lec13-203-11-fluidspost4-13-11

# lec13-203-11-fluidspost4-13-11 - F P A P gh liq disp B V g...

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Unformatted text preview: F P A P gh liq disp B V g m V Av const. 2 1 2 P v const. 13-0 - density = M kg V m 3 Pressure: P = = F force A area units: = Pa N m 2 or or lb in 2 lb ft 2 Atmospheric pressure 1 Atm = 14.7 1 Atm = 1.01 x 10 5 Pa lb in 2 13-1 Fluid - Statics 13-1a sample P Area, A 2 F 2 P A 2 F 1 PA 1 Pressure is transmitted through fluid without loss F 1 (Total) = F 1- P A 1 F 2 (Total) = F 2- P A 2 Piston at rest : 0 = F 1- PA 1 0 = F 2- P A 2 F 1 F 2 A 1 A 2 = P = F 2 = F 1 A 2 A 1 Pascal’s Principle 13-2 F s F L A s A L = P = F s = ( ) F s A L A s This factor can be made to be 1,000 or more. - Hydraulic Press - Hydraulic Jack Applications of Pascal’s Principle 13-3 P ATM h A Pressure Force per unit Area Any arbitrary cylinder V=Ah Pressure in a Liquid (or Fluid) vacuum F t =0 mg V=Ah mg= (Ah) g F h = P h A V F(total) = -F t-mg+ F h top 0 = -0- Ahg + P h A P h = hg vac fluid at rest Pressure due to weight of water on top If P o added at top (instead of vacuum) it is transmitted to bottom. P h = P o + hg 13-4 13-5 Note: h=0 at surface and all h’s are + #’s 13-5 Torricelli Barometer 13-6 A barometer compares the pressure due to the atmosphere to the pressure due to a column of fluid, typically mercury. The mercury column has a vacuum above it, so the only pressure is due to the mercury itself....
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lec13-203-11-fluidspost4-13-11 - F P A P gh liq disp B V g...

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