 Hello students, myself Ganesh B. Aghlavi working as an assistant professor in department of mechanical engineering, Valjean Institute of Technology, Sulapur. In this session of conduction, we will see composite cases, composite wall, composite cylinder and composite sphere. At the end, the students will be able to draw thermal network of the composite cases, that is composite wall, composite cylinder and composite sphere. First case is of composite wall. First case is of composite wall. If there are more than one wall is present, it is called composite wall. So, suppose this is the wall of material A and second wall is of material B. The surface temperature of wall on this side is T1, the interface temperature is T2 and outermost temperature is T3. Then, there thermal network, thermal network I will be drawing here. Now, temperatures of wall A, left-hand side temperature is T1, interface temperature is T2. Then, there will be conductive resistance. Then, there will be conductive resistance R1. The wall A has L thickness, wall B has also L thickness and the area normal to the heat flow is same in both the cases. So, conductive heat transfer is Q. Composite wall is in series. Now, the conductive heat, conductive heat Q will be transferring from wall A and B in equal, equal amount means QA will be equal to QB is equal to Q. Same rate of heat transfer is there through the both walls in series network. Wall B will oppose the thermal resistance and it works between temperatures T2 and T3 and resistance offered will be R2. Here, the first resistance which you can calculate separately that is R1 is equal to L by KAA. KAA is the thermal conductivity of A material and KB is the thermal conductivity of B material. Similarly, for second wall thermal network will be equal to L divided by KB into A. Here, the thicknesses may be different that is L1 and L2 whenever these are not same, I can substitute here their own notations. Now, in addition with conductive heat transfer, I can add the convective heat transfer. So, this will become combined mode of heat transfer. Suppose, the internal wall will be heated by means of the gases whose temperature is T i and convective heat transfer coefficient is H i. On the outermost side also, there will be fluid properties are there. Suppose, second fluid is passing through over this wall having the temperature T o and convective heat transfer coefficient is H o. Then, I have to add two more resistances. First resistance will be convective resistance. It will be working in between temperature difference T i and T1 and on outermost side, there will be one more convective resistance. This is R C 2 and suppose this is R C 1. Now, R C 1 will be equal to 1 by H i A and R C 2 is equal to 1 by H o A. So, rate of heat transfer, rate of heat transfer activation can be written as q is equal to q is equal to if I take the temperature T i and T o in consideration, then I have to write in the numerator T i minus T o divided by 4 resistances that is summation of resistances which will be equal to T i minus T o divided by this R C 1 plus R 1 plus R 2 plus R C 2. Either, you can find out the resistance R 1 R 2 R C 1 and R C 2 separately or can substitute their relations in this equation also. So, this will become equal to T i minus T o divided by 1 by H i into A plus L 1 by K A A plus L 2 K B into A plus 1 by H o A. Now, as area is same, I can take it common also and the equation can be written as q by A, q by A is nothing but heat flux which can be represented by small q. So, small q is equal to q A is equal to T i minus T o divided by 1 by H i plus L 1 by K A plus L 2 by K B plus 1 by H o 1 by H o. There are one more concept there is one more concept called overall heat transfer coefficient. So, the overall heat transfer coefficient q is equal to u A delta T. This overall heat transfer coefficient considers mixed mode of heat transfer. So, whenever such cases are there, so u A could be written as could be written as 1 by thermal resistance because this equation Fourier law of heat transfer in terms of Fourier law of heat transfer can be written as that is delta T divided by r thermal as there are two areas are available. So, overall heat transfer coefficient will be of two basis. So, first one is on inner side and second one overall heat transfer coefficient on other side. And one more concept is present that is called overall thermal contact resistance that is called thermal contact resistance. Now, second case we will see of composite cylinder. Suppose this is the pipe having r 1 and r 2 radius r 1 r 2 radius which is insulated with which is insulated with r 3. Then their temperatures are suppose T 1, T 2 and T 3. Then thermal network for the composite cylinder between temperature T 1 and T 2 will be first resistance between T 2 and T 3, T 3 there will be secondary resistance. This is for composite cylinder where this is the A material and this is the B material. So, you can write the convective heat transfer rate sorry conductive heat transfer rate which is equal to T i minus T 3 divided by r 1 plus r 2 r 1 plus r 2. Similarly, for sphere for sphere you could write also this is suppose hollow sphere which is insulated which is insulated the inner radius is r 1, the outer radius of the sphere is r 2 and insulation radius is r 3. Then the thermal network will be this is temperature T 1, temperature T 2, temperature T 3. So, this is the temperature T 1, temperature T 2, temperature T 3, this is for sphere hollow sphere composite case. Resistance first resistance will be r 1, second resistance will be r 2. I can add two more resistances of the convective case. So, this is suppose T i, this is the convective resistance first, this is the convective resistance second. I can add for composite cylinder also T i, this is the r c 1 convective resistances, r c 1 and r c 2 will be convective resistances. So, in the composite cases depending upon the number of walls, number of layers of insulations you can just go on adding the resistances you will be getting the rate of heat transfer. Similarly, here I can write q is equal to T i minus T o divided by r c 1 plus r 1 plus r 2 plus r c 2. You can calculate rate of heat conduction through the sphere with multiple layers and rate of heat transfer through hollow cylinder with multiple layers of the insulation. These are the different cases. For further study, you can refer fundamentals of heat and mass transfer by Incropera David. Thank you.