Setting Guidelines - ABB RELION 670 Series Applications Manual

Section 7
Impedance protection
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 Xd' = 0.2960 · 0.952 = 0.282 Ω
calculation Rs = 0.0029 · 0.952 = 0.003 Ω
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
particular with islanding. But for the two machine model as in Table 12, 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
reactance and resistance of only one power line.
• All the reactances and resistances must be referred to the voltage level where the Out-of-step
relay is installed; for the example case shown in Table 12, this is the generator nominal voltage
VBase = 13.8 kV. This affects all the forward reactances and resistances in Table 12.
106
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
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,
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)
ForwardX , ForwardR must therefore take into account the
1MRK 511 364-UUS A
GUID-CB86FCF6-8718-40BE-BDF2-028C24AB367D v6
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)
Application manual
2
= 0.038
2
= 0.004 Ω