 Hello students, I am Ganesh Biyaglavi working as an assistant professor in department of mechanical engineering, Walsh and Institute of Technology, Salafor. In this third session, we will see infinitely long fin. So, the learning outcome is at the end of this session students will be able to derive equation for infinitely long fin, general heat conduction equation for rectangular fin. Now, there are three cases of fins, first one is infinitely long fin. So, case first infinitely long fin, suppose this is the base of the fin, we are going to use the same notations, so this is the fin. The base temperature of the fin is T naught, the surrounding fluid temperature is T infinity. The length of fin is infinity, as we know the general heat conduction equation that is theta is equal to c 1 e raise to m x plus c 2 e raise to minus m x. In the last session, we have derived this general heat conduction equation. Now, for infinitely long fin, the boundary conditions, first boundary condition is at x is equal to 0, at x is equal to 0 theta is equal to theta naught. And second boundary condition at x is equal to infinity, now why this theta is equal to 0, for infinite long fin, the end temperature would become equal to surrounding temperature, so T infinity minus T infinity will become 0. Now, using first boundary condition, so we require to substitute these values in the general heat conduction equation, at x is equal to 0 theta is equal to theta 0. So, this becomes theta 0 is equal to c 1 e raise to 0 plus c 2 e raise to 0, so this becomes theta 0 is equal to c 1 plus c 2. Now, we will try to find the constant values. Using second boundary condition at x is equal to infinity, theta becomes 0, so I can write 0 is equal to c 1 e raise to m infinity plus c 2 e raise to minus m infinity, this is 0, this is 0. Now, here this cannot become 0, so c 1 will become 0, c 1 is the constant. If I substitute now c 1 is equal to 0 in the equation second, I will get theta naught is equal to c 2. Now these are the equations of constant c 1 and c 2, substituting these in equation number 1, if I substitute these constants in the first equation, then theta will become equal to c 2 is theta naught e raise to minus m x. I can take the ratio of theta that is theta by theta naught is equal to e raise to minus m x, which is nothing but t minus t infinity, this theta is nothing but t minus t infinity divided by theta naught is a t naught minus t infinity. This becomes temperature distribution equation, this is known as temperature distribution equation, means what? At any known length at any known length of the x at any x, suppose x 1, x 2 etcetera, we can find the unknown temperature at that location as t naught t infinity are easily measurable. Now after this for the same case, we will try to derive the rate of heat transfer rate of heat transfer. Here we can use the Fourier's law of heat conduction that is q is equal to minus k a d theta by d x, as we know in the fin heat enters from the base from the base, that is why this temperature gradient should be at x is equal to 0. Now here as we derived as we derived equation third, so I can differentiate equation third with respect to d x at x is equal to 0. So, this becomes equal to theta naught minus m e raise to 0 is 1, so this becomes equal to minus m theta naught. I can substitute now the equation for d theta by d x at x is equal to 0 in this equation. So, q will be equal to minus k a minus m theta naught minus minus plus, so this is k a into m into theta naught. Now please recall the m equation as we know m is nothing but square root of h p by k a into theta naught. This I can write as 2 times k a into k a into square root of h p by k a into theta naught. So, the one will get cancelled and this q will become equal to I can rearrange and can write h p k into a into theta naught, which is also equal to h p k a theta naught is, so it is theta naught minus t infinity. So, this becomes rate of heat transfer for infinitely long fin. Here I will introduce one new term known as efficiency of fin, which is equal to rate of heat transfer from the fin to the maximum possible heat transfer. Now this q max q max is nothing but h a theta h a theta naught, why? So, maximum possible heat transfer would be in the case in which the fin temperature would be at t naught. So, the rate of heat transfer would be by means of the convection from the fin surface to the surrounding that is why this becomes q convection becomes h a theta naught. So, this is h a theta naught and this q is as it is derived in this equation. So, I can write now the rate of heat transfer from actual q to the q max. So, this area is nothing but perimeter into length p l theta naught. So, this will be the fin efficiency. We can substitute this equation here and we can find the efficiency of fin. For further reference, you can have fundamentals of heat transfer, heat and mass transfer by Incoropera David. Thank you.