In [11]:

```
#Intialization of variables
xD = 0.90 # Distillate mole fraction
xB = 0.15# Reboiler mole fraction
xF = 0.50 #Feed mole fraction
F = 10. # mol/sec
dg = 3.1*10**-3 # g/cc
dl = 0.65 # g/cc
C = 0.11 # m/sec
from math import pi
#Calculations
D = ((xF*F)-(xB*F))/(xD-xB)
B = ((xF*F)-(xD*F))/(xB-xD)
L = 3.5*D
G = L+D
L1 = L+F
G1 = G
f = (L1/G1)*((dg/dl)**0.5) # flow parameter
vG = C*(((dl-dg)/dg)**0.5)#vapor velocity in m/sec
c = (22.4*10**-3)*340/373
d = (4*G1*c/(vG*pi))**0.5#m
#Results
print"The column diameter is m",round(d,1)
#Calculation mistake in textbook
```

In [1]:

```
#Initialization of variables
y1 = 0.9999
x0 = y1 # For a total condenser
y0 =0.58 + 0.42*x0 # The equilbirum line
LbyG = 0.75
yNplus1 = 0.99
A = LbyG/0.42
n= 1
from math import log
#Calculations
xN = (yNplus1-((1-LbyG)*y1))/LbyG
yN = 0.58 + 0.42*xN
N = (log((yNplus1-yN)/(y1-y0))/log(A))+n#, number of stages
#Results
print"the number of stages approximately is ",round(N,3)
```

In [13]:

```
#initialization of variables
x0 = 0.0082
xB = 10**-4
L = 1
from math import log
#Calculations
y0 = 36*x0
#There are two balancing equations , mole fraction balance , mole balance , from them G is
G0 = (xB-x0)*L/(xB-y0)
G = 3*G0
B = L-G
y1 = ((L*x0)-(B*xB))/G
yNplus1 = 36*xB
xN = (L*x0 - (G*(y1-yNplus1)))/L
yN = 36*xN
N = (log((yNplus1-yN)/(y1-y0)))/log((yNplus1-y1)/(yN-y0))
#Results
print"The number of stages are ",round(N)
#Answer might be wrong in textbook
```

In [14]:

```
#initialization of variables
yn = 0.84
ynplus1 = 0.76
ystarn = 0.874
GA = 0.14 # kg-mol/sec
Al = 0.04 # m^3
#Calculations
Murphree = (yn-ynplus1)/(ystarn-ynplus1)
Kya = GA/(Al*((1/Murphree)-1))
#results
print"the murphree efficiency is",round(Murphree,1)
print"\n the m.t.c along with the product with a is kg-mol/m^3-sec",round(Kya,1)
```

In [15]:

```
#Initialization of variables
R = 82. # atm-cm**3/gmol-K
T = 273 + 60 # Kelvin
pk = 1 # atm
a1 = 440. # sec**-1 (of gas)
a2 = 1.7 #sec**-1 (of liquid)
ck = 1.5/((0.47*(76.1))+(0.53*(158.7)))
x = 0.2
Vs = 10. # litres
GA = 59. # gmol/sec
m = 1.41
from math import exp
#Calculations
k = (R*T)/(pk*a1) + (m/(ck*a2))
Kya = (1/k)*1000 # gmol/l-sec
Murphree = 1 - exp(-Kya*Vs/(GA))
#Results
print"The murphree efficiency is ",round(Murphree,2)
```