MAGM 262

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MAGM 262
Hydraulic Fundamentals
Mr. Conrado
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Hydraulic Fundamentals

• Hydraulic systems are everywhere from
• Large excavation equipment
• Steering in your car
• Shocks
• Power trains

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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.

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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.

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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.

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Hydraulic Fundamentals

• Hydraulics doing work.
• Pascals 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.

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Hydraulic Fundamentals

• Pascals 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 A3.14x(2x2) or
  A 12.5 sq. in.

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Hydraulic Fundamentals

• Pascals 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

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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 Pascals law.
• For system pressure we use PFA
• So P50lps1sq.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.

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Hydraulic Fundamentals
40psi

• Cylinder One
• Solve for Force
• FP x A
• F 40psi x 5 in²
• Cancel out square inches to leave pounds and
  multiply
• F 200lbs.

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Hydraulic Fundamentals

• Cylinder One
• Solve for Force
• FP x A
• F 40psi x 5 in²
• Cancel out square inches to leave pounds and
  multiply
• F 200lbs.

200 pounds
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Hydraulic Fundamentals

• Cylinder Three
• Solve for Force
• FP x A
• F 40psi x 3in²
• Cancel out square inches to leave pounds and
  multiply
• F 120 pounds

40psi
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Hydraulic Fundamentals

• Cylinder Three
• Solve for Force
• FP x A
• F 40psi x 3in²
• Cancel out square inches to leave pounds and
  multiply
• F 120 pounds

120 pounds
40psi
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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

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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
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Homework

• Read chapter

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Lab

• Bottle Jack Lab