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

  

LAMPIRAN A

PERHITUNGAN NERACA MASSA

  Kapasitas produksi = 15737,084 ton/tahun = 1987,006 kg/jam

  Waktu Operasi = 330 hari Basis Perhitungan = 1 hari produksi (24 jam )

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

  1 (C H O ) 162 Abu 0,054

  6

  10 5 n

  2 Ca(CH COO) 158 Silika 0,036

  3

  2

  3 Ca(HCOO) 130 Lignin 0,181

  2 CaC O

  4 128 Pentosan 0,285

  2

  4 H O .2H O

  5 C 126 Derajat polimerisasi 600-1500

  2

  2

  4

  2 O

  6 H

  18 Selulosa 0,443

  2

  7 Ca(OH)

  74

  2

  8 O

  32

  2

  9 CaSO 136

4 SO

  10 H

  98

  2

  4

  11 CO

  44

  2 H O

  12 C

  90

  2

  2

  4 COOH

  13 CH

  60

  3

  14 CHOOH

  46 Satuan dalam kg/jam

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

  1 Alang-alang Alang-alang

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

  a. Selulosa = 44,28% x 1.987,006 = 879,846 kg/jam

  b. Abu = 5,42% x 1.987,006 = 107,696 kg/jam

  c. Silika = 3,60% x 1.987,006 = 71,532 kg/jam

  d. Lignin = 18,12% x 1.987,006 = 360,045 kg/jam

  e. Pentosan = 28,58% x 1.987,006 = 567,886 kg/jam

  2. Rotary Cutter Knife

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

  Fungsi :

  3 Alang-alang

  2

  1.987,006 larutan Ca(OH)

  2

  1 F

  F

  Alang-alang 1.987,006 1.987,006

  1.987,006 Memotong-motong alang-alang.

  2 Alang-alang

  1

  Keluar (kg/jam) F

  Alang-alang

  3 Alang-alang 1.987,006 1.987,006

  2 F

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

  Fungsi : Tempat terjadinya reaksi peleburan antara alang-alang dengan

  4. Reaktor Kalsium Oksalat

  3. Tangki Penyimpan Alang-alang

  1

2 Ca(OH)

  2 O

  7 H

  2 CaC

  

2

O

  6 Alang-alang

  2

  O

  2 O

  4

  CO

  2 Ca(HCOO)

  2 H

  50% Alang-alang

  2

  Ca(OH)

  5 Alang-alang

  2

  4 Ca(CH3COO) Ca(OH)

  2 O

  6 H

  2

  1

  2

  3

  4

  6

  7

  8

  9 H

  2 O

  Ca(HCOO)

  2

  5 komponen (C

  10 O

  Kondisi Operasi Temperatur Tekanan

  5

  )

  1050

  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

  1050CaC

  2 O+ 4200CO

  2

  2.980,508 0,443 1987,006

  Kadar selulosa Laju 1.987,006

  2

  3 COO)

  4

  2 O

  

2

  alang-alang = kg/jam Ca(OH)

  2

  1050

  5

  10 O

  6 H

• 1050Ca(CH
• 1050Ca(HCOO)
• 9450H

  0,003 1490,254

  0,003 3150

  3150 602,857

  2 Komposisi bahan masuk:

  50% : alang-alang = 1,5 : 1 Ca(OH)

  2

  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% 1.987,006

  4

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

  2

  879,846 879,846

  879,846 konversi 879,846 100%

  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 = 564,839 kg/jam

  7 F

  = mol bereaksi x 1050 x BM

  4

  = 0,003 x 1050 x 128 = 347,594 kg/jam

  7 F = mol bereaksi x 1050 x BM

  5

  = 0,003 x 1050 x 158 = 429,061 kg/jam

  7 F = mol bereaksi x 1050 x BM

  6

  = 0,003 x 1050 x 130 = 353,025 kg/jam

  6 F

  = mol bereaksi x 9450 x BM

  7

  = 0,003 x 9450 x

  18 = 439,923 kg/jam

  7 F = mol bereaksi x 4200 x BM

  8

  = 0,003 x 4200 x

  44 = 477,941 kg/jam

  7 F

  = Abu + Silika + lignin + Pentosan

  9

• = 107,696 71,532 360,045 567,886 = 1107,159 kg/jam

  Masuk (kg/jam) Keluar (kg) Komponen

  4

  5

  6

  7 F F F F

• Selulosa 879,846 - - Abu
• 107,696
• Silika 71,532 - -
• Lignin 360,045
• 567,886 - Pentosan - - Ca(OH) 1490,254 887,397

  2

• O

  564,839

• 347,594

• 353,025
• 5532,353 5532,353
• 564,839

  10 Alang-alang

  Memisahkan humus dengan CaC

  2

  H

  2 O

  Ca(CH

  3 COO)

  2 Ca(HCOO)

  2

  1930,177 429,061

  2980,508 5532,353 1490,254

  477,941 1107,159

  2 O

  Alang-alang

  4

  3 COO)

  2

  , (CH

  3 COO)

  2 Ca, Masuk (kg/jam) F

  9 keluar (kg/jam)

  1987,006

  Total

  Ca(OH)

  2

  , Ca(OH)

  4 CO

  Ca(CH

  1107,159

  5. TANGKI PENDINGIN

  Fungsi : Mendinginkan produk dari reaktor kalsium oksalat

  6. Vibrating Screen

  Fungsi : (HCOO)

  2 Ca dan H

  2 O

  887,397 347,594

  1930,177 429,061 353,025

  1107,159

  5054,412

  Humus

  Total

  5054,412

  2 Humus Komponen

  887,397

  F

  10 H

  2 O

  Ca(HCOO)

  2

  347,594 1930,177

  429,061 353,025

  2 CaC

  2 O

  4 CaC

  2 O

  9

  (COO) Ca

  2

  12

  11

  (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 CaC O

  4

  2

  4

  5 Ca(CH COO)

  3

  2

  6 Ca(HCOO)

  2

  7 H O

  2

  9 Humus Komposisi Bahan Masuk: Ca(OH)

  = 887,397 kg = 22,48%

  2 CaC O = 347,594 kg = 8,806%

  2

  4 Ca(CH COO) 429,061 kg = 10,870% =

  3

  2 Ca(HCOO) = 353,025 kg = 8,944%

  2 H O 1930,177 kg = 48,90% =

  2 Total

  3947,253 kg = 100,0%

  =

  Jumlah Humus 1107,159 kg/jam

  =

  Cake yang terikut pada humus = 2 % dari Humus = 2% x 1107,159 = 22,143 kg/jam

  11 Ca(OH) yang terikut di dalam humus F

  = Filtrat yang terikut x

  2

  2

  = 22,14 x 22,48% = 4,978 kg/jam

  11 CaC O yang terikut di dalam humus = Filtrat yang terikut x F

  2

  4

  4

  = 22,14 x 8,81% = 1,950 kg/jam

  11 Ca(CH COO) yang terikut di dalam humus = F

  Filtrat yang terikut x

  3

  2

  5

  = 22,14 x 8,81% = 1,95 kg/jam

  11 Ca(HCOO)

  F yang terikut di dalam humus =Filtrat yang terikut x

  2

  6

  = 22,14 x 10,87% = 2,41 kg/jam

  11 H O yang terikut di dalam cake F

  = Filtrat yang terikut x

  2

  7

  = 22,14 x 48,90% = 10,83 kg/jam

  11 Ca(OH) di dalam cake = F - Ca(OH) yang tinggal dalam cake

  2

  2

  2

• = 887,397 4,978

  

11

F4

  CaC - O di dalam cake = CaC O yang tinggal dalam cake

  2

  4

  2

  4

• = 347,594 1,950 = 345,644 kg/jam

  11 Ca(CH COO) di dalam cake = F - (CH COO) Ca dalam cake

  3

  2

  5

  3

  2

• = 429,061 1,950 = 427,111 kg/jam

  11 Ca(HCOO) di dalam filtrat F

  = - Ca (HCOO) dalam cake

  2

  6

  2

• = 353,025 2,407 = 350,618 kg/jam

  11 H O di dalam filtrat = F - H O dalam cake

  2

  7

  2

  = - 1.930,177 10,828 = 1.919,349 kg/jam

  2 Ca(HCOO)

  11

  3 COO)

  4 Ca(CH

  2 O

  16 CaC

  14

  17

  15

  887,397

  2

  2 H

  Ca(OH)

  Komponen

  345,644

  2 O

  4 H

  2 O

  4,978 CaC

  2 Komponen

• F

  2 O

  12 F

  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

  13 Ca(HCOO)

  5054,412 Keluar (kg/jam) F

  7. Rotary Vacuum Filter

  2

  H

  2 O

  4 H

  2 O

  2 CaC

  Ca(OH)

  2 O

  dan H

  , Ca(HCOO)

  x Jumlah solid masuk = x

  2

  3 COO)

  , Ca(CH

  2

  dengan Ca(OH)

  4

  2 O

  Fungsi : Memisahkan CaC

  2 O pencuci =

  Ca(CH

  5054,412 1129,272

  2 Humus 1107,159 1107,159

  5054,412 Masuk (kg) Total

  3925,140

  10,828 1,950 2,407

  1930,177 1,950

  429,061 353,025 347,594

  427,111 350,618

  345,644 1919,349

  

882,419

  4

  3 COO)

  2

  3 COO)

  7 Ca(CH

  6 Indeks

  5

  0,250 0,250

  2

  2 Ca(HCOO)

  2

• F

2 O total

  2

  CaC

  4

  = kg/jam Jumlah Filtra= F

  53,62% 24,65%

  11,93% 9,80%

  1930,177 345,644

  100,0% 3,46 11,93% 0,41

  86,411 2016,588

2 O

  1919,349 86,411

2 O = kg/jam =

  345,644 427,111 350,618

  2

  Ca(HCOO)

  2

  = kg/jam = kg/jam Jumlah Cake=

  Ca(OH)

  Ca(CH

  16

  = kg/jam H

  = F

  7

  = F

  7

  14

  7

  = + = kg/jam

  H

  Komposisi Bahan Masuk

  Cake:

  CaC

  4

  = kg/jam

  Filtrat :

  = kg/jam = = kg/jam =

  3 COO)

2 O

• F
• F
• F

  = x = kg/jam

  3,46 0,85 0,34

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

  = + + + = kg =

  14

  7

  14

  6

  2

  14

  3,46 9,80% 3,456

  3579,496 100,00%

  882,419 427,111 350,618 1919,349

  5

  14

  2

  Ca(OH)

  345,64 882,419

  Ca(CH

  Ca(HCOO)

  3 COO)

  2

  yang terikut di dalam cake =Filtrat yang terikut x F

  5

  14

  = x = kg/jam

  2

  24,65% 3579,496

  yang terikut di dalam cake= Filtrat yang terikut x F

  6

  14

  = x = kg/jam

  14

  yang terikut di dalam cake = Filtrat yang terikut x F

  2

2 O yang terikut di dalam cake

• Ca(OH)

  882,419

  14

• Ca(CH

  14

  7

  H

  dalam cake = kg/jam

  2

  5

  6

  F

  di dalam filtrat =

  2

  Ca(HCOO)

  dalam cake = - = kg/jam

  2

  3 COO)

  350,279 1.917,496

  14

  2

  di dalam filtrat = F

  Ca (CH

  53,62% 1,853

  3,46 881,567

  H

  = Filtrat yang terikut x F

  7

  14

  = x = kg/jam

  Ca(OH)

  2

  di dalam filtrat = F

  2

  14

  2

  yang tinggal dalam cake = - = kg/jam

3 COO)

  426,698

• Ca (HCOO)
• H

2 O di dalam filtrat = F

• H
• 427,111 0,412

  2 O Ca(CH COO) + H SO

  882,419 345,644

  1.919,349 1,85

  F

  15

  = - = kg/jam

  Masuk (kg/jam)

  0,85208 3925,140

  2 O dalam cake

  881,567 0,852

• 426,698 Ca(CH

  4

  2 O Total

  CaC

  2 O

  4 Keluar (kg/jam) F

  14

• 345,644 1919,349 86,411 2003,907 1,853

  86,411 349,100

  4011,551 4011,551

  2

  2 Ca(OH)

  427,111 0,412 Ca(HCOO)

• H
• CaSO

  2 H

  Fungsi : Untuk mereaksikan CaC

  2 SO

  4 .

  Reaksi yang terjadi adalah: CaC

  2 O

  4

  2 SO

  4 C

  2 O

  2

  4

  4

  3662,451

  F

  17 F

  16 Komponen

  8. Reaktor Asam Oksalat

  350,618 - 350,279 0,339

  3 COO)

  dengan H

  2CH COOH + CaSO

  3

  2

  2

  4

  3

  4 Ca(HCOO) + H SO

  2HCOOH + CaSO

  2

  2

  4

  4 Ca(OH) + H SO CaSO + 2H O

  2

  2

  4

  4

  2 H SO

  2

  4 CaC O

  2

  4 Ca(CH COO)

  18

  3 C H O

  2

  2

  4 Ca(HCOO)

  17

  2 CH COOH

  19

  3 H O

  2 HCOOH

  Ca(OH)

  2 CaSO

  4 H O

  2 Index Komponen

  Ca(OH)

  2

  2

  4 CaC O

  2

  4

  5 Ca(CH COO)

  3

2 Ca(HCOO)

  6

  2

  7 H O C H O

  10

  2

  2

  4

  11 CH COOH

  3

  12 HCOOH CaSO

  13

  4

  14 H SO

  2

4 Komposisi Bahan Masuk

  CaC O = 345,644 kg/jam

  2

  4 Ca(CH COO) = 0,412 kg/jam

  3

  2 Ca(HCOO)2 = 0,339 kg/jam

  H O = 1,853 kg/jam

  2 Ca(OH) = 0,852 kg/jam

  2 Asumsi: Konversi 100 % Reaksi 1

17 F = 345,644 kg/jam

  4

  = 2,700 kgmol

18 F bereaksi

  = mol H SO x BM H SO

  2

  4

  

2

  4

  = 2,700 x

  98 = 264,633 kg/jam

  19 F mol bereaksi x BM C H O

  =

  10

  2

  2

  4

  = 2,700 x

  90 = 243,031 kg/jam

  19 F terbentuk

  = mol bereaksi x BM CaSO

  4

  = 2,700 x 136 = 367,246 kg/jam

  Reaksi 2

  17 F = 0,412 kg/jam

  5

  = 0,003 kgmol

  18 F bereaksi

  = mol H SO x BM H SO

  2

  4

  

2

  4

  = 0,003 x

  98

  19 F = mol bereaksi x BM CH COOH

  11

  3

  = 0,003 x 2 x

  60 = 0,313 kg/jam

  19 F terbentuk

  mol bereaksi x BM CaSO =

  4

  = 0,003 x 136 = 0,355 kg/jam

  Reaksi 3

  17 F = 0,339 kg/jam

  6

  = 0,003 kgmol

  18 F bereaksi

  mol H SO x BM H SO =

  2

  4

  

2

  4

  = 0,003 x

  98 = 0,255 kg/jam

  19 F = mol bereaksi x BM HCOOH

  12

  = 0,003 x 2 x

  46 = 0,240 kg/jam

  F

2 SO

  4

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

  H

  2 SO

  4

  yang disuplai = x

  H

  2 SO

  4

  yang dibutuhkan = kg/jam

  H

  2 O pada H

  2 SO

  0,012 0,415

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

  2 SO

  1,200 136 136

  0,012 F

  2

  17

  0,003 0,354

  0,012 0,012

  0,255 0,852

  1,128 264,633 0,256 196 gr H

  

4

  /kg air 196

  1,128 319,527 266,273

  1630,242 319,527 0,196

  4

  yang dibutuhkan : H

  19

  98 = kg/jam

  terbentuk = mol bereaksi x BM CaSO

  4

  = x = kg/jam

  Reaksi 4

  = kg/jam = kgmol

  F

  18

  bereaksi = mol H

  4

  x BM H

  

2

SO

  4

  = x

  F

  4

  19

  terbentuk = mol bereaksi x BM CaSO

  4

  = x F

  7

  19

  terbentuk = mol Ca(OH)

  2

  x BM H

  2 O

  = x

  36 = kg/jam

  H

  2 SO

2 SO

  

Masuk (kg/jam) Keluar (kg)

Komponen

  17

  18

  19 F F F

  Ca(OH) 0,852 - -

2 CaC

• O 345,644

  2

  4

• Ca(CH 0,412 - COO)

  3

  2

• Ca(HCOO) 0,339

2 H O 1,853 1630,242 1632,510

  2

• C H O

  243,031

  2

  2

4 CH COOH - 0,313 -

  3

• HCOOH

  0,240 - H SO 319,527 - 53,255

  2

4 CaSO

  369,521 - -

  4

  349,100 1949,769

  Total 2298,869 2298,869

  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

  21

  20 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 Laju Filtrat Kg/jam

  H O 1632,510 84,61%

  2 C H O 243,031 12,60%

  2

  2

  4 CH COOH 0,313 0,02%

  3 HCOOH 0,240 0,01%

  H SO 53,255 2,76%

  2

  4

  Total 1929,348 100,000%

  Komponen Cake: Cake = CaSO

  4

  = 369,521 kg = 19,2%

  Filtrat

  21

• CaSO Jumlah filtra = F

  4

  2298,869 - = 369,521 = 1929,348 kg/jam

  Filtrat yang terikut pada cake= 1 % dari Cake = 1% x 369,521 = 3,695 kg/jam

  C H O keluar

  2

  2

  4

  20 Di dalam cake = Filtrat yang terikut x F

  = 3,695 x 12,60% = 0,465 kg/jam

  20

  22 F - F

  Di dalam filtrat =

  10

  10

• = 243,031 0,465 = 242,565 kg/jam

  CH COOH keluar

  3

  22 F

  Di dalam cake = Filtrat yang terikut x

  11

  = 3,695 x 0,02% = 0,001 kg/jam

  20

  22 Di dalam filtrat = F - F

  11

  11

• = 0,313 0,001 = 0,313 kg/jam

  HCOOH keluar

  22 Di dalam cake = Filtrat yang terikut x F

  12

  = 3,695 x 0,01% = 0,000 kg/jam

  20

22 F - F

  Di dalam filtrat =

  12

  12

• = 0,240 0,000 = 0,239 kg/jam

  H O keluar

  2

  20 F

  Di dalam cake = Filtrat yang terikut x

  7

  = 3,695 x 84,6% = 3,127 kg/jam

  20

22 Di dalam filtrat = F - F

  7

  7

• = 1.632,510 3,127 = 1.629,383 kg/jam

  H SO keluar

  2

  4

  20 Di dalam cake = Filtrat yang terikut x F

  = 3,695 x 2,76% = 0,102 kg/jam

  20

22 F - F

  Di dalam filtrat =

  14

  14

  = 53,255 0,102 - = 53,153 kg/jam

  Masuk (kg) Keluar (kg/jam) Komponen

  20

  21

  22 F F F

  H O 1632,510 1629,383 3,127

  2 C H O

  243,031 242,565 0,465

  2

  2

  4 CH COOH 0,313 0,313 0,001

  3 HCOOH 0,240 0,239 0,000

  H SO 53,255 53,153 0,102

  2

4 CaSO 369,521 369,521 -

  4

  1925,653 373,217

  Total 2.298,869 2.298,869

10. Evaporator

  3 COOH

  V 54,90% 15,39%

  V.Xf

  V 1.925,653

  V.Xv 15,39%

  L

  26 C

  2 H

  2 O

  4 CH

  HCOOH

  296,270 …………(1)

  24

  25 H

  2 O

  C

  2 H

  2 O

  4 CH

  3 COOH

  HCOOH

  539,653

  Fungsi : Mengurangi kandungan H

  2 O hingga konsentrasi larutan menjadi 30 o

  Komposisi bahan masuk : 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 F.Xf L

  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)

  = + F

  …………(1) 0,549

  539,653

  V F.Xf L.Xl 1925,653

  ………….(2) F L

  1925,653 L.Xl V = 1.386,000 kg H H O masuk H O uap

  O sisa= -

  2

  2

  2

• = 1629,383 1.386,000 = 243,384 kg

  Masuk (kg) Keluar (kg/jam) Komponen

  24

  25

  26 F F F

  H O 1629,383 1386,000 243,384

  2

• C H O 242,565 242,565

  2

  2

  4 CH 0,313 - COOH

  0,313

  3 HCOOH 0,239 0,239 -

  H SO

• 53,153 53,153

  2

  4

  1386,000 539,653

  Total 1.925,653 1.925,653

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 =

• massa

  2 H

  pelarut

  S BM dihidrat C m

  pelarut

  asam oksalat

  243,384

  36 F S C 243,384

  0,032 0,887

  539,653 0,113

  BM C

  2 O

  4

• 187,498
• 0,601 311,994 100 + 12,780 m

  173,954 242,565

  0,714 BM H C O

  .2H O 0,286 311,994

  243,384

  90 126

  243,384 0,887 100 + 12,780

  0,887 0,286 243,384

  89,141 154,242 0,887 215,078 0,100 0,633

  27,580 0,100 0,887 0,286

  539,653 12,780 0,449 m

  0,887 0,714

  asam oksalat

  2 O

  Neraca Massa di kristalizer : Feed masuk = +

  = + Neraca massa basis air :

  X

  air

  F = + BM H

  2 C

  2 O

  4

  .2H

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

  126 100 0,451 Larutan Kristal

  Neraca massa basis asam oksalat :

  X

  asam oksalat

  F = S + C

  = mpelarut + massaasam oksalat + (Geankoplis) x = S + C

  = S + C ..(2) Eliminasi persamaan (1) dan (2)

  = S + C ( x 0.113) = S + C ( x 0.887) = S + C = S + C = C

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

  = S + C = S + x = S + = S

  S = kg/jam (larutan) 0,286

  242,565 0,113

2 O = S

• 539,653 312,531 227,122 242,565 19,712

  HCOOH H

  0,113 173,954 19,712

  53,704 53,167 0,113

  C

  2 H

  2 O

  4 539,653 Total

  539,653

  53,167 C

  2 H

  2 O

  4 CH

  3 COOH

  28

  2 O

  4 .

  30 C

  2 H

  2 O

  4 CH

  3 COOH

  HCOOH H

  29 C

  2 H

  2 O

  4 CH

  3 COOH

  HCOOH H

  2 H2O

  2 O

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

  4

  = 1% x = kg/jam

  Larutan terdiri dari : H

  = x = kg/jam

  C

  2 H

  2 O

  4

  = S = x = kg/jam

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

  11. Centrifuge

  Fungsi : Memisahkan kristal C

  2 H

  

2

O

  .2H

  2 H

  0,887 173,954 53,704

  311,994 C

  Komponen

  154,242 Impurities 0,537 53,704

  0,887 154,242

  0,53704 311,994

  28 )

  2 O dari filtratnya

  28 ) Larutan (F

  27 Kristal (F

  F

  243,384

  2 O

  H

2 O

2 O

  Komposisi Bahan Masuk: H

  2 O (l)

  = kg/jam C

  2 H

  2 O

  311,994 154,242

  = kg/jam C

4.2 H

  2 O

  2H

  4

  kristal masuk) = - (0.01 x

  )

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

  = - 311,994

  19,712 0,537 312,531 227,122 539,653

  C

  2 H

  2 O 4.

  2 O dalam kristal yang keluar

  2 C

  2 O

  3,120 311,994 308,874

  0,532 0,532

  227,122 8,68% 0,17%

  23,41% 3,125 311,994 308,874

  0,537 0,537 0,537

  53,167 67,91% 99,83%

  6,251 312,531

  312,531

  2 O

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

  4

  2 C

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

  = kg/jam Jumlah filtrat = larutan

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

  = kg/jam Kristal yang lolos = 1 % x cake

  = 0.01 x = kg/jam dalam kristal = H

  2 C

  2 O

  4

  kristal yang masuk - (1% x H

  2 O

  kristal dalam cake = - = kg/jam

  4

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

  2 H

  2 O

  4

  kristal yang masuk - C

  2 H

  2 O

  4

  2 H

2 O dalam larutan yang keluar

2 O larutan yang masuk - H

  6,251 23,41% 19,170

  149,997 311,994

  0,005 0,537

  154,242 4,245 0,542

  0,532 C

  30 (larutan) Larutan

  29 (kristal) F

  Kristal Keluar (kg/jam) F

  .2H

  28 Komponen

  2 O

  0,542 1,463

  Kristal

  154,242 53,167 1,463

  Larutan

  6,251 6,251

  0,005 4,245

  0,542 1,463

  19,712 8,68%

  149,997

  67,91%

  4 Impurities

  2 H

  4,245 3,120 - 308,874 3,125

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

  = kg/jam 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

  4

  2 H

  C

  2 O

  4

  dalam larutan = x

  = kg/jam dalam filtrat = C H O lautan yang masuk - C H O larutan dalam kristal

  = - = kg/jam

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

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

  = - = kg/jam

  

Kristal Larutan

  H

  2 O

  51,704

• 51,704

  220,871

  539,653

  309,406

  312,531 19,170

  2 O Masuk (kg/jam) F

  4

  2 O

  2 H

  227,122 539,653

  Total

  6,251 19,712 53,167

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 = 309,406 kg/jam

  2

  2

  4

  2

  = C H O

  0,542 kg/jam

  2

  2

  4

  = H O 4,245 kg/jam

  2

  impurities = 1,463 kg/jam Recycle = 3,188 kg/jam

  Neraca Massa Overall di Ball Mill : P = B P (produk) = 318,845 kg

  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 = 318,845 ke pers (3) didapat P = 0,990 C 318,845 = 0,990 C C = 322,066 kg/jam A = 0,010 C A = 3,221 kg/jam

• B A = C 3,221 + B = 322,066

  B = 318,845 kg/jam C H O .2H O :

  2

  2

  4

  2

• Dari centrifuse = 309,406
• Recycle dari Vibrating Screen = 1% C = 1% x (P/0.99) = 1% x 309,406

  0,990 = 3,125 kg/jam

• Ke Vibrating Screen = C = (P/0.99) = 309,406

  0,990 = 312,531 kg/jam

  C H O :

  2

  2

  4

• Recycle dari Vibrating Screen = 1% C = 1% x (P/0.99) = 1% x 0,542

  0,990 = 0,005 kg/jam

• Ke Vibrating Screen = C = (P/0.99) = 0,542

  0,990 = 0,548 kg/jam

  H O :

  2

• Dari centrifuse = 4,245
• Recycle dari Vibrating Screen = 1% C = 1% x (P/0.99) = 1% x 4,245

  0,990 = 0,043 kg/jam

• Ke Vibrating Screen = C = (P/0.99) = 4,245

  0,990 = 4,288 kg/jam

  Impurities :

• Dari centrifuse = 1,463
• Recycle dari Vibrating Screen = 1% C = 1% x (P/0.99) = 1% x 1,463

  0,990 = 0,015 kg/jam

• Ke Vibrating Screen = C = 1,463

  0,990 = 1,478 kg/jam

  

Masuk (kg/jam) Keluar (kg)

Komponen

  31

  33

  32 F F F

  C H O .2H O 309,406 3,125 312,531

  2

  2

  4

  2 H O 4,245 0,043 4,288

  2 Impurities 1,463 0,015 1,478

  C H O 0,542 0,005 0,548

  2

  2

  4

  315,657 3,188

  Total 318,845 318,845

13. Vibrating Screen

  H O .2H O sesuai ukuran dengan Fungsi : Untuk memisahkan antara C

  2

  2

  4

  2 C H O .2H O yang tidak sesuai ukuran.

  2

  2

  4

  2

2 O

  318,845 312,531

  2 H

  4 CH

  2 O

  2 H

  C

  3 COOH BP-03

  4 CH

  2 O

  1% tidak normal C

  3 COOH

  Recycle ke BM

  HCOOH H

  3 COOH

  4 CH

  2 O

  2 H

  C

  312,531 4,288 312,531 3,125

  .2H

  HCOOH H

2 O

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

  Komposisi feed masuk : C H O

  .2H O = kg/jam C

  2 H

  2 O

  4

  = kg/jam Impurities = kg/jam H

  2 O = kg/jam

  3,188 0,548 1,478

  3

  2 C

  2 O

  4

  .2H

  2 O yang keluar :

  33

  34

  dalam H

  H

• Ke Ball Mill = 1 % x H
• Ke storage = H

  .2H

  2 O

  H

  2 O yang masuk -

  .2H

  4

  2 O

  2 C

  dalam H

  4

  2 C

  2 O

  = 0.01 x = kg/jam

  2 O yang masuk

  4

  2 O

  2 C

  4

  2 O

  

2

C

  2 C

  4

  2 O ke ball mill

  2 O yang keluar :

  = - = kg/jam

  2 C

  2 O

  4

  3,125 309,406

  Impurities dalam H

  2 C

  2 O

  dalam H

  4

  .2H

• Ke Ball Mill = 1 % x Impurities dalam H

  2 C

  2 O

  .2H

  2 O yang masuk

  = 0.01 x = kg/jam

  1,478 0,015

  4

• Ke Storage = Impurities dalam H C O .2H

  O yang masuk -

  2

  2

  4

  2 Impurities dalam H C O .2H O ke ball mill

  2

  2

  4

  2

  1,478 0,015 - = = 1,463 kg/jam

  H C O yang keluar :

  2

  2

4 C O yang masuk

• Ke Ball Mill = 1 % x H

  2

  

2

  4

  = 0.01 x 0,548 = 0,005 kg

  H C O C O d ke ball mill

• Ke Storage = yang masuk - H

  2

  2

  4

  2

  2

  4

  = 0,548 - 0,005 = 0,542 kg/jam

  H O yang keluar :

2 O yang masuk

• Ke Ball Mill = 1 % x H

  2

  = 0,043 kg/jam

• Ke storage = H O yang masuk - H O ke ball mill

  2

  2