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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clc
+//initialisation of variables
+psat= 143.3 //kPa
+R= 8.314 //J/mol K
+T= 110 //C
+m= 18.02 //gms
+pv= 150 //kPa
+v= 0.001052 //m^3/kg
+s= 10^-3
+//CALCULATIONS
+PL= psat+((R*(273.15+T)/(m*0.0010502))*log(pv/psat))
+D= (4*s/(PL-pv))*(75.64-13.91*(T/100)-3*(T/100)^2)*10^3
+//RESULTS
+printf (' equilibrium pressure= %.f kPa',PL-13)
+printf (' \n diameter of droplet= %.4f mm',D)