assignchap3-147xtracredit

assignchap3-147xtracredit - PROBLEM 3.147 KNOWN: Dimensions...

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Unformatted text preview: PROBLEM 3.147 KNOWN: Dimensions and materials of a finned (annular) cylinder wall. Heat flux and ambient air conditions. Contact resistance. FIND: Surface and interface temperatures (a) without and (b) with an interface contact resistance. SCHEMATIC: I 7,“ 7; k 50 W/m K l—r‘;:60mm f=2mm T T T L=320K L7. :66mm 59W"? h=100W/m2'K ' 45:70am” ._.,w_. Aluminum, vamm—J k=24OW/m-l< Rb 7;" 75.: 713: 7S I; u—f—F 2:10 W/m2 RC thr RW Rf ASSUMPTIONS: (1) One—dimensional, steady—state conditions, (2) Constant properties, (3) Uniform h over surfaces, (4) Negligible radiation. ANALYSIS: The analysis may be performed per unit length of cylinder or for a 4 mm long‘ section. The following calculations are based on a unit length. The inner surface temperature may be obtained from ' q’= Tillme = q’i'(27r 1;):105 W/m2x27rx0.06 m: 37,700 W/m tOt where Rgot = R; +R’LC + R’w + R’equiv; quuiv = (1/ R} +1/Ri3)_1. RE, Conduction resistance of cylinder wall: R, _ ln(r1/r1) _ ln (66/60) C _ 2:: k _ 23(50 W/m-K) 't C , Contact resistance: =3.034><1b‘4 mK/w R’LC : REC /277: 11210—4 m2 .K/WIza-rxon66 m = 2.411x10‘4 m-K/W R'W , Conduction resistance of aluminum base: R, 2 1n (rb In) : ln (70/66) W 27: k 27r><240 W/m-K Rf), Resistance of prime or unfinned surface: _ l _ 1 C hAb _ 100 W/m2 -K><0.5><27r(0.07 m) =3.902x10“5 m-KfW =454.7><10“4 m-K/W RE) R}, Resistance affins: The fin resistance may be determined from m Tb —T0° i 1 _ qi _ nrhA'r The fin efficiency may be obtained from Fig. 3.19, r2C 2 r0 + U2 2 0.096 m LC : L+t/2 2 0.026 m R'f Continued .. PROBLEM 3.147 (Cont.) AP = Lct =5.2><10“5 m2 rQC/r121.45 L3C/2 (h/kAp )“2 = 0.375 Fig. 3.19 —) "of z 0.88. The total fin surface area per meter length A} = 250%: (r3 — r13 )x2] : 250 m'1 [271' (0.0962 — 0.072)]162 = 6.78 m. c —1 Hence R} =[0.88><100 W/m2 -K><6.78 m] :16.8><10‘4 mK/W ungaquiV = (1/16.8x10_4 +1/454.7x10‘4)W/m-K = 617.2 W/m-K quuiv 216.2x10‘4 mK/W. Neglecting the contact resistance, R’tot =(3.034+0.390+16.2)10‘4 rn-K/W:19.6><10’4 m-KIW Ti 2 q’R'tot +r.,., = 37,700 W/mx19.6><10'4 m-K/W+320 K = 393.9 K < T1=Ti~q'R’w : 393.9 K—37,700 W/mx3034x10'4 m-K/W 2 382.5 K < Tb : T1 —q’RiJ = 382.5 K—37,700 W/m><3.902><10‘S m-K/W = 381.0 K. < Including the contact resistance, R’tot z (19.6x10“4 + 2.411x10‘4)m- W = 22.0><10‘4 m-K/w Ti = 37.700 W/mx22.0><10‘4 m-K/W+320 K = 402.9 K < TU = 402.9 K—37,700 W/mx3034x10‘4 m-K/W = 391.5 K < 71,0 2 391.5 K—37,700 WImx2.411x10'4 m-K/W = 382.4 K < Tb : 382.4 K—37,700 W/mx3902x10'5 m-K/W =380.9 K. < COMMENTS: (1) The effect of the contact resistance is small. (2) The effect of including the aluminum fins may be determined by computing Ti without the fins. In this case R'tot = R; +R'C0nv, where l 1 11273 I1 100 W/m2 -K 27: (0.066 m) R’Com, = = 241.i><10‘4 m-K/W. Hence, Rm 2 244.1x10"4 mm, and Ti : q’R’tot +7.... = 37,700 W/mx244ix10‘4 m-K/W+320 K = 1240 K. Hence, the fins have a significant effect on reducing the cylinder temperature. (3) The overall surface efficiency is no 217(A; /A§)(1—nf )=1—6.78 Ila/7.00 m(1—0.88) = 0.884. It follows that q'=nOhOA't€b = 37,700 W/m, which agrees with the prescribed value. ...
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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 Wisconsin.

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assignchap3-147xtracredit - PROBLEM 3.147 KNOWN: Dimensions...

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