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+
+// Theory and Problems of Thermodynamics
+// Chapter 4
+// Energy Analysis of Process
+// Example 11
+
+clear ;clc;
+
+//Given data
+V = 1                   // volume in m^3
+P0 = 0.2                // Initial Pressure in MPa
+Pf = 3                  // Final Pressure in MPa
+T = 350                // line carrying steam Temperature in C
+P = 3                  // line carrying steam Pressure in MPa
+P = P*1e3              // units conversion MPa tp kPa
+
+// at 2 MPa
+v_g = 0.8857            // v_g = v_0  units in m^3  
+h_g = 2706.7            // h_g = h_0  units kJ/kg
+
+m0 = V/v_g              // mass in kg
+u0 = h_g - P0*v_g       // units kJ/kg
+
+// at 3 MPa and 350 C
+h = 3115.3              // units kJ/kg
+
+// the first law of thermodynamics for the transient flow process
+// h1*(mf - m0) = mf*uf - m0*u0     (A)
+// the above equation convets after substituting the values
+// mf*(3115.3 -uf) = 661.3          (B)
+//the mass steam in final stage
+//mf = V/vf                         (C)
+
+// This can be solved by trail and error
+// Assume Tf
+Tf = 463                // temperature in C
+//at P = 3 MPa and T = 463 C
+vf = 0.1100             // data from steam tables 
+hf = 3373.25            // data from steam tables
+uf = hf-P*vf            // units kJ/kg
+mf = V/vf               // mass of steam in kg
+
+// substitute the values in LHS and RHS of equation B till both gets same
+LHS = mf*(h -uf)
+RHS = 661.3
+
+// Output Results
+mprintf('Mass of steam in the tank = %5.3f kg',mf)
+mprintf('\n Final temperature of steam in tank = %3.0f degree C',Tf)
+
+
+
+
+