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µ
15µ D φ D
.
1.0
15(3.0)(0.22)
.
µ M = C 1+
=
1+
= 16 cP or 3.87 lb/fth
.
φC
µC + µ D
0.78
10 + 3.0
. Exercise 8.30 (continued)
Analysis: (c) (continued)
Compute minimum rpm from Eq. (825), where, farright dimensionless group is, µ2 σ
387 2 (388,000)
.
M
=
= 3.42 × 10 −16
5
2
2
5
Di ρ M g ( ∆ρ)
(143) (49.3)(4.17 × 108 )(18.2) 2
.
2
N min ρ M Di
D
= 1.03 T
g∆ρ
Di 2 .76 2
Therefore, N min = 20.88 φ 0.106
D µ2 σ
M
5
Di ρ M g 2 ( ∆ρ) 2 0.084 = 1.03 4.3
1.43 2 .76 (0.22) 0.106 (3.42 × 10−16 ) 0.084 = 20.88 g∆ρ
(4.17 × 108 )(18.2)
= 20.88
= 9.9 × 107 (rph2)
ρ M Di
(49.3)(143)
. and Nmin = 9,950 rph or 166 rpm
(d) Determine the mixing vessel agitator Hp from Fig. 8.36 (b), which requires NRe.
Assume N = Nmin = 9,950 rph
From Eq. (822), N Re = Di2 Nρ M (143) 2 (9,950)(49.3)
.
=
= 2.6 × 105
µM
387
. 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,
P= N Po N 3 Di5ρ M (5.7)(9,950) 3 (143) 5 (49.3)
.
=
= 4.0 × 106 ftlbf/h or 2 Hp
8
gc
4.17 × 10 Compare this to the rule of thumb of 4 Hp/1,000 gallons.
Vessel volume = 62 ft3 or 464 gal.
Therefore, Hp = 4(464)/1,000 = 1.9 Hp, which is in very good agreement. Exercise 8.31
Subject: Droplet characteristics for extraction of acetic acid (A) from water (C) by isopropyl
ether (S) in a mixer unit.
Given: Conditions in Exercise 8.27.
Find: (a) Sauter mean drop size.
(b) Range of drop sizes.
(c) Interfacial area of emulsion.
Analysis:
(a) The Sauter (surfacemean) diameter, dvs , is given by Eq. (838) or (839), depending on the
Weber number. From Eq. (837),
3 N We 143 (9,950) 2 (45.3)
.
Di3 N 2ρC
=
=
= 33,800
σi
388,000 Since NWe > 10,000, Eq. (839) applies.
d vs = 0.39 Di N We
(b) −0 . 6 = 0.39(143)(33,900) −0.6 = 0.00107 ft or 0.32 mm
.
dmin = dvs /3 = 0.32/3 = 0.107 mm
dmax = 3dvs = 3(0.32) = 0.96 mm (c) From Eq. (836), the interfacial surface area per unit volume of emulsion is,
a= 6φ D 6(0.22)
=
= 1, 234 ft2/ft3
d vs 0.00107 Exercise 8.32
Subject: Mass transfer characteristics for extraction of acetic acid (A) from water (C) by
isopropyl ether (S) in a mixer unit.
Given: Conditions in Exercises 8.27 and 8.28. Diffusivity of acetic acid in each phase.
Distribution coefficient of acetic acid.
Find: (a)
(b)
(c)
(d) Dispersedphase masstransfer coefficient.
Continuousphase masstransfer coefficient.
Murphree dispersedphase efficiency.
Fraction of A extracted. Analysis:
(a) For the raffinate, which is the dispersed phase, Eq. (840 applies.
Diffusivity of A in C = DD = 1.3 x 109 m2/s and from Exercise 8.28, dvs = 0.32 mm = 0.00032
m
DD
13 × 10−9
.
= 6.6
= 26.8 × 10 −6 m/s or 0.32 ft/h
From a rearrangement of Eq. (840), k D = 6.6
d vs
0.00032
(b) For the extract, which is the continuous phase, Eq. (850) applies.
Diffusivity of A in S = DC = 2.0 x 109 m2/s or 7.75 x 105 ft2/h
(10)(2.42)
.
µC
In Eq. (844), N Sc =
=
= 690
ρC DC (45.3)(7.75 × 10 −5 )
In Eq. (850), N Re =
In Eq....
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 Spring '11
 Levicky
 The Land

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