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Unformatted text preview: L (0.00898)(0.143)(48)
=
= 262
[(0.35)(0.000672)]
ĀµL Take the surface tension of methanol at 154oF as Ļ = 17 dynes/cm = 0.001165 lbf/ft or
0.001165(32.2) = 0.0375 lbm/s2
2
u LĻ L d h ( 0.00898 ) (48)(0.143)
=
=
= 0.0148
Ļ
0.0375
2 Weber number = N WeL ,h ( 0.00898) = 0.0000175
u2
= L=
gd h 32.2(0.143)
2 Froude number = N FrL , h From (6136), ( ) (N aPh
ā1/ 2
= 1.5 ( ad h )
N Re L ,h
a ā0.2 = 1.5 [ (86.8)(0.0436) ] ā1/ 2 ) (N )
0.75 We L , h ā0.45 FrL , h ( 262 ) ( 0.0148) ( 0.0000175)
ā0.2 0.75 ā0.45 = 1.485 a
aPh Exercise 7.52 (continued)
Analysis: (c) at the top of the column (continued)
HL = 0.174 1
= 0.117 m
1.485 From Eq. (6132), for Montz, using SI units,
11
HL =
CL 12
= 0.0253 1/ 6 a
aPh 4hL Īµ
DL au L 1/ 2 uL
a a
1
1
=
1.165 12
aPh 1/ 6 4(0.0241)(0.930)
(3.0 Ć 10ā9 )(300)(0.00180) 1/ 2 0.00180
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 = Īµ
0.930
=4
= 0.0124 m = 0.0407 ft
a
300 uL d h Ļ L (0.00591)(0.0407)(48)
=
= 49.1
ĀµL
[(0.35)(0.000672)] 2
u LĻ L d h ( 0.00898 ) (48)(0.407)
=
= 0.00421
Ļ
0.0375
2 Weber number = N WeL ,h = ( 0.00898) = 0.0000615
u2
= L=
gd h 32.2(0.0407)
2 Froude number = N FrL ,h From (6136), ( aPh
ā1/ 2
= 1.5 ( ad h )
N Re L ,h
a
= 1.5 [ (300)(0.0124)] ) (N
ā0.2 ā1/ 2 HL = 0.0253 1
= 0.054 m
0.47 ) (N )
0.75 We L , h ā0.45 FrL , h ( 49.1) ( 0.00421) ( 0.0000615 )
ā0.2 0.75 ā0.45 = 0.47 a
aPh Exercise 7.52 (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 = 131 for water, and
V = 15.9 + 4( 2.31) + 611 = 312 for methanol
and molecular weights are 32 for methanol and 18 for water.
0.00143T 1.75
DV = DAB =
1/ 2
1/ 3
1/ 3 2
2
P
( V ) A + ( V )B
(1/ M A ) + (1/ M B ) 0.00143(350)1.75 = = 0.28 cm 2 /s 1/ 2 2
1/ 3
1/ 3 2
(13.1) + ( 31.2 )
(1/18) + (1/ 32)
Use the following additional properties and parameters, together with ĀµĻ = 1.1 x 102 cP and ĻV
uĻ
Āµ
= 0.0688 lb/ft3 , where from Eqs. (6134) and (6135), N ReV = V V and N ScV = V
aĀµV
ĻV DV
(1) CV
uV , m/s
(NSc)V
(NRe)V NOR PAC
0.322
3.96
0.36
4,560 Montz
0.422
2.56
0.36
850 From Eq. (6133), for NOR PAC, using SI units,
1
1/ 2 4Īµ
HG =
( Īµ ā hL )
CV
a4 1/ 2 (N ) (N )
ā3/ 4 ReV ā1/ 3 ScV 1
1/ 2 4(0.947)
=
( 0.947 ā 0.0162 )
0.322
86.84 1/ 2 uV a
DV aPh ( 4, 560 ) ā3/ 4 ( 0.36 ) ā1/ 3 (3.96)(86.8)
= 0.34 m
(0.28 Ć 10ā4 )(71.2) For Montz, 1
1/ 2 4(0.930)
HG =
( 0.947 ā 0.0241)
0.422
3004 1/ 2 ( 850 ) ā3/ 4 ( 0.36 ) ā1/ 3 (2.56)(300)
= 0.16 m
(0.28 Ć 10ā4 )(78) At the top of the column, vapor rate = V =1,495 lbmol/h and liquid rate = L = 727 lbmol/h.
Therefore, mV/L = (0.42)(1,495)/(727) = 0.864
mV
From Eq. (7.51), for NOR PAC, H OG = H G +
H L = 0.34 + (...
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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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