Design a heat exchanger for a factory that produces 10,000 liters of sulfuric acid per day starting at 70°C and needs to be cooled to 40°C for packaging. The available cooling water supply from the local municipal is at 20°C and is available at a flow rate of 1000 liters per minute. Design the heat exchanger to achieve the required temperature drop and calculate the heat transfer coefficient and the overall heat transfer coefficient.

Question

Design a heat exchanger for a factory that produces 10,000 liters of sulfuric acid per day starting at 70°C and needs to be cooled to 40°C for packaging. The available cooling water supply from the local municipal is at 20°C and is available at a flow rate of 1000 liters per minute. Design the heat exchanger to achieve the required temperature dro

Answer 1

To design a heat exchanger for this problem, we need to determine the heat transfer rate, the heat transfer coefficient, and the overall heat transfer coefficient. We will use the following equations:1. Heat transfer rate  Q  = mass flow rate  m  x specific heat capacity  c  x temperature change  T 2. Heat transfer coefficient  h  = Q /  A x T_lm 3. Overall heat transfer coefficient  U  = 1 /  1/h_h + 1/h_c Assumptions:- The specific heat capacity of sulfuric acid is 1.3 kJ/kgC.- The specific heat capacity of water is 4.18 kJ/kgC.- The density of sulfuric acid is 1.84 kg/L.- The density of water is 1 kg/L.- The heat transfer coefficients for the hot and cold sides are equal  h_h = h_c = h .First, let's calculate the heat transfer rate  Q :Q = m_acid x c_acid x T_acid = m_water x c_water x T_waterm_acid = 10000 L/day * 1.84 kg/L = 18400 kg/dayT_acid = 70C - 40C = 30Cm_water = 1000 L/min * 60 min/hour * 24 hours/day * 1 kg/L = 1440000 kg/dayT_water = T_out - 20CQ = 18400 kg/day * 1.3 kJ/kgC * 30C = m_water * 4.18 kJ/kgC * T_waterQ = 717600 kJ/day717600 kJ/day = 1440000 kg/day * 4.18 kJ/kgC * T_waterT_water = 0.122CNow, let's calculate the logarithmic mean temperature difference  T_lm :T1 = 70C - 20C = 50CT2 = 40C - 20.122C = 19.878CT_lm =  T1 - T2  / ln T1 / T2 T_lm =  50C - 19.878C  / ln 50C / 19.878C T_lm  32.7CNext, we will calculate the heat transfer coefficient  h :Q = h * A * T_lmA = heat transfer areaWe need to determine the area  A  of the heat exchanger. Let's assume a heat transfer coefficient  h  of 1000 W/mC, which is a typical value for water-to-water heat exchangers. We can adjust this value later if needed.h = 1000 W/mC = 1000 J/s*mCNow, we can calculate the area  A :A = Q /  h * T_lm A =  717600 kJ/day * 1000 J/kJ  /  86400 s/day * 1000 J/s*mC * 32.7C A  0.76 mFinally, we will calculate the overall heat transfer coefficient  U :Since we assumed h_h = h_c = h, the overall heat transfer coefficient  U  is equal to h:U = h = 1000 W/mCIn conclusion, the heat exchanger should have a heat transfer area of approximately 0.76 m and an overall heat transfer coefficient of 1000 W/mC to achieve the required temperature drop for the sulfuric acid. Note that these calculations are based on several assumptions, and the actual design may require adjustments based on specific conditions and requirements.