 I am Ganesh Biyagalave working as an assistant professor in Department of Mechanical Engineering Valchen Institute of Technology, Singapore. In this session of heat transfer, we will see the unit conduction in that general heat conduction equation in Cartesian coordinate system. Learning outcome, at the end of this session, students will be able to derive general heat conduction equation in Cartesian coordinate system. Consider this as an element of the system, of the system. Small part may be rectangular or square is taken of a big system, which is exposed to the heat transfer. Now, here the surface A, B, C, D through this face, Qx amount of heat is flowing in the x direction and the thickness in the x direction from the origin x is a dx. The face A dash, B dash, C dash, D dash is losing the Qx plus dx amount of heat that is in the x direction. Now, remaining two surfaces are also exposed to the heat transfer. So, in the y direction, Qy amount of heat enters through the bottom surface and the thickness in the y direction is dy. It flows, it conducts throughout this element and from the top surface it is losing Qy plus dy. Remember, this surface is exposed to the heat transfer in the conductive mode. Can you think for the z direction also? Yes, Qz amount of heat is entering from the back side and its thickness in the z direction is dz. Then this front surface will lose Qz plus dz amount of heat. So, this was the Cartesian coordinate system in which we have shown the small element of the big system. Now, we will try to see the different equations respectively x direction, y direction and in the z direction. Here we will be using the energy balance equation because what is going to happen? Whenever there is a heat transfer, the temperature gradient must be present that is low temperature and high temperature, heat always flows from high temperature to low temperature. Because of that, there may be some possibility of internal heat generation which will be denoted by small Qg and at the same time there might be internal rise in the temperature of the element. Now, we will go through the next step. So, here see here the Qx heat is entering into this surface which has the equation that is Fourier's law of heat conduction we can apply. So, Qx is equal to minus k dou t by dou x dy dz. Here we are going to consider all the three dimensions and for this equation we have considered only conductive heat in x axis. So, this was for the surface a, b, c, d. Now, from the opposite surface a dash, b dash, c dash, d dash the equation becomes dQx plus dx is equal to dQx plus dou by dou x of dQx dx this is the Taylor's expansion. Now, this can be simplified we can substitute dQx as minus k over here minus is taken common dou t by dou x plus dou by dou x as it is we are written here then what is the dQx it is k minus is taken common. So, k dou t by dou x into dx and remaining two axis will be dy dz here dx dy dz is nothing but the volume of the element. So, this equation these three equations we have seen along the x axis now we will consider the y axis through the y axis dQy amount of heat is entering and this top surface is leaving Qy plus dy amount of heat. So, similarly I can write for the y axis also dQy is equal to minus k dou t by dou y dx dz and through the top surface which is losing the heat is dQy plus dy is equal to minus is taken common. So, what is Qy it is minus k dou t by dou y plus dou by dou y as it is written here in the bracket k dou t by dou y bracket complete dy bracket complete dx dz this was y axis. Now, in the z axis similarly dQz is equal to minus k dou t by dou z dx dy and Taylor's expansion is used for dQz plus dz as minus k dou t by dou z plus dou by dou z k dou t by dou z bracket complete dz dx dy. So, this is regarding three equations we can we can write these three equations for three different axis in the Cartesian coordinate system this assume it as a first one then the equation can be simplified for the net amount of heat conducted net amount of heat conducted means dQz minus dQ sorry dQx minus dQx plus dx plus dQy minus dQy plus dy plus dQz minus dQz plus dz. So, that step is omitted over here and directly we have written dQx plus dQy plus dQz minus in another bracket dQx plus dx plus dQy plus dy plus dQz plus dz and we can get now see here we can get here dou by dou x in the bracket k dou t by dou x. Now, what is going to happen here we can write it also k dou square t by dou x square plus k dou square t by dou y square plus k dou square t by dou z square. So, all the three dimensions we have considered three axis we have consider over here and this is the volume dx dy dz. Now, this was regarding net amount of heat conducted ok now here we will consider the second part that is the possibility of internal heat generation inside the element. So, the internal heat generation will be equal to q dx q bar dx dy dz and the rate of change of internal energy is will be equal to m c p delta t m c p delta t. Now, rho is equal to mass per unit volume. So, m could be replaced by rho into v. So, what is v here dx dy dz. So, this equation becomes rho c p dou t by dou t this is the change in temperature over time dx dy dz. So, if we apply the energy balance equation that is net heat conducted plus internal heat generation is equal to change in internal energy of the system. The equation can be written as dou by dou x of k dou t by dou x plus dou by dou y k dou t by dou y y axis plus dou by dou z k dou t by dou z z axis plus internal heat generation is equal to rho c p dou t by dou t. This is known as generalize heat conduction equation in Cartesian coordinate system. Now, if the case is that there is no internal heat generation you can omit this and for steady state condition change in temperature over the change in temperature over the time will be 0. So, in that case for three dimensional heat flow without heat generation for steady state condition the equation will become equal to 0. In that case q will be 0 and this change in temperature will be 0 because it is a steady state condition we will try to see it. Now, for a homogeneous material homogeneous material in this here will be assuming the conductivity in x direction y direction and z direction is same that is k x is equal to k y is equal to k z is equal to k then k I can take common k in the bracket dou square t by dou x square plus dou square t by dou y square plus dou square t by dou z square plus q is equal to rho c p d t by d t. Now, here del operator we will consider. So, this is nothing but del square t plus q by k this will be dividing both sides by k. So, del square t plus q by k is equal to 1 by alpha dou t by dou t where del square is equal to this one. Now, here alpha alpha is nothing but thermal diffusivity which is a ratio of k by rho c p for highly conductive material the thermal diffusivity will be higher and for lower thermal conductive material like insulators the thermal diffusivity will be small. So, the unit of the thermal diffusivity is meter square per second. Now, just now we have written del square t is equal to 1 by alpha dou t by dou t this is for steady state condition. We can write this equation for steady state condition that is del square t is equal to 0 this will be general heat conduction equations Laplace star for transform without heat generation and for steady state condition. In this fashion you can use for a single dimension two dimensions etcetera etcetera the general heat conduction equation. For further study you can refer fundamentals of heat and mass transfer by in Cropera David. Thank you.