Unformatted text preview: NOR PAC and FP = 33 for Montz. Analysis: (b) (continued) Exercise 7.52 (continued) 2
From a rearrangement of the ordinate of Fig. 6.36a, uo = Y g ρH2 O (L)
FP
ρV 1
f {ρ L } f {µ L } 32.2
62.4
1
= 341 (ft/s)2 and uo = 18.5 ft/s
14
0.0689 16(0.8)
.
32.2
62.4
1
2
For Montz,
uo = 0.21
= 145 (ft/s)2 and uo = 12.0 ft/s
33 0.0689 16(0.8)
.
For fraction of flooding = f = 0.7, uV = uo f = 18.5(0.7) = 13.0 ft/s for NOR PAC,
uV = uo f = 12.0(0.7) = 8.4 ft/s for Montz.
2
For NOR PAC, uo = 0.21 4VM V
From Eq. (6103), column diameter = DT =
fu0 πρV
4(1495 / 3600)(30.9)
For NORPAC, DT =
13(3.14)(0.0689) 1/ 2 4VM V
=
uV πρV 1/ 2 1/ 2 = 4.3 ft.
1/ 2 4(1495 / 3600)(30.9)
For Montz,
DT =
= 5.3 ft
8.4(3.14)(0.0689)
Column diameter based on conditions at the bottom of the column,
where T = 207oF and P = 15.5 psia = 1.05 atm.
Use Figs. 6.36a, b, and c to compute flooding velocity, following Example 6.13.
From the ideal gas law, ρV = PMV/RT = (1.05)(18.1)/[0.730(207 + 460)] = 0.039 lb/ft3
For the liquid, which is mainly water, use ρL = 1.00 g/cm3 = 62.4 lb/ft3 and µ L = 0.27 cP.
LM L
The abscissa in Fig. 6.36a = X =
VM V ρV
ρL 1/ 2 2569.7(18.1) 0.039
=
1367.8(18.1) 62.4 1/ 2 = 0.047 From Fig. 6.36a, Y at flooding = 0.17; from Fig. 6.36b, for ρwater/ ρL = 1/1 = 1, f{ρL }= 1;
and from Fig. 6.36c, for µL = 0.27 cP, f{µL } = 0.77.
From Table 6.8, FP = 14 for NOR PAC and FP = 33 for Montz.
g ρH2 O (L)
1
2
From a rearrangement of the ordinate of Fig. 6.36a, uo = Y
FP
ρV
f {ρ L } f {µ L }
32.2 62.4
1
= 812 (ft/s)2 and uo = 28.5 ft/s
14
0.039 1(0.77)
32.2 62.4
1
2
For Montz,
uo = 0.17
= 345 (ft/s)2 and uo = 18.6 ft/s
33 0.039 1(0.77)
For fraction of flooding = f = 0.7, uV = uo f = 28.5(0.7) = 20.0 ft/s for NOR PAC,
uV = uo f = 18.6(0.7) = 13.0 ft/s for Montz.
2
For NOR PAC, uo = 0.17 4VM V
From Eq. (6103), column diameter = DT =
fu0 πρV 1/ 2 4VM V
=
uV πρV 1/ 2 Analysis: (b) (continued) Exercise 7.52 (continued) 4(1367.8 / 3600)(18.1)
For NOR PAC, DT =
20(3.14)(0.039) 1/ 2 = 3.4 ft.
1/ 2 4(1367.8 / 3600)(18.1)
For Montz,
DT =
= 4.2 ft
13.0(3.14)(0.039)
(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:
Ch
Packing
a, m2/m3
a, ft2/ft3
ε
NOR PAC
86.8
26.5
0.947 0.651
Montz
300
91.4
0.930 0.482
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 = (727/3600)(30.9)/ [48(3.14)(DT)2/4] = 0.166/(DT)2 ft/s
2
0166 / DT 48
.
33900
u Lρ L
From Eq. (698), N Re L =
=
=
2
µ L a (0.35)(0.000672)a
DT a
2
2
u L a ( 0.166 / DT ) a
a
=
=
= 0.000856 4
g
32.2
DT
2 From Eq. (699), N FrL Packing DT, ft a, ft2/ft3 uL, ft/s N Re L N FrL NOR PAC
4.3
0.00898
26.5
69.2 6.63 x 105
Montz
5.3
0.00591
91.4
13.2 9.91 x 105
0.
0
Since both values of N Re L > 5, use Eq. (6101), ah / a = 0.85Ch N Re25 N Fr.1
L
L
For...
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 Spring '11
 Levicky
 The Land

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