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Unformatted text preview: .0000631) HL = 0.24 −0.2 0.75 1
= 0.154 m
1.56 −0.45 = 1.56 a
aPh Exercise 7.53 (continued)
Analysis: (c) at the top of the column (continued)
Now estimate the value of HG from Eq. (6133). Estimate the gas diffusivity from Eq. (336),
where from Table 3.1,
.
V = 6(15.9 + 2.31) − 18.3 = 90.96 for benzene, and
V = 7(15.9 + 2.31) + 2.31 − 18.3 = 1115 for toluene
and molecular weights are 787 for benzene and 92 for toluene.
0.00143T 1.75
DG = DAB =
1/ 2
1/ 3
2
P
VA+
(1 / M A ) + (1 / M B ) 1/ 3 2
V B 0.00143(355)1.75 = 1/ 2 = 0.052 cm2 / s 2
1/ 3
90.96 + 1115
.
(1 / 78) + (1 / 92)
Use the following additional properties and parameters, together with µV = 0.0091 cP and ρV =
uρ
(7.84)(0.171)
0.171 lb/ft3 , where from Eq. (6134), N ReV = V V =
= 7,800
aµV (28.1)(0.0091)(0.000672)
µ
(0.0091)(0.000672)
From (6135) NScV = V =
= 0.639
ρV DV
0.052
(0.171)
(2.54) 2 (12) 2
From Eq. (6133), with CV = 0.408 from Table 6.8,
1/ 3 2 (1) 1
1/ 2 4ε
HG =
( ε − hL )
CV
a4 1/ 2 (N ) (N )
−3/ 4 ReV −1/ 3 ScV 1
1/ 2 4(0.977)
=
( 0.977 − 0.0273)
0.408
92.34 1/ 2 uV a
DV aPh ( 7,800 ) −3/ 4 ( 0.639 ) −1/ 3 (2.39)(92.3)
= 0.23 m
(0.052 × 10−4 )(144) At the top of the column, vapor rate = V =712 lbmol/h and liquid rate = L = 437 lbmol/h.
Therefore, V/L = (712)/(437) = 1.63 At the bottom of the column:
Near a mole fraction, xB , of 0.05, m = 2.1
Need an estimate of the liquid diffusivity of benzene in toluene at low concentrations of benzene.
At the bottom, temperature = 227oF = 382 K and liquid viscosity = 0.25 cP. This time, benzene
is component A and toluene is B. From Eq. (342),
T 1.29 (PB0.5 / PA0.5 )
3821.29 (24550.5 / 205.30.42 )
.
DL = DAB = 155 × 10−8
.
= 155 × 10−8
.
= 6.64 × 10−5
0.92 0.23
0.92
0.23
µB υ B
0.25 118.2 Exercise 7.53 (continued) Analysis: (c) bottom of the column (continued)
From Eq. (6132), using SI units 11
HL =
CL 12 1/ 6 1
1
=
1.168 12 4hL ε
DL au L 1/ 6 1/ 2 uL
a a
aPh 4(0.0312)(0.977)
(6.64 ×10 −9 )(92.3)(0.00911) 1/ 2 0.00911
92.3 a
a
= 0.26
aPh
aPh 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
0.977
=4
= 0.0423 m = 0.139 ft
a
92.3 uL d h ρ L (0.0299)(0.139)(48.7)
=
= 1205
[(0.25)(0.000672)]
µL Take the surface tension of toluene at 227oF as σ = 18 dynes/cm = 0.00123 lbf/ft or
0.00123(32.2) = 0.0397 lbm/s2
2
uL ρ L d h ( 0.0299 ) (48.7)(0.139)
=
= 0.152
0.0397
σ
2 Weber number = N WeL ,h = ( 0.0299 ) = 0.000200
u2
= L=
gd h 32.2(0.139)
2 Froude number = N FrL ,h From (6136), ( ) (N aPh
−1/ 2
= 1.5 ( ad h )
N Re L ,h
a −0.2 = 1.5 [ (92.3)(0.0423) ] −1/ 2 ) (N )
0.75 We L , h −0.45 FrL , h (1205 ) ( 0.152 ) ( 0.000200 ) HL = 0.26 −0.2 0.75 1
= 0.126 m
2.07 −0.45 = 2.07 Exercise 7.53 (continued) Analysis: (c) bottom of the column (continued) Now estimate the value of HG from Eq. (6133). Estimate the gas diffusivity from Eq. (336), by
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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