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// Theory and Problems of Thermodynamics
// Chapter 3
// Thermodynamic Properties of Fluids
// Example 10
clear ;clc;
//Given data
X = 0.8 // wet stream quality
P = 0.1 // Pressure in MPa
T = 300 // Temperature in C
h_f = 417.46 // Specific enthalpy of fluid
h_fg = 2258.0 // Specific enthalpy difference of gas and fluid
v_f = 0.001043 // Specific volume of liquid in m^3/kg
v_g = 1.6940 // Specific volume of vapor in m^3/kg
h2 = 3074.3 // specific enthalpy at 300C and 0.1MPa in kJ/kg
v2 = 2.639 // Specific volume at 300C and 0.1MPa in m^3/kg
P = P * 1e6 // Units conversion from MPa to Pa
// Calculations
h1 = h_f + X*h_fg
v1 = X*v_g + (1-X)*v_f
Q = h2 - h1 // heat interaction by the steam
W = P*(v2-v1) // Work done by the steam
W = W * 1e-3 // Units conversion from J to kJ
// Output Results
mprintf('heat interaction by the steam = %6.2f kJ/kg' ,Q)
mprintf('\n Work done by the steam = %6.2f kJ/kg' ,W)
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