 working as an assistant professor in Department of Mechanical Engineering, Vulture Institute of Technology, Solaford. In this session of heat exchangers, we will see few numericals by using LMTD approach. Learning outcome. At the end of this session, students will be able to design heat exchanger by LMTD method. Now, in this first numerical, So, given data, the hot fluid has flow rate 1000 kilogram per hour, specific heat 2.09 kilo joule per kilogram, THI 80 degree Celsius, THO 40 degree Celsius. Cold fluid has properties mass flow rate 1000 kilogram per hour, then the inlet temperature is 30 degree Celsius. So, the parallel flow heat exchanger is having such type of construction. This is hot fluid inlet temperature, hot fluid outlet temperature and the cold fluid inlet temperature is 30 degree Celsius, TCO is unknown. In the first part, we require to calculate whether the heat exchanger is of parallel flow or counter flow. If we will try to draw the temperature distribution along the heat exchanger, then the temperature profile of hot fluid will have such type of the factor and cold fluid like this one. But in the case of counter flow heat exchanger, we may get TCO greater than THO. This we will try to find now. So, we can have the rate of heat transfer for hot fluid and cold fluid. The heat transfer rate is same. So, Q is equal to for hot flow, hot fluid and for cold fluid. So, same rate of heat transfer will be there as the THI and THO are given. So, we can find out heat transfer rate is equal to M dot THCPH into bracket THI minus THO. This is 1000 into 2.09 into bracket 80 minus 40. So, we will get the answer 167200 kilojoule per hour. Which is also equal to cold fluid. So, 167200 is equal to M dot C CPC into bracket TCO minus TCI. So, substituting the values mass flow rate of cold fluid is 1000 as a water is used as a cold fluid. So, specific heat is 4.187 TCO minus TCI is 30. So, by solving this equation, we will get the outlet temperature of cold fluid as 50 degree Celsius. Now, see here the hot fluid inlet temperature given 80 degree Celsius and outlet temperature is given as 40 degree Celsius. Why? We got TCO 50 degree Celsius which is greater than hot fluid outlet temperature. Now, this case is not possible getting the outlet temperature of cold fluid greater than hot fluid outlet temperature is not possible with parallel flow. That is why the this type of heat exchanger must be of counter flow type of counter flow type. Now, second part in the second part they ask the surface area find out the surface area. Now, we know the equation Q is equal to UA delta Tm. We know the rate of heat transfer then we can find out delta Tm. Here delta Tm for counter flow for counter flow this is theta i this is theta o. So, theta o minus theta i by ln theta o by theta i. Substitute the values of theta o and theta i will get the log mean temperature difference as 18.20 degree Celsius. Then by using this equation Q is equal to UA delta Tm. And substituting the values of Q U delta Tm which we have calculated as 18.20. We will get the surface area of heat exchanger as 0.054 meter square. So, this in this way we may get the surface area of the counter flow heat exchanger means we can select any diameter of the tube any diameter of known tube. Then to satisfy this surface area we require to increase the length go on increasing the length till we get this surface area. Now, we will see next second numerical, second numerical saturated steam. Now, here as the condenser is given. So, remember this is the special type of the heat exchanger in which the condensation temperature of the steam will be same. And here if we consider this as the hot fluid. So, THI will be equal to THO while for the cold fluid the given inlet temperature is 20 degree Celsius. So, TCI having 20 degree Celsius is flowing through the condenser will absorb the heat and at the same time there will be rise in the temperature. So, rise in temperature is up to 90 degree Celsius. So, in the first part we require to find out the area. Now, for this condenser type of the heat exchanger this is theta i which is equal to THI minus TCI and theta O is equal to THO minus TCO. So, we may get theta i as 100 degree Celsius and theta O as 30 degree Celsius. So, LMTD is theta O minus theta i divided by LN theta O by theta i which is equal to 100 minus 30 divided by LN 100 by 30 which is equal to 58 degree Celsius. Now, since the rate of heat transfer for hot fluid and cold fluid is the same. So, we can write for the hot fluid as we can write for the cold fluid as M dot C CPC into bracket TCO minus TCI which is equal to this is 1000 multiplied by given data is 4.187. into this 90 minus 20 we will get the rate of heat transfer as 81.28 kW. Obviously, you require to convert this per hour into per second. Now, substituting the heat transfer value in this equation as UA delta TM. So, 81.28 into 10 raise to 3 this is the Watt is equal to the overall heat transfer coefficient is given as 1800 into A into this 58 delta TM. We will get the area as 0.774 meters square. So, this is the required answer. Now, in the second part we require to find out the rate of condensation. So, second to find the rate of condensation the heat given by the steam is equal to mass flow rate of steam into HFG. The rate is we know the rate of heat transfer 81.28 into 10 raise to 3 this is unknown into HFG is given 2200. So, this gives mass flow rate of steam this will give the mass flow rate of steam in kilogram per second. For further study you can refer heat and mass transfer by Incropera David. Thank you.