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+clear ;
+clc;
+// Example 6.5
+printf('Example 6.5\n\n');
+printf('Page No. 148\n\n');
+
+// given
+F = 1;// Weight of coal in kg
+//By analysis of coal in weight basis
+C = 0.74;// Carbon percentage - [%]
+H2 = 0.05;// Hydrogen percentage - [%]
+S = 0.01;// Sulphur percentage - [%]
+N2 = 0.001;// Nitrogen percentage - [%]
+O2 = 0.05;// Oxygen percentage - [%]
+H20 = 0.09;// Moisture percentage - [%]
+Ash = 0.05;// Ash percentage - [%]
+
+w_C = 12; // mol. weight of C
+w_H2 = 2; //mol. weight of H2
+w_O2 = 32; // mol. weight of O2
+w_S = 32; //mol. weight of S
+//Basis- Per kg of fuel
+mol_C = C / w_C;// kmol of C
+mol_H2 = H2 /w_H2;//kmol of H2
+mol_O2 = O2 /w_O2;//kmol of O2
+mol_S = S /w_S;//kmol of S
+//Calculation of excess air
+C_req = mol_C*1;//O2 required by entering C given by reaction C+O2->CO2 in kmol
+H_req = mol_H2*0.5;//O2 required by entering H2 given by reaction H2+(1/2)O2->H20 in kmol
+S_req = mol_S*1;//O2 required by entering S given by reaction S+O2->SO2 in kmol
+O2_req = (C_req + H_req + S_req) - mol_O2;// Total number of kmol of O2 required per kg of fuel in kmol
+m_O2 = O2_req*w_O2;// Mass of O2 required per kg of fuel
+printf('Mass of O2 required per kg of fuel is %3.2f kg \n',m_O2)
+//Calculation of air
+m_air = m_O2/0.232;// in kg
+printf('Mass of air required per kg of fuel is %3.1f kg \n',m_air')
+//Considering air as an ideal gas,calculating volume of air by ideal gas equation-P*V = n*R*T
+R = 8310;//Universal gas constant in J/kmol-K
+T = (273+0);// in K
+P = 1.013*10^5;// in N/m^2
+n = 1;// 1 kmol of air
+V_kmol = (n*R*T)/P;// In m^3/kmol
+M_air = 29;// Mol. weight of air
+V_kg = V_kmol/M_air;// in m^3/kg
+V_air = m_air*V_kg;// in m^3
+printf('Volume of air required is %3.1f m^3\n',V_air')
+
+
+
+
+