diff options
Diffstat (limited to '1775/CH1')
| -rwxr-xr-x | 1775/CH1/EX1.1/Chapter1_Example1.sce | 24 | ||||
| -rwxr-xr-x | 1775/CH1/EX1.10/Chapter1_Example10.sce | 25 | ||||
| -rwxr-xr-x | 1775/CH1/EX1.11/Chapter1_Example11.sce | 33 | ||||
| -rwxr-xr-x | 1775/CH1/EX1.12/Chapter1_Example12.sce | 40 | ||||
| -rwxr-xr-x | 1775/CH1/EX1.2/Chapter1_Example2.sce | 25 | ||||
| -rwxr-xr-x | 1775/CH1/EX1.3/Chapter1_Example3.sce | 25 | ||||
| -rwxr-xr-x | 1775/CH1/EX1.4/Chapter1_Example4.sce | 43 | ||||
| -rwxr-xr-x | 1775/CH1/EX1.5/Chapter1_Example5.sce | 48 | ||||
| -rwxr-xr-x | 1775/CH1/EX1.6/Chapter1_Example6.sce | 30 | ||||
| -rwxr-xr-x | 1775/CH1/EX1.7/Chapter1_Example7.sce | 55 | ||||
| -rwxr-xr-x | 1775/CH1/EX1.8/Chapter1_Example8.sce | 49 | ||||
| -rwxr-xr-x | 1775/CH1/EX1.9/Chapter1_Example9.sce | 52 |
12 files changed, 449 insertions, 0 deletions
diff --git a/1775/CH1/EX1.1/Chapter1_Example1.sce b/1775/CH1/EX1.1/Chapter1_Example1.sce new file mode 100755 index 000000000..12b509a48 --- /dev/null +++ b/1775/CH1/EX1.1/Chapter1_Example1.sce @@ -0,0 +1,24 @@ +//Chapter-1, Illustration 1, Page 15
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+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
+
+//OUTPUT
+mprintf('Minimum mass of air per kg of coal is %3.2f kg',m)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.10/Chapter1_Example10.sce b/1775/CH1/EX1.10/Chapter1_Example10.sce new file mode 100755 index 000000000..79466d5e7 --- /dev/null +++ b/1775/CH1/EX1.10/Chapter1_Example10.sce @@ -0,0 +1,25 @@ +//Chapter-1, Illustration 10, Page 23
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+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)
+
+//OUTPUT
+mprintf('Volume of air required for complete combustion is %3.3f (m^3)',v)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.11/Chapter1_Example11.sce b/1775/CH1/EX1.11/Chapter1_Example11.sce new file mode 100755 index 000000000..329cb1ff6 --- /dev/null +++ b/1775/CH1/EX1.11/Chapter1_Example11.sce @@ -0,0 +1,33 @@ +//Chapter-1, Illustration 11, Page 24
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+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
+
+//OUTPUT
+mprintf('Air-fuel ratio by volume is %3.3f:1',V)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.12/Chapter1_Example12.sce b/1775/CH1/EX1.12/Chapter1_Example12.sce new file mode 100755 index 000000000..d3d2ed7af --- /dev/null +++ b/1775/CH1/EX1.12/Chapter1_Example12.sce @@ -0,0 +1,40 @@ +//Chapter-1, Illustration 12, Page 24
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+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
+
+//OUTPUT
+mprintf('Volume of air required for complete combustion is %3.3f (m^3) \n Volume of CO2 per m^3 of gas fuel is %3.2f m^3/m^3 of gas fuel \n Volume of N2 per m^3 of gas fuel is %3.3f m^3/m^3 of gas fuel \n Volume of excess O2 per m^3 of gas fuel is %3.2f m^3/m^3 of gas fuel \n Total volume of dry combustion products is %3.3f m^3/m^3 of gas fuel \n Percentage volume of CO2 is %3.1f percent \n Percentage volume of N2 is %3.2f percent \n Percentage volume of O2 is %3.2f percent',v,vCO2,vN2,veO2,vT,pCO2,pN2,pO2)
+
+
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.2/Chapter1_Example2.sce b/1775/CH1/EX1.2/Chapter1_Example2.sce new file mode 100755 index 000000000..e9298ab9f --- /dev/null +++ b/1775/CH1/EX1.2/Chapter1_Example2.sce @@ -0,0 +1,25 @@ +//Chapter-1, Illustration 2, Page 16
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+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
+
+//OUTPUT
+mprintf('Theoretical volume of air at N.T.P per kg fuel is %3.2f (m^3)/kg of fuel',vth)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.3/Chapter1_Example3.sce b/1775/CH1/EX1.3/Chapter1_Example3.sce new file mode 100755 index 000000000..6b98548ab --- /dev/null +++ b/1775/CH1/EX1.3/Chapter1_Example3.sce @@ -0,0 +1,25 @@ +//Chapter-1, Illustration 3, Page 16
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+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
+
+//OUTPUT
+mprintf('Minimum quantity of air required for complete combustion is %3.2f kg \n Total mass of products of combustion is %3.3f kg',m,mt)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.4/Chapter1_Example4.sce b/1775/CH1/EX1.4/Chapter1_Example4.sce new file mode 100755 index 000000000..0e1b512ac --- /dev/null +++ b/1775/CH1/EX1.4/Chapter1_Example4.sce @@ -0,0 +1,43 @@ +//Chapter-1, Illustration 4, Page 17
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+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
+
+//OUTPUT
+mprintf('Mass of dry flue gases per kg of coal burnt is %3.1f kg',md)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.5/Chapter1_Example5.sce b/1775/CH1/EX1.5/Chapter1_Example5.sce new file mode 100755 index 000000000..4891b92d1 --- /dev/null +++ b/1775/CH1/EX1.5/Chapter1_Example5.sce @@ -0,0 +1,48 @@ +//Chapter-1, Illustration 5, Page 17
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+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
+
+//OUTPUT
+mprintf('Minimum air required to burn 1 kg of coal is %3.2f kg \n Mass of air actually supplied per kg of coal is %3.3f kg \n Amount of excess air supplied per kg of coal burnt is %3.3f kg',m,mact,me)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.6/Chapter1_Example6.sce b/1775/CH1/EX1.6/Chapter1_Example6.sce new file mode 100755 index 000000000..6b95d1115 --- /dev/null +++ b/1775/CH1/EX1.6/Chapter1_Example6.sce @@ -0,0 +1,30 @@ +//Chapter-1, Illustration 6, Page 19
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+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
+
+//OUTPUT
+mprintf('Mass of air to be supplied is %3.2f kg \n Mass of CO2 produced per kg of coal is %3.2f kg \n Mass of H2O produced per kg of coal is %3.2f kg \n Mass of SO2 produced per kg of coal is %3.2f kg \n Mass of excess O2 produced per kg of coal is %3.2f kg \n Mass of N2 produced per kg of coal is %3.2f kg \n',m,mCO2,mHw,mSO2,mO2,mN2)
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.7/Chapter1_Example7.sce b/1775/CH1/EX1.7/Chapter1_Example7.sce new file mode 100755 index 000000000..58929075d --- /dev/null +++ b/1775/CH1/EX1.7/Chapter1_Example7.sce @@ -0,0 +1,55 @@ +//Chapter-1, Illustration 7, Page 20
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+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
+
+//OUTPUT
+mprintf('Minimum mass of air required is %3.1f kg \n Total mass of dry flue gases per kg of fuel is %3.2f kg \n Percentage composition of CO2 by volume is %3.2f percent \n Percentage composition of SO2 by volume is %3.3f percent \n Percentage composition of O2 by volume is %3.1f percent \n Percentage composition of N2 by volume is %3.2f percent',m,md,d1,d2,d3,d4)
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.8/Chapter1_Example8.sce b/1775/CH1/EX1.8/Chapter1_Example8.sce new file mode 100755 index 000000000..d7be45407 --- /dev/null +++ b/1775/CH1/EX1.8/Chapter1_Example8.sce @@ -0,0 +1,49 @@ +//Chapter-1, Illustration 8, Page 21
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+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
+
+//OUTPUT
+mprintf('Mass of air actually supplied per kg of coal is %3.2f kg \n Percentage of excess air is %3.2f percent',mact,pe)
+
+
+
+
+
+
+
+//==============================END OF PROGRAM=================================
diff --git a/1775/CH1/EX1.9/Chapter1_Example9.sce b/1775/CH1/EX1.9/Chapter1_Example9.sce new file mode 100755 index 000000000..f35e7a7ae --- /dev/null +++ b/1775/CH1/EX1.9/Chapter1_Example9.sce @@ -0,0 +1,52 @@ +//Chapter-1, Illustration 9, Page 22
+//Title: Fuels and Combustion
+//=============================================================================
+clc
+clear
+
+//INPUT DATA
+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
+
+//OUTPUT
+mprintf('Mass of excess air supplied per kg of fuel burnt is %3.1f kg/kg of fuel \n Air-fuel ratio is %3.1f:1',me,mt)
+
+
+
+
+
+//==============================END OF PROGRAM=================================
+
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