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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 ksideinsulated 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 tallLi : 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 safetotoneh 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 2K 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  (10.2)1‘ttt 2pm)+(1 U'Uili Ila—31ml] when: at Jigr — 2 tun X 5800 K — [1,600 urnK, 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 [ﬂunkH 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) Steadystate. 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 Nslnil. k = 0.021
WliuK. 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 : pun11m : 0.470 kgmi" x 200 inJ’s X Eint/1.50 X 10 3 Nsi’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 Viifin". 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)
+Dl25mm +Di=25mm
——~DI=EEDmm thzsvam
+m=1mmm  +D1_100mrn In the upper lefthand 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
twosurface 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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This note was uploaded on 08/08/2008 for the course ME 364 taught by Professor Rothamer during the Fall '08 term at University of Wisconsin.
 Fall '08
 Rothamer
 Heat Transfer

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