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+

+clear;

+clc;

+printf("\t Example 6.6\n");

+            //wooden cloth is dried from 100 to 10 and then th efinal moisture content is changed to 16 percent from 10;

+

+Xcr=0.55;             //crtical moisture content

+X1=1;               //moisture content on dry basis intially

+X2=.1;              //moisture content on dry basis finally after drying

+Xbar=.06;            //equillibrium moisture 

+    //since eqn 1 is tobe divided by eqn 2 so let the value of Ls/A*Nc be = 1 as it will be cancelled 

+p=1;                  //let Ls/A*Nc be =p

+p=poly([0],'p');        //calc. of time 1

+tbar=1;                  //since the eqns are independent of tbar

+t1=roots(tbar- p*((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)))); //------eqn1

+X2bar=.16;    

+p=poly([0],'p');        //calc. of time 2

+t2=roots(tbar- p*((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2bar-Xbar))));//------eqn2

+

+    //let t1/t2 be = k

+k=t1/t2;

+ans=1/k-1;            //reduction in time for drying 

+printf("\n the reduction in time for drying is :%f percent",ans*100);

+//end
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