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- StudyBlue
- Alabama
- University of Alabama - Tuscaloosa
- Mechanical Engineering
- Mechanical Engineering 309
- Dent
- 4.34

Anonymous

PROBLEM 4.34 KNOWN: External corner of a two-dimensional system whose boundaries are subjected to prescribed onditions. c FIND: Finite-difference equations for these situations: (a) Upper boundary is perfectly insulated and side boundary is subjected to a convection process, (b) Both boundaries are perfectly insulated; ompare result with Eq. 4.43. c SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3) Constant roperties, (4) No internal generation. p ANALYSIS: Consider the nodal point configuration shown in the schematic and also as Case 4, Table 4.2. The control volume about the node – shaded area above of unit thickness normal to the page – has dimensions, (Δx/2)(Δy/2)⋅1. The heat transfer processes at the surface of the CV are identified as q 1 , q 2 ⋅⋅⋅. Perform an energy balance wherein the processes are expressed using the ppropriate rate equations. a (a) With the upper boundary insulated and the side boundary subjected to a convection process, the nergy balance has the form e (1,2) in out 1 2 3 4 E E 0 q q q q 0−= +++= () m-1,n m,n m,n-1 m,n m,n TT TT yxy k1 k1 h1TT 0 2x2y2 ∞ −− ΔΔΔ⎡⎤ ⎡⎤ ⎡⎤ ⋅+⋅+⋅− ⎢⎥ ⎢⎥ ⎢⎥ ΔΔ ⎣⎦ ⎣⎦ ⎣⎦ 0.+= L etting Δx = Δy, and regrouping, find m,n-1 m-1,n m,n hx 1hx TT T2 1T k2k ∞ ΔΔ⎡⎤ ++ − + ⎢⎥ ⎣⎦ 0.= (3) < (b) With both boundaries insulated, the energy balance of Eq. (2) would have q 3 = q 4 = 0. The same esult would be obtained by letting h = 0 in the finite-difference equation, Eq. (3). The result is r < m,n-1 m-1,n m,n T T 2T 0.+−= Note that this expression is identical to Eq. 4.43 when h = 0, in which case both boundaries are nsulated. i COMMENTS: Note the convenience resulting from formulating the energy balance by assuming that all the heat flow is into the node. Frank P. Incropera PROBLEM

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