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+// Example 32_27
+clc;funcprot(0);
+//Given data
+T=[0 2 3 6 8 12 14 15 17 23 24];// Time in hours 
+L=[1200 2000 3000 1500 2500 1800 2000 1000 500 800];//  Load in kW 
+
+//Calculation
+T_p=[0 0 2 2 3 3 6 6 8 8 12 12 14 14 15 15 17 17 23 23 24 24];// Time in hours for load curve
+L_p=[3200 1200 1200 2000 2000 3000 3000 1500 1500 2500 2500 1800 1800 2000 2000 1000 1000 500 500 800 800 200];//  Load in kW for load curve
+xlabel('Time (hours)');
+ylabel('LOAD (kW)');
+xtitle('Fig.32.27 Load curve');
+plot(T_p,L_p,'b')
+a=gca();
+a.x_ticks.labels=["6 A.M","8","10","12 NOON","2","4","6","8","10","12 NIGHT","2","4","6 A.M "];
+a.x_ticks.locations=[0;2;4;6;8;10;12;14;16;18;20;22;24];
+//(a)
+E_t=(L(1)*(T(2)-T(1)))+(L(2)*(T(3)-T(2)))+(L(3)*(T(4)-T(3)))+(L(4)*(T(5)-T(4)))+(L(5)*(T(6)-T(5)))+(L(6)*(T(7)-T(6)))+(L(7)*(T(8)-T(7)))+(L(8)*(T(9)-T(8)))+(L(9)*(T(10)-T(9)))+(L(10)*(T(11)-T(10)));// Total power generated in kW-hrs
+L_max=L(3);// Maximum load in kW
+LF=E_t/(L_max*24);// Load factor
+//(b)
+L_1=1200;// kW
+L_2=800;// kW
+L_3=2*500;// kW
+L_4=300;// kW
+//(c)
+Rc=1200;// Reserve capacity in MW
+Ic=L_1+L_2+L_3+L_4+Rc;// Installed capacity in kW
+CF=(E_t/(Ic*24))*100;// Plant capacity factor
+//(d)
+L_1=1200;// kW
+L_2=800;// kW
+L_3=500;// kW
+L_4=300;// kW
+E=(L_1*17)+(L_2*11)+(L_3*3)+(L_3*7)+(L_3*7)+(L_4*3);// The energy generated by the capacity of the plant in kW-hrs;
+UF=(E_t/E)*100;// Plant use factor
+printf('\n(a)Load factor=%0.3f \n(b)It is obvious from the load curve that the numberof sets required are 5 in number \n   One set of 1200 kW \n   One set of 800 kW \n   Two sets of 500 kW \n   One set of 300 kW \n(c)The reserve capacity of the plant=%0.0f kW \n   Capacity factor=%0.0f percentage \n(d)Plant use factor=%0.0f percentage',LF,Rc,CF,UF);