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// Theory and Problems of Thermodynamics
// Chapter 4
// Energy Analysis of Process
// Example 13
clear ;clc;
//Given data
P = 1 // Initial Pressure in MPa
T = 200 // line carrying steam Temperature in C
Pf = 1 // final Pressure in MPa
Pf = Pf*1e3 // units conversion MPa to kPa
// at P = 1 MPa T = 200 C
h1 = 2827.9 // units kJ/kg
// the first law of thermodynamics for the transient flow process
// h1*(mf - m0) = mf*uf - m0*u0 (A)
// This can be solved by trail and error
// Assume Tf
Tf = 250 // temperature in C
//at P = 1 MPa and T = 250 C
vf = 0.2327 // data from steam tables units m^3/kg
hf = 2942.6 // data from steam tables units kJ/kg
uf = hf-Pf*vf // units kJ/kg
// substitute the values in LHS and RHS of equation B till both gets same
//h1 = uf +Pf*vf/2
RHS = uf + (Pf*vf/2)
LHS = h1
// Output Results
mprintf('The approximate value equal to h1 = %6.1f kJ/kg',h1)
mprintf('\nFinal temperature of steam in tank = %4.0f degree C',Tf)
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