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+// Example 34_34
+clc;funcprot(0);
+//Given data
+P=[600 600 600 400];// Capacity of 4-generating sets in kW
+MD=1600;// kW
+F_l=0.45;// Load factor
+CC=10000;// Capital cost in Rs./kW
+Mc=60000;// Annual maintainence cost in rupees
+Oc=100000;// Operation cost in rupees
+Fc=7;// Fuel cost in Rs./kg
+Lc=40;// Lubricating oil cost in Rs./kg
+F=0.5;// Fuel consumed in kg/kWh
+O=0.0025;// Lubricating oil consumed in kg/kWh
+CV=42000;// kJ/kg
+n_g=0.92;// Generator efficiency
+
+//Calculation
+//(a)
+R_f3=P(1)/n_g;// Rating of first 3 sets in kW
+R_4=P(4)/n_g;// Rating of last set in kW
+//(b)
+AD=MD*F_l;// Average demand in kW
+E_g=AD*8760;// Energy generated/year in kWh
+//(c)(i)Fixed costs per year
+CC=((3*P(1))+(1*P(4)))*CC;
+Afc=.15*CC;// Annual fixed cost in rupees
+Tfc=Afc+Mc;// Total fixed cost in rupees
+//(i)Variable costs per year
+Fc=(E_g*F)*Fc;// Fuel cost in rupees
+Lc=(E_g*O)*Lc;// Lubricating oil cost in rupees
+Tvc=Fc+Lc+Oc;// Total variable cost in rupees
+Tc=Tfc+Tvc;// Total cost in rupees
+C=Tc/E_g;// Cost per kWh generated
+//(d)
+n_o=((E_g*3600)/(E_g*F*CV))*100;// Over all efficiency of the plant in %
+printf('\n(a)Rating of first 3 sets=%0.0f kW \n Rating of last set=%0.0f kW \n(b)Energy generated/year=%0.1e kWh \n(c)Cost of generation=Rs.%0.2f \n(d) Over all efficiency of the plant=%0.2f percentage',R_f3,R_4,E_g,C,n_o);
+// The answer vary due to round off error