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clear;
clc;
printf("\t Example 6_8_a\n");
Ls=1000; //mass of bone dry solid ais the drying surface
A=55; //both upper surafce and lower surface are exposed
v=.75; //velocity of air
Nc=.3*10^-3; //in kg/m^2*s
x2=.2; //moisture content on wet basis finally after drying
Xcr=0.125; //crtical moisture content
X1=0.15; //moisture content on dry basis intially
X2=0.025; //moisture content on dry basis finally after drying
Xbar=0.0; //equillibrium moisture
tbar=(Ls/(A*Nc))*((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
printf("\n the time for drying the sheets from .15 to .025 kg water /kg of dyr solid moisture under same drying conditions is :%f hour",tbar/3600);
//end
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