Example 5
Acetone is to be recovered from a 5 per cent acetone–air mixture by scrubbing with water
in a packed tower using countercurrent flow. The liquid rate is 0.85 kg/m2s and the gas
rate is 0.5 kg/m2s. The overall absorption coefficient KGa may be taken as 1.5 × 10?4 kmol/ [m3s (kN/m2) partial pressure difference] and the gas film resistance controls the process. What height of tower is required tower to remove 98 per cent of the acetone? The
equilibrium data for the mixture are:
Solution
straight line of slope ( ?y / ? x) m = 1.20.
y = 1.2 x
At the bottom of the tower:
M. wt %of air = mol% of air * M. wt / (mol % of air * M.wt + mol% of gas *M.wt )
= (0.95 * 29 0 / ( 0.95 * 29 + 0.05 *58 ) = 0.905
Gm = 0.5 Kmol / m2 S * 0.905 = 0.452 Kg / m2 S
( to convert from Kg / m2 S to Kmol / m2 S ) divided by M. wt of gas mixture)
Gm= 0.452 / 29 = 0.0155 kmol / m2 S
Lm = 0.85 / 18 = 0.0472 kmol / m2S
HOG = Gm / KGA PT = 0.0155 / 1.5 * 10-4 * 101.3 = 1.02
NOG=1/(1-?) ln??[ ( 1-?)y1/y2?+?] where ?=?mG ?_m/?L ?_m
The equilibrium data given are represented by a straight line of slope m = 1.20.
the equation for NOG may be integrated directly when the equilibrium line is given by
( y = 1.2 x ) ye = mx to give:
?=?mG ?_m/?L ?_m = 1.20(0.0164/0.0472) = 0.417
y1 /y2 = (0.05/0.001) = 50
The packed height = NOG × HOG = (5.80 × 1.02) = 5.92 m
Example 6
A solute gas is to be absorbed from an air – gas mixture in packed tower using free solute liquid. The equilibrium relationship is y=m .x
If 99% of the solute required used the liquid rate 1.75 time the min. , and the height of the transfer unit is 1.0 m .what is the packed height required ?
Solution :
y2 =(1-0.99)y1 = 0.01 y1
( L / G ) min = m ( 1- y2 / y1 ) = m ( 1- 0.01 y1 / y1 ) = 0.99( m )
(L / G ) act. = 1.75 * 0.99 m = 1.575 (m)
? = m G / L = m / 1.575 m = 0.577
NOG=1/(1-?) ln??[ ( 1-?)y1/y2?+? , NOG = 1 / (1-0.577) ln [(1-0.577)+ y1 / 0.01 y1 + 0.577 ] = 3.98
Z = HOG * NOG = 1 * 3.98 = 3.98 m
Example 7
Sulphur dioxide is recovered from a smelter gas containing 3.5 per cent by volume of SO2, by scrubbing it with water in a countercurrent absorption tower. The gas is fed into the bottom of the tower, and in the exit gas from the top the SO2 exerts a partial pressure of 1.14 kN/m2. The water fed to the top of the tower is free from SO2, and the exit liquor from the base contains 0.001145 kmol SO2/kmol water. The process takes place at 293 K, at which the vapour pressure of water is 2.3 kN/m2. The water flow rate is 0.43 kmol/s. If the area of the tower is 1.35 m2 and the overall coefficient of absorption for these conditions K La is 0.19 kmol SO2/sm3 (kmol of SO2/kmol H2O), what is the height of the column required?
The equilibrium data for SO2 and water at 293 K are:
Solution :
At the top of tower , Lm = 0.43 ( Kmol / S) / 1.35 m2 = 0.318 kmol / m2 S
PSO2 = y2 SO2 PT (Dalton law ) , y2SO2 = 1.14 / 101.3 = 0.013
y1 SO2 = 0.035
HOL = Lm / K La CT = Lm / K L a = 0.318 / 0.19 = 1.676
NOL = X1-X2 / ( Xe- X )lm
(Xe – X)lm = ( Xe1-X1) (Xe2- X2) / ln[ (Xe1-X1) / (Xe2 –X2)
From equilibrium cure , will be find the Xe1, Xe2
y1=0.035 , from eq. curve the Xe1 = 1.41 * 10-3 and y2 = 0.0113 , Xe2= 0.54*10-3
(Xe- X)lm = ( 1.4 *10-4) – ( 0.5 *10-3-0) / ln [ 2.65*10-4 / 0.5 810-3 ] = 3.86 * 10-3
NOL = X1-X2 / (Xe-X )lm = 0.001145-0 / 3.86 * 10 -3 = 2.96
Z = HOL * NOL = 2.96 * 1.671 = 4.96 ? 5 m
Example 8
It is desired to absorb 90% of the acetone in gas containing 1.0 mol% acetone in air in counter current stage tower . the total inlet gas flow rate to the tower is 30 kmol / hr , and the inlet pure water flow to the tower 90 kmol hr .the process is operated at 300K and 1 atm . the equilibrium relation for the acetone in the gas-liquid is y=2.53 X
Determined the No.of theoretical stage required .
If the overall tower efficiency is equal to 65% .
If the murphree efficiency ( Emv=0.75 ) , calculate the actual stage required .
Solution :
Y1= 1% =0.01 , Y2 = (1- 0.9 )Y1 = 0.001
Lm= 90 kmol / hr , Gm = 30 kmol / hr
N = log [( X2-A / B) /log ? ] , A = ( mX1- ? y1) / 1-? , B = y1 – A
? = m Gm / Lm
Y1= y1 / 1-y1 y1 = Y1 / 1+Y1 = 0.01 / ( 1+ 0.01) = 0.0099? 0.01
? = 2.53 *30 / 90 = 0.843
Gm( Y1-Y2) = Lm (X1-X2) X1 = Gm(Y1-Y2) / Lm = 30(0.01-0.001) / 90 = 3*10-3 , when X2 = 0
A= 2.53 (3*10-3 ) – 0.843 * 0.01 / (1-0.846 ) = - 5.35 810-3
B= y1-A= 0.01+5.35*10-3 = 0.01535
N = log [ 0.001+ 5. 35*10-3 / 0.01535 ] / log 0.846 = 5.1 ? 5
? (tray efficiency) = No.of theoretical tray / No.of actual tray
No. of tray ) act. = 5 / 0.65 ? 8 trays
C) y* act. = m X [ Emv (1-1/ ?) + 1 / ? ] = 2.53 X [ 0.75(1-1 / 0.846) = 1 / 0.846 ]
y* act = 2.647 X
? = mGm / Lm = 2.647 * 30 / 90 = 0.882
A = mX1-? y1 / 1-? = 2.64 ( 3*10-3 ) -0.882 (0.01) / 1-0.882 = -7.62*10-3
B= y1-A = 0.01 + ( 7.62 *10-3 ) = 0.0176
N = log [0.001 + 7.62 *10-3 / 0.0176] / log 0.882 = 5.68 ? 6 trays
Stripping:
Stripping is physical separation process where one or more component are removal from liquid stream by a pour (gas) stream
N=(log?[(?-S)/(1-S)])/(log?)-1 , S=(X2-X1)/(X2-(Y1/m))
Operating line : G(Y –Y1)= L(X-X1) , Y =L / G (X-X1)
Example : A 1% wt. solution of benzene in oil of M.wt 380 is to be stripped with steam at 1 atm and 100 C? to reduce the benzene content to0.05% (M.wt of benzene is 78)
Calculate the steam stage per 100 kg of feed
Calculate the number of theortical plate required for stripping tower when the steam consumptionis 1.8 times the minimum
If the tray effeiciency is equal to 65% calculate the actual number of trays, the equailibrium relation is Y= 1.842 X
Solution:
Y2=mX2, mol% = (wt.%/ M.wt ) / ?( wt% / M.wt) = (0.01 / 78 )/ (0.01 / 78 + 0.99/ 380)= 0.046 = X2
X1 = mol%= (0.05/78) / ( 0.05/78 +99.95 /180 )=2.43 X10-3
L(X2-X1)=G(Y2-Y1) , where Y1=0 AND Y2=1.842 x2 , THEN Y2=1.842 *(0.046) = 0.0847
G=L(X2-X1)/ Y2 = 100/ 380(0.046-2.43 X10-3 / 0.0847 = 1.59 KMOL / M2. S
G(Y2-Y1)=L(X2-X1) , (L/G)MIN =Y2-Y1 / X2-X1 , Y1=0 , THEN (L/G)MIN =Y2 / X2-X1 AND Y2=MX2
(L/G)MIN =MX2 / X2-X1 = 1.842*0.046 / 0.046-0.00243 =1.944
GMIN =L / 1.944=(100/380) / 1.944 = 0.135 KMOL
GACT =1.8 GMIN =1.8(0.135)=0.243KMOL
S=X2-X1 / X2-(Y1/ M) = 0.046-0.00243 / 0.046 = 0.947 , WHERE Y1/ M =0
?=MGM / LM =1.842*0.243 / (100/380)= 1.705
N= N=(log?[(?-S)/(1-S)])/(log?)-1 = [LOG (1.7-0.943 / 1-0.943) / LOG1.7] -1= 4.01?4 TRAYS
(No. OF TRAYS)ACT = ( No. OF TRAYS )THEO / TRAYS EFF. = 4/ 0.65 = 6.15 ?6 TRAYS
المادة المعروضة اعلاه هي مدخل الى المحاضرة المرفوعة بواسطة استاذ(ة) المادة . وقد تبدو لك غير متكاملة . حيث يضع استاذ المادة في بعض الاحيان فقط الجزء الاول من المحاضرة من اجل الاطلاع على ما ستقوم بتحميله لاحقا . في نظام التعليم الالكتروني نوفر هذه الخدمة لكي نبقيك على اطلاع حول محتوى الملف الذي ستقوم بتحميله .
الرجوع الى لوحة التحكم
|