Kapasitas produksi = ton/tahun =
Waktu Operasi = 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 1 162 2 158 3 130 4 128 5 C2H2O4.2H2O 126 6 H2O 18 7 Ca(OH)2 74 8 O2 32 9 CaSO4 136 10 H2SO4 98 11 CO2 44 12 C2H2O4 90 13 CH3COOH 60 14 46
Satuan dalam kg/jam
1. Gudang Penyimpanan Alang-Alang
Fungsi : Menyimpan persediaan alang-alang.
Adapun komponen alang-alang yang digunakan sebagai bahan baku adalah:
a. Selulosa = x = kg/jam b. = x = kg/jam c. = x = kg/jam d. = x = kg/jam e. = x = kg/jam 879,846 107,696 71,532 360,045 567,886 1.987,006 1.987,006 1.987,006 1.987,006 1.987,006 44,28% 18,12% Abu Silika Lignin CHOOH 5,42% 3,60% 28,58% Pentosan
LAMPIRAN A
PERHITUNGAN NERACA MASSA
Persentasi 0,054 0,036 0,181 0,285 600-1500 0,443 kg/jam Komposisi Abu Silika Lignin Pentosan Derajat polimerisasi Selulosa (C6H10O5)n Ca(CH3COO)2 Ca(HCOO)2 CaC2O4 1987,006 15737,084 330 Alang-alang 1 1 Alang-alang
2. Rotary Cutter Knife Fungsi :
3. Tangki Penyimpan Alang-alang Fungsi :
4. Reaktor Kalsium Oksalat
Fungsi : Tempat terjadinya reaksi peleburan antara alang-alang dengan Menyimpan alang-alang.
Komponen Masuk (kg/jam) Keluar (kg/jam)
F2 F3 Alang-alang 1.987,006 1.987,006 Alang-alang Keluar (kg/jam) F1 1.987,006 Memotong-motong alang-alang. F Masuk (kg/jam) Masuk(kg/jam) Keluar(kg/jam) Komponen Komponen Alang-alang 1.987,006 1.987,006 F1 F2 1.987,006 larutan Ca(OH)2 Ca(OH)2 CaC2O4 Ca(CH3COO)2 Ca(HCOO)2 H2O CO2 5 5 3 Alang-alang Ca(OH)2 50% Alang-alang 2 Alang-alang 1 4 7 H2O O2 6 3 Alang-alang 3 Alang-alang 2
: = 98 oC
= 1 atm
Reaksi yang terjadi adalah :
2(C6H10O5)1050 + 3150Ca(OH)2 + 6825O2
1050CaC2O4 + 1050Ca(CH3COO)2 + 1050Ca(HCOO)2 + 9450H2O+ 4200CO2 Komposisi bahan masuk:
alang-alang = kg/jam
Ca(OH)2 50% : alang-alang = 1,5 : 1
Ca(OH)2 50% = 1,5 x
= kg/jam
Ca(OH)2 yang dibutuhkan = x larutan Ca(OH)2
= kg/jam Derajat polimerisasi : 1050 Asumsi : Konversi 100% F14 = x = x = kg/jam F1 4 = x = x = kg/jam = kgmol
F25 yang bereaksi = mol bereaksi x x BM
= x x 74
= kg/jam
Kondisi Operasi Temperatur Tekanan 50% 1.987,006 Ca(OH)2 O2 879,846 879,846 879,846 konversi 879,846 100% 1 2 3 4 6 7 8 9 H2O Ca(HCOO)2 2.980,508 0,443 1987,006
Kadar selulosa Laju
1.987,006 5 komponen (C6H10O5)1050 CaC2O4 Ca(CH3COO)2 Indeks CO2 Humus 0,003 1490,254 0,003 3150 3150 602,857
F3 6 = mol bereaksi x x BM = x x 32 = kg/jam F47 = mol bereaksi x x BM = x x = kg/jam F5 7 = mol bereaksi x x BM = x x = kg/jam F67 = mol bereaksi x x BM = x x = kg/jam F76 = mol bereaksi x x BM = x x 18 = kg/jam F87 = mol bereaksi x x BM = x x 44 = kg/jam F97 = = + + + = kg/jam Ca(OH)2 Selulosa -1490,254 Silika Komponen
Abu + Silika + lignin + Pentosan
360,045 F4 -1107,159 0,003 0,003 0,003 4200 4200 F5 107,696 0,003 0,003 0,003 1050 9450 6825 6825 -1050 1050 1050 1050 1050 -Keluar (kg) F7 -887,397 -Lignin Pentosan Abu -879,846 107,696 567,886 71,532 9450 128 158 F6 -130 564,839 347,594 429,061 353,025 439,923 477,941 71,532 360,045 567,886 Masuk (kg/jam)
-5. TANGKI PENDINGIN
Fungsi : Mendinginkan produk dari reaktor kalsium oksalat
6. Vibrating Screen Fungsi : (HCOO)2Ca dan H2O 887,397 347,594 1930,177 429,061 353,025 1107,159 5054,412 Humus Total 1107,159 5054,412 887,397 F10 H2O Ca(HCOO)2 347,594 1930,177 429,061 353,025 -O2 CaC2O4 CaC2O4 CO2 Humus Komponen Ca(CH3COO)2 564,839 -347,594 -H2O Ca(CH3COO)2 Ca(HCOO)2 1930,177 429,061 2980,508 5532,353 1490,254 -- 353,025 477,941 1107,159
Memisahkan humus dengan CaC2O4, Ca(OH)2, (CH3COO)2Ca, Masuk (kg/jam) F9 keluar (kg/jam) -5532,353 5532,353 1987,006 Total Ca(OH)2 -564,839 Alang-alang 10 Alang-alang 9
Ca(OH)2 CaC2O4 H2O Ca(OH)2 = kg = CaC2O4 = kg = = kg = = kg = H2O = kg = = kg = Jumlah Humus =
Cake yang terikut pada humus =
= 2% x =
Ca(OH)2 yang terikut di dalam humus = Filtrat yang terikut x F211
= x
=
CaC2O4 yang terikut di dalam humus = Filtrat yang terikut x F411
= x = Ca(CH3COO)2 Ca(CH3COO)2 Indeks 2 4 5 6 kg/jam kg/jam kg/jam kg/jam Ca(HCOO)2 7 9 Humus
Komposisi Bahan Masuk:
1107,159 2 % dari Humus 429,061 8,944% 48,90% 100,0% 22,48% 8,806% 10,870% 3947,253 887,397 1,950 1930,177 22,14 22,14 8,81% 1107,159 22,48% 4,978 22,143 347,594 353,025 Komponen Humus Ca(HCOO)2 Total 13 11 12 (COO)2Ca Ca(CH3COO) Ca(HCOO)2 H2O Ca(OH)2 Humus (COO)2Ca Ca(CH3COO) Ca(HCOO)2 H2O Ca(OH)2
Ca(CH3COO)2 yang terikut di dalam humus = x F511
= x
=
Ca(HCOO)2 yang terikut di dalam humus =Filtrat yang terikut x F611
= x
=
H2O yang terikut di dalam cake = Filtrat yang terikut x F7 11
= x
=
Ca(OH)2 di dalam cake = F211 - Ca(OH)2 yang tinggal dalam cake
=
-= CaC2O4 di dalam cake = F4
11
-=
-=
Ca(CH3COO)2 di dalam cake = F511 - (CH3COO)2Ca dalam cake
=
-= Ca(HCOO)2 di dalam filtrat = F6
11
-=
-=
H2O di dalam filtrat = F711 - H2O dalam cake
= -= kg/jam kg/jam kg/jam kg/jam kg/jam kg/jam
CaC2O4 yang tinggal dalam cake
Ca (HCOO)2 dalam cake kg/jam kg/jam 10,828 887,397 4,978 882,419 429,061 1,950 1.930,177 427,111 353,025 2,407 347,594 1,950 10,87% 22,14 8,81% 1,95 22,14 2,41 350,618
Filtrat yang terikut
345,644 1.919,349 48,90% 10,83 22,14
7. Rotary Vacuum Filter
Fungsi : Memisahkan CaC2O4 dengan Ca(OH)2, Ca(CH3COO)2, Ca(HCOO)2 dan H2O
Ca(OH)2
CaC2O4
H2O
H2O pencuci = x Jumlah solid masuk
= x Ca(CH3COO)2 Ca(HCOO)2 0,250 0,250 5 6 Indeks 7 Ca(CH3COO)2 4 2 Humus 1107,159 1107,159 882,419 345,644 1919,349 427,111 350,618 429,061 353,025 347,594 1930,177 1,950 10,828 1,950 2,407 3925,140 5054,412 Masuk (kg) Total 5054,412 1129,272 5054,412 Keluar (kg/jam) F12 F13 Ca(HCOO)2 Komponen 4,978 CaC2O4 H2O 345,644 Komponen Ca(OH)2 -F11 887,397 15 17 14 16 CaC2O4 Ca(CH3COO)2 Ca(HCOO)2 H2O Ca(OH)2 H2O CaC2O4 Ca(CH3COO)2 Ca(HCOO)2 H2O Ca(OH)2 CaC2O4 Ca(CH3COO)2 Ca(HCOO)2 H2O Ca(OH)2
= kg/jam H2O total = F7 = F714 + F716
= +
= kg/jam
Komposisi Bahan Masuk Cake: CaC2O4 = kg/jam Filtrat : = kg/jam = = kg/jam = H2O = kg/jam = Ca(OH)2 = kg/jam = kg/jam
Jumlah Cake= CaC2O4 = kg/jam
Jumlah Filtrat= F2 14 + F5 14 + F6 14 + F7 14 = + + + = kg =
Filtrat yang terikut pada cake = 1% dari Cake
= 1% x
= kg/jam
Ca(OH)2 yang terikut di dalam cake = Filtrat yang terikut x F214
= x
= kg/jam
Ca(CH3COO)2 yang terikut di dalam cake =Filtrat yang terikut x F514
= x
= kg/jam
Ca(HCOO)2 yang terikut di dalam cake= Filtrat yang terikut x F614
= x = kg/jam Ca(CH3COO)2 24,65% 3579,496 345,64 882,419 Ca(HCOO)2 345,644 427,111 350,618 1919,349 86,411 86,411 2016,588 100,0% 3,46 11,93% 0,41 1930,177 345,644 3,46 0,85 0,34 11,93% 9,80% 53,62% 24,65% 3,46 9,80% 3,456 3579,496 100,00% 882,419 427,111 350,618 1919,349
H2O yang terikut di dalam cake = Filtrat yang terikut x F714
= x
= kg/jam
Ca(OH)2 di dalam filtrat = F214 - Ca(OH)2 yang tinggal dalam cake
=
-= kg/jam
Ca (CH3COO)2 di dalam filtrat = F514 - Ca(CH3COO)2 dalam cake
=
-= kg/jam
Ca(HCOO)2 di dalam filtrat = F614 - Ca (HCOO)2 dalam cake
= -= kg/jam H2O di dalam filtrat = F7 14 - H2O dalam cake = -= kg/jam
8. Reaktor Asam Oksalat
Fungsi : Untuk mereaksikan CaC2O4 dengan H2SO4. Reaksi yang terjadi adalah:
CaC2O4 + H2SO4 C2H2O4 + CaSO4 350,618 3662,451 -427,111 0,412 F17 F16 Komponen 426,698 - 345,644 1919,349 86,411 2003,907 1,853 350,618 - 350,279 0,339 - 426,698 Ca(CH3COO)2 427,111 0,412 Ca(HCOO)2 Ca(OH)2 882,419 345,644 1.919,349 1,85 F15 882,419 Masuk (kg/jam) 0,85208 3925,140 350,279 1.917,496 3,46 881,567 881,567 0,852 53,62% 0,339 1,853 -H2O Total CaC2O4 Keluar (kg/jam) F14 86,411 349,100 4011,551 4011,551
Ca(CH3COO)2 + H2SO4 2CH3COOH + CaSO4
Ca(HCOO)2 + H2SO4 2HCOOH + CaSO4 Ca(OH)2 + H2SO4 CaSO4 + 2H2O Ca(OH)2 CaC2O4 Ca(HCOO)2 H2O C2H2O4 CH3COOH HCOOH CaSO4 H2SO4
Komposisi Bahan Masuk
CaC2O4 = kg/jam = kg/jam Ca(HCOO)2 = kg/jam H2O = kg/jam Ca(OH)2 = kg/jam Asumsi: Konversi 100 % Reaksi 1 = kg/jam = kgmol F18 bereaksi = mol H2SO4 x BM H2SO4 = x = kg/jam 7 10 F4 17 14 Komponen 11 13 Ca(CH3COO)2 264,633 0,339 12 345,644 2,700 2,700 2 4 5 6 345,644 0,412 Ca(CH3COO)2 Index 1,853 0,852 98 18 17 19 C2H2O4 CH3COOH HCOOH CaSO4 H2O H2SO4 CaC2O4 Ca(CH3COO) Ca(HCOO)2 H2O Ca(OH)2
F1019 = mol bereaksi x BM C2H2O4
= x 90
= kg/jam
F19 terbentuk = mol bereaksi x BM CaSO4
= x = kg/jam Reaksi 2 = kg/jam = kgmol F18 bereaksi = mol H2SO4 x BM H2SO4 = x 98 = kg/jam F11 19
= mol bereaksi x BM CH3COOH
= x 2 x 60
= kg/jam
F19 terbentuk = mol bereaksi x BM CaSO4
= x = kg/jam Reaksi 3 = kg/jam = kgmol F18 bereaksi = mol H2SO4 x BM H2SO4 = x 98 = kg/jam
F1219 = mol bereaksi x BM HCOOH
= x 2 x 46 = kg/jam 2,700 136 136 0,003 0,003 F5 17 0,412 2,700 F617 0,003 0,240 0,339 0,003 0,003 0,255 0,003 0,003 0,256 0,313 0,355 367,246 243,031
F19 terbentuk = mol bereaksi x BM CaSO4 = x = kg/jam Reaksi 4 = kg/jam = kgmol F18 bereaksi = mol H2SO4 x BM H2SO4 = x 98 = kg/jam
F19 terbentuk = mol bereaksi x BM CaSO4
= x
= kg/jam
F7 19
terbentuk = mol Ca(OH)2 x BM H2O
= x 36
= kg/jam
H2SO4 yang dibutuhkan :
H2SO4 untuk reaksi = Reaksi (1+2+3+4)
= + + +
= kg/jam
H2SO4 yang disuplai = x H2SO4 yang dibutuhkan
= kg/jam H2O pada H2SO4 = 4 N = 2M = 2 x 98 = = = kg = kg/jam kg 0,012 0,415 1,200 136 136 1,566 0,012 F217 0,003 0,354 0,012 0,012 0,255 0,852 1,128 264,633 0,256 196 gr H2SO4/kg air 196 1,128 319,527 266,273 1630,242 319,527 0,196
9. Press Filter
Fungsi : Memisahkan CaSO4 dengan C2H2O4, CH3COOH, HCOOH, H2O dan H2SO4.
Kondisi operasi : Temperatur = 30oC Tekanan = 1 atm
Komposisi bahan yang masuk : - Bahan yang keluar dari reaktor
H2O C2H2O4 CH3COOH HCOOH H2SO4 Filtrat -Komponen Masuk (kg/jam)
F17 Total 1632,510 243,031 0,313 -H2O HCOOH H2SO4 0,240 -- -CaC2O4 -Ca(CH3COO)2 Ca(HCOO)2 -319,527 C2H2O4 -F18 0,412 0,339 1,853 1630,242 CH3COOH 345,644 Ca(OH)2 0,852 - -Keluar (kg) F19 0,01% 2,76% CaSO4 53,255 84,61% 12,60% 0,02% 2298,869 349,100 1949,769 -2298,869 53,255 369,521 Komponen Persentase Kg/jam 1632,510 243,031 0,313 0,240 Laju 21 CCH2H2O4 3COOH HCOOH H2O 20 22 C2H2O4 CH3COOH HCOOH CaSO4 H2O CaSO4
Cake = CaSO4
= kg =
Filtrat
Jumlah filtrat = F21 - CaSO4
=
= kg/jam
Filtrat yang terikut pada cake= 1 % dari Cake = 1% x
= kg/jam
C2H2O4 keluar
= Filtrat yang terikut x F1020
= x = kg/jam = F10 20 - F10 22 = -= kg/jam CH3COOH keluar
= Filtrat yang terikut x F1122
= x = kg/jam = F11 20 - F11 22 = -= kg/jam HCOOH keluar
= Filtrat yang terikut x F1222
= x = kg/jam Total 369,521 3,695 0,001 242,565 100,000% 0,313 Di dalam filtrat 19,2% Di dalam cake 12,60% 3,695 0,465 1929,348 0,001 2298,869 Di dalam cake 0,313 Komponen Cake: 369,521 Di dalam cake Di dalam filtrat 3,695 0,01% 0,000 243,031 0,465 3,695 0,02% 369,521 1929,348
= F1220- F1222
=
-= kg/jam
H2O keluar
= Filtrat yang terikut x F720
= x = kg/jam = F720- F722 = -= kg/jam H2SO4 keluar
= Filtrat yang terikut x F1420
= x = kg/jam = F14 20 - F14 22 = -= kg/jam 373,217 53,153 1.629,383 Di dalam cake Di dalam filtrat 0,240 369,521 Komponen F20 3,127 0,102 53,255 Di dalam filtrat 0,102 0,465 1629,383 242,565 0,102 1.632,510 H2SO4 3,127 CH3COOH H2O 0,313 Di dalam filtrat 3,695 C2H2O4 HCOOH Masuk (kg) 1632,510 3,127 2,76% Total 2.298,869 F22 F21 369,521 243,031 Keluar (kg/jam) 2.298,869 Di dalam cake 0,313 0,001 0,240 0,000 0,239 0,000 3,695 84,6% CaSO4 1925,653 0,239 53,153 -53,255
10. Evaporator
Fungsi : Mengurangi kandungan H2O hingga konsentrasi larutan menjadi 30oBe
Menghitung % larutan yang dipekatkan: Berdasarkan literatur :
Diketahui : 30oBe = 54,9oBrix Diuapkan sampai 54,9 oBrix = 54,9 % Solute
= 40,6 % air Komposisi bahan masuk :
Air = Solute =
= +
= +
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 0 …………(1) 0,549 539,653 V F.Xf L.Xl 1925,653 ………….(2) F L V F.Xf L 1925,653 L.Xl 296,270 …………(1) 539,653 V 54,90% 15,39% V.Xf V 1.925,653 V.Xv 84,61% 15,39% L 26 C2H2O4 CH3COOH HCOOH 24 25 H2O C2H2O4 CH3COOH HCOOH
V = kg H2O sisa= H2O masuk - H2O uap
=
-= kg
10. Kristalizer
Fungsi : Mengkristalkan asam oksalat anhidrat menjadi asam oksalat dihidrat Kondisi Operasi : Temperatur = 30 oC
Tekanan = 1 atm
Untuk mempermudah hitungan maka CH3COOH, HCOOH, H2SO4 disebut sebagai impurities.
Dasar Perhitungan :
1 . Kelarutan asam oksalat pada suhu 0-60oC ditunjukkan dengan persamaan : 3,42 + 0,168 t + 0,0048 t2
2 . Range suhu kristalisasi adalah 24-32oC
3 . Jenis kristalizer asam oksalat yang digunakan adalah "Cooling Crystalization", (Kirk Othmer vol 16 edisi 3)
Kelarutan asam oksalat pada 30oC adalah = kg/100 kg larutan Keluar (kg/jam) F26 Masuk (kg) 1629,383 242,565 0,313 F24 53,153 1386,000 -Total 539,653 1.925,653 12,78 0,313 0,239 1.925,653 0,239 CH3COOH 242,565 HCOOH 1386,000 1.386,000 F25 243,384 53,153 H2O H2SO4 -Komponen 243,384 C2H2O4 1629,383 1.386,000 C2H2O4 H2O C2H2O4 H2O impurities 27 28
Neraca Massa di kristalizer :
Feed masuk = +
= +
Neraca massa basis air :
Xair F = +
BM H2C2O4.2H2O (Geankoplis)
x = S + C
= S + C ..(1)
Neraca massa basis asam oksalat :
Xasam 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) 0,286 126 100 0,451 Larutan Kristal 0,887 0,714 -0,601 311,994 100 + 12,780 masam oksalat 539,653 12,780 0,449 mpelarut S BM dihidrat C mpelarut + massaasam oksalat
243,384 36 F S C 243,384 0,032 0,887 539,653 0,113 BM C2H2O4 -187,498 173,954 242,565 0,714 BM H2C2O4.2H2O 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 0,286 242,565 0,113
Total kristal = kg/jam
Impurities didalam kristal= 1% x impurities masuk = 1% x
= kg/jam
Larutan terdiri dari :
H2O = S = x = kg/jam C2H2O4 = S = x = kg/jam
Impurities = impurities yang masuk = 0.99 x
= kg/jam
11. Centrifuge
Fungsi : Memisahkan kristal C2H2O4.2H2O dari filtratnya
H2O 243,384 F27 Kristal (F28) Larutan (F28) 0,887 173,954 53,704 0,53704 311,994 Masuk (kg) 0,887 154,242 154,242 Impurities 53,704 0,537
Komponen Keluar (kg/jam)
311,994 C2H2O4.2H2O 0,113 173,954 19,712 53,704 53,167 0,113 C2H2O4 539,653 Total -- -539,653 312,531 227,122 242,565 19,712 539,653 53,167 C2H2O4 CH3COOH HCOOH H2O 28 30 C2H2O4 CH3COOH HCOOH H2O 29 C2H2O4 CH3COOH HCOOH H2O
Komposisi Bahan Masuk: H2O (l) = kg/jam C2H2O4.2H2O = kg/jam C2H2O4 (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 = 0.02 x
= kg/jam
Kristal yang lolos = 1 % x cake = 0.01 x
= kg/jam
dalam kristal = H2C2O4 kristal yang masuk - (1% x H2C2O4 kristal masuk)
= - (0.01 x )
= kg/jam
dalam filtrat = C2H2O4 kristal yang masuk - C2H2O4 kristal dalam cake
=
= kg/jam
Impuritiesdalam kristal yang keluar
dalam kristal = impurities kristal yang masuk -(1% x H2C2O4 kristal masuk)
= - (0.01 x )
= kg/jam
dalam filtrat = Impuritieskristal yang masuk - Impuritieskristal dalam cake
= -311,994 19,712 0,537 312,531 227,122 539,653
C2H2O4.2H2Odalam kristal yang keluar
311,994 154,242 3,120 311,994 308,874 0,532 0,532 312,531 227,122 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
= kg/jam H2O dalam larutan yang keluar
dalam kristal = Filtrat yang terikut kristal x H2O dalam larutan
= x
= kg/jam
dalam filtrat = H2O larutan yang masuk - H2O larutan dalam kristal
=
-= kg/jam
C2H2O4 dalam larutan yang keluar
dalam kristal = Filtrat yang terikut kristal x C2H2O4 dalam larutan
= x
= kg/jam
dalam filtrat = C2H2O4 lautanyang masuk - C2H2O4 larutan dalam kristal
=
-= kg/jam
Impurities dalam larutan yang keluar
dalam kristal = Filtrat yang terikut kristal x Impuritiesdalam larutan
= x
= kg/jam
dalam filtrat = Impuritieskristal yang masuk - Impuritieslarutan dalam kristal
= = kg/jam Kristal Larutan H2O C2H2O4 Impurities 149,997 311,994 0,005 0,537 Total 6,251 19,712 53,167 227,122 539,653 0,532 C2H2O4.2H2O Masuk (kg/jam) F28 Komponen 0,542 1,463 312,531 19,170 154,242 53,167 1,463 Larutan 6,251 6,251 0,005 4,245 0,542 1,463 -- -51,704 19,712 8,68% 149,997 309,406 51,704 220,871 4,245 3,120 -308,874 3,125 539,653 - - -67,91% Kristal 6,251 23,41% 19,170 Kristal Keluar (kg/jam) F29(kristal) F30 (larutan) Larutan 154,242 4,245 0,542
12. Ball Mill
Fungsi : Untuk menghaluskan kristal asam oksalat menjadi berukuran 200 mesh.
Komposisi bahan keluar dari centrifuse : C2H2O4.2H2O = kg/jam C2H2O4 = kg/jam H2O = kg/jam impurities = kg/jam Recycle = kg/jam Total = kg/jam
Neraca Massa Overall di Ball Mill :
P = B
P (produk) = 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 = C …………..(2)
P = 99% C = C …………..(3)
Substitusi P = ke pers (3) didapat
P = C = C C = kg/jam A = C A = kg/jam B + A = C B + = B = kg/jam 3,221 318,845 3,221 322,066 318,845 0,990 0,010 0,010 318,845 0,990 0,990 322,066 3,188 4,245 1,463 318,845 309,406 318,845 0,542 C2H2O4 CH3COOH HCOOH H2O C2H2O4 CH3COOH HCOOH H2O C2H2O4 CH3COOH HCOOH H2O 31 33 32
C2H2O4.2H2O :
- Dari centrifuse =
- Recycle dari Vibrating Screen = 1% C
= 1% x (P/0.99) = 1% x = kg/jam - Ke Vibrating Screen = C = (P/0.99) = = kg/jam C2H2O4 : - Dari Centrifuse =
- Recycle dari Vibrating Screen = 1% C
= 1% x (P/0.99) = 1% x = kg/jam - Ke Vibrating Screen = C = (P/0.99) = = kg/jam H2O : - Dari centrifuse =
- Recycle dari Vibrating Screen = 1% C
= 1% x (P/0.99) = 1% x = 0,043 kg/jam 309,406 0,542 0,542 0,990 0,542 0,990 309,406 0,990 0,990 4,245 4,245 0,990 309,406 3,125 312,531 0,548 0,005
- Ke Vibrating Screen = C = (P/0.99) = = kg/jam Impurities : - Dari centrifuse =
- Recycle dari Vibrating Screen = 1% C
= 1% x (P/0.99) = 1% x = kg/jam - Ke Vibrating Screen = C = (P/0.99) = = kg/jam 13. Vibrating Screen
Fungsi : Untuk memisahkan antara C2H2O4.2H2O sesuai ukuran dengan
C2H2O4.2H2O yang tidak sesuai ukuran. 4,288 0,015 1,478 Total 315,657 3,188 318,845 318,845 4,245 0,990 0,005 1,463 1,463 0,990 1,463 0,990
Komponen Masuk (kg/jam) Keluar (kg)
F31 F33 F32 C2H2O4.2H2O 309,406 3,125 312,531 H2O 4,245 4,288 Impurities 1,463 C2H2O4 0,542 0,548 0,015 1,478 0,043
Komposisi feed masuk :
C2H2O4.2H2O = kg/jam
C2H2O4 = kg/jam
Impurities = kg/jam
H2O = kg/jam
Feed yang tidak normal = 1% dari feed masuk = 0.01 x
= kg/jam
H2C2O4.2H2O yang keluar :
- Ke Ball Mill = 1 % x H2C2O4 dalam H2C2O4.2H2O yang masuk
= 0.01 x
= kg/jam
- Ke storage = H2C2O4 dalam H2C2O4.2H2O yang masuk
-H2C2O4 dalam H2C2O4.2H2O ke ball mill
=
= kg/jam
Impurities dalam H2C2O4.2H2O yang keluar :
- Ke Ball Mill = 1 % x Impurities dalam H2C2O4.2H2O yang masuk
= 0.01 x = kg/jam 3,125 309,406 3,188 0,548 1,478 318,845 312,531 312,531 4,288 312,531 3,125 1,478 0,015 C2H2O4 CH3COOH HCOOH H2O Recycle ke BM 1% tidak normal C2H2O4 CH3COOH BP-03 C2H2O4 CH3COOH HCOOH H2O 3 34 33
- Ke Storage = Impurities dalam H2C2O4.2H2O yang masuk
-Impurities dalam H2C2O4.2H2O ke ball mill
=
= kg/jam
H2C2O4yang keluar :
- Ke Ball Mill = 1 % x H2C2O4 yang masuk
= 0.01 x
= kg
- Ke Storage = H2C2O4 yang masuk - H2C2O4 d ke ball mill
=
= kg/jam
H2O yang keluar :
- Ke Ball Mill = 1 % x H2O yang masuk
= 0.01 x
= kg/jam
- Ke storage = H2O yang masuk - H2O ke ball mill
=
= kg/jam
Impurities yang keluar :
- Ke Ball Mill = 1 % x impurities yang masuk
= x
= kg/jam
- Ke Storage = Impurities yang masuk - Impurities ke ball mill
=
= kg/jam
Spesifikasi produk yang dihasilkan :
Total impurities dalam proses = Dalam kristal + yang ikut kristal
= + = kg/jam 0,548 4,2878 0,01 1,478 1,463 0,005 0,542 0,043 1,478 0,015 0,548 0,005 4,288 0,043 4,245 0,015 1,478 0,015 1,463 2,926 1,463 1,463
% Impurities pada produk yang dihasilkan = %
Total 318,845 315,657 3,188
318,845 1,836 Komponen Masuk (kg) Keluar (kg/jam)
F32 F34 F33
C2H2O4.2H2O 312,531 309,406 3,125
H2O 4,288 4,245 0,043
Impurities 1,478 1,463 0,015
Data Konstanta kapasitas panas Cp = A + BT + C/T2
Dimana :Cp = Kapasitas Panas (Kcal/Kmol K) A,B,C = Konstanta
T = Suhu (K)
Suhu reference = 25 oC = K
(Robert H Perry/Cecil H Chilton,Fifth Edition)
Cp = A + BT + C/T2
A,B,C = Konstanta
T = Suhu (K)
Kapasitas panas H2O(l)(Cp)
T Cp T Cp T
(Robert H Perry/Cecil H Chilton,Fifth Edition)
Selulosa (C6H10O5)x Asam Sulfat H2SO4 0oC K 25oC K 30oC K 50oC K 80oC K
PERHITUNGAN NERACA ENERGI
0,349 298,15
0,35 303,15
0,36 323,15
Senyawa Rumus Cp (kcal/kg K)
0,32
0,34 273,15
373,150 1,040 473,15 1,095
298,15 0,999 303,150 1,002 318,150 1,010
K kcal/kgK K kcal/kgK K kcal/kgK K kcal/kgK 0,00076 T Cp NaCl 58,5 10,79 0,0042 H2 2 4,97 Ca(OH)2 44 21,4 CaSO4 98 18,52 H2C2O4 136 0,259 LAMPIRAN B 298,15 Komponen BM A O2 16 8,27 0,000258 -187700 H2O(g) 18 8,22 0,00015 0,00000134 0,02197 -156800 CO2 32 10,34 0,00274 -195500 Cp 0,371 353,15 B C T
o
C
Asam Asetat CH3COOH (26-95
o
C)
Asam Formiat CHOOH 0oC K
o C K (20-100 oC) Asam Oksalat H2C2O4. oC K Dihidrat 2H2O oC K 0oC K 50oC K o C K
(Robert H Perry/Cecil H Chilton,Fifth Edition)
BM Calsium Oksalat CaC2O4
(Lange's,1999)
(Robert H Perry/Cecil H Chilton,Fifth Edition)
AHfo Beberapa komponen :
Selulosa (C6H10O5)x
Oksigen O2 (Perry 5 ed,1973)
Karbon Dioksida CO2 (Perry 5
ed
,1973)
Ca Hidroksida Ca(OH)2 (Perry 5 ed,1973)
Calcium Oksalat CaC2O4 (Lange's,1999)
Calsium Asetat Ca(CH3COO)2 (Lange's,1999)
Calcium Formiat Ca(COOH)2 (Lange's,1999)
Ca Karbonat CaCO3 (Lange's,1999)
Asam Sulfat H2SO4 (Perry 5 ed,1973)
Calsium Sulfat CaSO4 (Perry 5 ed,1973)
Asam Oksalat H2C2O4 (Lange's,1999)
Asam Asetat CH3COOH (Lange's,1999)
Asam Formiat CHOOH (Perry 5 ed,1973)
0,7690 0,5608 46 -97,8 -2126,1 90 -821,7 -2182,1 60 -486 -1935,9 98 -194 -1976,3 136 -339 -2490,7 130 1386,6 2549,27 100 -1207,6 -2886,2 128 -1360,6 -332 -2595,3 158 -1029 -1556,6 44 -94,1 -2137,5 74 -236 -3183,5 32 0 0 AHfo
kJ/mol kcal/m kcal/kg 15 9 10,53 14,44 Kalsium Asetat Kalsium Format (CH3COO)2Ca (HCOO)2Ca 158 130
Senyawa Rumus BM Σ atom AR rata-rata Cp ( kkal/kg K)
Senyawa Rumus Cp (kca/kg K)
128 152,8 0,285313 0,338 273,15 0,385 323,15 0,416 100 373,15 0,117 -200 73,15 0,239 -100 173,15 100 0,436 273,15 0,509 15,5 288,7 0,524 0,38 373,15 0,522 (162)x -4431,7 Rumus Senyawa BM Cp (J/mol K)
Asam Karbonat H2CO3 (Perry 5 ed ,1973) Air H2O (Perry 5 ed ,1973)
Steam yang digunakan adalah saturated steam dengan dengan temperatur dan tekanan 148 0C dan P= 4.7 bar
Hs = kj/kg o C = kcal/kgK hs = kj/kg o C = kcal/kgo(Geankoplis)K
Air pendingin yang digunakan adalah air cooling tower dengan suhu masuk25 oC dan keluar 45 oC
Cp H2O pada 25 oC = kcal/kgoC(Geankoplis)
Cp H2O pada45 oC = kcal/kgoC(Geankoplis)
Air pencuci yang digunakan adalah air proses dengan suhu masuk = 25 oC Cp H2O pada 25 o C = kcal/kgoC(Geankoplis) (Lange's.1999) (Lange's.1999) (Lange's.1999)
(Perry dan Green.1997)
Cp Komponen humus :
(Perry dan Green.1997)
(Kirk & Othmer.1954)
(Lange's.1999) Cp Komponen Humus : Cp = A+BT-C/T2 Senyawa Rumus B Abu 0,00333 A BM 40 5,31 Silika 0,316 Lignin 0,32 Pentosan 0,479 Ca -2860,743 -466,823 -18,95 0 Senyawa Cp (kkal/kg K) 0,999 1,010 0,9989 ΔHfo komponen humus : Senyawa ΔHfo (kkal/kg) 18 -57,8 -3211 2744,02 655,8365 623,572 149,0373 Silika Lignin Pentosan Abu 62 -167 -2696,6
1. Reaktor Kalsium Oksalat
Neraca Panas
H masuk = Panas yang terkandung dalam reaktan masuk H Keluar = Panas yang terkandung dalam produk Alang-alang masuk C6H10O5 = x = kg/jam = kg/jam x = kkal/jam Lignin = x = kg/jam = kg/jam x = kkal/jam Pentosan = x = kg/jam = kg/jam x kkal/kg K x = kkal/jam Silika = x = kg/jam = kg/jam x = kkal/jam Abu = x = kg/jam = kg/jam x kkal/kg 303,15-298,15 303,15-298,15 1360,087 113,021 1987,006 5,42% 107,696 107,696 1987,006 44,28% 879,846 879,846 0,32 1987,006 28,58% 567,886 567,886 1987,006 18,12% 360,045 360,045404 0,32 1987,006 3,60% 71,532 71,532 303,15-298,15 0,48 kkal/kg K x kkal/kg K x kkal/kg K x 303,15-298,15 0,32 1407,7537 576,0726 0,789 H feed T = 30 oC H Ca(OH)2 T = 30 oC H produk T = 98 oC
= kkal/jam
Maka H alang-alang masuk = C6H10O5 + Lignin + Pentosan + Silika + Abu
= + +
+ +
= kkal/jam
Data m ʃCp dT dan m Cp dT tiap komponen yang masuk :
Total massa bahan masuk = kg/jam
Total H bahan masuk = kkal/jam
Entalpi bahan keluar
Suhu bahan keluar = 98 oC = K Suhu referensi = 25oC = K ΔT = Tbahan keluar - T referensi
84,961 576,073 1360,087 5532,353 13713,699 2154,827 O2 564,839 0,980 553,444 371,15 298,15 Komponen 887,397 Q = m x ʃCp dT
kg/jam kkal/kg kkal/jam
Ca(OH)2 1490,254 1,446 ʃCp dT 21,1108 Ca(OH)2 m 18733,663 1407,754 H2O (l) 1490,254 5 1,002 7463,533 Komponen m ʃCp dT Q = m x ʃCp dT Komponen m T (K) Cp Q= m Cp dT Kg/jam kkal/kg K Q= m ʃCp dT
kg/jam kkal/kg kkal/jam
113,021 84,961 3541,895 ʃCp dT = 𝐴 ×𝑇𝑏𝑎ℎ𝑎𝑛 𝑚𝑎𝑠𝑢𝑘) + 𝐵2× 𝑇2⬚𝑏𝑎ℎ𝑎𝑛 𝑚𝑎𝑠𝑢𝑘 − 𝐴 ×𝑇𝑏𝑎ℎ𝑎𝑛 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒) + 𝐵2× 𝑇2⬚𝑏𝑎ℎ𝑎𝑛 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝐵𝑀
Entalpi Humus : Silika = m x Cp x dT = x x 73 K = Lignin = m x Cp x dT = x x 73 K = Abu = m x = x = Pentosan = m x Cp x dT = x x 73 K =
Maka Entalpi humus = Ca(CH3COO)2 146396,377 Q= m ʃCp dT Q= m Cp dT 477,941 1262,678 19857,276 Komponen m T (K) Cp (Kkal/kg K) Komponen 1930,177 73 m ʃCp dT 73 1650,105 Kg/jam kkal/kg K ʃCp dT 31180,72004 kkal/jam kkal/kg kkal/jam 567,886 kg/jam 0,479 kkal/kg kkal/jam 107,696 kg/jam 11,724502 kkal/jam H2O (l) 15,745 1,039 kkal/jam Q = m x ʃCp dT kg/jam Q = m.Cp.ΔT kkal/kg kkal/jam 0,316 kkal/jam Ca(HCOO)2 kkal/kg 71,532 kg/jam 8410,661 360,045 kg/jam 0,32 kkal/kg CO2 7525,230 429,061 0,769 73 Komponen 7239,627 24086,182 CaC2O4 347,594 0,285 m Cp ΔT kg/jam K 353,025 0,561 73 14452,264 ʃCp dT = 𝐴 ×𝑇𝑏𝑎ℎ𝑎𝑛 𝑚𝑎𝑠𝑢𝑘)+ 𝐵2× 𝑇2⬚𝑏𝑎ℎ𝑎𝑛 𝑚𝑎𝑠𝑢𝑘 −(𝑇𝑏𝑎ℎ𝑎𝑛 𝑘𝑒𝑙𝑢𝑎𝑟𝐶 ) − 𝐴 ×𝑇𝑏𝑎ℎ𝑎𝑛 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒) + 𝐵 2× 𝑇 2⬚ 𝑏𝑎ℎ𝑎𝑛 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 −(𝑇𝑒𝑓𝑒𝑟𝑛𝑐𝑒𝐶 ) 𝐵𝑀
Total massa bahan keluar = kg
Total H bahan keluar = kkal/jam
Reaksi yang terjadi dalam reaktor Kalsium Oksalat 2(C6H10O5)1050 + 3150 Ca(OH)2 + 6825 O2
1050(COO)2Ca + 1050 (CH3COO)2Ca + 1050 (HCOO)2Ca + 9450 H2O + 4200 CO2
ΔHf C6H10O5 = ΔHf o 25 = kkal/kg x kg/jam = kkal/jam ΔHf Ca(OH)2 = ΔHf o 25 = kkal/kg x kg/jam = kkal/jam ΔHf O2 = ΔHf o 25 = 0 kkal/kg x kg/jam = 0 kkal/jam ΔHf CaC2O4 = ΔHf o 25 = kkal/kg x kg/jam = kkal/jam ΔHf Ca(CH3COO)2 = ΔHf o 25 = kkal/kg x kg/jam = kkal/jam ΔHf Ca(HCOO)2 = ΔHf o25 = kkal/kg x kg/jam = kkal/jam ΔHf H2O = ΔHf o 25 = kkal/kg x kg/jam = kkal/jam ΔHf CO2 = ΔHf o25 = kkal/kg x kg/jam 899955,842 564,839 439,923 -3210,994 -1412590,396 -2137,545 5532,353 249614,065 477,941 -4431,719 879,846 -3899230,387 -3183,514 602,857 -1919204,939 -2595,313 347,594 -902113,763 -1556,563 429,061 -667859,917 2549,272 353,025
= kkal/jam = = ((-902113,763) + (-667859,917) + (899955,842) + (-1412590,396) + (-1021620,754)) - ((-3899230,387) + (-1919204,939) + (0)) = - ( ) = kkal/jam Perhitungan ΔHr o 25 Q = ΔHr + ΔHro25 + ΔHp = ΔHp+ ΔHro25 - ΔHr = + -= kkal/jam 2714206,338 7463,533 84,961 1360,087 576,073 13713,699 553,444 H2O 146396,377 Produk -1021620,754 ΔHr o25=ΔHfp- ΔHfr (m.Cp.dT)p (m.Cp.dT)r ΔHp = ΔHr o 25 + (m.Cp.dT)p - (m.Cp.dT)r kkal/jam Abu CaC2O4 Ca(CH3COO)2 Ca(HCOO)2 Jumlah Reaktan 2714206,338 19857,276 113,021 7239,627 Lignin 18733,663 CO2 ΔHr o25 ΔHf produk - ΔHf reaktan Komponen ΔHr o 98 2950106,703 Pentosan 14452,264 1407,754 2154,827 8410,661 1262,678 2950106,703 Σ 2714206,338 249614,065 Ca(OH)2 O2 24086,182 C6H10O5 H2O Silika 249614,065 13713,699 -3104228,988 -5818435,326 2714206,338 7525,230 1650,105
Q suplai = x Q = x = kkal/jam Q loss = 5% x Q dibutuhkan = 5% x = kkal/jam
Steam yang digunakan adalah saturated steam dengan T = 148 oC , P= 4,7 bar
Hs = kJ/kgoC = kkal/jam
hs = kJ/kgoC = kkal/jam
Q steam = m x (Hs - hs)
massa steam = kkal/jam
- kkal/kg
= kg/jam
Neraca Panas di reaktor 1
Q yang disuplai H2O Q loss 3111325,738 Jumlah Humus Panas Keluar 1,050 1,050 2950106,703 steam 146396,377 14452,264 24086,182 249614,065 31180,720 7525,230 18733,663 Ca(OH)2 CO2 ΔHr o 25 147505,335 3097612,038 3111325,738 147505,335 Jumlah Jumlah 2714206,338 3541,895 2154,827 553,444 7463,533 Panas Masuk Jumlah
kkal/jam 2744,02 655,837 623,572 149,037 3097612,038 Jumlah 2950106,703 H2O 13713,699 3097612,038 Alang-Alang Ca(HCOO)2 CaC2O4 2744,020 623,572 1460,829 7239,627 Ca(OH)2 O2 Jumlah Ca(CH3COO)2 kkal/jam
2. Tangki Pendingin
Entalpi bahan masuk
Suhu Bahan Masuk = 98 oC = K
Suhu Reference = 25 oC = K
ΔT = Suhu Bahan Masuk - Suhu Referensi
=
-= K
kkal/jam
Humus = kkal/jam
Total massa bahan masuk = kg/jam
Total H bahan masuk = kkal/jam
Entalpi bahan keluar
Cp 7239,627 24086,182 kkal/kg K kkal/jam Q= m ʃCp dT Q = m.Cp.ΔT 371,15 298,15 371,15 73 73 Komponen m (HCOO)2Ca 353,025 73 Ca(OH)2 18733,663 0,285313 429,061 0,7690 73 14452,264 Q= m Cp dT T (K) Cp Komponen kg/jam kkal/kg 887,397 21,1108 (Kkal/kg K) m Kg/jam 298,15 ΔT 31180,720 Komponen H2O (l) 1930,177 73 1,0390 m 0,5608 CaC2O4 146396,377 (CH3COO)2Ca 242088,835 347,594 5054,412 K ʃCp dT Q = m x ʃCp dT kg/jam H feed T = 98 oC H Pendingin out T =85 oC H Air pendingin in T = 25 oC H produk T = 35 o C
Suhu bahan keluar = 35 oC = K
= 25 oC = K
ΔT = Suhu Bahan keluar - Suhu Referensi
= -= 10 K Entalpi Humus : Silika = m x Cp x dT = x x 10 K = Lignin = m x Cp x dT = x x 10 K = kkal/jam Abu = m x = x = kkal/jam Pentosan= m x Cp x dT = x x 10 K = 298,15 3299,477 1979,762 kkal/kg K 298,15 2566,255 308,15 Cp ΔT ʃCp dT 107,696 kg/jam 1,580 kg/jam kg/jam (Kkal/kg K) K H2O (l) 1930,177 Kg/jam 308,15 Komponen m m ʃCp dT Q = m x ʃCp dT kg/jam Komponen Komponen m kg/jam kkal/kg kkal/kg kkal/jam Ca(OH)2 887,397 226,042 0,7690 360,045 (CH3COO)2Ca 429,061 kkal/jam (HCOO)2Ca 353,025 0,5608 10 991,730 kkal/jam Q = m.Cp.ΔT 19333,537 2,8919 1152,145 170,145 Cp CaC2O4 347,594 0,285313 10 kkal/kg 2720,175 567,886 kkal/jam T (K) Q= m ʃCp dT 0,479 kkal/kg Q= m Cp dT 0,32 10 10 1,0016 71,532 kg/jam 0,316 kkal/kg
Maka Entalpi humus =
Total massa bahan keluar = kg/jam
Total H bahan keluar = kkal/jam
Total H bahan masuk = kkal/jam
Q loss = 5% x H bahan masuk = 5% x
= kkal/jam
H bahan masuk - Q loss - H bahan keluar
= -
-= kkal/jam
Kebutuhan media pendingin
Digunakan air untuk mendinginkan bahan dengan suhu masuk = 25 oC dan suhu keluar 85 oC
Cp air pada 25 oC = kkal/kgK Cp air pada 85 oC = kkal/kgK
CP rata - rata = kkal/kgK
Q media pendingin (Q1) = m x Cp x ( Tout-Tin) massa air = Q media pendingin (Q1)
= =
= kg/jam
Neraca Panas di Tangki Pendingin
4268,507 19333,537 Humus 1,015 18733,663 Ca(OH)2 32439,269 Humus 12104,442 197545,124 197545,124 5054,412 32439,269 H2O 3242,567 kkal/jam 1979,762 242088,835 242088,835 CaC2O4 7239,627 CaC2O4
Ca(CH3COO)2 24086,182 Ca(CH3COO)2
14452,264 Ca(HCOO)2 3299,477 991,730 kkal/jam Jumlah Ca(HCOO)2 kkal/jam 4268,507327 H2O 31180,720 2566,255 146396,377 60,922 Ca(OH)2
Panas Masuk Jumlah Panas Keluar Q media pendingin = 242088,835 12104,442 197545,124 0,999 1,032 Cp x ( Tout-Tin) 1,015 x ( 85-25)
3.Rotary Vacuum Filter
Entalpi bahan masuk
Suhu bahan masuk = 35 oC = K
Total H bahan masuk = kkal/jam Entalpi bahan keluar
Suhu bahan keluar = T oC = T + 273,15 K
2551,859 12104,442 197545,124 Q loss 303,15 Cp ʃCp dT
Q yang diserap pendingin
242088,835 350,618 882,419 2005,760 0,999 4011,551 28824,311 20035,539 3284,482 986,166 0,769 Komponen m (kg//jam) kkal/kgK ΔT kkal/kg 0,285 Jumlah kkal/jam Q = m.ʃCp.dT 28824,311 10 Ca(HCOO)2 Ca(OH)2 H2O CaC2O4 Ca(CH3COO)2 345,644 427,111 Jumlah 0,561 1966,264 Q = m.Cp.dT 242088,835 Jumlah 10 10 10 2,892 H-215 H air pencuci T = 25 oC H filtrat T H cake T H feed T = 35 oC
(T-25) +
(T-25) +
Q loss = x H bahan masuk
= x
= kkal/jam
Suhu air pencuci = 25 oC = K H air pencuci = m.Cp.dT
= (25 % x jumlah solid masuk).Cp.dT
= x x (303.15-298,15) K
= kkal/jam
Neraca panas :
H bahan masuk = H cake + H filtrat - air pencuci + Q loss
= {(100,985 + 2536,868) (T25)} + {(0,852 + 881,567) ʃCp.dT} -431,579 + 1441,216
= (T-25) + ʃCp.dT
Perhitungan (T-25) dengan menggunakan trial and error
T = C = K H bahan masuk = (29,997-25) + x = + kkal/jam (T-25) (T-25) ʃCp dT (T-25) 0,339 0,561 0,852 0,190 0,317 1,853 1,004 (T-25) 328,131 2536,868 25403,111 2411,563 27814,674 2637,852 882,419 34,630 307,780 Ca(OH)2 (T-25) 196,437 2003,907 Q = m.ʃCp.dT (T-25) 100,985 0,852 0,999 86,411 ʃCp dT 28824,311 882,419 2,733 Jumlah 1,861 (T-25) (T-25) (T-25) ʃCp dT (T-25) Ca(CH3COO)2 (T-25) Jumlah 0,05 0,769 ʃCp dT 28824,311 298,15 H2O 431,579 0,852 H2O (T-25) Q = m.Cp.dT CaC2O4 Komponen 98,617 Cake 2012,300 2637,852 (HCOO)2 0,05 Filtrat Ca(CH3COO)2 426,698 0,769 (T-25) (T-25) Ca(HCOO)2 350,279 0,561 0,412 ʃCp dT 881,567 881,567 ʃCp dT m (kg//jam) ΔT kkal/kgK kkal/kg Ca(OH)2 881,567 ʃCp dT 1,004 345,644 0,285 1441,216 Cp
= kkal/jam
Jadi suhu bahan keluar rotary vacuum filter = oC = K H filtrat ke tangki =
= kkal/jam
H cake ke reaktor =
298,15)) x 0,852}
= kkal/jam
4. Reaktor Asam Oksalat
Entalpi bahan masuk
Cake yang kelur dari rotary vacuum filter = kkal/jam H2SO4 4 N pada suhu 30 oC = 303,15 K
- H2SO4 = kg/jam x kkal/kgK x 5 K
= kkal/jam
maka H2SO4 4 N adalah = kkal/jam
1441,216 974,835 - 298,15)) x 881,567} kkal/jam kkal/jam 974,835 28824,311 H filtrat 26839,840 34,63 26839,83952 974,8348855
Panas Masuk Jumlah Panas Keluar Jumlah
H umpan 29255,890 307,780 Q loss 27814,674 {2536,868 x (307,780 - 298,15)} + {(0,284 x (307,780
H air pencuci 431,579 H cake
Jumlah 29255,890 Jumlah 557,575 319,527 0,349 557,575 {100,985 x (307,780 298,15)} + {(0,284 x (307,780 -Q air pendingin T out = 45 o C T In = 25 oC H feed T =34,630 o C H H2SO4 T = 30 oC H produk T = 80 oC
- H2O = kg/jam x kkal/kgK x 5 K
= kkal/jam
maka H H2Oadalah = kkal/jam
Total entalpi bahan masuk = Cake yang keluar dari rotary vacuum filter + larutan H2SO4 4N
= + +
= kkal/jam
Entalpi bahan keluar
Suhu Bahan Keluar = 80 oC = K
Reaksi- reaksi yang terjadi pada reaktor asam oksalat :
Temperatur = 80 oC = K
Reaksi I : CaC2O4 + H2SO4 C2H2O4 + CaSO4 ΔHf CaC2O4 = ΔHf o 25 = ( kkal/kg x kg/jam ) = kkal/jam ΔHf H2SO4 = ΔHf o25 = ( kkal/kg x kg/jam ) = kkal/jam ΔHf C2H2O4 = = ( kkal/kg x kg/jam ) = kkal/jam 8164,624 8,993 75,224 92401,124 Jumlah 53,255 3614,228 kkal/kg Q= m.Cp.dT = m.ʃCp.dT (kkal/jam) H2O 0,240 2298,869 353,15 Komponen m kg/jam Cp kkal/kgK ΔT ʃCp.dT 6,905 C2H2O4 0,522 8164,624 8164,624 1630,242 1,002 557,575 1,029 0,522 0,524 1528,939 55 55 974,835 9,781 55 264,633 97635,412 9697,034 243,031 1632,510 0,310 ΔHf o 25 55 CH3COOH -1976,327 -523001,946 -2182,122 243,031 -530322,605 0,313 HCOOH H2SO4 CaSO4 345,644 369,521 353,15 -2595,313 -897053,110
ΔHf CaSO4 = ΔHf o25 = ( kkal/kg x kg/jam ) = kkal/jam ΔHr o 25 = ΔHf produk - ΔHf reaktan = (-523001,946)} = kkal/jam m.Cp.dT = ( x x 55 ) CaC2O4 = kkal/jam m.Cp.dT = ( x x 55 ) H2SO4 = kkal/jam m.ʃCp.dT = ( x ) C2H2O4 = kkal/jam m.ʃCp.dT = ( x ) CaSO4 = kkal/jam ΔHr o 80 (I) = + (m.Cp.dT)p - (m.Cp.dT)r = + -= kkal/jam 75,224 3591,975 3591,975 0,310 5423,915 5399,844 (m.Cp.dT)r Σ H2SO4 Produk CaSO4 C2H2O4 ΔHr o80 = ΔHr o 25 + (m.Cp.dT)p - (m.Cp.dT)r -2490,662 367,246 -914686,333 (kkal/jam) ΔHr o25 = ΔHfp -ΔHfr Reaktan CaC2O4 (m.Cp.dT)p ΔHf o 25 0,285 10823,759 3667,199 9,781 -32110,443 Komponen 75,224 3667,199 {(-530322,605) + (-914686,333)} - {(-897053,110) + 5423,915 5399,844 10823,759 345,644 264,633 243,031 367,246 -24953,883 0,371 -24953,883 -24953,883
Reaksi II Ca(CH3COO)2 + H2SO4 2CH3COOH + CaSO4 ΔHf Ca(CH3COO)2 = ΔHf o 25 = ( kkal/kg x kg/jam ) = kkal/jam ΔHf H2SO4 = ΔHf o25 = ( kkal/kg x kg/jam ) = kkal/jam ΔHf CH3COOH = ΔHf o 25 ( kkal/kg x kg/jam ) = kkal/jam ΔHf CaSO4 = ΔHf o 25 = ( kkal/kg x kg/jam ) = kkal/jam ΔHr o 25 = ΔHf produk - ΔHf reaktan = {(-606,408) + (-884,186)} - {(-641,969) + (-505,562)} = kkal/jam m.Cp.dT = ( x x 55 )
Ca(CH3COO)2 = kkal/jam
m.Cp.dT = ( x x 55 ) H2SO4 = kkal/jam mCp.dT = ( x x 55 ) CH3COOH = kkal/jam m.ʃCp.dT = ( x ) CaSO4 = kkal/jam ΔHr o 80 (II) = ΔHf o25 + (m.Cp.dT)p - (m.Cp.dT)r = + -= kkal/jam 0,412 0,256 0,371 0,313 0,522 -343,063 -641,969 8,993 3,472 -353,261 -343,063 12,465 22,663 -505,562 0,355 9,781 17,444 -1556,563 0,412 -1976,327 0,256 -1935,946 0,355 0,769 -606,408 -884,186 5,220 0,313 -2490,662
Reaksi III : Ca(HCOO)2 + H2SO4 2HCOOH + CaSO4 ΔHf Ca(HCOO)2 = ΔHf o 25 = ( kkal/kg x kg/jam ) = kkal/jam ΔHf H2SO4 = ΔHf o 25 = ( kkal/kg x kg/jam ) = kkal/jam ΔHf HCOOH = = ( kkal/kg x kg/jam ) = kkal/jam ΔHf CaSO4 = = ( kkal/kg x kg/jam ) = kkal/jam ΔHr o 25 = ΔHf produk - ΔHf reaktan = {(-509,408) + (-882,167)} - {(863,091) + (-504,408)} = kkal/jam m.Cp.dT = ( x x 55 ) Ca(HCOO)2 = 0,339 Σ 22,663 2549,272 -343,063 Komponen (m.Cp.dT)r (kkal/jam) Reaktan Ca(CH3COO)2 H2SO4 ΔHr o 80 = ΔHr o 25 + (m.Cp.dT)p - (m.Cp.dT)r ΔHr o25 = ΔHfp -ΔHfr (m.Cp.dT)p Produk CaSO4 CH3COOH -1976,327 5,220 0,255 -2126,087 0,240 -2490,662 0,354 ΔHf o 25 ΔHf o 25 12,465 17,444 863,091 -504,408 -509,408 -882,167 0,339 0,561 kkal/jam -1750,258 8,993 3,472 10,443
m.Cp.dT = ( x x 55 ) H2SO4 = kkal/jam mCp.dT = ( x x 55 ) HCOOH = kkal/jam m.ʃCp.dT = ( x ) CaSO4 = kkal/jam ΔHr o 80 (III) = ΔHf o25 + (m.Cp.dT)p - (m.Cp.dT)r = + -= kkal/jam
Reaksi IV : Ca(OH)2 + H2SO4 CaSO4 + 2H2O ΔHf Ca(OH)2 = ΔHf o25 = ( kkal/kg x kg/jam ) = kkal/jam ΔHf H2SO4 = ΔHf o 25 = ( kkal/kg x kg/jam ) = kkal/jam Komponen ΔHr o80 = ΔHr o 25 + (m.Cp.dT)p - (m.Cp.dT)r ΔHr o25 = ΔHfp -ΔHfr (m.Cp.dT)p (m.Cp.dT)r 6,905 (kkal/jam) Reaktan Ca(HCOO)2 H2SO4 -1976,327 1,128 Produk CaSO4 -1755,539 9,781 0,255 0,371 5,208 6,905 3,464 10,443 -1750,258 10,370 -1750,258 10,370 15,651 Σ 3,464 HCOOH -2712,613 -2230,151 15,651 5,208 -3183,514 0,852 0,240 0,524 0,354
ΔHf CaSO4 = ΔHf o25 = ( kkal/kg x kg/jam ) = kkal/jam ΔHf 2H2O = ΔHf o 25 = ( kkal/kg x kg/jam ) = kkal/jam ΔHr o 25 = ΔHf produk - ΔHf reaktan = {(-3900,346) + (-1331,041)} - {(-2712,613) + (2230,151)} = kkal/jam m.ʃCp.dT = ( x ) Ca(OH)2 = kkal/jam m.Cp.dT = ( x x 55 ) H2SO4 = kkal/jam m.ʃCp.dT = ( x ) CaSO4 = kkal/jam m.Cp.dT = ( x x 55 ) 2H2O = kkal/jam ΔHr o 80 (IV) = ΔHf o25 + (m.Cp.dT)p - (m.Cp.dT)r = + -= kkal/jam -288,623 38,779 CaSO4 23,462 1,566 -2490,662 -3210,994 0,415 36,578 Komponen ΔHr o 80 = ΔHr o 25 + (m.Cp.dT)p - (m.Cp.dT)r ΔHr o25 = ΔHfp -ΔHfr Reaktan (m.Cp.dT)p (m.Cp.dT)r (kkal/jam) -288,623 0,852 1,566 9,781 0,415 1,029 15,905 1,128 0,371 13,553 H2SO4 Produk -3900,346 -1331,041 13,553 15,317 23,026 23,462 23,026 Ca(OH)2 -286,422
ΔHr total = ΔHr o25(I) + ΔHr o25(II) + ΔHr o25(III) + ΔHr o25(IV)
= + + +
(kkal/jam)
= kkal/jam
Karena reaksi terjadi eksoterm maka diperlukan air pendingin
Air pendingin yang masuk pada suhu 25oC = K dan keluar pada
45 oC = K
Neraca Panas
Q pendingin = ΔHp + ΔHr o25 total -ΔHr
= (9697,034 + (-27335,827) -97635,412) = kkal/jam Q air pendingin = m.Cp.dT = m x x -m = x 20 m = kg/jam
Neraca Panas di Reaktor Asam Oksalat -343,063 36,578 2H2O 15,317 Jumlah -24953,883 Σ -288,623 -1750,258 298,15 Jumlah 974,835 H umpan H H2SO4 -27335,827 -60602,552 8164,624 9697,034 H H2O Q air pendingin 97635,412 -60602,552 -60602,552 1,002 (318,15 557,575 ΔH reaksi H produk 9697,034 -288,623 -60602,552 38,779 3008,654 1,007
Panas masuk Jumlah (kkal/jam) Panas keluar Jumlah (kkal/jam) 298,15) -27335,827
5.COOLER
Entalpi bahan masuk = Entalpi bahan keluar dari reaktor asam oksalat
= kkal/jam
Entalpi bahan keluar
suhu bahan keluar = 55 oC = K
Neraca Panas
Q loss = 5% x H bahan masuk = 5% x
= kkal/jam
H bahan masuk = H bahan keluar + Q media pendingin
= + Q media pendingin +
Q media pendingin = kkal/kg
0,360 30 97635,412 3990,832 40,262 49728,252 Q = m.Cp.dT 3,766 4,905 H2SO4 30 HCOOH 0,240 0,524 Q = m.ʃCp.dT 53,255 369,521 H2O 1632,510 97635,412 ΔT ʃCp dT kkal/kgK kkal/kg 30 0,166 38410,475 4881,771 CaSO4 Jumlah 4881,771 30 C2H2O4 243,031 1,015 0,360 30 2298,869 54343,167 54343,167 kkal/jam 575,149 CH3COOH 0,313 0,522 97635,412 328,15 Komponen m (kg//jam) Cp H air pendingin in T = 25 oC H produk T = 55 oC
H air pendingin out T = 45 oC H feed
Kebutuhan media pendingin
Digunakan air pendingin untuk mendinginkan bahan masuk dengan suhu
T in = 25 oC = K
T out = 45 oC = K
Q media Pendingin = m.Cp.ΔT
= m
massa media pendingin =
= kg/jam
6.FILTER PRESS
Entalpi bahan masuk = Entalpi bahan keluar dari Cooler I
= kkal/jam
Entalpi bahan keluar =
Suhu bahan keluar = T oC = T+273,15 K
C2H2O4 242,565 242,565 54343,167 Komponen m (kg//jam) Cp ΔT ʃCp dT Q = m.Cp.dT kkal/kgK kkal/kg Q = m.ʃCp.dT ʃCp.dT 298,15 318,15 Jumlah 97635,412 Filtrat Q loss Jumlah 97635,412 38410,475 38410,475 1,007 x ( 45-25) 2542,552 1,007 x ( 45-25)
Panas masuk Jumlah (kkal/jam) Panas keluar Jumlah (kkal/jam)
H umpan 97635,412 H produk 54343,167 Q media pendingin 38410,475 1654,088 H2O 4881,771 ʃCp.dT (T-25) 1629,383 1,015 (T-25) kkal/jam H feed T = 55 oC H produk T
(T-25) +
(T-25) +
Q loss = x H bahan masuk
= x
= kkal/jam
Neraca Panas
= +
H filtrat + H Cake = kkal/jam
H filtrat = (T-25) + ʃCp dT
H cake = (T-25) + ʃCp dT + ʃCp dT
H filtrat + H cake = (T-25) + +
ʃCp dT
Suhu bahan keluar dari Filter Press dihitung dengan trial and error
T = oC = K H Filtrat = (T-25) + ʃCp dT = (( x + [ x ((0,259 x 298,15) + (0,0008 x 298,15))] = + = kkal/jam CaSO4 369,521 54343,167 5% 5% 369,521 ʃCp dT 1674,096 242,565 3,213 0,465 ʃCp.dT 1674,096 (327,66-298,15)) 369,521 54,614 327,76 49576,815 39,656 0,000 369,52 ((0,259 x 327,76) + (0,0008 x 327,76)) - 1674,096 242,565 ʃCp dT 0,239 (T-25) (T-25) H2SO4 HCOOH 242,565 ʃCp dT +
H bahan masuk = H bahan keluar + Q loss
243,0 (T-25) 0,465 ʃCp.dT 54343,167 369,521 0,522 (T-25) Cake Jumlah 3,213 H2SO4 0,102 0,371 (T-25) 0,038 (T-25) 0,001 (T-25) ʃCp.dT 0,465 HCOOH 1674,096 CH3COOH 0,001 0,522 0,524 0,465 (T-25) 3,127 1,015 3,174 (T-25) 19,720 (T-25) 0,0002 242,565 Jumlah ʃCp.dT C2H2O4 0,125 (T-25) (T-25) (T-25) H filtrat + H cake 2717,158 H2O 53,153 0,371 CH3COOH 0,313 ʃCp.dT 0,163 (T-25) 2717,158 51626,008 49616,471 1677,309 0,524
H cake = (T-25) + ʃCp dT + ʃCp dT = (( x + x + = + + = kkal/jam
Neraca Panas pada Filter Press
7. EVAPORATOR
H bahan masuk = Entlapi bahan keluar dari filter press
= kkal/jam Panas masuk 0,465 ((18,52 x 327,76)+(0,022 x 327,76)+(-156800 x 327,76))-((18,52 x 298,15)+(0,022 x 298,15)+(-156800 x 298,15))x 567,093 ((0,259 x 327,76)+(0,0008 x327,76)-((0,259 x 298,15)+(0,0008x298,15) H umpan 54343,167 H produk H Cake 2009,537 0,076 54343,167 49616,471 3,213 0,465 Q loss 2717,158 Jumlah 54343,167 Jumlah Jumlah (kkal/jam) 95,144 1914,318 369,521 3,213 (327,66-298,15))
Panas keluar Jumlah (kkal/jam) 49616,471 2009,537 H vapour T = 100 o C H feed T = 54,61oC H Condensat T = 148 oC P = 4,7 bar H saturated steam in T = 148 oC P = 4,7 bar H produk T = 100 oC
Entalpi bahan keluar :
Suhu bahan keluar : oC = K
Hv air pada 100 oC = kkal/kg
Entalpi air yang menguap = m x Hv
= x = kkal/jam Neraca Panas Q = Hproduk - H umpan = (830473,421 +20571,982) -(49616,471) = kkal/jam Q suplai = 1,05 x Q Q Suplai = kkal/jam
Steam yang digunakan adalah saturated steam dengan T = 148 oC , P= 4,7 bar
Hs = kJ/kgoC = kkal/jam
hs = kJ/kgoC = kkal/jam
Q steam = m x (Hs - hs)
massa steam = kkal/jam
- kkal/kg = kg/jam Q loss = x Q = x = kkal/jam 801428,931 5% 5% 149,037 841500,378 841500,378 1506,875 9,398 12,240 98,085 18945,384 kkal/jam Q = m.ʃCp.dT Q = m.Cp.dT Jumlah 2744,0 623,6 655,837 149,037 801428,931 830473,421 1660,422 655,837 Komponen m (kg//jam) HCOOH 0,239 75 kkal/kg 0,524 C2H2O4 H2O 243,384 0,404 539,653 20571,982 1386,000 599,187 H2SO4 53,153 0,378 75 -CH3COOH 0,313 0,522 75 Cp ΔT ʃCp dT 242,565 kkal/kgK 100 373,150 1,038 75 599,187 40071,447
Neraca Panas di Evaporator
8.COOLER II
Entalpi bahan masuk = Entalpi bahan keluar dari evaporator
= kkal/jam
Entalpi bahan keluar
suhu bahan keluar = 55 oC = K
Neraca Panas
kkal/jam
0,000 7413,761 830473,421
Panas masuk Jumlah (kkal/jam) Panas keluar Jumlah (kkal/jam)
H umpan 49616,471 H produk 20571,982 Q Steam 841500,378 H Vapour kkal/kg Q = m.ʃCp.dT Q loss 40071,447 0,522 30 0,524 30 H2SO4 53,153 0,360 30 Jumlah 539,653 HCOOH 0,239 Jumlah 891116,849 Jumlah 891116,849 328,15 Komponen m (kg//jam) Cp ΔT ʃCp dT Q = m.Cp.dT kkal/kgK 4,896 574,048 3,759 H2O 243,384 1,015 30 C2H2O4 242,565 0,166 CH3COOH 0,313 20571,982 7996,464 H air pendingin in T = 25 oC H produk T = 55 oC
H air pendingin out T = 45 oC H feed
H bahan masuk = H bahan keluar + Q media pendingin
= + Q media pendingin
Q media pendingin = kkal/kg
Kebutuhan media pendingin
Digunakan air pendigin untuk mendinginkan bahan masuk dengan suhu
T in = 25 oC = K
T out = 45 oC = K
Q media Pendingin = m.Cp.ΔT
= m
massa media pendingin =
= kg/jam
9.CRYSTALIZER
Entalpi bahan masuk = Entalpi bahan keluar dari cooler = 7996,464 kkal/jam H umpan 20571,982 H produk 7996,464 12575,518 298,15 318,15 12575,518 1,007 x ( 45-30)
Panas masuk Jumlah (kkal/jam) Panas keluar Jumlah (kkal/jam) 12575,518 1,007 x ( 45-30) Q media pendingin 12575,518 Jumlah 20571,982 Jumlah 20571,982 20571,982 7996,464 832,427 H feed T = 55 o C H air pendingin in T = 25 oC H air pendingin out T = 45 o C H produk T = 30 o C
Entalpi bahan Pada 30 oC
Panas pendinginan sampai 30 oC = H bahan pada 55 oC - H bahan pada 30 oC
=
-= kkal/jam
H pengkristalan = Berat kristal x Panas Pengkristalan C2H2O4.2H2O
= Berat kristal x (-1 x Heat Solution C2H2O4.2H2O (Perry)
= x (-1x(-67,3413))
= kkal/jam
Q Kristalisasi = Panas untuk mendinginkan + Panas untuk mengkristalkan
= +
= kkal/jam
Q loss = x H bahan masuk
= x
= kkal/jam
Entalpi bahan keluar
Suhu bahan keluar = 30 oC = K
7996,464 5% 5% 399,823 -kkal/jam Q = m.ʃCp.dT Q = m.Cp.dT 6683,561 30043,163 303,15 0,366 5 570,949 Komponen m (kg//jam) Cp ΔT ʃCp dT kkal/kgK kkal/kg H2O - -Kristal C2H2O4 -0,350 5 93,983 0,000 Komponen m (kg//jam) Cp ΔT ʃCp dT kkal/kgK kkal/kg 446,133 C2H2O4 242,565 0,027 1312,903 6683,561 Impurities 53,704 1312,903 H2O 243,384 1,002 5 Impurities 0,537 0,350 C2H2O4.2H2O 5 C2H2O4.2H2O 311,994 Jumlah 539,653 7996,464 30043,163 36726,724 -1218,920 kkal/jam Q = m.ʃCp.dT Q = m.Cp.dT 0,940
Neraca Panas
H bahan masuk + Q kristalisasi = Q bahan keluar + Q loss + Q media pendingin
+ = +
+ Q media pendingin
Q media Pendingin = kkal/jam
Kebutuhan media Pendingin
Digunakan air pendingin untuk mendinginkan dengan suhu masuk 25 oC dan keluar 45 oC
Q media pendingin = m.Cp.dT
= x x
massa air pendingin =
= kg/jam
Neraca panas pada kristalizer
10. CENTRIFUGE 42885,418 42885,418 Jumlah (kkal/jam) (1,007 x 15) 2838,767 Q media pendingin 42885,418 Jumlah 44723,188 Jumlah 44723,188 1437,947 Q kristalisasi 36726,724 Q loss 399,823 7996,464 H produk massa air 1,007 (45-30) H umpan
Panas masuk Jumlah (kkal/jam) Panas keluar 19,712 42885,418 772,481 Jumlah 539,653 7996,464 C2H2O4 36726,724 1437,947 Impurities 53,167 0,350 5 0,027 399,823 1437,947 93,043 0,534 154,242 1,002 5 Larutan H2O
Entalpi bahan masuk = Panas yang keluar dari kristalizer
= kkal/jam
Entalpi Bahan keluar
Suhu bahan keluar = 30 oC = K
Neraca Panas di Centrifuge
Q = m.ʃCp.dT kkal/jam 0,015 565,240 21,259 0,027 C2H2O4 0,542 90,492 0,519 5,709 751,221 590,005 3,491 Jumlah 539,653 1437,947 H umpan 1437,947 H cake 590,005 Q larutan 847,942 Jumlah 1437,947 Jumlah 1437,947
Panas masuk Jumlah (kkal/jam) Panas keluar Jumlah (kkal/jam)
ʃCp dT Q = m.Cp.dT kkal/kgK kkal/kg C2H2O4.2H2O 3,120 0,366 5 Impurities 51,709 0,350 5 C2H2O4 19,170 0,027 315,657 Larutan H2O 149,997 1,002 5 Kristal H2O 4,245 1,002 5 C2H2O4.2H2O 308,874 0,366 5 Impurities 1,995 0,350 5 1437,947 303,15 Komponen m (kg//jam) Cp ΔT H feed T = 30 oC H filtrat T = 30 oC H cake T = 30 oC
1. GUDANG BAHAN BAKU ALANG-ALANG (GB-01)
Fungsi : Untuk menyimpan bahan baku alang-alang. Lama persediaan : 3 hari
Densitas alang-alang : fraksi Abu Silika Lignin Pentosan Selulosa
Maka densitas alang-alang = kg/m3
Laju bahan masuk = kg/jam
Faktor keloggaran = 2
Kapasitas penyediaan untuk 3 hari :
= kg/jam x 24 jam x 3 hari
= kg
Volume alang-alang :
Vgudang bahan baku = 2 x = m3
Asumsi 1 karung memiliki (p : 1 m;l : 0,5m; t : 0,3 m.) jumlah karung yang dibutuhkan
= 572 karung LAMPIRAN C
PERHITUNGAN SPESIFIKASI ALAT
Komponen Massa ρ ρc kg/jam massa (kg/m3) (kg/m3) 360,045 0,181 1600,000 289,920 567,886 0,286 1840,000 525,872 1667,6 1987,006 107,696 0,054 1150,000 62,330 71,532 0,036 2250,000 81,000 1987,006 1 hari 143064,399 879,846 0,443 1600,000 85,790 m3 Densitas 1667,602 171,581 V = Kapasitas = 143064,399 = 708,480 Jumlah 1987,006 1,000 8440,000 1667,602 571,937 85,790 0,150 = =
Gudang direncanakan : P = 1,5 l t = 8 m V = p x l x t = 1,5 l x l x 3 m = 1,5 l2 X 3 = 1,5 l2 l = m p = m
Maka design gudang
p = 8,0 m
l = 6 m
t = 8 m
2. ROTARY CUTTER KNIFE (RC-01)
Fungsi : Memotong alang-alang yang berasal dari gudang. Kapasitas : kg/jam x 1 jam / 3600 detik
: kg/s
Spesifikasi Alat : (Ulrich.1984 tabel 4-5)
Nama : Rotary Cutter
Fungsi : Memotong alang-alang
Tipe : Rotary Cutter Knife
Kapasitas maksimum : 50 kg/s Maks diameter feed masuk : 1 m
Max reduction ratio : 50
Bahan konstruksi : Carbon steel
Jumlah : 1 buah Temperatur design : 30 oC Power : 2,2 kW = 2,5 hp 1987,006 0,552 57,194 5,042 7,563 171,581
3. TANGKI PENAMPUNG ALANG-ALANG (BP-01)
Fungsi : Menampung alang alang setelah dipotong-potong Bentuk : Horizontal Silinder
Material : Carbon steel SA-283 Grade C
Jumlah : 1 buah
Menentukan Volume Bin
m = kg/jam = lb/jam
ρ = kg/m3 = lb/ft3
Laju padatan = ft3/jam
Dengan waktu tinggal 4 jam dimana volume solid mengisi 80 % volume bin digunakan 1 buah bin
Volume solid dalam bin = rate massa masuk x waktu tinggal
= x 4
= ft3
Volume solid = 80 % volume bin Volume bin = Volume solid dalam bin
=
= ft3
Menentukan dimensi tangki
Dirancang tangki berbentuk silinder tegak dengan volume
H/D =
Volume silinder (VS) = 1/4 π.D2.H = 0,25 π.D2.(1,5D) = 0,25π 1,5 D3
= D3
V tutup bawah (Ve) = 1
24 1987,006 4380,552 168,334 0,800 168,334 104,092 42,084 ρ solid 4380,552 104,092 1,178 πD3 1667,602 0,800 210,418 1,5
Volume total (Vt) = Vs + Ve = = D3 D = ft = in H = ft = in Volume Silinder Vs = ft3 Volume tutup Ve = ft3
Diameter dan tinggi tutup
Diameter tutup = diameter tangki = ft Rasio axis = 2:1
tinggi tutup =
Menentukan tekanan design (Pd) Pd = 1,05 x P hidrostatik
= 1,05 x ρ x (g/gc) x Hs
= x (105,334 x 1 x 14,041/144)
= psia = kPa
Menentukan tebal tangki 1. Tebal bagian silinder
Dipergunakan bahan konstruksi Carbon steel SA-283 grade C Efesiensi las , E =
f allowable = (Brownell & Young, 1959)
Faktor korosi = in (untuk 10 tahun )
8,351 100,218 203,227 7,191 210,418 1,178 D3 + 0,042 D3 210,418 1,219 5,568 5,568 = 1,392 ft 2 x 2 ts = Pd x ri + C fE - 0,6 Pd 1,050 6,339 43,706 0,85 12.650 0,125 5,568 66,812
dimana : ts : tebal bagian silinder (in) Pd : tekanan dalam bejana (lb/in2) ri : jari-jari dalam shell
f : allowable strees (lb/in2) E : faktor pengelasan C : Faktor korosi (in)
x (12650 x 0,85)-(0,6 x 6,339) distandarisasi menjadi 1/4 in OD = ID + 2ts = + = in = OD Standart = in = ft 2.Tebal tutup
Tutup tangki terbuat dari bahan yang sama dengan shell maka tebal tutup tangki standart yang digunakan adalah 1/4 Total tinggi tangki = Tinggi tangki +tebal tutup
= 100,218 + 1/4
= in
= ft
4. BELT CONVEYER (BC-01)
Fungsi : Mengangkut alang-alang dari bin-01 ke reaktor kalsium oksalat
Laju massa : kg/jam = ton/jam
Dari Perry & Green.1997, edisi 7. tabel 21-7, hlm 21-11 Untuk kapasitas 1,987 ton/jam, maka dipilih belt conyever :
67,437 72 6,00 100,468 8,372 + 0,125 ts = 0,127 in 66,812 0,625 ts = 6,339 2,784 1.987,006 1,987
Lebar belt = m
Kecepatan belt = 61 m/menit (untuk 32 ton/jam)
Jadi kecepatan belt = x = m/menit
Luas permukaan beban = m
Belt plies = 3 (min)
= 5 (maks)
Maksimum lump size = 2 in (sized material, 80% under) = 3 in (unsized material, not over 20%)
m = kg/jam = ton/jam = lbm/s
r = kg/m3 = lb/ft3
Rate Volumetrik = ft3/s
Alat yang direncanakan :
Jarak horizontal = 10 m = ft
Tinggi alat = 2 m = ft
Kemiringan : tga =
a =
Panjang belt (L) = (Jarak2 + tinggi2)0,5 = (102 + 22)0,5
= m = ft
Power :
Power vertikal (P1) = 0,34 hp/10 ft ( untuk 100 lb/ft3 material ) Untuk lbm/ft3material
P1 = hp
Power horizontal (P2) = 0,44 hp/10 ft ( untuk 100 lb/ft3 material )
P2 = hp
Power tambahan untuk tripped (P3) = 2 hp
Total Power (SP)= P1 + P2 + P3 = hp Efisiensi motor, hm = SP hm 3,79 32 0,01 1.987,006 1,99 1,22 0,35 1,987 61 10,198 33,452 104,108 104,108 0,012 32,808 6,5617 0,2 0,301 0,232 2,533 85% 1.667,602 Power motor = = 2,533 = 11,3 2,98 hp 0,85
