External faults, External faults -6 – Basler Electric BE1-87B User Manual

Page 24

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(summing point) on a per-phase basis. Three single-phase BE1-87B relays or one three-phase relay
provide complete protection of the bus. The relay will generate a trip output when the instantaneous
voltage applied across 5 and 7,or 3 and 7, or 1 and 7 of the three-phase model (A, B, and C respectively),
exceeds the voltage pickup setting (V

DIFF

) and the fault current is greater than the current sensitivity

setting.

External Faults

If the differential protection scheme is to perform satisfactorily, then it must not trip for faults external to
the zone of protection. For example, Figure 2-5 shows a one-line diagram for a three input differential
scheme. The BE1-87B must not operate for a fault at F1. Since the CTs in the faulted feeder (CT 3) will
see the most current, assume they will saturate completely, thus causing the magnetizing reactance to
drop to zero. The total current from the other CTs (CT 1 and 2) is forced through the parallel combination
of the high impedance relay (5,000

Ω) and the saturated CT secondary. The saturated CT secondary

winding resistance is in series with any resistance of the CT leads and connection cables (the total of
which presents a much lower resistance than the 5,000

Ω).

Therefore, nearly all the secondary fault current will flow through the saturated CT. A voltage drop V

R

caused by the flow of the fault current in this parallel path will appear across the BE1-87B relay. For this
fault, the highest voltage that could be developed at the relay would occur when the associated CT (CT3)
saturates completely, and the others (CT1 and CT2) did not saturate at all. When a CT with a distributed
toroidal winding (on the tap used) saturates completely, it produces no voltage and the impedance, as
seen at the secondary winding, is very nearly equal to the winding resistance (very small impedance).
Thus the highest peak voltage that can be developed across the relay during an external fault will be
equal to the voltage produced by the total secondary fault current flowing through the control cable
resistance plus the winding resistance of the CT associated with the faulted feeder. Refer to the example
case in Figure 2-5 while applying Equation (1):

)

2R

(R

)

(I

2

2

Vpeak

L

S

F

+

=

(EQUATION 1)

I

F

= rms symmetrical value of fault current in the fault CT in secondary amps.

R

S

= CT secondary winding resistance plus any lead resistance (at highest expected operating

temperature)

R

L

= Cable resistance from junction point to CT (at highest expected operating temperature)

Equation (1) above yields the peak voltage developed at the relay for a completely offset wave of current
having an rms symmetrical value of I

F

secondary amperes. Because the BE1-87B relay is calibrated in

symmetrical rms volts, Equation (2) below, which yields the rms voltage value, is used in the paragraphs
on Calculation of Settings.

)

2R

(R

I

V

L

S

F

R

+

=

(EQUATION 2)

The pickup voltage of the BE1-87B must be set above this value of rms voltage and above the rms value
of the other voltages obtained in a similar manner on all the circuits of the bus. Because the peak voltage
is proportional to the fault current, the highest possible value of expected fault current in rms symmetrical
amperes should be used in making the evaluation.

2-6

BE1-87B Application

9282300990 Rev P

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