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// Theory and Problems of Thermodynamics
// Chapter 5
//Second Law of Thermodynamcis
// Example 19
clear ;clc;
//Given data
T1 = 573.15 // entering temperature of superheated steam in K
P1 = 3.0 // entering pressure of superheated steam in MPa
P2 = 20 // leaving pressure of superheated steam in MPa
Ws = 1e3 // output power of turbine in KW
// For steam at 3.0 MPa and 573.15 K
h1 = 2993.5 // enthalphy for superheated steam kJ/kg
s1 = 6.539 // entrophy for superheated steam kJ/kg K
// at 20 kPa
hf = 251.4 // enthalphy kJ/kg
hfg = 2358.3 // enthalphy kJ/kg
sf = 0.832 // entrophy kJ/kg K
sfg = 7.9085 // entrophy kJ/kg K
// Reversible, adiabatic turbine: s1 = s2
//energy balance
deff('y=mass(X)', 'y = sfg*X + (1-X)*sf - s1')
X = fsolve(0,mass) // Mass of water vaporised in kg
h2 = hf + X * hfg // enthalphy for leaving steam
m = -Ws/(h2-h1) // flow rate of steam in kg/s
// Output Results
mprintf('The flow rate of steam = %4.2f kg/s', m)
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