Chapter 8 : Fuels

Example 8.1 Page No : 193

In [1]:
#Given:
HCV = 46900.      #Highest calorific value(HCV) of petrol in kJ/kg
pH2 = 14.4/100      #Composition of Hydrogen in petrol by mass
ufg = 2304.4      #Latent heat of evaporation for water in kJ/kg

#Solution:
#  2[H2] + [O2]  =  2[H2O]
H = 1      #Atomic mass of Hydrogen(H)
O = 16      #Atomic mass of Oxygen(O)
#Assume 1 kg of fuel consume
mH2 = 1*pH2      #Mass of Hydrogen in kg/kg of fuel
m_a = 2*(2*H+O)/(2*2*H)*mH2      #Mass of water produced in kg/kg of fuel
LCV = HCV-m_a*ufg      #Lowest calorific value in kJ/kg

#Results:
print " The LCV of petrol  =  %.0f kJ/kg"%(LCV)

 The LCV of petrol  =  43913 kJ/kg

Example 8.2 Page No : 198

In [2]:
from scipy.optimize import fsolve 
import math 

#Given:
pCO2 = 13./100      #Composition of Carbon di oxide in dry exhaust gas

#Solution:
p_v = 21./100      #Composition of Oxygen in air by volume
C = 12.      #Atomic mass of Carbon(C)
H = 1.      #Atomic mass of Hydrogen(H)
O = 16.      #Atomic mass of Oxygen(O)
#Combustion equation
#  [C8H18] + a[O2] + (1-p_v)/p_v*a[N2]  =  x[CO2] + y[H2O] + z[O2] + w[N2]
#On balancing the reaction
x = 8.
y = 9.      #Coefficients of combustion equation
def f(a):      #Defining the function, M of coefficient a for calculation of a
    z = a-x-y/2      #On balancing O
    w = (1-p_v)/p_v*a      #On balancing N
    return x/(x+z+w)-pCO2

#Since, Composition of CO2 calculated from the equation must be equal to the given composition of CO2
#Thus, function M solve for zero to get value of a
a = fsolve(f,1)
A_F_act = a/p_v      #Air fuel ratio by volume
#For chemically correct mixture
a = x+y/2      #Moles of air required
A_F_cc = a/p_v      #Chemically correct air fuel ratio
ratio = (1/A_F_act)/(1/A_F_cc)*100      #Ratio of actual to chemically correct fuel air ratio by volume

#Results:
print " The ratio by volume of fuel to air supplied  =  1/%.0f"%(A_F_act)
print " The volume fuel air ratio  =  %.1f percentage of chemically correct ratio"%(ratio)

 The ratio by volume of fuel to air supplied  =  1/66
 The volume fuel air ratio  =  90.1 percentage of chemically correct ratio

Example 8.3 Page No : 203

In [4]:
#Given:
pC = 84.
pH2 = 16.      #Percentage of Carbon, Hydrogen in fuel 
p_v = 20.9      #Percentage of Oxygen in air by volume

#Solution:
C = 12.      #Atomic mass of Carbon(C)
H = 1.      #Atomic mass of Hydrogen(H)
O = 16.      #Atomic mass of Oxygen(O)
N = 14.      #Atomic mass of Nitrogen(N)
m_f = 100.      #Mass of fuel (assume) in kg
#Combustion equation
#pC/C[C] + pH2/2[H2] + [a[O2] + (100-p_v)/p_v*a[N2]]   =   b[CO2] + d[O2] + e[N2] + f[H2O]
#Equating coefficients
b = pC/C
f = pH2/2
d = b/6
a = b+d+f/2      #Coefficients of combustion equation
m_a = a*2*O + (100-p_v)/p_v*a*2*N      #Mass of air supplied in kg
A_F = m_a/m_f      #Air fuel ratio
P_e = d/(a-d)*100      #Percentage excess air

#Results:
print " a)The air fuel ratio by mass, A_F  =  %.1f/1"%(A_F)
print " b)The percentage excess air supplied  =  %.1f percent"%(P_e)

 a)The air fuel ratio by mass, A_F  =  16.8/1
 b)The percentage excess air supplied  =  10.6 percent

Example 8.4 Page No : 208

In [5]:
#Given:
MS = 25.      #Mixture strength in percent
p = 23.1      #Percentage of oxygen in air by mass

#Solution:
C = 12.      #Atomic mass of Carbon(C)
H = 1.      #Atomic mass of Hydrogen(H)
O = 16.      #Atomic mass of Oxygen(O)
N = 14.      #Atomic mass of Nitrogen(N)
m_f = 1.      #Mass of fuel(C6H14) in kg
mC = (6*C)/((6*C)+(14*H))      #Mass of Carbon in kg
mH = (14*H)/((6*C)+(14*H))      #Mass of Hydrogen in kg
m_a = (2*O/C*mC+O/(2*H)*mH)*100/p      #Mass of air in kg
#For 25 percent rich mixture
m_f = m_f+m_f*MS/100      #Mass of fuel(C6H14) in kg
A_F = m_a/m_f      #Air fuel ratio
mO2 = p/100*A_F      #Mass of Oxygen available in kg
mO2_1 = O/(2*H)*mH      #Oxygen required for combustion of H to H2O in kg
mH2O = mH*(1+O/(2*H))      #Mass of H2O produced in kg
mO2_2 = O/C*mC      #Oxygen required for combustion of C to CO in kg
mCO = mC*(1+O/C)      #Mass of CO produced in kg
mO2_3 = mO2-(mO2_1+mO2_2)      #Mass of Oxygen remaining for combustion of CO to CO2
mCO_b = mO2_3*(C+O)/O      #Mass of CO burned to CO2 in kg
mCO2 = mCO_b*(1+O/(C+O))      #Mass of CO2 produced in kg
mCO_ub = mCO-mCO_b      #Mass of CO unburned in kg
nH2O = mH2O/(2*H+O)      #Moles of H2O
nCO2 = mCO2/(C+2*O)      #Moles of CO2
nCO = mCO_ub/(C+O)      #Moles of CO
mN2 = A_F*(1-p/100)      #Mass of Nitrogen (N2) in kg
nN2 = mN2/(2*N)      #Moles of N2
nT = nH2O+nCO2+nCO+nN2      #Total number of moles
pH2O = nH2O/nT
pCO2 = nCO2/nT
pCO = nCO/nT
pN2 = nN2/nT      #Composition of products

#Results:
print " The theoretical mass of air required, m_a  =  %.2f kg"%(m_a)
print " The composition of the products in percent\t H2O  =  %.2f\t CO2  =  %.2f\t CO  =  %.2f\t N2  =  %.2f"%(pH2O*100,pCO2*100,pCO*100,pN2*100)

 The theoretical mass of air required, m_a  =  15.30 kg
 The composition of the products in percent	 H2O  =  16.70	 CO2  =  5.25	 CO  =  9.07	 N2  =  68.98

Example 8.5 Page No : 213

In [6]:
import math 
from numpy import roots
from sympy import Symbol,solve

#C7H16 in Petrol engine
#Given:
p_v = 21.      #Percentage of Oxygen in air by volume
p_e = 50.      #Percentage of excess air supplied
#Solution:
m_f = 100.      #Mass of fuel (assume) in kg
C = 12.      #Atomic mass of Carbon(C)
H = 1.      #Atomic mass of Hydrogen(H)
O = 16.      #Atomic mass of Oxygen(O)
N = 14.      #Atomic mass of Nitrogen(N)
#a = poly(0,'a')      #Defining unknown number of moles of stoichiometric oxygen
a = Symbol('a')
#Combustion equation
#m_f/(7*C+16*H)[C7H16] + (1+p_e/100)*[a[O2] + (100-p_v)/p_v*a[N2]]   =   b[CO2] + d[O2] + e[N2] + f[H2O]
#Equating coefficients
b = m_f/(7*C+16*H)*7      #Moles of CO2 on balancing of C
f = m_f/(7*C+16*H)*16/2      #Moles of H2O on balancing of H
d = p_e/100*a      #Excess moles of oxygen
a = solve((1+p_e/100)*a-(b+d+f/2))[0]      #Balancing Oxygen of both sides
print a
m_a = a*2*O+(100-p_v)/p_v*a*2*N      #Mass of air supplied in kg
A_F = m_a/m_f      #Air fuel ratio
d = p_e/100*a      #Moles of Oxygen in products of combustion
e = (1+p_e/100)*(100-p_v)/p_v*a      #Moles of Nitrogen in products of combustion
nT = b+d+e+f      #Total number of moles in product of combustion
pH2O = f/nT*100
pCO2 = b/nT*100
pO2 = d/nT*100
pN2 = e/nT*100      #Percentage volumetric composition of the products of combustion

#Results:
print " a)The stoichiometric air fuel consumption by mass, A_F  =  %.2f:1"%(A_F)
print " b)The percentage volumetric composition of the products\t CO2  =  %.2f\t O2  =  %.2f\t N2  =  %.1f\t H2O  =  %.2f"%(pCO2,pO2,pN2,pH2O)

11.0000000000000
 a)The stoichiometric air fuel consumption by mass, A_F  =  15.11:1
 b)The percentage volumetric composition of the products	 CO2  =  8.48	 O2  =  6.66	 N2  =  75.2	 H2O  =  9.69

Example 8.6 Page No : 218

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

#Given:
pC = 85.
pH2 = 15.      #Percentage of Carbon, Hydrogen in fuel
A_F = 14.      #Air fuel ratio by mass
p_m = 23.2      #Percentage of oxygen in air by mass

#Solution:
m_f = 100.      #Mass of fuel (assume) in kg
m_a = A_F*m_f      #Mass of air in kg
C = 12.      #Atomic mass of Carbon(C)
H = 1.      #Atomic mass of Hydrogen(H)
O = 16.      #Atomic mass of Oxygen(O)
N = 14.      #Atomic mass of Nitrogen(N)
p_v = 20.9      #Percentage of Oxygen in air by volume
#Combustion equation
#pC/C[C] + pH2/2[H2] + [a[O2] + (100-p_v)/p_v*a[N2]]   =   b[CO2] + d[CO] + e[N2] + f[H2O]
#Equating coefficients
f = pH2/2      #Moles of H2O on balancing of H
a = m_a/(2*O+(100-p_v)/p_v*2*N)      #Balancing Oxygen of both sides
#On balancing C of both sides we get, b + d  =  pC/C     eq(1)
#On balancing O of both sides we get, b + d/2 + f/2  =  a     eq(2)
#Solving equations (1) and (2)
A = [[1,1],[1, 1./2]]
B = [[pC/C],[a-f/2]]
SOL = solve(A,B)      #Taking matrix A, B to get solution matrix, SOL  =  [b;d]
b = SOL[0]
d = SOL[1]      #Moles of CO2 and CO
e = (100-p_v)/p_v*a      #Moles of Nitrogen in products of combustion
mC = b/m_f*C      #Mass of carbon burning to CO2 in kg per kg of fuel
mCO2 = b/m_f*(C+2*O)      #Mass of CO2 produced in kg
mCO = d/m_f*(C+O)      #Mass of CO produced in kg
mN2 = e/m_f*(2*N)      #Mass of N2 produced in kg
mH2O = f/m_f*(2*H+O)      #Mass of H2O produced in kg
#Results:
print " a)The mass of the carbon burning to CO2  =  %.3f kg"%(mC)
print " b)The mass of each of the gases in the exhaust per kg of fuel\t CO2  =  %.2f kg\t CO  =  %.2f kg\t N2  =  %.2f kg\t H2O  =  %.2f kg"%(mCO2,mCO,mN2,mH2O)

 a)The mass of the carbon burning to CO2  =  0.685 kg
 b)The mass of each of the gases in the exhaust per kg of fuel	 CO2  =  2.51 kg	 CO  =  0.38 kg	 N2  =  10.75 kg	 H2O  =  1.35 kg

Example 8.7 Page No : 223

In [8]:
#Given:
pCO2 = 12./100
pCO = 2./100
pCH4 = 4./100
pH2 = 1./100
pO2 = 4.5/100      #Composition of Carbon di oxide(CO2), Carbon mono oxide(CO), Methane(CH4), Hydrogen(H2), Oxygen(O2) in dry exhaust gas

pN2 = 1-(pCO2+pCO+pCH4+pH2+pO2)      #Composition of Nitrogen(N2) in dry exhaust gas

#Solution:
C = 12.      #Atomic mass of Carbon(C)
H = 1.      #Atomic mass of Hydrogen(H)
p_v = 21.      #Percentage of Oxygen in air by volume
#Let X be the mass of the fuel per mole dry exhaust gas
#Let Y be the mole of O2 per mole dry exhaust gas
#Let 1 kg of fuel contain p kg of C and q kg of H2
#Combustion equation
#X*(p/C[C] + q/(2*H)[H2]) + Y[O2] + (100-p_v)/p_v*Y[N2]  =  pCO2[CO2] + pCO[CO] + pCH4[CH4] + pH2[H2] + pO2[O2] + a[H2O] + pN2[N2]
#Equating coefficients
Y = pN2/((100-p_v)/p_v)      #Nitrogen(N) balance
a = 2*(Y-(pCO2+pCO/2+pO2))      #Oxygen(O) balance
Xp = C*(pCO2+pCO+pCH4)      #Carbon(C) balance; X*p
Xq = (2*H)*(2*pCH4+pH2+a)      #Hydrogen(H) balance; X*q
p_q = Xp/Xq      #Ratio of C to H2 in fuel

#Results:
print " The proportion by mass of Carbon to Hydrogen in the fuel  =  %.2f/1"%(p_q)

 The proportion by mass of Carbon to Hydrogen in the fuel  =  7.36/1

Example 8.8 Page No : 228

In [11]:
#Given:
pCO2 = 7.5
pCO = 1.
pO2 = 9.4      #Percentage of Carbon di oxide(CO2), Carbon mono oxide(CO), Oxygen(O2) in dry exhaust gas
P = 1.02      #Pressure of the exhaust gas in bar
pO_v = 21.      #Percentage of Oxygen in air by volume
pN_v = 79.      #Percentage of Nitrogen in air by volume
M = 29.      #Molecular weight of air

#Solution:
C = 12.      #Atomic mass of Carbon(C)
H = 1.      #Atomic mass of Hydrogen(H)
#Let 100*x moles of air be used with fuel per 100 mole of dry exhaust products
pN2 = 100-(pCO2+pCO+pO2)      #Composition of Nitrogen(N2) in dry exhaust gas
#Combustion equation
#    a[C] + b[H2]) + pO_v*x[O2] + pN_v*x[N2]  =  pCO2[CO2] + pCO[CO] + pO2[O2] d[H2O] + pN2[N2]
#Equating coefficients
a = pCO2+pCO      #Carbon(C) balance
x = pN2/pN_v      #Nitrogen(N) balance
d = 2*(pO_v*x-(pCO2+pO2+pCO/2))      #Oxygen(O) balance
d = round(10*d)/10
b = d      #Hydrogen(H) balance
m_a = 100*x*M      #Mass of air in kg
m_f = a*C+b*2*H      #Mass of fuel in kg
A_F = m_a/m_f      #Air fuel ratio
pC = a*C/m_f*100      #Percentage of Carbon(C) in fuel
pH2 = 100-pC      #Percentage of Hydrogen(H2) in fuel
nT = pCO2+pCO+pO2+pN2+d      #Total number of moles in product of combustion
CO2 = pCO2/nT*100;O2 = pO2/nT*100;CO = pCO/nT*100;N2 = pN2/nT*100;H2O = d/nT*100      #Percentage volumetric composition of the products of combustion
PP = d/nT*P      #Partial pressure of H2O in bar
#From steam tables
if (PP == 0.0825):
    T = 42.8      #Saturation temperature in degreeC


#Results:
print " a)The air fuel ratio used, A_F  =  %.1f"%(A_F)
print " b)The mass analysis of the fuel\t Carbon  =  %.1f percent\t Hydrogen  =  %.1f percent"%(pC,pH2)
print " c)The wet products analysis in percent\t CO2  =  %.1f\t O2  =  %.2f\t CO  =  %.1f\t N2  =  %.2f\t Steam  =  %.1f"%(CO2,O2,CO,N2,H2O)
print " d)The minimum temperature to which the exhaust may be cooled before condensation occurs  =  %.1f degreeC"%(T)

 a)The air fuel ratio used, A_F  =  25.2
 b)The mass analysis of the fuel	 Carbon  =  85.3 percent	 Hydrogen  =  14.7 percent
 c)The wet products analysis in percent	 CO2  =  6.9	 O2  =  8.64	 CO  =  0.9	 N2  =  75.46	 Steam  =  8.1
 d)The minimum temperature to which the exhaust may be cooled before condensation occurs  =  42.8 degreeC

Example 8.9 Page No : 233

In [12]:
from numpy import sum,array

#Given:
pH2 = 49.4/100
pCO = 18./100
pCH4 = 20./100
pC4H8 = 2./100
pO2 = 0.4/100
pN2 = 6.2/100
pCO2 = 4./100      #Composition of Coal gas
MW = 20.      #Mixture weakness in percent

#Solution:
#Combustion equations for determining the moles of Oxygen used
#2[H2] + [O2] ---> 2[H2O]           #For Hydrogen
#2[CO] + [O2] ---> 2[CO2]           #For Carbon mono oxide
#[CH4] + 2[O] ---> [CO2] + 2[H2O]          #For Methane
#[C4H8] + 6[O2] ---> 4[CO2] + 4[H2O]         #For C4H8
nO2 = sum([pH2/2, pCO/2, 2*pCH4, 6*pC4H8 ,pO2])      #Moles of O2 required (error)
nCO2 = sum([pCO, pCH4 ,4*pC4H8 ,pCO2])      #Moles of CO2
nH2O = sum([pH2, 2*pCH4, 4*pC4H8])      #Moles of H2O
p_v = 21.      #Percentage of Oxygen in air by volume
n_a = nO2/p_v*100      #Moles of air required
n_f = 1.      #For 1 mole of fuel
A_F = n_a/n_f      #Air fuel ratio
#For weak mixture
A_F_act = A_F*(1+MW/100)      #Actual air fuel ratio
nN2 = (1-p_v/100)*A_F_act      #Moles of N2
nO2 = p_v/100*A_F_act-nO2      #Excess moles of Oxygen in products
nN2 = nN2+pN2      #Moles of Nitrogen in products
nT_d = nCO2+nO2+nN2      #Total dry moles of product
nT_w = nT_d+nH2O      #Total wet moles of product
p_d = array([nCO2, nO2 ,nN2])*100/nT_d      #Percentage volumetric composition of the dry products of combustion
p_w = array([nCO2, nH2O, nO2, nN2])*100/nT_w      #Percentage volumetric composition of the wet products of combustion

#Results:
print " The stoichiometric air fuel ratio used, A_F  =  %.1f/1"%(A_F)
print " The wet products analysis in percent\t CO2  =  %.0f\t H2O  =  %.2f\t O2  =  %.2f\t N2  =  %.2f"%(p_w[0],p_w[1],p_w[2],p_w[3])
print " The dry products analysis in percent\t CO2  =  %.2f\t O2  =  %.2f\t N2  =  %.2f"%(p_d[0],p_d[1],p_d[2])
#Answers in the book are wrong

 The stoichiometric air fuel ratio used, A_F  =  4.1/1
 The wet products analysis in percent	 CO2  =  9	 H2O  =  17.41	 O2  =  3.08	 N2  =  70.58
 The dry products analysis in percent	 CO2  =  10.82	 O2  =  3.73	 N2  =  85.45