Ch. 2 - Chapter 2 Fluid Power Basics Courtesy : Fluid Power...

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1 Chapter 2 Fluid Power Basics Courtesy : Fluid Power Circuits and Controls, John S.Cundiff, 2001 Introduction ± One of the underlying postulates of fluid mechanics is that, for a particular position within a fluid at rest, the pressure is the same in all directions. ± A second postulate states that fluids can support shear forces only when in motion. ± These two postulates define the characteristics of the fluid media used to transmit power and control motion.
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2 Introduction (contd. .) ± Traditional concepts such as static pressure, viscosity, momentum, continuity, Bernoulli’s equation, and head loss are used to analyze problems encountered in fluid power systems. ± The product of fluid mass density and the gravitational constant is the constant of proportionality between the depth of fluid in the container and the pressure acting at that depth in the fluid. ± P = ρ * g * h ² Where P = pressure ² ρ = fluid mass density ² h = depth of fluid ² g = gravitational constant Hydrostatic Pressure
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3 Hydrostatic Pressure (contd. .) ± The density of a fluid is often referred to in terms of specific gravity. ± Specific gravity, by definition, is the ratio of the specific weight of the fluid in question to that of water at standard conditions. Hydrostatic Pressure (contd. .) ± Modify the static pressure equation and introduce specific gravity as: ² P = S g ρ w gh ² Where S g = specific gravity ² ρ w = density of water ± Rearranging the terms we can solve for the specific gravity of fluid in the column ² S g = P / ρ w gh
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4 Hydrostatic Pressure (contd. .) ± Cavitation occurs when the pump does not completely fill with liquid. ± The incoming fluid is a mixture of gas and liquid. ± An important application of fluid power is the force multiplication that can be achieved with a hydraulic jack. Example Problem 2.1 ± The small cylinder of the jack shown below has a bore of 0.25” and the large cylinder has a bore of 4”. How much can be lifted if the jack handle is used to apply 10 lb f to the small cylinder (F s = 10 lb f )?
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5 Example Problem 2.1 (contd. .) ± Pressure developed: P s = F s / A s ² Where A s = area of small cylinder (in 2 ) ² A s = 0.25 2 π / 4 = 0.049 in 2 ² P s = 10 / 0.049 = 204 psi Example Problem 2.1 (contd. .) ± Pascal’s law: P l = P s ² Where P l = pressure on large cylinder (psi) ± Lift developed: ² F l = P l A l ³ Where A l = 4 2 π / 4 = 12.56 in 2 ³ F l = (204)(12.56) = 2560 lb f
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6 Example Problem 2.1 (contd. .) ± If the small cylinder bore is 0.125 in, how much can be lifted? ² A s = (0.125) 2 π / 4 = 0.0123in 2 ² P s = 10 / 0.0123 = 813 psi ² F l = (813)(12.56) = 10200 lb f ± A 10 lb f produces a 10,000-lb f lift. Fluid Statics
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Ch. 2 - Chapter 2 Fluid Power Basics Courtesy : Fluid Power...

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