# Chapter 9: Balances on Reactive Processes¶

## Example 9.9-1, page no. 443¶

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

ndot=2400.0 #mol/s
Hr1= -2878 #Kj/mol
HvWater=44.0 #Kj/mol
HvButane=19.2 #Kj/mol

#Calculations and print '%s %.3f' %ing :

print (" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
print ("Part 1")
E1= ndot/4
deltaH1=E1*Hr1
print '%s %.3E' %("enthalpy change (KJ/s) = ",deltaH1)
print ("part2")
Hr2=2*Hr1
E2=ndot/8
deltaH2=E2*Hr2
print '%s %.3E' %("Enthalpy change (kJ/s) = ",deltaH2)
print ("part 3")
Hr3=Hr1+5*HvWater+HvButane
deltaH3=E1*Hr3
print '%s %.3E' %("Enthalpy change (kJ/s) = ",deltaH3)
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
enthalpy change (KJ/s) =  -1.727E+06
part2
Enthalpy change (kJ/s) =  -1.727E+06
part 3
Enthalpy change (kJ/s) =  -1.583E+06
press enter key to exit

Out[1]:
''

## Example 9.1-2, page no. 445¶

In [2]:


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

deltaHr= -420.8 #kj/mol
R=8.314
T=298.0 #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("From reaction, only gaseous are counted")
left=1+2
right=1+1
deltaUr=deltaHr-R*T*(right-left)/math.pow(10,3)
print '%s %.3f' %("deltaUr (KJ/mol) = ",deltaUr)
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 reaction, only gaseous are counted
deltaUr (KJ/mol) =  -418.322
press enter key to exit

Out[2]:
''

## Example 9.3-1, page no. 447¶

In [3]:
#Initialization of variables
import math
import numpy
from numpy import linalg

HCO2= -393.5 #KJ/mol
HH2O= -285.84 #KJ/mol
HC5H12= -173. #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")
Hr=5*HCO2+6*HH2O-HC5H12
print '%s %.3f' %(" \n Heat of the rxn (KJ/mol) = ",Hr)
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 of the rxn (KJ/mol) =  -3509.540
press enter key to exit

Out[3]:
''

## Example 9.4-1, page no. 449¶

In [4]:
#Initialization of variables
import math
import numpy
from numpy import linalg

Hethane= -1559.9 #Kj/mol
Hethene= -1411 #Kj/mol
Hhydrogen= -285.84 #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")
Hr=Hethane-Hethene-Hhydrogen
print '%s %.3f' %(" \n Heat of the rxn (KJ/mol) = ",Hr)
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 of the rxn (KJ/mol) =  136.940
press enter key to exit

Out[4]:
''

## Example 9.5-1, page no. 453¶

In [5]:
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate
nNH3=100.0 #mol/s
nO2in=200.0 #mol/s
H1=8.470 #Kj/mol
H3=9.570 #Kj/mol
T1=25.0
T2=300.0
Hr= -904.7 #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")
def fun(T):
fun=29.50*math.pow(10,-3)+ T*0.8188*math.pow(10,-5) - math.pow(T,2) * 0.2925 *math.pow(10,-8) + math.pow(T,3) * 0.3652 * math.pow(10,-12)
return fun

H2, err=scipy.integrate.quad(fun,T1,T2)  #scipy.integrate.quad is an inbuilt function which can calculate definite integrals
E=nNH3/4.
nO2out=nO2in-nNH3*5./4.
nNO=nNH3
nH2O=nNH3*6./4.
deltaH=E*Hr+(nO2out*H1+nNO*H2+nH2O*H3)
Qdot=deltaH
print '%s %.3f' %(" \n Heat Transferred (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

Heat Transferred (kW) =  -19701.467
press enter key to exit

Out[5]:
''

## Example 9.5-2, page no. 454¶

In [6]:
#Initialization of variables
import math
import numpy
from numpy import linalg
import scipy
from scipy import integrate

NinMethane=100.0 #mol
NinOxygen=100.0 #mol
NinNitrogen=376.0 #mol
NoutMethane=60.0 #mol
NoutOxygen=50.0 #mol
NoutNitrogen=376.0 #mol
NoutFormal=30.0 #mol
NoutCarbon=10.0 #mol
NoutWater=50.0 #mol
H1= -74.85 #Kj/mol
H2= 2.235 #Kj/mol
H3= 2.187 #Kj/mol
H5=3.758#Kj/mol
H6=3.655 #Kj/mol
H8= -393.5+4.75 #Kj/mol
H9= -241.83+4.27 #Kj/mol
T1=25.0 #C
T2=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")
def fun1(T):
fun1=34.31*math.pow(10,-3)+ 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 fun1

H4= -74.85 + HoutMethane
def fun2(T):
fun2=34.28*math.pow(10,-3)+ T*4.268*math.pow(10,-5) - math.pow(T,3) * 8.694 * math.pow(10,-12)
return fun2

HoutFormal, err2=scipy.integrate.quad(fun2,T1,T2) #scipy.integrate.quadis an inbuilt function which can calculate definite integrals
H7= -115.90+ HoutFormal
deltaH=NoutWater*H9+NoutCarbon*H8+NoutFormal*H7+NoutNitrogen*H6+NoutOxygen*H5+NoutMethane*H4-NinNitrogen*H3-NinOxygen*H2-NinMethane*H1
Q=deltaH
print '%s %.3f' %(" \n Q (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

Q (KJ) =  -15296.233
press enter key to exit

Out[6]:
''

## Example 9.5-4, page no. 458¶

In [7]:
#Initialization of variables
import math
import numpy
from numpy import linalg

basis=150.0 #mol/s
x=0.9
HinEthanol= -212.19
HinEthanone= -147.07
HoutEthanol= -216.81
HoutEthanone= -150.9
HoutHydrogen=6.595
NinEthanol=135.0
NinEthanone=15.0
Q=2440.0 #KW

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
print("Carbon Balance")
print("basis*x *2 + basis* (1-x) *2=2 n1+2 n2")
print("Hydrogen Balance")
print("basis * x *6 + basis * (1-x)*4 = 6 n1+4 n2+2 n3")
print("Energy Balance")
print("Q= HoutEthanol n1  HoutEthanone n2 + HoutHydrogen n3 -NinEthanol * HinEthanol -NinEthanone* HinEthanone")
A=([[1, 1, 0],[3, 2, 1],[216.81, 150.9, -6.595]])
b=([[150],[435],[28412]])
C=numpy.dot(linalg.inv(A),b)
n1=C[0,0]
print '%s %.3f' %(" \n n1 (mol Ethanol/s) = ",n1)
n2=C[1,0]
print '%s %.3f' %(" \n n2 (mol Ethanone/s) = ",n2)
n3=C[2,0]
print '%s %.3f' %(" \n n3 (mol Hydrogen/s) = ",n3)
print("The solutions in the Text are Wrong")
fraction=(NinEthanol-n1)/NinEthanol
print '%s %.3f' %("Fractional conversion of Ethanol = ",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
Carbon Balance
basis*x *2 + basis* (1-x) *2=2 n1+2 n2
Hydrogen Balance
basis * x *6 + basis * (1-x)*4 = 6 n1+4 n2+2 n3
Energy Balance
Q= HoutEthanol n1  HoutEthanone n2 + HoutHydrogen n3 -NinEthanol * HinEthanol -NinEthanone* HinEthanone

n1 (mol Ethanol/s) =  91.957

n2 (mol Ethanone/s) =  58.043

n3 (mol Hydrogen/s) =  43.043
The solutions in the Text are Wrong
Fractional conversion of Ethanol =  0.319
press enter key to exit

Out[7]:
''

## Example 9.5-5, page no. 460¶

In [8]:
#Initialization of variables
import math
import numpy
from numpy import linalg

print("from tables,")
HfAcid= -1294. #Kj/mol
HfBase= -469.1 #Kj/mol
HfSalt= -1974. #Kj/mol
HfWater= -285.8 #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")
print("part 1")
Hr=HfSalt+3*HfWater-HfAcid-3*HfBase
print '%s %.3f' %("Hr of the rxn (KJ/mol) = ",Hr)
print("part 2")
deltaH=Hr*5./3.
print '%s %.3f' %("deltaH (KJ) = ",deltaH)
raw_input('press enter key to exit')

from tables,
All the values in the textbook are Approximated hence the values in this code differ from those of Textbook
part 1
Hr of the rxn (KJ/mol) =  -130.100
part 2
deltaH (KJ) =  -216.833
press enter key to exit

Out[8]:
''

## Example 9.5-6, page no. 461¶

In [9]:
import math
import numpy
from numpy import linalg

x=0.1
y=0.2
MAcid=98.1
MS=32.0
MSalt=142.0
MBase=40.0
MWater=18.0
MNa=46.0
basis=1000.0 #g
T2=35.0
T1=25.0
T3=40.0

#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 S balance, ")
m2=basis*x*MS*MSalt/(MAcid*MS)
print '%s %.3f' %(" \n m2  (g Na2SO4) = ",m2)
print("Using Na balance, ")
m1=2*MNa*m2*MBase/(y*MNa*MSalt)
print '%s %.3f' %(" \n m1  (g NaOH) = ",m1)
print("Total mass balance, ")
m3=basis+m1-m2
print '%s %.3f' %(" \n m3  (g H2O) = ",m3)
print '%s %.3f' %(" \n Mass of product solution = ",m2+m3)
m=m2+m3
Water=m2*2/MSalt
print '%s %.3f' %(" \n Water Formed in the reaction  (mol H2O) = ",Water)
print("H2SO4(aq):")
a1=basis*(1-x)/MWater
b1=basis*x/MAcid
rAcid=a1/b1
print '%s %.3f' %(" \n rAcid  (mol Water/mol Acid) = ",rAcid)
print("NaOH(aq):")
a2=m1*(1-y)/MWater
b2=m1*y/MBase
rBase=a2/b2
print '%s %.3f' %(" \n rBase  (mol Water/mol Base) = ",rBase)
print("Na2SO4(aq):")
a3=m3/MWater
b3=m2/MSalt
rSalt=a3/b3
print '%s %.3f' %(" \n rSalt  (mol Water/mol Salt) =",rSalt)
E=b1
print '%s %.3f' %(" \n Extent of reaction  (mol) = ",E)
nHAcid=basis*3.85*(T3-T1)/1000
nHSalt=m*4.184*(T2-T1)/1000
nHBase=0
HfSalt= -1384
HfAcid= -884.6
HfBase= -468.1
HfWater= -285.84
deltaHr=HfSalt+ 2*HfWater - HfAcid - 2*HfBase
print '%s %.3f' %(" \n Entahlpy change in the rxn  (KJ/mol) = ",deltaHr)
Q=E*deltaHr + (nHSalt-nHAcid-nHBase)
print '%s %.3f' %(" \n Q of the rxn  (KJ) = ",Q)
print("The answer in the 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
Using S balance,

m2  (g Na2SO4) =  144.750
Using Na balance,

m1  (g NaOH) =  407.747
Total mass balance,

m3  (g H2O) =  1262.997

Mass of product solution =  1407.747

Water Formed in the reaction  (mol H2O) =  2.039
H2SO4(aq):

rAcid  (mol Water/mol Acid) =  49.050
NaOH(aq):

rBase  (mol Water/mol Base) =  8.889
Na2SO4(aq):

rSalt  (mol Water/mol Salt) = 68.833

Extent of reaction  (mol) =  1.019

Entahlpy change in the rxn  (KJ/mol) =  -134.880

Q of the rxn  (KJ) =  -136.342
The answer in the Text is wrong.
press enter key to exit

Out[9]:
''

## Example 9.6-1, page no. 465¶

In [10]:
#Initialization of variables
import math
import numpy
from numpy import linalg

x=0.85
H1= 802 #KJ/mol
H2= 1428 #kJ/mol
M1=16.0
M2=30.0

#Calculations and printing :

print(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook")
y1=x*16
y2=(1-x)*30
xCH4=y1/(y1+y2)
HHVMethane=(H1+ 2*44.013)/M1
HHVEthane=(H2+ 3*44.013)/M2
HHV=xCH4*HHVMethane + (1-xCH4)*HHVEthane
print '%s %.3f' %(" \n HHV of Fuel (KJ/g) = ",HHV)
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

HHV of Fuel (KJ/g) =  54.725
press enter key to exit

Out[10]:
''