# Chapter 14 Extraction¶

## Example 14_3_1 pgno:412¶

In :
from math import pi,log
#initialization of variables
Rat1 = (6.5/3)*(1-0.47)# as Rat = x0/y0
m = 0.14
H = (6.5*10**3)/3600. # Extract flow in g/sec
L = (3*10**3)/3600.# Solvent flow in g/sec
d= 10 # cm
A = 0.25*pi*d**2 # cm**2
l = 65 # cm
#Calculations and Results
Kya = ((H/(l*A))*(1/(1-((m*H)/L)))*(log((1-0.14*Rat1)/(0.47))))*10**3# kg/m**3-sec
print"The value of Kya is   kg/m**3-sec",round(Kya,2)
Rat2 = (6.5/3)*(1-0.1)#For case B
l2 = l*(log(1/((1-0.14*Rat2)/(0.1))))/(log(1/((1-0.14*Rat1)/(0.47))))/100# m
print"\nThe length for 90 percent recovery is  m",round(l2,1)

The value of Kya is   kg/m**3-sec 0.29

The length for 90 percent recovery is  m 2.2


## Example 14_4_1 pgno:415¶

In :
#Initialization of variables
m = 0.018
H = 450. # litres/hr
L = 37. # litres/hr
Ynplus1byY1 = 100.
from math import log
#Calculations
E =m*H/L
nplus1 = log((Ynplus1byY1*((1/E)-1))+1)/log(1/E)
n = nplus1 -1
print"The number of ideal stages are ",round(n,1)
N = 0.60#Murphree efficienct
E1 = (m*H/L) + (1/N) - 1
nplus1 = log((Ynplus1byY1*((1/E1)-1))+1)/log(1/E1)
n=nplus1-1
print"\nThe number of stages required if Murphree efficiency is 60 percent is ",round(n)

The number of ideal stages are  2.9

The number of stages required if Murphree efficiency is 60 percent is  21.0


## Example 14_5_1 pgno:419¶

In :
#initialization of variables
F = 5. #kg feed
S = 2. # kg solvent
E = F-S # kg extract
W = 1 # kg waste
EG = 80. # ppm
y0 = (100-99)/100. # mole fraction of gold left
y1 = y0*EG*W/S # concentration in raffinate
from math import log
#Calculations
xN = (EG*W - y1*S)/E # solvent concentration
xNminus1 = ((xN*(E+S)) - EG*W)/F#feed stage balance
N = 1 + ((log((xN-xNminus1)/(y1))/log(F/S)))#numner of stages including feed stage
#Results
print"The number of stages including feed stage is ",round(N,2)

The number of stages including feed stage is  5.03