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+// Example 24_15
+clc;funcprot(0);
+//Given data 
+m_a=10;// kg/sec
+p_r=6;// Pressure ratio
+T_1=300;// K
+p_1=1;// bar
+T_6=1073;// K
+e=0.75;// The effectiveness of regenerator
+n_c=0.80;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of turbine
+C_pa=1;// kJ/kg.K
+r=1.4;// Specific heat ratio
+m=1;// kg
+
+//Calculation
+p_3=p_1*p_r;// bar
+p_2=sqrt(p_1*p_3);// bar
+T_2a=T_1*(p_2/p_1)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+//W_c=W_c1+W_c2=2*W_c1 (as intercooling is perfect)
+W_c=2*m*C_pa*(T_2-T_1);// kJ/kg
+// As T_3=T_1 and p_r=(p_2/p_1)=(p_3/p_2)
+T_4=T_2;// K
+T_7a=T_6/(p_3/p_1)^((r-1)/r);// K
+T_7=T_6-(n_t*(T_6-T_7a));// K
+W_t=C_pa*(T_6-T_7);// kJ/kg
+T_5=T_4+(e*(T_7-T_4));// K
+Q_s=m*C_pa*(T_6-T_5);// kJ/kg 
+W_n=W_t-W_c;// kJ/kg 
+P=m_a*W_n;//Power capacity of the plant in kW
+n_th=(W_n/Q_s)*100;// Thermal Efficiency in percentage
+printf('\nPower capacity of the plant=%0.0f kW\nThe thermal efficiency of the plant=%0.1f percentage',P,n_th);
+// The answer vary due to round off error