 Hello students, Myself Ganesh B. Aglade working as an assistant professor in Department of Mechanical Engineering, Walsh and Institute of Technology, Solopor. In this session of conduction, we will see the derivation of rate of heat transfer through the wall. Learning outcome, at the end of the session, students will be able to derive one-dimensional heat conduction equation through plane wall. Now, in last session, we have seen general heat conduction equation through Cartesian coordinate system. Now, for plane wall, we will try to derive the heat conduction equation. Suppose, this is the wall, plane wall, its left surface has temperature T1, right surface has temperature T2. These are the surface temperatures. The thickness of the wall is L and the conductive heat transfer is Q. As we know, the general rise heat conduction equation that is dou square T by dou x square plus dou square T by dou y square plus dou square T by dou z square plus Qg by k is equal to 1 by alpha dou T by dou T. Here, I can write for one-dimension d square T by dx square plus Qg by k is equal to 1 by alpha dT by dT. Now, as the internal heat generation is absent, that is the assumption. So, it will become 0 and under steady state condition, the temperature cannot change. So, what I can do now? I can write now, I can write d square T by dx square is equal to 0. If I take 2 times integration, then I will be getting the equation T is equal to c1 x plus c2. This is the equation number 1. Now, to find out the constants c1 and c2, we require to apply the boundary condition. We will be using the boundary condition. So, what are the boundary conditions? First boundary condition. At x is equal to 0, this is the x and this is the x axis, x is equal to 0 and x is equal to L. At x is equal to 0, the temperature, surface temperature is T1. T is equal to T1 and second boundary condition at x is equal to L, T is equal to T2. So, applying first boundary condition, applying first boundary condition and substituting in equation 1, I can write instead of T, it will be T1 which is equal to c1 into 0 plus c2. So, the constant c2 will have the value T1. Similarly, using second boundary condition, the equation 1 will become T2 is equal to c1 L plus what is c2? c2 is T1. So, I can write this as T2 minus T1 is equal to c1 L. So, c1 will become equal to T2 minus T1 by L. So, here we got the boundary condition for two constants. We got two constants c1 and c2. Now, substituting constant c1 and constant c2 in equation 1, we can have the equation 1 as T is equal to T2 minus T1 by L into x plus T1. Now, this can be further written as I can take T1 in the left hand side. So, T minus T1 will be equal to T2 minus T1 bracket complete L into x plus T1 plus T1. Now, I can take T2 minus T1 on the denominator side. So, the equation will be T minus T1 by T2 minus T1 is equal to x by L and this equation is known as temperature distribution equation. This is known as temperature distribution equation means what? By using thermocouples, we can measure the temperature T1 and T2. So, T1 T1 T2 temperatures can be measured length can be measured at any x at any x distance at any x distance. You can substitute here the x value you will be getting the T unknown temperature unknown temperature and the temperature distribution if you start finding at 1 mm, 2 mm, etcetera, etcetera the temperature distribution slope will be like this one, linear slope will be there. So, this was the first part. Now, in the second part we will find the rate of heat conduction rate of heat conduction by using the Fourier's law I can write q is equal to minus k A dT by Tx dT by dx. So, as we know the temperature T as we know temperature T, I will write temperature T is equal to T2 minus T1 by L into x plus T1. So, differentiating it with respect to x I can I can write dT by dx is equal to T2 minus T1 by L plus 0. So, in the equation 2 I can substitute this one. So, q will be equal to minus k A minus k A T2 minus T1 by L into x plus T1. So, minus T1 by L by L now the area here k is the thermal conductivity k is the thermal conductivity of this material. A is the area normal to heat flow A is the area normal to heat flow. So, this is the area this much area if I will try to draw the isometric sketch then the area will be L sorry here I will be letter B and C. So, here A will be equal to B into C. So, remember we should select the area normal to the heat flow. So, further I can substitute now or I can simplify q is equal to now can I remove the minus sign think over it yes. So, this can be written as k A T1 minus T2 divided by L this can further be written as here this can further be written as q is equal to q is equal to T1 minus T2 divided by L by k A. As you know the electrical analogy can you tell me the thermal resistance present over here in the equation number 3 which is the thermal resistance yes you are correct. So, thermal resistance will be equal to L by k A will be equal to L by k A. More and more thermal conductivity lower will be the thermal resistance lower will be the thermal conductivity that is the insulating materials higher will be the thermal resistance higher will be the thermal resistance. Now, here the area you have to keep it as a constant thickness you have to keep it as a constant you have to compare two different two different materials. Now, so in this in this video we have studied the temperature distribution equation we will summarize that is T minus T1 divided by T2 minus T1 is equal to x by L this is the equation A the temperature distribution equation and second equation that is q is equal to T1 minus T2 divided by L by k A this is the equation number B. So, today in this session we have derived these two temperature distribution equation and rate of heat conduction through the plane wall. For further study you can refer fundamentals of heat and mass transfer by Incropera David. Thank you.