LA.1 Mixer Liquid Feed (M-101)

(1)

LAMPIRAN A

PERHITUNGAN NERACA MASSA

Waktu operasi : 300 hari / tahun ; 24 jam / hari Basis perhitungan : 1 jam operasi

Satuan operasi : kilogram (kg)

Bahan baku : - Stirena oksida (C8H8O)

-- NaNattrriiuumm hhiiddrrookkssiiddaa ((NNaaOOHH))

-- MeMettaannooll ((CCHH33OOHH))

-- HiHiddrrooggeenn ((HH2)2)

-- KaKattaalliiss ((PPdd//CC 11%%)) Produk akhir : 2-Feniletanol (C8H10O) Kapasitas Produksi : 138,8889 kg/jam

LA.1 Mixer Liquid Feed (M-101)

Mixer Liquid Feed M-101 (4)

(6) (3)

(5)

NaOH

Stirena oksida

Metanol

NaOH

Stirena oksida Metanol

Dalam mixer ini terjadi pencampuran antara stirena oksida, natrium hidroksida, dan metanol.

Basis Perbandingan Umpan Stirena oksida : NaOH : Metanol (kg) = 1: 0,00026: 19

Neraca Massa Komponen:

Stirena oksida : F6Stirena oksida = F3Stirena oksida = 141,5894 kg/jam

NaOH : F6NaOH = 0,000026 × F3Stirena oksida = 0,00026 × 141,5894 = 0,0368 kg/jam F6NaOH = F4NaOH = 0,0368 kg/jam

(2)

F6Metanol = F5Metanol = 2690,1986 kg/jam

Neraca Massa Total :

F6 = F3 + F4 + F5 = 141,5894 + 0,0368 + 2690,1986 = 2931,8248 kg/jam

Tabel LA.1 Neraca Massa Mixer Liquid Feed (M-101)

Komponen

Alur Masuk Alur Keluar

Alur 3 Alur 4 Alur 5 Alur 6

F N F N F N F N

(kg/jam (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam)

C8H8O 141,5894 1,1784 - - - - 141,5894 1,1784

NaOH - - 0,0368 0,0009 0,0368 0,0009

CH3OH - - - - 2690,1986 83,9596 2690,1986 83,9596

Total 2831,8248 2831,8248

LA.2 Mixer Hydrogenation Feed (M-103)

Mixer Hydrogenation Feed

M-103

(10) (9)

(7) Hidrogen

NaOH

Stirena oksida Metanol

NaOH

Stirena oksida Metanol Hidrogen

Dalam mixer ini terjadi pencampuran antara stirena oksida, natrium hidroksida, metanol dan hidrogen.

Stirena oksida : F10Stirena oksida = F7Stirena oksida = F6Stirena oksida = 141,5894 kg/jam NaOH : F10NaOH = F7NaOH = F6NaOH = 0,0368 kg/jam

(3)

Neraca Massa Total:

F10 = F7 + F9 =2831,8248 + 2,8506 = 2834,6754 kg/jam

Tabel LA.2 Neraca Massa Mixer Hydrogenation Feed (M-103)

Komponen

Alur 7 Alur 9 Alur 10

F N F N F N

(kg/jam (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam)

C8H8O 141,5894 1,1784 - - 141,5894 1,1784

NaOH 0,0368 0,0009 - - 0,0368 0,0009

CH3OH 2690,1986 83,9596 - - 2690,1986 83,9596

H2 - - 2,8506 1,4141 2,8506 1,4141

Total 2834,6754 2834,6754

LA.3 Fixed Bed Reactor (R-101)

NaOH

Stirena oksida Metanol Hidrogen

Fixed Bed Reactor

R-101

(11) (10)

2-Feniletanol NaOH Metanol Hidrogen

Dalam reaktor ini terjadi reaksi hidrogenasi stirena oksida menghasilkan 2 feniletanol. Reaksi hidrogenasi berlangsung pada temperatur 40 0C dan tekanan 2,04 MPa (20 atm) dengan katalis Pd/C 1% sesuai dengan persamaan reaksi di bawah ini.

(4)

N10Stirena oksida = 𝐹𝑆𝑡𝑖𝑟𝑒𝑛𝑎 𝑜𝑘𝑠𝑖𝑑𝑎 10

𝑀𝑟𝑆𝑡𝑖𝑟𝑒𝑛𝑎 𝑜𝑘𝑠𝑖𝑑𝑎 = 120,1502 5894 , 141

= 1,1784 kmol/jam r1 = konversi × N10Stirena oksida = 1 × 1,1784 = 1,1784 kmol/jam

Reaksi yang terjadi di dalam reaktor:

C8H8O(l) + H2(g) → C8H10O(l)

M 1,1784 1,4141 -

B 1,1784 1,1784 1,1784

S - 0,2357 1,1784

Dimana:

M = jumlah mol senyawa mula-mula (kmol) B = jumlah mol senyawa yang bereaksi (kmol)

S = jumlah mol senyawa sisa setelah reaksi selesai (kmol)

Stirena Oksida: F10Stirena oksida = 141,5894 kg/jam Hidrogen : F10Hidrogen = N10Hidrogen × Mr Hidrogen

= 1,4141 × 2,0158 = 2,8506 kg/jam

F11Hidrogen = (N10Hidrogen – N10 Stirena oksida) × Mr Hidrogen = (1,4141 – 1,1784) × 2,0158

= 0,4751 kg/jam

2-Feniletanol : F112-Feniletanol = Freaksi2-Feniletanol = N10Stirena oksida × Mr 2-Feniletanol

= 1,1784 × 122,166

= 143,9649 kg/jam

NaOH : F11NaOH = F10NaOH = 0,0368 kg/jam Metanol : F11Metanol = F10Metanol = 2690,1986 kg/jam

Neraca Massa Total:

(5)

Tabel LA.3 Neraca Massa Fixed Bed Reactor (R-101)

Komponen

Alur 10 Alur 11

F N F N

(kg/jam (kmol/jam) (kg/jam) (kmol/jam)

C8H8O 141,5894 1,1784 - -

NaOH 0,0368 0,0009 0,0368 0,0009

CH3OH 2690,1986 83,9596 2690,1986 83,9596

H2 2,8506 1,4141 0,4751 0,2357

C8H10O - - 143,9649 1,1784

Total 2834,6754 2834,6754

LA.4 Knock Out Drum (FG-201)

Knock Out Drum FG-201 (11)

(12)

2-Feniletanol NaOH Metanol Hidrogen

2-Feniletanol NaOH Metanol (13)

Hidrogen

Pada drum ini digunakan untuk memisahkan gas hidrogen dari campuran. Semua gas hidrogen pada alur 12 akan terpisah pada knock-out drum ini.

Hidrogen : F12Hidrogen = F11Hidrogen = 0,4751 kg/jam

2-Feniletanol : F132-Feniletanol = F112-Feniletanol = 143,9649 kg/jam NaOH : F13NaOH = F11NaOH = 0,0368 kg/jam

Metanol : F13Metanol = F11Metanol = 2690,1986 kg/jam Neraca Massa Total:

(6)

F12 = F12Hidrogen = 0,4751 kg/jam F11 = F12 + F13

F13 = F11– F12 = 2690,1986 – 0,4751 = 2834,2003 kg/jam

Tabel LA.4 Neraca Massa Knock Out Drum (FG-201)

Komponen

Alur 11 Alur 12 Alur 13

F N F N F N

NaOH 0,0368 0,0009 - - 0,0368 0,0009

CH3OH 2690,1986 83,9596 - - 2690,1986 83,9596

H2 0,4751 0,2357 0,4751 0,2357 - -

C8H10O 143,9649 1,1784 - - 143,9649 1,1784

Total 2834,6754 2834,6754

LA.5 Mixing Point Hidrogen (M-102)

Mixing Point Hidrogen M-102

(14)

(8) (1)

Hidrogen

Hidrogen Hidrogen

Hidrogen : F8Hidrogen = F9Hidrogen = 2,8506 kg/jam F14Hidrogen = 0,4751 kg/jam

F1Hidrogen + F14Hidrogen = F8Hidrogen

F1Hidrogen = F8Hidrogen - F14Hidrogen = 2,8506 – 0,4751 = 2,3755 kg/jam

Tabel LA.5 Neraca Massa Mixing Point Hidrogen (M-102)

Komponen

Alur 1 Alur 14 Alur 8

F N F N F N

H2 2,3755 1,1784 0,4751 0,2357 2,8506 1,4141

(7)

LA.6 Tray Distillation Tower (TD-201)

Adapun kolom destilasi ini digunakan untuk memisahkan CH3OH (metanol) dari produk 2-Feniletanol berdasarkan perbedaan titik didih.

Kolom Destilasi

TD-201 2-Feniletanol

NaOH Metanol

2-Feniletanol NaOH Metanol

2-Feniletanol Metanol

(15)

(19)

(23)

Alur Umpan Masuk F15

N_metanol15 = Fmetanol15 Mrmetanol=

2690,1986 kg.jam-1

32,0416 kg.kmol-1 = 83,9596 kmol/jam NNaOH15 =

F_NaOH15 M_rNaOH=

0,0386 kg.jam-1

39,9969 kg.kmol-1=0,0009 kmol/jam

N_{2-Feniletanol}15 = F2-Feniletanol 15

Mr2-Feniletanol=

143,9649 kg.jam-1

122,166 kg.kmol-1=1,1784 kmol/jam X_metanol15 = Nmetanol15

N_metanol15 +N

2-Feniletanol

15 =

83,9596

83,9596+0,0009= 0,9862 X_air15 =1 -X_metanol15 = 1 – 0,9862= 0,0138

F15 = F15metanol + F152-Feniletanol

= 2690,1986 + 143,9649 = 2834,1635 kg/jam Feed = N15 = N15metanol + N152-Feniletanol

= 83,9596 + 1,1784 = 85,138 kmol/jam XF = 0,9862

(8)

Neraca mol total

F.XF = D.XD+ W.Xw

85,138 . 0,9862 = D. 0,9995 + (85,138 – D) . 0,0005

D = 84,001 kmol/jam

W = F – D = 1,137 kmol/jam

Alur Keluar Destilat F19

N_metanol19 = XD. D = 83,959 kmol/jam N_{2-Feniletanol}19 = (1 - XD) . D = 0,042 kmol/jam F_metanol19 = 83,959 kmol.jam-1. 32,0416 kg.kmol-1 = 2690,1804 kg/jam F_{2-Feniletanol}19 = 0,042 kmol.jam-1. 122,166 kg.kmol-1 = 5,131 kg/jam

Alur Keluar Bottom F23

N_metanol23 = XW. W = 0,0006 kmol/jam N_{2-Feniletanol}23 = (1 – XW) . W = 1,1364 kmol/jam F_metanol23 = 0,0006 kmol.jam-1. 32,0416 kg.kmol-1 = 0,0182 kg/jam F_{2-Feniletanol}23 = 1,1364 kmol.jam-1. 122,166 kg.kmol-1 = 138,8339 kg/jam

Tekanan Uap

Metanol : ln𝑃 𝑘𝑃𝑎 = 16,5785 +_𝑇_𝐶3628,27

+ 239,5 ... (1) (Smith, 2005)

2-Feniletanol : ln𝑃 𝑃𝑎 = 317,2 𝑅 −

89904,2 𝑅. 𝑇(𝐾)−

76,3 𝑅 ln⁡(

𝑇 𝐾

298,15) ... (2) R = 8,31447 J/K.mol

(Emel’yanenko, 2007)

Penentuan titik gelembung (bubble point) umpan :

Tekanan uap ditentukan dengan persamaan (1) dan (2).

(9)

P_{2-Feniletanol}sat = Psistem = 760 mmHg = 101,325 kPa T_{2-Feniletanol}sat = 490,0813 K

T_metanolsat = 337,8489 K b. Menghitung T rata-rata

Zat Xi T

2-Feniletanol (a) Metanol (b)

0,9862 0,0138

490,0813 337,8489 T rata-rata =



Xi . Ti = 339,956 K

c. Menghitung harga AB pada T rata-rata

Zat T Pi

339,956 339,956

0,1706 110,0141

AB= Pb Pa

= 0,001551 d. Menghitung P_bsat

P_bsat= P X_a.α_AB+X_b=

101,325

0,9862 x 0,001551+ 0,0138=102,745 kPa e. Menghitung T dari P_bsat dari persamaan (1).

Dari hasil iterasi, didapatkan bahwa harga T = 338,2033 K = 65,0533 oC

Penentuan titik gelembung (bubble point) bottom :

a. Menghitung temperatur jenuh masing masing komponen Asumsi P_metanolsat = Psistem = 760 mmHg = 101,325 kPa

T_metanolsat = 337,8489 K b. Menghitung T rata-rata

Zat Xi T

0,9995 0,0005

(10)

c. Menghitung harga AB pada T rata-rata

Zat T Pi

490,0052 490,0052

101,1224 5464,607

AB= Pb Pa

= 0,018505 d. Menghitung P_bsat

P_bsat= P X_a.α_AB+X_b=

101,325

0,9995 x 0,018505+ 0,0005=5334,098 kPa e. Menghitung T dari P_bsat dari persamaan (1).

Dari hasil iterasi, didapatkan bahwa harga T = 489,6257 K = 215,4757oC

Penentuan titik embun (dew point) destilat :

a. Menghitung temperatur jenuh masing masing komponen Asumsi P_metanolsat = Psistem = 760 mmHg = 101,325 kPa

T_metanolsat = 337,8489 K

a. Menghitung T rata-rata P

Pii..YYii == XXii.. PPsati K

Kaarreennaa ddiiaassuummssiikkaann PPii == PPsati _,_,_m_ma_ak_ka_a_X_Xi_i₌₌_Y_Yi_i

Zat Yi T

0,0005 0,9995

490,0813 337,8489 T rata-rata =



Yi . Ti = 337,9251 K

b. Menghitung harga AB pada T rata-rata

Zat T Pi

337,9251 337,9251

0,1489 101,6286

AB= Pb Pa

= 0,00146475

c. Menghitung P_asat

(11)

d. Menghitung T dari P_bsat dari persamaan (2)

Dari hasil iterasi, didapatkan bahwa harga T = 342,29 K = 69,14 oC

Umpan masuk pada temperatur 83,78 oC, sehingga 0< q <1, sebagian umpan berupa uap, dengan Yf = 0,9975 dari persamaan (1).

Perhitungan Reflux minimum

Rm = Xd - Yf Yf - Xf =

0,9995-0,9975 0,9975-0,9862 Rm =0,17634

Persamaan Fenske untuk mencari jumlah plat minimum

∝𝑏𝑣= 𝑌𝑓.

(1− 𝑋_𝑓)

𝑋𝑓.(1− 𝑌𝑓)

= 0,9975. (1−0,9862)

0,9862. (1−0,9975)= 5,60027

𝑛𝑚 + 1 =

log⁡ 𝑋𝑏_𝑋𝑎

𝑑 𝑋𝑎𝑋𝑏 𝑠 log∝_𝑏𝑣 =

log⁡ 0,9995_0,0005 0,9995_0,0005

log 5,60027 = 8,823237

Dan 1,2 < R/Rm < 1,5, sehingga dengan metode Gilliland, diperoleh hubungan rasio refluks dan jumlah plate sebagai berikut

Hubungan Rasio Refluks dan Jumlah Plate

R (R-Rm)/(R+1) [(n+1)-(nm+1)]/(n+2) n

0,2116 0,0291 0,6 7,5232

0,2293 0,043 0,58 7,6023

0,2469 0,0565 0,55 7,701

0,2645 0,0697 0,53 7,7556

D

Diippeerroolleehh :: R R == 00,,22111166 n

n == 77,,55223322 J

Juummllaahh ppiirriinngg tteeoorriittiiss == 77,,55223322 ++ 11 rreebbooiilleerr E

Effiissiieennssii pipirriinngg == 9900%%,, mamakkaa jjuummllaahh pipirriinngg yayanngg sseebbeennaarrnnyaya = =7,7,55223322//00,,99 == 88,,33559922 =

= 88 ppiirriinngg..

E

Essttiimmaassii ffeeeedd--ppllaatteeaattaauuppllaattuummppaannddeennggaannmmeettooddeeKKiirrkkbbrriiddee

logNe

N_s = 0,206 log XHF X_LF

W D

XLW X_HD

(12)

logNe Ns

= 0,206 log 0,0005 0,0005

0,5168 38,1823

0,00013 0,0019

2

logNe Ns

= -0,76658

N_e Ns

= 0,171165

Ne + Ns = 0,171165 Ns + Ns = N 1,171165 Ns = 8

Ns = 6,8308 ≈ 7

Ne + Ns = N Ne + 7 = 8

Ne = 1

Feed tray adalah tray ke-1 dari puncak kolom destilasi

Tabel LA.6 Neraca Massa Tray Distillation Tower (TD-201)

Komponen

Alur 15 Alur 19 Alur 23

F N F N F N

(kg/jam (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam)

C8H10O 143,9649 1,1784 5,1310 0,0420 138,8339 1,1364 CH3OH 2690,1986 83,9596 2690,1804 83,9590 0,0182 0,0006

NaOH 0,0368 0,0009 - - 0,0368 0,0009

Total 2834,2003 2834,2003

LA.7 Kondensor (E-202)

Kondensor

E-202 (17)

(19) (18)

Metanol 2-Feniletanol

Metanol 2-Feniletanol Metanol

2-Feniletanol

(13)

Vm = Lm - W = 102,9136 – 1,137 = 101,7766 kmol/jam Alur 18

F_metanol18 = X18metanol . Ln x Mr metanol

= (0,9995. 17,7757 kmol/jam) x 32,0416 kg/kmol = 589,2756 kg/jam

F_{2-Feniletanol}18 = (1 – X18metanol) . Ln x Mr 2-Feniletanol

= (0,0005. 17,7757 kmol/jam) x 122,166 kg/kmol = 1,0858 kg/jam

Alur 17

F_metanol17 = X17metanol . Vn x Mr metanol

= (0,9995 . 101,7766 kmol/jam) x 32,0416 kg/kmol = 2690,1804 kg/jam

F_{2-Feniletanol}17 = (1 – X17metanol) . Vn

= (0,0005 . 101,7766 kmol/jam) x 122,166 kg/kmol 1 = 5,131 kg/jam

Tabel LA.7 Neraca Massa Kondensor (E-202)

Komponen

Alur 17 Alur 18 Alur 19

F N F N F N

C8H10O 6,2168 0,0509 1,0858 0,0089 5,131 0,042 CH3OH 3259,4560 101,7268 589,2756 17,7668 2690,1804 83,9590

Total 3265,6728 3265,6728

LA.8 Reboiler E-203

Reboiler E-203 (22)

(23) (21)

2-Feniletanol Metanol NaOH

2-Feniletanol Metanol NaOH 2-Feniletanol

Metanol NaOH

Alur 21

(14)

F_NaOH21 = 𝐹𝑁𝑎𝑂𝐻23

(𝐹_{𝑚𝑒𝑡𝑎𝑛𝑜𝑙}23 +𝐹₂23_{−𝐹𝑒𝑛𝑖𝑙𝑒𝑡𝑎𝑛𝑜𝑙} )𝑥 (𝐹𝑚𝑒𝑡𝑎𝑛𝑜𝑙 21 ₊_𝐹

221−𝐹𝑒𝑛𝑖𝑙𝑒𝑡𝑎𝑛𝑜𝑙 ) = 0,0368/(0,0182 + 138,8339) x (3259,4742+145,0507) = 0,9026 kg/jam

Alur 22

F_metanol22 = F_metanol21 - F_metanol23 = 3259,4742 – 0,0182 = 3259,4560 kg/jam F_{2-Feniletanol}22 = F_{2-Feniletanol}21 - F_{2-Feniletanol}23 = 145,0507 – 138,8339 = 6,2168 kg/jam F_NaOH22 = 𝐹𝑁𝑎𝑂𝐻23

(𝐹_{𝑚𝑒𝑡𝑎𝑛𝑜𝑙}23 +𝐹₂23_{−𝐹𝑒𝑛𝑖𝑙𝑒𝑡𝑎𝑛𝑜𝑙} )𝑥 (𝐹𝑚𝑒𝑡𝑎𝑛𝑜𝑙 22 ₊_𝐹

222−𝐹𝑒𝑛𝑖𝑙𝑒𝑡𝑎𝑛𝑜𝑙 ) = 0,0368/(0,0182 + 138,8339) x (3259,4560 + 6,2168) = 0,8658 kg/jam

Tabel LA.8 Neraca Massa Reboiler E-203

Komponen

Alur 21 Alur 22 Alur 23

F N F N F N

(kg/jam (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam)

C8H10O 145,0507 1,1873 6,2168 0,0509 138,8339 1,1364 CH3OH 3259,4742 101,7263 3259,4560 101,7258 0,0182 0,0006 NaOH 0,9026 0,0226 0,8658 0,0216 0,0368 0,0009

Total 3405,4275 3405,4275

LA.9 Mixer Metanol (M-104)

Metanol

Mixing Point Metanol M-104

(20)

(5) (2)

Metanol

F5Metanol = 2690,1986 kg/jam F20Metanol = 2690,1804 kg/jam F2Metanol + F20Metanol = F5Metanol

(15)

Tabel LA.9 Neraca Massa Mixer Metanol (M-104)

Komponen

Alur 2 Alur 20 Alur 5

F N F N F N

(kg/jam (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam)

CH3OH 0,0182 0,0006 2690,1804 83,9590 2690,1986 83,9596

(16)

LAMPIRAN B

PERHITUNGAN NERACA PANAS

Kapasitas Produksi : 138,8889 kg/jam Basis perhitungan : 1 jam operasi

Waktu operasi : 300 hari / tahun ; 24 jam / hari Satuan operasi : kg/jam

Suhu referensi : 25oC (298,15 K)

Neraca panas ini menggunakan rumus-rumus perhitungan sebagai berikut: Persamaan untuk menghitung kapasitas panas (Reklaitis, 1983):

Cp = a + bT + cT2 + dT3

Jika Cp adalah fungsi dari temperatur maka persamaan menjadi:



       

Untuk sistem yang melibatkan perubahan fasa persamaan yang digunakan adalah:



2 



 



Perhitungan energi untuk sistem yang melibatkan reaksi:







B.1 Data Perhitungan Cp

Tabel LB.1 Nilai Konstanta a, b, c dan d untuk Perhitungan Cp Cairan

Komponen a b c d

CH3OH -2,5825E+02 3,3582 -1,1638E-02 1,4052E-05 H2O 1,8296E+01 4,7112E-01 -1,3387E-03 1,3142E-06 Sumber : (Reklaitis, 1983)

(17)

Tabel LB.2 Nilai Konstanta a, b, c, d dan e untuk Perhitungan Cp Gas

Komponen a b c d e

CH3OH 3,4492E+01 -2,9188E-02 2,8684E-04 -3,125E-07 1,0983E-10 H2 1,7639E+01 6,7005E-02 -1,3148E-04 1,0588E-07 -2,9180E-11 Sumber : (Reklaitis, 1983)

Cpg = a + bT + cT2 + dT3 + eT4 [J/mol. K]



         

T

T T e T T d T T c T T b T T a dT Cpg 2

1

) (

5 ) (

4 ) (

3 ) (

2 )

( ₂ ₁ ₂2 ₁2 ₂3 ₁3 ₂4 ₁4 ₂5 ₁5

B.2 Estimasi Cp Cairan dengan Metode Chueh dan Swanson

Tabel LB.3 Kontribusi Gugus Kapasitas Panas Cairan Gugus Harga (kJ/(kmol.K)

(ring) 22

(ring)

12

(ring)

18 (ring) 26 30,4 44,8 35 Sumber : (Reid ,dkk., 1987)

Cp C8H8O = 5 × (ring) + 1 × (ring) + 1 × (ring) + 1 × + 1 × (ring)

= 5 × 22 + 1 × 12 + 1 × 18 + 1 × 35 + 1 × 26 = 201 kJ/kmol.K

Cp C8H10O = 5 × (ring) + 1 × (ring) + 2 × + 1 × = 5 × 22 + 1 × 12 + 2 × 30,4 + 1 × 44,8

(18)

B.3 Perhitungan Estimasi ∆Hf(298) (kJ/mol)

P

Peerrhhiittuunnggaann esesttiimmaassii ∆H∆Hoof f (k(kJJ..mmooll --11

)

) dedennggaann mmeenngggguunnaakkaann memettooddee JoJobbaacckk d

deennggaann rurummuuss ::∆∆HHooff = = 6868,,2299 + + jjnnjj..∆∆HH ((RReeiidd,, dkdkkk;;,, 19198877)),, didi mamannaa kokonnttrriibbuussii g

guugguussnnyyaa ddaappaatt ddiilliihhaatt ppaaddaa ttaabbeell ddii bbaawwaahh iinnii.. T

Tababeell LLBB..44 KKoonnttrriibbuussii GGuugguuss PPaannaass PPeemmbbeennttuukkaann Gugus Harga (kJ/(mol)

(ring) 2,09

(ring)

46,43

(ring)

8,67 (ring) -26,8

-20,64 -208,04 -138,16 Sumber : (Reid ,dkk., 1987)

∆Hof(298) C8H8O = 68,29 + (5 × (ring) + 1 × (ring) + 1 × (ring) + 1 × + 1 × (ring)

= 68,29 + ( 5 × 2,09 + 1 × 46,43 + 1 × 8,67 + 1 × -138,16 + 1 × -20,64)

= -31,12 kJ/mol

∆Hof(298) C8H10O = 5 × (ring) + 1 × (ring) + 2 × + 1 × = 5 × 2,09 + 1 × 46,43 + 2 × -20,64 + 1 × -208,04 = -124,15 kJ/mol

B

B..44 NNiillaaiiPPaannaassLLaatteenn

T

Taabbeell LLBB..55 DDaattaa PPeerrhhiittuunnggaann NNiillaaii PPaannaass LLaatteenn

Senyawa w Tc (K) Pc (bar) T boiling

(°C)

T boiling (K)

CH3OH 0,564 512,6 80,97 65 338,15

C8H10O 0,603 751 44,7 216,93 490,08

(

(19)

R

R == 00,,000088331144 kkJJ//mmooll KK ∆

∆HHvvll == 11,,009933 RR .. TTcc [[TTbbrr((llnn[[PPcc//11,,0011332255]]--11))//((00,,9933--TTbbrr))]] T

Tbb == TT bbooiilliinngg ((KK)) (

(DDaauubbeerrtt,, 11998855))

LB.1 Heater (E-101)

Hidrogen 40oC, 20 atm Hidrogen

31,67oC, 20 atm

Saturated Steam 230oC, 27,61 atm

Kondensat 230oC, 27,61 atm

8 9

LB.1.1 Panas Masuk

Tabel LB.6 Perhitungan Panas Masuk Heater (E-101)

Komponen (i)

N8i



_Cp _dT 82 , 304

15 , 298

Qi = N8i .



Cp dT 82 , 304

15 , 298

(kmol/jam) _(kJ/mol) _(kJ/jam)

Hidrogen 1,4141 190,4253 269,285

Qin = 269,285

LB.1.2 Panas Keluar

Tabel LB.7 Perhitungan Panas Keluar Heater (E-101)

Komponen (i)

N9i



_Cp _dT 15 , 313

15 , 298

Qi = N9i .



Cp dT 15 , 313

15 , 298

Hidrogen 1,4141 429,0552 606,7373

Qout = 606,7373 Panas yang dibutuhkan (dQ/dT):

(20)

= 606,7373 – 269,285 = 337,4522 kJ/jam Massa steam yang diperlukan:

Steam yang digunakan adalah saturated steam dengan kondisi: Suhu : 230 oC

Tekanan : 27,61 atm λ (230oC) : 1811,7 kJ/kg m =

) 230 (

/ 0

C dt dQ



=

7 , 1811

4522 , 337

= 0,1863 kg/jam

LB.2 Heater (E-102)

Metanol Stirena Oksida

NaOH 40oC, 1 atm Metanol

Stirena Oksida NaOH 30oC, 1 atm

6 7

Komponen (i)

N6i



_Cp _dT 15 , 303

15 , 298

Qi = N6i .



Cp dT 15 , 303

15 , 298

Metanol 83,9596 406,5336 34132,3837

Stirena Oksida 1,1784 1005 1184,3288

NaOH 0,0009 235,85 0,2171

(21)

Komponen (i)

N7i



_Cp _dT 15 , 313

15 , 298

Qi = N7i .



Cp dT 15 , 313

15 , 298

Metanol 83,9596 1233,1418 103534,0433

Stirena Oksida 1,1784 3015 3552,9865

NaOH 0,0009 707,55 0,6512

Qout = 107087,6811

Panas yang dibutuhkan (dQ/dT): dQ/dT = Qout – Qin

= 107087,6811 – 32316,9296 = 71770,7514 kJ/jam

Massa steam yang diperlukan:

Tekanan : 27,61 atm λ (230o

C) : 1811,7 kJ/kg m =

) 230 (

/ 0

C dt dQ



=

7 , 1811

7514 , 71770

= 39,6151 kg/jam

LB.3 Fixed Bed Reactor (R-101)

Metanol Stirena Oksida

NaOH Hidrogen 40oC, 20 atm

NaOH Hidrogen 40o_{C, 20 atm} Air Pendingin

25o_C

Air Pendingin 40oC

10

(22)

Tabel LB.10 Perhitungan Panas Masuk Fixed Bed Reactor (R-101)

Komponen (i)

N10i



_Cp _dT 15 , 313

15 , 298

Qi = N10i .



Cp dT 15 , 313

15 , 298

Metanol 83,9596 1233,1418 103534,0433

Stirena Oksida 1,1784 3015 3552,9865

NaOH 0,0009 707,55 0,6512

Hidrogen 1,4141 429,0552 606,7373

Qin = 107694,4183

Tabel LB.11 Perhitungan Panas Keluar Fixed Bed Reactor (R-101)

Komponen (i)

N11i



_Cp _dT 15 , 313

15 , 298

Qi = N11i .



Cp dT 15 , 313

15 , 298

Metanol 83,9596 1233,1418 103534,0433

NaOH 0,0009 707,55 0,6512

2-Feniletanol 1,1784 3414 4023,1827

Hidrogen 0,2357 429,0552 101,1229

Qout = 107659,0002

LB.3.3 Panas Reaksi

Reaksi hidrogenasi di dalam Fixed Bed Reactor: C8H8O(l) + H2(g) → C8H10O (l) r1 = konversi x Nstirena oksida

r1 = 1 x 1,1784 = 1,1784 kmol/jam

Panas reaksi yang terjadi pada 25oC dan 1 atm: ∆Hr (25oC) = [∆Hof produk- ∆Hof reaktan]

= [∆Hof C8H10O - ∆Hof C8H8O –∆Hof H2] = [(-124,15) - (-31,12) - (0)]

(23)

∆Hr (40oC) = ∆Hr(25oC)+ _

  

  





15 , 313 15 , 298

tan

2 dT

Cp _Fenile _ol + _

  

  



,15

313 15 , 298

dT Cp_stirenaoks_ida -

    

  



,15

313 15 , 298

dT Cphidrogen

= -93030 + 3414 – 3015 – 429,0552 = -93060,0552 kJ/jam

Q reaksi = -93060,0552 kJ/jam x 1,1784 kmol/jam = -109665,3803 kJ/jam

Panas yang dibutuhkan (dQ/dT): dQ/dT = Qout – Qin + Qreaksi

= 107659,0002 – 107694,4183+ (-109665,3803) = -109700,7985 kJ/jam

Maka, massa air pendingin yang diperlukan: Air Pendingin masuk (25oC) ; H1 = 104,8 kJ/kg

Air Pendingin keluar (38oC) ; H2 = 159,1 kJ/kg (Smith dkk, 2005)

m =

1 2

/

H H

dt dQ

 

m =

3 , 54

) 7985 , 109700 (



m = 2020,2725 kg/jam

LB.4 Heater (E-201)

NaOH 83,78oC, 1 atm Metanol

2-Feniletanol NaOH 40oC, 1 atm

(24)

Komponen (i)

N14i



_Cp _dT 15 , 313

15 , 298

Qi = N14i .



Cp dT 15 , 313

15 , 298

Metanol 83,9596 1233,1418 103534,0433

2-Feniletanol 1,1784 707,55 0,6512

NaOH 0,0009 3414 4023,1827

Qin = 107557,8773

Komponen (i)

N15i



Cp dT 85 , 337 15 , 298

∆H_vl



CpvdT 9291 , 356

85 , 337

dT Cp



9291

, 356

15 ,

298 Qi

(kmol/jam) _(kJ/mol) _(kJ/mol) _(kJ/mol) _(kJ/mol) _(kJ/jam) Metanol 83,9596 3369,9298 33,7726 905,6462 361811,0127

2-Feniletanol 1,1784 2772,6191 15765,2706

NaOH 0,0009 13378,123 2,5519

dQ/dT = Qout – Qin

= 377578,8352 – 107557,8773 = 270020,9579 kJ/jam

C) : 1811,7 kJ/kg m =

) 230 (

/ 0

C dt dQ



=

7 , 1811

9579 , 270020

(25)

LB.5 Tray Distillation Tower

NaOH 83,78o_{C, 1 atm}

TD-201

Metanol 2-Feniletanol 69,14oC, 1 atm

Metanol 2-Feniletanol 215,4757oC, 1 atm E-202

E-203

15

16

18 19

21 22

23

Panas yang dibawa input feed (Qf) pada T = 83,78oC (356,9291 K)

Q15= N15metanol (



85 , 337 15 , 298

CpdT + ∆Hvl +



9291 , 356

85 , 337

CpvdT ) + N15 2-Feniletanol



9291 , 356

15 , 298

CpdT

Tabel LB.14 Neraca Panas pada Alur 15

Komponen (i)

N15i



Cp dT 85 , 337 15 , 298

∆Hvl



CpvdT 9291 , 356

85 , 337

dT Cp



9291

, 356

15 ,

298 Qi

(kmol/jam) _(kJ/mol) _(kJ/mol) _(kJ/mol) _(kJ/mol) _(kJ/jam)

Metanol 83,9596 3369,9298 33,7726 905,6462 361811,0127

2-Feniletanol 1,1784 2772,6191 15765,2706

NaOH 0,0009 13378,123 2,5519

Qtotal = 377578,8352

Panas Refluks keluar kondensor (QLo) pada T = 69,14oC (342,29K)

Q18= N18metanol



29 , 342 15 , 298

CpldT + N18 2-Feniletanol



29 , 342 15 , 298

CpdT

Tabel LB.15 Neraca Panas Pada Alur 18

Komponen (i)

N18i



_Cp _dT 8997 , 337

15 , 298

Qi (kmol/jam) _(kJ/mol) _(kJ/jam) Metanol 17,7668 3771,8017 67012,7144 2-Feniletanol 0,0089 10046,2640 89,2895

(26)

Panas destilat keluar kondensor (QD) pada T = 69,14oC (342,29K)

Q19= N18metanol



29 , 342 15 , 298



29 , 342 15 , 298

CpdT

Komponen (i)

N19i



_Cp _dT 8997 , 337

15 , 298

Qtotal = 317098,6088

Panas yang dibawa uap masuk kondensor (Qv) pada T = 69,14oC (342,29K)

Q16 = N16metanol x Hv + N162-Feniletanol x Hv

Komponen (i)

N20i

Hv Qi

(kmol/jam) _(kJ/mol) _(kJ/jam) Metanol 101,7258 33772,6444 3435547,7682 2-Feniletanol 0,0509 36189,3420 1841,6149

Qtotal = 3437389,3830 Panas hasil bawah (Qw) pada T = 215,4757oC (488,6257 K)

Q23= N23metanol



6949 , 489

15 , 298



6949 , 489

15 , 298

CpdT + + N23NaOH



6949 , 489

15 , 298

CpdT

Komponen (i)

N19i



_Cp _dT 6949 , 489

15 , 298

NaOH 0,0009 8984,7288 8,2696 Qtotal = 49289,6731

LB.5.1 Kondensor (E-202)

(27)

Qv = QLo + QD + QC

QC = 3437389,3830 - 67102,0038 - 317098,6088 = 3053188,704 kJ/jam

m =

1 2 H H

Qc 

m =

3 , 54

4 3053188,70

m = 56228,1542 kg/jam

LB.5.2 Reboiler (E-203)

Qinput = Qoutput

QF + QR = QD + QW + QC

QR = 317098,6088 + 49289,6731+ 3053188,704 - 377578,8352 = 3041998,2171 kJ/jam

C) : 1811,7 kJ/kg m =

) 230 ( 0C

Q_R



=

7 , 1811

71 3041998,21

(28)

LB.6 Cooler (E-204)

30oC, 1 atm Metanol

2-Feniletanol 69,14oC, 1 atm

Air 25oC

Air 40oC

19 20

Tabel LB.19 Perhitungan Panas Masuk Cooler (E-204)

Komponen (i)

N19i



_Cp _dT 8997 , 337

15 , 298

Qi = N19i .



Cp dT 8997 , 337

15 , 298

Metanol 83,9590 3771,8017 316676,6607 2-Feniletanol 0,042 10046,2640 412,9481

Qin = 317098,6088

Tabel LB.20 Perhitungan Panas Keluar Cooler (E-204)

Komponen (i)

N20i



_Cp _dT 15 , 303

15 , 298

Qi = N20i .



Cp dT 15 , 303

15 , 298

Metanol 83,9590 1233,1418 103533,3423

2-Feniletanol 0,042 1138 47,7966

= 103581,1388 – 317098,6088 = -213517,47 kJ/jam

(29)

m =

1 2

/

H H

dt dQ

 

m =

3 , 54

) 47 , 13517 2 (



m = 3932,1818 kg/jam

LB.7 Cooler (E-205)

NaOH 30oC, 1 atm Metanol

2-Feniletanol NaOH 215,4757oC, 1 atm

Air 25oC

Air 40oC

23 24

Tabel LB.21 Perhitungan Panas Masuk Cooler (E-205)

Komponen (i)

N23i



_Cp _dT 6949 , 489

15 , 298

Qi = N23i .



Cp dT 6949 , 489

15 , 298

Metanol 0,0006 25183,4183 14,3168

2-Feniletanol 1,1364 43352,2695 49267,0867

NaOH 0,0009 8984,7288 8,2696

Qin = 49289,6731 LB.7.2 Panas Keluar

Tabel LB.22 Perhitungan Panas Keluar Cooler (E-205)

Komponen (i)

N24i



_Cp _dT 15 , 303

15 , 298

Qi = N24i .



Cp dT 15 , 303

15 , 298

(30)

2-Feniletanol 1,1364 1138 268,0285

NaOH 0,0009 235,85 1,0474

= 269,3070 – 49289,6731 = -49020,3661 kJ/jam

m =

1 2

/

H H

dt dQ

 

m =

3 , 54

) 49020,3661 (



(31)

LAMPIRAN C

PERHITUNGAN SPESIFIK ALAT

LC.1 Tangki Penyimpanan Hidrogen (H2) (TT-101)

Fungsi : Menyimpan hidrogen untuk kebutuhan 10 hari Bahan Konstruksi : Low-Alloy Steel SA202, Grade B

Bentuk : Silinder vertikal dengan alas dan tutup spherical Jenis Sambungan : Double welded butt joints

Jumlah : 1 unit

Data Perhitungan:

Temperatur, T = 30oC

Tekanan, P = 30 atm

Kebutuhan perancangan, t = 10 hari

Laju alir massa, F = 2,3765 kg/jam

ρ hidrogen, ρ = 2,3713 kg/m3 (Green & Perry, 2008)

Perhitungan Ukuran Tangki:

1

1.. VVoolluummeeTTaannggkkii

V

V HHiiddrrooggeenn == ₃ / 3713 , 2

/ 24 10

/ 3765 , 2

m kg

hari jam hari

jam

kg  

=

= 224400,,66220033 mm33 V

V HHiiddrrooggeenn == 6633556644,,66772277 ggaall U

Unnttuukk ttaannggkkii ddeennggaann vvoolluummee lleebbiihh bbeessaarr ddaarriippaaddaa 1100..000000 ggaall,, mmaakkaa ddiigguunnaakkaann t

taannggkkii vveerrttiikkaall ((WWaallaass,, 11998888)).. F

Faakkttoorr kkeelloonnggggaarraann == 2200%% V

Voolluummee ttaannggkkii,, VVtt == 11,,22 ×× 224400,,66220033 == 228888,,77444444 mm33

2

2.. DDiiaammeetteerrddaannttiinnggggii sshheellll D

Diirreennccaannaakkaann:: T

Tiinnggggii sshheelllltatannggkkii :: ddiiaammeetteerr ttaannggkkii HHss :: DD == 55 :: 44 T

(32)

V

Voolluummee sshheellll ttaannggkkii ((VVss)) V

Vs s == ¼¼ π π DD22HHs s V

Vs s == 3 16

5

D



V

Voolluummee ttuuttuupp ttaannggkkii ((VVhh)) sspphheerriiccaall V

Vh h == 3 6 1

D 

V

Voolluummee ttaannggkkii ((VV)) V

V == VVs s ++ 22VVh h 2

28888,,77444444 == 3 16

5

D

 ++ 3

3D



2

28888,,77444444 == 3 48 31

D



M

Maakkaa ddiiaammeetteerr ttaannggkkii,, DD == 55,,22220022 mm == 220055,,55220044 iinn T

Tiinnggggii sshheelllltatannggkkii,, HHs s ==  6,5253

    

D D H_s

m m

T

Tiinnggggii ttuuttuupp ttaannggkkii,, HHhh ==  1,3051

    

D D H_h

m m T

Tiinnggggii ttaannggkkii,, HHtt == HHs s ++ 22HHh h == 99,,11335544 mm

3

3.. TTeebbaall sshheellll ttaannggkkiiddaannttuuttuuppttaannggkkii

U

Unnttuukk ttuuttuupp aattaass ttaannggkkii:: T

Teekkaannaann ooppeerraassii ddeennggaann ffaakkttoorr kkeeaammaannaann 2200%%:: P

Pooppeerraassii = = 33..003399,,7755 kkPPaa P

Pddeessaaiin n = = 11,,22 ×× 33..003399,,7755 kkPPaa == 33..664477,,77 kkPPaa

(33)

Tebal shell tangki :

t = nC

P SE

D P

 



2 , 1

2 (Peters, dkk., 2004)

t = 10 (1/80)

3647,7 2

, 1 85 , 0 46170 1 2

205,5204 3647,7

  

  



t

t == 33,,119966 iinn t

teebbaall sshheellllststaannddaarr yyaanng g ddiigguunnaakkaann == 33¼¼ iinn..

T

Teebbaallttuuttuuppttaannggkkiibbaawwaahhddaannaattaass::

t = nC

P SE

D P

 



4 , 0

t = 10 (1/80)

3647,7 4

, 0 85 , 0 146170 2

5204 , 05 2 3647,7

  

  



t

t == 33,,11559988 iinn t

teebbaall sshheellll ssttaannddaarr yyaanng g ddiigguunnaakkaann == 33¼¼ iinn..

LC.2 Tangki Penyimpanan Metanol (CH3OH) (TT-102)

Fungsi : Menyimpan hidrogen untuk kebutuhan 60 hari Bahan Konstruksi : Stainless Steel SA-240, Grade 340

Bentuk : Silinder vertikal dengan alas dan tutup ellipsoidal Jenis Sambungan : Double welded butt joints

Jumlah : 1 unit

Tekanan, P = 1 atm

ρ metanol, ρ = 784,7798 kg/m3 (Green & Perry, 2008)

1

(34)

V

V MMeettaannooll == ₃ / 7798 , 784

/ 24 60

/ 1,8864

m kg

hari jam hari

jam

kg  

=

= 33,,44661144 mm33 V

V MMeettaannooll == 991144,,33995555 ggaall U

Unnttuukk ttaannggkkii ddeennggaann vvoolluummee lleebbiihh kkeecciill ddaarriippaaddaa 11000000 ggaall,, mmaakkaa ddiigguunnaakkaann t

2

Tiinnggggii ttuuttuupp ttaannggkkii :: ddiiaammeetteerr ttaannggkkii HHh h :: DD == 11 :: 44

V

Vs s == 3 16

5

D



V

Voolluummee ttuuttuupp ttaannggkkii ((VVhh)) eelllliippssooiiddaall V

Vh h == D Hh

2 6



V

Vh h == 3 24D



V

V == VVs s ++ 22VVh h 4

4,,11553377 == 3 16

5

D

 ++ 3

12D



4

4,,11553377 == 3 48 19

D



M

    

D D H_s

m m

T

    

D D H_h

(35)

T

3

U

Pooppeerraassii = = 110011,,332255 kkPPaa P

Pddeessaaiinn = = 11,,22 ×× 110011,,332255 kkPPaa == 112211,,5599 kkPPaa

U

Unnttuukk sshheellll ttaangngkkii ddaann ttuuttuupp bbaawwaahh ttaannggkkii ::

Tinggi cairan dalam tangki, h = m m

m m

1797 , 2 6156 , 2 1537

, 4

4614 , 3

3 3

 

T

Teekkaannaann hhiiddrroossttaattiikk:: P

P == ρ ρ ×× gg ×× hh == 778844,,77779988 kkgg//mm33 ×× 99,,88 mm//ddeett22 ×× 22,,11779977 mm == 1166,,77663344 kkPPaa T

P ooppeerraassi i == 110011,,332255 kkPPaa ++ 1166,,77663344 kkPPaa == 111188,,00888844 kkPPaa P

P ddeessaaiin n == 11,,22 ×× 111188,,00888844 == 114411,,77006611 kkPPaa

Joint efficiency, E = 0,85 (Peters, dkk., 2004) Allowable stress, S = 128.900 kPa (Peters, dkk., 2004) Faktor korosi, C = 1/80 in (Peters, dkk., 2004) Umur alat, n = 10 tahun

t = nC

P SE

D P

 



2 , 1

t = 10 (1/80)

141,7061 2

, 1 85 , 0 128900 2

,8433 8 5

141,7061 _ _

   



t

t == 00,,11663311 iinn t

teebbaall sshheellllststaannddaarr yyaanng g ddiigguunnaakkaann ==33//116 6 iin.n.

T

Teebbaallttuuttuuppttaannggkkiibbaawwaahh::

t = nC

P SE

D P

 



2 , 0

(36)

t = 10 (1/80)

LC.3 Tangki Penyimpanan Stirena Oksida (C8H8O) (TT-103)

Fungsi : Menyimpan stirena oksida untuk kebutuhan 30 hari Bahan Konstruksi : Stainless Steel SA-240, Grade 340

Jumlah : 1 unit

Tekanan, P = 1 atm

ρ stirena oksida, ρ = 1019,4837 kg/m3 (Green & Perry, 2008)

(37)

U

Voolluummee ttaannggkkii,, VVtt == 11,,22 ×× 9999,,99996611 == 111199,,99995533 mm3 3

2

V

Vs s == 3 16

5

D



V

Vh h == D Hh

2 6



V

Vh h == 3 24D



V

V == VVs s ++ 22VVh h 1

11199,,99995533 == 3 16

5

D

 ++ 3

12D



1

11199,,99995533 == 3 48 19

D



M

    

D D H_s

m m

T

    

D D H_h

m m T

(38)

3

(39)

t

t == 00,,22991133 iinn t

teebbaall sshheellll ssttaannddaarr yyaanng g ddiigguunnaakkaann == 33//88 iin.n.

T

Teebbaallttuuttuuppttaannggkkiiaattaass: :

t = nC

P SE

D P

 



2 , 0

t = 10 (1/80)

5900 , 121 2 , 0 85 , 0 128900 2

5543 , 180 5900 , 121

  

  



t

t == 00,,22225522 iinn t

teebbaall sshheellll ssttaannddaarr yyaanng g ddiigguunnaakkaann == ¼¼ iinn..

LC.4 Tangki Penyimpanan 2-Feniletanol (C8H10O) (TT-201)

Fungsi : Menyimpan stirena oksida untuk kebutuhan 30 hari Bahan Konstruksi : Stainless Steel SA-240, Grade 340

Jumlah : 1 unit

Tekanan, P = 1 atm

ρ 2-feniletanol, ρ = 1003,2031 kg/m3 (Green & Perry, 2008)

1

V

V SSttiirreennaaookkssiiddaa == ₃

/ 2031 , 1003

/ 24 30

/ 8889 , 138

m kg

hari jam hari

jam

kg  

=

= 9999,,66880077 mm33 V

V SSttiirreennaaookkssiiddaa == 2266333322,,66552266 ggaall U

(40)

V

2

V

Vs s == ¼¼ π π DD2 2

H Hs s V

Vs s == 3 16

5

D



V

Vh h == D Hh

2 6



V

Vh h == 3 24D



V

V == VVs s ++ 22VVh h 1

11199,,66116688 == 3 16

5

D

 ++ 3

12D



1

11199,,66116688 == 3 48 19

D



M

    

D D H_s

m m

T

    

D D H_h

m m T

3

U

(41)

P

(42)

T

Teebbaallttuuttuuppttaannggkkiiaattaass: :

t = nC

P SE

D P

 



2 , 0

t = 10 (1/80)

5900 , 121 2 , 0 85 , 0 128900 2

3643 , 180 5900 , 121

  

  



t

t == 00,,22225511 iinn t

teebbaall sshheellll ssttaannddaarr yyaanng g ddiigguunnaakkaann == ¼¼ iinn..

L

LCC..55 GGuuddaannggPPeennyyiimmppaannaannNNaattrriiuummHHiiddrrookkssiiddaa((NNaaOOHH))ddaannKKaattaalliissPPdd//CC11% %

Fungsi : Menyimpan natrium hidroksida dan katalis Pd/C 1% dalam kemasan plastik selama 150 hari

Bahan Konstruksi : Dinding dari beton dan atap dari seng Bentuk : Prisma segi empat beraturan

Jumlah : 1 unit



 NNaattrriiuummHHiiddrrookkssiiddaa

Tekanan, P = 1 atm

Kebutuhan perancangan, t = 150 hari Laju alir massa, F = 0,0368 kg/jam

ρ natrium hidroksia, ρ = 2130 kg/m3 (Green & Perry, 2008) Kapasitas gudang = 0,0368 kg/jam × 24 jam/hari × 150 hari = 132,5277 kg. Natrium hidroksida dikemas dalam goni plastik dengan kapasitas 10 kg/goni. Maka goni yang dibutuhkan = 

goni kg

kg / 10

5277 , 132

13,2528 goni Asumsi:

Tebal 1 goni : 10 cm Lebar 1 goni : 25 cm Panjang 1 goni: 40 cm

Volume 1 goni = panjang x lebar x tebal = 0,4 x 0,25 x 0,1 = 0,01 m3/ goni

(43)



 KKaattaalliissPPdd//CC11% %

Tekanan, P = 1 atm

Kebutuhan perancangan, t = 150 hari Laju alir massa, F = 1,2743 kg/jam ρ katalis Pd/C 1%, ρ = 2016,813 kg/m3

(Green & Perry, 2008) Kapasitas gudang = 1,2743 kg/jam × 24 jam/hari × 150 hari = 4587,497 kg. Katalis Pd/C 1% dikemas dalam goni plastik dengan kapasitas 50 kg/goni. Maka goni yang dibutuhkan = 

goni kg

kg / 50

497 , 4587

91,7499 goni Tinggi gudang:

Asumsi tebal 1 goni plastik = 15 cm Maksimal tumpukan goni = 15 buah Faktor kelonggaran = 50%

Tinggi gudang yang dibutuhkan = 1,5 × 15 cm × 15 = 3,375 m = 3,5 m. Panjang gudang:

Direncanakan susunan goni = 3 goni × 2goni Dimana panjang 1 goni = 60 cm

Faktor kelonggaran = 30% Untuk jalan dalam gudang = 30%

Panjang gudang yang dibutuhkan = 1,6 × 60 cm × 3 = 2,88 m = 3 m. Lebar gudang :

Faktor kelonggaran = 30 % Dimana lebar 1 goni 45 cm

Lebar gudang yang dibutuhkan = 1,3 × 45 cm × 20 = 1,17 m = 1,5 m

L

LCC..66 VVeerrttiiccaall KKnnoocckkoouutt DDrruumm((FFGG--220011) )

Fungsi : Memisahkan fasa gas hidrogen dari produk keluaran reaktor Jenis : Vertical knockout drum

Bahan konstruksi : Stainless Steel SA-240, Grade 340

(44)

Data Perhitungan :

Temperatur = 40 oC

Tekanan = 19 atm

Lama Hold-up = 20 menit

Kondisi umpan gas masuk Knock Out Drum

Laju alir massa H2 = 0,4751 kg/jam = 1,0474 lbm/jam Densitas H2 = 0,041 kg/m3

Tabel LC.1 Kondisi umpan cair masuk Knock Out Drum Komponen Laju

massa

(kg/jam)

%

Berat

(x)

Densitas

(kg/m3)

ρ

campuran

(ρ . x)

CH3OH 2690,1986 0,94919 774,9601 735,5855 C8H10O 143,8649 0,05080 993,7333 50,4773

NaOH 0,0368 0,00001 2130 0,0213

Total 2834,2003 1 786,0905

Laju alir massa cairan =

jam lbm 3748 , 248 6 kg

1 lbm 2,2046 x

jam kg 2003 ,

2834 

Densitas gas = ₃ ₃

3 3

ft lbm 0026 , 0 ft

1 m 028317 ,

0 kg

1 lbm 2046 , 2 m

kg

0,4751 x x 

Densitas cairan = ₃ ₃

3 3

ft lbm 0745 , 9 4 ft

1 m 028317 ,

0 kg

1 lbm 2046 , 2 m

kg

786,0905 x x 

Laju alir volumetri gas (Qg) = 409,6498ft /jam 0,0026

0,4751 ρ

F_ _ 3

= 6,8275ft3/min = 0,1138 ft3/s

Laju alir volumetri cairan (Ql) = 127,3241ft /jam 49,0745

2834,2003 ρ

F_ _ 3

(45)

Vertical knockout drum yang digunakan mempunyai efisiensi sangat tinggi sehingga nilai k = 0,25 (Walas, 1988).

1 0026 , 0 0745 , 49 0,25 1 k

u   

g

 

= 34,634 ft/s

Diameter wire mesh yg dibutuhkan = Qg u     

4



= 34,634

4 1138 ,

0 

    

= 1,7597 ft

Diameter rancangan untuk vertical knockout drum adalah 2,75 ft = 33,003 in Ruang untuk uap minimum 4,5 ft.

Kedalaman cairan:Untuk hold-up 20 menit

=

 

2

3

ft 5625 , 7 π/4

menit .20 menit ft

1221 , 2

= 7,1427 ft Htotal = Hcairan + Hgas

= 7,1427 ft + 4,5ft = 11,6427 ft

H/D =11,6427ft/2,75 ft

= 4,2337 (H/D 3-5, memenuhi) D = 2,75 ft = 0,8382 m

Tinggi knock out drum = 11,6427 ft + 2 x 0,6875 = 13,0176 ft

Poperasi = 19 atm = 1925,175 kPa

Pdesign = 1,2 x Poperasi = 1,2 x 101,325 kPa = 121,59 kPa

Untuk shell tangki dan tutup bawah tangki :

Tinggi cairan dalam tangki, h = 7,1427 ft = 2,1771 m





2

cairan

4 / . H

D up hold waktu Q_l



(46)

Tekanan hidrostatik:

P = ρ × g × h = 768,0905 kg/m3 × 9,8 m/det2 × 2,1771 m = 16,7717 kPa Tekanan operasi dengan faktor keamanan 20%:

P operasi = 1925,175 kPa + 16,7717 kPa = 1941,9467 kPa P desain = 1,2 × 1941,9467 = 2330,3361 kPa

Joint efficiency, E = 0,85 (Peters, dkk., 2004) Allowable stress, S = 128.900 kPa (Peters, dkk., 2004) Faktor korosi, C = 1/80 in (Peters, dkk., 2004) Umur alat, n = 10 tahun

(47)

LC.7 Mixer I (M-101)

F

Fuunnggssii : : TTeemmppaatt ppeennccaammppuurraann uummppaann ccaaiirr bbeerruuppaa ssttiirreennaa o

okkssiiddaa,, mmeettaannooll,, ddaann nnaattrriiuumm hhiiddrrookkssiiddaa T

Tiippee : : TTaannggkkii BBeerrppeennggaadduukk B

Beennttuukk : : SSiilliinnddeerr vveerrttiikkaall ddeennggaann ttuuttuupp ddaann aallaass eelllliippssooiiddaall B

Baahhaann kkoonnssttrruukkssii : :SSttaaiinnlleessssSStteeeell SSAA--224400,, GGrraaddee 334400 Jenis Sambungan : Double welded butt joints

J

Juummllaahh : : 11 uunniitt

1

T

Taabbeell LLCC..22 KKoommppoossiissii UUmmppaann MMaassuukk kkee MMiixxeerr UUmmppaann MMaassuukk ((MM--110011)) K

Koommppoonneenn LaLajjuu aalliirr ((kkgg//jjaamm)) ρ ρ ((kkgg//mm33)) V V ((mm33/j/jaamm)) M

Meettaannooll 2.2.669900,,119999 787844,,77779988 3,3,4422779977 S

Sttiirreennaa OOkkssiiddaa 114411,,55889944 10101199,,44883377 0,0,1133888888 N

NaaOOH H 00,,00336688 21213300 0,0,0000000022 T

Toottaall 4.4.774466,,22992222 797966,,55332233 3,3,55666699

T

Taannggkkii ddiirraannccaanngg ddeennggaann ssppaaccee ttiimmee ((ττ),), ττ == 11 jjaamm

o o

V V

  

 

V = 1 jam × 3,5669 m3/jam = 3,5669 m3 Faktor kelonggaran = 20%

Volume tangki, Vt = 1,2 × 3,5669 = 4,2802 m3

2

2.. DDiiaammeetteerr((DD_ii))ddaannTTiinnggggiiRReeaakkttoorr((HHRR))

Asumsi Hs : Di = 1 : 1 Hh : Di = 1 : 4

Volume tangki (VR) = volume tutup dan alas + volume silinder 4,2802 m3 = 2 Di Di _H_s

                

4 24

2

3 _

(48)

4,2802 m3 = 2 i i i

D D D

                

4 24

2

3 _



4,2802 m3 = 1,0476 Di3

Di = 1,5987 m = 62,9391 in

Tinggi silinder (Hs) = 1,5987 m Tinggi tutup (h) = 0,3997 m

Tinggi tangki (H) = Hs + 2h = 2,398 m

3

3.. TTeebbaallssiilliinnddeerr((tts_s))ddaanntteebbaallhheeaadd((tthh))

U

Pooppeerraassii = = 110011,,332255 kkPPaa P

Pddeessaaiinn = = 11,,22 ×× 110011,,332255 kkPPaa == 112211,,5599 kkPPaa

U

Unnttuukk sshheellll ttaangngkkii ddaann ttuuttuupp bbaawwaahh ttaannggkkii ::

Tinggi cairan dalam tangki, h = m m

m m

9983 , 1 398 , 2 2802

, 4

5669 , 3

3 3

 

T

Teekkaannaann hhiiddrroossttaattiikk:: P

P == ρ ρ ×× gg ×× hh == 779966,,55332233 kkgg//mm33 ×× 99,,88 mm//ddeett22 ×× 11,,99998833 mm == 1155,,55998899 kkPPaa T

P ooppeerraassi i == 110011,,332255 kkPPaa ++ 1155,,55998899 kkPPaa == 111166,,99223399 kkPPaa P

P ddeessaaiin n == 11,,22 ×× 111166,,99223399 == 114400,,33008877 kkPPaa

Joint efficiency, E = 0,85 (Peters, dkk., 2004) Allowable stress, S = 128900 kPa (Peters, dkk., 2004) Faktor korosi, C = 1/80 in (Peters, dkk., 2004) Umur alat, n = 10 tahun

t = nC

P SE

D P

 



2 , 1

(49)

t = 10 (1/80)

(50)

L = panjang blade pada turbin W = lebar blade pada turbin J = lebar baffle

T

Taabbeell LLCC..33 VViisskkoossiittaass bbaahhaann mmiixxeerr uummppaann ccaaiirr ((MM--110011)) K

Koommppoonneenn LaLajjuu aalliirr ((kkgg//jjaamm)) N N ((kkmmooll)) XiXi μ μ ((ccPP)) lnln μ μ XiXi llnn μμ M

Meettaannooll 2690,1986 8383,,99559966 00,,9988661155 0,0,5555 --00,,55997788 -0-0,,558899555566 S

Sttiirreennaa OOkkssiiddaa 141,5894 1,1,11778844 00,,0011338844 0,0,3399 --00,,99441166 -0-0,,001133003333 N

Naattrriiuumm H

Hiiddrrookkssiiddaa 0,0368 0,0,00000011 00,,0000000011 11,,11 0,0,009933 00,,0000000011 T

Toottaall 28283311,,88224488 8585,,11338899 1 1 -0-0,,660022558888

V

Viisskkoossiittaass ccaammppuurraann ddaappaatt ddiihhiittuunngg ddeennggaann ppeerrssaammaaaann HHeerriicc--BBrreewweerr (

(GGrreeeenn && PPeerrrryy,, 22000088)) l

lnn μ μ == ΣΣ XXii llnn μ μ l

lnn μ μ == --00,,660022558888

μ

μ == eexxpp ((--00,,660022558888))

μ

μ == 00,,55447744 ccPP

Kecepatan pengadukan, N = 0,3 putaran / detik Bilangan Reynold (NRe) =   



 2

i

D N

123963,1

Dari fig 3.4-4 Geankoplis, 1997 dengan menggunakan kurva 1, untuk pengaduk jenis turbin impeller dengan daun enam dan empat baffle, diperoleh Np = 5.

Daya pengaduk = Np × ρ × N3 × Da5

Daya pengaduk = 5 × 796,5323 × 0,33 × (0,5329)5 = 4,6207 hp. = 5,7758 hp (efisiensi 80%)

LC.8 Reaktor Hidrogenasi (R-101)

F

Fuunnggssii : : TTeemmppaatt tteerrjjaaddiinnyyaa rreeaakkssii hhiiddrrooggeennaassii ssttiirreennaa ookkssiiddaa m

meemmbbeennttuukk 22--ffeenniilleettaannooll T

Tiippee : :SShheellll--TTuubbeeFFiixxeeddBBeeddRReeaaccttoorr

B

(51)

B

Baahhaann kkoonnssttrruukkssii : :SSttaaiinnlleessssSStteeeell SSAA--224400,, GGrraaddee 334400 Jenis Sambungan : Double welded butt joints

J

Juummllaahh : : 11 uunniitt

1

1.. VVoolluummeerreeaakkttoorr

T

Taabbeell LLCC..44 KKoommppoossiissii bbaahhaann mmaassuukk kkee RReeaakkttoorr HHiiddrrooggeennaassii ((RR--110011)) K

Koommppoonneenn LaLajjuu aalliirr ((kkgg//jjaamm)) ρ ρ ((kkgg//mm33) ) VVccaammppuurraan n ((mm33/j/jaamm)) M

Meettaannooll 2690,1986 777744,,99660011 33,,4477114400 S

Sttiirreennaa OOkkssiiddaa 141,5894 10101100,,99113355 00,,1144000066 N

Naattrriiuumm H

Hiiddrrookkssiiddaa 0,0368 21213300 00,,0000000022 H

Hiiddrroogegenn 2,8506 11,,55331122 11,,8866116677 T

Toottaall 28283344,,66775544 787855,,99885566 55,,4477331155

T

Taabbeell LLCC..55 KKoommppoossiissii bbaahhaann kkeelluuaarr kkee RReeaakkttoorr HHiiddrrooggeennaassii ((RR--110011)) K

Koommppoonneenn LaLajjuu aalliirr ((kkgg//jjaamm)) ρ ρ ((kkgg//mm33) ) VVccaammppuurraan n ((mm33/j/jaamm)) M

Meettaannooll 2690,1986 777744,,99660011 33,,4477114400 N

Naattrriiuumm H

Hiiddrrookkssiiddaa 0,0368 21213300 00,,0000000022 H

Hiiddrroogegenn 2,8506 11,,55331122 11,,8866116677 2

2--FFeenniilleettaannooll 141433,,99664499 999933,,77333333 00,,1144448877 T

Toottaall 28283344,,66775544 787855,,99559900 55,,4477779966

D

Daallaamm hhaall iinnii tteerrjjaaddii ppeerruubbaahhaann ddeennssiittaass ((ρ)ρ) kkaarreennaa ppeerruubbaahhaann jjuummllaahh mmooll sseellaammaa r

reeaakkssii,, yyaaiittuu ::

ρ

ρ CCaammppuurraannmmaassuuk k = = 778855,,99885566 kkgg//mm33

ρ

ρ CCaammppuurraannkkeelluuaarr = = 778855,,99559900 kkgg//mm33 p

peerruubbaahhaann ddeennssiittaass ssaannggaatt kkeecciill sseehhiinnggggaa ddaappaatt ddiiaabbaaiikkaann.. M

Maakkaa hhaarrggaa ffaakkttoorr vvoolluummee ((ε)ε) yyaaiittuu ε ε == 00.. R

Reeaakkttoorr ddiirraannccaanngg ddeennggaann ssppaaccee ttiimmee ((τ)τ),, τ τ == 22 jjaamm

o o

V V

  

(52)

V = 2 jam × 5,47315 m3/jam = 10,9463 m3 Faktor kelonggaran = 20%

Volume tangki, Vt = 1,2 × 10,9463 = 13,1356 m3

2

2.. DDiiaammeetteerrddaannttiinnggggii sshheellll Volume reaktor, Vt = 10,0871 m3 D

Diirreennccaannaakkaann,, HHs s :: DDii == 33 :: 11 H

Hhh :: DDi i == 11 :: 44 A

Addaa 77 ttuubbee V

Vs s == ¼¼ ππ DD2 2

HHss V

Vss == ¾¾ π π DD33 V

Vh h == 3 24D



V

V == VVs s ++ 22VVh h 1

133,,11335566 == 3 12 10

D



M

Maakkaa ddiiaammeetteerr ttaannggkkii,, D D == 11,,77111177 mm t

tiinnggggii sshheellll ttaannggkkii,, HHss ==  

    

D D H_s

5

5,,11335522 mm

t

tiinnggggii ttuuttuupp ttaannggkkii,, HHhh ==  

    

D D Hh

0

0,,4422779933 mm t

tiinnggggii ttaannggkkii HHtt == HHss ++ 22 HHhh == 55,,99991111 mm

V

Vss == ¾¾ π π DD33 == 1111,,88222200 mm33 == 1111,,88222200 mm33/7/7 == 11,,66888899 mm33/ /ttuubbee D

Diiaammeetteerr ttuubbee == VV ttuubbee xx 44 // π π .. HHss == 00,,66447700 mm == 2255,,44771133 iinn

3

U