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// Theory and Problems of Thermodynamics
// Chapter 9
// Air_water Vapor Mixtures
// Example 6
clear ;clc;
//Given data
V = 20*40*6 // dimensions of auditorium in cubic meters
RH = 0.6 // relative humidy
P = 101.325 // pressure of auditorium in kPa
T = 303.15 // Temperature auditorium in K
Ps_20 = 2.339 // saturation pressure of water at 20 degree C
Ps_25 = 3.169 // saturation pressure of water at 25 degree C
Ps_30 = 4.246 // saturation pressure of water at 30 degree C
M = 18 // molecular weight of water
R = 8.314 // gas constant
// Calculations
pw = RH*Ps_30 // partial pressure of water at 30 degree C
SH = 0.622*pw/(P-pw) // specific humidity of air
mw = pw*V*M/(R*T) // mass of water vapor in the auditorium
ps = pw // at dew point from steam table T = 21.3 degree C
// Output Results
mprintf('(a) Partial pressure of water at 30 degree C = %4.4f kPa' , pw);
mprintf('\n (b) Specific humidity of air = %4.3f kg H20/kg air' , SH);
mprintf('\n (c) Mass of water vapor in the auditorium = %4.2f kg' , mw);
mprintf('\n (d) Dew point = %4.4f kPa' , ps);
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