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Practice Final Problem 3 solution

Practice Final Problem 3 solution - velocities The key is...

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Name_________________________ Page 1 of 1 12/8/2005 ( Spring 2001) A U-tube of uniform area A contains a liquid of density ρ and, on one side, a float of area A/ 2, height h and density / 2, ρ so that the equilibrium state is as shown on the left below. Consider only quite small displacements of the liquid levels and the float from this equilibrium, as pictured on the right below. Neglect viscous effects, and assume the axis of the float remains vertical and centered. a. Relate displacement 1 y to displacements 2 y and z through an equation . The volume of fluid displaced from the left side of the tube must be equal to the volume of fluid displaced from the right side of the tube. If the float is raised a height z , a void of volume 2 A z occurs under the float. The void is supplied by fluid from both sides of the manometer, thus 2 1 2 2 A A z y y A = + . Simplifying gives 1 2 2 y z y = - b. Draw an approximate linear bond graph model of the system, labeling appropriate generalized
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Unformatted text preview: velocities. The key is to remember how compliance works, in particular for a floating mass. Recall problem 3.1 and its solution to the left. (The causality assignments shown below are there for part d not for this portion of problem.) c. Relate the moduli of your bond graph elements to the given physical parameters. 2 2 fl A I m h = = man I LA = 2 T = /2 2/ fl man C C gA = = 1/ man C gA = Note the moduli given on the cover page are associated with P and V , not F and x as in our case. What is the order of your model? Four (first order equations). fl I T z d 1 fl C /2 man C 2 y d 2 z y-d d man I 1 man C 1 y d surface of liquid x F gAx = L area A area A/ 2 area A h equilibrium level z y 2 y 1 displacements from equilibrium liquid, density float z 2 y 1 y water level static water level center of mass...
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