Buckingham Pi

V ? δp ሺl? ୟ ൬ m l ? ൰ ୠ ൬ l t ൰ ?

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v Δp ൌ ሺLሻ M L L T M L T ൰ ൌ M L T L: a – 3b + c - 1 = 0 M: b + 1 = 0 => b = -1 T: -c – 2 = 0 => c = -2 L: => a + 3 – 2 – 1 = 0 => a = 0 N ൌ D ρ ିଵ v ିଶ Δp ൌ Δp ρ v Repeat. Choose different variable from remaining list for N 3 : (arbitrarily put η in the denominator) N ൌ D ρ v η ିଵ ൌ ሺLሻ M L L T M L T ିଵ ൌ M L T L: a – 3b + c + 1 = 0 M: b - 1 = 0 => b = 1 T: -c + 1 = 0 => c = 1

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L: => a - 3 + 1 + 1 = 0 => a = 1 N ൌ D ρ v η ൌ Dρv η In conclusion, we have the following relationship: N ൌ fሺN , N D ൌ f ൬ Δp ρ v , Dρv η We could also write the more common form: N ൌ fሺN , N Δp ρ v ൌ f ൬ Dρv η , D Note that the function consists of three variables, as predicted by the following relation: Π = Z Y = 6 – 3 = 3
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• Spring '10
• CORDEIRO,WILLIAM
• Chemical Engineering, Buckingham π theorem, Buckingham Pi Theorem, Chemical Engineering 150A

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