Solutions for Exercises – Introduction to Fluid Mechanics 4th Edition by William Janna

1-1
CHAPTER 1
1.1 a) 3 teaspoons in 1 tablespoon
b) 236.5 ml in 1 cup
1.2 m = 6 slugs, g = 32.0 ft/s2, W = mg = 6(32) = 192 lbf
1.3 m = 46 kg, W = 450 N, g =
W
m
=
450
46
= 9.78 m/s2
1.4 128 fl oz · 2.957 x 10-5 = 3.78 x 10-3 m3
3.78 l · 1 x 10-3 = 3.78 x 10-3 m3
1 gallon · 3.785 x 10-3 = 3.78 x 10-3 m3 all equivalent
m
V
—
= ρ; m = ρV
— = 1 030(3.78 x 10-3 m3)
m = 3.9 kg
1.5 1/2 gallon · 3.785 x 10-3 = 1.89 x 10-3 m3
2 liter · 1 x 10-3 = 2 x 10-3 m3 (2 - 1.89)/2 = 0.055 = 5.5%
close enough to allow use of same container
1.6 V
— = 1 ft3; W = 62.4 lbf; W = mg; m =
W
g
=
62.4
32.2
a) m = 1.94 slug Wmoon = mg = 1.94(32.2/6) or
b) W = 10.4 lbf
1.7
BTU
hr
·
1 hr
3600 s
·
1054
745.7
= 3.926 x 10-4
HP
(BTU/hr)
1.8 1 gallon · 3.785 x 10-3 = 3.785 x 10-3 m3
m = 1 000 kg/m3(3.785 x 10-3 m3) = 3.785 kg
W = mg = 3.785(9.81) = 37.13 N = 8.35 lbf
1.9 Plot of data
1.10 ρ = 1 000 kg/m3 (SI); ρ =
1 000
16.01
= 62.5 lbm/ft3
ρ =
1 000
16.01(32.2)
= 1.94 slug/ft3 ρ =
1 000
1 000
= 1 g/cm3
s
m
t
b
9
8
@
g
m
a
i
l
.
c
o
m
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1-2
1.11 Sphere D = 8000 miles · 5280 ft/mile = 4.224 x 107 ft
V
— =
πD3
6
=
π(4.224 x 107)3
6
= 5.92 x 1022 ft3 ρ = 6 560 kg/m3 = 410 lbm/ft3
m = ρV
— = 410(5.92 x 1022) = 2.43 x 1025 lbm
m = 2.43 x 1025 · 4.535 x 10-1 = 1.1 x 1025 kg = m
1.12 Liquid weighs 1 - 1/2 = 1/2 lbf; ρg =
0.5 lbf
8 ounces
; using conversions,
ρg =
0.5 lbf
8 oz
·
4.448
2.957 x 10-5 = 9.4 x 103 N/m3 = SW
ρ =
9 400
9.81
= 958 kg/m3 = ρ ρ =
958
515.379
= 1.86 slug/ft3 = ρ
ρg = 1.86(32.2) = 59.9 lbf/ft3 = SW
1.13 V
— = 1 ft3; Carbon tetrachloride Appendix Table A-5, ρ = 1.59(1.94) slug/ft3;
W = mg = ρV
—g = 1.59(1.94)(1)(32.2) = 99.3 lbf; W = 99.3(4.448) or
W = 442 N
1.14 V
— = 5 ft3, Kerosene Appendix Table A-5, ρ = 0.823(1.94 slug/ft3);
m = ρV
— = 0.823(1.94)(5) = 7.99 slug; m = 7.99(14.59) or
m = 116.5 kg
1.15 minimum shear required is τo; movement impending when τ = τo
τo = 4 N/m2 =
force
area
=
mg
A
;
so m =
Aτo
g
=
0.5 m2 (4 N/m2)
9.81 m/s2 = 0.204 N·s2/m = 0.204 kg
so for movement to begin, m > 0.204 kg
1.16 m = 0.25 kg; W = mg = 0.25(9.81) = 2.45 N; τ =
F
A
=
2.45
0.5
= 4.9 N/m2
By definition, τ = τo + µo
∆V
∆y
; 4.9 = 4 + 4 x 10-3
∆V
0.005
∆V = 1.13 m/s
1.17 τ = µ
∆V
∆y
; τ =
W
A
=
W
0.5 m2
Table A-5 for turpentine, µ = 1.375 x 10-3 N·s/m2

1-3
∆V = 0.05 m/s; By substitution,
W
0.5
= 1.375 x 10-3
0.05
0.005
W = 0.007 N
1.18
F
A
= µ
∆V
∆y
; F = 0.89 N; A = 0.16 m2; ∆V = 0.12 m/s;
Table A-5, µ = 1.53 x 10-5 N·s/m2
∆y =
µ∆VA
F
=
1.53 x 10-5 (0.12)(0.16)
8.9
or
∆y = 3.3 x 10-5 m
1.19
F
A
= µ
∆V
∆y
; F = 0.025 kg(9.81 m/s2) = 0.245 N
∆y = 0.01 m; µ = 650 x 10-3 N·s/m2; A = 0.75 m2
∆V =
0.245(0.01)
0.75(650 x 10-3)
= 5.03 x 10-3 m/s
so V = 5.03 mm/s
1.20 Plot of data
1.21
F
A
= µ
∆V
∆y
; A = 2 sides ·
2.5 m2
side
= 5 m2; ∆V = 0.002 5 m/s; ∆y = 0.012 m
Table A-5 µ = 1.64 x 10-3 N·s/m2
so F =
5(1.64 x 10-3)(0.002 5)
0.012
or
F = 1.708 x 10-3 N
1.22
F
A
= µ
∆V
∆y
; F = 2 lbf; µ = 1.11 x 10-5 lbf·s/ft2 ; ∆V = 12 in./s = 1 ft/s
∆y = (0.05/12) ft; 2A =
F∆y
µ∆V
=
2(0.05/12)
1.11 x 10-5 (1)
= 750 ft2
A = 375 ft2
1.23 τ is the same on both sides. Given ∆y1 + ∆y2 = 1 - 0.1 or
∆y1 + ∆y2 = 0.9 cm = 0.009 m
τ = µ1
∆V
∆y1
= µ2
∆V
∆y2
so
µ1
∆y1
=
µ2
∆y2
From Table A-5, we get

1-4
µ1
µ2
=
16.2
42
=
∆y1
∆y2
or
0.009 m - ∆y2
∆y2
= 0.386 1.386∆y2 = 0.009
Solving,
∆y2 = 0.006 5 m = 0.65 cm ∆y1 = 0.9 - 0.65 or
∆y1 = 0.35 cm
1.24 The shear applied to both fluids is the same. Thus on the left,
τ = µ1
∆V
∆y1
and on the right, τ = µ2
∆V
∆y2
; so
µ1
∆y1
=
µ2
∆y2
From Table A-5, µ1 = 1.095 x 10-3 N·s/m2
Also, ∆y1 = 2∆y2 ; With µ2 = µ1
∆y2
∆y1
= 1.095 x 10-3 · (1/2) or
µ2 = 0.548 x 10-3 N·s/m2
1.25
200
100
0
0
2
4
6
8
10
dV/dy, 1/s
t,
N/m^2
Bingham Plastic

1-5
1.26
120
100
80
60
40
20
0
0.000
0.002
0.004
0.006
0.008
0.010
0.012
dV/dy, 1/s
t,
N/m^2
Pseudoplastic
1.27
60
50
40
30
20
10
0
0
10
20
dV/dy, 1/s
t,
lbf/ft^2
Newtonian (close enough)

1-6
1.28
1500
1000
500
0
0
200
400
600
800
1000
1200
dV/dy, 1/s
t,
N/m^2
Dilatant
1.29
Kraft Mayo
0
50
100
150
200
250
300
350
0 200 400 600 800 1000 1200
dV/dy in 1/s
shear
stress
in
Pa
pseudoplastic with an initial yield stress

1-7
1.30
Honey
0
100
200
300
400
500
600
700
800
900
1000
0 20 40 60 80 100
dV/dy in 1/s
shear
stress
in
Pa
Newtonian
1.31 A = 0.09 · 0.016 = 0.001 4 m2 ; V = 5 cm/s = 0.05 m/s
∆y = 2 mm = 0.002 m
a) τ =
F
A
=
0.07
0.001 4
= 48.6 N/m2
b)
dV
dy
=
∆V
∆y
=
0.05
0.002
= 25/s
c) µ =
τ
dV/dy
=
48.6
25
= 1.94 N·s/m2
1.32 Assume oil is Newtonian so τ = µ
dV
dy
; sleeve is stationary, shaft velocity is
V = 5 in./s = 0.417 ft/s; τ =
F
A
; A = area of contact. Surface area of shaft
does not equal surface of sleeve, so take an average. For shaft,
A = πDL = π(4/12)(12/12) = 1.047 ft2 ; for the sleeve,
A = π(4.01/12)(12/12) = 1.05 ft2 ; use A =
1.05 + 1.047
2
= 1.048 ft2
τ =
25
1.048
= 23.8 lbf/ft2 ;
dV
dy
=
∆V
∆y
=
5/12 - 0
0.005/12
= 1000
µ =
τ
dV/dy
=
23.8
1000
or µ = 23.8 x 10-3 lbf·s/ft2

1-8
1.33 Pseudoplastic τ = K




dV
dy
n
; substituting,
4.63 x 10-2 = K(25)n and 6.52 x 10-2 = K(50)n ; dividing gives
4.63
6.52
=




25
50
n
; which becomes 0.71 = (0.5)n ; ln(0.71) = n ln(0.5);
n = 0.494
4.63 x 10-2 = K(25)0.494 ; K =
4.63 x 10-2
4.91
K = 9.44 x 10-3
Check with second equation: 9.44 x 10-3 (50)0.494 = 6.52 x 10-2 which is OK.
τ = 9.44 x 10-3 (dV/dy)0.494 ; when τ = 7 x 10-2 ,
7 x 10-2 = 9.44 x 10-3




dV
dy
0.494
;
dV
dy
= (7.42)1/0.494 or
dV
dy
= 57.7 rad/s
1.34 τ = K




dV
dy
n
; 8.72 x 10-3 = K(20)n and 2.10 x 10-2 = K(40)n ; dividing,
8.72 x 10-3
2.10 x 10-2 =




20
40
n
; 0.415 = (0.5)n ; ln(0.415) = n ln(0.5)
n = 1.27
8.27 x 10-3 = K(20)1.27 ; K = 1.95 x 10-4
Check 1.95 x 10-4 (40)1.27 = 2.1 x 10-2 OK
τ = 1.95 x 10-4




dV
dy
1.27
; τ = 3 x 10-2
dV
dy
=




3 x 10-2
1.95 x 10-4
1/1.27
dV
dy
= 53.1 rad/s
1.35 µo = 0.029 lbf·s/ft2 ; τ = 2.7 lbf/ft2 ;
dV
dy
= 74.5 rad/s
τ = τo + µo
dV
dy
; 2.7 = τo + 0.029(74.5); τo = 2.7 - 0.029(74.5)
τo = 0.54 lbf/ft2

1-9
1.36 Table A-5 for acetone ρ = 0.787(1 000) = 787 kg/m3
µ = 0.316 x 10-3 N·s/m2 ν = µ/ρ = 4.015 x 10-7 m2/s
Using conversions from Table A-1,
ν =
4.015 x 10-7
9.290304 x 10-2 ;
ν = 4.32 x 10-6 ft2/s (Engineering system, Brit Grav & Brit abs)
ν =
4.015 x 10-7
(1 m/100 cm)2
ν = 4.015 x 10-3 cm-3/s (CGS)
1.37 At 0°C, Table A-4 µ = 1.787 x 10-3 N·s/m2
At 100°C, µ = 0.2818 x 10-3 N·s/m2
% change =
µ100 - µ0
µ0
=
0.2818 - 1.787
1.787
; % change = - 84%
At 0°C, ν = 1.787 x 10-6 m2/s
At 100°C ν = 0.2940 x 10-6
% change =
µ100 - µ0
µ0
=
0.2940 - 1.787
1.787
; % change = - 83.5%
1.38 µ = 8 cp · 1 x 10-3 = 8 x 10-3 N·s/m2 ; ρ = 59 lbm/ft3 · 16.01 = 945 kg/m3
ν =
µ
ρ
=
8 x 10-3
945
= 8.47 x 10-6 m2/s · (1002 cm2/m2)
ν = 8.47 x 10-2 cm2/s
1.39 pi - po =
2σ
R
; po = 101 300 N/m2 ; R = 0.001 m; σ = 23.1 x 10-3 N/m, so
pi = 101 300 +
2(23.1 x 10-3)
0.001
= 101 346 N/m2
1.40 pi - po =
2σ
R
; po = 70 000 N/m2 ; R = 250 x 10-6 m; σ = 72 x 10-3 N/m, so
pi = 70 000 +
2(72 x 10-3)
250 x 10-6 = 70 576 N/m2
1.41 pi - po =
2σ
R
; D = 1/16 in. = 0.0052 ft; R = 0.0026 ft
po = 14.7(144) = 2117 lbf/ft2 ; σ = 27.14 x 10-3 N/m (Table A-5); converting,

1-10
σ =
27.14 x 10-3
4.448
(0.3048) = 1.86 x 10-3 lbf/ft so
pi = 2117 +
2(1.86 x 10-3)
0.0026
or pi = 2118 lbf/ft2
1.42 Benzene σ = 28.2 x 10-3 N/m (Table A-9); D = 1 mm;
R = 0.5 mm = 0.000 5 m; po = 100 000 N/m2 ; pi - po =
2σ
R
;
pi = 100 000 +
2(28.2 x 10-3)
0.000 5
; solving,
a) pi = 1.001 x 105 N/m2 (benzene)
b) pi - po = 113 N/m2 =
2σ
R
for Hg; R =
2σ
113
; σ = 484 x 10-3 N/m (from
Appendix Table A-5 or A-9); R =
2(0.484)
113
or
b) R = 0.008 6 m = 8.6 mm
1.43 h =
2σ
ρRg
cos θ; ρ = 1 000 kg/m3, R = 0.002 m, θ = 0°
H2O at room temperature, Table A-5; σ = 71.97 x 10-3 N/m
h =
2(71.97 x 10-3)
1 000(0.002)(9.81)
; solving,
a) h = 0.007 337 m = 7.34 mm
Table A-5 for Hg, σ = 484 x 10-3 N/m; ρ = 13.6(1 000)
h =
2(484 x 10-3) cos 140°
13.6(1 000)(9.81)(0.002)
; and
b) h = - 2.8 x 10-3 m = - 2.8 mm

1-11
1.44 h =
2σ
ρRg
; R = 0.003 m; h =
2σ
ρRg
=
2σ
ρ(0.003)(9.81)
= 67.96
σ
ρ
Table A-4
T, °C σ, N/m ρ, kg/m3 h, m
0 75.6 x 10-3 0.9999(1 000) 0.00514
10 74.2 0.9997 0.00504
18 73.1 0.9986 0.00497
30 71.2 0.9957 0.00486
40 69.6 0.9922 0.00477
50 67.9 0.9881 0.00467
60 66.2 0.9832 0.00458
70 64.4 0.9778 0.00448
80 62.6 0.9718 0.00438
100 58.9 0.9584 0.00418
100
80
60
40
20
0
0.0040
0.0042
0.0044
0.0046
0.0048
0.0050
0.0052
T in deg C
h
in
m

1-12
1.45 σ =
xyθ
2
ρg ; σ = 71.97 x 10-3 N/m; ρ = 1 000 kg/m3 ; θ = 1° · 2π/360
θ = 0.017 45 rad; substituting,
71.97 x 10-3 = xy




0.017 45
2
(1 000)(9.81); xy = 8.048 x 10-4 m2
x y
1 8.408
2 4.2
4 2.1
8 1.05
8
6
4
2
0
0
2
4
6
8
10
x
y
1.46 h =
2σ
ρRg
cos θ; θ = 140° for Hg; Table A-5 for Hg, ρ = 13.6(1.94 slug/ft3)
D = 0.2 in. = 0.0166 ft; R = 0.00833 ft; h = - 0.052 in. = - 0.0043 ft
σ =
ρRhg
2 cos θ
=
13.6(1.94)(0.00833)(- 0.0043)(32.2)
2 cos 140°
σ = 0.0198 lbf/ft
1.47 θ = 0°; R = 0.002 5 m, ethyl alcohol, Table A-5; σ = 22.33 x 10-3 N/m;
ρ = 787 kg/m3 ; h =
2σ
ρRg
=
2(22.33 x 10-3)
787(0.002 5)(9.81)
h = 2.31 mm

1-13
1.48 θ = 0°; R = 0.002 m, benzene, Table A-5; σ = 28.18 x 10-3 N/m;
ρ = 876 kg/m3 ; h =
2σ
ρRg
=
2(28.18 x 10-3)
876(0.002)(9.81)
h = 3.28 mm
1.49 θ = 0°; R = 0.001 5 m, carbon tet, Table A-5; σ = 26.3 x 10-3 N/m;
ρ = 1 590 kg/m3 ; h =
2σ
ρRg
=
2(26.3 x 10-3)
1 590(0.001 5)(9.81)
h = 2.25 mm
1.50 θ = 0°; R = 0.001 25 m, glycerin, Table A-5; σ = 63.0 x 10-3 N/m;
ρ = 1 263 kg/m3 ; h =
2σ
ρRg
=
2(63.0 x 10-3)
1 263(0.001 25)(9.81)
h = 8.14 mm
1.51 θ = 0°; R = 0.001 m, octane, Table A-5; σ = 21.14 x 10-3 N/m;
ρ = 701 kg/m3 ; h =
2σ
ρRg
=
2(21.14 x 10-3)
701(0.001)(9.81)
h = 6.15 mm
1.52 θ = 140°; R = 0.005 m, Hg, Table A-5; σ = 484 x 10-3 N/m;
ρ = 13 600 kg/m3 ; h =
2σ
ρRg
cos θ =
2(484 x 10-3)
13 600(0.005)(9.81)
cos 140°
h = - 1.11 mm
1.53 m = 0.1 slug; ∆T = 25°F; Table A-6 for air, cp = 7.72 BTU/slug·°R
~
Q = mcp∆T = 0.1(7.72)(25)
~
Q = 19.3 BTU
1.54 CO2 Appendix Table A-6 cp = 876 J/(kg·K); m = 1.5 kg; ∆T = 25°C
~
Q = mcp∆T = 1.5(876)(25) = 3.28 x 104 J;
~
Q/m = 3.28 x 104/1.5 = 2.2 x 104 J/kg

1-14
1.55
~
Q = mcv∆T; Table A-6, cp = 523 J/(kg·K); cp/cv = 1.67; so we calculate
cv = 313 J/(kg·K);
~
Q = 8 kg(313 J/(kg·K))(∆T) = 50 000 J; solving,
∆T = 20.0 K increase in temperature
1.56 CO2 Appendix Table A-6 cp = 876 J/(kg·K); γ = 1.30;
cv =
876
1.3
= 674 J/(kg·K); ∆u = cv∆T = 674(50 - 25)
∆u = 1.68 x 104 J/kg
1.57 m = 1 kg; ∆T = 25°C
from Table A-6
He cp = 5 188 J/(kg·K)
H2 cp = 14 310 J/(kg·K)
~
Q = mcp∆T = 1(5 188)(25)
~
Q = 1.3 x 105 J for He
~
Q = mcp∆T = 1(14 310)(25)
~
Q = 3.6 x 105 J for H2
∆h =
~
Q
m
=
1.3 x 105
1
; ∆h = 1.3 x 105 J/kg for He
∆h =
~
Q
m
=
3.6 x 105
1
; ∆h = 3.6 x 105 J/kg for H2
Hydrogen requires more energy
1.58 Air loses
~
Q = mcv(120 - Tf)
H2 gains
~
Q = mcv(Tf - 60)
mair = 2mH2 ; Table A-6 gives
air cp = 7.72 BTU/slug·°R; cp/cv = 1.4
H2 cp = 110 BTU/slug·°R; cp/cv = 1.405
cvair =
7.72
1.4
= 5.51; cvH2 =
110
1.405
= 78.3
~
Q are equal so; mcv(120 - Tf)|air = mcv(Tf - 60)|H2
substituting, 2mH2(5.51)(120 - Tf) = mH2(78.3)(Tf - 60);
1322 - 11.02Tf = 78.3Tf - 4698; 6020 = 89.32Tf
T = 67.4°F

Solutions for Exercises – Introduction to Fluid Mechanics 4th Edition by William Janna

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