assignchap4-72 - PROBLEM 4.72 KNOWN: Straight fin of...

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Unformatted text preview: PROBLEM 4.72 KNOWN: Straight fin of uniform cross section with insulated end. FIND: (:1) Temperature distribution using finiteedifference method and validity of assuming one— dimensional heat transfer, (b) Fin heat transfer rate and comparison with analytical solution, Eq. 3.76, (c) Effect of convection coefficient on fin temperature distribution and heat rate. SCHEMATIC: Tm: OC _: h = 500 Wimz- K Tb = 100 00 —o 2 { LR = 50 Wim-K I W x L = 48 mm ASSUMPTIONS: (l) Steady-state COnditions, (2) Onedimensional conduction in fin, (3) Constant properties, (4) Uniform film coefficient. ANALYSIS: (3) From the analysis of Problem 4.45. the finite-difference equations for the nodal arrangement can be directly written. For the nodal spacing Ax = 4 mm, there will be 12 nodes. With E 3-) w representing the distance normal to the page, 2 LP .sz-o h” szzflAx2=———500W/m KX23 (4X10_3mm)=0.0533 kAo k-f'w kw 50W/m-Kx6x10_ m Node I: 100+T2 + 0.0533X30-(2+0.0533)T1= 0 or -2.053T. + '1‘; = -101.6 Node in: Tn+1+Tn_1+1.60A2.0533Tn : 0 or Tn_1-o2.053Tn +T,,_] = 71.60 Node I2: T1 1 + (0.0533/2)30—(0.0533/2+ 1)le = 0 or T11 —l.0267T;2 = —0.800 Using matrix notation, Eq. 4.52, where {A} [T] = [C]. the A-matrix is ttidiagonal and only the non-zero terms are shown below. A matrix inversion routine was used to obtain [T]. Tridiagonal Matrix A Cofumn Matrices Nonzero Terms Values Node C T a” 21].; 4.053 1 l —lOl.6 85.8 all an an 1 -2.053 1 2 -l .6 74.5 3.3.; 33.3 33.4 1 -2.053 1 3 -l a“ 34,4 34,5 1 -2.053 1 4 -l.6 58.6 314 355 35,5 1 l 5 ads 3“ am 1 -2.053 1 6 —l .6 48.8 37.5 31'"; 3.7.3 1 l 7 - 1 213‘? 33,3 3,3,9 1 l 8 -] 39,3 ago all") 1 l 9 -l.6 41.2 310.9 3mm am.“ 1 -2.053 I 10 -l .6 39.9 an‘go all,“ am; I I 312.11 312,12 al2.13 I - l .027 1 12 -0.8 38.9 The assumption of one-dimensional heat conduction is justified when Bi 5 h(wf2)fk < 0.1. Hence. with Bi = 500 wxm3.K(3 x 103 m)f50 Wlm-K = 0.03, the assumption is reasonable. PROBLEM 4.72 (Cont) (b) The fin heat rate can be most easily found from an energy balance on the control volume about Node 0. I f r T _T AX (if =QI+Qconv _k'“' D 1+h[2 ](T0 T00) Air 2 o —3 1 —85. c o q} 2SOW/m-K(6x10_3m)M+500W/m2-K 2-324“ (100730) c 4x10‘ in 2 q} =(1065+140)w/m =1205 w/m. < From Eq. 3.76. the fin heat rate is q = (thAC)UZ -9b-tanh mL. Substituting numerical values with P = 2(w + t): 2 e and A, = w- t , m = (hP/m)1’2 = 57.74 m'I and M = (hpmcim = 17.321? W/K. Hence. with at, = 70°C. q’ = 17.32 w/Kx70thanh(57.44x0.043) : 1203 win and the finite-difference result agrees very well with the exact (analytical) solution. (c) Using the IHT F infre-Difierence Equations Tao! Pad for ID, SS conditions. the fin temperature distribution and heat rate were computed for h = 10. 100, 500 and 1000 Wlmz-K. Results are plotted as follows. 100 1300 90 G 1500 g 80 m— n g 7° ,5 1200 g 60 E E 50 500 a 6-! '— 40 t E <6 600 30 g 300 + h: 10 waer 0 . , —e— n = 100 wirmnaK o 2 4 + h: 500 WImAZK 00 00 500 300 1000 E' h = 1000 WW3" Convection coefficient, htwtmnaK) The temperature distributions were obtained by first creating 3 Lookup Table consisting of 4 rows of nodal temperatures corresponding to the 4 values of h and then using the LO0KUPVAL2 interpolating function with the Explore feature of the IHT menu. Specifically, the function T_EVAL = LOOKUPVAL2(tO467. h, x) was entered into the workspace, where t0467 is the file name given to the Lookup Table. For each value of h, Explore was used to compute T(x), thereby generating 4 data sets which were placed in the Brewer and used to generate the plots. The variation of q’ with h was simply generated by using the Explore feature to solve the finite-difference model equations for values of h incremented by 10 from 10 to 1000 wmfis. Although q’f increases with increasing h, the effect of changes in h becomes less pronounced. This trend is a consequence of the reduction in fin temperatures, and hence the fin efficiency, with increasing h. For 10 S h S 1000 WimZ-K. 0.95 2 1-"; 0.24. Note the nearly isothermal fin for h = 10 Wlmz-K and the pronounced temperature decay for h = 1000 W/mE-K. ...
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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 University of Wisconsin Colleges Online.

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assignchap4-72 - PROBLEM 4.72 KNOWN: Straight fin of...

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