LAMPIRAN A PERHITUNGAN NERACA MASSA

  LAMPIRAN A PERHITUNGAN NERACA MASSA 1.

  Kapasitas Produksi Kapasitas produksi per tahun = 50.000 ton/tahun 1 tahun produksi = 330 hari Maka, kapasitas produksi = jam/hari24xhari/tahun330kg/ton1000xton/tahun50.000 = 6313,131 kg/jam 2.

  Spesifikasi Produk Kaprolaktam Kaprolaktam = minimal 99,5 % H

2 O = maksimal 0,3 %

  2 SO 4 = maksimal 0,2 % 3.

  2

  Reaksi samping:

  2 O (l)

  

5 COOH (l) + H

  6 H

  5 CH 3(l) + 3/2 O 2(g) C

  6 H

  C

  = 7,32 % mol (PT. Air Liquid Indonesia) 7. Reaktor – 01 (R-101) Reaksi yang terjadi :

  2

  = 90,52 % mol Ar = 2,16 % mol N

  ) (% mol) H

  Komposisi Bahan Baku Toluena Toluena = 98 % berat Benzena = 2 % berat

  2

  H

  = 21 % mol (Perry dkk., 1998) 6.

  2

  O

  2 = 79 % mol

  Komposisi Udara (% mol) N

  (www.alibaba.com) 5.

  3 = 35 % berat

  Komposisi Bahan Baku Oleum Asam sulfat = 65 % berat SO

  (PT Trans Pacific Petrochemical Indotama) 4.

  Komposisi Gas Hidrogen (H

  2 SO 4(l) + CO 2(g)

  6 H

  

11 ON.H

  6 H

  11 COOH (l) + NOHSO 4(l) C

  6 H

  C

  Reaktor R-301 Reaksi yang terjadi :

  (Sayari dkk., 2008) 12.

  (Chauvel dkk., 1989) 2. katalis masuk reaktor R-02 = 50 mg/2500 mg asam benzoat.

  Ketentuan : 1. konversi reaksi = 100 %

  11 COOH (l)

  6 H

  C

  2(g)

  5 COOH (l)

  C

  C

  2 keluar reaktor 4 % berat O 2 masuk 5.

  6 H

  5 CH 3(l) + O 2(g) C

  6 H

  5 CHO (l) + H

  2 O (l)

  Ketentuan : 1. konversi reaksi 30 % 2. selektivitas asam benzoat 90 % 3. selektivitas benzaldehida 10 % 4. gas O

  katalis Cobalt (Co) 0,22 lb/2000 lb toluena yang masuk (Chauvel dkk., 1989) 8.

  11. Reaktor R-201 Reaksi yang terjadi :

  Destilasi (T-101) Ketentuan : 99 % toluena terpisah sebagai hasil bawah.

  Semua benzena dan 99% air terpisahkan sebagai produk atas.

  9. Destilasi (T-102) Ketentuan : 99 % toluena terpisah sebagai produk atas.

  99% benzaldehida dan semua asam benzoat terpisah sebagai produk bawah.

  10. Destilasi (T-103) Ketentuan : Toluena, 99,9 % benzaldehida, dan 1 % asam benzoat terpisah sebagai produk atas.

  99,9 % asam benzoat terpisah sebagai produk bawah.

• 3 H

Ketentuan : 1. konversi reaksi 50 % 2.

  CHCA ekses 25 % (Marshall Sittiq, 1969) 3. kebutuhan oleum masuk R-03 adalah 3 mol / mol Kaprolaktam yang terbentuk.

  (Ullmann’s, 2005) 13.

  Tangki Pencampur M-301 Kelarutan kaprolaktam dalam air = 93, 5 % (Ullmann’s, 2005) 14.

  Reaktor Netralisasi R– 302 Reaksi yang terjadi :

  3

  2

  4

  4

  2

  4

  2 NH + H SO (NH ) SO Konversi 99 % 15.

  Evaporator FE– 301 Ketentuan : H 2 O dalam larutan pekat 0,1 % berat.

16. Titik Didih dan Berat Molekul Masing – Masing Zat

  Tabel A.1 Daftar Titik Didih dan Berat Molekul Masing-Masing Zat

  Nama Perusahaan Lokasi Kapasitas (ton/tahun)

  PT Indonesia Toray Tangerang, Banten 16.000 Synthetics Tangerang, Banten 28.000 PT Filamendo Sakti Purwakarta, Jabar 8.550 PT Indonesia Asahi Tangerang dan Cibitung 6.500 Chemical Ind. Rengasdengklok 7.000 PT Susila Indah Synthetic Cikupa, Banten 8.500 Ind.

  PT Sandang Utama Mulya PT Prima Rajuli Sukses

  Total kapasitas 74.750 No. Bahan Baku Harga US$

  1. Toluena 0,8 / L

  2. Asam Nitrosilsulfur 0,3/ kg

  3. Hidrogen 0,5/ kg

  4. Amonia 0,2 / L

  Tahun Jumlah ( ton ) Perkembangan (%)

• 2005 312.824,477 2006 580.086,042 >-
• 100 2008 13.191,585 100 2009 7.004,278 - 47 2010 31.756,655 353

  Rata-rata perkembangan (kenaikan) : +156 Komponen Masuk (kg/jam) Keluar (kg/jam)

  Arus 1 Arus 2 Arus 3 Toluena 5952,841 13009,479 18962,320 Benzena - 121,487 121,487

• Benzaldehida 6,621 6,621
• H2O 11,242 11,242

  Total 19101,67 19101,67

  Nama Perusahaan Lokasi Kapasitas 1.

  (ton/tahun)

  PT Indonesia Toray Tangerang, Banten 16.000 Synthetics Tangerang, Banten 28.000 PT Filamendo Sakti Purwakarta, Jabar 8.550 PT Indonesia Asahi Tangerang dan Cibitung 6.500 Chemical Ind. Rengasdengklok 7.000

  Neraca Massa di Tangki Pencampur M-101

  PT Susila Indah Synthetic Cikupa, Banten 8.500 M-101ToluenaBenzenaH2OToluenaBenzenaBenzaldehidaH2OToluenaBenzaldehida213 Gambar A.1 Tangki Pencampur (M-101) Fungsi : tempat untuk mencampur bahan baku dan bahan recycle untuk diumpankan ke dalam reaktor R-101.

  3

  1

  2 F = F + F .................................................................... (1)

  3

  1

  2 F Toluena = F Toluena + F Toluena ..................................... (2)

  F

  3 Benzena = F

  1 Benzena ............................................................ (3)

  F

  2 Benzaldehida = F

  3 Benzaldehida ..................................................... (4)

  2

  2

  3

  2 F H O = F H O ................................................................... (4a)

  2. Neraca Massa di Reaktor R-101

  7 662,057

  119,545 9837,562

  Gambar A.2 Reaktor (R-101)

  Fungsi : tempat mereaksikan toluena dan udara (oksigen) untuk menghasilkan asam benzoat. Reaksi : Konversi 30%, selektivitas asam benzoat 90%.

  1. Reaksi Utama : C

  6 H

  5 CH 3 + 3/2 O

  2 C

  6 H

  5 COOH + H

  2 O ................ r

  1

  2. Reaksi samping :

  6

  5

  3

  2

  6

  5

  

2

  2 C H CH + O C H CHO + H O .................. r

  6

  7

  3

  4

  5 F + F = F + F + F ................................................................... (5)

  6

• – F Toluena yang bereaksi................................. (6)

  3 F Toluena = F Toluena

  6 Toluena = F

  3 Toluena - toluena .r 1 .Mr toluena toluena .r

• F

  2 .Mr toluena ........ (7) 30% =

  σ σ ............. (8)

  6

  6

  5 C6H5COOH

  1 C6H5COOH

  F C H COOH = . r .Mr ................................. (9) σ

  Benzaldehida

2 enzaldehida

  F Benzaldehida reaksi 2 = . r . Mr B .............................. (10) σ

  F Benzaldehida reaksi 2 = 10 / 90 x N

  6 C

  6 H

  5 COOH x Mr benzaldehida .............. (11)

  F

  6 Benzaldehida = F

  

3 Benzaldehida + F Benzaldehida reaksi 2 ....... (12)

  7

  2

  5

  2 F O = 4% x F O ............................................................................ (13)

  7 O 2 = F

  5 O

  2 O2 .r 1 .Mr O2 O2 .r

• F

  2 .Mr O2 ................................. (14)

  σ σ F O

  2 reaksi 1 = 3 / 2 x N

  6 C

  6 H

  5 COOH x Mr O2 ............................................. (15)

  2 O2

  F O reaksi 2 = 1 / 1 x N Benzaldehida reaksi 2 x Mr .............................. (16)

  5

  2

  7

  2

  5

  2

  5

  2 N2

  F N = F N , di mana F N = 79 / 21 . N O . Mr ..................... (17) F

  3 Benzena = F

  

6 Benzena ........................................................................... (18)

  F

  4 Cobalt = 0,22 lb / 2000 lb F

  3 Toluena ................................................ (19)

  F

  4 Cobalt = F

  6 Cobalt .............................................................................. (20)

  6

  2

  6

  6

  5 H2O

  3 2 ...........

  F H O = ( N C H COOH + N Benzaldehida reaksi 2). Mr + F H O (21)

  3. Neraca Massa di Filter P -101

P-101C6H5COOHBenzaldehidaToluenaBenzenaCobaltH2OC6H5COOHBenzaldehidaToluenaBenzenaH2

OCobalt689

  Gambar A.3 Filter (P-101) Fungsi : untuk memisahkan katalis dari campuran produk dan umpan yang tidak bereaksi. Asumsi : cairan yang terikut pada padatan (katalis cobalt) dapat diabaikan.

  6

  8

  9 F = F + F ........................................................... (22)

  6

  9 F Toluena = F Toluena ........................................................ (23)

  F

  6 C

  6 H

  5 COOH = F

  9 C

  6 H

  5 COOH ................................................... (24)

  F

  6 Benzaldehida = F

  

9 Benzaldehida ............................................... (25)

  6

  9 F Benzena = F Benzena ........................................................ (26)

  6

  2

  9

  2 F H O = F H O .............................................................. (27)

  F

  6 Cobalt = F

  9 Cobalt ........................................................... (28)

  4. Neraca Massa di Kolom Destilasi T-101 T-101BenzenaH2O ToluenaBenzaldehidaC6H5COOHBenzena H2OToluenaH2OToluenaBenzaldehidaC6H5COOH91110

  Gambar A.4 Destilasi 1 (T-101) Fungsi : memisahkan air dan benzena dari komponen lainnya. Ketentuan : Hasil atas = Semua benzena, 99% air, dan 1% toluena. Hasil bawah = 1% air, 99% toluena, benzaldehida, dan asam benzoat.

  9

  10

  11 F = F + F .............................................................. (29)

  F

  10 Toluena = 0,99 x F

  9 Toluena .................................................. (30)

  F

  11 Toluena = 0,01 x F

  9 Toluena .................................................. (31)

  9

  6

  5

  10

  6

  5 F C H COOH = F C H COOH ...................................................... (32)

  9

  10 F Benzaldehida = F Benzaldehida ................................................... (33)

  F

  11 Benzena = F

  9 Benzena ................................................................... (34)

  9

  1 F Benzena = F Benzena ............................................................ (35)

  1

  1 F Benzena = 2% / 98% . F Toluena ........................................ (36)

  F

  11 H

  2 O = 0,99 x F

  9 H

  2 O ....................................................... (37)

  F

  10 H

  2 O = 0,01 x F

  9 H

  2 O ....................................................... (37a)

  5. Neraca Massa di Kolom Destilasi T-102

  T-102H2OToluenaBenzaldehidaC6H5COOHH2OToluenaBenzaldehidaToluenaBenzaldehidaC6H5C OOH10212 Gambar A.5 Destilasi 2 (T-102) Fungsi : memisahkan toluena dari komponen lainnya.

  Ketentuan : Semua air, 99 % toluena, dan 1 % benzaldehida terpisah sebagai hasil atas. F

  10 = F 2 + F 12 ............................................................................... (38)

  10

  2

  2

  2 F H O = F H O ............................................................................... (38a)

  F

  2 Toluena = 0,99 x F

  10 Toluena .............................................................. (39)

  F

  12 Toluena = 0,01 x F

  10 Toluena ............................................................. (39a)

  F

  10 C

  6 H

  5 COOH = F

  12 C

  6 H

  5 COOH .................................................................. (40)

  10

  2

  12 F Benzaldehida = F Benzaldehida + F Benzaldehida ................................ (41)

  Benzaldehida yang membentuk loop (F

  2 Benzaldehida) dijaga semininal mungkin. Untuk

  menghindari terjadinya akumulasi, benzaldehida yang dihasilkan (F

  12 Benzaldehida hasil reaksi 2) sama dengan benzaldehida yang dibuang (F

  Benzaldehida), maka : F

  10 Benzaldehida = F Benzaldehida Loop + F Benzaldehida hasil reaksi ...... (42)

  F Benzaldehida hasil reaksi = F

  12 Benzaldehida = 0,99 x F

  10 Benzaldehida....... (43)

  2

  10 F Benzaldehida Loop = F Benzaldehida = 0,01 x F Benzaldehida ....... (44)

  2

  3 F Benzaldehida = F Benzaldehida .................................................. (45)

  6. Neraca Massa di Kolom Destilasi T-103

  T-103ToluenaBenzaldehidaC6H5COOHToluenaBenzaldehidaC6H5COOHBenzaldehidaC6H5COOH 121413 Gambar A.6 Destilasi 3 (T-103) Fungsi : memisahkan asam benzoat (C

  6 H 5 COOH) dari komponen lainnya.

  Ketentuan : 99,9% benzaldehida, sisa toluena, dan 0,1% asam benzoat (C

  6 H

  5 COOH) terpisah sebagai produk atas.

  12

  13

  14 F = F + F ......................................................................... (46)

  F

  14 Toluena = F

  12 Toluena ................................................................... (46a)

  F

  12 C

  6 H

  5 COOH = F

  13 C

  6 H

  5 COOH + F

  14 C

  6 H

  5 COOH ................................ (47)

  13

  6

  5

  12

  6

  5 F C H COOH = 0,999 x F C H COOH ................................................ (48)

  14

  6

  5

  12

  6

  5 F C H COOH = 0,001 x F C H COOH ................................................ (49)

  14

  12 F Benzaldehida = 0,999 x F Benzaldehida .............................................. (50)

  F

  13 Benzaldehida = 0,001 x F

  12 Benzaldehida ............................................. (50a)

  7. Neraca Massa di Reaktor R-201

R-201BenzaldehidaC6H5COOHPaladiumH2ArN2CH4ArN2CH4CHCAPaladiumBenzaldehida131517181

  6 Gambar A.7 Reaktor 2 (R-201)

  6

  

5

Fungsi : tempat mereaksikan asam benzoat (C H COOH) dengan gas hidrogen untuk menghasilkan cyclohexane carboxylic acid (CHCA).

  Reaksi yang terjadi (stoikiometri):

  6

  5

  2 C H COOH + 3 H Cyclohexane Carboxilic Acid (CHCA)

  F

  13 + F 15 + F 16 = F 17 + F 18 ........................................................................... (51)

  F

  18 CHCA = F CHCA hasil reaksi ...................................................... (52)

  18

  13

  6

  5 CHCA

  F CHCA = N C H COOH x Mr ............................................. (53)

  16

  2

  13

  6

  5 H2

  F H = 3 / 1 x N C H COOH x Mr ....................................... (54) F

  16 Argon = F

  17 Argon = 2,16 / 90,52 x N

  16 H 2 x Mr Argon .................. (55)

  F

  16 N 2 = F

  17 N 2 ........................................................................... (56)

  16

  2

  16

  2 N2

  F N = 4,49 / 90,52 x N H x Mr ........................................... (57)

  15

  18 F Paladium = F Paladium ................................................................... (58)

  18

  19 F Paladium = F Paladium .................................................................... (59)

  F

  18 Benzaldehida = F

  13 Benzaldehida ............................................................ (59a)

  8. Neraca Massa di Filter P-201

  P-201CHCAPaladiumBenzaldehidaPaladiumBenzaldehidaCHCA181920 Gambar A.8 Filter 2 (P-201) Fungsi : memisahkan katalis paladium dari produk reaktor R-201.

  Asumsi : Benzaldehida terikut dalam cairan pada alur 19 dan diabaikan.

  19 F Paladium = recycle katalis Paladium 100%

  18

  19

  20 F = F + F .............................................................................. (60)

  18

  20 F CHCA = F CHCA ......................................................................... (61)

  F

  18 Paladium = F

  19 Paladium ...................................................................... (62)

  F

  19 Paladium = F

  15 Paladium ...................................................................... (63)

  19

  18 F Benzaldehida = F Benzaldehida ............................................................... (63a)

  9. Neraca Massa di Reaktor R-301

R-301CHCAOleumNOHSO4H2OCO2SO3CHCAKaprolaktam.H2SO4CHCAH2O222120242523

  Gambar A.9 Reaktor 3 (R-301) Fungsi : tempat mereaksikan CHCA, oleum, dan asam nitrosilsulfur (NOHSO

  4 ) untuk menghasilkan kaprolaktam sulfat.

  Ketentuan : CHCA ekses 25 % dan konversi 50 %. F

  20 + F 21 + F 22 + F 25 = F 23 + F 24 ............................................................. (64) masuk reaktor

  20

  25 F CHCA = F CHCA + F CHCA ..................................... (65) yang bereaksi

  N CHCA = %1250,5xCHCA) N +CHCA (N2520 N CHCA yang bereaksi = 0,4 (N

  20 CHCA + N

  25 CHCA) ........................... (66)

  N CHCA sisa (F

  25 CHCA) = 0,6 (N

  

20 CHCA + N

  25 CHCA) ........................... (67)

  23

  2

  4

  21

  2

  4 2 4hasil reaksi

  F H SO = F H SO + F H SO .......................... (68)

  21

  2

  4 F H SO = Asam Sulfat dalam Oleum 35 %

  = 65 % . N

  

21 kmol/jam . Mr H2SO4 ......................... (69)

  F H

  2 SO 4 hasil reaksi = 0,4 (N

  20 CHCA + N

  25 CHCA) kmol/jam. Mr H2SO4 ........... (70)

  21

  3

  24

  3 F SO = F SO ............................................................................... (71)

  21

  3

  3 F SO = SO dalam Oleum 35 % ..................................................... (72)

  24

  2

  20

  

25 CO2

  F CO = 0,4 (N CHCA + N CHCA) kmol/jam . Mr ............. (73) F

  22 NOHSO 4 = 0,4 (N

  20 CHCA + N

  25 CHCA) kmol/jam. Mr NOHSO4 .... (74) 4 yang bereaksi

  20

  25 NOHSO4

  F NOHSO = 0,4 (N CHCA + N CHCA) kmol/jam. Mr .... (75)

  23

  20

  25 Kaprolaktam

  F Kaprolaktam = 0,4 (N CHCA + N CHCA) kmol/jam. Mr (76) F

  23 H

  2 O = F

  22 H

  2 O .......................................................................... (77)

  F

  22 H

  2 O = 10% / 90% . F

  22 NOHSO 4 (kg/jam) .............................. (78)

  10. Neraca Massa di Tangki Pencampur (M-301)

  M-301Kaprolaktam.H2SO4CHCAH2OKaprolaktam.H2SO4CHCAH2OH2O262327 Gambar A.10 Tangki Pencampur (M-301) Fungsi : melarutkan kaprolaktam sulfat dengan air agar terpisah dari komponen lainnya.

  23

  26

  27 F + F = F ............................................................................... (79)

  23

  27 F CHCA = F CHCA ................................................................... (80)

  F

  23 Kaprolaktam = F

  27 Kaprolaktam ........................................................ (81)

  F

  23 H

  2 SO 4 = F

  27 H

  2 SO 4 .................................................................... (82)

  F

  27 H

  2 O = F

  23 H

  2 O + F

  26 H

  2 O ...................................................... (83)

  26

  2

  23 F H O = 100/ 93,5 x F Kaprolaktam ....................................... (84)

  11. Neraca Massa di Dekanter FL-301

  FL-301CHCAKaprolaktam.H2SO4CHCAH2OKaprolaktam.H2SO4H2O272528 Gambar A.11 Tangki Dekanter (FL-301) Fungsi : memisahkan kaprolaktam dari CHCA sebelum dinetralisasi di reaktor netralisasi R-302.

  Asumsi : larutan kaprolaktam yang terikut ke CHCA dapat diabaikan dengan pertimbangan perbedaan kelarutan dalam air dan perbedaan densitas yang besar. F

  27 = F 25 + F 28 ........................................................ (85)

  27

  25 F CHCA = F CHCA ..................................................... (86)

  27

  28 F Kaprolaktam = F Kaprolaktam .......................................... (87)

  F

  27 H

  2 SO 4 = F

  28 H

  2 SO 4 ..................................................... (89)

  F

  27 H

  2 O = F

  28 H

  2 O ......................................................... (90)

  12. Neraca Massa di Reaktor Netralisasi R-302 R-302Kaprolaktam.H2SO4H2ONH3H2OKaprolaktam (NH4)2SO4H2OH2SO4282930

  Gambar A.12 Reaktor Netralisasi (R-302) Fungsi : tempat netralisasi kaprolaktam dari asam sulfat dan pembentukan kristal ammonium sulfat.

  3

  2

  4

  4

  2

  4 Reaksi : 2 NH + Kaprolaktam.H SO (NH ) SO + Kaprolaktam

  30

  28

  29 F = F + F .................................................................... (91)

  F

  30 Kaprolaktam = F

  28 Kaprolaktam ..................................................... (92)

  F

  29 NH 3 = 2/1 . 0.99 . N

  28 H

  2 SO 4 (kmol/jam). Mr NH3 .............. (93)

  30

  4

  2

  4

  28

  2 4 (NH4)2SO4

  F (NH ) SO = 1/1 . 0.99 . N H SO (kmol/jam). Mr ....... (94)

  30

  2

  28

  2

  29

  2 F H O = F H O + F H O .................................................... (95)

  F

  29 H

  2 O = 0,5/ 99,5 x F

  29 NH 3 ................................................... (96)

  F

  30 H

  2 SO 4 = 0,01 . F

  28 H

  2 SO 4 ....................................................... (97)

  Di mana, F

  30 H

  2 SO 4 adalah asam sulfat sisa reaksi.

  13. Neraca Massa di Filter P-301

  P-301Kaprolaktam (NH4)2SO4H2OH2SO4 (NH4)2SO4H2OH2SO4Kaprolaktam H2OH2SO4303132 Gambar A.13 Filter (P-301) Fungsi : memisahkan kristal (NH

  4 )

  2 SO 4 (ammonium sulfat) dari campuran.

  Asumsi : cairan yang terikut ke kristal (NH

  

4 )

  2 SO 4 (ammonium sulfat) diabaikan.

  30

  31

  32 F = F + F ............................................................. (98)

  30

  32 F Kaprolaktam = F Kaprolaktam .............................................. (99)

  F

  30 (NH 4 )

  2 SO 4 = F 31 (NH 4 )

  2 SO 4 ................................................... (100)

  F

  30 H

  2 O = F

  32 H

  2 O + F

  31 H

  2 O ............................................ (101)

  31

  2

  31

  4

  2

  4 F H O = 0,8/98,5 . F (NH ) SO ................................... (102)

  30

  2

  4

  31

  2

  4

  32

  2

  4 F H SO = F H SO + F H SO ...................................... (103)

  F

  31 H

  2 SO 4 = 0,7/98,5 . F 31 (NH 4 )

  2 SO 4 ................................... (104)

  14. Neraca Massa di Evaporator FE-301 FE-301Kaprolaktam H2OH2SO4H2OH2SO4Kaprolaktam H2OH2SO4323433

  Gambar A.14 Evaporator (FE-301) Fungsi : memekatkan larutan kaprolaktam untuk dipadatkan di dalam spray dryer.

  F

  32 = F 33 + F 34 ............................................................. (105)

  32

  33 F Kaprolaktam = F Kaprolaktam .............................................. (106)

  32

  2

  33

  2

  34

  2 F H O = F H O + F H O ............................................ (107)

  F

  32 H

  2 SO 4 = F

  33 H

  2 SO 4 + F

  34 H

  2 SO 4 ...................................... (108) 15.

  Neraca Massa di Meeting Joint

  Kaprolaktam H2OH2SO4Kaprolaktam H2OH2SO4Kaprolaktam H2OH2SO4333940 Gambar A.15 Meeting Joint F

  40 = F 33 + F

  39 ...................................................

  .......... (109) F

  40 Kaprolaktam = F

  39 Kaprolaktam + F

  33 Kaprolaktam ............... (110)

  40

  2

  39

  2

  33 F H O = F H O + F

  H

  2 O ............................................

  (111)

  40

  2

  4

  39

  2

  4

  33 F H SO = F H SO + F

  2

  4 H SO ......................................

  (112) 16.

  Neraca Massa di Spray Dryer

  SR-301

  Gambar A.16 Spray Dryer (SR-301) Fungsi : membentuk kaprolaktam pekat menjadi padatan berbentuk butiran.

  40

  37

  35

  36 F +F = F + F ...................................................

  . (113) F

  40 Kaprolaktam = F

  35

  37 Kaprolaktam + F

  Kaprolaktam ...... (114) F

  37 Kaprolaktam = 0,009%

  x F

  40 Kaprolaktam

  ..................... (115)

  40

  2

  36

  2

  35

  2 F H O + F H O = F H O

• F

  37 H

  2 O

  ................................... (116) F

  36 H

  2 O + 0,009% F

  40 H

  2 O =

  37

  2 F H O

  .................................................... (117)

  36

  2

  37

  2 F O = F O ....................................................... (118)

  36

  2

  37

  2 F N = F N ....................................................... (119)

  F

  40 H

  2 SO 4 = F

  35 H

  2 SO 4 + 0,009% F

  40 H

  2 SO 4 .............. (120) 17.

  Neraca Massa di Screening S-301 Kaprolaktam H2OH2SO4Kaprolaktam H2OH2SO4Kaprolaktam H2OH2SO4353841

  Gambar A.17 Sreening (S-301) Fungsi : memisahkan butiran kaprolaktam (caprolactam prill) yang tidak sesuai spesifikasi.

  41

  35

  38 F = F - F ....................................................... (121)

  F

  41 Kaprolaktam = 0,9 x F

  35 Kaprolaktam ................................. (122)

  F

  41 H

  2 SO 4 = 0,9 x F

  35 H

  2 SO 4 ............................................ (123)

  41

  2

  35

  2 F H O = 0,9 x F H O ................................................ (124) 18.

  Neraca Massa di Melter ME-301 Gambar A.18 Melter (ME-301)

  Fungsi : meleburkan kembali butiran kaprolaktam yang ukurannya tidak sesuai dengan spesifikasi produk. F

  38 = F 39 ................................................................. (125)

  F

  38 Kaprolaktam = 0,1 x F

  35 Kaprolaktam ................................. (126)

  38

  39 F Kaprolaktam = F Kaprolaktam .......................................... (127)

  38

  2

  4

  35

  2

  4 F H SO = 0,1 x F H SO ............................................ (128)

  38

  2

  4

  39

  2

  4 F H SO = F H SO ..................................................... (129)

  F

  38 H

  2 O = 0,1 x F

  35 H

  2 O ................................................ (130)

  F

  38 H

  2 O = F

  39 H

  2 O ......................................................... (131) Penyelesaian Persamaan Neraca Massa Sebelumnya

  6

  11

  1. Neraca Massa Kaprolaktam (C H ON)

  Kapasitas produksi = 6313,131 kg/jam Kaprolaktam Spesifikasi Kaprolaktam : 99,5 % berat, maka :

41 Kaprolaktam

  ) = 0,995 x 6313,131 kg/jam Kaprolaktam (F

  = 6281,566 kg/jam

  2 O (F

  41 H

  2 O) = 0,3 % berat produk

  H = 0,003 x 6313,131 kg/jam = 18,939 kg/jam

  2 SO 4 (F

  41 H

  2 SO 4 ) = 0,2 % berat produk

  H = 0,002 x 6313,131 kg/jam = 12,626 kg/jam

41 Maka, F Kaprolaktam = 6281,566 kg/jam

  Melalui Pers. (123), (127), (116) dan (118), diperoleh :

  41

  35 F Kaprolaktam = 0,9 x F Kaprolaktam ; maka

  35 F Kaprolaktam = 6281,566 / 0,9

  = 6979,517 kg/jam

  F

  38 Kaprolaktam = 0,1 x F

  35 Kaprolaktam

  = 0,1 x 6979,517 = 697,952 kg/jam

  40 Kaprolaktam

  37 F Kaprolaktam = 0,009 % x F

  F

  40 Kaprolaktam = F

  35 Kaprolaktam + F

  37 Kaprolaktam F

  40 Kaprolaktam

  40

  = 6979,517 kg/jam + 0,009% F Kaprolaktam = 6979,517 / 0,99991 = 6980,146 kg/jam

  F

  37 Kaprolaktam = 0,009 % x F

  40 Kaprolaktam

  = 0,009% x 6282,131 = 0,628 kg/jam Melalui Pers. (128) dan (111), diperoleh :

  F

  39 Kaprolaktam = F

  38 Kaprolaktam = 697,952 kg/jam F

  40 Kaprolaktam = F

  39

  33 Kaprolaktam + F Kaprolaktam, maka :

• – 697,952) kg/jam = 6282,194 kg/jam Melalui Pers. (107), (100), (93), (89) dan (83) diperoleh :

  33 F Kaprolaktam = (6980,146

  33 F Kaprolaktam = 6282,194 kg/jam F

  32 Kaprolaktam = 6282,194 kg/jam F

  23 Kaprolaktam = 6282,194 kg/jam F

  27 Kaprolaktam = 6282,194 kg/jam

  28 F Kaprolaktam = 6282,194 kg/jam

  30 F Kaprolaktam = 6282,194 kg/jam

  2. Neraca Massa Air (H

  2 O)

  Melalui perhitungan spesifikasi produk, diperoleh :

  F

  41 H

  2 O = 18,939 kg/jam

  Melalui Pers. (125), (131), (117) dan (118), diperoleh :

  F

  35 H

  2 O = F

  41 H

  2 O / 0,9

  = 18,939 / 0,9 = 21,044 kg/jam Maka,

   F

  38 H

  2 O = 0,1 x F

  35 H

  2 O

  = 0,1 x 21,044 = 2,104 kg/jam

  40

  2

  36

  2

  

35

  2

  37 2 ;

  F H O + F H O = F H O + F H O di mana, F

  37 H

  2 O = F

  36 H

  2 O + 0,009% F

  40 H

  2 O

  Sehingga F

  40 H

  2 O + F

  36 H

  2 O = F

  35 H

  

2 O + F

  36 H

  2 O + 0,009% F

  40 H

  2 O

  40

  2

  35

  

2

  (100% - 0,009%) F H O = F H O

  40

  2

  99,991% F H O = 21,044 F

  40 H

  2 O = 21,044/0,99991 F

  40 H

  2 O = 21,046 kg/jam

  37 H

  2 O = 0,002 kg/jam

  Maka, F Melalui pers. (132) dan (112),

  39

  38 H

  2 O

  2 F H O = F

  = 2,104 kg/jam F

  40 H

  2 O = F

  39 H

  2 O + F

  33 H

  2 O F

  33 H

  2 O = 21,046

• – 2,104 = 18,941 kg/jam Pada Pers. (108), F

  32 H

  2 O = F

  33 H

  2 O + F

  34 H

  2 O

  32

  2

  34

  2 F H O = 18,941 + F H O ........................................................ (a)

  Dari Pers. (102) dan (96), diperoleh : F

  31 H

  2 O + F

  32 H

  2 O = F

  30 H

  2 O

  F

  30 H

  2 O = F

  28 H

  2 O + F

  29 H

  

2 O ............................................ (b)

  Subtitusi Pers. (91) dan (97), Di mana F

  27 H

  2 O = F

  28 H

  2 O dan F

  29 H

  

2 O = 0,5/99,5 . F

  29 NH 3, diperoleh :

  F

  31 H

  2 O + F

  32 H

  2 O = F

  30 H

  2 O

  27

  2

  29

  3

  = F H O + 0,5/ 99,5 x F NH kg/jam

  31

  2

  32

  2

  30

  2 F H O + F H O = F H O

  = F

  27 H

  2 O + (5,03.10 -3 x F

  29 NH

3 kg/jam) ............ (c)

  F

  27 H

  2 O dan F

  29 NH

3 belum diketahui, sehingga dicari terlebih dahulu.

  Melalui Pers. (85), (79), (80), dan (86) diperoleh :

  27

  2

  23

  2

  

26

  2 F H O = F H O + F H O

  23

  2

  22

  2

  di mana : F H O = F H O = 10/90 x F

  22 NOHSO 4 kg/jam

  dan F

  26 H

  2 O = 100/93,5 x F

  23 Kaprolaktam

  23

  26

  2 Karena F Kaprolaktam = 6282,194 kg/jam, maka F H O = 6718,924 kg/jam

  F

  27 H

  2 O = (10/90 x F

  22 NOHSO 4 kg/jam) + 6718,924 kg/jam ........................ (d)

  22

  4 Melalui Pers. (78) dan (76), F NOHSO dapat dihitung, yaitu :

  23

  20

  25 Kaprolaktam

  F Kaprolaktam = 0,4 (N CHCA + N CHCA) . Mr 0,4 (N

  20 CHCA + N

  25 CHCA) = F

  23 Kaprolaktam / Mr Kaprolaktam = 6282,194/113

  0,4 (N

  20 CHCA + N

  25 CHCA) = 55,595 kmol/jam

  22

  4

  20

25 NOHSO4

  F NOHSO = 0,4 (N CHCA + N CHCA) . Mr

  = 55,595. 127 = 7060,519 kg/jam

  23

  2

  22

2 Maka, F H O = F H O

  22

  4

  = 10/90 x F NOHSO kg/jam

   = 10/90 x 7060,519 kg/jam = 784,502 kg/jam. F

  27 H

  2 O

  = 784,502 + 6718,924 kg/jam

   = 7503,426 kg/jam

  Untuk F

  29 NH 3 , dapat dicari dengan substitusi Pers (94), (90), (84), (70), (71), dan (72) sehingga

  diperoleh :

  29

  3

  28

  2

  4

  3 F NH = 2 x 0,99 x N H SO kmol/jam . Mr NH

  N

  28 H

  2 SO 4 = N

  27 H

  2 SO 4 = N

  23 H

  2 SO 4 = N

  21 H

  2 SO 4 + N H

  2 SO 4 hasil reaksi ...... (e)

  Di mana, N

  21 H

  2 SO 4 = Asam sulfat yang terdapat dalam Oleum 35 % Kebutuhan Oleum 35 % = 3 mol/mol Kaprolaktam yang terbentuk.

  = 3 x 55,595 kmol/jam = 166,785 kmol/jam = 166,785 x Mr Oleum = 29687,536 kg/jam

  21

  2

4 F H SO = 65 % x 29687,536 kg/jam

  = 19296,898 kg/jam

  N

  21 H

  2 SO 4 = 196,907 kmol/jam N H

  2 SO 4 hasil reaksi

  20

  25

  = 0,4 (N CHCA + N CHCA) = 55,595 kmol/jam N

  21 H

  2 SO 4 dan N H

  2 SO

4 hasil reaksi dimasukkan ke Pers (e), diperoleh :

N

  28 H

  2 SO 4 = N

  27 H

  2 SO 4 = N

  23 H

  2 SO

  4

  = 196,907 + 55,595 = 252,502 kmol/jam

  28

  2

  4

  27

  2

  4

  23

  2

  4 F H SO = F H SO = F H SO = 24745,172 kg/jam

  29

  28 H

  2 SO 4 kmol/jam . Mr NH

  3

  3 Maka, F NH = 2 . N

  = 2. 227,251 . 17 = 8499,209 kg/jam

  27

  2

  29

3 F H O dan F NH masuk ke Pers. (c), sehingga diperoleh :

  31

  2

  32

  2

  30

  2

  27

  2

  29

  3 F H O + F H O = F H O = F H O + (0,5/ 99,5 x F NH kg/jam)

  F

  31 H

  2 O + F

  32 H

  2 O = F

  30 H