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MAGM 262

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

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  1. MAGM 262 Hydraulic Fundamentals Mr. Conrado

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

  3. 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.

  4. 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”.

  5. 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.

  6. 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.

  7. 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.

  8. 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

  9. 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.

  10. 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.

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. Homework • Read chapter

  17. Lab • Bottle Jack Lab

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