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+//Exa 4.11

+clc;

+clear;

+close;

+//given data

+k=40;// in W/m degree C

+d =12*10^-3;// in meter

+t=127;// in degree C

+t_i=877;// in degree C

+t_infinite=52;// in degree C

+h= 20;// in W/m^2 degree C

+rho=7800;// in W/m^2K

+C=600;// in J/kg K

+r=d/2;// in meter

+//l_s = V/A = r/3

+l_s =  r/3;

+Bi= h*l_s/k;

+// since Bi < 0.1 , hence lumped heat capacity analysis can be applied

+// (t-t_infinite)/(t_i-t_infinite)  =  %e^(-h*A*toh /(rho*V*C)) = %e^(-h*toh/(rho*l_s*C)) = %e^(-h*toh/(rho*C*l_s))

+toh = -log((t-t_infinite)/(t_i-t_infinite))*rho*C*l_s/h;// in sec

+disp("Time required for cooling process : "+string(toh)+" seconds or "+string(toh/60)+" minutes")
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