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# quiz_solutions - 1 2 hπr 3 R 2 = ln r 3/r 2 2 πk A R 3 =...

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Problem 2 Known: Plane wall with prescribed thermal conductivity, thickness, and surface tempera- tures Find: Heat flux, q , and the temperature gradient, d T/ d x , for the three different coordi- nate systems shown. Schematic: Assumptions: (1) One-dimensional heat flow, (2) steady state, (3) No internal generation, (4) Constant properties. Basic equation: q = - k d T d x (1) Simplified basic equations: d T d x = T ( L ) - T (0) L - 0 (2a) 1

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q = - k T ( L ) - T (0) L (2b) Computing equations: For parts (a) and (c): d T d x = T 2 - T 1 L = 600 - 400 0 . 1 = 2000 K / m (3a) q = - k T 2 - T 1 L = - 100 600 - 400 0 . 1 = - 200 kW / m 2 (3b) For part (b): d T d x = T 1 - T 2 L = 400 - 600 0 . 1 = - 2000 K / m (4a) q = - k T 1 - T 2 L = - 100 400 - 600 0 . 1 = 200 kW / m 2 (4b) 2
Problem 3 Assumptions: (1) One-dimensional radial conduction, (2) steady state, (3) constant properties. Basic Equations: R cond = ln( r 2 /r 1 ) 2 πLk R conv = 1 hA = 1 2 hπrL q = Δ T R tot ˙ E in - ˙ E out + ˙ E gen = d ˙ E st dt Simplified Basic Equations: R cond = ln( r 2 /r 1 ) 2 πk R conv = 1 2 hπr q = Δ T R tot ˙ E in - ˙ E out

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Unformatted text preview: 1 2 hπr 3 R 2 = ln( r 3 /r 2 ) 2 πk A R 3 = ln( r 2 /r 1 ) 2 πk B R 4 = 1 2 hπr 1 b) q i = T h-T i R 3 + R 4 q o = T h-T o R 1 + R 2 From conservation of energy: q o + q i = q h = 2 πr 2 q 00 h T h-T o R 1 + R 2 + T h-T i R 3 + R 4 = 2 πr 2 q 00 h T h ± 1 R 1 + R 2 + 1 R 3 + R 4 ² = T o R 1 + R 2 + T i R 3 + R 4 + 2 πr 2 q 00 h T h = ± 1 R 1 + R 2 + 1 R 3 + R 4 ²-1 ± T o R 1 + R 2 + T i R 3 + R 4 + 2 πr 2 q 00 h ² c) q o q i = ( T h-T o )( R 3 + R 4) ( T h-T i )( R 1 + R 2) 4 Computing equation in Algebraic Form: T h = ± 1 R 1 + R 2 + 1 R 3 + R 4 ²-1 ± T o R 1 + R 2 + T i R 3 + R 4 + 2 πr 2 q 00 h ² q o q i = ( T h-T o )( R 3 + R 4) ( T h-T i )( R 1 + R 2) 5...
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