User Manual

95
If reference on L3
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∑∑
−
=
−
=
⋅⋅+⋅−⋅=
1
0
1
0
1
11
1
02
1
3P
NechSec
n
NechSec
n
nAnU
NechSec
nAnU
NechSec
Note: The fundamental power factor is also called the displacement factor.
Total tangent
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[ ]
[ ]
3P
3Q
3Tan
1
1
=
16.1.6. ENERGIES
Energies excluding neutral – over Tint with refresh every second
16.1.6.1. Distribution system with neutral
Note: Tint is the power integration time in energy calculations; the beginning and end of this period are user-controlled.
a) Energies consumed (P[i][n] ≥ 0)
Consumed active energy of phase (i+1) with i ∈ [0; 2].
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[ ][ ]
∑
==
int
3600
0Wh0Ph
T
n
niP
ii
Consumed apparent energy of phase (i+1) with i ∈ [0; 2].
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[ ][ ]
∑
==
int
3600
0VAh0Sh
T
n
niS
ii
Consumed inductive reactive energy of phase (i+1) with i ∈ [0; 2].
(Non-active quantities broken down – Configuration >Calculation methods >var)
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[ ][ ]
∑
==
int
3600
0VARhL0hLQ
1
1
T
n
niQ
ii with Q
1
[i][n] ≥ 0
Consumed capacitive reactive energy of phase (i+1) with i ∈ [0; 2].
(Non-active quantities broken down – Configuration >Calculation methods >var)
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[ ][ ]
∑
−
==
int
3600
0VARhC0hCQ
1
1
T
n
niQ
ii with Q
1
[i][n] < 0
Consumed distortion energy of phase (i+1) with i ∈ [0; 2]
(Non-active quantities broken down – Configuration >Calculation methods >var)
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[ ][ ]
∑
==
int
3600
0VADh0Dh
T
n
niD
ii
Consumed non-active energy of phase (i+1) with i ∈ [0; 2]
(Non-active quantities not broken down – Configuration >Calculation methods >var)
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[ ][ ]
∑
==
int
3600
0VARh0Nh
T
n
niN
ii
Total consumed active energy
Ph[0][3] = Wh[0][3] =Ph[0][0] + Ph[0][1] + Ph[0][2]
Total consumed apparent energy
Sh[0][3] = VAh[0][3] = Sh[0][0] + Sh[0][1] + Sh[0][2]
Ph
Sh
hL
hC
Dh
Nh
Wh