JPS6238660B2 - - Google Patents

Description

[Detailed description of the invention]

The present invention relates to a method for locating a fault point on a power transmission line in the event of a one-line ground fault. Conventional methods for locating fault points on power transmission lines include methods for locating surges that occur at the fault instantaneously at the fault point based on the difference in time it takes to reach substations at both ends of the line; fire,
There are methods that determine the location based on the time it takes for this pulse to reflect from the failure point and return. However, the former method, which uses surges generated at the fault point, is likely to cause malfunctions due to induced surges caused by lightning, and the latter method, which fires pulses toward the fault point, is more likely to cause malfunctions due to the surge generated at the fault point. In the case of long distances, pulse attenuation may occur, making localization difficult. In addition, as a method other than the above, line voltage at the time of failure,
There is also a method of locating by measuring the line impedance from the substation at one end to the fault point using current, but this method is affected by uncertain factors such as the resistance at the fault point, resulting in a decrease in location accuracy. There are drawbacks. As mentioned above, the present invention is almost unaffected by fault point resistance, etc., and utilizes line voltage and current.
The purpose of this invention is to provide a locating method that can locate the fault point position with high accuracy in the event of a line-to-ground fault, and the details thereof will be explained next using the drawings. The principle will be explained by taking as an example a case where an a-phase one-line ground fault occurs at point F of a power transmission line L with a fault point resistance R _f as shown in FIG. Now, the ratio of the distance from substation A to point F to the total length of the line is n,
The positive sequence impedance of the entire length of the line is Z〓 ₁ , the zero-sequence impedance is Z〓 ₀ , the zero-sequence impedance of the neutral point grounding of A and B substations is Z〓 _0A , Z〓 _0B , and it flows out from A substation in the event of a failure. The symmetrical currents are I〓 ₀ , I〓 ₂ , I〓
₂ Let I = _L be the load current flowing between the substations at both ends of AB. In addition, if I = f is the current flowing to the ground through the fault point resistance R _f , that is, the fault point current, and if the positive sequence impedance Z = ₁ and the negative sequence impedance Z = ₂ of the entire line length are equal, then the A substation in this case The a-phase, that is, the ground fault-to-earth voltage V〓 _a at this point is obtained from the symmetrical equivalent circuit of Fig. 2 given by the symmetric coordinate method: V〓 _a =V〓 ₁ +V〓 ₂ +V〓 ₀ =I〓 _L nZ 〓 ₁ +I〓
₁ nZ〓 ₁ +
I〓 ₂ nZ〓 ₂ +I〓 ₀ nZ〓 ₀ +V〓 _1f +V〓 _2f +V〓 _0f where V〓 _1f , V〓 _2f , and V〓 _0f are symmetrical voltage divisions of the line fault voltage at the fault point F. However, since V〓 _1f +V〓 _2f +V〓 _0f is the a-phase voltage at the fault point F, this is R _f I _f in Fig. 1.
be equivalent to. Therefore, V〓 _a = (I〓 _L +I〓 ₁ +I〓 ₂ )nZ〓 ₁ +I〓 ₀ nZ
〓 ₀ +I〓 _f
R _f and I〓 _1f , I〓 _2f , and I〓 _0f are the symmetrical components of the current flowing out from the fault point, and I〓 _f = I〓 _1f + I〓 _2f + I〓 _0f , and I〓 _1
f = I〓 _2f
=I〓 _0f , so I〓 _f =3I〓 _0f . here,
I〓
If _a is the ground fault phase current of substation A, then I〓 _a =I〓 ₀ +
_{I 1 + I 2 + I _ _ _ _ _ _ 0+}
R _f I〓 _f (1). Therefore, by transforming equation (1), V〓 _a −(I〓 _a −I〓 ₀ ) nZ〓 ₁ −I〓 ₀ nZ〓 ₀
= I〓 _f R _f , and the fault point current I〓 _f is the reference vector (phase angle 0
), the right side of the equation is a real number, so the imaginary part (I _n ) on the left side is zero, that is, I _n [V〓 _a −(I〓 _a −I〓 ₀ )zZ〓 ₁ −I〓 ₀
nZ〓
₀ ]=0 Furthermore, when the fault point resistance R _f =0, the right side of the equation becomes 0, so the above equation holds true. From now on, the imaginary part I _n [V〓 _a ] of the a relative ground voltage V〓 _a at substation A is I _n [V〓 _a ]=I _n [(I〓 _a −I〓 ₀ )nZ〓 ₁ +I
〓 ₀ nZ〓
₀ ], and the ratio n of the distance from substation A to fault point F to the total line length, that is, the ratio of the distance from substation A to fault point F to the total line length n=I _n [V _a ]/I _n [(I _a −I ₀ )Z ₁ +I
₀ Z ₀ ](2). Therefore, detect the fault point current I〓 _f and the a-relative ground voltage V〓 _a1 at the A substation, the a-phase current I〓 _a , and use the fault point current I〓 _f as a reference vector to find the imaginary component of V _a and the known From the positive sequence impedance Z〓 ₁ of the entire line length and the zero sequence impedance Z〓 ₀ , (I〓 _a
−I〓 ₀ )Z〓 ₁ +I〓 ₀ Z〓 Take out the imaginary part of ₀ ,
(2)
By calculating n using the formula, it is possible to determine the distance from substation A to failure point F. However, in reality, the fault point current I〓 _f
It is impossible to detect this, and it is impossible to determine the imaginary component of the voltage relative to ground a by using _If as a reference vector. Therefore, in the present invention, V〓 _a , I〓 _a , I〓 ₀ and the known Z〓 which can be detected at substation A
₁ , Z = ₀ , find the value n corresponding to the distance from substation A to fault point F using the following method. First, using the zero _- sequence current I〓 ₀ having a phase close to the phase of the fault point current I〓 _f as a temporary reference vector, 〓
Difference between ₀ and a phase current and zero phase current (I〓 _a −I〓 ₀ )
From the positive-sequence impedance Z〓 ₁ and the zero-sequence impedance Z〓 ₀ , the imaginary part of V〓 _a shown in equation (2) and (I〓 _a −
I〓
₀ )Z〓 ₁ +I〓 ₀ Z〓 Find the imaginary component of ₀ , and find a temporary ratio n ₀ from that ratio. Therefore, assuming that the fault point occurs at the line length determined from n ₀ , the zero-sequence impedance of the circuit to the left of the hypothetical fault point is Z from the zero-sequence equivalent circuit given in Figure 3. 〓 _0A +n ₀ Z〓 ₀ , and that of the circuit on the right is Z〓 _0B
+(1-n ₀ )Z= ₀ . Therefore, the zero-sequence current I〓 ₀ ' detected from substation A is I〓' ₀ =Z _0B + (1-n ₀ )Z ₀ /Z _0A +
Z _0B +Z ₀ I〓' _0f where I〓' _0f is the zero-sequence component of the fault point current I' _f at the hypothetical fault point n ₀ . However, from Fig. 2, the symmetrical components of the fault current I〓 _1f , I〓 _2f , and I〓 _0f are equal, so I〓
′ _f =
3I〓' _0fFrom now on, the fault point current I〓' _f is I〓' _f =3Z _0A +Z _0B +Z ₀ /Z _0B + (1
−n ₀ )Z ₀ I〓′ ₀ (3). Therefore, now I〓′ ₀ , using the zero-sequence current I〓 ₀ detected at substation A, which does not have a large difference from this, and the known Z〓 _0A , Z〓 _0B , Z〓 ₀ and n ₀ , we can calculate I〓 ′ _f is obtained, a fault point current I〓 _f ′ having a value close to the actual fault point current I〓 _f is obtained, and its phase angle θ′ _f is obtained from this. This phase angle θ′ _f is subtracted from the respective phase angles of the a-relative earth voltage V〓 _a , the zero-sequence current I〓 ₀ , and the a-phase current I〓 _a detected at converter station A, and these are used as new detected values ( 2) Find n by putting it into the formula, and let the ratio be n ₁ , then n ₁ becomes n ₀
This is closer to the actual failure point. Therefore, by further adding n ₁ to equation (3), we obtain the known Z〓 _0A , Z〓 _0B , Z〓 ₀ and the zero-sequence current I〓 whose phase has been corrected by the previous θ′ _f .
_{0 ,} the phase angle θ″ _f of the fault point _{current is} determined again.
〓
₀ , I〓 Further subtract from the phase angle of _a , and again from equation (2), n
Find n ₂ , and then repeat this series of calculations until the difference between the current calculated value n _i+1 and the previous calculated value n _i becomes zero, or there is no practical problem. If the calculation is repeated until the ratio is sufficiently small, the calculated temporary ratio value will be corrected one by one, and finally the true orientation value n indicating the distance from substation A to fault point F will be obtained.
Or a value very close to this. Next, Table 1 shows an example of calculating the orientation value using the simulated track. In this example, the line constant is zero-sequence impedance Z ₀ in Figure 1.
= (30 + j72) Ω, positive sequence impedance Z ₁ = Z ₂ =
(12+j24) Ω, neutral point grounding zero-sequence impedance Z
_0A =Z _0B =300Ω, fault point resistance R _f =5Ω, load B

[Table] 20MW from substation to substation A (power factor 0.9)
This is the case where a one-wire ground fault occurs at the exit of substation B. In this example, the fault is at the exit of substation B, so the orientation value is 1.0, but the error at the first n ₀ is about 7%.
As n ₁ , n ₂ ... is corrected, it approaches 1.0,
At n ₆ , the error from the true orientation value of 1.0 is only 0.006%, a value extremely close to 1.0. As can be seen from the above explanation, since there is no term for the fault point resistance R _f in the calculation formula (2) for the orientation value n, the orientation is not affected by the fault point resistance that changes depending on the fault condition. Note that in the case of a one-wire ground fault in a high-resistance grounding system, I〓 ₀ =I〓 ₁ =I〓 ₂ , and the phase angle of I〓 ₀ is almost constant. On the other hand, the a-phase current I〓 _a is I〓 _a = I〓 _L +I〓 ₀ +I〓 ₁ +I〓 ₂ , and the load current I〓
_L is larger than the zero-sequence current I〓 ₀ , and I _n in equation (2)
The term [(I〓 _a −I〓 ₀ )Z〓 ₁ ] is affected by the power factor of the load. However, since the power factor during normal operation is approximately constant, it can be said that it has almost no effect on orientation. Although the above description has been made regarding a high-resistance grounding system, the present invention can also be applied to a direct grounding system. As is clear from the above explanation, according to the present invention, unlike the method of locating the fault point by emitting pulses or pulses generated at the fault point, the method of locating the fault point by receiving a surge induced by lightning or attenuation of the pulse is performed. It also has the advantage of being able to locate the fault point without being affected by fault resistance, unlike conventional methods that use system voltage and current, and is highly effective when used to locate the fault point in the event of a single-line ground fault. is obtained.

[Brief explanation of the drawing]

FIG. 1 is a power system example diagram for explaining the present invention, and FIGS. 2 and 3 are a symmetrical equivalent circuit diagram and a zero-phase equivalent circuit diagram at the time of a one-wire ground fault.

Claims (1)

[Claims] 1 When a one-wire ground fault occurs on a power transmission line, the ground fault phase voltage V〓 _a , the zero-sequence current I〓 ₀ , and the difference between the ground fault phase current I〓 _a and the zero-sequence current I〓 ₀ at one end substation of the line, I〓 _a −I〓 ₀
, the phase angle θ _v of the above V〓 _a with respect to the zero-sequence current I〓 ₀ , the phase angle θ ₀ of I〓 ₀ , and the phase angle θ _I of I〓 _a −I〓 ₀ are detected and memorized, From these, the positive sequence impedance Z〓 ₁ of the entire length of the line to be located, and the zero sequence impedance Z〓 ₀ , the imaginary part of the above V〓 _a can be calculated as Z〓 ₁ (I〓 _a −
I〓
₀ ) + Z〓 _{0 I〓} Tentative ratio of the distance from one end substation to the fault point to the total length of the line, divided by the imaginary part of 0 n ₀
After detecting the phase angle θ′ f of the temporary fault point current I〓 _f ′ from this n ₀ and the above zero-sequence current I〓 ₀ using the shunt ratio of the zero-phase circuit, the phase angle θ′ _f of the temporary fault point current _I〓 Phase angle θ _v , I〓 _a
−
I〓 _phase angle θ _I and I〓 ₀ phase angle θ ₀ to I〓
Using the phase angle obtained by subtracting the phase angle θ _f ′ of _f ′, the above
Using the same procedure as for finding n ₀ , find the provisional ratio n ₁ of the distance from one end substation to the fault point to the total length of the line, and
Hereinafter, in the same manner as above, we will calculate the temporary n using the phase angle of the temporary fault point current and the corrected value of the phase angle of each current based on the phase angle. A fault point locating method for a power transmission line, characterized in that the fault location is performed by determining the true ratio n of the distance from one end substation to the fault point to the total length of the line by locating the fault point until the difference becomes sufficiently small.