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clear;
clc;
printf("\t Example 6_3_a\n");
//part(i)
//table wt of wet slab,kg -- 5.0 4.0 3.6 3.5 3.4 3.06 2.85
// drying rate,kg/m^2s-- 5.0 5.0 4.5 4.0 3.5 2.00 1.00
// X,Dry basis -- 1.0 0.6 .44 0.4 .36 .224 0.14
// equillibrium relation is given under
p = [1.0 0.6 .44 0.4 .36 .224 0.14];
a = [5.0 5.0 4.5 4.0 3.5 2.00 1.00];
i=1; //looping for calc. of 1/N
while(i<8) //looping begins
t(i)=1/(a(i));
i=i+1;
end //as 1/N plot is needed
plot(p,a,"o-");
title("Fig.6.19(a) Example3 Drying Rate curve");
xlabel("X-- Moisture content, X(kg/kg) ---->");
ylabel("Y-- Drying Rate, N(kg/hr.m^2 ---->");
xset('window',1);
plot(p,t,"o-");
title("Fig.6.19(b) Example3 1/N vs X");
xlabel("X-- Moisture content, X(kg/kg) --->");
ylabel("Y-- 1/N, hr,m^2/kg --->");
//from X=0.6 to 0.44 ,falling rate is non linear and from X=.44 to .14 falling rate is linear
printf("\n from the graph we get critical moisture content as 0.6 kg moisture/kg dry solid");
//end
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