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authorpriyanka2015-06-24 15:03:17 +0530
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+clc;clear;
+//Example 5.13
+
+//given data
+V=6/1000;//in m^3
+Pgage=75;
+Patm=100;
+m1=1;
+Qind=0.5;//d stands for .
+t=30*60;//in s
+
+//calculation
+Pabs=Pgage+Patm;
+//from Table A-5, ths saturation temp
+T=116.04;
+disp(T,'the temperature at which cooking takes plac in C');
+//mass balance me=(m1-m2)cv
+//energy balance Qin - mehe = (m2u2 - m1u1)cv
+Qin=Qind*t;
+//from Table A-5
+he=2700.2;
+vf=0.001;
+vg=1.004;
+uf=486.82;
+ufg=2037.7;
+v1=V/m1;
+x1=(v1-vf)/(vg-vf);
+u1=uf+x1*ufg;
+U=m1*u1;
+//Qin = (m1 - V/v2)*he + (V/v2*u2 - m1*u1)
+//v2=vf + x2*(vg-vf)
+//u2=uf + x2*ufg
+//combining these equations we get
+//solved using EES
+x2=0.009;
+v2=vf + x2*(vg-vf);
+m2=V/v2;
+disp(m2,'amount of water left in kg')