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+// Example 7_7

+clc;funcprot(0);

+// Given data

+P_1=1*10^6;// Pa

+T_1=300+273;// K

+P_2=0.1*10^6;// Pa

+P_3=0.01*10^6;// Pa

+T_0=25+273;// K

+rho=1000;// kg/m^3

+

+// Calculation

+// From steam tables

+h_1=3051.2;// kJ/kg

+s_1=7.1237;// kJ/kg.K

+s_2=s_1;// kJ/kg.K

+// At P=0.1 MPa,

+x_2=0.96;// The quality of steam at state 2

+h_2=2587.3;// kJ/kg

+s_3=s_2;// kJ/kg.K

+// At P=0.01 MPa,

+x_3=0.86;// The quality of steam at state 3

+h_3=2256.9;// kJ/kg

+// The dead state for water is liquid at 25°C and 100 kPa

+h_f=104.9;// kJ/kg

+h_0=h_f;// kJ/kg

+s_f=0.3672;// kJ/kg.K

+s_0=s_f;// kJ/kg.K

+m_1=1;// kg

+m_2=0.10;// kg

+m_3=m_1-(10/100);// kg

+m_4=0.10;// kg

+s_4=0.6491;// kJ/kg

+h_4=191.8;// kJ/kg

+h_6=192.8;// kJ/kg

+X_2=m_2*[h_2-h_0-(T_0*(s_2-s_0))];// The availability at state 2 in kJ

+W_turb=(m_1*(h_1-h_2))+(m_3*(h_2-h_3));// kJ

+X_4=m_4*[h_4-h_0-(T_0*(s_4-s_0))];// The availability at state 4 in kJ

+W_pump=m_1*((P_1/10^3)-(P_2/10^3))/rho;// kJ

+Q_boil=m_1*(h_1-h_6);// kJ

+e_II=(X_2+W_turb)/(X_4+W_pump+([1-(T_0/T_1)]*Q_boil));// The second law effectiveness 

+printf("\nThe second law effectiveness,e_II=%0.2f",e_II);