How to do qn Qn3(d) and (e)?

QUESTION 3 (20 MARKS) A single-pass shell and tube heat exchanger is used to heat a polymer solution (p = 900 kg/m%). The polymer solution is flowing through the stainless-steel tubes with an inside diameter of 0.0428 m at a mass flow rate of 5.18 kg/s. The solution is heated from 30 C to 70 C. Water flows counter-currently in the shell side at a mass flow rate 3.5 kg/s and a temperature of 90 C. The overall heat transfer coefficient (U) for this heat exchanger is 3000 W/(m2. C). ) Calculate the a) heat transfer rate, q, in W, (2 marks) b) exit water temperature, (4 marks) c) log mean temperature difference, (4 marks) number of tubes used if the velocity of polymer solution inside the tube is 0.4 m/s, and (6 marks) e) required tube length. (4 marks) Useful data: Specific heat capacity of the polymer solution at 50 C, Cpc=2.3 kJ/(kg. C) Specific heat capacity of water at average temperature, Cph= 4.187 kJ/(kgC) Question 3 - 400 kg. di = 0.0428n AT 50C, Ope = 2300 goc Goh 4187 Try Tugit m = 5.18 kg me = 3.5kg V 3000 h 30% 706 904 7 (a) q = MCpc (Tco - Tea) = 15:18) (2.37.103) (10-30) > 476560 2 (6) Heat Balance min Gan (Tra- This inc. Cac (Too-Tha). (3.5)(4.18739?)( 90 - Thb) = (5.18)(23700 | 70-30) 1318405-146 544 Tht = 426563 Tho= 57.48C (c) Counter-cument flow AT, = 57.48-30 27:4 AT2 = 90-70 20*C AT = 27.48-20 23.54'c (a)