## Introduction to Hydraulic Principles

We all know that hydraulic principles are demonstrated when you use a liquid under controlled pressure to do work. There are laws that state the action of liquids under conditions of changing flows, increasing and decreasing pressures. The student must be able to state and to understand these laws in order to become successful as a heavy equipment technician.

## Using a Liquid

There are several advantages for using a liquid.

- Liquids conform to the shape of the container.

- Liquids are practically incompressible.

- Liquids apply pressure in all directions.

Illustration 1 | g01059986 |

## Liquids Conform to Shape

Liquids will conform to the shape of any container. Liquids will also flow in any direction through lines and through hoses of various sizes and various shapes.

## Practically Incompressible

Illustration 2 | g01059996 |

A liquid is practically incompressible. When a substance is compressed, there is less space. A liquid will occupy the same amount of space or volume even when under pressure. The space or volume that any substance will occupy is called displacement.

## Gas is compressible

Illustration 3 | g01060001 |

Gas is compressible. When the gas is compressed, it takes up less space, and the displacement becomes less. The space which has been occupied by the gas may be occupied by another object. Therefore, a liquid is best suited for the hydraulic system, because it continually occupies the same volume or the same displacement.

## Hydraulics Doing Work

Illustration 4 | g01060003 |

According to Pascal's Law, "Pressure exerted on a confined liquid is transmitted undiminished in all directions and acts with equal force on all equal areas." Therefore, a force exerted on any part of an enclosed hydraulic oil system transmits equal pressure in all directions throughout the system.

In the above example, a 227 kg (500 lb) force acting upon a piston with a 5.08 cm (2 inch) radius creates a pressure of approximately 280 kPa (40 psi) in a confined liquid. The same 280 kPa (40 psi) acting upon a piston with a 7.62 cm (3 inch) radius supports a 574 kg (1130 lb).

Illustration 5 | g01060005 |

Pascal's Law

A simple formula allows you to determine the Force, the Pressure, and the Area when two of the three are known. Understanding these terms is necessary in order to understand the fundamentals of hydraulics.

Force is the push or the pull that acts upon a body. Force is usually expressed in pounds (lbs) or in Newtons (N). Force is equal to the pressure times the area (F = P × A ).

Pressure is the force of a fluid per unit area,which is expressed in (psi) or in kilopascal (kPa) .

The area is a measurement of surface space. The area is calculated in square inches or in square centimeters. Sometime the surface area is referred to as the effective area. The effective area is the total surface that is used to create a force in the desired direction.

The surface area of a circle is calculated with the following formula:

- Area = Pi (3.14) times radius squared
- If the radius of the circle is 2 inches, see Illustration 4.
- A = Pi × R square
- A = 3.14 X (2 inch × 2 inch) or A = 3.14 × (5.08 cm × 5.08 cm)
- A = 12.5 in
^{2}or A = 81 cm^{2}

With the knowledge of the surface area, it is possible to determine how much system pressure it will take in order to lift a given weight. Pressure is the force per unit that is expressed in (psi) or (kPa).

If a force of 227 kg (500 lb) is acting upon an area of 81 cm^{2} (12.5 in^{2}), the pressure created would be 280 kPa (40 psi).

The pressure is calculated with the following formula:

- Pressure = Force divided by Area
- P = 500 lbs/12.5 in
^{2}or P = 227 kg/81 cm^{2} - P = 40 psi. or P = 2.8 kg/cm
^{2}or P = 280 kPa

Solving for the large piston in illustration 4 we find:

- Pressure × Area = Force
- 40 × (3 x 3) × 3.14 = Force or 2.8 × (7.62 × 7.62) × 3.14 = Force
- 40 × 28.26 = 1130 lbs. or 2.8 × 183 = 514 kg

## Mechanical Advantage

Illustration 6 | g01060006 |

Illustration 6 demonstrates how liquid in a hydraulic system provides a mechanical advantage.

All cylinders are connected. All of the areas must be filled before the system pressurizes.

Use the hydraulic formula and then calculate the items in the question. Cylinders are counted from left to right.

When the pressure in the system is calculated, use the two known values of the second cylinder from the left. The formula that is used, pressure equals force divided by area.

Pressure = Force/Area

Pressure = 50 lbs/1 in^{2} or Pressure = 22.7 kg/6.45 cm^{2}

Pressure = 50 psi or Pressure = 3.50 kg/cm^{2 } or Pressure = 350 kPa

When you find out the pressure in the system, you can calculate the force of the load for cylinder one and cylinder three, and the piston area for container four.

Calculate both cylinders with (?) on weight blocks using the formula, force equals pressure times area (Force = Pressure x Area) .

To calculate the cylinder piston area of the piston on the right, you use the following formula, area equals force divided by pressure (Area = Force / Pressure) .

## Orifice Effect

Illustration 7 | g01060007 |

When hydraulics is discussed, it is a common practice to use the term, pump pressure. However, the pump does not produce pressure. The pump produces flow. Pressure is produced when flow is restricted.

In Illustration 7 and Illustration 8, the pump flow through the pipe is 1 gpm.

In Illustration 7, there is no restriction to the flow through the pipe. Therefore, the pressure reading is zero for both gauges.

Illustration 8 | g01060008 |

## Orifice Offers Restriction

An orifice offers a restriction to the pump flow. When oil flows through an orifice, pressure is produced on the upstream side of the orifice.

In Illustration 8, there is an orifice in the pipe between the two gauges. The gauge up the stream of the orifice shows that a pressure of 207 kPa (30 psi) is needed to send a flow of 3.79 L/min (1 US gpm) through the orifice. There is no restriction to flow after the orifice. The gauge down the stream of the orifice shows 0 pressure.

## Oil Flow to Tank is Blocked

Illustration 9 | g01060009 |

When the end of either pipe is plugged, oil flow to the tank is blocked.

The positive displacement pump will continue to pump at 3.79 L/min (1 US gpm) and fills the pipe. When the pipe is filled, the resistance to any additional flow into the pipe produces pressure. The pressure reaction is the same as Pascal's Law which states that pressure exerted on a confined liquid is transmitted undiminished in all directions and acts with equal force on all equal areas. The two-gauge readings are the same.

The pressure will increase until the pump flow is diverted from the pipe to another circuit or to the tank. This diversion is usually done with a relief valve in order to protect the hydraulic system.

If total pump flow was not diverted from the pipe, pressure in the pipe would continue to rise and cause an eruption of the circuit.

## Restrictions In Series

Illustration 10 | g01060010 |

The two basic types of circuits are; the series and the parallel.

In Illustration 10, a pressure of 620 kPa (90 psi) is required to send 3.79 L/min (1 US gpm)1 gpm through either circuit.

In a hydraulic circuit, orifices or relief valves in series offer a resistance that is similar to an electrical circuit resistors in series. The oil, like the electrical current, must flow through each resistance. The total resistance is equal to the sum of each individual resistance.

## Restrictions In Parallel

Illustration 11 | g01060012 |

In a system with parallel circuits, pump oil follows the path of least resistance. In Illustration 11, the pump supplies oil to three parallel circuits. Circuit three has the lowest priority and circuit one has the highest priority.

When the pump oil flow fills the passage to the left of the three valves, pump oil pressure increases to 207 kPa (30 psi). The pump oil pressure opens the valve to circuit one and the oil flows into the circuit. When circuit one is filled, the pump oil pressure begins to increase. When the pump oil pressure increases to 414 kPa (60 psi) the circuit two valve will open. The pump oil pressure will not continue to increase until circuit two is filled. In order to open the valve to circuit three, the pump oil pressure must exceed 620 kPa (90 psi).

There must be a system relief valve in one of the circuits or at the pump in order to limit the maximum pressure in the system.