Question

Design a heat exchanger for a chemical plant which produces 1000 kg/hr of a chemical at a temperature of 350°C. The chemical needs to be cooled down to 50°C using water at an inlet temperature of 25°C. The cooling water flow rate should not exceed 5000 kg/hr. The overall heat transfer coefficient should be at least 500 W/m2K. Determine the surface area of the heat exchanger required for this process.

Answer 1

To design the heat exchanger, we first need to determine the heat duty  Q  required to cool the chemical from 350C to 50C. We will assume that the specific heat capacity  Cp  of the chemical and water are constant. Let's denote the specific heat capacity of the chemical as Cp_chem and the specific heat capacity of water as Cp_water.1. Calculate the heat duty  Q :Q = m_chem * Cp_chem *  T_chem_initial - T_chem_final where m_chem is the mass flow rate of the chemical  1000 kg/hr , T_chem_initial is the initial temperature of the chemical  350C , and T_chem_final is the final temperature of the chemical  50C .2. Calculate the heat capacity rates for the chemical and water:C_chem = m_chem * Cp_chemC_water = m_water * Cp_waterwhere m_water is the mass flow rate of the cooling water  5000 kg/hr .3. Calculate the temperature change of the cooling water:T_water = Q / C_water4. Calculate the logarithmic mean temperature difference  LMTD :T1 = T_chem_initial - T_water_inletT2 = T_chem_final -  T_water_inlet + T_water LMTD =  T1 - T2  / ln T1/T2 5. Calculate the required heat transfer area  A :A = Q /  U * LMTD where U is the overall heat transfer coefficient  500 W/m2K .Now, let's assume the specific heat capacity of the chemical  Cp_chem  is 2 kJ/kgK and the specific heat capacity of water  Cp_water  is 4.18 kJ/kgK. We can now calculate the required heat transfer area  A .1. Q = 1000 kg/hr * 2 kJ/kgK *  350C - 50C  = 600,000 kJ/hr = 166.67 kW2. C_chem = 1000 kg/hr * 2 kJ/kgK = 2000 kJ/hrK   C_water = 5000 kg/hr * 4.18 kJ/kgK = 20,900 kJ/hrK3. T_water = 166.67 kW / 20.9 kW/K = 7.98C4. T1 = 350C - 25C = 325C   T2 = 50C -  25C + 7.98C  = 17.02C   LMTD =  325C - 17.02C  / ln 325/17.02  = 54.24C5. A = 166.67 kW /  0.5 kW/m2K * 54.24K  = 6.14 m2Therefore, the required surface area of the heat exchanger is approximately 6.14 m2.