Lecture notes-chapter09-posted

Lecture notes-chapter09-posted - 1 Chapter 9 Fluids •...

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Unformatted text preview: 1 Chapter 9: Fluids • Introduction to Fluids • Pressure • Pascal’s Principle • Gravity and Fluid Pressure • Archimedes’ Principle • Continuity Equation • Bernoulli’s Equation 2 §9.1 Fluids A liquid will flow to take the shape of the container that holds it. A gas will completely fill its container. Fluids are easily deformable by external forces. A liquid is incompressible . Its volume is fixed and is impossible to change. 3 §9.2 Pressure Pressure arises from the collisions between the particles of a fluid with another object (container walls for example). There is a momentum change (impulse) that is away from the container walls. There must be a force exerted on the particle by the wall. 4 Pressure is defined as . A F P = The units of pressure are N/m 2 and are called Pascals (Pa). Note: 1 atmosphere (atm) = 101.3 kPa By Newton’s 3 rd Law, there is a force on the wall due to the particle. 5 Example (text problem 9.1): Someone steps on your toe, exerting a force of 500 N on an area of 1.0 cm 2 . What is the average pressure on that area in atmospheres? atm 49 Pa 10 013 . 1 atm 1 N/m 1 Pa 1 N/m 10 . 5 m 10 1.0 N 500 5 2 2 6 2 4- av = × × = × = = A F P 2 4 2 2 m 10 . 1 cm 100 m 1 cm . 1- × = 6 §9.3 Pascal’s Principle A change in pressure at any point in a confined fluid is transmitted everywhere throughout the fluid. (This is useful in making a hydraulic lift.) 7 Apply a force F 1 here to a piston of cross- sectional area A 1 . The applied pressure is transmitted to the piston of cross-sectional area A 2 here. 8 Mathematically, 1 1 2 2 2 2 1 1 A A A F A F 2 point at 1 point at F F P P = = ∆ = ∆ 9 Example: Assume that a force of 500 N (about 110 lbs) is applied to the smaller piston in the previous figure. For each case, compute the force on the larger piston if the ratio of the piston areas (A 2 /A 1 ) are 1, 10, and 100. F 2 1 500 N 10 5000 N 100 50,000 N 1 2 A A Using Pascal’s Principle: 10 The work done pressing the smaller piston (#1) equals the work done by the larger piston (#2). 2 2 1 1 d F d F = 11 Example: In the previous example, for the case A 2 /A 1 = 10, it was found that F 2 /F 1 = 10. If the larger piston needs to rise by 1 m, how far must the smaller piston be depressed? m 10 2 1 2 1 = = d F F d Using the result on the previous slide, Question 12 • A hydraulic lift is designed with an input cylinder that has 1/5 the diameter of the output cylinder. If a force F is applied to the input cylinder, the output cylinder can supply a force equal to _______. • A) 1/5 F • B) 1/25 F • C) 5 F • D) 25 F 13 §9.4 Gravity’s Effect on Fluid Pressure A cylinder of fluid FBD for the fluid cylinder P 1 A P 2 A w 14 Apply Newton’s 2 nd Law to the fluid cylinder: ( 29 gd P P gd P P gd P P g Ad A P A P w A P A P F ρ ρ ρ ρ + = =- ∴ =-- =-- =-- = ∑ 1...
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This note was uploaded on 12/23/2011 for the course PHY 2130 taught by Professor Rehse during the Fall '08 term at Wayne State University.

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Lecture notes-chapter09-posted - 1 Chapter 9 Fluids •...

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