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Assignment #8
PROBLEM 8.1
PROBLEM STATEMENT:
A thermodynamic cycle using air as the working fluid in a
closed system is outlined below. The specific heats of air can be assumed
constant throughout the cycle with values taken at the initial conditions. The four
processes that make up the cycle are:
Process 12: Air at standard conditions (State 1: 1 atm, 298 K) is heated at
constant volume to state 2, where
2
1
P =10P.
Process 23: The air at State 2 is expanded isobarically to State 3, where
3
2
v = 3v .
Process 34: The air is expanded isentropically until it returns to pressure P
1
at
State 4.
Process 41: The air is compressed isobarically to return it to State 1.
(a)
Sketch the P v
and
T s diagrams of this cycle. Pay special attention to the
relative slopes of the process lines.
(b)
Find the
numerical value
of the power cycle efficiency of this heat engine.
GIVEN:
Air, P
1
=
1 atm, T
1
=
298 K
Process 12: Constant volume to P
2
=10P
1
Process 23: Constant pressure (isobaric) to v
3
=3v
2
Process 34: Constant entropy (isentropic) to P
4
=P
1
Process 41: Constant pressure (isobaric) to v
1
FIND:
.
HE
P v
and
T s
diagram,
η
ASSUMPTIONS:
Ideal gas, constant specific heats.
GOVERNING RELATIONS:
1.
For constant specific heats,
(
29
(
29
∂
∂
∂
∂
V
P
V
P
T s
= T c
and
T s
= T c
2.
For an ideal gas, R
=
c
P
–
c
V
and
k
=
c
P
/
c
V
3.
Ideal gas equation, Pv =RT
4.
PT relation for isentropic process, constant specific heats:
(
29
(
29
(k1) k
b
a
b
a
T T = P P
5.
Thermal efficiency,
HE
NET
IN
= w
q
η
QUANTITATIVE SOLUTION:
a). The P

v diagram can be derived from the known processes. For the T  s
diagram, note that for the constant volume process,
∂
∂
V
V
T
T
=
s
c
and for the isobaric process,
∂
∂
P
P
V
T
T
T
=
=
s
c
R +c
Thus, the slope for the constant volume process (Process 12) is steeper than
the isobaric processes. The diagrams for this cycle are shown below.
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View Full Document Assignment #8
b). The temperature of each state can be found in terms of T
1
. Also, the thermal
efficiency can be expressed in terms of T
1
and k, where k is 1.4001 from Table
5s for T
=
25°C (298 K).
For State 2: (from the ideal gas equation)
P v
R
1
P v
=
T
R
2
2
1
1
1
2
P
T = T
=10T
P
T
⇒
For State 3:
P
v
R
3
P
=
T
v
R
3
3
2
1
1
2
2
v
T = T
= (10T ) 3 = 30T
v
T
→
×
For State 4:
k1
1.40011
k
1.4001
4
4
3
1
1
3
P
1
T = T
= 30T
=15.54T
P
10
Now apply the first law to determine the work and heat transfer for each
process:
Process 12:
,
,
(
)
IN 12
IN 12
2
1
V
2
1
w
0(constant volume)
q
u
u
c T
T
=
=

=

Process 23: Constant pressure
,
,
,
(
)
(
)
(
)
(
)
(
)
(
)
3
OUT 23
2
3
2
3
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This note was uploaded on 04/20/2010 for the course M E 320 taught by Professor Deinert during the Spring '08 term at University of Texas at Austin.
 Spring '08
 DEINERT

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