Kolom Destilasi

(1)

Kolom Destilasi

27 EG dan DEG

21 Konversi 99%

25.000kg\jam 402,787 Kmol

31

EG, EC dan DEG

Data (Backer,1985) Data (Kawabe,1998)

KomposisiX1

EG = 0,99 KomposisiX3EG=0,2482 KomposisiX2EG=0,99

Kompisoso X1

DEG =0,009 KomposisiX3DEG=0,6263 KomposisiX2DEG=0.01

Komposisi X1

EC= 0.001 KomposisiX3EC=0,1255

Neraca Massa Total – N1= N2 + N3

EG= X1 EGN1 = X3EGN3 + X2EGN2 0,99 N1_{= 0,2483 N}3 _{+ 0,99.25252,5253 kg/jam} 0,99 N1_{= 0,2483 N}3 _{+ 25000 kg/jam} 0,99 N1 _{= 0,2483 N}3_{+ 402,784 kmol/jam ... (1)} DEC= X1

DEC N1 = X1DEC N3 + X2DEC N2

0,009 N1_{= 0,6263 N}3_{+ 0,01 (25252,5253 kg/jam)}

0,009 N1_{= 0,6263 N}3_{+ 252,525253 kg/jam}

(2)

Eliminasi persamaan (1) dan (2) 0,99 N1_{- 0,2483 N}3_{= 402,784 x 0.009 0,00891 N}1_{- 0,0022 N}3_=3,625056 0,99 N1_{- 0,6263 N}3_{= 2,3823 x 0,99 0,00891 N}1_{– 0,6200 N}3_=2,358477 0,6222 N3_=1,266579 N3_{= 2,03564} 0,0089 N1_{– 0,0022 N}3_{= 3,625056} 0,0089 N1_{– 0,0022.2,3564 = 3,625056} 0,0089 N1_{– 0,00518408 = 3,625056} 0,0089 N1_{= 3,63024} N1_{= 407,8921} N2 = 407,8921 Kmol/jam N3 = 2,03564 Kmol/jam Alur 21 Total = N21_{= 407,8921 Kmol/jam} EG = N21_{. X}21 EG = 407,8921 x 0,99 = 403,81318 Kmol/jam DEG = N21_{. X}21 DEG = 407,8921 x 0,009 = 3,67103 Kmol/jam EC = N21_{. X}31 EC = 407,8921 x 0,001 = 0,4078 Kmol/jam Alur 31 Total = N31_{= 2,03564 Kmol/jam} EG = N31_{. X}31 EG = 2,03564 x 0,2482 = 0,50524 Kmol/jam DEG = N31_{. X}31 DEG = 2,03564 x 0,6263 = 1,27492 Kmol/jam EC = N31_.X31 EC = 2,03564 x 0,1255 = 0,25547 Kmol/ja

(3)

Alur27

N27 = 407,8921 - 2,03564 = 405,85647 Kmol/jam

EG = N27 . X27EG = 405,85647 x 0,99 = 401,7979053 Kmol/jam

DEG = N27 . X27DEG = 405,85647 x 0,01 = 4,017979 Kmol/jam

Kompon en

BM AlurMasuk (21) (27) Alurkeluar (31)

Kmol/jam Kg/jam Kmol/ja

m Kg/jam Kmol/ja m Kg/jam C2H6O2 62,068 403,81318 25063,8765 401,7979 24938,792 0,50524 31,35923 C4H10O 3 106 3,67103 389,12918 4,017979 425,9057 1,27492 135,14152 C3H4O3 88,06 0,4078 35,910868 - - 0,25547 22,4966882 Total 407,89201 25488,9165 5 405,8158 25364,6977 2,03653 188,997438 2

(4)

KONDENSOR

Table LA.2 Konstanta persamaan tekanan uap Antoine: In Psat = A-B/T+C ( P dalam Kpa T dalam Celcius )

Konstanta persamaan, tekanan uap Antoine (P dalam Kpa dan T dalam Kelvin )

Komponen A B C

C2H6O2 16,1847 4493,79 -82,1026

C4H10O3 17,6738 6034,08 -53,2122

C3H4O3 13,1897 3985,44 -68,9974

(sumber :Reklatis 1983) Menentukan kondisi umpan

Umpan masuk pada kondisi uap jenuh (trial umpan sampai syarat) P = 1 bar = 100 Kpa

(5)

Table LA.3 titik didih umpan masuk destilasi

Komponen Yif Pi Ki Xif=Yif\Ki Aif=Ki\khk

C2H6O2 0,9900 99,9012 0.9990 0.9910 4,0664

C4H10O3 0,0090 24,5675 0,2457 0,0366 1

C3H4O3 0,0010 1469,2572 14,6926 0,0001 59,8049

Total 1 1,0276

(sumber: Reklaitis 1983)

Menentukan kondisi operasi atas (kondensor total)

Untuk mengetahui suhu pada destilat, maka perlu perhitungan trial dew point sampai syarat Σyid/ki = 1 terpenuhi

P = 0,1 mPa =100 Kpa TDP = 195°c = 468,15 K Table LA.4 Dew Point Destilat

Komponen Yid Pi Ki Yid\K Aid

C2H6O2 0,9941 94,0839 0,9408 1,0566 4,1063

C4H10O3 0,0059 22,9122 0,2291 0,0257 1

C3H4O3 0 1407,3073 14,0731 0 61,4218

Total 1 1,0823

Menentukan Kondisi Operasi Bottom (reboiler)

Untuk mengetahui suhu Vb, maka perlu perhitungan train bubble point sampai syarat Σxy,ki = 1 terpenuhi

P = 0,141 Mpa = 141 Kpa TBP = 252°c = 525,15 K

Table LA.5 Boiling point produk bawah

Komponen Xib Pi Ki Xib.Ki aib

C2H6O2 0,2482 401,7360 3,9776 0,7073 3,1968 C4H10O3 0,6263 125,6666 1,2442 0,5582 1

C3H4O3 0,1255 82,6114 0,8178 0,0735 0,6574

(6)

Table LA.6 Omega point Destilat ɸ = 0,0001 dan =1,1Ɵ

Komponen Xif Alfa1 (alfa*Xif)(alfa

I-teta)

C2H6O2 0,9910 4,0664 1,3585

C4H10O3 0,0366 1,0000 -0,3663

C3H4O3 0,0001 59,8049 0,0001

Total 1,0276 6,1211 0,9921

Komponen Xi.d Alfai (alfa I*Xid)(alfa

I-teta)

C2H6O2 0,9941 4,1063 1,3579

C4H10O3 0,0059 1,0000 -0,0588

C3H4O3 0 0 0

Total 1,0000 1,2991

Mencari Refluks Minimum

Umpan masuk 468,15 k, sehingga q = 0 1 – q = Σ α1 – XF1 Rm + 1 = 1,2991 Rm = 0,2991 Rd = 1,5 Rm …...……….. (Geankopplis) Rd = 1,5 . 0,2991 = 0,4486 Rd = Ld/D Ld = Rd x D = 0,4486 x 405,85647

(7)

= 182,0672 Kmol/jam Alur LD (alur 26)

Total = Ld = 182,0672 kmol/jam

EG = LdEG = X19EG x Ld = 0,99 x 182,0672 = 180,24653

DEC = Lddec = X19DEC x Ld = 0,01 x 182,0672 = 1,820672

Alur 27 = 21

Total = F21_{= F}27_{= 405,85647 Kmol/jam}

EG = F21

EG = F27EG = 401,7979053 Kmol/jam

DEG = F21

DEG = F27DEG = 4,017979 Kmol/jam

Alur Vd (alur 25) F25_{= F}26_{+ F}27_{= 587,92368 Kmol/jam} EG = F26 EG + F27EG = 180,24653 + 401,7979053 = 582,0444 DEG = F26 DEG + F26DEG = 1,820672 + 4,017979 = 5,838651 Kompone n

BM Alur masuk (25/Vd) (26) alur keluar (27) Kmol/jam Kg/jam Kmol/jam Kg/jam Kmol/jam Kg/jam C2H6O2 62,068 582,0444 36126,332 180,24653 11187,5416 401,79790 24938,79206 C4H10O3 106 5,838651 618,897006 1,820672 192,991232 4,017979 425,905774 C3H4O3 18 - - - -Total 587,88305 1 36745,2290 1 182,06720 2 11380,5328 3 405,81587 9 25364,69783

(8)

REBOILER

Berdasarkan geankoplis (1997) untuk kondisi umpan dalam keadaan tidak jenuh q = 0 Lb = Vb + B atau Lb = Ld + (q+f) q = 0 Lb = Ld + (0 x f) Lb = Ld Lb = 182,0672 Kmol/jam Maka : Lb = VB + B VB = LB - B =182.0672 - 2,03564 = 180,03156 kmol/jam Komposisi : X31 EG = XvbEG = XLDEG = 0,2482 X31

DEG = XVBDEG = XLDDEG = 0,6263

X31 EC = XVBEC = XLDEC = 0,1255 Alur Lb Total Lb = B + Vb = 182,0672 kmol/jam Etilen glikol = LBEG = 182,0672 x XLDEG = 182,0672 x 0,2482 = 45,18907904 kmol/jam Dietilen glikol = LBDEG = 182,0672 x XLDDEG

=182,0672 x 0,6263 = 114,0286874 Kmol/jam Etilen karbonat = LBEC = 182,0672 x XLDEC

(9)

Alur Vb

Total Vb = Lb – B = 180,03156 Kmol/jam

Etilen glikol = VbEG = 180,03156 x XVBEG = 44,68383319 Kmol/jam

Dietilen glikol = VbDEG = 180,03156 x XVBDEG = 112,753766 Kmol/jam

Etilen Karbonat = VbEC = 180,03156 x XVBEC = 22,59396078 Kmol/jam

Kompone n

BM Alur masuk (LB) (31) alur keluar (VB) Kmol/jam Kg/jam Kmol/jam Kg/jam Kmol/jam Kg/jam C2H6O2 62,068 45,1890790 4 0,50524 44,6838331 9 C4H10O3 106 114,028687 4 1,27492 112,753766 C3H4O3 8 22,8494336 0,25547 22,5939607 8 Total 182,0672 2,03564 180,03156

(10)

FLASH DRUM

32 EG dan DEG

31

35 EC

Dimana dari perhitungan reboiler diperoleh perhitungan sebagai berikut: Alur 31 (Bottom = B) N31_{= 2,03564 Kmol/jam} N31 eg = 0,50524 Kmol/jam N31 deg = 1,27492 Kmol/jam N32 ec = 0,25547 Kmol/jam Menentukan Pi saturated

Konstanta Persamaan tekanan uap Antoine lnP = A - (B-\T+c) P dalam Kpa, T dalam K

Komponen A B C

C2H6O2 16,1847 4493,79 -82,1026

C4H10O3 17,6738 6034,08 -53,2122

C3H4O3 13,1897 3985,44 -68,9974

Tekanan uap jenuh komponen T = 250°c = 523,15 K

Komponen Pvp

C2H6O2 401,735991

C4H10O3 125,6666344

(11)

Menentukan Pbuble dengan Zi = Xi Pb = ΣXi . Pivp ………....… (Smith,2001) Pb = 185,5116329

Menentukan Pdew Dengan Zi = Xi Pdew = 1\Σyi\PIsat

P = 141 Kpa Pd < P < Pb

Karena P yang dihitung terletak antara Pb dan Pdew Ki = Pisat / P

K1 = 2,849191425

K2 = 0,891252726

K3 = 0,585896325

Data trial temperatur dan komposisi Flash Drum Kompone

n

Xi Pi Ki(Pi/P) XiPi KiXi 1/Σ(Yi/Pi)

C2H6O2 0,2353 401,7360 8,1179 94,5406 0,6705 0,0006

C4H10O3 0,6457 125,6666 0,7943 81,1424 0,5755 0,0051

C3H4O3 0,1190 82,6114 16,7944 9,8286 0,0697 0,0014

Σ 185,512 139,5847

Penentuan komposisi umpan dan bottom flash drum I mol umpan total (alur 31) N31_{= 2,05364 kmol/jam}

Vj+1 = Vj – f (Vj) F1 _(Vj)

J = 0,1,2,3 ... dst dilakukan iterasi Vj+1 = Vj (Smith, 2001) Iterasi

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Vo = 0,7240 F (Vo) = 0,0001 F1_{(Vo) = 1,1265} V1 = 0,7240 – ( 0,0001 ) = 0,7241 -1,1265 Zi . F = Xi . L + Yi . V . L Yi = Ki . Xi Zi F = Xi . L + Ki . Xi V Zi . F = Xi ( L + Ki . V ) Xi = Zif L + Ki . V

Basis F = 1 mol maka Xi = Zi L + Ki . V Maka V = 0,8

Table LA.12 Nilai V flash drum

Komponen Zi.Ki Ki-I

Sum(zi.ki\(1+v(ki-1)) EG 0,6705 1,8492 0,2704 DEG 0,5755 -0,1087 0,6303 EC 0,0697 -0,4141 0,1042 1,005 Dimana L = I – V = 0,2 Y1 = 0,27043 = 0 Y2 = 0,63031 Y4 = 0.10424 = 0 ΣY1 = 1.00499

(13)

Dari hasil diatas diperoleh, pada produk atas terdapat semua dietilen glikol dan etilen glikol ( temperature Flash drum 250°c (523,15 k ) jauh melebihi titik didih senyawa tersebut pada 1,41 atm (141 Kpa) sedangkan semua etilen karbonant

terdepat pada bagian bottom, karena tidak menguap kebagian destilat neraca massa komponen :

Etilen Glikol = F31

EG = F32EG + O x F35EG = 0,505241 Kmol/jam

Dietilen Glikol = F31

DEG = F32DEG + O x F35DEG = 1,27492 Kmol/jam

Etilen Karbonat = F31

EC = F32EC + O x F35EC

= 0,25547 Kmol/jam

Tabel LA.13 Tabel Neracamassa Flash Drum

Komponen BM Alur masuk N(Kmol/jam) F(kg/jam) C2H6O2 62,068 0,50524 31,359 C4H10O3 106 1,27492 135,14152 C3H4O3 88,06 0,25547 22,496 Komponen Alur Keluar

Alur Atas Alur Bawah

N (kmol/jam) F (kg/jam) N (kmol/jam) F (kg/jam)

C2H6O2 0,50524 31,359 0 0

C4H10O3 1,27492 135,14152 0 0

C3H4O3 0 0 0,25547 22,496

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Evaporator

19 EO EG DEG 18 20 EC EO H2O CO2

Asumsi efisiensi penguapan air pada evaporator = 100% alur keluar F20 = 407,89201 km01/jam = 25488,91655 Data (bhise : 1983) Komposisi X18 FG = 0,739 Komposisi X19H20 = 0,9485 Komposisi X18 H20 = 0,243 Komposisi X 19C20 = 0,0484 Komposisi X19 EO = 0,0031

Neraca massa total F18_{= F}19_{+ F}20

Neraca massa komponen Etilen glikol = N18_{= 0 x N}19_{= N} 20

0,739 x N18_{= N}20_{407,89201 kmol/jam ... (3)}

Air = F18

air = F19air + F20air

0,243 F18_{= 0,9485. F}19

air ... (4)

Eleminasi pers (3) & (4)

EG = 0,739 N18_{+ 0 N}19_{= 407,89201 x 0,243 0,179577 N}18_{+ 0 N}19_=99,117

(15)

0,7009 N19_{= 99,1177} N19_{= 141,414} N18_{= N}19_{= N}20 = 141,414 + 407,8921 = 549,3061Kmol/jam

Alur 19

Total = N19_{= 141,414 kmol/jam} Air = N19_{x X}19 air = 141,414 x 0,9485 =134,131179 kmol/jam CO2 = N19 x X19CO2 = 141,414 x 0,0484 = 6,8444376 kmol/jam EO = N19_{x X}19 EO = 141,414 x 0,0031 = 0,43838341 kmol/jam

Alur 18

Total N18_{= N}19_{+ N}20_{= 549,3061kmol/jam} EG = N18 EG = 403,81318kmol/jam DEG = N10 DEG = 3,67103kmol/jam EC = N18 DEG = 0,4078kmol/jam Air = N18 H2O = 134,131179 kmol/jam CO2 = N18CO2 = 6,8444376kmol/jam EO = N18 EO = 0,4383834kmol/jam

(16)

Neraca Massa Evaporator

Kompone n BM AlurMasuk 19 AlurKeluar 20 N (kmol/jam) F (kg/jam) N (kmol/jam) F (kg/jam) N (kmol/jam ) F (kg/jam) C2H6O2 62,06 8 403,81318 25063,8764 6 - - 403,81318 25063,87646 C4H10O3 106 3,67103 389,12918 - - 3,67103 389,12918 H2O 18 134,131179 2414,36122 2 134,131179 2414,36122 2 - -C3H4O3 88,06 0,4078 35,910868 - - 0,4078 35,910868 CO2 44 6,8444376 301,155254 1 6,8444376 301,155254 4 - -C2H4O 44,05 3 0,4383834 19,3121039 2 0,4383834 19,3121039 2 - -Total 549,3001 28223,7450 9 141,414 2734,82858 407,8921 25488,91655

(17)

Separator 11

Diasum

- Alur masuk F17_{= 549,3061}

kmol/jam

- Separatur dapat memisahkan CO2

sebanyak 90% Data (Bhise, 1983) Komposisi X15 EO = 0,018 Komposisi X15 CO2 = 0,3921

Menentukan Pisaturated

Tabel LA.15 Konstanta persamaan tekanan uap antoine In P = A (B/(T-C)) (P dalam Kpa, T dalam K) Sumber:Reklaitis, 1983 Komponen A B C C2H4O2 16,1847 4493,79 -82,1026 C4H10O3 17,6738 6034,08 -53,2122 H2O 16,5362 3985,44 -38,9974 C3H4O3 13,1897 3985,44 -68,9974 CO2 15,3768 1956,25 -2,1117 C2H4O 14,5116 2478,12 -33,1582

(18)

Tabel LA.16 tekanan uap jenuh komponen pada T = 148ºC = 421,15 K Komponen PVP C2H6O2 2,105678324 C4H10O3 0,305328832 H2O 100,4032116 C3H4O3 1,090108871 CO2 24451,16737 C2H4O 1370,369892

Menentukan Pbuble dengan Zi = Xi

Pb = ΣXi ,PiVp ... (Smith 2001) Pb = 9598,110797 Kpa

Menentukan P dew denganZi = Yi

P dew = 1 ... (Smith, 2001) ΣYi / Pisat

P dew = 3,683940099 kPa P = 250 kPaPd< P <Pb

Karena P yang dihitung terletak antara Pbuble dan Pdew Ki = Pisat / P K1 = 0,008422713 K2 = 0,001221315 K3 = 0,401612846 K4 = 0,004360435 K5 = 97,80466948 K6 = 5,48147957

(19)

Substitusi nilai K ke pers (10,16)

Σ[(Zi, Ki)/(1+V(Ki-1))] = 1 ... (Smith, 2001) V = 0,39

Tabel LA.17 nilai V separator 1

Komponen Zi . Ki Ki – 1 Sum(Zi,Ki/(1+V(Ki-1)) H2O 0,0257 -0,5984 0,0335 CO2 38,3524 96,8047 0,9629 C2H4O 0,0099 4,4815 0,0036 1 Dimana L = 1-V = 0,61 Denganpers (10,16) ... (Smith, 2001) Yi – ((Zi , Ki)/(1+V(Ki-1))) Y3 = 0,0335 Y5 = 0,9629 Y6 = 0,0036ΣYi = 1

Dengan pers (10,10) ... (Smith, 2001) Xi = Yi/Ki

X3 = 0,0834

X5 = 0,0101

X6 = 0,0007

ΣXi = 0,0941

(20)

CO2 F16CO2 = F16CO2 + F17CO2

0,3921 = 0,9896 N16_{+ 6,8444376 ... (5)}

EO F15

EO = F16EO + F17EO

0,0018 = 0,0036 N16_{+ 0,4383834 ...(6)}

Eliminasi persamaan 5 dan 6

CO2 0,3921 N15=0,9896 N16+6,8444376 x0,0018 7,0578x10-4N15=0,0017812 EO 0,0018 N15_{=0,0036 N}16_{+0,4383834 x0,3921 70578x10}-4_N15 N16_{= 431,6906828kmol/jam} _N15_{= N}16_{+ N}17 N15_{= 980,9967828kmol/jam} _{= 431,6906828 + 549,3061} = 980,9967828

Alur 16

Total = N16_{= 431,6906828kmol/jam} Air = 431,6906828 x 0,0335 = 14,46163787 CO2 = 431,6906828 x 0,9629 = 415,6749585 EO = 431,6906828 x 0,0036 = 1,554086458

Alur 15

Total N15_{= 980,9967828kmol/jam} EO: N15 _{= 0 + 403,81318 = 403,81318 kmol/jam} DEC: N15 DEC = 0 + 3,67103 = 3,67103 kmol/jam EC: N15 EC = 0 + 0,4078 = 0,4078 kmol/jam H2O:N15H20 = 14,46163787 + 134,131179 = 148,5928169 kmol/jam CO2: N15CO2 = 415,6749585 + 6,8444376 = 4235193961 kmol/jam

(21)

EO: N15 EO = 1,554086458 + 0,4383834 = 1,992469858 kmol/jam Kompone n BM AlurMasuk 15 16 AlurKeluar 17 N (kmol/jam) F (kg/jam) N (kmol/jam) F (kg/jam) N (kmol/jam) F (kg/jam) C2H6O2 62,06 8 403,81318 25063,87646 - - 403,81318 25063,87646 C4H10O3 106 3,67103 389,12918 - - 3,67103 389,12918 H2O 18 148,5928169 2674,670704 14,4616378 7 260,3094817 134,131179 2414,361222 C3H4O3 88,06 0,4078 35,910868 - - 0,4078 35,910868 CO2 44 422,5193961 18590,85343 415,674958 5 18289,69817 6,8444376 301,1552544 C2H4O 44,05 3 1,992469858 87,77427465 1,55408645 8 68,46217073 0,4383834 19,31210392 Total 980,9967828 46842,21492 431,690682 8 18618,46982 549,3061 28223,74509

(22)

Reaktor Hidrolis

6 Air EC EO 12 CO2 13 DEG Air EC EO CO2 EG

(23)

Data:

Dari ekspunder diperoleh data laju alir sebagai berikut: N13 _{= 980,9967828 N}13 EC = 0,4078 N13 EG = 403,81318 N13CO2 = 422,5193961 N13 DEG = 3,67103 N13EO = 1,992469858 N13 H20 = 148,5928169

Neraca massa total: F13_{= F}6 _{+ F}12 Reaksi 1 C3H4O3(l) + H2O(g) C2H6O2(l) + CO2(g) M = x B = 0,95 x S = 0,5 x Neraca komponen EG = N13_{= N}6 EG + N12EG + r1 r1 = 403,81318 kmol/jam Air = N13 air = N6air – r1 – r2 148,5928169 = N6 air – 403,81318 – r2 ... (7) EC = N13 EC = N12EC – r1 – 2 r1

(24)

0,4078 = N12 EC – 403,81318 – 2 r2 ... (8) CO2 = N13CO2 = N12CO2 + r1 + 2 r2 422,5193961 = N12 CO2 + 403,81318 – 2 r2 ... (9) Reaksi 2 2 C3H4O3(l) + H2O(g) C4H10O3(l) + 2 CO2(g) Neraca komponen DEG = N13

DEG = N6DEG + N12DEG + r2

T2 = N13DEG + (N6DEG – N12DEG) = 3,67103 + (0-0) = 3,67103 kmol/jam

EO = N13 EO = N12EO = 1,992469858 kmol/jam Air = 148,5928169 = N6 air – 403,81318 – 3,67103 N6 air = 148,5928169 + 403,81318 + 3,67103 = 556,0770269 kmol/jam EC = 0,4078 = N12 EC – 403,81318 – 2 (r2) N12 EC = 0,4078 + 403,81318 + 2 (3,67103) = 411,56304 kmol/jam CO2 = 422,5193961 = N12CO2 + 403,81318 – 2 (r2) N12 CO2 = 422,5193961 – 403,81318 + 2 (3,67103) = 26,0482761 kmol/jam

Neraca Massa Reaktor Hidrolisis

Kompone n BM AlurMasuk 15 6 AlurKeluar 12 N (kmol/jam) F (kg/jam) N (kmol/jam) F (kg/jam) N (kmol/jam) F (kg/jam) C2H6O2 62,06 8 403,81318 25063,87646 - - - -C4H10O3 106 3,67103 389,12918 - - - -H2O 18 148,5928169 2674,670704 556,077026 9 10009,38648 - -C3H4O3 88,06 0,4078 35,910868 - - 411,56304 36242,2413

(25)

CO2 44 422,5193961 18590,85343 - - 26,0482761 1146,124148 C2H4O 44,05 3 1,992469858 87,77427465 - - 1,99246985 8 87,7742746 5 Total 980,9967828 46842,21492 556,077026 9 - 439,603786 37476,1397 2

Separator 1

9 EC CO2 EO 8 EC CO2EO 10 EC EO

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CO2

 Dari data perhitungan di R Hidrolisis diperoleh laju alir 10 yaitu: 439,603786

kmol/jam  Data (bhise, 1983): Komposisi x8 co2 = 0,004 Komposisi x8 EO = 0,018

Menentukan Pi Saturated

Tabel konstanta persamaan tekanan uap antoine In P = A – (B/(T+C)) P dalam kPa T dalam K Komponen A B C CO2 15,3768 1956,25 -2,1117 C2H4O 14,5116 2478,18 -33,1582 C3H4O3 13,1897 3985,44 -68,9974 (Sumber: Reklaitis: 1983)

Tabel tekanan uap jenuh komponen pada T = 100ºC = 373,15K Komponen PVP

CO2 24451,17

C2H4O 1370,37

C3H4O3 1,090109

Menentukan P Buble denganZi = Xi Pb = ΣXi .PNP ... (Smith, 2001) Pb = 1116,851 kPa

Menentukan Pdew dengan Zi = Yi

P dew = 1 ... (Smith, 2001) ΣYi/Pisat

Pdew = 1,337 kPa P = 250 kPaPd< P <Pb

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Ki = Pisat / P K1 = 97,80467 K2 = 5,48148 K3 = 0,00436 Substitusi kepersamaan (10.16) Σ[(Zi.Ki)/(1+V(ki-1))] = 1 ... (Smith, 2001)

Tabel nilai V separator 1

Komponen Zi, Ki Ki-1 Sum [(Zi,ki/(1+V(ki-1))]

CO2 17,65525 96,8047 0,9581 C2H4O 0,0226 4,4815 0,0125 C3H4O3 0,0036 -0,9956 0,0043 0,9749 Dalam L = 1 – V = 1 – 0,18 = 0,82

Dengan persamaan (10,16) ... (Smith, 2001) Yi = ((Zi, Ki)//1+V(Ki-1))

Y1 = 0,9581

Y2 = 0,0125

Y3 = 0,0043

ΣYi = 0,9749

Dengan pers (10,10) ... (Smith, 2001) Xi = Yi/Ki

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X2 = 0,0023

X3 = 0,9934

ΣXi = 1,0855 Neraca total Massa

N8 = N9 _{+ N}10

Neraca Massa Komponen EC = N9 EC + N10EC ... (10) CO2 = 0,004 N8 = 0,9581 N9CO2 + 26,0482761 ... (11) EO = 0,018 N8 _{= 0,0125 N}9 EO + 1,992469858 ... (12)

Reaktor karbonasi

Co2 2 4 7 EO EO,DEG dan EG Data (bhisem,1983) Konversi reaksi = 99%

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N7_{= 466,4067896 kmol/jam} N7 ec = 411,6812602 kmol/jam N7 co2 = 52,38939594 kmol/jam N7 EO = 2,336133348 kmol/jam Reaksi : C2H4O+CO2 C C3H4O3 M= X B= 0,99X S=0,1 X

Neraca massa total: Alur7 EC=N7 ec=N2ec+N4ec-r r=N7 EC-(N2ec+N4ec)= 411,6812602-0=411,6812602 kmol/jam EO=N7 eo=N2eo + N4eo – r N4 eo= 2,33361333348+411,6812602=414,6812602 kmol/jam

CO2= N7co2= N2co2+N4co2 –r

N2

co2= 52,38939594-0 + 411,6812602=464,0706561

Kompo nen

BM Alur keluar 7 2 Alur masuk 4 N(Kmol/j am F(Kg/ja m) N(Kmol/ jam) F(Kg/ja m) N(Kmol/ jam) F(Kg/ja m) C3H4 O3 88,06 411,6812 602 36252,6 5177 - - - -CO2 44 52,38939 594 2305,13 3421 464,0706 561 20419,1 6887 - -C2H4o 44,05 30 2,334613 3348 102,913 6824 - - 414,0173 935 18238,7 0824 TOTA 466,4067 3860,69 464,0706 20419,1 414,0173 18238,7

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L 896 887 561 0887 935 0824

Eliminasi Nominasi Persamaan pers 11 & 12

0,004N8_{=0,9581 N}9_{+26,0482761 x0,018 7,2x10}-5_N8_=0,0172458N9_+0,468868969 0,018 N8_=0,0125N9_{+1,992469858 x0,004 7,2x10-5N =0,00005N9+0,007969879432} 0,0171958N9_=0,460899089 N9 _=26,80300358 N9 _{= 26,80300358 kmol/jam} N8_{= 466,4067896 kmol/jam} N10_{= 439,603786 kmol/jam}

Alur 9

Total N9 _{= 26,80300358 kmol/jam} EC = N9 _{x Y}9 ec = 27,49307989 x 0,0043 = 0,118220243 CO2= N9 x Y9Co2= 27,49307989 x 0,9581 = 26,341111984 EO=N9_{x Y}9 eo = 27,49307989 x o,0125 = 0,34366349

Alur 8

Total N8_{= 466,4067896 kmol/jam} EC= N8 ec = N9ec + N10 = 0,118220243 + 411,56304 = 411,6812602

CO2= N8co2= N9co2 + N10co2= 26,34111984+26,0482761 = 52,38939594

EO=N8

eo= N9eo + N10eo = 0,3436634951,992469858 = 2,336133348

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Kompon en

BM Alur masuk 8 9 Alur masuk 10 N(Kmol/ja m F(Kg/jam) N(Kmol/ja m) F(Kg/jam) N(Kmol/ja m) F(Kg/jam) C3H4O 3 88,06 411,68126 02 36252,65 177 0,1182202 43 10,41047 46 411,56304 36242,24 13 CO2 44 52,389395 94 2305,133 421 26,341119 84 1159,0092 73 26,048276 1 1146,1241 48 C2H4o 44,05 30 2,3346133 348 102,9136 824 0,3436634 9 15,13940 772 1,9924698 58 87,77427 465 TOTAL 466,40678 96 3860,698 87 26,803003 58 1184,5591 55 439,60378 6 37476,13 972