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// Theory and Problems of Thermodynamics
// Chapter 9
// Air_water Vapor Mixtures
// Example 10
clear ;clc;
//Given data
RH = 0.6 // relative humidity moist air
T1 = 313.15 // Temp of air entering cooling device in K
T2 = 293.15 // temperature at air required to leave in K
P = 101.325 // pressure of entering air in kPa
V = 300 // volumatric flow rate of air in m^3/min
M = 28.97 // molecular weight of air
R = 8.314 // gas constant
// data from psychometric chart
SH1 = 0.0284 // in kg H2O/kg air
pw1 = 4.4 // in kPa
h1 = 115 // in kJ/kg air
SH2 = 0.0148 // in kg H2O/kg air
h2 = 58 // in kJ/kg air
hw3 = 83.95 // in kJ/kg at T = 20`C
// at TDB = 40 `C; RH = 1
ma1 = (101.325-pw1)*1e3*V*M/(R*1e3*T1)
// masss balance for H2O: ma1*SH1 = ma2*SH2 + mw3
mw3 = ma1*(SH1-SH2) // amount of water condensed in kg/min
// Energy balance: ma1*h1 = ma2*h2 + mw3*hw3 + Qc
Qc = ma1*(h1-h2) - mw3*hw3 // in kJ/min
Qc = Qc*1e-3/60 // units conversion kJ/min to MW
// Output Results
mprintf('The amount of water condensed = %4.1f kg/min' , mw3);
mprintf('\n The rate of cooling required = %4.1f MW' , Qc);
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