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+// Example 8_6
+clc;funcprot(0);
+// Given data
+P_1=10;// kPa
+P_2=2;// MPa
+P_5=200;// kPa
+T_3=600;// °C
+h_2=192;// kJ/kg
+h_3=3690;// kJ/kg
+s_3=7.7032;// kJ/kg.K
+
+// Calculation
+// We have from Example 8.3 and the steam tables
+h_1=h_2;// kJ/kg
+h_7=505;// kJ/kg
+h_6=h_7;// kJ/kg
+h_4=2442;// kJ/kg
+h_5=(((7.7032-7.5074)/(7.7094-7.5074))*(2971-2870))+2870;// kJ/kg
+m_6=1;// kg
+m_5=((h_6-h_2)/(h_5-h_2))*m_6;// kg
+m_2=m_6-m_5;// kg
+w_T=(h_3-h_5)+((h_5-h_4)*m_2);// The work output from the turbine in kJ/kg
+q_B=h_3-h_7;// kJ/kg
+n_2=w_T/q_B;// The cycle efficiency
+n_1=0.3568;// The power cycle efficiency from example 8.3
+Pi=((n_2-n_1)/n_1)*100;// The percentage increase in efficiency
+printf(" The percentage increase in efficiency is %1.2f percentage.",Pi);
+