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clear;
clc;
printf("\t Example 6.17\n");
//moisture content reduces from 25 to 2
r=7.5*10^-5; //constant drying rate in kg/s
A1=.3*.3**2; // area of the sppecimen
Nc=r/A1; //drying rate
Xcr=.15/0.85; //.15 is the critical moisture content
Xo=.25/.75; //.25 is the initial moisture content
Xfinal=.02/0.98; //.02 is the final moisture content
Xbar=0; //equillibrium moisture content
A=1.2*.6*2; //area of the new solid
Ls=28.8; //bone dry weight of new solid
v1=.3*.3*.006; //volume of the old solid;
v2=.6*1.2*.012; //volume of the new solid
w2=1.8; //weight of the old solid
w3=864*10^-5*1.8*10^-5/54; //weight of the bone dry solid
//Nc is prporional to =(t-ts) = (G)^0.71---- whrere G is the mass flow rate
v1=3; //old velocity
Tg=52; //old dry bulb temperature
Tw=21; //wet bulb temperature
H=.002; //humidity
SH=0.015; //saturated humidity
vnew=5 //new velocity
Tgnew=66; //new DBT
Twnew=24; //new WBT
Hnew=.004; //new humidity
SH=.020; //new satuurated humidity
//hence drying rate of air under new condition
Nc=4.167*10^-4*((vnew/v1)*(273+Tg)/(273+Tgnew))^0.71 * ((.019-H)/(.015-H));//drying rate of air under new condition in kg/m^2*s
DT=Ls/(A*Nc) * ((Xo-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(Xfinal-Xbar)));
printf("\n the time for drying the sheets from 25 to 2 percent moisture under same drying conditions is :%f hours",DT/3600);
//end
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