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+//example 13
+//Calculating mass flow of steam in tank
+clear
+clc
+vf1=0.001725 //in m^3/kg
+vf2=0.0016 //in m^3/kg
+uf1=368.7 //in kJ/kg
+uf2=226 //in kJ/kg
+vg1=0.08313 //in m^3/kg
+vfg2=0.20381
+ug1=1341 //in kJ/kg
+ufg2=1099.7 //in kJ/kg
+Vf=1 //initial volume of liquid in m^3
+Vg=1 //initial volume of vapor in m^3
+mf1=Vf/vf1 //initial mass of liquid in kg
+mg1=Vg/vg1 //initial mass of vapor in kg
+m1=mf1+mg1 //initial mass of liquid in kg
+he=1461.1 //in kJ/kg
+V=2 //volume of tank in m^3
+disp('m1u1=mf1*uf1+mg1*ug1.If x2 is the quality,then m2=V/v2=2/(0.00160+0.20381*x2) and u2=uf2+x2*ufg2=226.0+1099.7*x2.')
+disp('Also,m2*(he-u2)=m1*he-m1u1.From this equation,we will get an equation for x2.')
+x2=((2*1461.1)-(2*226)-(0.00160*634706))/((634706*0.20381)+(2*1099.7)) //quality of ammonia
+v2=0.00160+(0.20381*x2) //final specific volume in m^3/kg
+m2=V/v2 //final mass of ammonia in kg
+m=m1-m2 //mass of ammonia withdrawn
+printf(" \n Hence,mass of ammonia withdrawn is m=%.1f kg. \n",m)
+\ No newline at end of file