 Hello students, myself Ganesh Biyagalave working as an assistant professor in department of mechanical engineering, Valchinist of technology, Sallapur. Now in this session of conduction, we will see the heat transfer rate equation for spherical coordinate system. Learning outcome, at the end of this session students will be able to derive equation for conduction, conductive heat transfer through spherical coordinate system. In previous sessions we have derived conductive heat transfer, conductive heat transfer through plane wall and through hollow cylinder. Now third case is through sphere, so I will be drawing the sphere, this is the sphere hollow sphere you consider and this is the small element of the sphere, small element of the sphere which is at radius r and thickness d r. The inner radius of the sphere is r i, inner temperature is t i, outer radius is r o, outer surface temperature is t o, this is the small element of the sphere. Now we know the general equation, we know the general equation that is dou square t by dou r square plus 2 by r dou t by dou r plus q g by k is equal to 1 by alpha dou t by dou r. Now here only radially heat transfer will be considered and without heat generation, assumption is that without heat generation under steady state condition. For steady state condition change in temperature with respect to time will be 0. Then this can be simplified to dou square t by dou r square plus 2 by r dou t by dou r is equal to 0. Now multiplying by multiplying by multiplying throughout by r square we get we get r square dou square t by dou r square plus 2 r dou t by dou r is equal to 0 as flow is in the radial direction instead of dou I can write d now r square d square t by d r square plus 2 r d t by d r is equal to 0. This is this equation is the differentiation of d by d r of r square d t by d r is equal to 0. Suppose this is the equation number 1. Now I can take the integration of equation number 1 then I can write d t is equal to c 1 c 1 by r square into d r. If I take second integration also then it will become t is equal to minus c 1 by r plus c 2 this is the equation number 2. Now in this equation we have two constants c 1 and c 2 so their relations are to be obtained for that purpose I will be using the boundary conditions. So boundary conditions are first boundary condition at r is equal to r i t is equal to t i and second boundary condition at r is equal to r o t becomes equal to t o because there are two boundary conditions are there and there are two temperatures are there. So using first boundary condition using first boundary condition equation 2 can be written as t i is equal to minus c 1 r i is equal plus c 2 and using second boundary condition it will become t o is equal to minus c 1 by r o plus c 2. I can subtract the equation 4 this equation 4 from third equation then I can write t i minus t o is equal to minus c 1 in the bracket 1 by r i plus 1 by r o c 2 c 2 will get cancelled. So t i minus t o I can divide it by I can divide it by I will simplify further minus c 1 minus c 1 in the bracket. So c 1 relation will become equal to t i minus t o divided by 1 by r o minus 1 by r i this is the constant c 1. Now we will move ahead for finding the constant c 2 for constant c 2 what I will do I will substitute c 1 in equation number 3. Now so t i in equation 3 I will be substituting. So t i is equal to minus t i minus t o by 1 by r o minus 1 by r i this is a constant c 1 multiplied by 1 by r i plus c 2. Further c 2 can be written as t i minus minus plus t i minus t o by 1 by r o minus 1 by r i into 1 by r i. So this is the constant c 2 substituting constant c 1 and c 2 in equation number 2 now in this equation we have to substitute constant c 1 and c 2. So t will become minus c 1 is t i minus t o by 1 by r o minus 1 by r i this is the c 1 multiplied by 1 by r and plus c 2 is a plus t i plus t i minus t o by 1 by r o minus 1 by r i this is the constant c 2 multiplied by 1 by r o minus 1 by r i into 1 by r i this t i I will take on the left hand side. So t minus t i will be there is equal to is equal to t i minus t o divided by 1 by r o minus 1 by r i can be taken common inside the bracket it will be 1 by r i minus 1 by r. So t minus t i divided by t i minus t o will become equal to will become equal to this r i r o will go in the numerator divided by r i minus r o in the bracket in the bracket 1 by r i minus 1 by r. So this is the temperature distribution equation this becomes temperature distribution equation first part. Now we will move to the second part rate of heat conduction we will move to the second part that is rate of heat conduction. So we have rate of heat conduction as we know Fourier's law of heat conduction we know that is minus k a d t by d r d t by d r. So here we will be substituting the t and t was t was equal to minus c 1 by r plus c 2 here c 1 and c 2 are to be substituted that we will do here. So which is equal to minus k a d by d r of t of t that is t i minus t o by 1 by r o minus 1 by r i minus 1 by r i into 1 by r plus c 2. So c 2 was equal to t i plus t i minus t o by 1 by r o minus 1 by r i into 1 by r i. If we differentiate now if we differentiate this will get cancelled this will also get cancelled because all r the constants are there. So I can write minus k a minus k a now taking the differentiation of r that will be minus 1 by r square in other bracket t i minus t o by r i minus r o divided by r i into r o and here the area. So this minus minus will become plus k 4 pi r square this will be 4 pi r square into 1 by r square into t i minus t o divided by r i minus r o into r i into r o. So this r square r square will get cancelled I can write directly q is equal to q is equal to 4 pi r i r o 4 pi r i r o k into t i minus t o divided by r i minus r o. Here this can be written in terms of the thermal resistance t i minus t o divided by r i minus r o by 4 pi r i r o k this is nothing but thermal resistance and here we are derived the rate of heat conduction. For further study you can refer fundamentals of heat and mass transfer by Incropera David. Thank you.