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Unformatted text preview: 7 lbmol/h, and V = L + D = 437 + 275 = 712
lbmol/h. At the bottom of the column, for constant molar overflow and 61.1 mol% feed vapor,
V = 712 − 0.611(450) = 437 lbmol / h, and L = V + B = 437 + 175 = 612 lbmol / h
Before computing liquid holdup, we need to compute column diameter.
(b) Column diameter based on conditions at the top of the column,
where T = 180oF and P = 14.7 psia = 1 atm.
Use Figs. 6.36a, b, and c to compute flooding velocity, following Example 6.13.
From the ideal gas law, ρV = PMV/RT = (1)(78.8)/[0.730(180 + 460)] = 0.171 lb/ft3
For the liquid, which is mainly benzene, use ρL = 50.6 lb/ft3 and µ L = 0.31 cP.
LM L
The abscissa in Fig. 6.36a = X =
VM V ρV
ρL 1/ 2 437(79.4) 0.171
=
712(78.8) 50.6 1/ 2 = 0.0376 From Fig. 6.36a, Y at flooding = 0.17; from Fig. 6.36b, for ρwater/ ρL = 62.4/50.6 = 1.23,
f{ρL } = 1.5; and from Fig. 6.36c, for µL = 0.31 cP, f{µL } = 0.8.
From Table 6.8, FP = 16 for 50mm Hiflow rings.
From a rearrangement of the ordinate of Fig. 6.36a,
g ρH2 O (L)
1
32.2 62.4
1
2
uo = Y
= 0.17
= 127 (ft/s)2 and uo = 11.2 ft/s
FP
ρV
f {ρ L } f {µ L }
16 0171 123(0.8)
.
.
For fraction of flooding = f = 0.7, uV = uo f = 11.2(0.7) = 7.84 ft/s.
From Eq. (6103), column diameter is,
4VM G
DT =
fu0 πρG 1/ 2 4VM G
=
uG πρG 1/ 2 4(712 / 3600)(78.8)
=
7.84(3.14)(0.171) 1/ 2 = 3.8 ft. Exercise 7.53 (continued) Analysis: (b) (continued)
Column diameter based on conditions at the bottom of the column,
where T = 227oF and P = 14.7 psia = 1 atm.
Use Figs. 6.36a, b, and c to compute flooding velocity, following Example 6.13.
From the ideal gas law, ρV = PMV/RT = (1)(90.6)/[0.730(227 + 460)] = 0.181 lb/ft3
For the liquid, which is mainly benzene, use ρL = 48.7 lb/ft3 and µ L = 0.25 cP.
ρV
ρL LM L
The abscissa in Fig. 6.36a = X =
VM V 1/ 2 612(90.6) 0.181
=
437(91.5) 48.7 1/ 2 = 0.085 From Fig. 6.36a, Y at flooding = 0.12; from Fig. 6.36b, for ρwater/ ρL = 62.4/48.7 = 1.28,
f{ρL } = 1.5; and from Fig. 6.36c, for µL = 0.25 cP, f{µL } = 0.78.
From Table 6.8, FP = 16 for 50mm Hiflow rings.
From a rearrangement of the ordinate of Fig. 6.36a,
g ρH2 O (L)
1
32.2 62.4
1
2
uo = Y
= 0.12
= 71 (ft/s)2 and uo = 8.43 ft/s
.
.
ρV
16 0181 150(0.78)
FP
f {ρ L } f {µ L }
For fraction of flooding = f = 0.7, uV = uo f = 8.43(0.7) = 5.9 ft/s.
From Eq. (6103), column diameter is,
4VM V
DT =
fu0 πρV 1/ 2 4VM V
=
uV πρV 1/ 2 4(437 / 3600)(90.6)
=
5.9(3.14)(0.181) 1/ 2 = 3.6 ft. (a) For liquid holdup estimates, assume the column operates in the preloading region.
Therefore, the holdup is independent of the gas rate. Follow Example 6.12. Pertinent packing
characteristics from Table 6.8 are a = 92.3 m2/m3 = 28.1 ft2/ft3, ε = 0.977, and Ch = 0.876
Use Eqs. (697) to (6101), which requires calculating the liquid Reynolds and Froude numbers. Liquid holdup based on conditions at the top of the column.
By the continuity equation, superficial liquid velocity is,
uL = LML /ρLS = (437/3600)(79.9)/ [50.6(3.14)(3.8)2/4] =0.0168 ft/s
0.0168 50.6
uρ
From Eq. (698), N Re L...
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 Spring '11
 Levicky
 The Land

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