Learn the basic operating principles of circuit breakers as well as the different types. Review the range of operating mechanisms with a focus on pneumatics and compressors.
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HV Circuit Breaker Operating Mechanisms  Pneumatic & Compressor Systems  First Chapter
Basic pneumatics for circuit breakers
1.1 What is pneumatics?
Pneumatics is the transmission and control of power, using gas as the fluid medium.
Power in a machine can be transmitted to the different parts of the machine by using electricity, mechanical connections, pneumatics or hydraulics. Generally a combination of the methods is used for power transmission. Fluids are used widely for transmission all over the world.
Fluid
A fluid is any material capable of flowing and it is usually a liquid or a gas.
Fluid power
Fluid power is the transmission and control of power using flowing fluids, under pressure, as the transmitting medium in a closed or restricted system.
1.1.1 Examples of uses of fluid power
Field Application
Electricity Supply Circuit Breakers
Manufacturing Handling, Bundling, Tying, Presses, Machine Tools, Mixing, Lifting
Mining Continuous miners, Pit props, Drills
Agriculture Farm Equipment
Aircraft Undercarriage, Flight controls, Cargo ramps, Deicing of wings
Marine Steering, Stabilizers, Winches, Cargo doors
Forestry Earth moving equipment, Tree loppers, Tree strippers
Transport Automatic doors of trains, Brakes and Lifting equipment
Entertainment Amusement park rides
Pneumatics has the following advantages over other means of power transmission.
Pneumatic fluid
The common pneumatic fluid used is air. Inert gases like nitrogen are used in special applications to reduce the risk of fire or explosion
Composition of air
Air contains a mixture of gases and other particles. The main constituents are nitrogen and oxygen. Air is made up of approximately 78% nitrogen, 20% oxygen and 2% of other gases.
Figure 1.1
Composition of air
Properties of air
Air is like any other fluid; it fills up the container it occupies. The desirable property of air is that, it is cheap and that it is compressible and can be made to perform a number of operations. The drawbacks of air are that it contains moisture and does not have the ability to lubricate moving parts.
1.2 Units of measurement
All the measurements in this manual are based on the standard base units of ‘Length’, ‘Mass’ and ‘Time’. S.I system (Systéme International) is the system of measurement used in Australia. It is based on the Metric system and makes use of the decimal system.
Base units
The base units of measurement are as follows
Quantity 
Unit of measurement 
Symbol 
Length 
Meter 
m 
Mass 
Kilogram 
kg 
Time 
Second 
s 
Table 1.1
Basic units of measurement
Measurements which are made up from combinations of base units are called derived units.
The following are derived units
Force
Definition of ‘Force’ as applied to pneumatics is ‘Any cause that tends to produce or modify motion’.
Force = mass × acceleration
= m a
Force is measured in Newtons. One Newton (N) is the force required to accelerate a mass of one kilogram at a rate of one meter per second per second.
Newton = kilogram × meters per second per second
N = kg × m/s^{2}
Acceleration due to gravity is commonly used for calculations and it is equal to 9.81 m/s^{2}.
Area
Area is the measure of the size of a surface. Area is measured in meters squared – m^{2}
Area of a rectangle A = l × b where l and b are the length and breadth of the rectangle
Area of a circle A = π × diameter × diameter
4
A = π D^{2}
4
Volume
Volume of a container is a measure of how much fluid it can accommodate inside it. The unit of measurement is cubic metre (m^{3})
Volume = Area × Length
V = A × l
Pressure
Pressure is the result of application of force on an area. The unit of pressure is Pascal. One Pascal is the pressure resulting from the application of 1 Newton force on an area of 1 square meter.
Pressure (P) = Force = F = N
Area A m^{2}
Figure 1.2
Relation between Force, Pressure and Area
Pressure is expressed in three different ways
Atmospheric pressure
Atmospheric pressure is the pressure exerted by the air around us. The pressure varies with the altitude. At sea level, the atmospheric pressure is approximately equal to 101.3 kilopascal. It is the force exerted by a column of air directly above an area of one square meter.
The atmospheric pressure at sea level is approximately equal to 760 mm of a mercury column. Refer to Figure 1.3 which shows that the atmospheric pressure is holding a 760 mm column of mercury inside the glass tube. The atmospheric pressure drops with elevation.
Figure 1.3
Atmospheric pressure
Gauge pressure
Gauge pressure is the reading obtained from a pressure gauge. Gauges can be calibrated to indicate any pressure required.
Absolute pressure
Absolute pressure is the total pressure obtained by the addition of gauge pressure and atmospheric pressure.
Prefixes
Prefix is a syllable or syllables which is placed at the beginning of a unit or word to indicate a multiple or submultiples of a unit
In S.I. system, the commonly used prefixes in pneumatics are
PREFIX

VALUE 
SYMBOL 
Mega

10^{6} 
M 
kilo

10^{3} 
k

centi

10^{–2} 
c

milli

10^{–3} 
m 
micro

10^{–6} 
μ 
Table 1.2
Prefixes used with the units in pneumatic system
The unit ‘Pascal’ is very small and hence, it is mostly used with a prefix. Pressures of 10 000 000 Pa are used quite often. Using prefixes, the same pressure can be mentioned in various ways like following
10 000 000 Pa
= 10 000 kPa (10 000 kilopascals)
= 10 MPa (10 Megapascals)
Another unit used commonly to denote pressure is ‘Bar’, even though it is not an Australian Standard unit. One ‘Bar’ is approximately equal to 100 kPa or 0.1 MPa.
10 000 kPa = 100 Bar
Temperature
Temperature is measured in degrees Centigrade or Celcius (˚^{C}). Absolute temperatures are used for the calculation of temperatures encountered during compression of air. The unit of absolute temperature is Kelvin (K).
Absolute zero on the Kelvin temperature scale equals –273˚^{C}.
One ˚^{C} is equal to 1 Kelvin unit and so to convert ˚^{C} to Kelvin, 273 is added to the centigrade temperature. To convert Kelvin to Centigrade, subtract 273 from the Kelvin units.
–100˚^{C} = 173 K
0˚^{C} = 273 K
100˚^{C} = 373 K
1.3 Pascal’s law
Pascal's law states that "Pressure applied to a confined static fluid is transmitted equally and undiminished in all directions throughout that fluid and acts with equal force on equal areas".
Figure 1.4
Pascal's Law
1.4 Force multiplication
The force acting on the piston in a large cylinder is greater than the force on the piston in a smaller cylinder for the same pressure due to the relationship formula shown below.
Force = Pressure × Area
For the same pressure, if the cross sectional area of the cylinder is doubled, then the force is also doubled. This principle is used in pneumatic systems where, much larger forces can be generated on a bigger sized piston by using the same pressure developed by a smaller sized piston.
Figure 1.5
Force amplification
For the smaller piston
Pressure = Force
Area
= 50 = 2.5
20
The same pressure gets transmitted to the bigger piston. The force acting on the bigger piston is given by
Force = Pressure × Area
= 2.5 × 400
= 1000 Newtons
The force is observed to have been amplified by the same proportion as the increase in the piston area. This principle is used in pneumatics to obtain much higher forces from lower values of applied forces.
1.5 Pressure intensification
Pressure intensification is used where a large pressure increase is required in a portion of the system. By using pressure intensifier it is possible to generate much higher forces, than what the pressure source is capable of developing. The Figure 1.6 shows the construction of a pressure intensifier.
Figure1.6
Construction of a pressure intensifier
Pressure intensifiers use two pistons having different areas and mechanically connected to each other. The inlet and outlet ports are connected to virtually separate systems.
When pressure is applied to the inlet port, the force on the bigger piston is directly transmitted to the smaller piston by the mechanical connection. The smaller piston then transmits the force to the fluid in the outlet port. Since the force is applied to a smaller area of the piston, the pressure is increased in direct proportion to the increase in the utilized area of the piston.
Figure1.7
Principle of operation of a pressure intensifier
For the large piston section
Force = Pressure × Area
= 900000 × 0.001256
= 1130.9 Newtons
For the small piston section
Pressure = Force
Area
= 1130.97
0.000078
= 14 400 000 Pa
= 14.4 MPa
In the above example of pressure intensifier, the pressure has been intensified from 900 kPa to 14.4 MPa.
1.6 Pressure acting on different areas
Application of same force on different piston areas produces different forces and this phenomenon is used effectively in some pneumatic systems. Refer to Figure 1.8.
Figure 1.8
Pressure on different sized pistons
When compressed air is introduced in the space between the two pistons, the forces developed on the two pistons are different due to the difference in the piston areas. In the calculation of the forces acting on pistons, the area of the shaft must be subtracted from the total area of the piston. The subtracted net area of the piston is called the annular area as shown in the Figure 1.9.
Figure 1.9
Annular area
The calculation of the forces acting on the two pistons can be calculated for the following case. Refer to Figure 1.10.
Figure 1.10
Forces on pistons having different areas
Large Piston,
Annular area of the piston = πD^{2} _ πd^{2}
4 4
= π × 0.04^{2} _ π × 0.01^{2}
4 5
= π × ( 0.04^{2} – 0.01^{2} )
4
= 1.17809 × 10^{–3}
Small Piston
Annulus area of the piston = πD^{2} _ πd^{2}
4 4
= π × 0.02^{2} _ π × 0.01^{2}
4 4
= π × ( 0.02^{2} – 0.01^{2} )
4
= 2.35619 × 10^{–4}
The force acting on the large piston is
F = p × A
= 3 000 000 × 0,00117809
= 3 534.2917 Newtons
= 3.53 kN
The force acting on the small piston is
F = p × A
= 3 000 000 × 0.000235619
= 706.85835 Newtons
= 0.71 Kn
The difference between the forces acting on the two pistons is
= 3.53 – 0.71
= 2.82 kN
Since the force acting on the large piston is more than the force acting on the smaller piston the piston will move towards left.
Figure 1.11 is an application of a pneumatic system with a piston, cylinder assembly.
Figure 1.11
Double acting linear actuator
The force on the piston from the side of Port 1 is more than the force acting on the piston from Port 2 side for the same pressure. This is due to the reduced area of the piston on account of the shaft connection.
1.7 Gas laws
Gas laws explain the behaviour of the gas when the pressure and temperature of the gas changes.
There are several laws and a few of them are given below.
Boyle’s law
When the temperature of an enclosed gas is kept constant, and the pressure is increased, the volume of the gas is decreased by the same proportion. The formula for Boyle’s law is
P_{1} × V_{1} = P_{2} × V_{2}
Absolute pressures have to be used while using the above formula.
Charles law
At constant pressure, the volume of gas varies in direct proportion to a change in temperature. The formula for Charles law is
V_{1} = V_{2}
T_{1} T_{2}
Absolute temperatures have to be used while using the above formula
GayLussac’s law
At constant volume, the pressure in a confined gas chamber is directly proportional to the absolute temperature of the gas.
The formula for GayLussac’s Law is
P_{1} = P_{2}
T_{1} T_{2}
Absolute temperatures have to be used while using the above formula
Combined gas law
The combined law is a combination of the gas laws of Boyle, Charles and GayLussac.
The formula for the Combined Gas Law is
P_{1} × V_{1} = P_{2 } × V_{2}
T_{1} T_{2}
Absolute temperatures and pressures have to be used while using the above formula.
Using this formula, it is possible to calculate the value of one unknown, if the values of the other five parameters are known. The gauge pressures have to be converted to absolute pressures and the temperatures in degrees centigrade have to be converted to Kelvin units.
Example
Calculation of final temperature of 5 m^{3 }of air at atmospheric pressure and a temperature of 20˚^{C, when it is compressed into a 1 }m^{3} volume at a gauge pressure of 500 kPa.
P_{1} X V_{1} = P_{2 } X V_{2}
T_{1} T_{2}
where P_{1 }= 101.3 kPa (absolute)
V_{1 } = 5 m^{3}
T_{1 } = 20 ˚^{C + 273 = 293 K (absolute)}
^{ P2 = 500 kPa + 101.3 (absolute)}
^{ = 601.3 kPa}
^{ V2 = 1 m3}
Therefore T_{2} = P_{2 } × V_{2 } × T_{1}
P_{1} × V_{1}
= ( 601.3 × 1 × 293 ) ÷ ( 101.3 × 5 )
= 347.8 K ( absolute )
= 74.8˚C
1.8 Pneumatic symbols
In a pneumatic circuit, the various pneumatic components are shown in the form of symbols.
Table 1.3 shows the pneumatic symbols and their description.
Symbol

Description


Pressure Line 
Exhaust Line, Control Line



Enclosing Line 

Mechanical Connection 

Vacuum Pump 

Lines Crossing 

Lines Connected 
Fixed Restriction 

Variable Restriction 


Fluid Flow Direction 

Pressure Gauge 

Pneumatic Compressor 

Linear Actuator 
Basic Envelope  Filters and Lubricators 


Filter  Manual Drain 

Filter Automatic Drain 
Lubricator ( Less Drain ) 


Lubricator ( Manual Drain ) 
Air Dryer 


Pressure Regulator  Adjustable Self Relieving 

Pressure Regulator  Adjustable Non Relieving 

Accumulator 

Check Valve


Valve  One Flow Path 

Valve Two Flow Paths 
Valve  Two Closed Ports 

Plugged line 

Shuttle Valve 

Silencer 

Cylinder Spring Return 


Pressure Actuated Electric Switch 

Shutoff Valve 
Flow Gauge 

Check Valve 


Check Valve  Spring Loaded 
Solenoid 


Cylinder Double Acting 
Table 1.3
Pneumatic circuit symbols
Table 1.4 below gives the conversion of the measurement units
Conversion Table


From

To 
Multiply by 
Inches 
mm 
25.4 
Feet 
Metres 
0.3048 
Square Feet 
Square Metres 
0.09290 
Cubic Feet 
Cubic Metres 
0.02832 
Pounds 
kgs 
0.4536 
PSI ( lb/inch^{2} ) 
Bar 
0.06895 
Table 1.4
Conversion table for units of measurement
1.9 Safety with pneumatics
Figure 1.12
Hazards of compressed air
Precautions when using compressed air
IDC Technologies © 1991  2019