Design a heat exchanger to cool down a mixture of water and ethylene glycol from 80°C to 20°C using cooling water from a nearby lake that enters at 15°C. The mixture flow rate is 10 L/min, and the desired outlet temperature is 20°C. The heat transfer coefficient for the heat exchanger is 250 W/(m².K) and the overall heat transfer coefficient is to be assumed as 350 W/(m².K). The heat capacity of the mixture is 3.8 kJ/(kg.K) and the density is 1030 kg/m³. Determine: a) the surface area of the heat exchanger required for this process,b) the volume flow rate of cooling water required, and c) the overall heat transfer rate in kW.

Question

Design a heat exchanger to cool down a mixture of water and ethylene glycol from 80°C to 20°C using coolin

Answer 1

a  To determine the surface area of the heat exchanger, we first need to calculate the heat transfer rate  Q  required for this process. The mass flow rate  m_dot  of the mixture can be calculated as follows:m_dot =  density * flow rate  / 1000m_dot =  1030 kg/m * 10 L/min  / 1000m_dot = 10.3 kg/minNow, we can calculate the heat transfer rate  Q  using the mass flow rate, heat capacity, and temperature difference:Q = m_dot * Cp * TQ = 10.3 kg/min * 3.8 kJ/ kg.K  *  80C - 20C Q = 10.3 * 3.8 * 60Q = 2354.4 kJ/minConvert Q to Watts  W :Q = 2354.4 kJ/min *  1000 W/kJ  *  1 min/60 s Q = 39240 W = 39.24 kWNow, we can calculate the surface area  A  of the heat exchanger using the overall heat transfer coefficient  U  and the logarithmic mean temperature difference  LMTD :LMTD = [ T1_hot - T1_cold  -  T2_hot - T2_cold ] / ln[ T1_hot - T1_cold  /  T2_hot - T2_cold ]LMTD = [ 80 - 15  -  20 - 15 ] / ln[ 80 - 15  /  20 - 15 ]LMTD =  65 - 5  / ln 65 / 5 LMTD  39.7CNow, we can calculate the surface area  A :A = Q /  U * LMTD A = 39240 W /  350 W/ m.K  * 39.7 K A  2.83 ma  The surface area of the heat exchanger required for this process is approximately 2.83 m.b  To determine the volume flow rate of cooling water required, we need to calculate the heat transfer rate  Q  for the cooling water. Since the heat exchanger is in a steady state, the heat transfer rate for the cooling water is equal to the heat transfer rate for the mixture:Q_cooling_water = Q_mixture = 39.24 kWAssuming the cooling water has a heat capacity  Cp_cooling_water  of 4.18 kJ/ kg.K  and a density  _cooling_water  of 1000 kg/m, we can calculate the mass flow rate  m_dot_cooling_water  of the cooling water:m_dot_cooling_water = Q_cooling_water /  Cp_cooling_water * T_cooling_water m_dot_cooling_water = 39.24 kW /  4.18 kJ/ kg.K  *  T2_cold - 15C  Since we don't know the outlet temperature  T2_cold  of the cooling water, we can't determine the exact volume flow rate of cooling water required. However, we can provide a range:For the minimum outlet temperature  T2_cold = 15C , the mass flow rate of cooling water would be infinite, which is not practical.For a more reasonable outlet temperature  e.g., T2_cold = 25C , we can calculate the mass flow rate:m_dot_cooling_water = 39.24 kW /  4.18 kJ/ kg.K  *  25C - 15C  m_dot_cooling_water  9.36 kg/sNow, we can calculate the volume flow rate  V_dot_cooling_water  of the cooling water:V_dot_cooling_water = m_dot_cooling_water / _cooling_waterV_dot_cooling_water = 9.36 kg/s / 1000 kg/mV_dot_cooling_water  0.00936 m/sb  The volume flow rate of cooling water required is approximately 0.00936 m/s  9.36 L/s  for an outlet temperature of 25C. The actual volume flow rate required will depend on the desired outlet temperature of the cooling water.c  The overall heat transfer rate in the heat exchanger is 39.24 kW.