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![Page 1: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/1.jpg)
Kiln kreasi
Dik : Mr CaCO3 = 100 kg/kmol SiO2Mr MgCO3 = 84 kg/kmol Al2O3
Mr CaO = 56 kg/kmol Fe2O3Mr MgO = 40 kg/kmolMr CO2 = 44 kg/kmolMr H2O = 18 kg/kmolMr H2 = 2 kg/kmolMr O2 = 32 kg/kmolBasis = 9752.9089 kg
CaCO3 = 0.998MgCO3 = 0.00054
SiO2 = 0.00016 1Al2O3 = 0.00016Fe2O3 = 0.00014 -0.00034H2O = 0.001
0.00053CaCO3 = 9733.4031 kg/jam 97.3340MgCO3 = 5.2666 kg/jam 0.0627
SiO2 = 1.5605 kg/jamAl2O3 = 1.5605 kg/jamFe2O3 = 1.3654 kg/jamH2O = 9.7529 kg/jam 0.5418
a. konversi = 98.0000 = 0.98000.95x (yang bereaksi) = 9538.7350 kg/jam= 95.3874
x = 97.3340 kmol/jam
CaCO3 CaO +mula-mula : x
reaksi : 0.95x 0.95xsisa : x - 0.95x 0.95x
mula-mula : 97.3340reaksi : 95.3874 95.3874sisa : 1.9467 95.3874
CO2 terbentuk = 4197.0434 kg/jamCaO terbentuk = 5341.6916 kg/jam
CaCO3 sisa = 194.6681 kg/jam= 9733.4031 kg/jam
b. konversi = 100.0000 = 1.0000MgCO3 = 0.0627 kmol/jam
MgCO3 MgO +mula-mula : 0.0627
Kandungan Limestone bahan baku :
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reaksi : 0.0627 0.0627sisa : 0.0000 0.0627
CO2 terbentuk = 2.7587 kg/jamMgO terbentuk = 2.5079 kg/jam
MgCO3 = 5.2666 kg/jam
Total CO2 = 4199.8021
c. konversi = 100.0000 = 1.0000
H2O = 0.5418 kmol/jam
H2O(l) H2O(g)mula-mula : 0.5418
reaksi : 0.5418 0.5418sisa : 0.0000 0.5418
H2) gas terbentuk = 9.7529 kg/jamO2 terbentuk = 0.0000 kg/jam
= 9.7529 kg/jam neraca masa baru
1 % dari material kiln akan terbawa ke aliran udara keluar
CaCO3 = 0.0000 kg/jamSiO2 = 0.0000 kg/jam
Al2O3 = 0.0000 kg/jamFe2O3 = 0.0000 kg/jamMgO = 0.0000 kg/jamCaO = 0.0000 kg/jam
jumlah = 0.0000 kg/jam
Komponen Massa masuk Massa keluarKiln Rotary Cooler ke kompresor buble
CaCO3 9733.4031 194.6681 0.0000MgCO3 5.2666
SiO2 1.5605 1.5605 0.0000Al2O3 1.5605 1.5605 0.0000Fe2O3 1.3654 1.3654 0.0000H2O(l) 9.7529H2O(g) 9.7529
MgO 2.5079 0.0000CO2 4199.8021CaO 5341.6916 0.0000 kemurnian CO2
sub jumlah 5543.3539 4209.55509752.9089 9752.9089
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rotary cooler
Komponen Massa masuk
CaCO3 194.6681 kg/jamMgO 2.5079 kg/jamSiO2 1.5605 kg/jam
Al2O3 1.5605 kg/jamFe2O3 1.3654 kg/jamCaO 5341.6916 kg/jam
5543.3539
S
Komponen
CaCO3 19.4668 kg/jamMgO 0.0025 kg/jamSiO2 0.0016 kg/jam
Al2O3 0.0016 kg/jamFe2O3 0.0015 kg/jamCaO 5.3417 kg/jam
jumlah 24.8156Komponen
yang keluar ke beltCaCO3
Komponen MgOSiO2
CaCO3 175.2013 kg/jam Al2O3MgO 2.5054 kg/jam Fe2O3SiO2 1.5589 kg/jam CaO
Al2O3 1.5589 kg/jam sub jumlahFe2O3 1.3639 kg/jam jumlahCaO 5336.3499 kg/jam
jumlah 5518.5383
SIKLON
Ke stack
furnace elektrik
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Ke rotari cooler 2
Komponen MASUK siklon
CaCO3 19.4668 kg/jamMgO 0.0025 kg/jamSiO2 0.0016 kg/jam
Al2O3 0.0016 kg/jamFe2O3 0.0015 kg/jamCaO 5.3417 kg/jam
24.8156Walas hal 616,1988, 1% material akan terbawa keluar dari siklon
Komponen
CaCO3 0.0195 kg/jamMgO 0.0000 kg/jamSiO2 0.0000 kg/jam
Al2O3 0.0000 kg/jamFe2O3 0.0000 kg/jamCaO 0.0059 kg/jam
0.0253
yang menuju belt
Komponen
CaCO3 19.4473 kg/jamMgO 0.0025 kg/jamSiO2 0.0016 kg/jam
Al2O3 0.0016 kg/jamFe2O3 0.0015 kg/jamCaO 5.3358 kg/jam
jumlah 24.7903
Belt
Komponen masukdari rotari cooler dari siklon jumlah
CaCO3 175.2013 19.4473 194.6486MgO 2.5054 0.0025 2.5079
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SiO2 1.5589 0.0016 1.5605Al2O3 1.5589 0.0016 1.5605Fe2O3 1.3639 0.0015 1.3654CaO 5336.3499 5.3358 5341.6857
sub jumlah 5518.5383 24.7903jumlah 5543.3286 5543.3286
konsentrai CaO 96.362423501364
Bila Cao yang digunakan sebanyak 6484.64812083708 Kg /jammaka butuh penampungan sementara sebab perjamnya dihaslnya Cao sebanyak
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60 kg/kmol102 kg/kmol160 kg/kmol
densitas2710 Kg/m3 Volume2958 Kg/m32650 Kg/m3
3965 Kg/m35250 Kg/m31000 Kg/m3
=kmol/jam =
=
kmol/jam
CO2
0.95x0.95x
95.387495.3874
CO2
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0.06270.0627
neraca masa barukpmonen mol masuk mol keluar mol produksi
CaCO3 97.334031077151 1.9467 -95.3874MgCO3 0.0626972716644 0.0000 -0.0627
SiO2 0.0260077571348 0.026007757 0.0000Al2O3 0.0152986806676 0.015298681 0.0000Fe2O3 0.0085337953099 0.008533795 0.0000H2O(l) 0.5418282736426 0.0000 -0.5418H2O(g) 0.5418 0.5418
MgO 0.0627 0.0627CO2 95.4500 95.4500CaO 95.3874 95.3874
sub jumlah 97.988396855571 193.4384446 95.4500477272727 sub jumlah
mol produksi= input -output
CO2 dari rekoverimake up
99.768314966668 %
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yang harus ddibuat dalam pabrik CO2 sebnayak4199.8021
umpan produksiklon belt
194.6681 19.4668 175.20132.5079 0.0025 2.50541.5605 0.0016 1.55891.5605 0.0016 1.55891.3654 0.0015 1.3639
5341.6916 5.3417 5336.349924.8156 5518.5383
5543.3539 5543.3539
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-1142.9624
maka butuh penampungan sementara sebab perjamnya dihaslnya Cao sebanyak
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3.5916620.00178
0.0005890.0003940.00026
0.0097533.604438 m3/jam3604.438 dm3/jam3604.438 liter/jam
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komponen massa in massa out massa produksi
CaCO3 9733.4031 194.6680621543 -9538.7350MgCO3 5.2666 0 -5.2666
SiO2 1.5605 1.560465428091 0.0000Al2O3 1.5605 1.560465428091 0.0000Fe2O3 1.3654 1.365407249579 0.0000H2O(l) 9.7529 0 -9.7529H2O(g) 0.0000 9.752908925566 9.7529MgO 0.0000 2.507890866574 2.5079CO2 0.0000 4199.8021 4199.8021CaO 0.0000 5341.691625514 5341.6916
sub jumlah 9752.9089 9752.9089 0.0000
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BM: NaCl 58.5 kg/kmol CO2 44.0H2O 18.0 kg/kmol NH3 17.0CaSO4 136.0 kg/kmol NaHCO3 84.0MgO 40.0000 CaO 56.0000
1, di MIX TankNaCl
BrineH2O
Basis 13910.7131 kg /jam garam industrikomponen garam - 99,5 % NaCl;
0,1 % CaSO4;, 0,4 % H2O
maka persentase kali basis untuk mendapatkan jumlah komponen
NaCl 13841.1595 kg /jam 236.6010 kmol/jamCaSO4 13.9107 kg /jam 0.1023 kmol/jamH2O 55.6429 kg /jam 3.0913 kmol/jam
kelarutan NaCl pada 30 C= 36,3 kg NaCl/100 kg H2O (Perry table-2-120)
maka 99,5 kg NaCl membutuhkan air sebanyak:
H2O yang dibutuhkan; 99,5/36,3*100 = 38129.9160
jika sudah ada kandungan air dalam garam industri sebesar 0,4 kg maka maka fress feed yang dibutuhkan:fress feed H2O; 274,1047 - 0,4= 38074.2732
Komponen masuk keluargaram industri fress feed Brine
NaCl 13841.1595 13841.1595CaSO4 13.9107 13.9107H2O 55.6429 38074.2732 38129.9160subjumlah 13910.7131 38074.2732jumlah 51984.9862 51984.9862
2. absorber
kimponenumpan make up recovery produk
NaCl 13841.1595 13841.1595CaSO4 13.9107 13.9107H2O 38129.9160 33733.3124NH3 4152.3479 0Co2 0 0NH4 OH 8548.9515
51984.9862 4152.3479 0.0000
MT
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56137.3341 56137.3341
2. Reaktor Buble
Brine NH4Cl + NaHCO3
NH3CO2
NH40H + CO2+ NaCl NaHCO3+
konversi NaCl menjadi NaHCO3 85% = 0.8000
NaCl yang terkonversi sebanyak ;0,85*99,5 11072.9276mol NACl mula2 99,5/58,4428mol H2O mula2 38129.9160 kg/jam / 18,01534=NH4OH mula2 244.255756089
co2 yang berubah menjadi 0.8 mula2 co2 harus ada agar na4oh dapat bereaksi seluruhnyaNH4OH + CO2 NH4HCO3
m 244.2558 244.2558berekasi 195.4046 195.4046 195.4046sisa 48.8512 48.8512 195.4046
NH4HCO3 + NACL NaHCO3+ NH4Cl
195.4046 236.6010189.2808 189.2808 189.2808 189.2808
6.1238 47.3202
sisa NH4HCO3 483.7795 kg
Mula2 :NH4OHmula2 8548.9515NaCl 1,7025*58,5 13841.1595
CO2 awal 1,3620*44 10747.2533jumlah 33137.3642
Sisa dan terbentukNH4OH sisa 1709.7903NaCl sisa 0,3405*58,5= 2768.2319NaHCO3 terbentuk= 1,3620*84= 15899.5884Nh4Cl terbentuk= 1,3620*53,5= 10126.5235
http://www.netl.doe.gov/publications/proceedings/04/carbon-seq/158.pdf
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CO2 sisa 2149.4507jumlah 32653.5848
komponen masuk keluar liiquid keluar gasNaCl 13841.1595 2768.2319CaSO4 13.9107 13.9107H2O 33733.3124 33733.3124NH4OH 8548.9515 1709.7903CO2 10747.2533 2149.4507NH4Cl 10126.5235NaHCO3 15899.5884NH4HCO3 483.7795jumlah 66884.5874 64735.1367 2149.4507
66884.5874
kristalizerkritaizing pada suhu 22 oC
komponenumpanNaCl(l) 2768.2319
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CaSO4(s) 13.9107H2O(l) 33733.3124NH4Cl(l) 10126.5235NaHCO3(l) 15899.5884NH4HCO3 483.7795NH4OH 1709.7903
64735.1367menurut paten 5275794
maka:
komponenUmpan output
NaCl(s) 2768.2319 2768.2319
CaSO4(s) 13.9107 13.9107
H2O(l) 33733.3124 33733.3124
NH4Cl(l) 10126.5235 10126.5235
NaHCO3(s) 15899.5884 15779.3481NaHCO3(aq) 120.2402
NH4HCO3 483.7795 483.7795Nh4OH 1709.7903 1709.7903total 64735.1367 64735.1367
RDVF
umpan cake
recovery
komponenumpan densitas kg/lt
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NaCl(l) 2768.2319 kg/jam 2.1650CaSO4(s) 13.9107 kg/jam 2.9600H2O(l) 33733.3124 kg/jam 0.9971NH4Cl(l) 10126.5235 kg/jam 1.5274NaHCO3(s) 15779.3481 kg/jamNaHCO3(aq) 120.2402 kg/jamNH4HCO3 483.7795 kg/jam 1.58nh4OH 1709.7903 kg/jam 0.88
64735.1367
Kebutuhan air pencuci pada RDF-301 = kandungan air yang terdapat dalam cake . (Mc.Cabe, 1985)voulme ruang kosong 0.01ml/gram Kirk Otmer , sehingga zat yang besrifat polar akan berada atau tercampur dalam cake
0.01 l/kg
H2O yang terdapat pada cake adalah 0.0100maka= 338.3210 lt/jam
NaCl Aq 12.7863 lt/jamNH4OH 19.42943514344 lt/jamNH4Cl 66.29909349345 lt/jamNaH4HCO3 3.0619 lt/jamCaSO4 0.046995652287 lt/jam
439.9447
komponen masuk keluarumpan air pencuci filter cake
NaCl 2768.2319 2740.5496 27.6823CaSO4(s) 13.9107 13.7716 0.1391H2O(l) 33733.3124 337.3331 33733.3124 337.3331NH4Cl(l) 10126.5235 10025.2583 101.2652NaHCO3(s) 15779.3481 0.0000 15779.3481NaHCO3(aq) 120.2402 120.2402 0.0000NH4HCO3 483.7795 478.9417 4.8378nh4OH 1709.7903 1692.6924 17.097902926227subjumlah 64735.1367 337.3331 48804.7662 16267.7036jumlah 65072.4698 65072.4698
4. furnace elektrik
umpan ke rotary calciner adalah cake dari RDVF
komponen umpancakeNaCl 27.6823 kg/jam 0.4732 kmol/jamCaSO4 0.1391 kg/jam 0.0010228 kmol/jamH2O 337.3331 kg/jam 18.7407 kmol/jamNaHCO3 15779.3481 kg/jam 187.8494 kmol/jamNH4OH 17.0979 kg/jam 0.4885 kmol/jam
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NH4HCO3 4.8378 kg 0.0612 kmol/jamNH4Cl(l) 101.2652 1.8928 kmol/jam
16267.7036 209.5069
Reaksi 1Konversi 0.9990 paten Us 5325606 dan menurut
BM 168.0000 106.0000 44.0000 18.00002NaHCO3 Na2CO3 + CO2(gas)+ H2O(gas)
M: 187.8494R; 187.6615 93.8308 93.8308 93.8308sisa 0.1878 93.8308 93.8308 93.8308
Na2CO3 terbentuk 0,6742*106 9946.0612Co2 gas terbentuk 0,6742*44 4128.5537H2Ogas terebentuk 0,6742*18 1688.9538NahCO3 sisa 15.7793
reaksi 3
aliran H2O dan CO2
NH3 9.3457H2O= 2026.0895
CO2= 4131.24826166.6834
produk padatan yang ke rotary coolerNa2CO3 9946.0612NaCl`` 27.6823CaSO4 0.1391nahCO3 15.7793NH4Cl 101.2652
jumlah 10090.9273air 10.0929
jumlah 10101.0202air keluar bersama produk 0.1% 0.9998
maka X = 0.1%*(D214+X)X=jumlah air yang terdapat dalam produk
mk:X= 0.001+(71.4466+X)0.999X= 10.0909
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X= 10.0929 0.5607 kmol
jumlah air yang bersama produk sebesar 0.0913 kg/ jam
maka air menguap 24.9568-0.0913 327.2402komponen umpan produk
cake steam gas padatanNaCl 27.6823 0.0000 27.6823CaSO4 0.1391 0.0000 0.1391H2O 337.3331 2026.0895 10.0929NaHCO3 15779.3481 15.7793Na2CO3 0.0000 9946.0612CO2 4131.2482NH4OH 17.097902926227NH3 9.3457NH4Cl 101.2652 101.2652NH4HCO3 4.8378sub jumlah 16267.7036 0.0000 6166.6834 10101.0202jumlah 16267.7036 16267.7036
5.rotary cooler
komponen masukdari furnace jumlah
NaCl 27.6823 27.6823CaSO4 0.1391 0.1391H2O 10.0929 10.0929Na2CO3 9946.0612 9946.0612nahco3 15.7793NH4Cl 101.2652jumlah 10101.0202
0.1% padatan akan terbawa udara/gas menuju siklon
NaCl 0.0277CaSO4 0.0001H2O 0.0101na2CO3 9.9461nahco3 0.0157793481145765nh4Cl 0.101265235401895
10.1010yang ke belt
NaCl 27.6546CaSO4 0.1390H2O 10.0829na2CO3 9936.1152nahCo3 15.7636nh4cl 101.1640
10090.9192
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neracakomponen masuk komponen keluar
dari furnace ke siklon 1 ke beltNaCl 27.6823 0.0277 27.6546CaSO4 0.1391 0.0001 0.1390H2O 10.0929 0.0101 10.0829Na2CO3 9946.0612 9.9461 9936.1152naHco3 15.7793 0.015779348115 15.7636nh4CL 101.2652 0.101265235402 101.1640sub jumlah 10101.0202 10.1010 10090.9192jumlah 10101.0202 10101.0202
7. siklon 1komponen masuk komponen keluar
ke stak ke beltNaCl 0.0277 2.7682319E-05 0.0277CaSO4 0.0001 1.39107131E-07 0.0001H2O 0.0101 1.00929458E-05 0.0101Na2CO3 9.9461 0.009946061246 9.9361naHco3 0.0157793481145765 1.57793481E-05 0.015763568766NH4Cl 0.101265235401895 0.000101265235 0.101163970166jumlah 10.1010 0.010101020201 10.0909
10.10108. belt
komponen masuk
dari siklon2 dari rotary cooler keluar beltNaCl 0.0276546367043917 27.6546 27.6823CaSO4 0.000138968023640159 0.1390 0.1391H2O 0.0100828528985004 10.0829 10.0929Na2CO3 9.9361 9936.1152 9946.0513nahCO3 0.0157635687664619 15.7636 15.7793nh4Cl 0.101163970166493 101.1640 101.2651
10.0909 10090.9192 10101.0101
5, Recovery di CSTRfiltrat RDVF
CaO
CaCl2
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komponen masuk produk
cairan gasNaCl 2740.5496NH4HCO3 478.9417H2O 33733.3124NH4Cl 10025.2583NaHCO3 120.2402nh4OH 1692.6924CaSo4 13.7716
CaCO3 194.6486MgO 2.5079SiO2 1.5605
Al2O3 1.5605Fe2O3 1.3654
CaO 5341.6857sub jumlah 48804.7662 5543.3286julmlah 54348.0948
conversi NH4Cl 98 %
rx1 2NH4Cl + CaO CaCl2+2NH3+H2ORX2 2NH4Cl + MgO MgCl2+2NH3+H2O
Rx1NH4Cl mula2 10025.2583 kg/jam= 187.3880 kmol/jamX= 0.9800NH4CL terkonversi 183.6402 kmol/jamCaO mula2 5341.6857 kg/jam 95.3872 kmol/jam
MK:2NH4Cl + CaO CaCl2 +2NH3 +H2O
M: 187.3880 95.3872R 183.6402 91.8201 91.8201sisa 3.7478 3.5671 91.8201
2NH4Cl sisa 4.0753*53.5= 200.5052 kg/jamcao sisa 103.2683*56= 199.7589 kg/jamCaCl2 trbntk 99.8457*111= 10192.0336 kg/jamNh3 grbntuk 199.6914*17= 3121.8842 kg/jamH2O trbntk 99.8457*18= 1652.7622 kg/jam
RX2 2NH4Cl + MgO MgCl2 +2NH3 +H2O
NH4Cl mula2 200.5052 kg/jam 3.7478 kmol/jam
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MgO mula 2.5079 kg/jam 0.0627 kmol/jamX MgO 0.8800 0.0552
M; 3.7478 0.0627R: 0.1103 0.0552 0.0552 0.1103sisa 3.6374 0.0075 0.0552 0.1103
NH4Cl sisa 194.6016 kg/jamMgO sisa 0.3009MgCl2 trbntk 5.2415NH3 trentuk 1.8759H2O terbentu 0.9931
RX3 NH4HCO3 NH3+ H2O CO2
m 6.06255312549r 6.06255312549 6.06255312549 6.06255312549 6.06255312548997
NH3 terbentuk 103.0634031333 kg/jamH2O terbentuk sebagai uap air 109.1259562588 kg/jamCo2 gas 266.7523375216 kg/jam
RX 4 NH4oH NH3 H20m 48.3626397056 48.36263970561 48.36263970561
NH3 terbentuk 822.1648749954 kg/jamH2O terbentuk sebagai uap air 870.527514701 kg/jam
dalam CSTR 80 oC terbentu 0.2 % H2O menjadi uap maka : 72.7334komponen masuk produk
cair gasCaSO4 13.7716 13.7716NaCl 2740.5496 2740.5496NH4HCO3 478.9417H2O 33733.3124 36293.9878 72.7334NH4Cl 10025.2583 194.6016
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NaHCO3 120.2402 120.2402NH4OH 1692.6924
CaCO3 194.6486 194.6486SiO2 1.5605 1.5605
Al2O3 1.5605 1.5605Fe2O3 1.3654 1.3654MgO 2.5079 0.3009CaO 5341.6857 199.7589
NH3 4048.9884MgCl2 5.2415CaCl2 10192.0336CO2 266.7523
49959.6207 4388.474154348.0948 54348.0948
NH3 yang dibutuhkan 4152.3479NH3 yang tersedia dari recoveri 4048.9884
jadi make up NH3 94.0138 kg/jam
6.Condenser Parsial kondisi 1 atm80 c
Komponen masukgas
H2O 2026.0895CO2 4131.2482NH3 9.3457Memisahkan CO2 dari campuran steam untuk umpan masuk ke reaktor. CO2 adalah komponen non condensable sehingga tidak mengembun.air merupakan komponen kondensable Nh3 juga akan menjadi liquid pada suhu 25 C dengan tekan 9,8 atm sehingga Nh3 juga bersifat non kondensableKirk Otmer keluaran dari Kiln kalsiner mengandung 99.7% kemurnian CO2 yang akan menjadi make upfeed pada reaktor.CO2 campuran dengan air 100/95*4438.0394 4143.6792air yang ikut menguap 77.6845air yang mengembun 1948.4049
komponen masuk produkgas cair gas
H2O 2026.0895 1948.4049 77.6845CO2 4131.2482 4131.2482NH3 9.3457 9.3457
1948.4049 4218.2785jumlah 6166.6834 6166.6834
gak usah diketik , gambarr lihat mbak upi or bob, 7,separator Hilangin aja massa separator di word, n kata separator di panas buang aja
![Page 23: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/23.jpg)
memisahkan komponen fasa cair dan fasa gas pada produk keluaran condenser parsial
Komponen masuk produkcair gas cair gas
H2O 1948.4049 77.6845 1948.4049 77.6845CO2 4131.2482 4131.2482sub jumlah 1948.4049 4208.9327 1948.4049 4208.9327jumlah 6157.3377 6157.3377
CO2 yang di recoveri sebayak 4398.0005CO2 yang dibutuhin 6349.2527
koNENDSWER PARSIAL
Umpan masuk KG/.JAM KMOL/JAM ZiH2O 2026.0895 112.5605257874 0.54376466777CO2 4131.2482 93.89200437739 0.453579567165NH3 9.3457 0.549749422536 0.002655765065
6166.6834 207.0022795874 1
PARAMETER TEKANAN UAP MURNI (YAWS)
A B C DH2O 29.8605 -3.15E+03 -7.30E+00 2.42E-09CO2 35.0187 -1.51E+03 -1.13E+01 9.34E-03NH3 37.1575 -2.03E+03 -1.16E+01 7.46E-03
T= 206.140455763024 K P= 11 Menghitung Buble point
komponen P'atm Ki=P'/P xi yi=Ki.xi
H2O 7.29606558656008E-06 7.29606559E-06 0.54376466777 3.967342679703E-06CO2 3.13452983559845 3.134529835598 0.453579567165 1.42175868609625NH3 0.135489105504655 0.135489105505 0.002655765065 0.0003598272331089Σ 1 1.42212248067204setelah di goal seek dapet T buble 206.140455763 K
-66.859544237 C
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2 menhitung dew point
T= 362.747259291524 K P 1komponen P'atm Ki=P'/P yi xi=yi/Ki
H2O 0.6813 0.6813 0.5438 0.7981CO2 220.2251 220.2251 0.4536 0.0021NH3 50.1044814707958 50.1044814708 0.002655765065 0.0001Σ 0.9973 0.8002setelah di goal seek dapet T dew 362.7472592915 k
89.74725929152 c
3 menhitung T kondensansi
T kondensasi harus berada di antara T buble dan T dew
misal T 35.8475832162885 C P= 1308.847583216288 K
misal persen umpan MENNGUAP(f) 0.477084413848
komponen P'atm Ki=P'/P Xf Xi
H2O 0.057752171 0.057752171 0.5438 0.9878CO2 80.1738 80.1738 0.4536 0.0117NH3 13.5694950858803 13.56949508588 0.002655765065 0.0004Σ 0.9973 0.9999
SEHINGGA t KONDENSASI PADA 35.84758321629 CDENGAN F 0.477084413848
KOMPONEN UAP DAPAT DIHITUNGA DENGAN
komponen yi=Ki.xi
H2O 0.0570CO2 0.9379NH3 0.0052Σ 1.0001
MAKAv= 98.757561222162 KMOL/JAML 108.244718365206 KMOL/JAM
∑i−1
Nc
xi=∑i−1
Ncxfi
f ( Ki−1)+1=1
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207.002279587368
KARENA KOMPONEN CO2 YANG TERDAPAT DALAM FRAKSI CAIR SANGAT KECIL MAKA DIASUMSIKAN SEMUA CO2 BERADA DALAM FASE GAS
KOPONEN H20 TERUAPKAN 4.315807422237 KMOL/JAM 77.6845336002655MISAL CO2 MENUAP SELURHNYA MK 4131.2481926051NH3 mengupa seluruhnya 9.3457
JUMLAH AIR MENGEMBUN 1948.404930573716166.6833969622
komponen L KMOL/JAM L KG/JAM v KMOL/JAM V. KG/JAM
H2O 108.244718365206 1948.404930574 4.315807422237 77.6845336002655CO2 0 0 93.89200437739 4131.2481926051NH3 0.549749422536 9.3457Σ 1948.404930574 98.75756122216 4218.27846638848
6166.6834
komponen INPUT OUPUTCAIR GAS
H2O 2026.0895 1948.404930574 77.68453360027CO2 4131.2482 4131.248192605NH3 9.3457 9.3457Σ 6166.6834 1948.4049 4218.2785
6166.6834
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kg/kmol NH4C 53.5 kg/kmolkg/kmol Na2C 106.0 kg/kmolkg/kmol CaCl 111.0 kg/kmol NH4HCO3 79.0000
MgCl 95.0 NH4OH 35.0000 kg/kmol
maka
Yang sudah dari recoveri
Mixing tank setelah penambahan dari recoveri
karena ada penambhan air dari recoveri maka fress feed yang dimasukansebanyak 72.7334 kg komponen
Komponen masuk keluargaram industri fress feed Brine
kg /jam NaCl 13841.1595 13841.1595CaSO4 13.9107 13.9107
jika sudah ada kandungan air dalam garam industri sebesar 0,4 kg maka maka fress feed yang dibutuhkan: H2O 55.6429 37913.8194 37969.4622kg /jam subjumlah 13910.7131 37913.8194
jumlah 51824.5325 51824.5325
Absorber dengan penambahan amoniak baru
kimponenumpan make up recovery
NaCl 13841.1595CaSO4 13.9107H2O 37969.4622 0.2829 72.7334NH3 94.0138 4058.3341Co2 0.0000 266.7523NH4 OH
51824.5325 94.2966 4397.819956316.6490
perbandingan berat bahan baku NH#/NaCl adalah maka NH3 yang dibuthkan pperbandingan umpan NH3 /NaCladalah 0.3jadi NH3 yang dibutuhkan 4152.3478535
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bila NH4OH dibutuhkan sebanyakmaka NH3 yang diserap dalam absorbsi sebanyak
###NH3+ H2O
mular 244.2558 244.2558NH3 yang diabsorb sebanyakair yang menyerap sebnayakjadi jumlah NH4OH
neraca masa di buble setelah penambahan CO2 baruNH4Cl
komponen absorber absorber top dari furnaceNaCl 13841.1595CaSO4 13.9107
kg/jam= ### kmol/jam H2O 33645.8750 9.7529### kmol/jam NH4OH 8548.9515### kmol/jam CO2 266.7523 4199.8021
NH4ClNaHCO3
mula2 co2 harus ada agar na4oh dapat bereaksi seluruhnya NH4HCO356049.8966 266.7523 4209.5550
66884.5874
33733.3124 6547.4512
CO2 dibuthkan di unit CACO3
kg/jamkg/jam
kpmonen mol masuk mol keluar mol produksikg/jam
NaCl 236.6010172941 47.320203459 -189.2808CaSO4 0.102284654979 0.102284655 0.0000
H2O 1874.072911145 1874.0729111 0.0000NH4OH 244.255756089 48.851151218 -195.4046
kg/jam CO2 244.2557564479 48.851151218 -195.4046kg/jam NH4Cl 0 189.28081384 189.2808kg/jam NaHCO3 0 189.28081384 189.2808
http://www.netl.doe.gov/publications/proceedings/04/carbon-seq/158.pdf
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kg/jam NH4HCO3 0 6.1237910358 6.1238
sub jumlah 2599.287725631 2403.8831204 -195.40460520.4713
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kelarutan sodium bikarbonate pada 25 oC adalah 0.1 g NaHCO3/100g air (A.Saberi at al,2008.precipitation kinetics of sodium bikarbonate in an industrial buble colum cristalizer. Wiley interscience)
persen berat NaHCO3 padatan dalam l 0.2456 100 air0.1 NaHCO3
total larutan 64735.1367
maka kristal yang terbentuk adalah :
C=
Dimana :C= berat kristal yang terbentukF = berat larutan% wt= % berat nAhco3 dalam larutanDimana,
%w= 0.2456
MAKA 15779.34811458 KG187.8493823164 kmol
any
Komponen ρi (kg/m3)NaCl 2165.0000CaSO4 2960.0000H2O 997.0800
dalam liter NaHCO3 (aq) 2159.0000
F x (% wt C6 H 4NH2 OH − % kelaru tan
1 − % kelaru tan )
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1278.6291 NaHCO3 (l) 2159.00004.6996 NH4Cl 1527.4000
33832.1021 NH4HCO3 1580.00006629.9093 NH4OH 880.0000
306.18961942.9435
43994.4732
Kebutuhan air pencuci pada RDF-301 = kandungan air yang terdapat dalam cake . (Mc.Cabe, 1985)voulme ruang kosong 0.01ml/gram Kirk Otmer , sehingga zat yang besrifat polar akan berada atau tercampur dalam cake
kg/jam337.333124
27.6823190217.09790293101.26523544.8377949180.139107131
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terdecomposisi pada 60 Cterdekomposisi pad 340.0000 C
reaksi 2titk didih NH4OH pada 27 oC sehingga pada STD ini engan suhu 120 C asumsi akan terurai menjadi NH3 dan uap h2O secara sempurana
BM 35.0000 17.0000 18.0000NH4OH NH3 + H2O
M: 0.4885R; 0.4885 0.4885 0.4885sisa 0.0000 0.4885 0.4885
kg/jam NH3 T 8.3047 kg/jamkg/jam H2O t 8.7932 kg/jamkg/jamkg/jam
CTTan:
NH4HCO3 NH3 CO2 H2OM: ###R; ### 0.0612 0.0612 0.0612sisa ### 0.0612 0.0612 0.0612
CO2 t 2.6945 kg/jamNH3 t 1.0410 kg/jamH2O t 1.1023 kg/jam
kpmonen mol masuk mol keluar mol produksigas padatan gas padatan
NaCl 0.4732 0.4732CaSO 0.0010 0.0010H2O 18.7407 112.5605 0.5607 94.3805NaHC 187.8494 0.1878 -187.6615Na2CO3 93.8308 93.8308
http://docs.google.com/viewer?a=v&q=cache:YQLS8RN43dgJ:www.ch.ntu.edu.tw/~genchem99/msds/exp21/Ammonium%2520hydroxide%2520water%2520solution.pdf+decomposition+of+ammonium+hydroxide&hl=id&gl=id&sig=AHIEtbTtel3g1iq0S3psR4tIpMq-xA-Vyw
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CO2 93.8920 93.8920NH4O 0.4885 0.0000 -0.4885NH3 0.5497 0.5497NH4C 1.8928 1.8928
kg/jam NH4H 0.0612 0.0000 -0.0612sub j 209.5069 207.0023 96.9464 94.4418jumlah
komp massa in massa out massa produksigas padatan gas padatan
NaCl 27.6823 27.6823CaSO 0.1391 0.1391H2O 337.3331 2026.0895 10.0929 1698.8493NaHC 15779.3481 15.7793 -15763.5688Na2CO3 9946.0612 9946.0612CO2 4131.2482 4131.2482NH4O 17.0979 -17.0979NH3 9.3457 9.3457NH4C 101.2652 101.2652NH4H 4.8378 -4.8378sub j 16267.7036 6166.6834 10101.0202 0.0000 0.0000jumlah 16267.7036
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bahan baku 80000tokemurnian produk10101.0101 kg/ja CaSO4 0.0014 %
H2O 0.0999 %NaCl 0.2741 %nahCO3 0.15621539013 %Na2CO3 98.4659 %nh4cl 1.0025 %
kondisi gas NH3
1,1 atm 30 C
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CaCl2 +2NH3 +H2O
183.6402 ###183.6402 ###
MgCl2 +2NH3 +H2O
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0.05520.0552
terdekomposisi pada suhu 60 oC Patnaik 2003,di dalam CSTR suhu 80 oCdianggapa terdekomposisi sempurna
kpmonen mol masuk mol keluar mol produksigas liwuid
CaSO4 0.1013 0.1013 0.0000NaCl 46.8470 46.8470 0.0000NH4HCO3 6.0626 0.0000 -6.0626H2O 1874.0729 4.0407 2016.3327 146.3005NH4Cl 187.3880 3.6374 -183.7506NaHCO3 1.4314 1.4314315189 0.0000NH4OH 48.3626 0.0000 -48.3626
CaCO3 1.9465 1.9465 0.0000SiO2 0.0260 0.0260 0.0000
Al2O3 0.0153 0.0153 0.0000Fe2O3 0.0085 0.0085 0.0000MgO 0.0627 0.0075 -0.0552CaO 95.3872 3.5671 -91.8201
NH3 238.1758 238.1758MgCl2 0.0552 0.0552CaCl2 91.8201 91.8201CO2 6.06255312549 6.0625531255sub jumlah 2261.7121 340.1544 2073.9207 152.3630jumlah
4.0407 kmol
48804.7662
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kadar NH3 yang ke 92.2641
kandungan NH3 makjadi make up sebenranya 94.2966sehingga kandungan air = 0.2829
Memisahkan CO2 dari campuran steam untuk umpan masuk ke reaktor. CO2 adalah komponen non condensable sehingga tidak mengembun.Nh3 juga akan menjadi liquid pada suhu 25 C dengan tekan 9,8 atm sehingga Nh3 juga bersifat non kondensable
keluaran dari Kiln kalsiner mengandung 99.7% kemurnian CO2 yang akan menjadi make up
kg/jamkg/jamkg/jam
gak usah diketik , gambarr lihat mbak upi or bob, Hilangin aja massa separator di word, n kata separator di panas buang aja
![Page 37: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/37.jpg)
memisahkan komponen fasa cair dan fasa gas pada produk keluaran condenser parsial
4.3158
kemurnian CO2 98.1543 %
E1.81E-067.76E-10
-9.58E-12
arm
YAWS
2.78973E-060.9997441890.0002530210.999746979
yi =ki*xi/Σkixi
log10 P=A+BT
+C log10T +DT +ET 2
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atm
0.99740.00260.00010.9999
atm
0.987920760.0116996290.000379611
1
77.6845
xi =(yi/Ki)/(Σ yi/Ki)
∑i−1
Nc
xi=∑i−1
Ncxfi
f ( Ki−1)+1=1
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KARENA KOMPONEN CO2 YANG TERDAPAT DALAM FRAKSI CAIR SANGAT KECIL MAKA DIASUMSIKAN SEMUA CO2 BERADA DALAM FASE GAS
KG/JAM 0.69
over all
c input out put
NaCl 13841.1595 2768.2319CaSO4 13.9107 13.9107H2O 38316.8312 38252.4856
CaCO3 9733.4031 194.6681MgCO3 5.2666
SiO2 1.5605 1.5605Al2O3 1.5605 1.5605Fe2O3 1.3654 1.3654
NH3 make up 94.0138Na2CO3 9946.0612NH4Cl 295.8668
NaHCO3MgO 0.3009CaO 199.7647
MgCl2 5.2415CaCl2 10192.0336
NaHCO3 136.019662009.0712 62009.0711
0.0000316
CaCO3MgOSiO2
Al2O3Fe2O3CaO
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karena ada penambhan air dari recoveri maka fress feed yang dimasukan
36.4534
2109.4146
pproduk 11390.8387top botom
0.0000 13841.15950.0000 13.9107
33645.8750
266.75230.0000 8548.9515
266.7523 56049.8966 38042.478656316.6490
pperbandingan umpan NH3 /NaCl
244.25575609
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222.894722887maka NH3 yang diserap dalam absorbsi sebanyak
NH4OH
244.25584152.3479 kg4396.6036 kg8548.9515
dari separator top produk botoom produk top produk2768.2319
13.910777.6845 33733.3124
1709.79034131.2482 2149.4507 2149.4507
10126.523515899.5884
483.77954208.9327 2149.4507 64735.1367 2149.4507
66884.587414.1477
CO2 dibuthkan di unit CACO34199.8021
komponen massa in massa out massa produksi
NaCl 13841.1595117 2768.23190234 -11072.9276CaSO4 13.9107130771 13.9107130771 0.0000
H2O 33733.3124006 33733.3124006 0.0000NH4OH 8548.95146311 1709.79029262 -6839.1612CO2 10747.2532837 2149.45065358 -8597.8026NH4Cl 10126.5235402 10126.5235NaHCO3 15899.5883622 15899.5884
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NH4HCO3 483.779491832 483.7795
sub jumlah 66884.5874 66884.5874 0.0000
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(A.Saberi at al,2008.precipitation kinetics of sodium bikarbonate in an industrial buble colum cristalizer. Wiley interscience)
komponen
NaCl(s)CaSO4(s)
H2O(l)
NH4Cl(l)
NaHCO3(s)
NaHCO3(aq)
NH4HCO3Nh4OH
total
( Chopey,N, Hal.108 F x (% wt C6 H 4 NH2 OH − % kelaru tan
1 − % kelaru tan )
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titk didih NH4OH pada 27 oC sehingga pada STD ini engan suhu 120 C asumsi akan terurai menjadi NH3 dan uap h2O secara sempuranahttp://docs.google.com/viewer?a=v&q=cache:YQLS8RN43dgJ:www.ch.ntu.edu.tw/~genchem99/msds/exp21/Ammonium%2520hydroxide%2520water%2520solution.pdf+decomposition+of+ammonium+hydroxide&hl=id&gl=id&sig=AHIEtbTtel3g1iq0S3psR4tIpMq-xA-Vyw
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komponen massa in massa outgas padatan
CaSO4 13.7716 13.7716NaCl 2740.5496 2740.5496NH4HCO3 478.9417 0.0000 0.0000H2O 33733.3124 72.7334 36293.9878NH4Cl 10025.2583 194.6016NaHCO3 120.2402 0.0000 120.2402NH4OH 1692.6924
CaCO3 194.6486 0.0000 194.6486SiO2 1.5605 1.5605
Al2O3 1.5605 1.5605Fe2O3 1.3654 1.3654MgO 2.5079 0.3009CaO 5341.6857 199.7589
NH3 4048.9884MgCl2 5.2415CaCl2 10192.0336CO2 266.7523sub jumala 54348.0948 4388.4741 49959.6207
54348.0948
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over all 2masuk
komponen garam air proses maake up NH3
NaCl 13841.1595CaSO4 13.9107H2O 55.6429 37913.8194 0.2829NH3 94.0138
CaCO3MgCO3
SiO2Al2O3Fe2O3
13910.7131 37913.8194 94.2966
62009.0712
keluarkomponen Belt 05 Siklon 902 condenser
NaCl 27.6823 2.7682319E-05CaSO4 0.1391 1.3910713E-07H2O 10.0929 1.0092946E-05 1948.4049Na2CO3 9946.0513 0.00994606125NH4Cl 101.2651 0.00010126524NH4HCO3
CaCO3SiO2
Al2O3Fe2O3MgOCaO
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MgCl2CaCl2
NAHCO3 15.7793 1.5779348E-05
10101.0101 0.0101 1948.4049
62009.0711
selisih -0.00002
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(A.Saberi at al,2008.precipitation kinetics of sodium bikarbonate in an industrial buble colum cristalizer. Wiley interscience)
masuk keluar massa terproduksi
47.32020345882930.102284654978625
1874.07291114514
189.280813835317
1.43143151894049 187.849382316377
189.280813835317
6.1237910358484648.8511512177914
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http://docs.google.com/viewer?a=v&q=cache:YQLS8RN43dgJ:www.ch.ntu.edu.tw/~genchem99/msds/exp21/Ammonium%2520hydroxide%2520water%2520solution.pdf+decomposition+of+ammonium+hydroxide&hl=id&gl=id&sig=AHIEtbTtel3g1iq0S3psR4tIpMq-xA-Vyw
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massa produksigas
-478.94172633.4088
-9830.65670.0000
-1692.69240.00000.00000.00000.0000
-2.2069-5141.92694048.9884
5.241510192.0336
266.75230.0000
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CaCO3 impurities air pencuci RDVF jumlah
13841.159513.9107
9.75290892556628 337.3331 38316.831294.0138
9733.40310771514 9733.40315.26657081980579 5.2666
1.5604654280906 1.56051.5604654280906 1.5605
1.36540724957928 1.36549752.9089 337.3331 62009.0712
CSTR S-901 Evaporator jumlah
2740.5496 2768.231913.7716 13.9107
36293.9878 0.0000 38252.48569946.0612
194.6016 295.86680.0000
194.6486 0.01946680621543 194.66811.5605 1.5604654281E-06 1.56051.5605 1.5604654281E-06 1.56051.3654 1.5019479745E-06 1.36540.3009 2.5078908666E-06 0.3009
199.7589 0.00587586078807 199.7647
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5.2415 5.241510192.0336 10192.0336
120.2402 136.0196
49959.6207 0.0253 0.0000 62009.0711
0.0000
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BM: NaCl 58.5000H2O 18.0000CaSO4 136.0000MgO 40.0000BM udara 28.9000
data Cp R=
komponen ANaCl 5.5260NaHCO3 5.1280CaCO3 12.5720Fe2O3 11.8120NH4Cl 5.9390CaCl2 8.6460CaO 6.1040
AMgO 10.4610MgCl2 21.8360Al2O3 -8.1210SiO2 2.4780komponen ANH3 22.6260H2O 8.7120NaCl 95.0160komponen AH2O 3.4700udara 3.3550NH3 3.5780CO2 5.4570udara 3.3550MgCO3
18.5200 0.0220Coeffs: -2.5700e+004 1.2248e+003 -4.4420e+000 7.6420e-003 -4.6090e-006Coeffs: 1.4001e+005 4.8571e+001 -1.6402e-003 0.0000e+000 0.0000e+000
Komponen A B C
-2570.0000 122.4800 0.0076
Data entalpi standar pada 25oC:ΔHf MgCO3 -261.9000 kcal/mol
-1095.7884 KJ/KMOLΔHf MgO -143.8100 kcal/mol
-601.7004 KJ/KMOLΔHf Caco3 -1206.9200 KJ/KMOL
CaSO4
Na2CO3(s)
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ΔHf CaO -635.0900 KJ/KMOLΔHf CO2 -393.5090 KJ/KMOLΔHf MgCO3 -1095.7884 KJ/KMOLΔHf MgO -601.7004 KJ/KMOLΔHf H20( l) -285.8300 KJ/KMOLΔHf H2O( g) -241.8180 KJ/KMOLΔHf NaHCO3(aq) -929.8888 Kj/kmolΔHf Na2CO3 -1130.6800 Kj/kmolΔHf CO2 -393.5090 Kj/kmolΔHf H2O -285.8300 Kj/kmolΔHf NaCl -411.1530 Kj/kmolΔHf NH4HCO3ΔHf NH4OH -361.2000 Kj/kmolΔHf NH4Cl -314.4000 Kj/kmol
Cp NH4HCO3 = 0.3600 kkal/kg.C119.8699 kj/kmol K
ΔHf NH4HCO3 = 599.3493 Kj/kmol ΔHf Cacl2 -595.8000 Kj/kmol
ΔHƒ° (MgCl2) -153.2800 kcal/mol-641.3228 Kj/kmol
ΔHf NH3 -46.1100 Kj/kmol
1. MT
T1 30.0000 C303.0000 K
Komponen masukgaram industri fress feed
NaCl 13841.1595CaSO4 13.9107H2O 55.6429 37913.8194subjumlah 13910.7131 37913.8194jumlah 51824.5325
a. Panas garam Input
umpan garam industri kg/jam kmol/jamNaCl 13841.1595 236.6010CaSO4 13.9107 0.1023H2O 55.6429 3.0913total 13910.7131 239.7946
b. Panas Fress Feed 37913.8194 2106.3233
c. Panas Pelarutan
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Panas kelarutan NaCl=
maka panas NaCL yang terlarut
Panas Total Input 2007975.8063
Bila Panas Input samadenga panas out put maka
d. Panas Out put
Tin misal 36.7281 c309.7281 K
output garam industri kg/jam kmol/jamNaCl 13841.1595 236.6010CaSO4 13.9107 0.1023H2O 37969.4622 2109.4146total 51824.5325 2346.1179
input outputQ in NaCl 61400.5486Q in H2O 794287.9338Q NaCl aq 2007975.8621
total Err:522 Err:522
2. Absorber
kimponenumpan make up
NaCl 13841.1595 0.0000CaSO4 13.9107 0.0000H2O 37969.4622 0.2829NH3 0.0000 94.0138Co2 0.0000 0.0000NH3(aq) 0.0000 0.0000total 51824.5325 94.2966
56316.6490
a. Umpan masuk dari larutan garam
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Tin 36.7281 C309.7281 K
umpan kg/jam kmol/jamNaCl 13841.1595 236.6010CaSO4 13.9107 0.1023H2O 37969.4622 2109.4146total 51824.5325 2346.1179
b. Umpan Make Up NH3
Tin 30.0000 C303.0000 K
umpan garam industri kg/jam kmol/jamNH3 94.0138 5.5302H2O 0.2829 0.0157Total 94.2966 5.5459
c.Umpan Recovery
Tin 30.0000 C303.0000 K
umpan kg/jam kmol/jam
72.7334 4.0407
4058.3341 238.7255
CO2 266.7523 6.0626
total 4397.8199 248.8288
d. Pelarutan NH3
panas kelarutan ammonia 27.1000 kcal/kmol113.3863 Kj/kmol
Banyaknya NH3 terlarut 4152.3479 Kg/jamMaka Pnas pelarutan 27695.2503 Kj
Maka total panas umpan 2075733.4715
e. Panas Out TOP
Tin 30.0000 C303.0000 K
output kg/jam kmol/jamCo2 266.7523 6.0626
f Panas ouput bottom
output kg/jam kmol/jam
H2O
NH3
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NaCl 13841.1595 236.6010CaSO4 13.9107 0.1023H2O 33645.8750 1869.2153NH3(aq) 8548.9515 244.2558total 56049.8966 2350.1743
Total Panas Keluar 803642.7592
g. Panas yang dilepaskan/diserap
Panas add = Pnas ouput-panas input-1272090.7123 Kj/jam
Bila Media pendingin adalah air maka banayak nya air yang dibutuhkan
Misal T nasuk 28.0000 C301.0000 K
Q=mCpdT CPdT air CPin=Cp outt
M=Q/(Cpout-Cpin)Massa air 991.0496 Kmol/jjam
17838.8937 Kg/jaminput output
Q in brine 2007975.8621Q in make up 847.8752Q recovery 39214.4839Q pelarutan NH3 27695.2503 0.0000Q out Top 1345.9520Q out Bootom 802296.8072Q pndingin in 224204.8123Q pndingin out 1496295.5246total 2299938.2838 2299938.2838
3. Reaktor Bubble
Kondisi Opersi isotermal T=30 C P= 4 Atmkomponen absorber botom absorber topNaCl 13841.1595 0.0000CaSO4 13.9107 0.0000H2O 33645.8750 0.0000NH3(aq) 8548.9515 0.0000CO2 0.0000 266.7523NH4Cl 0.0000 0.0000NaHCO3 0.0000 0.0000NH4HCO3 0.0000 0.0000total 56049.8966 266.7523
66884.5874
a. Umpan absorber bottom
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Tin 30.0000 C303.0000 K
umpan kg/jam kmol/jamNaCl 13841.1595 236.6010CaSO4 13.9107 0.1023H2O 33645.8750 1869.2153NH4OH 8548.9515 244.2558total 56049.8966 2350.1743
b. Umpan Top absorberumpan kg/jam kmol/jamCO2 266.7523 6.0626b1. umpan recyclerCO2 2149.4507 48.8512c. Umpan dari Pabri CaCO3 (BC 02)umpan kg/jam kmol/jamH2O 9.7529 0.5418CO2 4199.8021 95.4500total 4209.5550 95.9919
d. Umpan Kondenser separator
umpan kg/jam kmol/jamH2O 77.6845 4.3158CO2 4131.2482 93.8920total 4208.9327 98.2078
Total Energi masuk 858356.0111 KJ/jam
f. Panas OutToutput 30.0000 c
303.0000 Koutput kg/jam kmol/jamNaCl 2768.2319 47.3202CaSO4 13.9107 0.1023H2O 33733.3124 1874.0729NH4Cl 10126.5235 189.2808NaHCO3 15899.5884 189.2808NH4HCO3 483.7795 6.1238NH4OH 1709.7903 48.8512total 64735.1367 2355.0320CO2 2149.4507 48.8512total 66884.5874 2403.8831
e. Panas StandarRX1 NH4OH + CO2
195.4046 195.4046
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RX2 NH4HCO3 + NACL
195.4046 236.6010189.2808 189.2808
6.1238 47.3202
sisa NH4HCO3 483.7795 kg
RX1= 269100.7353000RX2= -272979.6810
g.menghitung panas yang dserap dan dilepas
T=
∆HR(T)=
Bila ∆HoR(T)= Kj
Reaksi 1
output kmol/jamNh3(aq) -1.0000 -22.6260CO2 -1.0000 -5.4570
NH4HCO3 1.0000 0.0000total -28.0830
CpdT -71.4389nCpdT= -13959.4983maka ∆HR(T)= 269029.2964Q= 52569563.3535Reaksi 2output kmol/jamNaCl -1.0000 -5.5260NH4Cl 1.0000 5.1280NaHCO3 1.0000 12.5720NH4HCO3 -1.0000 0.0000
ΔHof 25 = mol yang bereaksi x (ΔHF produk - ΔHF reaktan)
Qp=FaoX2.∆HR(303K)
Panas Reaksi ada 303 K,∆HR(T)∆HoR(TR)+∆CpdT
∆A
∆A
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
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total 2355.0320 12.1740
CpdT 89.1501 KjnCpdT 16874.3950maka ∆HR(T)= -272890.5309maka Q -51652941.7852 Kj
maka Q total= 916621.5683 KJ/mol
Q pendingin Qin-Qout-Qreaksi KJ-866135.5014
Bila air masuk ada suhu 30 C dan keluar pada suhu 50 C
Misal T nasuk 28.0000 C301.0000 K
Q=mCpdTCPdT air masuk 226.2296CPdT air keluar 1509.8088
m = 674.7815 mol
Q air dingin masuk 152655.5816 KJQair dingn keluar 1018791.0830 KJ
input Q reaksiQ in Bottom Abs 802296.8072 916621.5683Q in top ABS 1345.9520Q Pabrik CaCO3 21395.2588Q kndenser 22472.5122Q recycle 10845.4809Q pendinginin
Q pendingin outQ ProdukQproduk gastotal 858356.0111 916621.5683
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5. Evaporator
komponen masuk uapNaCl 0.0000 0.0000CaSO4 0.0000 0.0000H2O 0.0000 27443.7653NH4Cl 0.0000 0.0000NaHCO3 0.0000 0.0000NH4HCO3 0.0000 0.0000Total 0.0000 27443.7653
a. Panas aliran masuk
Tin`` 30.0000 c303.0000 K
umpan kg/jam kmol/jamNaCl 0.0000 0.0000CaSO4 0.0000 0.0000H2O 0.0000 0.0000NH4Cl 0.0000 0.0000NaHCO3 0.0000 0.0000
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NH4HCO3 0.0000 0.0000Total 0.0000 0.0000
b. panas aliran keluar uap
Tout 105.0000 C378.0000 K
ouput kg/jam kmol/jamH2O 0.0000 0.0000
d. panas aliran liquid
output kg/jam kmol/jamNaCl 0.0000 0.0000CaSO4 0.0000 0.0000H2O 0.0000 0.0000NH4Cl 0.0000 0.0000NaHCO3 0.0000 0.0000NH4HCO3 0.0000 0.0000total 0.0000 0.0000
Panas Penguapan (Hv) H2) pada Tc 647.3 K dan Tb 373 K=Panas penguapam (Hv)H2O pada 378 K adalah
Hv= 40.3996 Kj/Kmol
banyaknya H2O yang teruapkan sebesar= 0.0000sehingga Qv= 0.0000
d. Menghitung jumlah steam yang digunakan
Q steam=Qout+Qv-Qin
HV = HV ,b [ TC−T
TC−Tb ]0 ,38
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Qsteam= 0.0000 Kj/jam
Jka steam yang digunakan adalah team saturated dengan suhu 150C dengan tekanan P476 KpaHf= 632.1000 Kj/KgHv= 2745.4000 Kj/Kg
2113.3000 Kj/Kg
maka jumlah steam yang dibutuhkan
m= 0.0000 Kg/jam steam yang dibutuhkan
Komponen Fi Kg/jam Hf KJ/gsteam out 0.0000 632.1000
Komponen Fi Kg/jam HG KJ/gsteam in 0.0000 2745.4000
input outputQ in 0.0000Qstean in 0.0000Qsteanm out 0.0000Qv 0.0000Quap 0.0000Q liquid 0.0000
total 0.0000 0.0000
6 Kristalizer
komponen Kmol/jam
NaCl(l) 2768.2319CaSO4(s) 13.9107H2O(l) 33733.3124NH4Cl(l) 10126.5235NaHCO3(s) 15899.5884NH4HCO3 483.7795NH4OH 1709.7903total 64735.1367a. panas masukTin 30.0000 C
303.0000 k86.0000 F
Komponen Kmol/jamNaCl(l) 2768.2319 47.3202CaSO4(s) 13.9107 0.1023
λsteam=
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H2O(l) 33733.3124 1874.0729NH4Cl(l) 10126.5235 189.2808NaHCO3(s) 15899.5884 189.2808NH4HCO3 483.7795 6.1238NH4OH 1709.7903 48.8512total 64735.1367 2306.1808
b. Pnas keluar
T- 22.0000 C295.0000 K
71.6000 Kkomponen output kmol
NaCl(s) 2768.2319 47.3202CaSO4(s) 13.9107 0.1023H2O(l) 33733.3124 1874.0729NH4Cl(l) 10126.5235 189.2808NaHCO3(s) 15779.3481 187.8494NaHCO3(aq) 120.2402 1.4314NH4HCO3 483.7795 6.1238NH4OH 1709.7903 48.8512total 64735.1367 2355.0320c. Panas Kriatalisasipanas kristalisasi NaHCO3 = - panas kelarutan NaHCO3panas kelarutan NaHCO3= -17165.8800 kj/kmol (Perry’s, tabel 2-2)
maka panas kristalisasi= 17165.8800 Kj.kmol
maka panas kriatalisai bahan 3224599.9549 Kj/jam
d. Menghitng jumlah pendingin yang diperulkan
Qp= Qout-Qin-Qkristali-4660212.4272 Kj/jam
Bila air pendingin yang digunakan air maka
Bila air masuk ada suhu 25C dan keluar pada suhu50 C
Misal T nasuk 30.0000 C303.0000 K
86.0000 FQ=mCpdT CPdT air masuk 377.0969
CPdT air keluar 1509.8088
M air 4114.2078 Kmol/jam
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Q in pendingin 1551454.9064 KjQout pendingin 6211667.3336 Kj
Input Q kritalisasiQin 897996.5970 3224599.9549Qpedingin in 162987.2920
Qout pendinginQout produktotal 1060983.8891 3224599.9549
7. RDVF
Tin 22.0000 295.0000T ref 298.0000
a. Paas Mansukkomponen
umpan MolNaCl(s) 2768.2319 47.3202CaSO4(s) 13.9107 0.1023H2O(l) 33733.3124 1874.0729NH4Cl(l) 10126.5235 189.2808NaHCO3(s) 15779.3481 187.8494NaHCO3(aq) 120.2402 1.4314NH4HCO3 483.7795 6.1238NH4OH 1709.7903 48.8512total 64735.1367 2355.0320
a.1 Pnas air PencuciT=30C
umpan MolH2O(l) 337.3331 18.7407
b. Panas Keluar mother liqud
Tout 295.3499T ref 298.0000komponen KgNaCl(l) 2740.5496 46.8470CaSO4(s) 13.7716 0.1013H2O(l) 33733.3124 1874.0729NH4Cl(l) 10025.2583 187.3880NaHCO3(aq) 120.2402 1.4314NH4HCO3 478.9417 6.0626
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NH4OH 1692.6924 48.3626Total 48804.7662 2164.2658c. Panas Keluar outTout 295.3499T ref 298.0000komponen KgNaCl 27.6823 0.4732CaSO4(s) 0.1391 0.0010H2O(l) 337.3331 18.7407NaHCO3(s) 15779.3481 187.8494NH4Cl(l) 101.2652 1.8928NH4OH 17.0979 0.4885NH4HCO3 4.8378 0.0612total 16267.7036 209.5069
8 Kalsiner sodium bikarbonatekomponen Kg/jam Kmol/jamNaCl 27.6823 0.4732CaSO4(s) 0.1391 0.0010H2O(l) 337.3331 18.7407NaHCO3(s) 15779.3481 187.8494NH4Cl(l) 101.2652 1.8928NH4OH 17.0979 0.4885NH4HCO3 4.8378 0.0612total 16267.7036 209.5069a. Panas masukkomponen Kg/jam Kmol/jamNaCl 27.6823 0.4732CaSO4(s) 0.1391 0.0010H2O(l) 337.3331 18.7407NaHCO3(s) 15779.3481 187.8494NH4Cl(l) 101.2652 1.8928NH4OH 17.0979 0.4885NH4HCO3 4.8378 0.0612total 16267.7036 209.5069
b. Panas reaksi2NaHCO3 Na2CO3 +
187.6615 93.8308Panas reaksi mol reaktan yang bereaksi x HR(T)HR(T)= ∆Hf25 C+ CPdT
panas standar ∆Hf25 C= = mol yang bereaksi x (ΔHF produk - ΔHF reaktan)
b.1 panas standar -836.1182 Kj/b.2 CpdT
Tin 120.0000
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
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393.0000KomponenNaHCO3 -2.0000 -10.2560Na2CO3 1.0000 0.0000H2o 1.0000 3.4700co2 1.0000 5.4570total 1.0000 -1.3290CpdT= -1319.7261 Kj/KmolHR(T)= -2155.8443 Kj/kmolQ reaksi -404569.0419 Kj
Rx2 NH4HCO3
m 0.0612r 0.0612
∆Hf25 C= = mol yang bereaksi x (ΔHF produk - ΔHF reaktan) ∆Hf25 C= -1324.7983∆HR(T)= ∆HoR(TR)+∆CpdT
∆CpdT=Tin= 120.0000 c
393.0000 k
komponen ∆ANH4HCO3 -1.0000 0.0000NH3 1.0000 3.5780H2O 1.0000 8.7120CO2 1.0000 5.4570total 2.0000 17.7470
CpdT 1749.4216 Kj∆HR(T)= 424.6233 KjQ= 26.0030 Kj
RX4 NH4oHr 0.4885
∆Hf25 C= = mol yang bereaksi x (ΔHF produk - ΔHF reaktan) ∆Hf25 C= 29.2600Tin= 120.0000 c
393.0000 kkomponen ∆ANH4OH -1.0000 22.6260NH3 1.0000 3.5780h2o 1.0000 8.7120
1.0000 34.9160CpdT -984.7231∆HR(T)= -955.4631Q -466.7547
∆A
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Q reaks total -405009.7936
b.1 panas laten
Panas Penguapan (Hv) H2) pada Tc 647.3 K dan Tb 373 K=Panas penguapam (Hv)H2O pada 393 K adalah
Hv= 39.5293 Kj/Kmol
banyaknya H2O yang teruapkan sebesar= 327.2402sehingga Qv= 718.6425
Sehing
9. Rotary Cooler Na2CO3
a. Panas masukKomponen Kg/jam kmol/jamNaCl 27.6823 0.4732CaSO4 0.1391 0.0010H2O 10.0929 0.5607Na2CO3 9946.0612 93.8308
HV = HV ,b [ TC−T
TC−Tb ]0 ,38
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NaHCO3 15.7793 0.1878NH4Cl 101.2652 1.8928total 10101.0202 95.0536
b. Panas out ke beltTout 35.0000 C
308.0000 KKomponen Kg/jam kmol/jamNaCl 27.6546 0.4727CaSO4 0.1390 0.0010H2O 10.0829 0.5602Na2CO3 9936.1152 93.7369NaHCO3 15.7636 0.1877NH4Cl 101.1640 1.8909total 10090.9192 96.8494c. Panas out ke siklonTout 43.0207 C
316.0207 K
Komponen Kg/jam kmol/jamNaCl 0.0277 0.0005CaSO4 0.0001 0.0000H2O 0.0101 0.0006Na2CO3 9.9461 0.0938NaHCO3 0.0158 0.0002NH4Cl 0.1013 0.0019total 10.1010 0.0969d. Kebuthan udara keringQ pendinding -352100.3372 Kj/jam
Tin udara 30.0000 c303.0000 K
Tref 298.0000
Cp udara in 140.2427 GAK DIPKECp udara out 2105.3991
massa udara yang dibuthkan 179.17175178.0610
Komponen Fi Kmol/jam Cp (Kj/Kmol)udara masuk 179.1717 140.2427
Komponen Fi Kmol/jam Cp (Kj/Kmol)udara out 179.1717 2105.3991
input output
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Qin 387733.6825Qout siklon 66.7788Qout belt 35566.5665Q pendingin inQ pendingin outtotal 387733.6825 35633.3453
10. Condenser pasial n separator
Tekanan Condenser Parsial : 1atmSuhu Masuk Condenser Parsial 393.0000Suhu Keluar Condenser Parsial 308.8476Suhu Bubble Point 206.1405Suhu Dew Point 362.7473Tref 298.0000
a. Panas masukkomponen Kg/jam kmol/jamH2O 2026.0895 112.5605CO2 4131.2482 93.8920NH3 9.3457 0.5497Total 6166.6834 207.0023
b. Panas kelauar fase gas
komponen Kg/jam kmol/jamH2O 77.6845 4.3158CO2 4131.2482 93.8920NH3 9.3457 0.5497Total 4218.2785 98.7576
c. Pans keluar fase cairkomponen Kg/jam kmol/jamH2O 1948.4049 108.2447
Total 1948.4049 108.2447pansa sensibel 1162728.9403panas pengembinan 88589.5088d. menghtung jumlah pendingin yang diperlukan
Qp= 1251318.4492 Kj/jam
Bila air masuk ada suhu 25 C dan keluar pada suhu50 C
Misal T nasuk 28.0000 C301.0000 K
![Page 73: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/73.jpg)
Q=mCpdTCPdT air masuk 226.2296CPdT air keluar 1509.8088
M air 974.8666 Kmol/jam
Q in pendingin 220543.7201 KjQout pendingin 1471862.1693 Kj
Input Q pengembunanQin 1211694.2065 88589.5088Pnas pendingin in
Q fse gas 0.0000Pnas pendingiin outtotal 1211694.2065 88589.5088
11.Heater
menaikan suhu 20 c dari sentrifuge menjadi 80 cumpan ke CSTRa. Panas masuk mother liquorTin
295.3499 K
komponen Kg KmolNaCl(l) 2740.5496 46.8470H2O(l) 33733.3124 1874.0729NH4Cl(l) 10025.2583 187.3880NaHCO3(aq) 120.2402 1.4314NH4HCO3 478.9417 6.0626NH4OH 1692.6924 48.3626CaSO4 13.7716 0.1013Total 48804.7662 2164.2658b. Panas outTout 80.0000 c
353.0000 K
komponen Kg KmolNaCl(l) 2740.5496 46.8470H2O(l) 33733.3124 1874.0729NH4Cl(l) 10025.2583 187.3880NaHCO3(aq) 120.2402 1.4314NH4HCO3 478.9417 6.0626Nh4OH 1692.6924 48.3626CaSO4 13.7716 0.1013Total 48804.7662 2164.2658c. menghitung kebutuhan steam
Qp= 9447886.0781 Kj/jam
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misal steam yang akan digunakan steam dengan suhu 150C dengan P 476 Kpa
Hf 524.9900 kj/kg
Hg 2713.5000 k/kg
lamda steam 2188.5100 k/kg
maka jumlah steam pemanas yang dibutuhkan 4317.0404
Komponen Fi Kg/jam Hf KJ/gsteam pemanas out 4317.0404 524.9900
Komponen Fi Kg/jam HG KJ/gsteam pemanas in 4317.0404 2713.5000
input outputQin -430858.5899Qout 9017027.4883Q pemanas inQpemanas outtotal -430858.5899 9017027.4883
12. CSTR
a. Pnas moher liquor in
Tin 80.0000 c353.0000 k
komponen Kg KmolNaCl(l) 2740.5496 46.8470H2O(l) 33733.3124 1874.0729NH4Cl(l) 10025.2583 187.3880NaHCO3(aq) 120.2402 1.4314NH4HCO3 478.9417 6.0626NH4OH 1692.6924 48.3626CaSO4 13.7716 0.1013Total 48804.7662 2164.2658b. Panas masuk CaOTin 80.0000 c
353.0000 Kkomponen Kg Kmol
CaCO3 194.6486 1.9465MgO 2.5079 0.0627SiO2 1.5605 0.0260
Al2O3 1.5605 0.0153Fe2O3 1.3654 0.0085CaO 5341.6857 95.3872total 5543.3286 97.4463
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c. Panas reaksi
Rx1 2NH4Cl + CaO
m 187.3880 95.3872reaksi 183.6402 91.8201sisa 3.7478 3.5671
∆Hf25 C= = mol yang bereaksi x (ΔHF produk - ΔHF reaktan)
∆Hf25 C= 290.0400 Kj/jam
∆HR(T)=
∆CpdT=
Tin= 80.0000 c353.0000 k
komponen
NH4Cl -2.0000 -11.8780CaO -1.0000 -6.1040CaCl2 1.0000 8.6460Nh3 1.0000 3.5780h20 91.8201 799.9369total 794.1789
CpdT 45164.1753 Kj∆HR(T)= 45454.2153 KjQ= 8347223.2522 Kj
Rx2`` 2NH4Cl + MgO
m 3.7478 0.0627r 0.1103 0.0552s 3.6374 0.0075
∆Hf25 C= = mol yang bereaksi x (ΔHF produk - ΔHF reaktan)
∆Hf25 C= 430.8804 Kj
∆HoR(TR)+∆CpdT
∆A
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
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∆HR(T)=
∆CpdT=
Tin= 80.0000 c353.0000 k
komponenNH4Cl -2.0000 -11.8780MgO -1.0000 -10.4610MgCl2 1.0000 21.8360Nh3 1.0000 3.5780H2O 1.0000 8.7120total 0.0000 11.7870
CpdT -1125.5029 Kj∆HR(T)= -694.6225 KjQ= -76.6496 Kj
Rx 3 NH4HCO3 NH3+
m 6.0626r 6.0626 6.0626
∆Hf25 C= = mol yang bereaksi x (ΔHF produk - ΔHF reaktan)
∆Hf25 C= -1324.7983 Kj
∆HR(T)=
∆CpdT=
Tin= 80.0000 c353.0000 k
komponenNH4HCO3 -1.0000 0.0000NH3 1.0000 3.5780H2O 1.0000 8.7120CO2 1.0000 5.4570total 2.0000 17.7470
∆HoR(TR)+∆CpdT
∆A
∆HoR(TR)+∆CpdT
∆A
0.1103
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
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CpdT 999.9675 Kj
∆HR(T)= -324.8308 KjQ= -1969.3040
RX4 NH4oH NH3
r 48.3626 48.3626
∆Hf25 C= = mol yang bereaksi x (ΔHF produk - ΔHF reaktan)
∆Hf25 C= 29.2600
Tin= 80.0000 c353.0000 k
komponen
NH4OH -1.0000 22.6260NH3 1.0000 3.5780h2o 1.0000 8.7120
1.0000 34.9160CpdT -612.3657∆HR(T)= -583.1057Q -28200.5285
Qreaksi total 8316976.7701 Kj
d. panas out liquidTin 80.0000 c
353.0000 Kkomponen Kg Kmol
∆A
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CaSO4 13.7716 0.1013NaCl 2740.5496 46.8470H2O 36293.9878 2016.3327NH4Cl 194.6016 3.6374NaHCO3 120.2402 1.4314
CaCO3 194.6486 1.9465SiO2 1.5605 0.0260
Al2O3 1.5605 0.0153Fe2O3 1.3654 0.0085MgO 0.3009 0.0075CaO 199.7589 3.5671
MgCl2 5.2415 0.0552CaCl2 10192.0336 91.8201
total 49959.6207 2165.6948
e. Q out gas
Tin 80.0000 c353.0000 K
komponen Kg KmolH2O 72.7334 4.0407NH3 4048.9884 238.1758CO2 266.7523 6.0626total 4388.4741 248.2791
f. Menghitung pendingin yang dibutuhkan
Qp= -8202752.5076 Kj
Bila air pendingin yang digunakan air maka
Bila air masuk ada suhu 25 C dan keluar pada suhu50 C
Misal T nasuk 28.0000 C301.0000 K
Q=mCpdTCPdT air masuk 226.2296CPdT air keluar 1509.8088
M air 6390.5309 Kmol/jam
Q in pendingin 1445727.5480 KjQout pendingin 9648480.0556 Kj
Input Q reaksiQin mother liquor 9017027.4883 8316976.7701
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Qin Cao 238752.9008Qin pendinginQout pendinginQreaksiQout liquidQout gastotal 9255780.3890 8316976.7701
13. Kalsiner Kalsium Karbonatea. Panas CACO3 masukTin 30.0000 c
303.0000 k
komponen Kg KmolCaCO3 9733.4031 97.3340MgCO3 5.2666 0.0627
SiO2 1.5605 0.0260Al2O3 1.5605 0.0153Fe2O3 1.3654 0.0085H2O(l) 9.7529 0.5418total 9752.9089 97.9884
b. Pamas CaCO3 ke Rotari cooler
Tin 900.0000 c1173.0000 k
komponen Kg KmolCaCO3 194.6681 1.9467
SiO2 1.5605 0.0260Al2O3 1.5605 0.0153Fe2O3 1.3654 0.0085MgO 2.5079 0.0627CaO 5341.6916 95.3874total 5543.3539 97.4466
c. Panas gas out
Tin 900.0000 c1173.0000 k
komponen Kg KmolH2O(g) 9.7529 0.5418
CO2 4199.8021 95.4500total 4209.5550 95.9919
d. Qreaksi
RZ1 CaCO3
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m 97.3340r 95.3874s 1.9467
∆Hf25 C= = mol yang bereaksi x (ΔHF produk - ΔHF reaktan)
∆Hf25 C= 17117.0773 Kj
∆HR(T)=
∆CpdT=
Tin= 900.0000 c1173.0000 k
komponenCaCO3 -1.0000 -12.5720CaO 1.0000 6.1040CO2 1.0000 5.4570total 1.0000 -1.0110
CpdT -1394.7861 Kj
∆HR(T)= 15722.2912 Kj
Q= 1499707.6979 Kj
Rx2 MgCO3m 0.0627r 0.0627s 0.0000
∆Hf25 C= = mol yang bereaksi x (ΔHF produk - ΔHF reaktan)
∆Hf25 C= 752.5003 Kj
∆HR(T)=
∆CpdT=
∆HoR(TR)+∆CpdT
∆A
∆HoR(TR)+∆CpdT
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
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Tin= 900.0000 c1173.0000 k
komponenMgCO3 -1.0000 0.0000MgO 1.0000 10.4610CO2 1.0000 5.4570total 1.0000 15.9180
CpdT 43689.2285 Kj
∆HR(T)= 44441.7288 Kj
Q= 2786.3751 Kj
Reaksi 3 H2O(l)m 0.5418r 0.5418
∆Hf25 C= = mol yang bereaksi x (ΔHF produk - ΔHF reaktan)
∆Hf25 C= 23.8469 Kj
∆HR(T)=
∆CpdT=
Tin= 900.0000 c1173.0000 k
komponenH2O (l) -1.0000 -8.7120H2O(g) 1.0000 3.4700total 0.0000 -5.2420
CpdT -4332.4988 Kj
∆HR(T)= -4308.6519 Kj
Q= -2334.5494 Kj
Q total= 1500159.5236 Kj
∆A
∆HoR(TR)+∆CpdT
∆A
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
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e. Menghitung kebutuhan batubara
Qp=Qout-Qin-QrQp= 6775801.0452 Kj
Menurut mantell heat loss furnace terdur dari
radiasi 0.0170cooling 0.1000
maka total kehilangan panas= 0.8830
maka total kebutuhan panas adalah 7673613.8677
diketahui 1 kg batubara mengandung 29000.0000
maka kebuthan batubara adalah 264.6074
f.menghitung kebutuhan oksigen
C + O2 CO2mol batubara = mol oksigen yang digunakan jadi mol batubara adalah
jadi oksigen yang dibuthkan adalah 705.6197
massa tersebut diperloeh dari udara sebnayak
INPUT Q reaksiQin 40314.9479 1500159.5236QpemanasQ out gasQout padatanQ rQ losstotal 40314.9479 1500159.5236
14. Rotary cooler CaOa. Pamas InputTin 900.0000 c
1173.0000 K
komponen Kg KmolCaCO3 194.6681 1.9467
SiO2 1.5605 0.0260Al2O3 1.5605 0.0153Fe2O3 1.3654 0.0085MgO 2.5079 0.0627
![Page 83: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/83.jpg)
CaO 5341.6916 95.3874total 5543.3539 97.4466
b.Panas ke belt Out
Tin 80.0000 c353.0000 K
komponen Kg KmolCaCO3 175.2013 1.7520
SiO2 1.5589 0.0260Al2O3 1.5589 0.0153Fe2O3 1.3639 0.0085MgO 2.5054 0.0626CaO 5336.3499 95.2920total 5518.5383 97.1564
c. Pnas ke siklon out
Tin 114.7897 c387.7897 K
komponen Kg KmolCaCO3 19.4668 0.1947
SiO2 0.0016 0.0000Al2O3 0.0016 0.0000Fe2O3 0.0015 0.0000MgO 0.0025 0.0001CaO 5.3417 0.0954total 24.8156 0.2902
d. Kebuthan udara kering
Q pendinding -4213446.4189 Kj/jam
Tin udara 30.0000 c303.0000 K
Tref 298.0000
Cp udara in 140.2427 GAK DIPAKECp udara out 7736.9909
massa udara yang dibuthkan 554.6382209152.3159
Komponen Fi Kmol/jam Cp (Kj/Kmol)udara masuk 554.6382 140.2427
Komponen Fi Kmol/jam Cp (Kj/Kmol)udara out 554.6382 7736.9909
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input output
Qin 4452987.1020Qout siklon 1947.5398Qout belt 237593.1433Q pendingin inQ pendingin outtotal 4452987.1020 239540.6831
15. Coller pendingin gas dari CSTR ke ABSa.panas masukTin 80.0000 c
353.0000 K
komponen Kg KmolH2O 72.7334 4.0407NH3 4048.9884 238.1758CO2 266.7523 6.0626total 4388.4741 248.2791
b. panas out
Tin 30.0000 c303.0000 K
komponen Kg KmolH2O 72.7334 4.0407NH3 4048.9884 238.1758CO2 266.7523 6.0626total 4388.4741 248.2791
c. Menghitung jumlah pendingin yang dibuuthkan
Qp= -392800.7967 Kj
Bila air pendingin yang digunakan air maka
Bila air masuk ada suhu 30 C dan keluar pada suhu50 C
Misal T nasuk 28.0000 C301.0000 K
Q=mCpdTCPdT air masuk 226.2296CPdT air keluar 1509.8088
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M air 306.0199 Kmol/jam
Q in pendingin 69230.7774 KjQout pendingin 462031.5741 Kj
Input OutputQin 431931.5840Qpendingin in 69230.7774Qout 39130.7873Qpendingin out 462031.5741total 501162.3614 501162.3614
16. Kompresor
menghitung T capuran gas yang masuk
m1.T1 + m2.T2 + m3.T3 = Mc.Tctotal masa1 masuk yang berasal dari kondenser 4218.2785T1 308.8476massa 2 masuk berasal dari furnace calsium carbonate 4209.5550T2 1173.0000massa masuk 3 berasal dari absorber 266.7523T3 303.0000massa masuk recycle 2149.4507
303.0000jumlah massa campuran 10844.0365maka Tc adalah 643.0007 Kjadi suhu yang masuk kompresor adalah 643.0007
MENGHITUNG T OUT COMPRESOR
(y) dan rasio kompresi (Rc)
(Ludwig, 1963)
Jika Rc
= 3
4.0000
a. Menentukan jumlah stage
3,5 – 4 maka y (jumlah stage) yang digunakan benar. Digunakan 1 stage kompresor untuk mendapatkan rasio yang baik, dengan Rc adalah:
Rc=( Pout
Pin)1
y
¿
Rc=(41 )
11=4
![Page 86: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/86.jpg)
b.
Dimana untuk peristiwa kompresi,
Keterangan :
Temperatur keluar
Temperatur masukRc Rasio kompresi
m Polytropic Temperatur exponentEp Efisiensi politropik
ZX Compressibility function, XY Compressibility function, YR 8,314 kJ/kmol K
KOMPOSISI MASUK CP-501 F-201 Vurnace CACO3
H2O 77.6845 9.7529CO2 4131.2482 4199.8021total 4208.9327 4209.5550
b.1 kecepatan voumetrik
V = n x 22,4 x
3.6508
b.2 efisiensi politropikNilai Ep diperoleh dengan menghubungkan nilai lajuvolumetrik yang diperoleh pada gbr. 3.6 buku Coulson. R, vol.6, hal 75 untuk jenis kompresor sentrifugal.
Menentukan temperatur keluar (T out)
Menghitung suhu keluaran CP-101, Tout
Tout = Tin
Tin
Tout
1 kg gas ideal pada 0oC (273 K), 760 mmHg = 22,4 m3
3600
1 x
T
T
reff
in
[ Rc ]m
m =Z×RCp ( 1
Ep+ X)
![Page 87: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/87.jpg)
Diperolh Ep= 0.6800
b.3 m= dimana untuk gas polyatom 1.3000
m= 0.3394
Tout'=Tin [Rc]^m 1029.2726 K
T'mean= 836.1366 K
Pmean= 2.5000 atm
Diket Tc H2o 647.1000Pc H2O 217.6659Tc CO2 304.2000Pc CO2 72.8645
komponen total kg/jam kmol/jamH2O 87.4374 4.8576CO2 10747.2533 244.2558total 10834.6907 249.1134
Tr mean= 2.6895
Pr mean= 0.0330
18050.9495
X= 0.0010 (gbr 3.9 Coulson, 1983Y= 1.0000 (gbr 3.10 Coulson,1983)Z= 1.0000 (gbr 3.8 Coulson, 1983)
MAKA M =
M= 0.0007
Cpmix (pada T’ mean) = Sxi.Cpi
(γ - 1 )γ . Ep
T in+T out '
2
P in+Pout
2
T ' mean
∑ xi . Tc
P mean
∑ xi . Pc
m =Z×RCp ( 1
Ep+ X)
![Page 88: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/88.jpg)
Tout sebenarnya 643.6051 K
maka didapat Pin 1.0000kondisi opresai Tin= 643.0007
Pout 4.0000Tout 643.6051Tref 298.0000
maka neraca energi pada kompresor370.0007
1.Q inkomponen total kg/jam kmol/jamH2O 87.4374 4.8576CO2 10747.2533 244.2558total 10834.6907 249.1134
2. Qout
komponen total kg/jam kmol/jamH2O 87.4374 4.8576CO2 10747.2533 244.2558total 10834.6907 249.1134
Q kompresi 6986.9398 Kj
input Q kompresi
Q in 3942766.3005 6986.9398Q outQ kompresitotal 3942766.3005 6986.9398
3949753.240317. Cooler untukmenurunkan suhu dikompresor menjadi 30 C
a, Panas masuk
Tin 370.6051 c643.6051 K
komponen total kg/jam kmol/jam
H2O 87.4374 4.8576CO2 10747.2533 244.2558total 10834.6907 249.1134
b. panas keluar
![Page 89: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/89.jpg)
Tout 30.0000 c303.0000 K
komponen total kg/jam kmol/jamH2O 87.4374 4.8576CO2 10747.2533 244.2558total 10834.6907 249.1134
c. menghitng jumlah pendingin yangdiperlukan
Qp= -3893694.0365 Kj
Bila air pendingin yang digunakan air maka
Bila air masuk ada suhu 30 C dan keluar pada suhu50 C
Misal T nasuk 28.0000 C301.0000 K
Q=mCpdTCPdT air masuk 226.2296CPdT air keluar 1509.8088
M air 3033.4661 Kmol/jam
Q in pendingin 686259.9752 KjQout pendingin 4579954.0116 Kj
Input OutputQin 3949753.2403Qpendingin in 686259.9752Qout 56059.2038Qpendingin out 4579954.0116total 4636013.2155 4636013.2155
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kg/kmol CO2 44.0000 kg/kmolkg/kmol NH3 17.0000 kg/kmolkg/kmol NaHCO3 84.0000 kg/kmolkg/kmol CaO 56.0000 kg/kmol
8.3140 Kj/Kmol K
B C D E0.00200.01810.0026 -312000.00000.0097 -197600.00000.01610.0015 -30200.00000.0004 -104700.0000
B C D E0.1115 -0.00010.0144 0.00000.3869 -0.00030.1652 -0.0001
B C D E-0.1008 0.00020.0013 0.0000
-0.0311 0.0000 0.0000B C D E
0.0015 12100.00000.0006 -1600.00000.0030 -18600.00000.0010 -115700.00000.0006 -1600.0000
156800.0000 CP= A +BT+CT^2+DT^-2 Coeffs: -2.5700e+004 1.2248e+003 -4.4420e+000 7.6420e-003 -4.6090e-006Coeffs: 1.4001e+005 4.8571e+001 -1.6402e-003 0.0000e+000 0.0000e+000
D E
0.0000 0.0000 chem cad edisi 5
-1095788.3902
-601700.3757SMITH -1045716.0773
NH4CL
![Page 91: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/91.jpg)
SMITH -635090.0000 MGOSMITH -393509.0000 MGCl2PATNAIK 2003 -995961.8760 NH3PATNAIK 2003 -601700.3757 H2OSMITH -285830.0000SMITH -241818.0000
-855340.9457-1066492.5937
-393509.0000-285830.0000-411153.0000
-361200.0000-314400.0000
-485608.2647-858757.1616
-852411.5600-46110.0000
Tref 298.0000 K
keluarBrine
13841.159513.9107
37969.4622
51824.5325
CP ∆H (kj)254.2372 60152.7787802.2880 82.0618377.0969 1165.7081
1433.6221 61400.5486
377.0969 794287.9338
1 cal 4.1840
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1.1640 kg.cal/g mol perrys, 3-158 edsi 64870.1706 Kj/kmol
1152287.3239 Kj/Kmol
T ref 298.0000
CP ∆H (kj)596.9896 141248.3385983.4203 100.5888884.9028 1866626.9348
2465.3127 2007975.8621
Q pelarutan1152287.3239
pprodukrecovery top botom
0.0000 0.0000 13841.15950.0000 0.0000 13.9107
72.7334 0.0000 33645.87504058.3341 0.0000 0.0000
266.7523 266.7523 0.00000.0000 0.0000 8548.9515
4397.8199 266.7523 56049.896656316.6490
![Page 93: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/93.jpg)
Tref 298.0000
CP ∆H (kj)596.9896 141248.3385983.4203 100.5888884.9028 1866626.9348
2465.3127 2007975.8621
Tref 298.0000
CP ∆H (kj)152.2450 841.9487377.0969 5.9265529.3419 847.8752
Tref 298.0000
CP ∆H (kj)
377.0969 1523.7530
152.2450 36344.7789
222.0108 1345.9520
751.3527 39214.4839
patnaik 2003
244.2558 Kmol/jam
Kj/jam
Tref 298.0000
CP ∆H (kj)222.0108 1345.9520
CP ∆H (kj)
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254.2372 60152.7787802.2880 82.0618377.0969 704875.2391152.2450 37186.7277
1585.8671 802296.8072
Tout 45.0000 C318.0000 K
226.22961509.8088 1283.5792
Komponen Fi Kmol/jam Cp (Kj/Kmol)air pendingin in 991.0496 226.2296
Komponen Fi Kmol/jam Cp (Kj/Kmol)air pendingin out 991.0496 1509.8088
furnace dari separator recycle produk0.0000 0.0000 2768.23190.0000 0.0000 13.91079.7529 77.6845 33733.31240.0000 0.0000 1709.7903
4199.8021 4131.2482 2149.4507 0.00000.0000 0.0000 10126.52350.0000 0.0000 15899.58840.0000 0.0000 483.7795
4209.5550 4208.9327 2149.4507 64735.136766884.5874
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Tref 298.0000
CP ∆H (kj)254.2372 60152.7787802.2880 82.0618377.0969 704875.2391152.2450 37186.7277
1585.8671 802296.8072
CP ∆H (kj)222.0108 1345.9520
222.0108 10845.4809
CP ∆H (kj) aliran gas total umpan377.0969 204.3217 umpan 222.0108 21190.9370 H2O599.1076 21395.2588 CO2
total
CP ∆H (kj)377.0969 1627.4775222.0108 20845.0347599.1076 22472.5122
Trfe 298.0000
CP ∆H (kj)254.2372 12030.5557802.2880 82.0618377.0969 706707.0383448.0644 84809.9995439.8719 83259.3063599.3493 3670.2899152.2450 7437.3455
3073.1527 897996.5970222.0108 10845.4809 TOP produk
3295.1635 908842.0779
NH4HCO3195.4046
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NaHCO3+ NH4Cl
189.2808 189.2808
30.0000 C Tref 298.0000303.0000
∆B ∆C ∆D ∆E0.1008 -0.0002
-0.0013 0.0000 115700.00000.0000 0.0000
= 0.00000.0995 -0.0002 115700.0000
Q=delta H 255141.2370
∆B ∆C ∆D ∆E-0.0020 0.0000 0.00000.0181 0.0000 0.00000.0026 0.0000 0.00000.0000 0.0000 0.0000 0.0000
produk - ΔHF reaktan)
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
∫CpdT=R(A (T 2−T1 )+B2
(T 22−T 1
2)−D( 1T 2
− 1T1
))
![Page 97: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/97.jpg)
0.0188 0.0000 0.0000 0.0000
Q=delta H -256105.2860
misal amoniak yang akan digunakan steam dengan suhu -33 C dengan P 101.33 Kpa
Q total -964.0491 HVHflamda steam
51450.1159maka jumlah steam pemanas yang dibutuhka
Tref 298.0000Tout 45.0000 C Komponen
318.0000 K amoniak in
Kj/kmol KomponenKj/kmol amoniak out
12146.0671 Kg/jam
output Qpendingin in Q pendingin out152655.5816 1018791.0830
897996.597010845.4809
908842.0779 152655.5816 1018791.0830
![Page 98: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/98.jpg)
Liquid250.2330
12.57453049.3073
11213.434817606.1407
884.559433016.2497
Tref 298.0000
CP ∆H (kj)254.2372 0.0000802.2880 0.0000377.0969 0.0000448.0644 0.0000439.8719 0.0000
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599.3493 0.00003073.1527 0.0000
Tref= 298.0000
CP ∆H (kj)6061.7973 0.0000
CP ∆H (kj)4116.7562 0.00002855.0238 0.00006061.7973 0.00007570.7217 0.00007490.5974 0.00009589.5888 0.0000
37684.4853 0.0000
40.6830 Kj/Kmol
Kmol/jamKj
![Page 100: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/100.jpg)
Jka steam yang digunakan adalah team saturated dengan suhu 150C dengan tekanan P476 Kpa
Kg/jam steam yang dibutuhkan
Q amoniak in0.0000
Q amoniak in0.0000
Tref 298.0000
Cp H254.2372 12030.5557802.2880 82.0618
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377.0969 706707.0383448.0644 84809.9995439.8719 83259.3063599.3493 3670.2899152.2450 7437.3455
3073.1527 897996.5970
Tref 298.0000
Cp KJ/kmol C H
-152.3465 -7209.0660587.5421 60.0965
-226.1441 -423810.6107-267.2319 -50581.8702-262.1125 -49237.6778-262.1125 -375.1961-359.6096 -2202.1739
-87.1909 -4259.3771-1029.2060 -537615.8753
kj/kmol (Perry’s, tabel 2-2)
Tref 298.0000Tout 45.0000 C
318.0000 K113.0000 F
Kj/kmol Cp 4.1870 Kj/KgKj/kmol
74055.7405 Kgjam
![Page 102: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/102.jpg)
densitas air 1000.0000 kg/m374.0557 m3air yang dibutuhkan
Output Komponenair pendingin in
Komponen4823199.7192 air pendingin out-537615.87534285583.8440
C
Cp H Neraca panas sekitar RDF-301 :-152.3465 -7209.0660587.5421 60.0965
-226.1441 -423810.6107 maka Q masuk + Q air pncuci = Q liquid + Q cake-267.2319 -50581.8702-262.1125 -49237.6778 misal suhu produk liquid = cake maka-262.1125 -375.1961-359.6096 -2202.1739 Q liquid + Q cake=
-87.1909 -4259.3771-1029.2060 -537615.8753 Mentrial suhu keluaran
umpanNaCl(l)
Cp H CaSO4(s)377.0969 58755.6232 H2O(l)
NH4Cl(l)NaHCO3(s)NaHCO3(aq)NH4HCO3
C NH4OHtotal
Cp H-134.5842 -6304.8660596.9217 60.4454
-199.7709 -374385.1807-236.1238 -44246.7647-231.6094 -331.5330-317.6644 -1925.8571
Σ Q.
in = ΣQ.
out
![Page 103: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/103.jpg)
-77.0188 -3724.8337-599.8498 -430858.5899
C
Cp H-134.5842 -63.6855596.9217 0.6106
-199.7709 -3743.8518 Q in-231.6094 -43507.6878 Qout mother liquor-236.1238 -446.9370 Qout Cake
-77.0188 -37.6246 Qair pencuci-317.6644 -19.4531 total
30.9572 -47818.6293
Cp H-134.5842 -63.6855596.9217 0.6106
-199.7709 -3743.8518-231.6094 -43507.6878-236.1238 -446.9370
-77.0188 -37.6246-317.6644 -19.4531-599.8498 -47818.6293
CO2(gas)+ H2O(gas)93.8308 93.8308
= mol yang bereaksi x (ΔHF produk - ΔHF reaktan)
C Tref 298.0000
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
![Page 104: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/104.jpg)
k
-0.03630.0000 0.0000 0.00000.0015 0.0000 12100.00000.0010 0.0000 -115700.0000
-0.0338 0.0000 -103600.0000
NH3+ H2O CO2
0.0612 0.0612 0.0612
= mol yang bereaksi x (ΔHF produk - ΔHF reaktan)Kj
tref 298.0000
∆B ∆C ∆D0.0000 0.0000 0.0000
0.0030 0.0000 -18600.00000.0013 0.0000 0.00000.0010 0.0000 -115700.00000.0053 0.0000 -134300.0000
NH3 H200.4885 0.4885
tref 298.0000
∆B ∆C ∆D-0.1008 0.0002 0.00000.0030 0.0000 -18600.00000.0013 0.0000 0.0000
-0.0965 0.0002 -18600.0000
∆B ∆C ∆D
![Page 105: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/105.jpg)
40.6830 Kj/Kmol
kg/jam 18.1800 kmol/jamKj
Cp H4900.2763 2318.82073275.6980 3.35057205.0291 4039.99823863.1574 362483.0210
![Page 106: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/106.jpg)
9002.5882 1691.13069085.6337 17197.3613
37332.3826 387733.6825
Tref 298.0000
Cp H508.8824 240.5634936.8480 0.9573754.4309 422.6009352.4311 33035.8095883.5158 165.8019899.4763 1700.8335
3436.1082 35566.5665
Tref 298.0000
Cp H4591.1058 2.17251153.2931 0.00121360.2219 0.7627
644.2965 60.45481603.0615 0.30111630.5959 3.0864
10982.5747 66.7788
misalTout 100.0000 C373.0000 K
Kmol/jamKg/jam
Q udara25127.5152
Q udara377227.8524
Quadar pendingin in Quadara pendingin out
![Page 107: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/107.jpg)
135207.9140 487308.2512
135207.9140 487308.2512
kk 35.8476k -66.8595 ck 89.7473 c
Cp H7205.0291 811001.86064250.5488 399092.54882910.0478 1599.7971
14365.6257 1211694.2065
Cp H818.4188 3532.1378 0.0437481.9523 45251.4713 0.9507330.4362 181.6571 0.0056
1300.3711 48965.2662 0.9944
Cp H818.4188 88589.5088
818.4188 88589.5088
Tref 298.0000Tout 45.0000
318.0000
![Page 108: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/108.jpg)
Kj/kmol Cp 4.1870 Kj/KgKj/kmol
17547.5983 Kgjam
densitas air 1000.0000 kg/m317.5476 m3air yang dibutuhkan
Output Qpendigin in Qpendingin out220543.7201 1471862.1693
48965.2662
48965.2662 220543.7201 1471862.1693
CpKJ/Kmol C H-134.5842 -6304.8660-199.7709 -374385.1807-236.1238 -44246.7647-231.6094 -331.5330-317.6644 -1925.8571
-77.0188 -3724.8337596.9217 60.4454
-599.8498 -430858.5899
Tref 298.0000
Cp H2819.0497 132064.02344161.0466 7798104.72615112.8170 958080.58785112.8170 7318.64756592.8423 39969.45671680.4527 81271.12952161.8951 218.9174
25479.0254 9017027.4883
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Kg steam yang dibutuhkan
Q amoniakout2266403.0378
Q amoniak in11714289.1159
Q steam in Q stea out11714289.1159 2266403.0378
MRTref
Mr CaCO3Mr MgCO3
Cp H Mr CaO2819.0497 132064.0234 Mr MgO4161.0466 7798104.7261 Mr CO25112.8170 958080.5878 Mr H2O5112.8170 7318.6475 Mr H26592.8423 39969.4567 Mr O21680.4527 81271.1295 SiO22161.8951 218.9174 Al2O3
27640.9205 9017027.4883 Fe2O3MgCl2
Tref 298.0000 CaCl2
Cp H4785.0538 9314.04012104.0121 131.91572528.9009 65.77104632.1452 70.86565985.6366 51.08012401.9902 229119.2283
22437.7388 238752.9008
![Page 110: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/110.jpg)
CaCl2 +2NH3 +H2O
91.8201 183.640291.8201 183.6402
tref 298.0000
∆B ∆C ∆D
-0.0322 0.0000 0.0000-0.0004 0.0000 104700.00000.0015 0.0000 -30200.00000.0030 0.0000 -18600.00000.1148 0.0000 0.00000.0867 0.0000 55900.0000
MgCl2 +2NH3 +H2O
0.0552 0.1103 0.05520.0552 0.1103 0.0552
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
![Page 111: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/111.jpg)
tref 298.0000
∆B ∆C ∆D-0.0322 0.0000 0.0000-0.1115 0.0001 0.00000.0144 0.0000 0.00000.0030 0.0000 -18600.00000.0013 0.0000 0.0000
-0.1251 0.0001 -18600.0000
H2O CO2
6.0626 6.0626
tref 298.0000
∆B ∆C ∆D0.0000 0 0.00000.0030 0 -18600.00000.0013 -0.00000018 0.00000.0010 0 -115700.00000.0053 -0.00000018 -134300.0000
0.1103
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
![Page 112: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/112.jpg)
H20
48.3626
tref 298.0000
∆B ∆C ∆D
-0.1008 0.0002 0.00000.0030 0.0000 -18600.00000.0013 0.0000 0.0000
-0.0965 0.0002 -18600.0000
Tref 298.0000
Cp H
![Page 113: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/113.jpg)
2161.8951 218.91742819.0497 132064.02344161.0466 8390054.14425112.8170 18597.42735112.8170 7318.64754785.0538 9314.04012528.9009 65.77104632.1452 70.86565985.6366 51.08012104.0121 15.82992401.9902 8568.19381466.8213 80.92974050.0067 371872.1150
45160.2972 8938073.0676
Tref 298.0000
Cp H4161.0466 16813.73581680.4527 400243.14682453.5375 14874.70158295.0368 431931.5840
Tref 298.0000Tout 45.0000
318.0000
Kj/kmol Cp 4.1870 Kj/KgKj/kmol
115029.5560 Kgjam
densitas air 1000.0000 kg/m3115.0296 m3air yang dibutuhkan
Output Q pendingin in Q pendingin out1445727.5480 9648480.0556
![Page 114: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/114.jpg)
8938073.0676431931.5840
9370004.6516 1445727.5480 9648480.0556
Tref 298.0000
Cp H411.9188 40093.7142
11.7231 0.7350216.9407 5.6421397.8856 6.0871521.1856 4.4477377.0969 204.3217
1936.7507 40314.9479
Tref 298.0000
Cp H99074.4386 192866.289857290.6808 1490.002174530.6558 1140.2207
133701.5516 1140.981738531.4497 2415.816844596.4142 4253933.7909
447725.1908 4452987.1020
Tref 298.0000
Cp H69273.9362 37534.577340081.2145 3825753.8375
109355.1507 3863288.4147
CaO + CO2
![Page 115: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/115.jpg)
95.3874 95.387495.3874 95.3874
tref 298.0000
∆B ∆C ∆D-0.0026 0.0000 312000.00000.0004 0.0000 -104700.00000.0010 0.0000 -115700.0000
-0.0011 0.0000 91600.0000
MgO + CO2
0.0627 0.06270.0627 0.0627
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
![Page 116: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/116.jpg)
tref 298.0000
∆B ∆C ∆D0.0000 0.0000 0.00000.1115 -0.0001 0.00000.0010 0.0000 -115700.00000.1126 -0.0001 -115700.0000
H2O(g)
0.5418
tref 298.0000
∆B ∆C ∆D-0.0013 0.0000 0.00000.0015 0.0000 12100.00000.0002 0.0000 12100.0000
∫Tref
T
ΔC p dT=R(ΔA ( T - T R) + ΔB2
(T 2 -T2
R) + ΔC3
(T 3 -T3
R)−ΔD( 1T
- 1
TR))
![Page 117: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/117.jpg)
Kj/jam
KJ (Peray, Kurt E hal 62)
KG batubara
22.0506 mol
Kg
3360.0936 kg UDARA
OUTPUT Q loss Q hotgas897812.8225 7673613.8677
3863288.41474452987.1020
8316275.5167 897812.8225 7673613.8677
Tref 298.0000
Cp H99074.4386 192866.289857290.6808 1490.002174530.6558 1140.2207
133701.5516 1140.981738531.4497 2415.8168
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44596.4142 4253933.7909447725.1908 4452987.1020
Tref 298.0000
Cp H4785.0538 8383.47442528.9009 65.70534632.1452 70.79485985.6366 51.02402104.0121 131.78392401.9902 228890.3608
22437.7388 237593.1433
Tref 298.0000
Cp H8044.6689 1566.04014281.9160 0.11147824.3585 0.1197
10023.5129 0.09413522.4158 0.22083993.7548 380.9537
37690.6269 1947.5398
misalTout 300.0000 C573.0000 K
Kmol/jamKg/jam
Q udara77783.9461
Q udara4291230.3650
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Qpendingin in Q pendingin out248464.5198 4461910.9387
248464.5198 4461910.9387
Tref 298.0000
Cp H4161.0466 16813.73581680.4527 400243.14682453.5375 14874.70158295.0368 431931.5840
Tref 298.0000
Cp H377.0969 1523.7530152.2450 36261.0823222.0108 1345.9520751.3527 39130.7873
Tref 298.0000Tout 45.0000
318.0000
Kj/kmol Cp 4.1870 Kj/KgKj/kmol
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5508.3585 Kgjam
densitas air 1000.0000 kg/m35.5084 m3air yang dibutuhkan
Komponen Fi Kmol/jam Cp (Kj/Kmol)air pendingin in 306.0199 226.2296
Komponen Fi Kmol/jam Cp (Kj/Kmol)air pendingin out 306.0199 1509.8088
kg/jamkkg/jmKkg/jmK
Kg/jam370.0007
K 370.0007
) yang digunakan benar. Digunakan 1 stage kompresor untuk mendapatkan rasio yang baik, dengan Rc adalah:
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suhu rata
(Coulson, 1983)
Temperatur masuk
Polytropic Temperatur exponent
Compressibility function, XCompressibility function, Y
Compressibility factor
top ABS-201 TOP BUBLe total kmol/jam
87.4374 4.8576266.7523 2149.4507 10747.2533 244.2558266.7523 10834.6907 249.1134
m3/detik
Nilai Ep diperoleh dengan menghubungkan nilai lajuvolumetrik yang diperoleh pada gbr. 3.6 buku Coulson. R, vol.6, hal 75 untuk jenis kompresor sentrifugal.
3600
1 x
T
T
reff
in
[ Rc ]m
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KatmKatm
Xi. XiPc xi.Tc Cp10.0195 4.2444 12.6183 30923.48740.9805 71.4437 298.2682 17794.94701.0000 75.6881 310.8864 48718.4344
Kj/kmol
(gbr 3.10 Coulson,1983)(gbr 3.8 Coulson, 1983)
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370.6051 C
atmKatmKK
Cp masuk Q26556.4652 129001.633515613.8169 3813764.667042170.2822 3942766.3005
Cp masuk Q26603.9140 129232.122115641.4783 3820521.118242245.3923 3949753.2403
out put
3949753.2403
3949753.2403
terf 298.0000
Cp masuk Q
26603.9140 129232.122115641.4783 3820521.118242245.3923 3949753.2403
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Tref 298.0000
Cp masuk Q377.0969 1831.7992222.0108 54227.4046599.1076 56059.2038
Tref 298.0000Tout 45.0000
318.0000
Kj/kmol Cp 4.1870 Kj/KgKj/kmol
54602.3906 Kgjam
densitas air 1000.0000 kg/m354.6024 m3air yang dibutuhkan
Komponen Fi Kmol/jam Cp (Kj/Kmol)air pendingin in 3033.4661 226.2296
Komponen Fi Kmol/jam Cp (Kj/Kmol)air pendingin out 3033.4661 1509.8088
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NH4Cl 53.5000 kg/kmolNa2CO3 106.0000 kg/kmolCaCl2 111.0000 kg/kmol NH4HCO3 79.0000MgCl2 95.0000 kg/kmol NH4OH 35.0000
Cp /R at 25 C6.1110
10.53909.8480 solid
12.4800 Cp/R=A+BT+DT^-2 Smith10.7410
8.76205.0580
Cp at 25 C36.5900 solid26.2520 Cp=A+BT+CT^2 Yaws79.112043.1370
Cp /R at 25 C9.7180 Liquid9.0690 Cp/R= A+BT+CT^2 Smith
CP= A +BT+CT^2+DT^3 Yaws Cp /R at 25 C
4.03803.5090 gas4.46704.4670 Cp/R= A+BT+CT^2 +DT^-2 smith3.5090
perry pada 273 K
-314400.0000
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-601700.3757-852411.5600 27082.8384
-46110.0000-285830.0000
38.8157
752.5030
J
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yang dipakemisal amoniak yang akan digunakan steam dengan suhu -33 C dengan P 101.33 Kpa
HV 1417.4980 kj/kg
Hf 47.9006 k/kg
lamda steam 1369.5974 kj/kg
maka jumlah steam pemanas yang dibutu 928.8063
Komponen Fi Kg/jam HfKJ/gamoniak in 928.8063 47.9006
Komponen Fi Kg/jam Hv KJ/gamoniak out 928.8063 1417.4980
Q in224204.8123 input Qpelarutan
Q in brine 2007975.8621 27695.2503Q out Q in make up 847.8752
1496295.5246 Q recovery 39214.4839Q pndingin inQ out TopQ out Bootom
Q pndingin outtotal 2048038.2212 27695.2503
produk top
2149.4507
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mencari Tmasuk rata2 setelah penamabahan panas bahan masuk lain
kg/jam kmol/jam CP ∆H (kj)87.4374 4.8576 377.0969 1831.7992
10747.2533 244.2558 222.0108 54227.404610834.6907 249.1134 599.1076 56059.2038
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∫CpdT=R(A (T 2−T1 )+B2
(T 22−T 1
2)−D( 1T 2
− 1T1
))
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gak dipakemisal amoniak yang akan digunakan steam dengan suhu -33 C dengan P 101.33 Kpa
1417.4980 kj/kg 240.0000
47.9006 k/kg
1369.5974 kj/kg
maka jumlah steam pemanas yang dibutuhka 632.4015 Kg amoniak yang dibutuhkan Pery,3-125
Fi Kg/jam HfKJ/g Q amoniak in632.4015 47.9006 30292.4131
Fi Kg/jam Hv KJ/g Q amoniak in632.4015 1417.4980 896427.9145
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Pendingin yang digunakan = amonia pada T = -33.5 oC Hv ammonia pada -33,5 oC = 1417,498 kJ/kgHf ammonia pada -33,5 oC = 47,9006 kJ/kg λ (panas laten) = 1369,5973 kJ/kg
3402.6151 Kg amoni8506537.7013 rupiah
HV Komponen Fi KG/jam HV
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amoniak 3402.6148 1417.4980
Komponen Fi KG/jam Hfamoniak 3402.6148 47.9006
Fi Kmol/jam Cp (Kj/Kmol) Q in4114.2078 377.0969 1551454.9064
Fi Kmol/jam Cp (Kj/Kmol) Q out4114.2078 1509.8088 6211667.3336
Neraca panas sekitar RDF-301 :
297.2783maka Q masuk + Q air pncuci = Q liquid + Q cake
misal suhu produk liquid = cake maka
-478860.2521 Kj/jam Tin 22.3499 295.3499T ref 298.0000
Mol Cp H2768.2319 47.3202 -134.5842 -6368.5515
13.9107 0.1023 596.9217 61.055934070.6455 1892.8136 -199.7709 -378129.032510126.5235 189.2808 -236.1238 -44693.7017 trial buat dapretin suhu keluaran15779.3481 187.8494 -231.6094 -43507.6878
120.2402 1.4314 -231.6094 -331.5330483.7795 6.1238 -317.6644 -1945.3102
1709.7903 48.8512 -77.0188 -3762.458365072.4698 2373.7727 -831.4592 -478677.2191
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Input output-537615.8753
-430858.5899-47818.6293
58755.6232 -183.0329-478860.2521 -478677.2191
c. Q out put padatanTout 120.0000 C Tref
393.0000 Kkomponen Kg/jam kmol Cp H
NaCl 27.6823 0.4732 4900.2763 2318.8207CaSO4 0.1391 0.0010 3275.6980 3.3505H2O 10.0929 0.5607 7205.0291 4039.9982Na2CO3 9946.0612 93.8308 3863.1574 362483.0210NaHCO3 15.7793 0.1878 9002.5882 1691.1306NH4Cl 101.2652 1.8928 9085.6337 17197.3613total 10101.0202 96.9464 37332.3826 387733.6825
d. Q out gasTout 120.0000 C
393.0000 Kkomponen Kg/jam kmol Cp HH2O 2026.0895 112.5605 7205.0291 811001.8606CO2 4131.2482 93.8920 4250.5488 399092.5488NH3 9.3457 0.5497 2910.0478 1599.7971Total 6157.3377 207.0023 11455.5779 1211694.2065
e. Jumlah steam yang dibutuhkan
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Q p= Qout-Qin-QrQp= 2051537.6694 Kj/jam
INPUT Q reaksi Q laten OUTPUTQin -47818.6293 -405009.7936 718.6425Qsteam inQ out gas 1211694.2065Qout padatan 387733.6825Qsteam out
total -47818.6293 -405009.7936 718.6425 1599427.8890
misal steam yang akan digunakan steam dengan suhu 150C dengan P 476 Kpa
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Hf 524.9900 kj/kg
Hg 2713.5000 k/kg
lamda steam 2188.5100 k/kg
maka jumlah steam pemanas yang dibutu 937.4130 Kg steam yang dibutuhkan
Komponen Fi Kg/jam Hf KJ/g Q amoniakoutsteam pemanas out 937.4130 524.9900 492132.4376
Komponen Fi Kg/jam HG KJ/g Q amoniak insteam pemanas in 937.4130 2713.5000 2543670.1070
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menghitung massa udara yang dibutuhkan
Q= 352100.3372 Kj/jamCpu 29.1738 kj/kmol k
1.0095 Kj/kg KTG1 303.0000 KTG2 316.0207 K
Mu= 26787.7701 kg/jam10% dilebihkan 29466.5471 kg/jam
Qpendingin in 135207.9140 Kj/jamQpendingin out 487308.2512 Kj/jam
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Komponen Fi Kmol/jam Cp (Kj/Kmol) Q inair pendingin in 974.8666 226.2296 220543.7201
Komponen Fi Kmol/jam Cp (Kj/Kmol) Q outair pendingin out 974.8666 1509.8088 1471862.1693
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100.0000 kg/kmol84.0000 kg/kmol56.0000 kg/kmol40.0000 kg/kmol44.0000 kg/kmol18.0000 kg/kmol
2.0000 kg/kmol32.0000 kg/kmol60.0000 kg/kmol
102.0000 kg/kmol160.0000 kg/kmol
95.0000111.0000
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91.820191.8201
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Komponen Fi Kmol/jam Cp (Kj/Kmol) Q inair pendingin in 6390.5309 226.2296 1445727.5480
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Komponen Fi Kmol/jam Cp (Kj/Kmol) Q outair pendingin out 6390.5309 1509.8088 9648480.0556
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menghitung massa udara yang dibutuhkan
Q= 4213446.4189 Kj/jamCpu 29.1738 kj/kmol k
1.0095 Kj/kg KTG1 303.0000 KTG2 387.7897 K
Mu= 49226.4856 kg/jam
Q=M U CpU (T G2−T G1 )
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Q pendingin in 248464.5198 Kj/jamQout 4461910.9387 Kj/jam
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Q in69230.7774
Q out462031.5741
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127.0000
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x.Cpi602.9986
17447.950918050.9495
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Q in686259.9752
Q out4579954.0116
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kg/kmolkg/kmol
∫CpdT=R(A (T 2−T1 )+B2
(T 22−T 1
2)−D( 1T 2
− 1T1
))
∫CpdT=A (T 2−T 1 )+B2
(T 22−T 1
2 )+ C3
(T23−T 1
3 )
∫CpdT =R(A (T 2−T1 )+B2
(T 22−T 1
2)+ C3
(T 23−T 1
3))∫CpdT=A (T 2−T 1 )+
B2
(T 22−T 1
2 )+ C3
(T23−T 1
3 )+ D4
(T 24−T1
4 )
∫CpdT=R(A (T 2−T1 )+B2
(T 22−T 1
2)+ C3
(T 23−T 1
3)−D( 1T 2
− 1T1
))
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misal amoniak yang akan digunakan steam dengan suhu -33 C dengan P 101.33 Kpa
240.0000
Kg amoniak yang dibutuhkan Pery,3-125
Q amoniak in44490.3797
Q amoniak in1316581.0920
output Pendingin in pendingin out44490.3797 ###
0.00001345.9520
802296.8072
803642.7592 44490.3797 ###
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panas laten
1369.5974
Q in
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4823199.7192
Q out162987.2920
C
trial buat dapretin suhu keluaran
![Page 152: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/152.jpg)
298.0000
Tref 298.0000
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Q steam Qsteam out2543670.1070 492132.4376
2543670.1070 492132.4376
Menurut mantell heat loss furnace terdur dari
![Page 154: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/154.jpg)
radiasi 0.0170 GAK DIPAKE
cooling 0.1000
maka total kehilangan panas= 0.8830
maka total kebutuhan panas a 2323372.2190 Kj/jam
diketahui 1 kg batubara meng 29000.0000 KJ (Peray, Kurt E hal 62)
maka kebuthan batubara adal 80.1163 KG batubara
f.menghitung kebutuhan oksigen
C + O2 CO2mol batubara = mol oksigen yang digunakan jadi mol 6.6764 mol
jadi oksigen yang dibuthkan a 213.6434 Kg
massa tersebut diperloeh dari udara sebnayak ### kg UDARA
![Page 155: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/155.jpg)
Data MRKomponen ρi (kg/m3) vis (cp)NaCl 2165.0000 2.1000 Mr CaCO3 100.0000CaSO4 2960.0000 10.0000 Mr MgCO3 84.0000H2O 997.0800 0.8937 Mr CaO 56.0000NaHCO3 (aq) 2159.0000 0.9750 Mr MgO 40.0000NaHCO3 (l) 2159.0000 0.9750 Mr CO2 44.0000NH4Cl 1527.4000 1.2200 Mr H2O 18.0000NH4HCO3 1580.0000 0.0370 Mr H2 2.0000Total Mr O2 32.0000
SiO2 60.00001m3 35.3130 ft3 Al2O3 102.00001 ft3 7.4810 US gallon Fe2O3 160.00001 in 2.5400 cm MgCl2 95.00001 in 0.0254 m CaCl2 111.00001 in 0.0833 ft1 ft 0.3048 ft1 ft 12.0001 in1 ft 0.3048 mgc 32.1740 lbm ft/lbf.s2g 32.1740 ft/s21 rad= 0.1590 rps1hp 0.7450 kw1kg/m3 0.0624 lb/ft31cp= 2.4196 lb/ftjam1 kJ/(kg K) 0.2389 Btu/lb F1 w /m2K= 0.1761 Btu/ ft2 h oF1 kJ/kg = 0.4229 Btu/ lbm1 atm= 14.7000 psi1cp= 0.0010 kg/ms1 dyne 0.0000 kgm/s2
1dyne/cm 0.0010 kg/s2
1kj= 0.9478 BTU
1 lb 0.4536 kg1kw 1.3415 Hp1cp 2.4200 lb/ft hr
1 kJ/(kg K) 0.2389 Btu/(lbm oF) 1kJ 0.0003 kWh 1 W/m.K 0.5779 Btu/(ft h oF) 1 kcal/(kg oC) 1 Btu/(lbm oF)
4.1868 KJ/ (kg K)
1lb/ft2 0.0069 lb/in2
1 in2 0.0069 ft21 in2 0.0006 m2
∫CpdT=A (T 2−T 1 )+B2
(T 22−T 1
2 )+ C3
(T23−T 1
3 )+ D4
(T 24−T1
4 )
∫CpdT=R(A (T 2−T1 )+B2
(T 22−T 1
2)+ C3
(T 23−T 1
3)−D( 1T 2
− 1T1
))
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1 ft2 0.0925 m2
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kg/kmolkg/kmolkg/kmolkg/kmolkg/kmolkg/kmolkg/kmolkg/kmolkg/kmolkg/kmolkg/kmol
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coolerr cuy
fluida panas amonia
Tumpan= 36.7281 c T in309.7281 K
98.1106 FT produk Err:540 c T out
Err:540 KErr:540 F
laju alir 51824.5325 kg/jam laju alirErr:540 lb/jam
Beban panas cooler 1272090.7123 kj/jamErr:540 Btu/jam
Menghitung ΔT LMTD
fluidapanas air98.1 T tinggi -27.4
Err:540 Trendah -27.4Err:540 0.0
∆Tlmtd= Err:540 F
Tc = Tavg = (T1 + T2)/2 Err:540tc = tavg = (t1 + t2)/2 -27.4000
Menentukan UD dan NT
Dari tabel 4-15, ulrich, hal 165 diperoleh untuk amoniak dan air memiliki harga range UD =110-200dengan dirt factor 0.0010Asumsi : UD = 200.0000 Btu/jam.ft2.oF
Err:540 ft2
Karena A < 200 ft2, digunakan double pipe
anulus pipeIPS 3 1Sch. No 40 in 40OD 4.5 in 1.32ID 3.5 in 1.049a" 3.068 ft2/ft 0.344
temperatur kalorik, Tc dan tc
A=Q
U D×ΔT LMTD
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cek viskositas pada terminal dingin untuk tiap fluidabrine Tavg Err:540 Fammonia tavg -27.4000 F
anulus/brine1 flow area
D2= Err:540 ftD1 Err:540 ft
aa= Err:540 ft2
Diameter equivalent
De= Err:540 ft
2. Mass velocity
Gp[= Err:540 lb/ft2 hr
3. Pada temperatur Err:540 F
µ 1.3000 cp yawsErr:540 lb/ft hr
Re= Err:540
4. jH 110.0000
5. Pada temperatur Err:540 Fcp= 0.6900 Btu/lb Fk 0.2800 btu/hr/ft2(F/ft)
Err:540
aa=π ( D22−D
12 )/4
De=( D2
2−D12 )/ D1
Gp=Wa p
Rea=( De Ga )/ μ
( c . μk )
1/3
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6 Mencari ho
ho=
ho= Err:540 Btu/jam.ft2.oF
7Clean overall coefficient, Uc
Uc= Err:540
8.Ud koreksi
= Err:540
UD= Err:540
9 Required Surface Area
A= Err:540 ft2
External surface / lin ft, a'' = 0.9170 ft2 (Tabel.11 Kern, 1965)
Required length,
L= Err:540 ft
digunakan panjang hairpin = 20 ft, sehingga banyaknya hairpin yang diperlukan :
Hairpin = Err:540 harpin 4.0000
Actual length
La = 80.0000 ft
A aktual = L x a'' 73.3600 ft2= 6.8154
Actual Design Overall Coeffesient, UD act
UD akt= Err:540
Ud
1 RdUc
1
( c . μk )
1/3
jH ( κDe
) (cp. μκ )
1/3
( μμw
)0,14
U c=hio . ho
hio+ho
QUD . Δt
Aa} } } { ¿
¿¿
LLh
Lh×hairpin
QA×Δt
![Page 162: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/162.jpg)
Dirt Factor, RdRd= Err:540
anulus/brine
1.De untuk presure drop berbeda dengan De perpan
De= Err:540 ft
Re'= Err:540f= 0.00001 ft2 (Fig.26, Kern)s= 0.91 Btu/lb oF
56.8750 Lb/ft3
2). = Err:540
3).V= Err:540 ft/s
4) ∆Fi= Err:540 s
5) =
Err:540 psi
Err:540 psi
AlatKodeFungsiBentuk Double Pipe Heat ExchangerDimensi pipa Annulus:
IPS 3.0000
p=
QA×Δt
Uc−UdUc×Ud
De '=D2−D1
Rea '=( De ' Ga )/ μ
Δ Fa 4×f ×Ga2×L2×g×ρ2×De
Gaρ×3600
3 x (V 2
2 g )Δ Pa ( Δ Fa+ Δ Fi )×ρ
144
![Page 163: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/163.jpg)
Sch. No. 40OD 4.5000ID 3.5000
Inner pipe:IPS 1.0000Sch. No. 40OD 1.3200ID 1.0490
Jumlah hairpin 4.0000Panjang 1 pipa 20.0000∆P, annulus Err:540∆P, inner pipe Err:540
Bahan konstruksi Stainless Steel SA-240 A ISI tipe 316Jumlah 1 buah
![Page 164: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/164.jpg)
-33.0000 c240.0000 K-27.4000 F-33.0000 c240.0000 K-27.4000 F928.8063 kg/jam
Err:540 lb/jam
temp diff F Δt2/Δt1 ln Δt2/Δt1125.5 Δt2 Err:540 Err:540
Err:540 Δt1Err:540 Δt2 – Δt1
F fluida panasF air -33.0000
Dari tabel 4-15, ulrich, hal 165 diperoleh untuk amoniak dan air memiliki harga range UD =110-200
in
in inft2/ft
![Page 165: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/165.jpg)
karena kurang dari 1 cp maka Tc sadengan T avgµ 1.3000 cp sama dengan Tcµ 0.0720 cp sama dengan tc
pipeaimonia1 flow area
D= Err:540 ft
ap Err:540 ft2
2. Mass velocity
Ga= Err:540 lb/ft2 hr
3. Pada temperatur -27.4000 F
µ 0.0720 cpErr:540 lb/ft hr
Rea= Err:540
4. jH 390.0000
5. Pada temperatur -27.4000 Fcp= 0.0720 Btu/lb Fk 0.0950 btu/hr/ft2(F/ft)
Err:540
a p=πD2/ 4
Ga=Waa
( c . μk )
1/3
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6. Hio
ho= Err:540
m2
( c . μk )
1/3
![Page 167: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/167.jpg)
RESSURE DROPpipe/amoniak gas
1) Rep Err:540
f= 0.0002 ft3s= 1 Btu/lb oF
62.5000 lb/ft3
Err:540
2) = Err:540
=
Err:540 psi
Err:540 atm
in
p=
Δ Fp4×f ×G2×L
2×g×ρ2×De
Δ PpΔ Fp×ρ144
Δ Pp
![Page 168: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/168.jpg)
inin
in
inin
ftpsipsi
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Reaktor buble chimi
Atas dasar US Patent 4.291.002, reaksi antara ammonium hidroksida dan carbon dioksida berlangsung pada suhu 30 C dan tekanan 4 atm dengan residence time sampai 10 menit.dan menurut reksi pseodo orderreaksi
asumsi H2O berlebihdan NH3 berekasi secara sempurnasehingga waktu reaksi asumsi diabaikan
k1NH4OH + CO2 NACL + NH4HCO3
k2pada reaksi series untuk orde pseudo first order
misalNH4OH A NACLCO2 B NH4HCO3
untuk psedo firs order
A + Bmaka rA= .-k1CACBrD kiCACB.-K2CCCDrE k2CCCD
menurut frogmentrA + rC + r E = 0.0000jika keadaan tetap mka reaktan akan hilang dan pembentukan produk maka pendekatan akan ekivalen dengan :
.-Ra setara rE dan rC setara dengan nol.
sehingga mengitung volumV=Fao .xA /-rAV=FaO.x/k1CACB
mencari komposisi reaktan
![Page 170: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/170.jpg)
1. fase liquidumpan kg/jam kmol/jam
NaCl 13841.1595 236.6010CaSO4 13.9107 0.1023H2O 33645.8750 1869.2153NH4OH 8548.9515 244.2558total 56049.8966 2350.1743
BM campuran= 23.8493 Kg/kmoldensitas liquid
1039.1774
μ L = 1.1253
2. Komposisi Reaktan (Fase Gas)
umpan kg/jam kmol/jam
H2O 87.4374 4.8576CO2 10747.2533 244.2558total 10834.6907 249.1134
VolumeR : Konstanta gas ideal =
BM campuran 43.4930P= 4.0000 atmT= 30.0000 c
6.9971 kg/m3μ L = 0.0149
Volume gas dalam liquid 10.3565Volume campuran 1585.8920
Massa CO2 244.2558 mo/jam 538.4827
FAO= 244.2558 v1=
FBO= 244.2558 V2=
e^(wi*lnµ)
pG =
pG = e^(wi*lnµ)
Sehingga Densitas Campuran total =pG +pLpcampuran =
campurandensitas
totalmassa
ρL=1
∑ wi
ρi
BM×PR×T
![Page 171: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/171.jpg)
FCO 236.6010
CnolCO2 0.1540 kmol/m3 XA=NH4OH 0.1540 kmol/m3CnolNaCl brin so 0.1492 kmol/m3
r = (Derks,at all kinetic absrption of CO2 in aqueous ammonia solutin, 2009)
dimana : Yang mengikuti reaksi
CO2 + NH4OH + NaCL NaHCO3 + NH4Cl
CA C dari CO2CB C dari NH3
dari jurnal k2kh2o.k2/k-1kapp
QB= 1.0000QC= 0.9687e= -0.3333
h/d= 5-10
YA1 fraksi gas keluarYA2 fraksi gas masuk
Ya2 0.8000Ya1 0.0000
r=C A CB
1k2
+k−1
k2
1kC CC
h=Gf∫Y A 1
Y A 2 dY A
−r A
h=Gf∫Y A 1
Y A 2dYA
CA CB
1k2
+k−1
k2
1kC CC
∫Y A 1
Y A 2dYAC A CB
1k2
+k−1
k2
1kC CC
∫Y A 1
Y A 2
1k2
+k−1
k2
1kC CC
dYA
C A CB
∫Y A 1
YA 2
1k2
+k−1
k2
1
kC
CAO (ΘC−Y A)(1+ℓY A )
dYA
C AO(1−Y A)
(1+ℓY A ).C AO (ΘB−Y A )(1+ℓY A )
∫Y A 1
YA 2
1k2
+k−1
k2
1
kC .C AO(ΘC−Y A)(1+ℓY A )
dYA
CA
O2 (1−Y A)(ΘB−Y A)
(1+ℓY A )2
∫Y A 1
Y A 2 [1k2
+k−1
k2
1
kC .C AO (ΘC−Y A)(1+ℓY A ) ](1+ℓY A)2 dYA
CA
O2 (1−Y A)(ΘB−Y A)
∫Y A 1
Y A 2 [1k2
+k−1
k2
(1+ℓY A)kC .C AO (ΘC−Y A) ](1+ℓY A)2 dYA
CA
O2 (1−Y A)(ΘB−Y A)
![Page 172: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/172.jpg)
Volume liquid
HL
1. Perancangan Reaktor Bubble
a. Menentukan koefisien diffusivitas (DAL)
Proses difusi terjadi didalam fasa cair. Persamaan yang digunakan adalah
∫Y A 1
Y A 2dYAC A CB
1k2
+k−1
k2
1kC CC
∫Y A 1
Y A 2
1k2
+k−1
k2
1kC CC
dYA
C A CB
∫Y A 1
YA 2
1k2
+k−1
k2
1
kC
CAO (ΘC−Y A)(1+ℓY A )
dYA
C AO(1−Y A)
(1+ℓY A ).C AO (ΘB−Y A )(1+ℓY A )
∫Y A 1
YA 2
1k2
+k−1
k2
1
kC .C AO(ΘC−Y A)(1+ℓY A )
dYA
CA
O2 (1−Y A)(ΘB−Y A)
(1+ℓY A )2
∫Y A 1
Y A 2 [1k2
+k−1
k2
1
kC .C AO (ΘC−Y A)(1+ℓY A ) ](1+ℓY A)2 dYA
CA
O2 (1−Y A)(ΘB−Y A)
∫Y A 1
Y A 2 [1k2
+k−1
k2
(1+ℓY A)kC .C AO (ΘC−Y A) ](1+ℓY A)2 dYA
CA
O2 (1−Y A)(ΘB−Y A)
∫Y A 1
Y A 2 [1k2
+k−1
k2
1
kC .C AO (ΘC−Y A)(1+ℓY A ) ](1+ℓY A)2 dYA
CA
O2 (1−Y A)(ΘB−Y A)
![Page 173: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/173.jpg)
Keterangan :Φ : Association parameter =1M : Berat molekul larutan, kg/kmolT : Temperatur, Kμ : Viskositas larutan, kg/m.sVm : Volume molal zat terlarut, m3/kgmol
Berdasarkan Tabel 8.6 Coulson, diperoleh :
Vm CO2= 0.0340Vm H2O= 0.0189
Vmtotal= 0.0529
DAL= 8.99748108101236E-10
b. Menghitung diameter gelembung (dB)
Keterangan : db : Diameter gelembung, mdo : Diameter oriffice = σL : Tegangan muka cairang : Percepatan grafitasi, m/s2Δρ : Densitas (cairan-gas), kg/m3
Menentukan Δρρgas pada T=
P=
Δρ = ρ (cairan-gas)1032.1803
64.4080
Menghitung surface tension
Dimana:σL = surface tension, MJ/m2Pch = sudgen’s parachorρL = densitas cairan, kg/m3
maka ∆p=
DAL=117 ,3 .10−18 (φ×M )0,5×T
μ ×Vm0,6
db= ( 6×do×σ L
g×Δρ )1
3
σ L = [ Pch × ( ρL− ρv )M ]
4
× 10−12
![Page 174: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/174.jpg)
ρv = densitas saturated vapor, kg/m3M = berat molekul
Dari Coulson, 1983, hal 258 dapat dicari nilai Pch:
NaClCaSO4H2ONH4OH
σLNH4OH = 556.0254556.0254
Diperoleh surface tension (τ) campuran:
σLmix=σL1.x1+σL2.x2+….
σLmix= surface tension campuranσL1,σL2= surface tension komponenx1,x2, componen mol fraksi
σLmix= 133.14460.1331
133.1446
Jadi, diameter gelembung :
db= 0.00920.92439.2434
c. Menentukan koefisien transfer massa campuran (kL)
Diameter gelembung (db) 2 mm, sehingga persamaan yang digunakanPersamaan yang digunakan adalah :kL = kL (2 mm) 500 db
kL(2mm)= 0.0003Kl= 0.0012
d. Menentukan Bilangan Hatta
CAo = Konsentrasi liquid mula-mula
db= ( 6×do×σ L
g×Δρ )1
3
k L (2 mm )= 0,42 [ μL×g
ρL ]1
3 [ ρL×DAL
μL ]1
2
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Cao= 0.1540
M =
M= 1.92487994412095E-07
M< 0,3 = Reaksi membutuhkan volume bulk liquid yang besar.
Berdasarkan Perry’s Chemical engineering Hand’s Book, ada beberapa parameter design untuk reaktor gelembung yaitu : diameter gelembung (db), gas hold up (є), superficial velocity (usg), dan Interfacial area (α).1. Diameter gelembung (db)
Telah dihitung di atas diperoleh db =
2 Menentukan superficial gas velocity (usg)adalah laju volume aliran gas dibagi dengan luas area tiap lubang (treyball hal 143)
a. Lau volumetrik aliran gas
Q6/5= 18.6964
Q 11.4761
b. Menetukan diameter hole
Maka Luas tiap hole
A. Menghitung Parameter Design Reaktor Gelembung
k × CAo× DAB
kL2
Q6
5=d
b3×π×g3
5
8,268
D 0=db3 x ( ρL− ρg) xg
6 σ
![Page 176: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/176.jpg)
maka UsG= 14.61920.1462
3 Gas hold up (є)Untuk menghitung gas hold up(є), dapat digunakan persamaan :
e= 0.1983
4. Interfacial AreaUntuk menghitung interfacial area dapat digunakan persamaan sebagai berikut :
(α)=
(α)= 1.2870
Nilai (α) interfacial area yang diperoleh, masuk dalam range (α) di reaktor gelembung yaitu 40 – 600 m2/m3 sehingga memenuhi.
Berdasarkan Levenspiel (1999), untuk mengetahui apakah reaktor bubble sudah sesuai untuk reaksi tersebut, dapat dilihat dari faktor-faktor berikuta. Untuk bubble reaktor, jenis alirannya adalah plug flow untuk gas dan mixed flow untuk cairan.b. kG dan kL, untuk reaktor bubble, nilai kL harus lebih besar daripada kG, hal ini dibuktikan sebagai berikut:
kG = (Welty, 1976)
di mana H : Konstanta henry H258 atm.liter/mol
kG = 0.0212
B. Perhitungan Dimensi Reaktor
6×εdb
k L
H
ε=1 .2( μL .UsgσL )
1 /4
( Usg
( σL . gρL )
1/4 )1 /2
![Page 177: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/177.jpg)
D=
maka tinggi reaktor adalah
Berdasarkan Froment, hal 726 untuk ε ≤ 0,45; 0,03 < usg < 0,4 m/s; Dr > 0,15 m, maka rasio Z/dr adalah 0,3 < Z/Dr < 3.
a. Menentukan Sparger RingDr = 3.9624
7.0000ε = 0.1983usg = 0.1462Z= 11.8872z/DrDitetapkan diameter sparger ring, Ds = 40 % Dr
Ds= 1.92822.8000
Luas plate sparger (Ls) :
b. Menghitung diameter hole sparger
Berdasarkan Perry’s, 1997 diameter hole plate dapat ditentukan dengan persamaan
maka :Do= 1.0000
Jadi luas tiap hole :A hole= 0.7850
Direncanakan triangular pitch dengan jarak ke pusat
C = 1,5 × DoC= 1.5000
4
DsπLs
2
σ6,028
gρρdD GL
3b
o
o60sinC(h)Tinggi
![Page 178: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/178.jpg)
maka luas hole seluruhnya= rasio luas x luas plate spager
luas hole seluruhnya=
C. Penentuan Tebal Reaktor
(pers.13.1 Brownell and young,1959)
Keterangan :ts = Tebal shell, inP = Tekanan operasi, Psif = Allowable Stresss, PsiDr = Diameter reaktor, inE = Efisiensi PengelasanC = Tebal korosi (0,2 in)
Bahan konstruksi stain less stelll 167 grade 3, tipe 304, f = 18750 psi E = 80 % (double-welded butt joint) C = 0,125 in
Tekanan desain 5 -10 % di atas tekanan kerja normal/absolut (Coulson, 1988 hal. 637). Tekanan desain yang dipilih 10% diatasnya:Poperasi = 1 atm = 14,7 psia
Phid. =
ρmix = 65.2813
hL= Volume liquit/luas alas
hl= 16.853355.2931
maka
o60sinC(h)Tinggi
hC21segitigaLuas
pitchLuas
lubangLuasluasRatio
hole
holehole
tiapLuas
totalLuasJumlah
Pf
rPt i
s
6,0
144
hg
g. L
cmix
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Phid. = 25.0667
Tekanan desain 5 -10 % diatas tekanan normal Tekanan desain diambil 10 % diatasnya.
Pdesign = 1,1 (Poperating + Phid.)
Pdesain 92.25346.2757
Maka tebal shell :
ts= 0.5695
Diameter luar shell (ODs)= IDs + 2.ts
D. Menghitung Head Reaktor
Menentukan inside radius corner (icr) dan corner radius (rc).OD= 157.2500ts= 0.6250
maka icr 10.1250r 144.0000
1. Tebal head (th)
0.6250
th=P . rc .w
2fE−0,2P+C
![Page 180: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/180.jpg)
Dimana :
w= 1.6928
maka th= 1.2371
maka stnadar yang digunakan
maka untuk th=
2. Depth of dish (b)
Berdasarkan Brownel & Young hal.87 didapat :
a=ID/2 1.9812BC=rc-icr 133.8750AB=a-icr 67.8750
b= 28.60720.7266
3. Tinggi Head (OA)
OA = th + b + sf (Brownell and Young,1959.hal.87)
OA= 33.8572
0.86004. Tinggi reaktor
Tinggi reaktor = tinggi shell + (2 x tinggi head(OA))
tinggi reaktor 18.788761.6429
E. Merancang Koil Pendingin
Massa pendingin
th=P . rc .w
2fE−0,2P+C
w=14
.(3+√ rc
icr )
b=rc−√BC 2−AB2
![Page 181: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/181.jpg)
air suhu 28 CpµCpk
Pemilihan pipa standar (Tabel. 11, Kern) :
Dipilih tube :OD 1.3200ID 1.0490at” 0.8640a’ 0.3440NPS = 1.0000
Fluks Massa Pendingin Total (Gtot)
Gtot = 10169451.1836
Fluks Massa Tiap Set Koil
Gi =
Kecepatan medium pendingin di dalam pipa umumnya berkisar 1,5-2,5 m/s.
Dipilih : Vc = 2.5000
maka Gi= 535.30661927103.7469
Jumlah Set Koil (Nc)
5.2771
w
at'
ρc×vc
Nc=G c , tot
Gi
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Koreksi Fluks Massa Tiap Set Koil (Gi,kor)
1694908.5306
Cek Kecepatan Medium Pendingin (Vc,cek)
25963.15847.21202.1982
Koefisien transfer panas fluida sisi dalam tube :
Bilangan Reynold fluida dalam pipa adalah :
Nre =
Dari Gambar 24 (Kern, 1983), untuk Nre =dan faktor korksi tubeMaka hi adalah :
hi=
koefisien transfer panas pada pipa
transfer panas fluida sisi dalam koil
Koefisien transfer panas dari pipa ke luar pipa adalah
Hio = hi(Di/Do)
Untuk koil, harga hio dikoreksi dengan faktor koreksi sebagai barikut
Diketahui diameter spiral atau heliks koil = 0,7 – 0,8 Dt (Rase, 1977)
Dspiral koil= 10.4000
maka hio= 1129.5628
Koefisien transfer panas fluida sisi luar tube:
ho= 20274.5201
Gi , kor=G c , tot
Nc
V c , cek=Gi
ρc
DiGc , tot
μ
J H( kID )(C p μ
k )1/3
( μμw
)0,14
h io,koil=h io,pipa [1+3,5 (Dikoil
Dspiralkoil)]
ho=0 , 36 .k
De.( De .G
μ )0 ,55
.(Cp . μk )
13
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Uc = 1069.9521
untuk proses cooler, bahan pendinginnya berupa air-aquaueos solution pendingin maka nilai Ud adalah
UD 450.0000
Dirt factor, Rd
Rd =
Rd required 0,001
Perbedaan temperatur logaritmik rata-rata adalah
Hot Fluid F86.0000 High temperatur86.0000 low temperatur
0.0000
Menghitung luas transfer panas, A
A =
A= 26.1808
Jarak Antar Pusat Koil (Jsp)
Jsp = 2 x Odkoil
Panjang Satu Putaran Heliks Koil (Lhe)Lhe = ½ putaran miring + ½ putaran data
dhe=diameter spiral 10.4000
ho× h io
ho+hio
UC−U D
UC . U D
ΔTLMTD=(T 2 -t1)−(T 1 -t2)
ln (T 2 -t1
T 1 -t2)
QUD Δ t
Lhe=1/2 . π . rhe+1 /2 . π .dhe
Lhe=1/2 π (dhe2 +J
sp2 )1 /2+1/2π . dhe
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Lhe= 32.6597
Panjang Koil Tiap Set (Lci)
30.30199.2360
Jumlah Putaran Tiap Set Koil
0.9278
Tinggi Koil (Lc)
Lc = Jsp x Npc x Nc
Volume Koil (Vc)
Vc = ( (OD)2 Lci)
Cek Tinggi Cairan Setelah Ditambah Koil (hL)
Tinggi koil harus lebih kecil daripada tinggi cairan setelah ditambah koil agar seluruh koil tercelup dalam cairan.
16.854055.2953
hL = 55.2953 ft > Lc (
F. Desain Perpipaan dan NozzleF.1. Desain Perpipaan
Saluran dibuat dengan menggunakan bahan stainless steel. Diameter optimum tube yang bahannya terbuat dari stainless steel dihitung dengan menggunakan persamaan
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
Dimana : diopt = diameter optimum dalam tube (mm) G = kecepatan aliran massa fluida (kg/s) ρ = densitas fluida (kg/m3)
Pengecekan bilangan Reynolds (NRe)
Keterangan :
Lci=A
a t} } } } } {¿ ¿¿
¿¿
N pc=Lci
Lhe
π /4
hL=V reaktor+V koil
( π4
DR2 )
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• Saluran pemasukan bahan baku liquidLaju alir massa (G) = Densitas (ρmix) = Viskositas (μmix) =
Dari persamaan 5.15, hal. 161. Coulson, 1983 diperoleh :
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
Di opt=
Dipilih spesifikasi pipa (Peters and Timmerhaus, Tabel 13, 1980) :
Nominal pipe standar (NPS) = Schedule number = 40 ID = ODFlow area per pipe (A) =
Laju alir volumetrik (Fv) :
Kecepatan aliran, v :
Bilangan Reynold, NRe :
• Saluran pemasukan bahan baku gas
Laju alir massa (G) = Densitas (ρmix) = Viskositas (μmix) =
Dari persamaan 5.15, hal. 161. Coulson, 1983 diperoleh :
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
Di opt=
A
Fv v
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Dipilih spesifikasi pipa (Peters and Timmerhaus, Tabel 13, 1980) :
Nominal pipe standar (NPS) = Schedule number = 40 ID = ODFlow area per pipe (A) =
Laju alir volumetrik (Fv) :
Kecepatan aliran, v :
Bilangan Reynold, NRe :
• Saluran keluaran produk liquid
Laju alir massa (G) = Densitas (ρmix) = Viskositas (μmix) =
Dari persamaan 5.15, hal. 161. Coulson, 1983 diperoleh :
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
Di opt=
Dipilih spesifikasi pipa (Peters and Timmerhaus, Tabel 13, 1980) :
Nominal pipe standar (NPS) = Schedule number = 40 ID = ODFlow area per pipe (A) =
Laju alir volumetrik (Fv) :
Kecepatan aliran, v :
Bilangan Reynold, NRe :
• Saluran keluaran produk gas
A
Fv v
A
Fv v
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Laju alir massa (G) = Densitas (ρmix) = Viskositas (μmix) =
Dari persamaan 5.15, hal. 161. Coulson, 1983 diperoleh :
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
Di opt=
Dipilih spesifikasi pipa (Peters and Timmerhaus, Tabel 13, 1980) :
Nominal pipe standar (NPS) = Schedule number = 40 ID = ODFlow area per pipe (A) =
Laju alir volumetrik (Fv) :
Kecepatan aliran, v :
Bilangan Reynold, NRe :
• Saluran pemasukan pendingin
Laju alir massa (G) = Densitas (ρmix) = Viskositas (μmix) =
Dari persamaan 5.15, hal. 161. Coulson, 1983 diperoleh :
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
Di opt=
Dipilih spesifikasi pipa (Peters and Timmerhaus, Tabel 13, 1980) :
A
Fv v
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Nominal pipe standar (NPS) = Schedule number = 40 ID = ODFlow area per pipe (A) =
Laju alir volumetrik (Fv) :
Kecepatan aliran, v :
Bilangan Reynold, NRe :
• Saluran keluaran pendingin
Laju alir massa (G) = Densitas (ρmix) = Viskositas (μmix) =
Dari persamaan 5.15, hal. 161. Coulson, 1983 diperoleh :
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
Di opt=
Dipilih spesifikasi pipa (Peters and Timmerhaus, Tabel 13, 1980) :
Nominal pipe standar (NPS) = Schedule number = 40 ID = ODFlow area per pipe (A) =
Laju alir volumetrik (Fv) :
Kecepatan aliran, v :
Bilangan Reynold, NRe :
F.2. Desain NozzleBerdasarkan perhitungan saluran saluran pemasukan dan keluaran pada reaktor di
A
Fv v
A
Fv v
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atas maka dapat ditentukan jenis nozzle yang digunakan.
Gambar F.2. Desain opening pada Reaktor
• Nozzle Umpan liquidSpesifikasi nozzle standar (Brownel and Young, 1959, App. F ,hal.349) :Size = OD of pipe =Flange Nozzle thickness (n) = Diameter of hole in reinforcing plate (DR) = Length of side of reinforcing plate (L) = Width of reinforcing plate (W) = Distance, shell to flange face, outside (J) = Distance, shell to flange face, inside (K) = Distance from Bottom of tank to center of nozzle :
• Nozzle umpan gasSpesifikasi nozzle standar (Brownel and Young, 1959, App. F ,hal.349) :Size = OD of pipe =Flange Nozzle thickness (n) = Diameter of hole in reinforcing plate (DR) = Length of side of reinforcing plate (L) = Width of reinforcing plate (W) = Distance, shell to flange face, outside (J) = Distance, shell to flange face, inside (K) = Distance from Bottom of tank to center of nozzle :
• Nozzle keluaran produk (liquid)Spesifikasi nozzle standar (Brownel and Young, 1959, App. F ,hal.349) :Size = OD of pipe =Flange Nozzle thickness (n) = Diameter of hole in reinforcing plate (DR) = Length of side of reinforcing plate (L) =
OD
Q
J
T t
nWeld B
Weld B
Weld B
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Width of reinforcing plate (W) = Distance, shell to flange face, outside (J) = Distance, shell to flange face, inside (K) = Distance from Bottom of tank to center of nozzle :
• Nozzle keluaran gasSpesifikasi nozzle standar (Brownel and Young, 1959, App. F ,hal.349) :Size = OD of pipe =Flange Nozzle thickness (n) = Diameter of hole in reinforcing plate (DR) = Length of side of reinforcing plate (L) = Width of reinforcing plate (W) = Distance, shell to flange face, outside (J) = Distance, shell to flange face, inside (K) = Distance from Bottom of tank to center of nozzle :
• Nozzle masukan air pendinginSpesifikasi nozzle standar (Brownel and Young, 1959, App. F ,hal.349) :Size = OD of pipe =Flange Nozzle thickness (n) = Diameter of hole in reinforcing plate (DR) = Length of side of reinforcing plate (L) = Width of reinforcing plate (W) = Distance, shell to flange face, outside (J) = Distance, shell to flange face, inside (K) = Distance from Bottom of tank to center of nozzle :
• Nozzle keluaram air pendinginSpesifikasi nozzle standar (Brownel and Young, 1959, App. F ,hal.349) :Size = OD of pipe =Flange Nozzle thickness (n) = Diameter of hole in reinforcing plate (DR) =
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Length of side of reinforcing plate (L) = Width of reinforcing plate (W) = Distance, shell to flange face, outside (J) = Distance, shell to flange face, inside (K) = Distance from Bottom of tank to center of nozzle :
G. Penentuan ManholeManhole adalah lubang pemeriksaan yang diperlukan pada saat pembersihan atau pemeriksaan pada bagian dalam kolom. Direncanakan manhole di pasang pada kolom bagian atas reaktor dengan ukuran standar 24 in berdasarkan rekomendasi API Standard 12 C (Brownell and Young, Ap. F item 5, ), dengan spesifikasi :
Jumlah = Tebal shell = tinggi tnakiDiameter dalam manhole frame, ID = Maksimum diameter lubang, Dp = Diameter cover plate, DC = Diameter bolt circle, DB = Length of side, L = Lebar reinforcing plate, W Size of fillet weld A = Size of fillet weld B = (Brownel and Young, Ap.F item 3)
H. Perencanaan Flange pada Sambungan Head dengan ShellSambungan Head dengan Shell
Sambungan antara tutup bejana dengan bagian shell menggunakan sistem flange dan baut. Bahan konstruksi yang dipilih berdasarkan pada kondisi operasi. Data perancangan:Tekanan desain = Temperatur desain Material flange = SA-203, grade B (B & Y, 1959, Tabel 13.1)Bolting steel = SA-193 Grade B (B & Y, 1959, Tabel 13.1)Material gasket = Stainless steels Diameter luar shell = Diameter dalam shell = Ketebalan shell = Tegangan dari material flange = Tegangan dari bolting material = Tipe flange = Gambar 8.a, (B & Y, hal. 240)
Gasket
hG
t
hT
HG
HT
G
h
W
R hD Cgo
g1
g1/2
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Gambar F.3. Tipe flange dan dimensinya.
• Perhitungan Lebar Gasket
Keterangan:p = tekanan desain (psi)do = diameter luar gasket (in)di = diameter dalam gasket (in)y = yield stress, lb/in2 (Fig. 12.11)m = faktor gasket (fig. 12.11)
Dari fig 12.11 Brownell & Young, untuk material gasket Stainless steels y = 9000.0000m= 3.7500Sehingga :
1.0054
Asumsi bahwa diameter dalam gasket sama dengan diameter luar shell, yaitu
do 158.0949
Lebar gasket minimum (N) :
N = Keterangan :N = Lebar gasket minimum (in)do = Diameter luar gasket (in)di = Diameter dalam gasket (in)
jadi N=
Diameter gasket rata-rata (G) : G = di + lebar gasket
Dari Fig 12.12 B 7 Y, kolom 1, tipe 1.a, didapat
bo = 0.2500
jika bo ≤ 0,25 in maka b = b0 =
Digunakan gasket dengan lebar standar
Gasket
hG
t
hT
HG
HT
G
h
W
R hD Cgo
g1
g1/2
1)p(my
pmy
d
d
i
o
i
o
d
d
2
dd io
2
N
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Perhitungan Beban Baut (bolt)
a. Beban terhadap seal gasketBeban terhadap seal gasket :Wm2 = Hy = x b x G x y (Brownell and Young, pers. 12.88, 1959)
Wm2 =
b. Beban operasi total hp1HWm1
luas baut minimum 94.3184
digunakan baut ukuran 2 inroot area 2.3000BS 4.2500R 2.5000E 2.0000
jumlah baut minimum
BC 162.4328A 166.4328Ab aktual 64.4000
lebar gasket minimum
perhitungan MOMEN
W 1587183.9593hG 2.3727
Flange Moment (Ma) 3765859.7804
untuk kondisi operasi W = Wm1W 1886367.9187HD 1790742.9980
The lever armhD 2.5914
The moment
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MD 4640562.7245HG 85646.6705MG 203211.0708HT 9978.2501hT 2.4820MT 24766.4408MO 4868540.2361
Perhitungan tebal flange (B & Y, 1959, pers. 12.85) :
t = t = Ketebalan flange (in)A = Diameter luar flange (in)B = Diameter dalam flange (in) K =
K=
maka Y
t=
perhitungan PENYANGGA REAKTOR
Berat ShellID shell 3.9624OD shell 3.9942Tinggi shell 17.0688densitas carbon steelberat shell 58572.3116
Berat Head & BottomVhb 88.6577Berat Head & Bottom
Berat Coilberat coil 141.0300
Berat AksesorisNozzle umpan liquid
Bf
MY
a
max
B
A
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ukuran nozzle 3.5000berat nozzle 10.0000Nozzle umpan gasukuran nozzle 8.0000berat nozzle 39.0000Nozzle output gasukuran nozzle 0.7500berat nozzle 2.0000Nozzle output liquidukuran nozzle 3.5000berat nozzle 10.0000Nozzle input air pendinginukuran nozzle 1.5000berat nozzle 6.0000Nozzle output air pendinginukuran nozzle 1.5000berat nozzle 6.0000
berat nozzle total
Berat material dalam reaktorberat bahan baku 133769.1747
laju alir massa 66884.5874waktu tinggal 2.0000
berat air pendingin = volume koil x densitaspenidngin17.9073 lb
Total berat material dalam reaktor = 294928.5066Jadi, total berat reaktor = berat shell + berat head + berat coil + berat aksesoris + berat material dalam reaktorTotal= 397157.1412 lb
Sistem Penyanggaberat perancangan 476588.5695
Leg PlanningHeight 35.8214 ft
digunakan I-beam 12 in (Brownell and Young, App, G, item 2)dimensi I-beam : kedalaman f beam (h) = 12.0000
Lebar flange (b) = 5.0780Web thickness = 0.4280Ketebalan rata-rata flange 0.5440
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Area of section (A) = 10.2000Berat/ft = 35.0000
Peletakan dengan beban eksetrik (axis 1-1) : I 227.0000S 37.8000r 4.7200
Peletakan tanpa beban eksetrik (axis 2-2) :I 10.0000S 3.9000r 0.9900
Cek terhadap peletakan sumbu axis 1-1 maupun axis 2-2
axis 1-1, I/r 91.0714fc 12322.1995 psi
a = 0.5b + 1.5a 4.0390 iny = 0.5by 2.5390 inZ 89.4053 in3
Beban kompresi tiap legberat untuk perancangan 476588.5695P 79431.4282 lb
Hitung beban eksentrikfec 3588.4183 lb/in2f = fc- fecf 8733.7811 psiA 9.0947 in2
axis 2-2l/r 434.1992
lug planningP 79431.4282 lbPbolt 19857.8571 lbA bolt 1.6548 indigunakan baut dgn thread standar 1.5 in
Ketebalan plat horisontal (kompresi)My
6324.15831.3000
ln 7.19371.9732 in
b/l 1.5833
(P/4π)*((1+μ)*(ln(2l/πe)+(1-γ1)))P/4π1+ μln(2l/πe)ln(2l/πe)
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0.7890my 16223.3196 in lb
thp 2.3584 in2.2500
Ketebalan plat vertikal (gusset)tg 0.8438
Base plate planninghitung base plate areaberat 1 leg 1253.7501 lbbeban base plate 80685.1784base plate area 201.7129 in2
hitung tebal base plateAbp 201.7129di dapat nilai n 2.9825 inl 10.0275 inp 17.3651 in
Abp baru 174.1276 in2n baru 6.6513 inm baru 6.2705 in
tebal base platep 456.1679 psitbp 1.7399 in
1.6250
1-γ1
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T=
dan reaksi berlagsung sangat cpat /spontan
NH4HCO3NaHCO3+ NH4Cl
C NaHCO3 ED NH4CL F
Produk
jika keadaan tetap mka reaktan akan hilang dan pembentukan produk maka pendekatan akan ekivalen dengan : frogmant hal 29.
atau V=Fao .xC /rCV=FaO.xC/k2CCCD
53.9368
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wi ρ μkg/m3 cP
0.1007 2165.0000 2.1000 0.00000.0000 2960.0000 10.0000 0.00000.7954 997.0800 0.8937 0.00080.1039 880.0000 1.9750 0.00011.0000 7002.0800 14.9687 0.0010
FaO=v1o=
kg/m3 64.8447 lb/ft3
CaO=cp 0.0011 kg/ms
wi ρ μkg/m3 cP
0.0195 997.0800 0.0095 0.00000.9805 6.9971 0.0150 0.14011.0000 1004.0771 0.0245 0.1401
1518.24930.0821 m3.atm/kmol.K
Kg/Kmol
303.0000 K
0.4366 lb/ft3cp 0.0000 kg/ms
65.2813 lb/ft3 1046.1746 kg/m3m3/jamm3.jam
lb/jam
63.9325 m3/jam volumetrik CO2
5.4107 m3/jam volumetrik NH4OH
wi/ρ
wi/ρ
massa totaldensitas campuran
massa totaldensitas campuran
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4.0708 m3/jam
0.8000
(Derks,at all kinetic absrption of CO2 in aqueous ammonia solutin, 2009)
420.0000
7.5000 m3/mol.s 27000000 m3/kmoljam2.60E-06 m6/mol2.s 9360.0000 m6/kmol2jam
0.0021 /s 7.5600 /jam
1/k2 3.7037037037037E-081/(kh2o.k2/k-1)*CAO 0.000693670044660466Cao2 0.0237
f= 0.9686A 2.9186 m2D= 1.9282 m
0.0000
0.5000 3.8564 m12.6522 ft
standar 13.0000 ft , R3.9624 m
156.0016 in
[ 1k 2
+k−1
k 2
(1+ℓY A)kC . C AO(ΘC−Y A) ] C
AO2 (1−Y A )(ΘB−Y A ) [ 1
k2
+k−1
k2
(1+ℓY A )kC .C AO(ΘC−Y A) ](1+ℓY A )2
CA
O2 (1−Y A)(ΘB−Y A)
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(1+eYa)2
YA0.0000 0.000716149348900762 1.0000000 0.02370.2000 0.0008423139349305 0.8711111 0.01520.4000 0.00105722355063419 0.7511111 0.00850.6000 0.00150531176109305 0.6400000 0.00380.8000 0.00620236688218905 0.4946778 0.000287030327311352
dengan metode simpson rule
delta Y 0.2000
=
0.8076
4.3077h= 16.8533 m
55.2931 ftstandarisasi 56.0000 ft
17.0688 m
L/D 4.3702G= 85.3538f 0.9686
h
207.7171 m3
16.8533 m
215854.9677 kg
Proses difusi terjadi didalam fasa cair. Persamaan yang digunakan adalah
∫Y A 1
Y A 2 [1k2
+k−1
k2
1
kC .C AO (ΘC−Y A)(1+ℓY A ) ](1+ℓY A)2 dYA
CA
O2 (1−Y A)(ΘB−Y A)
=h3
[f ( X0 )+4 f ( X 1 )+2 f ( X2 )+4 f ( X3 )+ f ( X 4 )]
[ 1k 2
+k−1
k 2
(1+ℓY A)kC . C AO(ΘC−Y A) ] C
AO2 (1−Y A )(ΘB−Y A ) [ 1
k2
+k−1
k2
(1+ℓY A )kC .C AO(ΘC−Y A) ](1+ℓY A )2
CA
O2 (1−Y A)(ΘB−Y A)
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(Coulson 1983, vol 6 : 255)
m3/kmolm3/kmol
m3/kmol
m2/s
(Treyball, 1980 : 141)
10.0000 mm standard 0.0100 m
303.0000 K4.0000 atm
kg/m3 1.0322 g/cm3lb/ft3
(Coulson 1983, vol 6 : 258)
DAL=117 ,3 .10−18 (φ×M )0,5×T
μ ×Vm0,6
σ L = [ Pch × ( ρL− ρv )M ]
4
× 10−12
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Pch
112.2000
mJ/m2dyne/cm
(Coulson 1983, vol 6 : 258)
dyne/cmkg/s2g/s2
mcmmm
Diameter gelembung (db) 2 mm, sehingga persamaan yang digunakan
(Froment : 726)
m/s 0.0027 dm/sm/s
db= ( 6×do×σ L
g×Δρ )1
3
k L (2 mm )= 0,42 [ μL×g
ρL ]1
3 [ ρL×DAL
μL ]1
2
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kmol/m3
(Perry`s 1997,7thed : 23-42)
M< 0,3 = Reaksi membutuhkan volume bulk liquid yang besar.
Berdasarkan Perry’s Chemical engineering Hand’s Book, ada beberapa parameter design untuk reaktor gelembung yaitu : diameter gelembung (db),
0.0092 m0.9243 cm9.2434 mm
adalah laju volume aliran gas dibagi dengan luas area tiap lubang (treyball hal 143)
cm3/s 18.6962
0.0100 m1.0000 cm
10.0000 mm
0.7850 cm2
D 0=db3 x ( ρL− ρg) xg
6 σ
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cm/sm/s 526.2923 m/jam
Untuk menghitung gas hold up(є), dapat digunakan persamaan :
froment hal 725 pers 14,3,f-1
Untuk menghitung interfacial area dapat digunakan persamaan sebagai berikut :
(Perry, 1997)
/cm 128.6996 /m
Nilai (α) interfacial area yang diperoleh, masuk dalam range (α) di reaktor gelembung yaitu 40 – 600 m2/m3 sehingga memenuhi.
Berdasarkan Levenspiel (1999), untuk mengetahui apakah reaktor bubble sudah sesuai untuk
a. Untuk bubble reaktor, jenis alirannya adalah plug flow untuk gas dan mixed flow untuk cairan.b. kG dan kL, untuk reaktor bubble, nilai kL harus lebih besar daripada kG, hal ini dibuktikan sebagai berikut:
58.0000 atm dm3/mol
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1.9282 m tandarin 2.1336 m6.3261 ft 7.0000 ft
84.0009 inH= 0.0000 m standarin
0.0000 ft
Berdasarkan Froment, hal 726 untuk ε ≤ 0,45; 0,03 < usg < 0,4 m/s; Dr > 0,15 m, maka
mft
m/sm
3.0000(Peters and Timmerhause, 1991)
mft
2.9186 m26.1544 ft2
Berdasarkan Perry’s, 1997 diameter hole plate dapat ditentukan dengan persamaan
Keterangan :Do = Diameter hole, cmdb = Diameter bubble, cmρL = Densitas liquid, gr/cm3 ρG = Densitas gas, gr/cm3 σ = Tegangan permukaan liquid g = Percepatan gravitasi, 980 cm/det2
cm
cm2
cm
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1.2990 cm
0.9743 cm2
0.8057
4.9588 ft20.4588 m2
4588.1292 cm2
5844.7506 hole5845.0000 hole
(pers.13.1 Brownell and young,1959)
Bahan konstruksi stain less stelll 167 grade 3, tipe 304, f = 18750 psi (Brownell and Young, 1959, Tabel 13.2)
Tekanan desain 5 -10 % di atas tekanan kerja normal/absolut (Coulson, 1988 hal. 637).
P= 4.0000 atm58.8000 psi
lb/ft3 1046.1746 kg/m3
mft
hole
holehole
tiapLuas
totalLuasJumlah
144
hg
g. L
cmix
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psi
(Couldson, vol.6, 1983 : .637)
psiatm
in maka tebal shell satndar= 0.6250
3.9942 m157.2500 in
icr = r = OD = ID = b = OA = sf =
in distandaeisasi 168.0000 inin
in tabel 5.7 Brownel & Young didapat :in
th =
0.6250
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(Brownell and Young,1959,hal. 258)
w = stress-intensitication factor
1.2500 in
1.2500 in standar sf yang digunakan 4.0000 in0.3333 ft
m 78.0000 ininin
inm
OA = th + b + sf (Brownell and Young,1959.hal.87) ρ = 12.94 μ = 0.2609
in Cp = 1.2
m k = 0.354
mft
12146.0671 kg/jam
Sifat amoniak pada suhu rata-rata = -33,5 oC
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997.0800 lb/ft3 kg/m3 62.21780.8937 cp= cp 2.1624
377.0969 Btu/lboF Kj/Kmol K 20.94980.3300 Btu/jam.ft.oF Btu/jam.ft.oF
in 0.1100 ftin 0.0874 ftft2/ftin2/tube 0.0024 ft2/tubein
lbm/ft2.jam 2824.8476 lbm/ft2.s
Kecepatan medium pendingin di dalam pipa umumnya berkisar 1,5-2,5 m/s.
m/s = 8.2000 ft/s
lb/s.ft2 lb/jam.ft2
dibulatkan 6.0000 set koil
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lb/jam.ft2
ft/jamft/sm/s masuk dalam range
68518.5149
68518.5149 maka nilai jH = 25.00000.8600
1275.9094 Btu/ft2.hr.oF
1097.2821 Btu/ft2.hr.oF
872.0067 Btu/jam.ft2.o F
Untuk koil, harga hio dikoreksi dengan faktor koreksi sebagai barikut
Diketahui diameter spiral atau heliks koil = 0,7 – 0,8 Dt (Rase, 1977)
ft 3.1699 m
Btu/ft2.hr.oF
(Kern, 1950)
Btu/ jam.ft2.oF
h io,koil=h io,pipa [1+3,5 (Dikoil
Dspiralkoil)]
ho=0 , 36 .k
De.( De .G
μ )0 ,55
.(Cp . μk )
13
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Btu/hr ft2 oF
untuk proses cooler, bahan pendinginnya berupa air-aquaueos solution pendingin maka nilai Ud adalah
BTU ft2jamF
0.0013 T pendingin
cold fluid Temp. diff F Δt2/Δt1113.0000 27.0000 Δt2 7.5000
82.4000 3.6000 Δt130.6000 23.4000 Δt2 – Δt1
11.6135
Beban panas = 866135.5014 KJ/jam820934.4880 Btu/jam
Qc= 136822.4147 Btu/jam
0.2200 ft0.0671 m
ft
Lhe=1/2 π (dhe2 +J
sp2 )1 /2+1/2π . dhe
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ft
ftm
putaran 1.0000 putaran
1.3200 ft0.4023 m
0.2878 ft30.0082 m3
Tinggi koil harus lebih kecil daripada tinggi cairan setelah ditambah koil agar seluruh koil tercelup dalam cairan.
mft
1.3200 ft), berarti semua koil tercelup di dalam cairan.
Saluran dibuat dengan menggunakan bahan stainless steel. Diameter optimum tube yang bahannya terbuat dari stainless steel dihitung dengan menggunakan persamaan
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
G = kecepatan aliran massa fluida (kg/s)
Pengecekan bilangan Reynolds (NRe)
mix
mixRe μ
IDvρN
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ρmix = densitas campuran, kg/m3ID = diameter dalam tube, mµg = viskositas fluida, kg/m.jamv = kecepatan aliran, m/s
56049.8966 kg/jam 15.5694 kg/s1039.1774 kg/m3
0.0011 kg/ms
Dari persamaan 5.15, hal. 161. Coulson, 1983 diperoleh :
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
78.4157 mm7.8416 cm3.0872 in
Dipilih spesifikasi pipa (Peters and Timmerhaus, Tabel 13, 1980) :
3.5000 in
3.5480 in 0.0901 m4.0000 in 0.1016 m9.8900 in2 0.0064 m2
53.9368 m3/jam 0.0150
2.3578 m/s
196216.9658 (aliran turbulen)
10834.6907 kg/jam 3.0096 kg/s6.9971 kg/m30.0000 kg/ms
Dari persamaan 5.15, hal. 161. Coulson, 1983 diperoleh :
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
198.4463 mm19.8446 cm
7.8128 in
mixv ρ
GF
A
Fv v
mix
mixRe μ
IDvρN
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Dipilih spesifikasi pipa (Peters and Timmerhaus, Tabel 13, 1980) :
8.0000 in
7.9810 in 0.2027 m8.6250 in 0.2191 m
50.0000 in2 0.0321 m2
1548.4434 m3/jam 0.4301
13.3890 m/s
1277434.8481 (aliran turbulen)
64735.1367 kg/jam 17.9820 kg/s1084.0516 kg/m3
0.0009 kg/ms
Dari persamaan 5.15, hal. 161. Coulson, 1983 diperoleh :
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
83.0347 mm8.3035 cm3.2691 in
Dipilih spesifikasi pipa (Peters and Timmerhaus, Tabel 13, 1980) :
3.5000 in
3.5480 in 0.0901 m4.0000 in 0.1016 m9.8900 in2 0.0064 m2
59.7159 m3/jam 0.0166
2.6105 m/s
270267.3039 (aliran turbulen)
mixv ρ
GF
A
Fv v
mix
mixRe μ
IDvρN
mixv ρ
GF
A
Fv v
mix
mixRe μ
IDvρN
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2149.4507 kg/jam 0.5971 kg/s6.9971 kg/m30.0150 kg/ms
Dari persamaan 5.15, hal. 161. Coulson, 1983 diperoleh :
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
88.3891 mm8.8389 cm3.4799 in
Dipilih spesifikasi pipa (Peters and Timmerhaus, Tabel 13, 1980) :
3.5000 in
3.5480 in 0.0901 m4.0000 in 0.1016 m9.8900 in2 0.0064 m2
307.1895 m3/jam 0.0853
13.4287 m/s
564.5230 (aliran laminer)
12146.0671 kg/jam 3.3739 kg/s997.0800 kg/m3
0.0009 kg/ms
Dari persamaan 5.15, hal. 161. Coulson, 1983 diperoleh :
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
37.0356 mm3.7036 cm1.4581 in
Dipilih spesifikasi pipa (Peters and Timmerhaus, Tabel 13, 1980) :
mixv ρ
GF
A
Fv v
mix
mixRe μ
IDvρN
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1.5000 in
1.6100 in 0.0409 m1.9000 in 0.0483 m2.0360 in2 0.0013 m2
12.1816 m3/jam 0.0034
2.5867 m/s
124086.0430 (aliran turbulen)
12146.0671 kg/jam 3.3739 kg/s997.0800 kg/m3
0.0006 kg/ms
Dari persamaan 5.15, hal. 161. Coulson, 1983 diperoleh :
diopt = 226.G0^5.ρ^-0,35 (Pers 5.15, Coulson, 1983)
37.0356 mm3.7036 cm1.4581 in
Dipilih spesifikasi pipa (Peters and Timmerhaus, Tabel 13, 1980) :
1.5000 in
1.6100 in 0.0409 m1.9000 in 0.0483 m2.0360 in2 0.0013 m2
12.1816 m3/jam 0.0034
2.5867 m/s
191769.3392 (aliran turbulen)
Berdasarkan perhitungan saluran saluran pemasukan dan keluaran pada reaktor di
mixv ρ
GF
A
Fv v
mix
mixRe μ
IDvρN
mixv ρ
GF
A
Fv v
mix
mixRe μ
IDvρN
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Gambar F.2. Desain opening pada Reaktor
Spesifikasi nozzle standar (Brownel and Young, 1959, App. F ,hal.349) :3.5000 in4.0000 in
coupling inDiameter of hole in reinforcing plate (DR) = 4.1250 inLength of side of reinforcing plate (L) = 10.0000 in
12.6250 inDistance, shell to flange face, outside (J) = - inDistance, shell to flange face, inside (K) = - inDistance from Bottom of tank to center of nozzle :
Regular, Type H = 8.0000 inLow, Type G = 5.0000 in
Spesifikasi nozzle standar (Brownel and Young, 1959, App. F ,hal.349) :8.0000 in8.6250 in0.5000 in
Diameter of hole in reinforcing plate (DR) = 8.7500 inLength of side of reinforcing plate (L) = 20.2500 in
25.0000 inDistance, shell to flange face, outside (J) = 8.0000 inDistance, shell to flange face, inside (K) = 6.0000 inDistance from Bottom of tank to center of nozzle :
Regular, Type H = 13.0000 inLow, Type G = 10.1250 in
Spesifikasi nozzle standar (Brownel and Young, 1959, App. F ,hal.349) :3.5000 in4.0000 in
coupling inDiameter of hole in reinforcing plate (DR) = 4.1250 inLength of side of reinforcing plate (L) = 10.0000 in
OD
Q
J
T t
nWeld B
Weld B
Weld B
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12.6250 inDistance, shell to flange face, outside (J) = - inDistance, shell to flange face, inside (K) = - inDistance from Bottom of tank to center of nozzle : 0.0000
Regular, Type H = 8.0000 inLow, Type G = 5.0000 in
Spesifikasi nozzle standar (Brownel and Young, 1959, App. F ,hal.349) :3.5000 in4.0000 in
coupling inDiameter of hole in reinforcing plate (DR) = 4.1250 inLength of side of reinforcing plate (L) = 10.0000 in
12.6250 inDistance, shell to flange face, outside (J) = - inDistance, shell to flange face, inside (K) = - inDistance from Bottom of tank to center of nozzle :
Regular, Type H = 8.0000 inLow, Type G = 5.0000 in
Spesifikasi nozzle standar (Brownel and Young, 1959, App. F ,hal.349) :1.5000 in1.9000 in0.2000 in
Diameter of hole in reinforcing plate (DR) = 2.0000 inLength of side of reinforcing plate (L) = in
inDistance, shell to flange face, outside (J) = 6.0000 inDistance, shell to flange face, inside (K) = 6.0000 inDistance from Bottom of tank to center of nozzle :
Regular, Type H = 6.0000 inLow, Type G = 3.5000 in
Spesifikasi nozzle standar (Brownel and Young, 1959, App. F ,hal.349) :1.5000 in1.9000 in0.2000 in
Diameter of hole in reinforcing plate (DR) = 2.0000 in
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Length of side of reinforcing plate (L) = inin
Distance, shell to flange face, outside (J) = 6.0000 inDistance, shell to flange face, inside (K) = 6.0000 inDistance from Bottom of tank to center of nozzle :
Regular, Type H = 6.0000 inLow, Type G = 3.5000 in
Manhole adalah lubang pemeriksaan yang diperlukan pada saat pembersihan atau pemeriksaan pada bagian dalam kolom. Direncanakan manhole di pasang pada kolom bagian atas reaktor dengan ukuran standar 24 in berdasarkan rekomendasi
1.00000.6250 in
61.6429 ft24.0000 in ft26.5000 in32.7500 in30.2500 in53.5000 in64.0000 in
0.1875 in0.3750 in
Sambungan antara tutup bejana dengan bagian shell menggunakan sistem flange dan baut. Bahan konstruksi yang dipilih berdasarkan pada kondisi operasi. Data perancangan:
92.2534 psi303.0000 K
Material flange = SA-203, grade B (B & Y, 1959, Tabel 13.1)
157.2500 in156.0016 in
0.6250 in17500.0000 psi20000.0000 psi
Gasket
hG
t
hT
HG
HT
G
h
W
R hD Cgo
g1
g1/2
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Gambar F.3. Tipe flange dan dimensinya.
y = yield stress, lb/in2 (Fig. 12.11)
Dari fig 12.11 Brownell & Young, untuk material gasket Stainless steels
Asumsi bahwa diameter dalam gasket sama dengan diameter luar shell, yaitu
in
0.4375
0.4225 in
157.6875 in
in
0.2500 in
Digunakan gasket dengan lebar standar 0,25 in (Gambar 12.12, Brownell and Young,1959).
Gasket
hG
t
hT
HG
HT
G
h
W
R hD Cgo
g1
g1/2
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Wm2 = Hy = x b x G x y (Brownell and Young, pers. 12.88, 1959) Keterangan :Hy = Berat beban bolt maksimum (lb)b = Effective gasket (in)G = Diameter gasket rata-rata (in)
1114062.1875
85646.6705 lb1800721.2482 lb1886367.9187 lb
in2
in2ininin
41.0080 digunakan baut 28.0000
ininin2
0.1445 in
lbin
lb-in
lblb
in
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lb-inlblb-inlbinlb-inlb-in
166.4328 in156.0016 in
1.0669
25.0000 Brownell and Young,1959, fig. 12.22, hal. 238),
1.9283 in
m 13.0000 ft 156.0000m 13.1042 ft 157.2500m 56.0000 ft 672.0000
490.0000 lb/ft3lb
ft343442.2930 lb
lb
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inlb
inlb
inlb
inlb
inlb
inlb
73.0000 lb
kg 294910.5993 lbkg/jamjam
lbJadi, total berat reaktor = berat shell + berat head + berat coil + berat aksesoris + berat material dalam reaktor
lb
429.8572 in
inininin
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in2lb/ft
in4in3in
in4in3in
Cek terhadap peletakan sumbu axis 1-1 maupun axis 2-2
(< 120, OK)(<17500 psi, OK)
lb
e 0.5313
![Page 226: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/226.jpg)
γ 0.2110
dipakai 2.5 in
di ambil 3/8 in4.2272
6.3500
lb 26.8427
n Abp0.8100 73.98481.0000 81.23621.0100 81.62581.0200 82.01631.0300 82.40751.0400 82.79961.0500 83.19241.0600 83.58601.0700 83.98051.0800 84.37571.0900 84.7718
1.7500 in 1.0910 84.81141.0920 84.85111.0930 84.89081.0940 84.93042.0000 124.16102.1000 128.89342.3000 138.59842.5000 148.62342.6000 153.75582.7000 158.96832.8000 164.26082.9825 174.1276
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303.0000
viskositas NH4OH
![Page 228: buble,3](https://reader033.fdocuments.in/reader033/viewer/2022061604/55cf8ef6550346703b9775d7/html5/thumbnails/228.jpg)
C cpvolume(m3)/jam 0.0000 2.8000
30.0000 1.97500.0747 6.3931 40.0000 1.70000.0001 0.0047
-0.0894 33.74440.0707 9.71470.0561 49.8570
244.2558 mol/jam53.9368 m3/jam
4.5286 mol/m3
volume(m3)/jam
-0.0908 0.0877-4.1178 1535.9473-4.2086 1536.0350
asumsi waktu reaksi pertama sesudah co2 bereaksi dengan nh4oh tinggal bahan2 berikut dalam larutan
output kmol/jamNaCl 13841.1595 236.6010CaSO4 13.9107 0.1023H2O 33733.3124 1874.0729CO2 10747.2533 244.2558total 58335.6359 2355.0320
Vo faksi volum gasNaCl 809.2209 0.2373
wi*ln μ
wi*ln μ
massa totaldensitas campuran
massa totaldensitas campuran
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CaSO4 0.8133 0.0002H2O 1972.2121 0.5783CO2 628.3362 0.1842total 3410.5825 1.0000
NH4HCO3 15436.9638 195.4046
detikProduk kg/jam kmol/jam
outputNaCl 2768.2319 47.3202CaSO4 13.9107 0.1023H2O 33733.3124 1874.0729NH4Cl 10126.5235 189.2808NaHCO3 15899.5884 189.2808NH4HCO3 483.7795 6.1238NH4OH 1709.7903 48.8512total 64735.1367 2355.0320CO2 2149.4507 48.8512total 66884.5874 2403.8831
v NH4HCO3 0.4624FaNH4HCO3 6.1238maka C NH4HCO3 13.2427
NH
6.0000 ft72.0007 in
ρL=1
∑ wi
ρi
[ 1k2
+k−1
k2
(1+ℓY A )kC .C AO(ΘC−Y A) ](1+ℓY A )2
CA
O2 (1−Y A)(ΘB−Y A)
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0.03020.04830.09300.2538
10.6894
[ 1k2
+k−1
k2
(1+ℓY A )kC .C AO(ΘC−Y A) ](1+ℓY A )2
CA
O2 (1−Y A)(ΘB−Y A)
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Menghitung surface tension
sigma=A(1-T/Tc)^n (dyne/cm)
A T TcNaCl 201.3300 303.0000 3400.0000CaSO4 303.0000H2O 132.6740 303.0000 647.1300NH4OH 303.0000
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sigmaNaCl 175.0617 (dyne/cm)CaSO4 1.6530 (dyne/cm)H2O 72.5870 (dyne/cm)NH4OH
kL = kL (2 mm) 500 db
KL= 0.0012 m/s
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0.0580 atm dm3/kmol
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r 1.0668 m #REF!3.5000 ft
#REF! m56.0000 ft
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1640.4000
0.3750 1.5-30.3125 sama0.2500 1.5-2.5
Tebal insulattion Thermal conductivity zirconia (k1 ,2200 oC) = 0,002 kW/m.oC
k insulasi 0.0020 KW/mCT1 30.0000 CTref 25.0000 CQ 866135.5014 KJ/jamOdshell 168.0000 inOdinsulasi
Ashell 2461.7350 ft2
0.0095 m0.9467 cm0.3727 in
ΔX
qA
=kΔx
( ΔT )
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in 0.0159 m
tebal head, in inside corner radius, inradius of dish, in outside diameter, in inside diameter, in depth of dish, in overall dimension, in standard straight flange, in
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Tabel 5.8 Brownell and Young hal. 93
cp= 0.6315 lb/ft.jam
24292.1341 lb/jam
lb/ft3
Btu/lboF
Btu/jam.ft.oF
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lb/ft3lb/ftjam 0.0006 lb/ftsKj/Kg K 5.0049 Btu/lb F
mencari panas masuk rata2
total panas in buble
NaClCaSO4H2ONH4ClNaHCO3NH4HCO3
86.0000
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250-500 BTU ft2jamF kern
28.0000 c 25.0000301.0000 K
82.4000
ln Δt2/Δt12.0149
dari neraca panas
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AlatFungsiBentuk Bubble column dengan head berbentuk Torispherical head. yang dilengkapi dengan Coil pendingin dengan media pendingin berupa air pendinginKapasitasDimensi Diameter shell (D)
Tinggi shell (Z) Tebal shell (ts) Tebal head (th)
Sparger Triangular pitch doTekanan desainBahan konstruksi Carbon Steel SA 283 grade CJumlah 1 buah
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m3/s
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m3/s
komponen kg/jamNaCl 2768.2319CaSO4 13.9107H2O 33733.3124NH4Cl 10126.5235NaHCO3 15899.5884NH4HCO3 483.7795NH4OH 1709.7903
64735.1367
BM campuran=densitas liquid
μ L =
m3/s
ρL=1
∑ wi
ρi
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m3/s
μ L = 0.8500
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m3/s
μ L = 0.5500
m3/s
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buah
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ininin
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menentukan kinetika raksi2265000.0000
k= 0.4487 min-1 0.0075t 8.7179 menit
asumsi waktu reaksi pertama sesudah co2 bereaksi dengan nh4oh tinggal bahan2 berikut dalam larutanwi ρ μ
kg/m3 cP m3/jam0.1005 2165.0000 2.1000 0.0000 6.39310.0000 2960.0000 10.0000 0.0000 0.00470.7958 997.0800 0.8937 0.0008 33.83210.1037 1.8000 0.0150 0.0576 5970.69631.0000 6123.8800 13.0087 0.0585 6010.9262
wi/ρ
k=2. 265 x106exp (−13 . 512/ RT )
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0.8000 1580.0000 kg/m3 0.0005
17.1043 kg/m3
m3/jamkmol/jamkmol/m3
0.00 0.10 0.200.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0f(x) = NaN xR² = 0
t
1+εA X A
(1−X A )+ε A ln (1−X A )
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Yaws,
n sigma1.4978 175.0617 (dyne/cm)
1.6530 dyne/ cm0.9550 72.5870 (dyne/cm)
http://www.sciencedirect.com/
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in
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Thermal conductivity zirconia (k1 ,2200 oC) = 0,002 kW/m.oC
240593.1948 j/s 240.5932 KW13.9999 ft
354512.5271 in2227.7743 m2
standar 0.3750 in
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#REF!
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Bubble column dengan head berbentuk Torispherical head. yang dilengkapi dengan Coil pendingin dengan media pendingin berupa air pendingin#REF! m33.9942 m
18.7887 m0.6250 in1.2500 in1.0000 cm
92.2534 psi
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wi ρ μkmol/jam kg/m3 cP
47.3202 0.0201 2165.0000 2.1000 0.00000.1023 0.0000 2960.0000 10.0000 0.0000
1874.0729 0.7958 997.0800 0.8937 0.0008189.2808 0.0804 1527.4000 1.2200 0.0001189.2808 0.0804 2159.0000 0.9750 0.0000
6.1238 0.0026 1580.0000 0.0370 0.000048.8512 0.0207 880.0000 1.7000 0.0000
2355.0320 0.9793 11388.4800 16.9257 0.0009
27.4880 Kg/kmol
1084.0516 kg/m3 67.6448 lb/ft3
0.9436 cp 0.0009 kg/ms
wi/ρ
e^(wi*lnµ)
ρL=1
∑ wi
ρi
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cp 0.0009 kg/ms
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cp 0.0006 kg/ms
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1/(1-x) 50.0000Ln1/(1-x) 3.9120
/s 26.9241 /jam
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0.00 0.10 0.200.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0f(x) = NaN xR² = 0
t
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Bubble column dengan head berbentuk Torispherical head. yang dilengkapi dengan Coil pendingin dengan media pendingin berupa air pendingin
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0.01490.0001
-0.08940.0160
-0.0020-0.00860.0110
-0.0580
wi*ln μ