finalexamextraproblems

finalexamextraproblems - PROBLEM 12.122 KNOWN Shed...

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Unformatted text preview: PROBLEM 12.122 KNOWN: Shed roofol‘ weathered galvanized sheet metal exposed to solar insolation on a cool. clear spring day with ambient air at — IUDC and convection coefficient estimated by the empirical _ __. 7s 3 _ . . correlation h '— 10 ATM“ (W/in ‘K with temperature units of kelvins}. FIND: Temperature of the roof. '11.. (a) assuming the backside is well insulated. and (b) assuming the backside is exposed to ambient air with the same convection coefficient relation and experiences radiation exchange with the ground. also at the ambient air temperature. Comment on whether the roof will be a comfortable place for the neighborhood cat to snooze for these conditions. SCH ENIA'I‘IC: _ x .u _V_. u: . Ambient \— 5 (38 .L Tsky 5“ @333 T 8 0t (33 = 600 win2 4 5' ' S r00 = 10°C h : 1.0 area T5, sheet metal 6;}? T _ _ z _ e = 0.65 grd : m (a) _II -’_,z/-,z/'./{I - 01,8 — g _ I. _ w n my ‘ luv-"K ASSUMPTIONS: (l) Steady—state conditions. (2) The roof surface is diffuse. spectra“): selective. (1“ _} Sheet metal is thin with negligible thermal resistance. and (3) Roof is a small object compared to the large isothermal surroundings represented by the sky and the ground. ANALYSIS: (a) For the bac kside-insulated condition. the energy balance. represented schematically below. is Fl! Eliot : 0 41n— ask}. Eb(TSk}..)+ [ISGS — qg... — e Ehfl‘s) = 0 )4/3 __ MT: : 0 ask.,.arjkv + @568 — 1.0(Tb — Tm, With 0551“. = 5-? [see Comment 2} and 0' = 5.67 X Ill—8 ‘Wa’m2 -K4. [ind TR. 065 0(233 K)4 wran +0.8x600 meg 1.0(T5 —283 K)“3 w x m2 —[).65 or? : 0 T323125 K:39.5°c < fiEblTs) uskyEblTeky) OtsGs aSGS “°"\ \ z 5 M \ z (a) (b) Energybalances: backside condition- Cllw / P/ \ (a) insulated, {b} exposed to airlground sEb(Ts) ugrdEbUgrd) PROBLEM 12.122 (COHL) [ht With the backside exposed to convection with the ambient air and radiation exchange with the ground. the energy balance. represented schematically above. is E e timeb {Tskyt + r'tErdEthgrd + [JESUS — zqgv — 2.9 Eb[T5) = t) Substituting numerical values. recognizing that Tgrd 2'1)». and tall-Li : 5 (see ("ornment 2). l‘ind TS. _ ._ . 4 '3 _ _ ._ q 4 7' . , '3 0.65 0(233 K) W f m" +0.63 01283 K) Witn.‘ +0.8X000 W l in“ 413 l —-3x to ('1‘S —-283 K Win? -- 2x065 at}? r 0 TS : 299.5 K ; 26.5”(7 - < COMMENTS: (I) For the insulated—backside condition. the cut would find the roof quite warm remembering that 43°C represents a safe-to-toneh temperature. For the exposed—backside condition. the cat would [ind the roof comfortable. certainly compared to an area not exposed to the solar insolution (that is. exposed only to the ambient air through convection}. (2) For this spectrally selective surface, the absorptivin for the sky irradiation is equal to the emissivity. (151‘)- : E. since the sky irradiation and surface emission have the same approximate spectral regions. The same reasoning applies for the absomtivity of the ground irradiation. otgrd = e. PROBLEM 12.124 KNOWN: Opaque, speetrally—seleetive horizontal plate with electrical heater on backside is exposed to convection. solar irradiation and sky in‘adiation. FIND: Electrical power required to maintain plate at 60°C, SCHEMATIC: T =—40°C —[> Sky 1.0 —{> E=ZO°C _i> 600 W/mz pa h=10W 2-K Pia“ ' I Insufafion I ASSUMPTIONS: (1) Plate is opaque. diffuse and unifonn. (2) No heat lost out the backside of heater. ANALYSIS: From an energy balance on the plate—heater system. per unit area basis. 1/! '47 ltm -- l: = 0 out L131“; + as (is + 05G 5k). “glib (Ts )_ {lgonv : 0 where (3);“, = 0' d“. Fb 2 GT: ,andqzmw : hi'l's —T,,_.). The solar tibsomriviri' is ( as : [512,1 o Asa). fro Asa). = [:0 Ali “(2.. 5800 rod). 0 where (it‘s -r EU, (in. 5800 K). Noting that on = l — p,_. 0c) E;h‘1,(2.58001<)aa org - (1-0.2)1‘ttt 2pm)+(1 U'Uili Ila—31ml] when: at Jig-r — 2 tun X 5800 K —- [1,600 urn-K, find from Table 12_E__ Fm“: : 0941‘ (JCS 20.80X0941—h0'30_0_94]) : 0,771- The mmf. itemispfiei‘i'rsal eriiissii‘im‘ is . : _ j . _ . ' a (I (L) 1(072‘um) + (I 0.7)[1 rmwzflm) . At El" ‘— 2 pm X 333 ' ’- 666 K. tind [flunk-H r- 0.000; hence E — 0.30. The total. ht-’titi'.\‘pfierieuf t.a’h.\‘t)!‘f}t'tl'ti’_i for sky irradiation in; (1 E i 0.30 since the surface is gray for this‘ emission and irradiation process Substituling numerical values. qcicc : SGT: + 11 (T5 —Tml—(.XS GS - rrrrTji Y qfijlm. :(iatixniin nil +10 w s m2 -K{(i()—3(i)°C—(}.??1><(a()t) w 3m? —t.i.30xo-(233 K)‘ a 3 F) "1 tine. - 200.2 w m2 + 400.0 w s in“- 462.6 w m“ —50. l w. m -' . 00.5. w m3. < COMM ENTS: [l)l\'ote carefully why not " E for the sky irradiation. ..‘ l’Kl.)BLEM 12.125 KNOWN; (__'hor_t'l length and s ectral emissivity of wing. Ambient air temperature. sky temperature and solar iiTadiation for ground an in—flight conditions. Flight speed. FIND: Temperature oftop surface of Wing for (a) ground and (b) iii—flight conditions. SCHEMA'I‘IC: T5,“, = 270 or 235 K [Part a or D] " G = 800 1100 wrm2 Gsky Cleonv E S [Paafa or b} Wing, Ts (“Si “skip Tm = 40°C [PM D] um: 200 mis H— ' LC = 4 ri'i . ASSUMPTIONS: (l) Steady-state. t2) Negligible heat transfer from back of wing surface. (3) Diffuse surface behavior. (4) Negligible solar radiation for xi. > 3 turn (as : (1,15 3 pm 2 5,1 :1 3 pm 2 0.6). (5) Negligible sky radiation and surface emission for P. i 3 tan (ash. : onL > 3 pm : E»; > 3 jam 2 0.3 : s). to) Quiescent air for ground condition, (7) Air foil may be approximated as a flat plate. [8) Negligible viscous heating in boundary layer for inrfligbt condition. (0) The wing span W is much larger than the chord length L... i 10] Iii—flight transition Reynolds number is 5 x [0”. PROPERTIES: Part in). Table iii—4. air ['l'f 325 K): v = 1.84 x 10‘5 Fitz/S. 0c : 2.02 x 10'5 rn:/s. l; : 0.0202 VV!IT1'K.{JJ:U.(:]UBU7. Part (0). Given: p = 0.470 kgfrng. u = 1.50 x 10 5 N-slnil. k = 0.021 Wliu-K. Pr = 0.72. ANALYSIS: For both ground and iii—flight conditions. a surface energy balance yields 015k), ow. +03 (35 : .9ch: +E( i‘S —-'i;..,) (1) vi here (ISM. : E l 0.3 and (25 20.6. (a) For the ground condition. h may be evaluated from Eq. 9.30 or 9.31. where L = AJP :11: X Wl2 ELL. . . . . . 3 . .. - 2 . + W} : [cc/2 : 2m and RaL : gfi ( is — 13.) L Iva. Using the {HI software to solve in}. t1) and accounting for the effect of teinperature—depcndent properties. the surface temperature is is T 350.6 K : 77.6"C < _ 10 — 2 . . where RaL 2.32 X 10 and h : 6.2 W/ni -K. Heat transfer from the surface by cmtsston and convection is 257.0 and 3130 Winf. respectively. (b) For the iii—flight condition. ReL : pun-11m : 0.470 kgmi" x 200 inJ’s X Eint/1.50 X 10 3 N-si’rn; : 2.51 x ml. For mixed. laininarfturbulent boundary layer conditions (Section 7.2.3 of text} and a transition Reynolds number of ReM = 5 x 10". Nul . (0.0.irRefi’5— any?” 20s00 — L 0.03l W I m - KX 36.800 7 7 h :—Nu[_ : -—— =14i Wim- -l\ 1. 4m Stlelillllll’lg into liq. (l). a trial—and—error solution yields Ic : In? K — --3fi..i°(‘ < 3 Heat transfer from [lie surface by emission and L‘tlI‘HL'L‘Ilt'ill is now 54.3 and 057.6 Vii-fin". respectively. (TOMENTS: The temperature of the wing is stroneg iiilluenced bythe convection heat transfer _ coetliciciit. and the large COEiflClCIil associated with flight yields a surface temperature that is within D”(‘ of the air teiiipet'attu'e. PROBLEM 13.13 KNOWN: Parallel plates of infinite extent (1,?) having aligned opposite edges. FIND: View factor Fl: by using (at appropriate view factor relations and results for opposing parallel plates and (b) Hottel‘s string method described in Problem 13.12 SCHEMATIC: ASSUMPTIONS: [1) Parallel planes of infinite extent normal to page and (’2) Diffuse surfaces with uniform radiosity. ANALYSIS: From symmetry consideration [Fig : H14) and Eq. 13.5, it follows that F13 : (1/2)[Fl(2,3.4)‘1313] where A3 and A4 have been defined for convenience in the analysis. Each of these View factors can be evaluated by the first relation of Table 13.1 for parallel plates with midlines connected perpendicularly. Ill 1!? 113- H2 [(w, +w3 )1 +4] [(W: -‘ w1 )2 T4] [(2%)2 +4] {(2—33 +4] 1t1 = — 10.618 ' 2w] 2x2 Flam Wl=vtlfL—2 W(234):3w2f1:o 1/2 , 1:2 [(2%) +4] —[(()—2)‘+4] 11(114) — 2X2 Hence. find F12 : (If2)[0.944—0.618]=0.lo3. < (b) Using Hottel’s string method. a WFZ"? b n3 :(lllwl)[[ac+bd)e(ad +bc)l E q 1/: ac:(l+4“) =4.123 c “222,” d bd :l q .1/2 ad=(r+2-’) =2.23() he : ad 2 2.236 and substituting numerical values find PB 2 (t!2><’2)[(4.123+l)—(2.236+2.230)] = 0.163. < COMMENTS: Remember that Ilottel's string methodis applicable only to surfaces that are of infinite extenl in one direction and have unobstructed VICW'S of one another: PROBLEM 13.22 Kl‘i(’)\’\’.\l: Tubular heater radiates like hlaekhody at 1000 K. FIND: (11} Radiant power from the heater surface. As. intercepted by a disr. A]. at a prescribed loealion qq at: irradiation on the dial. (it: and (0) Compute and plot qs_,t and G] as a function of the eeparation distance L. for the range 0 S 1.1 E 200 mm for disk diameters D1 : 25. and 50 and 100mm. SCIIEMATIC: :i; 1 e e \ \ ‘. Ci :1 :i a :2 :s :s :s 6.— Insulation _ A4 ZEisk, 131 = 50 mm, A1 . - \ ' x ' s ‘ ~ - e - a - e ' e ii |-1—L =100mm—>{4—L =100mm 2 1 Heater surface T5 = 1000 K. A2 02 = 100 mm ASSUMPTIONS: (1} Heater surface behaves as blackbody with uniform temperature. ANALYSIS: (a) The radiant power leaving the inner surface of the tubular heater that is intercepted hr the disk i9 Qiral:(A2Eb2)FEl (l) where the heater is surface 2 and the dink is eurlaee 1. It follows from the reciprocity rule. Er]. [3.3. that Al F21 2 F122- (3) A“) Define now the hypothetical disks. A3 and A4. located at the ends of the tubular heater. By inspection. it follows that Fi4=F12+Fit 0r Fl2:fl4“F13 (3) where FM and F13, may be determined from Fig. 13.5. Substituting; numerical values, with D3 = D4 : I)“. _ 1. L1+14 200 r] 0312 10012 F1 1) 2 0.08 Wlth — :- — : : i : e- ' : —~— : ..._ ' x.- 13112 5012 L 1.] + L2 200 1. L 100 r- 1), 12 10012 PM = 0.20 with — 7 1 = ———— ~ 4 A .—. J— : --— = .- r. 7—0112 5012— L 1.I 100 Substituting liq. (3) into liq. t2) and then into Eq. (1). the result is Q24 : A1014 ’fiJlEh'z j . q: ,1¥[H(50X103) 012/4 1‘ (0.20 —0.08)><5.67 ><10‘8 w t m“ K4 (1000 K)“1 :13.4w < where 11h: : UTH . The irradiation (1] originating irom emisaion leaving the heater surface [5 I34 “1" _ . = r—q -— (1825 W / 1112. (4') < Al 740.050 in )‘~ 14 GI Tfirs—)1 (.‘ontinued PROBLEM 13.22 (Cont) lb) Using the foregoing equations in IH'I'alung with the Radiation Tani, View Hit-{ms fnr Coaxial Parallel Disks. GI and (159i were computed as a function of | .1 for selected values 0po The results are plotted below. so .. 400 “D g 300 RT P. n QJ 30 g . _— 200 a: i“ 20 5 E E‘ E 100 10 E n: n U . D 50 100 150 200 D 50 100 150 200 Separation tistance. L1 [mm] Separal'ron dis1ance. L1 (mm) +Dl-25mm +Di=25mm --——~DI=EEDmm thzsvam +m=1mmm - +D1_-100mrn In the upper left-hand plot. GI decreases with increasing separation distance. Fora given separation distance. the irradiation decreases with increasing diameter. With values of D] = 25 and 50 mm. the irradiation values are only slightly different, Which diminishes as l4 increases. In the upper right hand plet. the radiant power from the heater surface reaching the disk. qr"); decreases with increasing [.1 and decreasing D;. Note that while GI is nearly the same for D] : 25 and 50 min, their respective (1543 values are quite different. Why is this so? PROBLEM 13.27 KNOWN: Dimensions and temperature of a rectangular fin array radiating to deep space. FIND: Expression for rate ofradiation transfer per unit length from a unit section of the array. SCHEMA'I'IC: Deep space f A:3 = W- T3 : 0 K! 83 : 1 IXA‘Z:2L ASSUMPTIONS: (1') Surfaces may be approximated as blackbodies. t2) Surfaces are isothermal. (3) Length of aiTay (normal to page) is much larger than W and L. ANALYSIS: Deep space may be represented by the hypothetical surface A3. which acts as a blackbody at absolute zero temperature. The net rate of radiation heat transfer to this surface is therefore equivalent to the rate of heat rejection by a unit section of the array. q} = A] 1:13 0(1‘14 7- T.f]+ A’2 F23 0(11'? \Vitl] Air = A3; Fng 2 Ai Fl}. T1: T2 =T and [1:1’70, quai (53+r13)ar4:wg'r4 < Radiation from a unit section of the array corresponds to emission from the base. Hence. if blackbody behavior can. indeed, be maintained. the fins do nothing to enhance heat rejection. ('(..)l\'IMENTS: ti} The foregoing result should come as no surprise since the surfaces of the unit section form an isothermal blackbody cavity for which emission is proportional to the area of the opening. (2) Because surfaces. I and 2 have the same temperature. the problem could be treated as a two-surface enclosure consisting of the combined ( l. 2) and 3. It follows that q} L (1’03 )3 2 AU?) 11(13):; all"; I A3173“ 2)UT4 —— \VGT4. (3) [f blackbody behaviorcannot be achieved [t—‘l . 6‘3 < I). enhancement would be afforded by the fins. ...
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finalexamextraproblems - PROBLEM 12.122 KNOWN Shed...

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