Intro to Pneumatics Modified

7/31/2019 Intro to Pneumatics Modified

1/95

Introduction toPneumatics


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2

Air Production System Air Consumption System


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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


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4

Properties of compressed air

Availability

Storage

Simplicity of design and control

Choice of movement

Economy


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5

Properties of compressed air

Reliability

Resistance to Environment

Environmentally clean.

Safety


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What is Air?

NitrogenOxygen

Carbon Dioxide

Argon

Nitrous 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.


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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/m

3

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 80 100 120 140 160 180

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

HUMIDITY & DEWPOINT


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Pressure and Flow

Sonic Flow

Qn (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


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Air Treatment


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10

Compressing Air

One cubic foot of air

7.8 cubic feet of free air

One cubic foot of

100 psig

compressed air(at Standard conditions)with 7.8 times the

moisture and dirt

compressor

CFM vs SCFM

psig + 1 atm

1 atm

Compression

ratio=

Compressed air is always related at Standard conditions.


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11

Relative Humidity

Compressor

1 ft3 @100 psig

1950 F

100% RH

57.1

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

56.37

grams of

H20

.72

grams of

H20

Adsorbtion DryerCompressor

Exit

Reservoir

TankAirline

Drop


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Air Mains

Ring

Main

Dead-End

Main


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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 unitof 100,000 Pa.

100,000 Pa = 100 kPa = 1 bar

Atmospheric Pressure =14.696 psi =1.01325 bar =1.03323 kgf/cm2.

h i h ( l )


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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 V1V2

V2 = P1 x V1P2

Example P2 = ?

P1 = Pa (1.013bar)

V1 = 1m V2 = .5m

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 T2T1

T2 = T1 x V2V1

Example V2 = ?

V1 = 2m

T1 = 273K (0C) T2 = 303K (30C)

V2 = 2 x 303

273 = 2.219m

10


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

P1 x P2

T1 x T2

P2 = P1 x T2T1

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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400

2000

20000

250

500

1000

1500

2500

4000

5000

10000

15000

25000

4000050000

10000025 30

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

10 7 5(bar)p :

(mm)

F

(N

)

1250

12500

5

4

2.5

10

15

202530

40

50

100

500

1000

250

2.5 4 6 8 10 12 2016 (mm)

F(N)

125

150

200

400

300

12.5


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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%~ 1.5:1

Vertical.50%~ 2.0:1


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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) 7 1.5 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

50400 -- - - - 4 79.9

200 - _ 2 40100 (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 78.1500 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


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Speed control

The speed of a cylinder is define by theextra 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 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?


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24

b c d

x

A

B

h

G

y

R a

F=G F= G W =m/2 va2

F=G (sin + cos )

a db c

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

X axes, (horizontal lifting force).cos x x MTotal force = Y + X

=friction coefficients

E l


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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


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_____

F = ________ (N)

______kg

= __

Force Y = sin x M =

Force X = cos x x M =

Total Force = Y + X =


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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


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Relative humidity (r.h.) = actual water content X100% saturated quantity (dew point)

Example 1

T = 25C r.h = 65%

V = 1m

From table 3.7 air at 25C contains

23.76 g/m

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

13


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29

Relative HumidityExample 2

V = 10m

T1= 15C

T2= 25C

P1 = 1.013bar

P2 = 6bar

r.h = 65%

? H0will condense out

From 3.17, 15C = 13.04 g/m

13.04 g/m x 10m = 130.4 g

130.4 g x .65 r.h = 84.9 g

V2 = 1.013 x 10 = 1.44 m

6 + 1.013

From 3.17, 25C = 23.76 g/m

23.76 g/m x 1.44 m = 34.2 g

84.9 - 34.2 = 50.6 g

50.6 g of water will condense out

13


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V = __________m

T1= __________CT2= __________C

P1 =__________bar

P2 =__________barr.h =__________%

? __________H0

will condense out

l


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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


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Sonic / Subsonic flow

Example

P1 = 7bar

P2 = 6.3bar

S = 12mm

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


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P1 = _________bar

P2 = _________bar

S = _________mm

Q = ____?_____l/min


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34

Receiver sizing

Example

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

22


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37 30


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3100

4"


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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

0.7

0.80.9

0.25

1.75

2.5

2.25

pkPa / m

= bar /100 mPipe Length

2

1

0.5

0.1

1.5

0.2

0.3

0.4

0.01

0.050.04

0.03

0.02

0.015

0.15

0.025

90

80

70

60

50

40

30

20

15

25

35

Inner Pipe Dia. ,

mm

Reference

Line

X

3"

2.5"

2"

1.5"

1.25"

1"

3/4"

1/2"

3/8"

Q (m /s3n

33


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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 7.1180* 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 9.4

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

34


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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

3100

4"


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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

0.7

0.80.9

0.25

1.75

2.5

2.25

pkPa / m

= bar /100 mPipe Length

2

1

0.5

0.1

1.5

0.2

0.3

0.4

0.01

0.050.04

0.03

0.02

0.015

0.15

0.025

90

80

70

60

50

40

30

20

15

25

35

Inner Pipe Dia. ,

mm

Reference

Line

X

3"

2.5"

2"

1.5"

1.25"

1"

3/4"

1/2"

3/8"

Q (m /s3n

33


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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/s

Using the ring main example on page 29 size for the

following requirements:


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Auto

Drain

1

2

34

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 Air

Aftercooler

d

a

b

c

Auto

Drain

39


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Example

P = 7 bar (700,000 N/m) D = 63mm (.063m)

d = 15mm (.015m)

F = x (D -d) x P4

F = 3.14 x (.063 - .015) x 700,0004

F = 3.14 x (.003969 - .0.000225) x 700,0004

F = .785 x .003744 x 700,000

F = 2057.328 N

54


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400

2000

20000

250

500

1000

1500

2500

4000

5000

10000

15000

25000

40000

50000

10000025 30

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

10 7 5(bar)p :

(mm)

F

(N

)

1250

12500

5

4

2.5

10

15

202530

40

50

100

500

1000

250

2.5 4 6 8 10 12 2016 (mm)

F(N)

125

150

200

400

300

12.5

E l


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Example

M = 100kg

P = 5bar

= 32mm

= 0.2

F = /4 x Dx 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/s , therefor

185.7 N = 185.7 kg m/s

divide mass into remaining force

m/s = 185.7 kg m/s

100kg

= 1.857 m/s


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M = _______kg

P = _______bar

= _______mm

= 0.2

F = /4 x Dx P = 401.9 N

Air Flow and Consumption


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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 10004


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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.

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 0.99350 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 Flow


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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 theoreticalvolume 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 10004

Working Pressure in bar


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52

g

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 1.39150 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 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.


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Formulae comparison


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 D ( ) (P P ) t k ( ) # t k / i 1000


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54


= _______mm

stroke = _______mm

s/min = _______ x 2

P =_______bar

I ti


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Inertia

Example 1

m = 10kg

a = 30mm

j = ___?

J= m (kg) x a (m)

12

J= 10 x .03

12

J= 10 x .000912

J = .00075

a


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Inertia

Example 2

m = 9 kg

a = 10mm

b = 20mm

J = ___?

J = ma x a + mb x b

3 3

J = 3 x .01 + 6 x .02

3 3

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

J = .0001 + .0008

J = .0009

a b


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57

a b

m = ________ kg

a = _________mm

b = _________mm

J = _________?


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58

Valve identification

A(4) B(2)

EA P EB(5) (1) (3)


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59

Valve Sizing

The Cv factor of 1 is a flow capacity ofone US Gallon of water per minute, witha pressure drop of 1 psi.

The kv factor of 1is a flow capacity ofone liter of water per minute with apressure drop of 1 bar.

The equivalent Flow Section S of avalve is the flow section in mm2 of anorifice in a diaphragm, creating thesame 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 +

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

Fl m l


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61

Flow example

S = 35

P1 = 6 bar

P2 =5.5 bar

= 25C

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 273298

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


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62

Cv = ________between 1 -5

P1 = ________bar

P2 = ________5 bar

= ________C

Flow capacity formulae transposed


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63

p y p

Cv = Q

400 x (P2 + 1.013) x P

Kv = Q

27.94 x (P2 + 1.013) x P

S = Q22.2 x (P2 + 1.013) x P

Fl it l


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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 = 75022.2 x 9.113 x .9

S = 750

22.2 x 2.86

S = 750 S = 11.8163.49


65/95

65

Q = _________ l/min

P1 = _________ bar

P = _________%

Cv = _________ ?

O ifi i i ti


66/95

66

Orifices in a series connection S total = 1

1 + 1 + 1S1 S2 S3

Example

S1 = 12mm

S2 = 18mm S3 = 22mm

S total = 11 + 1 + 1

12 18 22

S total = 11 + 1 + 1

144 324 484

Stotal

= 1 = 1.00694 + .00309 + .00207 .0121

S total = 9.09


67/95

67

Cv = _________

Cv = _________

Cv = _________

Cv total = ________


68/95

68

Tube Length in

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


69/95

69

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. fittings4 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 10.0026.1 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 25.0512 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


70/95

70

Average piston speed in mm/s

dia. mm 50 100 150 200 250 300 400 500 750 10008,10 0.1 0.1 0.15 0.2 0.25 0.3 0.4 0.5 0.75 1

12,16 0.12 0.23 0.36 0.46 0.6 0.72 1 1.2 1.8 2.420 0.2 0.4 0.6 0.8 1 1.2 1.6 2 3 4

25 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 11

40 0.85 1.7 2.6 3.4 4.3 5 6.8 8.5 12.8 17

50 1.4 2.7 4 5.4 6.8 8.1 10.8 13.5 20.3 27

63 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 81 108

125 8.4 16.8 25.2 33.6 42 50.4 67.2 84 126 168

140 10.6 21.1 31.7 42.2 52.8 62 84.4 106 158 211

160 13.8 27.6 41.4 55.2 69 82.8 110 138 207 276

Equivalent Flow Section in mm2

Table 7.31Equivalent Section S in mm2 for the valve and the tubing, for6 bar working pressure and a pressure drop of 1 bar (Qn Conditions)

Flow Amplification


71/95

71

p

Signal Inversion


72/95

72

g

Selection


73/95

73

greenred

Memory Function


74/95

74

greenred

Memory Function


75/95

D l d it hi ff


76/95

76

Delayed switching off

Pulse on switching on


77/95

77

Pulse on switching on

P l l i l


78/95

78

Pulse on releasing a valve


79/95

79

Direct Operation and Speed Control

Control from two points: OR Function


80/95

80

Shuttle Valve

Safety interlock: AND Function


81/95

81

Safety interlock: AND Function


82/95

82

1

3

2


83/95

83

Inverse Operation: NOT Function


84/95

84

P

AB

Direct Control


85/95

85

P

ABHolding the end positions


86/95

86

Cam valve

Semi Automatic return of a cylinder


87/95

87

Repeating Strokes

2 4


88/95

88

3

1 2

4

Sequence Control


89/95

89Commands

Signals Start

A+ B+ A- B-

b0b1 a0a1

b1

A+ B+

b0 a1

A- B-

aostart

ISO SYMBOLS for AIR TREATMENT EQUIPMENT

Air Cleaning and Drying


90/95

90

PressureRegulator

Regulatorwith relief

WaterSeparator

Filter

Auto Drain Air Dryer

Filter /Separator

Filter /Separator

w. Auto Drain

Multi stageMicro Filter

Lubricator

Air

Heater

Heat

Exchanger

AirCooler

BasicSymbol

DifferentialPressure

Regulator

PressureGauge

FRL Unit, detailed

FRL Unit,

simplified

Refrigerated

Air Dryer

AdjustableSetting

Spring

Pressure Regulation

Units


91/95

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


92/95

92

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

ManualOperation

ClosedInput

Inputconnected to

OutputReturnSpring

ManualOperation

ClosedInput

Inputconnected to

OutputReturnSpring


93/95

93

Air SupplyExhaust

Manually Operated,

Normally Open 3/2 valve(normally passing)

with Spring

OR

MechanicalOperation

Inputconnected to

Output

Input closed,Output

exhaustedReturnSpring

Air Supply Exhaust

Mechanicallynormally closed 3/2

(non-passing)

Valve with Spring Return

OR

MechanicalOperation

Inputconnected to

Output

Input closed,Output

exhaustedReturnSpring

Manually operated Valvesdetent, must correspond with valve position

no pressure pressureno pressure pressure


94/95

94

3/2, normally closed/normally openbistable valves: both positions possible

3/2, normally closed 3/2, normally openmonostable 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

an1an1

All valves with index "0" are operated.

an 0

pressure

an 0

no pressure

First stroke of the cycleA B

Last stroke of the cycleC


95/95

POWER Level

LOGIC Level

SIGNAL INPUT Level

Start

Memories,AND's, OR's,

Timings etc.

A+ A- B+ B- C

Codes: a , a , b , b , c and c .1010 10

Documents

Intro to Pneumatics Modified