ME 330 PS6 Solutions Fall 2010

ME 330 PS6 Solutions Fall 2010 - (a The relevant variables...

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Unformatted text preview: (a) The relevant variables for this problem are i. Geometry: D ii. Material properties: ρ, σ iii. External effects: Q, g Hence, Q = f(D, ρ, σ, g). The number of Pi terms required is k (=5) – r (=3) = 2. There are two options to express in terms of dimensionless variables Π1 = φ ( Π 2 ) , ⎛ ρ gD 2 ⎞ Q =φ⎜ 1. If we choose D, ρ and σ as the repeating variable, ⎟ D Dσ / ρ ⎝σ⎠ ⎛ ρ gD 2 ⎞ Q = φ2 ⎜ 2. If we choose D, ρ and g as the repeating variable, 2 ⎟ D Dg ⎝σ⎠ It does not matter which set of Pi terms you have chosen. At the same Π 2 location, Q will be the same value for both cases (1 & 2). However, since the following questions require conditions under different gravitational effects, it is wise to choose case 1 in order to save the trouble of worrying about the gravitational term appearing in both axes. (b) Yes, you can achieve reduced gravitational effects by changing the value for Π 2 . Knowing that ρ ρ gD 2 Π2 = , decreasing is the same effect as decreasing g since we are ultimately σ σ interested in decreasing the nondimensionalized Π 2 . (c) Yes. Given that we have collected data for Π1 = φ ( Π 2 ) , we can move along the Π 2 axis (equivalent to changing gravity) and find the corresponding Π1 value. Since Π1 = Q , D Dσ / ρ if the fluid properties and the diameter of the tube are known, we can determine the flow rate Q. (d) According to the problem statement, we are keeping ρ, σ and Q constant. Looking into ⎛ ρ gD 2 ⎞ Q =φ⎜ ⎟ , we can see this equation as D Dσ / ρ ⎝σ⎠ ρ g moon C1 Q = f ( C2 D 2 ) where the constants are C1 = and C2 = . Looking at the σ DD σ /ρ Π1 = φ ( Π 2 ) , which is ( )( ) equation, C1 = D D f C2 D 2 which means that the value of constant C1 is purely dependent on the diameter D. Depending on the functional relationship, it may both be possible or impossible to achieve the same flow rate Q. (eg. D may be cancelled depending on the ( )( ) function and in this case, C1 = D D f C2 D 2 becomes C1 = f ( C2 ) . Since C1 and C2 are constants, the equality may not hold depending on the value of the two constants, thus resulting in no solution.) Note: If students have assumed a constant D, then it will be impossible to achieve the flow rate Q. ...
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This note was uploaded on 09/05/2011 for the course ME 205 taught by Professor Koen during the Spring '07 term at University of Texas.

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