ABB Relion 670 Series Applications Manual page 120

Section 7
Impedance protection
1. Remain stable for normal steady state load.
2. Distinguish between stable and unstable rotor swings.
3. Locate electrical centre of a swing.
4. Detect the first and the subsequent pole-slips.
5. Prevent stress on the circuit breaker.
6. Distinguish between generator and motor out-of-step conditions.
7. Provide information for post-disturbance analysis.
7.2.3 Setting guidelines
The setting example for generator protection application shows how to calculate the most
important settings
Table 12: An example how to calculate values for the settings ForwardR, ForwardX, ReverseR, and ReverseX
Generator
Turbine
200 MVA
(hydro)
GEN
CT 1
to OOS relay
ANSI11000090 V2 EN-US
Generator
Data VBase = Vgen = 13.8 kV
required
IBase = Igen = 8367 A
Xd' = 0.2960 pu
Rs = 0.0029 pu
1-st step in ZBase = 0.9522 Ω (generator)
calculation Xd' = 0.2960 · 0.952 = 0.282 Ω
Rs = 0.0029 · 0.952 = 0.003 Ω
2-nd step in
calculation
3-rd step in 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)
calculation 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 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)
resulted 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
Settings
• 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
the post-disturbance configuration of the simplified power system. This is not always easy, in
114
ForwardR , ForwardX , ReverseR , and ReverseX .
13.8 kV
CT 2
Step-up transformer
V1 = 13.8 kV
usc = 10%
V2 = 230 kV
I1 = 12 551 A
Xt = 0.1000 pu (transf. ZBase)
Rt = 0.0054 pu (transf. ZBase)
ZBase (13.8 kV) = 0.6348 Ω
Xt = 0.100 · 0.6348 = 0.064 Ω
Rt = 0.0054 · 0.635 = 0.003 Ω
ForwardR , ForwardX , ReverseR , and ReverseX .
ForwardX , ForwardR , ReverseX and ReverseR must, if possible, take into account,
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Double circuit power line
Transformer
230 kV, 300 km
300 MVA
Y
Single power line
Vline = 230 kV
Xline/km = 0.4289 Ω/km
Rline/km = 0.0659 Ω/km
Xline = 300 · 0.4289 = 128.7 Ω
Rline = 300 · 0.0659 = 19.8 Ω
(X and R above on 230 kV basis)
Xline= 128.7 · (13.8/230)
0.463 Ω
Rline = 19.8 · (13.8/230)
Ω
(X and R referred to 13.8 kV)
1MRK 511 407-UUS Rev. J
GUID-CB86FCF6-8718-40BE-BDF2-028C24AB367D v7
Equivalent
power
system
ANSI11000090_2_en.vsd
Power system
Vnom = 230 kV
SC level = 5000 MVA
SC current = 12 551 A
φ = 84.289°
Ze = 10.5801 Ω
Xe = Z · sin (φ) = 10.52 Ω
e
Re = Z · cos (φ) = 1.05 Ω
e
(Xe and Re on 230 kV basis)
2
= Xe = 10.52 · (13.8/230)
Ω
2
= 0.071 Re = 1.05 · (13.8/230)
(X and R referred to 13.8 kV)
Phasor measurement unit RES670
Application manual
2
= 0.038
2
= 0.004 Ω