Compressed Air Distribution

2

Air Production System Air Consumption System

What can Pneumatics do?

3

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

What can Pneumatics do?

4

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

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

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 80 100 120 140 160 180

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

What is Air?

8

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.

The weight of a

one square inch

column of air

(from sea level

to the outer atmosphere,

@ 680 F, & 36% RH)

is 14.69 pounds.

Pressure and Flow

9

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

Air Treatment

10

Compressing Air

11

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.

Relative Humidity

12

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 Dryer Compressor

Exit

Reservoir

Tank Airline

Drop

Air Mains

13

Ring

Main

Dead-End

Main

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 (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 V1 P2

Example P2 = ?

P1 = Pa (1.013bar)

V1 = 1m

V2 = .5m

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 = 2m

T1 = 273K (0C)

T2 = 303K (30C)

V2 = 2 x 303 273

= 2.219m

16 10

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 T2 T1

T2 = T1 x P2 P1

Example P2 = ?

P1 = 4bar

T1 = 273K (OC)

T2 = 298K (25C)

P2 = 4 x 298 273

= 4.366bar

17

18

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

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.

19

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

20

21

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 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 78.1

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 control

The speed of a cylinder is defined by the extra force behind the piston, above the force opposed by the load

The lower the load ratio, the better the speed control.

22

23 29

24 29

25 30

26

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

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

Pressure and Flow

28

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

29

Compression Ratio

30

Cylinder Diameter

31

Load Ratio

32

Relative Humidity

33

Receiver Sizing