Introduction to Pneumatics

Air Production System Air 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 air Availability Storage Simplicity of design and control Choice of movement Economy

Properties of compressed air Reliability Resistance to Environment Environmentally clean. Safety

What is Air? The weight of a 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.

HUMIDITY & DEWPOINT

Pressure and Flow Example P1 = 6bar  P = 1bar P2 = 5bar Q = 54 l/min (1 Bar = 14.5 psi) P1 P2

Air Treatment

with 7.8 times the moisture and dirt 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.

Relative Humidity 100% RH .73 grams of H20 100% RH .01 grams of H20 Compressor Exit Reservoir Tank Adsorbtion Dryer Airline Drop Compressor 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 1950 F 100% RH 57.1 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

Air Mains Dead-End Main Ring 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.

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

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

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 = 273°K (O°C) T2 = 298°K (25°C) P2 = 4 x 298 273 = 4.366bar

P1 = ________bar T1 = _______°C ______°K T2 = _______°C ______°K

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.

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

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.

Angle of Movement 1. 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 M X axes, (horizontal lifting force)….cos x  x M Total force = Y + X  = friction coefficients

Example F = ________ (N) Force Y = sin  x M = .642 x 150 = 96.3 N  = .01 F = ________ (N) 150kg 40° 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

Force Y = sin  x M = Force X = cos  x  x M = Total Force = Y + X =  = __ ______kg _____° Force Y = sin  x M = Force X = cos  x  x M = Total Force = Y + X = F = ________ (N)

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Relative humidity (r. h. ) = actual water content X 100% 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

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

V = __________m³ T1= __________°C T2= __________°C P1 =__________bar P2 =__________bar r.h =__________% ? __________H²0 will condense out

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

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

P1 = _________bar P2 = _________bar S = _________mm² Q = ____?_____l/min

Receiver sizing V = 505.84 liters If Example Q = 5000 P1 = 9 bar Pa = 1.013 V = 5000 x 1.013 9 + 1.013 V = 5065 10.013 V = 505.84 liters Example V = capacity of receiver Q = compressor output l/min Pa = atmospheric pressure P1 = compressor output pressure V = Q x Pa P1 + Pa 22

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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 = m³/s 30 = .24 kPa/m 125 16800 x .00001667 = 0.28 m³/s chart lines on Nomogram 31

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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

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P = .5 bar (_________kPa) L = 200 m pipe length P = kPa/m L Using the ring main example on page 29 size for the following requirements: Q = 20,000 l/min P1 = 10 bar (_________kPa) P = .5 bar (_________kPa) L = 200 m pipe length P = kPa/m L l/min x .00001667 = m³/s

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Example F = 2057.328 N P = 7 bar (700,000 N/m²) D = 63mm (.063m) 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

Example 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 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

M = _______kg P = _______bar  = _______mm  = 0.2 F = /4 x D²x P = 401.9 N

Air Flow and Consumption Air 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 = D² (m) x  x (P + Pa) x stroke(m) x # strokes/min x 1000 4

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.

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

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.

Formulae comparison Q = 473.54 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

 = _______mm stroke = _______mm s/min = _______ x 2 P =_______bar 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

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

Inertia J = ma x a² + mb x b² 3 3 Example 2 J = 3 x .01² + 6 x .02² 3 3 J = 3 x .0001 + 6 x .0004 3 3 J = .0001 + .0008 J = .0009 Example 2 m = 9 kg a = 10mm b = 20mm J = ___? a b

m = ________ kg a = _________mm b = _________mm J = _________? a b

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

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.

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 + 

S = 35 P1 = 6 bar P2 =5.5 bar  = 25°C Flow example Q = 1405.383 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 S = 35 P1 = 6 bar P2 =5.5 bar  = 25°C

Cv = ________between 1 -5 P1 = ________bar P2 = ________5 bar  = ________°C

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

Q = 750 l/min P1 = 9 bar P = 10% S = ? S = Q 22.2 x (P2 + 1.013) x P 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 = 750 22.2 x 9.113 x .9 S = 750 22.2 x 2.86 S = 750 S = 11.81 63.49

Q = _________ l/min P1 = _________ bar P = _________% Cv = _________ ?

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

Cv = _________ Cv total = ________

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)

Flow Amplification

Signal Inversion

Selection

Memory Function

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

Safety interlock: AND Function

Safety interlock: AND Function 3 2 1

Inverse Operation: NOT Function

Direct Control

Holding the end positions

Semi Automatic return of a cylinder

Repeating Strokes

Sequence Control