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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 25 o C dengan Satuan Kkal/Kg o C yaitu: Senyawa A B (10 -3 ) C (10 -6 ) D (10 -9 ) Methane (CH 4 ) 1,702 9,081 2,164 -2,703 Carbon dioxide (CO 2 ) 5,457 1,045 13,381 -1,157 Hydrogen Sulfide (H 2 S) 3,931 1,490 5,631 -1,000 Water (H 2 O) 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 (C 2 H 7 NO) 17,817 3,988 -86,931 Sumber : (Lamp. A Sherwood & Prousnitz hal: 627) 1. ABSORBER

Baru Neraca Panas

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Page 1: Baru Neraca Panas

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

Page 2: Baru Neraca Panas

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)

Page 3: Baru Neraca Panas

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

Page 4: Baru Neraca Panas

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

Page 5: Baru Neraca Panas

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

Page 6: Baru Neraca Panas

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

Page 7: Baru Neraca Panas

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

Page 8: Baru Neraca Panas

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)

Page 9: Baru Neraca Panas

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)

Page 10: Baru Neraca Panas

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

Page 11: Baru Neraca Panas

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

Page 12: Baru Neraca Panas

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

Page 13: Baru Neraca Panas

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

Page 14: Baru Neraca Panas

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

Page 15: Baru Neraca Panas

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

Page 16: Baru Neraca Panas

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

Page 17: Baru Neraca Panas

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

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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 .

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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 :

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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

Page 21: Baru Neraca Panas

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

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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

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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

Page 24: Baru Neraca Panas

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

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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)

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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

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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

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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

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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 :

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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

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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

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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

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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)

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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

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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

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Total 154372,7877 154372,7877