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authorpriyanka2015-06-24 15:03:17 +0530
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+
+clear;
+clc;
+printf("\t Example 5.2\n");
+ //dry bulb temperature=25 and wet bulb temperature=22
+Tg=25; //dry bulb temperature=50
+To=0; //refrence temperature in degree celcius
+Mb=28.84; //average molecular weight of air
+Ma=18; //average molecular weight of water
+
+//part(i)
+hum=.0145 //0.0145 kg of water/kg of dry air
+printf("\n the saturation humidity(from chart) is :%f percent",hum);
+
+//part(ii)
+humper=57; //humidity percentage
+printf("\n the percentage humidity is \t\t:%f percent",humper);
+
+//part(iii)
+pt=1; //total pressure in atm
+sathum=0.0255; //molal humidity =pa/(pt-pa)
+pa1=sathum*pt*(28.84/18)/(1+(sathum*(28.84/18)));
+//the vopour pressure of water(steam tables)at 25 = .0393*10^5 N/m^2
+pt=1; //total pressure in atm
+molhum=0.0145; //molal humidity =pa/(pt-pa)
+pa2=molhum*pt*(28.84/18)/(1+(molhum*pt*(28.84/18)));
+//the vopour pressure of water(steam tables)at 25 = .0393*10^5 N/m^2
+relhum=(pa2/pa1)*100; //percentage relative humidity =partial pressure/vapour pressure
+printf("\n the percentage relative humidity is \t :%f ",relhum);
+
+//part(iv)
+dewpoint=19.5; //dew point temperature in degree celcius
+printf("\n the dew point temperature \t :%f degree celcius",dewpoint);
+
+//part(v)
+Ca=1005;
+Cb=1884;
+ybar=.0145; // humidity inkg water /kg dry air
+Cs=Ca+Cb*ybar; //humid heat in j/kg dry air degree celcius
+d=2502300; //latent heat in j/kg
+H=Cs*(Tg-0)+ybar*d; //enthalpy for refrence temperature of 0 degree
+printf("\n we get Humid heat H as \t :%f j/kg",H);
+//the actual answer is 62091.3 bt in book it is given 65188.25(calculation mistake in book)
+//end \ No newline at end of file