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+// Example 22_30
+clc;funcprot(0);
+//Given data
+p_a=10;// bar
+p_b=0.2;// bar
+p_1=40;// bar
+T_1=400;// °C
+T_2=40;// °C
+m_s=500;// kg/sec
+T_sa=515.5;// °C
+T_sb=277.3;// °C
+h_fa=72.33// kJ/kg
+h_fb=38.35;// kJ/kg
+h_ga=363.0;// kJ/kg
+h_gb=336.55;// kJ/kg
+s_fa=0.1478;// kJ/kg-K
+s_fb=0.0967;// kJ/kg-K
+s_ga=0.5167;// kJ/kg-K
+s_gb=0.6385;// kJ/kg-K
+v_fa=80.9*10^-6;// m^3/kg
+v_fb=77.4*10^-6;// m^3/kg
+v_ga=0.0333;// m^3/kg
+v_gb=1.163;// m^3/kg
+
+//Calculation
+//From h-s chart:
+h_1=3230;// kJ/kg
+h_2=2120;// kJ/kg
+// From steam tables
+h_3=167.5;// kJ/kg
+h_4=h_3;// kJ/kg
+// s_a=s_b
+x_b=(s_ga-s_fb)/(s_gb-s_fb);
+h_b=h_fb+(x_b*(h_gb-h_fb));// kJ/kg
+h_c=38.35;// kJ/kg
+h_d=h_c;// kJ/kg
+//(a)
+h_a=h_ga;// kJ/kg
+m_Hg=(h_1-h_4)/(h_b-h_c);// kg/kg of steam
+n_Hg=((h_a-h_b)/(h_a-h_d))*100;// The efficiency of the mercury cycle in %
+n_H2o=((h_1-h_2)/(h_1-h_3))*100;// The efficiency of the steam cycle in %
+n_o=(((m_Hg*(h_a-h_b))+(1*(h_1-h_2)))/(m_Hg*(h_a-h_c)))*100;// The over all efficiency of the plant in %
+P=((m_s/60)*((m_Hg*(h_a-h_b))+(1*(h_1-h_2))))/1000;// Total power generated in the system in MW
+printf('\nMass of mercury required to generate one kg of steam=%0.2f kg/kg of steam \nThe efficiency of the mercury cycle=%0.1f percentage \nThe efficiency of the steam cycle=%0.2f percentage \nThe over all efficiency of the plant=%0.1f percentage \nThe power generating capacity of the plant=%0.2f MW',m_Hg,n_Hg,n_H2o,n_o,P);
+// The answer vary due to round off error