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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} Steadystate 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‘sni, Ir.
= 3.066? “IrmaIi, 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 kgfsm
SinceﬂieﬂuwisﬁﬂlydeuelepedmduubiﬂennuseﬂieﬂimEBueltﬁ'cmrdaﬁon,
Nun = 9.1123 11.635 Pro3 = 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, = MATL) 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 mtgK, k = 186 meK; ”5" M , Air (Tam: 362 K): v = 2.221 x 10'5 m‘ﬂ's, l: = 0.03069 WinnK,
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 Jtf 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]m3x9251fkgK 22545 M
tf:—[ _1 211545 3M 80—25
1I4K1M 2.2496W/m2K 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: 683 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 temperaturetime 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/ﬂ12K andradiativeexchangebecomm animportant [access PROBLEM 9.69
MOW: Thinwalled 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 WImK 5.0
T3: ASSUI'tIPTIONS: (1} Steadystate 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 meK, 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 115.151x1z1“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/m2Kf45—15)K=2o?w/m2K < Zovmﬁspwr puﬁ'lurlov: Therequiredheatﬂuxoons'l . gr 1 iDIllS
=(6.90+6_ﬂl)W/m2K(45—15}K=38TW/m2K < (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 [III2 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 WaltonK and from 3.3 to 3.3 WinfK, respectively. ...
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 Summer '08
 Rothamer
 Heat Transfer

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