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clear;
clc;
printf("\t Example 6.12\n");
d=.22; //density of dry pulp in g/cc;
Ls=1.125*10^-2*.22*10^3; //mass of bone dry solid ais the drying surface
A=1.5*1.5*2; //both upper surafce and lower surface are exposed
v=1.5*1.5*.5; //volume of material
Nc=1.4; //in kg/m^2*hr
x2=.2; //moisture content on wet basis finally after drying
Xcr=0.46; //crtical moisture content
X1=0.15; //moisture content on dry basis intially
X2=0.085; //moisture content on dry basis finally after drying
Xbar=0.025; //equillibrium moisture
//tbar=(Ls/(A*Nc))*((Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
// but initial moisture is more than Xcr, so there is constant rate drying period and only falling rate peroid is observed
tbar=(Ls/(A*Nc))*((Xcr-Xbar)*log((X1-Xbar)/(X2-Xbar)));
printf("\n the time for drying the sheets from 15 to 8.5 percent moisture under same drying conditions is :%f min",tbar*60);
//end
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