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diff --git a/599/CH6/EX6.6/example6_6.sce b/599/CH6/EX6.6/example6_6.sce
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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
+\ No newline at end of file