Unformatted text preview: ued) Exercise 7.41 (continued) (b) From Eq. (742), Eo = 13.3  66.7 log µ
Take the viscosity as that of the feed = 0.34 cP
Eo = 13.3  66.7 log (0.34) = 44.6%
This is poor agreement with the performance data.
(c) From Eq. (743), Eo = 50.3(αµ)0.226
At the feed composition, x = 0.36 and y = 0.71. Therefore, from Eqs (219) and (221 combined,
the relative volatility is, α = ( y/x)/[(1  y)/(1  x)] = (0.71/0.36)/(0.29/0.64) = 4.4 Exercise 7.41 (continued)
Analysis: Eo = 50.3[(4.4)(0.34)]0.226 = 45.9% Now correct for length of liquid path from Fig. 7.5. Column diameter = 6 ft. Assume length of
liquid path = 70% of column diameter = 0.7(6) = 4.2 ft. From Fig. 7.5, correction to be added =
10%. Therefore corrected Eo = 45.9 + 10 = 55.9%
This also appears to be low.
(d) From Eq. (656), NOG =  ln (1  EOV). Therefore, EOV = 1  exp(NOG)
Use Eqs. (662, (664), (666), and (667) as in Example 6.7. Carry out the calculations at the
bottom tray based on methanol diffusion. Conditions are: Molar flow rate, kmol/h
Molecular weight
Density, kg/m3 Gas
630
18.5
0.657 Liquid
1,192
18.0
940 Estimate liquid diffusivity of methanol in water at 212oF = 373 K. From Perry's Handbook, DL =
1.6 x 105 cm2/s at 25oC. The viscosity of water at 212oF = 0.25 cP. Using Eq. (339) to correct
for temperature and viscosity, DL = 1.6 x 105 (373/298)(1/0.25) = 8 x 105 cm2/s
Estimate gas diffusivity of methanol in water vapor from Eq. (336). T = 373 K
MAB = 2/[(1/32) + (1/18)] = 23. Using Table 3.1,
= 15.9 + 2.31(4) + 611 = 31.3,
.
= 131
.
V
V
methanol water 0.00143(373)1.75
DV =
= 0.30 cm2/s
1/ 2
1/ 3
1/ 3 2
(151 / 14.7)23 [31.3 + 131 ]
.
.
Required tray dimensions: DT = 6 ft,
A = 3.14(6)2/4 = 28.3 ft2 = 2.63 m2
Aa = 0.91 A = 0.91(2.63) = 2.39 m2 = 23,900 cm2, Lw = 42.5 in. = 1.08 m
Tray conditions: φε = 1  0.617 = 0.383,
qL = 47,300/(60)(8.33) = 94.6 gpm = 5,970 cm3/s
From Eq. (654), Cl = 0.362 + 0.317 exp[3.5(2)] = 0.362
From Eq. (651),
f = 0.40, 94.6
hl = 0.383 2.0 + 0.362
(42.5)(0.383) From Eq. (664), 2/3 = 1.21 in. = 3.08 cm2 t L = (3.08)(23,900)/5,970 = 12.3 s Analysis: (d) (continued)
The continuity equation is,
mV = U a Aa ρV , so U a , ft/s = From Eq. (665), tG = From below Eq. (667), Exercise 7.41 (continued) mV
630(18.5)(2.205) / 3600
=
= 6.8 ft/s = 2.07 m/s
Aa ρV 28.3(0.91)(0.657 /16.02) (0.617)(3.08)
= 0.024 s
(0.383)(6.8)(2.54)(12) 0.5
F=Ua ρV = 2.07(0.657)0.5 = 1.68 (kg/m)0.5/s From Eq. (667), k L a = 78.8(8 × 10 −5 ) 0.5 (168 + 0.425) = 1.48 s1
.
From Eq. (666), k G a = 1,030(0.30)1/ 2 0.40 − 0.842(0.40) 2
3.081/ 2 = 85.3 s1 From Eq. (663), N L = k L at L = 148(12.3) = 18.2
.
From Eq. (662), N G = k G at G = 85.3(0.024) = 2.05
From the vaporliquid equilibrium data at the bottom of the column,
Kmethanol = 0.156/0.0246 = 6.34
Absorption factor = KV/L = (6.34)(630)/1,192 = 3.35
1
1 ( KV / L)
1
3.35
=
+
=
+
= 0.488 + 0.184 = 0.672
N OG N G
NL
2.05 18.2
NOG = 1/0.672 = 1.488 From Eq. (661), Thus, the gasphase resistance is more importa...
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 Spring '11
 Levicky
 The Land

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