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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)
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