tpk-2 minggu 3
TRANSCRIPT
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TEKNIK PERMESINAN KAPAL II(Minggu – 3)
LS 1329 ( 3 SKS)Jurusan Teknik Sistem Perkapalan
ITS Surabaya
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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=η
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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 engineOtto cycleDiesel cycle
Gas turbines Brayton cycle
RefrigerationReversed 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
2π
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 (andrejection) process;
Assuming constant specific heat:
QQ - 1 =
QQ - Q =
H
L
H
LHthη
T C m = Q vin ∆
1-TTT
1 - TTT
-1 =)T - T( C m)T - T( C m - 1 =
2
32
1
41
23v
14vthη
T C m = Q v ∆Rej
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For an isentropic compression (and expansion) process:
where: γ = Cp/Cv
Then, by transposing,
TT =
VV =
VV =
TT
4
3
3
41-
2
11-
1
2
γγ
TT =
TT
1
4
2
3
Otto Cycle Derivation
TT-1 =
2
1thηLeading 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 ratioand is equal to the expansion ratio in an ottocycle 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 γ
γη
)r( = VV =
TT -1
v1
2-1
2
1 γγ
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
LHthη T C m = Q v ∆
1-TTT
1 - TTT
-1 =
2
32
1
41
thη
)r( = VV =
TT -1
v1
2-1
2
1 γγ
)r(1 - 1 = )r( - 1 = 1-v
-1vth γ
γη
TT =
TT
1
4
2
3
2
11TT th −=η
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 – 1The air at the beginning of the compression stroke of an air-standard Otto cycle is at 95 kPa and 22°C 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 −− −=−=η
585.0=η
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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