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TERMODINAMIKA
dan PEMBAKARAN
Satworo Adiwidodo, S.T., M.T
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TIK
Mengetahui siklus tabel termodinamika, tabel sifat-sifat gas dan diagram uap.
Mengetahui manfaat dan aplikasi siklus pada AC, otomotif dan pembakaran bahan bakar.
Pokok BahasanSifat-sifat termodinamika, hukum Termodinamika 1dan 2, aplikasi termodinamika padarefrigerasi, siklus daya ideal dan pembakaran bahan bakar.
Kepustakaan
1. Rayner Joel, Basic Engineering Thermodynamics in SI Unit, Longman
2. Gordon Van Willen, Clasical Thermodynamics, John Willey and Son Inc.3. Moran, M.J., Shapiro, H.N., Fundamentals of Engineering Thermodynamics, John
Willey and Son Inc.
4. Incropera, F.P, Dewitt, D.P., Fundamental of Heat Transfer, John Willey and Son Inc.
5. Irawan, B., Termodinamika Idan II, Jurusan Teknik Mesin Politeknik Negeri Malang.
TERMODINAMIKA dan PEMBAKARAN
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R efrigeration&
Air Conditioning
A plikasi termodinamika pada:
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Aplikasi Refrigerasi dan Pengkondisian Udara
Penghangatan,
pengaturan kelembaban
dan kualitas udara
Refrigerasi industri,
meliputi pengawetan
makanan, kimia dan
proses industri
Pendinginan dan
pengurangan
kelembaban pada
pengkondisian
udara
Pengkondisian Udara Refrigerasi
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Aplikasi Refrigerasi dan Pengkondisian Udara
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Refrigerator dan Pompa Kalor (heat pump)
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Refrigerator dan heat pump
Refrigerator dan heat pumps pada dasarnya merupakan peralatan
yang sama.
Refrigerator dan heat pumps berbeda hanya pada tujuannya saja. Tujuan dari refrigerator adalah mengambil kalor (QL) dari
medium bersuhu rendah (mempertahankan ruang pendingin
tetap dingin)
Tujuan dari heat pump adalah mensuplai kalor (QH) ke
medium bersuhu tinggi (mempertahankan ruang pemanastetap panas)
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COP : Refrigerator and heat pump
Unjuk kerja (prestasi) refrigerator dan heat pump dinyatakan dalam
coefficient of performance (COP), yang didefinisikan sebagai:
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Reverse Carnot Cycle = Carnot Heat Pump
T-s Diagram
T
s
P 2
P 1
1
23Q H
4
Q L
COLD medium at T L
Q L
WARM medium at T H
Q H
W in
C ondenser
Evaporator
C ompressor Turbine
3 2
41
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Reverse Carnot Cycle = Carnot Heat Pump
T-s Diagram
T
s
P 2
P 1
1
23Q H
4
Q L
Proses yang terbentuk dari daur:
1-2 : kompresi adiabatik reversible
(isentropik)
2-3 : pelepasan kalor isothermal
3-4 : ekspansi adiabatik reversible
(isentropik)
4-1 : pemasukan kalor isothermal
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Siklus carnot dibalik
(The R eversed Carnot Cycle)
R efrigerator atau heat pump yang bekerja berdasarkan siklus Carnot
yang dibalik (reversed Carnot cycle) disebut refrigerator Carnot atau
Pompa Kalor Carnot (a Carnot heat pump)
COP ±nya adalah :
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Gas Ideal : Isentropik
(specific heats constans)
konstan !1-k vT
konstank)/k -(1 !TP
konstan!k v P
Proses Isentropik (specific heats constans) pada gas ideal berlaku :
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Gas ideal : Isentropik
(variable specific heats)
P
P2
1
¨
ª©¸
º¹s=konstan
r2
r1
=P
P
r1
r2
konstan=s1
2 =
v
v
v
v¹¹
º
¸©©
ª
¨
Proses Isentropik (specific heats variable) pada gas ideal berlaku :
Hanya untuk udara
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Contoh Perhitungan
Refrigerasi dengan Daur carnot
1. Sebuah sistem pendingin beroperasi berdasarkan siklus gas idealdengan fluida kerja udara. Udara masuk kompresor pada temperatur 12 oC dan tekanan 50 kPa. Udara masuk ke turbin pada temperatur 47oC dan tekanan 250 kPa. Laju aliran masa udara adalah 0,08 kg/s.
Dengan menggunakan asumsi panas jenis udara adalah fungsitemperatur tentukan
Laju pendinginan
Kebutuhan daya kompresor
COP
2. Selesaikan soal no 1 dengan menggunakan asumsi panas jenis konstan.
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PenyelesaianPenyelesaian
No 1 No 1
47 o
C, 250 kPa
0,08 kg/s.12 oC, 50 kPa
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Penyelesaian
T
s
Q H
Q L
1
4
3
2
12 oC
50 kPa
47
o
C250 kPa
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Asumsi udara gas ideal dg panas jenis variable
1584,1
285,14 28527312
1
1
22o
1
!
!!!
r
Atabel
P
kJ/kg h K C T
K T
kJ/kg h P P P P
Atabel
r r
36,450
17,452 5,7921,1584x50250
2
2
22
1
1
22
!
!!!!
7375,1
320,29 32027347
3
3
22o
3
!
!!!
r
Atabel
P
kJ/kg h K C T
K T
kJ/kg h P P
P P
Atabel
r r
84,200
200,813475,01,7375x250
50
4
4
22
3
3
44
!
!!!!
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Asumsi panas jenis variable
skJ hhmQ L /7461,641 !!
skJ hhW co p /3624,1312 !!
skJ hhW ¡
urbin /5581,943 !!
skJ W W W turbinco¢ pinnet /8043,3
, !!
773,1,
!!innet
L
W
QCOP
Laju Pendinginan
Kebutuhan Daya Kompresor
Daya output Turbin
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No. 2 Asumsi panas jenis konstan
Proses 1-2 : Isentropik
v
k !
1)/k -(k
1
212 ¹¹
º ¸©©
ª¨! P P T T
1)/1,4-(1,4
50
250285 2 ¹
º
¸©ª
¨! T
K T 4,451 2 !
1)/k -(k
12
1)/k -(k
21 P T P T !
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Asumsi panas jenis konstan
Proses 3-4 : Isentropik
1)/1,4-(1,4
250
50320 4 ¹
º
¸©ª
¨! T
K T 202 4 !
1)/k -(k
3
434 ¹¹
º ¸©©
ª¨! P P T T
1)/k -(k
34
1)/k -(k
43 P T P T !
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Asumsi panas jenis konstan
4141 T T C mhhmQ p L !!
1212 T T C mhhmW com p !!
4343 T T C mhhmW P Turbin !!
turbincompinnet W W W !,
innet
L
W
QCOP
,
!
Laju Pendinginan
Kebutuhan Daya Kompresor
Daya output Turbin
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Perbaikan Siklus Carnot
Siklus Kompresi Uap Ideal
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1 ± 2 : Isentropic compression
2 to 3 : Constant pressure condensation
3 to 4 : Isenhalpic expansion
4 to 1 : Constant pressure evaporation
Siklus Kompresi Uap Ideal
dalam P-h Diagram
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The ordinary household refrigerator is a good example of the application of this cycle.
COP Q
W
h h
h h
COP Q
W
h h
h h
R L
net in
H P H
net in
! !
! !
,
,
1 4
2 1
2 3
2 1
h1-h4 : dampak refrigerasi
Kapasitas refrigerasi
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Evaporator incooled space
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Evaporator incooled space
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Fantastic Fridges in the Real World
Compressor
pumps andcompressesrefrigerant
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C
ompressorspump andcompressrefrigerant
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Condenserrejects heat
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Condensersreject heat
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ExpansionValve Reducepressure
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Basic AC System
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8/9/2019 1Thermodinamika II (Siklus Refrijerasi I)
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35OC
- OC
Ja a :
h3=h4 h h
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Property Refrigeran
T = - oC, h = 46 J/ g
vg = , 652 m3/ g
sg = ,9424 J/ g.
p = 3,5484 ar
TCond = 35oC, sg = ,9424 J/ g.
p2 = 3.536 ar
h2 = 279,656 J/ g
TABELA7
TABELA9
T3 = 35oC, h3 = 88, 25 J/ g
h4 = h3 = 88, 25 J/ g TABEL A7
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8/9/2019 1Thermodinamika II (Siklus Refrijerasi I)
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PERUMUSAN Dampak refrigerasi :h1-h4
Laju alir massa( ) :Ql /(h1-h4)
Daya Kompressor (Wcomp) : . (h2-h
1)
COP : Ql /Wcomp
Laju alir Volume isap komp : . Vg
Daya Komp per KW refrigerasi :Wcomp /Ql
Suhu Buang Komp : Tabel Interpolasi + extrapolasi
m
m
m
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RESULT Dampak refrigerasi :
h1-h4 = (246- 88,0025) KJ/Kg = 157,9975 KJ/Kg
Laju alir massa ( ) :
Ql /(h1-h4) = 50 KW/157,9975 KJ/Kg = 0,316 Kg/det
Daya Kompressor (Wcomp) :(h2-h1) = 0,316 Kg/det.(279,656 ± 246)KJ/Kg = 10,635 KW
COP :
Ql /Wcomp = 50 KW/10,635 KW = 4,70
Laju alir Volume isap komp :
. Vg = 0,316 Kg/det . 0,0652 m3 /Kg = 20, 6032 l/det Daya Komp per KW refrigerasi :
Wcomp /Ql = 10,635 KW/50 KW = 0,213 KW/KW
Suhu Buang Komp :
Interpolasi + extrapolasi = 56,97
o
C
m
m
m
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Example 2
Refrigerant- 34a is the orking fluid in an ideal compression refrigeration cycle. The
refrigerant leaves the evaporator at -20oC and has a condenser pressure of 0.9 M a.
The mass flo rate is 3 kg/min. Find COPR and COPR, Carnot for the same T max andT min , and the tons of refrigeration.
Using the Refrigerant- 34a Ta les, e have
12
2 2
11 2
2 11
3
3
21
238.41278.23
900200.9456 43.79
0.94561.0
3
900
0
s
so
o
s
s
StateState kJ
kJ h C om pressor exit hC om pressor inlet kg
kg P P k PakJ T C s T C kJ
kg £
s s xkg
£
St ate
C ondenser exit
P k Pa
x
¾¾ ® ±! ®±± ± !±± ±±
! !¿ ¯ ¿ ¯! ±± ±±! !°±± ± ! !! °À ±À
!
!
3 4
44 13
4 3
4101.61 0.358
0.4053200.3738
.0
o
St atekJ h x
T hrottle exit kg kJ
skJ T T C s kg
¤
kg ¤
h h
¾¾ ®! !®±±±
±± ±±¿ ¯ ¿ ¯
!! ! ±± ±±! °±± ± !°À À
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1 4 1 4
, 2 1 2 1
( )
( )
(238.41 101.61)
(278.23 238.41)
3.44
L R
net in
Q m h h h hCOP
W m h h h h
kJ
kg
kJ
kg
! ! !
!
!
& &
& &
The tons of refrigeration, often called the cooling load or refrigeration effect, are
1 4( )
13 (238.41 101.61)
min211
min
1.94
LQ m h h
kg kJ Ton
kJ kg
Ton
!
!
!
& &
,
( 20 273)
(43.79 ( 20))
3.97
L R Carnot
H L
T COP T T
K
K
!
!
!
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Another measure of the effectiveness of the refrigeration cycle is ho much input
po er to the compressor, in horsepo er, is required for each ton of cooling.
The unit conversion is 4.715 hp per ton of cooling.
, 4.715
4.715
3.44
1.37
net in
L R
W
Q COP
hp
Ton
hpTon
!
!
!
&
&
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Actual Vapor-Compression Refrigeration Cycle
22s
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Heat Pump Systems
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1 Ton of R efrigeration = Kalor yang diambil dari 1 ton
(2,000 lb) air yang bersuhu 32 F sehingga
menjadi es pada 32 F selama 24 jam
1 Ton = 12,000 Btu/h = 3.517 kW