Spring 2006
1
ME 201
Thermodynamics
Homework #5 Solutions
1. (5 pts) Calculate the entropy change for N
2
as it goes from 250 K and 1000 kPa to
1300 K and 60 kPa.
Solution:
Substance Type: Ideal Gas (N
2
)
Problem Type: Process
State 1
State2
T
1
= 250 K
T
2
= 1300 K
P
1
= 1000 kPa
P
2
= 60 kPa
Using our air tables the entropy change will be given by
Δ
s =

 R
ln
P
P
2
1
2
1
φ
φ
⋅
°
±
²
³
´
µ
From the N
2
tables we have
φ
2
= 236.831/ 28.013 = 8.454 kJ / kg
K
⋅
φ
1
= 186.370 / 28.013 = 6.653 kJ / kg
K
⋅
The gas constant for N
2
is R = 0.2968 kJ/kg
⋅
K, so that upon substituting we get
Δ
s = 8.454  6.653  0.2968)ln
60
1000
=
2.636 kJ
/
kg
K
(
°
±
²
³
´
µ
⋅
.
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ME 201 Thermodynamics
Spring 2006
2
2. (20 pts) For the two processes given below, determine the final temperature,
pressure, specific volume, and the changes in internal energy, enthalpy, and
entropy.
a. Air at 8400 R and 66 psia goes isothermally to 1.7 psia.
Solution:
Substance Type: Ideal Gas (Air)
Problem Type: Process (Isothermal)
State 1
State2
T
1
= 8400 R
T
2
=
8400 R
P
1
= 66 psia
P
2
= 1.7 psia
Bold
values are calculated
We note that to fix state 2 we need another property, but since the process is
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 Spring '06
 SOMERTON
 Thermodynamics, Enthalpy, Energy

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