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4.4.4.
Heuristic power control
Note: The function is only available when using
GPC model 3PH, with separate control of three
individual single-phase loads.
This type of control is used to limit the delivery of total
power to the loads in order to avoid input peaks from the
single-phase power line.
An example of this type of input occurs during the start
phases when the machine is cold; the demand for heating
power is almost 100% until temperature values near the
threshold are reached.
It is useful to avoid simultaneous conduction even when
there is ON-OFF modulation for temperature maintenance.
The cycle time must be unique or identical for all modules;
the power percentage of each module is limited to that
necessary to maintain current within set limits.
This function acts by enabling the control to search for the
most appropriate input combinations.
EXAMPLE
We have three single-phase loads supplied with 380 V:
•
GPC-M = 32 A,
•
GPC-E1 = 16 A,
•
GPC-E2 = 25 A.
The maximum current, in the case of simultaneous conduc-
tion, is 73 A.
The current limit value set with I.HEU is 50 A.
Bearing in mind that combinations without repetitions are
equal to
it is seen that the possible combinations of conduction can
be:
•
I1 + I2
= 48 A
•
I1 + I3
= 57 A
•
I2 + I3
= 41 A
•
I1 + I2 + I3 = 73 A
The combinations corresponding to current values below
the current limit value I.HEU are therefore:
•
I1 + I2 = 48 A
•
I2 + I3 = 41 A
Of these, the one with the lowest current is the sum of
GPC-E1 and GPC-E2.
In the single cycle for the enabled modules, the delivery
of power may be reduced to comply with the maximum
current limit.
The time distribution of module activation is calculated at
the start of each cycle:
Ou.P
= Ou.P_1 + (Ou.P_2* or Ou.P_3**)
tot
(*) if Ou.P_2 > Ou.P_3
For GPC-E1 and GPC-E2, simultaneous operation is per-
missible because their sum is less than the set current limit.
The percentage of conduction time of the x-module during
the cycle is calculated as
% conduction time module-x = Px 100%
n!
k! (n-k)!
(**) if Ou.P_3 > Ou.P_2
P
tot
81901 "MSW_GPC-40/600A"_03-2021_ENG_page 140
Case 1
If Ou.P_1 = 100%, Ou.P_2 = 100% and Ou.P_3 = 100% we
have Ou.P
= 200%.
tot
As Ou.P
> 100% we therefore have:
tot
•
% time Ou.P_1
= 0.50 × 100% = 50%.
•
% time Ou.P_2
= 0.50 × 100% = 50%.
•
% time Ou.P_3
= 0.50 × 100% = 50%.
A
50
M +
E1 +
32 A
E2
0
A
50
E2
0
A
50
E1
0
A
50
M
32 A
0
50%
Case 2
If Ou.P_1 = 100%, Ou.P_2 = 50% and Ou.P_3 = 0% we get
Ou.P
= 150%.
tot
As Ou.Ptot > 100% rounding up, we there have:
•
% time Ou.P_1
= 0.66 × 100% = 67%.
% time Ou.P_2
•
= 0.66 × 50% = 33%.
•
% time Ou.P_3
= 0.66 × 0% = 0%.
A
50
M +
E1 +
32 A
E2
0
A
50
E2
0
A
50
E1
0
A
50
M
32 A
0
67%
41 A
(16 A + 25 A)
25 A
16 A
50%
Cycle time
16 A
16 A
33%
Cycle time
I.HEU
I.HEU
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