ELEKTROGAS – TECHNICAL MANUAL
7-12
When the flow read on the diagram is referred to operating pressure instead of standard
conditions, the pressure drop
∆
p read on the diagram must be multiplied for the factor:
(1+ relative pressure in bar)
Example:
In the 2” solenoid valve with an air flow of 80 Nm
3
/h there is a pressure drop
∆
p = 5 mbar.
If we consider that 80 m
3
/h is the flow at 200 mbar of inlet pressure, then the pressure drop to
be consider is:
∆
p = 5x(1+0,2) = 6 mbar
Normally, pressure drop and flow rate for the valves are read from the gas flow diagram.
However, the valves can also be chosen in accordance with the characteristic ”Kvs value” which
is shown in table 2.
The selection of the valve requires the calculation of the Kv under the operating conditions.
Considering only subcritical pressure drops:
2
1
p
p
<
∆
Kv can be calculated with the formula:
(
)
2
273
514
p
p
t
V
Kv
⋅
∆
+
=
ρ
where
V
= flow rate [Nm
3
/h]
Kv = flow factor [m
3
/h]
ρ
= density [Kg/m
3
]
p
1
= absolute inlet pressure [bar]
p
2
= absolute outlet pressure [bar]
∆
p = differential pressure p
1
-p
2
[bar]
t
= media temperature [°C]
To the Kv value calculated from operating conditions we add an allowance of 20%, to obtain the
minimum Kvs value which the valve should have:
Kvs > 1,2 Kv
Valve must be selected considering the following:
-
Pressure drops
∆
p
≤
0,1p
1
are recommended and
∆
p > p
1
/2 are always unadvisable
-
Flow velocities w
≤
15 m/s are recommended and w > 50 m/s are always unadvisable.
