Unformatted text preview: (850), N Fr = Di2 NρC (143) 2 (9,950)(45.3)
.
=
= 381,000
µC
(2.42) Di N 2 (143)(9,950) 2
.
=
= 0.34
4.17 × 108
g In Eq. (850), Di /dvs = 1.43/(0.00032/0.3048) = 1,360
In Eq. (850, dvs/DT = (0.00032/0.3048)/4.3 = 2.44 x 104
In Eq. (850, N Eo = 2
ρ D d vs g (63.5)(0.00032 / 0.3048) 2 (4.17 × 108 )
=
= 0.0752
σ
388,000 From Eq. (850), N Sh C kd
= C vs = 1237 × 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 (690)1/ 3 (381,000) 2 / 3 (0.22) −1/ 2 (0.34) 5/12 (1,360) 2 (2.44 × 10 −4 )1/ 2 0.0752 5/ 4 = 889 Exercise 8.32 (continued)
Analysis: (b) (continued)
kC = ( NSh )C DC
d vs = 889(7.75 ×10 −5 )
= 65.6 ft/h
(0.00032 / 0.3048) (c) To compute EMD from Eq. (833), KOD must be computed from Eq. (828).
That equation requires the slope of the equilibrium curve for acetic acid, m = dcC/dcD .
From Perry's Handbook, Section 15, the equilibrium constant under dilute conditions in massfraction composition units is KD = 0.429. Converting this to concentration units, mass/unit
volume, m =KD (ρS/ρC) = 0.429(45.3/63.5) = 0.306.
1
1
=
= 0.315 ft/h
From Eq. (828), KOD =
1
1
1
1
+
+
k D mk C 0.32 0.306(65.6)
Thus, mass transfer is controlled by the dispersed phase.
In Eq. (833), need a, V, and QD . From Exercise 8.28, a = 1,234 ft2/ft3 , V = 62 ft3, and
QD = 331 ft3/h. From Eq. (833),
KOD aV / QD
0.315(1,234)(62) / 331
E MD =
=
= 0.986 or 98.6%
1 + KOD aV / QD 1 + 0.315(1,234)(62) / 331
(d) To compute the fraction of solute extracted, the relation in Example 8.7 can not be used
because in that example, the extract is the dispersed phase. Here, the raffinate is the dispersed
phase. Therefore, derive a new equation. By material balance on the solute,
QD cD , in − cD , out = QC cC , out or QD 1 − cD , out
c
= QC C , out
cD , in
cD , in Therefore, f extracted = 1 − E MD = cD , in − cD , out
*
cD , in − cD , out cD , out QC cC , out
=
cD , in QD cD , in
1− = (1)
cD , out
cD , in cD , in − cD , out
=
cC , out
1 cC , out
cD , in −
1−
m
m cD , in (2) Combining Eqs. (1) and (2) and solving for fextracted ,
E MD
0.986
f extracted =
=
= 0.511 or 51.1% extracted
0.986 331
E MD QD
1+
1+
0.306 1,148
m QC Exercise 8.33
Subject: Design of mixer unit for extraction of benzoic acid (A) from water (C) by toluene (S).
Given: Conditions in Example 8.4. Properties of raffinate and extract phases. 6flatblade
impeller in a closed mixer unit with baffles. Extract phase is dispersed.
Find: (a) Minimum rpm of impeller for complete and uniform dispersion.
(b) Power requirement of agitator.
(c) Sauter mean diameter.
(d) Interfacial area.
(e) KOD.
(f) NOD.
(g) EMD.
(h) Fractional extraction of A.
Analysis: From Example 8.4, DT = 7.9 ft. Use Di = DT/3 = 7.9/3 = 2.63 ft.
Assume QR = QF = 500 gpm, and QE = QS = 750 gpm. Therefore, total Q = 500 + 750 = 1,250
gpm. Assume phase volume holdups in the mixing vessel are in proportion to the volumetric
flow rates. Therefore, with the extract phase dispersed, φ∆...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details