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clear;
clc;
printf("\t Example 6_5_a\n");
//table 6.5.1
//S.NO. Time (Hr) weight of wet material(kg)
// 0.0 5.314
// 0.4 5.238
// 0.8 5.162
// 1.0 5.124
// 1.4 5.048
// 1.8 4.972
// 2.2 4.895
// 2.6 4.819
// 3.0 4.743
// 3.4 4.667
// 4.2 4.524
// 4.6 4.468
// 5.0 4.426
// 6.0 4.340
// infinite 4.120
w=[5.314 5.238 5.162 5.124 5.048 4.972 4.895 4.819 4.743 4.667 4.524 4.468 4.426 4.340 4.120]
t=[0.0 0.4 0.8 1.0 1.4 1.8 2.2 2.6 3.0 3.4 4.2 4.6 5.0 6.0]
//part(i)
x=4.120; //weight of the dried material
printf("\n moisture content (dry basis) ");
i=1; //looping starts
while(i<16) //calculation of moisture content
p(i)=(w(i)-x)/x;
printf("\n :%f",p(i));
i=i+1;
end
printf("\n \n Drying rate kg/hr*m^2");
i=2;
while(i<15)
a(i)=(p(i-1)-p(i))*4.12/(t(i)-t(i-1));
printf("\n :%f ",a(i));
i=i+1;
end
a(1)=.19;
a(15)=0;
printf("\n\n from the above data it is clear that critical moisture content Xcr=0.11");
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 ---->");
//end
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