Introduction to Pneumatics

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Introduction to Pneumatics
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Air Production System
Air Consumption System
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What can Pneumatics do?

• Operation of system valves for air, water or
  chemicals
• Operation of heavy or hot doors
• Unloading of hoppers in building, steel making,
  mining and chemical industries
• Ramming and tamping in concrete and asphalt
  laying
• Lifting and moving in slab molding machines
• Crop spraying and operation of other tractor
  equipment
• Spray painting
• Holding and moving in wood working and furniture
  making
• Holding in jigs and fixtures in assembly
  machinery and machine tools
• Holding for gluing, heat sealing or welding
  plastics
• Holding for brazing or welding
• Forming operations of bending, drawing and
  flattening
• Spot welding machines
• Riveting
• Operation of guillotine blades
• Bottling and filling machines
• Wood working machinery drives and feeds
• Test rigs
• Machine tool, work or tool feeding

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Properties of compressed air

• Availability
• Storage
• Simplicity of design and control
• Choice of movement
• Economy

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Properties of compressed air

• Reliability
• Resistance to Environment
• Environmentally clean.
• Safety

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What is Air?
The weight of a one square inch column of
air (from sea level to the outer atmosphere, _at_
680 F, 36 RH) is 14.69 pounds.
In a typical cubic foot of air --- there are
over 3,000,000 particles of dust, dirt, pollen,
and other contaminants. Industrial air may be 3
times (or more) more polluted.
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HUMIDITY DEWPOINT
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Pressure and Flow
Example P1 6bar ? P 1bar P2 5bar Q 54
l/min (1 Bar 14.5 psi)
P1
P2
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Air Treatment
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Compressing Air
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Relative Humidity
Reservoir Tank
Adsorbtion Dryer
Compressor Exit
Airline Drop
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Air Mains
Dead-End Main
Ring Main
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Pressure

• It should be noted that the SI unit of pressure
  is the Pascal (Pa)
• 1 Pa 1 N/m2 (Newton per square meter)
• This unit is extremely small and so, to avoid
  huge numbers in practice, an agreement has been
  made to use the bar as a unit of 100,000 Pa.
• 100,000 Pa 100 kPa 1 bar
• Atmospheric Pressure
• 14.696 psi 1.01325 bar 1.03323 kgf/cm2.

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Isothermic change (Boyles Law)with constant
temperature, the pressure of a given mass of gas
is inversely proportional to its volume

• P1 x V1 P2 x V2
• P2 P1 x V1 V2
• V2 P1 x V1 P2

• Example P2 ?
• P1 Pa (1.013bar)
• V1 1m³
• V2 .5m³
• P2 1.013 x 1 .5
• 2.026 bar

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Isobaric change (Charles Law)at constant
pressure, a given mass of gas increases in volume
by 1 of its volume for every degree C in
temperature rise. 273

• V1 T1
• V2 T2
• V2 V1 x T2 T1
• T2 T1 x V2 V1

• Example V2 ?
• V1 2m³
• T1 273K (0C)
• T2 303K (30C)
• V2 2 x 303 273
• 2.219m³

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Isochoric change Law of Gay Lussac at constant
volume, the pressure is proportional to the
temperature

• P1 x P2 T1 x T2
• P2 P1 x T2 T1
• T2 T1 x P2 P1

• Example P2 ?
• P1 4bar
• T1 273K (OC)
• T2 298K (25C)
• P2 4 x 298 273
• 4.366bar

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P1 ________bar T1 _______C ______K T2
_______C ______K
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Force formula transposed

• D 4 x FE
  ? x P

• Example
• FE 1600N
• P 6 bar.
• D 4 x 1600 3.14 x 600,000
• D 6400 1884000
• D .0583m
• D 58.3mm
• A 63mm bore cylinder would be selected.

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Load Ratio

• This ratio expresses the percentage of the
  required force needed from the maximum available
  theoretical force at a given pressure.
• L.R. required force x 100
  max. available theoretical force
• Maximum load ratios
• Horizontal.70 1.51
• Vertical.50 2.01

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Speed control

• The speed of a cylinder is define by the extra
  force behind the piston, above the force opposed
  by the load
• The lower the load ratio, the better the speed
  control.

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Angle of Movement
1. If we totally neglect friction, which cylinder
diameter is needed to horizontally push a load
with an 825 kg mass with a pressure of 6 bar
speed is not important. 2. Which cylinder
diameter is necessary to lift the same mass with
the same pressure of 6 bar vertically if the load
ratio can not exceed 50. 3. Same conditions
as in 2 except from vertical to an angle of 30.
Assume a friction coefficient of 0.2. 4. What is
the force required when the angle is increased to
45?
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Y axes, (vertical lifting force).. sin? x M X
axes, (horizontal lifting force).cos? x ? x
M Total force Y X ? friction coefficients
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Example
? .01
F ________ (N)
150kg
40?
Force Y sin ? x M .642 x 150
96.3 N Force X cos ? x ? x M .766 x .01
x 150 1.149 N Total Force Y X
96.3 N 1.149 N 97.449 N
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? __
______kg
_____?
Force Y sin ? x M Force X cos ? x ? x
M Total Force Y X
F ________ (N)
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Relative humidity (r.h.) actual water content
X 100
saturated quantity (dew point)

• Example 1
• T 25C
• r.h 65
• V 1m³

• From table 3.7 air at 25C contains 23.76 g/m³
• 23.76 g/m³ x .65 r.h 15.44 g/m³

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Relative Humidity Example 2

• V 10m³
• T1 15C
• T2 25C
• P1 1.013bar
• P2 6bar
• r.h 65
• ? H²0 will condense out

• From 3.17, 15C 13.04 g/m²
• 13.04 g/m² x 10m³ 130.4 g
• 130.4 g x .65 r.h 84.9 g
• V2 1.013 x 10 1.44 m³ 6 1.013
• From 3.17, 25C 23.76 g/m²
• 23.76 g/m² x 1.44 m³ 34.2 g
• 84.9 - 34.2 50.6 g
• 50.6 g of water will condense out

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V __________m³ T1 __________C T2
__________C P1 __________bar P2
__________bar r.h __________ ?
__________H²0 will condense out
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Formulae, for when more exact values are required

• Sonic flow P1 1.013 gt 1.896 x (P2 1,013)
• Pneumatic systems cannot operate under sonic flow
  conditions
• Subsonic flow P1 1.013 lt 1.896 x (P2
  1,013)
• The Volume flow Q for subsonic flow equals
• Q (l/min) 22.2 x S (P2 1.013) x ? P

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Sonic / Subsonic flow

• Example
• P1 7bar
• P2 6.3bar
• S 12mm²
• l/min

• P1 1.013 ? 1.896 x (P2 1.013)
• 7 1.013 ? 1.896 x (6.3 1.013)
• 8.013 ? 1.896 x 7.313
• 8.013 lt 13.86 subsonic flow.
• Q 22.2 x S x (P2 1.013) x ?P
• Q 22.2 x 12 x (6.3 1.013) x .7
• Q 22.2 x 12 x 7.313 x .7
• Q 22.2 x 12 x 5.119
• Q 22.2 x 12 x 2.26
• Q 602 l/min

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P1 _________bar P2 _________bar S
_________mm² Q ____?_____l/min
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Receiver sizing

• If
• Q 5000
• P1 9 bar
• Pa 1.013
• V 5000 x 1.013 9 1.013
• V 5065 10.013
• V 505.84 liters

• Example
• V capacity of receiver
• Q compressor output l/min
• Pa atmospheric pressure
• P1 compressor output pressure
• V Q x Pa P1 Pa

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Sizing compressor air mains

• Example
• Q 16800 l/min
• P1 9 bar (900kPa)
• ?P .3 bar (30kPa)
• L 125 m pipe length
• ?P kPa/m L
• l/min x .00001667 m³/s

• 30 .24 kPa/m 125
• 16800 x .00001667 0.28 m³/s
• chart lines on Nomogram

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Sizing compressor air mains

• Example 2
• Add fittings to example 1
• From table 4.20
• 2 elbows _at_ 1.4m 2.8m
• 2 90 _at_ 0.8m 1.6m
• 6 Tees _at_ 0.7m 4.2m
• 2 valves _at_ 0.5m 1.0m
• Total 9.6m
• 125m 9.6 134.6m
• 135m

• 30kPa 0.22kPa/m 135m
• Chart lines on Nomogram

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Using the ring main example on page 29 size for
the following requirements
Q 20,000 l/min P1 10 bar (_________kPa) ?P
.5 bar (_________kPa) L 200 m pipe
length ?P kPa/m L l/min x .00001667 m³/s
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Example

• P 7 bar (700,000 N/m²)
• D 63mm (.063m)
• d 15mm (.015m)
• F ? x (D² -d²) x P 4
• F 3.14 x (.063² - .015²) x 700,000 4
• F 3.14 x (.003969 - .0.000225) x 700,000 4
• F .785 x .003744 x 700,000
• F 2057.328 N

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Example

• Calculate remaining force
• 401.9 x 48.8 (.488) 196N 100
• assume a cylinder efficiency of 95
• 196 x 95 185.7 N 100
• Newtons kg m/s² , therefor
• 185.7 N 185.7 kg m/s²
• divide mass into remaining force
• m/s² 185.7 kg m/s²
  100kg
• 1.857 m/s²

• M 100kg
• P 5bar
• ? 32mm
• ? 0.2
• F ?/4 x D²x P 401.9 N
• From chart 6.16
• 90KG 43.9 Lo.
• To find Lo for 100kg
• 43.9 x 100 48.8 Lo. 90

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M _______kg P _______bar ? _______mm ?
0.2 F ?/4 x D²x P 401.9 N
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Air Flow and ConsumptionAir consumption of a
cylinder is defined aspiston area x stroke
length x number of single strokes per minute x
absolute pressure in bar.
Q D² (m) x ? x (P Pa) x stroke(m) x
strokes/min x 1000 4
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• Example.
• ? 80
• stroke 400mm
• s/min 12 x 2
• P 6bar.

• From table 6.19... ?80 at 6 bar 3.479
  (3.5)l/100mm stroke
• Qt Q x stroke(mm) x of extend
  retract strokes 100
• Qt 3.5 x 400 x 24 100
• Qt 3.5 x 4 x 24
• Qt 336 l/min.

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Peak Flow

• For sizing the valve of an individual cylinder
  we need to calculate Peak flow. The peak flow
  depends on the cylinders highest possible speed.
  The peak flow of all simultaneously moving
  cylinders defines the flow to which the FRL has
  to be sized.
• To compensate for adiabatic change, the
  theoretical volume flow has to be multiplied by a
  factor of 1.4. This represents a fair average
  confirmed in a high number of practical tests.

Q 1.4 x D² (m) x ? x (P Pa) x stroke(m) x
strokes/min x 1000 4
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• Example.
• ? 80
• stroke 400mm
• s/min 12 x 2
• P 6bar

• From table 6.20... ?80 at 6 bar 4.87
  (4.9)l/100mm stroke
• Qt Q x stroke(mm) x of extend
  retract strokes 100
• Qt 4.9 x 400 x 24 100
• Qt 4.9 x 4 x 24
• Qt 470.4 l/min.

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Formulae comparison

• Q 1.4 x D² (m) x ? x (P Pa) x stroke(m) x
  strokes/min x 1000 4
• Q 1.4 x .08² x .785 x ( 6 1.013) x .4 x
  24 x 1000
• Q 1.4 x .0064 x .785 x 7.013 x .4 x 24 x
  1000
• Q 473.54

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Q 1.4 x D² (m) x ? x (P Pa) x stroke(m) x
strokes/min x 1000 4

? _______mm stroke _______mm s/min
_______ x 2 P _______bar
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Inertia

• Example 1
• m 10kg
• a 30mm
• j ___?

• J m (kg) x a² (m) 12
• J 10 x .03² 12
• J 10 x .0009 12
• J .00075

a
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Inertia

• J ma x a² mb x b²
  3 3
• J 3 x .01² 6 x .02²
  3 3
• J 3 x .0001 6 x .0004
  3 3
• J .0001 .0008
• J .0009

• Example 2
• m 9 kg
• a 10mm
• b 20mm
• J ___?

a b
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m ________ kg a _________mm b
_________mm J _________?
a b
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Valve identification
A(4) B(2)
EA P EB (5) (1) (3)

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Valve Sizing

• The Cv factor of 1 is a flow capacity of one US
  Gallon of water per minute, with a pressure drop
  of 1 psi.
• The kv factor of 1 is a flow capacity of one
  liter of water per minute with a pressure drop of
  1 bar.
• The equivalent Flow Section S of a valve is the
  flow section in mm2 of an orifice in a diaphragm,
  creating the same relationship between pressure
  and flow.

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Q 400 x Cv x (P2 1.013) x ?P x
273 273 ? Q 27.94 x kv x (P2
1.013) x ?P x 273 273 ? Q
22.2 x S x (P2 1.013) x ?P x
273 273 ?
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Flow example

• Q 22.2 x S x (P2 1.013) x ?P x 273
  273 ?
• Q 22.2 x 35 x (5.5 1.013) x .5 x
  273 273 25
• Q 22.2 x 35 x 6.613 x .5 x 273
  298
• Q 22.2 x 35 x 6.613 x .5 x 273
  298
• Q 22.2 x 35 x 1.89 x .957
• Q 1405.383

• S 35
• P1 6 bar
• P2 5.5 bar
• ? 25C

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Cv ________between 1 -5 P1
________bar P2 ________5 bar ?
________C
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Flow capacity formulae transposed

• Cv Q 400 x (P2 1.013) x ?P
• Kv Q 27.94 x (P2 1.013) x
  ?P
• S Q 22.2 x (P2 1.013) x ?P

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Flow capacity example

• Q 750 l/min
• P1 9 bar
• ?P 10
• S ?

• S Q 22.2 x (P2 1.013) x ?P
• S 750 22.2 x (8.1 1.013) x .9
• S 750 22.2 x 9.113 x .9
• S 750 22.2 x 2.86
• S 750 S 11.81 63.49

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Q _________ l/min P1 _________ bar ?P
_________ Cv _________ ?
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Orifices in a series connection

• S total 1 1 1 1
  S1² S2² S3²
• Example
• S1 12mm²
• S2 18mm²
• S3 22mm²

S total 1 1 1 1
12² 18² 22² S total
1 1 1 1 144 324
484 S total 1 1
.00694 .00309 .00207 .0121 S total
9.09
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Cv _________ Cv _________ Cv
_________ Cv total ________
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Table 7.31 Equivalent Section S in mm2 for the
valve and the tubing, for 6 bar working pressure
and a pressure drop of 1 bar (Qn Conditions)
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Flow Amplification
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Signal Inversion
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Selection
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Memory Function
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Delayed switching on
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Delayed switching off
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Pulse on switching on
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Pulse on releasing a valve
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Direct Operation and Speed Control
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Control from two points OR Function
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Safety interlock AND Function
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Safety interlock AND Function
3
2
1
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Inverse Operation NOT Function
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Direct Control
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Holding the end positions
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Semi Automatic return of a cylinder
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Repeating Strokes
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Sequence Control
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