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clear;
clc;
printf("\t Example 6.2\n");
//table X*100,(kgmoisture/kg dry solid) N*100 (kg moisture evaporated /hr*m^2)
// 35 30
// 25 30
// 20 30
// 18 26.6
// 16 23.9
// 14 20.8
// 12 18
// 10 15
// 9 9.7
// 8 7
// 7 4.3
// 6.4 2.511111
Ls=262.5; //mass of bone dry solid ais the drying surface
A=262.5/8; //both upper surafce and lower surface are exposed
Nc=0.3; //in kg/m^2*hr
x2=.06; //moisture content on wet basis finally after drying
x1=.25; //moisture content on wet basis finally after drying
Xcr=0.20; //crtical moisture content
X1=x1/(1-x1); //moisture content on dry basis intially
X2=x2/(1-x2); //moisture content on dry basis finally after drying
Xbar=0.025; //equillibrium moisture
t1=Ls/(A*Nc) *(X1-Xcr); //so for constant rate period
//for falling rate period we find time graphically
p = [.20 .18 .16 .14 .12 .10 .09 .08 .07 .064];
a = [3.3 5.56 6.25 7.14 8.32 10.00 11.11 12.5 14.29 15.625];
plot(p,a,"o-");
title("Fig.6.18 Example2 1/N vs X for fallling rate period");
xlabel("X-- Moisture content, X(kg/kg)");
ylabel("Y-- 1/N, hr,m^2/kg");
Area=1.116; //area under the curve
t2=Area *Ls/A; //falling rate period we find time graphically
ttotal=t1+t2; //total time for drying
printf("\n the total time for drying the wet slab on wet basis is :%f min",ttotal);
//end
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