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Unformatted text preview: VM V ρ L
2
From Fig. 6.36, Y = 0.18. Therefore, uo = 1/ 2 (514 + 18.5)(18) 0.0734
=
(312)(28.3)
62.4 1/ 2 Yg ρH 2O
(0.18)(32.2) 62.4
=
= 88
FP ρV
56
0.0734 Therefore, flooding velocity = uo = 9.4 ft/s
For 50% of flooding, f = 0.5
4GM G
From Eq. (6103), DT =
fuo πρG 1/ 2 = 0.0373 4(312 / 3, 600)(28.3)
=
(0.5)(9.4)(3.14)(0.0734) 1/ 2 = 3.0 ft. Exercise 6.38 (continued)
Analysis: (continued)
Assume gas phase controls mass transfer rate. Then, from Eq. (689), lT = HOGNOG
From Eq. (693), for xin = 0, yin = 0.06, and yout = 0.00065,
ln 92.3
N OG = A −1
1
+
A
A
A −1
A (1) For the absorption factor, A = L/KV. Take L = 514 lbmol/h. Take V = 312 lbmol/h.
From Fig. 6.49, K = 0.82. Therefore, A = (514)/(0.82)((312) = 2.0
From Eq. (1),
2 −1 1
ln 92.3
+
2
2
= 7.7
N OG =
2 −1
2
Masstransfer data for 2inch metal Pall rings for the absorption of NH3 from air into water are
presented in Figs. 6.42 and 6.43. In the solutions to Exercise 6.34, these data are fitted to the
equation,
(3)
kGa = 43.4 F0.7 uL0.45 , s1
where the gas capacity factor = F = uV(ρV)0.5 = f uo (ρV)0.5 in kg1/2s1m1/2 and from the
continuity equation, uL = m/SρL in m/s or m3/m2s. Here, S = 3.14(3)2/4 = 7.07 ft2
For our case, F = (0.5)(9.4)(0.0734)0.5 = 1.27 lb1/2s1ft1/2 or 3.10 kg1/2s1m1/2
and uL = [(532.5)(18)/3,600)]/(7.07)(62.4) =0.006 ft/s = 0.00182 m3/m2s
Substitution of these values into Eq. (3) gives, kGa = 43.4(3.10)0.7(0.00182)0.45 = 5.6 s1
Note that this kGa is in concentration units for the driving force.
Therefore, from Table 6.7 by analogy to the liquid phase case,
HOG = HG = VMV/kGaSρV = (312/3,600)(28.3)/(5.6)(7.07)(0.0734) = 0.84 ft
Now, we must correct this value to 1inch metal Pall rings.
From Table 6.8, using interpolation when necessary, we have the following characteristics:
Packing
1.0inch Pall rings
2.0inch Pall rings a, m2/m3
223.5
112.6 ε
0.954
0.951 Ch
0.719
0.784 CV
0.336
0.410 Analysis: (continued) Exercise 6.38 (continued) From Eqs. (6136) to (6140), aPh/a is proportional to ε1/2/a . Therefore, aPh / a for 1.0inch Rings
0.954
=
aPh / a for 2.0inch Rings
0.951 1/ 2 112.6
= 0.505
223.5 From Eq. (6133), if we ignore holdup in the term (ε  hL), then, H G is proportional to Therefore, 1
a
1.25
CV a
aPh 1
0.954
H G for 1.0inch Rings 0.336 223.51.25
=
1
0.951
H G for 2.0inch Rings
0.410 112.61.25 This ratio should be about the same for HOG .
Therefore, for the 1.5inch rings, HOG = 0.84(1.03) = 0.87 ft.
From Eq. (2), column height = 7.7 (0.87) = 6.7 ft. 1
= 1.03
0.505 Exercise 6.39
Subject: Absorption of SO2 from air by water in a packed column.
Given: Mass velocity of entering gas = 6.90 lbmol/hft2 on SO2free basis, containing 80 mol%
air and 20 mol% SO2. Water enters at a mass velocity of 364 lbmol/hft2. Gas exits with 0.5
mol% SO2. Tower operates at 30oC and 2 atm. Equilibrium yx equation given for SO2. Tower
is packed with Montz B1200 metal structured packing.
Assumptions: No stripping of water. No absorption of air.
Find: (a) Molar material balance operating line.
(b) NOG
Analysis: (a) A material balance for the solute, SO2, from the top of the column to an
intermediate point gives:
Yn +1V ′ + X in L′ = YoutV ′ + X n L′ (1) Solving Eq. (1), the operating line is given by,
Y= L′
L′
X + Yout − X in
V′
V′ (2) From the given data, L'/V' = 364/6.90 = 52.75. Xin = 0.0 and Yout = 0.005/0.995 = 0.005025
Rearrange Eq. (2), letting Y = y/(1y) and X = x/(1x), which is approximately x be...
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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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