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// Theory and Problems of Thermodynamics
// Chapter 4
// Energy Analysis of Process
// Example 14
clear ;clc;
//Given data
V = 1 // Volume of tank in m^3
P0 = 1 // Initial Pressure in MPa
Pf = 0.8 // final Pressure in MPa
P0 = P0*1e3 // units conversion MPa to kPa
Pf = Pf*1e3 // units conversion MPa to kPa
// at P = 1 MPa
v0 = 0.19444 // units m^3/kg
h0 = 2778.1 // units kJ/kg
m0 = V/v0 // mass in kg
u0 = h0-P0*v0 // units kJ/kg
mf = 4.243 // assume mf in kg
vf = V/mf // units in m3/kg
//at 0.8 MPa
vf_sat = 0.001115 // data from steam tables units m^3/kg
vg = 0.2404 // data from steam tables units m^3/kg
hf_sat = 721.11 // data from steam tables units kJ/kg
hg = 2769.1 // data from steam tables units kJ/kg
hfg = 2048.97 // data from steam tables units kJ/kg
// Calculations
X = (vf - vf_sat)/(vg - vf_sat) // Fraction of steam
hf = hf_sat + X * hfg // units kJ/kg
uf = hf-Pf*vf // units kJ/kg
h2 = (h0 + hg) * 0.5
// substitute the values in LHS and RHS of equation D till both gets same
//(m0-mf)*h2 = m0*u0 - mf*uf
LHS = (m0-mf)*h2
RHS = m0*u0 - mf*uf
mass_steam = m0-mf // mass of the steam in kg
// Output Results
mprintf('Amount of steam withdrawn from tank = %3.1f kg', mass_steam)
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