Unformatted text preview: lbf/ft
uµ ρ
Assume o C C = 0.01
σ i ∆ρ
(0.000924)(63.5 − 45.3)
σ ∆ρ
Therefore, uo = 0.01 i
= 0.01
= 0.177 ft/s
(0.000021)(45.3)
µ C ρC
U D mD / ρ D 21,000 / 63.5
=
=
= 0.288
The flooding correlation of Fig. 8.39 is too difficult
U C mC / ρC 52,000 / 45.3
to read for low values of UD/UC . Instead, use Eqs. (862) and (859).
(1)
From Eq. (862), with UC/UD = 1/0.289 = 3.46,
φ Df = 1+ 8 UC / U D 1/ 2 4 UC / U D − 1 −3 = 1 + 8 3.46 1/ 2 4 3.46 − 1 −3 = 0.239 From Eq. (859), using φD = (φD)f = 0.239,
UD
UC
UD
UC
+
=
+
= uo 1 − φ D = 0.177(1 − 0.239)
φ D 1 − φ D 0.239 1 − 0.239
which simplifies to, 4.184U D + 1314U C = 0.1347
.
(2)
Solving Eqs. (1) and (2) simultaneously, UC = 0.0534 ft/s and UD = 0.0154 ft/s
Therefore, at flooding, (UD + UC) = 0.0154 + 0.0534 = 0.0688 ft/s
Size the column for 50% of flooding. Therefore, (UD + UC) = 0.0688/2 = 0.0344 ft/s
Total volumetric flow rate = Q = 21,000/63.5 + 52,000/45.3 = 1,480 ft3/h = 0.411 ft3/s
Column crosssectional area = AC = Q/(UD + UC) = 0.411/0.0344 = 11.9 ft2
Column diameter = DC = (4AC/π)1/2 = [4(11.9)/3.14]1/2 = 3.9 ft Exercise 8.35
Subject: Diameter of a Karr column for extraction of benzoic acid (A) from water (C) by
toluene (S).
Given: Conditions in Exercise 8.33.
Find: Column diameter.
Analysis:
Volumetric flow rate of feed/raffinate = 500 gal/min
Volumetric flow rate of solvent/extract = 750 gal/min
Extract is dispersed.
The following properties are given in Exercise 8.30.
ρR = ρD = 0.860(62.4) = 53.7 lb/ft3 and ρE = ρC = 0.995(62.4) = 62.1 lb/ft3
µC = 0.95 cP or 0.000020 lbfs/ft2
Interfacial tension = σI = 22 dyne/cm or 0.0015 lbf/ft
uµ ρ
Assume o C C = 0.01
σ i ∆ρ
(0.0015)(62.1 − 53.7)
σ ∆ρ
= 0.01
= 0101 ft/s
.
Therefore, uo = 0.01 i
µ C ρC
(0.000020)(62.1)
U D QD 750
=
=
= 150
.
U C QC 500 From Fig. 8.39, U D + UC f uo = 0.27 Therefore, (UD +UC)f = 0.27 uo = 0.27(0.101) = 0.0273 ft/s
Size the column for 50% of flooding. Therefore, (UD + UC) = 0.0273/2 = 0.0137 ft/s
Total volumetric flow rate = Q =(500 + 750)(60)/7.48 = 10,030 ft3/h = 2.79 ft3/s
Column crosssectional area = AC = Q/(UD + UC) = 2.79/0.0137 = 204 ft2
Column diameter = DC = (4AC/π)1/2 = [4(204)/3.14]1/2 = 16.1 ft
From Table 8.2, maximum Karr diameter is 1.5 m or 4.9 ft. Therefore, need multiple Karr
πD 2 314(4.9) 2
.
columns in parallel of maximum crosssectional area = Ac max =
=
= 18.8 ft2
4
4
Therefore, need 204/18.8 = 11 units in parallel of 1.5 m diameter each.
Would probably be better to specify one RDC column of 16.1 ft diameter or 5 m , which is less
than the maximum diameter of 8 m. Exercise 8.36
Subject: HETS of an RDC column for extraction of acetic acid (A) from water (C) by isopropyl
ether (S).
Given: Conditions in Exercises 8.28, 8.30, and 8.34.
Find: HETS.
Analysis: From Fig. 8.40, for an interfacial tension of 13.5 dyne/c from Exercise 8.30,
HETS
= 4 with HETS and DT in inches.
1
DT/ 3
From Exercise 8.34, for the conditions of Exercise 8.28, DT = 2.8 ft or 33.6 inches.
Therefore, HETS = 4(33.6)1/3 = 12.9 in...
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 Spring '11
 Levicky
 The Land

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