hw8_sol - PROBLEM 13.24 KND‘WN Circular plate(A1...

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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) Temp-eratm'e efthe conical surfiaoe T2 and (b) Electric powerrequiredmmaintain plate at 600 K. SCHELIATIC: ASSMTIDNS: (l) Steady-state 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[Ti-Tfi) 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]n-1;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 diffuse-gray 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: =11-0.1.05,1.0 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 Sin-face 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—0-05 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_8me2-K1r[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. COLE-EMS: 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'e-cten radiation exchange with interior objects, which increases with increasing £1. PRUBLEI’tI 13.63 KNOWN: Cylindlieal-shaped, 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) Stu-faces have uniform radidsigr or emissive pm, [2) Upper and insulated surfaea are difliJse-gray, [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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