1 hari produksi (24 jam ) Tabel LA-1 Tabel LA-2 Data Nilai Berat Molekul (Kgmol) Komposisi Alang-alang No Rumus Molekul BM

  Kapasitas bahan baku = ton/tahun =

  129,235 85,839

  0,054 0,036 0,181 0,285

  

LAMPIRAN A

PERHITUNGAN NERACA MASSA

Persentasi

  CHOOH 5,42% 3,60%

  Abu Silika Lignin

  44,28% 18,12%

  432,054 2.384,407 2.384,407 2.384,407 2.384,407

  d. = x = kg/jam 1055,815

  Komposisi

  c. = x = kg/jam

  b. = x = kg/jam

  a. Selulosa = x = kg/jam

  Adapun komponen alang-alang yang digunakan sebagai bahan baku adalah:

  1. Gudang Penyimpanan Alang-Alang Fungsi : Menyimpan persediaan alang-alang.

  46 Satuan dalam kg/jam

  600-1500 0,443 kg/jam

  Abu Silika

  60

  3 COO)

  Alang-alang

  2384,407 18884,501 330

  4

  2 O

  2 CaC

  2 Ca(HCOO)

  Ca(CH

  Lignin Pentosan

  n

  )

  5

  10 O

  6 H

  (C

  Derajat polimerisasi Selulosa

  14

  3 COOH

  Waktu Operasi = hari Basis Perhitungan = 1 hari produksi (24 jam )

  2 O

  2

  7 Ca(OH)

  18

  2 O

  6 H

  126

  .2H

  8 O

  4

  2 O

  2 H

  5 C

  1 162 2 158 3 130 4 128

  Tabel LA-1 Tabel LA-2 Data Nilai Berat Molekul (Kg/mol) Komposisi Alang-alang No Rumus Molekul BM

  74

  2

  13 CH

  2

  90

  4

  2 O

  2 H

  12 C

  44

  11 CO

  32

  98

  4

  2 SO

  10 H

  136

  4

  9 CaSO

1 Alang-alang

  e. = x = kg/jam

  1

  3 Alang-alang

  2. Rotary Cutter Knife

  2

  2.384,407 larutan Ca(OH)

  2

  1 F

  F

  Alang-alang 2.384,407 2.384,407

  F Masuk (kg/jam) Masuk(kg/jam) Keluar(kg/jam) Komponen Komponen

  2.384,407 Memotong-motong alang-alang.

  Keluar (kg/jam) F

  2 Alang-alang

  Pentosan

  681,463 2.384,407 28,58% Alang-alang

  3 Alang-alang 2.384,407 2.384,407

  2 F

  Komponen Masuk(kg/jam) Keluar(kg/jam) F

  Fungsi : Tempat terjadinya reaksi peleburan antara alang-alang dengan Menyimpan alang-alang.

  4. Reaktor Kalsium Oksalat

  Fungsi :

  3. Tangki Penyimpan Alang-alang

  Fungsi :

  1

2 Ca(OH)

  2 O

  7 H

  2 CaC

  4 Ca(CH3COO)

  6 Alang-alang

  2

  O

  2 O

  4

  CO

  2 Ca(HCOO)

  2 H

  50% Alang-alang

  2

  Ca(OH)

  5 Alang-alang

  

2

  

2

O

  1

  5

  2

  3

  4

  6

  7

  8

  9 H

  2 O

  Ca(HCOO)

  2

  5 komponen (C

  6 H

  10 O

  )

  Kondisi Operasi Temperatur Tekanan

  1050

  Ca(OH)

  2 O

  2 CaC2O

  4 Ca(CH

  )

  3 COO)

  2 Indeks

  CO

  Reaksi yang terjadi adalah : 2(C

  C = 1 atm

  o

  98

  : =

2 Humus

• 3150Ca(OH)
• 6825O

  1

  

2

  3.576,610 0,443 2384,407

  Kadar selulosa 2.384,407

  2 O+ 4200CO

  2

  2

  3 COO)

  4

  2 O

  1050CaC

  2

  alang-alang = kg/jam Ca(OH)

  1050

  5

  10 O

  6 H

• 1050Ca(CH
• 1050Ca(HCOO)
• 9450H

  0,003 1788,305

  Laju 0,003

  3150 3150

  723,429

  2 Komposisi bahan masuk:

  2

  4

  F

  = x = x = kg/jam

  F

  1

  4

  = kgmol F

  2

  5

  yang bereaksi = mol bereaksi x x BM = x x 74 = kg/jam

  50% 2.384,407

  50% : alang-alang = 1,5 : 1 Ca(OH)

  1055,815 1055,815

  1055,815 konversi 1055,815 100%

  = kg/jam Derajat polimerisasi : 1050 Asumsi : Konversi 100%

  2

  yang dibutuhkan = x larutan Ca(OH)

  2

  Ca(OH)

  50% = 1,5 x = kg/jam

  2

  = x = x = kg/jam

  6 F = mol bereaksi x 6825 x BM

  3

  = 0,003 x 6825 x

  32 = 677,807 kg/jam

  7 F

  = mol bereaksi x 1050 x BM

  4

  = 0,003 x 1050 x 128 = 417,112 kg/jam

  7 F = mol bereaksi x 1050 x BM

  5

  = 0,003 x 1050 x 158 = 514,873 kg/jam

  7 F = mol bereaksi x 1050 x BM

  6

  = 0,003 x 1050 x 130 = 423,630 kg/jam

  6 F

  = mol bereaksi x 9450 x BM

  7

  = 0,003 x 9450 x

  18 = 527,908 kg/jam

  7 F = mol bereaksi x 4200 x BM

  8

  = 0,003 x 4200 x

  44 = 573,529 kg/jam

  7 F

  = Abu + Silika + lignin + Pentosan

  9

• = 129,235 85,839 432,054 681,463 = 1328,591 kg/jam
• 85,839

  Pentosan Abu

  1055,815 129,235

• Masuk (kg/jam)

  432,054 681,463

• 677,807

  O

  1788,305

  5

  F

  1064,876 Lignin

  6

  4

• Alang-alang

  573,529 1328,591

  2 O

  9

  417,112

  Total

  10 Alang-alang

  2 CaC

  2 O

  4

  H

  Ca(CH

  3576,610 6638,824 1788,305

  3 COO)

  2 Ca(HCOO)

  2 CO

  2 Humus

  677,807

  9 keluar (kg/jam)

  Masuk (kg/jam) F

  2316,213 514,873

• 423,630
• 6638,824 6638,824 F

  2384,407

  F

  417,112 2316,213

  Fungsi : Mendinginkan produk dari reaktor kalsium oksalat 1064,876

  417,112 2316,213

  514,873 423,630

  1328,591

  6065,295

  Humus

  Total

  1328,591

  6065,295

  CaC

  2 O

  4 Ca(HCOO)

  2

  Keluar (kg/jam)

  7

  Selulosa

  Komponen

  Silika

  Komponen F

  10 H

  2 O

  Ca(CH

  3 COO)

  2 Ca(OH)

  2

  514,873 423,630

  Ca(OH) 1064,876

  F

5. TANGKI PENDINGIN

6. Vibrating Screen

  Fungsi : Memisahkan humus dengan CaC O , Ca(OH) , (CH COO) Ca,

  2

  4

  2

  3

  2

  (HCOO) Ca dan H O

  2

  2

  (COO) Ca

  2

  11

  12

  (COO) Ca Ca(CH COO)

  2

  3 Ca(CH COO)

  Ca(HCOO)

  3

  2 Ca(HCOO)

  H O

  2

  2

  Ca(OH)

  13 H O

  2

  2 Ca(OH)

  Humus

  2 Humus

  Indeks Komponen

  2 Ca(OH)

  2

  4 CaC O

  2

  4

  5 Ca(CH COO)

  3

  2

  6 Ca(HCOO)

2 H O

  7

  2

  9 Humus Komposisi Bahan Masuk: Ca(OH) 1064,876 kg = 22,48%

  =

  2 CaC O = 417,112 kg = 8,806%

  2

  4 Ca(CH COO) = 514,873 kg = 10,870%

  3

  2 Ca(HCOO) 423,630 kg = 8,944% =

  2 H O = 2316,213 kg = 48,90%

  2 Total = 4736,703 kg = 100,0%

  Jumlah Humus

  = 1328,591 kg/jam

  Cake yang terikut pada humus = 2 % dari Humus = 2% x 1328,591 = 26,572 kg/jam

  11 Ca(OH) yang terikut di dalam humus = Filtrat yang terikut x F

  2

  2

  = 26,57 x 22,48% = 5,974 kg/jam

2 O

3 COO)

2 O yang terikut di dalam cake = Filtrat yang terikut x F

• Ca(OH)

• = - = kg/jam

  Ca(HCOO)

  H

  

11

  6

  di dalam filtrat = F

  2

  2 Ca dalam cake

  = - = kg/jam

  

11

  3 COO)

  

11

  5

  F

  di dalam cake =

  7

  CaC

  2 O dalam cake

  = - = kg/jam

  2 O

  4

  yang tinggal dalam cake Ca (HCOO) dalam cake

  420,741 Filtrat yang terikut

  8,81% 2,34

  414,772 1.064,876 5,974 1.058,902

  514,873 2,340 512,533 423,630 2,888

  12,993 2.303,219

  417,112 2,340 2.316,213

  26,57 2,89

  12,99 26,57 48,90% 26,57 8,81% 2,340

  10,87% 26,57

2 O

  2

  Ca(CH

  

11

  2

  2

  Ca(HCOO)

  = x = kg/jam

  11

  5

  yang terikut di dalam humus = x F

  Ca(CH

  6

  = x = kg/jam

  11

  4

  yang terikut di dalam humus = Filtrat yang terikut x F

  4

  CaC

  di dalam cake = F4

  yang terikut di dalam humus =Filtrat yang terikut x F

  11

  F

  4

  CaC

  yang tinggal dalam cake = - = kg/jam

  2

  

11

  = x = kg/jam

  2

  di dalam cake =

  2

  Ca(OH)

  = x = kg/jam

  11

  7

  H

• (CH

3 COO)

• = - = kg/jam
• H

2 O di dalam filtrat = F

  2 Ca(HCOO)

  1064,876

  3 COO)

  4 Ca(CH

  2 O

  

16

CaC

  14

  17

  15

  11

  2 H

  2

  Ca(OH)

  2 O Komponen

  4 H

  2 O

  5,974 CaC

  13

  12 F

• F

  2 O

  6065,295 1355,127

  Ca(OH)

  Ca(OH)

  2 O

  2 H

  2 Ca(HCOO)

  3 COO)

  4 Ca(CH

  2 O

  2 CaC

  2 O

  Ca(OH)

  2 H

  2 Ca(HCOO)

  3 COO)

  4 Ca(CH

  2 O

  CaC

  2 O

  2 H

  6065,295 Keluar (kg/jam) F

  6065,295 Masuk (kg/jam) Total

  7. Rotary Vacuum Filter

  2

  H

  2 O

  4 H

  2 O

  2 CaC

  Ca(OH)

  2 O

  dan H

  , Ca(HCOO)

  = x = kg/jam

  2

  3 COO)

  , Ca(CH

  2

  dengan Ca(OH)

  4

  2 O

  Fungsi : Memisahkan CaC

  2 O pencuci = x Jumlah solid masuk

  Ca(CH

  4710,168

  2 Humus 514,873 423,630 417,112

  12,993 2,340 2,888

  4 2,340

  414,772

  2 Komponen

  512,533 420,741 Ca(HCOO)

  414,772 2303,219

  1.058,902

  2316,213 1328,591 1328,591

  103,693 Indeks

  3 COO)

  0,250 0,250

  2

  3 COO)

  7 Ca(CH

  6

  5

  2

  2 Ca(HCOO)

  2

• F

2 O total = F

2 O

  yang terikut di dalam cake= Filtrat yang terikut x F

  7

  2 O yang terikut di dalam cake = Filtrat yang terikut x F

  H

  = x = kg/jam

  14

  6

  Ca(HCOO)

  2

  414,77 Ca(CH

  = x = kg/jam

  14

  5

  yang terikut di dalam cake =Filtrat yang terikut x F

  2

  14

  2 Ca(HCOO)

  3 COO)

  103,693 4,15 9,80%

  1058,902 512,533 420,741 2303,219

  512,533 420,741

  4,15 11,93% 0,49 4,15

  53,62% 24,65% 100,0%

  11,93% 9,80%

  2419,906 1,02 0,41

  414,772 2303,219

• F
• F
• F

  4295,396 100,00%

  4295,396 1058,902

  414,772 24,65%

  4,148 2316,213

  2

2 O = kg/jam =

2 O

  Ca(CH

  14

  = x = kg/jam

  Komposisi Bahan Masuk

  = kg/jam = = kg/jam =

  Filtrat :

  = kg/jam

  4

  CaC

  Cake:

  = + = kg/jam

  Ca(OH)

  16

  7

  14

  7

  = F

  7

  H

  H

  2

  2

  14

  yang terikut di dalam cake = Filtrat yang terikut x F

  2

  Ca(OH)

  Filtrat yang terikut pada cake = 1% dari Cake = 1% x = kg/jam

  = + + + = kg =

  14

  7

  6

  = kg/jam = kg/jam Jumlah Cake=

  14

  5

  14

  2

  = kg Jumlah Filtra= F

  4

  CaC

3 COO)

• Ca(OH)
• Ca(CH

3 COO)

  dalam cake = - = kg/jam

  Ca(HCOO)

  2

  di dalam filtrat = F

  6

  

14

  2

  H

  2

  = F

  7

  

14

  1.057,880 H

  1.058,902 4,15

  512,038 2,22

  dalam cake = - = kg/jam

  3 COO)

  53,62% 0,406 2,224

  2

  420,335 2.300,995

  420,741 4710,168

  = x = kg/jam

  Ca(OH)

  2

  di dalam filtrat = F

  

14

  

14

  2

  yang tinggal dalam cake = - = kg/jam

  Ca (CH

  2

  di dalam filtrat = F

  5

• Ca (HCOO)
• H

2 O di dalam filtrat

  4

  8. Reaktor Asam Oksalat

  1058,902 - 0,495

  Ca(HCOO)

  2

  420,741 - 420,335 0,406 512,533 0,495

  F

  16 F

  17 Ca(CH

  3 COO)

  2

  512,533 - 512,038

  Komponen

  = - = kg/jam

  dengan H

  2 O dalam cake

  2 O

  CaC

  2 O

  4 Keluar (kg/jam) F

  14 F

  15

  414,772 2303,219

  103,693 418,920 103,693 2404,688 2,224

  Masuk (kg/jam)

  1,02250 2.303,219

  Fungsi : Untuk mereaksikan CaC

  1057,880 1,022

  2

• H
• CaSO

  4 Ca(CH

  2 SO

  4 .

  Reaksi yang terjadi adalah: CaC

  2 O

  4

  2 SO

  4 C

  2 H

  2 O

  2 O

• H

  3 COO)

  4394,941 Ca(OH)

  2

  2 SO

  4

  2CH

  3 COOH + CaSO

  4 Ca(HCOO)

• H
• H

  2

  2 SO

  4

  2HCOOH + CaSO

  4

  414,772 - -

  Total

  4

• 2H

  11 Index

  4 CaC

  2 SO

  H

  2 O

  4 H

  HCOOH CaSO

  3 COOH

  4 CH

  2 O

  Ca(OH)

  3 COO)

  2

  14 Komponen

  2

  4 Ca(CH

  4

  5 414,772

  2 H

  12

  13 F

  19 C

  17

  6

  7

  10 317,560

  0,406 2,224 1,022

  414,772 0,495

  Ca(CH

  3 COO)

  2 O

  3 COO)

  18

  4 CH

  2

  2 SO

  4 CaSO

  4

  2 O

  Ca(OH)

  2 CaC

  2 O

  4 Ca(HCOO)

  2 H

  2 O

  C

  2 H

  2 O

  3 COOH

  Ca(HCOO)

  HCOOH CaSO

  4 H

  2 SO

  4 Komposisi Bahan Masuk

  CaC

  4

  = kg/jam = kg/jam

2 O

  Ca(HCOO)2 = kg/jam H

  = kg/jam Ca(OH)

  2

  Ca(OH)

  2 O

  2 H

  2

2 O

  98

  

2

SO

  2

  = kg/jam Asumsi: Konversi 100 %

  = kg/jam = kgmol

  F

  18

  bereaksi = mol H

  2 SO

  4

  x BM H

  4

  17

  = x = kg/jam

  F

  10

  19

  = mol bereaksi x BM C

  2 H

  2 O

  3,240 3,240

  4

  4 Ca(CH

  = 3,240 x

  90 = 291,637 kg/jam

  19 F terbentuk

  = mol bereaksi x BM CaSO

  4

  = 3,240 x 136 = 440,696 kg/jam

  Reaksi 2

  17 F = 0,495 kg/jam

  5

  = 0,003 kgmol

  18 F bereaksi

  mol H SO x BM H SO =

  2

  4

  

2

  4

  = 0,003 x

  98 = 0,307 kg/jam

  19 F = mol bereaksi x BM CH COOH

  11

  3

  = 0,003 x 2 x

  60 = 0,376 kg/jam

  19 F terbentuk

  mol bereaksi x BM CaSO =

  4

  = 0,003 x 136 = 0,426 kg/jam

  Reaksi 3

  17 F = 0,406 kg/jam

  6

  = 0,003 kgmol

  18 F bereaksi

  = mol H SO x BM H SO

  2

  4

  

2

  4

  = 0,003 x

  98 = 0,306 kg/jam

  19 F

  = mol bereaksi x BM HCOOH

  12

  = 0,003 x 2 x

  46 = 0,288 kg/jam

  19 F terbentuk

  = mol bereaksi x BM CaSO

  4

  = x = kg/jam

2 SO

  = 4 N = 2M = 2 x 98 = = = kg/ja = kg/jam kg/jam

  H

  2 SO

  4

  yang disuplai = x H

  2 SO

  4

  yang dibutuhkan = kg/jam

  H

  2 O pada H

  2 SO

  4

  0,306 1,022

  136 136

  4

  0,014 1,354 0,014 1,879 0,014 0,497

  1,200 317,560 0,307

  196 196 gr H

  

4

  /kg air F

  2

  17

  0,003 0,425

  0,014 1,354

  383,433 319,527

  1956,290 383,433 0,196

  untuk reaksi = Reaksi (1+2+3+4) = + + + = kg/jam

  2 SO

  Reaksi 4

  terbentuk = mol bereaksi x BM CaSO

  = kg/jam = kgmol

  F

  18

  bereaksi = mol H

  4

  x BM H

  

2

SO

  4

  = x

  98 = kg/jam

  F

  19

  4

  yang dibutuhkan : H

  = x = kg/jam

  F

  7

  19

  terbentuk = mol Ca(OH)

  2

  x BM H

  2 O

  = x

  36 = kg/jam

  H

  2 SO

  4

2 SO

  

Masuk (kg/jam) Keluar (kg/jam)

Komponen

  17

  18

  19 F F F

  Ca(OH) 1,022 - -

  2 O - CaC 414,772 -

  2

  4 Ca(CH

• COO) 0,495

  3

  2 Ca(HCOO)

• 0,406

  2 H O 2,224 1956,290 1959,012

  2 C H O - - 291,637

  2

  2

  4 CH

  0,376 - - COOH

  3

• HCOOH

  0,288 H SO 383,433 63,905 -

  2

4 CaSO

  443,426 - -

  4

  418,920 2339,723

  Total 2758,643 2758,643

9. Press Filter

  Fungsi : Memisahkan CaSO dengan C H O , CH COOH, HCOOH, H O dan

  4

  

2

  2

  4

  3

  2 H SO

  2 4. o

  Kondisi operasi : Temperatur =

  30 C Tekanan = 1 atm

  C H O C H O

  2

  2

  4

  2

  2

  4

  20

  21 CH COOH

3 CH COOH

  3 HCOOH

  HCOOH CaSO

  H O

  4

  2

  22 H O

  2 CaSO

  4 Komposisi bahan yang masuk :

• Bahan yang keluar dari reaktor

  Komponen Persentase Filtrat Kg/jam

  H O 1959,012 84,61%

  2 C H O

  291,637 12,60%

  2

  2

  4 CH COOH 0,376 0,02%

  3 HCOOH 0,288 0,01%

  H SO 63,905 2,76%

  2

4 Total 2315,217 100,000%

  Komponen Cake: Cake = CaSO

  4

  = 443,426 kg = 19,2%

  Filtrat

  21

• CaSO Jumlah filtra = F

  4

  = 2758,643 443,426 - = 2315,217 kg/jam

  Filtrat yang terikut pada cake= 1 % dari Cake = 1% x 443,426 = 4,434 kg/jam

  C H O keluar

  2

  2

  4

  20 Di dalam cake = Filtrat yang terikut x F

  10

  = 4,434 x 12,60% = 0,559 kg/jam

  20

  22 Di dalam filtrat = F - F

  10

  10

• = 291,637 0,559 = 291,078 kg/jam

  CH COOH keluar

  3

  22 Di dalam cake = Filtrat yang terikut x F

  11

  = 4,434 x 0,02% = 0,001 kg/jam

  20

  22 F - F

  Di dalam filtrat =

  11

  11

• = 0,376 0,001 = 0,375 kg/jam

  HCOOH keluar

  22 F

  Di dalam cake = Filtrat yang terikut x

  12

  = 4,434 x 0,01% = 0,001 kg/jam

  20

22 F - F

  Di dalam filtrat =

  12

  12

  = - 0,288 0,001

  = 0,287 kg/jam

  H O keluar

  2

  20 Di dalam cake = Filtrat yang terikut x F

  7

  = 4,434 x 84,6% = 3,752 kg/jam

  20

22 F - F

  Di dalam filtrat =

  7

  7

  = 1.959,012 3,752 - = 1.955,260 kg/jam

  H SO keluar

  2

  4

  20 Di dalam cake = Filtrat yang terikut x F

  14

  = 4,434 x 2,76% = 0,122 kg/jam

  20

22 F - F

  Di dalam filtrat =

  14

  14

  = - 63,905 0,122 = 63,783 kg/jam

  Masuk (kg/jam) Keluar (kg/jam) Komponen

  20

  21

  22 F F F

  H O 1959,012 1955,260 3,752

  2 C H O 291,637 291,078 0,559

  2

  2

  4 CH COOH 0,376 0,375 0,001

  3 HCOOH 0,288 0,287 0,001

  H SO 63,905 63,783 0,122

  2

  4

  443,426 443,426 - CaSO

  4

  2310,783 447,860

  Total 2.758,643 2.758,643

10. Evaporator

  3 COOH

  2.310,783 F L

  15,39%

  V L L

  V.Xv 84,61% 15,39%

  2310,783

  26 C

  2 H

  2 O

  4 CH

  HCOOH

  355,523 F.Xf L.Xl

  24

  25 H

  2 O

  C

  2 H

  2 O

  4 CH

  3 COOH

  HCOOH

  2310,783 ………….(2)

  Fungsi : Mengurangi kandungan H

  2 O hingga konsentrasi larutan menjadi 30 o

  Komposisi bahan masuk : Air = Solute =

  Be Menghitung % larutan yang dipekatkan: Berdasarkan literatur :

  Diketahui :

  30

  o

  Be = 54,9

  o

  Brix Diuapkan sampai 54,9

  o

  Brix = 54,9 % Solute = 40,6 % air

  = + = +

  V V F.Xf

  Dimana : Xf = (Total filtrat dalam feed/ total feed)x100 % Xl = Filtrat dalam liquid V = Vapor L = Umpan ke evaporator

  Neraca massa (untuk Solute) = + = +

  Neraca Massa Komponen (untuk Solute) = +

  ……..(2) = x L + = L

  L = kg/jam Substitusi ke persamaan (1)

  = + 0,549

  647,584 …………(1)

  647,584

  V.Xf

  V F

  54,90% L.Xl V = 1.663,199 kg/jam H H O masuk H O uap

  O sisa= -

  2

  2

  2

  1955,260 1.663,199 - = = 292,060 kg/jam

  Masuk (kg/jam) Keluar (kg/jam) Komponen

  24

  25

  26 F F F

  H O 1955,260 1663,199 292,060

  2

• C H O 291,078 291,078

  2

  2

4 CH COOH 0,375 - 0,375

  3

• HCOOH 0,287 0,287
• H SO 63,783 63,783

  2

  4

  1663,199 647,584

  Total 2.310,783 2.310,783

10. Kristalizer

  Fungsi : Mengkristalkan asam oksalat anhidrat menjadi asam oksalat dihidrat

  o

  Kondisi Operasi : Temperatur =

  30 C Tekanan = 1 atm

  C H O

  2

  2

  4 C H O

  27

  2

  2

  4

  

28

H O

  2 H O

  2

  impurities Untuk mempermudah hitungan maka CH COOH, HCOOH, H SO disebut sebagai

  

3

  2

  4 impurities.

  Dasar Perhitungan :

o

  1 . Kelarutan asam oksalat pada suhu 0-60 C ditunjukkan dengan persamaan :

  2

• 0,0048 t 3,42 + 0,168 t

  o

  2 . Range suhu kristalisasi adalah 24-32 C 3 . Jenis kristalizer asam oksalat yang digunakan adalah "Cooling Crystalization", (Kirk Othmer vol 16 edisi 3)

  o

  Kelarutan asam oksalat pada 30 12,78 kg/100 kg larutan C adalah =

  Neraca Massa di kristalizer :

  Larutan Kristal F S C

• massa

  100 BM C

  2 H

  2 O

  4

  126 0,451

  0,286 374,393 m

  pelarut

  S BM dihidrat C 0,887

  647,584 m

  pelarut

  asam oksalat

  m

  asam oksalat

  0,449 647,584 12,780 0,887

  126 292,060

  36

  374,393 292,060 100 +

  90 0,286

  292,060 0,887 100 + 12,780

  0,887 0,286 292,060

  106,969 185,091 0,887

  208,745 291,078 0,113 0,714

  12,780 374,393

• (Geankoplis) x = S + C = S + C ..(2)

  258,094 0,100 0,633 33,096 0,100 0,032

  0,887 0,286 291,078 0,113

  .2H

  2 O

  4

  Feed masuk = + = +

  Neraca massa basis air :

  X

  air

  F = +

  BM H

  2 C

  2 O

  4

  .2H

  2 O

  (Geankoplis) x = S + C = S + C ..(1)

  Neraca massa basis asam oksalat :

  X

  asam oksalat

  F = S + C =

  mpelarut + massaasam oksalat

  Eliminasi persamaan (1) dan (2) = S + C ( x 0.113) = S + C ( x 0.887) = S + C = S + C = C

  C = kg (kristal) Substitusi C ke pers (1)

  = S + C = S + x = S + = S

  S = kg/jam (larutan) Total kristal = kg/jam

  0,714 BM H

  2 C

  2 O

  292,060

• 224,998 -0,601

2 O

  63,801

  4 CH

  2 O

  2 H

  647,584 375,038 C

  Masuk (kg)/jam) 647,584

  67,91%

  647,584

  HCOOH H

  272,546 291,078 23,655

  64,445 0,64445

  64,445 63,801

  0,113 208,745 23,655

  2 H2O

  3 COOH

  2 O

  Impurities didalam krist = 1% x impurities masuk = 1% x = kg/jam

  28

  30 C

  2 H

  2 O

  4 CH

  3 COOH

  HCOOH

  2

  29 C

  2 H

  2 O

  4 CH

  3 COOH

  HCOOH H

  4 .

  2 O

  2 H

  0,644

  2 H

  2 O

  4

  = S = x = kg/jam

  Impurities = impurities yang masuk = 0.99 x = kg/jam

  Fungsi : Memisahkan kristal C

  2 H

  

2

O

  4

  .2H

  2 O dari filtratnya

  Larutan terdiri dari : H

  374,393 C

  F

  = S = x = kg/jam

  27 Kristal (F

  28 ) Larutan (F

  28 )

  Impurities H

  2 O 292,060 Total

  185,091 185,091 0,887

  C

  2 H

  2 O

  4

  0,113 64,445

  Komponen Keluar (kg/jam)

  C

11. Centrifuge

2 O

  Komposisi Bahan Masuk: H

  185,091 0,887 208,745

  2 O (l) = kg C

  2 H

  2 O

  99,83% 7,501

4.2 H

  2 H

  2H

  4

  kristal masuk) = - (0.01 x

  )

  = kg/jam dalam filtrat = Impurities kristal yang masuk - Impurities kristal dalam cake = - = kg/jam

  374,393 374,393 23,655 0,644 375,038 272,546

  0,006 0,644

  C

  2 H

  2 O 4.

  2 O dalam kristal yang keluar

  2 C

  3,744 0,638

  3,750 63,801

  2 O = kg

  375,038 375,038

  374,393 370,649

  374,393 370,649 0,644 0,644

  375,038 272,546

  647,584 0,638

  8,68% 0,17%

  23,41%

  2 O

  Impurities dalam kristal yang keluar dalam kristal = impurities kristal yang masuk - (1% x H

  2 O

  kristal yang masuk - (1% x H

  4

  (l) = kg

  Impuritis (s) = kg Impuritis (l) = kg Total Solid = kg Total Liquid = kg kg Jumlah kristal = Kristal

  = kg/jam Jumlah filtrat = larutan

  = kg/jam Filtrat yang terikut kristal = 2 % x cake

  = 0.02 x = kg/jam

  Kristal yang lolos = 1 % x cake = 0.01 x = kg/jam dalam kristal = H

  2 C

  2 O

  4

  2 C

  kristal dalam cake = - = kg/jam

  2 O

  4

  kristal masuk) = - (0.01 x ) = kg/jam dalam filtrat = C

  2 H

  2 O

  4

  kristal yang masuk - C

  2 H

  2 O

  C

  4

2 O dalam larutan yang keluar

2 O larutan yang masuk - H

  Keluar (kg/jam) F

  Impurities dalam larutan = x

  = kg/jam dalam filtrat = Impurities kristal yang masuk - Impurities larutan dalam kristal = - = kg/jam

  H

  2 O

  C

  185,091 5,094 0,651 23,655

  30 (larutan) Larutan

  29 (kristal) F

  Kristal

  larutan dalam kristal = - = kg/jam

  5,094 7,501 23,41%

  Komponen

  0,651 1,756

  179,997

  67,91%

  63,801 23,004

  5,094 - - - 375,038

  28

  Masuk (kg/jam) F

  Impurities dalam larutan yang keluar = kg dalam kristal = Filtrat yang terikut kristal x

  4

  8,68% 179,997

  4

  H

  dalam kristal = Filtrat yang terikut kristal x H

  2 O dalam larutan

  = x = kg/jam dalam filtrat = H

  2 O larutan dalam kristal

  = - = kg/jam

  C

  2 H

  2 O

  dalam larutan yang keluar dalam kristal = Filtrat yang terikut kristal x C

  2 O

  2 H

  2 O

  4

  dalam larutan = x

  = kg/jam dalam filtrat = C

  2 H

  2 O

  4

  lautan yang masuk - C

  2 H

  185,091 63,801 1,756 7,501 7,501

  4

  647,584

  2 H

  2 O

  0,006 0,644 62,045

• 62,045

  265,045

  23,004

  23,655 63,801 0,638

  3,750 7,501 272,546

  .2H

  Total

  0,651 1,756

  370,649 374,393 3,744 - C

  2 H

  371,287

  2 O

  2 O

  4 Impurities 647,584

12. Ball Mill Fungsi : Untuk menghaluskan kristal asam oksalat menjadi berukuran 200 mesh.

  C H O

  2

  2

  4 CH COOH

  3 C H O

  HCOOH

  2

  2

  4

  31

  32 CH COOH

  H O

  3

  2 HCOOH

  C H O

  2

  2

  4

  33 H O

  2 CH COOH

  3 HCOOH

  H O

2 Komposisi bahan keluar dari centrifuse :

  C H O .2H O = 371,287 kg/jam

  2

  2

  4

  2

  = C H O

  0,651 kg/jam

  2

  2

  4

  = H O 5,094 kg/jam

  2

  impurities = 1,756 kg/jam Recycle = 3,826 kg/jam Total = 382,614 kg/jam

  Neraca Massa Overall di Ball Mill : P = B P (produk) = 382,614 kg/jam

  Neraca massa di Vibrating Screen : C = P + A

  …………..(1)

  Asam Oksalat yang ukurannya tidak sesuai spesifikasi (dikembalikan ke ball mill = 1% A = 1% C = 0,010 C

  …………..(2)

  P = 99% C = 0,990 C

  …………..(3)

  Substitusi P = 382,614 ke pers (3) didapat P = 0,990 C 382,614 = 0,990 C C = 386,479 kg/jam A = 0,010 C A = 3,865 kg/jam

• B A = C 3,865 + B = 386,479

  B = 382,614 kg/jam C H O .2H O :

  2

  2

  4

  2

• Dari centrifuse = 371,287
• Recycle dari Vibrating Screen = 1% C = 1% x (P/0.99) = 1% x 371,287

  0,990 = 3,750 kg/jam

• Ke Vibrating Screen = C = (P/0.99) = 371,287

  0,990 = 375,038 kg/jam

  C H O :

  2

  2

  4

• Dari Centrifuse = 0,651
• Recycle dari Vibrating Screen = 1% C = 1% x (P/0.99) = 1% x 0,651

  0,990 = 0,007 kg/jam

• Ke Vibrating Screen = C = (P/0.99) = 0,651

  0,990 = 0,658 kg/jam

  H O :

  2

• Dari centrifuse = 5,094
• Recycle dari Vibrating Screen = 1% C = 1% x (P/0.99) = 1% x 5,094

  0,990 = 0,051 kg/jam

• Ke Vibrating Screen = C = (P/0.99) = 5,094

  0,990 = 5,145 kg/jam

  Impurities :

• Dari centrifuse = 1,756
• Recycle dari Vibrating Screen = 1% C = 1% x (P/0.99) = 1% x 1,756

  0,990 = 0,018 kg/jam

• Ke Vibrating Screen = C = (P/0.99) = 1,756

  0,990 = 1,774 kg/jam

  Keluar (kg/jam) Masuk (kg/jam) Komponen

  31

  33

  32 F F F

  C H O .2H O 371,287 3,750 375,038

  2

  2

  4

  2 H O 5,094 0,051 5,145

  2 Impurities 1,756 0,018 1,774

  C H O 0,651 0,007 0,658

  2

  2

  4

  378,788 3,826

  Total 382,614 382,614

13. Vibrating Screen

2 O

  .2H

  C

  2 O yang keluar :

  33

  34

  dalam H

  3

  HCOOH H

  3 COOH

  .2H

  4 CH

  2 O

  2 H

  3 COOH BP-03

  4

  4 CH

  2 O

  2 H

  1% tidak normal C

  Recycle ke BM

  HCOOH H

  3 COOH

  4 CH

  2 O

  2 H

  C

2 O

  375,038 5,145 375,038 3,750

  2 O

  3,826 0,658 1,774

  Komposisi feed masuk : C

  Fungsi : Untuk memisahkan antara C

  2 H

  2 O

  4

  .2H

  2 O sesuai ukuran dengan

  C

  2 H

  2 O

  4

  .2H

  2 O yang tidak sesuai ukuran.

  2 H

  2 C

  2 O

  4

  .2H

  2 O = kg/jam

  C

  2 H

  2 O

  4

  = kg/jam Impurities = kg/jam H

  2 O = kg/jam

  Feed yang tidak normal = 1% dari feed masuk = 0.01 x = kg/jam

  H

  382,614 375,038

• Ke Ball Mill = 1 % x H

  4

  H

  4

  2 O

  2 C

  dalam H

  4

  2 O

  2 C

  H

  = 0.01 x = kg/jam

  2 O yang masuk

  4

  2 O