Chapters_15

# Chapters_15 - The physics of fluid Balloon Hydraulic lift...

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1 The physics of fluid Human circulatory system Hydraulic lift: an application of Pascal’s principle Balloon These pictures show the phenomena and wonderful applications of the physics of fluid (fluid mechanics)

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2 Chapter 15: Fluid mechanics gh P P ρ + = 0 fluid in immersed volume object fluid Buoy gV F ρ = 2 2 1 1 υ υ A A = t cons gy P tan 2 1 2 = + + ρυ ρ Outline ---- 1. Pressure in fluids and Pascal’s principle: --- 1) pressure P : definition and units --- 2) pressure in fluids: --- 3) Pascal’s 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) Bernoulli’s equation: 4. Ability to solve real problems
3 1. Pressure in fluids and Pascal’s 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 mm-Hg = 133 Pa. 1 atm = 760 mm-Hg, d) mm-Hg , (2) pressure P in fluids: (i) fluids exerts pressure in all directions. (ii) the pressure P in fluids depends on ρ fluid (fluid density) and h (depth) – (at the same depth, the pressure is the same) F A , 0 Ahg g m A P PA F fluid fluid ρ = = - = , 0 gh P P fluid ρ + = , 0 gh P P fluid ρ + = definition: the force F over the contact area A . h P = P 0 + ρ fluid gh fluid air P 0 Note: for water, ρ H2O = 1.0 × 10 3 kg/m 3 , if h = 10 m , ρ 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 = 0 , P 0 is the pressure outside the fluid, and h counts from the surface of the fluid, i.e. h is the depth under the fluid ---

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4 Application examples Example 1: (3) Pascal’s principle: a change in pressure to a confined fluid applies throughout everywhere in the fluid by the same amount. Example 2: Example 3: . 0 gh P P P P fluid B A ρ + = = = 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 π π . 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 (let’s 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 0 0 y H g P gh P P O H O H - + = + = ρ ρ [ ] , ) ( 2 0 dy w y H g P PdA dF O H - + = = ρ Solution: The tubing shown here on the left is actually a manometer, with one end open the atmosphere and the other end is sealed with some gas. It is filled with liquid (density ρ fluid ).
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