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Introduction to PneumaticsPowerPoint Presentation

Introduction to Pneumatics

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Air Consumption System

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
- Component and material conveyor transfer
- Pneumatic robots
- Auto gauging
- Air separation and vacuum lifting of thin sheets
- Dental drills
- and so much more… new applications are developed daily

Properties of compressed air

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

Properties of compressed air

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

What is Air?

The weight of a

one square inch

column of air

(from sea level

to the outer atmosphere,

@ 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.

ratio

=

psig + 1 atm

One cubic foot of air

1 atm

compressor

One cubic foot of

100 psig

compressed air

(at Standard conditions)

with 7.8 times the moisture and dirt

7.8 cubic feet of free air

CFM vs SCFM

Compressed air is always related at Standard conditions.

Compressing Air1 ft3 @100 psig

-200 F

1 ft3 @100 psig

770 F

1 ft3 @100 psig

770 F

1 ft3 @100 psig

1950 F

100% RH

57.1

grams of H20

0.15% RH

.01

grams of H20

100% RH

.73

grams of H20

100% RH

.01

grams of H20

.72

grams of H20

56.37

grams of H20

Relative HumidityReservoir

Tank

Adsorbtion Dryer

Compressor

Exit

Airline

Drop

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.

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

Isothermic change (Boyle’s Law)with constant temperature, the pressure of a given mass of gas is inversely proportional to its volumeV1 = T1

V2 T2

V2 = V1 x T2 T1

T2 = T1 x V2 V1

Example V2 = ?

V1 = 2m³

T1 = 273°K (0°C)

T2 = 303°K (30°C)

V2 = 2 x 303 273

= 2.219m³

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

10

P1 x P2 T1 x T2

P2 = P1 x T2 T1

T2 = T1 x P2 P1

Example P2 = ?

P1 = 4bar

T1 = 273°K (O°C)

T2 = 298°K (25°C)

P2 = 4 x 298 273

= 4.366bar

Isochoric change Law of Gay Lussacat constant volume, the pressure is proportional to the temperatureD = 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.

Force formula transposedLoad 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%~ 1.5:1
- Vertical…….50%~ 2.0:1

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.

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°?

Y axes, (vertical lifting force)….. sin x M

X axes, (horizontal lifting force)….cos x x M

Total force = Y + X

= friction coefficients

= .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

______kg

_____°

Force Y = sin x M =

Force X = cos x x M =

Total Force = Y + X =

F = ________ (N)

T = 25°C

r.h = 65%

V = 1m³

From table 3.7 air at 25°C contains 23.76 g/m³

23.76 g/m³ x .65 r.h = 15.44 g/m³

Relative humidity (r.h.) = actual water content X 100% saturated quantity (dew point)13

T1= 15°C

T2= 25°C

P1 = 1.013bar

P2 = 6bar

r.h = 65%

? H²0 will condense out

From 3.17, 15°C = 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, 25°C = 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

Relative Humidity Example 213

T1= __________°C

T2= __________°C

P1 =__________bar

P2 =__________bar

r.h =__________%

? __________H²0 will condense out

Formulae, for when more exact values are required

- Sonic flow = P1 + 1.013 > 1.896 x (P2 + 1,013)
- Pneumatic systemscannot operate under sonic flow conditions
- Subsonic flow = P1 + 1.013 < 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

16

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 < 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

Sonic / Subsonic flow16,17

V = capacity of receiver

Q = compressor output l/min

Pa = atmospheric pressure

P1 = compressor output pressure

V = Q x Pa P1 + Pa

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

Receiver sizing22

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

Sizing compressor air mains31

Add fittings to example 1

From table 4.20

2 elbows @ 1.4m = 2.8m

2 90° @ 0.8m = 1.6m

6 Tees @ 0.7m = 4.2m

2 valves @ 0.5m = 1.0m

Total = 9.6m

125m + 9.6 = 134.6m

=135m

30kPa = 0.22kPa/m 135m

Chart lines on Nomogram

Sizing compressor air mains31

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

39 following requirements:

Example following requirements:

- 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

54

M = 100kg following requirements:

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

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

ExampleM = _______kg following requirements:

P = _______bar

= _______mm

= 0.2

F = /4 x D²x P = 401.9 N

Air Flow and Consumption following requirements:Air consumption of a cylinder is defined as:piston 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

Example. following requirements:

= 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.

Peak Flow following requirements:

- 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

Example. following requirements:

= 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.

Formulae comparison following requirements:

- 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

Q = 1.4 x D² (m) x following requirements: x (P + Pa) x stroke(m) x # strokes/min x 1000 4

= _______mm

stroke = _______mm

s/min = _______ x 2

P =_______bar

Example 1 following requirements:

m = 10kg

a = 30mm

j = ___?

J= m (kg) x a² (m) 12

J= 10 x .03² 12

J= 10 x .0009 12

J = .00075

Inertiaa

Example 2 following requirements:

m = 9 kg

a = 10mm

b = 20mm

J = ___?

J = ma x a² + mb x b²3 3

J = 3 x .01² + 6 x .02²3 3

J = 3 x .0001 + 6 x .00043 3

J = .0001 + .0008

J = .0009

Inertiaa b

Valve Sizing following requirements:

- 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 1is 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.

Q = 400 x Cv x (P2 + 1.013) x following requirements: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 +

S = 35 following requirements:

P1 = 6 bar

P2 =5.5 bar

= 25°C

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

Flow exampleCv = ________ following requirements:between 1 -5

P1 = ________bar

P2 = ________5 bar

= ________°C

Flow capacity formulae transposed following requirements:

- 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

Q = 750 following requirements: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 = 750S = 11.81 63.49

Flow capacity exampleOrifices in a series connection following requirements:

- 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

Table 7.31 following requirements: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)

Flow Amplification following requirements:

Signal Inversion following requirements:

Selection following requirements:

Memory Function following requirements:

Delayed switching on following requirements:

Delayed switching off following requirements:

Pulse on switching on following requirements:

Pulse on releasing a valve following requirements:

Direct Operation and Speed Control following requirements:

Control from two points: OR Function following requirements:

Safety interlock: AND Function following requirements:

Inverse Operation: NOT Function following requirements:

Direct Control following requirements:

Holding the end positions following requirements:

Semi Automatic return of a cylinder following requirements:

Repeating Strokes following requirements:

Sequence Control following requirements:

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