hw11soln - ME 375 HOMEWORK #11 Solution Fall 2009 PROBLEM 1...

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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 12 () d A xA x 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: 11 22 R qpp R qp p = = From the expression for x above: A x p k = Thus, 2 1 2 A Ax p q q p p p p kR R == = . Rewriting, this gives 2 21 A p pp p R += + ± 2 1 RA p p k + ± To find the equivalent fluid capacitance, we need the following relationship: f qA x C p p ± Substituting x ± from above, we get 2 f A C k =
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2 (b). Given : Find : Differential equation relating P 3r to P s Solution : Continuity: 1c2 c12 QQQ =+⇒=− Compatibility: r1 12 23 3cr PPPP 0 +++ = r1 12 23 3Rr 0 = Elemental Equation: r1 1r s 1 12 1 1 23 3cr c 3Rr 2 2 f PPP dQ PR Q , PI dt 1 PQ d
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This note was uploaded on 12/22/2011 for the course ME 375 taught by Professor Meckle during the Fall '10 term at Purdue University-West Lafayette.

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hw11soln - ME 375 HOMEWORK #11 Solution Fall 2009 PROBLEM 1...

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