A shell-and-tube heat exchanger, shown below in the link, is required to cool 500kg/hour of milk from 28DegC to 15DegC by using 800kg/hour of water at a inlet temperature of 10DegC.
i) The heat rate need to be removed from the milk is Q = m_milk ? Cp_milk ? ?T_milk = m_milk ? Cp_milk ? (T_milk,in -T_milk,out) = 500kg/hr ? 4200J/kgK ? (28 - 15)K = 27.3×10?J/hr = 7583J/s = 7583W ii) Cooling water absorbs the same amount of heat: Q = m_water ? Cp_water ? (T_water,out -T_water,in) Hence: T_water out = T_water,in + Q/(m_water ? Cp_water) = 10°C + 8.125°C = 18.125°C iii) The temperature difference between the two media at right hand side of the heat exchanger is: ?T_A = T_milk,in - T_water,out = (28 - 18.125)K = 9.875K The temperature difference between the two media at left hand side of the heat exchanger is: ?T_B = T_milk,out - T_water,in = (15 - 10)K = 5K So the log mean temperature difference of the heat exchanger is LMTD = (?T_A - ?T_B)/ln(?T_A/?T_B) = (9.875K - 5K)/ln(9.875K/5K) = 7.163K iv) Specified diameter of heat exchanger pipes is commonly the internal diameter. d_i = 20mm = 0.02m => d_o = 20mm + 2?3mm = 26mm = 0.026m Overall heat transfer coefficient, with respect to internal pipe surface is: U = 1/(1/h_i + d_i?ln(d_o/d_i)/(2?k) + d_i/(h_o?d_o) = 1/(1/1800W/m?K + 0.02m?ln(0.026/0.02)/(2?800W/mK) + 0.026m/(1800W/m?K?0.02m) = 780.6W/m?K Because Q = U?A?LMTD the total exchanger area needed is A = Q/(U?LMTD) = 7583W / (780.6W/m?K ? 7.163K) = 1.3562m? The internal surface of a single pipe is A_pipe = π·d_i·L = π · 0.02m · 1.5m = 0.09425m? So the total number of pipes needed is: n = A/A_pipe = 1.3562m?/0.09425m? = 14.39 ≈ 15
