88 / 218
Pressure booster
Where large forces are required, for example for cutting and/or bending operations, cylinders with large diameters
– or Tandem cylinders can be used. In both cases, problems with spacemay be an issue, a possible solution is to
increase the supply pressure, this can be achieved through use of the pressure booster.
The pressure booster has a similar function to two opposing cylinders joined by the piston rod; the movement of
these cylinders is oscillating. The central part of the Pressure Booster is fed through a pressure regulator, the two
sides by a 5/2-way control valve. The oscillating movement is determined automatically by the internal limit
switches that invert the state of the 5/2 - control valve once the piston has reached the end position.
Thepressure increase is obtained, as the sumof areas on the thrust side is greater than that of the resistant surface,
therefore the volume of the resistant chamber continues to decrease and as a result the pressure increases.
Figure 41
Pos. A
: the pressure booster is fed by compressed air, the left piston is in the positive end position, the right piston
is at its negative end position, and its relative chamber is exhausted. The pressure regulator feeds compressed air
into the central part. The two generated Forces, being in opposite directions, cancel out each other.
Pos.B
: the left piston activates the limit switch integrated in the central block that reverses the5/2-way valve, the
right chamber is pressurized, while the left is exhausted. The pressure in the negative chamber of the piston to the
right increases, when this pressure exceeds the spring setting of the integrated unidirectional valve, this air exits
through the outlet
U
. The unidirectional valves prevent the exhaust through the regulator.
Pos. C
: the right piston has reached the positive end position and activates the limit switch that reverses the
5/2-way control valve.
Pos. D
: the 5/2-way control valve has switched, the left chamber is pressurized, while the right is exhausted.
The pressure in the negative chamber of the piston to the left increases, when this pressure exceeds the spring
setting of the integrated unidirectional valve, the air exits through the outlet
U
. The unidirectional valves prevent
the exhaust through the regulator. The pistonmoves towards the right since the thrust surfaces are greater than the
surface on the resistant side.When the piston arrives at the stroke end, the cycle repeats automatically.
The ratio between the inlet pressure and the outlet is normally 1:2 even if different values are possible.
Assuming that
: the surface of a piston on the outer side is
S
e
= 20
cm
2
that of a piston on the rod side
S
s
=16
cm
2
working pressure
p
=6
bar
Calculate the value of the outgoing pressure
p
u
:
F
=
(
S
e
+
S
s
) *
p
F
= (20+16) * 6
F
=
216
kg
p
u
=
F
p
u
=
216
=13,6
p
u
=
13,6
bar
S
s
16
U
U
U
U
P=6bar
S=20 cm+ 16cm= 36 cm
F=S xP= 36 x 6=216Kg (2160N) P=F : S= 216 : 16= 13,5bar
2
2
2
AIR
A
B
C
D
Fig. 41
3
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