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ECM5_09WINTER_Quiz#2Soln

ECM5_09WINTER_Quiz#2Soln - 2 An experiment is conducted in...

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ECM 5 Quiz #2 February 11, 2009 Total Test Time – 20 minutes 1. For a fluid flowing through a packed bed of particles, careful experimentation has shown that the pressure drop across bed can be calculated using " P = K μ 0.15 H # 0.85 V 1.85 d 1.2 $ % & ( ) where Pressure Drop: Δ P [=] ML -1 T -2 Viscosity: µ [=] ML -1 T -1 Column Height: H[=] L Density: ρ [=] ML -3 Velocity: V [=] LT -1 Particle Diameter: d [=] L Experimental Constant: K What dimensions does the constant K have? We can start this problem off by isolating the constant K. K = " P d 1.2 μ 0.15 H # 0.85 V 1.85 $ % & ( ) Now we’ll substitute in the given dimensions for each of the variables. K [ = ] M L T 2 L 1.2 ( M L T ) 0.15 L ( M L 3 ) 0.85 ( L T ) 1.85 " # $ $ $ % & We’ll then raise each of the dimensions to the powers that they are being raised to. K [ = ] M L T 2 L 1.2 1 L 0.15 T 0.15 M 0.15 1 L L 2.55 M 0.85 T 1.85 L 1.85 We then combine each dimension’s powers to find the answer.
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K [ = ] M 1 " 0.15 " 0.85 L 1.2 + 0.15 + 2.55 " 1 " 1 " 1.85 T 0.15 + 1.85 " 2 K [ = ] M 0 L 0.05 T 0 [ = ] L 0.05
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Unformatted text preview: 2. An experiment is conducted in which four variables are varied. These variables are velocity(v), gravity(g), area(A), and diameter(D). (a) Determine the total number of dimensionless variables that can be formed from these variables. We know that each of the above variables has the following dimensions: v[=] L T-1 g[=] L T-2 A[=] L 2 D[=] L We can easily count that there are 4 variables and 2 dimensions. This leads us to find there are 2 dimensionless variables as a result of: G = NV – ND = 4 – 2 = 2 (b) Find as many dimensionless variables as you determined there were in part (a) We can easily find many dimensionless variables for this group of variables, but we can only define 2 at a time that will be independent . Two such variables are listed below. A D 2 , gD v 2...
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