hw8 SOL - -- a-uvuuuLVJ. - II V" ‘7 U 301W:...

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Unformatted text preview: -- a-uvuuuLVJ. - II V" ‘7 U 301W: (CV15 M5114: Far/f 2009 KNOWN: Temperature 3 moving through a. tube and mass flow rate of various liquid of prescribed diameter. Mercury, wafer; 0:- engine oil ‘ 7 11': = 0. 03kg/s Tm :27; a" ASSUMPTIONS: Constant properties. PROPERTIES: (T = 300K) «De 7r(0.025m),u —— p: “UL RED = 4d: = 4 x 0.03kg/s '_ 1.53kg/s'111 ' 0. Q) - .12, 1 m JC Hence, even for water (,0 = 0.855x10'3 N's/m2), Rep < 2300 and the flow is laminar. (7‘!j % From qu. 8.3 and 8.23 it follows that ' .( A —.3 . p. ‘ —3 Hence: Liquid “m0” /-9} 31rd,“ 7”) mtfm) <} Oil 0.069 0.0030 25.2 Mercury 0.0045 1.257 0.031 Water 0.061 2.234 13.02 COMNTS: Note the efi‘e ' effect of Pr on the thermal entry length. PROBLEM neg KNOWN: Flow rate and inlet temperature of engine oil in a tube of prescribed length, diameter. and surface temperature. Total heat transfer and oil outlet temperature with and without the assumption of fully developed flow. SCHEMATIC: I vi: = 05k 9/5 Tmi=25°C/V K,’ i}- "1 '§;%ai.=5aétat£;;;ais }-'-.'Er . ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic energy, potential energy and flow work changes, (3) Constant properties. @301: $152.! £0135 PROPERTIES: Table 21.5, Engine oil, (Tm = 340K): p=_860‘kg/m3, (razors J/kg-K, u: samtc'z’Eg/s-m, k: ouaewlm-K, Pr=393T 575‘ 5:59 5.72. {at (1751. 13 ANALYSIS: From Eqs. 3:43!» and 89?, “5-21.96; .1} rating-at; .sa‘. '( 1'i}E21‘5p’1‘trtE-{11firewaiw'giétiiigétélifi film" an _ ._ _ 41:21:,— = a _ a (u 1:0.025mx10m a) “rm—T, m Tm,i)exp( mp 11) 100 C 75 Cexp ___H~.~0.5kglsx2076mg_l{ h P D ‘ — :2 r Tm = 100°C - 75°C exp(— momma) L4: 0 105—qu _ q=n'lc (T - )=05kg/s>c2g(a}'ifi?fkg-K(T 45°C) I ‘5 filigm r 1: mm 111.! - _ mun . , Una With Rep == 4 Iii/qu = 4(O.5 kg/s)/rt(0.025 m)0.0531 kg/s-m = Wthe flow is laminar. flow Considering the thermal entry (re) region, it follows from Eq. fithat 5% [3mm] cw [awn-2.61m] =mwm. 1+0.04[(nn.)a= mm 0.025 m 14034940)” ' D a; i I a Tmm, = 100°C — 75°Cexp(-0.000757x92.5) = 100°C — 693°C =3§19C Q .3 “- ii ‘5'! 51 Q “'5’” t qt“) =1033wm<30.1-25)°c=529w‘t Q If fully developed (fd) conditions are assumed for the entire tube, — 1: 0.139 W/m-K 2 a _. _ = _......— . = 20. - h D366 0.025!“ 366 3W/m K med, = 100°C — 739°C = 25.1°C Q qm = 1038W/K(26.1—25)°C = 1190w . Q CONIMENTS: The assumption of fully developed fonditions throughout the tube leads to a large error in the calculation of h and hence q. Note that Liairm = 0.05 D ReD Pr = 0.05(0.025 m)480(793) = 476 m, which is much_1arger than the tube length. The calculations should be repeated with properties evaluated atTm = 300K. PROBLEM 1:63 KNOWN: Tubing with glycerin Welded to transformer lateral surface to remove dissipated power. Maximum allowable temperature rise of coolant is 6°C. FIND: (a) Required coolant rate m, tube length L and lateral spacing S between turns, and (b) Effect of flowrate on outlet temperature and maximum power. SCHEMATICL ' . Transformer. 1000 W (:alycerint T J: 24 0c, . T T Ts = 47 °C ‘ H = 500 mm '5 i l D = 20 mm k——>l—D =aoo mm ASSUMPTIONS: (l) Steady-state conditions, (2) All heat dissipated by transformer transferred to glycerin, (3) Fully-developed flow (part a), (4) Negligible kinetic and potential energy changes, (5) Negligible tube wall thermal resistance. PROPERTIES: Table A5. Glycerin (Tm = 300 K): p = 1259.9 kgrmi. cp = 2427 Mtg-K, a = 79.9 x 10'1 N-s/ml. k = 286 x 10'3 W/rn-K. Pr = 6730. ANALYSIS: (a) From an overall energy balance assuming the maximum temperature rise of the glycerin coolant is 6°C, find the flow rate as .q=ch(Tm'n— mJ) rh=q/cp(Tm‘n—- m)=1000W/2427I/kg-K(6K)=6.87x10'2kg/s < From Eqrflsfi', the length of tubing can be determined, fly- : eitp(—PLh/r'ncp) Ts "Tami where P = TED. For the tube flow, find RED_ 4:11 = 4x6.87x10 'kg/s =54? ' itDu it x 0.020m x799 x 10'2 N - s/m1 which implies laminar flow, and if fully developed, _ -3 Nunzflgzififi E=3.66x286><10 W/m Kzsmw/mLK k 0.020m ' %‘%:_¥% = exp[—('J't(0.020m) x 52.3W/m3 - K xL)/(6.87 x10"2 kg/s x 2427J/kg - K)] L = 15.3 m. < The number of turns of the tubing. N, is N # L/(er) = (15.3 m)!rt(0.3 in) = 16.2 and hence the spacing S will be s =HIN=500 min/16.2=30.8 mm. ‘ < continued... PROBLEM 8.26 (Cont) - 053. Ema i, 1355?: W“ Cad, ¢_7LC__ (b) Parametric calculations Were‘performed using the-H-HLGarrehrrizfiooipad—based-oa—Eq: 8 26 (a then‘nal entry length condition), and the following results were obtained. 35 - E y ' u g :5, - a1 S E E E E 29 fl — =- . E 7 Z------- “II-I... gt a, 25 . . . 0.05 0.09 0.13 0.17 0.21 0.25 0.05 . ' . 0.1T . 0.25 Mass llowrata. mdaltkgls) Mass flowrata. mdollkgis) With T, maintained at 47°C, the maximum allowable transforrne power (heat rate) and glycerin outlet temperature increase and decrease. respectchly, with increasin m. The increase in q is due to an increase in 'fifin (and hence h) with increasing Reg. The va a of NED increased from 5.3 to 9.4 with increasing In from 0.05 to 0.25 kg/s. CONflVIENTS: Since G2},l = (L/DVReD Pr = (15.3 .02 m)l(5.47 x 6780) = 0.0206 < 0.05. entrance length effects are significant, and Eq. 8.56 should be sed. to determine my. Nw aim 1/040WWQR60’ “43.04 EDI!) Edyafi PROBLEM KNOWN: Flow rate, inlet temperature and desired outlet temperature of water passing through a tube of prescribed diameter'and surface temperature. FIND: (a) Required tube length. L, for prescribed conditions, (b) Required length using tube diameters ovor the range 30 S D s 50 mm with flow rates In = 1. 2 and 3 kg/s; represcnt this design information graphically. and (c) Pressure gradient as a function of tube diameter for the three flow rates assuming the ' tube wall is smooth. SCHEMATIC: m = 2 kg/s /. Tm'i = 00/7 . ASSUMPTIONS: (l) Steady-state conditions, (2) Negligible potential energy, kinetic energy and flow ' work changes, (3) Constant preperties. PROPERTIES: Table A6, Water (Tm = 323 K): on = 4181 J/kg-K. it = 547 x 10'6 N-s/mz, 1: = 0.643 W/m-K. P: = 3.56. ANALYSIS: (a) From Eq. 8.6. the Reynolds number is R6D=iri=_‘l§3k_gii___=msx105. ' (1) Hence the flow is turbulent, and assuming fully developed conditions throughout the tube, it follows from the Dittus-Boelter correlation, Eq‘ 8.60. i - k 0.643W/m- K ___ 4,5 M_ 5 415 0.4” 2_ 11.130.023RsD Pr .. 004m 0.023(1.16x10 ) (3.56) _6919w/m K (2) From Eq. 3.4221, we then obtain .. —' 2 AT AT 2k‘s41811k-Kfn25T75‘C - L: mcp n( _D/ 0:“ g/( /E ) ( q / )=10I6m' < non 1t(0.04m)6919W/m*- K Tube 15 glh. L (m) equations, the required length - . flow rate is computed and pi Tuba diameter. D (mm) —8— Flow rate. rndot=1 kgls -— mdot = 2 kgls —¢— mdct = 3 kgfs Continued... ...
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This note was uploaded on 09/08/2010 for the course ME 114 at San Jose State University .

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hw8 SOL - -- a-uvuuuLVJ. - II V&amp;quot; ‘7 U 301W:...

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