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MAE 301: Engineering Thermodynamics
Spring 2010
HOMEWORK #6
(
Solutions
)
Problem 1.
Problem 5.41 from the book.
Solution:
This is a steadyflow process since there is no change with time. Nitrogen is
an ideal gas with variable specific heats.
Potential energy changes are negligible.
The
device is adiabatic and thus heat transfer is negligible.
There are no work interactions.
The molar mass of nitrogen is
M
= 28 kg/kmol (Table A1). The enthalpies are (Table A
18)
kJ/kmol
8580
K
295
=
C
22
kJ/kmol
8141
K
280
=
C
7
2
2
1
1
=
→
°
=
=
→
°
=
h
T
h
T
(a
) There is only one inlet and one exit, and thus
&&&
mmm
12
=
=
. We take diffuser as the
system, which is a control volume since mass crosses the boundary. The energy balance
for this steadyflow system can be expressed in the rate form as
2
2
0
0)
pe
W
(since
/2)
+
(
)
2
/
(
2
1
2
2
1
2
2
1
2
2
1
2
2
2
2
2
1
1
V
V
M
h
h
V
V
h
h
Q
V
h
m
V
h
m
−
+
−
=
−
+
−
=
≅
Δ
≅
≅
=
+
&
&
&
&
,
Substituting,
()
( )
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
+
−
=
2
2
2
2
2
/s
m
1000
kJ/kg
1
2
m/s
200
kg/kmol
28
kJ/kmol
8141
8580
0
V
It yields
V
2
= 93.0 m/s
(
b
)
The ratio of the inlet to exit area is determined from the conservation of mass
relation,
or,
(
)
(
)
0.625
=
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
=
⎯→
⎯
=
m/s
200
kPa
K/85
295
m/s
93.0
kPa
K/60
280
/
/
/
/
1
1
1
2
2
2
1
1
2
1
1
2
2
2
1
1
1
2
2
1
2
1
1
1
1
2
2
2
V
V
P
T
P
T
A
A
V
V
P
RT
P
RT
V
V
A
A
V
A
V
A
v
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Problem 2.
Problem 5.49 from the book.
Solution:
Air is an ideal gas.
The flow is steady.
The gas constant of air is
R
= 0.287
kPa
⋅
m
3
/kg
⋅
K (Table A1).
The specific volumes of air at the inlet and outlet are
/kg
m
2506
.
0
kPa
1000
K)
273
K)(600
/kg
m
kPa
287
.
0
(
3
3
1
1
1
=
+
⋅
⋅
=
=
P
RT
v
/kg
m
3575
.
1
kPa
100
K)
273
K)(200
/kg
m
kPa
287
.
0
(
3
3
2
2
2
=
+
⋅
⋅
=
=
P
RT
The mass flow rate is
kg/s
11.97
=
=
=
/kg
m
0.2506
m/s)
)(30
m
1
.
0
(
3
2
1
1
1
V
A
m
&
The outlet area is
2
m
1.605
=
=
=
m/s
10
/kg)
m
75
kg/s)(1.35
97
.
11
(
3
2
2
2
V
m
A
&
Problem 3.
Problem 5.51 from the book.
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 Spring '08
 Hassan

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