hw11_sol - PROBLEM 3.3’9 KNOW Temperaun'e and value of...

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Unformatted text preview: PROBLEM! 3.3’9 KNOW: Temperaun'e and value of flaw between allel ates of bribed surface temperamre and separatien "Iliickngs loeaiim of plalepianrsert. Pl PIE FIND: Heatfluxtoflieplales (a)udfl10utand{b)wiflithe insert SCHENIATIC: ASSUMPTIONS: {1} Steady-state Dundilions, (2) Negligible radiation, (3] Gas has properties of atmospheric air, (4) Plate: are of infinite width W, (5} Fully,r devekiped flew. . 3 —".-‘ PROPERTIES: Table A4, Airfl aim, Tm = 1000K): p = £1348 kgr’m , p = 424.4 x 10 kgI‘s-ni, Ir. = 3.066? “Irma-Ii, Pr = 0326. ANALYSIS: {a} Based upon the hydraulic diameter D11, the Reynalds number is uh =49c 19 =4{H.w)12(H +w)=2H =summ p um 11h _ 9.34s kgxm3 {69 mf5)fl.03 n1 9 _ 424.4x 111—3r kgfs-m SincefliefluwisfifllydeuelepedmduubiflennuseflieflimE-Bueltfi'cmrdafion, Nun = 9.1123 11.635 Pro-3 = 9.923 { 39.3%}“5 [9.926)”'3 = 99.1 h ziNuD zwwl 282.6 9.9612 -K Dh 13.08 in q" =11{T111 :1; ) =92.6 w1m2.K{1999—359)K =53,?99 9.9612. 6: (b]an1cen1in11ity, m=(p 111,191];i =[p umfijb umjb =umL1 [ijaf [ijb =69 6115(49120) =120m15. Foremhofiheraulfingchannels, Dh=0.02n1and p um [:11 _ 9.34s kgxm3 {129 1615) 9.92 1:11 ReDh = =39,360. Re 2 = 1 9,6911. Db 9 424.4x 164mg. :6 Since the flew is still mrl'sflent, 56.9 9.9669 1.91 -K Nun = 9.923(1 9,6811)“5 {fi.?26)fl'3 = 56.9 h = w = 189.8 9.91112 -K 9.92 111 q" =199s‘919m2 -K {1999—359)K =123,4119 1191612. <Z CORIMENTS: From the Dilms—Boelter equation, (12 (13 hhfh =(uni,b fume) [Dha’lnhb) Henne,heattrausferenhauemnentduemfl1enismispriruarilyaresultoftbeinnreaseinqnand seenndarilyarault eftbedeereasein D11. = [2)” [4)”-2 =1.?4 91.32 =2.39. PROBLEM 9.12 KNOWN: Temperature dependence of free convection coefiicient, h = CL‘LTIM, for a solid suddenly submerged in a quiacent fluid. FIND: [a] Expression for cooling time, tr, [b] Considering a plate ofprficribed geometry and thermal conditions, the time required to reach 30°C using the appropriate con'elation from Problem 9.10 and (c) Plot the temperature—time historyI obtained from part (b) and compare with resulm using a constant Ho fiomanappropriatecorrelationbased uponan average surface temperature T=(Ti+Tf)/2. SCHIl'rIA'IIC: m 2024 plane. '- uiescant _ 150 1150 mm. fluid, Ta tram, = MAT-L) 5 mm minimum 5 : snarl: J1“- TED} = 225 ”C 4 new: A. l THEFT! 1;, = 25 “6 ASSUMPTIONS: (l) Linnped capacitance approximation is valid, (2) Negligible radiation, [3] Constant properties. PROPERTIES: ““5 “-1 , Aluminum alloy 2024 [T = (Ti +Tf);'2 = 4110K]:p= 2?”!0 kgmfi, c1, = 925 mtg-K, k = 186 me-K; ”5" M , Air (Tam: 362 K): v = 2.221 x 10'5 m‘fl's, l: = 0.03069 Winn-K, 0:: 3.18? x10‘5m’rs,Pr= 0.0970, B = 1! Tfilm. ANALYSIS: (3) Apply an energy balance to a eontrol surface about the object, ‘Eout: 2 Est , and substitute the convection rate equation, with h = CATIM, to find 43A: [T _T.,,,)5M = did; [chT)_ (1) Separating variables and integrating, find and: = —(CA5fch)(T— T...)5M _ Tr Tf dT :_ CA5 J-tf dt —4(T—T..,) IM‘ :_CA5 tf Ti (T _ Too )Si‘l- pV—e 0 Ti ch 1'4 1f=Effie—1.0T“—(Ti—T..)—”4]= —4PV‘ 1:4 [ii—iml —1 . (2).: 5 CA5 (Ti —TW) 1" _ on (1]) Considering the almninumplate, initially at T(D)=225°C, and suddenlyexposedtoaniaientair atzTm3 = 25°C: from Problem 9.10 the convection coeflicienthas the form - M M - 1M hi =1.40 f h.- =CnT _ 1.-4_ 1x4_ 2 3"!4 - whereC—IAOJL —1.40l(0.15l]) —2.2496W m -K . U51ngEq.(2),find Continued... PROBLEM 9.12 (Cont) 4X2??Okg m3(0.1503xu_005]m3x9251fkg-K 22545 M tf:—[ _1 211545 3M 80—25 1I4K1M- 2.2496W/m2-K x2x(fl.150n1)2(225—25) (c) For the vertical plate, Eq. 9.2? is an appropriate correlation. Evaluating properties at: THE =(T5+Tw)/2=((Ti+Tf)/2+Tw)/2=362 K where TS 2426K , the average plate temperame, find gfifis —TW]L3 = QBmI/sz (H362KH426—298)K(fl_150m)3 7' RaL= =1_652x10 v0: 2.221x1CI—5mE/sx3_18?x1fl_5m2/s r 0670 1652 10? 1M _ 14 _ (_ x ) NuL =0.68+LL4!9 =U_68+ 4:9 :34 1+(0 49 9,!16 93’16 _ are] 1+(u_492jn_59?5) ED =5fiL = Wflfix—t: 6-83 tar/m? -K L {lljflm Frem Eq.5.l5,theternperamre—timehistorywithaconstantconvection eoeficientis T(t)=TW+[Ti—Tm)exp[—(hoAsfch)t] (3) where [ts/V: 2L3/(Lxwa} = 2/“: 2400.114. The temperature-time histories forthe h=car1"4 and E0 analyses are shown in plot below. Phhtinmm. Twp} — 011m mnmwsswrm‘ax + Va'labte construed, h =1254Ts -TIIT‘|]25 COMIMENTS: (1] The times toreachTfltn) = 80°C were 1154 311112125 forthe variable and constant coefficient analysis, respectively, a difi‘erenee of 5%. Fer convenience, it is reasonable to evaluate the convection eoeflicient as described in part (b). {2) Note that Ra; < 109 so indeed the exprasion selected from Problem 9.10 was the appropriate one. (3) Recognize that if the emissivity of the plate were unity, the average linearized radiation coeflicient using Eq. (1.9] is brad =ll_GW/fl12-K andradiativeexchangebecomm animportant [access PROBLEM 9.69 MOW: Thin-walled tube mounted herizaontalljr in quiescent air and wrapped with an electrical tape passing hot fluid in an experimental loop. FED: (a) Heat flux q; from the heating tape required to prevent heat loss fi'onlthe hot fluid when (a) neglecting and {11) including radiation exchange with the surroundings, (e) Effect of insulation on :1; and convectionfradiation rates. SCHILBJIA'I'IC : 1;,=150c Thin heater, q"? 3,, = I195 Insulation. Dé féflfl mm a, = 0.00. k,- = 0.050 WIm-K 5.0 T3: ASSUI't-IPTIONS: (1} Steady-state conditions, (2} Ambient air is quiescent and extensive, (3] Sun'oundings are large compared to the tube. PROPERTIES: ”“5 M, Air (Tr: {I,+ To, 1112: (45 +15]°C1'2 = 303 K, 1 mu): 1»! =16.19 x 111*1 111215, 11 = 22.9 x 104‘ m1f5,k = 26.5 x 111'3 me-K, Pr = 0.303, B = 111.}. ANALYSIS: (ah) To prevent heat losses from the hot fluid, the heating tape telnperamre must he maintained at Tm, hence Tgi = Tm. From a surface energy balance, 111:. =0Em+0iad =lEDi+hrllTai—TW) wherethelinearized radiation coefficient, Eq. 1.9,is hI =ED' (T5,i+Tm)(Tii+T;),or 11r = -:1.515x5.151rx111‘a 111/1112 4:4 (313+ 2331(3132 +2333}:3 = solar/m3 -K. “51’5“” ”“1“” Forthe 11332311131 cylinder, Eq. 934333135 3 35(Ts,i —Tm}Di = Siam/53 (113031;) (45—15)K((1.1:-2-:1m)3 R31:- 2 = 20.9110 WI 115.151x1-z1“5 m2 5X22.9X10_6 ml/s 2 _ h— _D. ngfihlffi Nun: D]: ‘= (mm—D 121 [1+ (0.559fpr)9”5]fi PROBLEM 9.69 (Cont) HIS — - 0.385 20 900 hm =W 060.4 flow/mag Q0213 ' J 2? m [1+ (0.5 Swami)“ ‘5]3 Hence, neglecting radiation, the required heat flux is q;=snow/m2-Kf45—15)K=2o?w/m2-K < Zovmfispwr pufi'lurlov: Therequiredheatfluxoons'l . gr 1- iDIll-S =(6.90+6_fll)W/m2-K(45—15}K=38TW/m2-K < (c) “.ch insulation, the surface energy balance must be modified to account for an increase in the outer diameter from I}1 to 1),: Di + 2t and for the attendant thermal resistance associated with conduction across the insulation. From an energy balance at the inner surface of the insulation, , 2mg (Tm —T5 0) *1; (’3 Di) : qoond : —' '11 (DofDi) and from an energy balance at the outer surface, ‘1’ch =Qibonv +‘1i'ad =EDo (EDIn +hr)(Ts,o _ Too) The foregoing expressions maybeusedtaodetermine T5,, and q; as a functionoft, 1with the [HT 1” ”FEM“ W and HMHWW Tool Pafi used to evaluate EDo . The desired results are plotted as follows. E *3 a E 'I 2 3 I] 111104 [“133 [“112 0315 [III-2 Insulal'lm Ii'lioknas. lint} I] 111134 [LUDH- D.D12 0.015 [LUZ . TDH + mm II'ELIEliCIn I'IlDlfl'lEE. tlm} —.— REIdiEIIlDf‘I By adding 21) mm ofinsulation, the required power dissipation is reduced by a factor ofapproxirnately 3. Convection and radiation heat rates at the outer sufioe are conmarable. COMltrIENTS: Over the range of insulation thickness, I'm decreases from 45°C to 20°C, while ED“, and III decrease from 6.9 to 3.5 Walton-K and from 3.3 to 3.3 Winf-K, respectively. ...
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