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+clear;

+clc;

+disp('Example 8.23');

+

+//  aim : To determine

+// (a) the actual quantity of air supplied/kg of fuel

+// (b) the volumetric efficiency of the engine

+

+// given values

+d = 300*10^-3;// bore,[m]

+L = 460*10^-3;// stroke,[m]

+N = 200;// engine speed, [rev/min]

+

+C = 87;//  %age mass composition of Carbon in the fuel

+H2 = 13;//  %age mass composition of H2 in the fuel

+

+mc = 6.75;// fuel consumption, [kg/h]

+

+CO2 = 7;// %age composition of CO2 by volume

+O2 = 10.5;// %age composition of O2 by volume

+N2 = 7;// %age composition of N2 by volume

+

+mC = 12;// moleculer mass of CO2,[kg/kmol]

+mH2 = 2;// moleculer mass of H2, [kg/kmol]

+mO2 = 32;// moleculer mass of O2, [kg/kmol]

+mN2 = 28;// moleculer mass of N2, [kg/kmol]

+

+T = 273+17;// atmospheric temperature, [K]

+P = 100;// atmospheric pressure, [kn/m^2]

+R =.287;// gas constant, [kJ/kg k]

+

+// solution

+// (a)

+// combustion equation by no. of moles

+// 87/12 C + 13/2 H2 + a O2+79/21*a N2 = b CO2 + d H2O + eO2 + f N2

+// equating  coefficient

+b = 87/12;// [mol]

+a = 22.7;// [mol]

+e = 10.875;// [mol]

+f = 11.8*b;// [mol]

+// so fuel side combustion equation is

+// 87/12 C + 13/2 H2 +22.7 O2 +85.5 N2

+mair = ( 22.7*mO2 +85.5*mN2)/100;// mass of air/kg fuel, [kg]

+mprintf('\n (a) The mass of actual air supplied per kg of fuel is  =  %f  kg\n',mair);

+

+// (b)

+m = mair*mc/60;// mass of air/min, [kg]

+V = m*R*T/P;// volumetric flow of air/min, [m^3]

+SV = %pi/4*d^2*L*N/2;// swept volume/min, [m^3]

+

+VE = V/SV;// volumetric efficiency

+mprintf('\n (b) The volumetric efficiency of the engine is  =  %fpercent\n',VE*100);

+

+//  End