Leakage protector leapfrog trip reason

There are several possibilities for the leakage protector to trip over the leap:

First, it may be that the rear-stage leakage protector is larger than the front-stage leakage protector, causing the rear-stage protector not to trip and the front-stage tripping;

Second, it may be that the rear leakage protector is damaged, resulting in failure of protection;

Third, it is possible that the power of the front-stage leakage protector is too small, and the power consumption of the latter stage is greater than the rated current that the front-end protector can withstand, thereby causing the front-stage protector to trip.

In general, the third possibility is even greater. In the case of ensuring that the rear-stage leakage protector is normal, it may be appropriate to increase the front-end leakage protector appropriately.

Solution for leakage protection of the leakage protector:

(1) Improper setting of protection settings, especially if the protection values ​​of the upper and lower levels are improperly matched, when the lower level fails, the protection of this level does not work or the protection of the upper and lower levels simultaneously;

(2) The upper and lower protection time limits are improperly matched. When the fault occurs, the lower protection time limit is not reached and the upper time limit is reached to make the upper protection action; the setting of the relay protection of the incoming and outgoing lines and the time limit are very important, otherwise it is easy to occur. The leapfrog trip. In order to ensure the stable operation of the power system, the power supply department has higher requirements for the relay protection of the incoming line of the user. The quick-break and over-current of the incoming line must meet the requirements of the upper-level power grid, and the shorter the time, the better. This gives a certain difficulty in the protection setting of the line switch. In some places, the quick-break of the incoming and outgoing lines of the substation of the substation only depends on the action current. There is no time difference in the quick-break. When the short-circuit capacity of the grid is large, it is easy to cooperate with the action current. A leapfrog trip occurred. A short-circuit fault occurs on the high-voltage side of the transformer, and the short-circuit current is substantially equal to the busbar. If the quick-break has no time to match, the over-level trip is easy to occur, or tripped at the same time. When the low-voltage side outlet of the transformer fails, the short-circuit current calculation is performed. If the quick-break current setting value is too small, when the power grid capacity is large, when the low-voltage side outlet of the transformer has an accident, it is easy to cause a step-by-step trip.

The overcurrent of the incoming and outgoing lines is based on the overcurrent and time difference to ensure the selectivity of the relay protection. The time difference of overcurrent matching should generally be less than 0, 5 seconds. Although the high voltage switch now uses a vacuum circuit breaker, its inherent operating time is relatively small, but the inherent operating time of the switch, the relay protection exit time, the operation time of the intermediate relay, and the operation. The operating time of the organization has a certain relationship with the relay protection setting time. Therefore, the time difference between the incoming line and the outgoing line overcurrent protection is too small, and the over-the-counter trip is also prone to occur.

(3) The wiring of the relay protection circuit is incorrect. For example, if the current relay is connected in series to be connected in parallel, the protection setting value is doubled, and the current relay is connected in series to double the protection setting to protect the secondary circuit. If the wiring is incorrect, the quick disconnection will be connected to the overcurrent, and the overcurrent will be connected to the quick disconnect. When the loop is faulty and the setting value is correct, it will not operate correctly.