# Chapter 6: Multiphase Systems¶

## Example 6.1-1, page no. 244¶

In :
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

T2=15.4+273.2  #K
T1=7.6+273.2  #K
P1=40.0 #mm of Hg
P2=60.0 #mm of Hg
T=42.2+273.2 #K
R=8.314 #J/mol.k

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
print("let deltaHv/R = S")
S= - (T1*T2* math.log(P2/P1))/(T1-T2)
deltahv=S*R
print '%s %d' %(" \n Latent Heat of Vaporization (J/mol) =",deltahv)
B=math.log(P1) + S/T1
print '%s %.3f' %("\n B=",B)
P=math.exp(-S/T + B)
print '%s %.3f %s %.3f' %("\n P* at",T," K is (mm Hg)= ",P)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook
let deltaHv/R = S

Latent Heat of Vaporization (J/mol) = 35023

B= 18.691

P* at 315.400  K is (mm Hg)=  207.400
press enter key to exit

Out:
''

## Example 6.3-1, page no. 250¶

In [ ]:
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

P=760.0 #mm of Hg
Pstar=289.0 #mm of Hg

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
y=Pstar/P
print '%s %.3f %s %.3f' %(" \n Molar composition of Water is",y," and Air is",1-y)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook

Molar composition of Water is 0.380  and Air is 0.620


## Example 6.3-2, page no. 251¶

In :
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

T=100.0+273.2 #K
PT=5260.0 #mm of Hg
y=0.1 #by volume
basis= 100 #mol of feed gas

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
P=y*PT
if(P<760):
print("The Vapour is Super heated")
elif(P==760):
print("The vapour is At Dew point")
else:
print("The vapour is not Super heated")

print("From tables Tdp=90 C")
print("Superheat = 100-90=10 C ")
print("Using Raoult law at the outlet")
y1=355/PT
print '%s %.3f' %("y1=",y1)
print("Balance on Dry Air")
n2=basis*(1-y)/(1-y1)
print '%s %.3f' %("n2 (mol) = ",n2)
print("Total mole balance")
n1=basis-n2
print '%s %.3f' %("n1 (mol) = ",n1)
print '%s %.3f' %(" \n Percentage condesation=",n1*100/(y*basis))
Psaturation=760/y
print '%s %.3f %s' %("\n Any increase in pressure above", Psaturation," mm of Hg must cause condensation ")
print(" \n For the next part of the problem use the same code by modifying PT to be 8500 mm of Hg")
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook
The Vapour is Super heated
From tables Tdp=90 C
Superheat = 100-90=10 C
Using Raoult law at the outlet
y1= 0.067
Balance on Dry Air
n2 (mol) =  96.514
Total mole balance
n1 (mol) =  3.486

Percentage condesation= 34.862

Any increase in pressure above 7600.000  mm of Hg must cause condensation

For the next part of the problem use the same code by modifying PT to be 8500 mm of Hg
press enter key to exit

Out:
''

## Example 6.3-3, page no. 254¶

In :
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

T=75.0 +273 #K
P75=289.0 #mm of Hg
hr=0.3
Porig=825.0 #mm of Hg
PorigBar=1.1 #bar
Vdot=1000.0 #M^3/h
R=0.0831

#Calculations and printing :

print (" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
P=hr*P75
y=P/Porig
ndot=PorigBar*Vdot/(R*T)
ndotWater=ndot*y
print '%s %.3f' %(" \n Molar flowrate of Water (Kmol/h) = ",ndotWater)
ndotBDA=ndot*(1-y)
print '%s %.3f' %(" \n Molar flowrate of Dry Air (Kmol/h) = ",ndotBDA)
ndotO2=ndotBDA*0.21
print '%s %.3f' %(" \n Molar flowrate of Oxygen (Kmol/h) = ",ndotO2)
hm=P/(Porig-P)
ha=hm*18/29
hmdot=P75/(Porig-P75)
hp=100*hm/hmdot
print '%s %.3f' %(" \n Molal Humidity (mol water/mol BDA) = ",hm)
print '%s %.3f' %(" \n Absolute Humidity (kg water/kg BDA) = ",ha)
print '%s %.3f' %(" \n Percentage Humidity=",hp)
print("\n From table B.3, Tdp=48.7 C")
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook

Molar flowrate of Water (Kmol/h) =  3.997

Molar flowrate of Dry Air (Kmol/h) =  34.040

Molar flowrate of Oxygen (Kmol/h) =  7.148

Molal Humidity (mol water/mol BDA) =  0.117

Absolute Humidity (kg water/kg BDA) =  0.073

Percentage Humidity= 21.780

From table B.3, Tdp=48.7 C
press enter key to exit

Out:
''

## Example 6.4-1, page no. 255¶

In :

#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

basis=100.0 #lb-mole/h
x=0.45
PH2O=31.6 #mm of Hg
PSO2=176.0 #mm of Hg
P=760.0 #mm of Hg
y=2.0
M1=64.0
M2=18.0

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
yH2O=PH2O/P
ySO2=PSO2/P
yAir=1-yH2O-ySO2
print("Using Air balance, ")
nG2=(1-x)*basis/yAir
print '%s %.3f' %("nG2 (lbm/h) = ",nG2)
xSO2=y/102
xH2O=1-xSO2
print("Using SO2 balance, ")
nL2=(basis*x-nG2*ySO2)*M1/(xSO2)
print '%s %.3f' %("nL2 (lbm/h) = ",nL2)
print("Using H2O balance, ")
nL1=nG2*yH2O*M2 + nL2*xH2O
print '%s %.3f' %("nL1 (lbm H2O/h) = ",nL1)
SO2Absorbed=nL2*xSO2
SO2Fed=basis*x*M1
Fraction=SO2Absorbed/SO2Fed
print '%s %.3f' %(" \n Fraction SO2 absorbed (lbm SO2 absorbed/lbm SO2 fed) = ",Fraction)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook
Using Air balance,
nG2 (lbm/h) =  75.670
Using SO2 balance,
nL2 (lbm/h) =  89683.186
Using H2O balance,
nL1 (lbm H2O/h) =  87981.325

Fraction SO2 absorbed (lbm SO2 absorbed/lbm SO2 fed) =  0.611
press enter key to exit

Out:
''

## Example 6.4-2, page no. 258¶

In :
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

basis=100.0 #lb-mole/h
x=0.45
PH2O=31.6 #mm of Hg
PSO2=176.0 #mm of Hg
P=760.0 #mm of Hg
y=2.0
M1=64.0
M2=18.0

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
yH2O=PH2O/P
ySO2=PSO2/P
yAir=1-yH2O-ySO2
print("Using Air balance, ")
nG2=(1-x)*basis/yAir
print '%s %.3f' %("nG2 (lbm/h) = ",nG2)
xSO2=y/102
xH2O=1-xSO2
print("Using SO2 balance, ")
nL2=(basis*x-nG2*ySO2)*M1/(xSO2)
print '%s %.3f' %("nL2 (lbm/h) = ",nL2)
print("Using H2O balance, ")
nL1=nG2*yH2O*M2 + nL2*xH2O
print '%s %.3f' %("nL1 (lbm H2O/h) = ",nL1)
SO2Absorbed=nL2*xSO2
SO2Fed=basis*x*M1
Fraction=SO2Absorbed/SO2Fed
print '%s %.3f' %(" \n Fraction SO2 absorbed (lbm SO2 absorbed/lbm SO2 fed) = ",Fraction)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook
Using Air balance,
nG2 (lbm/h) =  75.670
Using SO2 balance,
nL2 (lbm/h) =  89683.186
Using H2O balance,
nL1 (lbm H2O/h) =  87981.325

Fraction SO2 absorbed (lbm SO2 absorbed/lbm SO2 fed) =  0.611
press enter key to exit

Out:
''

## Example 6.5-1, page no. 264¶

In :
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

basis=150.0 #kg feed
S100=0.905 #g AgNO3/g
S20=0.689 #g AgNO3/g
inputx=0.095 #kg water/kg
outputx=0.311 #kg water/kg

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
print("\n Composition of Filter cake,")
print("\n m2=4m3")
print("\n Water balance around the crystallizer,")
print '%s' %("\n basis*inputx* kg H2O = outputx m1 + outputx m1 \n")
print("Mass balance around crystallizer, \n")
print '%d %s' %(basis,"=m1+m2+m3")
A=([[0, 1, -4],[outputx, 0, outputx],[1, 1, 1]])
b=([,[basis*inputx],[basis]])
C=numpy.dot(linalg.inv(A),b)
#Here we solved two linear equations simultaneously
m1=C[0,0]
print '%s %.2f' %(" \n m1 (Kg) = ",m1)
m2=C[1,0]
print '%s %d' %(" \n m2 (Kg) = ",m2)
m3=C[2,0]
print '%s %d' %(" \n m3 (Kg) = ",m3)
print("\n Overall AgNO3 balance,")
m5=(1-inputx)*basis - (1-outputx)*m1
print '%s %d' %("\n m5 (kg AgNO3 crystals recovered) = ",m5)
percentage=m5*100/(basis*(1-inputx))
print '%s %.3f' %(" \n Percentage recovery=",percentage)
print("\n Overall mass balance")
m4=basis-m1-m5
print '%s %d' %("\n m4 (Kg water removed in the Dryer) = ",m4)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook

Composition of Filter cake,

m2=4m3

Water balance around the crystallizer,

basis*inputx* kg H2O = outputx m1 + outputx m1

Mass balance around crystallizer,

150 =m1+m2+m3

m1 (Kg) =  19.77

m2 (Kg) =  104

m3 (Kg) =  26

Overall AgNO3 balance,

m5 (kg AgNO3 crystals recovered) =  122

Percentage recovery= 89.963

Overall mass balance

m4 (Kg water removed in the Dryer) =  8
press enter key to exit

Out:
''

## Example 6.5-2, page no. 266¶

In :
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

inputx=0.6
basis=100.0 #kg Feed
S=63.0 #Kg KNO3/100 Kg H2O

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
outputx=S/(S+100)
print '%s %.3f' %("x (Kg KNO3/Kg) = ",outputx)
print("Water balance")
m1=basis*(1-inputx)/(1-outputx)
print '%s %.3f' %(" \n m1 (Kg) = ",m1)
print("Mass balance")
m2=basis-m1
print '%s %.3f' %(" \n m2 (kg) = ",m2)
percentage=m2*100/(basis*inputx)
print '%s %.3f' %(" \n Percentage of KNO3 in the feed that crystallizes is ",percentage)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook
x (Kg KNO3/Kg) =  0.387
Water balance

m1 (Kg) =  65.200
Mass balance

m2 (kg) =  34.800

Percentage of KNO3 in the feed that crystallizes is  58.000
press enter key to exit

Out:
''

## Example 6.5-3, page no. 267¶

In :
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

basis=1.0 #Tonne Epsom salt produced/h
inputx=0.301 #Tonne MgSO4/tonne
outputx=0.232 #Tonne MgSO4/tonne
M=120.4
M1=246.4

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
print("Total mass balance")
print("m1=1+m2")
print("MgSO4 balance")
print(" \n inputx m1 = bassis* M/ M1 + m2 outputx")
A=([[1, -1],[inputx, -outputx]])
b=([,[basis*M/M1]])
C=numpy.dot(linalg.inv(A),b)
#Here we solved two linear equations simultaneously
m1=C[0,0]
m2=C[1,0]
print '%s %.3f' %(" \n m1 (Tonne/h) = ",m1)
print '%s %.3f' %(" \n m2 (Tonne/h) = ",m2)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook
Total mass balance
m1=1+m2
MgSO4 balance

inputx m1 = bassis* M/ M1 + m2 outputx

m1 (Tonne/h) =  3.719

m2 (Tonne/h) =  2.719
press enter key to exit

Out:
''

## Example 6.5-4, page no. 270¶

In :
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

m1=5.0 #g of solute
m2=100.0 #g of Water
P=1.0 #atm
Tf=100.421 #C
Ti=25.0 #C
R=8.314 #J/mol.K

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
x=(Tf-100.)*40656./(R*373.16*373.16)
y=m2/18.016
Ms=m1*(1-x)/(y*x)
print '%s %.3f' %(" \n Ms (g/mol) =",Ms)
deltaTm=R*273.16*273.16*x/6009.5
Tms=0-deltaTm
print '%s %.3f' %(" \n Tms (C)=",Tms)
Pstar=(1-x)*23.756
print '%s %.3f' %(" \n Solvent Vapour pressure (mm Hg) = ",Pstar)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook

Ms (g/mol) = 60.028

Tms (C)= -1.526

Solvent Vapour pressure (mm Hg) =  23.405
press enter key to exit

Out:
''

## Example 6.6-1, page no. 272¶

In :
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

V1=200.0 #CC Acetone
x=0.1 #Wt acetone
V2=400.0 #CC chloroform
DA=0.792 #g/cc
DC=1.489 #g/cc
DW=1.0 #g/cc

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
Dbar=DA*DW/(x*DW + (1-x)*DA)
mass1=V1*Dbar
mass2=V2*DC
print("C balance")
m4=mass2
print '%s %.3f' %(" \n m4=",m4)
print("W balance")
m2=(1-x)*mass1
print '%s %.3f' %(" \n m2=",m2)
print("A balance")
print("m1+m3=x * mass1")
print("Distribution Cooefficient ,K=m3*(m1+m2)/m1*(m3+m4)")
print("On solving, ")
m1=2.7
m3=16.8
percentage=m3*100/(x*mass1)
print '%s %.3f' %(" \n m1=",m1)
print '%s %.3f' %(" \n m3=",m3)
print '%s %.3f' %(" \n percentage of acetone transferred to chloroform=",percentage)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook
C balance

m4= 595.600
W balance

m2= 175.394
A balance
m1+m3=x * mass1
Distribution Cooefficient ,K=m3*(m1+m2)/m1*(m3+m4)
On solving,

m1= 2.700

m3= 16.800

percentage of acetone transferred to chloroform= 86.206
press enter key to exit

Out:
''

## Example 6.6-2, page no. 274¶

In :
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

basis=1000.0 #Kg of solution
inputxA=0.3 #Wt. fraction of acetone
outputxA1=0.05
outputxM1=0.02
outputxA2=0.1
outputxM2=0.87

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
print("Mass balance")
print("ms + basis = mE + mR")
print("acetone balance")
print("inputxA *basis = outputxA2 *mE + outputxA1* mR")
print("Water balance")
print(" \n 1-inputxA *basis = 1-outputxA2-outputxM2 *mE + 1-outputxA1-outputxM1* mR")
A=([[1, 1, -1],[outputxA2, outputxA1, 0],[1-outputxA2-outputxM2, 1-outputxA1-outputxM1, 0]])
b=([[basis],[inputxA*basis],[(1-inputxA)*basis]])
C=numpy.dot(linalg.inv(A),b)
#Here We solved three linear equations simultaneously
mE=C[0,0]
mR=C[1,0]
mS=C[2,0]
print '%s %.3f' %(" \n mE (Kg) = ",mE)
print '%s %.3f' %(" \n mR (Kg) = ",mR)
print '%s %.3f' %(" \n mS (Kg MIBK) = ",mS)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook
Mass balance
ms + basis = mE + mR
acetone balance
inputxA *basis = outputxA2 *mE + outputxA1* mR
Water balance

1-inputxA *basis = 1-outputxA2-outputxM2 *mE + 1-outputxA1-outputxM1* mR

mE (Kg) =  2666.667

mR (Kg) =  666.667

mS (Kg MIBK) =  2333.333
press enter key to exit

Out:
''

## Example 6.7-1, page no. 276¶

In :
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

V=50.0 #L
P=1.0 #atm
T=34.0+273.2 #K
y=0.3
xF=0.001
R=0.08206
Pstar=169.0 #mm of Hg
Pmm=760.0 #mm of Hg

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
n=P*V/(R*T)
print '%s %.3f' %(" \n No.of moles (mol) = ",n)
y0=y*Pstar/Pmm
print '%s %.3f' %(" \n Y0= (mol CCl4/mol) = ",y0)
Pfinal=xF*Pmm
b=0.096*Pfinal
Xstar=0.794*b/(1+b)
print '%s %.3f' %(" \n Mass of CCl4 adsorbed to Carbon at equilibrium (g CCl4 ads/g C) = ",Xstar)
Mads=(y0*n- xF*n)*154
print '%s %.3f' %(" \n Mass of CCl4 adsorbed (g) = ",Mads)
Mc=Mads/Xstar
print '%s %.3f' %(" \n Mass of carbon Required (g) = ",Mc)
raw_input('press enter key to exit')

 All the values in the textbook are Approximated hence the values in this code differ from those of Textbook

No.of moles (mol) =  1.983

Y0= (mol CCl4/mol) =  0.067

Mass of CCl4 adsorbed to Carbon at equilibrium (g CCl4 ads/g C) =  0.054

Mass of CCl4 adsorbed (g) =  20.071

Mass of carbon Required (g) =  371.750
press enter key to exit

Out:
''