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Introduction to Pneumatics
*Air Production SystemAir 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 airAvailability
Storage
Simplicity of design and control
Choice of movement
Economy
*Properties of compressed airReliability
Resistance to Environment
Environmentally clean.
Safety
*What is Air?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 inchcolumn of air(from sea levelto the outer atmosphere,@ 680 F, & 36% RH)is 14.69 pounds.
*HUMIDITY & DEWPOINT
Temperature C
0
5
10
15
20
25
30
35
40
g/m3n *(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/m3n (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
*Pressure and FlowExampleP1 = 6bar P = 1barP2 = 5barQ = 54 l/min(1 Bar = 14.5 psi)
P1P2
Sonic Flow Range
Q
n
(54.44 l /
min)
S = 1 mm
2
0
20
40
80
100
120
60
10
9
8
7
6
5
4
3
2
1
(dm
/min)
3
n
Q
p
(bar)
*Air Treatment
*Compressing Air
*Relative HumidityAdsorbtion DryerCompressorExitReservoir TankAirline Drop
*Air MainsRing MainDead-End Main
*PressureIt 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 (Boyles Law)with constant temperature, the pressure of a given mass of gas is inversely proportional to its volumeP1 x V1 = P2 x V2
P2 = P1 x V1 V2
V2 = P1 x V1 P2Example P2 = ?P1 = Pa (1.013bar)V1 = 1mV2 = .5m
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 = T1V2 T2
V2 = V1 x T2 T1T2 = T1 x V2 V1
Example V2 = ?V1 = 2mT1 = 273K (0C)T2 = 303K (30C)
V2 = 2 x 303 273= 2.219m10
*Isochoric change Law of Gay Lussac at constant volume, the pressure is proportional to the temperature P1 x P2 T1 x T2P2 = P1 x T2T1T2 = T1 x P2P1Example P2 = ?P1 = 4barT1 = 273K (OC)T2 = 298K (25C)
P2 = 4 x 298 273= 4.366bar
*P1 = ________bar
T1 = _______C ______K
T2 = _______C ______K
*
_915341748.unknown
*Force formula transposed
D = 4 x FE x P
ExampleFE = 1600NP = 6 bar.D = 4 x 1600 3.14 x 600,000D = 6400 1884000D = .0583mD = 58.3mmA 63mm bore cylinder would be selected.
*Load RatioThis 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 ratiosHorizontal.70%~ 1.5:1 Vertical.50%~ 2.0:1
*
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
40
100
(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 controlThe 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 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?
*Y axes, (vertical lifting force).. sin x MX axes, (horizontal lifting force).cos x x MTotal force = Y + X = friction coefficients
a
b
c
d
b
c
d
x
A
B
h
G
y
a
a
R
a
F
=
G
F
=
G
W
=
m
/
2
v
a
2
F
=
G
(sin
a
+ cos
a
)
*Example40F = ________ (N) 150kg = .01Force 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
*_____F = ________ (N)______kg = __Force Y = sin x M =
Force X = cos x x M =
Total Force = Y + X =
*13
Temperature C
0
5
10
15
20
25
30
35
40
g/m3n *(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/m3n (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
*Relative humidity (r.h.) = actual water content X 100% saturated quantity (dew point)
Example 1
T = 25Cr.h = 65%V = 1mFrom table 3.7 air at 25C contains 23.76 g/m
23.76 g/m x .65 r.h = 15.44 g/m13
*Relative Humidity Example 2V = 10mT1= 15CT2= 25CP1 = 1.013barP2 = 6barr.h = 65%? H0 will condense outFrom 3.17, 15C = 13.04 g/m13.04 g/m x 10m = 130.4 g130.4 g x .65 r.h = 84.9 gV2 = 1.013 x 10= 1.44 m 6 + 1.013From 3.17, 25C = 23.76 g/m 23.76 g/m x 1.44 m = 34.2 g84.9 - 34.2 = 50.6 g
50.6 g of water will condense out13
*V = __________mT1= __________CT2= __________CP1 =__________barP2 =__________barr.h =__________%? __________H0 will condense out
*Formulae, for when more exact values are requiredSonic 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 flowExample
P1 = 7barP2 = 6.3barS = 12mml/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 7.3138.013 < 13.86 subsonic flow.Q = 22.2 x S x (P2 + 1.013) x PQ = 22.2 x 12 x (6.3 + 1.013) x .7Q = 22.2 x 12 x 7.313 x .7Q = 22.2 x 12 x 5.119Q = 22.2 x 12 x 2.26Q = 602 l/min16,17
*P1 = _________bar
P2 = _________bar
S = _________mm
Q = ____?_____l/min
*Receiver sizingExampleV = capacity of receiverQ = compressor output l/minPa = atmospheric pressureP1 = compressor output pressureV = Q x PaP1 + PaIf Q = 5000P1 = 9 barPa = 1.013
V = 5000 x 1.013 9 + 1.013V = 506510.013V = 505.84 liters22
*29
_915341790.unknown
*29
_915341789.unknown
*30
_915341788.unknown
*Sizing compressor air mainsExampleQ = 16800 l/minP1 = 9 bar (900kPa)P = .3 bar (30kPa)L = 125 m pipe lengthP = kPa/m Ll/min x .00001667 = m/s 30 = .24 kPa/m 12516800 x .00001667 = 0.28 m/schart lines on Nomogram31
*33
_915341785.unknown
*34
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
7.1
180* 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
*Sizing compressor air mainsExample 2Add fittings to example 1From table 4.202 elbows @ 1.4m = 2.8m2 90 @ 0.8m = 1.6m6 Tees @ 0.7m = 4.2m2 valves @ 0.5m = 1.0mTotal = 9.6m125m + 9.6 = 134.6m=135m
30kPa = 0.22kPa/m 135mChart lines on Nomogram31
*33
_915341785.unknown
*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/sUsing the ring main example on page 29 size for the following requirements:
*39
Sub-micro Filter
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
Odor Removal Filter
Adsorbtion Air Dryer
Aftercooler
d
a
b
c
Auto
Drain
*ExampleP = 7 bar (700,000 N/m)D = 63mm (.063m)d = 15mm (.015m)F = x (D -d) x P4F = 3.14 x (.063 - .015) x 700,000 4F = 3.14 x (.003969 - .0.000225) x 700,000 4F = .785 x .003744 x 700,000F = 2057.328 N54
*
_915341748.unknown
*ExampleM = 100kgP = 5bar = 32mm = 0.2
F = /4 x Dx P = 401.9 N
From chart 6.1690KG = 43.9% Lo.To find Lo for 100kg43.9 x 100= 48.8 % Lo. 90Calculate remaining force401.9 x 48.8 (.488) = 196N 100assume a cylinder efficiency of 95%196 x 95 = 185.7 N 100Newtons = kg m/s , therefor185.7 N = 185.7 kg m/sdivide mass into remaining forcem/s = 185.7 kg m/s 100kg
= 1.857 m/s
*M = _______kg
P = _______bar
= _______mm
= 0.2
F = /4 x Dx P = 401.9 N
*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 1000 4
_915341743.unknown
*Example.
= 80stroke = 400mms/min = 12 x 2P = 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 100Qt = 3.5 x 400 x 24 100Qt = 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
0.993
50
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 FlowFor 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 D (m) x x (P + Pa) x stroke(m) x # strokes/min x 1000 4
*Example.
= 80stroke = 400mms/min = 12 x 2P = 6barFrom table 6.20... 80 at 6 bar = 4.87 (4.9)l/100mm stroke
Qt= Q x stroke(mm) x # of extend + retract strokes 100Qt = 4.9 x 400 x 24 100Qt = 4.9 x 4 x 24
Qt = 470.4 l/min.
Working Pressure in bar
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
1.391
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
Table 6.20 Air Consumption of double acting cylinders in liters per 100 mm stroke corrected for losses by adiabatic change
*Formulae comparisonQ = 1.4 x D (m) x x (P + Pa) x stroke(m) x # strokes/min x 1000 4
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 x D (m) x x (P + Pa) x stroke(m) x # strokes/min x 1000 4 = _______mm
stroke = _______mm
s/min = _______ x 2
P =_______bar
*InertiaExample 1
m = 10kga = 30mmj = ___?
J= m (kg) x a (m) 12J= 10 x .03 12J= 10 x .0009 12J = .00075
a
*InertiaExample 2
m = 9 kga = 10mmb = 20mmJ = ___?
J = ma x a + mb x b 3 3J = 3 x .01 + 6 x .02 3 3J = 3 x .0001 + 6 x .0004 3 3J = .0001 + .0008
J = .0009a b
*a bm = ________ kg
a = _________mm
b = _________mm
J = _________?
*Valve identification A(4) B(2) EA P EB (5) (1) (3)
*Valve SizingThe 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 1bar. 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 +
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
*Flow exampleS = 35P1 = 6 barP2 =5.5 bar = 25CQ = 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
*Cv = ________between 1 -5
P1 = ________bar
P2 = ________5 bar
= ________C
*Flow capacity formulae transposedCv =Q 400 x (P2 + 1.013) x P
Kv =Q 27.94 x (P2 + 1.013) x PS =Q 22.2 x (P2 + 1.013) x P
*Flow capacity exampleQ = 750 l/minP1 = 9 barP = 10%S = ?S =Q22.2 x (P2 + 1.013) x PS =75022.2 x (8.1 + 1.013) x .9S =750 22.2 x 9.113 x .9S =750 22.2 x 2.86S = 750S = 11.81 63.49
*Q = _________ l/min
P1 = _________ bar
P = _________%
Cv = _________ ?
*Orifices in a series connectionS total = 1 1 + 1 + 1 S1 S2 S3
Example S1 = 12mmS2 = 18mmS3 = 22mmS total = 1 1 + 1 + 1 12 18 22S total = 1 1 + 1 + 1 144 324 484S total = 1= 1 .00694 + .00309 + .00207 .0121S total = 9.09
*Cv = _________
Cv = _________
Cv = _________
Cv total = ________
*
S mm
3
4
6
7.5
9
2
2
0.2
0.02
Tube Length in m
0
10
20
30
40
50
60
0.1
1
5
10
0.5
0.05
*
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. fittings
4 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
10.00
26.1
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
25.05
12 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
*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)
2
Equivalent Flow Section in mm
13.8
27.6
41.4
55.2
69
82.8
110
138
207
276
160
10.6
21.1
31.7
42.2
52.8
62
84.4
106
158
211
140
8.4
16.8
25.2
33.6
42
50.4
67.2
84
126
168
125
5.4
10.8
16.2
21.6
27
32.4
43.2
54
81
108
100
3.4
6.8
10.2
13.6
17
20.4
27.2
34
51
68
80
2.1
4.2
6.3
8.4
10.5
12.6
16.8
21
31.5
42
63
1.4
2.7
4
5.4
6.8
8.1
10.8
13.5
20.3
27
50
0.85
1.7
2.6
3.4
4.3
5
6.8
8.5
12.8
17
40
0.55
1.1
1.7
2.2
2.8
3.7
4.4
5.5
8.5
11
32
0.35
0.67
1
1.3
1.7
2
2.7
3.4
5
6.7
25
0.2
0.4
0.6
0.8
1
1.2
1.6
2
3
4
20
0.12
0.23
0.36
0.46
0.6
0.72
1
1.2
1.8
2.4
12,16
0.1
0.1
0.15
0.2
0.25
0.3
0.4
0.5
0.75
1
dia. mm
50
100
150
200
250
300
400
500
750
1000
8,10
Average piston speed in mm/s
*Flow Amplification
*Signal Inversion
*Selection
green
red
*Memory Function
green
red
*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
Shuttle Valve
*Safety interlock: AND Function
*Safety interlock: AND Function132
*Inverse Operation: NOT Function
*Direct Control
P
A
B
*Holding the end positions
P
A
B
*Semi Automatic return of a cylinder
Cam valve
*Repeating Strokes
*Sequence Control
3
1
2
4
2
4
*
_915341647.unknown
b
1
A+
B+
b
0
a
1
A-
B-
a
o
start
*
Pressure
Regulator
Regulator
with
relief
ISO SYMBOLS for AIR TREATMENT EQUIPMENT
Water
Separator
Filter
Auto Drain
Air Dryer
Filter /
Separator
Filter /
Separator
w. Auto Drain
Multi stage
Micro Filter
Lubricator
Air
Heater
Heat
Exchanger
Air
Cooler
Basic
Symbol
Differential
Pressure
Regulator
Pressure
Gauge
FRL Unit, detailed
FRL Unit,
simplified
Refrigerated
Air Dryer
Adjustable
Setting
Spring
Air Cleaning and Drying
Pressure Regulation
Units
*
_978162463.doc
Single Acting Cylinder,
Spring retract
Single Acting Cylinder,
Spring extend
Double Acting Cylinder
Double Acting Cylinder with
adjustable air cushioning
Double Acting Cylinder,
with double end rod
Rotary Actuator,
double Acting
*
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
_978162460.doc
_978162461.doc
_978162459.doc
*
with Spring Return
Normally Open 3/2 valve
(normally passing)
Manually Operated,
to
to
to
to
Valve with Spring Return
normally closed 3/2
(non-passing)
Mechanically Operated,
Manual
Operation
Closed
Input
Input
connected
Output
Return
Spring
Air Supply
Exhaust
Manual
Operation
Closed
Input
Input
connected
Output
Return
Spring
OR
Mechanical
Operation
Input
connected
Output
Input closed,
Output
exhausted
Return
Spring
Air Supply
Exhaus
t
OR
Mechanical
Operation
Input
connected
Output
Input closed,
Output
exhausted
Return
Spring
*
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
pressure
no pressure
No valve with index "1" is operated.
All valves with index "0" are operated.
Mechanically operated Valves
no pressure
an
0
pressure
an
0
an
1
an
1
no pressure
pressure
*
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 .
1
0
1
0
1
0
**