ME 375 HOMEWORK #11 Solution Fall 2009PROBLEM 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: mxpAkx=−±±With the plate mass small, this leads to pApAkxxk=⇒=. Continuity requires that 12()dAxAxqqdt==−±, where q1is the flow through the left resistor, and q2is the flow through the right resistor. The elemental equations for the resistors are given by: 1122RqppRqpp=−=−From the expression for xabove: Axpk=±±Thus, ()()2121211AAxpqqppppkRR==−=−−−±±. Rewriting, this gives ()21221AppppkRR+=+±()212122RAppppk+=+±To find the equivalent fluid capacitance, we need the following relationship: fqAxCpp==±±±Substituting x±from above, we get 2fACk=
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