 Myself, Ganesh Bhagalai working as an assistant professor in Department of Mechanical Engineering Valchin Institute of Technology, Sallapur. In this session of conduction, we will see rate of heat conduction through cylinder. Learning outcome at the end of this session, students will be able to derive heat conduction through a cylindrical coordinate system. In previous sessions, we have derived heat conduction through plane wall. Now, in the cylindrical coordinate system, I will be drawing here cylinder. This is the isometric sketch. In this cylinder, the hollow cylinder, small elemental area is selected, this one. It has DA area. The cylinder has internal radius R i, outer radius R o, internal temperature T i, and surrounding temperature is T o, surrounding in the sense surface temperature, surrounding temperature T o. Now, as we know the thermal resistance, thermal resistance for this hollow cylinder could be represented by electrical applying electrical analogy. So, here this is the inner temperature, this is the outer temperature. The rate of heat conduction is denoted by Q. At the end, we will be writing here the thermal resistance for the cylinder. As we have we have the equation dou square T dou square T by dou R square plus 1 by R dou T by dou R is equal to 0. This is general heat conduction equation in radial direction without heat generation under steady state condition. This could be written as dou by dou R of R dou T by dou R is equal to 0. If I take the integration, then this can be written as R dou T by dou R is equal to C 1. Now, here it will become D T. This can be simplified to D T by D R is equal to C 1 by R. Now, in terms of temperatures, this can be reduced to D T is equal to C 1 by R into D R. Now, if I integrate on both sides, on both sides, if I integrate on both sides, this can be written as T is equal to C 1 ln R ln R ln R ln R plus C 2. This is suppose equation number 1. Now, here to find out the constants C 1 and C 2, I should apply the boundary condition. So, boundary conditions are first boundary condition at R is equal to R i, T is equal to T i. At radius R i, inner radius temperature is T i. These are the inner radius and inner temperatures. Second boundary condition at R is equal to R o, T will be equal to T o. Now, using first boundary condition, the equation 1 becomes T i is equal to C 1 ln R i plus C 2. Similarly, I can use equation 2. Then this will become T o is equal to C 1 ln R o plus C 2. I can take the difference of this equation 2 and equation 3. Then if I subtract equation 3.2 from 2, then I can write here T i minus T o is equal to C 1 ln R i minus C 1 ln R o. This C 2, C 2 will get cancelled. Now, so T i minus T o is equal to C 1 ln R i by R o. C 1 can be written as T i minus T o by ln R i by R o. This is the constant C 1. This is the constant C 1. Now, I can use the relation of C 1 in equation number 2. So, this can be written as T i is equal to, instead of C 1, I will be substituting the relation T i minus T o by ln R i by R o into R i plus C 2 plus C 2. So, C 2 will become T i minus T o by ln R i by R o into R i. This is the relation for C 2. Now, we will substitute C 1 and C 2 in equation number 1. So, substituting constant C 1 and constant C 2 in equation 1. Equation number 1 is T is equal to C 1 T i minus T o by ln R i by R o into ln R plus T i. This is the C 2 now. C 2 relation plus T i minus T i minus T o by ln R i by R o into R i. Here, I can take T i on the left hand side. So, T minus T i will be equal to, from these two equations, I will be taking common T i minus T o. In the bracket, it will be ln R. Denominator is also same. So, it can be taken as a common ln R i R o minus R i minus R i. ln is, sorry, ln is forgotten. So, this is ln R i. Now, further simplification is there. T minus T i divided by T i minus T o will be equal to ln R by R i by ln R i by R o. This is known as, this is known as temperature distribution equation. This is known as temperature distribution equation, where T i T o R i R o can be measured easily. Whereas, at any radius, at any radius, at any radius unknown temperature T can be calculated. So, that is the use of temperature distribution equation. So, next part is, that is part b to find out rate of heat conduction, heat conduction through hollow cylinder denoted by q. Now, think over the equation, which will be required for deriving this. Yes, it is Fourier's law. It is Fourier's law, which has q is equal to minus k a, minus k a, minus k a, minus k a, minus k a, minus d t by d r. Now, because in the radial direction, we are finding the equation for the conductive rate of heat transfer. Since we have d by d r, this equation, since we have d by d r of r, d t by d r is equal to 0. If I take integration, then I can write d t by d r, d t by d r is equal to c 1 and that r in the denominator, c 1 by r. So, here this d t by d r relation can be substituted in Fourier's law of heat conduction. So, q will become equal to minus k a into c 1 by r. So, what is c 1? Just now we got the c 1. So, q is equal to minus k into a, c 1 is t i minus t o by l n r i by r o into 1 by r into 1 by r. Now, here the area, here the area is 1 by r. Minus k pi d l surface area. So, 2 pi r l, this is the surface area t i minus t o by l n r i by r o into 1 by r. This r r will get cancelled. We will be having the equation minus k t i minus t o by a into 2 pi k l. This k I can remove it. So, that I can write 2 pi k l l n r i by r o, which can you can be written as t i minus t o divided by l n r o by r i divided by 2 pi k l. So, this is the rate of heat transfer through the heat through the plane wall, where the thermal resistance, where the thermal resistance for the hollow cylinder is equal to l n r o by r i divided by 2 pi k l. So, in this way you can solve the numerical for further study, you can refer fundamentals of heat and mass transfer by in Krupera David. Thank you.