Solutions Manual for Engineering Fluid Mechanics 9th Edition 2026/2027 | Complete Worked Solutions

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PROBLEM 2.1
Situation: An engineer needs density for an experiment with a glider.
Local temperature = 74.3 ◦ F = 296.7 K.
Local pressure = 27.3 in.-Hg = 92.45 kPa.
Find: (a) Calculate density using local conditions.
(b) Compare calculated density with the value from Table A.2, and make a recom-
mendation.
J
Properties: From Table A.2, Rair = 287 kg· K
, ρ = 1.22 kg/ m3 .

APPROACH
Apply the ideal gas law for local conditions.

ANALYSIS

a.) Ideal gas law
p
ρ =
RT
92, 450 N/ m2
=
(287 kg/ m3 ) (296.7 K)
= 1.086 kg/m3


ρ = 1.09 kg/m3 (local conditions)
b.) Table value. From Table A.2

ρ = 1.22 kg/m3 (table value)


COMMENTS

1. The density difference (local conditions versus table value) is about 12%. Most
of this difference is due to the effect of elevation on atmospheric pressure.

2. Answer ⇒ Recommendation—use the local value of density because the effects
of elevation are significant.




1

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,h including
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PROBLEM 2.2
Situation: Carbon dioxide is at 300 kPa and 60o C.
Find: Density and specific weight of CO2 .
Properties: From Table A.2, RCO2 = 189 J/kg·K.

APPROACH
First, apply the ideal gas law to find density. Then, calculate specific weight using
γ = ρg.

ANALYSIS
Ideal gas law

P
ρCO2 =
RT
300, 000
=
189(60 + 273)
= 4.767 kg/m3



Specific weight
γ = ρg
Thus


γ CO2 = ρCO2 × g
= 4.767 × 9.81
= 46.764 N/m3




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PROBLEM 2.3
Situation: Methane is at 500 kPa and 60o C.
Find: Density and specific weight.
J
Properties: From Table A.2, RMethane = 518 kg· K
.

APPROACH
First, apply the ideal gas law to find density. Then, calculate specific weight using
γ = ρg.

ANALYSIS
Ideal gas law

P
ρHe =
RT
500, 000
=
518(60 + 273)
= 2.89 kg/m3

Specific weight
γ = ρg
Thus


γ He = ρHe × g
= 2.89 × 9.81
= 28.4 N/m3




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PROBLEM 2.4
Situation: Natural gas (10 ◦ C) is stored in a spherical tank. Atmospheric pressure is
100 kPa.
Initial tank pressure is 100 kPa-gage. Final tank pressure is 200 kPa-gage.
Temperature is constant at 10 ◦ C.
Find: Ratio of final mass to initial mass in the tank.

APPROACH
Use the ideal gas law to develop a formula for the ratio of final mass to initial mass.

ANALYSIS
Mass
M = ρV (1)
Ideal gas law
p
ρ= (2)
RT
Combine Eqs. (1) and (2)

M = ρV−
−
= (p/RT )V

Volume and gas temperature are constant so
M2 p2
=
M1 p1
and

M2 300 kPa
=
M1 200 kPa
= 1.5




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