Troubleshoot the pump trip failure 1. Symptom: Since a power plant 125 mw unit has been put into production since the pump occasionally a closing or tripping problem, without any signal relay card. After excluding the switchgear failure, according to the conventional method of checking the cable, the secondary circuit wiring and the relay and its rating are normal, start again often successful. Suspected to be dcs system soft failure caused, but the change in the control panel operation, there will still be this phenomenon.
2. Test to find the reasons: In order to find out the reason for this phenomenon, observe the changes in the process of closing the switch to confirm what is the reason for the pump trip. Test voltage meter which monitors the microcomputer trip circuit, mA watch differential relay 1cj, 2cj action, ammeter monitoring thermal protection circuit. Pick up a good meter, start to the pump, after a period of testing, and finally once the start of the pump trip, while observing the mA ammeter deflection of the watch a bit, the other watch does not respond, the new xjl- 0025/31 integrated block signal relay 1xj also action card, indicating that the differential protection action led to the pump trip.
3 root cause analysis: differential protection action, first of all suspect that there is a fault within the protected equipment. Through routine inspection, the pump motor and its cable are normal, the differential relay is calibrated, and the polarity of the current transformer is connected correctly. After eliminating the causes of equipment failure and wiring errors, differential protection operates during motor start-up, indicating that the differential current of the differential circuit exceeded the differential relay setting during this process. Under normal circumstances caused by differential current loop differential current are two main reasons. First, both sides of the motor current transformer ratio error is different, there is a small difference current, the difference current is less than 5% of the motor rated current id. Second, both ends of the current transformer secondary load difference will cause the difference between the ratio, resulting in a poor current. In the feed pump motor differential protection circuit current transformer load difference is only the difference between the length of the secondary cable, about 50m, and the rated current, the differential relay power consumption is not greater than 3va, the secondary load is not heavy. Check found to pump motor differential protection with the end of the current transformer models are lmzbj-10, b-level 15 times the rated current, the ratio of 600/5, capacity 40va, fully able to meet the requirements of the secondary load.
The above pump trip analysis is based on the normal operation of the condition, the motor starts, the situation is different. When the motor starts, the current is very large. The current transformer on both sides of the motor may be saturated. In this case, the secondary differential current may be very large due to the inconsistent magnetization characteristics of the current transformers. According to the relay factory in acheng lcd-12 differential relay setting instructions, the relay operating current setting izd = △ i1 × kk × in / n = 0.06 × 3 × 356/120 = 0.534a (where: △ i1- First, the end of the current transformer maximum error during normal operation, 0.04 ~ 0.06; kk-reliability factor, 2 to 3; in- motor rated current; n- current transformer ratio). Should be set at 1.0a position. The use of b-level transformer case, the differential relay operating current set at 1.5a, the braking factor of 0.4, the differential protection is still occasionally activated when the motor starts, is due to b-level current transformer magnetization saturation point Lower, anti-saturation capacity is low, can not meet the requirements of the differential relay. Usually require differential protection circuit current transformer using d-level, d-level transformer saturation point higher, not so easy to saturation, can reduce the difference between the current flowing through the differential loop when the motor starts. In the replacement of d-level current transformer, while the differential relay operating current set at 1.0a, the braking factor of 0.4, then there is no switch over a switch that is the fault of the pump trip.
2. Test to find the reasons: In order to find out the reason for this phenomenon, observe the changes in the process of closing the switch to confirm what is the reason for the pump trip. Test voltage meter which monitors the microcomputer trip circuit, mA watch differential relay 1cj, 2cj action, ammeter monitoring thermal protection circuit. Pick up a good meter, start to the pump, after a period of testing, and finally once the start of the pump trip, while observing the mA ammeter deflection of the watch a bit, the other watch does not respond, the new xjl- 0025/31 integrated block signal relay 1xj also action card, indicating that the differential protection action led to the pump trip.
3 root cause analysis: differential protection action, first of all suspect that there is a fault within the protected equipment. Through routine inspection, the pump motor and its cable are normal, the differential relay is calibrated, and the polarity of the current transformer is connected correctly. After eliminating the causes of equipment failure and wiring errors, differential protection operates during motor start-up, indicating that the differential current of the differential circuit exceeded the differential relay setting during this process. Under normal circumstances caused by differential current loop differential current are two main reasons. First, both sides of the motor current transformer ratio error is different, there is a small difference current, the difference current is less than 5% of the motor rated current id. Second, both ends of the current transformer secondary load difference will cause the difference between the ratio, resulting in a poor current. In the feed pump motor differential protection circuit current transformer load difference is only the difference between the length of the secondary cable, about 50m, and the rated current, the differential relay power consumption is not greater than 3va, the secondary load is not heavy. Check found to pump motor differential protection with the end of the current transformer models are lmzbj-10, b-level 15 times the rated current, the ratio of 600/5, capacity 40va, fully able to meet the requirements of the secondary load.
The above pump trip analysis is based on the normal operation of the condition, the motor starts, the situation is different. When the motor starts, the current is very large. The current transformer on both sides of the motor may be saturated. In this case, the secondary differential current may be very large due to the inconsistent magnetization characteristics of the current transformers. According to the relay factory in acheng lcd-12 differential relay setting instructions, the relay operating current setting izd = △ i1 × kk × in / n = 0.06 × 3 × 356/120 = 0.534a (where: △ i1- First, the end of the current transformer maximum error during normal operation, 0.04 ~ 0.06; kk-reliability factor, 2 to 3; in- motor rated current; n- current transformer ratio). Should be set at 1.0a position. The use of b-level transformer case, the differential relay operating current set at 1.5a, the braking factor of 0.4, the differential protection is still occasionally activated when the motor starts, is due to b-level current transformer magnetization saturation point Lower, anti-saturation capacity is low, can not meet the requirements of the differential relay. Usually require differential protection circuit current transformer using d-level, d-level transformer saturation point higher, not so easy to saturation, can reduce the difference between the current flowing through the differential loop when the motor starts. In the replacement of d-level current transformer, while the differential relay operating current set at 1.0a, the braking factor of 0.4, then there is no switch over a switch that is the fault of the pump trip.
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