Several methods for troubleshooting of the pump tripping by Baili Biological Editor:
Trouble shooting method for pump tripping
1: Failure phenomenon: Since the 125 mw unit of a power plant has been put into production, the feed water pump occasionally has the problem of closing and tripping without any signal relay. After troubleshooting the switching mechanism, check the cables, secondary circuit wiring, and relays and their settings according to conventional methods, and it is often successful to start again. Later it was suspected that it was caused by a soft fault in the dcs system, but this phenomenon still occurs even if it is operated on the control panel.
2: Test to find the cause: In order to find out the cause of this phenomenon, observe the changes of the meters in the closing process of the switch to confirm what caused the pump to trip. In the test, the voltmeter monitors the trip circuit of the microcomputer, the mA meter monitors the operation of the differential relays 1cj and 2cj, and the ammeter monitors the thermal protection circuit. After connecting the meter, start the feed water pump. After a period of testing, the pump tripped once at the start. At the same time, the pointer of the mA meter was deflected. Other monitoring meters did not respond. The 0025/31 integrated block signal relay 1xj has also been deactivated, indicating that the differential protection action caused the pump to trip.
3: Root cause analysis: differential protection action, first suspect that there is a fault inside the protected equipment. Through routine inspection, the feedwater pump motor and its cable are normal, the differential relay is verified normally, and the polarity of the current transformer is connected correctly. After eliminating the cause of equipment faults and wiring errors, the differential protection operates during the motor starting process, indicating that the differential current of the differential circuit exceeds the setting value of the differential relay during this process.
Under normal circumstances, there are two main reasons for the differential current of the differential circuit: one is that the ratio error of the current transformer on both sides of the motor is different, there is a small difference current, this difference current is less than 5% of the motor rated current id . Second, the difference in the secondary load of the current transformers on both sides of the head and tail will also cause the difference in its transformation ratio, so there is a difference current. The load difference of the current transformer in the differential protection circuit of the feed water pump motor is only the difference of the length of the secondary cable, which is about 50 m, and at the rated current, the power consumption of the differential relay is not more than 3 va, the secondary load is not weight.
The inspection found that the current transformer models used for the differential protection of the feed water pump motor are all lmzbj-10, class b 15 times rated current, transformation ratio 600/5, capacity 40 va, which can fully meet the requirements of secondary load. The above analysis of the pump trip is based on the conditions of normal operation, and the situation is different when the motor is started. When the motor starts, the current is very large, and the current transformers on both sides of the head and tail may be saturated. At this time, due to the inconsistent magnetization characteristics of the current transformers, the secondary differential current may be large. According to the setting instructions of lcd-12 differential relay of Acheng Relay Factory, the setting value of the operating current of the relay is izd = △ i1 × kk × in / n = 0.06 × 3 × 356/120 = 0.534a (where: △ i1— The maximum error of the first and the tail current transformers during normal operation, 0.04 ～ 0.06; kk—reliability coefficient, 2 ～ 3; in—rated motor current; n—current transformer ratio). It should be set at 1.0a. In the case of using a class b transformer, the operating current of the differential relay is set at 1.5a, and when the braking coefficient is 0.4, the differential protection still occasionally operates when the motor is started, due to the saturation point of the magnetization characteristic of the class b current transformer Low, low saturation resistance, can not meet the requirements of differential relays.
It is usually required that the current transformer of the differential protection circuit adopts class d, and the saturation point of the class d transformer is higher, which is not so easy to saturate, which can reduce the differential current flowing through the differential circuit when the motor is started. After replacing with the current transformer of class d, and setting the operating current of the differential relay at 1.0a and the braking coefficient at 0.4, the failure of the switch to close, that is, the pump trip, has not occurred.
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