Down the product of this new variable and those in

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down the product of this new variable and those in step 3 in the form: N = x 1 x 2 a x 3 b x 4 c where x 1 is the new variable and x 2 , x 3 , and x 4 were selected in step 3 (we have assumed in this example that Y =3). 5. Solve algebraically for the exponents a, b, and c that make N dimensionless. 6. Repeat steps 4 and 5 until Z-Y dimensionless groups (N’s ) have been formed.
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Chemical Engineering 150A Buckingham Pi Example Problem: Pipe flow 1. Z = 6 (D, ρ , v, l , Δ p, η ) 2. Y = 3 (L, M, T) 3. Choose D, ρ , v Check that they do not form a dimensionless group: D ሾൌሿ L , ρ ሾൌሿ M L , v ሾൌሿ L T Contains all 3 dimensions (L, M, T) Does not form dimensionless group by themselves since only ρ contains M, only v contains T 4. N ൌ D ρ v ℓ ൌ ሺLሻ M L L T ሺLሻ ൌ M L T (For N 1 to be dimensionless, exponents must be zero) 5. L: a – 3b + c + 1 = 0 M: b = 0 => b = 0 T: -c = 0 => c = 0 L: => a + 1 = 0 => a = -1 N ൌ D ିଵ ρ v ℓ ൌ D 6. Repeat. Choose different variable from remaining list for N 2 : N ൌ D ρ
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  • Spring '10
  • CORDEIRO,WILLIAM
  • Chemical Engineering, Buckingham π theorem, Buckingham Pi Theorem, Chemical Engineering 150A

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