# Chapter 8: Balances on Nonreactive Processes¶

## Example 8.1-1, page no. 362¶

In [2]:
#Initialization of variables
import math
basis=100. #mol/s
ninAcv=66.9 #mol/s
noutAcv=3.35 #mol/s
noutAcl=63.55 #mol/s
ninN2=33.1 #mol/s
noutN2=ninN2
HinAcv=35.7 #Kj/mol
HoutAcv=32.0 #Kj/mol
HinN2=1.16 #Kj/mol
HoutN2= -0.10 #Kj/mol
#Calculations and printing:
deltaH=noutAcv*HoutAcv+ noutN2*HoutN2-ninN2*HinN2-ninAcv*HinAcv
Q=deltaH
print '%s %.2f' %("Heat transferred from condenser to achieve required cooling and condensation at the rate of (kW) ",Q)
raw_input('press enter key to exit')

Heat transferred from condenser to achieve required cooling and condensation at the rate of (kW)  -2322.84
press enter key to exit

Out[2]:
''

## Example 8.3-1, page no. 367¶

In [3]:
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate
mass=200.0 #kg
Ti=20.0 #C
Tf=150.0 #C

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \n ")
def Cv(T):
Cv=0.855+ T*9.42*math.pow(10,-4)
return Cv

Ucap, err= scipy.integrate.quad(Cv,Ti,Tf)  #integrate.quad is an inbult function used for definite integration
Q=mass*Ucap
print '%s %.3f' %("Heat Required (KJ) = ",Q)
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

Heat Required (KJ) =  24311.820
press enter key to exit

Out[3]:
''

## Example 8.3-2, page no. 369¶

In [4]:
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate
T1=20.0 #C
T2=100.0 #C
T3=90.0 #C
T4=30.0 #C
P=3.0 #bar
V=5.0 #L
R=0.08314 #L.bar/mol.K
T=363.0 #K
ndot=100.0 #mol/min

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \n ")
print("part 1")
def fun1(T):
fun1=0.02900+ T*0.2199*math.pow(10,-5) + math.pow(T,2) * 0.5723 *math.pow(10,-8) - math.pow(T,3) * 2.871 * math.pow(10,-12)
return fun1

Qdot=ndot*deltaH
print '%s %.3f' %("Heat Transferred (KJ/min) = ",Qdot)
print("part2")
def fun2(T):
fun2=0.02900+ T*0.2199*math.pow(10,-5) + math.pow(T,2) * 0.5723 *math.pow(10,-8) - math.pow(T,3) * 2.871 * math.pow(10,-12)-8.14* math.pow(10,-3)
return fun2

n=P*V/(R*T)
Q=n*deltaU
print '%s %.3f' %("Heat transferred (KJ) = ",Q)
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

part 1
Heat Transferred (KJ/min) =  233.238
part2
Heat transferred (KJ) =  -0.627
press enter key to exit

Out[4]:
''

## Example 8.3-3, page no. 371¶

In [5]:
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate
T1=430.0 #C
T2=100.0 #C
ndot=15.0 #Kmol/min

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \n ")
def fun(T):
fun=0.02894 + T* 0.4147 *math.pow(10,-5) + math.pow(T,2) * 0.3191 * math.pow(10,-8) - math.pow(T,3) * 1.965 * math.pow(10,-12)
return fun

Qdot=ndot*deltaH*math.pow(10,3) /60.
print '%s %.3f' %(" \n Rate of heat removal (KW) = ",Qdot)
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

Rate of heat removal (KW) =  -2494.904
press enter key to exit

Out[5]:
''

## Example 8.3-4, page no. 373¶

In [6]:
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate
x=0.6
T1=0.0
T2=400.0
ndot=150.0 #mol/h

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \n ")
def fun1(T):
fun1=0.04937 + T*13.92*math.pow(10,-5) - math.pow(T,2) *5.816*math.pow(10,-8) + math.pow(T,3) *7.280 * math.pow(10,-12)
return fun1

def fun2(T):
fun2=0.06803 +T*22.59*math.pow(10,-5) - math.pow(T,2) *13.11*math.pow(10,-8) + math.pow(T,3) *31.71 * math.pow(10,-12)
return fun2

def fun(T):
fun=x*fun1(T)+ (1-x)*fun2(T)
return fun

deltaH , err=scipy.integrate.quad(fun,T1,T2) #scipy.integrate.quad is an inbult function used for definite integration
print '%s %.3f' %(" \n Heat capacity of Mixture (KJ/mol) = ",deltaH)
Qdot=ndot*deltaH
print '%s %.3f' %(" \n Heat Required (KJ/h) = ",Qdot)
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

Heat capacity of Mixture (KJ/mol) =  34.890

Heat Required (KJ/h) =  5233.495
press enter key to exit

Out[6]:
''

## Example 8.3-5, page no. 374¶

In [7]:
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate
x=0.1 #CH4
T1=20.0
T2=300.0
Vdot=2000.0 #L/min

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \n ")
ndot=Vdot/22.4
def fun(T):
fun=0.03431 + T*5.469*math.pow(10,-5) + math.pow(T,2) *0.3661*math.pow(10,-8) + math.pow(T,3) *11*math.pow(10,-12)
return fun

print '%s %.3f' %("H1 (Kj/mol) = ",H1)
print("From Tables H2= -0.15 Kj/mol   ,   H3=8.17 Kj/mol")
H2= -0.15
H3=8.17
Qdot=(ndot*x*H1+ ndot*(1-x)*(H3-H2))/60.
print '%s %.3f' %("Heat Input (KW) = ",Qdot)
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

H1 (Kj/mol) =  12.112
From Tables H2= -0.15 Kj/mol   ,   H3=8.17 Kj/mol
Heat Input (KW) =  12.945
press enter key to exit

Out[7]:
''

## Example 8.4-1, page no. 379¶

In [9]:
#Initialization of variables
import math
import numpy
from numpy import linalg

mdot=1500.0 #g/min
M=32.0 #g/mol
deltaHv=35.3 #Kj/mol

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \n ")
Qdot=mdot*deltaHv/(M*60.)
print '%s %.3f' %("Rate of Heat transfer (KW) = ",Qdot)
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

Rate of Heat transfer (KW) =  27.578
press enter key to exit

Out[9]:
''

## Example 8.4-2, page no. 379¶

In [10]:
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate
deltaHv=28.85  #Kj/mol at 69 C
T1=25.0 #C
T2=69.0 #C
Cp=0.2163 #Kj/mol C
V=1.0 #L
P=7.0 #bar
D=0.659 #KG/L
M=86.17 #Kg
ndot=100.0 #mol/h
T3=300.0 #C

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
deltaHA=Cp*(T2-T1) + V*(1.013-P)*M/(D*math.pow(10,4))
print '%s %.3f' %(" \n deltaHA (Kj/mol) =",deltaHA)
deltaHD=deltaHv
print '%s %.3f' %(" \n deltaHD (Kj/Kg) =",deltaHD)
def fun1(T):
fun1=0.13744 + T*40.85*math.pow(10,-5) - math.pow(T,2) *23.92*math.pow(10,-8) + math.pow(T,3) *57.66*math.pow(10,-12)
return fun1

print '%s %.3f' %(" \n deltaHG (KJ/mol) = ",deltaHG)
Qdot=ndot*(deltaHA+deltaHD+deltaHG)/3600
print '%s %.3f' %(" \n rate of Heat supply (KJ/mol) = ",Qdot)
print(" \n In this problem we neglected V*deltaP as it is negligible ")
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

deltaHA (Kj/mol) = 9.439

deltaHD (Kj/Kg) = 28.850

deltaHG (KJ/mol) =  47.149

rate of Heat supply (KJ/mol) =  2.373

In this problem we neglected V*deltaP as it is negligible
press enter key to exit

Out[10]:
''

## Example 8.4-3, page no. 382¶

In [11]:
#Initialization of variables
import math
import numpy
from numpy import linalg

T1=337.9 #K
T2=473.0 #K
Tc=513.2 #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("using Trouton rule,")
deltaHvT1=0.109*T1
print("In this case, trouton rule gives a better estimate")
print("using Watson correction")
deltaHvT2=36.8*math.pow(((Tc-T2)/(Tc-T1)),0.38)
print '%s %.3f' %("Estimated value using Trouton rule (Kj/mol) = ",deltaHvT1)
print '%s %.3f' %(" \n Estimated value using watson correction (Kj/mol) = ",deltaHvT2)
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 Trouton rule,
In this case, trouton rule gives a better estimate
using Watson correction
Estimated value using Trouton rule (Kj/mol) =  36.831

Estimated value using watson correction (Kj/mol) =  21.029
press enter key to exit

Out[11]:
''

## Example 8.4-4, page no. 383¶

In [1]:
#Initialization of variables
import math
import numpy
from numpy import linalg

basis=1.0 #mol feed
x=0.684 #mole fraction Of B
y=0.4

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
A=([[1, 1],[x ,y]])
b=([[basis],[basis/2]])
C=numpy.dot(linalg.inv(A),b)
#Here We solved two linear equations simultaneously
nV=C[0,0]
nL=C[1,0]
H1=5.332
H2=6.340
H3=37.52
H4=42.93
Q=nV*x*H1 + nV*(1-x)*H2 + nL*y*H3 + nL*(1-y)*H4
print '%s %.3f' %(" \n Heat transferred (KJ) = ",Q)
print("The answer for this problem in Text is wrong")
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

Heat transferred (KJ) =  28.401
The answer for this problem in Text is wrong
press enter key to exit

Out[1]:
''

## Example 8.4-6, page no. 390¶

In [13]:
#Initialization of variables
import math
basis=1. #lbm dry air
Tin=80. #F Inlet temperature
Tout=51. #F Outlet temperature

#Calculations and printing:
print("For 80F and 80 percent RH, ha=0.018 lbm H20/lbm DA H1=38.8 Btu/lbm DA")
ha=0.018 #lbm H20/lbm DA
H1=38.8 #Btu/lbm DA
m1=basis*ha
print("\n For 51F and saturated, ha=0.0079 lbm H20/lbm DA H2=20.9 Btu/lbm DA")
ha2=0.0079 #lbm H20/lbm DA
H2=20.9 #Btu/lbm DA
m2=basis*ha2
print("\n Balance on H2O")
m3=m1-m2
Fraction=m3/m1
print '%s %.3f' %("\n Fraction H2O condensed = ",Fraction)
H3=basis*(Tout-32.)
Q=basis*H2+m3*H3-basis*H1
print("\n From psychrometric chart, Vh=13.0 ft^3/lbm DA")
Vbasis=basis*13.0
Qdot=Q*1000./Vbasis
print '%s %.2f' %("Rate at which heat must be removed (Btu/min) = ",Qdot)
raw_input('press enter key to exit')

For 80F and 80 percent RH, ha=0.018 lbm H20/lbm DA H1=38.8 Btu/lbm DA

For 51F and saturated, ha=0.0079 lbm H20/lbm DA H2=20.9 Btu/lbm DA

Balance on H2O

Fraction H2O condensed =  0.561

From psychrometric chart, Vh=13.0 ft^3/lbm DA
Rate at which heat must be removed (Btu/min) =  -1362.16
press enter key to exit

Out[13]:
''

## Example 8.5-1, page no. 397¶

In [14]:
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate
M1=35.6 #g/mol
M2=18.0 #g/mol
x=0.2
mdot=1000.0 #kg/h
T1=25.0 #C
T2=100.0 #C
T3=40.0 #C

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
nHCl=mdot*x*math.pow(10,3) /M1
nH2O=mdot*(1-x)*math.pow(10,3) /M2
def fun(T):
fun=29.13*math.pow(10,-3) - T*0.1341*math.pow(10,-5) +math.pow(T,2) *0.9715*math.pow(10,-8) - math.pow(T,3) *4.335*math.pow(10,-12)
return fun

r=nH2O/nHCl
print("From table B.11, deltaHa= -67.4 Kj/mol HCl")
deltaHa= -67.4
y=nHCl/(nHCl+nH2O)
Cp=0.73*mdot*4.184/nHCl
deltaHb=Cp*(T3-T1)
H2=deltaHa+deltaHb
Qdot=nHCl*(H2-H1)
print '%s %.3E' %("Heat (kJ/h) = ",Qdot)
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
From table B.11, deltaHa= -67.4 Kj/mol HCl
Heat (kJ/h) =  -3.451E+05
press enter key to exit

Out[14]:
''

## Example 8.5-2, page no. 400¶

In [15]:
#Initialization of variables
import math
import numpy
from numpy import linalg

x=0.05
y=0.4
mdot=1000.0 #lbm/h
Hf=10.0 #Btu/lbm
Hl= -17 #Btu/lbm
Hv=1138.0 #Btu/lbm

#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 sulphuric acid balance")
m2=x*mdot/y
print("\n Total Mass balance")
m1=mdot-m2
Qdot=m1*Hv+m2*Hl-mdot*Hf
print '%s %.3f' %(" Rate of Heat transfer (Btu/h) = ",Qdot)
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

sulphuric acid balance

Total Mass balance
Rate of Heat transfer (Btu/h) =  983625.000
press enter key to exit

Out[15]:
''

## Example 8.5-5, page no. 405¶

In [16]:
#Initialization of variables
import math
import numpy
from numpy import linalg

basis=100.0 #lbm/h
Hv=728.0 #Btu/lbm
Hl=45.0 #Btu/lbm
HF=100.0 #Btu/lbm
T=120.0 #F
xF=0.30

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
print("From figure 8.5-2, ")
xL=0.185
xV=0.89
mL=basis*((xV-xF)/(xV-xL))
mV=basis-mL
Qdot=mV*Hv + mL*Hl - basis*HF
print '%s %.3f' %("Rate of heat transfer (Btu/h) =",Qdot)
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
From figure 8.5-2,
Rate of heat transfer (Btu/h) = 5641.135
press enter key to exit

Out[16]:
''