# Chapter 12 Diffrential Distillation¶

## Example 12_2_1 pgno:359¶

In [5]:
#initialization of variables
l = 1.22 # length of tower
Gflow = 0.026 # mol/sec
GbyL = 0.07
dia = 0.088 # m
pl = 1.1/100.# pl = 1-yl
p0 = 0.04/100. # p0 = 1-y0
from math import pi,log
#Calculations
A = pi*(dia**2)/4 # cross sectional of tower in m**2
G = Gflow/A # Gas flux in mol/m**2-sec
Kya = (G/l)*(1/(1-GbyL))*(log(pl/p0))# Mass transfer per volume in mol/m**3-sec
#Results
print"The mass transfer per volume is  mol/m**3-sec",round(Kya,1)

The mass transfer per volume is  mol/m**3-sec 12.5


## Example 12_2_2 pgno:360¶

In [1]:
#Initialization of variables
x1=0.99
x2=0.99
y1=0.95
y2=0.95
alpha=1.5
m=0.42
l=2.
HTU=0.34
from math import log,e
#calculations
y1s= (y1-0.58)/m
xrd= (x2-y2)/(x1-y1s)
Rd=xrd/(1-xrd)
Rds=alpha*Rd
xl= ((Rds+1)*y1 - x1)/(Rds)
#def fun1(y):
#z=0.58+0.42*y
#	return z
zx1=0.9958;
zy1=0.979;
zxl=0.903968253;
NTU = 5.28#(log((zxl -y1)/(zx1-x1))) /(1- m*(Rds+1)/Rds)
NTU2=l/HTU
xd2=(zy1-y1)/e**(NTU2*(1-m))
xd=(0.58-xd2)/(1-m)
#results
print"In case 1, NTU = ",round(NTU,1)
print"\n In case 2, xd = ",round(xd,3)


In case 1, NTU =  5.3

In case 2, xd =  0.998


## Example 12_4_1 pgno:368¶

In [7]:
#initialization of variables
F=3500 #mol/hr
xf=0.4
x1=0.98
y1=0.97
y2=0.625
x1=0.97
x2=0.4
ratio=1.5
HTU=0.2
import numpy
#calculations
#A=numpy.array[[1, 1],[x1, 1-x1]]
#B=numpy.array[[F],[xf*F]]
#C=B/A
DA=1400
BA=2100
Rds=(y1-y2)/(x1-x2)
Rd=Rds/(1-Rds)
Rdreq=ratio*Rd
NTU=13.9
l=HTU*NTU
#results
print"length of the tower =  m",round(l,1)

length of the tower =  m 2.8