Tabel LA.1 Komposisi Sabut Kelapa Sawit NO Komposisi

  

LAMPIRAN A

PERHITUNGAN NERACA MASSA

  Kapasitas produksi = 15000 ton/tahun Satuan massa = kg/jam Waktu operasi pertahun = 330 hari Sehingga kapasitas produksi : ton kg

  1 tahun 1 hari = 15000 x 1000 x x 1.893,9394 kg/jam

  =

  tahun ton 330 hari 24 jam

  

Tabel LA.1 Komposisi Sabut Kelapa Sawit

NO Komposisi %

  1 Selulosa 59,6 0,596

  2 Lignin 28,5 0,285

  3 Lemak 1,9 0,019

  4 Protein 3,6 0,036

  5 Abu 5,6 0,056

  6 Impuritis 0,8 0,008 Untuk basis bahan baku 1000 kg/jam dihasilkan glukosa sebanyak 715,4037 kg/jam, maka untuk kapasitas produksi 15000 ton/tahun di peroleh kapasitas bahan baku sebanyak 2.647,3716 kg/jam. Dengan faktor pengalinya:

  1.893,9394 kg/jam Faktor pengali : 1000 / = 2.647,3716 kg/jam

  x kg jam

  715,4037 kg/jam Berikut ini adalah perhitungan neraca massa pada setiap peralatan proses.

  LA.1 Hammer Mill (HM-01)

  1

  2 F F

  Selulosa H O

  2 Lignin

1 Lemak

  

2

Protein

  Abu Impuritis

  3 F

  Selulosa Lignin

  

3

Lemak

  Protein Abu H O

  2 Impuritis

  1

  2

3 Neraca massa total : F + F = F

  Neraca massa komponen :

  a. Selulosa

  1

  1 F Selulosa = 0,596 x F

  = 0,596 x 2.647,3716 kg/jam = 1.577,8335 kg/jam

  3

  1 F Selulosa = F Selulosa

  = 1.577,8335 kg/jam

  b. Lignin

  1

  1 F Lignin = 0,285 x F

  = 0,285 x 2.647,3716 kg/jam = 754,5009 kg/jam

  3

  1 F Lignin = F Lignin

  = 754,5009 kg/jam

  c. Lemak

  1

  1 F Lemak = 0,019 x F

  = 0,019 x 2.647,3716 kg/jam = 50,3001 kg/jam

  3

  1 F Lemak = F Lemak

  = 50,3001 kg/jam d. Protein

  1

  1 F Protein = 0,036 x F

  = 0,036 x 2.647,3716 kg/jam = 95,3054 kg/jam

  3

  1 F Protein = F Protein

  = 95,3054 kg/jam

  e. Abu

  1

  1 F Abu = 0.056 x F

  = 0,056 x 2.647,3716 kg/jam = 148,2528 kg/jam

  3

  1 F Abu = F Abu

  = 148,2528 kg/jam

  f. Impuritis

  1

  1 F Impuritis = 0,008 x F

  = 0,008 x 2.647,3716 kg/jam = 21,1790 kg/jam

  3

  1 F Impuritis = F Impuritis

  = 21,1790 kg/jam

  g. H O

  2

  2

  1 Air yang masuk pada alur 2 (F ) adalah sebanyak 1:2 dari alur 1 (F ).

  2

1 F H O = ½ x F

  2

  = ½ x 2.647,3716 kg/jam = 1.323,6858 kg/jam

  3

2 F H O = F H O

  2

  2

  = 1.323,6858 kg/jam

2 O - 1.323,6858 1.323,6858

  F

  Komponen Alur Masuk Alur Keluar Alur 1 (kg/jam) Alur 2 (kg/jam) Alur 3 (kg/jam)

  Selulosa 1.577,8335 - 1.577,8335 Lignin 754,5009 - 754,5009

  Lemak 50,3001 - 50,3001 Protein 95,3054 - 95,3054

  Abu 148,2528 - 148,2528 Impuritis 21,1790 - 21,1790

  H

  

Sub Total 2.647,3716 1.323,6858 3.971,0574

Total 3.971,0574 3.971,0574 LA.2 Reaktor Hidrolisa (RH-01)

  Tabel LA.2 Neraca Massa Hammer Mill (HM-01)

  4 HCl 37%

  H

2 O 63%

  2 O n(C

  5 C

  2 O

  H

  Protein Abu

  Lignin Lemak

  3 Selulosa

  2 O F

  H

  Abu Impuritis

  Lemak Protein

  6 Lignin

  12 O

  6 H

  3 F

  6 H

  4

  5

  HCl 37% Reaksi : (C

  6 H

  10 O

  5 )n + H

  4

  3

  Konversi : 99.5% Neraca massa total : F

  6 )

  12 O

  Impuritis

5 Neraca massa komponen :

• F

  3 Lignin

  5 Lignin = F

  a. Lignin F

  = F

  = 754,5009 kg/jam b. Lemak

  5

  3 F Lemak = F Lemak

  = 50,3001 kg/jam

  c. Protein

  5

  3 F Protein = F Protein

  = 95,3054 kg/jam

  d. Abu

  5

  3 F Abu = F Abu

  = 148,2528 kg/jam

  e. Impuritis

  5

  3 F Impuritis = F Impuritis

  = 21,1790 kg/jam

  f. HCl

  4

  1 HCl yang ditambahkan pada alur 4 (F ) sebanyak 10% dari alur 1 (F ).

  4

1 F = 0,1 x F

  = 0,1 x 2.647,3716 kg/jam = 264,7372 kg/jam

  4

  4 F HCl = 37% x F

  = 0,37 x 264,7372 kg/jam = 97,9527 kg/jam

  5

4 F HCl = F HCl

  = 97,9527 kg/jam

  g. Selulosa Mencari BM Selulosa : (C H O )n = (C H O )

  6

  10

  5

  6

  10 5 1000

  Derajat polimerisasi selulosa (n) = 1.000 (Weilen, tanpa tahun) Maka : C = 6 x 1.000 x 12 = 72.000

  H = 10 x 1.000 x 1 = 10.000 O = 5 x 1.000 x 16 = 80.000

   Total = 162000 kg/kmol

  Sehingga BM selulosa = 162.000 kg/kmol

  2 F Selulosa

3 N Selulosa =

  BM Selulosa

  1.577,8335 kg/jam =

  162.000 kg/kmol = 0,00974 kmol/jam

2 X.N Selulosa

  r = ∂

  0,995 x 0,00974 kmol/jam =

  1 = 0,00969 kmol/jam

  Banyaknya air yang masuk ke dalam reaktor hidrolisa :

  4

  4

4 F H O = F – F HCl

  2

  = 264,7372 kg/jam – 97,9527 kg/jam = 166,7844 kg/jam

  4

  3 Total air yang masuk = F H O + F H O

  2

  2

  = 166,7844 kg/jam + 1323,6858 kg/jam = 1490,4702 kg/jam = 82,8039 kmol/jam

  Air yang bereaksi = 0,0097 x 1000 = 9,6910 kmol/jam HCl 95%

  (C H O )n + H O n(C H O )

  6

  10

  5

  2

  6

  12

  6 M 0,00974 82,8039

  R 0,00969 9,6910 9,6910 S 4,86986E-05 73,1129 9,6910

5 F Selulosa = 4,86986E-05 kmol/jam x BM Selulosa

  = 4,86986E-05 kmol/jam x 162000 kg/kmol = 7,8892 kg/jam

5 F H O = 73,1129 kmol/jam x BM H O

  2

  2

  = 73,1129 kmol/jam x 18 kg/kmol = 1.316,0319 kg/jam

  5 F C H O = 9,6910 kmol/jam x BM C H O

  6

  12

  6

  6

  12

  6

  = 9,6910 kmol/jam x 180 kg/kmol = 1.744,3825 kg/jam

  Tabel LA.3 Neraca Massa Reaktor Hidrolisa (RH-01)

  Alur Masuk Alur Keluar Komponen Alur 3 (kg/jam) Alur 4 (kg/jam) Alur 5 (kg/jam)

• Selulosa 1577,8335 7,8892
• Lignin 754,5009 754,5009 Lemak - 50,3001 50,3001
• Protein 95,3054 95,3054
• Abu 148,2528 148,2528
• Impuritis 21,1790 21,1790 H O 1323,6858 166,7844 1316,0319

  2

• C H O

  1744,3825

  6

  12

  6

• HCl 97,9527 97,9527

  

Sub Total 3971,0574 264,7372 4235,7945

Total 4235,7945 4235,7945

LA.3 Cooler (C-01)

  5

  6 F F

  C H O Selulosa

  6

  12

  6 Lignin

  Lignin Lemak

  Lemak Protein

  6 Protein

  5 Abu

  Abu Impuritis

  Impuritis H O H O

  2

  2 HCl

  HCl C H O

  Selulosa

  6

  12

  6 .

  HMF

  5

  6 Neraca massa total : F = F

  Neraca massa komponen :

  a. Selulosa

  6

  5 F Selulosa = F Selulosa

  = 7,8892 kg/jam

  b. Lignin

  6

  5 F Lignin = F Lignin

  = 754,5009 kg/jam

  c. Lemak

  6

  5 F Lemak = F Lemak

  = 50,3001 kg/jam

  d. Protein

  6

  5 F Protein = F Protein

  = 95,3054 kg/jam

  e. Abu

  6

  5 F Abu = F Abu

  = 148,2528 kg/jam

  f. Impuritis

  6

  5 F Impuritis = F Impuritis

  = 21,1790 kg/jam

  g. H O

  2

  6

  5 F H O = F H O

  2

  2

  = 1.316,0319 kg/jam

  h. HCl

  6

  5 F HCl = F HCl

  = 97,9527 kg/jam i. HMF (Hidroksi Metil Furfural)

  Hidroksi metil furfural terbentuk beberapa saat setelah reaksi terjadi yang berasal dari 1% glukosa yang terbentuk. Terbentuknya HMF menyebabkan perubahan warna pada produk.

  6

6 F HMF = 1% x F C H O

  6

  

12

  6

  = 0,01 x 1.744,3825 kg/jam = 17,4438 kg/jam j. C H O

  6

  12

  6

  6

  5

  

6

F C H O = F C H O – F HMF

  6

  12

  6

  6

  12

  6

  = 1.744,3825 kg/jam – 17,4438 kg/jam = 1.726,9387 kg/jam

  6 F

  

Total 4235,7945 4235,7945

LA.4 Filter Press (FP-01)

  

FP - 102

  8

  7

  HCl C

  6 H

  12 O

  6 HMF F

  6 Selulosa

  Lignin Lemak Protein

  Abu Impuritis

  H

  HMF - 17,4438

  8 H

  1744,3825 1726,9387 HCl 97,9527 97,9527

  6

  12 O

  6 H

  C

  2 O 1316,0319 1316,0319

  Impuritis 21,1790 21,1790 H

  Protein 95,3054 95,3054 Abu 148,2528 148,2528

  Selulosa 7,8892 7,8892 Lignin 754,5009 754,5009 Lemak 50,3001 50,3001

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

  Tabel LA.4 Neraca Massa Cooler (C-01)

  2 O

2 O

  Abu Impuritis

  6 HMF

  Asumsi : Semua bahan padatan dapat dipishkan dari cairan, dan cairan yang terikut bersama padatan sebanyak 0.1%. Neraca massa total : F

  6

  = F

  7

  12 O

  H

  HCl C

  6 H

  

7

Selulosa

  Lignin Lemak Protein

  6 HMF F

  

12

O

  6 H

  HCl C

  

2

O

8 Neraca massa komponen :

• F

  a. Selulosa F

  6 Selulosa

  = 7,8892 kg/jam

  7 Selulosa = F b. Lignin

  7

  6 F Lignin = F Lignin

  = 754,5009 kg/jam

  c. Lemak

  7

  6 F Lemak = F Lemak

  = 50,3001 kg/jam

  d. Protein

  7

  6 F Protein = F Protein

  = 95,3054 kg/jam

  e. Abu

  7

  6 F Abu = F Abu

  = 148,2528 kg/jam

  f. Impuritis

  7

  6 F Impuritis = F Impuritis

  = 21,1790 kg/jam

  g. H O

  2

  7

  6 F H O = 0,1% x F H O

  2

  

2

  = 0,001 x 1.316,0319 kg/jam = 1,3160 kg/jam

  8

  6

  7 F H O = F H O – F H O

  2

  2

  

2

  = 1.316,0319 kg/jam – 1,3160 kg/jam = 1.314,7159 kg/jam

  h. HCl

  7

  6 F HCl = 0,1% x F HCl

  = 0,001 x 97,9527 kg/jam = 0,0980 kg/jam

  8

  6

  7 F HCl = F HCl – F HCl

  = 97,9527 kg/jam – 0,0980 kg/jam = 97,8548 kg/jam i. HMF

  7

  6 F HMF = 0,1% x F HMF

  = 0,001 x 17,4438 kg/jam

  = 0,0174 kg/jam

  8

  6

  7 F HMF = F HMF – F HMF

  = 17,4438 kg/jam – 0,0174 kg/jam = 17,4264 kg/jam j. C H O

  6

  12

  6

  7

  6 F C H O = 0,1% x F C H O

  6

  12

  6

  6

  12

  6

  = 0,001 x 1.726,9387 kg/jam = 1,7269 kg/jam

  8

  6

  7 F C H O = F C H O – F C H O

  6

  12

  6

  6

  12

  6

  6

  12

  6

  = 1.726,9387 kg/jam – 1,7269 kg/jam = 1.725,2118 kg/jam

  Tabel LA.5 Neraca Massa Filter Press (FP-01)

  

Alur Masuk Alur Keluar

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

• Selulosa 7,8892 7,8892
• Lignin 754,5009 754,5009 Lemak 50,3001 - 50,3001
• Protein 95,3054 95,3054
• Abu 148,2528 148,2528
• Impuritis 21,1790 21,1790 H O 1.316,0319 1,3160 1.314,7159

  2 C H O 1.726,9387 1,7269 1.725,2118

  6

  12

  6 HCl 97,9527 0,0980 97,8548

  HMF 17,4438 0,0174 17,4264

  Sub Total 4.235,7945 1.080,5857 3.155,2089 Total 4.235,7945 4.235,7945

  LA.5 Reaktor Netralisasi (RN-01)

9 F

  NaOH 37% H O 63 %

  2

  8 F

  9 H O

  2

  8 HCl

  C H O

  6

  12

  6 HMF

  10 F

  H O

  2

  10 NaCl

  C H O

  6

  12

  6 HMF

  Reaksi : HCl + NaOH NaCl + H O

  2 Konversi : 100% terhadap HCl

  8

  9

10 Neraca massa total : F + F = F

  Neraca massa komponen :

  a. C H O

  6

  12

  6

  10

  8 F C H O = F C H O

  6

  12

  6

  6

  12

  6

  = 1.725,2118 kg/jam

  b. HMF

  10

  8 F HMF = F HMF

  = 17,4264 kg/jam

  c. NaOH

  10

  8 F HCl = F HCl – (r.

  ԏ.M) = 97,8548 kg/jam – (r x 1 x 36,458) r = 2,6840 kmol/jam

  10

  9 F NaOH = F NaOH – (r.

  ԏ.M)

  9

  = F NaOH – (2,6840 kmol/jam x 1 x 40 kg/kmol)

9 F NaOH = 107,3617 kg/jam

  9

  9 F NaOH = 37% x F

  9

  107,3617 kg/jam = 0,37 x F

9 F = 290,1667 kg/jam

  d. NaCl

  10

8 F NaCl = F NaCl + (r.

  ԏ.M) = 0 + (2,6840 kmol/jam x 1 x 58,45 kg/kmol) = 156,8822 kg/jam

  e. H O

  2

  10

  8

9 F = F + F

  = 3.155,2089 kg/jam + 290,1667 kg/jam = 3.445,3756 kg/jam

  10

  10

  10

  10

  10 F H O = F – (F NaCl + F C H O + F HMF)

  2

  6

  12

  6

  = 3.445,3756 kg/jam – (156,8822 kg/jam + 1.725,2118 kg/jam + 17,4264 kg/jam) = 1.545,8551 kg/jam

  9

  9

9 F H O = F – F NaOH

  2

  = 290,1667 kg/jam – 107,3617 kg/jam = 182,8050 kg/jam

  Tabel LA.6 Neraca Massa Reaktor Netralisasi (RN-01)

  Alur Masuk Alur Keluar Komponen Alur 8 (kg/jam) Alur 9 (kg/jam) Alur 10 (kg/jam)

  H O 1.314,7159 182,8050 1.545,8551

  2 C H - O 1.725,2118 1.725,2118

  6

  12

  6 HCl 97,8548 - -

• HMF 17,4264 17,4264
• NaOH 107,36
• NaCl

  156,8822

  Sub Total 3.155,2089 290,1667 3.445,3756

Total 3.445,3756 3.445,3756

  LA.6 Dekanter (DK–01)

  10 F

  12 F

  H O

  2 H O

  2 NaCl

  12

  10 C H O

  6

  12

  6 C H O

  6

  12

  6 HMF

  HMF

  

11

  11 F

  H O

  2 NaCl

  C H O

  6

  12

  

6

HMF

  Asumsi : Semua NaCl dapat dipisahkan, tetapi C H O H O dan HMF terikut

  6 12 6,

  2 sebanyak 0.1% bersama NaCl.

  10

  11

12 Neraca massa total : F = F + F

  Densitas dari : O = 1 gr/cm3

  2

  ρH = 1.29 gr/cm3

  ρHMF H O = 1.54 gr/cm3

  ρC

  6

  12

  6

  = 2.16 gr/cm3 ρNaCl

  Neraca massa komponen :

  a. C H O

  6

  12

  6

  11

  10 F C H O = 0,1% x F C H O

  6

  12

  6

  

6

  12

  6

  = 0,001 x 1.725,2118 kg/jam = 1,7252 kg/jam

  12

  10

  11 F C H O = F C H O – F C H O

  6

  12

  6

  6

  12

  6

  6

  12

  6

  = 1.725,2118 kg/jam – 1,7252 kg/jam = 1.723,4866 kg/jam

  b. H O

  2

  11

  10 F H O = 0,1% x F H O

  2

  

2

  = 0,001 x 1.545,8551 kg/jam = 1,5459 kg/jam

  12

  10

  11 F H O = F H O – F H O

  2

  2

  2

  = 1.545,8551 kg/jam - 1,5459 kg/jam = 1.544,3093 kg/jam

  c. HMF

  11

  10 F HMF = 0,1% x F HMF

  = 0,001 x 17,4264 kg/jam = 0.0174 kg/jam

  12

  10

  11 F HMf = F HMF – F HMF

  = 17,4264 kg/jam - 0.0174 kg/jam = 17,4090 kg/jam

  d. NaCl

  11

10 F NaCl = F NaCl

  = 156,8822 kg/jam Tabel LA.7 Neraca Massa Tangki Dekanter (DK-01)

  Alur Masuk Alur Keluar Komponen Alur 10 (kg/jam) Alur 11 (kg/jam) Alur 12 (kg/jam)

  H O 1.545,8551 1,5459 1.544,3093

  2 C H O 1.725,2118 1,7252 1.723,4866

  6

  12

  6 HMF 17,4264 0,0174 17,4090

  NaCl 156,8822 156,8822 -

  

Sub Total 3.445,3756 160,1707 3.285,2048

Total 3.445,3756 3.445,3756 LA.7 Tangki Decolorizing (TDL-01)

13 F

• Karbon Aktif

  13

  12

  14 F F

  H O -H O

  2

  2

  12

  14 C H O -C H O

  6

  12

6 TDL -101

  6

  12

  6 HMF -Karbon Aktif Asumsi : HMF yang menyebabkan warna, habis terserap oleh karbon aktif.

  12

  13

14 Neraca massa total : F + F = F

  Neraca massa komponen :

  a. C H O

  6

  12

  6

  14

  12 F C H O = F C H O

  6

  12

  6

  6

  12

  6

  = 1.723,4866 kg/jam

  b. H O

  2

  14

  12 F H O = F H O

  2

  2

  = 1.544,3093 kg/jam

  c. Karbon aktif

  13 Karbon aktif yang ditambahkan pada alur 13(F ) adalah 2.2% dari alur

  1 1(F ).

  13

  1 F Karbon aktif = 2,2% x F

  = 0,022 x 2647,3716 kg/jam = 58,2422 kg/jam

  14

  13

  12 F Karbon aktif = F Karbon aktif + F HMF

  = 58,2422 kg/jam + 17,4090 kg/jam = 75,6511 kg/jam

  Tabel LA.8 Neraca Massa Tangki Decolorizing (TDL-01)

  

Alur Masuk Alur Keluar

Komponen Alur 12 (kg/jam) Alur 13 (kg/jam) Alur 14 (kg/jam)

• H2O 1.544,3093 1.544,3093 C H O - 1.723,4866 1.723,4866

  6

  12

  6

• HMF 17,4090
• Karbon aktif 58,2422 75,6511

  Sub Total 3.285,2048 58,2422 3.343,4470 Total 3.343,4470 3.343,4470

  LA.8 Filter Press (FP-02)

14 F

  16 F

  H O

  2

  16

  14 H O

  2 C H O

  6

  12

  6 C H O

  6

  12

  6 Karbon Aktif _{FP - 102}

  15

  15 F

  H O

  2 C H O

  

6

  12

  6 Karbon aktif

  Asumsi : Semua karbon aktif dapat dipisahkan, tetapi cairan C H O dan H O

  6

  12

  6

  2 terikut sebanyak 0.1% bersama karbon aktif.

  14

  15

16 Neraca massa total : F = F + F

  Neraca massa komponen:

  a. C H O

  6

  12

  6

  15

  14 F C H O = 0,1% x F C H O

  6

  12

  6

  6

  12

  6

  = 0,001 x 1.723,4866 kg/jam = 1,7235 kg/jam

  16

  14

  15 F C H O = F C H O – F C H O

  6

  12

  6

  6

  12

  

6

  6

  12

  6

  = 1.723,4866 kg/jam- 1,7235 kg/jam = 1.721,7631 kg/jam

  b. H O

  2

  15

  14 F H O = 0,1% x F H O

  2

  2

  = 0,001 x 1.544,3093 kg/jam = 1,5443 kg/jam

  16

  14

  

15

F H O = F H O – F H O

  2

  2

  2

  = 1.544,3093 kg/jam – 1,5443 kg/jam = 1.542,7650 kg/jam

  c. Karbon aktif

  15

14 F Karbon aktif = F Karbon aktif

  = 75,6511 kg/jam Tabel LA.9 Neraca Massa Filter Press (FP-02)

  Alur Masuk Alur Keluar Komponen Alur 14 (kg/jam) Alur 15 (kg/jam) Alur 16 (kg/jam)

  H O 1.544,3093 1,5443 1.542,7650

  2 C H O 1.723,4866 1,7235 1.721,7631

  6

  12

  6 Karbon aktif 75,6511 - 75,6511

Sub Total 3.343,4470 78,9189 3.264,5281

Total 3.343,4470 3.343,4470

  LA.9 Evaporator (EV-01) Fungsi : Untuk memekatkan larutan C H O .

  

6

  12

  6

  17 F

  Uap air

  17

  16

  18 F F

  16 H O

  18 H O

  2

  2 C H O C H O

  6

  12

  6 _{EV - 101}

  6

  12

  6 Kepekatan : 78%

  16

  17

18 Neraca massa total : F = F + F

  Neraca massa komponen :

  a. C H O

  6

  12

  6

  18

16 F C H O = F C H O

  6

  12

  6

  6

  12

  

6

  = 1.721,7631 kg/jam

  b. H O

  2

  18

  16 F H O = 22% x F H O

  2

  2

  = 0.22 x 1.542,7650 kg/jam = 339,4083 kg/jam

  17

  16

  

18

F Uap Air = F H O – F H O

  2

  2

  = 1.542,7650 kg/jam – 339,4083 kg/jam = 1.203,3567 kg/jam

  CR-101

  C

  18 F

  19 H

  2 O

  C

  6 H

  12 O

  6 H

  2 O F

  18 H

  2 O

  6 H

  Tabel LA.10 Neraca Massa Evaporator (EV-01)

  12 O

  18

  6

  

Sub Total 3.264,5281 1.203,3567 2.061,1714

Total 3.264,5281 3.264,5281 LA.10 Crystallizer (CR-01)

  1.721,7631 - 1.721,7631 Uap air - 1.203,3567 -

  6

  12 O

  6 H

  H2O 1.542,7650 - 339,4083 C

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

  19

19 Neraca massa komponen :

  6

  6

  F

  19 H

  2 O = F

  18 H

  2 O – Komposisi H

  2 O dalam C

  6 H

  12 O

  6

  .H

  2 O

  = 339,4083 kg/jam – 172,1763 kg/jam = 167,2320 kg/jam

  2 O =

  6

  .H

  6

  12 O

  6 H

  C

  2 O

  .H

  6

  12 O

  6 H

  a. C

  = F

  Neraca massa total : F

  = 1.893,9394 kg/jam – 1.721,7631 kg/jam = 172,1763 kg/jam

  

12

O

  18

  Komposisi H

  6

  

12

  6

  2

  6

  12

  6 O H C xF

  O H C BM O .H O H C BM = kg/jam 1 x1.721,763

  180 198

  = 1.893,9394 kg/jam

  b. H

  2 O

  2 O dalam C

  6 H

  6 H

  12

  6

  .H

  

2

O :

  = F

  19 C

  6 H

  12 O

  6

  .H

  2 O – F

  18 C

  12 O Tabel LA.11 Neraca Massa Crystallizer (CR-01)

  Alur Masuk Alur Keluar Komponen Alur 18 (kg/jam) Alur 19 (kg/jam)

  H O 339,4083 167,2320

2 C H O - 1.721,7631

  6

  12

  6

• C H O .H O 1.893,9394

  6

  12

  6

2 Total 2.061,1714 2.061,1714

  LA.11 Rotry Dryer (RD-01)

  20 F

  H O

  2

  20

  21

  19 F F

  H O H O

  19 _{RD - 101}

  21

  2

  2 C H O .H O

  C H O

  6

  12

  6

  2

  6

  12

  6 Pengurangan air sampai 89%

  19

  20

  21 Neraca massa total : F = F + F

  Neraca massa komponen :

  a. C H O .H O

  6

  12

  6

  2

  21

19 F C H O .H O = F C H O .H O

  6

  12

  6

  2

  6

  12

  6

  2

  = 1.893,9394 kg/jam

  b. H O

  2

  20

  

19

F H O = 86% x F H O

  2

  2

  = 0.86 x 167,2320 kg/jam = 143,8195 kg/jam

  21

  19

  20 F H O = F H O – F H O

  2

  2

  2

  = 167,2320 kg/jam – 143,8195 kg/jam = 23,4125 kg/jam

  Tabel LA.12 Neraca Massa Rotary Dryer (RD-01)

  Alur Masuk Alur Keluar Komponen Alur 19 (kg/jam) Alur 20 (kg/jam) Alur 21 (kg/jam)

  H2O 167,2320 143,8195 23,4125

• C H O .H O 1.893,9394 1.893,9394

  6

  12

  6

2 Sub Total 2.061,1714 143,8195 1.917,3519

  Total 2.061,1714 2.061,1714

  LA.12 Rotary Cooler (RC-01)

  21

  22 F F

  21

  22 H O

  H O

  2 RC-01

  2 C H O .H O C H O .H O

  6

  12

  6

  2

  6

  12

  6

  2

  21

  22 Neraca massa total : F = F

  Neraca massa komponen :

  a. C H O .H O

  6

  12

  6

  2

  22

  21 F C H O .H O = F C H O .H O

  6

  12

  6

  2

  6

  12

  

6

  2

  = 1.893,9394 kg/jam

  b. H O

  2

  22

  21 F H O = F H O

  2

  2

  = 23,4125 kg/jam Tabel LA.13 Neraca Massa Rotary Cooler (RC-01)

  Alur Masuk Alur Keluar Komponen Alur 21 (kg/jam) Alur 22 (kg/jam)

  H2O 23,4125 23,4125 C H O .H O 1.893,9394 1.893,9394

  6

  12

  6

2 Total 1.917,3519 1.917,3519

  

LAMPIRAN B

PERHITUNGAN NERACA PANAS

  Kapasitas produksi = 15.000 ton/tahun Basis perhitungan = 1 hari produksi (24 jam)

  o

  Suhu refrensi = 25 C = 298 K Basis perhitungan = kJ/jam

  Neraca panas ini menggunakan rumus-rumus perhitungan sebagai berikut :

• Perhitungan panas yang masuk dan keluar

  T ₁ Q = n..Cp.dT (Smith, 2005) ∫

  T ₂ T ₁ F.Cp.dT

  Q = (Reklaitis,1999)

  ∫ T ₂

• Perhitungan panas penguapan

  Q = n.H (Smith, 2005)

  VL

• Untuk sistem yang melibatkan perubahan fasa persamaan yang digunakan adalah

  :

  T T T _{1 b 1}

• Cp.dT = Cp dT Cp dT (Reklaitis, 1983) ΔH

1 VL v

  ∫ ∫ ∫ T T T _{2 1 b}

• Perhitungan energy untuk sistem yang melibatkan reaksi:

  T T _{2 2}

  dQ = + r (T) N Cp.dT − N Cp.dT (Reklaitis, 1983)

• dT

  Δ

  r out out ∫ ∫

  T T _{1 1} LB.1 KAPASITAS PANAS LB.1.1 Kapasitas Panas Gas dan Cairan

  2

  3

  4 Cp = A + BT + CT + DT + ET T ₂

    b c d d

• Cp.dT = a(T − T )

  2

  1

   

  2

  

2

  3

  3

  4

  4

  5

  5 ∫

  2(T − T ) 3(T − T ) 4(T − T ) 5(T − T )

  T

  2

  

1

  2

  1

  2

  1

  2

  1 ₁ 

   Dalam hubungan ini : Cp = Kapasitas panas cairan, J/mol, K T = Suhu, K A,B,C,D = Konstanta

  Tabel LB.1 Kapasitas Panas Gas dan Cairan

  ∑ =

  Na 26,19

  H 7,56 S 12,36 N 18,74

  Nilai Konstribusi Atom ∆E (J/mol.K) C 10,89 O 13,42

  = Jumlah unsur atom i dalam senyawa = Nilai dari kontribusi unsur atom i pada Tabel LB.2 Tabel LB.2 Kontribusi Unsur Atom dengan Metode Hurst dan Harrison

  N i

  S = Kapasitas panas padatan pada 298,15 K (J/mol.K) n = Jumlah unsur atom yang berbeda dalam senyawa

  Dimana : C p

  Cps

  Ni Δ

  n 1 i Ei

  =

  (J/mol.K), menggunakan metode Hurst dan Harrison dengan rumus :

  Senyawa A B C D E

  s

  Perhitungan estimasi kapasitas panas padatan, Cp

  LB.1.2 Estimasi Cp s Dengan Metode Hurst dan Harrison

  17,7227 9,04261E-01 -5,64496E-03 1,13383E-05 - (Reklaitis, 1983)

  (l)

  18,2964 4,72118E-01 -1,3387E-03 1,31424E-06 - HCl

  2 O (l)

  30,3088 -7,60900E-03 1,32608E-05 -4,33363E-09 - H

  (g)

  34,0471 -9,65064E-03 3,29983E-05 -2,04467E-08 4,30228E-12 HCl

  2 O (g)

  H

  (Perry & Green, 1999)

• Selulosa (C
• Glukosa (C
• Lignin (C

  Cp

  ∆E

  H + 3.

  ∆E

  C + 6.

  ∆E

  s = 6.

  6 O 3 )

  Cp

  6 H

  = 247,44 J/mol.K

  s

  = 108,9 + 98,28 + 40,26 Cp

  s

  = 10 (10,89) + 13 (7,56) + 3 (13,42) Cp

  O

  s

  ∆E O Cp

  ∆E H + 1.

  s

  = 26,19 + 7,56 + 13,42 Cp

  s

  = 1 (26,19) + 1 (7,56) + 1 (13,42) Cp

  s

  ∆E O Cp

  ∆E Na + 1.

  = 6 (10,89) + 6 (7,56) + 3 (13,42) Cp

  s = 1.

  Cp

  = 150,96 J/mol.K

  s

  = 65,34 + 45,36 + 40,26 Cp

  s

  s

  ∆E H + 3.

  = 47,17 J/mol.K

  ∆E H + 5.

  s

  = 65,34 + 75,6 + 67,1 Cp

  s

  = 6 (10,89) + 10 (7,56) + 5 (13,42) Cp

  s

  ∆E O Cp

  ∆E C + 10.

  6 H

  s = 6.

  Cp

  5 )

  10 O

  6 H

  Berdasarkan rumus diatas maka didapatlah kapsitas panas padatan T = 298,15.K :

  = 208,04 J/mol.K

  12 O 6 )

  ∆E C + 13.

  = 65,34 + 90,72 + 80,52 Cp

  s = 10.

  Cp

  10 O 2 (OCH 3 ))

  9 H

  = 236,58 J/mol.K

  s

  s

  Cp

  = 6 (10,89) + 12 (7,56) + 6 (13,42) Cp

  s

  ∆E O Cp

  ∆E H + 6.

  ∆E C + 12.

  s = 6.

• Hidroksi Metil Furfural (C
• Natrium Hidroksida (NaOH)

  LB.1.3 Perhitungan Cp dengan menggunakan metode Chuch dan Swanson

  Perhitungan estimasi kapasitas panas (Cp) dengan menggunakan metode Chueh and Swanson, dimana kontribusi elemen atomnya dapat dilihat pada tabel berikut :

  Tabel LB.3 Nilai Gugus pada Perhitungan Cp dengan Metode Chueh and Swanson

Gugus Cp (J/mol.K)

• CH2- 3>CH- 20,92
• COOH 7
• NH

  2 58,58

  (Perry & Green, 1999)
• Protein Cp = 1 ( -NH
• Kapasitas panas abu
• 5
• CH2- -20,64
• OH- -208,04
• CH- 29,89

  Berdasarkan gugus diatas maka didapatlah kapsitas panas T = 298,15.K :

  2 ) +16 ( -CH2-) + 1 ( -COOH) + 1 (-CH-) Cp = 1 (58,58) + 16 (30,38) + 1 (79,91) + 1 (20,92) Cp = 58,58 + 468,08 + 79,91 + 20,92 Cp = 627,49 J/mol.K

  Cp = 0,180 + 7,78 x 10

  T (Reklaitis, 1983) Untuk nilai kapasitas panas (Cp) komponen yang lainnya, dapat dilihat pada tabel dibawah ini :

  Tabel LB.4 Kapasitas panas (Cp), dalam kJ/kg.K

  Komponen Cp (kJ/kg.K)

  Lemak 2,0408 Impuritis 1,9958

  NaCl 0,8698 Karbon aktif 0,7029

• CH

• -O- -132,22

• ∆H

  6 H

  o

  f = 68,29 + 5 (-CH-) + 3 (-OH-) + 2 ( -O-) + 1 ( -CH2-) ∆H

  o

  f = 68,29 + 5 (29,89) + 3 (-208,04) + 2 (-132,22) + 1 (-20,64) ∆H

  o

  f = 68,29 + 149,45 – 624,12 – 264,44 – 20,64 ∆H

  o

  f = -691,46 kJ/kmol.K

  12 O 6 )

  5 )

  ∆H

  o

  f = 68,29 + 6 (-OH-) + 1 (COH) + 4 (-CH- ) + 1 ( -CH2-) ∆H

  o

  f = 68,29 + 6 (-208,04) + 1 (2,09) + 4 (29,89) + 1 (-20,64) ∆H

  o

  f = 68,29 – 1248,24 + 2,09 + 119,56 – 20,64 ∆H

  o

  ∆H

  10 O

  f = 1.078,94 kJ/kmol.K

  Hi = nilai dari unsur i Nilai gugus fungsi dapat dilihat pada tabel di bawah ini :

  LB.1.4 Perhitungan estimasi ∆H f o (kJ. mol -1 ) dengan menggunakan metode Joback yang didasarkan pada kontribusi gugusnya dengan rumus :

  ∆H

  o f

  = 68,29

  ∑ = n

  1 i Ei

  Ni Δ

  Keterangan : n = nilai atom yang terdapat dalam molekul Ni = nilai grup atom i yang terdapat dalam molekul

  Tabel LB.5 Nilai Gugus pada Perhitungan ∆H f

  6 H

  o

  dengan Metode Joback

  Gugus H o f 298,15 (kJ/kmol.K)

  O

  2,09

  (Perry & Green, 1999) Berdasarkan rumus dan gugus diatas maka didapatlah kapsitas panas padatan

  T = 298,15.K :

  o f Selulosa (C

• Glukosa (C

  Untuk nilai panas pembentukan ( 298) dapat dilihat pada tabel di bawah

  f

  ΔH ini :

  o

  Tabel LB.6 Panas Pembentukan ( f 298) ∆H

  o

Komponen f298 (kJ/kmol)

∆H

  H2O -285,8509 HCl -167,4437

  NaOH -469,4448 NaCl -407,1116

  LB.2 PERHITUNGAN NERACA PANAS LB.2.1 Hammer Mill (HM-01) ₁

₂

Q Q ² H O Selulosa P = 1 atm

  Lignin T = 30 ⁰C Lemak

  Protein Abu

  

2

1 Impuritis

  T = 30 ⁰ P = 1 atm ₃ Q Selulosa

  Lignin HM-01 Lemak T = 30

  ⁰C Protein

  

3

P = 1 atm Abu ₂ H O Impuritis T = 30 ⁰C P = 1 atm

  Panas masuk

  Tabel LB.7 Perhitungan Panas Masuk pada Hammer Mill (HM-01)

  

303

  Cp.dT (kJ/kg)

  Alur Komponen F (kg/jam) Q (kJ/jam)

∫

  

298

  Selulosa 1.577,8335 6,4210 1.013,2492 Lignin 754,5009 6,8354 5.157,2846

  1 Lemak 50,3001 10,2038 513,2518 Protein 95,3054 9,1002 867,2980 Tabel LB.7 Perhitungan Panas Masuk pada Hammer Mill (HM-01)…..(Lanjutan) Abu 148,2528 4,2575 631,1914

  Impuritis 21,1790 9,9788 211,3416 H2O 1.323,6858 20,8180 27.556,4878

  Total 45.068,1042 Panas keluar

  Tabel LB.8 Perhitungan Panas Keluar pada Hammer Mill (HM-01)

  Alur Komponen F (kg/jam) ∫

  303 298

  Cp.dT (kJ/kg)

  Q (kJ/jam)

  3 Selulosa 1.577,8335 6,4210 1.013,2492 Lignin 754,5009 6,8354 5.157,2846 Lemak 50,3001 10,2038 513,2518

  Protein 95,3054 9,1002 867,2980 Abu 148,2528 4,2575 631,1914

  Impuritis 21,1790 9,9788 211,3416 H