1 / 17

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

lacy
Download Presentation

MAGM 262

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related