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+//Exa 4.8
+clc;
+clear;
+close;
+//given data
+k=45;// in W/m degree C
+d =0.1;// in meter
+l =0.30;// in meter
+t=800;// in degree C
+t_i=100;// in degree C
+t_infinite=1200;// in degree C
+h= 120;// in W/m^2 degree C
+alpha=1.2*10^-5;// in meter
+rhoC= k/alpha;
+V=%pi/4*d^2*l;// in m^3
+A= %pi*d*l + 2*%pi/4*d^2;// in m^2
+// l_s= V/A = (%pi/4*d^2*l)/(%pi*d*l + 2*%pi/4*d^2) = d*l/(4*l+2*d^2)
+l_s = d*l/(4*l+2*d^2);
+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/(rhoC*l_s))
+toh = -log((t-t_infinite)/(t_i-t_infinite))*rhoC*l_s/h;// in sec
+
+// So, the velocity of ingot passing through the furnace
+FurnaceLength = 8*100;// in cm
+time = toh;
+Velocity = FurnaceLength/time;// in cm/sec
+disp(Velocity,"Maximum speed in cm/sec")