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Unformatted text preview: .247) 1/ 2 0.247 1
= 5.23 ft
33.9 2.02 uL
0.247
=
= 0.0234 s1
H L ( aPh / a ) 5.23 ( 2.02 ) This is 35% of the suggested value of 0.067 s1
Coefficient kGa is obtained from the gasphase analog of (1), kG a = uV
H G ( aPh / a ) (2) with HG obtained from (6133), using dimensionless groups from (6134) and (6135).
From Exercise 6.28, uV = 1.59 ft/s, ρV = 0.148 lb/ft3, µV = 0.018 cP = 1.21 x 105 lb/fts
From Eq. (6134), N ReV = uV ρV
(1.59)(0.148)
=
= 574
(33.9)(1.21× 10−5 )
aµV From Exercise 6.23, the diffusivity of benzene in air = DV = 0.086 cm2/s at 70oF and 15 psia.
Using Eq. (336) to correct for temperature and pressure, DV = 0.086 15
(2)(14.7) From Eq. (6135), NScV = 77 + 460
70 + 460 1.75 = 0.045 cm2/s = 4.84 x 105 ft2/s µV
1.21× 10−5
=
= 1.69
ρV DV (0.148)(4.84 × 10 −5 ) Exercise 6.29 (continued)
Analysis: (continued)
From Eq. (6133) 1
1/ 2 4ε
HG =
( ε − hL )
CV
a4 1
4(0.919)
( 0.919 − 0.177 )1/ 2
0.368
33.94
From (2), kG a = 1/ 2 1/ 2 (N ) (N )
−3/ 4 ReV ScV ( 574 )−3 / 4 (1.69 )−1/ 3 −1/ 3 uV
DV a
aPh 1.59 4.84 ×10−5 1.59
= 1.73 s1
0.455 ( 2.02 ) This is approximately twice the suggested value of 0.80 s1. = 1
= 0.455 ft
2.02 Exercise 6.30
Subject: Absorption of NH3 from air into water in a packed tower.
Given: Column operation at 68oF and 1 atm. Inlet gas mass velocity = 240 lb/hft2. Inlet water
mass velocity = 2,400 lb/hft2. Tower packed with 1.5inch ceramic Berl saddles. Henry's law
applies for NH3 with p (atm) = 2.7x.
Assumptions: No stripping of water. No absorption of air. Dilute system.
Find: (a) Packed height for 90% absorption of NH3.
(b) Minimum water mass velocity for 98% absorption of NH3.
(c) KGa, pressure drop, maximum liquid rate, KLa, packed height, column diameter, HOG,
and NOG for 1.5inch ceramic Hiflow rings.
Analysis: (a) Assuming Dalton's law, the equilibrium equation for 1 atm is,
p = yP = 2.7x atm or with P = 1 atm, y = Kx = 2.7x (1) Because the system is dilute, Eq. (689) can be used to compute NOG .
Then the packed height is given by Eq. (685),
lT = NOGHOG (2)
The absorption factor from Eq. (615) is, A = L/KV = (2,400/18)/2.7(240/29) = 6.0
Entering and exiting ammonia mole fractions are: yin = 0.02, yout = 0.002, xin = 0.0
The ammonia mole fraction in the exiting liquid is obtained by an overall ammonia balance:
240
2,400
240
2,400
yin +
xin =
yout +
xout
29
18
29
18 (3) Solving Eq. (3), xout = 0.00112.
From Eq. (689),
N OG = ln ( A − 1) / A ( yin − Kxin ) / ( yout − Kxin ) + (1 / A)
( A − 1) / A = ln (6 − 1) / 6 0.02 / 0.002 + (1 / 6)
(6 − 1) / 6 = 2.57 To obtain an estimate of HOG, the correlations of Billet and Schultes in Chapter 6 or literature
data can be used. The latter is found on page 1638 of Perry's Chemical Engineers' Handbook,
6th edition or on page 663 of "EquilibriumStage Separation Operations in Chemical
Engineering" by Henley and Seader, as shown below. Analysis: (a) (continued) Exercise 6.30 (continued) AirAmmoniaWater system with...
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 Spring '11
 Levicky
 The Land

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