Question

Design a heat exchanger that can transfer heat from a 10,000 L/h hot water stream at 80°C to a 5,000 L/h cold water stream at 20°C with a maximum temperature difference of 10°C. Calculate the required heat transfer area and select appropriate materials for the heat exchanger components based on the fluids' properties and operating conditions.

Answer 1

To design a heat exchanger, we first need to determine the heat transfer rate, the overall heat transfer coefficient, and the logarithmic mean temperature difference  LMTD . Then, we can calculate the required heat transfer area and select appropriate materials for the heat exchanger components.1. Heat transfer rate  Q :We need to find the heat transfer rate from the hot water stream to the cold water stream. We can use the formula:Q = m * Cp * Twhere m is the mass flow rate, Cp is the specific heat capacity of water, and T is the temperature difference.For the hot water stream:m_hot =  10,000 L/h  *  1 kg/L  *  1 h/3600 s  = 2.78 kg/sCp_hot = 4.18 kJ/ kgK T_hot = 80C - 70C = 10C  since the maximum temperature difference is 10C For the cold water stream:m_cold =  5,000 L/h  *  1 kg/L  *  1 h/3600 s  = 1.39 kg/sCp_cold = 4.18 kJ/ kgK T_cold = 30C - 20C = 10CNow, we can calculate the heat transfer rate:Q = m_hot * Cp_hot * T_hot = m_cold * Cp_cold * T_coldQ = 2.78 kg/s * 4.18 kJ/ kgK  * 10 K = 116.2 kW2. Logarithmic mean temperature difference  LMTD :LMTD =  T1 - T2  / ln T1/T2 where T1 is the temperature difference at one end of the heat exchanger and T2 is the temperature difference at the other end.T1 = 80C - 20C = 60CT2 = 70C - 30C = 40CLMTD =  60C - 40C  / ln 60/40  = 49.2C3. Overall heat transfer coefficient  U :The overall heat transfer coefficient depends on the materials and the design of the heat exchanger. For a typical shell and tube heat exchanger with water as the fluid, U is approximately 800 W/ mK .4. Required heat transfer area  A :Now, we can calculate the required heat transfer area using the formula:A = Q /  U * LMTD A = 116.2 kW /  800 W/ mK  * 49.2 K  = 2.98 m5. Material selection:For the heat exchanger components, we need materials that have good thermal conductivity, corrosion resistance, and mechanical strength. Stainless steel  e.g., SS316  is a common choice for heat exchanger components due to its excellent corrosion resistance and good thermal conductivity.In conclusion, we need a heat exchanger with a heat transfer area of 2.98 m, and stainless steel  e.g., SS316  is an appropriate material for the heat exchanger components.