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Unformatted text preview: 10−5 N Sc
.
DC 1/ 3 N Re 2/3 φ −1/ 2
D N Fr 5/12 Di
d vs 2 d vs
DT 1/ 2 N Eo 5/ 4 .
.
= 1.237 × 10 −5 (434)1/ 3 (904,000) 2 / 3 (0.60) −1/ 2 (0148) 5/12 (1,520) 2 (2.19 × 10−4 )1/ 2 01058
N Sh DC 5/ 4 = 1,052 1,052(8.53 × 10−5 )
= 518 ft/h
.
d vs
0.00173
The slope of the equilibrium curve for dilute conditions is given by Eq. (829) as m = dcC/dcD =
cC/cD = 1/21 = 0.0476
1
1
1
From Eq. (828), K OD =
=
=
= 0.202 ft/h
1
1
1
1
4.55 + 0.405
+
+
k D mkC 0.22 0.0476(51.8)
where, it is seen that the dispersed phase controls the rate of mass transfer.
kC = C = Exercise 8.33 (continued)
Analysis: (continued)
(f) From Eq. (832, N OD = K OD aV (0.202)(2, 080)(334)
=
= 26.3
QD
(750)(60) / 7.48 (g) From Eq. (833), E MD = N OD
26.3
=
= 0.963 or 96.3%
1 + N OD 1 + 26.3 (h) Because the extract is the dispersed phase, Eq. (3) in Example 8.7 applies, f extracted 750(0.963)
QD E MD
1
500
QC m
21
=
=
= 0.968 or 96.8% extracted
QD E MD
750(0.963)
1+
1+
1
QC m
500
21 Exercise 8.34
Subject: Diameter of an RDC column for extraction of acetic acid (A) from water (C) by
isopropyl ether (S).
Given: Conditions in Exercises 8.28 and 8.30.
Find: Column diameter.
Analysis:
Conditions of Exercise 8.28:
Flow rate of feed/raffinate = 12,400 lb/h. Flow rate of solvent/extract = 24,000 lb/h
Assume raffinate is dispersed.
The following properties are given in Exercise 8.30.
ρR = ρD = 63.5 lb/ft3 and ρE = ρC = 45.3 lb/ft3
µC = 1.0 cP or 0.000021 lbfs/ft2
Interfacial tension = σI = 13.5 dyne/cm or 0.000924 lbf/ft
uµ ρ
Assume o C C = 0.01
σ i ∆ρ
σ ∆ρ
(0.000924)(63.5 − 45.3)
Therefore, uo = 0.01 i
= 0.01
= 0.177 ft/s
µ C ρC
(0.000021)(45.3)
U D mD / ρ D 12,400 / 635
.
=
=
= 0.369
The flooding correlation of Fig. 8.39 is too difficult
U C mC / ρC 24,000 / 45.3
to read for low values of UD/UC . Instead, use Eqs. (862) and (859).
From Eq. (862), with UC/UD = 1/0.369 = 2.71,
(1)
φ Df = 1+ 8 UC / U D 1/ 2 4 UC / U D − 1 −3 = 1 + 8 2.71 1/ 2 4 2.71 − 1 −3 = 0.257 From Eq. (859), using φD = (φD)f = 0.257,
UD
UC
UD
UC
+
=
+
= uo 1 − φ D = 0.177(1 − 0.257)
φ D 1 − φ D 0.257 1 − 0.257
which simplifies to, 3891U D + 1346U C = 01315
.
.
.
(2)
Solving Eqs. (1) and (2) simultaneously, UC = 0.0473 ft/s and UD = 0.0175 ft/s
Therefore, at flooding, (UD + UC) = 0.0175 + 0.0473 = 0.0648 ft/s
Size the column for 50% of flooding. Therefore, (UD + UC) = 0.0648/2 = 0.0324 ft/s
Total volumetric flow rate = Q = 12,400/63.5 + 24,000/45.3 = 725 ft3/h = 0.201 ft3/s
Column crosssectional area = AC = Q/(UD + UC) = 0.201/0.0324 = 6.2 ft2
Column diameter = DC = (4AC/π)1/2 = [4(6.2)/3.14]1/2 = 2.8 ft Exercise 8.34 (continued)
Analysis: (continued)
Conditions of Exercise 8.30:
Flow rate of feed/raffinate = 21,000 lb/h. Flow rate of solvent/extract = 52,000 lb/h
Raffinate is dispersed.
The following properties are given in Exercise 8.30.
ρR = ρD = 63.5 lb/ft3 and ρE = ρC = 45.3 lb/ft3
µC = 1.0 cP or 0.000021 lbfs/ft2
Interfacial tension = σI = 13.5 dyne/cm or 0.000924...
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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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