hw11soln

# hw11soln - ME 375 HOMEWORK#11 Solution Fall 2009 PROBLEM...

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ME 375 HOMEWORK #11 Solution Fall 2009 PROBLEM 1 (40%) (a). Given : Prob. 7.30 in Palm textbook. Solution : In the absence of hydrostatic pressure, the gage pressure at the bottom of the plate is p . Thus, Newton’s law gives: mx pA kx = ±± With the plate mass small, this leads to pA pA kx x k = = . Continuity requires that 1 2 ( ) d Ax Ax q q dt = = ± , where q 1 is the flow through the left resistor, and q 2 is the flow through the right resistor. The elemental equations for the resistors are given by: 1 1 2 2 Rq p p Rq p p = = From the expression for x above: A x p k = ± ± Thus, ( ) ( ) 2 1 2 1 2 1 1 A Ax p q q p p p p k R R = = = ± ± . Rewriting, this gives ( ) 2 1 2 2 1 A p p p p k R R + = + ± ( ) 2 1 2 1 2 2 RA p p p p k + = + ± To find the equivalent fluid capacitance, we need the following relationship: f q Ax C p p = = ± ± ± Substituting x ± from above, we get 2 f A C k =

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