Rt = 0.0054 pu (transf. ZBase)
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
36, 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 must be referred to the voltage level where the
Out-of-step relay is installed; for the example case shown in Table
36, this is the
generator nominal voltage VBase = 13.8 kV.
This affects all the forward
reactances and resistances in Table
36.
• All reactances and resistances must be finally expressed in percent of ZBase,
where ZBase is for the example shown in T
able
36 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.
1MRK 504 163-UUS A Section 8
Impedance protection
Transformer protection RET670 2.2 ANSI 485
Application manual
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