300MW unit desulphurization booster fan analysis of differential protection misoperation

2016-02-03 10:01:46

Author: Guo Xuefeng
Abstract: the booster fan as an important area of thermal power plant desulfurization equipment, there are some problems in the two circuit design, caused by the misoperation of differential protection. Check the transformer due to limit error (10%) a lot of work, the reasons for misoperation are discussed, and puts forward a new solution for the design, debug.
Keywords: differential protection, booster fan, malfunction analysis, solution
introduction
Huaneng Group power plant desulphurization booster island #2 unit fan motor power is 2700KW, voltage 6KV, differential transformer protection 400/5 5P20 30VA parameters. During the test run, booster fan differential protection escape starting current, improve the setting value of the protection device itself is invalid, the commissioning quality and the two circuit wiring and differential polarity correctness check. Booster fan protection device installed in 6kV cabinet, neutral point long cable, about 250 m, 2.3 ohm resistance, load has exceeded the CT secondary winding capacity, resulting in current numerical error is bigger, cause the mal operation of differential protection of the main reason.
1 study on the problems of research and testing
Determination of 1.1CT two winding overload
The booster fan startup process, see Figure 1-1 graphic recorder:                                                                   Figure 1-1
1L1 cabinet A 1L2 cabinet B phase current, phase current, phase current, 1L3 cabinet C 2L1 A neutral point phase current and phase current, 2L2 neutral B 2L3 C neutral point current
From book wave current graphics can be seen, the startup process and phase current distortion of the neutral side of phase A and C can judge neutral side of phase A and C phase of secondary winding of the transformer load is too large, cause the saturation of the transformer, caused by the distortion of the current waveform.
1.2CT two winding capacity calculation
The selection and ratio of two winding capacity CT and the accuracy limit. "According to the DL/T866 current and voltage transformer selection and calculation of guide" in the P type current transformer performance calculation:
1, the current transformer rated accuracy limit primary current Ipal should be greater than the current Ipcf fault checking protection. The booster fan rated current is 306A, the starting current ratio of 6.5.
(306 * 6.5) /20=99.5A, 100A
According to this method results, design selection of transformer 400/5 to meet the application requirements. But the project side transformer neutral point switch cabinet far distance, rated two current by 5A. Therefore, according to the time limit of two rated EMF accurately checking.
2, according to the rated two limit electric type checking
A) P type current transformer rated two limit electromotive force (Es1) for
Es1=Kalf * Isn * =20 * 5 * (Rct+Rbn) (4.1+1.2) =530V
Type: Kalf-- accuracy limit factor, 20 times
Isn-- two times rated current, 5A
The two winding resistance Rct-- current transformer, 4.1.
Rbn-- current transformer rated load two, 1.2.
B) two EMF relay performance verification requirements (Es) for
Es=K * Kpcf * Isn * (Rct+Rb) =20 * 1 * 5 * (4.1+2.3) =640V
Type: Kpcf-- protection coefficient, take 20 times, with the same accuracy limit factor
K -- given transient coefficient 1
Isn-- two times rated current, 1A
The two winding resistance Rct-- current transformer, 4.1.
Rb-- of current transformer two load, 2.3.
C) two EMF current transformer rated two limit electromotive force should be greater than checking protection requirements, i.e.:
Es1 is more than or equal to Es but the calculation results of 530V = 640V
So the transformer can not choose from starting current.
The calculation of 1.3 booster fan CT two connection line increase effective cross section after error
Two load 0.77 will be pressurized neutral point CT two connection line to increase the effective section after real measured
Es=K * Kpcf * Isn * (Rct+Rb) =20 * 1 * 5 * (4.1+0.77) =487V
530V = 487V qualified
In the rectification, start the booster fan again, differential protection during startup without misoperation.
The influence of current numerical error of 2CT on differential protection and the solutions
From the above calculation shows that, the original secondary connection line measured secondary load for 2.3 Omega, current transformer saturation, distortion in secondary current, and cabinet in the current transformer secondary side waveform inconsistent, resulting in large imbalance current, resulting in differential protection misoperation.
Solution: in view of this phenomenon, because the neutral point current transformer two, increase the cable cross section is not economic, the follow-up units are changed two times for current transformer 1A.
Conclusion 3
Because of the booster fan neutral point long cable, a core section is too small, secondary circuit connection line impedance is wide and beyond transformer secondary load allowed, in booster fan start instant start current greatly, resulting in neutral point side transformer deep saturation, caused a serious imbalance in the differential protection circuit, resulting in differential protection misoperation. According to the actual situation of #2 machine to increase the use of cable section two practice transformer, follow-up all units to replace the two current transformer 1A. Practice has proved that did not happen during the startup of differential protection misoperation phenomenon.