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solutions manual Fluid Mechanics by Russell C. Hibbeler 1st Edition provides a comprehensive and well-illustrated introduction to the theory and application of Fluid Mechanics (14 chapter with answers). The text presents a commitment to the development of student problem-solving skills and features many of the same pedagogical aids unique to Hibbeler texts.
Chapter 1 Fundamental Concepts
Chapter 2 Fluid Statics
Chapter 3 Kinematics of Fluid Motion
Chapter 4 Conservation of Mass
Chapter 5 Energy of Moving Fluids
Chapter 6 Fluid Momentum
Chapter 7 Differential Fluid Flow
Chapter 8 Dimensional Analysis and Similitude
Chapter 9 Viscous Flow Within Enclosed Surfaces
Chapter 10 Analysis and Design for Pipe Flow
Chapter 11 Viscous Flow Over External Surfaces
Chapter 12 Turbomachinery
Chapter 13 Open Channel Flow
Download Fluid Mechanics by Russell C. Hibbeler 1st Edition solutions manual pdf
1. 1
1-1. Represent each of the following quantities with combi-
nations of units in the correct SI form, using an appropriate
prefix: (a) GN • mm, (b) kg / mm, (c) N / ks2
, (d) kN / ms.
SOLUTION
a) GN • mm = (109
)N(10-6
)m = 103
N • m = kN • m Ans.
b) kg / mm = (103
)g / (10-6
)m = 109
g / m = Gg / m Ans.
c) N / ks2
= N / (103
s)2
= 10-6
N / s2
= mN / s2
Ans.
d) kN / ms = (103
)N / (10-6
)s = 109
N/s = GN / s Ans.
Ans:
a) kN • m
b) Gg / m
c) pN / s2
d) GN / s
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2. 2
1-2. Evaluate each of the following to three signifi-
cant figures, and express each answer in SI units using
an appropriate prefix: (a) (425 mN)2
, (b) (67 300 ms)2
,
(c) [ 723(106
) ]1/2
mm.
SOLUTION
a) (425 mN)2
= [ 425(l0-3
) N ]2
= 0.181 N2
b) (67 300 ms)2
= [67.3(103
)(10-3
) s]2
= 4.53(103
) s2
c) [723(106
) ]1/2
mm = [723(106
) ] 1/2
(10-3
) m = 26.9 m
2
Ans.
2 2 2
Ans.
Ans.
Ans:
a) 0.181 N2
b) 4.53(103
) s
c) 26.9 m
2
3. 3
1-3. Evaluate each of the following to three signifi-
cant figures, and express each answer in SI units using
an appropriate prefix: (a) 749 mm
/63 ms, (b) (34 mm)
(0.0763 Ms) /263 mg, (c) (4.78 mm)(263 Mg).
SOLUTION
a) 749 mm / 63 ms = 749(10-6
) m / 63(10-3
) s = 11.88(10-3
) m / s
= 11.9 mm/s Ans.
b) (34 mm)(0.0763 Ms)/263 mg = [34(10-3
) m] [0.0763(106
) s]/[263(10-6
)(103
) g]
= 9.86(106
) m • s / kg = 9.86 Mm• s / kg Ans.
c) (4.78 mm)(263 Mg) = [4.78(10-3
) m] [263(106
) g]
4. 4
= 1.257(106
) g • m = 1.26 Mg • m Ans.
Ans:
a) 11.9 mm/s
b) 9.86 Mm•
s/kg
c) 1.26 Mg •
m
5. 5
*1-4. Convert the following temperatures: (a) 20°C to
degrees Fahrenheit, (b) 500 K to degrees Celsius, (c) 125°F
to degrees Rankine, (d) 215°F to degrees Celsius.
SOLUTION
a) TC = 5
(TF - 32)
20°C = 55 (TF - 32)
TF = 68.0°F
b) TK = TC + 273
500 K = TC + 273
TC = 227°C
c) TR = TF + 460
TR = 125°F + 460 = 585°R
d) TC = 5
(TF - 32)
TC = 5 (215°F - 32) = 102°C
Ans.
Ans.
Ans.
Ans.
6. 6
1-5. Mercury has a specific weight of 133 kN / m3
when the
temperature is 20°C. Determine its density and specific
gravity at this temperature.
SOLUTION
g = Pg
133(103
) N/m3
= pHg(9.81 m/s2
)
pHg = 13 558 kg / m3 = 13.6 Mg / m3 Ans.
P
Hg 13 558 kg/m3
SH = — =----------------- = 13.6
Hg Pw 1000 kg/m3
Ans.
Ans:
p
Hg =
^Hg =
13.6 Mg/m3
13.6
7. 7
1-6. The fuel for a jet engine has a density of 1.32 slug / ft3
.
If the total volume of fuel tanks A is 50 ft3
, determine the
weight of thefuel when the tanks are completely full.
SOLUTION
The specific weight of thefuel is
g = pg = (1.32 slug/ft3
)(32.2 ft / s2
) = 42.504 lb/ft3
Then, theweight of the fuel is
W = gV = (42.504 lb/ft3
)(50 ft3
) = 2.13(103
) lb = 2.13 kip Ans.
Ans:
g = 42.5 lb/ft3
W = 2.13 kip
8. 8
1-7. If air within the tank is at an absolute pressure of
680 kPa and a temperature of 70°C, determine the weight
of the air inside the tank. The tank has an interior volume
of 1.35 m3
.
SOLUTION
From the table in Appendix A, the gas constant for air is R = 286.9 J / kg • K.
P = PRT
680(103
) N/m2
= p(286.9 J/kg • K)(70° + 273) K
p = 6.910 kg / m3
The weight of the air in the tank is
W = pg V =
= 91.5 N (6.910 kg/m3
)(9.81 m / s2
)(1.35 m3
)
Ans.
Ans:
91.5 N
9. 9
*1-8. The bottle tank has a volume of 1.12 m3
and contains
oxygen at an absolute pressure of 12 MPa and a temperature
of 30°C. Determine the mass of oxygen in the tank.
SOLUTION
From the table in Appendix A, the gas constant for oxygen is R = 259.8 J / kg • K.
p = pR
T
12(106
) N/m2
= p(259.8 J/kg • K)(30° + 273) K
p = 152.44 kg / m3
The mass of oxygen in the tank is
m = pV = (152.44 kg/m3
)(0.12 m3
)
= 18.3 kg Ans.
10. 10
1-9. The bottle tank has a volume of 0.12 m3
and contains
oxygen at an absolute pressure of 8 MPa and temperature
of 20°C. Plot the variation of the temperature in the tank
(horizontal axis) versus the pressure for 20°C < T < 80°C.
Report values in increments of AT = 10°C.
SOLUTION
TC (°C) 20 30 40 50 60 70 80
p(MPa) 8.00 8.27 8.55 8.82 9.09 9.37 9.64
p(MPa)
10 -
From the table in Appendix A, the gas constant for oxygen is R = 259.8 J / (kg • K).
For T = (20°C + 273) K = 293 K,
p = pRT
8(106
) N/m2
= p [ 259.8 J/(kg • K) ] (293 K)
p = 105.10 kg/m3
Since themass and volume of theoxygen in the tank remain constant, its density will
also be constant.
p = pRT
p = (105.10 kg/m3
) [259.8 J/(kg • K) ] (TC + 273)
p = (0.02730 TC + 7.4539)(106
) Pa
p = (0.02730TC + 7.4539) MPawhere TC is in °C. (a)
The plot of p vs TC is shown in Fig. a.
0 10 20 30 40 50 60 70 80
TC(°C)
Ans:
p = (0.0273 Tc + 7.45) MPa, where Tc is in C°
11. 11
1-10. Determine the specific weight of carbon dioxide
when the temperature is 100°C and the absolute pressure
is 400 kPa.
SOLUTION
From the table in Appendix A, the gas constant for carbon dioxide is R = 188.9 J / kg • K.
p = pRT
400(103
) N/m2
= p(188.9 J/kg • K)(100° + 273) K
p = 5.677 kg / m3
The specific weight of carbon dioxide is
g = pg = (5.677 kg / m3
)(9.81 m / s2
)
= 55.7 N/m3
Ans.
Ans:
55.7 N / m3
12. 12
1-11. Determine the specific weight of air when the
temperatureis 100°F and the absolute pressureis 80 psi.
SOLUTION
From the table in Appendix A, the gas constant for the air is R = 1716 ft • lb / slug • R.
p
12 in. 2p = pRT
2 12 in. 2
#
lb/in' (if J = P(1716 ft • l80 lb/in2
( ] = p(1716 ft • lb/slug • R)(100° + 460) R
p = 0.01200 slug/ft3
The specific weight of theair is
g = pg = (0.01200 slug/ft3
) (32.2 ft/s2
)
= 0.386 lb / ft3
Ans.
Ans:
0.386 lb / ft3
13. 13
*1-12. Dry air at 25°C has a density of 1.23 kg / m3
. But if it
has 100% humidity at the same pressure, its density is
0.65% less. At what temperature would dry air produce this
same density?
SOLUTION
For both cases, the pressures are the same. Applyingtheideal gas
with p1 = 1.23 kg/m3
, p2 = (1.23 kg/m3
)(1 - 0.0065) = 1.222005 kg/m3
T1 = (25°C + 273) = 298 K,
p = p1 RT1 = (1.23 kg / m3
) R (298 K) = 366.54 R
Then
p = p2RT2; 366.54 R = (1.222005 kg / m3
)R(Tc + 273)
TC = 26.9°C
law
and
Ans.
14. 14
1-13. The tanker carries 1.5(106
) barrels of crude oil in its
hold. Determine the weight of the oil if its specific gravity is
0.940. Each barrel contains 42 gallons, and there are
7.48 gal/ft3
.
SOLUTION
The specific weight of theoil is
go = Sogw = 0.940(62.4 lb/ft3
) = 58.656 lb/ft3
Weight of one barrel of oil:
1 ft3
Wb = goV = (58.656 lb/ft3
)(42 gal/bl) (7-48gal J
= 329.4 lb / bl
Totalweight:
W = 1.5(106
) bl(329.4 lb/bl)
17. 17
1-14. Water in the swimming pool has a measured depth
of 3.03 m when the temperature is 5°C. Determine its
approximate depth when the temperature becomes 35°C.
Neglect losses due to evaporation.
9 m
4 m
SOLUTION
From Appendix A, at Tj = 5°C, (pw) j = 1000.0 kg/m3
. The volume of the water is
V = Ah. Thus, Vj = (9 m)(4 m)(3.03 m). Then
m „„„„ „ , , ^ m
V ;
(
Pw)
1 =JT ; 1000
.0 k
g/m3 = o. 2
36 m2(3.03 m)
m = 109.08(103
) kg
At T2 = 35°C, (pw)2 = 994.0 kg/m3
. Then
m , 109.08(103)
(Pw)2 " U 994 0 kg/m “ H6m‘)h_
h = 3.048 m = 3.05 m Ans.
Ans:
3.05 m
18. 18
1-15. The tank contains air at a temperature of 15°C and
an absolute pressure of 210 kPa. If the volume of the tank is
5 m3
and the temperature rises to 30°C, determine the mass
of air that must be removed from the tank to maintain the
same pressure.
SOLUTION
For T1 = (15 + 273) K = 288 K and R = 286.9 J/kg • K for air, the ideal gas
law gives
p1 = pRT; 210(103
) N / m2
= p1(286.9 J / kg • K)(288 K)
p1 = 2.5415 kg/m3
Thus, the mass of air at T1 is
mi = piV = (2.5415 kg / m3
)(5 m3
) = 12.70768 kg
For T2 = (273 + 30) K = 303 K and R = 286.9 J/kg #
K
p2 = p1RT2; 210(103
) N/m2
= p2(286.9 J/kg • K)(303 K)
p2 = 2.4157 kg / m3
Thus, the mass of air at T2 is m2 = p2V = (2.4157 kg / m3
) (5 m3
) = 12.07886 kg
Finally, the mass of air that must be removed is
Am = m1 — m2 = 12.70768 kg — 12.07886 kg = 0.629 kg Ans.
Ans:
0.629 kg
19. 19
*1-16. The tank contains 2 kg of air at an absolute pressure
of 400 kPa and a temperature of 20°C. If 0.6 kg of air is added
to the tank and the temperature rises to 32°C, determine the
pressurein the tank.
SOLUTION
For T1 = 20 + 273 = 293 K, p1 = 400 kPa and R = 286.9 J / kg • K for air, the ideal
gas law gives
p1 = p1RT1; 400(103
) N/m2
= p1(286.9 J/kg • K)(293 K)
p1 = 4.7584 kg / m3
Since thevolume is constant. Then
m1 m2 m2V
= = ; p
2 = p
1
p1 p2 m1
Here m1 = 2 kg and m2 = (2 + 0.6) kg = 2.6 kg
P2 = (gggg) (4.7584 kg / m3
) = 6.1859 kg / m3
Again applyingthe ideal gas law with T2 = (32 + 273) K = 305 K
p2 = p2RT2 = (6.1859 kg/m3
)(286.9 J/kg • k)(305 K) = 541.30(103
) Pa
= 541 kPa Ans.
20. 20
1-17. The tank initially contains carbon dioxide at an
absolute pressure of 200 kPa and temperature of 50°C. As
more carbon dioxide is added, the pressure is increasing at
25 kPa / min. Plot the variation of the pressure in the tank
(vertical axis) versus the temperature for the first 10 minutes.
Report thevalues in increments of two minutes.
SOLUTION
From the table in Appendix A, the gas constant for carbon dioxide is
R = 188.9 J/(kg • K). For T = (50°C + 273) K = 323 K,
p = pRT
200(103
) N/m2
= p [ 188.9 J / (kg • K) ] (323 K)
p = 3.2779 kg / m3
Since the mass and the volume of carbon dioxide in the tank remain constant, its
density will also be constant.
p = pRT
p = (3.2779 kg/m3
)[188.9 J/(kg • K)] (TC + 273) K
p = (0.6192 TC + 169.04)(103
) Pa
p = (0.6192 TC + 169.04) kPa where TC is in °C
The plot of p vs TC is shown in Fig. a
Ans:
p = (0.619 Tc + 169) kPa, where Tc is in C°
p(kPa)
200
225 250 275 300 325
TC(°C) 50.00 90.38 130.75 171.12 211.50 251.88
p(kPa)
21. 21
1-18. Kerosene has a specific weight of gk = 50.5 lb / ft3
and benzene has a specific weight of gb = 56.2 lb/ft3
.
Determine the amount of kerosene that should be mixed
with 8 lb of benzene so that the combined mixture has a
specific weight of g = 52.0 lb/ft3
.
SOLUTION
The volumes of benzene and kerosene are given by
The specific weight of mixture is
Wk = 20.13 lb = 20.1 lb
56.2 lb / ft
.3
50.5 lb / ft3
8 lb
¥b = 0.1423 ft3
¥k = 0.019802 Wk k
52.0 lb / ft
3 Wk + 8 lb
0.1423 ft3
+ 0.019802 Wk
Ans.
24. 24
*1-20. Kerosene is mixed with 10 ft3
of ethyl alcohol so
that the volume of the mixture in the tank becomes 14 ft3
.
Determine the specific weight and the specific gravity of the
mixture.
SOLUTION
From AppendixA,
Pk = 1.58 slug / ft3
Pea = 1.53 slug /
ft3
The volume of kerosene is
Vk = 14 ft3
- 10 ft3
= 4 ft3
Then thetotal weight of the mixture is therefore
W = Pkg Vk + Pea g Vea
= (1.58 slug/ft3
)(32.2 ft/s2
)(4 ft3
) + (1.53 slug/ft3
)(32.2
ft/s2
)(10 ft3
)
= 696.16 lb
The specific weight and specific gravity of the mixture are
solut ion- m anual- for- fluid- mechanics- 1st- edit ion-by-russell- c-hibbeler
gm =
e =
m
W
V
gm
gw
696.16 lb _
14 ft3
=
49.73 lb/ft3
62.4 lb / ft3
49.73 lb/ft3
= 49.7 lb / ft3
= 0.797
Ans.
Ans.
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solution-manual/
25. 25
Download full verion of Fluid Mechanics Hibbeler 1st Edition solutions manual
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