ALL 15 CHAPTERS COVERED
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.
1
,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
2
, 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
3
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.
1
,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
2
, 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
3
