Treatment of tripping of feed water pump

1. Fault Phenomenon Since the 125 MW unit of Meixian Power Plant was put into operation, the feed pump occasionally trips or fails to close, with no relay signal being triggered. After conventional checks—such as inspecting the cable, secondary wiring, relays, and their settings—everything appears normal. The system usually restarts successfully. Initially, a suspected DCS soft fault was considered, but even after switching to manual control on the panel, the issue still occurred. 2. Test and Diagnosis To identify the root cause, we monitored the meter readings during the switch closing process to determine what triggered the trip. A voltmeter was connected to the microcomputer trip circuit, a milliammeter to the differential relays (1CJ, 2CJ), and an ammeter to the thermal protection loop. During testing, the feed pump was started and ran for some time before tripping. At that moment, the milliammeter showed a deflection, while the other meters remained unaffected. We replaced the new XJL and found that the -0025/31 integrated block signal relay 1XJ also activated, confirming that the trip was caused by the differential protection. 3. Root Cause Analysis The first assumption was that there was an internal fault in the protected equipment. A routine inspection confirmed that the feed pump motor and its cables were in good condition, and the differential relay was functioning properly. The current transformer polarity was correctly connected. After eliminating possible equipment failures and wiring errors, the differential protection operated during the motor start-up, indicating that the differential current exceeded the relay’s set value at that time. There are two main reasons for differential current under normal conditions: - First, the current transformers on both sides of the motor may have different ratio errors, resulting in a small difference current, typically less than 5% of the motor's rated current. - Second, differences in the secondary load of the current transformers on the motor’s input and output can also lead to a difference in the ratio, causing a difference current. In this case, the secondary load difference is due to the length of the secondary cable, which is about 50 meters. The power consumption of the differential relay is within 3 VA at rated current, so the secondary load is not excessive. The current transformers used for the differential protection of the feed pump motor are LMZBJ-10, B-class, with a ratio of 600/5 and a capacity of 40 VA, meeting all requirements. However, during motor start-up, the current is very high, and the current transformers on both ends may become saturated. This can lead to inconsistent magnetization characteristics, resulting in a large secondary differential current. According to the LCD-12 differential relay settings from Acheng Relay Factory, the operating current is calculated as: IZD = △I1 × KK × IN / N = 0.06 × 3 × 356 / 120 = 0.534A (Where △I1 is the maximum error of the current transformers, 0.04–0.06; KK is the reliability coefficient, 2–3; IN is the motor’s rated current; N is the current transformer ratio). It should be set to 1.0A. When using B-class current transformers, the differential relay is set to 1.5A with a braking coefficient of 0.4. Due to the lower saturation resistance of B-class CTs, they may not meet the relay’s requirements, leading to occasional tripping during motor start-up. Typically, D-class current transformers are recommended for differential protection because they have a higher saturation point and are less likely to saturate, reducing the differential current during motor startup. After replacing the B-class CTs with D-class ones and adjusting the relay settings to 1.0A with a braking coefficient of 0.4, the switch closure no longer resulted in any faults.

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