 Hello students, I am Ganesh Vyaghlalai working as an assistant professor in department of mechanical engineering, Wiltian Institute of Technology, Solopur. In this session of extended surfaces or fins, we will see generalized one-dimensional heat conduction equation through rectangular fin. Learning output. At the end of this session, students will be able to derive general heat conduction equation in one-dimension for rectangular fin. Consider this as a base surface. On this base surface, this is the rectangular fins. Base surface has temperature T naught. In fluid temperature T infinity, the width of this fin is suppose W. The length of the fin is suppose L. Now, what we will do? We will consider elemental area dA. This is the elemental area. On x axis, the distance is suppose x. This area thickness is dx. Now, as we know, the transfer of heat occurs from base surface to the fin by means of conduction. So from the base, the Q amount of heat is getting conducted. On this surface, on left hand side, the conductive heat transfer is suppose Qx. And from this right hand side surface, this surface, the rate of conductive heat transfer is Qx plus dx. Now, to this elemental area, if I want to apply the energy balance equation, then Qx will be equal to Qx plus dx. This is the conductive heat transfer. And from this elemental area to the fluid, the heat loss would be by means of convection. So I can write here the energy balance equation as Qx is equal to Qx plus dx plus Q convection. Suppose this is the equation 1. Now here, Qx is the conductive heat transfer where it is equal to minus kA dt by dx. Then I can write for Qx plus dx as minus kA dt by dx plus minus kA dt by dx square into dx. And for convection, Q convection is equal to, in first session we have seen HA Ts minus T infinity. Now here, the conductive heat transfer occurs from the surface area. So I can write, H area is nothing but a perimeter into dx. This is nothing but HA and T minus T infinity. As this is the general equation, this is the fin temperature. This stands for fin temperature minus surrounding fluid temperature. Now substituting these relations in equation number 1, what will happen? This Qx minus kA minus kA dt by dx, this term will get cancelled. Just substitute this here and this one here. So what will remain? Zero will be equal to minus kA dt by dx square into dx plus HP dx T minus T infinity. If I bring it on the left hand side, I can write the equation kA d square T by dx square is equal to HP T minus T infinity. Now I can simplify this equation as d square T by dx square is equal to HP by k into bracket T minus T infinity. Now here we will consider new term, let m square equal to HP by kA and theta is equal to T minus T infinity. So I can write the equation d square theta by dx square is equal to instead of HP by kA, I can write m square T minus T infinity. If I bring this right hand term to the left side, this becomes d square theta by dx square minus m square T minus T infinity is equal to zero. So suppose this is the equation number second. This is the second order LDE whose general solution we can write. This is second order linear differential equation as this is the general equation whose solution, general solution we can write theta is equal to C1 e raise to mx plus C2 e raise to minus mx. This equation is known as general heat conduction equation considering one-dimensional heat transfer for rectangular fin. Now depending upon the cases, fin cases, we can get the constant value that is C1 and C2. By using C1 C2, we may get temperature profile and rate of heat transfer equations. These are three cases in next part we will see. For further study, you can refer fundamentals of heat transfer by Incropara David. Thank you.