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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
- 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

- For system pressure we use P=F÷A
- 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

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.

- Solve for Force

- 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.

- Solve for Force

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 pounds

- Solve for Force

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

- Solve for Force

120 pounds

40psi

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

- Solve for Area

- Cylinder four
- Solve for Area
- A = F ÷ P
- A = 100 pounds ÷ 40 psi
- Cancel pounds to get in² and divide
- A = 2.5 in²

- Solve for Area

2.5 in²

Homework

- Read chapter

Lab

- Bottle Jack Lab

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