Design a heat exchanger for a process where hot air (80°C) needs to be cooled down to 30°C using water (10°C) as the cooling medium. The flow rate of hot air is 2 kg/s and the desired outlet temperature of the cooled air is 35°C. The maximum allowable pressure drop for air side is 50 kPa and for water side is 100 kPa. Design the heat exchanger using the LMTD (logarithmic mean temperature difference) method and calculate the required area of heat transfer.

Question

Design a heat exchanger for a process where hot air (80°C) needs to be cooled down to 30°C using water (10°C) as the cooling medium. The flow rate of hot air is 2 kg/s and the desired outlet temperature of the cooled air is 35°C. The maximum allowable pressure drop for air side is 50 kPa and for water side is 100 kPa. Design the heat exchanger using the LMTD (logarithmic mean tem

Answer 1

To design the heat exchanger using the LMTD method, we first need to determine the heat transfer rate  Q  and the overall heat transfer coefficient  U . Then, we can calculate the required area of heat transfer  A .1. Determine the heat transfer rate  Q :Q = m_air * C_p_air *  T_inlet_air - T_outlet_air Assuming the specific heat capacity  C_p  of air is approximately 1.005 kJ/kgK, we can calculate the heat transfer rate:Q = 2 kg/s * 1.005 kJ/kgK *  80C - 35C Q = 2 * 1.005 * 45Q = 90.45 kJ/s2. Determine the inlet and outlet temperatures of the cooling water:Since we know the heat transfer rate  Q  and the flow rate of hot air, we can calculate the temperature increase of the cooling water. Assuming the specific heat capacity  C_p  of water is approximately 4.18 kJ/kgK, we can calculate the mass flow rate  m_water  of the cooling water:Q = m_water * C_p_water *  T_outlet_water - T_inlet_water Rearranging the equation to solve for m_water:m_water = Q /  C_p_water *  T_outlet_water - T_inlet_water  Assuming the outlet temperature of the cooling water is 30C:m_water = 90.45 kJ/s /  4.18 kJ/kgK *  30C - 10C  m_water = 90.45 /  4.18 * 20 m_water = 1.08 kg/s3. Calculate the LMTD  logarithmic mean temperature difference :LMTD =  T_1 - T_2  / ln T_1/T_2 Where T_1 is the temperature difference between the hot air inlet and the cooling water outlet, and T_2 is the temperature difference between the hot air outlet and the cooling water inlet:T_1 = 80C - 30C = 50CT_2 = 35C - 10C = 25CLMTD =  50 - 25  / ln 50/25 LMTD  35.6C4. Determine the overall heat transfer coefficient  U :The overall heat transfer coefficient  U  depends on the materials and design of the heat exchanger. For this problem, we will assume a typical value for a shell and tube heat exchanger, which is around 1000 W/mK  or 1 kJ/smK .5. Calculate the required area of heat transfer  A :Now that we have the heat transfer rate  Q , the LMTD, and the overall heat transfer coefficient  U , we can calculate the required area of heat transfer:A = Q /  U * LMTD A = 90.45 kJ/s /  1 kJ/smK * 35.6C A  2.54 mSo, the required area of heat transfer for the heat exchanger is approximately 2.54 m.