Unformatted text preview: 0.864)(0.117) = 0.44 m = 1.5 ft
L Exercise 7.52 (continued)
Analysis: (c) at the top of the column (continued)
mV
For Montz, H OG = H G +
H L = 0.16 + (0.864)(0.054) = 0.21 m = 0.68 ft
L
At the bottom of the column:
Near a mole fraction, xB , of 0.005, m = 2.5
Estimate HL from Eq. (6123), using the following properties and parameters:
CL
hL , m3/m3
ε , m3/m3
uL, ft/s
a, m2/m3
ah, m2/m3 NOR PAC
1.080
0.0269
0.947
0.0228
86.8
91.7 Montz
1.165
0.0341
0.930
0.0150
300
162 Need an estimate of the diffusivity of methanol in water at low concentrations of methanol.
From Example 3.7, the diffusivity of methanol in water at infinite dilution at 25oC is 1.5 x 105
cm2/s. Use Eq. (339) to correct this for temperature to 207oF or 370 K and viscosity. For water,
viscosity at 25oC (298 K) = 0.90 cP and at 370 K, viscosity = 0.26 cP. Therefore,
370 0.90
DMeOH = 150 × 10 −5
.
= 6.4 × 10−5 cm2/s
298 0.26
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 a
1
1
=
aPh
1.08 12 1/ 6 4(0.0269)(0.947)
(6.4 ×10 −9 )(86.8)(0.0228) 1/ 2 0.0228
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).
= 0.456 d h = packing hydraulic diameter = 4 Reynolds number = N Re L , h = ε
0.947
=4
= 0.0436 m = 0.143 ft
a
86.8 u L d h ρ L (0.0228)(0.143)(62.4)
=
= 1120
µL
[(0.27)(0.000672)] a
aPh Exercise 7.52 (continued)
Analysis: (c) at the bottom of the column (continued)
Take the surface tension of water at 207oF as σ = 60 dynes/cm = 0.00411 lbf/ft or 0.00411(32.2)
= 0.132 lbm/s2
Weber number = N WeL ,h ( 0.0228 ) (62.4)(0.143) = 0.035
u 2ρ d
= L L h=
σ
0.132 Froude number = N FrL , h ( 0.0228 ) = 0.000113
u2
= L=
gd h 32.2(0.143) 2 2 From (6136), ( ) (N aPh
−1/ 2
= 1.5 ( ad h )
N Re L ,h
a −0.2 = 1.5 [ (86.8)(0.0436)] −1/ 2 HL = 0.456 ) (N )
0.75 We L , h −0.45 FrL , h (1120 ) ( 0.035 ) ( 0.000113)
−0.2 0.75 −0.45 = 0.92 1
= 0.50 m
0.92 From Eq. (6132), for Montz, using SI units,
11
HL =
CL 12
= 0.0595 1/ 6 a
aPh 4hL ε
DL au L 1/ 2 uL
a a
1
1
=
aPh
1.165 12 1/ 6 4(0.0341)(0.930)
(6.4 × 10 −9 )(300)(0.0150) 1/ 2 0.0150
300 m 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).
d h = packing hydraulic diameter = 4 Reynolds number = N ReL , h = ε
0.930
=4
= 0.0124 m = 0.0407 ft
a
300 u L d h ρ L (0.0150)(0.0407)(62.4)
=
= 210
µL
[(0.27)(0.000672)] a
aPh Exercise 7.52 (continued)
Analysis: (c) at the bottom of the column (continued) ( 0.0150 ) (62.4)(0.407) = 0.0433
u 2ρ d
= L L h=
σ
0.132
2 Weber number = N WeL ,h 2
( 0.0150 ) = 0.000172
uL
=
=
gd h 32.2(0.0407)
2 Froude number = N FrL , h ( aPh
−1/ 2
= 1.5 ( ad h )
N Re L ,h
a From (6136), ) (N
−0.2 = 1.5 [ (300)(0.0124)] −1/ 2 HL = 0.0...
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 Spring '11
 Levicky
 The Land

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