1

Treat 500 SCFM of air containing 14%mol acetone to remove 95% of the acetone by absorption in liquid water in a packed bed operating at 80°F and 1 atm, with 1 in rasching rings. The feed water contains 0.02% acetone and the flowrate is 1.1 times the minimum. The partial pressure of acetone over an aqueous solution at 80°F can be calculated from pA = PA°µAxA where ln µA = 1.95(1-x)^2 where PA° = 0.33 atm at 80°F. V2 L2 2.67As designer, you must select the following: Y2 0.0081 X2 0.0002

flowrate of water

diameter of tower height of packing Jawab G 500 SCFM y1 0.14 Recover 0.95 x2 0.0002 V1 1.39 L1 L2 1.1 Lmin Y1 0.1628 X1 0.0810pA = PA°µAxA where ln µA = 1.95(1-x)^2 PA° 0.33 atm P 1 atm 1. Merubah SCFM menjadi lbmol/min V (STP) 22.4 ltr/gmol 359 ft3/lbmol V1 1.39 lbmol/min 2. Menentukan Laju alir minimum 2.1 Menggambar kurva kesetimbangan Dalam bentuk rasio mol

x µA pA y* = pA/P 0 7.0287 0.0000 0.0000

0.01 6.7612 0.0223 0.0223 0.02 6.5064 0.0429 0.0429 0.03 6.2636 0.0620 0.0620 0.04 6.0322 0.0796 0.0796 0.05 5.8117 0.0959 0.0959 0.06 5.6014 0.1109 0.1109

2

0.07 5.4008 0.1248 0.1248 0.08 5.2095 0.1375 0.1375 0.09 5.0269 0.1493 0.1493 0.1 4.8525 0.1601 0.1601

2.2 Membuat garis operasi Merubah dalam bentuk rasio mol X = x / (1-x) Y = (L/V) X + Y2 = 1.91 X + 0.00814 x2 0.0002 --> X2 0.0002 V1 y1 + L x = V y + L1 x1 y1 0.1400 --> Y1 0.1628 V' (y1/(1-y1)) + L' (x/(1-x)) = V'(y/(1-y)) + L'(x1/(1-x1)) Y2 0.00814 Kurva kesetimbangan (EC) berada di bawah garis operasi (OL). Pembuktian rumus Selanjutnya, minimum slope tercapai ketika OL menyinggung EC. V' = V(1-y2) Akhirnya, X1 bisa dibaca pada OL. L' = L(1-x2) X1 = 0.081 y2 0.008074

1 1-y2 0.991926 (L'/V')min =(Y1-Y2)/(X1-X2) 1.914 1-x2 0.9998 maka Lmin 2.666 lbmol/min jadi V' hampir sama V

Atau 18 lbmol/min 8.330 lb/gal 5.76 gal/min jadi L' hampir sama L

X Y* Y

0.0000 0.0000 0.0081 0.0101 0.0228 0.0274 0.0204 0.0449 0.0471 0.0309 0.0661 0.0672 0.0417 0.0865 0.0877 0.0526 0.1061 0.1087 0.0638 0.1247 0.1301 0.0753 0.1425 0.1519 0.0870 0.1595 0.1742 0.0989 0.1755 0.1970 0.1111 0.1907 0.2204

3

Jika diinginkan x1 = 0.07 V2 L2 0.00 Hitung berapa nilai NOG ? y2 0.00807 x2 0.0002

Jawab 1. Menghitung slope OL L [x1/(1-x1) - x2/(1-x2)] = V [y1/(1-y1) - y2/(1-y2)] L [X1 - X2] = V [Y1 - Y2]

y1 0.14 Y1 0.1628 y2 0.00807 Y2 0.0081 x1 0.07 X1 0.0753 V1 1.39 L1 x2 0.0002 X2 0.0002 y1 0.1400 x1 0.0700

Y1-Y2 -0.1547 X1-X2 -0.0751 L/V 2.060178 slope y* = 0.33 * exp (1.95*(1-x2)^2) * x 2. Menghitung NOG Menggunakan metoda Wiegand NOG = INTEGRAL y1-y2 [dy/(y*-y)] atau dibalik NOG = INTEGRAL y2-y1 [dy/(y-y*)]

y X x y* 1/(y-y*) Integrasi0.00807 0.0002 0.0002 0.0005 131.467 2.0885 0.022 0.0072 0.0071 0.0161 168.393 2.6133 0.036 0.0144 0.0142 0.0311 204.932 3.0153 0.05 0.0218 0.0213 0.0456 225.819 2.8986

0.063 0.0289 0.0281 0.0585 220.119 2.6900 0.076 0.0362 0.0349 0.0708 193.725 2.2949 0.089 0.0437 0.0418 0.0827 159.343 1.8611 0.102 0.0514 0.0489 0.0941 126.987 1.4787 0.115 0.0593 0.0560 0.1051 100.508 1.1737 0.128 0.0675 0.0632 0.1155 80.063 0.8741 0.14 0.0753 0.0700 0.1248 65.615 20.988

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3. Menghitung ketinggian tower, jika Kga 0.04 lbmol/s m3 DT 1 m Jawab R 0.5 m S 0.7854 m2 dari soal diatas V= 1.39 lbmol/min 0.023167 lbmol/s G 0.0295 lbmol/s m2 HOG 0.7374 m HOG = V/S/Kga Z 15.48 m Z=HOG x NOG

5

A gas stream containing a valuable hydrocarbon (MW = 44) is to be scrubed with na nonvolatile oil (MW=300) in a tower packed with 1-in Raschig rings. The entering gas analyzes 20 mole percent hydrocarbon and 95% percent of this hydrocarbon is to be recovered. The inert gas molecular weight is 29. The gas stream enters the column at 5000 lb/hr ft2 and hydrocarbon-free oil enters the top of the column at 10000 lb/hr/ft2. Determine the NOG for this operation. Jawab V2 126.6 L2 33.3 Dari soal y2 0.0123 x2 0

MW HC 44 lb/lbmol

MW Gas Inert 29 lb/lbmol MW Oil 300 lb/lbmol GV 5000 lb/hr/ft2 GL 10000 lb/hr/ft2 y1 0.2 fraksi mol Recover 0.95 L2 33.3 lbmol/hr ft2 x2 0 fraksi mol V1 156.25 L1 63.0 y1 0.20 x1 0.471 1. Menghitung V1 Basis 1 lbmol gas masuk Menghitung MW gas % in V MW

0.2 8.8 0.8 23.2 cara cepat menghitung y2

32 Y1 0.25 Maka Y2 0.0125 V1 156.25 lbmol/hr ft2 y2 0.0123457 2. Menghitung V2

Menghitung y2

Gas bebas solute masuk = Gas bebas solute keluar = 0.8 lbmol/mol gas masuk HC keluar 0.01 lbmol y2 0.01235 V2 126.6 lbmol/hr ft2 = 0.8*V1/(1-Y2)

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3. Menghitung x1 Lb mol HC di L1 0.19 lbmol/mol V1 HC di L1 29.7 lbmol Oil bebas solute 33.3 lbmol Total mole 63.0 lbmol x1 0.47 fraksi mol 4. Membuat operating line V1 y1 + L x = V y + L1 x1 V' = V(1-y) non difusing component L' = L(1-x) V' (y1/(1-y1)) + L' (x/(1-x)) = V'(y/(1-y)) + L'(x1/(1-x1)) V' = 125 L' = 33.33

31.25 33.33 *(x/(1-x)) = 125 *(y/(1-y))+ 29.6875 (x/(1-x)) = 3.75 *(y/(1-y))+ -0.04687 x (x/(1-x)) (y/(1-y)) y

0 0.0000 0.0125 0.0123 0.0307 0.0317 0.0209 0.0205 0.1325 0.1527 0.0532 0.0505

0.271 0.3717 0.1116 0.1004 0.383 0.6207 0.1780 0.1511 0.472 0.8939 0.2509 0.2006

5. Membuat equilibrium line Membuat pers EC (karena di soal dalam bentuk grafik) y y* (y/(1-y)) (x/(1-x)) x

0.0124 0 0.0126 0.0002 0.0002 0.03 0.003 0.0309 0.0691 0.0646 0.05 0.0117 0.0526 0.1505 0.1308 0.08 0.03 0.0870 0.2792 0.2183

0.1 0.0475 0.1111 0.3698 0.2700 0.14 0.0945 0.1628 0.5636 0.3604 0.15 0.106 0.1765 0.6149 0.3808 0.18 0.1445 0.2195 0.7763 0.4370

0.2 0.173 0.2500 0.8906 0.4711 6. Menghitung NOG (menggunakan metoda penurunan)

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y* dibaca dari kurva (1-y)lm = (1-y)-(1-y*)/ln((1-y)/(1-y*)) y interval dari y2 ke y1 A B C D Metoda Wiegand

y (y/(1-y)) (x/(1-x)) x y* 1-y 1-y* (1-y)lm y*-y B/(A*C) Integrasi -D 1/(y-y*) Integrasi 0.0124 0.0126 0.0002 0.0002 0.0003 0.9876 0.9997 0.9936 -0.0121 -83.2 1.0610 83.2 82.65 1.0520

0.03 0.0309 0.0691 0.0646 0.0029 0.9700 0.9971 0.9835 -0.0271 -37.4 0.6336 37.4 36.89 0.6232 0.05 0.0526 0.1505 0.1308 0.0107 0.9500 0.9893 0.9695 -0.0393 -26.0 0.7021 26.0 25.43 0.6862 0.08 0.0870 0.2792 0.2183 0.0308 0.9200 0.9692 0.9444 -0.0492 -20.8 0.4092 20.8 20.31 0.3983 0.1 0.1111 0.3698 0.2700 0.0488 0.9000 0.9512 0.9254 -0.0512 -20.1 0.8399 20.1 19.52 0.8174

0.14 0.1628 0.5636 0.3604 0.0931 0.8600 0.9069 0.8832 -0.0469 -21.9 0.2251 21.9 21.34 0.2193 0.15 0.1765 0.6149 0.3808 0.1056 0.8500 0.8944 0.8720 -0.0444 -23.1 0.7864 23.1 22.51 0.7686 0.18 0.2195 0.7763 0.4370 0.1452 0.8200 0.8548 0.8373 -0.0348 -29.3 0.6706 29.3 28.73 0.6583 0.2 0.2500 0.8906 0.4711 0.1731 0.8000 0.8269 0.8134 -0.0269 -37.7 5.3279 37.7 37.11 5.2233

Jadi NOG = 5.3279 menggunakan trapeziodal rule Bisa menggunakan metoda W