Unformatted text preview: nsity
difference computes to be 0.106, which is greater than 0.10, the required residence time in the
mixer should not exceed 5 minutes, Therefore, should be able to handle a feed rate of the acetic
acidwater mixture of (27/5)(6,660 + 23,600) = 163,400 lb/h or 27/5 = 5.4 times that in Fig. 8.1.
Next, consider the mixer agitator Hp factor.
The vessel volume in gallons = 5,872. Therefore, using the criterion of 4 Hp/1,000 gal, need
4(5,872)/1,000) = 23.5 Hp. This is greater than the given 20 Hp. So reduce allowable feed rate
of acetic acid in water to 20/23.5(163,400) = 139,000 lb/h. This increases residence time in
mixer to 5(23.5)/20 = 5.9 minutes.
Now, check settler disengaging factor.
Assume that the total volumetric flow leaving the settler is equal to that entering the mixer.
Therefore, Q = (27/5)(5/5.9)(217.1) = 994 gpm. Using the settling criterion of 5 gpm/ft2 of DTL,
Need DTL = 994/5 = 200 ft2. The available settler has DTL = 10(40) = 400 ft2, which is twice that
needed.
Therefore, a feed rate of acetic acid + water of 139,000 lb/h is safe. This is 4.6 times that in Fig.
8.1. Exercise 8.30
Subject: Sizing an agitator for extraction of acetic acid (A) from water (C) by isopropyl ether
(S) at 25oC.
Given: For one of the mixers, flow rates and properties of raffinate and extract are given.
Raffinate is dispersed phase in mixer, with residence time of 2.5 minutes.
Find: (a)
(b)
(c)
(d) Dimensions of closed, baffled mixing vessel.
Diameter of flatbladed impeller.
Minimum rate of rotation of impeller in rpm for complete and uniform dispersion.
Power of agitator at minimum rpm. Analysis: First, compute the volumetric flow rates of the extract and raffinate.
Extract flow rate = 52,000/45.3 = 1,148 ft3/h or 1,148(7.48)/60 = 143 gpm
Raffinate flow rate = 21,000/63.5 = 331 ft3/h or 331(7.48)/60 = 41 gpm
Total flow rate = Q = 1,148 + 331 = 1,479 ft3/h or 143 + 41 = 184 gpm
Assume this total flow also enters and leave the mixer.
(a) For a residence time of 2.5 minutes,
volume of mixing vessel = V = Qtres = (1,479/60)(2.5) = 62 ft3
Assume H = DT , then V = πDT2H/4 = πDT3/4
Solving, DT3 = 4(62)/3.14 = 79 ft3 and DT = H = 4.3 ft.
(b) Impeller diameter = Di = (1/3)DT = 4.3/3 = 1.43 ft.
(c) To determine the minimum impeller rpm, assume that the volume fractions of the two phases
in the mixer are in the same ratio as for the flow rates of the phases leaving the settler.
Therefore, the volume fractions are: φΕ = 1,148/1,479 = 0.78 and φR = 1  0.78 = 0.22. Because
the raffinate volume fraction is less than 0.26, the raffinate is probably the dispersed phase.
Therefore, φD = 0.22 and φC = 0.78. We are given the following properties, Phase
Dispersed (raffinate)
Continuous (extract) Density, lb/ft3
63.5
45.3 Viscosity, cP
3.0
1.0 Also, interfacial tension = σI = 13.5 dyne/cm = 388,000 lb/h2
Density difference = ∆ρ = 63.5  45.3 = 18.2 lb/ft3
Phase mixture density from Eq. (823) = ρM = 45.3(0.78) + 63.5(0.22) = 49.3 lb/ft3
From Eq. (824), mixture viscos...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details