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7.11 Harmonic Measurement Specifications
Determination of Measurement Functions during Harmonic Measurement
Measurement Functions during
Harmonic Measurement
Voltage U( ) [V]
Current I( ) [A]
Active power
Apparent power
(TYPE3)*
Reactive power
(TYPE3)*
Power factor
Phase difference φ( ) [°]
Phase difference with respect to U(1)
φU(
) [°]
Phase difference with respect to I(1)
φI(
) [°]
Impedance of the load circuit
Z(
) [Ω]
Series resistance of the load circuit
Rs(
) [Ω]
Series reactance of the load circuit
Xs(
) [Ω]
Parallel resistance of the load circuit
Rp(
) [Ω] (= 1/G)
Parallel reactance of the load circuit
Xp(
) [Ω] (= 1/B)
* For details on the type of S and Q equations, see section 5.9 in the User's Manual IM WT3001E-01EN.
7-42
Characters/Numbers inside the parentheses of
dc
(when k = 0)
U(dc) =U
I(dc) = I
P(dc) = U
(0)
P( ) [W]
r
S( ) [VA]
S(dc) = P(dc)
Q( ) [var]
Q(dc) = 0
P(dc)
λ( )
λ(dc) =
S(dc)
U(dc)
Z(dc) =
I(dc)
P(dc)
Rs(dc) =
I(dc)
Q(dc)
Xs(dc) =
U(dc)
Rp(dc) =
P(dc)
U(dc)
Xp(dc) =
Q(dc)
Note
•
Variables k, r, and j denote the harmonic order, real part, and imaginary part, respectively.
•
Variables U(k), Ur(k), Uj(k), I(k), Ir(k), and Ij(k) are expressed using rms values.
•
min denotes the minimum order. You can select 0 (DC component) or 1 (fundamental signal
component) for the minimum order. For details, see section 7.5.
•
Variable max is the upper limit of measured order. The upper limit is determined
automatically (maximum is 100) by the frequency of the PLL source.
Method of Determination, Equation
measurement functions
1
(when k = 1)
U(k) =
(0)
r
(0)
I(k) =
r
I
(0)
P(k) =
U
•
r
S(k) =
Q(k) =
U
φ(k) = ATAN2{P(k), Q(k)}
Regarding ATAN2{x,y} in the equation above:
φU(k) =
φI(k) = Phase difference of I(k)
Rs(k) =
2
Xs(k) =
2
I(dc)
2
Rp(k) =
2
Xp(k) =
k
(When k = 2 to max)
2
2
U
(k)
+
U
(k)
r
j
2
2
I
(k)
+
I
(k)
r
j
(k)
I
(k) +
U
(k)
I
(k)
r
•
r
j
•
j
2
2
P(k)
+ Q(k)
(k)
I
(k) –
U
(k)
I
(k)
r
•
j
j
•
r
P(k)
λ(k) =
S(k)
{ }
y
–1
When x > 0,
tan
x
When x < 0 and y > 0,
tan
When x < 0 and y < 0,
tan
Phase difference of U(k)
with respect to U(1)
with respect to I(1)
U(k)
Z(k) =
I(k)
P(k)
2
I(k)
Q(k)
2
I(k)
2
U(k)
P(k)
2
U(k)
Q(k)
(Continues on the next page)
(Table 1/3)
Total {No ( )}
max
U
2
U =
(k)
k = min
max
I
2
I =
(k)
k = min
max
P
P =
(k)
k = min
2
2
S =
P
+ Q
max
Q
Q =
(k)
k = min
P
λ =
S
φ = ATAN2{P, Q}
{ }
y
–1
+ 180°
x
{ }
y
–1
- 180°
x
IM WT3001E-51EN
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