Unformatted text preview: 996(0.02645) = 7.924 x 104 mol/s
Increase in water content of the gas = rate of mass transfer of water =
nA = 7.924 x 104  2.426 x 104 = 5.5 x 104 mol/s
(b) From Eqs. (3211) and (3218), noting that there is no resistance in the liquid phase, KG = k p = ( NA
pAi − pAb ) =
LM nA
AP yAi − yAb ( ) (1)
LM Exercise 3.38 (continued)
Analysis: (b) (continued)
A = πDL = 3.14(1.46)(82.7) = 379 cm2
At 25oC, the interface mole fraction of water vapor = yA i = y A i − yA b LM = yA i − yA in − yA i − yA out
ln yA i − yA in
yA i − yA out = PAs 0.46 psi
=
= 0.0313
P 14.7 psi 0.0313 − 0.00825 − 0.0313 − 0.02645
= 0.0117
0.0313 − 0.00825
ln
0.0313 − 0.02645 From Eq. (1), KG = 5.5 × 10−4
= 1.24 × 10−4 mol/scm2atm
(379)(1)(0.0117) Note that the correction for bulk flow amounts to only about 2.5% because the average
mole fraction of air between the interface and the bulk is 0.975. Exercise 3.39
Subject: Absorption of ammonia (A) from air (B) into 2 N H2SO4 in a wettedwall column, with
countercurrent flow.
Given: Column inside diameter = D = 0.575 inch (0.0479 ft) and length = L = 32.5 in. (2.71 ft)
Air flow rate = n = 0.260 lbmol/h at 1 atm. Partial pressures of ammonia in the air are 0.0807
atm entering and 0.0205 leaving. Air enters at 77oF and exits at 84oF. Aqueous acid enters at
76oF and exits at 81oF. Change in acid strength is negligible.
Assumptions: Reaction of the basic ammonia with the sulfuric acid is instantaneous such that
the partial pressure, pA i , of ammonia at the gasliquid interface is zero. Therefore, the only
mass transfer resistance is in the gas phase. Air flow rate refers to the total entering gas.
Find: Masstransfer coefficient, kp , for the gas phase. Negligible bulk flow effect.
Analysis: From a rearrangement of Eq. (3207), kp = nA
A pAb − pAi ( ) =
LM nA
A pA b − 0 ( ) =
LM nA
A pAb () (1)
LM The rate of mass transfer is obtained by material balance for the decrease in partial pressure of
the ammonia in the air. First compute the flow rate of just the air in the gas.
P − pA in 10 − 0.0807
.
= 0.239 lbmol/h
P
1.0
pA in
0.0807
Ammonia in entering gas =
0.239 = 0.021 lbmol/h
nB =
10 − 0.0807
.
pBin
nB = n Ammonia in exiting gas = pA out
pBout nB = = 0.260 0.0205
0.239 = 0.005 lbmol/h
10 − 0.0205
. Rate of mass transfer of ammonia = nA = 0.021  0.005 = 0.016 lbmol/h
Area for mass transfer = A = πDL = 3.14(0.0479)(2.71) = 0.408 ft2
pA b LM = () From Eq. (1), k p = nA / A pAb pA in − pA out 0.0807 − 0.0205
=
= 0.0439 atm
pA in
0.0807
ln
ln
0.0205
pA out LM = 0.016 /(0.408)(0.0439) = 0.893 lbmol/hft2atm The bulk flow effect amounts to a correction of about 5%. Exercise 3.40
Subject: Overall masstransfer coefficient for a packed cooling tower, where water (A) is
evaporated into air (B).
Given: Crosssectional area for air flow = S = 0.5 ft2. Height over which data were taken = H =
7 ft. Air rate = n = 0.401 lbmol/h. Water rate = 20 lbmol/h. At the bottom, P = 14.1 psia, T =
120oF, PA...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details