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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 steadystate 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. SEEIEMATIC:
_ . LA zoom
unseat: , Eta—30 ‘3 LanignaISM
2A a20W/wk‘ . kc Jaw/mg
En as? C
haZSW 2K m T” 7;; Ea ‘ . ._—“'b
I E £1. £9“ 9" =— MEL
an kAA kBA itch ASSUMPTIONS: [it Steadyslate ennditions. [2) le'tetlitnensiomll 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}\‘.""HaK ‘iﬂ' W35” _ “31211. 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 notforprofit 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/mzK. Assume p is 1 g/cm3 and Cp is 1 J/gmK. (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 :_~ ‘ZLAMAQJ' 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/mK), 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. Onedimensional 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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