Chapter 7:Liquid-Liquid Extraction¶
Example 7.1,Page number:429¶
In [42]:
from pylab import *
#For water phase:
Cw=[1.23,1.29,1.71,5.10,9.8,16.90] #Chloroformm concentration in wt %
Aw=[15.80,25.6,36.0,49.30,55.7,59.60] #Acetone concentration in wt %
#For Chloroform phase
Cc=[70.0,55.7,42.9,28.4,20.4,16.9] #Chloroformm concentration in wt%
Ac=[28.70,42.10,52.70,61.30,61.00,59.6] #Acetone concentration in wt %
for i in range(0,6):
Cw[i]=Cw[i]/100 #Weight fraction
Aw[i]=Aw[i]/100 #Weight fraction
Cc[i]=Cc[i]/100 #Weight fraction
Ac[i]=Ac[i]/100 #Weight fraction
a1=plot(Cw,Aw)
a2=plot(Cc,Ac)
X=linspace(1,0)
Y=linspace(0,1)
a3=plot(X,Y)
xlabel("$Weight fraction of chloroform$")
ylabel("$Weight fraction of acetone$")
title("Equilibrium for water-chloroform-acetone at 298 K and 1 atm.\n\n")
a4=plot([Cw[1],Cc[1]],[Aw[1],Ac[1]],label='$Tie line$')
a5=plot([(Cw[2]+Cw[3])/2,(Cc[2]+Cc[3])/2],[(Aw[2]+Aw[3])/2,(Ac[2]+Ac[3])/2],label='$Tie line$')
a6=plot([Cw[5],Cc[3]],[Aw[5],Aw[1]],label='$Conjugate line$')
legend(loc='upper right')
Out[42]:
Example 7.2,Page number:433¶
In [1]:
#Variable declaration
# 'b'-solvent 'f'-feed 'r'-raffinate 'e'-extract 'c'-one of the # component in feed
F = 50 # [feed rate, kg/h]
S = 50 # [solvent rate, kg/h]
xcf = 0.6
xbf = 0
ycs = 0
ybs = 1.0
# The equilibrium data for this system can be obtained from Table 7.1 and # Figure 7.6
# Plot streams F (xcF = 0.6, xBF = 0.0) and S (yes = 0.0, yBs = 1.0). After # locating streams F and S, M is on the line FS its exact location is found # by calculating xcm from
#Calculation
xcm = (F*xcf+S*ycs)/(F+S)
# From figure 7.8
xcr = 0.189
xbr = 0.013
yce = 0.334
ybe = 0.648
M = F+S # [kg/h]
# From equation 7.8
E = M*(xcm-xcr)/(yce-xcr) # [kg/h]
R = M-E # [kg/h]
#Result
print"The extract and raffinate flow rates are",round(E,2),"kg/h and",round(R,2)," kg/h respectively\n"
print"The compositions when one equilibrium stage is used for the separation is",xcr,"and",xbr," in raffinate phase for component b and c respectively and",yce,"and",ybe,"in extract phase for component b and c respectively"
Example 7.4,Page number:439¶
In [1]:
#Variable declaration
# C-acetic acid A-water
# f-feed r-raffinate s-solvent
fd = 1000 # [kg/h]
xCf = 0.35 # [fraction of acid]
xAf = 1-xCf # [fraction of water]
# Solvent is pure
xAr = 0.02
yCs = 0
import math
from pylab import *
from numpy import *
print "Solution 7.4(a)\n"
# Solution(a)
# From Figure 7.15
xCMmin = 0.144
# From equation 7.11
Smin = fd*(xCMmin-xCf)/(yCs-xCMmin) # [kg/h]
#result
print"The minimum amount of solvent which can be used is",round(Smin),"kg/h"
print "Solution7.4(b)"
# Solution(b)
S = 1.6*Smin # [kg/h]
# From equation 7.11
xCM = (fd*xCf+S*yCs)/(fd+S)
# Data for equilibrium line
# Data_eqml = [xCeq yCeq]
Data_eqml =matrix([[0.0069,0.0018],[0.0141,0.0037],[0.0289,0.0079],[0.0642,0.0193],[0.1330,0.0482],[0.2530,0.1140],[0.3670,0.2160],[0.4430,0.3110],[0.4640,0.3620]])
# Data for operating line
# Data_opl = [xCop yCop]
Data_opl =matrix([[0.02,0],[0.05,0.009],[0.1,0.023],[0.15,0.037],[0.20,0.054],[0.25,0.074],[0.30,0.096],[0.35,0.121]])
a1=plot(Data_eqml[:,0],Data_eqml[:,1],label='$Equilibrium line$')
a2=plot(Data_opl[:,0],Data_opl[:,1],label='$Operating line$')
legend(loc='upper right')
xlabel("wt fraction of acetic acid in water solutions, xC")
ylabel("wt fraction of acetic acid in ether solutions, yC")
title('Mc-Cabe thiele Diagram')
# Now number of theoritical stages is determined by drawing step by step # stairs from xC = 0.35 to xC = 0.02
# From figure 7.16
# Number of theoritical stages 'N' is
N = 8
show(a1)
show(a2)
print"\nThe number of theoretical stages if the solvent rate used is 60 percent above the minimum is ",N
Example 7.5,Page number:444¶
In [3]:
#Variable declaration
# C-nicotine A-water B-kerosene
# F-feed R-raffinate S-solvent
F = 1000 # [feed rate, kg/h]
xAF = 0.99 # [fraction of water in feed]
# Because the solutions are dilute therefore
xCF = 0.01 # [fraction of nicotene in feed, kg nicotene/kg water]
xCR = 0.001 # [fraction of nicotene in raffinate, kg nicotene/kg water ]
m = 0.926 # [kg water/kg kerosene]
import math
print "Solution 7.5(a) "
# Solution(a)
yCS = 0 # [kg nicotene/kg water]
# Because, in this case, both the equilibrium and operating lines are # straight,if the minimum solvent flow rate Bmin is used, the concentration # of the exiting extract, yCmax, will be in equilibrium with xCF. Therefore
yCmax = m*xCF # [kg nicotene/kg kerosene]
A = F*xAF # [kg water/h]
# From equation 7.17
Bmin = A*(xCF-xCR)/(yCmax-yCS) # [kg kerosene/h]
#Result
print"The minimum amount of solvent which can be used is",round(Bmin),"kerosene/h"
print"\nSolution 7.5(b)"
# Solution(b)
B = 1.2*Bmin # [kg kerosene/h]
EF = m*B/A
Nt = math.log((xCF-yCS/m)/(xCR-yCS/m)*(1-1/EF)+1/EF)/math.log(EF)
#Result
print"The number of theoretical stages if the solvent rate used is 20 percent above the minimum is",round(Nt,2)
print "Solution7.5(c)"
# Solution(c)
Eme = 0.6 # [Murphree stage efficiency]
# from equation 7.20
Eo = math.log(1+Eme*(EF-1))/math.log(EF) # [overall efficiency]
Nr = Nt/Eo # [number of real stages]
#Result
print"The number of real stages required is",round(Nr)
Example 7.6,Page number:449¶
In [5]:
# C-styrene A-ethylbenzene B-diethylene glycol
F = 1000 # [kg/h]
XF = 0.6 # [wt fraction of styrene]
XPE = 0.9
XN = 0.1
# All above fractions are on solvent basis
# Equilibrium Data for Ethylbenzene (A)-Diethylene Glycol (B)-Styrene (C) at 298 K
# Data_eqm = [X Y]
# X - kg C/kg (A+C) in raffinate solution
# Y - kg C/kg (A+C) in extract solution
import math
from scipy.optimize import fsolve
from pylab import*
from numpy import*
Data_eqm=matrix([[0,0],[0.087,0.1429],[0.1883,0.273],[0.288,0.386],[0.384,0.48],[0.458,0.557],[0.464,0.565],[0.561,0.655],[0.573,0.674],[0.781,0.863],[0.9,0.95],[1,1]])
#Illustration 7.6(a)
# Solution(a)
# Minimum theoretical stages are determined on the XY equilibrium distribution diagram, stepping them off from the diagonal line to the equilibrium curve, beginning at XPE = 0.9 and ending at XN = 0.1
Data_opl=matrix([[0,0],[0.09,0.09],[0.18,0.18],[0.27,0.27],[0.36,0.36],[0.45,0.45],[0.54,0.54],[0.63,0.63],[0.72,0.72],[0.81,0.81],[0.90,0.90],[1,1]])
a1=plot(Data_eqm[:,0],Data_eqm[:,1],label='$Equilibrium line$')
a2=plot(Data_opl[:,0],Data_opl[:,1],label='$Operating line$')
legend(loc='upper left')
title('Equilibrium-distribution diagram and minimum number of stages')
xlabel("$X,kg C/kg (A+C) in raffinate solution$")
ylabel("$Y,kg C/kg (A+C) in extract solution$")
show(a1)
show(a2)
# Figure 7.20
Nmin = 9 # [number of ideal stages]
print"Ans.(a)The minimum number of theoretical stages are ",Nmin
#Illustration 7.6(b)
# Solution(b)
# Since the equilibrium-distribution curve is everywhere concave downward# ,the tie line which when extended passes through F provides the minimum
# reflux ratio
# From figure 7.19
NdeltaEm = 11.04
NE1 = 3.1
# From equation 7.30
# Y = R_O/P_E, external reflux ratio
Ymin = (NdeltaEm-NE1)/NE1 # [kg reflux/kg extract product]
print"Ans.(b)The minimum extract reflux ratio is",round(Ymin,3),"kg reflux/kg extract product"
#Illustration 7.6(c)
# Solution(c)
Y = 1.5*Ymin # [kg reflux/kg extract product]
# From equation 7.30
NdeltaE = Y*NE1+NE1
# From figure 7.19
NdeltaR = -24.90
# From figure 7.21
N = 17.5 # [number of equilibrium stages]
# From figure 7.19
# For XN = 0.1 NRN = 0.0083
NRN = 0.0083
# Basis: 1 hour
# e = [P_E R_N]
# Solution of simultaneous equation
def G(e):
f1 = F-e[0]-e[1]
f2 = F*XF-e[0]*XPE-e[1]*XN
return(f1,f2)
# Initial guess:
e = [600,300]
y = fsolve(G,e)
P_E = y[0] # [kg/h]
R_N = y[1] # [kg/h]
R_O = Y*P_E # [kg/h]
E_1 = R_O+P_E # [kg/h]
B_E = E_1*NE1 # [kg/h]
E1 = B_E+E_1 # [kg/h]
RN = R_N*(1+NRN) # [kg/h]
S = B_E+R_N*NRN # [kg/h]
print"Ans(c.)The number of theoretical stages are",N
print"The important flow quantities at an extract reflux ratio of 1.5 times the minimum value are\n\n"
print"PE =",round(P_E),"kg/h\n","RN =",round(R_N),"kg/h\n","RO =",round(R_O),"kg/h\n","E1 =",round(E_1),"kg/h\n","BE =",round(B_E)," kg/h\n","E1 =",round(E1),"kg/h\n","RN =",round(RN),"kg/h\n","S =",round(S),"kg/h\n"
Example 7.7,Page number:454¶
In [5]:
#Variable declaration
Ff = 1.89 # [cubic m/min]
Fs = 2.84 # [cubic m/min]
t = 2 # [min]
print "Solution 7.7(a)\n"
# Solution(a)
import math
Q = Ff+Fs # [total flow rate, cubic m/min]
Vt = Q*t # [cubic m]
# For a cylindrical vessel H = Dt
Dt = (4*Vt/math.pi)**(1.0/3.0) # [m]
H = Dt # [m]
#Result
print"The diameter and height of each mixing vessel is",round(Dt,1)," m and",round(H,1)," m respectively"
print"Solution 7.7(b) "
# Solution(b)
# Based on a recommendation of Flynn and Treybal (1955),
P = 0.788*Vt # [mixer power, kW]
#Result
print"The agitator power for each mixer is",round(P,2),"kW"
print"\nSolution 7.7(c)"
# Solution(c)
# Based on the recommendation by Ryan et al. (1959), the disengaging area # in the settler is
# Dt1*L1 = Q/a = Y
a = 0.2 # [cubic m/min-square m]
Y = Q/a # [square m]
# For L/Dt = 4
Dt1 = (Y/4)**0.5 # [m]
L1 = 4*Dt1 # [m]
#Result
print"The diameter and length of a settling vessel is",round(Dt1,2)," m and",round(L1,2)," m respectively"
print"Solution 7.7(d)"
# Solution(d)
# Total volume of settler
Vt1 = math.pi*Dt1**2*L1/4 # [cubic m]
tres1 = Vt1/Q # [min]
#Result
print"The residence time in the settling vessel is",round(tres1,1),"min"
Example 7.8,Page number:456¶
In [1]:
#Variable declaration
Ff = 1.61 # [flow rate of feed, kg/s]
Fs = 2.24 # [flow rate of solvent, kg/s]
t = 2*60 # [residence time in each mixer, s]
df = 998 # [density of feed, kg/cubic m]
uf = 0.89*10**-3 # [viscosity of feed, kg/m.s]
ds = 868 # [density of solvent, kg/cubic m]
us = 0.59*10**-3 # [viscosity of solvent, kg/m.s]
sigma = 0.025 # [interfacial tension, N/m]
g = 9.8 # [square m/s]
import math
Qf = Ff/df # [volumetric flow rate of feed, cubic m/s]
Qs = Fs/ds # [volumetric flow rate of solvent, cubic m/s]
# Volume fractions in the combined feed and solvent entering the mixer
phiE = Qs/(Qs+Qf)
phiR = 1-phiE
print"\nSolution7.8(a)\n"
# Solution(a)
import math
Q = Qf+Qs # [total flow rate, cubic m/s]
Vt = Q*t # [vessel volume, cubic m]
# For a cylindrical vessel, H = Dt
# Therefore,Vt = math.pi*Dt**3/4
Dt = (4*Vt/math.pi)**(1.0/3.0) # [ diameter, m]
H = Dt # [height, m]
Di = Dt/3 # [m]
#Result
print"The height and diameter of the mixing vessel are",round(Dt,3)," m and",round(H,3)," m respectively.\n"
print"The diameter of the flat-blade impeller is",round(Di,3)," m"
print"\nSolution (b)\n"
# Solution(b)
# For the raffinate phase dispersed:
phiD = phiR
phiC = phiE
deltad = df-ds # [kg/cubic m]
rowM = phiD*df+phiC*ds # [kg/cubic m]
uM = us/phiC*(1 + 1.5*uf*phiD/(us+uf)) # [kg/m.s]
# Substituting in equation 7.34
ohm_min=math.sqrt(1.03*phiD**0.106*g*deltad*(Dt/Di)**2.76*(uM**2*sigma/(Di**5*rowM*g**2*deltad**2))**0.084/(Di*rowM))*60 # [rpm]
#Result
print"The minimum rate of rotation of the impeller for complete and uniform dispersion.is",round(ohm_min),"rpm."
print"\nSolution 7.8(c)"
# Solution(c)
ohm = 1.2*ohm_min # [rpm]
# From equation 7.37
Re = ohm/60*Di**2*rowM/uM # [Renoylds number]
# Then according to Laity and Treybal (1957), the power number, Po = 5.7
Po = 5.7
# From equation 7.37
P = Po*(ohm/60)**3*Di**5*rowM/1000 # [kW]
# Power density
Pd = P/Vt # [kW/cubic m]
#Result
print"The power requirement of the agitator at 1.20 times the minimum rotation rate is",round(P,3)," kW."
print"Power density is",round(Pd,3),"kW/m^3"
Example 7.9,Page number:460¶
In [7]:
#Variable declaration
# From example 7.8
Di = 0.288 # [m]
sigma = 0.025 # [N/m]
ohm = 152.0*1.2/60.0 # [rps]
ds = 868.0 # [kg/cubic m]
phiD = 0.385
#Calculation
import math
# Therefore from equation 7.49
We = Di**3*ohm**2.0*ds/sigma # [Weber number]
# From equation 7.50
dvs = Di*0.052*(We)**-0.6*math.exp(4*phiD) # [m]
# Substituting in equation 7.48
a = 6*phiD/dvs # [square m/cubic m]
dvs=dvs*1000 #[mm]
#Result
print"The Sauter mean drop diameter and the interfacial area is",round(dvs,3)," mm and",round(a)," square m/cubic m respectively."
Example 7.10,Page number:461¶
In [8]:
#Variable declaration
Dd = 1.15*10**-9 # [molecular diffusivity of furfural in water, square m/s]
Dc = 2.15*10**-9 # [molecular diffusivity of furfural in toluene, square m/s]
m = 10.15 # [equilibrium distribution coefficient, cubic m raffinate/cubic m extract]
print"Solution7.10(a)\n"
# Solution(a)
# From example 7.8 and 7.9
dvs = 3.26*10**-4 # [m]
Shd = 6.6 # [sherwood number for dispersed phase]
# From equation 7.52
kd = Shd*Dd/dvs # [dispersed phase mass transfer coefficient, m/s]
#Result
print"The dispersed-phase mass-transfer coefficient is",round(kd,7)," m/s"
print"\n Solution 7.10(b) "
# Solution(b)
dd = 998
dc = 868 # [density of continuous phase, kg/cubic m]
uc = 0.59*10**-3 # [viscosity of continuous phase, kg/m.s]
ohm = 182.2 # [rpm]
g = 9.8 # [square m/s]
Di = 0.288 # [m]
sigma = 0.025 # [N/m]
phiD = 0.385
Dt = 0.863 # [m]
Scc = uc/(dc*Dc)
Rec = Di**2*ohm/60*dc/uc
Fr = Di*(ohm/60)**2/g
Eo = dd*dvs**2*g/sigma
# From equation 7.53
Shc=1.237*10**-5*Rec**(2.0/3.0)*Scc**(1.0/3.0)*Fr**(5.0/12.0)*Eo**(5.0/4.0)*phiD**(-1.0/2.0)*(Di/dvs)**2*(dvs/Dt)**(1.0/2.0)
# Therefore
kc = Shc*Dc/dvs # [continuous phase mass transfer coefficient, m/s]
#Result
print"The continuous-phase mass-transfer coefficient is",round(kc,5),"m/s"
print"Solution 7.10(c)"
# Solution(c)
a = 7065 # [square m/cubic m]
Vt = 0.504 # []
Qd = 0.097/60 # [cubic m/s]
Qc = 0.155/60 # [cubic m/s]
# From equation 7.40
Kod = kd*kc*m/(m*kc+kd) # [m/s]
# From equation 7.45
N_tod = Kod*a*Vt/Qd
# From equation 7.46
Emd = N_tod/(1+N_tod)
#Result
print"The Murphree dispersed phase efficiency is ",round(Emd,3)
print"Solution 7.10(d)"
# Solution(d)
# From equation 7.57
fext = Emd/(1+Emd*Qd/(m*Qc))
#Result
print"The fractional extraction of furfural is",round(fext,3)
Example 7.11,Page number:466¶
In [9]:
#Variable declaration
# Preliminary Design of an RDC
T = 293 # [K]
F1 = 12250 # [flow rate for dispersed organic phase, kg/h]
F2 = 11340 # [flow rate for continuous aqueous phase, kg/h]
d1 = 858 # [kg/cubic m]
d2 = 998 # [kg/cubic m]
n = 12 # [Equilibrium stages]
#Calculation
import math
Qd = F1/d1 # [cubic m/h]
Qc = F2/d2 # [cubic m/h]
# Assume that based on information in Table 7.5
# Vd+Vc = V = 22 m/h
V = 22 # [m/h]
# Therefore column cross sectional area
Ac = (Qd+Qc)/V # [square m]
# Column diameter
Dt = math.sqrt(4*Ac/math.pi) # [m]
# Assume that based on information in Table 7.5
# 1/HETS = 2.5 to 3.5 m^-1
# Therefore
HETS = 1.0/3.0 # [m/theoritical stages]
# Column height
Z = n*HETS # [m]
#Result
print"The height and diameter of an RDC to extract acetone from a dilute toluene-acetone solution is",Z," m and",round(Dt,2),"square m respectively"