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Unformatted text preview: 1 The physics of fluid Human circulatory system Hydraulic lift: an application of Pascals principle Balloon These pictures show the phenomena and wonderful applications of the physics of fluid (fluid mechanics) 2 Chapter 15: Fluid mechanics gh P P + = fluid in immersed volume object fluid Buoy gV F = 2 2 1 1 A A = t cons gy P tan 2 1 2 = + + Outline  1. Pressure in fluids and Pascals principle:  1) pressure P : definition and units  2) pressure in fluids: 3) Pascals principle: a change in pressure to an enclosed fluid applies to everywhere inside the liquid with the same amount. 2. Buoyancy and Archimedes principle : 3. Fluid in motion:  1) equation of continuity:  2) Bernoullis equation: 4. Ability to solve real problems 3 1. Pressure in fluids and Pascals principle (1) pressure P : , A F P = units: (a) N/m 2 or Pa , 1 Pa = 1 N/m2, b) atm , 1 atm = 1.013 10 5 Pa, 1 bar = 1.000 10 5 Pa, (1 atm = 1.013 bar) c) bar , e) torr , 1 torr = 1 mmHg = 133 Pa. 1 atm = 760 mmHg, d) mmHg , (2) pressure P in fluids: (i) fluids exerts pressure in all directions. (ii) the pressure P in fluids depends on fluid (fluid density) and h (depth) (at the same depth, the pressure is the same) F A , Ahg g m A P PA F fluid fluid = = = , gh P P fluid + = , gh P P fluid + = definition: the force F over the contact area A . h P = P + fluid gh fluid air P Note: for water, H2O = 1.0 10 3 kg/m 3 , if h = 10 m , H2O gh = 1.0 10 3 9.8 10 1 atm.  hi, this means each 10 m in depth under water adds a pressure of ~ 1 atm ! So, if d = , P is the pressure outside the fluid, and h counts from the surface of the fluid, i.e. h is the depth under the fluid  4 Application examples Example 1: (3) Pascals principle: a change in pressure to a confined fluid applies throughout everywhere in the fluid by the same amount. Example 2: Example 3: . gh P P P P fluid B A + = = = Solution: The car is lifted up in a workshop for repair. The left side cylinder tubing has a radius r 1 = 5 cm , and the right side tubing r 2 = 25 cm . Oil is filled between. If the car is 4000 lb , how much is the force F 1 on the left side needed to lift it up ? Solution: , 2 1 P P = (the oil level on both sides is essentially the same, negligible in difference) , 2 2 1 1 A F A F = , 25 1 25 5 2 2 2 1 2 2 2 1 2 1 2 1 = = = = = r r r r A A F F . 160 25 / 4000 25 / 2 1 lb F F = = = hi, the force needed to lift up a car is really small ! This is a water dam. How much is the force from the water (lets not consider the speed of the water flow for simplicity)? (note, w dam width, H water depth) , PA F = , dy w P PdA dF = = ), ( 2 2 y H g P gh P P O H O H + = + = [ ] , ) ( 2 dy w y H g P PdA dF O H + = = Solution: The tubing shown here on the left is actually a manometer, with one end open the atmosphere...
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This note was uploaded on 07/13/2008 for the course PHYS 6C taught by Professor Staff during the Spring '01 term at UCLA.
 Spring '01
 staff
 mechanics

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