PNEUMATIC AND HYDRAULIC EQUIPMENT

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MODULE TITLE : APPLICATIONS OF PNEUMATICS AND

HYDRAULICS

TOPIC TITLE : PNEUMATIC AND HYDRAULIC EQUIPMENT

LESSON 4 : BASIC HYDRAULIC PRINCIPLES

APH - 2 - 4

Published by Teesside University Open Learning (Engineering)

School of Science & Engineering

Teesside University

Tees Valley, UK

TS1 3BA

+44 (0)1642 342740

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

In this lesson we deal with the fundamental principles behind the transmission

of power using liquid in a confined system. We will consider the production of

force, motion and power by a hydraulic system, and discuss the causes of

power losses in a hydraulic power transmission system, indicating ways by

which such losses can be reduced.

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YOUR AIMS ________________________________________________________________________________________

On completion of this lesson you should be able to:

• explain the advantages to be gained from using hydraulic power

transmission systems

• calculate the force, torque, and speed developed by a hydraulic motor

or cylinder for a given hydraulic power input

• calculate the amount of hydraulic power available from given

pressure and flow data

• calculate the size of a suitable system prime mover

• list and explain the prime causes of excessive fluid friction in a

hydraulic system and understand the consequences of such losses.

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THE FUNDAMENTAL PRINCIPLES OF HYDRAULICS ________________________________________________________________________________________

As far as we are concerned here, hydraulics is that branch of engineering

concerned with the transmission and control of power by means of a liquid in a

confined system.

Power is supplied to the hydraulic system by a prime mover which is usually

an electric motor, diesel or petrol engine. The mechanical energy supplied is

converted to hydraulic power by the pump, controlled by the valves of the

hydraulic system, then reconverted into mechanical work in a more usable

form by the hydraulic actuator at the end of the system.

FIGURE 1 illustrates these features that are common to every hydraulic power

transmission system.

FIG. 1 Power flow and energy conversion in a hydraulic system

Hydraulic control system

Output device

Prime mover

Hydraulic pump

Hydraulic actuator

Hydraulic energy

Mechanical energy

Hydraulic energy

Mechanical energy

Power flow:

Energy conversion:

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PRESSURE AND FORCE

In a hydraulic system pressure is created as resistance to flow is encountered.

If fluid is supplied to a hydraulic cylinder, then the resistance to flow imposed

by the piston will increase the pressure on the fluid, and result in a force being

applied to the piston. The size of this force will depend upon the maximum

pressure that can be generated in the system and the area of the piston.

For example, in the system shown in FIGURE 2, fluid at a pressure of 50 bar is

supplied to a hydraulic cylinder of diameter 200 mm. Calculate the force

exerted on the piston in newtons.

FIG. 2

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Force (newtons) = pressure (newtons metre–2) × area (metre2)

To convert bars to N m–2 multiply by 100 000

Calculate the force generated on the piston, if fluid at a pressure of 75 bar is applied to

a piston of 60 mm diameter.

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force (N) = pressure N m area m

= 75

–2 2( ) × ( )

× 105(( ) × ×( )

× × ×

π 0 06 4

10 2 83 10

21 208 5

5 3

–= 75

= N

= 11 2. kN

∴ ×( ) × ×( ) force (N) = 50 N m m–2100 000 0 2 2π . 22

= N

= kN

5000 000 0 03142

157 100

157 1

× .

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TORQUE

If fluid is supplied to a rotary or semi-rotary actuator, then this will result in a

turning force or torque being generated. The amount of torque generated can

be calculated by using the following formula:

For example, a hydraulic motor has a displacement of 200 cm3/rev. If the

pressure drop across the motor is 85 bar calculate the theoretical torque

generated by the motor.

It can be seen from the above examples that increasing system pressure will

increase the maximum amount of force or torque that can be produced by the

actuator. The maximum pressure that can be generated in a hydraulic system

must be closely controlled as excessive pressure in the system can cause

serious damage to pipework and components.

motor torque = pressure N m motor displac–2( ) × eement/rev m

( )

×( ) × ×( ) ×

T 85 10 200 10

2 3

5 6–

..142

270= N m

motor torque = pressure N m motor disp

–2

T ( ) × llacement/rev m

3( ) π

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What do you think would be the detrimental effects of running a system at an

excessively high pressure?

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The detrimental effects of running a system at an excessively high pressure are:

• failure of hoses, pipework and fittings, leading to high fluid leakage and potentially

dangerous occurrences

• premature failure of components

• high fluid temperatures, leading to rapid deterioration of hydraulic fluid

• power loss resulting in high operational costs.

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PISTON VELOCITY

If the pressure of the fluid acting on the actuator is responsible for providing

the force output, then the amount or volumetric flowrate of oil supplied to the

actuator will govern the speed at which it will operate.

The rate at which fluid flows into an actuator will control its velocity.

To calculate the velocity that a piston will travel for a given flowrate of oil, we

have to use the formula given below:

For example, suppose we want to calculate the velocity of a piston of diameter

50 mm, if supplied with fluid at a flowrate of 16 litres min–1.

From the above

Do note that this is a theoretical value which assumes that there is no leakage

of fluid across the piston seals.

flowrate = min = m s

–1 3 –1Q( ) 16 16 60 000

0 0

� /

. 0002666

0 052

m s

the piston area =

3 –1

A( ) ×( )π .

0 00196

= m

the piston velocity

∴ ( v)) ( )= m s = m s m

–1 3 –1

A v

0 0002666 0 00196

. .

00 136. m s–1

piston velocity m s = flowrate m s

–1 3

v Q( ) ( )

––1

2piston area mA( )

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If we supply the same amount of oil into the rod end of the cylinder, it will retract at a

higher velocity than it extends. Why?

...................................................................................................................................................

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The piston rod reduces the volume that has to be filled by the fluid to retract the rod,

therefore for the same flowrate the piston will return at a higher velocity.

To calculate the rotational speed of a motor, we must divide the flowrate (Q)

by the motor displacement per revolution, expressed in the same system of

units.

For example, suppose we want to calculate the rotational speed of a hydraulic

motor with a displacement of 23 cm3 rev–1, if it is supplied with fluid at a

flowrate (Q) of 15 � min–1.

This is also a theoretical speed, as it assumes no internal leakage. In practice

internal leakage will always be present and, dependant upon motor condition,

will vary with pressure drop across the motor and the viscosity of the fluid

used.

motor speed = fluid flowrate

motor displacementt per revolution

fluid flowrate = miQ( ) 15 � nn = cm min

displacement per revolu

–1 3 –115 000

ttion = cm s

the motor speed = c

3 –123

15 000 ∴

mm min cm rev

= rev min 3 –1

3 –1 –1

23 652

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So, if we increase flow into the motor, its speed will increase and, if we

increase the pressure of the fluid, we can use the motor to generate more

torque. As both flow and pressure can be infinitely variable up to the

maximum operating pressure and flow of the system, we can infinitely control

the force or torque and speed of an actuator, and therefore its power output.

This stepless control of power is one of the major advantages offered by a

hydraulic power transmission system.

What other advantages do you think can be gained by the use of hydraulic power

transmission systems?

...................................................................................................................................................

________________________________________________________________________________________

Other advantages to be gained by the use of hydraulic systems include:

• actuators (cylinders or motors) can be rapidly reversed

• the system is protected against overload

• actuators can be safely stalled, and will start up again immediately as soon as the load

is reduced, without any resetting of equipment.

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HYDRAULIC POWER

As we have seen, to allow the hydraulic cylinder or motor to do the work for

which it has been designed, it must be supplied with sufficient hydraulic power

in terms of pressure and flow. This power requirement will need to be

calculated, if we are to be able to correctly size the drive motor for the system.

Flowrate is measured in litres per minute

Example

A hydraulic pump delivers 90 litres min–1 of fluid against a pressure of 55 bar.

Calculate the hydraulic power developed and the size of the electric drive

motor required, if the overall pump efficiency is 65%.

(1 min = m s

hydraulic power (

–1 3 –1� 1 60 000/ )

wwatts) = (bar) ( min

= (

–1p Q

× ×100 000 60 000

� )

bbar) ( min

hydraulic power (

–1× ×

∴

10 6

Q � )

kkW) = (bar) ( min

= (bar)

–1p Q

p Q

× × ×

10 6 1000

� )

( min–1� ) 600

Hydraulic power = pressure flowrate

power is m

eeares in watts watt = Nm s

pressure i

–1( )1 1

ss measured in bar (1 bar = N m –2100 000 )

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Solution

Electric power of drive motor required (power input to system)

= ouput power required

overall efficiency

= 8 2. 55 0 65

12 7

= kW

Hydraulic power (kW) = bar min–1p Q( ) × ( )�

600

= kW

55 90 600

4950 600

8 25

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FRICTION AND FLOW

From the above example, it would seem that all of the hydraulic energy created

by the pump in the system is supplied to the actuator without any losses.

Unfortunately this is not the case in the real world: hydraulic energy cannot be

transmitted through pipelines without losses. Friction within the fluid (internal

friction) and between the fluid and the solid boundary (external friction) occurs

in all devices and lines in a hydraulic system through which liquid passes.

This pressure drop (∆p, delta p) is caused by friction in the pipe, which converts hydraulic energy into thermal energy, raising the temperature of the

fluid. System designers and manufacturers strive to keep the causes of this

flow friction to a minimum.

The amount of pressure drop is a function of the following factors:

• flow velocity dependent on the cross-section area of the pipeline and the

piston velocity or flowrate

• type of flow: laminar or turbulent

• viscosity of the oil dependent on temperature and pressure

• pipeline length and diameter

• internal surface condition of pipelines

• type and number of cross-sectional reductions in the system of pipelines

such as throttles and orifices

• pipeline arrangement: type and number such as bends and junctions.

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The further the fluid has to travel around the system the higher will be the

amount of friction and heat generated, and the greater the loss of hydraulic

power. The length of pipe runs should therefore be kept to a minimum, with

power-packs being sited as close as possible to the actuators using the

hydraulic power supplied.

Valves and fittings in the system will also act as points of constriction.

Bernoulli's principle states that, if the flowrate of the fluid is constant, the sum

of the kinetic energy of the flowing fluid, by virtue of its velocity and mass and

its potential energy in the form of its pressure and height, must also remain

constant. This means simply that, if one increases, the other must decrease in

proportion.

Therefore, as the fluid enters the restricted flow path created by the valve or

fitting, the fluid velocity will increase to maintain the flowrate of fluid around

the system. This will result in a drop in pressure of the fluid as it passes

through the constriction. On leaving the constriction, the fluid will return to its

original velocity and, although the pressure of the fluid will rise to a value

close to what it was prior to it entering the restriction, it will always be less due

to the power loss, caused by the increased friction and heat generated by the

increased velocity of the fluid passing through the constriction. FIGURE 3

below illustrates this effect.

FIG. 3 Pressure Changes due to Constriction

Q in

Q out

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If the outlet of the restriction shown in FIGURE 3 were closed and all flow through the

constriction ceased, what effect would this have upon the readings of the pressure

gauges?

...................................................................................................................................................

________________________________________________________________________________________

When the outlet of the restriction in FIGURE 3 is closed, there will be no flow in the

system, which becomes a static case. Pascal's law states that in a static system, pressure is

transmitted undiminished in all directions, and acts with equal force on equal areas, and at

right angles to them. This will therefore result in the pressure gauges in FIGURE 3 all

having the same maximum pressure reading.

Fluid velocity also affects the amount of pressure drop experienced by the fluid

in its path around the system. When fluid travels through a straight pipe at low

velocity, the fluid particles move in straight and parallel flow paths, and

friction losses are minimal. This condition is known as laminar flow and is

illustrated in FIGURE 4.

FIG. 4 Laminar Flow in a Pipe System

Gradual changes in cross-section

do not upset streamline flow

Gradual changes in direction do not upset

streamline flow

Low velocity flow in straight pipe is streamlined

Q in

Q out

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Poor flow conditions, created by sharp bends, rough internal pipe surfaces and

high flow velocity, will cause cross currents to be formed, which will result in

high energy losses and turbulent flow as illustrated in FIGURE 5.

FIG. 5 Turbulent Flow in a Pipe System

To avoid turbulence, the velocity at which the fluid travels through the

pipelines must be closely controlled. Typical fluid velocities are given below:

Suction lines 0.5 to 1.2 m s–1

Return lines 2 to 3 m s–1

Pressure lines (up to 50 bar) 4 m s–1

(50 to 100 bar) 4 to 5 m s–1

(100 to 200 bar) 5 to 6 m s–1

(over 200 bar) 7 m s–1

When sizing hydraulic pipework, it is common practice to use a nomogram of

the type shown in FIGURE 6.

Sudden changes in direction

will cause turbulence

Abrupt changes in cross-section will cause turbulence

Flow may start out streamlined

Q in

Q out

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FIG. 6 Hydraulic Pipe-sizing Nomogram

L min–1 mm m s–1

F lo

w -r

a te

P ip

e in

te rn

a l

d ia

m et

400

300

200

150

100

200

150

125

100

80 70 60 50

40 35 30

6 5

0.3

0.4

0.5

0.6 0.7 0.8 0.9 1.0

1.2

1.5

2.0

2.5

5 6 7

7 8 9 10

7 8 9

8 9

F lo

w v

el o ci

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Example

A hydraulic system has a flow rate (Q) of 50 � min–1 and a maximum working

pressure of 120 bar. Use the nomogram to obtain a pipe size suitable for the

suction, pressure and return lines of the system.

Solution

To use the nomogram to obtain suction line size, join a point, which coincides

with a flow rate of 50 � min–1 on the left-hand line of the nomogram, to a point

that coincides with 0.8 m s–1 flow velocity (mid-range value) on the vertical

line on the right-hand side of the nomogram. Where the line crosses the centre

axis, this will indicate a pipe diameter of 35 mm. The nomogram is used in the

same way to obtain pipe sizes for the pressure and return lines.

Use the nomogram to find pipes of a suitable size for the pressure and return lines

indicated in the example given above.

...................................................................................................................................................

________________________________________________________________________________________

Return line size using a flow velocity of 2.5 m s–1 is 20 mm diameter. Pressure line size

using a flow velocity of 5 m s–1 is 15 mm diameter. These are the recommended pipe sizes:

it may not be possible in practice to obtain a pipe of exactly that size. In such a case it is

usual to use the closest larger commercially-available pipe.

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Fluid of too high a viscosity will provide increased resistance to flow: then a

lot of energy will be expended by the system in just circulating the fluid. It is

therefore very important that fluid of the correct viscosity is used, and that the

temperature of the system is closely controlled, to ensure that the fluid remains

as close as possible to its desired viscosity during operation of the system.

How is fluid viscosity affected by change in temperature in a hydraulic system?

...................................................................................................................................................

________________________________________________________________________________________

As the operational temperature of the system rises, the viscosity of the fluid will reduce,

resulting in thinner oil. This will reduce flow resistance but increase internal leakage. If the

temperature falls, the oil will tend to thicken and flow resistance increase. This can have the

effect of increasing the pump suction pressure, resulting in cavitation and pump damage. It

is usual to try to maintain an operating temperature between 30°C and 60°C, but ideally it

should be kept as close as possible to 40°C.

As a summary, the four main causes of excessive fluid friction are:

• excessive length of pipelines

• excessive number of bends and fittings

• high fluid velocity

• high viscosity of fluid.

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FIGURE 7 shows the way that the fluid pressure reduces in its passage around

the system. The system shown is a poor example of a hydraulic system, with

excessive use being made of pipe fittings with tight bends; but it is not unusual

to come across such systems in practice.

Note that it is only the pressure drop of the fluid, as it passes through the

actuator, that will result in useful work; the remaining power is lost in just

pushing the fluid through the system.

FIG. 7 Poor Design of Hydraulic System

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. What is the function of the prime mover in a hydraulic system?

2. If fluid with a pressure of 80 bar is supplied to a hydraulic cylinder with a

piston diameter of 40 mm, how much force will be exerted on the piston?

3. A hydraulic motor is supplied with fluid at a flowrate of 35 litres min–1.

If the motor has a displacement per revolution of 41 cm3 rev–1, calculate

the theoretical speed of the motor.

4. If a hydraulic system provides flow to an actuator at a rate of

1500 litres min–1 at a maximum pressure of 200 bar, how much hydraulic

power is being provided by the system?

5. List the four main causes of excessive fluid friction in a hydraulic system.

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________________________________________________________________________________________

NOTES ________________________________________________________________________________________

...................................................................................................................................................

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. The function of the prime mover in a hydraulic system is to provide the

mechanical energy to the pump for conversion into hydraulic power.

Hydraulic power (kW) = pressure bar flowrat( ) × ee min

= 200

600

= kW

–1�( )

600

1500

500

Motor speed = flowrate cm min

motor displ

3 –1( ) aacement cm rev

= cm min

3 –1

( )

×35 1000 41 rev

= rev min

–1

–1853 66.

Force N = pressure N m area m

–2 2( ) ( ) × ( )

×80 105(( ) × ×( )

× × ×

3 142 0 04

80 10 1 257 10

10056

5 3

. .

. –=

= NN

= 10.056 kN

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5. The four main causes of excessive fluid friction in a hydraulic system are:

• excessive length of pipelines

• excessive number of bends and fittings

• high fluid velocity

• excessive viscosity of the fluid.

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

In this lesson we have looked at the way power is produced in a hydraulic

system, and the advantages to be gained from using hydraulic power

transmission. We also noted the causes of power loss in a hydraulic system,

and ways in which this power loss can be minimised.

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