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Figure 58
Tube Sizing
For the selection of the tubes it is necessary to check the flow characteristics according to their length and
diameter (see
Fig. 58
). Incorrect sizing will consequentially increase
∆
p
with repercussions on travel time of the
cylinders, and therefore on the cycle. When several components are aligned in sequence, the resulting flow rate
is not of the component with the lowest value. To calculate the value it is necessary to introduce a new variable
called “
conductance
”.
The conductance is the ratio between the flow rate value in free flowQ and the value of the absolute pressure
.
C
=
Q [Nl
/
min]
P
rel
[bar]
+
P
atm
[bar]
Taking as reference the valve with
Q
n
≅
400
Nl
/
min
we see that the value of flow in free air at 6
bar
is
Q
≅
610
Nl
/
min
, fromwhich the conductance is equal to:
C
=
Q
C
=
610
C
=
87,14
Nl / min * bar
P
rel
+
P
atm
6+1
If the supply pressurewere 5
bar
:
C
=
520
C
=
87,14
Nl / min * bar
5+ 1
If the supply pressurewere 4
bar
:
C
=
435
C
=
87,14
Nl / min * bar
4+ 1
We conclude that the value of conductance does not vary by changing the supply pressure.
Generally in catalogues, the nominal flow rate
Q
n
is indicated, calculated according toUNI ISO8778with a supply
pressure of 6
bar
and
∆
p
1
bar.
Taking the measured values on the graph, we observe that the free air delivery
with a supply pressure of 6
bar
is:
Q
≅
610
Nl
/
min
The conductance of this valve as in previous calculations is:
C
=
87,14
Nl
/
min
*
bar
Also from the graphwe observe that with a supply pressure of 6
bar
and
∆
p
1
bar
the flow rate is:
Q
n
≅
400
Nl
/
min
Taking the value of
Q
n
and dividing it by the value of
C
we obtain a coefficient
X
that experimental tests have
determinedwith good approximation, to be valid for all the sections traversed by a fluid.
X
=
Q
n
X
=
400
X
≅
4,6
C
87,14
Another way to calculate the conductance of an element is:
C
=
Q
n
X
4
122
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