Pembuatan Silikon Karbida (SiC) dari Pasir Silika (SiO2) dan Karbon (C) dengan Kapasitas 30.000 Ton/Tahun

  = 82,1869 kmol/jam Massa impuritis

  = 3.295,3842 kg/jam Mol SiC murni =

   SiO

  = 121,5447 kg/jam  Na

  

40

3.295,3842

  

0962 ,

  =

  SiC Mr SiC F

  = 3.787,8788 kg/jam Produk Akhir = Silikon Karbida (SiC) dengan kemurnian 87 % Kapasitas produksi = 3.787,8788 kg/jam Massa SiC Murni = 87 % x 3.787,8788 kg/jam

  (4,93%) = 186,8109 kg/jam  C (3,21%)

  1 1000 kg

  24 hari 1 ton

  330 hari tahun 1 . jam

  

tahun

ton

.

  1 hari = 24 jam Kapasitas produksi tiap jam = 30.000 .

  Kapasitas produksi Silikon Karbida = 30.000 ton/tahun, dengan kemurnian 87% (% berat) dengan ketentuan sebagai berikut: 1 tahun = 330 hari kerja

  

LAMPIRAN A

NERACA MASSA

A.1 PERHITUNGAN PENDAHULUAN A.1.1 Menghitung Kapasitas Produksi

  2

2 O (2,60%) = 98,5025 kg/jam

   FePO

  4

  (2,26%) = 85,6365 kg/jam

  A.1.2 Menghitung Kapasitas Feed

  Reaksi : SiO

  2

• 3 C → SiC + 2 CO
• Pereaksi pembatas :SiO

  2

• Konversi SiO

  2

  sebesar 96 % Massa SiC murni = 3295,3842 kg/jam Mol SiC murni = 82,1869 kmol/jam Mol SiO

  2

  = 85,2960 kmol/jam Massa SiO

  =

96 SiC N

  2

  2 O) = 6,5%

  .60H

  2

  dalam 10Na

  2

  murni/ jam = x kg/jam Jumlah SiO

  2

  Basis Jumlah bahan baku SiO

  ) = 1% (Lowe, 1958)

  4

  4. Besi Fosfat ( FePO

  .60H

  = N SiO

  2

  %

  2. Karbon (C) = 36%

  ) = 56,5%

  2

  1. Pasir Silika (SiO

  Bahan baku dan Rasio (%wt)

  = 85,2960 x 60,0864 = 5.125,1287 kg/jam

  2

  x Mr SiO

  2

  3. Larutan Natrium Silikat (10Na

2 O.30SiO

2 O.30SiO

  O H SiO O Na massa x O H SiO O Na Mr SiO Mr x _{2 2 2 2 2 2} ² 60 .

  36  

  

2

O/jam

  =

  1  

  56

  = _jam ^kg _jam ^kg 5025 , 98 4.838,4606 5 ,

  4

  FePO

  6  

  56 5 ,

  2 O = 4.838,4606 556 6371 , _jam ^kg _jam ^kg 5 ,

  .60H

  2

  2 O.30SiO

  10Na

  56

  10 60 . 30 .

  . 082 9129 , 3 4.838,4606

5 ,

  murni = 4.838,4606 kg/jam C = _jam ^kg _jam ^kg

  2

  = 4.838,4606 kg/jam Jumlah bahan baku : SiO

  x

  total = 5.125,1287 kg/jam x kg/jam + 0,0592x kg/jam = 5.125,1287 kg/jam

  2

  = 0,0592 x kg/jam Jumlah bahan baku SiO

  5 , 51  

  56 5 , 6 %

  SiO murni massa ₂ 5 ,

  =

  30

  10

  30 .

  A.2 PERHITUNGAN NERACA MASSA A.2.1 Mixer (M-101)

  Fungsi: Tempat pencampuran semua bahan baku

  10Na O.30SiO .60H _{2 2 2} O

  5 SiO ₂ C SiO ₂

  4

  6 FePO ₄ C

  10Na O.30SiO .60H _{2 2 2} FePO ₄ O Neraca massa komponen: Alur 4

  (4)

  F SiO = 4.838,4606 kg/jam

  2 (4)

  F FePO 85,6365 kg/jam

  =

  4 (4)

  F C 3.082,9129 kg/jam

  = Alur 5 (5)

  F

  10Na O.30SiO .60H O 556,6371 kg/jam

  =

  2

  2

  2 Alur 6 (6)

  F FePO 85,6365 kg/jam

  =

  4 (6)

  F C 3.082,9129 kg/jam

  = (6) _{( 4 ) ( 5 )}

  30  Mr SiO ₂ F SiO =

  2

  5 F SiO  F ₂ 10 Na O . ₂ 30 SiO . ₂

  30 H O  ₂ 10 . 30 .

  30 Mr Na O SiO H O _{2 2 2} =

  .125,1287 kg/jam

  (6) _{( 5 )  Mr Na O}

  10 ₂ F Na O =

2 F 10 Na O .

  ₂ 30 SiO . ₂

  30 H O  ₂ 10 . 30 .

  30 Mr Na O SiO H O _{2 2 2} = 98,5025 kg/jam

  (6) _{( 5 )  Mr H O}

  60 F H O ² =

  2 10 . 30 .

  30 F Na O SiO H O  _{2 2 2} Mr 10 Na O . ₂ 30 SiO . ₂

  30 H O ₂ = 171,4665 kg/jam

• 4.838,4606 5.125,1287 C - 3.082,9129 3.082,9129
• 85,6365 85,6365 Na

  H

  10Na ^{2 O.30SiO 2 .60H 2 O}

  30 ^o

  7

  6

  8

  30

^o

  C, 1 atm

  Fungsi : Mengubah dan membentuk slurry menjadi pellet

  Total 8.563,6471 8.563,6471 A.2.2 Pelletizing Machine (PL-102)

  Subtotal 556,6371 8.007,0100 8.563,6471

  2 O - - 171,4665

  2 O - - 98,5025

  10Na ^{2 O.30SiO 2 .60H 2 O} SiO ² C FePO ⁴

  4

  FePO

  2

  SiO

  2 O 556,6371 - -

  .60H

  2

  2 O.30SiO

  10Na

  Komponen Masuk (kg/jam) Keluar (kg/jam) Alur 5 Alur 4 Alur 6

  Tabel A.1 Neraca massa pada Tangki Mixer

  Neraca massa total :

C, 1 atm

  F

  (6)

  Neraca massa komponen: Alur 6

  F

  (6)

  SiO

  2 =

  C, 1 atm H ^{2 O}

  40 ^o

  4.838,4606 kg/jam F

  (6)

  4 = 85,6365 kg/jam

  2 O = 171,4665 kg/jam SiO ² C FePO ⁴

  H

  (6)

  98,5025 kg/jam F

  =

  Na

  (6)

  3.082,9129 kg/jam F

  =

  C

  FePO

2 O

  Alur 7

  Dari Tabel 20.44 Perry Handbook, moisture requirements untuk mengubah dan membentuk slurry menjadi pellet berkisar antara 13,0 O.

• – 13,9 % H

  2 Misalkan, jumlah total = X kg/jam

  = 8.563,6471 + 0,139 X

   X X = 9.747,0158 kg/jam (7)

  F H O (0,139 x 9.747,0158)

  =

  2

• – 171,4665 1.183,3687 kg/jam

  = Alur 8 (8)

  F SiO 5.125,1287 kg/jam

  =

  2 (8)

  F FePO = 85,6365 kg/jam

  4 (8)

  F C = 3.082,9129 kg/jam

  (8)

  F Na O 98,5025 kg/jam

  =

  2 (8)

  F H O 1.354,8352 kg/jam

  2 = Neraca massa total :

  Tabel A.2 Neraca massa pada Pelletizing Machine (L-102)

  Masuk (kg/jam) Keluar (kg/jam) Komponen Alur 6 Alur 7 Alur 8

• SiO 5.125,1287 5.125,1287

  2

• C 3.082,9129 3.082,9129
• FePO 85,6365 85,6365

  4

• Na O 98,5025 98,5025

  2 H O 171,4665 1.183,3687 1.354,8352

  2 Subtotal 8.563,6471 1.183,3687 9.747,0158 Total 9.747,0158 9.747,0158

• Komposisi gas alam (alur 22) :

  X

  C

  4 H

  10

  = 1,25 % (Speight, dkk., 2006)

  X

  (12)

  O

  2

  = 21 %

  (12)

  X

  N

  2

  = 79 %

  4

  2 →

  CO

  2

  2 O

  Konversi CH

  4 ≈ 100%

  (10)

  = 1,25 %

  = 90 %

  X

   Kiln Preheater (B-102) Udara ^E-139 Gas Alam ^B-101

  30 ^{o C, 2 atm 30 o C, 2 atm 863 o C, 1 atm} O ₂ N ² CO ₂ H ^{2 O} FC ¹³ ₁₀

  12 ¹¹ Dimana :

  X

  (10)

  CH

  4

  8

  (10)

  A.2.3 Burner (B-101)

  C

  2 H

  6

  = 7,5 %

  X

  (10)

  C

  3 H

  Fungsi : Tempat pembakaran gas alam sebagai sumber panas Rotary

• Komposisi Udara :
• Reaksi :

1. CH

• 2O
• 2H
• 2

• 3H
• 5O

• 4H

  2

  4CO

  13 O 2 →

  10

  4 H

  C

  2 O = 4 4.

  2

  = 3 σ H

  = -5 σ CO

  2

  = -1 σ O

  8

  σ C

  2 H 6 ≈ 100%

  Konversi C

  

2

O

  Konversi C

  2 H 6 ≈ 100%

  = 4 σ H

  C, maka jumlah gas alam yang dibutuhkan adalah 600 kg/jam dengan kebutuhan udara (excess 20%) sebesar 16.958,4138 kg/jam.

  o

  C sampai 863

  kiln preheater dari 30 o

  Berdasarkan energi yang dibutuhkan untuk menaikkan suhu rotary

  Karena pembakaran dengan menggunakan oksigen berlebih dari udara, maka reaksi pembakaran gas alam mempunyai konversi yang mendekati 100%.

  2

  σ C

  σ CO

  13

  2

  = -

  2

  = -1 σ O

  10

  

2

O

  2

  3CO

  C

  

2

O

  2

  2CO

  7 O 2 →

  6

  2 H

  2 O = 2 2.

  2 H

  = 1 σ H

  2

  = -2 σ CO

  2

  = -1 σ O

  4

  σ CH

  →

  Konversi C

  6

  ≈ 100% σ C

  2

  8

  3 H

  3. C

  2 O = 3

  = 2 σ H

  2

  σ CO

  7

  2

  = -

  2

  = -1 σ O

  6

  2 H

3 H

• 2
• 5H

4 H

2 O = 5

  Perhitungan neraca massa :

  (% CH  MrCH )  (% C H  MrC H )  (% C H  MrC H )  (% C H  MrC H ) _{4 4 2 6 2 6 3 8 3 8 4 10 4 10} Mr gas alam = 100 % ( , 9  16 , 0425 )  ( , 075  30 , 07 )  ( , 0125  44 , 096 )  ( , 0125  58 , 124 )

  =

  100 %

  = 17,9712 kg/kmol

  Alur 10 (10)

  F = 600 kg/jam ₁₀

  (10) ^kg 600

  N = F = jam = 33,3867 kmol/jam _kg

  Mr gas alam 17 , 9712 _kmol (10) (10)

  N CH = 0,9 x N = 30,0480 kmol/jam

  4 (10) (10)

  F CH = N CH x Mr CH = 482,0449 kg/jam

  4

  4

  4 (10) (10)

  N C H = 0,075 x N = 2,5040 kmol/jam

  2

  6 (10) (10)

  F C H = N C H x Mr C H = 75,2953 kg/jam

  2

  6

  2

  6

  2

  6 (10) (10)

  N C H = 0,0125 x N = 0,4173 kmol/jam

  3

  8 (10) (10)

  F C H = N C H x Mr C H = 18,4027 kg/jam

  3

  8

  3

  8

  3

  8 (10) (10)

  N C H = 0,0125 x N = 0,4173 kmol/jam

  4

  10 (10) (10)

  F C H = N C H x Mr C H = 24,2571 kg/jam

  4

  10

  4

  10

  4

  10 Alur 12 (12)

  F = 16.958,4138 kg/jam

  (12) ^{12 kg} 16.958,413

  8 N = F = jam = 587,8065 kg/jam _kg 28 , 8503

  Mr gas alam kmol (12) (12) (12)

  N O = X O x N = 123,4394 kmol/jam

  2

  2 (12) (12)

  F O = N O x Mr O = 3.949,9115 kg/jam

  2

  2

  2 (12) (12) (12)

  N N = X N x N = 464,3671 kmol/jam

  2

  2 (12) (12)

  F N = N N x Mr N = 13.008,5023 kg/jam

  2

  2

  2

• – (1 x 30,0480) =
• – (1 x 2,5040) =
• – (1 x 0,4173) =
• – (1 x 0,4173) =

  2

  (13)

  O

  2

  x Mr O

  2

  = 1.592,9880 kg/jam F

  (13)

  N

  = F

  2

  (12)

  N

  2

  = 13.008,5023 kg/jam N

  (13)

  CO

  2

  = N

  = N

  O

  O

  4 H

  6

  )

  3 H

  8

  )

  2

  13

  x r C

  10

  (13)

  ) = 123,4394

  2

  7

  x (1 x1,6693)

  2

  13

  x(1 x 0,4173) = 49,7800 kmol/jam

  F

  (12)

  2 – (2 x r

  x r C

  ) + (3 x r C

  CO

  2

  x Mr CO

  2

  = 1.671,2871 kg/jam N

  (13)

  H

  4

  2 H

  = N

  

6

  ) + (4 x r C

  3 H

  8

  ) + (5 x r C

  4 H

  10

  ) = (2 x (1 x 30,0480) + (3 x 1 x 2,540) + (4 x (1 x 0,4173)

  (13)

  2

  CH

  3 H

  4

  )

  

2

  

7

  x r C

  2 H

  6

  )

  8

  CO

  )

  2

  13

  x r C

  4 H

  10

  ) = (123,4394 x (1 x 30,0480) + (2 x (1 x 2,540) + (3 x (1 x 0,4173)

  F

  (13)

  2 H

  

7

  (10)

  C

  2 H 6 – r C

  2 H

  6

  = N

  (10)

  C

  2 H 6 – (konversi x N (10)

  2 H

  (10)

  6

  ) = 2,5040

  N

  (13)

  C

  3 H

  8

  = N

  C

  = N

  C

  (10)

  CH

  4

  = N

  (10)

  CH

  4 – r CH

  4

  = N

  CH

  6

  4 – (konversi x N (10)

  CH

  4

  ) = 30,0480

  N

  (13)

  C

  2 H

  Alur 13

  3 H 8 – r C

  N

  (13)

  C

  4 H 10 – (konversi x N (10)

  C

  4 H

  10

  ) = 0,4173

  N

  O

  = N

  2

  = N

  (12)

  O

  2 – (2 x r

  CH

  4

  )

  (10)

  10

  3 H

  ) = 0,4173

  8

  = N

  (10)

  C

  3 H 8 – (konversi x N (10)

  C

  3 H

  8

  N

  4 H

  (13)

  C

  4 H

  10

  = N

  (10)

  C

  4 H 10 – r C

  (13)

• – (
• – (5 x r C
• – (
• – (2 x (1 x 20,0320) – (
• – (5 x (1 x 0,4173 – (
• – (
• – (5 x r C
• – (
• (4 x (1 x 0,4173) = 37,9773 kmol/jam

  

2

2 O = (2 x r CH

• (5 x (1 x 0, 4173)

  = 71,3640 kmol/jam

  (13) (13)

  F H O = N H O x Mr H O = 1.285,6364 kg/jam

  2

  2

2 Neraca massa total:

  Tabel A.3 Neraca massa pada Burner (B-101)

  Masuk (kg/jam) Keluar (kg/jam) Komponen Alur 10 Alur 12 Alur 13

  CH 482,0449 - -

  4

• C H 75,2953

  2

6 C H

• 18,4027

  3

  6

• C H 24,2571

  4

  8

• O

  3.949,9115 1.592,9880

2 N - 13.008,5023 13.008,5023

  2

• CO

  1.671,2871

  2 H O - 1.285,6364 -

  2 Subtotal 600,0000 16.958,4138 17.558,4138 Total 17.558,4138 17.558,4138

  A.2.4 Rotary Kiln Preheater (B-102)

  Fungsi: Pemanas awal bahan baku sampai suhu 617

  C, sebelum dikirim ke Electric Furnace (B-103)

  O ₂ N ₂ CO _{2 625}

_{C, 1 atm}

H O _{2 14} SiO ₂ C _{35 C, 1 atm} FePO ₄ H O _{2 9}

  10Na O.30SiO .60H O _{2 2 2} SiO _{2 15} C FePO _{4 617 C, 1 atm} Na O ₂ O _{2 13} N ₂ CO _{2 863 C, 1 atm} H O ₂ Dimana :

  Asumsi oksigen (O ) tidak bereaksi dengan pasir silika (SiO ) dan Karbon (C)

  2

  2 Neraca massa komponen Alur 9 :

  Massa masuk alur 9 Rotary Kiln Preheater (B-102) = Massa keluar alur 8 Pelletizing

  Machine (L-101) (9)

  F SiO = 5.125,1287 kg/jam

  2 (9)

  F FePO 85,6365 kg/jam

  =

  4 (9)

  F C 3.082,9129 kg/jam

  = (9)

  F Na O = 98,5025 kg/jam

  2 (9)

  F H O 1.354,8352 kg/jam

  =

  2

  Alur 13 :

  Massa masuk alur 13 Rotary Kiln Preheater (B-102) = Massa keluar alur 13 Burner (B-101)

  (13)

  F CO = 1.671,2871 kg/jam

  2 (13)

  F N = 13.008,5023 kg/jam

  2 (13)

  F O = 1.592,9880 kg/jam

  2 (13)

  F H O = 1.285,6364 kg/jam

  2 Alur 15 : (15)

  F SiO 5.125,1287 kg/jam

  =

  2 (15)

  F FePO 85,6365 kg/jam

  =

  4 (15)

  F C = 3.082,9129 kg/jam

  (15)

  F Na O = 98,5025 kg/jam

  2 Alur 14 : (14) (13)

  F CO = F CO = 1.671,2871 kg/jam

  2

  2 (14) (13)

  F N N 13.008,5023 kg/jam

  = F =

  2

  2 (14) (13)

  F O O 1.592,9880 kg/jam

  2 = F 2 = (14) (9) (13)

  F H O = F H O + F H O

  2

  2

  2 = (1.354,8352 + 1.285,6364) kg/jam = 2.640,4716 kg/jam

  Neraca massa total

  Tabel A.4 Neraca massa pada Rotary Kiln Preheater (B-102)

  Masuk (kg/jam) Keluar (kg/jam) Komponen Alur 9 Alur 13 Alur 15 Alur 14

• SiO 5.125,1287 5.125,1287 -

2 C - 3.082,9129 3.082,9129 -

• FePO 85,6365 85,6365 -

  4

• Na O 98,5025 98,5025 -

2 O - - 1.592,9880 1.592,9880

  2

  13.008,5023 - 13.008,5023 - N

  2

  1.671,2871 - 1.671,2871 - CO

  2 H

• O 1.354,8352 1.285,6364 2.640,4716

  2 Subtotal 9.747,0158 17.558,4138 8.392,1806 18.913,2490 Total 27.305,4296 27.305,4296

  A.2.5 Electric Furnace (B-103)

  Fungsi: Tempat reaksi reduksi dimana terjadinya pembentukan SiC pada suhu 1600 C

  N

  2

  17 CO _o

  2 1400

C, 1 atm

  

16

_o Udara

  30 C, 1,2 atm o _o

  1600

  C, 1 atm 617

C, 1 atm

  SiC SiO ₂

  18 SiO ₂

  15 C C FePO ₄ FePO ₄ Na O ₂ Na O ₂ Reaksi :

• SiO + 3 C

  2 → SiC + 2 CO

  Konversi SiO sebesar 96 %

  2

  = -1

  2

  σ SiO = -3

  σ C = 1

  σ SiC = 2

  σ CO

  1

• CO + O

  2

  2 2 → CO

  Konversi CO ≈ 100% = -1

  σ CO

  1

  = - σ O

  2

  2

  = 1 σ CO

  2

  85,2960

  Alur 18

  = C Ar C F ^{) 15 (}

  = _{kmol kg} ^{jam kg}

  0107 ,

  12 3.082,9129

  = 256,6805 kmol/jam F

  (15)

  Na

  N

  (15)

  (18)

  SiO

  2 = N

  (15)

  SiO

  2 – r SiO

  2 =

  C

  3.082,9129 kg/jam N

  Alur 15

  0864 ,

  F

  (15)

  SiO2 = 5.125,1287 kg/jam N

  (15)

  SiO2

  = ₂ ^{2 ) 15 (} SiO Mr SiO F

  = _{kmol kg} ^{jam kg}

  60 5.125,1287

  =

  = 85,2960 kmol/jam F

  (15)

  FePO

  4 =

  85,6365 kg/jam F

  (15)

  C

2 O = 98,5025 kg/jam

• – (0,96 x 85,2960)
• – 3 x r SiO
• – 3 x (0,96 x 85,2960)

  3,1090 kmol/jam F

  2 =

  2 =

  0,96 x 85,2960

  =

  82,1869 kmol/jam F

  (18)

  SiC = N

  (18)

  SiC x Mr SiC = 82,1869 x 40,0962

  = 3.295,3842 kg/jam Alur 16

  N

  (16)

  O

  2

  (18)

  1

  x r CO + r S =

  2

  1

  x (1x(2 x r SiO

  2

  ))

  =

  2

  1

  x (1x(2 x 0,96 x 85,2960))

  = 82,1869 kmol/jam

  SiC = r SiO

  121,5447 kg/jam N

  (18)

  (15)

  SiO

  2 = N

  (18)

  SiO

  2

  x Mr SiO

  2 = 3,1090 x 60,0864 = N

  =

  (18)

  C

  =

  N

  C

  =

  2 =

  256,6805

  = 10,1197 kmol/jam

  F

  (18)

  C

  =

  N

  (18)

  C x Ar C

  

=

  10,1197 x 12,0107

  186,8109 kg/jam N

  2 =

  1x(2 x 0,96 x 85,2960)

  F

  (17)

  4

  F

  (16)

  N

  2 =

  8.661,1680 kg/jam N

  (17)

  CO

  2 = r CO = 1x(2 x r SiO

  2

  )

  =

  =

  8.661,1680 kg/jam

  2 = 164,3739 x 44,0962 =

  2

  SiO

  Komponen Masuk (kg/jam) Keluar (kg/jam) Alur 15 Alur 16 Alur 17 Alur 18

  Tabel A.5 Neraca massa pada Electric Furnace (B-103)

  Neraca massa total:

  7.234,1854 kg/jam

  x Mr CO

  164,3739 kmol/jam F

  2

  CO

  (17)

  2 = N

  CO

  (17)

  Alur 17

  2 = 309,1795 x 28,0134 =

  FePO

  O

  N

  (16)

  N

  2 = 82,1869 x 31,9988 = 2.629,8837 kg/jam

  x Mr O

  2

  (16)

  2 ) 16 (

  N

  2 =

  O

  (16)

  F

  85,6365 - - 85,6365 Na

  2 =

  %

  x Mr N

  (16)

  2

  N

  (16)

  N

  2 =

  N

  309,1795 kmol/jam F

  21 %

  =

  79 

  21 %

  82,1869 %

  =

  79 O N 

  5.125,1287 - - 186,8109 C 3.082,9129 - - 121,5447

2 O 98,5025 - - 98,5025

  O

• 2.629,8837 - - N

  2

• 8.661,1680 8661,1680 - CO

  2

• 7234,1854 - SiC - - - 3.295,3842

  2

  Subtotal 8.392,1806 11.291,0517 15.895,3534 3.787,8788

  Total 19.683,2323 19.683,2323

  A.2.6 Mixing Point (M-102) _{FC FC} M-102 O

  2 O

  2 N

  2 N

  2

  22

  2 CO

  24 CO

  2 _o ^o 1031

  2 H O

  C, 5 atm H

  2 O 625

C, 5 atm

  _FC O

  2

23 N

  2 _o 1400

C, 5 atm Neraca massa komponen Alur 22

  (22)

  F CO 1.671,2871 kg/jam

  =

  2 (22)

  F N 13.008,5023 kg/jam

  2 = (22)

  F O = 1.592,9880 kg/jam

  2 (22)

  F H O 2.640,4716 kg/jam

  =

  2 Alur 23 (21)

  F N 8.661,1680 kg/jam

  =

  2 (21)

  F CO 7234,1854 kg/jam

  =

  2 Alur 24 (24) (22) (23)

  F CO F CO + F CO

  =

  2

  2

  

2

  8.905,4725 kg/jam

  = (24) (22) (23)

  F N = F N + F N

  2

  2

  2

  21.669,6703 kg/jam

  = (24)

  F O 1.592,9880 kg/jam

  =

  2 (24)

  F H O = 2.640,4716 kg/jam

  2

• 1.592,9880
• 2.640,4716

2 O

  90 ^o

  2 N

  2 CO

  2 H

  2 O H

  2 O H

  2 O 600 ^o

  

^o

  2 H

  C, 148 atm 100 ^o

  C, 1 atm

  25

  26

  27

  28

  2 O O

  2 CO

  2

  2 N

  Tabel A.6 Neraca Massa pada Mixing Point (M-102)

  Komponen Masuk (kg/jam) Keluar (kg/jam) Alur 22 Alur 23 Alur 24

  O

  2

  1.592,9880

  N

  13.008,5023 8661,1680 21.669,6703 CO

  2

  1.671,2871 7234,1854 8.905,4725 H

  2.640,4716

  Subtotal 18.913,2490 15895,3534 34.808,6025

  

Total 34.808,6025 34.808,6025

A.2.7 Steam Boiler (E-201)

  Fungsi: Memanaskan boiler feed water untuk menghasilkan superheated

  steam

  Neraca massa total

C, 1atm

• PC

  TC O

C, 148 atm 565

  Neraca massa komponen Alur 25 = Alur 27 (24) (27)

  F CO = F CO = 8.905,4725 kg/jam

  2

  2 (24) (27)

  F N F N = 21.669,6703 kg/jam

  =

  2

  2 (24) (27)

  F O F O = 1.592,9880 kg/jam

  2 =

  2 (24) (27)

  F H O = F H O = 2.640,4716 kg/jam

  2

  2 Alur 26 = Alur 28 (26) (28)

  F H O = F H O = 24.016,2058 kg/jam

  2

2 Neraca massa total

  Tabel A.7 Neraca massa pada Steam Boiler (E-201)

  Masuk (kg/jam) Keluar (kg/jam) Komponen Alur 24 Alur 26 Alur 28 Alur 27

• O 1.592,9880 1.592,9880

  2

• N 21.669,6703 21.669,6703

  2

• CO 8.905,4725 8.905,4725

  2 H O 2.640,4716 24.016,2058 24.016,2058 2.640,4716

  2 Subtotal 34.808,6025 24.016,2058 24.016,2058 34.808,6025 Total 58.824,8083 58.824,8083

  

LAMPIRAN B

PERHITUNGAN NERACA PANAS

  Basis perhitungan : 1 jam operasi Satuan Operasi : kJ/jam Temperatur referensi :

  25 C (298 K) Kapasitas : 30.000 ton/tahun Perhitungan neraca panas menggunakan rumus sebagai berikut: Perhitungan beban panas pada masing-masing alur masuk dan keluar. _T

  Q = H = (Smith dan Van Ness, 2001)

  n x Cp x dT  T _ref

  Persamaan umum untuk menghitung kapasitas panas adalah sebagai berikut:

  

2

  3 Cp  a  bT  cT  dT _{x , T}

  Jika Cp adalah fungsi dari temperatur maka persamaan menjadi : _{T T 2 2}

  2

  3

  ( )

  CpdT  a  bT  CT  dT dT _{T T T}   _{1 2 1} b c d

  2

  

2

  3

  3

  4

  4

  ( ) ( ) ( ) ( )

  CpdT  a T  T  T  T  T  T  T  T

  2

  1

  2

  1

  2

  1

  2

  1  _{T 1}

  2

  3

  4 Untuk sistem yang melibatkan perubahan fasa persamaan yang digunakan adalah : _{T T T 2 b 2} CpdT  Cp dT   H  Cp dT _{l Vl v T T T}    _{1 1 b} Perhitungan energi untuk sistem yang melibatkan reaksi : _{T T 2 2}

  dQ

  ( )  r  H T  N CpdT  N CpdT _{r out in}

    dt _{T T 1}

₁

• dT
• eT

2 O 1,82964E+01 4,72118E-01 -1,33878E-03 1,31424E-06 0,00000E+00

• dT
• eT

  4 3,8387E+01 -2,3664E-02 2,9098E-04 -2,6385E-07 8,0068E-11 C

  2 H

  6 3,3834E+01 -1,5518E-02 3,7689E-04 -4,1177E-07 1,3889E-10 C

  3 H

  8 4,7266E+01 -1,3147E-01 1,1700E-03 -1,6970E-06 8,1891E-10 C

  4 H

  10 6,6709E+01 -1,8552E-01 1,5284E-03 -2,1879E-06 1,0458E-09

  Kapasitas Panas Padatan, Cp

  (Perry, 2007) Tabel B.3 Kapasitas Panas Padatan (s)

  s

  = a + bT + cT-

  2

  (kal/mol K) Komponen a b c T range (K)

  SiO

  2

  10,87 0,0087 -241.200 273

  CH

  2 1,9022E+01 7,9629E-02 -7,3707E-05 3,7457E-08 -8,1330E-12 H

  2 2,9412E+01 -3,0068E-03 5,4506E-05 5,1319E-09 -4,2531E-12 CO

  4 (J/mol K) Komponen a b c d e

  B.1 Data-Data Kapasitas Panas, Panas Perubahan Fasa, dan Panas Reaksi Komponen

  Tabel B.1 Data Kapasitas Panas Komponen Cair ( J/mol K)

  Kapasitas Panas Cairan, Cp l

  = a + bT + CT

  2

  3

  H

  2 2,9883E+01 -1,1384E-02 4,3378E-05 -3,7006E-08 1,0101E-11 N

  (Perry, 2007) Tabel B.2 Data Kapasitas Panas Komponen Gas ( J/mol K)

  Kapasitas Panas Gas, Cp g

  

= a + bT + CT

  2

  3

  4 (J/mol K) Komponen a b c d e

  O

2 O 3,4047E+00 -9,6506E-03 3,2998E-05 -2,0447E-08 4,3023E-12

• – 848 10,95 0,0055 - 848
• – 1.873 SiC 8,89 0,0029 -284.000 173
• – 1.629 C 2,637 0.0026 -116.900 273
• – 1.373 (Perry, 2007)

Tabel B.4 Data Panas Reaksi Pembentukan P

  f,

  25 C) anas Reaksi Pembentukan (∆H

  Komponen ( kJ/kmol)

  CH -78.451,6774

  4 C H -84.684,0665

  2

  6 C H -103.846,7654

  3

  8 C H -126.147,4607

  4

  10 H O -241.834,9330

  2 CO -393.504,7656

  2 CO -110.541,1580

  SiO -851.385,7800

2 SiC -117.230,4000

  (Perry, 2007)

  B.2 Perhitungan Neraca Panas B.2.1 Pelletizing Machine

  Fungsi : Memperbesar ukuran bahan menjadi bentuk pellet, untuk memperbesar porositas bahan. Asumsi : Selama proses terjadi kenaikan suhu bahan menjadi 40 C.

  10Na O.30SiO .60H _{2 2 2} O

  5 SiO ₂ C SiO ₂

  4

  6 FePO ₄ C

  10Na O.30SiO .60H _{2 2 2} FePO ₄ O

313 303

dQ out in

   N CpdT  N CpdT _{s s}

  

 298   298

dT

a. Menghitung Panas Masuk

  303

  6 SiO : Qi = N . Cp dT

  2 SiO2 SiO2 _{SiO 2} 

  298

  = 85,2960 (kmol/jam). 1840,5824 (J/mol) = 156.994,2867 (kJ/jam)

  303

  6 C : Qi = N .

  Cp dT C C C

   298

  = 256,6805 (kmol/jam). 529,4370 (J/mol) = 135.896,1639 (kJ/mol)

  303

  6 FePO : Qi = N . Cp dT

  4 FePO4 FePO4 _{FePO 4} 

  298

  = 0,5678 (kmol/jam). 474,5705 (J/mol) = 269,4703 kJ/jam

  303

  6 Na O : Qi = N . Cp dT

  2 Na2O Na2O _{Na O 2} 

  298

  = 1,5893 (kmol/jam). 346,4530 (J/mol) = 550,6137 kJ/jam

  303

  7 H O : Qi = N .

  Cp dT

  2 H2O H2O H O ₂ 

  298

  = 9,5179 (kmol/jam). 374,6878 (kJ/mol) = 3.566,2332 kJ/jam

b. Menghitung Panas Keluar

  313

  8 SiO : Qo = N .

  Cp dT

  2 SiO2 SiO2 SiO ₂ 

  298

  = 85,2960 (kmol/jam). 2.301,5507 (J/mol) = 196.313,0313 (kJ/jam)

  313

  8 C : Qo = N . Cp dT C C _C 

  298

  = 256,6805 (kmol/jam). 621,9829 (J/mol) = 159.650,8941 (kJ/mol)

  313

  8 FePO : Qo = N .

  Cp dT

  4 FePO4 FePO4 FePO ₄ 

  298

  = 0,5678 (kmol/jam). 1.461,4185 (J/mol) = 830,4358 kJ/jam

  313

  8 Na O : Qo = N . Cp dT

  2 Na2O Na2O _{Na O 2} 

  298

  = 1,5893 (kmol/jam). 1.050,3585 (J/mol) = 1.668,7526 kJ/jam

  313

  8 H O : Qo = N . Cp dT

  2 H2O H2O _{H O 2} 

  298

  = 9,5179 (kmol/jam). 1.125,7408 (kJ/mol) = 10.714,6635 kJ/jam

  Tabel B. 5 Neraca Energi pada Pelletizing Machine (L-102)

  Komponen Masuk (kJ) Keluar (kJ)

  SiO 156.994,2867 196.313,0313

2 C 135.896,1639 159.650,8941

  Na O 269,4703 1.668,7526

  2 FePO 550,6137 830,4358

  4 H O 3.566,2332 10.714,6635

  2 Jumlah 297.276,7679 369.177,7774

• ∆Hr
• Q 71.901,0095

  Total 369.177,7774 369.177,7774

  B.2.2 Bucket Elevator (C-110)

  Fungsi : Mengangkut bahan baku dari pelletizing machine ke rotary kiln pre-heater . Asumsi : terjadi penurunan suhu bahan menjadi 35 C, selama pengangkutan. _o H ₂ H O ₂

  35 C, 1 atm SiO

_{2 SiO}

_o O ²

  40 C, 1 atm C C

  8

  9 FePO _{4 FePO 4}

  10Na O.30SiO .60H _{2 2 2}

  10Na O.30SiO .60H _{2 2 2} O O

a. Menghitung Panas Masuk

  Panas masuk bucket elevator sama dengan panas keluar pelletizing machine pada alur 8, yaitu = 369.177,7774 kJ/jam.

  Menghitung Panas Keluar b.

  308

  9 SiO : Qo = N .

  2 SiO2 SiO2 Cp dT _{SiO 2} 

  298

  = 85,2960 (kmol/jam). 2.069,7463 (kJ/mol) = 176.541,0481 (kJ/jam)

  308

  9 C : Qo = N .

  Cp dT C C SiO ₂

   298

  = 256,6805 (kmol/jam). 575,1541 (J/mol) = 147.630,8532 (kJ/mol)

  308

  9 FePO : Qo = N . Cp dT

  4 FePO4 FePO4 _{SiO 2} 

  298

  = 0,5678 (kmol/jam). 961,7110 (J/mol) = 546,0781 kJ/jam

  308

  9 Na O : Qo = N .

  Cp dT

  2 Na2O Na2O SiO ₂ 

  298

  = 1,5893 (kmol/jam). 696,5830 (J/mol) = 1.107,0712 kJ/jam

9 H2O .

8 H

  830,4358 546,0781 H

  4

  FePO

  Na

  H

  2

  196.313,0313 176.541,0481 C 159.650,8941 147.630,8532

  Komponen Masuk (kJ/jam) Keluar (kJ/jam) H