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6.3 Sample application
Adjusting the flow limitation (medium: water)
when the pressure loss across the plant is
known.
Known:
− Type 46-6, DN 15, K
The regulator is to limit the flow rate in
the plant to 0.63 m³/h.
− The pressure drop across the plant
) is 0.4 bar.
(Δp
plant
− Determine the adjustment values using
the adjustment diagram Fig. 9 on
page 28.
To be determined:
What is the limit value of the differential
pressure set point and how many turns of the
set point screw are necessary?
Solution:
Sequence: points A to E in diagram (Fig. 9).
The calculation is based on the pressure
drop Δp in the plant, therefore, this value
must be known.
Δp = 0.4 bar is specified in the example
and corresponds with point A in the dia-
gram. The differential pressure at the restric-
tion (Δp
) assumed to be 0.2 bar in the
restriction
example, must be added.
A line representing this value is drawn from
A across to the right and results in point B.
Point B is situated on the same straight line
for the differential pressure Δp = 0.6 bar.
Differential pressure to be set: Δp = 0.6 bar.
EB 3130 EN
= 1
VS
A vertical line is drawn from point B until it
reaches the limiting curve for the flow rate
(in the example, it is 0.63 m³/h). This is
point C.
The horizontal line is drawn from point C
across to the curve relevant for the nominal
size (DN) or K
coefficient used; this is point
VS
D.
When a line is drawn vertically upwards
from point D, this results in point E which in-
dicates how many turns of the set point
screw are required.
The example shows that around 6 turns are
required.
Based on a closed restriction, turn the set
point screw (17) six turns counterclockwise
to open the restriction.
Operation
27
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