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+// Example 25_4
+clc;funcprot(0);
+//Given data 
+W_t=100*10^3;// kW
+W_g=(60/100)*W_t;// kW
+T_1=300;// K
+p_1=1;// bar
+T_3=1000+273;// K
+p_r=8;// Pressure ratio
+n_c=0.85;// Isentropic efficiency of compressor
+n_t=0.90;// Isentropic efficiency of both turbines
+n_com=0.95;// Combustion efficiency
+Gc=2500;// Rs./ton
+T_7=600+273;// K
+T_6=200+273;// K
+p_9=0.05;// bar
+C_pa=1;// kJ/kg.K
+C_pg=1.1;// kJ/kg.K
+r_a=1.4;// Specific heat ratio for air
+r_g=1.33;// Specific heat ratio for gases
+CV=40*10^3;// kJ/kg
+dT=10;// °C
+C_pw=4.2;// kJ/kg°C
+
+// Calculation
+// Considering compressor
+p_2=p_1*p_r;// bar
+T_2a=T_1*(p_r)^((r_a-1)/r_a);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+// Considering turbine
+T_4a=T_3/(p_r)^((r_g-1)/r_g);// K
+T_4=T_3-((T_3-T_4a)*n_t);// K
+T_5=T_3;// K
+// Considering heat balance in CC-I
+function[X]=mass(y);
+    X(1)=(y(1)*CV*n_com)-((y(2)+y(1))*C_pg*(T_3-T_2));
+    X(2)=(W_g)-(((y(2)+y(1))*C_pg*(T_3-T_4))-(y(2)*C_pa*(T_2-T_1)));
+    X(3)=(y(3)*CV*n_com)-((y(2)+y(1)+y(3))*C_pg*(T_5-T_4));
+endfunction
+y=[1 100 1];
+z=fsolve(y,mass);
+m_a1=z(2);// kg/sec
+m_f1=z(1);// kg/sec
+m_f2=z(3);// kg/sec
+AF_1=m_a1/m_f1;
+m_f=m_f1+m_f2;// kg/sec
+Q_s=(m_f*CV);// kW
+n=(W_t/Q_s)*100;// Efficiency of the plant in %
+// From h-s chart:
+h_7=3610;// kJ/kg
+// From steam table
+h_9=32.6;// kJ/kg
+m_s=((m_a1+m_f1+m_f2)*C_pg*(T_5-T_6))/(h_7-h_9);// kg/sec
+Afsf=m_a1/m_s;// Air flow to steam flow ratio
+Cf=((m_f*3600)/1000)*Gc;// Cost of fuel per hour in rupees
+E_g=W_t;// Energy generated per hour kWh
+Cg=Cf/E_g;// Cost of generation in rupees/kWh
+// From h-s chart:
+h_8=2220;// kJ/kg
+m_w=(m_s*3600*(h_8-h_9))/(C_pw*dT*1000);// Quantity of cooling water required in tons/hr
+printf('\n(i)Overall efficiency of the plant=%0.1f percentage \n(ii)Air flow to steam flow ratio=%0.2f \n(iii)Cost of generation=%0.2f rupees/kWh \n(iv)Quantity of cooling water required=%0.0f tons/hr',n,Afsf,Cg,m_w);
+// The answer provided in the textbook is wrong