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+//Exa 4.4
+clc;
+clear;
+close;
+//given data
+k=8;// in W/mK
+alpha=4*10^-6;// in m^2/s
+h=50;// in W/m^2K
+D=6*10^-3;// in m
+R=D/2;
+T=0.5;// where T = (t-t_infinite)/(t_i-t_infinite)
+//l_s= V/A = R/3
+l_s= R/2;// in m
+Bi= h*l_s/k;
+// since Bi < 0.1 , hence lumped heat capacity analysis can be applied
+// toh= rho*V*C/(h*A) = rho*C*l_s/h = k*l_s/(h*alpha)
+toh= k*l_s/(h*alpha);// in seconds
+disp(toh,"time constant in seconds");
+// It is given that (t-t_infinite)/(t_i-t_infinite) = 0.5 = %e^(-h*A*c /(rho*V*C)) = %e^(-h*c/(rho*l_s*C)) = %e^(-h*alpha*c/(l_s))
+// or (t-t_infinite)/(t_i-t_infinite) = %e^(-h*alpha*c/(l_s);
+c= -log(T)*l_s/(h*alpha);// in sec
+disp(c,"The time required to temperature change to reach half of its initial value in seconds");
+