0 Intro to Pneumatics Modified

Introduction to Pneumatics

*Air Production SystemAir 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 airAvailability

Storage

Simplicity of design and control

Choice of movement

Economy

*Properties of compressed airReliability

Resistance to Environment

Environmentally clean.

Safety

*What is Air?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.The weight of aone square inchcolumn of air(from sea levelto the outer atmosphere,@ 680 F, & 36% RH)is 14.69 pounds.

*HUMIDITY & DEWPOINT

Temperature C

0

5

10

15

20

25

30

35

40

g/m3n *(Standard)

4.98

6.99

9.86

13.76

18.99

25.94

35.12

47.19

63.03

g/m3 (Atmospheric)

4.98

6.86

9.51

13.04

17.69

23.76

31.64

41.83

54.11

Temperature C

0

5

10

15

20

25

30

35

40

g/m3n (Standard)

4.98

3.36

2.28

1.52

1.00

0.64

0.4

0.25

0.15

g/m3 (Atmospheric)

4.98

3.42

2.37

1.61

1.08

0.7

0.45

0.29

0.18

Temperature F

32

40

60

80

100

120

140

160

180

g/ft3 *(Standard)

.137

.188

.4

.78

1.48

2.65

4.53

7.44

11.81

g/ft3 (Atmospheric)

.137

.185

.375

.71

1.29

2.22

3.67

5.82

8.94

Temperature F

32

30

20

10

0

-10

-20

-30

40

g/ft3 (Standard)

.137

.126

.083

.053

.033

.020

.012

.007

.004

g/ft3 (Atmospheric)

.137

.127

.085

.056

.036

.023

.014

.009

.005

*Pressure and FlowExampleP1 = 6bar P = 1barP2 = 5barQ = 54 l/min(1 Bar = 14.5 psi)

P1P2

Sonic Flow Range

Q

n

(54.44 l /

min)

S = 1 mm

2

0

20

40

80

100

120

60

10

9

8

7

6

5

4

3

2

1

(dm

/min)

3

n

Q

p

(bar)

*Air Treatment

*Compressing Air

*Relative HumidityAdsorbtion DryerCompressorExitReservoir TankAirline Drop

*Air MainsRing MainDead-End Main

*PressureIt 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.

*Isothermic change (Boyles Law)with constant temperature, the pressure of a given mass of gas is inversely proportional to its volumeP1 x V1 = P2 x V2

P2 = P1 x V1 V2

V2 = P1 x V1 P2Example P2 = ?P1 = Pa (1.013bar)V1 = 1mV2 = .5m

P2 = 1.013 x 1 .5= 2.026 bar

*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 = T1V2 T2

V2 = V1 x T2 T1T2 = T1 x V2 V1

Example V2 = ?V1 = 2mT1 = 273K (0C)T2 = 303K (30C)

V2 = 2 x 303 273= 2.219m10

*Isochoric change Law of Gay Lussac at constant volume, the pressure is proportional to the temperature P1 x P2 T1 x T2P2 = P1 x T2T1T2 = T1 x P2P1Example P2 = ?P1 = 4barT1 = 273K (OC)T2 = 298K (25C)

P2 = 4 x 298 273= 4.366bar

*P1 = ________bar

T1 = _______C ______K

T2 = _______C ______K

*

_915341748.unknown

*Force formula transposed

D = 4 x FE x P

ExampleFE = 1600NP = 6 bar.D = 4 x 1600 3.14 x 600,000D = 6400 1884000D = .0583mD = 58.3mmA 63mm bore cylinder would be selected.

*Load RatioThis 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 ratiosHorizontal.70%~ 1.5:1 Vertical.50%~ 2.0:1

*

Cyl.Dia

Mass (kg)

60

45

30

0.01

0.2

0.01

0.2

0.01

0.2

0.01

0.2

25

100

4

80

50

2.2

40

25

(87.2)

(96.7)

71.5

84.9

50.9

67.4

1

20

12.5

51.8

43.6

48.3

35.7

342.5

25.4

33.7

0.5

10

32

180

-

-

-

-

-

4.4

-

90

-

-

-

-

2.2

43.9

45

-

(95.6)

-

78.4

(93.1)

55.8

73.9

1.1

22

22.5

54.9

47.8

53

39.2

46.6

27.9

37

0.55

11

40

250

3.9

78

125

(99.2)

2

39

65

72.4

(86)

51.6

68.3

1

20.3

35

54.6

47.6

52.8

39

46.3

27.8

36.8

0.5

10.9

50

400

--

-

-

-

4

79.9

200

-

_

2

40

100

(87)

(96.5)

71.3

84.8

50.8

67.3

1

20

50

50

43.5

48.3

35.7

42.4

25.4

33.6

0.5

0

63

650

4.1

81.8

300

1.9

37.8

150

(94.4)

82.3

(91.2)

67.4

80.1

48

63.6

0.9

18.9

75

47.2

41.1

45.6

33.7

40.1

24

31.8

0.5

9.4

80

1000

3.9

78.1

500

2

39

250

(97.6)

85

(94.3)

69.7

82.8

49.6

65.7

1

19.5

125

48.8

42.5

47.1

34.8

41.4

24.8

32.8

0.5

9.8

100

1600

4

79.9

800

2

40

400

(87)

(96.5)

71.4

84.4

50.8

67.3

1

20

200

50

43.5

48.3

35.7

42.2

25.4

33.6

0.5

10

Table 6.16 Load Ratios for 5 bar working pressure and friction coefficients of 0.01 and 0.2

*Speed controlThe 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 Movement1. 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 MX axes, (horizontal lifting force).cos x x MTotal force = Y + X = friction coefficients

a

b

c

d

b

c

d

x

A

B

h

G

y

a

a

R

a

F

=

G

F

=

G

W

=

m

/

2

v

a

2

F

=

G

(sin

a

+ cos

a

)

*Example40F = ________ (N) 150kg = .01Force 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

*_____F = ________ (N)______kg = __Force Y = sin x M =

Force X = cos x x M =

Total Force = Y + X =

*13

Temperature C

0

5

10

15

20

25

30

35

40

g/m3n *(Standard)

4.98

6.99

9.86

13.76

18.99

25.94

35.12

47.19

63.03

g/m3 (Atmospheric)

4.98

6.86

9.51

13.04

17.69

23.76

31.64

41.83

54.11

Temperature C

0

5

10

15

20

25

30

35

40

g/m3n (Standard)

4.98

3.36

2.28

1.52

1.00

0.64

0.4

0.25

0.15

g/m3 (Atmospheric)

4.98

3.42

2.37

1.61

1.08

0.7

0.45

0.29

0.18

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

Example 1

T = 25Cr.h = 65%V = 1mFrom table 3.7 air at 25C contains 23.76 g/m

23.76 g/m x .65 r.h = 15.44 g/m13

*Relative Humidity Example 2V = 10mT1= 15CT2= 25CP1 = 1.013barP2 = 6barr.h = 65%? H0 will condense outFrom 3.17, 15C = 13.04 g/m13.04 g/m x 10m = 130.4 g130.4 g x .65 r.h = 84.9 gV2 = 1.013 x 10= 1.44 m 6 + 1.013From 3.17, 25C = 23.76 g/m 23.76 g/m x 1.44 m = 34.2 g84.9 - 34.2 = 50.6 g

50.6 g of water will condense out13

*V = __________mT1= __________CT2= __________CP1 =__________barP2 =__________barr.h =__________%? __________H0 will condense out

*Formulae, for when more exact values are requiredSonic flow = P1 + 1.013 > 1.896 x (P2 + 1,013)Pneumatic systems cannot 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

*Sonic / Subsonic flowExample

P1 = 7barP2 = 6.3barS = 12mml/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 7.3138.013 < 13.86 subsonic flow.Q = 22.2 x S x (P2 + 1.013) x PQ = 22.2 x 12 x (6.3 + 1.013) x .7Q = 22.2 x 12 x 7.313 x .7Q = 22.2 x 12 x 5.119Q = 22.2 x 12 x 2.26Q = 602 l/min16,17

*P1 = _________bar

P2 = _________bar

S = _________mm

Q = ____?_____l/min

*Receiver sizingExampleV = capacity of receiverQ = compressor output l/minPa = atmospheric pressureP1 = compressor output pressureV = Q x PaP1 + PaIf Q = 5000P1 = 9 barPa = 1.013

V = 5000 x 1.013 9 + 1.013V = 506510.013V = 505.84 liters22

*29

_915341790.unknown

*29

_915341789.unknown

*30

_915341788.unknown

*Sizing compressor air mainsExampleQ = 16800 l/minP1 = 9 bar (900kPa)P = .3 bar (30kPa)L = 125 m pipe lengthP = kPa/m Ll/min x .00001667 = m/s 30 = .24 kPa/m 12516800 x .00001667 = 0.28 m/schart lines on Nomogram31

*33

_915341785.unknown

*34

Type of Fitting

Nominal pipe size (mm)

15

20

25

30

40

50

65

80

100

125

Elbow

0.3

0.4

0.5

0.7

0.8

1.1

1.4

1.8

2.4

3.2

90* Bend (long)

0.1

0.2

0.3

0.4

0.5

0.6

0.8

0.9

1.2

1.5

90* Elbow

1.0

1.2

1.6

1.8

2.2

2.6

3.0

3.9

5.4

7.1

180* Bend

0.5

0.6

0.8

1.1

1.2

1.7

2.0

2.6

3.7

4.1

Globe Valve

0.8

1.1

1.4

2.0

2.4

3.4

4.0

5.2

7.3

9.4

Gate Valve

0.1

0.1

0.2

0.3

0.3

0.4

0.5

0.6

0.9

1.2

Standard Tee

0.1

0.2

0.2

0.4

0.4

0.5

0.7

0.9

1.2

1.5

Side Tee

0.5

0.7

0.9

1.4

1.6

2.1

2.7

3.7

4.1

6.4

Table 4.20 Equivalent Pipe Lengths for the main fittings

*Sizing compressor air mainsExample 2Add fittings to example 1From table 4.202 elbows @ 1.4m = 2.8m2 90 @ 0.8m = 1.6m6 Tees @ 0.7m = 4.2m2 valves @ 0.5m = 1.0mTotal = 9.6m125m + 9.6 = 134.6m=135m

30kPa = 0.22kPa/m 135mChart lines on Nomogram31

*33

_915341785.unknown

*Q = 20,000 l/minP1 = 10 bar (_________kPa)P = .5 bar (_________kPa)L = 200 m pipe length

P = kPa/m L l/min x .00001667 = m/sUsing the ring main example on page 29 size for the following requirements:

*39

Sub-micro Filter

Auto

Drain

1

2

3

4

5

6

7

Refrigerated

Air Dryer

Compressor

Tank

a

a

a

a

a

b

b

b

c

d

Micro Filter

Odor Removal Filter

Adsorbtion Air Dryer

Aftercooler

d

a

b

c

Auto

Drain

*ExampleP = 7 bar (700,000 N/m)D = 63mm (.063m)d = 15mm (.015m)F = x (D -d) x P4F = 3.14 x (.063 - .015) x 700,000 4F = 3.14 x (.003969 - .0.000225) x 700,000 4F = .785 x .003744 x 700,000F = 2057.328 N54

*

_915341748.unknown

*ExampleM = 100kgP = 5bar = 32mm = 0.2

F = /4 x Dx P = 401.9 N

From chart 6.1690KG = 43.9% Lo.To find Lo for 100kg43.9 x 100= 48.8 % Lo. 90Calculate remaining force401.9 x 48.8 (.488) = 196N 100assume a cylinder efficiency of 95%196 x 95 = 185.7 N 100Newtons = kg m/s , therefor185.7 N = 185.7 kg m/sdivide mass into remaining forcem/s = 185.7 kg m/s 100kg

= 1.857 m/s

*M = _______kg

P = _______bar

= _______mm

= 0.2

F = /4 x Dx P = 401.9 N

*Air Flow and ConsumptionAir 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 = D (m) x x (P + Pa) x stroke(m) x # strokes/min x 1000 4

_915341743.unknown

*Example.

= 80stroke = 400mms/min = 12 x 2P = 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 100Qt = 3.5 x 400 x 24 100Qt = 3.5 x 4 x 24

Qt = 336 l/min.

Working Pressure in bar

Piston dia.

3

4

5

6

7

20

0.124

0.155

0.186

0.217

0.248

25

0.194

0.243

0.291

0.340

0.388

32

0.319

0.398

0.477

0.557

0.636

40

0.498

0.622

0.746

0.870

0.993

50

0.777

0.971

1.165

1.359

1.553

63

1.235

1.542

1.850

2.158

2.465

80

1.993

2.487

2.983

3.479

3.975

100

3.111

3.886

4.661

5.436

6.211

Table 6.19 Theoretical Air Consumption of double acting cylinders from 20 to 100 mm dia,

in liters per 100 mm stroke

*Peak FlowFor 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 D (m) x x (P + Pa) x stroke(m) x # strokes/min x 1000 4

*Example.

= 80stroke = 400mms/min = 12 x 2P = 6barFrom table 6.20... 80 at 6 bar = 4.87 (4.9)l/100mm stroke

Qt= Q x stroke(mm) x # of extend + retract strokes 100Qt = 4.9 x 400 x 24 100Qt = 4.9 x 4 x 24

Qt = 470.4 l/min.

Working Pressure in bar

Piston dia.

3

4

5

6

7

20

0.174

0.217

0.260

0.304

0.347

25

0.272

0.340

0.408

0.476

0.543

32

0.446

0.557

0.668

0.779

0.890

40

0.697

0.870

1.044

1.218

1.391

50

1.088

1.360

1.631

1.903

2.174

63

1.729

2.159

2.590

3.021

3.451

80

2.790

3.482

4.176

4.870

5.565

100

4.355

5.440

6.525

7.611

8.696

Table 6.20 Air Consumption of double acting cylinders in liters per 100 mm stroke corrected for losses by adiabatic change

*Formulae comparisonQ = 1.4 x D (m) x x (P + Pa) x stroke(m) x # strokes/min x 1000 4

Q = 1.4 x .08 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 D (m) x x (P + Pa) x stroke(m) x # strokes/min x 1000 4 = _______mm

stroke = _______mm

s/min = _______ x 2

P =_______bar

*InertiaExample 1

m = 10kga = 30mmj = ___?

J= m (kg) x a (m) 12J= 10 x .03 12J= 10 x .0009 12J = .00075

a

*InertiaExample 2

m = 9 kga = 10mmb = 20mmJ = ___?

J = ma x a + mb x b 3 3J = 3 x .01 + 6 x .02 3 3J = 3 x .0001 + 6 x .0004 3 3J = .0001 + .0008

J = .0009a b

*a bm = ________ kg

a = _________mm

b = _________mm

J = _________?

*Valve identification A(4) B(2) EA P EB (5) (1) (3)

*Valve SizingThe 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 1bar. 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 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 +

1 Cv =

1 kv =

1 S =

The normal flow Qn for other various flow capacity units is:

981.5

68.85

54.44

The Relationship between these units is as follows:

1

14.3

18

0.07

1

1.26

0.055

0.794

1

*Flow exampleS = 35P1 = 6 barP2 =5.5 bar = 25CQ = 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

*Cv = ________between 1 -5

P1 = ________bar

P2 = ________5 bar

= ________C

*Flow capacity formulae transposedCv =Q 400 x (P2 + 1.013) x P

Kv =Q 27.94 x (P2 + 1.013) x PS =Q 22.2 x (P2 + 1.013) x P

*Flow capacity exampleQ = 750 l/minP1 = 9 barP = 10%S = ?S =Q22.2 x (P2 + 1.013) x PS =75022.2 x (8.1 + 1.013) x .9S =750 22.2 x 9.113 x .9S =750 22.2 x 2.86S = 750S = 11.81 63.49

*Q = _________ l/min

P1 = _________ bar

P = _________%

Cv = _________ ?

*Orifices in a series connectionS total = 1 1 + 1 + 1 S1 S2 S3

Example S1 = 12mmS2 = 18mmS3 = 22mmS total = 1 1 + 1 + 1 12 18 22S total = 1 1 + 1 + 1 144 324 484S total = 1= 1 .00694 + .00309 + .00207 .0121S total = 9.09

*Cv = _________

Cv = _________

Cv = _________

Cv total = ________

*

S mm

3

4

6

7.5

9

2

2

0.2

0.02

Tube Length in m

0

10

20

30

40

50

60

0.1

1

5

10

0.5

0.05

*

Tube

Material

Length

Fittings

Total

Dia.

1 m

0.5 m

Insert type

One Touch

0.5 m tube +

(mm)

straight

elbow

straight

elbow

2 strt. fittings

4 x 2.5

N,U

1.86

3.87

1.6

1.6

1.48

5.6

4.2

3.18

6 x 4

N,U

6.12

7.78

6

6

3.72

13.1

11.4

5.96

8 x 5

U

10.65

13.41

11

(9.5) 11

6.73

18

14.9

9.23

8 x 6

N

16.64

20.28

17

(12) 16

10.00

26.1

21.6

13.65

10 x 6.5

U

20.19

24.50

35

(24) 30

12.70

29.5

25

15.88

10 x 7.5

N

28.64

33.38

30

(23) 26

19.97

41.5

35.2

22.17

12 x 8

U

33.18

39.16

35

(24) 30

20.92

46.1

39.7

25.05

12 x 9

N

43.79

51.00

45

(27) 35

29.45

58.3

50.2

32.06

Table 7.30 Equivalent Flow Section of current tube connections

*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)

2

Equivalent Flow Section in mm

13.8

27.6

41.4

55.2

69

82.8

110

138

207

276

160

10.6

21.1

31.7

42.2

52.8

62

84.4

106

158

211

140

8.4

16.8

25.2

33.6

42

50.4

67.2

84

126

168

125

5.4

10.8

16.2

21.6

27

32.4

43.2

54

81

108

100

3.4

6.8

10.2

13.6

17

20.4

27.2

34

51

68

80

2.1

4.2

6.3

8.4

10.5

12.6

16.8

21

31.5

42

63

1.4

2.7

4

5.4

6.8

8.1

10.8

13.5

20.3

27

50

0.85

1.7

2.6

3.4

4.3

5

6.8

8.5

12.8

17

40

0.55

1.1

1.7

2.2

2.8

3.7

4.4

5.5

8.5

11

32

0.35

0.67

1

1.3

1.7

2

2.7

3.4

5

6.7

25

0.2

0.4

0.6

0.8

1

1.2

1.6

2

3

4

20

0.12

0.23

0.36

0.46

0.6

0.72

1

1.2

1.8

2.4

12,16

0.1

0.1

0.15

0.2

0.25

0.3

0.4

0.5

0.75

1

dia. mm

50

100

150

200

250

300

400

500

750

1000

8,10

Average piston speed in mm/s

*Flow Amplification

*Signal Inversion

*Selection

green

red

*Memory Function

green

red

*Delayed switching on

*Delayed switching off

*Pulse on switching on

*Pulse on releasing a valve

*Direct Operation and Speed Control

*Control from two points: OR Function

Shuttle Valve

*Safety interlock: AND Function

*Safety interlock: AND Function132

*Inverse Operation: NOT Function

*Direct Control

P

A

B

*Holding the end positions

P

A

B

*Semi Automatic return of a cylinder

Cam valve

*Repeating Strokes

*Sequence Control

3

1

2

4

2

4

*

_915341647.unknown

b

1

A+

B+

b

0

a

1

A-

B-

a

o

start

*

Pressure

Regulator

Regulator

with

relief

ISO SYMBOLS for AIR TREATMENT EQUIPMENT

Water

Separator

Filter

Auto Drain

Air Dryer

Filter /

Separator

Filter /

Separator

w. Auto Drain

Multi stage

Micro Filter

Lubricator

Air

Heater

Heat

Exchanger

Air

Cooler

Basic

Symbol

Differential

Pressure

Regulator

Pressure

Gauge

FRL Unit, detailed

FRL Unit,

simplified

Refrigerated

Air Dryer

Adjustable

Setting

Spring

Air Cleaning and Drying

Pressure Regulation

Units

*

_978162463.doc

Single Acting Cylinder,

Spring retract

Single Acting Cylinder,

Spring extend

Double Acting Cylinder

Double Acting Cylinder with

adjustable air cushioning

Double Acting Cylinder,

with double end rod

Rotary Actuator,

double Acting

*

Return Spring (in fact not an operator, but a built-in element)

Mechanical (plunger):

Roller Lever:

one-way Roller Lever:

Manual operators: general:

Lever:

Push Button:

Push-Pull Button:

Detent for mechanical and manual operators (makes a monostable valve bistable):

Air Operation is shown by drawing the (dashed) signal pressure line to the side of the square; the direction of the signal flow can be indicated by a triangle:

Air Operation for piloted operation is shown by a rectangle with a triangle. This symbol is usually combined with another operator.

Direct solenoid operation

solenoid piloted operation

_978162460.doc

_978162461.doc

_978162459.doc

*

with Spring Return

Normally Open 3/2 valve

(normally passing)

Manually Operated,

to

to

to

to

Valve with Spring Return

normally closed 3/2

(non-passing)

Mechanically Operated,

Manual

Operation

Closed

Input

Input

connected

Output

Return

Spring

Air Supply

Exhaust

Manual

Operation

Closed

Input

Input

connected

Output

Return

Spring

OR

Mechanical

Operation

Input

connected

Output

Input closed,

Output

exhausted

Return

Spring

Air Supply

Exhaus

t

OR

Mechanical

Operation

Input

connected

Output

Input closed,

Output

exhausted

Return

Spring

*

Manually operated Valves

detent, must correspond with valve position

no pressure

3/2, normally closed/normally open

pressure

bistable valves: both positions possible

3/2, normally closed

no pressure

pressure

3/2, normally open

monostable valves never operated

Solenoids are never operated in rest

Air operated valves may be operated in rest

Electrically and pneumatically operated Valves

pressure

no pressure

No valve with index "1" is operated.

All valves with index "0" are operated.

Mechanically operated Valves

no pressure

an

0

pressure

an

0

an

1

an

1

no pressure

pressure

*

POWER Level

LOGIC Level

SIGNAL INPUT Level

First stroke of the cycle

Start

Memories,

AND's, OR's,

Timings etc.

A

A+

A-

B

B+

B-

Last stroke of the cycle

C

C

Codes: a , a , b , b , c and c .

1

0

1

0

1

0

**