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Problem 8_23 - PROBLEM 8.23 KNOWN Ethylene glycol flowing...

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Unformatted text preview: PROBLEM 8.23 KNOWN: Ethylene glycol flowing through a coiied, t weilustirred water bath maintained at a constant ternper 8.23 Ethylene giyco 1 flows at 0 0i kg/s through a 3-mm diameter. thin-walled tube, The tube is coiicd and submerged in a wail-stirred water bath maintained at 25°C, if the fluid enters the tube at 85°C. what heat ante and lube iength are required for the fluid to leave a: 350?? Neglect heat Hamster enhance,“m associated With the coiling FIND: Heat rate and required tube length for prescribed conditions C C: . S HEMATI $m=001kgls T 7;”.1” =850C Thin—walled 1tube, D=5mm Weiiws‘i‘irred wafer bath, L=25°c ASSUMPTIONS: (1.) Steady-state conditions, (2) Tube wali thermal resistance negligible, (3) Convection coefficient on water side infinite; cooling process approximates l "Jam-15) constant wall surface temperature distribution, (4) KB, 93 and flow work changes negligibie, (5) Constant properties, (6) Negiigible heat transfer enhancement associated with the coiling“ M PROPERTIES: Table A~5, Ethylene glycol (Tm =(85+35) ° 0/2 a 60 ° C = 333K): c? 22562 J/kg»K., n=o.522x10—2 N“s/m2, k =0,26{} W/rn'K, Pr msle. ANALYSIS: From an overali energy balance on the tube, qmm, = filcpiTm,o“Tm,i) m 0.01kg/S x 2562J/kg(35—85) ° C m —1281 W , (1) 4 For the constant surface temperature condition, from the rate equation, A: : Cicenv [E ATfm AT are“, mm'romanwn ' ° = [(35—25)°o ~(85—25)“C]/€n AT, Find the Reynoids number to determine flow conditions, 4th 4 x 0.01 kg/s (2) 35”“25 227.9"0 ,. (3) 85~25 (4) Hence, the flow is laminar and, assuming the flow is fully developed, the appropriate correlation is min i: w Nun = .——— 2 366, ii = Nu—— 2 355 x transom/0.003171 = 317 “1/le , (5) k D IIVK From Eq. (2), the required area, As, and tube length, L, are A, =1281 W/317W/m2«K. x 273 ° C e 0.1443 m2 L = A,/7rD = 0.1448m2/fi'(0.003m) 2 15.4m . <1 COMMENTS: Note that for fuily developed laminar flow conditions, the requirement is satisfied: Ga—1 : (L/D)/R€g Pr = (15.3/0,003)/(813 x 51.3) m 0122 > 0,05 . Note also the sign of the heat rate gem... when using Eqs, (1) and (2) ...
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