Vinayak Bindu Basic Pneumatic Syste

8/7/2019 Vinayak Bindu Basic Pneumatic Syste

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Introduction toPneumatics

By- Prof. S.B. CHIKALTHANKAR


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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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6

What is Air?

Nitrogen

Oxygen

Carbon Dioxide

Argon

Nitrous Oxide

Water 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.

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been corrupted. Restart your computer,and then open the file again. If the red xstill appears, you may have to delete theimage and then insert it again.

The weight of a

one square inch

column of air(from sea level

to the outer atmosphere,

@ 680F, & 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/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 80 100 120 140 160 180

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

Q n (54.44 l / min)

S = 1 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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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 anddirt

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 HumidityCompressor

1 ft3

@100 psig1950 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 SIunit 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 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. 27

3

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

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 20 16 (mm)

F(N)

125

150

200

400

300

12.5


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Force formula transposed

D = 4 x FE

T 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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C y l.D ia M a ss (k g ) o 60 45 30 m

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 40100 (87) (96.5) 71.3 8 4.8 5 0.8 6 7.3 1 20

50 50 43.5 4 8.3 3 5.7 4 2.4 2 5.4 3 3.6 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 1 600 4 79.9

800 2 40

400 (87) (96.5) 71.4 8 4.4 5 0.8 6 7.3 1 20

200 50 43.5 48.3 35.7 42.2 25.4 33.6 0.5 10

Tab le 6 .16 Load Ra tios for 5 bar working pressure an d fr iction coefficients of 0.01 and 0.2


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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 Movement1. If we totally neglect friction, which cylinder diameter is needed to

horizontally push a load with an 825 kg mass with a pressure of6 bar;speed is not important.

2. Which cylinder diameter is necessary to lift the same mass with the

same pressure of6 bar vertically if the load ratio can not exceed 50%.

3. Same conditions as in #2 except from vertical to an angle of30.

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

h

y

E

E

R a

= = = / 2 v a 2

= (s in E + cosE )

a db c

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

X axes, (horizontal lifting force).cosE x Q x M

Total force = Y + X

Q = friction coefficients


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Example

40E

F = ________ (N)150kg

Q = .01

Force Y = sin E x M = .642 x 150 = 96.3 N

Force X = cos E x Q 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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_____E

F = ________ (N)

______kg

Q = __

Force Y = sin E x M =

Force X = cos E x Q 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/m

3(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 X 100%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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Relative Humidity Example 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 m6 + 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


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Formulae, for when more exact values are required

Sonic flow = P1 + 1.013 > 1.896 x (P2 + 1,013)

Pneumatic systems cannotoperate 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 = 506510.013

V = 505.84 liters

22


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35 29


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36 29


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37

The Water remains

in the Pipe

The Water runs into the

Auto Drain

a b

30


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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 = m/s

30 = .24 kPa/m125

16800 x .00001667 = 0.28 m/s

chart lines on Nomogram

31


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2

3

4

5

6

7

8

9

10

11

12

Line

ressure

(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

p

k

a /

= bar /100

ipe Length

2

1

0.5

0.1

3

1.5

0.2

0.3

0.4

0.01

0.05

0.04

0.03

0.02

0.015

0.15

0.025

100

90

80

70

60

50

40

30

20

15

25

35

Inner

ipe

ia. ,

eference

Line

4"

3"

2.5"

2"

1.5"

1.25"

1"

3/4"

1/2"

3/8"

Q ( /s3n

33


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Type f Fitti g N i l pipe ize ( )

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 4.20Equivalent Pipe Lengths for the main fittings

34


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2

3

4

5

6

7

8

9

10

11

12

Line

ressure

(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

p

k

a /

= bar /100

ipe Length

2

1

0.5

0.1

3

1.5

0.2

0.3

0.4

0.01

0.05

0.04

0.03

0.02

0.015

0.15

0.025

100

90

80

70

60

50

40

30

20

15

25

35

Inner

ipe

ia. ,

eference

Line

4"

3"

2.5"

2"

1.5"

1.25"

1"

3/4"

1/2"

3/8"

Q ( /s3n

33


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Q = 20,000 l/min

P1 = 10 bar (_________kPa)

(P = .5 bar (_________kPa)L = 200 m pipe length

(P = kPa/mL

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

3

4

5

6

7

Refrigerated

Air 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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46

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 :

(

)

(

)

1250

12500

5

4

2.5

10

15

202530

40

50

100

500

1000

250

2.5 4 6 8 10 12 20 16 (

)

(

)

125

150

200

400

300

12.5


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Example

M = 100kg P = 5bar

= 32mm

Q = 0.2

F = T/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) = 196N100

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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48

M = _______kg

P = _______bar

= _______mm

Q = 0.2

F = T/4 x Dx P = 401.9 N


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Air Flow and ConsumptionAir consumption of a cylinder is defined as:piston area x stroke length x number ofsingle strokes per minute x absolute pressure in bar.

a

b c

Q = D (m) xTx (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.

W rking Press re in r

Pist n di . 3 6

0.124 0.155 0.1 6 0.217 0.24

0.1 4 0.243 0.2 1 0.340 0.3

3 0.31 0.3 0.477 0.557 0.6360.4 0.622 0.746 0. 70 0. 3

0.777 0. 71 1.165 1.35 1.553

63 1.235 1.542 1. 50 2.15 2.465

1. 3 2.4 7 2. 3 3.47 3. 75

3.111 3. 6 4.661 5.436 6.211

Tab le 6 .19Theoret ical

ir onsu

pt ion of double a ct ing cyl inders fro

20 to 100

dia,

in l i ters per 100

stroke


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Peak Flow

For sizing the valve of an individual cylinder we need to

calculate Pea flow. The peak flow depends on thecylinders highest possible speed. The peak flow of all

simultaneously moving cylinders defines the flow to which

the FRL has to be sized.

To compensate foradiabatic 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) xTx (P + Pa) x stroke(m) x # strokes/min x 10004


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Wor ing ress re in ar

iston ia. 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

Tab le 6 .20 i r onsu p t ion o f doub le act ing cy l inders in l ite rs

per 100 stroke corrected for losses by adiaba t ic chan ge

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


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54


= _______mm

stroke = _______mm

s/min = _______ x 2

P =_______bar


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55

Inertia

Example 1

m = 10kg

a = 30mm

j = ___?

J= m (kg) x a (m)12

J= 10 x .03

12

J= 10 x .0009

12

J= .00075

a


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Inertia

Example 2

m = 9 kg

a = 10mm

b = 20mm

J = ___?

J = ma x a + mb x b3 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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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 of one USGallon of water per minute, with a pressure drop

of 1 psi.

The kv factor of 1 is a flow capacity of one liter ofwater per minute with a pressure drop of 1 bar.

The equivalent Flow Section S of a valve is the

flow section in mm

2

of an orifice in a diaphragm,creating the same relationship between pressure

and flow.


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60

Q = 400 x Cv x (P2 + 1.013) x (P x 273

273 + U

Q = 27.94 x kv x (P2 + 1.013) x (P x 273273 + U

Q = 22.2 x S x (P2 + 1.013) x (P x 273273 + U

1 Cv = 1 kv = 1 S =

The normal lo Qn or other various lo capacity units is: 981.5 68.85 54.44

The elationship bet een these units is as ollo s: 1 14.3 18

0.07 1 1.26

0.055 0.794 1


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61

Flow example

S = 35 P1 = 6 bar

P2 =5.5 bar

U = 25C

Q = 22.2 x S x (P2 + 1.013) x (P x 273

273 + U

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

U = ________C

Fl it f l t d


63/95

63

Flow capacity formulae transposed

Cv = Q

400 x (P2 + 1.013) x (P

Kv = Q27.94 x (P2 + 1.013) x (P

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


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64

Flow capacity example

Q =7

50 l/min P1 = 9 bar

(P = 10%

S = ?

S = Q22.2 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 = _________ ?


66/95

66

Orifices in a series connection S total = 1

1 + 1 + 1

S1 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

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

S total = 9.09


67/95

67

Cv = _________

Cv = _________

Cv = _________

Cv total = ________


68/95

68

Tube Le ng th in

0

1 0

2 0

3 0

4 0

5 0

6 0

0 .1 1 5 1 0 0 .50 .05 0 .0 2 0 .2 2

2

9

7 .5

6

4

3

S m m


69/95


70/95

70

A ve r a g e p i t p eed i /

dia . 5 0 10 0 1 50 2 0 0 25 0 30 0 40 0 5 0 0 7 5 0 1 0 0 0

8,1 0 0.1 0.1 0.1 5 0.2 0.2 5 0.3 0.4 0.5 0.7 5 11 2 ,1 6 0.1 2 0.2 3 0.3 6 0.4 6 0.6 0.7 2 1 1 .2 1 .8 2 .4

20 0.2 0.4 0.6 0.8 1 1 .2 1 .6 2 3 4

25 0.3 5 0.6 7 1 1 .3 1 .7 2 2 .7 3 .4 5 6.7

32 0.5 5 1 .1 1 .7 2 .2 2 .8 3 .7 4 .4 5.5 8.5 1 1

40 0.8 5 1 .7 2 .6 3 .4 4 .3 5 6.8 8.5 1 2 .8 1 7

50 1 .4 2 .7 4 5.4 6.8 8.1 10.8 1 3 .5 2 0.3 2 7

63 2 .1 4 .2 6.3 8.4 10.5 12 .6 16.8 2 1 3 1 .5 4 2

80 3 .4 6.8 1 0.2 1 3 .6 1 7 20 .4 27.2 34 5 1 6 8

1 0 0 5 .4 10.8 1 6.2 2 1 .6 2 7 32 .4 43 .2 5 4 8 1 1 0 8

1 2 5 8 .4 16.8 2 5.2 3 3 .6 4 2 50 .4 67.2 8 4 1 2 6 1 6 8

1 4 0 10 .6 21 .1 3 1 .7 4 2 .2 52 .8 6 2 84 .4 1 06 1 5 8 2 1 1

1 6 0 13 .8 27.6 4 1 .4 5 5.2 6 9 82 .8 11 0 1 38 2 0 7 2 7 6 Eq uiva len t Fl ect i n in 2

Table7.31 Equivalent ecti n in 2 f rthevalveand the tubing, f r6bar rkingpressureandapressuredropof1 bar( n Conditions)

l A lifi i


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71

Flow Amplification

Si l I i


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72

Signal Inversion

S l i


73/95

73

greenred

Selection


74/95

74

greenred

Memory Function


75/95


76/95


77/95

77

Pulse on switching on


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79/95

79

Direct Operation and Speed Control

C t l f t i t OR F ti


80/95

80

Sh ut tle Va lve

Control from two points: OR Function

Safety interlock: AND Function


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81

Safety interlock: AND Function


82/95


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

Ca va lve

Semi Automatic return of a cylinder


87/95

87

Repeating Strokes

2 4


88/95

88

3

1 2

4

2 4

Sequence Control


89/95

89Commands

Signals Start

A+ B+ A- B-

b 0b 1 a 0a1

b1

A + B +

b0 a1

A - B -

ao

s t a r t

IS O S Y M B O L S f o r A IR TR EATM E T EQ IPM E T

A i C l i d i


90/95

90

Pressureegulator

egulatori th rel ief

aterSeparator

Fi l t e r

Auto rain A ir rye r

Fi l t e r /Separator

Fi l t e r /Separator. Auto rain

Multi stageMicro Fi l t e r

Lubricator

Aireate r eat Exchang e rA irCoole r

BasicSy bo l

if fe rentia lPressure

R egulator

PressureG auge

FR L nit, de tailed

FR L nit ,

s i p li fi ed

R e frige ra tedA ir rye r

AdjustableSe tt ing

Spring

A ir Cleaning and ry ing

Pressur e R egulat ion

nits


91/95


92/95

92

Return pring (in act not an

operator, but a built-in element)echanical (plunger):

Roller ever: one- ay Roller ever:

anual operators: general: ever:

ush utton: ush- ull utton:

etent or mechanical and manual operators (makes a monostable valve bistable):

ir peration is sho n by dra ing the (dashed) signal pressure line to the side o

the square; the direction o the signal lo can be indicated by a triangle:

ir peration or piloted operation is sho n by a rectangle ith a triangle. This

symbol is usually combined ith another operator.

irect solenoid operation solenoid piloted operation

Manual Operation

ClosedInput

Inputconnected to

OutputReturnSpring

Manual Operation

ClosedInput

Inputconnected to

OutputReturnSpring


93/95

93

Operation Input Output Spring

Air SupplyExhaust

Man ua l ly Opera ted ,

Norm a l ly Ope n 3 /2 va l ve

(no rma l ly pass ing )

wi th Spr ing

Operation Input Output Spring

O R

MechanicalOperation

Inputconnected to

Output

Input closed,Output

exhaustedReturnSpring

A ir Supp ly E xhaust

Mechan i ca l l y no rma l l y c losed 3 /2

(non -pass ing )

Va l ve w i th Sp r i ng Re tu rn

O R

MechanicalOperation

Inputconnected to

Output

Input closed,Output

exhaustedReturnSpring

M a n u a l l y o p e r a t e d V a lv e sdetent, must co rrespond with valve posit ion


94/95

94

no pressure

3/2, normal ly c losed/normal ly open

pressure

bis table v alves: both po s i tions po ss ib le

3/2, normal ly c losed

no pressure pressure

3/2, normal ly ope n

monostable valves n e v e r operated

So len o id s ar e n ev er o p er at ed in re s t

A i r o p er at ed va lv es m ay b e o p er at ed in re s t

El ec t r i c a l l y an d p n eu m a t i c a l l y o p e ra ted Va lv es

p ressu reno p ressu re

pressure

M e c h a n i c a l l y o p e r a t e d V a lv e s

No v a l ve w i t h i n d e x " 1 " i s o p e ra t e d .no pressure

a n 1a n 1

A l l va l ve s w i t h i n d e x " 0 " a re o p e ra t e d .

a n 0

pressure

a n 0

no pressure

First s t roke of the cycle

A B

Last s t roke of the cycle

C


95/95

P O W E R L e v el

L O G IC L e v e l

SIG N A L IN P U T L e v e l

Start

Memor ies ,

AND 's , OR 's ,

Timings etc.

A + A - B + B - C

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

Documents

Vinayak Bindu Basic Pneumatic Syste