ME410-2_HW6_FALL2010_Solutions

ME410-2_HW6_FALL2010_Solutions - ME 410-2 Name: SfiLoTlJN...

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Unformatted text preview: ME 410-2 Name: SfiLoTlJN Homework #6 (due October 18, 2010) Homework Format 1. 2. 3. 9°.\‘9‘.U':‘> Use this page as a cover sheet. One problem per page. Always state what is given, your assumptions, provide a sketch, and analysis. Circle your answer. Include units. Write legibly. No late homework accepted. Work independently. Use paper with four straight edges. Staple together your work. Atmospheric air is in parallel flow (11.. = 20 m/s, T... = 15°C) over a flat heated surface that is to be maintained at a temperature of 140°C. The heater surface area is 0.5 m2, and the airflow is known to induce a drag force of 0.25 N on the heater. What is the electrical power needed to maintain the prescribed surface temperature. Assume the bottom surface of the plate to be adiabatic. Engine oil at 100°C and a velocity of 0.1 m/s flows over both surfaces of a l—m long smooth flat plate maintained at 20°C. Determine the following: a. The velocity and thermal boundary layer thickness at the trailing edge. b. The local heat flux and surface shear stress at the trailing edge e. The total drag force and heat transfer per unit width of the plate. Consider the wing of an aircraft as a flat plate of 2.5 m length in the flow direction. The plane is moving at 100 m/s in air that is at a pressure of 0.7 bar and a temperature of —10°C. The top surface of the wing absorbs solar radiation at a rate of 800 W/mz. Assume the wing to be of solid construction and to have a single, uniform temperature. Estimate the steady—state temperature of the wing. 1.. KNOWN: Q‘Q Flaw Cmo17‘qm§_mn 0236 F024: fimasfi‘lm born} 9 Hearse of _ mescmeso scam-)5; fr am Meal, QN‘O-szi’fipxrzéo aemm 9MB? SCt—SEMQ'fic‘. a ~—-—-> M.» w :: 20m]; TL, = IS‘°C p a law assump'nouc : s; 6.on _.9U€F66ér ts noxma'n c1 QEYNGLOS Humoév \s appumaw onPérrnEg; 9w; (Tymww) ;_ F: GHQS‘k‘D/mg} C? = \ooq \Tfiaqxj Pr = 0700. L). Qwfiwms: "me. fluembe Wits? m0 $1216ch Cowman nae _ a M 733‘ E2: 62‘; '5 \U/m'z A 0.?‘5"‘m‘2 " 5; =51; ..=. Wm" _. .. 'L z {3% (am; In?) as misfit/'2. 2 933 “9’3 WM mg. (eewom fiwwéy, —. “ _" —Zg goLlee E . . . . 5,“ g = 03% Wm; Wm(s)_(_woq9.43x:6 >(m3 : n3 W/mZ—K Z (6: (Ts 97¢) :_ \Bmfi (140- 6306 it 3.3% kw ’ ’2. Known: T and \/ ac away/3e ocl. Tau! \flng‘Ha r6 PM 914% 1a) V€(OC\.+S C‘flé “NEW/hove \owv‘ Amrh )6; er Ha‘ékmss 4+ +mt'lM‘g £461 fl “3 “66* Ohm a“; Méce ck“: J WAR 9&6»: C) To” dvas Que aué. [4953‘ {muster ‘06? um" (J‘AH flcsuwghfixz Q6)“: smog, «Now m {7,9 1; devm swat/co Profevhégi Engu‘xp 0H Ca) 833|Z= Tit: p: 5’04 FOO/W3) y: suwa mz/s, k: 0.140 W/mk, Pr= may- A“alfig;s‘- a) Q : ligL: th/gx \m ‘91. V W2] =‘ M" "- \av‘mi‘av PJXD W S @ rtl .1/ ~ (g = 5L QeLQ ; 50",) ("£15 "a: 0.14‘7M 4 5;,— SPr‘V3: 0.:47m (mm #5: 0.0|43m \o) V2 '/ \U _ 1 3 am My \ 1 h; f 0.332 Q1 Pr : fix. 0.332(m.n’2(.0m/3 M : 10.2; 'U/thK [I 0 50x: hL (TS-RA : 1(42: w/mfig (20-!003L: 4300 [$12 a} -'/2 4 k / 2 2 J4 115$: - —2_ 01474 (2‘51. 8!” '3 m (O'I‘M/SBO-ééqowu SIL : 0.0 342 kco /m_g'2 =0.og41 ww— C) 9’: 2mg,L -l/z B 7— ;: 2L V 2 ’32ng = 20.“) (8M k‘o/m3 (aim/s3 D (to “#2) L x hheUIuH/z :- 0073 N/m T‘MIC CB, 2 2. L‘AL( 5’ 9L\— " ’2( I 32,; W 1v? K (2' 40023 “g? 00 W/ m 3| __3« £@E.L_59§1EE_AEK_ME9_WD HBSarzeeo Same” FLUX Foe m mamas-r wnmc, 6F Peasanqu maqu 4m) 59% he FIND: smmv— smug T 6? Iowb E:— q 6800 was r . .N- mam..- ._—.—~......____-..____.._____ — “2 g - ' fififixfiyi umva wuuc 12m», neeuame manna: E; __.m?_§9aq§§:_~95_@_La§~21nfléjmzzegmwmfifiw% 3 kn:- _______.__..____j}.1.9.:Z$§._,gi_;§_1m§f_vri/s K 0.”? V _____Qu&}‘t_$t_$._E__.__F‘_29M. énmr/__léwe_w_w~_a_ _.______‘.,__._S~1M.(=___Q£é_.§2§m!;=_. ‘Eflééflg 5‘ “A . _.zsLo7____ 11.14 Xlo‘b WI 2/5 __..___A. 12319..~ 94¢ ..=__ . £fl.9§~_,_ ._me___flés~_a__m_ge_e@mmg0 63 “meantm 0935...... 66.15. ' ...
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ME410-2_HW6_FALL2010_Solutions - ME 410-2 Name: SfiLoTlJN...

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