Question

Oil of unknown properties is heated in a shell-and-tube heat exchanger with one shell pass and 20 tube passes. The oil flows through the shell, and hot water flows inside the single copper tube that has an inner diameter of 20 mm, a wall thickness of 2 mm, and a length of 3 m per pass. The water enters at 360 K at a mass flow rate of 0.2 kg/s and leaves at 300 K. The inlet and outlet temperatures of the oil are 280 K and 310 K respectively. Calculate the following: a) The Reynolds number of the flow inside the tube. Verify that the tube is long enough for the flow to be both hydrodynamically and thermally developed. b) The average convective heat transfer coefficient of the flow inside the tube. c) The rate of heat transfer between the water and the oil. d) The heat capacity rate of the oil. e) The maximum rate of heat transfer between the water and the oil and use this to estimate the heat exchanger effectiveness. f) The number of transfer units, NTU. g) The overall heat transfer coefficient times the area, UA. h) The average convection heat transfer coefficient of the tube outer surface using your estimate of UA. Assume there is no fouling and that the thermal resistance of the copper tube is negligible.

Answer 1

applying every balare b/w oil and wate> (MCP) w. (87) wr - Miceland (OT)ois (MCP) w (360-300) = (Mcploie ( 310 -280) (Mcp)w (310-280) (micplois = (360-300) - Capacity Rote Radio . - and (micp) w c (in Cebie Maximum Rate of heat transfer Omar: micp) small. (AT) Big – Mi Co) w - (Twin - To, in) - 012 84.18 (360-2001 Tomax = 66.88 Kw] and Heat exchanger offechienen! - Mi Calw (Tw, in- Tw, out) Pactual (mi(e) (Tw; in - To, in) Oman E = (360 - 300) 40 2x 4.18 5016 360 - 280) x 0:2X4.18 16.88 Je=0.75 l 560 Now water 310 as To, out > Tw, out (30) > (300) so He Is Counter flow shell and AT; - 360-310 z 50'c o lote 280 to be HE. Ате 300-28 20°c

LMTD (DIM) . ATI-ATE 50-20 en Cor e a (%) O im = 32.741 (MCP Jw (AT) w 2 UA DIm 0.2% 4180* (360-300) = UA * 32.741 UA = 1532.038 W NTV2 UA (mcp) small - UA mcpw = 1532-038 024186 INTU – 1188258 | A . gingen mw = 0.2 ks PAV: 012 1037 (0.02)? XV = 0.2 Reynalde 0.63662> 0.02 Reynold N. V= 0.63662 m/s 7 for water at ang semp of 330k cinematic Viscosity V- U-967x100 mg Re - VIP - 0163662x 0.02 4.967 x 102 TRE = 25633.975) Re > Reer (2000) 80 turbulent How KO

Thermal Gonductivity of water at 33° K Cany. Semb.) K-0.638 W/mK 14 Pr = 3.207 for furbulent how Using Dittes Belter en No hirde 0.023 (R. Jo (Px 20.3 for cosuty of water 1 Tix. 002 0.023 ( 28633.97839*(3-23,0-3 01638 hi = 3502.684 W/m2 i total surface aree Ar 200 ADL 20% ZX 0.02 X 3 37699 m2 and UA2 1532-038 W 1532-038 3.7699 Over all He J. Coeff Uz 40613857 W/m2K - this the 06.3853= 5502.004 to ho=459-7233 W/m x 7 Rate of Meat transfer b/w oil and water. Pacton: (Mcp)w (Allw . 02 * 4:18 (360-309 Cachel = 50.16 kw) ©

@ 999999 Heat Capacity rate of one (mi (p)one (To, our - To, in): (mcpw (DJ) w m'cp)ovie (310-280) - 0.2% 4-18 * (360-300) Į (mice Joie = 1.632 kg @ as Re- 25633.975 (turbulent flow) a hydrodynamic entrance length (where how developed . July ) La, torb. = 4:40. (Reje = 4:4 * 0.02 x (266331975) Yo In: 014778 m] So bipe is long enough for How to be developed thermal endance longth hydrodynamiently for Jurbulent Now, M : 100 ata 10X0:02 Far-0.2m] so pipe is long to be Pact aman 66.88 enough for How developed thermally - E. 2016 x LE :0.75 ] Effectivenes