Solutions Manual – Fox and McDonald’s Introduction to Fluid Mechanics 10th Edition | PDF

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SOLUTION MANUAL

,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.




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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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