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Unformatted text preview: y) For the rectifying section, the limits on y are y = 0.37 at x = 0.1 to y = 0.80.
Using the trapezoidal method to solve the integral, ( NOG ) RS = ∆y
(0.42 − 0.37) (0.665 − 0.42) ( 0.80 − 0.665 )
=
+
+
= 12.4
0.084
0.063
0.017
( y *−y) (c) For 15.8 equilibrium stages at 80% tray efficiency, need 15.8/0.8 = 19.8 or 20 trays. If they
are on 18inch spacing, column height = 19(1.5) + 4 + 10 = 42.5 ft.
(d) For HOG = 1.2 ft, packed height = 1.2(2.4 + 12.4) = 18 ft. Add 14 ft for disengagement at the
top and sump at the bottom. Need 18 + 14 = 32 ft column.
(e) We would need data on individual values of HL and HG. From Eq. (6132),
1
H L is proportional to
DL 1/ 2 . Therefore, can ratio with the liquid diffusivities. From Eq. (6133), H G is proportional to ρV
µV −3 / 4 µV
ρV DV −1/ 3 . Therefore, we can ratio off these properties. Then from Eq. (7.52),
HOG = HG + mH L , where m = dy / dx , which varies over the equilibrium curve. Exercise 7.52
Subject: Doubling plant capacity for distillation of methanolwater by adding a packed column.
Given: Conditions to duplicate those of Exercise 7.41. Design for 70% of flooding. Two
packings to be considered: 1. 50mm plastic NOR PAC rings and 2. Montz metal B1300.
Assumptions: Constant molar overflow. Partial reboiler is an equilibrium stage.
Find: (a) Liquid holdup.
(b) Column diameter.
(c) HOG.
(d) Packed height.
(e) Pressure drop.
Advantages of packed column over trayed column. Preferred packing.
Analysis: First, convert the performance data for feed and product flow rates and compositions
from mass units into mole units, using molecular weights of 32.04 for methanol and 18.02 for
water. The results are as follows:
Component
Methanol
Water
Total: Flow rate, lbmol/h:
Feed
Distillate
709.1
702.7
1260.8
65.3
1969.9
768.0 Bottoms
6.4
1195.5
1201.9 Mole fraction:
Feed
Distillate
0.360
0.915
0.640
0.085
1.000
1.000 Bottoms
0.0053
0.9947
1.0000 The results of Exercise of 7.41 show that the observed reflux and boilup ratios (0.947 and 1.138,
respectively) are consistent with the separation achieved. Therefore, at the top of the column, L
= RD = 0.947(768) = 727 lbmol/h, and V = L + D = 727 + 768 = 1,495 lbmol/h.
At the bottom of the column,
V = VB B = 1138(1,2019) = 1,367.8 lbmol / h, and L = V + B = 1,367.8 + 1,2019 = 2,569.7 lbmol / h
.
.
.
Before computing liquid holdup, we need to compute column diameter.
(b) Column diameter based on conditions at the top of the column,
where T = 154oF and P = 14.7 psia = 1 atm.
Use Figs. 6.36a, b, and c to compute flooding velocity, following Example 6.13.
From the ideal gas law, ρV = PMV/RT = (1)(30.9)/[0.730(154 + 460)] = 0.0689 lb/ft3
For the liquid, which is mainly methanol, use ρL = 0.77 g/cm3 = 48 lb/ft3 and µ L = 0.35 cP.
LM L
The abscissa in Fig. 6.36a = X =
VM V ρV
ρL 1/ 2 727(30.9) 0.0689
=
1495(30.9)
48 1/ 2 = 0.0184 From Fig. 6.36a, Y at flooding = 0.21; from Fig. 6.36b, for ρwater/ ρL = 1/0.77 = 1.3, f{ρL } = 1.6;
and from Fig. 6.36c, for µL = 0.35 cP, f{µL } = 0.8.
From Table 6.8, FP = 14 for...
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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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