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Unformatted text preview: PROBLEM 13.24 KND‘WN: Circular plate (A1] maintained at 600 K positioned evtiaaiall},r with a conical shape (A2)
whosebaekside is insulated. Plate andcone are blacksurfaces and locatedin large, insulated
enclesme at 300 K FIND: (a) Temperatm'e efthe conical surﬁaoe T2 and (b) Electric powerrequiredmmaintain plate at
600 K. SCHELIATIC: ASSMTIDNS: (l) Steadystate conditions, [2) Plate and cone are black, (3) Cone behaves as
insulated, Ieradiating surface, (4) Smoundings are large Dmnpared to plat]: and cone. ANALYSIS: (:1) Recognizing that the plate, Done, and surrelmdings Train a three{hlack] surface
enclasure, perform a radiation balance on the cone. I12 2021123 +q21 =A2E30(Ti'T§)+A2F210[TiTﬁ) where the View factor F31 canbe determined frurn the “ﬂimhmms ““5 relation (Table 13.2 or
Fig. 13.5)mmm2rm5m500 = 0.5,01 = 0.5, s =1+(1+0.f]n1;1 =1+(1 + 0.55:0.522000, aﬂdmﬁﬂg FT121321=
2 2 f2 2 U2
51:05 s—[s —4(rj.l'ri) J1 =0.5:l§—[6 —4[ﬂ_5!0_5)2] }=0_1?2_ For the enclesure, the summation rule presides, E3 = 1 ‘FZ'I = 1 411?? = 0.323. Hence,
0.023(T51 — 3004 ) = 0 +0.1?2(T§' — 6004) T2 =413 K. <1
(b) The pnwer required 11: maintain the plate at T2 follews than a radiation balance,
4 4 4
'11 =I3112+lll3 =A11izU(T1 —T2 )+A1}‘130(T14—T3 )
WhCIC 1:12 =A2’F2’1 .I'rﬂl =51 20.1?2 and 1713 21—1312 20.323, ql =(.1r0_52 £4)m20 [0.102(000“ — 4134 )K“ + 0.020(0004 — 3.00“ )K4] ql =1312 w. <1 PROBLEM 13.53
MOW: Emissivities, diameters and temperatures of ooneenlric sphera. FIND: {a} Radiation transfer rate for black surfaces. (11] Radiation transfer rate for diffusegray
surfaces, (:2) Effeeﬁ of increasing the diameter and assuming blackhody hehmrior for the outer sphere.
(d) Ef'iact ofemissivities on net radiation exchange. scmﬂﬂﬁ. _
F12" $=12mnr2€lm
$1 '25:? 2  300 r:
=110.1.05,1.0 0054.22.14: J1 J2
El“ W 5m
”was! 1:11. 1 I  t:
31141 “11:12 .5242 ASSUMPTIONS: (l)Bla1:'.ld:mdj.r or difﬁlse—gray surface behavior.
ANALYSIS: (:1) Assuming blackbody behavior, it follows from Eq. 13.13 ‘112 = AlFucr (T12 —T3) =a'(0_3 n1)2 (1)5.67x10'3w1m2 ~K4[(4ﬂ{l K)4 —(300 K)‘1 ] = 1905 w. <1
(h) For diffusegray Sinface behavior, it follows from Eq. 13.26 0:11 [T14 43*) 5.6?xlﬂ_sWIm2 I~:‘*.1r(0_sm)2 [4004 —300“]1~:4
‘112 = = =191w_ <2 L+1€2 r_1 n+1—005 E 2
£1 £2 {2 0.5 0.05 0.6 (c) Willi D2 = 21] m, it ﬁollows from Eq. 132:5 5.ﬁ?x10_8me2K1r[0.8m)2[[401] K)‘1 —(300 KT‘] 1 +1—0.05 0.4 2
0.5 0.05 10 912: =933w. <2 With 22 = 1, instead ofllﬂi Eq. 13.26 reduces to Eq. 13.2? and £112 =0A191 (T14 424) = ammo—3101012 K41r(0_8m)2 0.5 [(400 K)4 —(300 K11]: 9923 W. <3 PROBLEM 13.53 (Cent) (:1) Using the ‘HT ”“1““ Tool Pad, the following results were obtained Minn mm. mm +
+3:
+m Net radiatinn exchange increases with 21 and £2, and the trends are due to increases in emission from and
absurptinn by surfaces 1 and 2, respectively. COLEEMS: Frnm part (c) it is evidentthat the actual surface emissivity efa 3"” 5 enclosure has a
small effect nnradiatien exchange with small surfaca inthe enclosure. Waking with 22 = 1.0 instead (if
92 = 0.05, the value of I112 is increased by 42ml},r (998 — 983F933 = 1.5%. In centrast, fmm the resulm of (d) it is e'eident that the surface emissivity £2 ofa a“ “33‘ enclosure has a large eﬂ'ecten radiation
exchange with interior objects, which increases with increasing £1. PRUBLEI’tI 13.63
KNOWN: Cylindliealshaped, three surface encloslne with lateral surface insulated. mu: Temperatlm efdle lewer plate T1 and insulated side surface T3.
SCHEL'IATIC: L! 0.2 m L\\\‘l\l‘l\\\\l\“l ASSUMPTIONS: (1) Stufaces have uniform radidsigr or emissive pm, [2) Upper and insulated
surfaea are diﬂiJsegray, [3) Negligible eonvectinn. ANALYSIS: Find the temperamre afﬁne 1mm: plate T1 from Eq. 13.31]
4 4
0 (T1 —T2 )
[1) I11 = —1 .
—1
I{1—91)I“€12”L1+[AIFL2 +[113A1F13)+(IIA2F23)] ] +(1—92F92A2
FromTable 13.2forparallel coaxial disks,
R1=r1IL=0.1!0.2=0.5 R2 =IEIL=0.1I0.2=0.5 s=1+(1+R§]iR12 =1+(1+0.52)£0.52 20.0 F12 =1I2{s—[s2 4(12 I11)2]1ﬂ}=1r2l0—[62 —4(0.5!0.5)2]m]=0.1?2. Usingtbe summation rule forthe euclosme, F13=1—F12 = l — 0.1?2 = 0.323, and from symmety, F23 = F13. With A1 = A2 = JIDQM = 1110.2 m)2f4 = 0.03142 n12 and substituting nmnetical values into
Eq. (1), obtain 0031421112 x50?x10‘3trwm2 K"' [T14 —4?34)K4
10.000 w: 1 1
0+[0.1?2+[(1I0.323)+(1!0.1?2)]’] +(1—0.3)£0.3 10,000 =4.540x109[1'1" —4?34) T1 =1225 K. {I The tenpemture ofthe insulated side surface can be determined from the radiation balance, E]. 13.31,
withA1=A2,
J1—13 I3 42 _ “513 ”1:23
where 11: 01'14 and J; can be evaluated ﬁ'omEq. 13.151= 0 [2) PRUBLERI 13.63 (Cunt) 5.6?x10‘3WIm2 K“ (473 K)‘i 42
[l—ﬂ.8)f[ﬂ_3xﬂ.ﬂ3l42m2) 154:2 42 =— —lU,UUUW=
(1—92 )J'lngl '12 ﬁnd :2 = 32,495 Win12. Substitmzing mmical values into Eq. {2), 5.6?){10—anm2 K4 (1225 K)4 —13 13 42,405 mez _
1mm 1mm 0 ﬁnd 13 = 105,043 Win12. Hence, for this insulated, re—radiating (adiabatic) surface, E413 =ch§ =105,u43 Win12 T3 =11ﬁ7 K. ...
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 Summer '08
 Rothamer
 Heat Transfer

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