# Chapter 1 : Fuels and Combustion¶

## Example 1.1 Page no : 15¶

In [1]:
# Variables
C = 0.91;			#Percentage composition of Carbon
H = 0.03;			#Percentage composition of Hydrogen
O = 0.02;			#Percentage composition of Oxygen
N = 0.008;			#Percentage composition of Nitrogen
S = 0.008;			#Percentage composition of Sulphur

# Calculations
m = (11.5*C)+(34.5*(H-(O/8)))+(4.3*S);			#Mass of air per kg of coal in kg

# Results
print 'Minimum mass of air per kg of coal is %3.2f kg'%(m)

Minimum mass of air per kg of coal is 11.45 kg


## Example 1.2 Page no : 16¶

In [2]:
# Variables
C = 0.86;			#Percentage composition of Carbon
H = 0.12;			#Percentage composition of Hydrogen
O = 0.01;			#Percentage composition of Oxygen
S = 0.01;			#Percentage composition of Sulphur
v = 0.773;			#Specific volume of air at N.T.P in (m**3)/kg

# Calculations
m = (11.5*C)+(34.5*(H-(O/8)))+(4.3*S);			#Theoretical mass of air per kg of coal in kg
vth = m*v;			#Theoretical volume of air at N.T.P per kg fuel in (m**3)/kg of fuel

# Results
print 'Theoretical volume of air at N.T.P per kg fuel is %3.2f m**3)/kg of fuel'%(vth)

Theoretical volume of air at N.T.P per kg fuel is 10.85 m**3)/kg of fuel


## Example 1.3 Page no : 16¶

In [3]:
# Variables
C = 0.78;			#Percentage composition of Carbon
H = 0.06;			#Percentage composition of Hydrogen
O = 0.078;			#Percentage composition of Oxygen
N = 0.012;			#Percentage composition of Nitrogen
S = 0.03;			#Percentage composition of Sulphur

# Calculations
m = (11.5*C)+(34.5*(H-(O/8)))+(4.3*S);			#Minimum quantity of air required in kg
mt = ((11*C)/3)+(9*H)+(2*S)+(8.32+N);			#Total mass of products of combustion in kg

# Results
print 'Minimum quantity of air required for complete combustion is %3.2f kg  \
\nTotal mass of products of combustion is %3.3f kg'%(m,mt)

Minimum quantity of air required for complete combustion is 10.83 kg
Total mass of products of combustion is 11.792 kg


## Example 1.4 Page no : 17¶

In [2]:
# Variables
C = 0.84;			#Percentage composition of Carbon
H = 0.09;			#Percentage composition of Hydrogen
CO2 = 0.0875;			#Volumetric composition of CO2
CO = 0.0225;			#Volumetric composition of CO
O2 = 0.08;			#Volumetric composition of Oxygen
N2 = 0.81;			#Volumetric composition of Nitrogen
M1 = 44.;			#Molecular mass of CO2
M2 = 28.;			#Molecular mass of CO
M3 = 32.;			#Molecular mass of O2
M4 = 28.;			#Molecular mass of N2

# Calculations
c1 = CO2*M1;			#Proportional mass of CO2
c2 = CO*M2;	    		#Proportional mass of CO
c3 = O2*M3;		    	#Proportional mass of O2
c4 = N2*M4;			    #Proportional mass of N2
c = c1+c2+c3+c4;			#Total proportional mass of constituents
m1 = c1/c;		    	#Mass of CO2 per kg of flue gas in kg
m2 = c2/c;		    	#Mass of CO per kg of flue gas in kg
m3 = c3/c;		    	#Mass of O2 per kg of flue gas in kg
m4 = c4/c;		    	#Mass of N2 per kg of flue gas in kg
d1 = m1*100;			#Mass analysis of CO2
d2 = m2*100;			#Mass analysis of CO
d3 = m3*100;			#Mass analysis of O2
d4 = m4*100;			#Mass analysis of N2
m = ((3*m1)/11)+((3*m2)/7.);			#Mass of carbon in kg
md = C/m;			    #Mass of dry flue gas in kg

# Results
print 'Mass of dry flue gases per kg of coal burnt is %.f kg'%(md)

# note : rounding off error

Mass of dry flue gases per kg of coal burnt is 19 kg


## Example 1.5 Page no : 17¶

In [5]:
# Variables
C = 0.624;			#Percentage composition of Carbon
H = 0.042;			#Percentage composition of Hydrogen
O = 0.045;			#Percentage composition of Oxygen
CO2 = 0.13;			#Volumetric composition of CO2
CO = 0.003;			#Volumetric composition of CO
O2 = 0.06;			#Volumetric composition of Oxygen
N2 = 0.807;			#Volumetric composition of Nitrogen
M1 = 44;			#Molecular mass of CO2
M2 = 28;			#Molecular mass of CO
M3 = 32;			#Molecular mass of O2
M4 = 28;			#Molecular mass of N2
mw = 0.378;			#Mass of H2O in kg

# Calculations
m = (11.5*C)+(34.5*(H-(O/8)));			#Minimum air required in kg
c1 = CO2*M1;			#Proportional mass of CO2
c2 = CO*M2;			#Proportional mass of CO
c3 = O2*M3;			#Proportional mass of O2
c4 = N2*M4;			#Proportional mass of N2
c = c1+c2+c3+c4;			#Total proportional mass of constituents
m1 = c1/c;			#Mass of CO2 per kg of flue gas in kg
m2 = c2/c;			#Mass of CO per kg of flue gas in kg
m3 = c3/c;			#Mass of O2 per kg of flue gas in kg
m4 = c4/c;			#Mass of N2 per kg of flue gas in kg
d1 = m1*100;			#Mass analysis of CO2
d2 = m2*100;			#Mass analysis of CO
d3 = m3*100;			#Mass analysis of O2
d4 = m4*100;			#Mass analysis of N2
mC = ((3*m1)/11)+((3*m2)/7);			#Mass of carbon in kg
md = C/mC;			#Mass of dry flue gas in kg
mact = (md+mw)-(C+H+O);			#Actual air supplied per kg of fuel in kg
me = mact-m;			#Mass of excess air per kg of fuel in kg

# Results
print 'Minimum air required to burn 1 kg of coal is %3.2f kg  \
\nMass of air actually supplied per kg of coal is %3.3f kg  \
\nAmount of excess air supplied per kg of coal burnt is %3.3f kg'%(m,mact,me)
#rounding-off errors

Minimum air required to burn 1 kg of coal is 8.43 kg
Mass of air actually supplied per kg of coal is 11.521 kg
Amount of excess air supplied per kg of coal burnt is 3.090 kg


## Example 1.6 Page no : 19¶

In [6]:
# Variables
C = 0.78;			#Percentage composition of Carbon
H = 0.03;			#Percentage composition of Hydrogen
O = 0.03;			#Percentage composition of Oxygen
S = 0.01;			#Percentage composition of Sulphur
me = 0.3;			#Mass of excess air supplied

# Calculations
m = (11.5*C)+(34.5*(H-(O/8)))+(4.3*S);			#Mass of air per kg of coal in kg
mec = me*m;			#Excess air supplied per kg of coal in kg
mact = m+mec;			#Actual mass of air supplied per kg of coal in kg
mCO2 = (11*C)/3;			#Mass of CO2 produced per kg of coal in kg
mHw = 9*H;			#Mass of H2O produced per kg of coal in kg
mSO2 = 2*S;			#Mass of SO2 produced per kg of coal in kg
mO2 = 0.232*mec;			#Mass of excess O2 produced per kg of coal in kg
mN2 = 0.768*mact;			#Mass of N2 produced per kg of coal in kg

# Results
print 'Mass of air to be supplied is %3.2f kg  \
\nMass of CO2 produced per kg of coal is %3.2f kg  \
\nMass of H2O produced per kg of coal is %3.2f kg\
\nMass of SO2 produced per kg of coal is %3.2f kg  \
\nMass of excess O2 produced per kg of coal is %3.2f kg  \
\nMass of N2 produced per kg of coal is %3.2f kg '%(m,mCO2,mHw,mSO2,mO2,mN2)

Mass of air to be supplied is 9.92 kg
Mass of CO2 produced per kg of coal is 2.86 kg
Mass of H2O produced per kg of coal is 0.27 kg
Mass of SO2 produced per kg of coal is 0.02 kg
Mass of excess O2 produced per kg of coal is 0.69 kg
Mass of N2 produced per kg of coal is 9.90 kg


## Example 1.7 Page no : 20¶

In [3]:
# Variables
C = 0.9;			#Percentage composition of Carbon
H = 0.033;			#Percentage composition of Hydrogen
O = 0.03;			#Percentage composition of Oxygen
N = 0.008;			#Percentage composition of Nitrogen
S = 0.009;			#Percentage composition of Sulphur
M1 = 44;			#Molecular mass of CO2
M2 = 64;			#Molecular mass of SO2
M3 = 32;			#Molecular mass of O2
M4 = 28;			#Molecular mass of N2

# Calculations
m = (11.5*C)+(34.5*(H-(O/8)))+(4.3*S);			#Minimum mass of air per kg of coal in kg
mCO2 = (11*C)/3;			#Mass of CO2 produced per kg of coal in kg
mHw = 9*H;			#Mass of H2O produced per kg of coal in kg
mSO2 = 2*S;			#Mass of SO2 produced per kg of coal in kg
mt = 11.5*1.5;			#Total mass of air supplied per kg of coal in kg
me = mt-m;			#Excess air supplied in kg
mO2 = 0.232*me;			#Mass of excess O2 produced per kg of coal in kg
mN2 = 0.768*mt;			#Mass of N2 produced per kg of coal in kg
mtN2 = mN2+N;			#Total mass of Nitrogen in exhaust in kg
md = mCO2+mSO2+mO2+mtN2;			#Total mass of dry flue gases per kg of fuel in kg
CO2 = (mCO2/md)*100;			#Percentage composition of CO2 by mass in percent
SO2 = (mSO2/md)*100;			#Percentage composition of SO2 by mass in percent
O2 = (mO2/md)*100;			#Percentage composition of O2 by mass in percent
N2 = (mN2/md)*100;			#Percentage composition of N2 by mass in percent
c1 = CO2/M1;			#Proportional volume of CO2
c2 = SO2/M2;			#Proportional volume of SO2
c3 = O2/M3;			#Proportional volume of O2
c4 = N2/M4;			#Proportional volume of N2
c = c1+c2+c3+c4;			#Total proportional volume of constituents
m1 = c1/c;			#Volume of CO2 in 1 (m**3) of flue gas
m2 = c2/c;			#Volume of SO2 in 1 (m**3) of flue gas
m3 = c3/c;			#Volume of O2 in 1 (m**3) of flue gas
m4 = c4/c;			#Volume of N2 in 1 (m**3) of flue gas
d1 = m1*100;			#Volume analysis of CO2
d2 = m2*100;			#Volume analysis of SO2
d3 = m3*100;			#Volume analysis of O2
d4 = m4*100;			#Volume analysis of N2

# Results
print 'Minimum mass of air required is %3.1f kg  \
\nTotal mass of dry flue gases per kg of fuel is %3.2f kg  \
\nPercentage composition of CO2 by volume is %3.2f percent  \
\nPercentage composition of SO2 by volume is %3.3f percent  \
\nPercentage composition of O2 by volume is %3.1f percent  \
\nPercentage composition of N2 by volume is %3.2f percent'%(m,md,d1,d2,d3,d4)

# note : rounding off error

Minimum mass of air required is 11.4 kg
Total mass of dry flue gases per kg of fuel is 17.93 kg
Percentage composition of CO2 by volume is 12.69 percent
Percentage composition of SO2 by volume is 0.048 percent
Percentage composition of O2 by volume is 7.2 percent
Percentage composition of N2 by volume is 80.08 percent


## Example 1.8 Page no : 21¶

In [5]:
# Variables
C = 0.88;			#Percentage composition of Carbon
H = 0.036;			#Percentage composition of Hydrogen
O = 0.048;			#Percentage composition of oxygen
CO2 = 0.109;			#Volumetric composition of CO2
CO = 0.01;			#Volumetric composition of CO
O2 = 0.071;			#Volumetric composition of Oxygen
N2 = 0.81;			#Volumetric composition of Nitrogen
M1 = 44.;			#Molecular mass of CO2
M2 = 28.;			#Molecular mass of CO
M3 = 32.;			#Molecular mass of O2
M4 = 28.;			#Molecular mass of N2

# Calculations
m = (11.5*C)+(34.5*(H-(O/8)));			#Theoretical air required in kg
c1 = CO2*M1;			#Proportional mass of CO2
c2 = CO*M2;			#Proportional mass of CO
c3 = O2*M3;			#Proportional mass of O2
c4 = N2*M4;			#Proportional mass of N2
c = c1+c2+c3+c4;			#Total proportional mass of constituents
m1 = c1/c;			#Mass of CO2 per kg of flue gas in kg
m2 = c2/c;			#Mass of CO per kg of flue gas in kg
m3 = c3/c;			#Mass of O2 per kg of flue gas in kg
m4 = c4/c;			#Mass of N2 per kg of flue gas in kg
mC = ((3*m1)/11)+((3*m2)/7);			#Mass of carbon in kg
md = C/mC;			#Mass of dry flue gas in kg
hc = H*9;			#Hydrogen combustion in kg of H2O
mair = (md+hc)-(C+H+O);			#Mass of air supplied per kg of coal in kg
me = mair-m;			#Excess air per kg of coal in kg
mN2 = m4*md;			#Mass of nitrogen per kg of coal in kg
mact = mN2/0.768;			#Actual mass of air per kg of coal in kg
pe = (me/m)*100;			#Perccentage excess air in percent

# Results
print 'Mass of air actually supplied per kg of coal is %3.2f kg  \
\nPercentage of excess air is %.f percent'%(mact,pe)

# note : rounding off error

Mass of air actually supplied per kg of coal is 18.20 kg
Percentage of excess air is 60 percent


## Example 1.9 Page no : 22¶

In [1]:
# Variables
C = 0.84;			#Percentage composition of Carbon
H = 0.14;			#Percentage composition of Hydrogen
O = 0.02;			#Percentage composition of oxygen
CO2 = 8.85;			#Volumetric composition of CO2
CO = 1.2;			#Volumetric composition of CO
O2 = 6.8;			#Volumetric composition of Oxygen
N2 = 83.15;			#Volumetric composition of Nitrogen
M1 = 44.;			#Molecular mass of CO2
M2 = 28.;			#Molecular mass of CO
M3 = 32.;			#Molecular mass of O2
M4 = 28.;			#Molecular mass of N2
a = 8/3.;			#O2 required per kg C
b = 8.; 			#O2 required per kg H2
mair = 0.23;			#Mass of air

# Calculations
c = C*a;			#O2 required per kg of fuel for C
d = H*b;			#O2 required per kg of fuel for H2
tO2 = c+d+O;			#Theoreticcal O2 required in kg/kg of fuel
tm = tO2/mair;			#Theoretical mass of air in kg/kg of fuel
c1 = CO2*M1;			#Proportional mass of CO2 by Volume
c2 = CO*M2;			#Proportional mass of CO by Volume
c3 = O2*M3;			#Proportional mass of O2 by Volume
c4 = N2*M4;			#Proportional mass of N2 by Volume
c = c1+c2+c3+c4;			#Total proportional mass of constituents
m1 = c1/c;			#Mass of CO2 per kg of flue gas in kg
m2 = c2/c;			#Mass of CO per kg of flue gas in kg
m3 = c3/c;			#Mass of O2 per kg of flue gas in kg
m4 = c4/c;			#Mass of N2 per kg of flue gas in kg
mC = ((m1*12)/M1)+((m2*12)/M2);			#Mass of carbon per kg of dry flue gas in kg
md = C/mC;			#Mass of dry flue per kg of fuel in kg
p = (4*m2)/7;			#Oxygen required to burn CO in kg
meO2 = md*(m3-p);			#Mass of excess O2 per kg of fuel in kg
me = meO2/mair;			#Mass of excess air in kg/kg fuel
mt = tm+me;			#Total air required per kg fuel

# Results
print 'Mass of excess air supplied per kg of fuel burnt is %3.1f kg/kg of fuel  \
\nAir-fuel ratio is %3.1f:1'%(me,mt)

# rounding off error

Mass of excess air supplied per kg of fuel burnt is 6.0 kg/kg of fuel
Air-fuel ratio is 20.7:1


## Example 1.10 Page no : 23¶

In [10]:
# Variables
H2 = 0.27;			#Percentage composition of H2 by volume
CO2 = 0.18;			#Percentage composition of CO2 by volume
CO = 0.125;			#Percentage composition of CO by volume
CH4 = 0.025;			#Percentage composition of CH4 by volume
N2 = 0.4;			#Percentage composition of N2 by volume

# Calculations
v = (2.38*(H2+CO))+(9.52*CH4);			#Volume of air required for complete combustion in (m**3)

# Results
print 'Volume of air required for complete combustion is %3.3f m**3)'%(v)

Volume of air required for complete combustion is 1.178 m**3)


## Example 1.11 Page no : 24¶

In [3]:
# Variables
H2 = 0.5;			#Percentage composition of H2 by volume
CO2 = 0.1;			#Percentage composition of CO2 by volume
CO = 0.05;			#Percentage composition of CO by volume
CH4 = 0.25;			#Percentage composition of CH4 by volume
N2 = 0.1;			#Percentage composition of N2 by volume
pCO2 = 8;			#Percentage volumetric analysis of CO2
pO2 = 6;			#Percentage volumetric analysis of O2
pN2 = 86;			#Percentage volumetric analysis of N2

# Calculations
v = (2.38*(H2+CO))+(9.52*CH4);			#Volume of air required for complete combustion in (m**3)
vN2 = v*0.79;			#Volume of nitrogen in the air in m**3
a = CO+CH4+CO2;			#CO2 formed per m**3 of fuel gas burnt
b = vN2+N2;			#N2 formed per m**3 of fuel gas burnt
vt = a+b;			#Total volume of dry flue gas formed in m**3
ve = (pO2*vt)/(21-pO2);			#Excess air supplied in m**3
V = v+ve;			#Total quantity of air supplied in m**3
afr = V/1

# Results
print 'Air-fuel ratio by volume is %3.3f'%(afr)

Air-fuel ratio by volume is 5.055


## Example 1.12 Page no : 24¶

In [12]:
# Variables
H2 = 0.14;			#Percentage composition of H2 by volume
CO2 = 0.05;			#Percentage composition of CO2 by volume
CO = 0.22;			#Percentage composition of CO by volume
CH4 = 0.02;			#Percentage composition of CH4 by volume
O2 = 0.02;			#Percentage composition of O2 by volume
N2 = 0.55;			#Percentage composition of N2 by volume
e = 0.4;			#Excess air supplied
# Calculations
v = (2.38*(H2+CO))+(9.52*CH4)-(4.76*O2);			#Volume of air required for complete combustion in (m**3)
ve = v*e;			#Volume of excess air supplied in m**3
vtN2 = v-(v*0.21);			#Volume of N2 in theoretical air in m**3
veN2 = ve-(ve*0.21);			#Volume of N2 in excess air in m**3
vt = vtN2+veN2;			#Total volume of N2 in air supplied in m**3
vCO2 = CO+CH4+CO2;			#CO2 formed per m**3 of fuel gas
vN2 = vt+N2;			#N2 formed per m**3 of fuel gas
veO2 = ve*0.21;			#Volume of excess O2 per m**3 of fuel gas
vT = vCO2+vN2+veO2;			#Total volume of dry combustion products
pCO2 = (vCO2*100)/vT;			#Percentage volume of CO2
pN2 = (vN2*100)/vT;			#Percentage volume of N2
pO2 = (veO2*100)/vT;			#Percentage volume of O2

# Results
print 'Volume of air required for complete combustion is %3.3f m**3)  \
\nVolume of CO2 per m**3 of gas fuel is %3.2f m**3/m**3 of gas fuel  \
\nVolume of N2 per m**3 of gas fuel is %3.3f m**3/m**3 of gas fuel  \
\nVolume of excess O2 per m**3 of gas fuel is %3.2f m**3/m**3 of gas fuel  \
\nTotal volume of dry combustion products is %3.3f m**3/m**3 of gas fuel \
\nPercentage volume of CO2 is %3.1f percent  \
\nPercentage volume of N2 is %3.2f percent  \
\nPercentage volume of O2 is %3.2f percent'%(v,vCO2,vN2,veO2,vT,pCO2,pN2,pO2)

Volume of air required for complete combustion is 0.952 m**3)
Volume of CO2 per m**3 of gas fuel is 0.29 m**3/m**3 of gas fuel
Volume of N2 per m**3 of gas fuel is 1.603 m**3/m**3 of gas fuel
Volume of excess O2 per m**3 of gas fuel is 0.08 m**3/m**3 of gas fuel
Total volume of dry combustion products is 1.973 m**3/m**3 of gas fuel
Percentage volume of CO2 is 14.7 percent
Percentage volume of N2 is 81.25 percent
Percentage volume of O2 is 4.05 percent