Unformatted text preview: NOR PAC, ah / a = 0.85(0.651)(69.2) 0.25 (6.63 × 10 −5 ) 0.1 = 0.61
For Montz, ah / a = 0.85(0.482)(13.2) 0.25 (9.91 × 10 −5 ) 0.1 = 0.31
From Eq. (697), the fractional liquid holdup is hL = 12
6.63 × 10 −5
For NOR PAC, hL = (12)
69.2 1/ 3 1/ 3 ( 0.61) 2/3 N FrL 1/ 3 N Re L = 0.0162 m3/m3 9.91× 10−5
2/3
For Montz,
hL = (12)
( 0.31) = 0.0241 m3/m3
13.2
These liquid holdups at the top of the column are quite small. ah
a 2/3 Exercise 7.52 (continued)
Analysis: (a) (continued)
Liquid holdup based on conditions at the bottom of the column.
By the continuity equation, superficial liquid velocity is,
uL = LML /ρLS = (2570/3600)(18.1)/ [62.4(3.14)(DT)2/4] = 0.264/(DT)2 ft/s
2
0.264 / DT 62.4
90800
u Lρ L
From Eq. (698), N Re L =
=
=
2
µ L a (0.27)(0.000672)a
DT a
2
2
u L a ( 0.264 / DT ) a
a
=
=
= 0.00216 4
g
DT
32.2
2 From Eq. (699), N FrL Packing DT, ft uL, ft/s a, ft2/ft3 N Re L NOR PAC
Montz 3.4
4.2 0.0228
0.0150 26.5
91.4 296
56.3 N FrL 4.28 x 104
6.39 x 104 0.
0
Since both values of N Re L are > 5, use Eq. (6101), ah / a = 0.85Ch N Re25 N Fr.1
L
L .
For NOR PAC, ah / a = 0.85(0.651)(296) 0.25 (4.28 × 10 −4 ) 0.1 = 1057
For Montz, ah / a = 0.85(0.482)(56.3) 0.25 (6.39 × 10 −4 ) 0.1 = 0.54
From Eq. (697), the fractional liquid holdup is hL = 12
4.28 × 10 −4
For NOR PAC, hL = (12)
69.2 1/ 3 1/ 3 (1.057 ) 2/3 N FrL
N Re L 1/ 3 ah
a 2/3 = 0.0269 m3/m3 6.39 × 10 −4
2/3
For Montz,
hL = (12)
( 0.54 ) = 0.0341 m3/m3
56.3
These liquid holdups at the bottom of the column are quite small.
mV
(c) From Eq. (7.52), HOG = HG +
H L , where we must use the slope, m, of the equilibrium
L
curve instead of a Kvalue because the equilibrium curve is curved.
At the top of the column:
Near a mole fraction, xD , of 0.9, m = (0.9780  0.9359)/0.10 = 0.42
Estimate HL from Eq. (6132), using the following properties and parameters:
NOR PAC
Montz
CL
1.080
1.165
3
3
hL , m /m
0.0162
0.0241
0.947
0.930
ε , m3/m3
uL, ft/s
0.00898
0.00591
2
3
a, m /m
86.8
300
ah, m2/m3
52.9
93 Exercise 7.52 (continued) Analysis: (c) at the top of the column (continued) Need an estimate of the diffusivity of methanol in water at high concentrations of methanol.
From Example 3.7, the diffusivity of methanol in water at a mole fraction of 0.8 and 25oC is 1.5
x 105 cm2/s. Use Eq. (339) to correct this for temperature to 170oF or 350 K and viscosity. For
methanol, viscosity at 25oC (298 K) = 0.55 cP and at 350 K, viscosity = 0.32 cP. Therefore,
350 0.55
DMeOH = 150 × 10−5
.
= 3.0 × 10−5 cm2/s
298 0.32
From Eq. (6132), for NOR PACK, using SI (instead of American Engineering) units,
11
HL =
CL 12 1/ 6 4hL ε
DL au L 1/ 2 uL
a 1/ 6 a
1
1
=
1.08 12
aPh 4(0.0162)(0.947)
(3.0 × 10−9 )(86.8)(0.00274) 1/ 2 0.00274
86.8 a
m
aPh
To compute (aPh /a), we need for the liquid phase, the Reynolds, Weber, and Froude
numbers from equations (6138), (6139), and (6140), respectively, based on the packing
hydraulic diameter from (6137).
= 0.174 d h = packing hydraulic diameter = 4 Reynolds number = N ReL , h = ε
0.947
=4
= 0.0436 m = 0.143 ft
86.8
a u L d h ...
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 Spring '11
 Levicky
 The Land

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