Fluid Power Helpful Information

1. Hydraulics is a means of transmitting power. It may be used to multiply force or modify motions.

2. PASCAL'S LAW: Pressure exerted on a confined fluid is transmitted undiminished in ALL directions, and acts with EQUAL FORCE on all EQUAL AREAS and at right angles to them.

3. To find the area of a circle (eg. a piston), square the diameter and multiply by .7854. So, we have:

4. The force (pounds) exerted by a piston can be determined by multiplying the piston area (square inches) by the pressure applied (PSI). So, we have:

5. To determine the VOLUME (cubic inches) required to move a piston a given distance, multiply the piston area (sq.in.) by that distance (inches). Convert volume (cu.in.) to U.S.Gallons by dividing by 231. Often-used conversion factor: 231 cubic inches = 1.0 Gallon.

6. WORK is force acting through a distance: WORK = FORCE × DISTANCE. Example: Work (in-lbs) = Force (lbs) ×Distance (in.)

7. POWER is the rate of doing work or another way: 

8. One HORSEPOWER = 550 ft.lbs/sec. Thus, we have: Power = 33,000 ft.lbs/min. In other words, 33,000 lbs raised one foot in one minute. Other conversion factors: 1.0 H.P. (elec) = 746 watts; 1.0 HP 42.4 BTU/minute.

9. HYDRAULIC OIL serves as a lubricant and is practically NON-compressible. It will compress only about $\text{0.4%}$ at 1,000 PSI and $\text{1.1%}$ at 3,000 PSI at a temperature of 120 degrees F.

10. The weight of hydraulic oil may vary with a change in VISCOSITY ; however, 55 to 58 lbs/cu.ft. (density) covers the viscosity range from 150 SSU to 900 SSU at 100 degrees F.

11. Pressure at the bottom of a one-foot column of oil will be approximately 0.4 PSI. To find the approximate pressure at the bottom of any column of oil, multiply the height in feet by 0.4.

12. Atmospheric pressure equals 14.7 PSIA at sea level. Pressure gauge readings ("PSIG") do NOT include atmospheric pressure unless marked "PSIA", and are really 14.7 PSI less than the absolute pressure being exerted on the system. 

13. There must be a pressure drop (pressure difference) across an orifice (restriction) to cause fluid flow through it. Conversely, if there is no flow, there is no pressure drop. Now this does not mean that there is no pressure! It does mean that there is no DIFFERENCE in pressure, and Pascal's law is in effect.

14. A fluid is PUSHED, not drawn, into a pump. Although we use the term "suction" in regard to the inlet side of a pump or compressor, the working fluid enters due to a pressure pushing it.

15. A pump does not create pressure; it creates FLOW. Pumps used to transmit power are usually "positive displacement" types.

16. Pressure is caused by RESISTANCE TO FLOW. A pressure gauge indicates the workload at any given moment.

17. Fluid always takes the path of LEAST resistance.
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18. The SPEED of a cylinder is dependent upon 2 parameters: its SIZE (piston area) and the RATE OF FLOW into it. Notice the "pressure" exerted on the cylinder is NOT even mentioned here.

19. Fluid velocity through a pipe varies inversely as the SQUARE of the inside diameter. Doubling the I.D. increases the area 4 times, but velocity is cut to 1/4th of its original value.

20. Friction losses (pressure drop) of a fluid in a pipe varies directly with velocity. The faster the fluid is moving, the more the pressure loss due to friction there is (more work for pump!).
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21. To find the actual area of a pipe (and thus, size or diameter) needed to handle a given flow, use the formula:

Or

Then, knowing area, find D (diameter) from #3 above.

22. In standard pipe, the actual inside diameter is usually larger than the nominal size quoted. A standard conversion chart should be used when selecting pipe sizes.

23. Steel and copper tubing size indicates the outside diameter. To find the actual inside diameter (fluid carrying capacity), subtract 2 times the wall thickness from the tube size quoted.

24. Hydraulic hose sizes are usually their nominal inside diameters. With some exceptions among manufacturers, this is given by a dash number showing the number of sixteenth-inch increments in their inside diameters.
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25. To find the H.P. required for a given flow rate (GPM) at a known pressure (PSI) in a system, use the formula:

26. The H.P. (motor size) needed to drive (input H.P.) a pump will be somewhat higher than computed system H.P.requirements since the pump is not 100% efficient. So we modify the above formula (#26) to account for pump efficiency to read:

27. It is often desirable to convert back and forth from horsepower to torque without computing pressure and flow. This can be done for any rotating equipment by using the general torque-power formulas:

or

Note that torque in these formulae must be in "inch-pounds".
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28. To find the pump capacity (GPM) required to extend a cylinder (piston area square inches) a given distance (inches) in a specific time (seconds), use the formula:

( Note how the units cancel in the above equation. This is called again, "Factor Label".)

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