Intro to Pneumatics Modified_2
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Transcript of Intro to Pneumatics Modified_2
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I ntroduction toPneumatics
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Air Production System Air Consumption System
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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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Properties of compressed air
Availability
Storage
Simplicity of design and control
Choice of movement
Economy
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Properties of compressed air
Reliability
Resistance to Environment
Environmentally clean.
Safety
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What is Air?
NitrogenOxygen
Carbon Dioxide
Argon
Nitrous OxideWater 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 aone square inch
column of air(from sea level
to the outer atmosphere,@ 680 F, & 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/m
3
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 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
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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 and dirt
compressor
CFM vs SCFM
psig + 1 atm
1 atm
Compression
ratio=
Compressed air is always related at Standard conditions.
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Relative Humidity
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 DryerCompressor
Exit
Reservoir
TankAirline
Drop
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Air Mains
Ring
Main
Dead-End
Main
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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 unitof 100,000 Pa.
100,000 Pa = 100 kPa = 1 bar
Atmospheric Pressure =14.696 psi =1.01325 bar =1.03323 kgf/cm2.
h i h ( l )
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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 V1V2
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. 273
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
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Isochoric change Law o f 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
4000050000
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
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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.
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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
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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) 7 1.5 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
50400 -- - - - 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 78.1500 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 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
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Speed control
The speed of a cylinder is define by theextra 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 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?
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b c d
x
A
B
h
G
y
R a
F=G F= G W =m 2 va2
F=G (sin + cos )
a db c
Y axes, (vertical lifting force).. sin x M
X axes, (horizontal lifting force).cos x x MTotal force = Y + X
=friction coefficients
E l
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Example
40
F = ________ (N)150kg
= .01
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
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_____
F = ________ (N)
______kg
= __
Force Y = sin x M =
Force X = cos x 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/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
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Relative humidity (r.h.) = actual water content X100% 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
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Relative HumidityExample 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 m
6 + 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
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V = __________m
T1= __________CT2= __________C
P1 =__________bar
P2 =__________barr.h =__________%
? __________H0
will condense out
l
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Formulae, for when more exact values are required
Sonic flow = P1 + 1.013 > 1.896 x (P2 + 1,013)
Pneumatic systemscannotoperate 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
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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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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 = 5065
10.013
V = 505.84 liters
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37 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
3100
4"
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2
3
4
5
6
7
8
9
10
11
12
Line
Pressure(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
kPa / m= bar /100 mPipe Length
2
1
0.5
0.1
1.5
0.2
0.3
0.4
0.01
0.050.04
0.03
0.02
0.015
0.15
0.025
90
80
70
60
50
40
30
20
15
25
35
Inner Pipe Dia. ,
mm
Reference
Line
X
3"
2.5"
2"
1.5"
1.25"
1"
3/4"
1/2"
3/8"
Q (m /s3n
33
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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 7.1180* 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 4.20 Equivalent Pipe Lengths for the main fittings
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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
3100
4"
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2
3
4
5
6
7
8
9
10
11
12
Line
Pressure(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
kPa / m= bar /100 mPipe Length
2
1
0.5
0.1
1.5
0.2
0.3
0.4
0.01
0.050.04
0.03
0.02
0.015
0.15
0.025
90
80
70
60
50
40
30
20
15
25
35
Inner Pipe Dia. ,
mm
Reference
Line
X
3"
2.5"
2"
1.5"
1.25"
1"
3/4"
1/2"
3/8"
Q (m /s3n
33
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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/s
Using the ring main example on page 29 size for the
following requirements:
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Auto
Drain
1
2
34
5
6
7
RefrigeratedAir 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
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Example
P = 7 bar (700,000 N/m) D = 63mm (.063m)
d = 15mm (.015m)
F = x (D -d) x P4
F = 3.14 x (.063 - .015) x 700,0004
F = 3.14 x (.003969 - .0.000225) x 700,0004
F = .785 x .003744 x 700,000
F = 2057.328 N
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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 2016 (mm)
F
(N)
125
150
200
400
300
12.5
E l
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Example
M = 100kg
P = 5bar
= 32mm
= 0.2
F = /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) = 196N
100
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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M = _______kg
P = _______bar
= _______mm
= 0.2
F = /4 x Dx P = 401.9 N
Air Flow and Consumption
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Air Flow and ConsumptionAir consumption of a cylinder is defined as:piston area x stroke length x number of single strokes per minute x absolu te pressur e in bar .
Q = D (m) x x (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.
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 0.99350 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 Flow
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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 foradiabatic change, the theoreticalvolume 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 10004
Working Pressure in bar
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g
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 1.39150 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 litersper 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) 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 D (m) x x (P + Pa) x stroke(m) x # strokes/min x 10004
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 D ( ) (P + P ) t k ( ) # t k / i 1000
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Q = 1.4 x D (m) x x (P + Pa) x stroke(m) x # strokes/min x 10004
= _______mm
stroke = _______mm
s/min = _______ x 2
P =_______bar
I ti
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Inertia
Example 1
m = 10kg
a = 30mm
j = ___?
J= m (kg) x a (m)
12
J= 10 x .03
12
J= 10 x .000912
J = .00075
a
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Inertia
Example 2
m = 9 kg
a = 10mm
b = 20mm
J = ___?
J = ma x a + mb x b
3 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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a b
m = ________ kg
a = _________mm
b = _________mm
J = _________?
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Valve identification
A(4) B(2)
EA P EB(5) (1) (3)
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Valve Sizing
The Cv factor of 1 is a flow capacity ofone US Gallon of water per minute, witha pressure drop of 1 psi.
The kv factor of 1is a flow capacity ofone liter of water per minute with apressure drop of 1 bar.
The equivalent Flow Section S of avalve is the flow section in mm2 of anorifice in a diaphragm, creating thesame relationship between pressure
and flow.
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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
Flo e ample
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Flow example
S = 35
P1 = 6 bar
P2 =5.5 bar
= 25C
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 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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Cv = ________between 1 -5
P1 = ________bar
P2 = ________5 bar
= ________C
Flow capacity formulae transposed
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p y p
Cv = Q
400 x (P2 + 1.013) x P
Kv = Q
27.94 x (P2 + 1.013) x P
S = Q22.2 x (P2 + 1.013) x P
Fl it l
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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 = 75022.2 x 9.113 x .9
S = 750
22.2 x 2.86
S = 750 S = 11.8163.49
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Q = _________ l/min
P1 = _________ bar
P = _________%
Cv = _________ ?
O ifi i i ti
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Orifices in a series connection S total = 1
1 + 1 + 1S1 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
Stotal
= 1 = 1.00694 + .00309 + .00207 .0121
S total = 9.09
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Cv = _________
Cv = _________
Cv = _________
Cv total = ________
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Tube Length in
0
10
20
30
40
50
60
0.1 1 5 100.50.050.02 0.2 2
2
9
7.5
6
4
3
S mm
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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. fittings4 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 10.0026.1 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 25.0512 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
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Average piston speed in mm/s
dia. mm 50 100 150 200 250 300 400 500 750 10008,10 0.1 0.1 0.15 0.2 0.25 0.3 0.4 0.5 0.75 1
12,16 0.12 0.23 0.36 0.46 0.6 0.72 1 1.2 1.8 2.420 0.2 0.4 0.6 0.8 1 1.2 1.6 2 3 4
25 0.35 0.67 1 1.3 1.7 2 2.7 3.4 5 6.732 0.55 1.1 1.7 2.2 2.8 3.7 4.4 5.5 8.5 11
40 0.85 1.7 2.6 3.4 4.3 5 6.8 8.5 12.8 17
50 1.4 2.7 4 5.4 6.8 8.1 10.8 13.5 20.3 27
63 2.1 4.2 6.3 8.4 10.5 12.6 16.8 21 31.5 4280 3.4 6.8 10.2 13.6 17 20.4 27.2 34 51 68
100 5.4 10.8 16.2 21.6 27 32.4 43.2 54 81 108
125 8.4 16.8 25.2 33.6 42 50.4 67.2 84 126 168
140 10.6 21.1 31.7 42.2 52.8 62 84.4 106 158 211
160 13.8 27.6 41.4 55.2 69 82.8 110 138 207 276
Equivalent Flow Section in mm2
Table 7.31Equivalent Section S in mm2 for the valve and the tubing, for6 bar working pressure and a pressure drop of 1 bar (Qn Conditions)
Flow Amplification
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p
Signal Inversion
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g
Selection
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greenred
Memory Function
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greenred
Memory Function
Delayed switching on
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Delayed switching on
D l d it hi ff
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Delayed switching off
Pulse on switching on
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Pulse on switching on
P l l i l
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Pulse on releasing a valve
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Direct Operation and Speed Control
Control from two points: OR Function
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Shuttle Valve
Safety interlock: AND Function
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Safety interlock: AND Function
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1
3
2
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Inverse Operation: NOT Function
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P
AB
Direct Control
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P
ABHolding the end positions
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Cam valve
Semi Automatic return of a cylinder
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Repeating Strokes
2 4
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3
1 2
4
Sequence Control
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89Commands
Signals Start
A+ B+ A- B-
b0b1 a0a1
b1
A+ B+
b0 a1
A- B-
aostart
ISO SYMBOLS for AIR TREATMENT EQUIPMENT
Air Cleaning and Drying
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PressureRegulator
Regulatorwith relief
WaterSeparator
Filter
Auto Drain Air Dryer
Filter /Separator
Filter /Separator
w. Auto Drain
Multi stageMicro Filter
Lubricator
Air
Heater
Heat
Exchanger
AirCooler
BasicSymbol
DifferentialPressure
Regulator
PressureGauge
FRL Unit, detailed
FRL Unit,
simplified
Refrigerated
Air Dryer
AdjustableSetting
Spring
Pressure Regulation
Units
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Single Acting Cylinder,Spring retract Single Acting Cylinder,Spring extend
Double Acting Cylinder Double Acting Cylinder withadjustable air cushioning
Double Acting Cylinder,with double end rod
Rotary Actuator,double Acting
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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
ManualOperation
ClosedInput
Inputconnected to
OutputReturnSpring
ManualOperation
ClosedInput
Inputconnected to
OutputReturnSpring
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Air SupplyExhaust
Manually Operated,
Normally Open 3/2 valve(normally passing)
with Spring
OR
MechanicalOperation
Inputconnected to
Output
Input closed,Output
exhaustedReturnSpring
Air Supply Exhaust
Mechanicallynormally closed 3/2
(non-passing)
Valve with Spring Return
OR
MechanicalOperation
Inputconnected to
Output
Input closed,Output
exhaustedReturnSpring
Manually operated Valvesdetent, must correspond with valve position
no pressure pressureno pressure pressure
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3/2, normally closed/normally openbistable valves: both positions possible
3/2, normally closed 3/2, normally openmonostable valves never operated
Solenoids are never operated in rest
Air operated valves may be operated in rest
Electrically and pneumatically operated Valves
pressureno pressure
pressure
Mechanically operated Valves
No valve with index "1" is operated.no pressure
an1an1
All valves with index "0" are operated.
an 0
pressure
an 0
no pressure
First stroke of the cycleA B
Last stroke of the cycleC
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POWER Level
LOGIC Level
SIGNAL INPUT Level
Start
Memories,AND's, OR's,
Timings etc.
A+ A- B+ B- C
Codes: a , a , b , b , c and c .1010 10