Latihan Soal Packed Bed Absorber

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  As designer, you must select the following: Y2 0.0081 X2

2.67 0.0002

flowrate of water diameter of tower height of packing Jawab G 500 SCFM 0.14 y1 Recov ecover er 0.95 0.95 0.0002 x2 L2 1. 1.1 Lmin pA = PA°µAxA where ln µA = 1.95(1-x)^2 PA° 0.33 0.33 atm atm P 1 atm

V1 Y1

1.39 0.1628

L1 X1

0.0810

1. Merubah Merubah SCFM SCFM menjadi lb mol/mi n V (STP (STP)) 22.4 22.4 ltr/ ltr/gm gmol ol 359 359 ft3/ ft3/lb lbmo moll 1.39 1.39 lbmo lbmol/ l/mi min n V1 2. Menentukan Laju alir minimum 2.1 Menggambar kurva keseti mbangan Dalam bentuk rasio mol x 0 0.01 0.02 0.03 0.04 0.05 0.06

µA 7.0287 6.7612 6.5064 6.2636 6.0322 5.8117 5.6014

pA 0.0000 0.0223 0.0429 0.0620 0.0796 0.0959 0.1109

y* = pA/P 0.0000 0.0223 0.0429 0.0620 0.0796 0.0959 0.1109 1

0.07 0.08 0.09 0.1

5.4008 5.2095 5.0269 4.8525

0.1248 0.1375 0.1493 0.1601

0.1248 0.1375 0.1493 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  jadi L' hampir sama  Atau 18 lbmol/min 8.330 lb/gal 5.76 gal/min L X 0.0000 0.0101 0.0204 0.0309 0.0417 0.0526 0.0638 0.0753 0.0870 0.0989 0.1111

Y* 0.0000 0.0228 0.0449 0.0661 0.0865 0.1061 0.1247 0.1425 0.1595 0.1755 0.1907

Y 0.0081 0.0274 0.0471 0.0672 0.0877 0.1087 0.1301 0.1519 0.1742 0.1970 0.2204

2

Jika diinginkan x1 = 0.07 Hitung berapa nilai NOG ?

V2 y2

0.00807

L2 x2

0.00 0.0002

V1 y1

1.39 0.1400

L1 x1

0.0700

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 y2 x1 x2

0.14 0.00807 0.07 0.0002 Y1-Y2 X1-X2 L/V

Y1 0.1628 Y2 0.0081 X1 0.0753 X2 0.0002 -0.1547 -0.0751 2.060178 slope

y* = 0.33 * exp (1.95*(1-x2)^2) * x 2. Menghitung NOG Menggu nakan metoda Wiegand NOG = INTEGRAL y1-y2 [dy/(y*-y)] atau dibalik NOG = INTEGRAL y2-y1 [dy/(y-y*)] y 0.00807 0.022 0.036 0.05 0.063 0.076 0.089 0.102 0.115 0.128 0.14

X 0.0002 0.0072 0.0144 0.0218 0.0289 0.0362 0.0437 0.0514 0.0593 0.0675 0.0753

x 0.0002 0.0071 0.0142 0.0213 0.0281 0.0349 0.0418 0.0489 0.0560 0.0632 0.0700

y* 0.0005 0.0161 0.0311 0.0456 0.0585 0.0708 0.0827 0.0941 0.1051 0.1155 0.1248

1/(y-y*) Integrasi 131.467 2.0885 168.393 2.6133 204.932 3.0153 225.819 2.8986 220.119 2.6900 193.725 2.2949 159.343 1.8611 126.987 1.4787 100.508 1.1737 80.063 0.8741 65.615 20.988 3

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

m m2 1.39 lbmol/min 0.023167 lbmol/s lbmol/s m2 m HOG = V/S/Kga m Z=HOG x NOG

4

 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 Dari soal MW HC MW Gas Inert MW Oil GV GL y1 Recover L2 x2

44 29 300 5000 10000 0.2 0.95 33.3 0

V2 y2

126.6 0.0123

L2 x2

33.3 0

V1 y1

156.25 0.20

L1 x1

63.0 0.471

lb/lbmol lb/lbmol lb/lbmol lb/hr/ft2 lb/hr/ft2 fraksi mol lbmol/hr ft2 fraksi mol

1. Menghit ung V1 Basis 1 lbmol gas masuk Menghitung MW gas % in V MW 0.2 8.8 0.8 23.2 32 Maka V1 156.25 lbmol/hr ft2

cara cepat menghitung y2 Y1 0.25 Y2 0.0125 y2 0.0123457

2. Menghit ung 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)

5

3. Menghit ung x 1 Lb mol HC di L1 HC di L1 Oil bebas solute Total mole x1

0.19 29.7 33.3 63.0 0.47

lbmol/mol V1 lbmol lbmol lbmol fraksi mol

4. Membuat operating lin e 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))+ (x/(1-x)) = 3.75 *(y/(1-y))+ -0.04687 x 0 0.0307 0.1325 0.271 0.383 0.472

(x/(1-x)) 0.0000 0.0317 0.1527 0.3717 0.6207 0.8939

(y/(1-y)) y 0.0125 0.0209 0.0532 0.1116 0.1780 0.2509

0.0123 0.0205 0.0505 0.1004 0.1511 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 0.03 0.003 0.0309 0.0691 0.05 0.0117 0.0526 0.1505 0.08 0.03 0.0870 0.2792 0.1 0.0475 0.1111 0.3698 0.14 0.0945 0.1628 0.5636 0.15 0.106 0.1765 0.6149 0.18 0.1445 0.2195 0.7763 0.2 0.173 0.2500 0.8906 6. Menghi tung NOG

29.6875

0.0002 0.0646 0.1308 0.2183 0.2700 0.3604 0.3808 0.4370 0.4711

(menggunakan metoda penurunan) 6

y* dibaca dari kurva y interval dari y2 ke y1 y 0.0124 0.03 0.05 0.08 0.1 0.14 0.15 0.18 0.2 Jadi NOG =

(y/(1-y)) 0.0126 0.0309 0.0526 0.0870 0.1111 0.1628 0.1765 0.2195 0.2500

(1-y)lm = (x/(1-x)) 0.0002 0.0691 0.1505 0.2792 0.3698 0.5636 0.6149 0.7763 0.8906

x 0.0002 0.0646 0.1308 0.2183 0.2700 0.3604 0.3808 0.4370 0.4711

(1-y)-(1-y*)/ln((1-y)/(1-y*)) A y* 0.0003 0.0029 0.0107 0.0308 0.0488 0.0931 0.1056 0.1452 0.1731

5.3279 menggunakan trapeziodal rule

1-y 0.9876 0.9700 0.9500 0.9200 0.9000 0.8600 0.8500 0.8200 0.8000

1-y* 0.9997 0.9971 0.9893 0.9692 0.9512 0.9069 0.8944 0.8548 0.8269

B

C

D

(1-y)lm 0.9936 0.9835 0.9695 0.9444 0.9254 0.8832 0.8720 0.8373 0.8134

y*-y -0.0121 -0.0271 -0.0393 -0.0492 -0.0512 -0.0469 -0.0444 -0.0348 -0.0269

B/(A*C) -83.2 -37.4 -26.0 -20.8 -20.1 -21.9 -23.1 -29.3 -37.7

Metoda Wiegand Integrasi 1.0610 0.6336 0.7021 0.4092 0.8399 0.2251 0.7864 0.6706 5.3279

-D 1/(y-y*) Integrasi 83.2 82.65 1.0520 37.4 36.89 0.6232 26.0 25.43 0.6862 20.8 20.31 0.3983 20.1 19.52 0.8174 21.9 21.34 0.2193 23.1 22.51 0.7686 29.3 28.73 0.6583 37.7 37.11 5.2233 Bisa menggunakan metoda

7