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# me309test - ME 309 Fluid Mechanics Examination 1 Name 13...

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Unformatted text preview: ME 309 Fluid Mechanics Examination # 1 Name ______________________ 13 Feb. 2008 PURDUE UNIVERSITY School of Mechanical Engineering ME 309 Examination 1 Wednesday February 2008 7:00 – 9:00 PM FRNY G140 Closed Book INSTRUCTIONS This examination consists of four problems, each of which is worth 25 points. The breakdown of specific parts may be given on some problems. Work neatly and completely to receive full credit . Use proper format with coordinate system(s), control volume(s), governing equation(s), assumptions, and answer(s) (with units) labeled clearly. State all numerical answers to three significant figure precision. Begin each problem in the space provided on the examination sheets and/or the following blank pages. Request additional paper if the number of pages provided for any problem is insufficient. Work on one side of each sheet only. Write your name on all sheets you hand in . If you remove the staple from the packet handed out or include additional sheets place the problem solutions in order prior to turning in. Place all solution sheets for any given problem after the examination sheet for that problem. Hand in test papers with this sheet on top. Please identify your section number by placing an X in the appropriate box below: Section Section 1 (Merkle) Section 2 (Fleeter) PERFORMANCE 1 2 3 Problems 4 Total ME 309 Fluid Mechanics Examination # 1 Name ______________________ 13 Feb. 2008 Problem # 1 GIVEN: Two mercury manometers are used to measure the pressure in the same vessel. The manometers are identical except for the diameter of the leg that is open to atmosphere. The diameter of the atmospheric leg of Manometer, L, on the left is d R , while the diameter of the atmospheric leg of Manometer, R , on the right is d R . Both manometers have precisely the same volume of mercury. The density of mercury is hg ρ . The density of the working fluid is fl ρ . Find: An expression for the mercury levels, 1 L Z and 2 L Z , in Manometer L as a function of the diameter ratio R L d d / and the corresponding manometer readings, 1 R Z and 2 R Z , in Manometer R. It is sufficient to express the solution for 1 L Z and 2 L Z as a two-by-two matrix system for the unknowns. It is not necessary to solve for the variables explicitly. z L1 z R1 Reference Manometer (Note: Corners are to be taken as rounded.) Mercury Working Fluid P p d L d R z 0 z 0 z L2 z R2 ME 309 Fluid Mechanics Examination # 1 Name ______________________ 13 Feb. 2008 Solution: Manometer R measures the pressure in the vessel as: ( ) ( ) atm R fl R R hg p p z z g z z g p +- +- = 1 1 2 ρ ρ Manometer L measures the pressure in the vessel as: ( ) ( ) atm L fl L L hg p p z z g z z g p +- +- = 1 1 2 ρ ρ Equate the two pressures, cancel atmospheric terms and eliminate g : ( ) ( ) ( ) ( ) 1 1 2 1 1 2 z z z z z z z z L fl L L hg R fl R R hg- +- =- +- ρ ρ ρ ρ Use conservation of mass to equate the volume of mercury (after cancelling pi/4): L L L L R R R R z d z d z d z d 1 2 2 2 1 2...
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me309test - ME 309 Fluid Mechanics Examination 1 Name 13...

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