Download - TPK-2 Minggu 3
TEKNIK PERMESINAN KAPAL II(Minggu – 3)
LS 1329 ( 3 SKS)Jurusan Teknik Sistem Perkapalan
ITS Surabaya
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
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
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.
Gas cycles have many engineering applications
Internal combustion engineOtto cycleDiesel cycle
Gas turbines Brayton cycle
RefrigerationReversed Brayton cycle
Some nomenclature before starting internal combustion engine cycles
More terminology
Terminology
Bore = d Stroke = s Displacement volume =DV = Clearance volume = CV Compression ratio = r
4ds
2π
CVCVDVr +
=TDC
BDC
VV
=
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
−=
The net work output of a cycle is equivalent to the product of the mean effect pressure and the displacement volume
Real Otto cycle
Real and Idealized Cycle
Otto Cycle P-V & T-s Diagrams
Pressure-Volume Temperature-Entropy
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
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
Differences between Otto and Carnot cycles
T
s
1
2
3
4
T
s
1
2
3
4
2
3
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
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
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
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
Effect of compression ratio on Otto cycle efficiency
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.
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
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!
Get T3 with first law:
( )23v23 TTmcQ −=Solve for T3:
2v
3 TcqT += K705.6
kgkJ0.855
kg6.29x10kJ8.6 3
+=−
K2304.7T3 =
Thermal Efficiency
11.41k 911
r11 −− −=−=η
585.0=η
Sample Problem – 2
Solution P
v
1
2
3
4
Diesel Cycle P-V & T-s Diagrams
Sample Problem – 3
Gasoline vs. Diesel Engine