LAMPIRAN A PERHITUNGAN NERACA MASSA

  

LAMPIRAN A

PERHITUNGAN NERACA MASSA

  Waktu operasi : 330 hari / tahun ; 24 jam / hari Basis perhitungan : 1 jam operasi Satuan operasi : kilogram (kg) Bahan baku dan berat molekul yang digunakan (Wikipedia, 2011; Perry, 1999)

• 1,3 Butadiene (C H ) = 54,09 kg/kmol

  4

  6

• Asam Asetat (CH COOH) = 60,05 kg/kmol

  3

• Katalis Amberlyst 15 (C H ) .(C H O S) = 314,403 kg/kmol

  10 10 n

  8

  8 3 m

• Katalis Nikel = 28 kg/kmol
• Hidrogen (H ) = 2,02 kg/kmol

2 Produk akhir : n-butyl asetat (C H O )

  6

  12

  2 Kapasitas Produksi : 883,8384 kg/jam

  Perbandingan Reaktan masuk ke reaktor adalah Butadiene : asam asetat : katalis = 700 : 3600 : 85 (Gracey, BP dan Norbat, WJK, 2002) Basis perhitungan = 1121,8237 kg/jam Butadiene di Reaktor Neraca Massa untuk alat- alat sebelum reaktor :

  LA.1 Mixer (M-101)

  Fungsi : Untuk membuat larutan HCl 1 N HCl 38%

  7

  9

8 HCl 1N

  H O

  2

  7

  8

9 Neraca Massa Total : F + F = F

  Aktivasi Katalis dengan rasio 100 gram katalis amberlyst : 500 ml HCl 1N (Shakoor, Zaidoon M, et all). Katalis yang digunakan adalah (85/700) x 1121,8237 kg/jam = 136,2215 kg/jam.

  3

  3 Diketahui ρ HCl 38% : 1,18 gr/cm ,ρ air : 1 gr/cm dan ρ HCl 1 N = 1,0134 total

  F = 136,2215 kg/jam x (500ml/100ml) x 1,0134

  HCl 1N

  = 690,2343 kg/jam atau Volume HCl 1 N yang digunakan adalah 681,1075 L

  9 total

  16 F = F HCl 1N HCl 1N – F HCl 1N

  = 690,2343 kg/jam

• – 686,7831 kg/jam = 3,4512 kg/jam

  Didalam Mixer ini, akan dibuat HCl 1N. HCl yang tersedia adalah HCl 38% (12,67 N), dari perhitungan maka

  V N =

  V N

  1

  1

  2

  2 V . 12,67 = 3,4512 . 1

1 V = 0,2724 L HCl

  1

  7 F = 0,2724 L x 1,0134 kg/L = 0,2761 kg / jam HCl 38%

  Untuk membuat larutan HCl 1 N maka ke dalam Mixer-101 dimasukkan HCl 38 % 0,2761 kg kemudian diencerkan dengan air hingga 3,4512 kg. Berarti air yang ditambahkan adalah sebanyak = 3,4512 kg/jam

• – 0,2761 kg /jam = 3,1751 kg/jam

  8 F = 635,7564 kg/jam H2O

  Tabel LA.1 Neraca massa di Mixer 101

  Alur Masuk Alur Keluar Alur 7 Alur 8 Alur 9 Komponen N F N F N F (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam)

  HCl 38% 0,00756

• 0,2761
• 0,1764 3,1751 - - H O

  2

  0,0945 - - - HCl 1N 3,4512 -

  Total 3,4512 3,4512

  LA.2 Mixer (M-102)

  Fungsi : Sebagai pengaktivasi dan juga regenerisasi katalis amberlyst 15

  (C H ) .(C H O S) _{10 10 n 8 8 3 m} (C H ) .(C H O S) _{10 10 n 8 8 3 m}

  5

  10 HCl 1 N HCl 1N

  9

  

22

(C H ) .(C H O _{10 10 n 8 8 3} S)m

  5

  9

  16

  22

  10 Neraca Massa Total = F + F + F + F = F

  22 F = 136,2215 kg/jam (C10H10)n.(C8H8O3S)m 5 total

  22 F F (C10H10)n.(C8H8O3S)m = (C10H10)n.(C8H8O3S)m – F (C10H10)n.(C8H8O3S)m

  = (136,2215 -136,2215) kg/jam = 0 kg/jam

  9 F HCl 1N = 0,5% x 690,2343 kg/jam = 3,4512 kg/jam

  16 F HCl 1N = 99,5% x 690,2343kg/jam = 686,7831 kg/jam

  10 F = (136,2215 +3,4512+690,2343) kg/jam total

  = 826,4558 kg/jam Tabel LA.2 Neraca Massa di Mixer 102

  Alur Masuk Alur Keluar Alur 5 Alur 9 Alur 16 Alur 22 Alur 10 Komponen _{(C H ) .(C H O S) 10 10 n 8 8 3 m kmol/jam kg/jam (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) kmol/jam (kg/jam) kmol/jam (kg/jam)} N F N F N F N F N F

• 0,4333 136,2215 0,4333 136,2215
• HC
• 0,0945 3,4512 18,816 686,7831 18,9105 690,2343 - - total 826,4558 826,4558

  LA.3 Filtrasi (P-101)

  Fungsi = Memisahkan katalis (C H ) .(C H O S) dari larutan HCl dan tempat

  10 10 n

  8

  8 3 m

  pencucian katalis dengan air Efisiensi = 99,5 % Massa air pencuci = 3x Massa larutan HCl

  11 H

  2 O

  10 (C H ) .(C H O S) _{10 10 n 8 8 3 m} H

  2 O

  12 HCl 1N (C H ) .(C H O S) _{10 10 n 8 8 3 m}

  HCl 1N

16 H O

  17

  2 HCl 1 N (C H ) .(C H O S)

  10

  11

  16

  

12

  17 ^{10 10 n 8 8 3 m}

  Neraca Massa Total = F + F + F = F + F

  10 F = 136,2215 kg/jam (C10H10)n.(C8H8O3S)m

  10 F = 690,2343 kg/jam HCl 1N

  11 F = 2070,7029 kg/jam H2O

  Alur 12

  12 F = 136,2215 (C10H10)n.(C8H8O3S)m

  12 F = 0.005 x 2070,7029 kg/jam = 10,3535 kg/jam H2O

  Alur 16

  16

  10 F = 0,995 x F HCl 1N

  = 0,995 x 690,2343 kg/jam = 686,7831 kg/jam Alur 17

  17

  11

  12 F = F H2O H2O – F H2O

  = (2070,7029

• – 10,3535) kg/jam = 2060,3494 kg/jam

  17

  10 F = F = 3,4512 kg/jam HCl 1N HCl 1N Tabel LA.3 Neraca massa di Filtrasi

  Alur masuk Alur Keluar Alur 10 Alur 11 Alur 12 Alur 16 Alur 17 Komponen N F N F N F N F N F

(kmol/ (kmol/ (kmol/ (kmol/ (kmol/

(kg/jam) (kg/jam) (kg/jam) (kg/jam) (kg/jam) _{(C H ) .(C H O 10 10 n 8 8 3}

jam) jam) jam) jam) jam)

  0,4333 136,2215 0,4333 136,2215 - - _m

• )

• - - - HCl 1N 18,9105 690,2343 2,1844 686,7831 -

  0,0946 3,
• 115,0391 2070,702 - 0,5752 10,3535 103,5351 2060,3494

  H O

  2 total 2897,1587 2897,1587

  LA.4 Rotary Dryer (D-101)

  13

  14 (C H ) .(C H O S) _{10 10 n 8 8 3 m} (C H ) .(C H O S) _{10 10 n 8 8 3 m} H

  2 O

  2 O H

  18 H

  2 O H

  2 O

  13

  14

  18 Neraca Massa Total = F = F + F

  Alur Masuk 13

  13 F = 136,2215 kg/jam (C10H10)n.(C8H8O3S)m

  13 F = 10,3535 kg/jam H2O

  Alur Keluar Efisiensi di disc centrifuge adalah 99,5 %

  14

  13 F = F (C10H10)n.(C8H8O3S)m (C10H10)n.(C8H8O3S)m

  = 136,2215 kg/jam

  18 F = 0,995 x 10,3535 H2O

  = 10,3017

  18

  12

  14 F = F - F H2O H2O H2O

  = (10,3535 - 10,3017) kg/jam = 0,0518 kg/jam

  2

  Komponen Alur Masuk Alur Keluar Alur 13 Alur 18 Alur 14 N F N F N F (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (C ₁₀ H ₁₀ ) _n. (C ₈ H ₈ O ₃ S) _m 0,4333 136,2215 0,4333 136,2215 H ₂ O 0,5752 10,3535 0,5723 10,3017 0,00288 0,0518 Total 146,575 146,575

  LA.5 Mix Point 1 (MP-101)

4 F

  Neraca Massa total = F

  2

  20

  6

  4 H

  6 C

  4 H

  6 C

  4 H

  20 C

  4

• F

  = (1121,8237

  Komponen Alur Masuk Alur Keluar Alur 2 Alur 20 Alur 4 N F N F N F (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) C ₄ H ₆ 15,3271 829,0406 5,4129 292,7831 20,7339 1121,8237 total 1121,8237 1121,8237

  20

  = F

  2

  = 292,7831 kg/jam F

  20

  = 1121,8237 kg/jam F

  4

  = F

  4

• – F
• – 292,7831) = 829,0406kg/jam Tabel LA.5 Neraca Massa di Mix Point 1

  LA.6 Mix Point 1I

  

32 C H O

  

2

  4

  2

  1 C H O

  2

  4

  2

  3 C

2 H

  

4 O

  2

  1

  32

  3 Neraca Massa total = F + F = F

  3 F = 1121,8237 x (3600/700) kg/jam (Gracey, BP dan Norbat, WJK, 2002)

  = 5769,3792 kg/jam

  32 F = 4841,3912 kg/jam

  1

  3

  32 F = F

• – F = (5769,3792
• – 4841,3912) kg/jam = 927,9879 kg/jam

  Tabel LA.6 Neraca Massa di unit Mix Point 2

  Alur Masuk Alur Keluar Alur 1 Alur 32 Alur 3 Komponen N F N F N F (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) C H O 15,4536 927,9879 80,6227 4841,3912 96,0763 5769,3792 _{2 4 2} total 5769,3792 5769,3792

  LA.7 Reaktor (R-101)

  Fungsi : Mereaksikan antara asam asetat (CH COOH) dengan butadiene (C H )

  3

  4

  6

  untuk menghasilkan n-butenyl asetat (C H O ) dan sec-butenyl asetat (C H O )

  6

  

10

  2

  6

  10

  2 (C

  10 H 10 ) n .(C

  8 H

  8 O

  3 S)

  15 CH

  3 COOH

  3 P = 1 P = 1 atm atm T = 60 T = 60 CH

  3 COOH o o C

  4 H

  4 H

  6

  4 n-C

  6 C

  6 H

  10 O

  2

  19 sec-C H O

  6

  10

  2 (C H ) .(C H O S)

  10 10 n

  8

  8 3 m

  3

  4

  15

  19 Neraca Massa total = F + F + F = F

  Perbandingan reaktan masuk (Gracey, Benjamin Patrick dan Norbat, WJK, 2002)

  4 F = 1121,8237 kg/jam C4H6

  3 F = 5769,3792 kg/jam CH3COOH

15 F = 136,2215 kg/jam

  _{(C10H10)n.(C8H8O3S)m} Konversi reaksi adalah : 0,7377 (Gracey, Benjamin Patrick dan Norbat, WJK, 2002) Reaksi:

  2C H + 2C H O n-C H O + sec- C H O

  4

  6

  2

  4

  2

  6

  10

  2

  6

  10

  2

• Mula : 20,7399 96,0763 Reaksi: 15,2999 15,2999 7,6499 7,6499 Sisa : 5,4401 80,7764 7,6499 7,6499 Konversi= 0,7377

  3 Maka N yang bereaksi = 0,7377 x 20,7399 kmol/jam C4H6

  = 15,2999 kmol/jam

  3 N yang bereaksi = 15,2999 kmol/jam C2H4O2

  Alur 19

  19

  4 N = N C4H6 C4H6 – N C4H6 bereaksi

  = (20,7399

• – 15,2999) kmol/jam = 5,4401 kmol/jam

  19

  3 N = N C2H4O2 C2H4O2 – N C2H4O2 bereaksi

  = (96,0763

• – 15,2999) kmol/jam = 80,7764 kmol/jam

  19 N = (1/2) x 15,2999 kmol/jam n-C6H10O2

  = 7,6499 kmol/jam

  19 N = (1/2) x 15,2999 kmol/jam sec-C6H10O2

  = 7,6499 kmol/jam Tabel LA.7 Neraca Massa di unit Reaktor

  Alur masuk Alur Keluar Alur 4 Alur 3 Alur 15 Alur 19 Komponen N F N F N F N F (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) C H 20,7399 1121,8237 _{4 6}

  5,4401 294,2544

C H O 96,0763 5769,3792 80,7764 4850,6227

_{2 4 2} n-C H O _{6 10 2} 7,6499 873,1629 Sec-C H O _{6 10 2}

  7,6499 873,1629 (C ^{10 H 10 ) n .(C 8 H 8 O 3 S) m}

  0,4333 136,2215 0,4333 136,2215 total 7027,4244

  7027,4244 LA.8 Knock Out drum (FG-201)

  Fungsi : Untuk memisahkan gas butadiene dari keluaran reaktor untuk direcycle kembali.

  C

  4 H

  6

  20 CH

  3 COOH

  16

  19 C H

  4

  6 CH

  6

  10

  3 COOH n-C H O

  2 C

  4 H

  6 H

  10 O

  6 sec-C

  2 n-C

  6 H

  10 O

  2 (C10H10)n.(C8H8O3S)m sec-C

  6 H

  10 O

  2 (C10H10)n.(C8H8O3S)m

  21

  19

  20

21 Neraca massa total = F = F + F

  Efisiensi Knock Out drum adalah 99,5 % (Walas, 1988)

  20 Alur 20 = 0,995 x F = 0,995 x 294,2544 kg/jam = 292,7831kg/jam C4H6

  Alur 21

  21

  19 F = F = 4850,6277 kg/jam CH3COOH CH3COOH

  21

  19

  20 F = F C4H6 C4H6 – F C4H6

  = (294,2544 = 1,4713 kg/jam

• – 292,7831) kg/jam

  21

  19 F = F = 873,1629kg/jam n-C6H10O2 n-C6H10O2

  21

  19 F = F = 873,1629 kg/jam sec-C6H10O2 sec-C6H10O2

  21

  19 F (C10H10)n.(C8H8O3S)m = F = 136,2215 kg/jam (C10H10)n.(C8H8O3S)m Tabel LA.8 Neraca massa di Knock Out Drum

   Alur Masuk Alur Keluar Alur 19 Alur 20 Alur 21 Komponen N F N F N F (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) C H _{4 6} 5,4401 294,2544 5,4129 292,7831 0,0272 1,4713

  C H _{2 4 2}

• 80,7764 4.850,6277 80,7764 4.850,6277
• n-C H O 7,6499 873,1629 7,6499 873,1629
_{6 10} 2<
• O
• sec-C H O 7,6499 873,1629 7,6499 873,1629
_{6 10} 2<
• (C H )n.(C H O S)m 0,4333 136,2215 0,4333 136,2215
_{10 10 8 8 3} Total 7.027,4244 7.027,4244 LA.9 Disc Centrifuge (FF-201)

  Fungsi : Untuk memisahkan (C10H10)n.(C8H8O3S)m (Katalis Amberlyst 15) dari larutan, sehingga (C10H10)n.(C8H8O3S)m dapat digunakan kembali.

  CH

  3 COOH C

  4 H

  6

  21 n-C H O

  6

  10

  2 sec-C H O

  6

  10

  2 (C10H10)n.(C8H8O3S)m CH COOH ₃ n-C

  6 H

  10 O

  2

  23

  24 sec-C

  6 H

  10 O

  2 C

4 H

  6 (C10H10)n.(C8H8O3S)m

  

22

(C10H10)n.(C8H8O3S)m

  21

  22

  23

  24 Neraca Massa Total = F = F + F + F

  Katalis Amberlyst yang bisa direcovery pada alur 22 adalah sebesar 99%, sedangkan sisanya yang berupa larutan blood C H dan katalis amberlyst

  4

  6

  (C10H10)n.(C8H8O3S)m akan dialirkan ke unit utilitas pengolahan limbah. (Moore, W.P , 1964). Alur 21 dapat dilihat pada tabel LA.6 Alur 22

  25

  21 F = 0,99 x F (C10H10)n.(C8H8O3S)m (C10H10)n.(C8H8O3S)m

  = 0,99 x 136,2215 kg/jam = 134,8592 kg/jam

• – F
• – 134,8592) kg/jam = 1,3622 kg/jam F
• F
• F
• – 1,3622 – 134,8592) kg/jam = 0 kg/jam F
• – F
• – 1,4713) kg/jam = 0 kg/jam F

  4 H

  21 n-C6H10O2

  = 873,1629 kg/jam F

  24 sec-C6H10O2

  = F

  21 sec-C6H10O2

  = 873,1629 kg/jam Tabel LA.9 Neraca Massa Disc centrifuge

  Komponen

Alur Masuk Alur Keluar

Alur 21 Alur 23 Alur 22

   Alur 24 N F N F N F N F (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam)

  C

  6 0,0272 1,4713 0.0272 1.4713 - - - -

  24 n-C6H10O2

  C

  2 H

  4 O

  2 80,7764 4850,6277 - - - - 80,7764 4850,6277

  n-C

  6 H

  10 O

  2 7,6499 873,1629 - - - - 7,6499 873,1629

  sec-C

  6 H

  = F

  = 4850,6277 kg/jam F

  Alur 23 F

  =F

  26 (C10H10)n.(C8H8O3S)m

  = F

  24 (C10H10)n.(C8H8O3S)m

  25 (C10H10)n.(C8H8O3S)m

  = (136,2215

  26 C4H6

  = F

  24 C4H6

  = 1,4713 kg/jam Alur 24 F

  24 (C10H10)n.(C8H8O3S)m

  21 (C10H10)n.(C8H8O3S)m

  21 CH3COOH

  23 (C10H10)n.(C8H8O3S)m

  24 (C10H10)n.(C8H8O3S)m

  = (136,2215

  24 C4H6

  = F

  24 C4H6

  26 C4H6

  = (1,4713

  24 CH3COOH

  = F

  10 O 2 7,6499 873,1629 - - - - 7,6499 873,1629

(C10H10)n.(C8H8O3S)m 0,4333 136,2215 0.0043 1.3622 0.4289 134,8592 - -

Total 6.734,6413 6.734,6413

  LA.10 Kolom Destilasi (D-201)

  Fungsi : Untuk memisahkan C H O (asam asetat) dari n-C H O (n-butenyl asetat)

  2

  4

  2

  6

  10

  2 dan sec- C H O (sec-butenyl asetat) berdasarkan perbedaan titik didih.

  6

  10

  2

26 CH COOH

  _{3 E-303} CH COOH ₃ n-C

  6 H

  10 O

  2 n-C

  6 H

  10 O

  2 sec-C H O sec-C H O

  6

  10

  2

  6

  10

  2

  28

  31

  25 _D-301

  30 CH COOH ₃ n-C

  6 H

  10 O

  2 _E-302 sec-C

  6 H

  10 O

  2

  29

  33 Laju Alir Mol Masuk

  25 N = 80,7764 kmol/jam C2H4O2

  25 N = 7,6499 kmol/jam n-C6H10O2

  25 N = 7,6499 kmol/jam sec-C6H10O2

  Fraksi mol produk bawah(Bottom) yang diinginkan:

  33 N = 0,01 C2H4O2

  33 N = 0,495 n-C6H10O2

  33 N = 0,495 sec-C6H10O2

  Fraksi mol produk atas(Destilat) yang diinginkan:

  31 N = 0,999 C2H4O2

  31 N = 0,0005 n-C6H10O2

  31 N = 0,0005 sec-C6H10O2

  Dari persamaan F = D + W dan X .F = Y .D + X .W diperoleh:

  

if id ib

W = 15,3729 kmol/jam dan D = 80,7034 kmol/jam.

  Maka:

  33

  31

  31 N =. N .N C2H4O2 C2H4O2

  = 0,01 x 15,3729 kmol/jam = 0,1537 kmol/jam

  33

  33

  33 N = N . N n-C6H10O2 n-C6H10O2

  = 0,495 x 15,3729 kmol/jam = 7,6096 kmol/jam

  33

  33

  33 N = N . N sec-C6H10O2 sec-C6H10O2

  = 0,495 x 15,3729 kmol/jam = 7,6096 kmol/jam

31 Laju Alir Mol Produk Atas (N )

  31

  25

  33 N = N C2H4O2 C2H4O2 – N C2H4O2

  = 80,7764 kmol/jam

• – 0,1537 kmol/jam = 80,6227 kmol/jam

  31

  25

  33 N = N - N n-C6H10O2 n-C6H10O2 n-C6H10O2

  = 7,6499 kmol/jam

• – 7,6096 kmol/jam = 0,0404 kmol/jam

  31

  25

  33 N = N - N sec-C6H10O2 sec-C6H10O2 sec-C6H10O2

  = 7,6499 kmol/jam

• – 7,6096 kmol/jam = 0,0404 kmol/jam

  LA.10 Neraca Massa Kolom Destilasi

   Alur Masuk Alur Keluar Alur 25 Alur 33 Alur 31 Komponen N F N F N F (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam)

C H O 80,7764 4.850,6227 0,1537 9,2314 80,6227 4.841,3912

_{2 4 2}

n-C H O 7,6499 873,1629 7,6096 868,5572 0,0404 4,6057

_{6 10 2}

sec-C H O 7,6499 873,1629 7,6096 868,5572 0,0404 4,6057

_{6 10 2} Total 6.596,9485 6.596,9485

  LA.10.1 Kondensor (E-203) CH COOH ₃

  26 n-C H O

  6

  10

  2 E-303 sec-C

  6 H

  10 O

  2 CH COOH ₃ n-C

  6 H

  10 O

  2

  28

  31 sec-C

  6 H

  10 O

  2 Tekanan uap komponen, dapat dihitung berdasarkan persamaan Antoine: .......................... (Yaws, 2007)

  Keterangan: P = Tekanan (mmHg)

  A, B, C = Konstanta Antoine

  o

  T = Temperatur (

  C)

  LA.11 Tabel konstanta Antoine komponen A B C

  C H O 7.8152 1800.0300 246.8940

  2

  4

  2

  n-C H O 8.0620 2156.3500 272.8900

  6

  10

  2

  sec-C H O 8.1003 2188.3200 295.6800

  6

  10

  2

  (Sumber: Yaws, 2007) Suhu Umpan Masuk Kolom Destilasi I

  o

  T = 119,732 C

  Trial

  P = 760 mmHg Tabel LA.12 Suhu Umpan Masuk Kolom Destilasi I

  Ki.Xif

  Komponen Xif Pa Ki(Pa/P) αiF

  C H O 0.8408 804.4198 1.0584 0.8899 2.1660

  2

  4

  2

  n-C H O 0.0796 371.3828 0.4887 0.0389 1.0000

  6

  10

2 Sec-C H O 0.0796 679.9383 0.8947 0.0712 1.8308

  6

  10

2 Total 1.0000 1.0000

  Dari hasil perhitungan diperoleh harga = 1, maka trial T dapat diterima. Penentuan Titik Embun Destilat

  o

  T = 117,921 C

  Trial

  Tabel LA.13 Titik Embun Kolom Destilasi I Yid/Ki

  Komponen Yid Pa Ki(Pa/P) αid(Ki/Kj)

  C H O 0,9990 760,5191 1,0007 0,9983 2,1714

  2

  4

  2

  n-C H O 0,0005 350,2445 0,4608 0,0011 1,0000

  6

  10

  2

  sec-C H O 0,0005 644,7683 0,8484 0,0006 1,8409

  6

  10

  2 Total 1,0000 1,0000 Dari hasil perhitungan diperoleh harga = 1, maka trial T dapat diterima.

  Penentuan Titik Gelembung Destilat

  o Trial T = 132,097 C

  Tabel LA.14 Titik Gelembung Kolom Destilasi I Xib.Ki

  Komponen Xib Pa Ki(Pa/P) αib(Ki/Kj)

  C H O 0,0100 1163,2356 1,5306 0,0153 2,1289

  2

  4

  2

  n-C H O 0,4950 546,3967 0,7189 0,3559 1,0000

  6

  10

  2

  sec-C H O 0,4950 965,4670 1,2704 0,6288 1,7670

  6

  10

  2 Total 1,0000 1,0000 Dari hasil perhitungan diperoleh harga = 1, maka trial T dapat diterima.

  Refluks Minimum Destilat Umpan masuk berupa cairan yang berada pada titik didihnya, maka q = 1 Sehingga

  Trial

  θ = 1,04403 Tabel LA.15 Omega Point Kolom Destilasi I

  (αiF.Xif)/(αiF- Komponen Xif αiF αiF.Xif αiF-θ θ)

  C H O 0.8408 2.1660 1.8211 1.1220 1.6231

  2

  4

  2

  n-C H O 0.0796 1.0000 0.0796 -0.0440 -1.8084

  6

  10

  2

  sec-C H O 0.0796 1.8308 0.1458 0.7868 0.1853

  6

  10

  2 Total 1.0000

  0.0000 Oleh Karena ; sehingga trial θ = 1,04403 dapat diterima.

  Tabel LA.16 Perhitungan R

  Dm Komponen Yid αid αid.Yid αid-θ (αid.Yid)/(αid-θ)

  C H O 0,9990 2,1714 2,1692 1,1274 1,9242

  2

  4

  2

  n-C H O 0,0005 1,0000 0,0005 -0,0440 -0,0114

  6

  10

  2

  sec-C H O 0,0005 1,8409 0,0009 0,7969 0,0012

  6

  10

  2 Total 1,0000

  1,9128 = 1,9128

  R = 0,9128

  Dm

  R = L /D

  D D

  L = R x D

  D D

  = 0,9128 x 80,7034 kmol/jam = 74,0948 kmol/jam

  31 Laju Alir Mol Destilat (N )

  31 N = 80,6227 kmol/jam C2H4O2

  31 N = 0,0404 kmol/jam n-C6H10O2

  31 N = 0,0404 kmol/jam sec-C6H10O2

  28 Laju Alir Mol Refluks (N )

  28 N =. Yid x Ld C2H4O2

  = 0,999 x 74,0948 kmol/jam

  = 73,5922 kmol/jam

  28 N = Yid x Ld n-C6H10O2

  = 0,0005 x 74,0948 kmol/jam = 0,0368 kmol/jam

  28 N = Yid x Ld sec-C6H10O2

  = 0,0005 x 74,0948 kmol/jam = 0,0368 kmol/jam

26 Laju Alir Mol Masuk Kondensor (N )

  26

  28

  31 N = N + N C2H4O2 C2H4O2 C2H4O2

  = 73,5922 kmol/jam + 80,6227 kmol/jam = 154,2149 kmol/jam

  26

  28

  31 N = N + N n-C6H10O2 n-C6H10O2 n-C6H10O2

  = 0,0368 kmol/jam + 0,0404 kmol/jam = 0,0772 kmol/jam

  26

  28

  31 N = N + N sec-C6H10O2 sec-C6H10O2 sec-C6H10O2

  = 0,0368 kmol/jam + 0,0404 kmol/jam =0,0772 kmol/jam

  Tabel LA. 16 Neraca Massa Kondensor 1

  

Alur Masuk Alur Keluar

Alur 26 Alur 28 Alur 31 Komponen N F N F N F (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam)

  C H O 154,2149 9260,6026 73,5922 4419,2114 80,6227 4841,3912

  2

  4

  2

  n-C H O 0,0772 8,8099 0,0368 4,2041 0,0404 4,6057

  6

  10

  2

  sec-C H O 0,0772 8,8099 0,0368 4,2041 0,0404 4,6057

  6

  10

2 Total 9278,2223 9278,2223

  N

  CH ₃ COOH n-C

  30 sec-C6H10O2

  = Xib x Vb = 0,4950 x 154,7982 kmol/jam = 76,6251 kmol

  Laju Alir Mol Masuk Kondensor (N

  29

  ) N

  29 C2H4O2

  = N

  30 C2H4O2

  31 C2H4O2

  = 1,5480 kmol/jam + 0,1537 kmol/jam

  6 H

  = Xib x Vb = 0,4950 x 154,7982 kmol/jam = 76,6251 kmol

  10 O

  2 sec-C

  6 H

  10 O

  2 CH ₃ COOH n-C

  6 H

  10 O

  2 sec-C

  6 H

  10 O

• (q x f) ..................................................(Geankoplis,1997) q = 1 L
• – 15,3729 kmol/jam = 154,7982 kmol/jam

  30 n-C6H10O2

  2

  33 sec-C6H10O2

  LA.10.2 Reboiler (E-204) E-302

  30

  29

  33 Laju Alir Mol Keluar sebagai bottom (N

  33

  ) N

  33 C2H4O2

  = 0,1537 kmol/jam N

  33 n-C6H10O2

  = 7,6096 kmol/jam N

  = 7,6096 kmol/jam Laju alir massa yang dikembalikan ke kolom destilasi L

  N

  B

  = L

  D

  B

  = 74,0948 kmol/jam + (1 x 96,0763 kmol/jam) = 170,1711kmol/jam

  V B = L

  B – W

  = 170,1711 kmol/jam

  Laju alir mol keluar Reboiler N

  30 C2H4O2

  =. Xib x Vb = 0,01 x 154,7982 kmol/jam = 1,548 kmol

• N

  = 1,7017 kmol/jam

  29

  30

  31 N = N + N n-C6H10O2 n-C6H10O2 n-C6H10O2

  = 76,6251 kmol/jam + 7,6096 kmol/jam = 84,2347 kmol/jam

  29

  30

  31 N = N + N sec-C6H10O2 sec-C6H10O2 sec-C6H10O2

  = 76,6251 kmol/jam + 7,6096 kmol/jam = 84,2347 kmol/jam

  Tabel LA.17 Neraca Massa Reboiler 1

  

Alur Masuk Alur Keluar

Alur 29 Alur 30 Alur 33 Komponen N F N F N F (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam)

  C H O 1,7017 102,1877 1,5480 92,9563 0,1537 9,2314

  2

  4

  2

  n-C H O 84,2347 9614,5456 76,6251 8745,9884 7,6096 868,5572

  6

  10

  2 sec-C H O _{6 10 2 84,2347 9614,5456 76,6251 8745,9884 7,6096 868,5572}

  Total 19331,2788 19331,2788

  LA.11 Reaktor Hidrogenasi (R-201)

  Fungsi : Mereaksikan antara n-C H O dan sec-C H O dengan H (hidrogen)

  6

  10

  2

  6

  10

  2

  2 untuk menghasilkan n-C H O (n-butyl asetat) dan sec-C H O (sec-butyl asetat).

  6

  12

  2

  6

  12

  2 Ni

  33 CH COOH ₃ n-C

  6 H

  10 O

  2 sec-C

  6 H

  10 O

  2

  

35

CH COOH ₃ n-C

  6 H

  12 O

  2 sec-C

  6 H

  12 O

  2 H ₂

6 Reaktor Hidrogenasi dilengkapi katalis Nikel dengan berat 0,05% umpan masuk(Perry, 1999).

  = 0,05% x 1746,3458 kg/jam = 0,8732 kg/jam

  5

  36

  37 Neraca Massa Total = F + F = F

  33 Laju alir pada F dapat dilihat pada tabel LA.16 Konversi = 100% Berdasarkan reaksi

  (i) n-C + H O H n-C H O

  6

  10

  2

  2

  6

  12

  2

  mula 7,6096 7,6096 - reaski 7,6096 7,6096 7,6096 sisa

  7,6096

• (ii) sec-C H O H sec-C H O

  6

  10

  2

  2

  6

  12

  2

  mula 7,6096 - 7,6096 reaski 7,6096 7,6096 7,6096 sisa

  7,6096 Berdasarkan reaksi, maka jumlah H yang dibutuhkan adalah 2 x 7,6096 kmol/jam

2 Atau setara dengan 15,2192 kmol/jam, sehingga

  6 F = 30,7427 kg/jam H2

  33 F = 7,6096 kmol/jam x 116.16 kg/kmol n-C6H12O2

  = 883,9286 kg/jam

  35 F = 7,6096 kmol/jam x 116.16 kg/kmol sec-C6H12O2

  = 883,9286 kg/jam LA.18 Neraca Massa Hidrogenasi

   Alur masuk Alur keluar Alur masuk 33 Alur masuk 6 Alur keluar 35 Komponen N F N F N F (Kmol/jam) (kg/jam) (Kmol/jam) (kg/jam) (Kmol/jam) (kg/jam) C H _{2 4 6}

• O 0,1537 9,2314 0,1537 9,2314
_{6 10} 2<
• n-C H O 7,6096 868,5572 sec-C H - - O
_{6 10} 2<
• 7,6096 - 868,5572
₂ - - H 15,2192 30,7427 - - _{6 12} 2<
• n-C H O

  7,6096 883,9286 _{6 12 2}

• sec-C - H O

  7,6096 883,9286 Total 1.777,0885 1.777,0885

  LA.12 Kolom Destilasi (D-202)

  Fungsi : Untuk memisahkan n-C H O (n-butyl asetat) dan sec- C H O (sec-butyl

  6

  12

  2

  6

  12

  2 asetat) berdasarkan perbedaan titik didih.

  36 CH COOH _{3 E-303} n-C

  6 H

  12 O

  2 CH COOH ₃ sec-C

  6 H

  12 O

  2 n-C

  6 H

  12 O

  2

  38

  41 sec-C

  6 H

  12 O

  2

  35 _D-301

  40 CH COOH _{3 E-302} n-C

  6 H

  12 O

  2 sec-C H O

  6

  12

  2

  39

  42

  35

  41

42 Neraca Massa Total = F = F + F

  Laju Alir Mol Masuk

  35 N = 0,1537 kmol/jam C2H4O2

  35 N = 7,6096 kmol/jam n-C6H12O2

  35 N = 7,6096 kmol/jam sec-C6H12O2

  Fraksi mol produk bawah(Bottom) yang diinginkan:

  42 N = 0 C2H4O2

  42 N = 0,995 n-C6H12O2

  42 N = 0,005 sec-C6H12O2

  Fraksi mol produk atas (Destilat) yang diinginkan:

  41 N = 0,0198 C2H4O2

  41 N = 0,005 n-C6H12O2

  41 N = 0,9752 sec-C6H12O2

  Dari persamaan F = D + W dan X .F = Y .D + X .W diperoleh:

  if id ib

  W = 7,6088 kmol/jam dan D = 7,7641 kmol/jam

  Maka:

  42

  42

  35 N =. N .N C2H4O2 C2H4O2

  = 0 x 7,6088 kmol/jam = 0 kmol/jam

  42

  42

  35 N = N . N n-C6H12O2 n-C6H12O2

  = 0,995 x 7,6088 kmol/jam = 7,5708 kmol/jam

  42

  42

  35 N = N . N sec-C6H12O2 sec-C6H12O2

  = 0,005 x 7,6088 kmol/jam = 0,0380 kmol/jam

  Laju Alir Mol Produk Atas

  41

  35

  42 N = N C2H4O2 C2H4O2 C2H4O2

• – N = 0,1537 kmol/jam
• – 0 kmol/jam = 0,1537 kmol/jam

  41

  35

  42 N = N - N n-C6H12O2 n-C6H12O2 n-C6H12O2

  = 7,6096 kmol/jam

• – 7,5708 kmol/jam = 0,0388 kmol/jam

  41

  35

  42 N = N - N sec-C6H12O2 sec-C6H12O2 sec-C6H12O2

  = 7,6096 kmol/jam

• – 0,0380 kmol/jam = 7,5715 kmol/jam

  Tabel LA.19 Neraca Massa destilasi 2

  Alur Masuk Alur Keluar Alur 35 Alur 41 Alur 42 Komponen N F N F N F (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) (kmol/jam) (kg/jam) C H O 0,1537 _{2 4 2}

• 9,2314 - 0,1537 9,2314 n-C H O 7,6096 883,9286 0,0388 4,5094 7,5708 879,4192
_{6 12 2}

sec-C H O 7,6096 883,9286 7,5715 879,5094 0,0380 4,4192

_{6 12 2} Total 1.777,0885 1.777,0885

  LA 12.1 Kondensor (E-207)

36 E-303

CH COOH

  ₃ CH COOH ₃ n-C

  6 H

  12 O

  2

  

38

41 n-C

  6 H

  12 O

  2 sec-C

  6 H

  12 O 2 sec-C

  6 H

  12 O

  2 Tekanan uap komponen, dapat dihitung berdasarkan persamaan Antoine: .......................... (Yaws, 2007)

  Keterangan: P = Tekanan (mmHg)

  A, B, C = Konstanta Antoine

  o

  T = Temperatur (

  C) LA.20 Tabel konstanta Antoine

  komponen A B C

  C H O 7,8152 1800,0300 246,8940

  2

  4

  2

  n-C H O 7,2350 1515,7600 222,0770

  6

  12

  2

  sec-C H O 7,3235 1539,5600 234,6040

  6

  12

  2

  (Sumber: Yaws, 2007) Suhu Umpan Masuk Kolom Destilasi II

  o Trial T = 118,315 C

  P = 760 mmHg Tabel LA.21 Suhu Umpan masuk kolom destilasi II

  Komponen Xif Pa Ki(Pa/P) Ki.Xif αiF

  C H O 0,0100 769,8980 1,0130 0,0101 1,2718

  2

  4

  2

  n-C H O 0,4950 605,3637 0,7965 0,3943 1,0000

  6

  12

  2

  sec-C H O 0,4950 914,4063 12032 0,5956 1,5105

  6

  12

2 Total 1,0000 1,0000 Dari hasil perhitungan diperoleh harga = 1, maka trial T dapat diterima.

  Penentuan Titik Embun Destilat

  2 H

  0,9752 739,4636 0,9730 0,9681 1,0000 sec-C

  6 H

  12 O

  2

  0,0050 1104,8575 1,4538 0,0073 1,4941 Total 1,000 1,0000 Dari hasil perhitungan diperoleh harga = 1, maka trial T dapat diterima.

  Refluks Minimum Destilat Umpan masuk berupa cairan yang berada pada titik didihnya, maka q = 1 Sehingga

  Trial

  θ = 1,195135 Tabel LA.24 Omega Point Kolom Destilasi II

  Komponen Xif αiF αiF.Xif αiF-θ (αiF.Xif)/(αiF-θ)

  C

  4 O

  12 O

  2

  0,0100 1,2718 0,0127 0,0767 0,1659 n-C

  6 H

  12 O

  2

  0,4950 1,0000 0,4950 -0,1951 -2,5367 sec-C

  6 H

  12 O

  2

  0,4950 1,5105 0,7477 0,3154 2,3709 Total 1,0000

  2

  6 H

  Trial T = 112,058 o

  0,0050 499,6087 0,6574 0,0076 1,0000 sec- C

  C Tabel LA.22 Titik Embun Kolom Destilasi II

  Komponen Yid Pa Ki(Pa/P) Yid/Ki αid(Ki/Kj)

  C

  2 H

  4 O

  2

  0,0050 631,7109 0,8312 0,0060 1,2644 n-C

  6 H

  12 O

  2

  6 H

  0,0198 946,6764 1,2456 0,0247 1,2802 n-C

  12 O

  2

  0,9900 762,7843 1,0037 0,9864 1,5268 Total 1,0000 1,0000 Dari hasil perhitungan diperoleh harga = 1, maka trial T dapat diterima.

  Penentuan Titik Gelembung Destilat

  Trial T = 125,09 o

  C Tabel LA.23 Titik Gelembung Destilat

  Komponen Xib Pa Ki(Pa/P) Xib.Ki αib(Ki/Kj)

  C

  2 H

  4 O

  2

  0,0000 Oleh Karena ; sehingga trial θ = 1,195135 dapat diterima. Tabel LA.25 Perhitungan R

  Dm Komponen Yid αid αid.Yid αid-θ (αid.Yid)/(αid-θ)

  C H O 0,0050 1,2644 0,0063 0,0693 0,0913

  2

  4

  2

  n-C H O 0,0050 1,0000 0,0050 -0,1951 -0,0256

  6

  12

  2

  sec-C H O 0,9900 1,5268 1,5115 0,3316 4,5578

  6

  12

2 Total 1,0000

  4,6234 = 4,6234

  R = 3,6234

  Dm

  R = L /D

  D D

  L = R x D

  D D

  = 3,6234 x 7,7641 kmol = 28,1326 kmol

  41 Laju Alir Mol Destilat (N )

  41 N = 0,1537 kmol/jam C2H4O2

  41 N = 0,0388 kmol/jam n-C6H12O2

  41 N = 7,5715 kmol/jam sec-C6H12O2

  38 Laju Alir Mol Refluks (N )

  38 N =. Yid x Ld C2H4O2

  = 0,005 x 28,1326 kmol/jam = 0,1407 kmol/jam

  38 N = Yid x Ld n-C6H12O2

  = 0,005 x 28,1326 kmol/jam = 0,1407 kmol/jam

  38 N = Yid x Ld sec-C6H12O2

  = 0,99x 28,1326 kmol/jam = 28,1045 kmol/jam

• N
• N
• N

  40

  12 O

  2

  0,1795 20,8488 0,1407 16,3394 0,0388 4,5094 sec-C

  6 H

  12 O

  2

  35,6760 4.144,1283 28,1045 3264,61

  9 7,5715

  879,509

  4 Total 4182,6554 4182,6554

  LA. 12.2 Reboiler (E-208) E-302

  39

  0,2944 17,6782 0,1407 8,4468 0,1537 9,2314 n-C