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// Theory and Problems of Thermodynamics
// Chapter 5
//Second Law of Thermodynamcis
// Example 20
clear ;clc;
//Given data
P1 = 0.5 // entering pressure of air at nozzle in MPa
T1 = 500 // entering temperature of air at nozzle in K
V1 = 2 // velocity of air enters at nozzle
P2 = 0.1 // leaving pressure of air from nozzle in MPa
r = 1.4 // ratio of specific heats of air
R = 8.314 // gas constant
T2 = T1*(P2/P1)^((r-1)/r) // enthalphy for superheated steam kJ/kg
// from law of thermodynamics
//(V2^2)/2 = h1 + (V1^2)/2 - h2
//(V2^2)/2 = cp*(T1-T2) + (V1^2)/2
deff('y=mass(V2)', 'y = (V2^2)/2 - (R*r/(r-1))*(T1-T2) + (V1^2)/2')
V2 = fsolve(0,mass) // Mass of water vaporised in kg
// Output Results
mprintf('The exit velocity of air = %5.2f m/s', V2)
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