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How can a heat exchanger be designed for a specific industrial process involving the heating and cooling of a liquid stream with a flow rate of 500L/min and a specific heat capacity of 4.2 kJ/kg K? The temperature of the hot and cold streams are 120°C and 25°C respectively, and the process requires a heat transfer rate of at least 200 kW. What should be the surface a

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To design a heat exchanger for this specific industrial process, we need to determine the required surface area  A  and overall heat transfer coefficient  U  to meet the process requirements. We can use the following equation for heat transfer in a heat exchanger:Q = U * A * LMTDwhere Q is the heat transfer rate, LMTD is the log mean temperature difference, U is the overall heat transfer coefficient, and A is the surface area.First, let's calculate the LMTD:LMTD =  T1 - T2  / ln T1/T2 where T1 is the temperature difference between the hot and cold streams at one end of the heat exchanger, and T2 is the temperature difference at the other end.In this case, we know the temperature of the hot stream  Th  is 120C and the temperature of the cold stream  Tc  is 25C. We also know the heat transfer rate  Q  required is 200 kW. However, we need to determine the outlet temperatures of both streams to calculate the LMTD.To find the outlet temperatures, we can use the energy balance equation:Q = m * Cp *  Tout - Tin where m is the mass flow rate, Cp is the specific heat capacity, Tout is the outlet temperature, and Tin is the inlet temperature.We are given the flow rate  500 L/min  and specific heat capacity  4.2 kJ/kg K . We need to convert the flow rate to mass flow rate  kg/min  by multiplying it by the density of the liquid   . Assuming the liquid is water, the density is approximately 1000 kg/m.m = 500 L/min *  1 m/1000 L  * 1000 kg/m = 500 kg/minNow, we can use the energy balance equation to find the outlet temperature of the cold stream  Tc_out :200 kW =  500 kg/min  *  4.2 kJ/kg K  *  Tc_out - 25C Solving for Tc_out:Tc_out =  200,000 kJ/min  /  500 kg/min * 4.2 kJ/kg K  + 25C = 95CNow we can calculate the LMTD:T1 = 120C - 25C = 95CT2 = 120C - 95C = 25CLMTD =  95C - 25C  / ln 95C/25C   58.5CNow we can use the heat transfer equation to find the product of U and A:200 kW = U * A * 58.5CWe need to determine the overall heat transfer coefficient  U  for the heat exchanger. This value depends on the materials used, the type of heat exchanger, and the flow conditions. Assuming a typical value for a shell and tube heat exchanger with water as the medium, we can estimate U to be around 1000 W/m K  which is equal to 1 kJ/m K min .Now we can solve for the surface area  A :A = 200,000 kJ/min /  1 kJ/m K min * 58.5C   3420 mSo, the required surface area of the heat exchanger is approximately 3420 m, and the overall heat transfer coefficient is 1 kJ/m K min  or 1000 W/m K  to meet the process requirements.

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