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# MAGM 262 - PowerPoint PPT Presentation

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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Hydraulic Fundamentals

• Hydraulic systems are everywhere from:

• Large excavation equipment

• Steering in your car

• Shocks

• Power trains

• 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.

• 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”.

• 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.

• 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.

• 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.

• 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

• 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.

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.

• 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

• 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

• 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

• Cylinder four

• Solve for Area

• A = F ÷ P

• A = 100 pounds ÷ 40 psi

• Cancel pounds to get in² and divide

• A = 2.5 in²

• 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