Introduction to
Pneumatics
Air Production System
Air Consumption System
2
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
3
Properties of compressed air
• Availability
• Storage
• Simplicity of design and control
• Choice of movement
• Economy
4
Properties of compressed air
• Reliability
• Resistance to Environment
• Environmentally clean.
• Safety
5
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.
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.
6
HUMIDITY & DEWPOINT
Temperature °C
0
4.98
5
6.99
10
9.86
15
13.76
20
18.99
25
25.94
30
35.12
35
47.19
40
63.03
4.98
6.86
9.51
13.04
17.69
23.76
31.64
41.83
54.11
0
4.98
–5
3.36
–10
2.28
–15
1.52
–20
1.00
–25
0.64
–30
0.4
–35
0.25
–40
0.15
g/m (Atmospheric)
4.98
3.42
2.37
1.61
1.08
0.7
0.45
0.29
0.18
Temperature °F
g/ft3 *(Standard)
g/ft3 (Atmospheric)
Temperature °F
g/ft3 (Standard)
g/ft3 (Atmospheric)
32
.137
.137
32
.137
.137
40
.188
.185
30
.126
.127
60
80
.78
.71
10
.053
.056
100
1.48
1.29
0
.033
.036
120
2.65
2.22
-10
.020
.023
140
4.53
3.67
-20
.012
.014
160
7.44
5.82
-30
.007
.009
180
11.81
8.94
–40
.004
.005
3
g/m n *(Standard)
3
g/m (Atmospheric)
Temperature °C
3
g/m n (Standard)
3
.4
.375
20
.083
.085
7
Pressure and Flow
10
p (bar)
9
S = 1 mm 2
Example
8
P1 = 6bar
7
P1
P2
6
 P = 1bar
5
P2 = 5bar
4
Q = 54 l/min
3
(1 Bar = 14.5 psi)
2
Sonic Flow
Range
1
0
20
40
Q
60
80
n (54.44 l / min)
100
120
3
Q (dm n /min)
8
Air Treatment
9
Compressing Air
One cubic foot of air
Compression
ratio
=
psig + 1 atm
1 atm
compressor
One cubic foot of
100 psig
compressed air
7.8 cubic feet of free air
CFM
vs
SCFM
(at Standard conditions)
with 7.8 times the
moisture and dirt
Compressed air is always related at Standard conditions.
10
Relative Humidity
Compressor
Exit
Reservoir
Tank
Adsorbtion Dryer
Airline
Drop
Compressor
1 ft3 @100 psig
1950 F
1 ft3 @100 psig
770 F
1 ft3 @100 psig
-200 F
1 ft3 @100 psig
770 F
100% RH
57.1
grams of
H2 0
100% RH
.73
grams of H20
100% RH
.01
grams of
H20
0.15% RH
.01
grams of
H20
56.37
grams of
H2 0
.72
grams of
H2 0
11
Air Mains
Dead-End
Main
Ring
Main
12
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.
13
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
14
Isobaric change (Charles Law)
…volume
at constant pressure, a given mass of gas increases in volume by 1 of its
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
15
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
16
P1 = ________bar
T1 = _______°C ______°K
T2 = _______°C ______°K
17
ø (mm)
4
6
8
10
12
16
20
25
30
100000
500
400
50000
40000
300
250
200
25000
20000
p : (bar) 10
150
7
5
15000
125
12500
100
10000
50
40
5000
4000
30
25
20
2500
2000
1500
15
12.5
1250
10
1000
5
4
500
400
2.5
32
40
50
63
80
100
125 140
160
200
250
250
300
ø (mm)
18
F (N)
F (N)
2.5
1000
Force formula transposed
• Example
• D=
4 x FE
 x P
• 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
Cyl.Dia
Mass (kg)

60°
µ
µ 0.2
0.01
–
–
–
–
(87.2) (96.7)
43.6
48.3
µ
0.01
–
–
71.5
35.7
45°
µ 0.2
55.8
27.9
–
73.9
37
4.4
2.2
1.1
0.55
43.9
22
11
–
(99.2)
51.6
27.8
–
–
68.3
36.8
3.9
2
1
0.5
78
39
20.3
10.9
-
4
79.9
_
–
2
40
50.8
25.4
67.3
33.6
1
0.5
20
0
48
24
–
–
63.6
31.8
4.1
1.9
0.9
0.5
81.8
37.8
18.9
9.4
49.6
24.8
–
–
65.7
32.8
3.9
2
1
0.5
78.1
39
19.5
9.8
50.8
25.4
–
–
67.3
33.6
4
2
1
0.5
79.9
40
20
10
100
50
25
12.5
–
–
–
51.8
32
180
90
45
22.5
54.9
40
250
125
65
35
–
–
–
54.6
50
400
--
-
-
200
-
–
–
100
50
–
50
71.3
35.7
84.8
42.4
650
300
150
75
–
–
(94.4)
47.2
67.4
33.7
–
–
80.1
40.1
1000
500
250
125
–
–
(97.6)
48.8
69.7
34.8
–
–
82.8
41.4
1600
800
400
200
–
–
–
50
71.4
35.7
–
–
84.4
42.2
63
80
100
(95.6)
47.8
53
78.4
39.2
(93.1)
46.6
–
–
–
47.6
–
–
–
52.8
–
–
72.4
39
–
–
(86)
46.3
(96.5)
48.3
82.3
41.1
–
–
(91.2)
45.6
85
42.5
–
–
(94.3)
47.1
(87)
43.5
–
–
(96.5)
48.3

–
–
84.9
342.5
25
(87)
43.5
30°
µ 0.2
µ
0.01
–
–
50.9
25.4
µ 0.2
–
–
67.4
33.7
µ
0.01
4
2.2
1
0.5
80
40
20
10
Table 6.16 Load Ratios for 5 bar working pressure and friction coefficients of 0.01 and 0.2
21
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.
22
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°?
23
F = G · (sin  + µ · cos )
F=G
F = µ ·G
W a =m /2 · v
R
2
B
x
a

c
b
a
b

d
c
h
y
G
A
d
Y axes, (vertical lifting force)….. sin x M
X axes, (horizontal lifting force)….cos x  x M
Total force = Y + X
 = friction coefficients
24
Example
 = .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
25
 = __
______kg
_____°
Force Y = sin  x M =
Force X = cos  x  x M =
Total Force = Y + X =
F = ________ (N)
26
Temperature °C
3
g/m n *(Standard)
0
4.98
5
6.99
10
9.86
15
13.76
20
18.99
25
25.94
30
35.12
35
47.19
40
63.03
4.98
6.86
9.51
13.04
17.69
23.76
31.64
41.83
54.11
0
4.98
–5
3.36
–10
2.28
–15
1.52
–20
1.00
–25
0.64
–30
0.4
–35
0.25
–40
0.15
g/m (Atmospheric) 4.98
3.42
2.37
1.61
1.08
0.7
0.45
0.29
0.18
3
g/m (Atmospheric)
Temperature °C
3
g/m n (Standard)
3
27
13
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
28
13
Relative Humidity Example 2
•
•
•
•
•
•
•
•
•
•
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
• ? H²0
will condense out
– 50.6 g of water will condense out
V = 10m³
T1= 15°C
T2= 25°C
P1 = 1.013bar
P2 = 6bar
r.h = 65%
29
13
V = __________m³
T1= __________°C
T2= __________°C
P1 =__________bar
P2 =__________bar
r.h =__________%
?
__________H²0
will condense out
30
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
31
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
32
16,17
P1 = _________bar
P2 = _________bar
S = _________mm²
Q = ____?_____l/min
33
Receiver sizing
• If
Example
– Q = 5000
V = capacity of receiver
– P1 = 9 bar
Q = compressor output l/min
– Pa = 1.013
Pa = atmospheric pressure
P1 = compressor output
• V = 5000 x 1.013
pressure
9 + 1.013
• V = Q x Pa
• V = 5065
P1 + Pa
10.013
•
•
•
•
•
• V = 505.84 liters
34
22
35
29
36
29
37
30
Sizing compressor air mains
•
•
•
•
•
•
Example
• 30 = .24 kPa/m
125
Q = 16800 l/min
• 16800 x .00001667 = 0.28 m³/s
P1 = 9 bar (900kPa)
• chart lines on Nomogram
P = .3 bar (30kPa)
L = 125 m pipe length
P = kPa/m
L
• l/min x .00001667 = m³/s
38
31
3
4"
2
1.5
2
3.0
2.5
3
2.25
2.0
1.75
1.5
3"
1
80
70
2.5"
0.5
0.4
100
90
60
2"
50
0.3
4
7
1.0
0.9
0.8
0.7
0.6
8
0.5
9
0.4
5
6
10
11
12
0.2
0.15
0.1
1.5"
40
1.25"
35
30
0.05
0.04
1"
25
0.03
0.025
0.02
0.3
0.25
0.015
3/4"
20
0.2
Line
Press ure
(bar)
0.01
1/2"
0.15
15
3/8"
X
²p
kPa / m
= bar /100 m
Pipe Length
3
Q (m n
Reference
Line
/s
Inner Pipe Dia. ,
mm
39
33
Type of Fitting
Elbow
90* Bend (long)
90* Elbow
180* Bend
Globe Valve
Gate Valve
Standard Tee
Side Tee
Nominal pipe size (mm)
15
20
25
30
40
50
65
0.3
0.4
0.5
0.7
0.8
1.1
1.4
0.1
0.2
0.3
0.4
0.5
0.6
0.8
1.0
1.2
1.6
1.8
2.2
2.6
3.0
0.5
0.6
0.8
1.1
1.2
1.7
2.0
0.8
1.1
1.4
2.0
2.4
3.4
4.0
0.1
0.1
0.2
0.3
0.3
0.4
0.5
0.1
0.2
0.2
0.4
0.4
0.5
0.7
0.5
0.7
0.9
1.4
1.6
2.1
2.7
Table 4.20 Equivalent Pipe Lengths for the main fittings
80
1.8
0.9
3.9
2.6
5.2
0.6
0.9
3.7
100
2.4
1.2
5.4
3.7
7.3
0.9
1.2
4.1
125
3.2
1.5
7.1
4.1
9.4
1.2
1.5
6.4
40
34
Sizing compressor air mains
• Example 2
• Add fittings to example 1
• From table 4.20
–
–
–
–
–
–
–
• 30kPa = 0.22kPa/m
135m
• Chart lines on Nomogram
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
41
31
3
4"
2
1.5
2
3.0
2.5
3
2.25
2.0
1.75
1.5
3"
1
80
70
2.5"
0.5
0.4
100
90
60
2"
50
0.3
4
7
1.0
0.9
0.8
0.7
0.6
8
0.5
9
0.4
5
6
10
11
12
0.2
0.15
0.1
1.5"
40
1.25"
35
30
0.05
0.04
1"
25
0.03
0.025
0.02
0.3
0.25
0.015
3/4"
20
0.2
Line
Press ure
(bar)
0.01
1/2"
0.15
15
3/8"
X
²p
kPa / m
= bar /100 m
Pipe Length
3
Q (m n
Reference
Line
/s
Inner Pipe Dia. ,
mm
42
33
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
43
Aftercooler
Tank
1
Refrigerated
Air Dryer
Auto
Drain
Compressor
a
2
Auto
Drain
3
a Micro Filter
a
b Sub-micro Filter
c Odor Removal Filter
d Adsorbtion Air
a
b
b
c
d
b
Dryer
a
a
4
5
6
7
44
39
Example
•
•
•
•
P = 7 bar (700,000 N/m²)
D = 63mm (.063m)
d = 15mm (.015m)
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
45
54
ø (mm)
4
6
8
10
12
16
20
25
30
100000
500
400
50000
40000
300
250
200
25000
20000
p : (bar) 10
150
7
5
15000
125
12500
100
10000
50
40
5000
4000
30
25
20
2500
2000
1500
15
12.5
1250
10
1000
5
4
500
400
2.5
32
40
50
63
80
100
125 140
160
200
ø (mm)
250
250
300
46
F (N)
F (N)
2.5
1000
Example
– Calculate remaining force
•
•
•
•
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
• 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
47
M = _______kg
P = _______bar
 = _______mm
 = 0.2
F = /4 x D²x P = 401.9 N
48
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
49
Working Pressure in bar
Piston dia.
3
4
5
6
7
0.124
0.155
0.186
0.217
0.248
20
0.194
0.243
0.291
0.340
0.388
25
0.319
0.398
0.477
0.557
0.636
32
0.498
0.622
0.746
0.870
0.993
40
0.777
0.971
1.165
1.359
1.553
50
1.235
1.542
1.850
2.158
2.465
63
1.993
2.487
2.983
3.479
3.975
80
3.111
3.886
4.661
5.436
6.211
100
Table 6.19 Theoretical Air Consumption of double acting cylinders from 20 to 100 mm dia,
in liters per 100 mm stroke
• 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)
100
Qt = 3.5 x 400 x 24
100
Qt = 3.5 x 4 x 24
x
# of extend + retract strokes
• Qt = 336 l/min.
50
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
51
Working Pressure in bar
Piston dia.
3
4
5
6
7
0.174
0.217
0.260
0.304
0.347
20
0.272
0.340
0.408
0.476
0.543
25
0.446
0.557
0.668
0.779
0.890
32
0.697
0.870
1.044
1.218
1.391
40
1.088
1.360
1.631
1.903
2.174
50
1.729
2.159
2.590
3.021
3.451
63
2.790
3.482
4.176
4.870
5.565
80
4.355
5.440
6.525
7.611
8.696
100
Table 6.20 Air Consumption of double acting cylinders in liters
per 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)
100
Qt = 4.9 x 400 x 24
100
Qt = 4.9 x 4 x 24
x
# of extend + retract strokes
• Qt = 470.4 l/min.
52
Formulae comparison
• 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
53
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
54
Inertia
• Example 1
a
• m = 10kg
• a = 30mm
• j = ___?
• J= m (kg) x a² (m)
12
• J= 10 x .03²
12
• J= 10 x .0009
12
• J = .00075
55
Inertia
• Example 2
a
•
•
•
•
b
m = 9 kg
a = 10mm
b = 20mm
J = ___?
• J = ma x a² + mb x b²
3
3
• J = 3 x .01² + 6 x .02²
3
3
• J = 3 x .0001 + 6 x .0004
3
3
• J = .0001 + .0008
• J = .0009
56
m = ________ kg
a
b
a = _________mm
b = _________mm
J = _________?
57
Valve identification
A(4)
EA P
(5) (1)
B(2)
EB
(3)
58
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
59
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
The normal flow Qn for other various flow capacity units is:
The Relationship between these units is as follows:
273
273 + 
1 Cv =
1 kv =
981.5
68.85
1S=
54.44
1
0.07
0.055
14.3
1
0.794
18
1.26
1
60
Flow example
•
•
•
•
S = 35
P1 = 6 bar
P2 =5.5 bar
 = 25°C
• 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
61
Cv = ________
between 1 -5
P1 = ________bar
P2 = ________5 bar
 =
________°C
62
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
63
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
63.49
S = 11.81
64
Q = _________ l/min
P1 = _________ bar
P = _________%
Cv = _________ ?
65
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
.00694 + .00309 + .00207
=
1
.0121
S total = 9.09
66
Cv = _________
Cv = _________
Cv = _________
Cv total = ________
67
2
S mm
60
9
50
40
7.5
30
6
20
10
4
0
0.02
3
0.05
0.1
0.2
0.5
1
2
5
10
Tube Length in
m
68
Tube
Dia.
(mm)
4 x 2.5
6x4
8x5
8x6
10 x 6.5
10 x 7.5
12 x 8
12 x 9
Material
N,U
N,U
U
N
U
N
U
N
Length
1m
0.5 m
1.86
6.12
10.65
16.64
20.19
28.64
33.18
43.79
3.87
7.78
13.41
20.28
24.50
33.38
39.16
51.00
Fittings
Insert type
straight
elbow
1.6
1.6
6
11
17
35
30
35
45
One Touch
straight
elbow
5.6
4.2
13.1
11.4
18
14.9
26.1
21.6
29.5
25
41.5
35.2
46.1
39.7
6
(9.5) 11
(12) 16
(24) 30
(23) 26
(24) 30
(27) 35
58.3
50.2
Table 7.30 Equivalent Flow Section of current tube connections
Total
0.5 m tube +
2 strt. fittings
1.48
3.18
3.72
5.96
6.73
9.23
10.00
13.65
12.70
15.88
19.97
22.17
20.92
25.05
29.45
32.06
69
dia. mm
8,10
12,16
20
25
32
40
50
63
80
100
125
140
160
50
0.1
0.12
0.2
0.35
0.55
0.85
1.4
2.1
3.4
5.4
8.4
10.6
13.8
100
0.1
0.23
0.4
0.67
1.1
1.7
2.7
4.2
6.8
10.8
16.8
21.1
27.6
Average piston speed in mm/s
150
200
250
300
400
500
0.15
0.2
0.25
0.3
0.4
0.5
0.36 0.46
0.6
0.72
1
1.2
0.6
0.8
1
1.2
1.6
2
1
1.3
1.7
2
2.7
3.4
1.7
2.2
2.8
3.7
4.4
5.5
2.6
3.4
4.3
5
6.8
8.5
4
5.4
6.8
8.1
10.8 13.5
6.3
8.4
10.5 12.6 16.8
21
10.2 13.6
17
20.4 27.2
34
16.2 21.6
27
32.4 43.2
54
25.2 33.6
42
50.4 67.2
84
31.7 42.2 52.8
62
84.4 106
41.4 55.2
69
82.8 110
138
Equivalent Flow Section in mm2
750
0.75
1.8
3
5
8.5
12.8
20.3
31.5
51
81
126
158
207
1000
1
2.4
4
6.7
11
17
27
42
68
108
168
211
276
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)
70
Flow Amplification
71
Signal Inversion
72
Selection
red
green
73
Memory Function
red
green
74
Delayed switching on
75
Delayed switching off
76
Pulse on switching on
77
Pulse on releasing a valve
78
Direct Operation and Speed Control
79
Control from two points: OR Function
Shuttle Valve
80
Safety interlock: AND Function
81
Safety interlock: AND Function
3
1
2
82
Inverse Operation: NOT Function
83
B
Direct Control
A
P
84
Holding the end positions
B
A
P
85
 
Cam valve
Semi Automatic return of a cylinder
86
 
Repeating Strokes
87
2
4
Sequence Control
3
4
1
2
88
A+
B+
A-
b0
B-
a1
start
ao
b1
Signals
a1
Start
Commands
A+
b1
B+
a0
A-
b0
B-
89
ISO SYMBOLS for AIR TREATMENT EQUIPMENT
Air Cleaning and Drying
Auto Drain
Air Cooler
Water
Separator
Filter
Refrigerated
Air Dryer
Filter /
Separator
Air Dryer
Air
Heater
Filter /
Multi stage
Separator
Micro Filter
w. Auto Drain
Heat
Exchanger
Lubricator
Pressure Regulation
Basic
Symbol
Adjustable
Setting
Spring
Pressure
Regulator
Regulator
with relief
Differential
Pressure
Regulator
Pressure
Gauge
Units
FRL Unit, detailed
FRL Unit,
simplified
90
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
91
Return Spring (in fact not an
operator, but a built-in element)
Roller Lever:
one-way Roller Lever:
Manual operators: general:
Lever:
Push Button:
Push-Pull Button:
Mechanical (plunger):
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
92
Manual
Operation
Closed
Input
Input
connected to Return
Output
Spring
Manual
Operation
Closed
Input
Input
connected to Return
Output
Spring
OR
Manually Operated,
Normally Open 3/2 valve
(normally passing)
with Spring
Return
Air Supply
Exhaust
Mechanical
Operation
Input
connected to
Output
Input closed,
Output
exhausted
Return
Spring
Mechanical
Operation
Input
connected to
Output
Input closed,
Return
Output
exhausted Spring
OR
Mechanically
normally closed 3/2
Operated,
(non-passing)
Valve with Spring Return
Air Supply
Exhaust
93
Manually operated Valves
detent, must correspond with valve position
no pressure
3/2, normally closed
no pressure
pressure
3/2, normally closed/normally open
3/2, normally open
monostable valves never operated
pressure
bistable valves: both positions possible
Electrically and pneumatically operated Valves
Air operated valves may be operated in rest
pressure
no pressure
Solenoids are never operated in rest
Mechanically operated Valves
No valve with index "1" is operated.
no pressure
pressure
a…n1
a…n1
All valves with index "0" are operated.
pressure
a…n
0
no pressure
a…n
94
0
Last stroke of the cycle
First stroke of the cycle
B
A
A+
A-
C
B+
B-
C
POWER Level
LOGIC Level
Memories,
AND's, OR's,
Timings etc.
Start
SIGNAL INPUT Level
Codes: a 0 , a1 , b0 , b , c0
1
and c1 .
95