Refresh The basic upon which all hydraulic systems operate is the relationship between area, force and pressure. P= F/A In hydraulics, pressure is commonly.

Refresh

• The basic upon which all hydraulic systems operate is the relationship between area, force and pressure.

• P= F/A

• In hydraulics, pressure is commonly expressed in pounds (of force) per square inch (of area), abbreviated “psi”.

Pascal’s Law

• The operation of all closed hydraulic systems using fluid under pressure is based upon a principle of physics known as Pascal’s Law.

• Pascal’s Law states that pressure on a confined fluid is transmitted equally and perpendicular to the entire surface of its container

Cylinder Construction

• To extend the piston rod, the oil entering the cap end port is pressurized.

• This oil pushes against the face of piston, and the piston rod extends with a force determined by pressure of the oil and the area of the piston face. (F = PA)

• To retract the piston rod (pull it back), oil under pressure is pumped into the rod end port.

• Now the oil has less area to push against, because the rod itself takes up part of the area of the piston.

• For any given pressure then, the force exerted by the piston rod will be less when it is retracted than when extended.

What force is exerted when retracting a 3” diameter piston

having a ½” diameter rod, if the pressure of the hydraulic

fluid is150psi?

• Area of the piston, Ap = 0.785(3)2 = 7.065 sq. in.

• Area of the rod, Ar = 0.785(0.5)2 = 0.196 sq. in.

• The area A, which hydraulic fluid actually presses against, is found by, A = Ap – Ar

• A = 6.869 sq. in.• Thus, F = PA

F = 150(6.869) = 1030.35 lbs.

Hydraulic Leverage

• A force applied to a surface develops a pressure, which depend on the amount of force and the size of the surface.

• If the surface now made to push against a confined fluid, the fluid is said to be pressurized.

• The pressure of the fluid will be the same as that of the surface.

• If the pressurized fluid is then made to push against a second surface, a second force will be developed which depends upon the amount of pressure and the size of the second surface.

• We can conclude, the ratio of the two forces is exactly the same as the ratio of the two areas,

• F2/F1= A2/A1

f1 f2

d1 d2

• By selecting appropriate sizes of hydraulic cylinders, we are able to develop hydraulic leverage and convert a given force to the level needed to perform some needed function.

• But how fast a piston rod would extend?• This shows to us, the rate of piston

extension or retraction is independent of pressure.

• Extension rate is normally expressed in inches per minute, which we write as ipm or in./min

• To determine how fast the rod extends we need to know the flow rate at which hydraulic oil is directed into the cylinder port.

• This is expressed in gallons per minute (GPM) and is usually, but not always, determine by the capability of the pump.

• We determine the rate at which it extends or retracts using calculation based upon volume, capacity and fluid flow.

Capacity

• A container is said to have a capacity of one gallon, or one ounce, or one litter, and it need have no specific shape.

• Each unit of capacity, however, corresponds to a volume which can be expressed in cubic units.

• One gallon = 231 cubic inches

• Volume, V = Area x Height

What will happened, if we doubling the piston

area?• Doubling the piston area doubles the force

exerted, but reduce the length of the rod by half.

Extension Rate

• Extension rate depends only upon the cylinder diameter and the flow rate (pumping) of the incoming oil.

• R = 231(g)/0.785d2

where R = Rod extension rate

g = oil flow rate (GPM)

d = cylinder diameter

Retraction Rate

• R = 231(g)/A

A = 0.785(dcyl)2 – 0.785(drod)

2

where R = Rod extension rate

g = oil flow rate (GPM)

A = Cylinder Area – Piston Area

Actuation Time

To find the time required for a piston rod to

extend or retract a specific distance, use a

three step approach

1. Solve for the extension or retraction rate in inches per minute (R)

2. Divide this figure by 60 to convert inches per second (Rs)

3. Solve for time in seconds using the formula:

T = D/Rs