quiz+2+key+2009 - PTFE 3210 QUIZ#2 Spring 2009 Name Problem...

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Unformatted text preview: PTFE 3210 QUIZ #2 Spring 2009 Name Problem Maximum Score Score 1 30 2 49 3 39 Total 100 PTFE 3210 QUIZ #2 Spring 2009 Name 1. Find kg. The composite wall of an oven consists of three materials, two of which are of known thermal conductivity, k A = 20 W/m - K and kc = 50 W/m ° K, and known thickness, LA = 0.30 m and LC = 0.15 In. The third material, B, which is sandwiched between materials A and C, is of known thick- ness, LB = 0.15 m, but unknown thermal conductivity k3. Under steady-state operating conditions, measurements reveal an outer surface temperature of T5", = 20°C, an inner surface temperature of T5,,- = 600°C, and an oven air temperature of T5,, = 800°C. The inside convection coefficient k is known to be 25 W/m2 - K. What is the value of k3? AK / / 25$ _, .\ /V6645c2- fl/rD/fltWCW’ if (3‘9 Pam/'05) PRUBLEM 3.9 Ki‘fln’t'fi': ‘l'itittkneaxm “flinch: nulleriziis uhich Farm :1 cnl'npusitv wall and thermal conductiviticn of urn of the Innteriztln, inner and outer Marleen: tempemtnrea of the composite: also. lempemture and convection coefficient ilfii‘li'lL'iLllL‘ti with adjoining gas“. FIND: \‘nlne of unknown thermal conductivity. I»; g. SEE-IEMATIC: _ . LA zoom unseat: , Eta—30 ‘3 LanignaISM 2A a20W/wk‘ . kc Jaw/mg En as? C haZSW 2-K m T” 7;; Ea ‘ . ._—“'b I E £1. £9“ 9" -=— MEL an kAA kBA itch ASSUMPTIONS: [it Steady-slate ennditions. [2) le'te-tlitnensiomll conduction. [3} Constant. moment-:22. Ml Negliglhlc contact rcxlsmncm (:3) Negligible radiation et‘fcctg. ANAIJ’SSS: Rct'cs‘ring tn the thermal eil‘euiL the heat flux may he expressed an [-3 TM Tum m {1.1m emit: l ' ltA AE_LE' (1.3111 LtHSm 0.15111 RA ‘ kl! ‘ kg 111 “7911vi V. RB ‘::E}\‘."-"Ha-K ‘ifl' W35” _ “3-1211. 1!} IEJJlSiHJD-‘kg The heat flux may be obtained From Z ’ S q"=h('l'..,: 41;“): 5 Wimg -K{_3()()—(a(l(l]c C (2} (160% Wflnz. Substilnling for [he heal Iiux from Eq. (2) inlo Eq. (I), find 353=§wmsz f O *0.0l8=w /,//4i 2 kB Q 'J[ of )3; RB 2% W/an- K. < COMMENTS: Radiation ell‘ects ate likely to have :1 significant influence on the net heat flux at the inner surface ol‘the oven. 3 O P0/M75l fl Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted, Any other reprodlcfion or translation of this work beyond that permitted by Sections 107 or I 08 ofrhe 1976 United States Copyright Act without the permission ofrhe copyright owner is unlawful. 2. A rectangular prism (see figure below) of material is to be cooled in a fluid bath that can be treated as an infinite bath. The length d is 0.01 m, and the thermal conductivity of the material is 1 W/m—K. The average heatwtransfer coefficient for the cooling process is 10 W/mz-K. Assume p is 1 g/cm3 and Cp is 1 J/gm-K. (40 points) Note: all sides are perpendicular to each other. 2d If the prism is at a uniform temperature of 400 K and the temperature of the bath is 300K, how long does it take for the surface temperature of the prism to reach 350 K? A?) r 2f V r gm»; fl :_~ ‘ZLAMA-QJ' 462W M5 2/4255: #— dud/mi; 4W / g :3 1 MO . ZT%:&wm 2225727 5 1 ”7’ - *WM/ B 1'1 z. 00 W/fi’ffi ((9,0an .2 (9, 632 Fr / %%-/< 5% 5». 4 [1/ WSW- SECQF Esr— 7" he" 14 f?“ # lo mam Consider a plane wall that can be treated as one dimensional with wall thickness L= 1.0 m. There' IS uniform and constant heat generation per unit volume of 1,000 W/ni3 in the wall. The thermal conductivity is constant (K= 10 W/m-K), and steady state conditions exists. The wall on the left hand side (x I 0) is insulated. The temperature of the wall on the right hand side (x = 1.0 m) is 300 K. e eat 1 8mm equation 15. 8x 6x 8y 6y 62 6—2 9': PC a: The following assumptions allow simplifying the governing equation: A1. One-dimensional in x directions A2. Steady state conditions exist A3. Constant properties The boundary conditions are: BC. #1 dT/dx = 0 at the insulated wall B.C.#2 T=300Katx=L a) Using the assumptions, simplify the heat diffusion equation to its simplest form. (5 points) 0) / Q4 ,4; kéj\t_214f<:”)44g:§591 ;( K; 07/“; “’2 W W // Jx LL a; A? o ,//\. 1“ Cum/W C5 #31449 M) #45 {rléc— /’o’/V“ S, S aildd: M flg/fi b) Solve for the temperature distribution in the wall, that is, find T(x). (20 points) ii Bgzz :3300g+ 3K W L g H :.30QK+ @%l§§giflya é (a?) /z9/%JT<\ 7/2? : EGO‘VJF WM- tfl ...
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