8- 95

8-73
Air at 96 kPa, 27 ºC enters an eight-cylinder, four-stroke Otto cycle that operates at 3,000 RPM.
Each
cylinder has a bore of 9 cm and a stroke of 8.5 cm.
At top dead center, the volume is 15% of the cylinder
volume at bottom dead center.
The maximum temperature in the cycle is 2200 K. Using an air-standard
analysis, determine:
a. the net work per cycle per cylinder (in kJ)
b. the cycle thermal efficiency
c. the power developed (in kW and hp).
Approach:
For a cycle net work is equal to net heat transfer.
Apply conservation of energy to the heat addition and
heat rejection processes to determine the heat transfer.
Evaluate properties using the ideal gas tables.
Assumptions:
1.
Neglect potential and kinetic energy effects.
2.
All processes are ideal.
3.
Air is the working fluid and is an ideal gas.
Solution:
a)
The net work per cycle
23
41
net
net
W
= Q
Q
Q
=
−
Applying the closed system energy equation to these processes, neglecting potential and kinetic energy effects,
and noting that no work occurs during processes 2-3 and 4-1:
(
)
23
3
2
Q
m u
u
=
−
and
(
)
41
4
1
Q
m u
u
=
−
Evaluating the properties from the air tables and using the ideal gas equation to determine the mass:
1
1
1
PV M
m
RT
=
Volume 1 needs to be determined.
From the given geometry
(
)(
) (
)
2
2
4
1
2
4
4
0.09m
0.085m
5.41 10
m
V
V
B S
π
π
−
−
=
=
=
×
3
1
V
We are also given
.
Therefore, combining these two equations we obtain
2
0.15
V
=
and
4
3
1
6.36
10
m
V
−
=
×
4
3
2
0.954
10
m
V
−
=
×
The compression ratio is
1
2
6.67
V
r
V
V
=
=
(
)(
)
(
)
(
)(
)
2
4
3
4
1
1
1
96kN m
6.36
10
m
28.97kg kmol
7.09
10
kg
8.314kJ
kmol K
300K
PV M
m
RT
−
−
×
=
=
=
×
From the air table, Appendix A-9:
(
)
1
300K
214.02kJ kg
u
=
1
621.2
r
v
=
(
)
(
)
2
1
2
1
621.2 1 6.67
93.18
r
r
v
v
V
V
=
=
=
by interpolation
2
457.1kJ kg
u
=
(
)
3
2200 K
1872.4kJ kg
u
=
3
2.012
r
v
=
(
)
(
)
4
3
4
3
4
2.012 6.67
13.41
959.9kJ kg
r
r
v
v
V
V
u
=
=
=
→
=
Therefore,
(
)
(
)
4
23
7.09
10
kg
1872.4
457.1 kJ kg
1.003kJ
Q
−
=
×
−
=
Answer
(
)
(
)
4
41
7.09
10
959.9
214.07
0.529kJ
Q
−
=
×
−
=
Answer
1.003
0.529
0.475kJ
net
W
=
−
=
Answer
Note that these results are per cylinder
b)
Cycle thermal efficiency
23
0.475kJ
0.473
1.003kJ
net
net
cycle
in
W
W
=
=
Q
Q
η
=
=
Answer
c) Power developed can be calculated following the procedure given in Example 8-15.
(
)
kJ
rev
1min
1kWs
0.475
8cylinders
3000
cylinder-powerstroke
min
60s
1 kJ
rev
2
powerstroke
net
net
W
n N
W
X
⎛
⎞
⎛
⎞
⎛
⎞⎛
⎞
⎜
⎟
⎜
⎟
⎜
⎟⎜
⎝
⎠⎝
⎠
⎝
⎠
⎝
⎠
=
=
±
⎟
95kW
127hp
=
=
Answer
8- 96

8-74
An eight-cylinder, four-stroke Otto cycle runs at 5000 rpm.
The compression ratio is 12.2, and the engine
displacement is 396 in
3
.
Air enters the engine at 14.4 psia, 70 ºF. The maximum temperature in the engine
is 4000 ºF. Using an air-standard analysis, determine:
a. the net work per cycle per cylinder (in Btu)
b. the cycle thermal efficiency
c. the net power output (in kW and hp)
d. the airflow through the engine (in ft
3
/min).
Approach:
The approach used in Example 8-15 can be followed.
Assumptions:
1.
Neglect potential and kinetic energy effects.

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