Unformatted text preview: 47. Therefore, the particle density
39
ρ
from a rearrangement of Eq. (154) is ρ p = b =
= 73.6 lb gel/ft3 of particles.
1 − ε b (1 − 0.47)
When the partial pressure of H2O = 0.0277 atm, we have 0.1264/100 = 0.001264 lb H2O/ft3 gas.
Therefore, K = 15(73.6)(0.0277)/0.001264 = 24,200. The equilibrium adsorption isotherm is
now, q in lb H2O/ft3 silica gel particles = 24,200 c in lb H2O/ft3 gas
From above, k = 1/[0.0292 K] = 1/[0.0292(24,200)] = 1.42 x 103 s1 or 0.0849 min1
From Eq. (2), k c = The Klinkenberg Eq. (15114) is:
where, from Eq. (15115), ξ = c1
≈ 1+
cF 2 kKz 1 − ε b
u
εb τ− ξ+ 1
8τ + 1 (3) 8ξ
(4) u = actual (interstitial velocity) = superficial velocity/εb = 100/0.47 = 213 ft/min = 1.08 m/s
1
z
1 − 0.47
0.0292
From Eq. (4), ξ =
(5)
= 35.8z (in meters)
108
.
0.47 Exercise 15.26 (continued)
Analysis: (continued)
From Eq. (15116),
z
z
τ=k t−
= 0.0849 t −
= 0.0849t (in minutes) − 0.00131z (in meters)
(6)
u
1.08(60)
The maximum value of z, the distance through the bed, is the bed height = 5 ft = 1.524 m.
At breakthrough, c/cF = 0.05 at z = 1.524 m and, therefore, ξ = 35.8(1.524) = 54.6
and τ = 0.0849t  0.00131(1.524) = 0.0849t (in minutes)  0.00200
Must solve Eq. (1) for t at c/cF = 0.05 and ξ= 54.6. The time will be less than the ideal
equilibrium time above of 679 minutes. Can do this by trial and error with a spreadsheet or with
a nonlinear solver that can handle the error function, erf. From a spreadsheet, breakthrough time
= 451 minutes.
To obtain the width of the MTZ, solve Eq. (1) for z at c/cF = 0.95 and t = 451 minutes. By trial
and error from a spreadsheet, at breakthrough, the MTZ extends from z = 0.718 m to z =1.524 m.
Therefore, the width of the MTZ = 1.524  0.718 = 0.81 m = 2.6 ft.
Breakthrough curves similar to Figs. 15.30 and 15.31, but in terms of distance, z, through the bed
and time, instead of the dimensionless groups, ξ and τ , are shown on the next two pages. They
are computed with a spreadsheet from Eq. (3) above for gas concentration and Eq. (15119) for
adsorbent loading, together with Eqs. (5) and (6) to convert from ξ and τ to z and t, respectively.
With the spreadsheet, the error function, erf, must be used carefully, noting that erf(x) = erf(x).
Thus, when the argument of the error function is negative, the argument should be made positive
and the result should be multiplied by 1.
To compute the average loading of the bed at the breakthrough time of 451 minutes, use the
equation in Example 15.11:
1.524 qavg = qdz , where z is in meters.
(7)
1.524
A spreadsheet is used with Eq. (15119) to compute q as a function of z for t = 451 minutes
0 (breakthrough time) and q * = 0.440 lb H2O/lb silica gel from above (value in equilibrium with
F
the feed. The results are:
q , lb/lb
0.440 0.440 0.433 0.382 0.260 0.125 0.042 0.018
z, meters
0.2
0.4
0.6
0.8
1.0
1.2
1.4 1.524
Solving Eq. (7) numerically by the trapezoid method,
qavg = [0.440(0.2) + 0.440(0.2) + 0.4365(0.2) + 0.4075(0.2) + 0.321(0.2) + 0.1925(0.2) +
0.0835(0.2)...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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