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Unformatted text preview: = 750/1,250 = 0.60 and φC = 0.40.
(a) We are given the following properties, Phase
Dispersed (extract)
Continuous (raffinate) Density, lb/ft3
53.7
62.1 Viscosity, cP
0.59
0.95 Diffusivity of A, cm/s
1.5 x 105
2.2 x 105 Also, interfacial tension = σI = 22.0 dyne/cm = 633,000 lb/h2
Density difference = ∆ρ =62.1  53.7 = 8.4 lb/ft3
Phase mixture density from Eq. (823) = ρM = 53.7(0.60) + 62.1(0.40) = 57.0 lb/ft3
From Eq. (824), mixture viscosity is,
µ
15µ D φ D
.
0.95
15(0.59)(0.60)
.
µ M = C 1+
=
1+
= 2.65 cP or 6.41 lb/fth
φC
µC + µ D
0.40
0.95 + 0.59
Compute minimum rpm from Eq. (825), where, farright dimensionless group is, µ2 σ
6.412 (633,000)
M
=
= 2.96 × 10 −19
Di5ρ M g 2 ( ∆ρ) 2
(2.63) 5 (57.0)(4.17 × 108 )(8.4) 2
2
N min ρ M Di
D
= 103 T
.
g∆ρ
Di 2 .76 φ 0.106
D µ2 σ
M
Di5ρ M g 2 ( ∆ρ) 2 0.084 = 103
. 7.9
2.63 2 .76 (0.60) 0.106 (2.96 × 10−19 ) 0.084 = 1004
. Exercise 8.33 (continued)
Analysis: (a) (continued)
g∆ρ
(4.17 × 108 )(8.4)
2
Therefore, N min = 1004
.
= 1004
.
= 235 × 106 (rph2)
.
ρ M Di
(57.0)(2.63)
and Nmin =4,840 rph or 81 rpm
(b) Determine the mixing vessel agitator Hp from Fig. 8.36 (b), which requires NRe.
Assume N = Nmin =4,840 rph
From Eq. (822), N Re = Di2 Nρ M (2.63) 2 (4,840)(57.0)
=
= 2.98 × 105
µM
6.41 From Fig. 8.36 (b), for a mixing vessel with baffles (thus, no vortex), Npo = 5.7
From Eq. (821), the power is given by,
N Po N 3 Di5ρ M (5.7)(4,840) 3 (2.63) 5 (57.0)
.
P=
=
= 1110 × 106 ftlbf/h or 5.6 Hp
8
4.17 × 10
gc
Compare this to the rule of thumb of 4 Hp/1,000 gallons.
Vessel volume = 334 ft3 or 2,500 gal.
Therefore, Hp = 4(2,500)/1,000 = 10 Hp, which is 80% higher.
(c) The Sauter (surfacemean) diameter, dvs , is given by Eq. (838) or (839), depending on the
Weber number. From Eq. (837),
3 N We = 2.63 (4,840) 2 (57.0)
Di3 N 2ρC
=
= 41,800
633,000
σi Since NWe > 10,000, Eq. (839) applies.
d vs = 0.39 Di N We −0 . 6 = 0.39(2.63)(41,800) −0.6 = 0.00173 ft or 0.53 mm (d) From Eq. (836), the interfacial surface area per unit volume of emulsion is,
a= 6φ D 6(0.60)
=
= 2, 080 ft2/ft3
d vs 0.00173 Exercise 8.33 (continued)
Analysis: (continued) 1 (e) Eq. (828) applies, where, KOD = 1
1
+
k D mk C
For the extract, which is the dispersed phase, Eq. (840 applies.
D
15 × 10 −5
.
= 188 × 10−3 cm/s or 0.22 ft/h
.
From a rearrangement of Eq. (840), k D = 6.6 D = 6.6
d vs
0.0527
For the raffinate, which is the continuous phase, Eq. (850) applies, with DC = 2.2 x 105 cm2/s or
8.53 x 105 ft2/h.
µC
(0.95)(2.42)
=
= 434
In Eq. (844), N Sc =
ρC DC (62.1)(8.53 × 10−5 )
In Eq. (850), N Re = In Eq. (850), N Fr = Di2 NρC (2.63) 2 (4,840)(62.1)
=
= 904,000
µC
(0.95)(2.42) Di N 2 (2.63)(4,840) 2
=
= 0.148
g
4.17 × 108 In Eq. (850), Di /dvs = 2.63/(0.00173 = 1,520
In Eq. (850, dvs/DT = 0.00173/7.9 = 2.19 x 104
In Eq. (850, N Eo = 2
ρ D d vs g (53.7)(0.00173) 2 (4.17 × 108 )
=
= 0.1058
633,000
σ From Eq. (850), N Sh C kd
= C vs = 1237 ×...
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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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