Siklus Carnot , Siklus Otto dan Siklus Diesel
OLEH :
JURUSAN PENDIDIKAN FISIKA FAKULTAS MIPA UNIVERSITAS PENDIDIKAN GANESHA SINGARAJA 2011
Siklus Carnot, Otto Diesel Suardika
Page 1
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1.1 Siklus Carnot
Secara eksperimen usaha dapat diubah seluruhnya menjadi kalor. Berdasarkan Hukum I Termodinamika, Q = ΔU + W , pada proses isothermal untuk gas ideal maka Q = W. Jadi persamaan pada proses ekspansi isothermal, seluruh kalor dapat diubah menjadi usaha luar. Akan tetapi secara kontekstual, hal ini mustahil karena kita menghendaki perubahan kalor menjadi usaha luar secara terus menerus selama sistem diberi kalor, atau dengan kata lain volume tabung akan terus membesar yang ditunjukkan dengan pergerakan tutup tabung sampai tak terhingga. Secara praktis hal ini tidak mungkin bisa dilakukan karena kita harus menyediakan silinder dengan volume yang tak terbatas. Perhatikan gambar berikut.
Agar tanpa memerlukan volume yang tak berhingga tetapi konversi dapat berjalan terusmenerus, maka digunakan rangkaian proses. Rangkaian proses ini adalah siklus, yakni rangkaian proses sedemikian rupa sehingga keadaan sistem pada akhir proses sama dengan keadaan awalnya sehingga proses dapat diulang. Contoh proses siklus pertama dikemukakan oleh Nicolas Leonard Sadi Carnot. Siklus Carnot terdiri dari 4 proses yaitu dua proses isothermal yaitu A adiabatik yaitu B C dan DA.
B dan C D, dan dua proses
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Secara konkret, mesin Carnot dinyatakan dengan gambar (3) berikut.
I
Gas
II
q2
R 2
III
Gas
Gas
T 2
IV
q1 R1
Gas
T 1
terisolasi
terisolasi
Gambar (3) I.
Sistem dikontakkan dengan reservoir suhu tinggi ( R2) yang bertemperatur T 2, sehingga sejumlah kalor ( q2) masuk sistem dan menyebabkan sistem berekspansi, temperatur sistem dipertahankan sebesar T 2. Pada gambar (2) ditunjukkan pada proses A→ B.
II. Sistem diisolasi dan dibiarkan berekspansi menyebabkan temperatur sistem turun dari T 2 menjadi T 1, pada gambar (2) ditunjukkan dengan proses B→C . III. Sistem dikompresi dan dikontakkan dengan reservoir suhu rendah ( R1), sehingga sejumlah kalor ( q1) keluar sistem. Temperatur sistem dipertahankan sebesar T 1. Pada gambar (2) ditunjukkan dengan proses C → D. IV. Sistem kembali diisolasi dan dikompresi, menyebabkan keadaan sistem kembali seperti semula. Pada gambar (2) ditunjukkan dengan proses D→ A. Kalau diterapkan Hukum I Termodinamika, yaitu: u q w , karena dalam proses siklus u 0 , sehingga: wq w q2 q1 ....................................................................................................................(1)
Diagram alir mesin kalor ditunjukkan seperti gambar (4).
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Gambar (4) Berdasarkan penjelasan di atas dapat disimpulkan bahwa tidak seluruh kalor dapat diubah menjadi usaha, karena ada sebagian kalor yang terbuang. η) didefinisikan sebagai perbandingan antara kerja yang dihasilkan Efisiensi mesin kalor ((η (w) dengan jumlah kalor yang diserap ( q2). Secara matematis dapat dinyatakan dengan persamaan :
w
q1 q 2
q2
q2
…………………………………………………………………….. (2 (2))
Bila sistem yang dikaji adalah gas ideal, bagaimanakah persamaan efisiensi mesin Carnot? Oleh karena itu, dapat dikaji kembali gambar (2). Persamaan umum gas ideal: p v R T , maka p
R T v
Pada proses A→ B, sistem menjalani proses isotermal, sehingga besarnya usaha : b
wab p dv a
Substitusikan nilai p ke persamaandi atas, maka diperoleh : b
R T a v dv
wab
RT merupakan konstanta, maka dapat dikeluarkan dari integrasi sehingga didapatkan: b
dv
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wab R T 2 ln va b
wab R T 2 ln vb ln va wab R T 2 ln
vb va
…………………………………………………………………(3 )
Sedangkan untuk proses dari B→C , sistem menjalani proses adiabatik sehingga dq 0 , dengan menerapkan hukum I Termodinamika, maka diperoleh. đw = -du đw = -cvdT Sehingga diperoleh usaha pada proses B→C adalah sebagai berikut. c
wbc cv dT b
cv merupakan konstanta, maka dapat dikeluarkan dari integrasi sehingga didapatkan : c
wbc cv
dT b
wbc cv T b c
wbc cv T c T b wbc cv T b T c
Berdasarkan gambar (2), temperatur di b adalah T2 sedangkan temperatur di c adalah T1 sehingga persamaan di atas menjadi : wbc cv T 2 T 1 .......... .......... .......... .......... .......... .......... .......... .......... ......... 4
Pada proses C → D, sistem menjalani proses isotermal, sehingga besarnya usaha: d
wcd p dv c
Substitusikan nilai p ke persamaandi atas, maka diperoleh : d
R T c v dv
wcd
RT merupakan konstanta, maka dapat dikeluarkan dari integrasi sehingga didapatkan :
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d
dv
wcd R T
v
c
Berdasarkan gambar (2), temperatur di D adalah T1 sedangkan temperaturdi A adalah T2 maka, ...............................................................................(5) wda cv (T 1 T 2 ) ....................
Bila sistem yang dikaji adalah gas ideal, maka pada proses isotermal tidak ada perubahan energi dalam sistem atau u 0 , sehingga berdasarkan gambar (2) pada proses akan berlaku : A B, q 2 wab C D, q1 wcd
Berdasarkan persamaan (2) diperoleh : w
q2 wab wbc wcd wda
q2
Karena nilai wbc wda sehingga didapatkan:
wab wda wcd wda q2 wab wcd q2
Substitusikan
persamaan
R T 2 ln
vb va
R T 1 ln
R T 2 ln
R T 2 ln
R T 2 ln
vb va vb va
(3) vd vc
vb va
R T 1 ln R T 2 ln
vd vc vb va
dan
(5)
ke
persamaan
di
atas
sehingga
diperoleh:
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T 1 ln 1
T 2 ln
vd vc
................................................................................................6
vb va
pada B→C pada B→C merupakan merupakan proses adiabatik sehingga berlaku: T v
1
c
1
T b vb
T c vc
1
Berdasarkan gambar (2), temperatur di b adalah T 2 sedangkan temperatur di c adalah T 1 sehingga persamaan di atas menjadi : T 2 vb T 2 T 1
1
vc
vb
T 1 vc
1
1 1
v c T 1 vb
T 2
1
............................................................................................................7
Pada D→ A merupakan proses adiabatik sehingga berlaku : T v
1
T d vd
c
1
T a va
1
Berdasarkan gambar (2), temperatur di D adalah T 1 sedangkan temperatur di A adalah T 2 sehingga persamaan di atas menjadi : T 2 v a T 2 T
1
T 1 vd
vd
1
1
1
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T 2 T 1
vc vb vc vb vc v d
T 2 T 1 1
v d v a
v d
vb
1
va va
Berdasarkan persamaan (6), didapatkan sebagai berikut. T 1 ln 1
T 2 ln
Karena nilai
ln
v d vc
1
T 2 ln
vc vb va
ln
T 1 ln
1
vd
vc v d
, maka diperoleh :
vc vd vb va
T 1 T 2
T 2 T 1 T 2
.................................................................................................9
Efisiensi siklus Carnot hanya tergantung pada temperatur kedua reservoir. Berdasarkan
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A B T2
D C T1 Gambar (5)
Diagram mesin pendingin ditunjukkan pada gambar (6) berikut.
Gambar (6)
Pada mesin kalor, kerja dihasilkan oleh mesin sedangkan pada mesin pendingin, kerja harus diberikan pada mesin (sistem dikenakan kerja) supaya mesin itu dapat beroperasi. Daya guna refrigrator ( E ) Carnot didefinisikan sebagai perbandingan antara kalor yang
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Substitusikan nilai q1 wcd dan q 2 wab ke persamaan di atas, maka diperoleh : E
wcd wcd wab
Substitusikan persamaan (3) dan (5) ke persamaan di atas sehingga didapatkan : T 1 ln E T 1 ln
Karena nilai
ln
v d vc
vd vc
vd vc
T 2 ln
ln
vc v d
vb va
, maka diperoleh :
v T 1 ln c vd E v v T 1 ln c T 2 ln b vd va T 1 ln E T 2 ln
E
vb va
T 1 T 2 T 1
vc vd
T 1 ln
vc vd
.......... .......... .......... .......... .......... .......... .......... .......... .......... ...... 11
Daya guna mesin pendingin Carnot mungkin lebih besar dari 100% jika kalor yang diserap oleh sistem lebih besar dari kerja yang diberikan pada sistem sedangkan efisiensi mesin kalor Carnot, selalu lebih kecil dari 100% karena kalor yang diserap tidak seluruhnya dapat
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p c
q2
d b
p0
e a v1
q1
v2
ad ad
v
Gambar (7)
Siklus Otto seperti ditunjukkan pada gambar (7), terdiri atas enam proses, dengan empat gerakan piston, sehingga mesin yang menggunakan siklus Otto diberi nama dengan mesin empat tak . Keenam proses tersebut adalah sebagai berikut.
1.
e→a :
merupakan intake stoke, pada proses tersebut udara masuk ke dalam silinder pada tekanan udara luar ( po), volume spesifik sistem berubah dari nol menjadi v2. Persamaan yang berlaku pada titik a adalah : p o v 2 R T a , dengan T a merupakan temperatur udara luar.
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q2 cv T c T b .
4.
c→d :
Proses ekspansi adiabatik, yaitu sistem mengembang secara adiabatik. Pada cabang ini usaha dihasilkan, tekanan dan temperatur sistem turun atau berkurang. Persamaan yang berlaku pada proses ini adalah : T c vc
5.
d→a :
1
1
T d vd
.
Proses pendinginan isokorik. Pada proses ini, sejumlah kalor keluar
sistem
sehingga
energi
dalam
sistem
berkurang,
sedangkan temperatur sistem juga berkurang. Persamaan yang berlaku pada proses ini adalah: q1 cv T a T d . 6.
a→e :
Proses isobarik, seluruh gas sisa pembakaran dibuang dan diganti dengan udara baru untuk menjalani siklus berikutnya.
Proses pertama dan keenam saling meniadakan sehingga untuk menentukan efisiensi ( η) hanya diperhatikan empat proses saja yaitu proses 2, 3, 4, dan 5. Berdasarkan definisi efisiensi diperoleh :
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1
T d vd
T a va
1
T c vc
1
T b vb
1
Berdasarkan gambar (7) diperoleh bahwa v a v d v 2 dan vb vc v1 . Substitusikan nilai tersebut ke persamaan di atas, maka didapatkan : T d v2
1
T a v2
T d T a v2 T d T a T c T b
1
1
1
T c v1
1
T b v1
1
T c T b v1
1
v1 v2
1
v 1 T c T b v2
T d T a
1
Dengan mensubstitusikan nilai
v2 v1
r v yaitu nilai rasio kompresi, maka didapatkan hubungan
antara efisiensi mesin Otto dengan rasio kompresinya adalah : 1
T d T a T c T b
1
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p b
c q2
d
p0
q1
e
ad
a ad vb
vc
va
Gambar (8)
v
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sehingga sejumlah kalor masuk sistem. Persamaan yang berlaku pada proses ini adalah : q 2 n c p T c T b dan 4.
c→d :
T c vc
T b vb
.
Proses ekspansi adiabatik, yaitu sistem mengembang secara adiabatik. Pada cabang ini, usaha dihasilkan sedangkan tekanan dan temperatur sistem turun atau berkurang. Persamaan yang berlaku pada proses ini adalah :
T c vc
1
1
T d vd
dan
pc vc pd vd .
5.
d→a :
Exhaust. Proses isokorik. Pada proses ini, sejumlah kalor keluar sistem. Persamaan yang berlaku pada proses ini adalah : q1 n cv T a T d dan
6.
a→e :
T a p a
T d p d
.
Proses isobarik, pada proses ini seluruh gas sisa pembakaran dibuang dan diganti dengan gas baru untuk menjalani siklus selanjutnya.
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1
1 T a T d
1
1
T c T b
1 T a T d T a T b
T T T b c b T a 1 T a T d T a 1
T b T c
.................................................... . 14
T b 1
Pada proses a→b : T a va
1
T b vb
1
T a vb 1 vb 1 T b va va Dengan :
va vb
1
r v , maka diperoleh :
T a vb T b va
1
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pa va pb vb
Pada proses c→d : pc vc pd vd
Berdasarkan kedua persamaan tersebut didapatkan : pc vc
pb vb
pd vd p a va
Berdasarkan gambar (8) diperoleh hubungan : v a v d dan p c pb sehingga diperoleh :
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DAFTAR PUSTAKA Rapi,Ni Ketut. 2099. Buku Ajar Termodinamika. Singaraja : FMIPA UNDIKSHA.