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+
+clear;
+clc;
+printf("\t Example 6.10\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
+
+//Ls/ A*Nc is unknown;
+
+Xcr=0.16; //crtical moisture content
+X1=.33; //moisture content on dry basis intially
+X2=0.09; //moisture content on dry basis finally after drying
+Xbar=.05; //equillibrium moisture
+tbar=7; //time needed to dry from 33 to 9 percent on bone dry basis
+
+ //let Ls / A*Nc be = p
+p=poly([0],'p'); //calc. of Ls / A*Nc be = p value
+x=roots(tbar-p * ((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar))));
+
+//new X1 AND X2 are now given as follows
+X1=0.37; //new moisture content on dry basis intially
+X2=0.07; //new moisture content on dry basis finally after drying
+tbar=x * ((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
+printf("\n the time for drying the sheets from 33 to 9 percent moisture under same drying conditions is :%f hr",tbar);
+
+//end \ No newline at end of file