ME 106 - Saturday Professor Philip S Marcus ME 106...

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Saturday, December 12, 2009 Professor Philip S. Marcus ME 106 Solution to Final Examination 1a) The relevant “input” parameters are: ρ , R outer , Ω inner , H , R inner , Ω outer , and ν . Specifcally, we do not include g , and the air pressure. We choose units oF { Length } ≡ R outer ; { Time } ≡ 1 / Ω inner ; and { Mass } ≡ ρ ( R outer ) 3 . These choices are arbitrary, and you are free to choose any other units that make sense . With these units, an expression For the period τ oF the ±ow is τ = τ * { Time } =1 / Ω inner ˆ f ± ρ * ,R * outer , Ω * inner , Ω * outer ,H * ,R * inner * ² (1) Expressing the dimensionless Form (with asterisks) oF the input parameters in the units chosen above, eq. (1) becomes τ =1 / Ω inner ˆ f ± 1 , 1 , 1 , Ω outer / Ω inner ,H/R outer ,R inner /R outer ,ν/ [( R 2 outer Ω inner ] ² (2) and dropping the “slots” flled with numerical values rather than dimensionless groupings oF parameters, we obtain τ =1 / Ω inner f ± Ω outer / Ω inner ,H/R outer ,R inner /R outer ,ν/ [( R 2 outer Ω inner ] ² (3) where f is a dimensionless Function oF dimensionless “slots”. Note that the 4 dimensionless grouping in the 4 slots oF f are independent oF each other, as required. b) You must make all Four oF the expressions in the slots oF f the same in the small model and in the Etcheverry design; otherwise, the physics will di²er, and the ±ows will di²er and not be similar. In Fact, unless all Four oF the slots have the same numerical values For the small apparatus and For the Etcheverry apparatus, it is possible that the ±ow in Etcheverry won’t be periodic in time. By your design requirement that all lengths are 10 times bigger in the Etchev- erry version, H/R outer and R inner /R outer are same in the Etcheverry and in the small apparati. In addition, you must keep Ω outer / Ω inner 0 in both apparati.
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Thus, keeping the values in the slots the same requires that Ω outer = 0 in the Etcheverry apparatus and ν/ [( R 2 outer Ω inner ] be the same in both apparati. Or, ν Etch ( R Etch outer ) 2 Ω Etch inner = ν small ( R small outer ) 2 Ω small inner . (4) Or ν Etch ν small = ± R Etch outer R small outer ² 2 Ω Etch inner Ω small inner = 100 Ω Etch inner Ω small inner . (5) Note that to enforce the correct period of 25 seconds, we use eq. (3), remember- ing that we will have the same numerical value of f in both apparati (because the numerical values in the 4 slots of f are the same in both apparati) τ =1 / Ω inner f, (6) so Ω small inner Ω Etch inner = τ Etch τ small 25 / 10 = 2 . 5 . (7) Using this ratio of the inner cylinder speeds in eq. (5), we obtain ν Etch ν small = 40 . (8) Equations (5) and (8) along with the parameter values in the small apparatus give Ω Etch inner =0 . 4 radians/second and ν Etch =0 . 4 cm 2 /sec. No experiments are needed.
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ME 106 - Saturday Professor Philip S Marcus ME 106...

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