Question

can you solve only last 4 questions

Q3. 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 Kat a mass flow rate of 0.2 kg/s and leaves at 300 K. The inlet and outlet temperatures of the oil are 280 Kand 310 K respectively. Calculate the following: a) The Reynolds number of the flow inside the tube. Verify that the tube is long [3marks) enough for the flow to be both hydrodynamically and thermally developed. b) The average convective heat transfer coefficient of the flow inside the tube. [3marks) c) The rate of heat transfer between the water and the oil. [2marks) d) The heat capacity rate of the oil. [2marks) e) The maximum rate of heat transfer between the water and the oil and use this to [3marks) estimate the heat exchanger effectiveness. f) The number of transfer units, NTU. [2marks] g) The overall heat transfer coefficient times the area, UA. [2marks] 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 [3marks] the copper tube is negligible.

Answer 1

a Answer Thi/360k 012 Thi-Teo water Mzoo2 2 kg/s (360-310) = so To=3 lok Tho -3ook Ozc (Tho-Teo 022 300-280 Ti = 280k = 20 Our Oz 50-20 LMTD 2 In(o) en(590) LMTD 32.74 Heat exchange (mc) watert (Thi-Tho) = aque between water & oil (Mc) or (Teo-Tai) 0.27 U180x (360-300) = (me) oil x(310-289) : أنه (W 1672 KI/K Cat Cwater mc) unter (water 0-2x4180 836 kJ/k Emin = [min (Coil Cuater) | Emin 836 KJ /

Maximum rate of heat transfer C Qmax = (min [ Tmax - Tmin ((mm) [Thi-Te 836 [ 360 – 280] Qmax 66880 W 66.88 KW Ane To determine the NTU first find effectiveness of heat exchanger Thi Tho Thi-Tei = مع 360-300 - 360 - 280 Ez 0.75 satio C (c) eat capacity Cmin (maxi 836 1672 =0.5 effectiveney can also be written like in terms of NTO

-NTUCIRC) Ez 1-e -NTUCI-3) -NTU (1-0,5) rce 6675 = l-é -NTU (1-05) 1-6.5xe NTO -1.83 Am g over all heat transfer coefficient N TU = UA Cmin where us over all heat transfer coefficient A= Surface area Tdi LX No. of tube KX010233 x 20 3.76 m2 1083= UX3.76 836 UE 406.88 W W/mk UANTUX Cmin 1.83X 836 UAZ 1529.88 W/

10 کا The average convection heat transfor Coefficient of the tube outer surface 1 I + 1 hi Ai ho to UA Tvalue of thie Ai inner tube surface area Ao z outer tube surface area Average convective heat transfer hi take from part (b) coefficient of the flow inside the tube Aiz Idi e TX 0.02x3 0.1884 m2 Ai Ź {do= ditzt t- thickner) Ad = ac di +2x+) Xh 7/000242200002) X3 Ao= 0.2261 m2 - 1 - 1 I t Thi A ho ho UA A I + hox0.2261 1529,8% hix 0.1884 1 I hixo. 1894 15 29.8% Note hox0.2261 Deer, students put the value of hi, which you have already find out in part (b) & get the value of ho (of outer surface coefficient). -
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