20
E
N
G
L
IS
H
P
N
:
P
N
= U
N
x I x PF (cos
ϕ)
(single-phase system);
P
N
= š3 x U
N
x I x PF (cos
ϕ)
(three-phase, three-wire system);
P
N
= 3 x U
N
x I x PF (cos
ϕ)
(three-phase, four-wire system);
where:
U
N
= the real phase-neutral voltage of the electrical system being
measured;
I = the maximum phase current of the electrical system being measured;
Power factor (cos
ϕ)
= the average power factor (cos
ϕ)
of the electrical
system to be measured;
P
I
:
P
I
= U
I
x I
I
x VT (ratio) x CT (ratio) (single-phase system);
P
I
= š3 x U
I
x I
I
x VT (ratio) x CT (ratio) (three-phase, three-wire system);
P
I
=
3 x U
I
x I
I
x VT (ratio) x CT (ratio) (three-phase, four-wire system);
where:
U
I
= rated input voltage of the transducer; it changes depending on the
model (see table on page 19);
I
I
= rated input current of the transducer; it changes depending on the
model (see table on page 19);
VT (ratio) = the value of the voltage transformer ratio (P-S parameter in
the "in.U" menu);
CT (ratio) = the rated value of the current transformer ratio (P-S parameter
in the "in.C" menu);
Example 5:
model AV3.3 (3-wire system)
U
N
= 6kV (Ð voltage)
I
= 265A (single phase current)
Cos
ϕ
= 0.85 (system's power factor)
U
I
= 100V;
I
I
= 5A
VT (ratio) = 6kV = 60
CT (ratio) = 300A = 60
100V
5A
P
N
becomes: š 3 x U
N
x I x cos
ϕ =
š 3 x 6000 x 265 x 0.85 = 2.33MW
P
I
becomes: š 3 x U
I
x I
I
x VT (ratio) x CT (ratio) = š 3 x 100 x 5 x 60 x 60
= 3.12MW, the ratio P
N
is: 2.33 = 0.75
P
I
3.12
Example 6:
model AV3.3 (4-wire system)
U
N
= 6kV / š 3;
I
= 265A
Power factor (Cos
ϕ)
= 0.85
U
I
= 100V / š 3
I
I
= 5A
VT (ratio) = 6kV / š 3 = 60
CT (ratio) = 300A = 60
100V / š 3
5A