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Unformatted text preview: . The value of FLV is obtained first,
using the conditions in Fig. 7.48.
MV = [187.6(58.1) + 176.4(72.2) + 82.5(86.2)]/446.5 = 68.9
ML = [112.4(58.1) + 223.6(72.2) + 217.5(86.2)]/553.5 = 74.8
From a simulation program, ρV = 1.13 lb/ft3, compared to 0.965 from the ideal gas law
and ρL = 33.2 lb/ft3. Also given, V = 446.5 lbmol/h and L = 553.5 lbmol/h
In Fig. 6.24, FLV LM L ρV
=
VMV ρ L 1/ 2 = 553.6(74.8) 113
.
446.5(68.9) 33.2 1/ 2 = 0.248 From Fig. 6.24, CF = 0.27 ft/s. From Eq. (6.42), C = (1)(1)(1)0.27 = 0.27 ft/s.
From Eq. (640),
ρ − ρV
Uf =C L
ρV 1/ 2 33.2 − 113
.
= 0.27
113
. 1/ 2 = 144 ft/s
. From Eq. (644),
4VMV
DT =
fU f πρV 1/ 2 4(446.5 / 3600)(68.9)
=
0.85(1.44)(314)(113)
.
. 1/ 2 = 2.81 ft Assume a liquid residence time, t, of 5 minutes (0.0833 h), half full
.
2 LM L t 2(553.5)(74.8)(0.0833)
=
= 208 ft3
From Eq. (744), vessel volume = VV =
ρL
33.2
4V
4(208)
Now compute the height, H, from Eq. (745). H = V2 =
= 33.6 ft
πDT 314(2.81) 2
.
Thus, H/D = 33.6/2.81 = 12.0. This is too large, so redimension the volume to give H/D = 4.
From Eq. (746),
V
DT = V
π 1/ 3 208
=
3.14 From Eq. (746), H = 4DT = 4(4.05) = 16.2 ft 1/ 3 = 4.05 ft Exercise 7.46
Subject: Sizing of a horizontal reflux drum. Given: Flow conditions leaving reflux drum in Fig. 7.49.
Find: Drum length and diameter.
Analysis:
Liquid rate leaving drum = (3 + 1)120 = 480 lbmol/h
Assume properties of pure nhexane. From Fig. 2.4, boiling point of nC6 at 1 atm = 150oF.
From Fig. 2.3, ρL = 0.63 g/cm3 = 39 lb/ft3
Assume a liquid residence time in the reflux drum of 5 minutes (0.0833 h), half full.
From Eq. (744),
2 LM L t 2(480)(86.2)(0.0833)
VV =
=
= 177 ft3
ρL
39
From Eq. (746),
V
DT = V
π 1/ 3 177
=
3.14 1/ 3 = 3.83 ft From Eq. (746), H = 4DT = 4(3.83) = 15.3 ft
Might use a drum of 4 ft diameter and 16 ft long. Exercise 7.47
Subject: Sizing of a sievetray distillation column separating methanol and water. Given: Column top and bottom conditions in Fig. 7.50.
Assumptions: Column feed is a bubblepoint liquid. Constant molar overflow.
Find: (a) Column diameters at top and bottom. Should column be swaged?
(b) Length and diameter of horizontal reflux drum.
Analysis: First, compute the liquid and vapor flow rates at the top and bottom of the column.
Molecular weight of distillate = 0.9905(32.04) + 0.0095(18.02) = 31.91
Distillate flow rate = D = 462,385/31.91 = 14,490 lbmol/h
Molecular weight of bottoms = 0.0101(32.04) + 0.9899(18.02) = 18.16
Bottoms flow rate = B = 188,975/18.16 = 10,410 lbmol/h
From these values and the given compositions, the overall material balance is:
Lbmol/h:
Component
Feed
Distillate Bottoms
Methanol
14,457
14,352
105
Water
10,443
138
10,305
Total: 24,900
14,490
10,410
From the reboiler duty, assuming the bottoms is pure water, can compute the boilup rate.
Heat of vaporization of water at 262.5oF = 937 Btu/lb = 16,885 Btu/lbmol
Boilup rate = V = 442,900,000/16,885 = 26,230 lbmol/h
Assume constant molar overflow with saturated liquid feed. T...
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 Spring '11
 Levicky
 The Land

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