Power supply: The power supply is connected to terminals 15 and 16, with no polarity.
Grounding: The casing must be separately grounded to the ground grid with a copper wire of not less than 12 AWG; It is recommended that each device has independent wiring.
Insulation test warning: Before conducting the withstand voltage test, the connecting plug must be unplugged, otherwise high voltage may damage the internal semiconductor devices.
4. Setting calculation (core steps)
The purpose of setting BE1-87B is to ensure reliable non operation in case of faults outside the zone and reliable operation in case of minimum faults within the zone. The setting is divided into three steps: voltage setting, current setting, and sensitivity verification.
4.1 Voltage setting (based on the maximum voltage of external faults)
When there is a fault outside the area, the fault line CT may saturate, causing its secondary impedance to drop to the winding resistance. The current of other non fault CTs will all flow through the fault CT circuit, generating voltage at both ends of the differential circuit. The voltage must be less than the relay voltage setting value.
Calculation formula:
VDIFF=1.25×(RS+P⋅RL)×IFNV DIFF=1.25×(R S +P⋅R L)× NI FVDIFFV
DIFF: Minimum required voltage setting value (taking the next step greater than the calculated value, with a step size of 50V).
RSR S: Fault CT secondary winding DC resistance+lead resistance (converted based on maximum operating temperature).
RLR L: one-way cable resistance from differential junction to fault CT (also calculated based on maximum temperature).
PP: Phase number coefficient, set to 1 for three-phase short circuits and 2 for single-phase grounding (because the current passes through a two core cable circuit during single-phase).
IFI F: Maximum out of zone fault current (primary symmetrical effective value), usually taken as the maximum breaking current of the circuit breaker or the maximum short-circuit current of the system.
NN: CT ratio (e.g. 1200/5, then N=240).
1.25 is the reliability coefficient.
Simplified method (compliance): directly take the maximum breaking current of all circuit breakers as
IFI F, Take the cable resistance of the farthest CT as RLR L and set P=2. Calculate once to obtain the maximum value
VDIF FV DIFF, Then take a higher level. This method is simple and does not require recalculating with system changes.
Accurate method: Calculate the three-phase and single-phase faults at each feeder end separately, take the corresponding IFI F and RLR L, calculate them separately, take the maximum value, and then take the higher level. This method may result in lower setting values and improve sensitivity.
Example: A 1200/5 CT, RS=0.524 Ω R S=0.524 Ω, farthest CT cable one-way resistance 0.493 Ω, maximum short-circuit current 12500A, N=240. then
VDIFF=1.25×(0.524+2×0.493)×12500/240≈98.3V
V DIFF=1.25 × (0.524+2 × 0.493) × 12500/240 ≈ 98.3V, select the next level of 100V.
4.2 Current setting
The current setting needs to consider the following factors:
Avoiding CT circuit induced noise: Due to the high impedance of the differential circuit, external faults may induce voltage, but the induced current is extremely small. To prevent misoperation, the current setting is generally not less than 0.5A.
Combined with CT testing function: When using a CT testing source, the test current under an unhealthy CT circuit (short circuit) is approximately
Vtest/100V
Test/100 (100 Ω series resistance) is usually around 0.3A, so the current setting should be greater than this value, such as 0.5A.
If there is a lightning arrester, the current setting must be set to 2.5A (highest level) to avoid the surge current during the operation of the lightning arrester.
If high sensitivity is desired, 0.25A or 0.5A can be set when there is no lightning arrester and the system is impedance grounded.
4.3 Minimum internal fault sensitivity verification
It is necessary to ensure that the relay can reliably operate in the event of minimal internal faults on the busbar (such as single-phase grounding). Sensitivity verification requires the peak characteristics corrected by the excitation characteristic curve of CT.
Steps:
Find the inflection point voltage based on the conventional excitation curve (effective value) of CT
EsE s and inflection point current IeI e.
Calculate points on double logarithmic coordinates
V=7Es
V=7E s,I=5Ie
I=5I e, Draw a straight line with a slope of 1/2 across this point (voltage corresponds to two ten times the current for every ten times the current), and form a corrected peak curve at the intersection with the extension of the lower straight line of the original curve.
Calculate the voltage required for the operation of relay voltage components
VS=22×VDIFF
V S=22 × V DIFF (in this case, VS=2.83 × 100=283VV S=2.83 × 100=283V).
Read the excitation current Ie'E '(approximately 0.05A in this case) corresponding to the voltage from the correction curve.
The current of the relay itself at the critical point of action is IR=(2 × VDIFF)/5000 Ω=0.04AI
R=(2 × V DIFF)/5000 Ω=0.04A (internal impedance of the relay is approximately 5000 Ω).