motor bakar minggu-2
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thanksTRANSCRIPT
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MOTOR BAKAR
( 3 SKS) Jurusan Teknik Mesin
Sekolah Tinggi Teknologi Angkatan Laut (STTAL)
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Thermodynamic Principles All internal combustion Open cycle, heated engine
Gasoline (Otto) engine Spark ignition Compresses air-fuel mixture
Diesel engine Compressed ignition Compresses air only
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INTERNAL COMBUSTION ENGINE:
AN ENGINE THAT PRODUCES POWER BY BURNING FUEL INSIDE A COMBUSTION CHAMBER WITHIN THE ENGINE
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Gas Cycles
Carnot Cycle
T2
T1
s1 s2
Work W
1
2 3
4
1-2 - ADIABATIC COMPRESSION (ISENTROPIC) 2-3 - HEAT ADDITION (ISOTHERMAL) 3-4 - ADIABATIC EXPANSION (ISENTROPIC) 4-1 - WORK (ISOTHERMAL)
Heat Q
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Carnot Cycle
Carnot cycle is the most efficient cycle that can be executed between a heat source and a heat sink.
However, isothermal heat transfer is difficult to obtain in reality--requires large heat exchangers and a lot of time.
2
1
TT-1=
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Carnot Cycle
Therefore, the very important (reversible) Carnot cycle, composed of two reversible isothermal processes and two reversible adiabatic processes, is never realized as a practical matter.
Its real value is as a standard of comparison for all other cycles.
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Gas cycles have many engineering applications
Internal combustion engine Otto cycle Diesel cycle
Gas turbines Brayton cycle
Refrigeration Reversed Brayton cycle
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Some nomenclature before starting internal combustion engine cycles
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More terminology
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Terminology
Bore = d Stroke = s Displacement volume =DV = Clearance volume = CV Compression ratio = r
4ds
2
CVCVDVr +=
TDC
BDC
VV
=
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Mean Effective Pressure
Mean Effective Pressure (MEP) is a fictitious pressure, such that if it acted on the piston during the entire power stroke, it would produce the same amount of net work.
minmax VVWMEP net
=
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The net work output of a cycle is equivalent to the product of the mean effect pressure and the displacement volume
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Real Otto cycle
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Real and Idealized Cycle
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Otto Cycle P-V & T-s Diagrams
Pressure-Volume Temperature-Entropy
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Otto Cycle Derivation
Thermal Efficiency: For a constant volume heat addition (and
rejection) process; Assuming constant specific heat:
QQ - 1 =
QQ - Q =
H
L
H
LHth
T C m = Q vin
1-TTT
1 - TTT
-1 =)T - T( C m)T - T( C m - 1 =
2
32
1
41
23v
14vth
T C m = Q v Rej
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For an isentropic compression (and expansion) process:
where: = Cp/Cv
Then, by transposing,
TT =
VV =
VV =
TT
4
3
3
41-
2
11-
1
2
TT =
TT
1
4
2
3
Otto Cycle Derivation
TT-1 =
2
1thLeading to
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Differences between Otto and Carnot cycles
T
s
1
2
3
4
T
s
1
2
3
4
2
3
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The compression ratio (rv) is a volume ratio and is equal to the expansion ratio in an otto cycle engine.
Compression Ratio
VV =
VV = r
3
4
2
1v
1 + vv = r
vv + v =
volume Clearancevolume Total = r
cc
sv
cc
ccsv
where Compression ratio is defined as
Otto Cycle Derivation
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Then by substitution,
)r(1 - 1 = )r( - 1 = 1-v
-1vth
)r( = VV =
TT -1
v1
2-1
2
1
The air standard thermal efficiency of the Otto cycle then becomes:
Otto Cycle Derivation
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Summarizing
QQ - 1 =
QQ - Q =
H
L
H
LHth T C m = Q v
1-TTT
1 - TTT
-1 =
2
32
1
41
th
)r( = VV =
TT -1
v1
2-1
2
1
)r(1 - 1 = )r( - 1 = 1-v
-1vth
TT =
TT
1
4
2
3
2
11TT th =
where
and then
Isentropic behavior
Otto Cycle Derivation
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Heat addition (Q) is accomplished through fuel combustion
Q = Lower Heat Value (LHV) BTU/lb, kJ/kg
Q AF m =Q fuelain cycle
Otto Cycle Derivation
T C m = Q vin also
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Effect of compression ratio on Otto cycle efficiency
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Sample Problem 1 The air at the beginning of the compression stroke of an air-standard Otto cycle is at 95 kPa and 22C and the cylinder volume is 5600 cm3. The compression ratio is 9 and 8.6 kJ are added during the heat addition process. Calculate: (a) the temperature and pressure after the compression and heat addition process (b) the thermal efficiency of the cycle
Use cold air cycle assumptions.
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Draw cycle and label points
P
v
1
2
3
4
T1 = 295 K
P1 = 95 kPa
r = V1 /V2 = V4 /V3 = 9
Q23 = 8.6 kJ
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Carry through with solution
kg 10 x 29.6RT
VPm 3-1
11 ==
Calculate mass of air:
Compression occurs from 1 to 2:
ncompressio isentropic VVTT
1
2
112
=
k
( ) ( ) 11.42 9K 27322T +=K 705.6T2 = But we need T3!
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Get T3 with first law:
( )23v23 TTmcQ =Solve for T3:
2v
3 TcqT += K705.6
kgkJ0.855
kg6.29x10kJ8.6 3+=
K2304.7T3 =
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Thermal Efficiency
11.41k 911
r11 ==
585.0=
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Sample Problem 2
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Solution P
v
1
2
3
4
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Diesel Cycle P-V & T-s Diagrams
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Sample Problem 3
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Gasoline vs. Diesel Engine
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MOTOR BAKARThermodynamic PrinciplesSlide Number 3Gas CyclesSlide Number 5Carnot CycleCarnot CycleGas cycles have many engineering applicationsSome nomenclature before starting internal combustion engine cyclesMore terminologyTerminologyMean Effective PressureThe net work output of a cycle is equivalent to the product of the mean effect pressure and the displacement volumeReal Otto cycleReal and Idealized CycleOtto Cycle P-V & T-s DiagramsOtto Cycle DerivationOtto Cycle DerivationDifferences between Otto and Carnot cyclesOtto Cycle DerivationOtto Cycle DerivationOtto Cycle DerivationOtto Cycle DerivationEffect of compression ratio on Otto cycle efficiencySample Problem 1Draw cycle and label pointsCarry through with solutionGet T3 with first law:Thermal EfficiencySlide Number 30SolutionSlide Number 32Slide Number 33Slide Number 34Slide Number 35Slide Number 36Slide Number 37Slide Number 38Slide Number 39Gasoline vs. Diesel Engine