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Fig.5.
Equivalent phase circuit of proposed device
There symbols
S
,
k
and
sec
on
u
denote capacity, the short-circuit impedance and
secondary voltage of the transformer respectively. The Eq.(3) is calculated at
2000 kVA
.
TSC consists of 3 channels which capacity of 150kVar, 300kVar and 600kVar
respectively. In order to avoid terrible 5
th
harmonic current there is a reactance ratio of
6% of the TSC. AS shown in figure5 the compensation capacitor of every channel
was denoted by
C
1
,
C
2
and
C
3,
corresponding smoothing reactor is
L
1
,
L
2
and
L
3.
There is a single tuning filter with the capacity of 3600kVar in PPF based on
analyzing harmonic current and original compensation capacitor. According to
classical calculation method the parameters of the PPF can be given.
Considering the economic benefits of the project and space constraint there is
remained fixed capacitor which capacity of 1800kVar. The symbol
C
F
denotes
equivalent capacitor of the remained FC.
The harmonic source denotes total harmonic current that caused by corresponding
rectifier and converter device of the secondary side of the transformer. Detail
harmonic currents were illustrated by TABLE I. According to Thevenin and Norton
theorem the repercussion effect of every harmonic current need to analyze.
As shown in Figure5 due to the equivalent inductance fall into a part of the 6.3kV
bus, FC and PPF could be switched individually or in a flotilla, and there are 3
channel of TSC, so their manner of working may be expressed as
,
and
sec
S
=
u
=
6300 V
k
=
9.06%
N
ond
(
)
1
2
C
,
CC
+
and
2
2
(
)
1
2
3
respectively. The total combination of equivalent inductance, TSC, FC
and PPF can be theoretically is presented as,
(
CC C
++
3
3
3
)(
)
11
2
1
2
3
= + + + =
(4)
Equation (4) denotes that there are 21 network topologies of proposed device. Every
circuit topology need to analyze in order to avoid the resonance occurs. Equivalent
impedance of every above part is defined as
nCCCCC C
21
1
1
2
2
3
3
3
(
)
Z
Z
Z
TSC
Zi
=
1, 2, 3
,
,
and
Ls
Fc
PPF
Z
=
ω
Ls
which are illustrated with Figure5. Above symbols are written as:
,
Ls
( )
( )
( ) (
)
Z
=−
1.0 /
ω
C
,
Z
=−
ω
L
1.0 /
ω
C
Z
=−
ω
L
1.0 /
ω
C
i
=
.
1, 2,3
Fc
F
PPF
5
5
TSCi
i
i
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