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+

+clear;

+clc;

+printf("\t Example 6.9\n");

+

+            //determine the drying condition of sample 0.3*0.3 size.sheet lost weight at rate of 10^-4 kg/s until the moisture fell to 60 percent

+            

+x1=.75;                //moisture content on wet basis

+xbar=0.1;                //equilllibrium moisture on dry basis

+xcr=0.6;                //critical moisture content

+Ls=0.90;                //mass of bone dry solid ais the drying surface

+A=0.3*0.3*2;            //both upper surafce and lower surface are exposed

+//A*Nc=10^-4;

+x2=.2;                //moisture content on wet basis  finally after drying

+Xcr=0.6/0.4;          //crtical moisture content

+X1=3;                 //moisture content on dry basis  intially

+X2=0.25;              //moisture content on dry basis  finally after drying

+Xbar=0.1/0.9;        //equillibrium moisture 

+tbar=Ls/(10^-4) * ((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));

+printf("\n the time for drying the sheets from 75 to 25 percent moisture under same drying conditions is :%f hr",tbar/3600);

+

+//end
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