assignchap4-46 - 3: _ l 5 PROBLEM 4.46 KNOWN: Nodal...

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Unformatted text preview: 3: _ l 5 PROBLEM 4.46 KNOWN: Nodal temperatures from a steady—state, finite-difference analysis for a one—eighth symmetrical section of a square channel. FIND: (a) Beginning with properly defined control volumes, derive the finite-difference equations for nodes 2, 4 and 7, and determine T3, T4 and T7. and (b) Heat transfer loss per unit length from the channel. I q . SCHEMATIC: run: 300 K, h = 50 Wlmz- K _d_11q_'1_ _2T9'g _ 3Tq_'3_ _4lq’4 Node WC) 1 430 i Symmetry adiabat 3 394 5 t, 7 / 3‘ 6 492 l C><olr=1 Wim-K VI 8’9 (’00 r I Ax = a .01 m I 6. qu \é 6| y: X _?_ ASSUMPTIONS: (l) Steady~state conditions, (2) TWO-dimensional conduction, (3) No internal volumetric generation, (4) Constant properties. ANALYSIS: (a]Def'1nc control volumes about the nodes 2, 4, and 7, taking advantage of synunetry where appropriate and performing energy balances, Em e Bout = 0 , with Ax = Ay, Node 2: q; +q’rJ +012; WEI =0 T3 — To hAx(T,,°—T2)+k(Ay/2) A “+kAxT6_T2 x T3 {0511+ 0.5T3 + T6 + (naming/[2+ (hAx/k)] T3 =l:0.5x430+0.5><394+492+ (SOW/mz-KXODIm/lW/m-K)300]K/[2+0.50] < Node 4: q; + qij +qé = 0 T3 —T4 Ax T4 = [T3 + (hAx/k)r,,, ]/[1+ (hAxfkfl T4 =[394+0.5x3001K/[1+0.5]:363K < =0 h(Ax/2)(Tw —T4)+O+k(Ay/2) PROBLEM $4.46 (Cont) K Symmetry plane Node 7: From the first schematic, recognizing that the diagonal is a symmetry adiabat, we can treat node 7 as an interior node. hence T7 = 0.25(T3 +T3 + T6 +T6 ) :0.2s(394+394+ 492 + 492)K = 443K < (b) The heat transfer loss from the upper surface can be expressed as the sum of the convection rates from each node as illustrated in the first schematic. CIEV =cti+ci'2+q:2+q'4 qgv = h(Ax/2)(T1-Tm)+ 1mm}; —rm)+ 1mm “Too )+ h (Ax/2)(T4 4m) qgv = sow/m2 -Kx0.1m[(430~300)/2+(422— 300)+ (394—300)+ (363—300)/2] K qgv =156W/m < COMMENTS: (1) Always look for symmetry conditions which can greatly simplify the writing of the nodal equation as was the case for Node 7. (2) Consider using the IHT Tool, F inite-Diflcrence Equations, for Steady—State, Two-Dimensional heat transfer to determine the nodal temperatures T. - T; when only the boundary conditions T3. T9 and (Tmh) are specified. ...
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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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assignchap4-46 - 3: _ l 5 PROBLEM 4.46 KNOWN: Nodal...

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