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+//Exa 4.6

+clc;

+clear;

+close;

+//given data

+k=385;// in W/mK

+h=100;// in W/m^2K

+delta =2*10^-3;// thickness of plate in meter

+A=25*25;// area of plate in square meter

+rho=8800;// kg/m^3

+C=400;// J/kg-K

+// l_s= V/A= L*B*delta/(2*L*B) = delta/2

+l_s= delta/2;// in meter

+Bi= h*l_s/k;

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

+

+// Part(i)

+// toh= rho*V*C/(h*A) = rho*C*l_s/h

+toh= rho*C*l_s/h;// in second

+disp(toh,"Time constant in seconds");

+

+// Part(ii)

+t_i=400;// in degree C

+t=40;// in degree C

+t_infinite=25;// in degree C

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

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

+disp(toh,"The time required for the plate to reach the temperature of 40 degree C in seconds");

+