Contoh Otto Cycle
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Transcript of Contoh Otto Cycle
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Contoh 1 Siklus otto ideal mempunyai rasio kompresi 8. Pada permulaan langkah kompresi udara pada 100 kPa dan 170C. Kalor ditransfer sebanyak 800 kJ/kg ke dalam sistem.
Tentukan: a). temperatur dan tekanan maksimum yang mungkin
terjadi
b). kerja netto
c). efisiensi termal
Asumsi : fluida kerja adalah udara dengan kalor jenis konstan.
Penyelesaian:
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T1 = 290K qc = 800 kJ/kg
P1 = 100 kPa udara k = 1.4
rc = 8 = V1/V2 cp = 1.005 kJ/kgK
cv = 0.718 kJ/kgK
a). Temperatur dan tekanan maksimum terjadi pada posisi 3.
Langkah 1 – 2 isentropis
k
TT
VV −
=
11
2
1
2
1 KTVVT
k
25.666290.8. 4.01
1
2
12 ==
=
−
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kk
TT
PP −
=
1
2
1
2
1 kPaPTTP
kk
97.183725.666
290.100.4.0
4.1
1
1
2
12 =
=
=
−−
( ) KcqTTTTcq
v
cvc 97.1780
718.080025.6662323 =+=+=⇒−=
Langkah 2 – 3 isokhorik
kPaxTTPP
TP
TP 53.4910
25.66697.178097.1937.
2
323
3
3
2
2 ===⇒=
Temp. max = 1780.97 K ; Tek. Max = 4910.53 kPa
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b) Karena wnet = qc – qe, sedangkan qe = cv(T4 – T1) harus dicari harga T4.
Langkah 3 – 4 isentropis
KTVVT
VVT
kk
21.77597.1780.81..
4.0
3
1
1
23
1
4
34 =
=
=
=
−−
( ) kgkJTTcq ve /38.348)29021.775(718.014 =−=−=⇔
∴wnet = 800 – 348.38 = 451.62 kJ/kg
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c) Efisiensi termal
%4.56564.0800
82.451⇒===
c
netth q
wη
atau
%4.56564.081111 4.01 ⇒=−=−= −k
cth r
η
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Contoh 2 Siklus otto bekerja berdasarkan siklus udara standard dengan jumlah silinder 4, compression ratio 8.6 dan volume langkah piston total 1000 cc. Keadaan awal langkah kompresi 100 kPa, 180C. Jumlah energi yang disupplai tiap siklus 135 J.
Hitung: a). Efisiensi termal mesin
b). Tekanan dan temperatur akhir langkah supplai energi (P3, T3)
c). e.m.p mesin
Penyelesaian:
T1 = 180C=291K Qc = 135 J/cycle
P1 = 100 kPa jumlah silinder = 4
rc = 8.6 = V1/V2 total PD volume = 1000 cc
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a. Untuk udara standard k = 1.4
( )%7.57
6.81111 4.01 =−=−= −k
cth r
η
Proses 1 – 2 kompresi isentropis
( ) K
VVTTVTVT
kkk
5.6886.8291 14.1
1
2
112
122
111
==
=⇒=
−
−
−−
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cc r
VVVVr 1
22
1 =⇒=
PDrVVPDVV
c
+=⇒+= 1121
410
6.81111
3
11
−
=
−⇒=
− VPDV
rc
331 1028.0 mxV −=∴
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Persamaan gas ideal PV=mRT m=PV/RT
Pada titik 1:
kgxmx
xxmRT
VPm
3
3
1
11
10334.0
291287.01028.0100
−
−
=
=⇒=
Supplai energi/cycle (proses 2-3):
( )( )( )( )
)4.979(5.1252
5.688./107165.010334.01350
3
333
23
CKTKTKkgJxkgxJ
TTmcQ vc
=
−=
−=−
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Proses (2-3) berlangsung pada volume konstan:
=⇒=
2
323
3
3
2
2
TTPP
TP
TP
Sehingga:
( )kc
kkk rP
VVPPVPVP 1
2
1322211 =
=⇒=
Dari proses (1-2) isentropik:
( )
kPa
TTrPP k
c
3700 5.6885.1252)6.8)(100( 4.1
2
313
=
=
=
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e.m.p mesin Supplai energi/siklus/silinder:
%7.57135
==
th
c JQη
Kerja/siklus:
silinderJxWi /9.77135577.0 ==
Volume langkah piston: PD = 10-3 m3
( )kPaPaxpme
mJJpmePDWpme i
6.311106.311..
/4
109.77....
3
33
==
=⇒= −
100 kPa
311.6 kPa