Exam2_S11_solution_rev1_marked

Exam2_S11_solution_rev1_marked - <Solution for Problem...

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<Solution for Problem 1> (a) Air at 350 K ν air = 20.92 × 10 6 m 2 / s , Pr air = 0.7, k air = 30 × 10 3 W / m K Re D = V D = 4 m / s × 0.04 m × 10 6 m 2 / s = 7646.49 Nu D = 0.3 + 0.62 Re D 1 2 Pr 1 3 1 + 0.4 / Pr ( ) 2 3 1 4 + Re D 28,2000 5 8 4 5 = 46.04 h = k D Nu D = × 10 3 W / m K m (46.04) = 34.53 W / m 2 K h = 34.53 W / m 2 K (b) Bi = h L c k g = h R k g = (34.53 W / m 2 K ) 0.02 m ( ) 1.38 W / m K = 0.5 For Bi = 0.5, ς 1 = 0.9408 and C 1 = 1.1143 cylinder ( ) Fo = α t R 2 = 0.8 × 10 6 m 2 / s ( ) 120 s ( ) m ( ) 2 = 0.24 > 0.2 θ 0 * = T r = 0, t = 120 s T T i T = C 1 exp ( 1 2 Fo ) = 1.1143 exp ( 0.9408 2 0.24) = 0.901046 T r = 0, t = 120 s = T + 0 * T i T ( ) = 300 K + 90.1 K = 390.1 K T r = 0, t = 120 s = 390.1 K
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2 (c)
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3 <Solution for Problem 2> (a) q electric = q conv + q evap q conv = h A T s T ( ) q evap = ˙ m h fg = n A " ρ A , sat T s ( ) A , T ( ) [ ] A h fg where ˙ m = n A " A = h m A A , sat T s ( ) A , T ( ) [ ] Overall , q electric = ˙ m T s T ( ) A , sat T s ( ) A , T ( ) c p Le 2 / 3 + ˙ m h fg q
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This note was uploaded on 02/09/2012 for the course ME 352 taught by Professor Staff during the Fall '08 term at Purdue University-West Lafayette.

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Exam2_S11_solution_rev1_marked - &lt;Solution for Problem...

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