 Hello everyone, I am Ganesh B. Aglawe working as an assistant professor in department of mechanical engineering, Valchian Institute of Technology, Singapore. So in this session we will see of the heat exchangers numericals. So here we will be using LMTD approach, learning outcome. At the end of the session students will be able to design heat exchanger by LMTD method or approach. So in the first numerical the statement is like this. And oil cooler for lubrication system has to cool 1000 kilogram per hour of oil, whose specific heat is given as 2.09 kilo joule per kilogram Kelvin, from 80 degree Celsius to 40 degree Celsius by using a cooling water flow of 1000 kilogram per hour at 30 degree Celsius. Do your choice for a parallel flow or counter flow heat exchanger with reasons. Also calculate the surface area of heat exchanger if the overall heat transfer coefficient is 24 watt per meter Kelvin. Take Cp that is specific heat of water as 4.18 kilo joule per kilogram Kelvin. In this numerical the hot fluid is given as oil. The mass flow rate of hot fluid is 1000 kilogram per hour. Specific heat of the hot fluid is 2.09 kilo joule per kilogram Kelvin. The inlet temperature of hot fluid is 80 degree Celsius, outlet temperature as 40 degree Celsius. Now to cool the hot fluid water is used. So here the cold fluid is water. Now the mass flow rate of cold fluid is also 1000 kilogram per hour and the cold fluid inlet temperature is given as 30 degree Celsius. Now here they are asked to find the type of the heat exchanger. Is it the parallel flow or counter flow? As we know for the parallel flow heat exchanger the temperature profile has such type of shape in which THR is the hot fluid inlet temperature, THO hot fluid outlet temperature. For parallel flow in the same direction the cold fluid also flows while flowing it against the heat and the living temperature will be TCO. Remember the temperature of cold fluid is less than hot fluid. Now since the rate of heat exchange between the hot fluid and cold fluid is same. Here QH will be equal to m dot THCPH into bracket THI minus THO. After substituting the values in this equation we will get here THI is 80 degree and THO is 40 degree. Now if you do the calculation we will get some value of QH. Similarly I can write as QH is equal to QC the relation for QC which is m dot C CPC into bracket TCO minus TCI. Now we can substitute the values of m dot C CPC and in this equation we note TCI. So if these values are substituted then I will get TCO which is equal to 50 degree Celsius. Now can you think over the obtained TCO with THO of the given? So TCO here is greater than THO how much is TCO 50 degree Celsius which is greater than THO which has a value of 40 degree Celsius. Now this is not possible by using parallel flow heat exchanger means the type of the heat exchanger obviously will be counter flow, counter flow heat exchanger. So here we have given the answer of the type of the heat exchanger either parallel flow or counter flow. Now in the second part of this we required to find the surface area of the heat exchanger also calculate the surface area of heat exchanger. Now rate of heat transfer is also equal to UA delta Tm where delta Tm is equal to log main temperature difference which is the ratio of theta 1 minus theta 2 by ln theta 1 by theta 2. Here for the counter flow heat exchanger theta 1 will be equal to THI minus TCO and theta 2 is equal to THO minus TCI. If we will substitute theta 1 theta 2 in this equation U is also given overall heat transfer coefficient is 24. So the Q value is already known overall heat transfer coefficient value is given this surface area is to be calculated and delta Tm lmtd by using this equation we can find. So after substituting the values we will get the surface area as 535.9 centimeter square or 0.5359 meter square in this fashion we can get the surface area of heat exchanger. Now we will see the second numerical in the second numerical the statement is like this saturated steam at 120 degree Celsius is condensing on the outer tube surface of a single pass heat exchanger the type of the passes and the heat exchanger is of single pass heat exchanger. The overall heat transfer coefficient on outer side is 1800 watt per meter square Kelvin determine the surface area of a heat exchanger capable of heating 1000 kilogram per hour of water from 20 degree Celsius to 90 degree Celsius. So compute the rate of the heat exchanger rate of condensation of steam and here they have given the latent heat as 2200 kilo joule per kg. In this numerical the given data is saturated steam the steam is saturated has temperature 120 degree Celsius is condensing on outer tube surface of a single pass heat exchanger. The overall heat transfer coefficient is given as 1800 watt per meter square Kelvin. Determine the surface area surface area is to be determined of a heat exchanger capable of heating 1000 kilogram of water. So mass flow rate of the fluid is given as 1000 kilogram per hour the initial temperature as 20 degree Celsius and outlet temperature as 90 degree Celsius. Now steam will get condensed by rejecting heat to the cold fluid. So this will be the cold fluid inlet temperature cold fluid outlet temperature. This is the special type of the heat exchanger. In the special type of the heat exchanger the phase change will takes place at constant temperature and that constant temperature of hot fluid will be inlet temperature will be equal to outlet temperature and that is given as 120 degree Celsius. Meanwhile the fluid cold fluid water flowing has a temperature 20 degree Celsius inlet temperature outlet temperature 90 degree Celsius. So we can write the equation rate of heat transfer equation Q is equal to UA delta T m where delta T m delta T m is equal to theta 1 minus theta 2 by ln theta 1 by theta 2. Remember here theta 1 is T h i minus T c i theta 2 as T h o minus T c o. So we can get the delta T m value since the rate of heat transfer is equal to mass of cold fluid specific heat of cold fluid outlet temperature of cold fluid minus inlet temperature. After substituting we will get the rate of heat flow as 81394.4 Watt. Now I can substitute this Q in this equation 1 Q is substituted U is also substituted area is to be calculated delta T m after substituting these values we will get delta T m then by solving this equation we will get the surface area as 0.78 meter square this is the required surface area for the heat exchanger. Now how to find the rate of condensation this amount of heat will be rejected by the steam. So Q will be equal to mass flow rate of steam into its latent heat. So Q value is calculated they have given the latent heat. So mass of steam which is condensed will be obtained as 133.2 kilogram per margals. For further reference or study you can refer fundamentals of heat and mass transfer by