Schweitzer Engineering SEL-387-0, Instruction Manual

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10.29
Date Code 20170601 Instruction Manual SEL-387-0, -5, -6 Relay
Testing and Troubleshooting
Acceptance Testing
Step 3. Purpose: Connect two current test sources to one phase of one
winding input.
Method:
a. Connect a current source to the IAW1 input.
b. Connect a second current source in parallel with the
first source to the
IAW1
input as shown in Figure 10.7 .
Step 4. Purpose: Apply fundamental current to pick up the 87R
element.
Method: Turn on the first current test source connected to the
Winding 1 input (
IAW1
) equal to the TAP1 setting multiplied by
the connection constant A shown in Table 10.2 . The
87R
LED
will illuminate once current is applied to the relay.
The following test applies to a single-harmonic injection at a
time, i.e., only the second or the fourth harmonic, not both.
a. Set E87W1 = E87W2 = Y1, and HRSTR = Y.
b. Set the second current source for second-harmonic
current (120 Hz for NFREQ = 60 and 100 Hz for
NFREQ = 50).
c. Turn on the second current test source connected to the
Winding 1 input ( IAW1 ).
d. Starting at zero current, slowly increase the magnitude
of applied current until the 87R element deasserts,
causing the 87R LED to extinguish completely. Note
the value of the applied current from the second test
source. The general equation to calculate the percentage
of harmonic content for a single slope is: (±5% ±0.10
A (5 A relay) or ±5% ±0.02 A (1 A relay)):
Equation 10.12
where f(SLP) is the value of the slope for the chosen
value of IRT1
and
Equation 10.13
For inrush conditions, current normally is applied to
one side of the transformer, and the equation simplifies
to the following:
Equation 10.14
For example
SLP1 = 50 percent
PCT2 = 20 percent
Equation 10.15
where
PCT2 = second-harmonic setting in percent
SLP1 = slope 1 setting
I1F2 IOP1 IRT1
–
f SLP
 
•
 
PCT
100
----------- • =
% harmonic
I1F2
IOP1
-------------
100 percent
• =
% harmonic PCT2 1
SLP1
200
-------------- –
 
 
• =
% harmonic 20 1
50
200
--------- –
 
 
•
15 percent
= =