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Unformatted text preview: eight of carbon bed for a timetobreakthrough of 8 h, for the same
conditions of flow rate, temperature, and pressure as the lab data.
Analysis: Use a continuity equation, similar to Eq. (643) to compute the bed diameter:
πD 2 Q
(1)
m = u s A ρ = Qρ o r A =
=
4
us
At operating conditions, Q = 12,000(560/520) = 12,900 ft3/min. us = 60 ft/min
From Eq. (1), A = 12,900/60 = 215 ft2. D = [4(215)/3.14]1/2 = 16.5 ft.
Use Eq. (15122) to compute the ideal bed length, LES:
cQt
c (12,900)(8 × 60)
c
LES (in ft) = F F B = F
= 960 F
(2)
q F ρb A
q F (30)(215)
qF
Want cF in lb EA/ft3 inlet gas. MW of EA = 88.1. One scf of gas contains 379 lbmol.
Therefore, have 12,000/379 = 31.66 lbmol/min. The mole fraction of EA = 0.005.
Concentration of EA in the entering gas of 0.005(31.66)(88.1)/12,900 = 0.00108 lb EA/ft3 gas.
From the given equilibrium data, for pEA = 0.005 atm, qF = 0.270 lb EA/lb C.
From Eq. (2),
0.00108
LES = 960
= 385 ft
.
0.270
From Eq. (15120), the required bed length taking into account the MTZ is,
LB = LES + LUB, where LUB is given by Eq. (15121).
The breakthrough data for Le = 2 ft are plotted on the next page as yEA out against time in
minutes. The data are almost straight in the mole fraction range of 0.00475 (c/cF = 0.95), where
time = te = 160 minutes, to 0.00025 (c/cF = 0.05), where time = tb = 75 minutes. Therefore, the
midpoint time, ts = (160 + 75)/2 = 117.5 minutes. Alternatively, ts can be evaluated by
integration, using Eq. (15123).
From Eq. (15121), LUB = (ts  tb)Le/ts = (117.5  75)2/117.5 = 0.72 ft.
Therefore, LB = 3.85 + 0.72 = 4.57 ft. So the bed is 16.5 ft in diameter by 4.57 ft high. This is a
poor bed heighttodiameter ratio. Might increase the breakthrough time by a factor of 4 to get a
bed height approximately equal to the bed diameter. Exercise 15.28 (continued)
Analysis: (continued) 0.0050 Mole fraction of EA in gas effluent 0.0045 Lab Breakthrough Data 0.0040 te 0.0035
0.0030
0.0025
0.0020
0.0015 ts tb 0.0010
0.0005
0.0000
0 20 40 60 80 100 Time, minutes 120 140 160 180 Exercise 15.29
Subject: Desorption of benzene from silica gel using air under isothermal, isobaric conditions.
Given: Values are given in a table below for dimensionless benzene concentration, φ, and
loading, ψ, profiles in a bed following adsorption. Bed is 2 feet in diameter and 30 ft high. Bed
contains silica gel adsorbent. Desorption by air at 1 atm and 145oF at an interstitial velocity, u,
of 98.5 ft/min. Bed porosity = εb = 0.5. During desorption, masstransfer coefficient = k = 0.206
min1 and Henry's law equilibrium adsorption constant = 1,000 in Eq. (1) on p. 837.
Assumptions: Plug flow with a constant interstitial velocity equal to the given inlet value.
Negligible axial dispersion. Desorption air is passed up the bed, opposite in direction to the
initial flow of benzeneair feed mixture during adsorption.
Find: Desorption time required to remove 90% of the benzene from the bed.
Analysis: First, compute the initial amount of benzene adsorbed in the bed.
Bed volume = V...
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 Spring '11
 Levicky
 The Land

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