exam 1 solution

# exam 1 solution - Name Spring Zﬂﬂﬂ ME 3304 Test 1...

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Unformatted text preview: Name Spring Zﬂﬂﬂ ME 3304: Test 1 Open beak and 1 review sheet allawed. Ne ether materials permitted. Instrtteter: Viekﬁ Nichole RyIander Chris Rylat‘tder Fitltat] Prtthlem 1 : 250% A long wire of radius rl = {1.25 em and thermal conductivity kw at 25 Wire—“C. is heated due to reaiatanee heating that generates heat unifurmly in the wire at a rate of r} = 2t} erm3_ The wire it; pntteetet] by a 1 13.25 cm thick layer at inaulatiun material whose thermal enttdueliviiy is 1-11 =1 WitnT. Under mead}.r ennditinne. the enter surface temperature a!" the insulation was measured to he T3 = 35°C. [a] Determine the temperature between the Interface at the wire and the insulation layer. T1: __ I—I‘I “a. “a. . I? _ .l'. .I" 1' 1' "': Isl _ -}‘I It. _ :9,- frt - hematite. « J-ié m I Tr _ Fri-I 1 .r_ ." I ( r? I‘lll “x - J:— {jq irﬂ .r- J v—‘-'-_-'_“-' —F -w"""i :- .ﬁ-I i [[1] Determine the temperature ef the center at the wire. Ta = [Wino/61' W/ Herr? Gauwm‘ﬂm I Name ‘ _ Spring 2008 ME 3304: Test 1 Open book and 1 review sheet allowed. No other materials permitted. Instructor: Vick_ Nichole Rylander Chris Rylander Ekkad Problem 1: 25% A long wire of radius r1 = 0.25 cm and thermal conductivity kw of 25 W/m—°C is heated due to resistance heating that generates heat uniformly in the wire at a rate of q = 20 W/cm3 . The wire is protected by a t = 0.25 cm thick layer of insulation material whose thermal conductivity is ki =1 W/m°C. Under steady conditions, the outer surface temperature of the insulation was measured to be T2 = 35°C. (a) Determine the temperature between the interface of the wire and the insulation layer. T1=j€'3‘2f€/ (b) Determine the temperature of the center of the wire. To = ‘ 157 C” 1 ’L Q C; To :: 7" ’“f‘ m 1.: C Q»! (whiz, H.322: N“ umﬁ. ‘ > moan Bag: mmmwam mm 8383 E8505. 3:599. a? H“; B“ UHoHBr wnwoo chEO Mm Boga ﬂow :5 \$38 loan. 155 “wage—=0 o». 90 can om 9o 9&qu mm HaHGQO. .55 9339. M5. HmBngm BE .5»: Hangman 80399: m8 HEHMoO BE EHH o “SEN? 1:6 8:53 RmeSboo 33:35 :5 mmmﬁaa E5 :5 30». mm W33. .25 mmmwoﬁ on: Go Bogoon 8 mb 5:38 mu. my BBQ 90 3g” Hemm Egan: mmmcoﬁ 35. mm,” 3 u 9% W 49% mi?” .‘x\ H Show 5 Wmmowxvﬂm. E 3:; :5 Havana—no M: 8m. om mmmuonm :83. Hﬁhiém .9: “WM. All H AUG 7 5. my % \$ 7 £5” am.\,\m um um W9. qmlxﬂmo “0 NW. . Am onb 8 35m :5 Bow HmbaBgm mace—mom—q gnaw. : 53¢ Av wwlw do a“ s u E W W+nl 3 5 can: 8 Raga :5 :9: 5mm \$.95 Ea moon 599.8 \$3099. 50 mozosmbm 633\$on £53m so 523me Eu 3 moowammma Ag». ﬁx in mahmn+ Azmv Pug owv cu|w w|v {HIM Problem 3: 20% Consider two—dimensional, steady heat transfer in the region shown. The top surface is exposed to convection while the left surface experiences a uniform heat ﬂux, q”s. Convection \ (Lagging a) Derive in symbolic form (no numbers) the finite difference equation for node 1 assdming a uniform grid, Ax = Ay. Show your derivation. a “km W 33% i , . . {2W « N gin g; 1 ,r We semi“: V I ’ A \ I, D m S Q r“ i; v; ‘ W sins ﬁg”. a ,4 W M m. R mm A mm in am Ti 2“ if? m, Mggfc‘gmmm we? r I ‘ Wme a «an r. ‘ inmum»imammvmwwmwmm razvrawmwmwzm‘rgmxm 42% b) For the special case where q”5=0, h =0, and T2 = T3 = 100°C, determine T1 = 5‘ t3} {ﬁeld in to em i& e o WQJ it’d e23 mutate “1/55” E MWMWWW m «aw ﬁrwa Wm Mr B gmammwxmwsamammemmmmmw;imimem‘mmwmfm9K ,. ‘ s it mg“? grits 5 Warmmmmwmwwﬁrtemmmwwﬁ'wﬁug§ My! , Humor—n5 a” m3» > £5“ 23. Egg? Unm BB. Hum? ﬁumo BB. Emmy Buou Wm“ Ea mwaowmo ramp ounmoo HEWHA mm Emma 9550 cam 85 3% (\$63 H3 "300. \$5256 EN: :5 ram» Hangman Bowman?“ om mu. mm wuwoo ﬁQBNW man 90 505\$ nonucnmSQ om 9m Emma Wm wuqm \$3.75 13.5 Swan mm Emo 9683 8 m :2 395305 ram» max om pauroQo 4,33”. we? 85.0030: mum Shanna: m3 manic:de \$me on :8 wage Ba Em ’58. 3555 no 092038 B. mam—non 05 9n 3% ow .90 Ewan. 155 2:6 :mm mm BEE ﬂnﬁwoﬁgo om HWHNODO. >mm§m m 3:63 \$39858 Semi. A3 USE m mmmmnma \$5436 50 3589 5an 395. Dwain Ea :63me 835 3:58 9% 5232 m: 38H 9&5qu Boaramam 333% 5 90 @8295. \$18 9% 98a cannon mmgvomném 3 835m 0m 90 miﬁ: 6335088. m nw 19L W 4 no) ,\ Mm+ H mu). 1. oF++ Maw) a .3?an u Alp b.» 148>< 30\$. n A.”.PP>M I. TDUGAII Aunvov 3n “MPH... n «\DJO VJ «Afar I 73.....QbJ AS mum En mg: 438 Ragga—Ho m: mﬁommw mg? 1? n _ m A 09 ©u+£< ablﬂ m n0 . Q 1—! u n~.v§L.LaIn—Do)’\ o in.» km .. rbmﬁﬁuﬁuov rbmnwdvbu 4ybm ﬁnd. .. II- I {a 1D..." 9:??? LII—HOB m H... u E .T u,me kw .ruupmmﬁn Wu. A3 wrong H3 52% 38a. wroﬁ gm 5on 15. as. am gm :5 8833659 m» can :50 83?? .3. 06qu Banana H n a n 050 \$80 8853. Ev Ugandan 50 ER Haawoamﬁﬁo 35305 .5 mundcomo moms mm m @530: om 33? H3 n H 3mm A? 3 - {m Saab n5. 90\$“ T}.M Q." All-1+" ME... 4... .2) H nu... EM \$ 0 :4. ml" ﬁd. "n >1 9.. 13.113" 570.16 \JIH 3.90 .9)... ...
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