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authorpriyanka2015-06-24 15:03:17 +0530
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+
+clear;
+clc;
+printf("\t Example 5.8\n");
+pa1=4.24 //data: vapour pressure of water at 30degree = 4.24 kpa
+pa2=1.70 // vapour pressure of water at 30degree = 1.70 kpa
+
+//part(i)
+pt=100; //total pressure
+ys_bar=pa1/(pt-pa1); //kg water vapour/kg dry air
+rh=.8; //relative humidity
+pa3=rh*pa1; //partial pressure
+y_bar=pa3*(18/28.84)/(pt-pa3); //molal humidity
+printf("\n the molal humidity:%f kg/kg dry air",y_bar);
+
+//part(ii)
+//under these conditions the air will be saturated at 15 degree as some water is condensed
+pa=1.7;
+pt=200;
+ys=pa/(pt-pa);
+ys_bar=ys*(18/28.84);
+printf("\n the molal humidity if pressure doubled and temp. is 15 :%f kg/kg dry air",ys_bar);
+
+//part(iii)
+Ma=18; //molecular weight
+Mb=28.84; //molecular weight
+Tg=30; //temperature of mixture
+rh=(pa/pa2)*100; //relative humidity in pecentage
+pt=10^5; //total pressure in pascal
+VH=8315*[(1/Mb)+(y_bar/Ma)]*[(Tg+273)/pt]; //humid volume in m^3mixture/kg of dry air
+printf("\n we get humid volume VH as \t :%f m^3/kg of dry air",VH);
+w=100/VH; //100 m^3 of original air
+wo= w*y_bar; //water present in original air
+wf= w*ys_bar; //water present finally
+wc=wo-wf; //water condensed from 100m^3 of original sample
+printf("\n the weight water condensed from 100m^3 of original sample:%f kg",wc);
+
+//part(iv)
+Tg=15; //temperature of mixture
+pt=2*10^5; //total pressure in pascal
+VH=8315*[(1/Mb)+(ys_bar/Ma)]*[(Tg+273)/pt]; //humid volume in m^3mixture/kg of dry air
+vf=VH*110.6; //final volume of mixture
+printf("\n we get VH final volume of mixture as \t :%f m^3",vf);
+
+//end \ No newline at end of file