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// Theory and Problems of Thermodynamics
// Chapter 9
// Air_water Vapor Mixtures
// Example 9
clear ;clc;
//Given data
V1 = 2 // volume of air conditioner delivers in cubic meter
T1 = 278.15 // Temp of air completely saturates in K
V2 = 5 // volume of air mixed in cubic meter
T2 = 308.15 // Temp of air mixed in K
RH = 0.4 // relative humidity in room
R = 8.314 // gas constant
M = 28.97 // molecular weight of air
// data from psychometric chart
SH1 = 0.0056 // in kg H2O/kg air
pw1 = 0.8 // in kPa
// at TDB = 5 degree C; RH = 1
SH2 = 0.0138 // in kg H2O/kg air
pw2 = 2.2 // in kPa
// at TDB = 5 `C; RH = 1
ma1 = (101.325-pw1)*1e3*V1*M/(R*1e3*T1)
ma2 = (101.325-pw2)*1e3*V2*M/(R*1e3*T2)
//ma1/ma2 = (SH2-SH3)/(SH3-SH1)
deff('y=SH(SH3)', 'y = ma1/ma2 - (SH2-SH3)/(SH3-SH1)')
SH3 = fsolve(0.01,SH)
// Join the states 1 and 2 by a straight line and read TDB = 26 degree C and
// RH3=0.55 at the intersection of the line with SH3=0.0113 kg H20/kg air
// Output Results
mprintf('Specific humidity of air after mixing = %4.4f kg H20/kg air' , SH3);
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