Here's a precise problem for a chemistry student to solve related to designing a heat exchanger for a specific process:A chemical plant produces a product with a flow rate of 500 liters per hour, and it needs to be cooled from its initial temperature of 120°C to its final temperature of 80°C before it can be packaged for shipment. The cooling medium that will be used is water, which is available at a flow rate of 400 liters per hour and an initial temperature of 20°C. Design a heat exchanger to cool the product and determine the required surface area of the heat exchanger if the overall heat transfer coefficient is 500 W/m2K.

Question

Here's a precise problem for a ch

Answer 1

To design the heat exchanger, we need to determine the required surface area. We can use the following equation to calculate the heat transfer rate  Q  and then find the surface area  A  using the overall heat transfer coefficient  U .Q = m_product  C_p,product   T_initial,product - T_final,product where m_product is the mass flow rate of the product, C_p,product is the specific heat capacity of the product, T_initial,product is the initial temperature of the product, and T_final,product is the final temperature of the product.First, we need to convert the flow rate of the product from liters per hour to kg/s. Assuming the product has a density  _product  similar to water  1000 kg/m , we can calculate the mass flow rate  m_product  as follows:m_product =  500 L/h  1000 kg/m  /  3600 s/h  = 0.1389 kg/sNext, we need to find the specific heat capacity  C_p,product  of the product. Assuming it is similar to water, we can use the specific heat capacity of water, which is approximately 4.18 kJ/kgK  4180 J/kgK .Now, we can calculate the heat transfer rate  Q :Q = 0.1389 kg/s  4180 J/kgK   120C - 80C  = 23,102.8 WThe heat transfer rate  Q  can also be expressed as:Q = U  A  T_lmwhere U is the overall heat transfer coefficient, A is the surface area of the heat exchanger, and T_lm is the log mean temperature difference.To find T_lm, we need to calculate the temperature difference at both ends of the heat exchanger:T_1 = T_initial,product - T_initial,water = 120C - 20C = 100CT_2 = T_final,product - T_final,waterTo find T_final,water, we can use the energy balance equation:m_product  C_p,product   T_initial,product - T_final,product  = m_water  C_p,water   T_final,water - T_initial,water Assuming the specific heat capacity of water  C_p,water  is 4180 J/kgK and the mass flow rate of water  m_water  is:m_water =  400 L/h  1000 kg/m  /  3600 s/h  = 0.1111 kg/sWe can now solve for T_final,water:0.1389 kg/s  4180 J/kgK   120C - 80C  = 0.1111 kg/s  4180 J/kgK   T_final,water - 20C T_final,water = 61.6CNow, we can find T_2:T_2 = T_final,product - T_final,water = 80C - 61.6C = 18.4CNext, we can calculate the log mean temperature difference  T_lm :T_lm =  T_1 - T_2  / ln T_1 / T_2  =  100C - 18.4C  / ln 100C / 18.4C  = 54.3CNow, we can find the required surface area  A  of the heat exchanger:A = Q /  U  T_lm  = 23,102.8 W /  500 W/mK  54.3C  = 0.846 mTherefore, the required surface area of the heat exchanger is approximately 0.846 m.