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+// Theory and Problems of Thermodynamics

+// Chapter 5

+//Second Law of Thermodynamcis

+// Example 23

+

+clear ;clc;

+

+//Given data

+V = 1                   // volume of insulated tank in m^3

+r = 1.4                 // ratio of specific heat

+P0 = 0.1*1e6            // pressure of air in insulated tank in Pa

+T0 = 300                // temperature of air in insulated tank in K

+Pi = 1*1e6              // pressure of high pressure line in Pa

+Ti = 600                // temperature of high pressure line in K

+Pf = 1*1e6              // final pressure of air in the tank (Pa)

+R = 8.314               // universal gas constant

+// The first law expresion for a control volume (the tank) ignoring the changes//in KE and PE gives

+// mi * hi = dE/dt  

+// Cp*Ti*(Nf-N0)=Cv*(Nf*Tf-N0*T0)                       (A)

+// Nf = Pf*V/(R*Tf);    N0 = P0*V/(R*T0)

+// Equation A can be written as

+deff('y=temp(Tf)', 'y = r*Ti*(Pf/Tf-P0/T0) - Pf + P0') 

+Tf = fsolve(10,temp)

+

+N0 = P0*V/(R*T0)                    //mass of air in tank(mol)

+Nf = Pi*V/(R*Tf)                    //mass of air entered from line(mol)

+m = Nf-N0                           //mass of air entered tank(mol)

+

+// 40.093 mol air changed from 0.1 MPa, 300K to 1 MPa, 711.87 K.

+// The entrophy change of this air is del S1

+S1 = N0*R*{r/(r-1)*log(Tf/T0)-log(Pf/P0)}  

+

+// 128.869 mol air changed from 1 MPa, 600K to 1 MPa, 711.87 K.

+// The entrophy change of this air is del S2

+S2 = m*R*r/(r-1)*log(Tf/Ti)

+

+S = S1 + S2                                // total entropy change

+ 

+

+// Output Results

+mprintf('Mass of air entered the tank = %5.3f mol', m)

+mprintf('\n Entrophy change associated with the process = %5.3f J/K', S)