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Unformatted text preview: quid mole fraction, as given by y* = Kxb .
Solving Eq. (4) for the driving vapor and liquid phase driving forces,
Adding Eqs. (5) and (6) to eliminate yi and solving for r and equating to the last term in Eq. (4),
r= yb − y* = 1
K
+
kG acG k L acL Solving Eq. (7) for KGa,
KG a = yb − y *
1
K G acG 1
KcG
1
+
kG a k L acL The total gas concentration is obtained from the ideal gas law, (7) (8) Analysis: (d) continued) xercise 6.26 (continued)
cG = P
2(101.3)
=
= 81.8 mol/m3
−3
RT 8.314 × 10 (298) ρ L 106 g / m3
=
= 55,500 mol/m3
ML
18.02
1
1
KG a =
=
= 0.22 s1
1
150(81.8)
1.25 + 3.29
+
0.80 0.067(55,500)
This is with a concentration driving force.
cL = The liquid phase controls with a relative resistance of 3.29 compared to 1.25 for the gas phase.
To obtain KGa with a partial pressure driving force, we write the rate of mass transfer as,
r = K G acG ( yb − y* ) = ( K G ) p aP ( yb − y* ) (9) where KGa is with the concentration driving force = 0.22 s1 and (KG)pa is with the partial
pressure driving force. Solving Eq. (9), and applying the ideal gas law in the form, cG=P/RT,
(K G ) p a = K G a cG K G a
0.22
=
=
= 0.089 mol/kPam3s
P
RT (8.314 × 10−3 )(298) (e) From Table 6.7, with a partial pressure driving force, H OG = G
(K G ) p aPS However, the cross sectional area of the tower, S, is not known. Therefore, can not compute HOG
From Eq. (689), lT = HOGNOG . Therefore, can not compute the height because HOG is
unknown. However, we can compute the packed volume.
Packed volume = lTS = NOGHOGS = N OG G
(149)(454)
= (14.2)
= 14.8 m3
(K G ) p aP
0.089(3, 600)(2)(101.3) Exercise 6.27
Subject: Absorption of GeCl4 from air into dilute caustic solution in an existing packed column.
Given: Column operates at 25oC (77oF) and 1 atm. Gas enters at 23,850 kg/day containing 288
kg/day of GeCl4 and 540 kg/day of Cl2. Dissolved GeCl4 and Cl2 react with the caustic so that
neither has a vapor pressure. Packed tower is 2ft diameter with 10 ft of 1/2inch ceramic
Raschig ring packing of given characteristics. Liquid rate is to give 75% of flooding. Equation
is given for estimating Kya.
Assumptions: No stripping of water. No absorption of air.
Find: (a) Entering dilute caustic flow rate.
(b) Required packed height. Which controls, GeCl4 or Cl2? Is 10ft height adequate?
(c) % absorption of GeCl4 and of Cl2 for 10 ft of packing. If necessary, select an
alternative packing.
Analysis: (a) Determine entering dilute caustic flow rate from the 75% of flooding
specification, using the flooding curve of Fig. 6.36(a), with the correction factors of Figs.
6.36(b,c). The air rate in the entering gas = 23,850  288  540 = 23,022 kg/day.
MW of Cl2 = 71. MW of GeCl4 = 214.6. MW of air = 29.
23,022 288 540
Molar gas rate = V =
+
+
= 803 kmol/day or 73.7 lbmol/h or 2189 lb/h
29
214.6 71
Average molecular weight of gas = 23,850/803 = 29.7
Tower cross sectional area for 2ft diameter = 3.14(2)2/4 = 3.14 ft2
The continuity equation for flow through the tower based on the gas s...
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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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