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SI Engine Cycle vs Thermodynamic Otto Cycle
A
I
R
Combustion
Products
Ignition
Intake
Stroke
FUEL
Fuel/Air
Mixture
Air TC
BC
Compression
Stroke
Power
Stroke
Exhaust
Stroke
Qin Qout
Compression
Process
Const volume
heat addition
Process
Expansion
Process
Const volume
heat rejection
Process
Actual
Cycle
Otto
Cycle
SIKLUS OTTO
Process 0 1 intake
Process 1 2 Isentropic compression
Process 2 3 Constant volume heat addition (isovolumetric)
Process 3 4 Isentropic expansion
Process 4 1 Constant volume heat rejection (isovolumetric)
Process 1 0 Exhaust
Qout
Qin
3
4
2
1
v
v
v
vrv
Analysis of Otto Cycle
12 Isentropic Compression, entropy (s) constant
m
W
m
Quu in )( 12
k
vrv
v
T
T
P
P
2
1
1
2
1
2
AIR
)( 12 uum
Win
1
1
2
1
1
2
k
v
k
rv
v
T
T)()( 1212 TTcuu
m
Wv
in
1
2
1
2
r
r
v
v
v
v Jika s1= s2
cold air standar analysis constant k
air standar analysis from tabel
23 Constant Volume Heat Addition
m
W
m
Quu in )()( 23
)( 23 uum
Qin
2
3
2
3
T
T
P
P
AIR Qin
TC
Analysis of Otto Cycle
air standar analysis from tabel
)( 23 TTcm
Qv
in
cold air standar analysis constant k
3 4 Isentropic Expansion
AIR m
W
m
Quu out )( 34
)( 43 uum
Wout
krv
v
T
T
P
P 1
4
3
3
4
3
4
1
1
4
3
3
4 1
k
k
rv
v
T
T
Analysis of Otto Cycle
air standar analysis from tabel
cold air standar analysis constant k
)( 43 TTcm
Wv
out
3
4
3
4
r
r
v
v
v
v Jika s3= s4
4 1 Constant Volume Heat Removal
AIR Qout m
W
m
Quu out )( 41
)( 41 uum
Qout 1
1
4
4
T
P
T
P
BC
Analysis of Otto Cycle
air standar analysis from tabel
cold air standar analysis constant k
)( 41 TTcm
Qv
out
23
1243
uu
uuuu
Q
W
in
cycle
th
23
14
23
1423 1uu
uu
uu
uuuu
Cycle indicated thermal efficiency:
Net cycle work:
1243)( uumuumWWW inoutcycle
1
2
1
23
14 111
)(
)(1
k
v
vth
rT
T
TTc
TTc
Analysis of Otto Cycle
Air standar analysis:
Cold Air standar analysis:
Work per cycle is represented in terms of a mean effective
pressure and the displacement.
p
V
MEP = Mean effective pressure
X = Displacement
thin
thincycle
r
r
u
mQ
kr
r
VP
Q
P
imep
VV
Wimep
1
/
1
1
1 111121
Indicated mean effective pressure is:
Indicated mean effective pressure
Efek Kompresi rasio terhadap effisiensi thermal
1
11
kconst cth
rV
Traditional
SI engines
9 < r < 11
k = 1.4
• Spark ignition engine compression ratio limited by T3 (autoignition)
and P3 (material strength), both ~rk
• For r = 8 the efficiency is 56%
Efek Specific Heat Ratio terhadap Efisiensi Thermal
1
11
kconst cth
rV
Specific heat
ratio (k)
Cylinder temperatures vary between 300K and 2000K so 1.2 < k < 1.4
k = 1.3 most representative
The net cycle work of an engine can be increased by either:
i) Increasing the r (1’2)
ii) Increase Qin (23”)
P
V2 V1
Qin Wcycle
1
2
3
(i)
4
(ii)
Faktor yang berpengaruh terhadap kerja per siklus
1’
4’
4’’
3’’
thincycle
r
r
V
Q
VV
Wimep
1121
Otto Cycle
9.1 An air-standard Otto cycle has a compression ratio of 8.5. At
the beginning of compression, p1= 100 kPa and T1 = 300 K.
The heat addition per unit mass of air is 1400 kJ/kg. Determine
(a) the net work, in kJ per kg of air.
(b) the thermal efficiency of the cycle.
(c) the mean effective pressure, in kPa.
(d) the maximum temperature in the cycle, in K.
9.2 Solve Problem 9.1 on a cold air-standard basis with specific
heats evaluated at 300 K.
Latihan
9.3 At the beginning of the compression process of an air standard Otto
cycle, p1 1 bar, T1 290 OK, V1 400 cm3. The maximum temperature in the
cycle is 2200 OK and the compression ratio is 8. Determine
(a) the heat addition, in kJ.
(b) the net work, in kJ.
(c) the thermal efficiency.
(d) the mean effective pressure, in bar.
9.4 Solve Problem 9.3 on a cold air-standard basis with specific heats
evaluated at 300 OK.
Latihan