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PS2_sol_S08

# PS2_sol_S08 - Problem Set 2(2.006 Spring 08 Problem 1(a...

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Problem Set 2 (2.006, Spring 08) Problem 1 (a) There are 8 important parameters. There are 4 different dimensions appearing in the table. Try j = 4 and ρ , μ , D , and k f as repeating parameters. ( ) ( ) ( ) ( ) 3 1 1 3 1 0 0 0 0 a b d c a b c d f D k ML ML T L MLT M L T ρ μ θ θ ⎤ = = : 0 : 3 0 : 3 0 : 0 M a b d L a b c d T b d d θ + + = + + = = = 0 a b c d = = = = Now non-dimensionalize h , U , c p , and k s with the repeating variables. h: ( ) ( ) ( ) ( ) 3 1 3 1 1 3 1 0 0 0 0 a b d c a b c d f h D k MT ML ML T L MLT M L T ρ μ θ θ θ ⎤ = = : 1 0 : 3 0 : 3 3 0 : 1 0 M a b d L a b c d T b d d θ + + + = + + = = − − = 0, 1, 1 a b c d = = = = − 1 f hD k Π = U: ( ) ( ) ( ) ( ) 1 3 1 1 3 1 0 0 0 0 a b d c a b c d f U D k LT ML ML T L MLT M L T ρ μ θ θ ⎤ = = : 0 : 1 3 0 : 1 3 0 : 0 M a b d L a b c d T b d d θ + + = + + = − − = = 1, 1, 0 a c b d = = = − = 2 UD ρ μ Π = c p : ( ) ( ) ( ) ( ) 2 2 1 3 1 1 3 1 0 0 0 0 a b d c a b c d p f c D k L T ML ML T L MLT M L T ρ μ θ θ θ ⎤ = = : 0 : 2 3 0 : 2 3 0 : 1 0 M a b d L a b c d T b d d θ + + = + + = = − − = 0, 1, 1 a c b d = = = = − 3 p f c k μ Π = k s must have the dimension of k f . ' 4 s f k k Π = Parameter h U ρ μ D f k p c s k Dimension 3 1 MT θ 1 LT 3 ML 1 1 ML T L 3 1 MLT θ 2 2 1 L T θ 3 1 MLT θ

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(b) 1 Nu f hD k Π = = , 2 Re UD ρ μ Π = = , 3 Pr p f f p c k k c μ μ ρ ν ρ α Π = = = = 1 ' 4 4 Bi s f f s hD k hD k k k Π Π = = = = Π Nu and Bi look similar but they have different thermal conductivity. ( ) Nu Re, Pr, Bi f = (c) 0 10 20 30 40 50 0 10000 20000 30000 40000 h (W/m 2 K) v (m/s) Water Therminol 60 Air It seems that each case follows a power function with different coefficient and exponent. (d) Air (Pr=0.700) Therminol 60 (Pr=81.1) Water (Pr=5.93) Re Nu Re Nu Re Nu 63.3 4.1 15.6 10.9 574.4 26.1 316.4 9.0 78.0 24.0 2872.0 58.1 632.8 12.4 156.0 33.3 5743.9 81.8 3164.0 27.3 779.8 74.4 11487.9 139.1 6328.0 38.6 1559.6 108.5 17231.8 176.8 12655.9 65.9 3119.1 155.0 22975.8 210.3 18983.9 82.4 4678.7 186.0 28719.7 241.4 25311.8 101.1 6238.2 209.3 34463.7 270.0 31639.8 116.1 7797.8 232.6 40207.6 297.9 10 100 1000 10000 10 100 Nu Re Therminol 60 (Pr=81.1) Water (Pr=5.93) Air (Pr=0.700) (e) Since Bi # is not important, the functional form reduces to ( ) Nu Re, Pr f = . By observing linear log-log relationship between Nu and Re for a given Pr=constant, we infer that the slope of the linear plot doesn’t change with Pr. However, y-intersect of the linear log-log plots is shifted when Pr number changes. ( ) ( ) (Pr) (Pr) ' Nu Re (Pr) Re 10 Nu Re 10 Re (Pr) m A m A m Log m Log A Log A = + = => = =
where A(Pr) is y-intersect which is a function of Pr. By letting ' (Pr) Pr n A = Nu Re Pr m n = Do linear regressions on ( ) ( ) ( ) Log Nu Log Re Log Pr m n = + for each case. Air Therminol 60 Water Pr 0.700 81.1 5.93 m: slope 0.537994 0.497356 0.585716 nLog(Pr): y-intersect -0.39554 0.439558 -0.23637 Calculated m’s are not the same, but very close to each other. Take the average of the 3 values as a representative number. ( ) 1 0.5380 0.4974 0.5857 0.5404 3 m + + = As suggested in the problem, Nu Pr Re n m = is plotted as a function of Pr. By taking logarithm, the plot shows nLogPr vs LogPr and the slope of the linear fit is n. You should set the y-intersect zero when you do the fitting. Using m=0.5404 for all cases, 1 10 100 1 Nu/Re m Pr Achieved n= 0.1456 .

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PS2_sol_S08 - Problem Set 2(2.006 Spring 08 Problem 1(a...

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