This preview shows page 1. Sign up to view the full content.
Unformatted text preview: = 1.75, can apply
Eq. (529), which was derived for extraction but applies to absorption using:
Yout
=
Yin 1
N En where E = L′
18.23
=
= 1.356
K ′G ′ (175)(7.679)
. n=0 Therefore, Yout 0.000912
=
= 0.0295 =
Yin
0.03093 1
N
n =0 1356
. Solving, N = 7.5 equilibrium stages
n Exercise 6.37 (continued) Analysis: (c) (continued)
Could also solve this using Kremser's equation or by a graphical construction like Fig. 6.12.
(d) Because operating line and equilibrium line are straight, use Eq. (695) with A = E = 1.356
N OG = N t ln(1/ A)
ln(1/1.356)
= 7.5
= 8 .7
(1 − A) / A
(1 − 1.356) /1.356 (e) From Eq. (6141), lT = HOGNOG
From Table 6.7, H OG = V′
7.679
0.64
=
=
, where S = cross sectional area of tower in ft2
KY aS (12.0) S
S The allowable gas superficial velocity = uV = 2.4 ft/s. Take this at the bottom of the tower where
gas rate is the highest. From the continuity equation, Q = uV S = gas volumetric rate
Therefore, S = Q/uV = (50/60)/2.4 = 0.347 ft2. Therefore, HOG = 0.64/0.347 = 1.84 ft
Packing height = (1.84)(8.7) = 16 ft.
(f) To obtain packed height as a function of L'/V' with V' and HOG constant, the only change will
be the value of NOG . Use Eq. (693), which for xin = 0 and yin/yout = 0.03/0.000911 = 32.9, is: ln 32.9
N OG = A −1
1
+
A
A
A −1
A (2) where for a dilute system, we can take A = L'/K'V' = (L'/V')/K' = (L'/V')/1.75
The results obtained from Eq. (2) are as follows: L'/V'
1.75
2.0
2.5
3.0
5.0
10.0 A
1.00
1.143
1.429
1.714
2.857
5.714 1/ A
1.00
0.875
0.700
0.583
0.350
0.175 NO'G Height, ft 12.9
7.86
6.38
4.74
4.01 23.6
14.5
11.7
8.72
7.38 ∞ ∞ Exercise 6.38
Subject: Absorption of NH3 from air into water with a packed column.
Given: Column operating at 68oF and 1 atm. Gas enters at 2,000 ft3/min at 68oF and 1 atm,
containing 6 mol% NH3 in air. Entering water rate is twice the minimum. Absorb 99% of the
NH3. Gas velocity at 50% of flooding velocity.
Assumptions: No stripping of water. No absorption of air.
Find: Tower diameter and packed height.
Analysis: Entering gas rate is 2,000 60 + 460
60 = 312 lbmol /h
379 68 + 460 Compute material balance. NH3 in entering gas = 0.06(312) = 18.72 lbmol/h
NH3 in exiting gas = 0.01(18.72) = 0.19 lbmol/h
Air in entering and exiting gas = V' = 312  18.72 = 293.28 lbmol/h
NH3 in exiting water = 18.72  0.19 = 18.53 lbmol/h
Therefore, Xin = 0.0
Yin = 18.72/293.28 = 0.06383
Yout = 0.19/293.28 = 0.000648
At the minimum water rate, the exiting liquid is in equilibrium with the entering gas.
From Fig. 6.49, X*out = 0.072 for Yin = 0.06383. Therefore, L'min = 18.53/0.072 = 257 lbmol/h
Operating liquid rate = 2 times minimum = 2(257) = 514 lbmol/h
Now, compute column diameter using Fig. 6.36. Corrections for liquid density and viscosity are
neglected. From the ideal gas law on the entering gas, with an average molecular weight = 28.3,
ρV =PM/RT = (1)(28.3)/(0.7302)(68+460) = 0.0734 lb/ft3
From Table 6.8, packing factor for 1inch metal Pall rings = 56 ft2/ft3.
LM L ρV
The abscissa in Fig. 6.36 is, X =...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details