Solutions Manual for Fluid Mechanics – 3rd Edition by R.C. Hibbeler

,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.
2 3 2 -6 2 2
c) N>ks = N>(10 s) = 10 N>s = mN>s Ans.
3 -6 9
d) kN>ms = (10 )N>(10 )s = 10 N>s = GN>s Ans.




Ans:
a) kN
m
b) Gg>
m
c) µN>
s2
d) GN>
s

,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) 3 723( 106 ) 4 1>2 mm.




SOLUTION
a) (425 mN)2 = 3425(10-3 ) N 4 2 = 0 .181 N2 Ans.
2
b) (67 300 ms) = 367.3(10 )(10 ) s 4
3 -3 2
= 4.53(10 ) s
3 2
Ans.
c) 3723(106 ) 41>2 mm = 3723(106 ) 41>2(10-3 ) m = 26.9 m Ans.




Ans:
a) 0.181 N2
b) 4.53(103 ) s2
c) 26.9 m

,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 = 334(10 ) m4 30.0763(10 ) s4 > 3263(10-6 )( 103 ) g4
-3 6


= 9.86(106 ) m # s>kg = 9.86 Mm # s>kg Ans.
c) (4.78 mm)(263 Mg) = 34.78(10 ) m 4 3263(10 ) g 4
-3 6


= 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 #

,*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
5
a) TC
(TF - 32)
= 9
5
20°C = - 32)
(T
9 F
TF = 68.0°F Ans.
b) TK = TC + 273
500 K = TC +
273
TC = 227°C Ans.
c) TR = TF + 460
TR = 125°F + 460 = 585°R Ans.
5
d) TC (T - 32)
9 F
=
5
(215°F - 32) = 102°C Ans.
TC 9
=

,1–5. Mercury has a specific weight of 133 kN>m 3 when
the temperature is 20°C. Determine its density and
specific gravity at this temperature.




SOLUTION
g rg
133(103 ) N>m 3 = rHg(9.81 m>s2 )
rHg = 13 558 kg>m 3 = 13.6 Mg>m 3 Ans.
rHg
SHg 13 558 = 13.6 Ans.
rw =
= kg>m 3
1000
kg>m 3




Ans:
rHg = 13.6 Mg>m 3
SHg = 13.6

,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 the fuel when the tanks are
completely full.

A




SOLUTION
The specific weight of the fuel is
g = rg = ( 1.32 slug>ft3 )( 32.2 ft>s 2 ) = 42.504

lb>ft 3 Then, the weight of the fuel is
W = gV = (42.504 lb>ft 3 )( 50 ft3 ) = 2.13(103 ) lb = 2.13 kip Ans.




Ans:
g = 42.5 lb>ft 3
W = 2.13 kip

,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 = rRT
680(103 ) N>m 2 = r(286.9 J>kg # K)(70° + 273) K
r = 6.910
kg>m 3 The weight of the air in the tank is
W = rg V = ( 6.910 kg>m 3 )( 9.81 m>s2 )( 1.35 m3 )
= 91.5 N Ans.




Ans:
91.5 N

,*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 = rRT
12( 10 ) N>m 2 = r(259.8 J>kg # K)(30° + 273) K
6


r = 152.44 kg>m 3
The mass of oxygen in the tank is
m = r V = ( 152.44 kg>m 3 )( 0.12 m3 )
= 18.3 kg
Ans.

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