AD Automation partnering with Enerpac, believe that hydraulic safety is a core value, and wish to help educate and show our customers safe ways of working. Dealing with hydraulic pressures of 700 bar can be dangerous, so we have some pages to help guide you as you develop your system.
This information is found the Yellow Pages section of the current Enerpac catalogue. Please feel free to contact us, and we can get one to you, or sit down and discuss your task.
|Flow||A hydraulic pump produces flow.||Pressure||Pressure occurs when there is resistance to flow.|
|Pascal's Law||Pressure applied at any point upon a confined liquid is transmitted undiminished in all directions (Fig.1).
||This means that when more than one hydraulic cylinder is being used, each cylinder will lift at its own rate, depending on the force required to move the load at that point (Fig. 2).
Cylinders with the lightest load will move first, and cylinders with the heaviest load will move last (Load A), as long as the cylinders have the same capacity.
To have all cylinders operate uniformly so that the load is being lifted at the same rate at each point, either control valves (see Valve section) or Synchronous Lift System components (see Cylinder section) must be added to the system (Load B).
|Force||The amount of force a hydraulic cylinder can generate is equal to the hydraulic pressure times the “effective area” of the cylinder (see cylinder selection charts).||Example 1
An RC-106 cylinder with 2.24 in2 effective area operating at 8,000 psi will generate what force?
Force = 8,000 psi x 2.24 in2 = 17,920 lbs.
An RC-106 cylinder lifting 14,000 lbs will require what pressure?
Pressure = 14,000 lbs ÷ 2.24 in2 = 6,250 psi.
An RC-256 cylinder with 5.15 in2 effective area is required to produce a force of 41,000 lbs. What pressure is required?
Pressure = 41,000 lbs ÷ 5.15 in2 = 7961 psi.
Four RC-308 cylinders each with 6.49 in2 effective area are required to produce a force of 180,000 lbs. What pressure is required?
Pressure = 180,000 lbs ÷ (4 x 6.49 in2) = 6933 psi.
Remember, since four cylinders are used together, the area for one cylinder must be multiplied by the number of cylinders used.
A CLL-2506 cylinder with 56.69 in2 effective area is going to be used with a power source that is capable of 7,500 psi. What is the theoretical force available from that cylinder?
Force = 7,500 psi x 56.79 in2 = 425,925 lbs.
Use this formula to determine either force, pressure or effective area if two of the variables are known.
|Cylinder Oil Capacity||The volume of oil required for a cylinder (cylinder oil capacity) is equal to the effective area of the cylinder times the stroke*.||Example 1
An RC-158 cylinder with 3.14 in2 effective area and an 8 in. stroke will require what volume of oil?
Oil Capacity = 3.14 in2 x 8 in = 25.12 in3
An RC-5013 cylinder has an effective area of 11.05 in2 and a stroke of 13.25 in. How much oil will be required?
Oil Capacity = 11.05 in2 x 13.25 in = 146.41 in3
An RC-10010 cylinder has an effective area of 20.63 in2 and a stroke of 10.25 in. How much oil will it require?
Oil Capacity = 20.63 in2 x 10.25 in = 211.46 in3
Four RC-308 cylinders are being used, each with an effective area of 6.49 in2 and stroke of 8.25 in. How much oil will be required?
Oil Capacity = 6.49 in2 x 8.25 in = 53.54 in3 for one cylinder. Multiply by four to obtain the required capacity: 214.17 in3
|* Note: these are theoretical examples and do not take into account the compressibility of oil under high pressure.