i. Module 4 Fluids and Thermodynamics ... - Workforce 3 One

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You can determine the different values for Work, Force, Pressure, and Distance forhydraulics at HyperPhysics: http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html#hpressHyperPhysics Home Pagehttp://hyperphysics.phyastr.gsu.edu/hbase/hframe.htmlHydraulic Presshttp://hyperphysics.phyastr.gsu.edu/hbase/pasc.html#hpressClick on Calculation or Scroll downYou can enter values in either U.S. or metricdimensions:r: Radius of the pistonsA: Areas of the pistonsIt will calculate the force multiplicationThen enter values for the input force (F 1 )It will calculate the Pressure (P) and theoutput force (F 2 )Enter the distance for the output distance (d 2 )It will calculate the input distance (d 1 )Note if you scroll down the page, an explanationof the Automobile Hydraulic Lift if provided.Module 4 Student Guide Fluids & Thermodynamics Page 16
How does a Hydraulic Jack Work?Hydraulic jacks work on the basis of Pascal's Principle.Pressure is described mathematically by a Force divided by Area. Thereforeif you have two cylinders connected together, a small one and a large one,and apply a small Force to the small cylinder, this would result in a givenpressure. By Pascal's Principle, this pressure would be the same in thelarger cylinder, but since the larger cylinder has more area, the forceemitted by the second cylinder would be greater. This is represented byrearranging the pressure formula P = F/A, to F = PA. The pressure stayedthe same in the second cylinder, but Area was increased, resulting in alarger Force. The greater the differences in the areas of the cylinders, thegreater the potential force output of the big cylinder.A hydraulic jack is simply two cylinders connected as shown above.Assume a small piston (one squareinch area) applies a weight of 1 lb. toa confined hydraulic fluid.That provides a pressure of 1 lb. persquare inch throughout the fluid.If another larger piston with an areaof 10 square inches is in contact withthe fluid, that piston will feel a forceof 1 lbs/square inch x 10 squareinches = 10 lbs.So we can apply 1 lb. to the small piston and get 10 lbs. of force to lift a heavy object withthe large piston. Is this 'getting something for nothing'? Unfortunately, no. Just as a leverprovides more force near the fulcrum in exchange for more distance further away, thehydraulic lift merely converts work (force x distance) at the smaller piston for the SAMEwork at the larger one. In the example, when the smaller piston moves a distance of 10inches it displaces 10 cubic inches of fluid. The 10 cubic inches displaced at the 10 squareinch piston moves it only 1 inch, so a small force and larger distance has been exchangedfor a large force through a smaller distance.You will have to operate the handle 10 times to lift the heavy object 10 inches.Module 4 Student Guide Fluids & Thermodynamics Page 17
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