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Unformatted text preview: mass transfer coefficient for Cl2 is higher and the corresponding HG is lower.
Therefore, the assumption that GeCl4 controls is correct.
(c) If a packed height of 10 ft rather than the computed 8.3 ft is used, then for GeCl4 ,
NG = lT /HG = 10/1.8 = 5.55 rather than 4.61.
y
From above, N G = ln in = 5.55
yout
y
Solving, in = 258
Therefore, yout = 0.00162/258 = 0.0000063
yout
% absorption of GeCl4 = (1  0.0000063/0.00162) = 0.996 or 99.6% compared to 99% specified
For Cl2, from above, HG = HG of GeCl4 (DGeCl4/DCl2)2/3 = 1.8(0.000006/0.000013)2/3 = 1.08
Therefore, NG = 10/1.08 = 9.26 = = ln yin
yout Solving, yin
= 10,500
yout % absorption of Cl2 = (1  1/10,500) = 0.9999 or 99.99% compared to 99% specified Exercise 6.28
Subject: Packed tower diameter and pressure drop for conditions of Exercise 6.26.
Given: Suggested tower diameter of 0.80 m and pressure drop of 500 N/m2m of packed height
(0.612 in. H2O/ft).
Find: (a)
(b)
(c)
(d) Fraction of flooding if FP = 24 ft2/ft3.
Pressure drop at flooding.
Pressure drop at operating conditions using GPDC chart.
Pressure drop at operating conditions using BilletSchultes correlation. Analysis: (a) Using data from Exercise 6.26 and Fig. 6.36,
LML = 600(8.33) = 5,000 lb/min
VMV = (1,000/379)(29) = 76.5 lb/min
From ideal gas law, ρV = PM/RT = (2)(29)/(0.7302)(460+77) = 0.148 lb/ft3
ρL =62.3 lb/ft3
1/ 2 1/ 2 LM L ρV
5, 000 0.148
From Fig. 6.36, X =
=
= 3.19
VM V ρ L
76.5
62.3
From Fig. 6.36, at flooding, Y = 0.0053
For the actual operation, by the continuity equation, m = uoSρ
Area for flow = S = πD2/4 = (3.14)(0.8/0.3048)2/4 = 5.41 ft2
Superficial velocity = uo = m/Sρ = (76.5/60)/(5.41)(0.148) = 1.59 ft/s
From Eq. (6102), noting that for the dilute aqueous solution at 25oF, f{ρL} = 1 and f{µL} = 0.98
for a liquid viscosity of 0.95 cP at 25oC,
2
uo FP ρV
Y=
g ρH 2O (1.59) 2 (24) 0.148
(1.0)(0.98)=0.0044
f {ρ L }f {µ L }=
32.2
62.3
1/ 2 u
Y
0.0044
Fraction of flooding = o =
=
uflood
Yflood
0.0053
This is too high. Should increase tower diameter. 1/ 2 = 0.91 or 91% (b) From Eq. (6104), ∆Pflood = 0.115 FP0.7 = 0.115(24)0.7 = 1.06 in. H2O/ft of packed height
(c) From Fig. 6.36, ∆P at design = 1.50 in. H2O/ ft of packed height (seems high)
(d) From Table 6.8, for 2inch (50mm) plastic Pall rings, the packing characteristics are:
a = 111.1 m2/m3 or 33.9 ft2/ft3 , ε = 0.919 m3/m3 , Ch = 0.593 , Cp = 0.698, Cs = 2.816
The pressure drop per unit height of packed bed is given by Eq. (6115),
∆P
ε
=
∆Po
ε − hL 3/ 2 exp 1/ 2
13300
N
3/ 2 ( FrL )
a (1) Exercise 6.28 (continued)
Analysis: (d) (continued)
where, hL is given by (697) and is in m2/m3 and N FrL is given by (699).
From Eq. (6110), ∆Po
a u2ρ 1
= Ψo 3 V V
lT
ε 2 KW (2) 1− ε
1 − 0.919
=6
= 0.0143 ft
a
33.9
1
2 1 DP
2
1
0.0143
From Eq. (6111),
= 1+
= 1+
= 1045
.
KW
3 1 − ε DT
3 1 − 0.919 2.62
Therefore, KW = 0.957 .
From Eq. (6112), DP = 6 From Perry's Handbook, µ for air at 25oC = 0.018 cP, and uo = uV = 1.59 ft/s
From Eq. (6114), N ReV = uV DP ρV...
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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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