Question

Question 6ACLO5 War T-50 Ashall d e la change with 2-shell passes and be passis wed to how ethyl alcoholic - 2670 kg in the tubes fro 25°C to Came of 21 kgs. The huaring is done by water -4190 kg the enters the shell side 95C and love 1 45°C. Thew of the heat exchanger is A -25°C tube passes) -10 m andh. - 35 W TC Determine the logarithmic mean temperature difference =- ATT -T AT, T.-T ...... Ilear Trade ATW AT,AT.) AT-AT Determine the correction Factor Fuse Fig. 6) R=1 Correction factor F 017111 0 0.1 0.2 03 04 05 06 07 08 09 10 P (b) Two-shell passes and 4. &. 12. etc. (any multiple of 4), tube passes El Correction factor Echart for common shall-and-tube heat exchangers Determine the overall heal trasfer coefficit ater fouling with a fouling factor of R-0.0016 C/W the outer surface of the tubes: m 4 Determine the rate of heat transfer in the heat exchanger if the son of the heat exchanger A4 Heel Track

Answer 1

- Assumptions, t. Steady operating conditions exist. 2. The heat enchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid ei equal to the heat transfer to the cold fleid. (3 change in the kinetic and potential energia of fluud streams are negligible. ® There is not in properties of fleid. Posperties - The specific heats of water and ethyl alcohol are given to be 4190 I lugar and 2670 7/19°C Respectively. I water Ti = 956 t 70' er Ethyl Alcohul 250 2. kgls. t (8 tube ) Je zA5C

2 The logarithmic mean temperature difference for Counter - flow arrangment and the correction factor are. F D oT, = Thin-Thout = 95C - 706 = 25°C. AT₂ = Thout - Tc, in a 45°C -25°C = 20°C A Timcs - - im, CE Oti-ate ln (st/otel 25-20 - ln (25) Alim, cf = 22.4° C 2 Determine the correction free for F :- P= tar ti - 45-95 25-gg = R = F=0.77. I-Tz tant 25-70 -zuig 45-95 ③ The overall heat transfer :- ti + + Ry

U = It is +0.0006 160 tu= 21.34 W/ m² 4 The rate of heat enchenger ett transfer in the heat = UAF & Timice 2 21.34 x 4 x 0.774 22,4'c Q = 1472.6 W