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Unformatted text preview: herefore, overhead vapor rate also
is equal to 26,230 lbmol/h. The reflux rate = V  D = 26,230  14,490 = 11,740 lbmol/h
A more accurate calculation by CHEMCAD with an overall energy balance gives a reflux rate of
15,800 lbmol/h and an overhead vapor rate of 30,300 lbmol/h. But use the constant molar
overflow values.
(a) Column diameter at the bottom:
From a simulation program, at 262.5oF and 40 psia, ρV = 0.0994 lb/ft3 and ρL =57.9 lb/ft3.
V = 26,230 lbmol/h and L = 10,410 + 26,230 = 36,640 lbmol/h
In Fig. 6.24,
FLV LM L ρV
=
VMV ρ L 1/ 2 = 36,640(18.2) 0.0994
26,230(18.29) 57.9 1/ 2 = 0.058 From Fig. 6.24, for 24inch plate spacing, CF = 0.27 ft/s.
For Eqs. (640) and (642), use FHA = 1.0 and FF = 1.0. For water, σ = 51 dyne/cm
Therefore, FST = (51/20)0.2 = 1.21. From Eq. (6.42), C = 1.21(1)(1)0.37 = 0.45 ft/s. Analysis: (a) (continued) Exercise 7.47 (continued) ρ − ρV
From Eq. (640), U f = C L
ρV 1/ 2 57.9 − 0.0994
= 0.45
0.0994 1/ 2 = 10.9 ft/s Since FLV < 0.1, from below Eq. (644), Ad /A = 0.1. Assume f = 0.80.
4VM V
From Eq. (644), DT =
fU f π(1 − Ad / A)ρV 1/ 2 4(26, 230 / 3600)(18.2)
=
0.80(10.9)(3.14)(1 − 0.1)(0.0994) 1/ 2 = 14.7 ft Column diameter at the top:
From a simulation program, at 189oF and 33 psia, ρV = 0.1566 lb/ft3 and ρL =45.3 lb/ft3.
V = 26,230 lbmol/h and L = 11,740 lbmol/h
In Fig. 6.24,
FLV LM L ρV
=
VMV ρ L 1/ 2 = 11,740(34) 01566
.
26,230(34) 45.3 1/ 2 = 0.026 From Fig. 6.24, for 24inch plate spacing, CF = 0.38 ft/s
For Eqs. (640) and (642), use FHA = 1.0 and FF = 1.0. For methanol, σ = 17 dyne/cm
Therefore, FST = (17/20)0.2 = 0.97
From Eq. (6.42), C = 0.97(1)(1)0.38 = 0.37 ft/s.
From Eq. (640),
ρ − ρV
Uf =C L
ρV 1/ 2 = 0.37 45.3 − 01566
.
0.1566 1/ 2 = 6.28 ft/s Since FLV < 0.1, from below Eq. (644), Ad /A = 0.1. Assume f = 0.80.
From Eq. (644),
4VM V
DT =
fU f π(1 − Ad / A)ρV 1/ 2 4(26, 230 / 3600)(34)
=
0.80(6.28)(3.14)(1 − 0.1)(0.1556) 1/ 2 = 21.2 ft Therefore, column would be swaged.
(b) Sizing of reflux drum.
Assume a liquid residence time in the reflux drum of 5 minutes (0.0833 h), half full.
From Eq. (744),
VV = 2 LM L t 2(26,230)(34)(0.0833)
=
= 3,280 ft3
ρL
45.3 From Eq. (746),
V
DT = V
π 1/ 3 3, 280
=
3.14 From Eq. (746), H = 4DT = 4(10.1) = 40.4 ft 1/ 3 = 10.1 ft Exercise 7.48
Subject:
Given: Tray hydraulics for methanolwater separation'
Data from Exercise 7.41. Find: (a)
(b)
(c)
(d)
(e) Percent of flooding.
Tray pressure drop in psi.
Weeping potential.
Entrainment rate.
Downcomer froth height. Analysis: Use the material balance from Exercise 7.41. Make calculations at the top and
bottom trays.
(a) At the top of the column, the reflux rate is 0.947(768) = 727 lbmol/h. The overhead vapor
rate = 727 + 768 = 1,495 lbmol/h. Use the entrainment flooding correlation of Fig. 6.24, with
densities from a simulation program, where the abscissa is,
FLV LM L ρV
=
VM G ρ L 1/ 2 = 727(30.9) 0.0703
1,495(30.9) 47.2 1/ 2 = 0.0188 (1) From Fig. 6.24, for 24inch tray spacing, CF = 0.38 ft/s. Because FLV < 1, Ad /A = 0...
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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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