 1 74
1 74 1 72
1 72
100 1 2%
. . . .
. .
.
.
%error =
θ
π
δ
π
δ
δ
π
π
δ
δ δ
π
δ
δ
δ
= −





 = − − + −





 = − + =
∫sin sin cos sin . .
y y
dy
y y
2
1
2
2
2 2 2
2
0 137
0
0
term
∴ = × =
−
× =
∞ ∞
θ
ν ν
. . . .
. .
.
. .
137 4 79 0 654
654 644
644
100 1 6%
x
U
x
U
%error =

151. 192
8.111 a) δ
ν
= =
× ×





 =
∞
−
4 65 4 65
1 6 10 20
12
0 0759
4
1 2
. .
.
. .
/
x
U
ft
b) τ ρ
ν
0
2 2
4
1 2
5
323 323 0024 12
1 6 10
20 12
9 11 10
= = × ×
×
×





 = ×
∞
∞
−
−
. . .
.
. .
/
U
xU
psf
c) Drag =
2
1
20 15 1.29
2
U
LU
ν
ρ ∞
∞
× × ×
1 / 2
4
2
1 1.6 10
.0024 12 300 1.29 0.0546 lb.
2 20 12
−
 
×
= × × × × =
 
 
×
 
d) δ
∂
∂ δ δ
δ
x
u
x
U
y y d
dx
=
−
∞
=
× ×
= = − +






10
4
2
3
4
4 65
1 6 10 10
12
0 0416
3
2
3
2
.
.
. .
ft.
∴ = −
×
+
×






×
×
= − +
−
∂
∂
u
x
y y
y y
12
3
2 0416
3
2 0416
4 65
2
1 6 10
10 12
27 9 16140
2
3
4
4
3
. . . .
. .
. .
∴ = − = × − × =
∫
v
u
x
dy
∂
∂
δ
27 9
2
0416
16140
4
0416 0 0121
2 4
0
.
. . . .
fps
8.112 a) δ
ν
= =
× ×





 =
∞
−
4 65 4 65
15 10 6
4
00221
5 1 2
. .
.
. .
/
x
U
m
b) τ ρ
ν
0
2 2
5 1 2
0323 323 122 4
15 10
6 4
0 00498
= = × ×
×
×





 =
∞
∞
−
. . .
.
. .
/
U
xU
Pa
c) Drag =
1
2
129
1
2
122 4 6 5 129
15 10
6 4
0299
2 2
5 1 2
ρU Lw
v
LU
∞
∞
−
× = × × × × ×
×
×





 =
. . .
.
. .
/
N
d)
3
2 4
3 3
2
2
u y y d
U
x dx
∂ δ
∂ δ δ
∞
 
= − +
 
 
 
2 3
2 5 4 5 2
3 4 3 4
4 .
2
2 4.65 1.5 10 3 4.65 (1.5 10 3)
y d
y
dx
δ
− −
 
×
= − +
 
× × × × × × ×
 
 
[ ]
∴ = − + ×
×
= − + ×
−
∂
∂
u
x
y y y y
4 6166 253 10
465
2
15 10
4
1
3
641 2 63 10
7 3
5
5 3
.
. .
. . .
∴ = − = × −
×
× =
∫
v
u
x
dy
∂
∂
δ
641
2
0156
2 63 10
4
0156 0 00391
2
5
4
0
.
.
.
. . ,
m / s
where δx=
−
=
× ×
=
3
5
465
15 10 3
4
01560
.
.
. m.

152. 193
8.113 a) δ
ν
= =
× ×





 =
∞
−
5 5
1 5 10 2
10
0 00866
5
1 2
x
U
.
. .
/
m Use τ ρ
ν
0
2
332
= ∞
∞
. .
U
xU
Drag = τ 0
2
5
0
332 1 22 10
1 5 10
10
2
1 2
4 0 561
wdx
L
= × ×
×
× =
−
∫ . .
.
/
. .
N
b) δ = ×
×
×





 =
−
.
.
. .
.
38 2
1 5 10
10 2
0 0453
5 2
m
Drag =
1
2
074
1
2
1 22 10 2 4 074
1 5 10
10 2
2 15
2
2
2
5
2
ρ
ν
U Lw
U L
∞
∞
−
×





 = × × × × ×
×
×





 =
. . .
.
. .
. .
N
8.114 a) δ τ
= ×
×
×





 = = × × ×
×
×






− −
.
.
. . . .
.
. .
38 6
1 5 10
20 6
0 0949
1
2
1 22 20 059
1 5 10
20 6
5 2
0
2
5 2
m
= . .
6 Pa
b) δ τ
= ×
×





 = = × × ×
×






− −
. . . .
. .
38 6
10
20 6
0 0552
1
2
1000 20 059
10
20 6
6 2
0
2
6 2
m
= 286 Pa.
8.115 u y U
u
y
U y
u
y
U
y
( ) . . / .
/ /
= = = =
∞ ∞
− −
=
∞
δ
∂
∂
δ
∂
∂
δ
δ
1
7
1
7
6 7 1 7
∂
∂ δ
u
y y=
should be zero. Thus, this condition is not satisfied.
τ µ
∂
∂
µ δ
0
0
1 7
1
7
1
0
= = = ∞
=
∞
−
u
y
U
y
/
. Thus, this is unacceptable and
∂
∂
u
y
at, and near, the wall is not valid.
u U
y y
= −






∞
3
2
1
2
3
3
δ δ
.
u U
y
=






∞
δ
1 7
/
.
8.116 a) Drag =
1
2
0024 20 12 15 074
1 58 10
20 12
1060
1 58 10
20 12
0 31
2
4 2 4
× × × ×
×
×





 −
×
×














=
− −
. ( ) .
. .
. .
.
lb
cubic
turb (power-law)
U
y
u
u

153. 194
b) Drag =
1
2
0024 20 12 15 074
1 58 10
20 12
1700
1 58 10
20 12
0 27
2
4 2 4
× × × ×
×
×





 −
×
×














=
− −
. ( ) .
. .
. .
.
lb
c) Drag =
1
2
0024 20 12 15 074
1 58 10
20 12
2080
1 58 10
20 12
0 25
2
4 2 4
× × × ×
×
×





 −
×
×














=
− −
. ( ) .
. .
. .
.
lb
8.117 a) Drag =
1
2
1000 1 2 1 2 074
10
1 2 1
1060
10
1 2 1
5 21
2
6 2 6
× × × ×
×





 −
×














=
− −
. ( ) .
. .
. .
.
N
b) Drag =
1
2
1000 1 2 1 2 074
10
1 2 1
1700
10
1 2 1
4 44
2
6 2 6
× × × ×
×





 −
×














=
− −
. ( ) .
. .
. .
.
N
c) Drag =
1
2
1000 1 2 1 2 074
10
1 2 1
2080
10
1 2 1
3 99
2
6 2 6
× × × ×
×





 −
×














=
− −
. ( ) .
. .
. .
.
N
8.118 U∞ = = = ×
×
×





 =
60
1000
3600
16 67 38 100
16 67 10
235
5
2
. .
.
.
.
m /s. 000
1.5 10
m
-5
δ
τ ρ
0
2 2
5
5
2
1
2
1
2
1 22 16 67 059
1 5 10
16 67 10
0 0618
= = × × ×
×
×














=
∞
−
U c f . . .
.
.
. .
.
Pa
b) τ ρ
0
2 2
5
5
2
1
2
1
2
1 22 16 67
455
06
16 67 10
1 5 10
0 151
= = × ×
×
×












=
∞
−
U c
n
f . .
.
.
.
.
. .
l
Pa
∴ = = ∴ =
×
+ ∴ =
−
u n
τ
δ
δ
.
.
.
.
.
.
.
.
. . .
151
1 22
351
16 67
351
2 44
351
1 5 10
7 4 585
5
m /s. m
l
Both (a) and (b) are in error, however, (b) is more accurate. 0.
p
x
∂
∂
<
8.119 a) 5 =
uτ ν
δ
ν
(See Fig. 8.24 b). ∴ =
× ×
= ×
−
−
δν
5 1 5 10
351
2 14 10
5
4
.
.
. .
m
b) δ
µ
δ δ
τ τ
δ
δ
δ
δ
δ
ν
d
U
U u dy
U
n
y
dy
u
U
n
y
dy
= − = −





 + −
∞
∞
∞ ∞
∫
∫
∫
1
2 5 2 44 3 74
15
15
0
( ) . . .
.
.
l l
( )
= − − −





 − −












∞
= =
u
U
y n
y
y y n
y
y
v
τ
ν
δ
δ δ
δ δ
δ δ
2 5 15 2 44 3 74
15 87 8
87 8
585
. . . .
. .
.
l l
= + − + − =
.
.
[ . ] . .
351
16 67
219 620 008 2188 951 43 7 m
Note: We cannot use zero as a lower limit since the ln-profile does not go to the

154. 195
wall. Hence, we use δν ; the lower limit provides a negligible
contribution to the integral.
8.120 a) Use Eq. 8.6.40: c
n
f
=
×
×












=
−
.
.
.
. .
455
06
300 20
1 58 10
0 00212
4
2
l
b) τ ρ
0
2 2
1
2
1
2
0024 300 00212 0 229
= = × × × =
∞
U c f . . . .
psf uτ = =
.
.
.
229
0024
9 77 fps.
c) δ
ν
ν
τ
= = × × = ×
− −
5
5 1 58 10 9 77 8 09 10
4 5
u
. / . . .
ft
d)
300
9 77
2 44
9 77
1 58 10
7 4 0 228
4
.
.
.
.
. . . .
=
×
+ ∴ =
−
ln
δ
δ ft
8.121 a) τ ρ
0
2 2
6
2
1
2
1
2
1000 10
455
06
10 3
10
110
= = × ×
×












=
∞
−
U c
n
f
.
.
.
l
Pa
∴ = = ∴ = =
×
= ×
−
−
u
u
τ ν
τ
δ
ν
110
1000
332
5 5 10
332
1 51 10
6
5
.
.
. .
m /s. m
b) u u
= = × =
5 5 332 1 66
τ . . .
m /s
c) y =. .
15δ — Do part (d) first! ∴ = × =
y . . . .
15 0333 0 005 m
d)
10
332
2 44
332
10
7 4 0 0333
6
.
.
.
. . . .
= + ∴ =
−
ln
δ
δ m
8.122 Assume flat plates with dp dx
n
f
/ .
.
.
. .
= =
×












=
−
0
523
06
10 100
10
00163
6
2
C
l
∴Drag = 2
1
2
1000 10 10 100 00163
2
× × × × × × =
. .
163 000 N
To find δ τ
max we need u .
τ τ
0
2
6
2
1
2
1000 10
455
06
10 100
10
70 9
70 9
1000
0 266
= × ×
×












= ∴ = =
−
.
.
.
.
.
ln
u
Pa. m /s.
10
266
2 44
266
10
7 4 0 89
6
.
.
.
. . . .
= + ∴ =
−
ln
δ
δ m
max
8.123 a) Assume a flat plate of width πD. 8
5
15 600
Re 6 10 .
1.5 10
UL
ν −
×
= = = ×
×

155. 196
2 8 1/5 2
1 1
drag 0.073(6 10 ) 1.2 15 600 100 32600 N
2 2
f
C U L D
ρ π π
−
= = × × × × × × × =
power 32600 15 489000 W or 655 hp or 164 hp/engine
D
F U
= × = × = .
b) 3
helium
100
0.167 kg/m .
2.077 288
p
RT
ρ = = =
×
air helium
B
F W W V
ρ
= − = ∆ × 2 7
(1.2 0.167) 9.8 50 600/2 2.38 10
π
= − × × × × = ×
payload = 6 6 6
23.8 10 9.8 1.2 10 12 10 N
B
F W
− = × − × × = ×
8.124 u
y
u
x x y
v
x
u
y y
u
y y
= = = − = =
∂ψ
∂
∂
∂
∂ ψ
∂ ∂
∂ψ
∂
∂
∂
∂ ψ
∂
∂
∂
∂ ψ
∂
, , , , .
2 2
2
2
2
3
3
Substitute into Eq. 8.6.45 (with dp dx
/ ):
= 0
∂ψ
∂
∂ ψ
∂ ∂
∂ψ
∂
∂ ψ
∂
ν
∂ ψ
∂
y x y x y y
2 2
2
3
3
− = .
8.125 We also have
∂ψ
∂
∂ψ
∂φ
∂φ
∂
∂ψ
∂η
∂η
∂
∂ ψ
∂ ∂
∂ ∂ψ ∂
∂φ
∂φ
∂
∂ ∂ψ ∂
∂η
∂η
∂
x x x x y
y
x
y
x
= + = +
,
( / ) ( / )
2
Recognizing that ∂φ ∂ ∂φ ∂ ∂η ∂ ν
/ , / , / / ,
x y x
y
U x
= = = − ∞
1 0
2
3
and
∂η ∂ ν
/ / ,
y U x
= ∞
∂ψ
∂ νφ
∂ψ
∂η
∂ψ
∂
∂ψ
∂φ ν
∂ψ
∂η
y
U
x
y U
x
= = −
∞ ∞
,
2 3
∂ ψ
∂ ∂ νφ
∂ ψ
∂φ∂η νφ
∂ψ
∂η νφ
∂ ψ
∂η νφ
2 2
3
2
2 3
2
x y
U U U y U
= + −






∞ ∞ ∞ ∞
-
1
2
∂ ψ
∂ νφ
∂ ψ
∂η νφ
∂ ψ
∂ νφ
∂ ψ
∂η νφ
2
2
2
2
3
3
3
3
y
U U
y
U U
=





 =
∞ ∞ ∞ ∞
,
Equation 8.6.47 then becomes, using U y
∞ =
/ / ,
νφ η
η ∂ψ
∂η
η ∂ ψ
∂φ∂η
η ∂ψ
∂η
η ∂ ψ
∂η
∂ψ
∂φ
η ∂ψ
∂η
η ∂ ψ
∂η
y y yx yx x y
2 2 2
2
2
2
2
2
2 2 2
− −





 − −












= ∞
ν
ν
η ∂ ψ
∂η
U
x y
3
3
Multiply by y2 2
/η and Eq. 8.6.49 results:
−





 + − = ∞
1
2
2 2 2
2
3
3
φ
∂ψ
∂η
∂ ψ
∂φ∂η
∂ψ
∂η
∂ψ
∂φ
∂ ψ
∂η
ν
∂ ψ
∂η νφ
U

156. 197
8.126 u
y
U x
dF
d y
U x F
U
x
U F
= = = =
∞ ∞
∞
∞
∂ψ
∂
ν
η
∂η
∂
ν η
ν
η
'( ) '( ).
We used Eq. 8.6.50 and Eqs. 8.6.48.
( )
v
x x
U x F
U
x
F U x
F
x
= − = − = − −
∞
∞
∞
∂ψ
∂
∂
∂
ν
ν
ν
∂
∂η
∂η
∂
1
2
= − − −






∞
∞
∞ −
1
2
1
2
3 2
U
x
F U x F y
U
x
ν
ν
ν
' /
= − + = −
∞ ∞ ∞ ∞
1
2 2
1
2
U
x
F
y U
x
U
x
F
U
x
F F
ν
ν
ν ν
η
' ( ' ).
8.127 The results are shown in Table 8.5.
8.128 a) τ 0
2
5
0 332 1 22 5
1 5 10
2 5
0 0124
= × ×
×
×
=
−
. .
.
. .
Pa
b) δ =
× ×
=
−
5
15 10 2
5
0 0122
5
.
. .
m
c) v
U
x
F F
max
max
m /s
= −






=
× ×
× =
∞
−
ν
η
1
2
1 5 10 5
2
8605 0 00527
5
( ' )
.
. . .
d) Q udy U
dF
d
dy U
dF
d
d
vx
U
= = =
∞ ∞
∞
∫ ∫
∫ 0 0
0
δ δ
δ
η η
η
= − =
× ×
× =
∞
∞
−
U
x
U
F F
ν
δ
[ ( ) ( )]
.
. . /
0 5
1 5 10 2
5
3 28 0 04
5
m s / m
2
8.129 a) τ 0
2
4
4
332 0024 15
1 6 10
6 15
2 39 10
= × ×
×
×
= ×
−
−
. .
.
. .
psf
b) δ =
× ×
=
−
5
1 6 10 6
15
0 04
4
.
. ft.
c) v
U
x
F F
max
max
fps
= −






=
× ×
× =
∞
−
ν
η
1
2
1 6 10 15
6
8605 0 0172
4
( ' )
.
. . .
d) Q udy U
x
U
F
= = =
× ×
× =
∞
∞
−
∫
ν
δ
δ
( )
.
. . / .
15
1 6 10 6
15
3 28 0 394
4
0
ft sec /ft
2
8.130 At x = 2m, Re = 5 × 2/10-6 = 107. ∴Assume turbulent from the leading edge.
a) τ ρ
0
2
2
1
2
0 455
0 06
= ∞
U
n x
.
( . Re )
l

157. 198
= × ×
×
=
1
2
1000 5
0 455
0 06 10
32 1
2
7 2
.
( . )
.
ln
Pa
b) uτ τ ρ
= = =
0 32 1 1000 0 1792
/ . / . m /s
5
0 1792
2 44
0 1792
10
7 4 0 0248
6
.
.
.
. . .
= + ∴ =
−
ln m or 24.8 mm
δ
δ
c) Use the 1/7 the power-law equation:
Q y dy
= =
∫5 0 0248 0 109
1 7
0
0 0248
( / . ) .
/
.
m /s /m
3
8.131 From Table 8.5 we would select η = 6:
a)
5
1.5 10 2
6 6 0.0147 m
5
x
U
ν
δ
−
∞
× ×
= = =
b)
5
15.8 10 6
6 6 0.047 ft or 0.57 in.
15
x
U
ν
δ
−
∞
× ×
= = =
8.132 From Table 8.5 we interpolate for F' .
= 0 5 to be
η =
−
−
− + =
0 5 0 3298
0 6298 0 3298
2 1 1 1 57
. .
. .
( ) .
=
× ×
∴ =
−
y y
5
1 5 10 2
0 00385
5
.
. . m or 3.85 mm
( )
v
U
x
F F
=





 −
∞
ν
η
1
2
' =
× ×
=
−
1 5 10 5
2
0 207 0 00127
5
.
( . ) . m / s
u v
y x
∂ ∂
τ µ
∂ ∂
= + 2
"
F U
xU
ν
ρ ∞
∞
 
=
 
 
5
2 1.5 10
0.291(1.2)5 0.011 Pa
2 5
−
×
= =
×
8.133
If v y v y v y
= = > = <
0 10 0 0
at and at then
δ δ ∂ ∂
, / and continuity demands that
/ 0. The component, for must then be greater than ,
u x u y U
∂ ∂ δ
> > 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
= =
δ. ,
If at large say
0
y = 10δ , 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 = δ
v
v
y
y = δ
v = 0
v
(a) (b) (c)
U
v
v = 0
u > U

158. 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 u U
y y
= −






∞
3
2
1
2
3
3
δ δ
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
δD
δΒ
δC
δΑ
δE

159. 200
CHAPTER 9
Compressible Flow
9.1
Btu ft-lb lbm ft-lb
0.24 778 32.2 6012
Btu slug
lbm- R slug- R
p
c = =
o o
c c R
v p
= − = − = =
6012 1716 4296 4296
1
778
1
32 2
ft -lb
slug- R
ft -lb
slug- R
Btu
ft - lb
slug
lbm
o o
.
Btu
0.171
lbm- R
= o
9.2 c c R c kc c
c
k
R c
k
R
p v p v p
p
p
= + = ∴ = + −





 =
. . .
or 1
1
∴ = −
c Rk k
p /( ).
1
9.3 If ∆s = 0, Eq. 9.1.9 can be written as
c n
T
T
R n
p
p
n
T
T
n
p
p
p
c R
p
l l l l
2
1
2
1
2
1
2
1
=





 =






or
It follows that, using c c R c c k
p v p v
= + =
and / ,
T
T
p
p
p
p
R c
k
p
2
1
2
1
2
1
1
1
=





 =






−
/
.
Using Eq. 9.1.7,
T
T
p
p
p
p
p
p
k
k
2
1
2 1
2 1
2
1
1
1
1
2
2
1
1
= =





 =






− −
ρ
ρ
ρ
ρ
or
/
.
Finally, this can be written as
p
p
k
2
1
2
1
=






ρ
ρ
.
9.4 Substitute Eq. 4.5.18 into Eq. 4.5.17 and neglect potential energy change:
& &
&
~ ~ .
Q W
m
V V p p
u u
S
−
=
−
+ − + −
2
2
1
2
2
2
1
1
2 1
2 ρ ρ

160. 201
Enthalpy is defined in Thermodynamics as h u pv u p
= + = +
~ ~ / .
ρ Therefore,
& &
&
.
Q W
m
V V
h h
S
−
=
−
+ −
2
2
1
2
2 1
2
Assume the fluid is an ideal gas with constant specific heat so that ∆ ∆
h c T
p
= .
Then
( )
& &
&
.
Q W
m
V V
c T T
S
p
−
=
−
+ −
2
2
1
2
2 1
2
Next, let c c R k c c c R k k
p v p v p
= + = = −
and so that
/ / ( ).
1 Then, with the ideal gas
law T p R
= / ,
ρ the first law takes the form
& &
&
.
Q W
m
V V k
k
p p
S
−
=
−
+
−
−






2
2
1
2
2
2
1
1
2 1 ρ ρ
9.5 Differentiate p c d xy ydx xdy
k
ρ−
= = +
using ( ) :
ρ ρ ρ
− − −
− =
k k
dp pk d
1
0.
Rewrite:
dp
d
k
p
ρ ρ
= .
9.6 The speed of sound is given by
c dp d
= / .
ρ
For an isothermal process TR p K K
= =
/ ,
ρ where is a constant. This can be
differentiated:
dp Kd RTd
= =
ρ ρ.
Hence, the speed of sound is
c RT
= .
9.7 Eq. 9.1.4 with & &
Q W
V
c T
S p
= = + =
0
2
2
is: cons't.
V
c T
V V
c T T
V V V V
c T c T
p p p p
2 2 2 2
2 2
2
2
+ =
+
+ + =
+ +
+ +
( )
( )
( )
.
∆
∆
∆ ∆
∆
2
2 ( )
0
2 2
V V V
∆ ∆
∴ = + . .
p p
c T V V c T h
+ ∆ ∴− ∆ = ∆ = ∆
We neglected (∆V)2
. The velocity of a small wave is V c h c V
= ∴ = −
. .
∆ ∆
9.8 For water
ρ
ρ
dp
d
= ×
2110 106
Pa
Since ρ = 1000 kg / m we see that
3
,

161. 202
c dp d
= / ρ
= × =
2110 10 1000 1453
6
/ m / s
9.9 For water c
p dp
d
= ≅ =
×
=
∆
∆ρ ρ
2110 10
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
∆t
d
c
= = =
10
1450
0 0069
. seconds
9.11 a) M = =
× ×
=
V
c
200
1 4 287 288
0 588
.
. .
b) M 600/ 1.4 1716 466 0.567.
= × × =
c) M = × × =
200 1 4 287 223 0 668
/ . . .
d) M 600/ 1.4 1716 392 0.618.
= × × =
e) M = × × =
200 1 4 287 238 0 647
/ . . .
9.12 c kRT d ct
= = × × = ∴ = = × =
1 4 287 263 256 256 1 21 309
. . .
m / s. m
9.13 a) Assume T = 20°C:
c kRT
= = × × =
1 4 287 293 343
. m /s.
d c t
= = × =
∆ 343 2 686 m
b) Assume T = 70°F:
c kRT
= = × × =
1 4 1716 530 1130
. fps.
d c t
= = × =
∆ 1130 2 2260 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
= × × = = =
1 4 287 263 256
1
. sin .
m / s. α
1000
sin 0.256. tan 0.2648 . 3776 m
L
L
α α
= ∴ = = ∴ =
∆t = =
3776
1000
3 776
. .
s
1000 m
V
L

162. 203
9.15 Use Eq. 9.2.13:
a)
c
V
V
= =
× ×
=
sin
.
sin
α or m /s
1 4 287 288
22
908
o
b)
c
V
V
= =
× ×
=
sin
.
sin
α or fps
1 4 1716 519
22
2980
o
9.16 Eq. 9.2.4: ∆
∆ ∆
V
p
c
p
kRT
= − = − = −
× ×
= −
ρ ρ
0 3
00237 1 4 1716 519
0 113
.
. .
. .
fps
Energy Eq:
V
c T
V V
c T T V V c T
p p p
2 2
2 2
0
+ =
=
+ + ∴ = +
( )
( ). .
∆
∆ ∆ ∆
∴ = − = − × × − × × =
∆ ∆
T c V cp
/ . ( . ) /( . . ) . .
1 4 1716 519 113 0 24 778 32 2 0 021o
F
Note: cp = × × = × ×
. . . . .
24 778 32 2 24 778 32 2
Btu
lbm - F
ft - lb
Btu
lbm
slug
ft - lb
slug - F
o o
Then
ft
ft - lb /(slug - F)
ft lb - sec F
sec ft - lb - ft
F.
2 2 2
2
/sec
2
o
o
o
=
− −
−
= (units can be a pain!)
9.17 a) ρ ρ ρ ρ ρ ρ ρ ρ ρ
AV AV AdV AVd Ad dV VdA dAdV Vd dA d dAdV
= + + + + + + +
Keep only the first order terms (the higher order terms—those with more
than one differential quantity—will be negligible):
0 = + +
ρ ρ ρ
AdV AVd VdA
Divide by ρAV:
dV
V
d dA
A
+ + =
ρ
ρ
0
b) Expand the r.h.s. of Eq. 9.3.5 (keep only first order terms):
V k
k
p V VdV k
k
p dp
d
2 2
2 1
2
2 1
+
−
=
+
+
−
+
+
ρ ρ ρ
.
Hence,
0
2
2 1
= +
−
+
+
−






VdV k
k
p dp
d
p
ρ ρ ρ
= +
−
+ − −
+






VdV
k
k
p dp p pd
d
1 2
ρ ρ ρ ρ
ρ ρ ρ
= +
−
−






VdV
k
k
dp pd
1 2
ρ ρ
ρ
where we neglected ρ ρ ρ
d compared to 2
. For an isentropic process Eq. 9.2.8
gives ρ ρ
dp kpd
= , so the above becomes

163. 204
0
1 2
= +
−
−
VdV
k
k
kpd pd
ρ ρ
ρ
= +
−
−
VdV
k
k
k pd
1
1
2
( ) ρ
ρ
= +
VdV k
p
d
ρ
ρ
2
But d dV V dA A
ρ ρ
/ / /
= − − so that the above equation is
0 = + − −






VdV k
p dV
V
dA
A
ρ
which can be written as
dA
A
V
kp
dV
V
= −






2
1
ρ
.
Since c kp
2
= / ,
ρ and M = V/c, this is put in the form
dA
A
V
c
dV
V
= −






2
2
1 or ( )
dA
A
dV
V
= −
M2
1
c) Substituting in V c c kRT R c k k
p
= = = −
M and we find
, , / ( ) / ,
2
1
T
T
V
c T
c
c T
kRT
c T
p p p
0
2 2 2 2
2
1
2
1
2
1
= + = + = +
M M
=
−
+ = +
−
M
M
2
2
1
2
1 1
1
2
k k
k
k
( )
.
d) &
/( )
/
m p
k
TR
A
p
k
T
k
k
R
A
k k
= =
+
−






+
−






−
−
M
M
M
M
0
2
1
0
2
1 2
1
1
2
1
1
2
= +
−






+
−
p
k
RT
A
k
k
k
0
0
2
1
2 1
1
1
2
M M
( )
At the critical area A* *
, .
M = 1 Hence,
& .
*
( )
m p
k
RT
A
k
k
k
=
+






+
−
0
0
1
2 1
1
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
2 1
0
0
1
2 1
1
1
2
1
2
M M
+
−





 =
+






+
−
+
−
( )
*
( )
or
A
A
k
k
k
k
*
( )
( )
=
+ −
+






+
−
1 2 1
1
2
1
2 1
M
M

164. 205
9.18 a) atm 10 69.9 10 79.9 kPa abs.
s
p p
= + = + =
1 69.9 kPa abs.
p =
From 1 → s :
V p p p
p
s
s
s
s
k
1
2
1
1
1
1
1 1 1 4
2
906
79 9
69 9
0 997
+ = =





 =





 =
ρ ρ
ρ ρ
. .
.
.
. .
/ / .
kg /m3
∴ + = ∴ =
V
V
1
2
1
2
69 79
77 3
900
.906
900
.997
m / s
. . .
b) p p
s = + = =
26 4 10 36 4 26 4
1
. . .
kPa abs. kPa abs.
From 1 → s :
V p p p
p
s
s
s
s
k
1
2
1
1
1
1
1 1 1 4
2
0 412
36 4
26 4
0 518
+ = =





 =





 =
ρ ρ
ρ ρ
. .
.
.
. .
/ / .
kg /m3
V
V
1
2
1
2
26 36
111
+ = ∴ =
400
.412
400
.518
m /s
. .
9.19 a)
1/ 1/1.4
2
3
1 1
1
1 1
105
. 1.22 1.254 kg/m .
2 101
k
s s
s
s
p p
V p
p
ρ ρ
ρ ρ
   
+ = = = =
   
 
 
2
1
1
1.4101 000 105 0001.4
. 81.3 m/s.
2 .4 1.22 1.254 .4
V
V
+ = ∴ =
b)
2
1
1
4000 81.3-81
. 81.0 m/s. % error = 100 0.42%.
2 1.22 81.3
V
V
= ∴ = × =
9.20 Is p p
r < × = kPa.
. ? . .
5283 0 5283 200 105 7
0
a) p p V kRT p
r e e e e
< ∴ ∴ = ∴ = =
choked flow. M kPa.
. . . . .
5283 1 105 7
0
2
1000 298
1 4 287
2
1000 248 1 315 8
× =
×
+ ∴ = =
.
. . .
T
T T V
e
e e e
K, m / s.
ρ π
e m
=
×
= ∴ = × × × =
105 7
287 248 1
1 484 1 484 01 315 8 0 1473
2
.
. .
. . & . . . . .
kg / m kg / s
3
b) p p
V
r e
e
> ∴ < × + =






M 1000 298 =
000
e
2
e
. . .
.
.
.
.
.
5283 1
2
1 4
4
130 130
200 2 338
0
1 4
ρ
ρ
ρ ρ
0
200
287 298
2 338 1 7187 257 9
=
×
= ∴ = ∴ =
.
. . . . .
kg / m m / s.
3
e e
V
∴ = × × × =
& . . . . .
m 1 7187 01 257 9 0 1393
2
π kg /s
9.21 Is p p
r < × = psia.
. ? . .
5283 0 5283 30 15 85
0
a) p p V kRT
r e e e
< ∴ = = =
15 choked flow and M psia.
. . , . .
85 1 15 85 2
0 24 530
1 4 1716
2 778 32 2
0 24 441 7 1030
.
.
( . )
. . .
× =
× ×
×
+ ∴ = =
T
T T V
e
e e e
R, fps.
o
V
s
1
Vs=0

165. 206
3
15.85 144
0.003011 slug/ft .
1716 441.7
e
ρ
×
= =
×
2
.5
.003011 1030 0.01692 slug/sec.
12
m π
 
∴ = × × =
 
 
&
b) 15.85. M 1, and 20 psia.
r e e
p p
> ∴ < =
3
0
30 144
.00475 slug/ft .
1716 530
ρ
×
= =
×
1/1.4
3
20
.00475 .003556 slug/ft .
30
e
ρ
 
∴ = =
 
 
2
1.4 20 144
0.24 530(778 32.2) .
2 .4 .003556
e
V ×
× × = +
∴ = ∴ = ×





 × =
V m
e 838 9 003556
5
12
838 9 0 01627
2
. & .
.
. . .
fps. slug /sec
π
(Note:
ft-lb ft-lb
0.24 Btu/lbm- R=0.24 778 0.24 778 32.2 .)
lbm- R slug- R
cp = × = × ×
o
o o
9.22 a) p p p T
r e e e
< ∴ = ∴ = × = = × =
M kPa. K.
. . . . . . .
5283 1 5283 200 105 7 8333 298 248 3
0
ρe e
V
=
×
= = × × =
105 7
287 248 3
1 483 1 4 287 248 3 315 9
.
. .
. . . . .
kg / m m / s.
3
∴ = × × × =
& . . . . .
m 1 483 01 315 9 0 1472
2
π kg /s
b) p p p
p
p
T T
r e
e
e e
> ∴ = = ∴ = =
kPa, M
. . . . . , .
5283 130 0 65 81 884
0
0
0
ρe e
V
=
×
= = × × =
130
287 263 4
1 719 81 1 4 287 263 4 263 5
. .
. , . . . .
kg / m m / s.
3
∴ = × × × =
& . . . . .
m 1 719 01 263 5 0 1423
2
π kg /s
9.23 a) p p p
r e e
< ∴ = ∴ = × =
M psia.
. . . . .
5283 1 5283 30 15 85
0
Te = × =
. .
8333 530 441 6o
R.
∴ =
×
×
= = × × =
ρe e
V
15 85 144
1716 441 6
003012 1 4 1716 441 6 1030
.
.
. . . .
slug
ft
fps.
3
& .
.
. .
m = ×





 × =
003012
5
12
1030 0 01692
2
π slug /sec
b) p p p
p
p
T T
r e
e
e e
> ∴ = = = ∴ = =
psia. M
. . . . . . . .
5283 20
20
30
6667 785 0 890
0
0
0
∴ =
×
×
= = × × =
ρ0
20 144
1716 472
00356 785 1 4 1716 472 836
. . . . fps.
Ve
∴ =





 × =
& .
.
. .
m 00356
5
12
836 0 01664
2
π slug /sec

166. 207
9.24 p T
e e
= × = = × =
. . . . .
5283 400 211 3 8333 303 252 5
kPa abs K.
V m
e = × × = ∴ =
×
× × =
1 4 287 252 5 318 5
211 3
287 252 5
05 318 5 7 29
2
. . . &
.
. .
. . . .
m / s. kg /s
π
9.25 p p p T
e e
= = ∴ = = × =
. . . . . .
5283 101 191 2 8333 283 235 8
0 0
kPa kPa abs K.
V m
e = × × = ∴ =
×
× × =
1 4 287 235 8 307 8
101
287 235 8
03 307 8 1 30
2
. . . &
. .
. . . .
m / s. kg /s
π
p p p T
e e
0 0
2 191 2 382 4 5283 202 0 235 8
= × = = = =
. . . . .
kPa abs. kPa abs. K.
V m
e e
= = ∴ =
×
× × =
307 8 1
202
287 235 8
03 307 8 2 60
2
. . &
. .
. . . .
m / s since M kg / s
π
9.26 p p p T
e e
= = ∴ = = × =
. . . . . . .
5283 14 7 27 83 8333 500 416 6
0 0
psia psia R.
o
1.4 1716 416.6 1000 fps.
e
V = × × =
3
0.3203 kg/m and 199.4 kPa abs.
e e
p
ρ
∴ = =
0 0
2 27.83. 0.5283 29.4 psia, 416.6 R, 1000 fps.
e e e
p p p T V
= × = = = =
o
∴ =
& . .
m 0 202 slug / sec
9.27 Treat the pipeline as a reservoir. Then, p p
e = =
. .
5283 264 5
0 kPa abs.
M and m /s.
e e
V
= = × × =
1 1 4 287 8333 283 307 8
. (. ) .
&
.
. (. )
. .
m =
× ×
× × × =
−
264 5
287 8333 283
30 10 307 8 3 61
4
kg /s.
∆
∆
− = =
× ×
× ×
=
V
m t
& .
. / (. . )
.
ρ
361 6 60
2645 287 8333 283
333 m3
9.28 5193 300
1 667 2077
2
5193 225 200
225
300
1 667
667
× =
×
+ ∴ = ∴ =






.
.
.
.
K.
T
T T p
e
e e e
=97.45 kPa.
Next, T p V
t t t
= = ∴ = × × =
225 97 45 1 667 2077 225 882 6
K, kPa; m /s.
. . .
3 2 2
97.45
= 0.2085 kg/m . 0.2085 × × .03 × 882.6 = 0.075
2.077 225
t e e
V
ρ π ρ π
= ×
×
5193 300
2
1 667
667
200
200 2 077 300
1330
2 1 667
1 667
× = + =
×





 =
V p
p
e e
e
e
e
e
.
.
.
/ .
.
.
ρ
ρ
ρ kPa.
= + × × −
V
V
e
e
2
3 667
2
3324 10 9 54
. .
.
or 3 116 10 63 420 10 91 8
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.
e e
p
ρ
∴ = =

167. 208
9.29 ρ ρ
1
1
1
2
1 1 4
300 100
287 293
4 757 4 757
340
400
4 236
= =
+
×
= =





 =
p
RT .
. . . . .
/ .
kg / m kg /m
3 3
V V V V
1
2
2
2
2 1
4 757 10 4 236 5 4 492
× × = × × ∴ =
. . . . .
V k
k
p V k
k
p V V
1
2
1
1
2
2
2
2
1
2 2
1
2
2 1 2 1 2
1 4
4
400 4 492
2
1 4
4
340
+
−
= +
−
+ = +
ρ ρ
.
.
.
. .
.
.
000
4.757
000
4.236
∴ =
V1 37 35
. .
m
s
∴ = = × × × =
& . . . . .
m A V
ρ π
1 1 1
2
4 757 05 37 35 1 395 kg /s
9.30 ρ1
1
1
45 14 7 144
1716 520
0 009634
= =
+
×
=
p
RT
( . )
. .
slug
ft3
slug /ft3
ρ2
1 1 4
009634
50 7
59 7
008573
=





 =
.
.
.
. .
/ .
V V V V
1
2
2
2
2 1
009634 4 008573 2 4 495
× × = × × ∴ =
. . . . .
V V
V
1
2 2
1
2
1
2
1 4
4
59 7 144 4 495
2
1 4
4
50 7 144
121 9
+
×
= +
×
∴ =
.
.
. . .
.
.
. .
.009634 .008573
fps.
∴ = × × =
& . ( / ) . . .
m 009634 2 12 121 9 0 1025
2
π slug /sec
9.31 Energy 0 → 2: 1000 303
2
1000 3
2
2
2 2 2
× = + =
V
T V kRT
.
2 1.627 20 32.5 kPa.
p
∴ = × =
l
1.4
.4 3
2 2
107.9 5.39
200 5.390 kPa. 0.1740 kg/m .
303 .287 107.9
p ρ
 
∴ = = = =
 
×
 
Energy 0 → 1: 1000 303
2
1000
1 4 287
318 4 252 3
1
2
1 1
× = +
×
∴ = =
V
V T
V
m /s, K.
1
2
.
. . .
p1
1 4
4
1
200
252 3
303
105 4
105 4
287 252 3
1 455
=





 = =
×
=
.
.
.
. .
. .
.
.
kPa. kg /m3
ρ
Continuity: 1 455 05 318 4 174
4
3 1 4 287 107 9 0 2065
2 2
2
2
. . . . . . . . .
π π
× × = × × × ∴ =
d
d m
9.32 V kRT
T
T T V
t t
t
t t t
2
1000 293
1 4 287
2
1000 244 0 313 1
= × =
×
+ ∴ = =
.
.
. . .
K. m / s.
∴ =





 = ∴ =
×
=
pt t
500
244
293
263 5
263 5
287 244
3 763
1 4
4
.
.
.
.
.
. .
kPa abs. kg /m3
ρ
0 1 2

168. 209
2
2 2
1.4 1.4
1.4 263 500
1000 293 . 3.763 .025 313.1 .075 .
2 .4 3.763
e e e
e e
e e
V p p
V
π ρ π
ρ ρ
× = + × × × = × =
2
6 .4
293 000= 1.014 10 . Trial-and-error: 22.2 m/s, 659 m/s.
2
e
e e
V
V V
−
∴ + × =
∴ = ∴ =
ρe e
p
5 897 0 1987 494 2 4 29
. , . . . , . .
kg / m kPa kPa abs
3
9.33
*
0
9. 0.997 from Table D.1. 500 .997 498.5 kPa.
e e
e
A p
p
p
A
= ∴ = ∴ = × =
and
0
0.00855 from Table D.1. 4.28 kPa abs.
e
e
p
p
p
= ∴ =
9.34 M psia, R.
t = ∴ = × = = × =
1 5283 120 63 4 8333 520 433 3
. . . . .
p T
t t
o
∴ =
ρt . .
01228
slug
ft3
& . . . . . .
m
d
d
t
t
= = × × ∴ =
1 01228
4
1 4 1716 433 3 0 319
2
π
ft
p
p
T V
e e
0
15
120
125 2 014 552 520 287 2 014 1 4 1716 287
= = ∴ = = × = = × ×
. . . , . . .
M R,
e
o
= 684 fps.
A
A
d
d
e
e
*
. . .
.
. . .
= ∴ =
×
∴ =
1 708
4
1 708
319
4
0 417
2 2
ft
π π
9.35
*
M 4. 10.72, .006586 2000 13.17 kPa, .2381 293 69.76 K.
e e e
A
p T
A
= = = × = = × =
For
A
A
p p
e
*
. , . . . . . .
= = ∴ = = × =
10 72 0584 9976 9976 2000 1995 2
0
M kPa abs
e
9.36 Let 1
150
M 1. Neglect viscous effects. M 0.430.
1.4 287 303
t = = =
× ×
2
2
1
*
.05
1.5007. . 0.0816 m or 8.16 cm.
1.5007 1.5007 4
t
t t
d
A
A
A d
A
π
π ×
∴ = ∴ = = = ∴ =
9.37 p T
e es
= × = = × =
. . . . .
5283 400 211 3 8333 303 252 5
kPa abs.
303
.96 . 254.5 K. 1.4 287 254.5 319.8 m/s.
303 252.5
e
e e
T
T V
−
= ∴ = ∴ = × × =
−
2
211.3
.05 319.8 7.27 kg/s.
.287 254.5
m π
∴ = × × =
×
&

169. 210
9.38 Isentropic flow. Since k = 1.4
for nitrogen, the isentropic table
may be used.
M = =
3 4 235
: . .
*
A
A
3
100
3 1.4 297 373 1181 m/s. .9027 kg/m .
.297 373
i i
V ρ
= × × = = =
×
∴ = =
×
= ∴ = =
A
m
V
A
i
i i
t
&
.
. .
.
.
. .
ρ
10
9027 1181
0 00938
00938
4 235
0 00221
m m
2 2
At M = = =
3 3571 02722
0 0
, . , . .
T T p p
∴ = = = = = =
T T p p
e e
0 0
373
3571
1044
100
02722
3670
.
.
.
.
K or 772 C kPa
o
9.39 Isentropic flow. Since k = 1.4
for nitrogen, the isentropic table
may be used.
M = =
3 4 235
: . .
*
A
A
Vi i
= × × = =
×
×
=
3 1 4 1776 660 3840
15 144
1776 660
001843
. . .
fps.
slug
ft3
ρ
A A
i t
=
×
= ∴ = =
.
.
. .
.
.
. .
2
001843 3840
0283
0283
4 235
0 00667
ft ft
2 2
At M = = =
3 3571 02722
0 0
, . , . .
T T p p
∴ = = = = = =
T T p p
e e
0 0
660
3571
1848
15
02722
551
.
.
.
.
o o
R or 1388 F psia
9.40 Assume pe e
= =
×
=
101
101
189 1273
4198
kPa. Then kg / m3
ρ
.
. .
Momentum: 2 2 2
80 000 9.81
. .4198 .25 .
6
F mV AV V
ρ π
×
= = = ×
&
1260 m/s.
V
∴ =
9.41 F mV AV
= = =
×
=
& .
.
. .
ρ ρ
2 101
287 873
403 kg / m (Assumegases are air.)
3
4 2
100 9.81 .403 200 10 . 349 m/s.
V V
−
× = × × ∴ =
9.42 M M
t e
= = ∴ = =
1 4 2 94 02980 0
. ; . , . .
*
A
A
p p
e
e
0
.3665 .3665 300 109.95 K,
e
T T
= = × =
0 0
100 .0298 . 3356 kPa abs.
e
p p p
= = ∴ =
i t
e
Ve = 0
Mt = 1
M > 1
M < 1
~
i t
e
Ve = 0
Mt = 1
M > 1
M < 1
~
Ve
FB
p0
A0

170. 211
∴ = × × =
Ve 2 94 1 4 287 109 95 618
. . . m / s.
∴ =
−
×
× × + × =
FB
100
287 109 95
05 618 3 2
2 2 2
. .
. . .
π π
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
= +
−






−
M
Using k = 1.4 this gives M or M
2
00424 0 206
= =
. . .
For standard conditions V c
= = × × =
M m / s
0206 14 287 288 70
. . .
9.44 a) 2 2 2 2
0.9850 1000 . 80 000 0.985 1000( 1000)
V p V
ρ
× = − = × −
V p
2
2 2
2
2
1
1000
2
1 4
4
287 283 0
80
287 283
9850
−
+ − ×





 = =
×
=






.
.
.
.
. .
ρ
ρ kg /m3
2 2
2 2
2
1000 1.4
( 985 1 065 000) 284 300 = 0
2 2 .4 985
V V
V
− + − + −
∴ − + ∴ = =
3 3784 784 261 3 774
2
2
2 2 2
V V V
300 = 0. m /s. kg / m3
ρ . .
Substitute in and find p2 808
= kPa.
M K or 473 C
1 2
1000
1 4 287 283
2 966
808
287 3 774
746
=
× ×
= =
×
=
.
. .
. .
.
T o
M2
261
1 4 287 746
0 477
=
× ×
=
.
. .
b) M M 0.477. kPa
1 2 2 1
1000 1 4 287 283 2 97 10 12 809 6
= × × = ∴ = = =
/ . . . . . .
p p
T2 2
2 644 283 748
809 6
287 748
3 771
= × = ∴ =
×
=
. .
.
.
. .
K or 475 C kg / m3
o
ρ
9.45 a) ρ ρ
1 2 2
12 144
1716 500
002014 002014 3000
=
×
×
= × =
. . . .
slug
ft3
V
Momentum: 2 2
12 144 .002014 3000( 3000).
p V
× − = × −
2 2
2 2
2
3000 1.4
1716 500 0.
2 .4
V p
ρ
 
−
+ − × =
 
 
V
V
V
2
2 2 2
2
6
3000 7
6 042
19 854 6 042 6 006 10
− +





 − − ×
.
( , . ) . = 0.
∴ − + × ∴ = =
6 23 15 10 833 0 00725
2
2
2
6
2 2
V V V
, . .
000 =0. fps.
slug
ft3
ρ
p2 102 9
= . .
psia
M R or 731 F
1 2
3000
1 4 1716 500
2 74
102 9 144
1716 00725
1191
=
× ×
= =
×
×
=
.
. .
.
.
.
T o o

171. 212
M2
833
1 4 1716 1191
0 492
=
× ×
=
.
. .
b) 1 2 2
M 3000/ 1.4 1716 500 2.74. M 0.493. 8.592 12 103.1 psia.
p
= × × = ∴ = = × =
T2 2
2 386 500 1193
103 1 144
1716 1193
0 00725
= × = ∴ =
×
×
=
. .
.
. .
o o
R or 733 F slug / ft3
ρ
9.46
[ ]
ρ
ρ
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 2 2
1
2 1
= =
− +
+
+
+
−





 − +
=
+
+ −
p
p
T
T
k k
k
k
k
k k
k
k
M M
M M
M
M
( ) ( )
( )
.
M1
2 2
1
1
2
1
2
=
+
+
−
k
k
p
p
k
k
. (This is Eq. 9.4.12). Substitute into above:
ρ
ρ
2
1
2
1
2
1
2
1
2 2
1
1 1 1
4 1 1 1
1 1 1
1 1 1
=
+ + + −






+ − + + −






=
+ + + −






+ + − +
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( ) ( )( )
.
k k
p
p
k
k k k
p
p
k
k k
p
p
k
k k k
p
p
=
− + +
+ + −
k k p p
k k p p
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
>> =
+
−
, .
ρ
ρ
9.47 Assume standard conditions: T1 1
15 101
= =
o
C, kPa.
ρ
∴ = × × =
V1 2 1 4 287 288 680
. m / s.
1 2 2
M 2. M .5774. 1.688 288 486 K.
T
= ∴ = = × =
2 4.5 101 454 kPa.
p = × =
2 .5774 1.4 287 486 255 m/s.
V
∴ = × × =
induced 1 2 680 255 425 m/s.
V V V
∴ = − = − =
The high pressure and high induced velocity cause extreme damage.
9.48 If M then M m /s
2 1 1
5 2 645 2 645 1 4 287 293 908
= = ∴ = × × =
. , . . . . .
V
p2 2
8 00 200 1600
1600
287 2 285 293
8 33
= × = =
× ×
=
. .
. ( . )
. .
kPa abs kg /m3
ρ
9.49 If M then M fps
2 1 1
5 2 645 2 645 1 4 1716 520 1118
= = ∴ = × × =
. , . . . . .
V
p2 2
8 00 30 240
240 144
1716 2 285 520
0 01695
= × = =
×
× ×
=
. .
( . )
. .
psia slug /ft3
ρ
V1
V2
stationary
shock

172. 213
9.50 p T
1 1 1
2615 101 26 4 223 3 1000 1 4 287 223 3 3 34
= × = = = × × =
. . . / . . . .
kPa. K. M
∴ = = × = = × =
M kPa. K.
2 2 2
4578 12 85 26 4 339 3 101 223 3 692 5
. . . . . . .
p T
For isentropic flow from ‚ → €: For M = .458, p = .866 p0 and
T T p T
= ∴ = = = =
. . /. . . /. .
960 339 866 391 692 5 960 721
0 0 0
kPa abs K or 448 C
o
9.51 After the shock M kPa abs.
2 2
4752 10 33 800 8264
= = × =
. , .
p
For isentropic flow from ‚ → €: For M = .475, p = .857 p0 .
∴ = =
p0 8264 857 9640
/. .
kPa abs
9.52
A
A
p p p
e
*
. . . . /. . .
= ∴ = = ∴ = =
4 147 985 101 985 102 5
0 0
M kPa abs
e
M kPa. K.
t = = × = = × =
1 5283 102 5 54 15 8333 298 248 3
. . . . . .
p T
t t
∴ =
×
= = × × =
ρt t
V
5415
287 2483
7599 14 287 248 3 3159
.
. .
. . . . .
kg / m m / s.
3
∴ = × × × =
& . . . . .
m 7599 025 315 9 0 471
2
π kg /s If throat area is reduced, Mt
remains at 1, ρ π
t m
= = × × × =
. & . . . . .
7599 7599 02 315 9 0 302
2
kg /m and kg /s
3
9.53 p p
A
A
p p
e = = ∴ = =
101 4 2 94 9 918
2 1 2 1
kPa = M and
. . . , / . .
*
∴ = = = =
p p p
1 1 0
101 9 918 10 18 2 94 0298
/ . . . , / . .
kPa. At M
∴ = =
p0 10 18 0298 342
. /. .
kPa abs
M kPa abs K.
t = = × = = × =
1 5283 342 181 8333 293 244 1
, . . . .
p T
t t
∴ = × × =
Vt 1 4 287 244 1 313
. . .
m /s
M kPa abs K.
1 1 1
2 94 10 18 3665 293 107 4
= = = × =
. , . . . .
p T
∴ = × × =
V1 2 94 1 4 287 107 4 611
. . . .
m /s
M kPa K.
2 2
4788 101 2 609 107 4 280 2
= = = = × =
. , . . . .
p T T
e e
∴ = × × =
V2 4788 1 4 287 280 2 161
. . . .
m /s
9.54 p p
A
A
p p
e = = ∴ = =
14 7 4 2 94 9 918
2 1 2 1
. . . . , / . .
*
psia = M and
∴ = = = =
p p p
1 1 0
14 7 9 918 1 482 2 94 0298
. / . . . , / . .
psia. At M
∴ = =
p0 1 482 0298 49 7
. /. . .
psia
M psia R.
t = = × = = × =
1 5283 49 7 26 3 8333 520 433 3
, . . . . . .
p T
t t
o
∴ = × × =
Vt 1 4 1716 433 3 1020
. . .
fps
M psia R.
1 1 1
2 94 1 482 3665 520 190 6
= = = × =
. , . . . .
p T o
∴ = × × =
V1 2 94 1 4 1716 190 6 1989
. . . .
fps

173. 214
M psia R.
2 2
4788 14 7 2 609 190 6 497 3
= = = = × =
. , . . . . .
p T T
e e
o
∴ = × × =
V2 4788 1 4 1716 497 3 523
. . . .
fps
9.55 M kPa K.
t t t
p T
= = × = = × =
1 5283 500 264 8333 298 248 3
. . . . .
A
A
p
1
2
2 1 1
8
5
2 56 2 47 0613 500 30 65
*
. . . , . . .
= = ∴ = = × =
M
T V
1 1
451 298 134 4 2 47 1 4 287 134 4 574
= × = ∴ = × × =
. . . . . .
K. m /s
M kPa K.
2 2 2
516 6 95 30 65 213 2 108 134 4 283 3
= = × = = × =
. , . . . . . .
p T
After the shock it’s isentropic flow. At M = =
. , . .
*
516 1 314
A
A
p A
02
2
511 500 255 5
04
1 314
003825
= × = =
×
=
. .
.
.
. .
*
kPa. m2
π
A
A
p p
e
e r
*
.
.
. . . . . . . .
=
×
= ∴ = × = =
π 05
003825
2 05 940 255 5 240 298
2
kPa abs = Me
T V
e e
=





 = ∴ = × × =
283 3
213
240
273 8 298 1 4 287 273 8 99
2857
. . . . . .
.
K. m /s
9.56 p p T
t t
= = × = =





 =
. .
. / .
546 546 1200 655 673
655
1200
585
0
3 1 3
kPa. K.
∴ =
×
= = × × = =
ρt t
V
655
462 585
2 42 1 3 462 585 593 1
.
. . . ( .)
kg / m m /s. M
3
t
& . . . . .
m A V
d
d
t t t
t
t
= ∴ = ×
×
× ∴ =
ρ
π
m or 6 cm
4 2 42
4
593 0 060
2
Te e
=





 = ∴ =
×
=
673
101
1200
380 2
101
462 380 2
575
3 1 3
. / .
.
. .
. .
K kg /m3
ρ
Ve
2
2
1872 380 2 1872 673
+ × = ×
. . (Energy from € → e .) (cp = ⋅
1872 J / kg K)
∴ = ∴ = × ∴ =
V
d
d
e
e
e
1050 4 575
4
1050 0 092
2
m /s. m or 9.2 cm
. . . .
π
9.57 M kPa.
e = = = × =
1 546 546 1000 546
0
. . .
p p
e
Te e
=





 = ∴ =
×
=
623
546
1000
542
546
462 542
2 18
3
1 3
.
.
.
. .
K.
kg
m3
ρ
V
d
d
e
e
e
= × × = = × ∴ =
1 3 462 542 571 15 2 18
4
571 0 124
2
. . . . .
m / s. m or 12.4 cm
π

174. 215
9.58 M psia. R.
e = = × = =





 =
1 546 150 81 9 1160
81 9
150
1009
3
1 3
. . .
.
.
.
p T
e e
o
∴ =
×
×
= = × × =
ρe e
V
81 9 144
2762 1009
00423 1 3 2760 1009 1903
.
. . .
slug
ft
fps.
3
. . . . .
25 00423
4
1903 0 199
2
= × ∴ =
πd
d
e
e ft. or 2.39"
9.59 M kPa. K.
t = = × = =





 =
1 546 1200 655 673
655
1200
585
3 1 3
. .
. / .
p T
t t
∴ = × × = =
×
=
Vt t
1 3 462 585 593
655
462 585
2 42
.
.
. .
m / s. kg / m3
ρ
∴ = × × × =
& . . .
m 2 42 0075 593 0 254
2
π kg /s per nozzle
Te =





 =
673
120
1200
396
3 1 3
. / .
.
K
9.60 M1
800
1 4 287 303
2 29
=
× ×
=
.
. .
From Fig. 9.15, β = 46 79
o o
, .
a) 1n
46 . M 2.29sin46 1.65.
β = ∴ = =
o o
2n 2
M .654 M sin(46 20 ).
∴ = = −
o o
∴ =
M2 1 49
. .
p T
2 2
3 01 40 120 4 1 423 303 431
= × = = × =
. . . .
kPa abs K.
V2 1 4 287 431 1 49 620
= × × × =
. . .
m /s
b) β = ∴ = = ∴ = = −
79 2 29 79 2 25 541 79 20
2
o o o o
. . sin . . . sin( ).
M M M
1n 2n
∴ =
M2 0 631
. .
p T
2 2
5 74 40 230 1 90 303 576
= × = = × =
. . .
kPa abs K.
V2 1 4 287 576 631 303
= × × × =
. . .
m /s
c)
V1
V2
= 20o
V1
= 35o
a detached
shock

175. 216
9.61 β θ
1 40 10
= ∴ =
o o
. .
M M M M =1.58.
1n 2n
= = ∴ = = − ∴
2 40 1 29 791 40 10
2 2
sin . . . sin( ).
o o o
If θ β
2 2
10 1 58 51 1 58 51 1 23
= = = = ∴ =
o o o
then, with M M M
2n 2n
. , . . sin . . .
∴ = = − ∴ = = − = − =
M M M
3n . sin( ). . . .
824 51 10 1 26 10 51 10 41
3 3 2
o o o
β β
9.62 M M K.
1n 2n
= = ∴ = = × =
3 5 35 2 01 576 1 696 303 514
2
. sin . . . . .
o
T
M2 1 2 2
576
35 20
2 26 20 47
=
−
= = = ∴ =
.
sin( )
. . . .
o o
o o
θ θ β
M M M M
2n 3n 3
= = ∴ = = − ∴ =
2 26 47 1 65 654 47 20 1 44
3
. sin . . . sin( ). . .
o o o
T V kRT
3 3 3 3
1 423 514 731 1 44 1 4 287 731 780
= × = = = × × =
. . . .
K. M m /s
9.63 M M R.
1n 2n
= = ∴ = = × =
3 5 35 2 01 576 1 696 490 831
2
. sin . . . . .
o o
T
M2 1 2 2
576
35 20
2 26 20 47
=
−
= = = ∴ =
.
sin( )
. . . .
o o
o o
θ θ β
M M M M
2n 3n
= = ∴ = = − ∴ =
2 26 47 1 65 654 47 20 1 44
3 3
. sin . . . sin( ). . .
o o o
T V kRT
3 3 3 3
1 423 831 1180 1 44 1 4 1716 1180 2420
= × = = = × × =
. . . .
o
R. M fps
9.64 M M M
1n 2n
1 1
3 10 28 3 28 1 41 736
= = ∴ = = = ∴ =
, . . sin . . . .
θ β
o o o
∴ = × =
p2 2 153 40 86 1
. . kPa.
M kPa
2 3
736
28 10
2 38 6 442 86 1 555
=
−
= ∴ = × =
.
sin( )
. . . . .
o o
p
( ) . .
p3 10 33 40 413
normal kPa
= × =
9.65 At M1 1 1
3 49 8 19 47
= = =
, . , . .
θ µ
o o
(See Fig. 9.18.)
θ θ
1 2 2
49 8 25 74 8 4 78
+ = + = ∴ =
. . . . .
o
M
From isentropic flow table: p p
p
p
p
p
2 1
0
1
2
0
20
1
02722
002452 1 80
= = × × =
.
. . .
kPa
0 2
2 1 2
1 0
1
253 .1795 127K or 146 C. 12.08 .
.3571
T T
T T
T T
µ
= = × × = − =
o o
V2 4 78 1 4 287 127 1080 90 25 70 53 12 08 32 4
= × × = = + − − =
. . . . . . .
m /s α o
9.66 θ θ
1 26 4 4 65 8
= = =
. . , . .
o o
For M (See Fig. 9.18.)
∴ = − =
θ 65 8 26 4 39 4
. . . .
o
T T
T
T
T
T
V
2 1
0
1
2
0
2
273
1
5556
2381 117 4 1 4 287 117 867
= = × = ∴ = × × =
.
. . .
K. m /s
T2 156
= − o
C.

176. 217
9.67 θ θ
= = =
26 4 4 65 8
. . , . .
o o
For M ∴ = − =
θ 65 8 26 4 39 4
. . . .
o
0 2
2 1
1 0
1
490 .2381 210 R or 250 F.
.5556
T T
T T
T T
= = × = −
o o
V2 4 1 4 1716 210 2840
= × × =
. .
fps
9.68 a) θ θ
1 2 2
39 1 39 1 5 44 1 2 72 20
1
0585
04165
= = + = ∴ = = ×
. . . . . . .
.
.
o o
Mu u
p
= 14.24 kPa.
For θ β
= = = = = ∴ =
5 2 5 27 2 5 27 1 13 889
o o o
and M M M
1n 2n
. , . . sin . . . .
∴ = × =
p2 1 32 20 26 4
l . . .
kPa
M M
l o o
= =
−
=
2
889
27 5
2 37
.
sin( )
. .
b) M M M
1n 2n
= = ∴ = = = =
2 72 5 25 2 72 25 1 15 875
. , . . . sin . , . .
θ β
o o o
M2u =
−
=
.
sin( )
. .
875
25 5
2 56
o o
For M = 36.0 For M2
= = + = =
2 37 36 5 41 2 58
. , . , . .
θ θ
o o
l
c) Force on plate = ( . . ) .
26 4 14 24 1000
− × × =
A F
C
F
V A
A
A
L = =
× ×
× × ×
=
cos . .
. .
. .
5
1
2
12 2 996 1000
1
2
1 4 2 5 20 000
0 139
1 1
2 2
o
ρ
d) C
F
V A
A
A
D = =
× ×
× × ×
=
sin . .
. .
. .
5
1
2
12 2 1000 0872
1
2
1 4 2 5 20 000
0 0122
1 1
2 2
o
ρ
9.69 β = = = ∴ = × = =
19 4 19 1 30 1 805 20 36 1 786
2
o o
. sin . . . . . .
M kPa. M
1n 2n
p
M M
2 1 2 3
786
19 5
3 25 54 36 59 36 3 55
=
−
= = = ∴ =
.
sin( )
. . . . . . . .
o o
θ θ
p p
p
p
p
p
3 2
0
2
3
0
36 1
1
0188
0122 23 4
= = × × =
.
.
. . kPa.
C
A A
V A
D =
−






=
×
× × ×
=
36 1
2
23 4
2
5
1
2
6 35 0872
1
2
1 4 4 20
0 0025
1
2 2
. . sin
. .
.
. .
o
ρ
Lift
Drag
F
Airfoil
surface
M1
M2
M3
p2
p3
shock

177. 218
9.70 If θ β
= = →
5 4
1
o o
with M then Fig. 9.15 =18
, .
1n 2n
M 4sin18 1.24. M .818.
= = ∴ =
o
2 1.627 20 32.5 kPa.
p
∴ = × =
l
∴ =
−
=
M2l o o
.
sin( )
. .
818
18 5
3 64
At M At M2u u
1 1 2 1
0 2
0
4 65 8 75 8 4 88 20
002177
006586
= = = = =
, . . . . .
.
.
θ o o
p p
p
p
p
p
= 6.61 kPa.
C
V A
A A A
A
L = =
− × − × ×
× × ×
=
Lift
1
2
325 5 20 2 6 61 2 10
1
2
14 4 20
0 0854
1
2 2
ρ
. cos / . / cos
.
. .
o o
C
V A
A A
A
D = =
− × ×
× × ×
=
Drag
1
2
32 5 5 6 61 2 10
1
2
1 4 4 20
0 010
1
2 2
ρ
. sin . / sin
.
. .
o o
M1
M2u
M2l
shock
shock