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+
+clear;
+clc;
+printf("\t Example 6.14\n");
+
+d=450; //density of dry pulp in kg/m^3;
+thickness=0.05; //thickness in m^2
+Ls=d*thickness; //mass of bone dry solid ais the drying surface
+A=1; //area in m^2
+v=1*5*10^-2; //volume of material
+Nc=4.8; //in kg/m^2*hr
+xcr=.2;
+xbar=0.02;
+x1=.45; //new moisture content on wet basis
+x2=0.05; //new moisture content on wet basis
+X1=x1/(1-x1); //new moisture content on dry basis intially
+X2=x2/(1-x2); //new moisture content on dry basis finally after drying
+Xbar=xbar/(1-xbar); //crtical moisture content
+Xcr=xcr/(1-xcr); //equillibrium moisture
+
+//tbar=(Ls/(A*Nc))*((Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
+ // but initial moisture is more than Xcr, so there is constant rate drying period and only falling rate peroid is observed
+tbar=Ls/(A*Nc) * ((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
+printf("\n the time for drying the sheets from 45 to 5 percent moisture under same drying conditions is :%f min",tbar);
+
+//end \ No newline at end of file