Table 31: An example how to calculate values for the settings ForwardR, ForwardX, ReverseR, and ReverseX
Turbine
(hydro)
Generator
200 MVA
Transformer
300 MVA
Double power line
230 kV, 300 km
Equivalent
power
system
CT 1
To OOS relay
CT 2
13.8 kV
IEC10000117-2-en.vsd
to OOS relay
IEC10000117 V2 EN-US
Generator
Step-up transformer
Single power line
Power system
Data
required
UBase = Ugen = 13.8 kV
IBase = Igen = 8367 A
Xd' = 0.2960 pu
Rs = 0.0029 pu
U1 = 13.8 kV
U2 = 230 kV
usc = 10%
I1 = 12 551 A
Xt = 0.1000 pu (transf. ZBase)
Rt = 0.0054 pu (transf. ZBase)
Uline = 230 kV
Xline/km = 0.4289 Ω/km
Rline/km = 0.0659 Ω/km
Unom = 230 kV
SC level = 5000 MVA
SC current = 12 551 A
φ = 84.289°
Ze = 10.5801 Ω
1-st step in
calculation
ZBase = 0.9522 Ω (generator)
Xd' = 0.2960 · 0.952 = 0.282 Ω
Rs = 0.0029 · 0.952 = 0.003 Ω
ZBase (13.8 kV) = 0.6348 Ω
Xt = 0.100 · 0.6348 = 0.064 Ω
Rt = 0.0054 · 0.635 = 0.003 Ω
Xline = 300 · 0.4289 = 128.7 Ω
Rline = 300 · 0.0659 = 19.8 Ω
(X and R above on 230 kV
basis)
Xe = Z
e
· sin (φ) = 10.52 Ω
Re = Z
e
· cos (φ) = 1.05 Ω
(Xe and Re on 230 kV basis)
2-nd step in
calculation
Xd' = 0.2960 · 0.952 = 0.282 Ω
Rs = 0.0029 · 0.952 = 0.003 Ω
Xt = 0.100 · 0.6348 = 0.064 Ω
Rt = 0.0054 · 0.635 = 0.003 Ω
Xline= 128.7 · (13.8/230)
2
=
0.463 Ω
Rline = 19.8 · (13.8/230)
2
=
0.071 Ω
(X and R referred to 13.8 kV)
Xe = 10.52 · (13.8/230)
2
= 0.038
Ω
Re = 1.05 · (13.8/230)
2
= 0.004
Ω
(X and R referred to 13.8 kV)
3-rd step in
calculation
ForwardX = Xt + Xline + Xe = 0.064 + 0.463 + 0.038 = 0.565 Ω; ReverseX = Xd' = 0.282 Ω (all referred to gen. voltage 13.8 kV)
ForwardR = Rt + Rline + Re = 0.003 + 0.071 + 0.004 = 0.078 Ω; ReverseR = Rs = 0.003 Ω (all referred to gen. voltage 13.8 kV)
Final
resulted
settings
ForwardX = 0.565/0.9522 · 100 = 59.33 in % ZBase; ReverseX = 0.282/0.9522 · 100 = 29.6 in % ZBase (all referred to 13.8 kV)
ForwardR = 0.078/0.9522 · 100 = 8.19 in % ZBase; ReverseR = 0.003/0.9522 · 100 = 0.29 in % ZBase (all referred to 13.8 kV)
Settings
ForwardR, ForwardX, ReverseR, and ReverseX.
•
A precondition in order to be able to use the Out-of-step protection and construct a
suitable lens characteristic is that the power system in which the Out-of-step protection
is installed, is modeled as a two-machine equivalent system, or as a single machine –
infinite bus equivalent power system. Then the impedances from the position of the Out-
of-step protection in the direction of the normal load flow can be taken as forward.
•
The settings
ForwardX, ForwardR, ReverseX and ReverseR must, if possible, take into
account, the post-disturbance configuration of the simplified power system. This is not
always easy, in particular with islanding. But for the two machine model as in Table
, the
most probable scenario is that only one line is in service after the fault on one power line
has been cleared by line protections. The settings
ForwardX, ForwardR must therefore
take into account the reactance and resistance of only one power line.
•
All the reactances and resistances (ForwardX, ForwardR, ReverseX and ReverseR) must be
referred to the voltage level where the Out-of-step relay is installed; for the example case
shown in Table
, this is the generator nominal voltage UBase = 13.8 kV. This affects all
the forward reactances and resistances in Table
•
All reactances and resistances must be finally expressed in percent of ZBase, where ZBase
is for the example shown in Table
the base impedance of the generator, ZBase = 0.9522
Ω. Observe that the power transformer’s base impedance is different, ZBase = 0.6348 Ω.
Observe that this latter power transformer ZBase = 0.6348 Ω must be used when the
power transformer reactance and resistance are transformed.
Section 7
1MRK 505 343-UEN B
Impedance protection
392
Application manual
Содержание Relion 670 series
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