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+
+// Theory and Problems of Thermodynamics
+// Chapter 5
+//Second Law of Thermodynamcis
+// Example 22
+
+clear ;clc;
+
+//Given data
+Ph = 0.1                // pressure of hot air flow in MPa
+Th = 400                // temperature of hot air flow in K
+Pc = 0.1                // pressure of cold air flow in MPa
+Tc = 200                // temperature of cold air flow in K
+Pf = 0.5                // pressure of air flow fed in MPa
+Tf = 300                // temperature of air flow fed in K
+R = 8.314               // universal gas constant in J/mol/K
+Cp = 3.5 * R            // specific heat of ideal gas
+
+// Suppose 1 mol air at 0.5 MPA, 300 K (State i) enters the device
+// The device gives 0.5 mol air at 0.1 MPa, 400K(state e1) and
+// 0.5 mol air at 0.1 MPa, 200K(state e2)
+// The first law of thermodynamics gives  he-hi=0
+ h12 = Pf*Cp*{Th-Tf+Tc-Tf}  // h12 = (he1 - hei)+(he2-hei)
+// The Second law of thermodynamics for a flow process gives se-s1>=0
+se1i = Pf*{Cp*log(Th/Tf)-R*log(Ph/Pf)}   // se1i = se1-si (J/K)
+se2i = Pf*{Cp*log(Tc/Tf)-R*log(Pc/Pf)}   // se2i = se2-si (J/K)
+sei  = se1i + se2i                      // se1i = se1-si (J/K)
+
+// Output Results
+mprintf('The sum of difference in he1 and hi and difference in he2 and hi = %1.0f J/mol', h12)
+mprintf('\n The difference in se1 and si = %6.4f J/mol K', se1i)
+mprintf('\n The difference in se2 and si = %7.4f J/mol K', se2i)
+mprintf('\n The difference in se  and si = %6.4f J/mol K', sei)
+mprintf('\n The device satisfies the laws of thermodynamics. It is Theoritically feasible')