problem4-47 - PROBLEM 4.47 KNOWN: Outer surface...

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PROBLEM 4.47 KNOWN: Outer surface temperature, inner convection conditions, dimensions and thermal onductivity of a heat sink. c F IND: Nodal temperatures and heat rate per unit length. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) Two-dimensional conduction, (3) Uniform outer surface emperature, (4) Constant thermal conductivity. 20 t ANALYSIS: (a) To determine the heat rate, the nodal temperatures must first be computed from the corresponding finite-difference equations. From an energy balance for node 1, () 51 21 1 TT hx / 2 1T T ky / 2 1 kx 1 0 xy Δ⋅ −+ + Δ = ΔΔ 12 5 3T T 2 kk ⎛⎞ + + + = ⎜⎟ ⎝⎠ 0 ( 1 ) With nodes 2 and 3 corresponding to Case 3 of Table 4.2, 3 6 2 T2 2T T2 T T 0 + + + = ( 2 ) 23 7 T T T −++ + = 0 0 , ( 3 ) where the symmetry condition is invoked for node 3. Applying an energy balance to node 4, we obtain ( 4 ) 45s 2T T T 0 + = T he interior nodes 5, 6 and 7 correspond to Case 1 of Table 4.2. Hence, ( 5 ) 14 56s TT 4 TT T 0 +− ++= ( 6 ) 25 67s TT4 TTT0 ( 7 ) 367 s T4 +−+ = where the symmetry condition is invoked for node 7. With s T5 0 C , T 2 0 C = °=
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This note was uploaded on 12/07/2010 for the course MAE Heat Trans taught by Professor Lee,j.s. during the Spring '10 term at Seoul National.

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problem4-47 - PROBLEM 4.47 KNOWN: Outer surface...

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