Post on 20-Jul-2016
Lampiran B
NERACA PANAS
Berdasarkan hasil perhitungan Neraca Massa dapat dibuat Neraca Panas yaitu :
Basis waktu : 1 jam
Operasi : 24 jam per hari
Satuan : kkal
Suhu reterensi : 25º
Diketehui data kapasitas panas tiap senyawa gas pada suhu standar 25oC dengan
Satuan Kkal/Kg oC yaitu:
Senyawa A B (10-3) C (10-6) D (10-9)
Methane (CH4) 1,702 9,081 2,164 -2,703
Carbon dioxide (CO2) 5,457 1,045 13,381 -1,157
Hydrogen Sulfide (H2S) 3,931 1,490 5,631 -1,000
Water (H2O) 3,470 1,450 -3,450 0,121
Sumber : (HM Spencer & KK Kelley vol: 40)
Data kapasitas panas Untuk larutan yaitu:
senyawa A B (10-3) C (10-6)
MEA (C2H7NO) 17,817 3,988 -86,931
Sumber : (Lamp. A Sherwood & Prousnitz hal: 627)
1. ABSORBER
MEA 30 % (T = 40ºC)
T = 30ºC T = 60ºC
T = 60ºC
Treff = 25ºC (298 K)
T2 = 30ºC (303 K)
AB-01
B-2
Panas Masuk
Q = m ∫T 1
T 2
cpdt cp = (a + bt + ct2 + dt3) dtkkal/kg ºc
QC H 4=158592kg ∫250C
300C
(1,702+9,081 .10−3T−2,164 .10−6T2−2,703 .10−9T 3 )dT kkal /kg kg0C
= 15,8592 kg
(1,702 (30−25 )30o+ 9,081−10−3
2(302−252 )℃−2,146 .10−6
3(302−252 )℃−2,703 .10−9
4(304−254 )℃)
kkal/kg ºC
=154,62897kkal.
QCO2 = 43,6128 kg ∫25℃
30℃
(5,457+1,045 .10−3T−13,38 .10−6T 2+1,57 .10−9¿T3)¿ dt
kkal/kg ºc
¿43,6128 kg ¿
= 1194,03653kkal
Q H2S=0,21kg ∫250C
300C
(3,931+1,490 .10−3T−15,716 .10−6T2+0,232 .10−9T3 )dt kkal /kg0C
= 94,52 kg
(3,931 (30−25 )℃+ 1,490−10−3
2(302−252 )℃−15,716 .10−6
3( 303−253 )℃+ 0,232 .10−9
4(304−254 )℃)kkal /kg℃
= 1649,32716kkal
Total Qin = QCH4 + QCO2 + QH2S + QH2O
= 154,62897 + 1194,03653 + 4,15807 + 1649,32716
= 3002,15073kkal
Panas larutan penyerap masuk absorber (panas larutan MEA 30 % berat)
B-3
yang keluar dari stripper yaitu = 88568,274 kkal
Panas penyerap QºR
MEA + CO2 + H2O MEA CO2
Diketahui data entalphi pembentukan standar pada suhu 25ºc
(298 ºk) :ΔHºt :
ΔHºt MEA = - 62520 kkal/kg mol
ΔHºt CO2 = - 94052 kkal/kg mol
ΔHºt H2O = - 68317 kkal/kg mol
ΔHºt MEA CO2 = - 229710 kkal/kg mol
Sehingga =
Entalphi reaksi penyerapan standar pada suhu 25ºc (298k)
ΔHºR = ΔHºt produk - ΔHºtreaktan
= [ (−229710 ) ]−(−62520 )+(−94052 )+(−68317 )
= - 4821 kkal/kg mol
Maka panas reaksi QºR
QºR = kg mol MEA CO2 terbentuk X ΔHºR
¿ 121,7957kg123 kg/kgmol
X (−4821kkal /kgmol )
= 4773,79732kkal
Panas keluar :
Komponen gas
Q=m∫T1
T2
CpdT Cp = (a + bt + a2 + dt3) dt
Tref = 25ºc = 298 k
B-4
T2 = 60ºc = 333 k
QC H 4=15,8592kg ∫250C
600C
(1,702+9,081 .100T−2,164 .10−6T 2−2,703 .10−9T 3 )dt kkal /kg0C
= 15,8592 kg
(1,702 (60−25 )℃+ 9,081−10−3
2(602−252 )℃−2,164 .10−6
3(603−253 )℃−2,703 .10−9
4( 604−254 )℃)
kkal/kg ºc
= 1156,53146kkal
QCO2=43,6128 kg ∫250C
600C
(5,457+1,045.10−3T−13,30 .10−6T 2+1,157 .10−9T 3)dt kkal /kg0C
= 43,6128 kg
(5,457 (60−25 )℃+ 1,045 .10−3
2( 602−252 )℃−13,38 .10−6
3(603−253 )℃+1,157 .10−9T 3
4(604−254 )℃)
kkal/kg ºc
= 8358,80203 kkal
Q H2S=0,21 kg ∫250C
600C
(3,931+1,490.10−3T−15,716 .10−6T2+0,232 .10−9T3 )dt kkal /kg0C
= 0,21 kg
(3,931 (60−25 )℃+ 1,490 .10−3
2(602−252 )℃−15,716 .10−6
3(603−253 )℃+ 0,232 .10−9
4(604−254 )℃)
kkal/kg ºc
= 29,138012 kkal
B-5
Maka QGas = QCH4 + QCO2 + QH2S
= 1156,53146 + 8358,80203 + 29,138012
= 9544,471502 kkal
Komponen cairan
Q = m ∫T 1
T 2
cpdt
QMEA = 12,08056 kg ∫25oC
60oC
(17,8174+3,988 .10−3T−86,93 .10−6T 2) dt kkal/kg ºc
= 12,08056 kg
(17,8174 (60−25 )℃+ 3,988 .10−3
2(602−252 )℃−86,93 .10−6
3(603−253 ))
kkal/kg ºc
= 7535,06749 kkal
QMEA. CO2 = 12,7957 kg X 0,523 kkal/kgºc (35ºc) = 2229,4703 kkal
QH2O = m ∫T 1
T 2
cpdt
94,052= ∫250C
600C
(3,470+1,450 .10−3T−3,450 .10−6T 2+0,721.10−9T 3)dt kkal /kg0C
¿94,052 kg¿
= 11603,84046 kkal
Maka QLiquid = QMEA + QMEA.CO2 + QH2O
= 7535,06749 + 2229,4703 + 11603,84046
= 21368,37825kkal
Qout = Qg + Qi
H - 01
B-6
= 9544,471502 + 21368,37825
= 30912,84975kkal
Qsteam = Qout - Qin
= 30912,84975 – 3002,15073
= 27910,69902kkal
Neraca Panas Total (AB-01)
Masuk (kkal) Keluar (kkal)
QCH4 154,62897 QCH4 1156,53146
QCO2 1194,03653 QCO2 8358,80203
QH2S 4,15807 QH2S 29,138012
QH2O 1649,32716 QMEA 7535,06749
Qsteam 27910,69902 QMEA CO2 2229,4703
QH2O 116603,84046Total 30912,84975 Total 30912,84975
2. HEATER
Steam = 1500C
(AB-01) (ST-01)T = 600C T = 900C
Kondensat steam
B-7
Panas larutan masuk heater I (H - 01) = panas larutan keluar dari bottom
AB – 01
= 21368,37825 kkal
Panas larutan keluar heater I (H - 01)
Treff = 2590
T2 = 9090
Q = m.cp.∆T
QMEA=12,08056 kg ∫250C
900C
(17,8174+3,988 .10−3T−86,93 .10−6T 2 )dt kkal /kg0C
¿12,08056 kg¿
= 14165,46351 kkal
QMEA CO2 = 121.7357 x0,523kkal/kg ℃ .(90−25) = 4140,444822 kkal
QH 2O=94,052 ∫250C
900C
¿¿
¿94,052 kg¿
= 217.62757 kkal.
Maka :
Qout = QMEA + Q MEA CO2 + QH2O
= 14165,46351 + 4140,444822 + 21721,62757
= 40027, 5359 kkal
Panas yang dibutuhkan dari pemanas
Qs = Qout – Qin
= 40027,5359 – 21368,37825
B-8
= 18659,15765kkal
Sebagai pemanas pada heater (H-01) digunakan saturated steam (uap jenuh) pada
kondisi suhu 150°C dan tekanan 4,6 kg/cm2 dari tabel steam diketahui data
entalphi :
Cair jenuh Hf = 151.2 kkal /kg
Laten kondensasi N = 504,9 kkal /kg
Uap jenuh Hg = 656,1 kkal /kg
Maka jumlah steam pemanas yang dibutuhkan (Ms)
Ms=QsNs
=18659,15765 kkal504,9 kkal /kg
= 36,95615 Kg
Sehingga :
Panas steam masuk heater (H-01)
Qin = ms x Hg
= 36,95615 Kg x 656,1 kkal/ kg
= 24246,92679kkal
Panas steam keluar heater (H – 01)
Qout = ms x Hf
= 36,95615 Kg x 151,2 kkal / kg
= 5587,76988kkal
Neraca Panas Total Heater (H-01)
ST - 01
B-9
Masuk (kkal) Keluar (kkal)
QMEA 7535,06749 QMEA 14165,46351QMEA.CO2 2229,4703 QMEA.CO2 4140,444822QH2O 1103,84046 QH2O 21721,62757Qsteam in 24246,92679 Q Steam out 5587,76988
Total 45615,3057 Total 45615,3057
3. Strippel (ST-01)
T = 1500C
(H-01) Q Steam
T=900C
Panas masuk streapper = panas laurat keluar heater (H-01)
= 40027,5359kkal
Panas pelucutan Q oR
MEA.CO2 MEA + CO2 H2O
Diketahui entalpi pembentukan standar komponen pada suhu 25 C
H f MEA = -62520 kkal / kg
H f CO2 = -94052 kkal / kg
H f H2O = -68317 kkal / kg
H f MEA.CO2 = 229710 kkal /kg
(sumber perrs, edisi 7)
Entalphi reaksi standar pada 25 (298 k)
B-10
∆ H °R= H f produk - H f reakta
= [ (∆ H° f MEA+∆H ° f CO2+∆H ° f H 2O )−∆H °MEACO2 ]
= [ (−62520 )−(−94052 )+(−68317 ) ] [(−229710) ]
= 4821 kkal / kg mol
Panas reaksi standar
Q° R = Kg mol MEA CO2 terlucuti x ∆H0R
¿ 121,7957 kg123 kg/kgmol
x ∆0 R=4773,797315 kkal
Panas keluar stripper (ST-01)
Panas yang keluar terdiri atas panas gas (Qg) keluar stripper dan panas
larutan (Q1) yang keluar bottom stripper.
Panas komponen uap (Qg)
Q = m∫T 1
T 2
cpdt
Diketahui = Tref = 25 C
T2 = 95 C
Maka :
QCO2 = 43,56924 kg
∫25 °C
95 °C
(5,457+1,045.10−3T−13,38. 10−6T−2+1,157 .10−9T−3)dt kkal /kg°C
¿43,56924 kg¿
¿16670,77369 kkal
B-11
Q H2O=116,3682 kg ∫250C
950C
(3,470+1,145 .10−3T−3,450 .10−6T2+0,121 .10−9T3 )dT kkal /kg0C
¿116,3682kg¿
= 28862.15904 kkal
Maka Qg = QCO2 + QH2O
= 16670,77369 + 28862,15904
= 45532,93273kkal
Panas komponen cairan (Q1)
Q = m ∫T 1
T 2
cpdt
QMEA = 72,48337 kg ∫25℉
95℃
¿¿ dt
kkal/kg
¿72,48337 kg¿
= 89848,68209 kkal
QH2O = 52,7595 kg ∫25℉
95℃
¿¿
0,121. 10−9T3 ¿dT kkal/kg
= 52,7595 kg ¿(952-252)
℃−3,450.10−6
3(953−253 )℃+ 0,121.10−9
4(954−254 )℃ kkal /kg℃
= 13085,64608 kkal
Maka : Ql = QMEA + QH2O
= 89848,68209 + 13085,64608
B-12
= 102934,3282 kkal
Sehingga :
Qtotal = Qg + Ql
= 45532,93273 + 102934,3282
= 148467,2609 kkal
Panas yang dibutuhkan pada reboiler stripper
Q1 + Q4 = Q0R + Q3 atau Q4 = Q0R + Q3 . Q1
Atau Qmasuk + Qreboiler = Qreaksi + Qtotal
Q4 = 4773,797315 + 148467,2609 - 40027,5359
= 113213,5233 kkal
Sebagai pemanas pada reboiler stripper digunakan saturated steam atau uap
tanah pada kondisi suhu 1500C dan tekanan 4,9 kgf / cm2 dari tabel steam (A.II.2
Hal.604 stoichiometry) Diketahui entalphi :
Cairan jenuh Hf = 151,2 kkal/kg
Laten kondensasi N = 504,9 kkal/kg
Uap jenuh Hg = 656,1 kkal/kg
Maka jumlah steam pemanas yang dibutuhkan :
ms=Q 4
N s
ms=113213,5223 kkal504,9 kkal /kg
=224,2295946 kg
Sehingga :
panas steam yang masuk :
Qin= ms x Hg
= 224,2295946 x 656,1 kkal/kg
= 147117, 037 kkal
Panas steam yang keluar
B-13
= Ms x Hf
= 224,2295946 kg x 151,2 kkal / kg
= 33903,5147kkal
Neraca Panas Total Stripper (ST-01)Masuk (kkai) Keluar (kkai)
QMEA 14165,46351 QMEA 89848,68209
QMEA.CO2 4140,444822 QH2O(l) 13085,64608
QH2O 21721,62757 QH2O(g) 28862,15904
QCO2 16670,77369
Qsteam 147117,037 Q Steam 33903,5147
Q R4773,797315
Total 187144,5729 Total 187144,5729
4. Cooler (C-01)
Fungsi ; untuk menurunkan suhu gas keluar dari stripper sebelum masuk ke
tangki pengencer (TP -02)
tair = 300C (3030K)
T = 400C (313 0K)
T = 106 0C
(379 K)
tair = 37 0C (310 0K)
Panas larutan MEA masuk cooler (C-01) = panas larutan MEA keluar dari
Botton stripper (ST-01)
= 102934, 3282 kkal
Panas larutan MEA keluar Cooler I (C-01)
Q=m∫T1
T2
cpdr
C - 01
B-14
Tnf = 25oC
T2 = 40oC
QMEA = 72,48337 kg
∫25oC
40oC
❑(17,8174+3,988.10−3T−86,93. 10−6T2 ) Kkal /kgoC
= 72,48337 kg (17,8174 ¿ Kkal/kgoC
= 19411,64656kkal
QH2O = 52,7595 kg ∫25oC
40oC
❑(3,470 + 1,450.
10−3T−3,450.10−6T2+0,121. 10−9T3 ¿dT Kkal ¿kgoC
= 52,7595 kg (3,470 (40-25) + 1,450.10−3
2(402−252 )oC –
3,450.10−6
3(403−253 )oC 0,121.10−9
2( 404−254 )oC ¿Kkal /kgoC
= 2780,494732 kkal
Maka :
Q1 = QMEA +QH2O
= 19411,64656 + 2780,494732
= 22192,14129kkal
Panas yang diserap pendingin Qserap
Qserap = Qmasuk – Q1
= 102934,3282-22192,14129
= 80742,18691kkal
Sebagai pendingin pada cooler I (c-01)digunakan air
B-15
- Suhu air masuk cooler t1= 30oC (303ok)
- Suhu air keluar cooler t2 = 37oC (310ok)
Maka jumlah air pendingin yang dibutuhkan :
m = Qserap
(CpH 2O )dT
m =
80742,18691 kkal
∫30oC
37oC
❑( 3,470+1,450.10−3T−3,450. 10−6T2+0,121. 10−9T3 )dT kkal /kgoC
m =
80742,18691kkal
(3,470 (37−30 )oC+ 1,450.10−3
2(372−302 )oC−3,450. 10−6
3(373−303 )oC ¿+0,121.10−9
4(404−254 )oC)Kkal/kgo
= 80742,18691kkal24,60285624 kg = 3281,821676 kg
Sehingga panas air pendingin masuk cooler
Qin = 3281,821676 kg
∫25oC
300C
❑ (3,470+1,450.10−3T−3,450. 10−6T 2+0,121.10−9T 3 )dT
= 3281,821676 kg ( 3,470(30−25 )oC+ 1,450.10−3
2(302−252)oC –
3,450.10−6
3(303−253 )oC+ 0,121.10−9
4(304−254 )oC) kkal/kgoC
= 57551,03058kkal
Panas air pendingin yang keluar
Qout = 3281,821676 kg
B-16
∫25oC
370C
❑ (3,470+1,450.10−3T−3,450. 10−6T 2+0,121.10−9T 3 )dT
= 3281,821676 kg ( 3,470(37−25 )oC+ 1,450.10−3
2(372−252 )oC –
3,450.10−6
3(303−253 )oC+ 0,121.10−9
4(374−254 )oC) kkal/kgoC
= 138293,2175kkal
Neraca Panas Total Cooler (C-01)
Komponen Masuk (kkal) Keluar (Kkal)
QMEA 89848,68209 19411,64656
QH2O 13085,64608 2780,494732
QPendingin 57551,03058 1323,2175
Total 160485,3588 160485,3588
5. Tangki Pengencer (TP-02)
H2O Proses (T = 40 ºC)
MEA dari (C-01) T=40ºC (Larutan campuran MEA 30%)
T=40 ºC
Panas masuk tangki pelarutan (TP-01)
Panas larutan MEA keluar dari cooler 1 (C-01)
Q=m∫T1
T2
Cp .dT
Tref = 25 ºC
T2 = 40 ºC
QMEA=72,48337 kg ∫250C
400C
(17,8174+3,998 .10−3T−86,93 .10−6T 2 )dt
= 72,48337 kg
TP-02
B-17
(17,8174 ( 40−25 )℃+ 3,998 .10−3
2( 40−25 )℃−86,93 .10−6
3(40−25)℃)
kkal/kg ºc
= 19411,64656kkal
Panas H2O pelarut yang masuk tangki pelarutan (TP-01)
Q=∫T 1
T 2
Cp .dT
Q H 2O=116,3682 kg ∫250C
400C
(3,470+1,450 .10−3T−3,450 .10−6+0,121 .10−9T3 )dT
= 116,3682 kg
(3,470 (40−25 )℃+ 1,450 .10−3
2(402+252 )℃−3,450 .10−6
3( 403+253 )℃+ 0,121.10−9
3(403+253 )℃)
= 6132,756509 kkal
Qin = QMEA + QH2O
= 19411,64656 + 6132,756509
= 25544,40307kkal
Panas larutan MEA 30 % keluar tangki pengencer
Neraca panas tangki pengencer
Qin = Qout
= 25544,40307 kkal
Neraca Panas Total (TP-02)
Komponen Masuk (kkal) Keluar (kkal)
QMEA 19411,64656 19411,64656QH2O proses 6132,756509
B-18
QH2O 6132,756509Total 2544,40307 2544,40307
6. Compressor Gas (CG-01)
Fungsi : menaikkan tekanan gas sintetik yang keluar absorber (AB-01).
Sebelum masuk tangki penampung gas (gas hader)
(air, T=30ºc) air (T=30ºc)P1 = 1 atmT1 = 60ºc
P2 = 46 atm
Air, T = 45 0C air, T = 45 0C T2 = ........0C
Data – data yang diketahui :
Tekanan gas masuk P1 = 1 atm
Tekanan gas keluar P2 = 46 atm
Rasio spesifik panas γ = 1,31
Laju air massa gas masuk compressor
Komponen Kg BM Kmol x i. fraksi mol BM x i
CH4 15,8592 16 0,9912 0,13741 2,19856CO2 43,6128 44 0,9912 0,13741 6,04604H2S 0,21 34 0,00617647 8,56 . 10−4 0,0291H2O 94,052 18 5,2251 0,72433 13,0379Total 153,734 112 7,21367647 1,000006 21,3116
BM Campuran gas = ∑ BM . xi
= 21,3116 kg/mol
Penentuan jumlah stage :
RasioCompressor (RC )=P2
P1= 46atm
1atm=46atm
Karena rasio kompresi compressor melebihi (RC) maksimun = 4 (Ulrich .
B-19
hal 160) maka digunakan multi stage adiabatic centrifugal compressor. Digunakan
3 stage compressor, dengan (RC) pada masing-masing stage :
Rcn=RC
1n
¿4613=3,5830
(Rcn<Rc maksimun = 4, maka jumlah stage dapat digunakan :
a. Compressor Stage 1
P1 = atm
P1 = 60 ºc
T2
P2
Suhu gas masuk compressor T1 = 60 ºC
Tekanan gas masuk compressor P1 = 1 atm
Tekanan gas keluar compressor P2 ?
P2 = Rc X P1
= 3,5830 X 1 atm
= 3,5830 atm
Suhu gas keluar compressor T2 :
T 2=T 1( P2
P1)γ−1 /γ
......................................(pers. 3-9 smith – van ness hal-70)
T 2=600C( 3,5830atm1atm )
1,31−1 /1,31
= 81,5024 ºC
Tenaga yang dibutuhkan compressor :
B-20
-ws = γ
γ−1 (R .T 1
BM )[(P2
P1 )γ−¿γ
−1]= 1,31
1,31−1 ( 8,314 J /mol ° k X333 ° k17,70192862 kg/mol ) [(3,5830atm
1atm )1,31−1/1,31
−1]= 893,597275 J /kg
Diambil efesiensi compressor η = 90 %
Power compressor P= ws X m
η
= 893,5972751 γ /kg X110,1646 kg / jam X1 jam /36005
0,90
= 30,38357γ /s
= 30,38357 X 1,34 X 10−3 HP
= 0,041 HP
Digunakan motor dengan efisiensi η = 90 %
Power motor compressor :
BHP = Pη =
0,041HP0,90
= 0,04524 HP
Neraca panas compressor stage 1
a. Panas gas masuk compressor stage 1 = panas gas yang keluar dari Absorber
(AB-01)
= 9544,471502 kkal
b. Panas gas yang keluar compressor stage 1
Q = m ∫T 1
T 2
cpdt
QCH4 = 15,8592 kg
B-21
∫25℃
81℃
(1,702+9,081 .10−3T−2,164 .10−6T−2−2,703 .10−9T−3 )dt
= 15,8592 kg
(1,702 (81−25 )℃+ 9,081 .10−3
2(812−252 )℃− 2,164 .10−6
3( 812−252 )−2,703 .10−9
4(814−254 ))kkal /kg℃
= 1932,657667 kkal
QCO2 = 43,6128 kg
∫25℃
81℃
(5,437+1,045 .10−3T−13,38 .10−6T 2+1,157 .10−9T 3 )dt
= 43,6128 kg
(5,437 (81−25 )℃+1,095 .10−3
2( 812−252 )℃−13,38 .10−6
3(813−253 )℃+ 1,157 .10−9
4( 814−254 )℃)kkal /kg℃
= 13314,35013 kkal
QH2S = 0,21 kg ∫25℃
81℃
(3,931+1,490 .10−3T−15,716 .10−6T 2+0,232.10−9T 3) dt
¿0,21 kg(3,931 (81−25 )0C+ 1,490 .10−3
2(812−252)℃−15,716 .10−6
3(813−253 )℃+ 0,232 .10−9
4(814−254 )℃)
= 46,590727 kkal
Qtotal = QCH4 + QCO2 + QH2S
= 1932,657667 + 13314,35013 + 46,590727
= 15293,59853 kkal
Panas compressor gas:
Qc = Qtotal + QIn
= 15293,59853 – 9544,471502
= 5749,127026 kkal
Neraca Panas Total Compressor Stage 1 (CP-01)
Komponen Masuk (kkal) Keluar (kkal)QCH4 1156,53146 1932,657667
B-22
QCO2 8358,80203 13314,35013QH2S 29,138012 46,590727Qcompressor 5749,127026Total 15293,59852 15293,59852
7. Inter Cooler (IC-01)
Fungsi : untuk menurunkan suhu gas yang keluar dari compressor stage
1 sebelum masuk ke compressor stage II
T = 30 ºc (air)
Cp – 01 Cp - 02
T = 81 ºC T = 50 ºC
Air T = 450C
Panas gas masuk inter cooler (1C–01) = panas gas yang keluar dari compressor
stage 1
= 15293, 59852 Kkal
Panas gas keluar inter cooler yaitu :
Q=m∫t 1
t 2
CpdT
QCH4 = 15,859 2kg (1,702 + 9,081. 10 -3- 2,164. 10-6 T2- 2,703. 10 -9T3) dT
= 15,8592 kg ( 1,702 (50-25) 0C + 9,081.10−3
2 (502 – 252 )0C
9,081.−2 ,164.10−6
2 (503.253)0C −2,703.10−9
2 ( 504-254)0 C )
= 808,5112491Kkal
Q CO2 = 43,6128 kg ∫250C
500C
❑(5,457 + 1,045. 10-3 – 13,38.10-6T2 =1,157.10-9T3) dT
IC -01
B-23
= 43,6128 kg(5,457(50−25)0C+ 1,045.10−3
2 (502-252)oC−13,38.10−6
2
(503-253)oC + 1,157.10−9
9 (504-254)0C ) Kkal/Kg oC
= 5971,402202 Kkal
QH2S = 0,21 kg ∫250C
500C
❑(3,931 + 1,490 . 10-3T – 15,716.10-6T2 + 0,232.10-9T3) dT
= 0,21 kg (3,931 (50-25)oC + 1,490.10−3
2 (502-252)o C −15,716.10−6
3
(503-253)oC + 0,232.10−9
4 (504-254)oC ) Kkal/kgoC
= 20,81083949Kkal
Maka QTotal = QCH4 + QCo2 + QH2S
= 808,5112491 + 5971,402202 + 20,81083949
= 6800,724291 Kkal
Panas yang diserap pendingin :
Qserap = Qin – Qtotal
= 15293,59852 – 6800,724291
= 8492,874229 Kkal
Sebagai pendingin pada inter cooler (1C-01 ) digunakan air :
Suhu air masuk t1 = 30o C
Suhu air keluar t2 = 45o C
Jumlah air pendingin yang dibutuhkan :
m = QserapcpdT
B-24
m =
8492,874229Kkal
∫300C
450C
(3,470+1,450. 10−3¿T−3,450. 10−6T−2+0,121.10−9T 3)dT ¿
= 8514,680484 Kkal
¿¿¿
+0,121.10−9
4( 454−304 )oC) Kkal/kgoC
= 8492,874229Kkal
52,79198079 Kkal/Kg
= 160,8743241 Kg
Sehingga :
Panas air pendingin masuk intercooler :
Qin = 160,8743241 Kg
∫250C
300C
❑ (3,470+1,450.10−3T−3,450. 10−6T 2+0,121. 10−9T 3 )dT
= 160,8743241 Kg ( 3,470
(30−25)oC 1,450.10−3
2(30¿¿2−252)oC−3,450. 10−6
3¿
(30¿¿3−253)oC+ 0,121.10−9
4(30¿¿4−254)oC ¿Kkal /Kg oC ¿
= 2821,141445 Kkal
Panas Air Pendingin yang keluar
Qout = 160,8743241 Kg
∫250C
450C
❑ (3,470+1,450. 10−3T−3,450.10−6T 2+0,121.10−9T 3 )dT
= 160,8743241 Kg ( 3,470
B-25
(45−25)oC 1,450. 10−3
2(45¿¿2−252)oC −3,450. 10−6
3(45¿¿3−253)oC+ 0,121. 10−9
4(45¿¿ 4−254)oC ¿Kkal /Kg oC ¿¿
= 11314,01567Kkal
Neraca Panas Total Inter Cooler (IC-01)
Komponen Masuk (kkal) Keluar (kkal)
QCH4 1932,65767 808,5112491
QCO2 13314,35013 5971,402202
QH2S 46,590727 20,81083949
Qin 2821,141445
Qout 11314,01567
Total 18114,73997 18114,73997
8. Compresor Stage II (CP-02)
P1 = 3,5830 atm
T1 = 50 0C
T2
P1
Suhu gas masuk compressor , T, = 50oC
Tekanan gas yang masuk P1 = 3,5830atm
Tekanan gas yang keluar P2
P2 = Rc x P1
= 3,5830 x 3,5830 atm
= 12,837889 atm
Suhu keluar compressor, T2
T 2=T 1( P2
P1)α−1 /α
(pers. 3-9 Smith – Van Ness Hal. 70)
B-26
T2 = 50oC ( 12,8378893,5830
)1,311,31
= 323oK x 1,3584
= 438,755oK (165,755oC)
Tenaga yang dibutuhkan oleh compressor :
-Ws = ∝∝−1 ( RTBM
¿( P2P1
)∝−1
∝−1
= 1,31
1,31−1 (8,314∝/molo K .323oK21,3116 kg/Mol
) ( 12,837889atm3,5830atm
)1,31−1/1,31
−1
= 723,1454∝/kg
Diambil efesiensi compressor ƞ = 90%
Maka :
Power Compressor ρ = -Ws x m/ƞ
= 723,1454 ∝
kgx153,734 kg
jamx1 jam
3600s
0,90
= 34,3123∝/s
= 34,3123 x 1,34.10-3Hp
= 0,0459Hp
Digunakan motor compressor BHP
BHP = Pn
BHP = 0,0459Hp
0,90 = 0,051 Hp
Neraca panas compressor stage II
B-27
a. Panas gas yang masuk compressor stage II = panas gas sintetik yang keluar
intercooler (1c-01)= 6800,724291
b. Panas gas keluar compressor stage II
Q=m∫t 1
t 2
CpdT
QCH4 = 15,8592 kg ∫250C
1660C
❑(1,702 + 9,081.10−3T -2,164.10−6T 2
−2,703. 10−9T3 ¿dT15,8592 kg (1,702 (166oC−25oC ¿+
9,081.10−3
2(1662−252)oC −2,164. 10−6
3(1663−253)oC −2,703.10−9
4(1664−254)oC
) kkal/kgoC
= 5684,9063kkal
QCO2 = 43,6128 kg ∫250C
1660C
❑(5,457 + 1,045.10−3+13,38.10−6T 2
+1,157.10−9T 3)dT
= 43,6128 kg (5,457
(166−25)oC+1,045. 10−3
2(166¿¿2−252)oC −13,38.10−6
3(166¿¿3−253)oC+ 1,157. 10−9
4(166¿¿4−254)oC ¿¿¿
) kkal/kgoC
= 33293,8327kkal
QH2S = 0,21 kg ∫250C
1660C
❑ (3,931 + 1,490.
10−3T−15,716.10−6T 2+0,232. 10−9T3)dT
= 0,21 kg (3,91
(166−25)oC+1,490. 10−3
2(1662−252)oC−15,716.10−6
3(1663−253)oC
+0,232.10−9
4(1664−254)oC) kkal/kgoC
= 115,60458kkal
B-28
Maka :
Q total = QCH4 + QCO2+ QH2S
= 5684,9063 + 33293,8327+115,60458
= 39094,34358 kkal
Panas compressi gas
Qc = Qtotal – Qm
= 39094,34358 – 6800,724291
= 32293,61929kkal
Neraca Panas Total Compressor Stage II (CP-02)
Komponen Masuk (kkal) Keluar (kkal)
QCH4 808,5112491 5684,9063
QCO2 5971,402202 33293,8327
QH2S 20,81083949 115,60458
QCompressor 32293,61929
Total 39094,34358 39094,34358
9. Inter Cooler II (IC-02)
Fungsi : menurunkan suhu gas yang keluar dari compressor stage II
Air (T = 300C)
Cp – 02 Cp – 03 (T = 500C)
T = 1660C
Air (T = 450C)
a. Panas gas yang masuk ke (IC-02) = panas yang keluar dari (cp-02)
= 39094,34358
IC-02
B-29
b. Panas gas yang keluar dari (1c-02)
Q = m t1ʃt2CpdT
QCH4 = 15,8592 kg ∫250C
500C
❑(1,702 + 9,081.10−3T 2−2,703.10−9T 3 ¿dT
= 808,5112491kkal
QCO2 = 43,6128 kg ∫250C
500C
❑ (5,457 + 1,045.
10−3T−13,38.10−6T 2+0,232. 10−9T3 ¿dT
= 5971,402202 kkal
QH2S = 0,21 kg ∫250C
500C
❑¿3,931+ 1.490.
10−3T−15,716.10−6T 2+0,232 .10−9T3 ¿dT
= 20,81083949kkal
Q total = CH4 + QCO2+ QH2S
= 808,5112491 + 5971,402202 + 20,81083949
= 6800,724291 kkal
Panas yang diserap pendingin
Qs = 39094,34358-6800,724291
= 32293,61929kkal
Sebagai pendingin pada intercooler II (1C-02) digunakan air
- suhu air masuk t1 = 30oC
- suhu air keluar t2 = 45oC
Jumlah air pendingin yang dibutuhkan :
B-30
M=QS
CpdT
¿ 32292,61929 kkal
∫300C
450C
(3,470+1,450 .10−3T−3,450 .10−6T 2+0,121 .10−3 )dT= 32293,61929 kkal
52,79198079 kkal /kg
= 611,714484 kg
Maka panas air pendingin masuk inter cooler II (1c-02)
Qin = 611,714484 kg ∫250C
300C
❑ (3,470 + 1,450.
10−3T−3,450.10−6T2+0,121. 10−9T3 ¿dT
= 10727,21264kkal
Panas air pendingin yang keluar
Qout = 611,714484 kg ∫250C
450C
❑ (3,470 + 1,450.
10−3T−3,450.10−6T2+0,121. 10−9T3 ¿dT
= 43020,83193 kkal
Neraca Panas Total Inter Cooler II (IC-02)
Komponen Masuk (kkal) Keluar (kkal)
Q CH4 5684,9063 808,5112491
Q CO2 33293,8327 5971,402202
Q H2S 115,60458 20,81083949
Q in 10727,21264
Q out 43020,83193
Total 49821,55622 49821,55622
10. Compressor Stage III
P1 : 12,8379 atm
T1 : 50oC
B-31
T2
P2 = 46 atm
Suhu gas masuk compressor T1 : 50oC
Tekanan gas masuk P1 : 12,8379 atm
Tekanan gas keluar P2
P2 : Rc x P1
: 3,5830 x12,8379 atm
: 46 Atm
Suhu gas keluar T2
T2 : T1 (P2
P1)∝−1 /∝
T2 : 323ok x 1,3583
: 630,296ok (357,296oC)
Tenaga yang dibutuhkan compressor :
-Ws = ∝∝−1
( R .T 1BM
)( P2P1
)∝−1/∝
-1
¿ 1,311,31−1 ( 8,314 α /mol0K .3230 K
21,3116 kg /mol )( 46atm12,8379atm )
1,31−1/1,31
−1
= 1064,97∝/kg
Diambil efesiensi compressor n :90 %
Power compressor = -ws x m/n
¿ 1064,97α /kg x 153,734 kg / jam x1 jam/3600S0,90
= 50,5315 α/s x 1,34.10−3 Hp
= 0,0677 Hp
B-32
Digunakan motor dengan efesiensi n :90 %
BHP = pn =
0,0677 Hp0,90 = 0,0752Hp
Neraca Panas Compressor Stage III
a. Panas gas masuk compressor stage III = panas yang keluar dari inter cooler II
(1C-02)
= 6800,724291 kkal
b. Panas gas yang keluar
Q=m∫T1
T2
CpdT
QCH4=15,8592 kg ∫250C
3570C
(1,702+9,081. 10−3T−2,164.10−6T 2−2,703. 10−9T3)dT
¿15,8592 kg¿
= 17399,49943kkal
QCO2 = 43,6128 kg ∫250C
3570C
❑(5,457 + 1,045.
10−3T−13,38.10−6T 2+1,157.10−9T 3 ¿dT
= 43,6128 kg (5,457 ((357−25)oC+1,045.10−3
2(357¿¿2−252)o¿C -
−13,38.10−6
3(357 ¿¿3−253)oC+ 1,157.10−9
4(357¿¿4−254)oC ¿¿)
kkal/kgoC
= 73262,10465 kkal
QH 2S=0,21 kg ∫250C
3570C
(3,931+1,490. 10−3T−15,716.10−6T 2+1,157.10−9T 3 )dT
C - 02
B-33
¿0,21 kg¿
= 1162,24408kkal
Qtotal = QCH4 + QCO2 + QH2S
= 17399,49943 + 73262,10465 + 1162,24408
= 91823,84816
Panas kompresi gas
Qc = Qtotal – Qin
= 91823,84816 – 6800,724291
= 85023,12387 kkal
Neraca Panas Total Compressor Stage III (CP-03)
Komponen Masuk (kkal) Keluar (kkal)
QCH4 5684,9063 17399,49943
QCO2 33293,8327 73262,10465
QH2S 115,60458 1162,24408
Qcompresor 85023,12387
Total 91823,84816 91823,84816
11. Cooler II (C-02)
Fungsi : Menurunkan suhu gas keluar dari compressor stage III (cp-03)
sebelum masuk ke tangki penampang gas (T-01)
Air (t = 300C)
Cp – 03 T – (T = 400C)T = 357 0C
Air (t = 370C)
B-34
a. Panas gas yang masuk cooler II (c-02) = panas gas yang keluar dari
compressor stage III
= 91823,84816kkal
b. Panas yang keluar cooler II (c-02)
Q = m t1ʃt2CpdT
QCH4 = 15,8592 kg ∫250C
400C
❑¿+ 9,081.
10−3T−2,164. 10−6T2−2,703.10−9T 3¿dT
= 474,5172084kkal
QCO2 = 43,6128 kg ∫250C
400C
❑¿+ 1,045.10−3T - 13,38.
10−6T 2+1,157.10−9T 3¿dT
= 3582,761536 kkal
QH2S = 0,21 kg ∫250C
400C
❑(3,931 + 1,490.10−3 T – 15,716.
10−6T 2+0,232.10−9T 3¿ dT
= 12,48199687kkal
Qtotal = QCH4 + QCO2 + QH2S
= 474,5172084 + 3582,761536 + 12,48199687
= 4069,760741 kkal
Panas yang diserap pendingin :
Qs = Qin – Qout
= 91823,84816 + 4069,760741
= 87754,08742kkal
B-35
Sebagai pendingin yang dibutuhkan pada cooler II digunakan air
- Suhu air masuk t1 = 30oC
- Suhu air keluar t2 = 37oC
Jumlah air pendingin yang dibutuhkan :
m= QsCpdT
m=87754,08742 kkal ∫300C
370C
(3,470+1,450 .10−3T−3,450 .10−6T 2+0,12110−9T 3 )dT
¿ 87754,08742 kkal24,285624 kkal /kg
= 3566,825192 kg
Maka panas air pendingin masuk cooler II (c-02)
Qin = 3566, 825192 kg (3,470 + 1,450.10−3T−3,450.10−6T6+0,121. 10−9T 3
= 62548,93957kkal
Panas air pendingin keluar cooler II (c-02)
Qout=3566,825192kg ∫300C
370C
(3,470+1,450. 10−3T−3,450.10−6T 2+0,121. 10−9T3 )dT
= 150303,027 kkal
Neraca Panas Total Cooler II (C-02)
Komponen Masuk, (kkal) Keluar, (kkal)
Q CH4 17399,49943 474,5172084
Q CO2 73262,10465 3582,761536
Q H2S 1162,24408 12,48199687
Q in 62548,93957
Q out 150303,027
B-36
Total 154372,7877 154372,7877