 I am Ganesh Bhaiaglava working as an assistant professor in department of mechanical engineering waltzy new stuff technology solopour. In this session of the heat exchangers, we will see LMTD approach for counter flow, learning outcome. At the end of the session students will be able to derive LMTD relation for counter flow. So, we are going to continue the LMTD approach now the flow will be counter flow. So, as usual the tube in tube type of the structure is there. So, this is the inner tube, this is the outer tube. In the counter flow, the cold fluid direction will change, this is suppose a T H I, T H O. Now, the flow of cold fluid will be in the reverse direction, counter direction opposite direction. This is T C I, this is T C O. We can plot the temperature distribution for this counter flow. If this is T H I as the heat transfer is taking place, there will be decrease in the temperature of the hot fluid and the outlet temperature will be T H O. The let the cold fluid inlet temperature is T C I. As it flows in the counter direction it will gain that heat and that is why there will be rise in the temperature and the living temperature is T C O. Now, the counter flow heat exchanger gives the T C O temperature depending upon the length of the heat exchanger may be equal to T H O less than T H O and if the length of the heat exchanger is much more, we may get T C O greater than T H O. Keep in mind and also try to see such type of the flows in day to day life. That means in the actual practice. Now, we will write the heat transfer rate equation which is equal to minus m dot T H C P H into bracket T H I minus T H O. Now, as we have seen already in terms of heat rate will be equal to C H into del T H. So, for D T H we can write Q divided by minus C H. Similarly, for cold fluid it will be equal to minus m dot C C P C T C O minus T C I. For heat rate it will be minus C C into D T C. So, for D T C we may write minus Q by C C. Since, the temperature difference is theta between the hot and cold fluid. So, it is a T H minus T C which is also equal to del T taking the differentiation of D theta in differentiation for which will be equal to D T H minus D T C. So, so string their relation for D T H and D T C. So, D theta is equal to minus Q divided by C H minus Q divided by C C. Now, here minus minus will become plus. Here, we can take Q common minus Q into bracket 1 by C H minus 1 by C C. Suppose, this is the equation number first. Since, Q is equal to U D A del T which is equal to U D A theta because del T is equal to theta here also we can write theta. Now, substituting Q in equation number first then we can write Q is equal to minus U D A minus U D A into theta bracket C H minus 1 by C C. Now, let this as B. So, Q is equal to minus U D A theta into B. So, this relation can be written as. So, this will be theta this is theta D theta. So, D theta by theta is equal to minus U B into D A. Now, taking integration from theta 1 to theta 2 and this is from 0 to A. So, after taking the integration L N theta O by theta I which is equal to minus U B into A. Now, substituting the relation for B. So, this will be minus U A into bracket 1 by C H minus 1 by C C. Since, from this relation, relation of Q which is equal to minus m dot H C P H into bracket T H minus T H O. So, this C H is nothing but m H C P H we can write 1 by m dot H C P H is equal to T H I minus T H O by Q. Similarly, can be written for 1 by m dot C C P C is equal to T C O minus T C I by Q. Then the equation 2 will be will become L N theta O by theta I is equal to as it is minus U A we have written. Then instead of 1 by C H we can write T H I minus T H O by Q minus 1 T C O minus T C I divided by Q. We can take a Q common here and we can simplify it like L N theta O by theta I which is equal to minus U A Q is taken common into bracket it will be T H I minus T H O minus T C O minus minus plus T C I. Now, rearranging the temperatures T H I T C O and minus U A by Q into bracket T H I minus T C O minus T H O minus T C I. Now, this theta I is T H I minus T C O and theta O here is T H O minus T C I. So, we can write here minus U A by Q instead of T H I minus T C O T H I minus T C O theta I into bracket theta I instead of T H O minus T C I theta O. So, this is theta O. Now, we can take minus sign here. So, it will become minus minus plus U A by Q into bracket theta O minus theta I and in the left hand side it is L N theta O by theta I. So, for heat transfer rate we can write Q is equal to U A into theta O minus theta I by L N theta O by theta I. We can also rearrange the terms and can get theta I minus theta O and here theta I by theta O there is no problem. So, this is similar with Q U A del T M remember this del T M is nothing but L M T D. So, L M T D approach for counter flow the relation obtained is del T M is equal to theta O minus theta I by L N theta O by theta I which is also equal to theta I minus theta O by L N theta I by theta O. So, if inlet and outlet temperatures of both fluids are given or we can easily measure it also by using the thermocouples then we can use this equation for finding the L M T D log mean temperature difference. We can substitute it here the Q will be known to us U we can take from the table and in this fashion we can find out the surface area of the heat exchanger. Then this helps us for deciding the size of the heat exchanger. For the references you can have fundamentals of heat and mass transfer by Frank and David the Wille publication.