siklus diesel
TRANSCRIPT
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DIESEL CYCLE
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Proses Siklus DieselProses Siklus Diesel
1-2 - KOMPRESI ADIABATIK (ISENTROPIC)
2-3 - PEMASUKAN PANAS PADA TEKANAN KONSTAN
3-4 - KOMPRESI ADIABATIK (ISENTROPIC)
4-1 - PENGELUARAN PANAS PADA VOLUME KONSTAN
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Efisiensi SiklusEfisiensi Siklus
in
net
q
w
Thermal Efficiency:
Berarti harus tahu qin dan Wnet
Bagaimana menerapkan hukum pertama, dalam sistem tertutup?
q - w = u
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Langkah kompresi 1 ke 2 (second law) Isentropic s1 = s2 (adiabatic and reversible) Langkah kompresi isentropis dari 1 ke 2
0q12 121212 uuwq 0uuw 2112
1212in uuww
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Pemasukan panas pd tekanan konstan, 2 ke3232323 uuwq
2323 hhqqin )()( 2323223 uuvvPq
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Langkah ekspansi isentropik, S3 = S4, 3 ke 4
q34 = 0 (adiabatic and reversible)
343434 uuwq
4334out uuww
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Pengeluaran panas pada volume konstan
414141 uuwq
1441out uuqq so,0Pdvwbut 41
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Cycle PerformanceCycle Performanceoutinnet qqw Get net work from energy balance of cycle:
Efficiency is then:
Substituting for qin and qout: netw )uu()hh( 1423
in
net
q
w
23
1423
hh
)u(u)h(h
23
14, hh
uu1
Dieselth
subtitusikan kerja dan pamasukan panas:
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KarenaKarena Constant volume: V1 = V4 ,
2
1
r
r
s2
1
2
1
v
v
V
Vr
V
V
efisiensi dengan asumsi udara dalam keadaan dingin (cold efisiensi dengan asumsi udara dalam keadaan dingin (cold air cycle assumptions)air cycle assumptions)
Jika kita mengasumsikan nilai kapasitas panas konstan:
)T(TC
)T(TC1
hh
uu1
23p
14v
23
14
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Cycle performance with Cycle performance with cold air cycle assumptionscold air cycle assumptionsBecause we’ve got two isentropic (adiabatic and reversible) processes in the cycle, T1 can be related to T2, and T3 can be related to T4 with our ideal gas isentropic relationships….
1
1
2
1
1
2
1
1
2
k
kk
k
rV
V
P
P
T
T
1k
4
3
1
4
3
3
4
V
V
P
P
T
T
k
k
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Cycle performance with Cycle performance with cold air cycle assumptionscold air cycle assumptionsPada tekanan konstan (isobaris) berlaku hubungan :
3
33
2
22
T
VP
T
VP
2
3
2
3
T
T
V
V
Proses isobaris P2 = P3 :
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)1/(
)1/(1
232
141
TTkT
TTT)Tk(T
)T(T1
)T(TC
)T(TC1
23
14
23p
14v
Cycle PerformanceCycle Performance Jika : k
C
C
v
P
)1(
111
1
1,c
kc
kDieselth rk
r
r
Kita sekarang mendefinisikan nilai baru, cutoff ratio rc
What are the limitations for this expression?
2
3
V
Vrc
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Contoh soalContoh soalSebuah mesin dengan siklus diesel ideal bekerja pada ratio tekanan 18 dan cutoff ratio 2. Pada awal langkah kompresi tekanan udara adalah 14,7 psi, 80oF dan 117 in3. Dengan asumsi keadaan udara dingin standar {R=0,3704 Psi.ft3/(lbm.R); Cp=0,240 Btu/(lbm.R); Cv=0,171 Btu/(lbm.R) }hitung :(a) temperature pada setiap proses(b) Efesiensi thermal siklus
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Penyelesaian :a) temperature setiap proses
Proses 1-2 isentropis
Proses 2-3 isobaris
Proses 3-4 isentropis
314
3323
33
12
117
135.62
5.618
117
inVV
ininVrV
inin
r
VV
c
Rin
inR
V
VTT
K
1425117
133432
14.1
3
31
4
334
RRV
VTT 343221716
2
323
RRV
VTT
K
171618540 14.1
1
2
112
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b) efesiensi Massa udara
Pemasukan panas pada tekanan konstan
Pengeluaran panas pada volume konstan
Sehingga : Maka efisiensi :
lbm
in
ft
RRlbmftpsi
inpsi
RT
VPm 00498.0
1728
1
540./.3707.0
1177.143
3
3
3
1
11
BtuTTmCq pin 051.2)17163432(240,0.004980,0)( 23
BtuTTmCq vin 758,0)5401425(171,0.004980,0)( 14
Wnet = qin-qout = 2,051-0,171=1,293 btu
63,0051,2
293,1
in
net
q
W
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Dual cycle
p
v
1
2
x 3
4
qin
qout
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PRINSIP KERJAMOTOR DUAL FUELED
Bahan Bakar Gas disemprotkan ke dalam intake manifold, bercampur dengan udara masuk ke dalam silinder dan dikompresikan
Bahan bakar solar disemprotkan sesaat sebelum TMA, untuk memulai pembakaran (solar hanya untuk memulai pembakaran)
Pembakaran selanjutnya terjadi pada campuran udara dan bahan bakar gas
Sisa pembakaran dibuang melalui saluran buang oleh dorongan langkah buang torak
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)T(TC)T(TC
)T(TC1
x3p2xv
14v
Performance dual cyclePerformance dual cycle Jika : k
C
C
v
P
)()(
)(1
32
14,
xxDualth TTkTT
TT