Introduction to Pneumatics

2

Air Production System Air Consumption System

3

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

4

Properties of compressed air

• Availability

• Storage

• Simplicity of design and control

• Choice of movement

• Economy

5

Properties of compressed air

• Reliability

• Resistance to Environment

• Environmentally clean.

• Safety

6

What is Air?NitrogenOxygenCarbon DioxideArgonNitrous OxideWater Vapor

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 inch

column of air(from sea level

to the outer atmosphere,@ 680 F, & 36% RH)

is 14.69 pounds.

7

Temperature °C 0 5 10 15 20 25 30 35 40

g/m3

n *(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/m3

n (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 180g/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 –40g/ft3

(Standard) .137 .126 .083 .053 .033 .020 .012 .007 .004g/ft3 (Atmospheric) .137 .127 .085 .056 .036 .023 .014 .009 .005

HUMIDITY & DEWPOINT

8

Pressure and Flow

Sonic FlowRange

Q n (54.44 l / min)

S = 1 mm 2

0 20 40 80 100 12060

10

9

8

7

6

5

4

3

2

1

(dm /min)3

nQ

p (bar)

Example

P1 = 6bar

P = 1bar

P2 = 5bar

Q = 54 l/min

(1 Bar = 14.5 psi)

P1

P2

9

Air Treatment

10

Compressing Air

One cubic foot of air

7.8 cubic feet of free air

One cubic foot of100 psig

compressed air

(at Standard conditions) with 7.8 times the moisture and dirt

compressor

CFM vs SCFM

psig + 1 atm

1 atm

Compressionratio

=

Compressed air is always related at Standard conditions.

11

Relative Humidity

Compressor

1 ft3 @100 psig1950 F

100% RH57.1

grams of H20

1 ft3 @100 psig 770 F

100% RH.73

grams of H20

1 ft3 @100 psig -200 F

100% RH.01

grams of H20

1 ft3 @100 psig 770 F

0.15% RH.01

grams of H20

56.37grams of H20

.72grams of H20

Adsorbtion DryerCompressorExit

Reservoir Tank

Airline Drop

12

Air Mains

Ring Main

Dead-End Main

13

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.

14

Isothermic change (Boyle’s 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

15

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 = 273°K (0°C)

• T2 = 303°K (30°C)

• V2 = 2 x 303 273

• = 2.219m³

10

16

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 T2T1

• T2 = T1 x P2P1

• Example P2 = ?

• P1 = 4bar

• T1 = 273°K (O°C)

• T2 = 298°K (25°C)

• P2 = 4 x 298 273

• = 4.366bar

17

P1 = ________bar

T1 = _______°C ______°K

T2 = _______°C ______°K

18

400

2000

20000

250

500

1000

1500

2500

40005000

10000

15000

25000

4000050000

10000025 30

32 40 50 63 80 100 125 140 160 200 250 300

10 7 5(bar)p :

ø (mm)

F (

N)

1250

12500

54

2.5

10

15

202530

4050

100

500

1000

250

2.5 4 6 8 10 12 2016ø (mm)

F (

N)

125150

200

400

300

12.5

19

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.

20

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%~ 1.5:1 – Vertical…….50%~ 2.0:1

21

Cyl.Dia Mass (kg) 60° 45° 30° µ

0.01µ 0.2 µ

0.01µ 0.2 µ

0.01µ 0.2 µ

0.01µ 0.2

25 100 – – – – – – – 4 80

50 – – – – – – – 2.2 4025 – (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 43.945 - (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 40100 – (87) (96.5) 71.3 84.8 50.8 67.3 1 2050 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 40400 – (87) (96.5) 71.4 84.4 50.8 67.3 1 20200 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

22

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.

23

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°?

24

b c d

x

A

B

h

Gy

R a

F = G F = µ ·G W =m /2 · va2

F =G · (sin + µ · cos)

a db c

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

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

Total force = Y + X

= friction coefficients

25

Example

40°

F = ________ (N) 150kg

= .01

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

26

_____°

F = ________ (N)

______kg

= __

Force Y = sin x M =

Force X = cos x x M =

Total Force = Y + X =

27

Temperature °C 0 5 10 15 20 25 30 35 40

g/m3

n *(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/m3

n (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

13

28

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

• Example 1

• 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

13

29

Relative Humidity Example 2

• V = 10m³• 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

13

30

V = __________m³T1= __________°CT2= __________°CP1 =__________barP2 =__________barr.h =__________%? __________H²0

will condense out

31

Formulae, for when more exact values are required

• Sonic 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

32

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 < 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

16,17

33

P1 = _________bar

P2 = _________bar

S = _________mm²

Q = ____?_____l/min

34

Receiver sizing• Example

• V = capacity of receiver

• Q = compressor output l/min

• Pa = atmospheric pressure

• P1 = compressor output pressure

• V = Q x PaP1 + Pa

• If

– Q = 5000

– P1 = 9 bar

– Pa = 1.013

• V = 5000 x 1.013 9 + 1.013

• V = 506510.013

• V = 505.84 liters

22

3529

3629

37

The Water remains in the Pipe

The Water runs into the Auto Drain

a b

30

38

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

31

39

2

3

4

5

6

7

8

9

10

11

12

Line Pressure

(bar)

3.0

1.0

2.0

1.5

0.5

0.4

0.3

0.2

0.15

0.6

0.70.80.9

0.25

1.75

2.5

2.25

²p kPa / m

= bar /100 m Pipe Length

2

1

0.5

0.1

3

1.5

0.2

0.3

0.4

0.01

0.050.04

0.03

0.02

0.015

0.15

0.025

100

90

80

70

60

50

40

30

20

15

25

35

Inner Pipe Dia., mm

Reference Line

X

4"

3"

2.5"

2"

1.5"

1.25"

1"

3/4"

1/2"

3/8"

Q (m /s3n

33

40

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 3.290* 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 7.1180* 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 9.4Gate 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 1.5Side 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

34

41

Sizing compressor air mains• Example 2• 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

31

42

2

3

4

5

6

7

8

9

10

11

12

Line Pressure

(bar)

3.0

1.0

2.0

1.5

0.5

0.4

0.3

0.2

0.15

0.6

0.70.80.9

0.25

1.75

2.5

2.25

²p kPa / m

= bar /100 m Pipe Length

2

1

0.5

0.1

3

1.5

0.2

0.3

0.4

0.01

0.050.04

0.03

0.02

0.015

0.15

0.025

100

90

80

70

60

50

40

30

20

15

25

35

Inner Pipe Dia., mm

Reference Line

X

4"

3"

2.5"

2"

1.5"

1.25"

1"

3/4"

1/2"

3/8"

Q (m /s3n

33

43

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 = m³/s

Using the ring main example on page 29 size for the following requirements:

44

AutoDrain

1

2

3

4

5

6

7

RefrigeratedAir Dryer

Compressor

Tank

a

a

a

a

a

b

b

b

c

d

Micro Filter

Sub-micro Filter

Odor Removal Filter

Adsorbtion AirDryer

Aftercooler

d

a

b

c

AutoDrain

39

45

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

54

46

400

2000

20000

250

500

1000

1500

2500

40005000

10000

15000

25000

4000050000

10000025 30

32 40 50 63 80 100 125 140 160 200 250 300

10 7 5(bar)p :

ø (mm)

F (

N)

1250

12500

54

2.5

10

15

202530

4050

100

500

1000

250

2.5 4 6 8 10 12 2016ø (mm)

F (

N)

125150

200

400

300

12.5

47

Example

• 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

– 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

48

M = _______kg

P = _______bar

= _______mm

= 0.2

F = /4 x D²x P = 401.9 N

49

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.

a

b c

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

50

• 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.

Working Pressure in barPiston dia. 3 4 5 6 7

20 0.124 0.155 0.186 0.217 0.24825 0.194 0.243 0.291 0.340 0.38832 0.319 0.398 0.477 0.557 0.63640 0.498 0.622 0.746 0.870 0.99350 0.777 0.971 1.165 1.359 1.55363 1.235 1.542 1.850 2.158 2.46580 1.993 2.487 2.983 3.479 3.975100 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

51

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

52

Working Pressure in barPiston dia. 3 4 5 6 7

20 0.174 0.217 0.260 0.304 0.34725 0.272 0.340 0.408 0.476 0.54332 0.446 0.557 0.668 0.779 0.89040 0.697 0.870 1.044 1.218 1.39150 1.088 1.360 1.631 1.903 2.17463 1.729 2.159 2.590 3.021 3.45180 2.790 3.482 4.176 4.870 5.565100 4.355 5.440 6.525 7.611 8.696

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

• 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.

53

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

54

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

55

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

56

Inertia• Example 2

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

• 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

a b

57

a b

m = ________ kg

a = _________mm

b = _________mm

J = _________?

58

Valve identification

A(4) B(2)

EA P EB

(5) (1) (3)

59

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.

60

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 14.3 180.07 1 1.260.055 0.794 1

61

Flow example

• S = 35• 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

62

Cv = ________between 1 -5

P1 = ________bar

P2 = ________5 bar

= ________°C

63

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

•

64

Flow capacity example

• Q = 750 l/min

• P1 = 9 bar P = 10%• S = ?

• S = Q22.2 x (P2 + 1.013) x P

• S = 75022.2 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

65

Q = _________ l/min

P1 = _________ bar

P = _________%

Cv = _________ ?

66

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

67

Cv = _________

Cv = _________

Cv = _________

Cv total = ________

68

Tube Length inm

0

10

20

30

40

50

60

0.1 1 5 100.50.050.02 0.2 2

2

9

7.5

6

4

3

S mm

69

Tube Material Length Fittings TotalDia. 1 m 0.5 m Insert type One Touch 0.5 m tube +

(mm) straight elbow straight elbow 2 strt. fittings4 x 2.5 N,U 1.86 3.87 1.6 1.6 1.48

5.6 4.2 3.186 x 4 N,U 6.12 7.78 6 6 3.72

13.1 11.4 5.968 x 5 U 10.65 13.41 11 (9.5) 11 6.73

18 14.9 9.238 x 6 N 16.64 20.28 17 (12) 16 10.00

26.1 21.6 13.6510 x 6.5 U 20.19 24.50 35 (24) 30 12.70

29.5 25 15.8810 x 7.5 N 28.64 33.38 30 (23) 26 19.97

41.5 35.2 22.1712 x 8 U 33.18 39.16 35 (24) 30 20.92

46.1 39.7 25.0512 x 9 N 43.79 51.00 45 (27) 35 29.45

58.3 50.2 32.06Table 7.30 Equivalent Flow Section of current tube connections

70

Average piston speed in mm/sdia. mm 50 100 150 200 250 300 400 500 750 1000

8,10 0.1 0.1 0.15 0.2 0.25 0.3 0.4 0.5 0.75 112,16 0.12 0.23 0.36 0.46 0.6 0.72 1 1.2 1.8 2.4

20 0.2 0.4 0.6 0.8 1 1.2 1.6 2 3 425 0.35 0.67 1 1.3 1.7 2 2.7 3.4 5 6.732 0.55 1.1 1.7 2.2 2.8 3.7 4.4 5.5 8.5 1140 0.85 1.7 2.6 3.4 4.3 5 6.8 8.5 12.8 1750 1.4 2.7 4 5.4 6.8 8.1 10.8 13.5 20.3 2763 2.1 4.2 6.3 8.4 10.5 12.6 16.8 21 31.5 4280 3.4 6.8 10.2 13.6 17 20.4 27.2 34 51 68

100 5.4 10.8 16.2 21.6 27 32.4 43.2 54 81 108125 8.4 16.8 25.2 33.6 42 50.4 67.2 84 126 168140 10.6 21.1 31.7 42.2 52.8 62 84.4 106 158 211160 13.8 27.6 41.4 55.2 69 82.8 110 138 207 276

Equivalent Flow Section in mm2

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)

71

Flow Amplification

72

Signal Inversion

73

greenred

Selection

74

greenred

Memory Function

75

Delayed switching on

76

Delayed switching off

77

Pulse on switching on

78

Pulse on releasing a valve

79

Direct Operation and Speed Control

80

Shuttle Valve

Control from two points: OR Function

81

Safety interlock: AND Function

82

Safety interlock: AND Function

1

3

2

83

Inverse Operation: NOT Function

84

P

AB

Direct Control

85

P

ABHolding the end positions

86

Cam valve

Semi Automatic return of a cylinder

87

Repeating Strokes

88

3

1 2

4

2 4

Sequence Control

89Commands

Signals Start

A+ B+ A- B-

b0b1 a0a1

b1

A+ B+

b0 a1

A- B-

aostart

90

PressureRegulator

Regulatorwith relief

ISO SYMBOLS for AIR TREATMENT EQUIPMENT

WaterSeparator

Filter

Auto Drain Air Dryer

Filter /Separator

Filter /Separator

w. Auto Drain

Multi stageMicro Filter

Lubricator

AirHeater

Heat Exchanger

Air Cooler

BasicSymbol

DifferentialPressureRegulator

PressureGauge

FRL Unit, detailed

FRL Unit,simplified

RefrigeratedAir Dryer

AdjustableSettingSpring

Air Cleaning and Drying

Pressure Regulation

Units

91

Single Acting Cylinder,Spring retract

Single Acting Cylinder,Spring extend

Double Acting Cylinder Double Acting Cylinder withadjustable air cushioning

Double Acting Cylinder,with double end rod

Rotary Actuator,double Acting

92

R e tu rn S p rin g ( in fa c t n o t a no p e ra to r , bu t a b u ilt - in e le m e n t )

M e c h a n ic a l (p lu ng e r) :

R o lle r L e v e r : o n e -w a y R o lle r L e v e r :

M a n u a l o p e ra to rs : g e n e ra l: L e ve r :

P u sh B u t to n : P u sh -P u ll B u t to n :

D e te n t fo r m e c h a n ic a l a nd m a n u a l o p e ra to rs (m a k e s a m o no sta b le va lv e b is ta b le ) :

A ir O p e ra t io n is sh o w n b y d ra w ing th e (d a sh e d ) s ig na l p re ssu re lin e to th e s id e o fth e sq u a re ; the d ire c t io n o f the s ig n a l flo w c a n b e in d ic a te d by a t r ia ng le :

A ir O p e ra t io n fo r p ilo te d o p e ra t io n is sh o w n by a re c ta n g le w ith a t r ia n g le . T h issym bo l is u su a lly c o m bin e d w ith a n o th e r o p e ra to r .

D ire c t so le n o id o p e ra t io n so le no id p ilo te d o p e ra t io n

93

ManualOperation

ClosedInput

Inputconnected to

OutputReturnSpring

Air SupplyExhaust

Manually Operated,Normally Open 3/2 valve (normally passing)

with SpringReturn

ManualOperation

ClosedInput

Inputconnected to

OutputReturnSpring

OR

MechanicalOperation

Inputconnected to

Output

Input closed,Output

exhaustedReturnSpring

Air Supply Exhaust

MechanicallyOperated,normally closed 3/2

(non-passing)Valve with Spring Return

OR

MechanicalOperation

Inputconnected to

Output

Input closed,Output

exhaustedReturnSpring

94

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

pressureno pressure

pressure

Mechanically operated Valves

No valve with index "1" is operated.no pressure

a…n1 a…n1

All valves with index "0" are operated.

a…n0

pressure

a…n0

no pressure

95

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 .1010 10