Treatment of tripping of feed water pump

1 Fault phenomenon
Since the 125 mw unit of Meixian Power Plant has been put into production, the feed pump occasionally has a problem of tripping or tripping, and there is no signal relay to be dropped. After the failure of the switching mechanism is eliminated, the cable, the secondary circuit wiring, and the relays and their settings are normal according to the conventional method, and the restart is often successful. After the suspected dcs system soft fault, but changed to operate on the control panel, this phenomenon will still occur.
2 test find reasons
In order to find out the cause of this phenomenon, observe the changes of the meters in the process of closing the switch to confirm what caused it to trip.
In the test, the voltmeter monitors the microcomputer trip circuit, the milliampere monitors the differential relays 1cj, 2cj, and the ammeter monitors the thermal protection loop. After the meter is connected, the feed pump is started. After a period of test, the feed pump is tripped once, and the pointer of the milliampere meter is deflected. The other monitors do not respond. The new xjl is replaced. The -0025/31 type integrated block signal relay 1xj also acts off, indicating that it is tripped by the differential protection action.
3 root cause analysis
The differential protection action first suspects that there is a fault inside the protected device. Through routine inspection, the feed pump motor and its cable are normal, the differential relay is verified normally, and the current transformer polarity is connected correctly. After eliminating the cause of equipment failure and wiring error, the differential protection operates during the motor starting process, indicating that the differential current of the differential circuit exceeds the differential relay setting value during this process.
There are two main reasons for the differential current difference current under normal conditions:
First, the current transformer on both sides of the motor has different ratio error, and there is a small difference current, which is less than 5% of the rated current id of the motor.
Second, the difference in the secondary load of the current transformer on both sides of the head and tail will also cause the difference in the ratio, so that there is a difference current. The current transformer load difference in the differential protection circuit of the feed pump motor is only the difference of the length of the secondary cable, which is about 50 m, and the power consumption of the differential relay is not more than 3 va at the rated current. The secondary load is not weight. Check that the first and last side current transformers used for the differential protection of the feed pump motor are lmzbj-10, b level 15 times rated current, ratio 600/5, capacity 40 va, fully meet the requirements of secondary load.
The above analysis is based on normal operating conditions, 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 first and last ends may be saturated. At this time, the secondary differential current may be large due to the inconsistent magnetization characteristics of the current transformers.
According to the lcd-12 differential relay setting of Acheng Relay Factory, the operating current setting value of the relay is izd=△i1×kk×in/n=0.06×3×356/120=0.534a (in the formula: △i1— The maximum error of the first and last current transformers during normal operation, 0.04 ~ 0.06; kk - reliability coefficient, 2 ~ 3; in - motor rated current; n - current transformer ratio). Should be set at 1.0a.
In the case of using the b-class transformer, the differential relay operating current is set at 1.5a, and the braking coefficient is 0.4. The differential protection will occasionally operate when the motor starts, due to the saturation characteristic of the b-level current transformer magnetization characteristic. Lower, low saturation resistance, can not meet the requirements of differential relays. Generally, the current transformer of the differential protection circuit is required to adopt d-stage, and the saturation point of the d-stage transformer is higher, which is not so easy to be saturated, and can reduce the differential current flowing through the differential circuit when the motor starts. After replacing the current transformer of class d, and setting the operating current of the differential relay to 1.0a, and the braking coefficient is 0.4, there is no fault of the switch being closed when the switch is closed.

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