PS_6%20soln - ChBE 421 Problem Set #6 Solution Fall 2009 1....

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Unformatted text preview: ChBE 421 Problem Set #6 Solution Fall 2009 1. The variables involved are: , , , D , v , e Using the Buckingham theorem, we will get 3 different dimensionless groups. Since the problem gives relations for , and , well use these as the core variables. Hence, 1 2 3 a b c a b c a b c D v e = = = Solving for each a , b and c and applying the numbers gives 1 2 2 3 2 D v e = = = These terms should be equal for the case of water and the actual application. If we denote 2 H O D xD = , 2 H O v yv = , 2 H O e ze = , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 H O H O H O H O H O H O 2 2 2 2 H O H O H O H O H O H O H O H O H O H O H O H O H O H O 2 2 2 2 H O H O 1.8 2.1 2.1 1.8 1.8 2.1 xD D D v yv v ze e e = = = = = = Solving for x, y, z, we get 2 2 2.1 2.45 1.8 1.8 0.857 2.1 2.1 2.45 1.8 x y z = = = = = = and therefore, 2 H O 2.452....
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PS_6%20soln - ChBE 421 Problem Set #6 Solution Fall 2009 1....

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