Schweitzer Engineering Laboratories SEL-751 - Page 206

4.80
SEL-751 Relay Instruction Manual Date Code 20170927
Protection and Logic Functions
Group Settings (SET Command)
EXAMPLE 4.9 Convert Maximum Loads to Equivalent Secondary
Impedances
Start with maximum forward load:
800 MVA • (1/3) = 267 MVA per phase
230 kV • (1/ 3) = 132.8 kV line-to-neutral
267 MVA • (1/132.8 kV) • (1000kV/MV) = 2010 A primary
2010 A primary • (1/CT ratio) = 2010 A primary • (400) = 5.03 A
secondary
132.8 kV • (1000 V/kV) = 132800 V primary
132800 V primary • (1/PT ratio) = 132800 V primary •
(2000) = 66.4 V secondary
Now, calculate the equivalent secondary impedance:
This secondary value can be calculated more expediently with the following
equation:
Again, for the maximum forward load:
To provide a margin for setting ZLF, multiply by a factor of 0.9:
For the maximum reverse load:
Again, to provide a margin for setting ZLR:
EXAMPLE 4.10 Convert Power Factors to Equivalent Load Angles
The power factor (forward load) can vary from 0.90 lag to 0.95 lead.
Setting PLAF := cos
–1
(0.90) = 26°
Setting NLAF := cos
–1
(0.95) = –18°
The power factor (reverse load) can vary from 0.80 lag to 0.95 lead.
Setting PLAR := 180
°
– cos
–1
(0.95) = 180
°
– 18
°
= 162
°
Setting NLAR := 180
°
+ cos
–1
(0.80) = 180
°
+ 37
°
= 217
°
66.4 V secondary
5.03 A secondary
------------------------------------------ 13.2  secondary=
line-line voltage in kV
2
CT ratio•
3-phase load in MVA PT ratio•
----------------------------------------------------------------------------------------
230
2
400•
800 2000•
--------------------------- 13.2  secondary=
ZLF 13.2  secondary 0.9• =
11.90  secondary=
230
2
400•
500 2000•
--------------------------- 21.1  secondary=
ZLR 21.1 secondary 0.9• =
19.00  secondary=