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Unformatted text preview: 595 ) (N )
0.75 We L , h −0.45 FrL , h ( 210 ) ( 0.0433) ( 0.000172 )
−0.2 0.75 −0.45 = 1.25 1
= 0.048 m
1.25 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.
DV = DAB = 0.00143T 1.75
2
P
(1/ M A ) + (1/ M B ) 1/ 2 ( ) 1/ 3 V A +( 0.00143(370)1.75 = 1/ 2 ) 1/ 3 V 2 B = 0.31 cm 2 /s 2
1/ 3
1/ 3 2
(13.1) + ( 31.2 )
(1/18) + (1/ 32)
Use the following additional properties and parameters, together with µV = 1.2 x 102 cP and ρV
uρ
µ
= 0.0370 lb/ft3 , where from Eqs. (6134) and (6135), N ReV = V V and N ScV = V
aµV
ρV DV
(1) CV
uV , m/s
(NSc)V
(NRe)V
From Eq. (6133), for NOR PAC, NOR PAC
0.322
6.10
0.65
3,470 Montz
0.422
3.96
0.65
652 Exercise 7.52 (continued) Analysis: (c) at the bottom of the column (continued)
From Eq. (6133), for NOR PAC, using SI units,
1
1/ 2 4ε
HG =
( ε − hL )
CV
a4
= 1/ 2 (N ) (N )
−3/ 4 ReV −1/ 3 ScV 1
1/ 2 4(0.947)
( 0.947 − 0.0269 )
0.322
86.84 1/ 2 uV a
DV aPh ( 3, 470 ) −3/ 4 ( 0.65 ) −1/ 3 (6.10)(86.8)
= 0.69 m
(0.34 × 10−4 )(44.2) For Montz, HG = 1
1/ 2 4(0.930)
( 0.947 − 0.0341)
0.422
3004 1/ 2 ( 652 ) −3/ 4 ( 0.65) −1/ 3 (3.96)(300)
= 0.073 m
(0.34 × 10−4 )(207) At the bottom of the column, vapor rate = V =1,368 lbmol/h and liquid rate = L = 2,570 lbmol/h.
Therefore, mV/L = (2.50)(1,368)/(2,570) = 1.33
mV
From Eq. (7.51), for NOR PAC, H OG = H G +
H L = 0.69 + (1.33)(0.50) = 1.36 m = 4.4 ft
L
mV
For Montz, H OG = H G +
H L = 0.073 + (1.33)(0.048) = 0.14 m = 0.45 ft
L
(d) The packed height is given by Eq. (6127), lT = HOGNOG dy
y −y
Apply this equation to the rectifying section and to the stripping section, using the McCabeThiele diagrams developed in Exercise 7.41. Calculate NOG for each section, using from Table 6.7 for EM diffusion, N OG = * For the rectifying section, the limits on y from the feed to the distillate are 0.64 to 0.915. The
difference, y*y is almost constant at an average value of approximately 0.065. Therefore,
(NOG)RS = ∆y/( y*y) = (0.9150.64)/0.065 = 4.2. Therefore,
Packed height for NOR PAC = (1.5)(4.2) = 6.3 ft
Packed height for Montz = (0.68)(4.2) = 2.9 ft
For the stripping section, after accounting for the partial reboiler, the limits on y from the
bottom to the feed are 0.05 to 0.68. The difference, y*y varies from 0.12 at y = 0.05, to 0.23 at y
= 0.17, to 0.19 at y = 0.40, to 0.09 at y = 0.59, to 0.03 at y = 0.68 Using the trapezoidal method to
solve the integral,
∆y
(0.17 − 0.05) (0.40 − 017) 0.59 − 0.40 (0.68 − 0.59)
.
N OG RS =
=
+
+
+
= 4.6
y *−y
0175
.
0.21
014
.
0.06 Analysis: (d) (continued) Exercise 7.52 (continued) Packed height for NOR PAC = (4.4)(4.6) = 20.2 ft
Packed height for Montz = (0.45)(4.6) = 2.1 ft
The total packed height for both sections are:
6.3 + 20.2 = 26.5 ft for NOR PAC
2.9 + 2.1 = 5.0 ft for Montz
However, these he...
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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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