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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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