 like working as an assistant professor in Department of Mechanical Engineering, Volcane Instruct Technology. So, in this session of the heat exchangers we will try to use analysis LMTD approach, learning outcome. At the end of this session students will be able to describe overall heat transfer coefficient and derive LMTD for parallel flow heat exchanger. Now, before proceeding we will see overall heat transfer coefficient capital U. In the heat exchanger there will be tube in tube type of the condition in which the through the inner tube hot fluid is flowing and through the annular space cold fluid is flowing. So, there will be heat transfer through the wall then we required to use one relation means for mixed heat transfer the capital U is used instead of small h that is convective heat transfer coefficient. So, this is a enlarge view in which wall is shown the inner surface is at temperature T i outer temperature is at temperature T o the wall thickness is this much. So, by means of convective heat transfer there will be transfer of heat from the fluid to the wall through the wall there will be conductive heat transfer and from wall to the annular fluid or the cold fluid convective heat transfer will takes place. Now, we will try to see the relation for the overall heat here there are three resistances R 1, R 2, R 3. Consider this radius as R 1, this radius as R 2 the inner temperature is T i outer temperature is T o inside convective heat transfer coefficient is H i outer is H o. Now, the total resistance will be equal to R 1 plus R 2 plus R 3 R 1 is 1 by H i A i R 2 is L n R 2 by R 1 divided by 2 pi k L plus 1 by H o A o. As we know total heat transfer rate is equal to del T by R here it is total which is equal to U A del T. Now, both if we will compare these two equations then here U A is nothing but 1 by R total which can be written as U is equal to 1 by R total into A i and this capital U for the mixed flow heat transfer is known as overall heat transfer coefficient which is having two different types. If we will consider the inner surface it will be U i which will be equal to 1 by R total into A i and for the outer surface overall heat transfer coefficient will be U o which is equal to 1 divided by R into A k L. Now, the typical values are 100 0 to 350 and similarly if the steam condenser is there and water is flowing in the tube the value is 1000 to 6000. So, you will find that for phase change you try to think about the unit of the overall heat transfer coefficient and also compare it with the unit of the convective heat transfer coefficient. Now, there are two methods of heat exchanger analysis in the first method L M T approach is used and in the second method N T approach is used number of transparently initially we will see the L M T approach. If we know all inlet and outlet temperatures of the heat exchanger then and then only we can adopt the L M T approach you can use the L M T approach. Now, there are two types of the flows are there parallel flow and for the parallel flow L M T D for parallel flow this is the inner tube this is the outer tube hot fluid inlet temperature hot fluid outlet temperature this is cold fluid inlet temperature cold fluid outlet temperature. For the parallel flow the direction of both fluids is the same then here we can plot the temperature profile this is T i j as it flows its temperature will go on decreasing cold fluid inlet temperature is somewhere here as it flows it will be gaining the heat. So, the directions of the both fluids is the same. Now, let D Q is equal to heat transfer heat then for hot fluid it is m dot h C p h into bracket T h i minus T h o which is equal to minus C h where capital C h is heat capacity rate into bracket T h i minus T h o which can be written as minus C h into del T h. So, del T h which is equal to D T h will be equal to D Q by C h. Similarly, for cold fluid we can write D T c is equal to D Q by C c since theta is equal to T h minus T c. So, this theta is the temperature difference of hot fluid and cold fluid which is equal to del T. Now, differentiating D Q D theta. So, D theta is equal to D T h minus D T c. Now, we can substitute the values of D T h and D T c then we can write it as minus D Q by C h minus D Q by C c. Here I can take D Q common. So, it into into the bracket it will be C h plus 1 by C c. If we will simplify this then we can write D theta is equal to instead of del Q it will be minus U D a then theta into let this will be B. Now, this theta can be taken on left hand side. So, D theta by theta is equal to minus U D a into B taking the integration from theta 1 to theta 2 and for this integration 0 to a because U b are constants. So, I can write I can get L n theta o by theta i which is equal to minus U b into a. Now, substituting the relation for B which is equal to minus U a into bracket 1 by C h plus 1 by C c. Now, instead of this capacity rate I can write bracket directly minus U a into bracket 1 by m dot h C p h plus 1 by m dot c C p c. Since heat transfer by hot fluid is equal to Q which is also same for the cold fluid which is equal to m dot h C p h into bracket T h i minus T h o. So, you can write instead of m dot h C p h Q by T h i minus T h o or 1 by m C h C p h is equal to T h i minus T h o by Q. So, we will get L n theta o by theta i which is equal to minus U a into bracket T h i minus T h o by Q plus T c o minus T c i by Q. By simplifying this equation we will get L n theta o by theta i which is equal to U a by Q into bracket theta o minus theta i. So, for heat transfer rate we can write the equation as U a theta o minus theta i by L n theta o by theta i. Now, this is also equal to U a delta T m. Now, comparing these two equations U a U a is same then del T m will be equal to theta o minus theta i by L n theta o by theta i. So, this is known as for further study.