 I am Ganesh B. Aglai working as an assistant professor in Department of Mechanical Engineering Walsh and Instruct Technology, Sallapur. In this session of heat exchangers, we will see LMTD approach for counter flow heat exchanger learning outcome. At the end of this session students will be able to describe LMTD method for counter flow and design the heat exchanger. In the analysis of heat exchanger, we have seen LMTD approach for parallel flow. Now second type of the flow is counter flow. Now for counter flow, the direction of one fluid we have to change. Then the flow will become counter. Now see here this is the hot fluid flow direction, I have changed the cold fluid direction. In the opposite direction it will start flowing, there will be change in the notations. For parallel flow this was the inlet cold fluid temperature, for counter it becomes right hand side as the cold fluid inlet temperature. As the fluid start passing through the heat exchanger, there will be heat transfer between the fluids. This is the temperature profile for the hot fluid, THI is the hot fluid inlet temperature at is as it flows through the heat exchanger, it loses the heat. Because of it the temperature of that hot fluid goes on decreasing and at THO or THE it will leave the heat exchanger. While the cold fluid inlet temperature is TCI along the flow in the heat exchanger, cold fluid is gaining the heat and because of that the temperature of cold fluid goes on increasing. And the increase in the heat transfer responsible for rise in the cold fluid temperature gives TCO. You can think over the relationship between TCE cold fluid exit temperature and hot fluid exit temperature. For parallel flow heat exchanger TCE will always be less than THE for limited length of the heat exchanger while for counter flow heat exchanger of that same length or size of the heat exchanger we may get TCE greater than THE. Here the notations TH1 or I represents the difference in hot and cold fluid temperatures while THE is the difference in the temperatures of hot and cold fluid. To do the analysis we will consider the small element of the heat exchanger that is DA which is B into DX. Now we will consider the relations. Rate DQ is the rate of heat transfer from the hot fluid is equal to rate of heat transfer from the cold fluid. Both these rate of heat exchanger rates will be same means del Q is equal to del QH is equal to del QC. Del Q considering overall heat transfer coefficient into bracket TH-TC into BDX will be equal to minus m dot H CPH DTH where minus sign is there because of negative slope of the hot fluid which can be written for the cold fluid minus m dot C CPC DTC. Here also minus sign is there because both slopes has negative. So this minus minus notations are taken. Here del T is the temperature difference between hot and cold fluid. Now I can differentiate this relation then I can obtain D del T which is equal to DTH minus DTC. I can replace DTH and DTC in the form of del Q by mH. So here see here DTH will be equal to minus DQ by m dot H CPH and DTC will be equal to minus DQ by m dot C CPC. So after replacing I can get the equation minus into bracket 1 by m dot H CPH minus 1 by m dot C CPC into DQ. Here rate of heat transfer is also equal to U del T BDX. So that is written here. So DQ is replaced by U del T B into DX. This is nothing but area. D del T by I can take it on the left hand side. So this becomes del T which is equal to minus into bracket 1 by m dot H CPH minus 1 by m dot C CPC bracket complete U as it is B is also taken and DX is also taken. Now this was for this small elemental area. If I want to consider the total area of the heat exchanger I should integrate it from del T I to del T E and the area varies from 0 to A. See here for analysis purpose we consider small area. Now to cover the entire heat exchanger surface entire heat exchanger surface may have single pass one shell as we have seen in the classification two pass one shell or two shells four passes. Here the number of passes depends upon the amount of heat exchange to be carried away between the flutes. For the open type of the heat exchanger means what? Only one fluid either cold fluid or hot fluid you have to pass through the tube. The outer surface by means of the fins will get heated or cold depending upon the application. So the surface area is very important in the heat exchangers as it decides the size and depending upon the availability of the size the heat exchangers are selected or I can attach the fins on outer surface of the single tube. For the shell and tube heat exchangers if you try to recall on inner side of the shell we have inserted the baffles ok. So here what we will do we will cover the entire surface entire surface of the heat exchanger. So for that I required to integrate this equation for from del T 1 to del T 2 and here area 0 to a then I will get ln del T e by del T i which is equal to minus 1 by m dot h C p h minus 1 by m dot c C p c bracket complete into u a. Here del T i del T e is del T h e minus T c i because this is the counter flow and del T is equal to del T h minus T c e then these temperatures gets changed for the counter flow remember it. So ln del T by del T e and i is equal to minus 1 by q into bracket I can rearrange these temperatures for getting del T i and del T e. So I can write q is equal to u a as it is del T i minus del T e by ln del T i by del T e. Now this ratio is nothing but l m T d del T i minus del T e by ln del T i by del T e though the relation this l m T d relation will be same for counter flow and parallel flow but the del T i for the counter flow is not same as that of the parallel flow keep in mind. So we got the l m T d for counter flow heat exchanger which can be used in the equation in the equation u delta T into del T m ok. So overall heat transfer between the heat exchanger fluids will be equal to u a delta T m log mean temperature difference. For further study you can refer fundamentals of heat and mass transfer by Incropera and David. Thank you.