Hydraulic Fundamentals

Hydraulic Fundamentals• Hydraulic systems are

everywhere from:– Large excavation

equipment– Steering in our car– Shocks– Power trains

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 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 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 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 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 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 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 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.

40psi

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 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 pounds40psi

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 pounds40psi

120 pounds

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• $#%&^&(*^$@!@#$)*&%