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AVENTICS | LU1/LU2 | 1829927007–BDL–001–AD | English
14 Appendix
Control examples
6
Pneumatic-only control
Ensure that the S1 and S2 pneumatic signals are strong enough (see
1 to 4).
Example 1:
Pneumatic-only control with 5/3 way valve, center position 1 connected with 4 and 2.
Actuation of locking unit via shuttle valve.
Example 2:
Pneumatic-only control with two 3/2 way valves, normally open, i. e. normal position
3 connected with 2. Actuation of locking unit with shuttle valve.
Example 3:
Pneumatic-only control with 5/3 way valve, center position 1 connected with 4 and
2. Actuation of locking unit with additional 5/2 way valve.
Example 4:
Pneumatic-only control with two 3/2 way valves, normally open, i.e. normal position
3 connected with 2. Actuation of locking unit with additional
5/2 way valve.
7
Electro-pneumatic control
Configure the electrical control ensuring that – parallel to energizing
the solenoid for the retracting and extending cylinder movements – the solenoid
(y2a) for the locking unit valve is also energized. (See
Example 5:
Electro-pneumatic control with 5/3 way valve, center position 1 connected with 4
and 2. Actuation of locking unit with additional 3/2 way valve.
Example 6:
Electro-pneumatic control with two 3/2 way valves, normally open, i. e. normal
position 3 connected with 2. Actuation of locking unit with additional 3/2 way valve.
Example 7:
Electro-pneumatic control with 5/3 way valve, center position 1 connected with 4
and 2. Actuation of locking unit with additional 5/2 way valve.
Example 8:
Electro-pneumatic control with two 3/2 way valves, normally open, i. e. normal
position 3 connected with 2. Actuation of locking unit with additional 5/2 way valve.
Example for actuation at 2nd pressure level
8
9
(Example: 9 – 12)
Design-related differences in cylinder surface areas can result in force differences.
You can compensate for these force differences by applying a reduced pressure to
the base of the cylinder.
The following examples apply to pneumatic and electro-pneumatic controls (see
8
9
also
and
).
Piston surface area. piston end
Piston surface area, rod end
(with piston rod Ø 20
AST = 3.1 cm
Working pressure
Calculation of 2nd pressure level
(reduced pressure)
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6
7
: Examples 5 to 8).
2
A1 = 19.6 cm
2
A2 = 16.5 cm
2
)
(P
)P
= 6 bar
2
e
P
xA
2
2
P
=
------------- -
1
A
1
·
6x 16
5 ,
P
=
-------------------- -
1
19 6 ,
P
= 16,05 bar
1
: Examples
8
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