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MAGM 262. Hydraulic Fundamentals. Mr. Conrado. Hydraulic Fundamentals. Hydraulic systems are everywhere from: Large excavation equipment Steering in your car Shocks Power trains. Hydraulic Fundamentals. Using liquids to transfer force They conform to their container

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MAGM 262

Hydraulic Fundamentals

Mr. Conrado


Hydraulic fundamentals
Hydraulic Fundamentals

  • Hydraulic systems are everywhere from:

    • Large excavation equipment

    • Steering in your car

    • Shocks

    • Power trains


Hydraulic fundamentals1
Hydraulic Fundamentals

  • Using liquids to transfer force

    • They conform to their container

    • Practically incompressible

    • Apply pressure in all directions

    • Flow in any direction through lines and hoses.


Hydraulic fundamentals2
Hydraulic Fundamentals

  • Liquids for all practical purposes are incompressible.

    • When a substance is compressed it takes up space. A liquid does not do this even under large pressures.

    • The space any substance occupies is called “displacement”.


Hydraulic fundamentals3
Hydraulic Fundamentals

  • Gases are compressible

    • When a gas is compressed it takes up less space and its displacement is less. For this reason liquids are best used for hydraulic systems.


Hydraulic fundamentals4
Hydraulic Fundamentals

  • Hydraulics doing work.

    • Pascal’s law – “ Pressure exerted on a confined liquid is transmitted undiminished in all directions and acts as a equal force on all equal areas.”

    • Thus a force exerted on any part of a confined liquid the liquid will transmit that force (pressure) in all directions within the system.

      • In this example a 500 pound force acting upon a piston with a 2 inch radius creates a pressure of 40 psi on the fluid.

      • This same liquid with a pressure of 40 psi acting on a piston with a 3 inch diameter can support 1130 pounds.


Hydraulic fundamentals5
Hydraulic Fundamentals

  • Pascal’s Law

    • To understand how this works we must understand a very simple but fundamental formula.

    • To find one of the three areas two of the others must be known.

      • Force – The push or pull acting on a body usually expressed in pounds.

      • Pressure – The force of the fluid per unit area. Usually expressed in pounds per square inch or psi.

      • Area – A measure of surface space. Usually calculated in square inches.

      • To calculate the area of a circle use the formula Area = Pi (3.14) x radius squared.

        • Ex: For a 2” diameter piston A=3.14x(2”x2”) or A= 12.5 sq. in.


Hydraulic fundamentals6
Hydraulic Fundamentals

  • Pascal’s Law

    • With the knowledge of the surface area it is possible to determine how much system pressure will be required to lift a given weight.

    • The pressure needed for a 500 pound given weight is calculated with the formula

      • Pressure = Forced ÷ Area

      • P = 500lbs ÷ 12.5 Sp. In.

      • P = 40 psi


Hydraulic fundamentals7
Hydraulic Fundamentals

  • Mechanical Advantage

    • Here we see and example of how a hydraulic system can create a mechanical advantage.

    • We can calculate the items in question by using the systems known items and Pascal’s law.

      • For system pressure we use P=F÷A

        • So P=50lps÷1sq.in (cylinder #2)

        • P= 50psi

    • Now we know the system pressure we can calculate the load force for cylinders 1 & 3 and the piston area for 4. Do so on a separate piece of paper and wait for instructions.


Hydraulic fundamentals8
Hydraulic Fundamentals

40psi

  • Cylinder One

    • Solve for Force

      • F=P x A

      • F= 40psi x 5 in²

      • Cancel out square inches to leave pounds and multiply

      • F = 200lbs.


Hydraulic Fundamentals

  • Cylinder One

    • Solve for Force

      • F=P x A

      • F= 40psi x 5 in²

      • Cancel out square inches to leave pounds and multiply

      • F = 200lbs.

200 pounds


Hydraulic fundamentals9
Hydraulic Fundamentals

  • Cylinder Three

    • Solve for Force

      • F=P x A

      • F = 40psi x 3in²

      • Cancel out square inches to leave pounds and multiply

      • F = 120 pounds

40psi


Hydraulic Fundamentals

  • Cylinder Three

    • Solve for Force

      • F=P x A

      • F = 40psi x 3in²

      • Cancel out square inches to leave pounds and multiply

      • F = 120 pounds

120 pounds

40psi


Hydraulic fundamentals10
Hydraulic Fundamentals

  • Cylinder four

    • Solve for Area

      • A = F ÷ P

      • A = 100 pounds ÷ 40 psi

      • Cancel pounds to get in² and divide

      • A = 2.5 in²


Hydraulic Fundamentals

  • Cylinder four

    • Solve for Area

      • A = F ÷ P

      • A = 100 pounds ÷ 40 psi

      • Cancel pounds to get in² and divide

      • A = 2.5 in²

2.5 in²


Homework
Homework

  • Read chapter


Lab

  • Bottle Jack Lab


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