 Hello students, Myself Ganesh Biajlawe working as an assistant professor in Department of Mechanical Engineering, Vultrin Institute of Technology, Solopore. So, in the this fourth session of Finns, we will see fin with insulated end, learning outcome. At the end of this session, students will be able to derive equation for fin with insulated end, this is our second. This is the second case, fin with insulated end, so this is the base surface of the fin, the base temperature T naught surrounding temperature T infinity. In a few applications, so what is observed in comparison with the length of the fin, the end is very very thin, so we can assume the end to be adiabatic means insulated, insulated. So, in this case, the boundary conditions will be at x is equal to 0, theta is equal to theta naught and second boundary condition at x is equal to L, d theta by dx would be 0 as the temperature gradient is 0. Now, try to recall the general heat conduction equation, yes, so it is theta is equal to c 1 e raise to m x plus c 2 e raise to minus m x. In the previous case, already we got the constant that is c 1 by substituting theta naught is equal to c 1 e raise to m x after substituting it becomes 1 plus c 2, suppose this is the first equation. Now, what I will do now here, I will differentiate the general heat conduction equation. So, it becomes d theta by dx is equal to m c 1 e raise to m x minus m c 2 e raise to minus m x. Now, if I put the condition that is d theta substituting the second boundary condition, applying the second boundary condition that is d theta by dx at x is equal to L will become equal to m c 1 e raise to m L minus m c 2 e raise to minus m L. Already, we know this equation the temperature gradient at L x is equal to L is 0. So, I can write this as 0 is equal to m c 1 e raise to m L minus m c 2 e raise to minus m L. I can take this on left hand side m m will get cancelled. So, I can write a c 1 e raise to m L is equal to c 2 e raise to minus m L. I can get the constant c 1 which is equal to c 2 into e raise to minus m L divided by e raise to m L which will become equal to c 2 e raise to minus 2 times m L. So, this is the c 1 constant. If I now substitute the c 1 in equation A substituting c 1 in equation A general heat conduction equation general heat conduction equation. So, theta will be equal to the c 1 is c 2 e raise to minus 2 m L. This is the replacement of c 1 into e raise to m x plus c 2 e raise to minus m x. So, this theta would become equal to c 2 e raise to minus 2 m L plus m x plus c 2 e raise to minus m x. Now, I can write this as theta can take c 2 common into bracket substituting c 1 in equation 1. I can write this is c 1 substituting in this equation 1 I can write theta naught is equal to c 1 plus c 2. So, what is the c 1? c 1 is c 2 e raise to minus 2 m L plus c 2. So, this theta naught will become c 2 into bracket e raise to minus 2 m L plus 1. So, c 2 is nothing but theta naught by e raise to minus 2 m L plus 1. Now, here I can divide numerator and denominator by e raise to minus m L. So, c 2 dividing the numerator by e raise to minus m L and denominator e raise to minus 2 m L plus 1 by e raise to minus m L. The simplification will be theta naught into e raise to m L divided by e raise to minus m L plus e raise to plus m L. This will be the relation for c 2. Now, we have c 2 I can substitute this c 2 in relation for c 1. So, since c 1 is equal to c 2 into e raise to minus 2 m L and c 2 is nothing but theta naught into e raise to m L divided by e raise to minus m L plus e raise to m L. This is nothing but c 2 into e raise to minus 2 m L. So, this is equal to theta naught into e raise to minus m L divided by e raise to minus m L plus e raise to m L. So, we got the equations of constant that is c 2 and c 1. Now, we can substitute these constants in equation 1. So, equation 1 becomes theta naught is equal to c 1 is theta naught into e raise to minus m L to be substituted in general heat conduction equation. So, theta will become equal to theta naught into e raise to minus m L divided by e raise to m L plus e raise to minus m L into e raise to m X. This is c 1 then c 2 is plus c 2 is theta naught e raise to m L by plus m L into minus m L into e raise to minus m X. Now, so this theta I can take theta naught common will be equal to now e raise to m into bracket X minus L by e raise to m L plus e raise to minus m L plus e raise to m L minus X by e raise to m L plus e raise to minus m L. If you simplify this, if you simplify this then we will get the temperature distribution equation as theta by theta naught is equal to hyperbolic cos of m L minus X by hyperbolic cos of m L. So, this is the temperature distribution equation for fin with insulated end. Similarly, if you try for the rate of heat conduction rate of heat conduction for fin with insulated end q becomes equal to square root of h p k a theta naught into hyperbolic tan of m L. So, these are the two equations temperature distribution equation and rate of heat transfer from the fin with in for further study you can refer fundamentals of heat and mass transfer by Ncropera David. Thank you.