JPH0136594B2 - - Google Patents

Description

[Detailed description of the invention]

The present invention relates to a method for detecting disconnection of an electric line that can reliably and easily distinguish between a disconnection of an electric line supplying power to a three-phase load and an open load at a substation. Distribution lines are generally ungrounded at the neutral point, and the system is dendritic. Accidents such as ground faults and overcurrents on distribution lines are detected using ground fault relays, overcurrent relays, etc., but in order to improve the reliability of distribution lines and ensure safety, The use of insulated wires for railway lines is progressing. However, even if the insulated wire breaks and falls to the ground, the insulated wire has a covering, so
Since the internal conductor does not contact the ground, zero-sequence current does not flow, and there are cases where a ground fault cannot be detected at the substation, posing a problem in terms of safety and power supply. In order to solve the above-mentioned problems, the present invention has been developed to solve the above problem. Find the absolute value of the ratio of the vector change in the line current and the vector difference in the vector change in the line current of the B phase and C phase, and then calculate the remaining 2
The absolute values of the two phases, ie, the B phase and the C phase, are determined in the same manner as above. A wire breakage detection method for detecting a wire breakage by utilizing the fact that the sum of the three absolute values obtained in this way or the absolute value of each of them has a different relationship between when one line is broken and when the load is released. was provided. Hereinafter, the disconnection detection method of the present invention will be explained with reference to the drawings. FIG. 1 is a diagram showing a line in which a load L ₁ and a load L ₂ are connected to a substation T ₁ , and the line current I ₀ in the substation is in a loaded state or in a line connected only to that load. The line current I ₁ for a load L ₁ which does not change and the load whose load condition or line condition changes
This is the vector sum of L ₂ and the line current I ₂ , but the line impedance between the substation and each load is ignored as it is smaller than the impedance of each load. Furthermore, each line voltage has a phase difference of 120°, and even if the line condition of load L ₂ changes, the phase difference between each line voltage and each line current at point P shown in Figure 1 does not change. shall be taken as a thing. Therefore, the change ΔI ₀ in the line current at the substation depends only on the change ΔI ₂ due to the change in the line condition of the load L ₂ ,
Since the change in load L ₁ is ΔI ₀ =0, the load L ₁ will not be considered in the following explanation. Therefore, the change in the line state of the load _L2 can be shown by the equivalent circuits shown in FIGS. 2A to 2D. However, Figure 2A is the equivalent circuit when one wire of A phase is disconnected, Figure B is the equivalent circuit when the BC phase single-phase load is opened, and Figure C is the equivalent circuit when the AB/CA phase V load is opened. Figure D is an equivalent circuit when the three-phase load is released. In FIGS. 2A to 2D, A in the steady state
The line currents of phase, B phase and C phase are Ia, Ib and Ic
Let the voltage between each line be Vab, Vbc and Vca, and the current of the load connected between each line Iab, Ibc and Ica,
If their absolute values are iab, ibc, and ica, and the power factor angles of each load are α, β, and τ, then each line current is Ia=Iab−Ica=iabe ^j 〓−icae ^j 〓・a ……(1) Ib=Ibc−Iab=ibce ^j 〓・a ² −iabe ^j 〓 ...(2) Ic=Ica−Ibc=icae ^j 〓・a−ibce ^j 〓・a ² ...(3). However, a=ej2/3π and a ² =aj4/3π. Here, when the A-phase line is disconnected as shown in Figure 2A, each line current after the disconnection is Ia', Ib' and
Ic′ and the load currents Iab′, Ibc′ and Ica′, Ia′=0 Ib′=Ibce ^j 〓・a ² +
1/1/iabe ^j 〓+1/icae ^j 〓・a ² Ic′=−Ib′, and the changes in each line current ΔIa, ΔIb, and ΔIc
is as follows. ΔIa=I′a−Ia=−(iabe ^j 〓−icae ^j 〓・a)……(4
) ΔIb＝ΔI′b−Ib＝iabe ^j 〓／iabe ^j 〓＋icae ^j 〓(iabe ^j
〓−icae ^j 〓． a)=−iabe ^j 〓/iabe ^j 〓+icae ^j 〓・Δ
Ia……(5) ΔIc＝I′c−Ic＝iabe ^j 〓／iabe ^j 〓＋icae ^j 〓(iabe ^j
〓−icae ^j 〓・a)=−icae ^j 〓/iabe ^j 〓+icae ^j 〓・Δ
Ia...(6) From equations (5) and (6), ΔIb/ΔIc=iab/icaej(α−τ)...(7) From equations (4) and (7), ΔIa=(iabe ^j 〓−icae ^j 〓・a) =−ica(iab/icae ^j 〓・a) =−ica(ΔIb/ΔIce ^j 〓−e ^j 〓・a) =−ica/ΔIc(ΔIbe ^j 〓−ΔIce ^j 〓・a) ……(8) Transforming equation (8), ΔIc/ica=−(ΔIb−ΔIc・a)/ΔIa・e ^j 〓……(
9), and taking the absolute value of the right-hand side, the following formula becomes the formula for determining the A-phase disconnection index STa. STa=|ΔIb−ΔIc·a/ΔIa| (10) Similarly, the formulas for determining the disconnection index at the time of B-phase and C-phase disconnection are as follows. STb=|ΔIc・a ² −ΔIa/ΔIb| =|ΔIc−ΔIa・a/ΔIb| ...(11) STc=|ΔIa・a−ΔIb・a ² /ΔIc| =|ΔIa−ΔIb・a/ΔIc ｜ ...(12) Next, using equations (10) to (12) above, the values of the breakage indices STa, STb, and STc at the time of one wire breakage and when switching various loads are shown in Figures 3 to 7. Find it while referring to. Figures A to D show the case when one wire of the A phase is disconnected and the single phase load of the BC phase is opened under a three-phase balanced load, respectively.
A vector diagram of the line current in the substation when the AB/CA phase V load is opened and the 3-phase load is opened,
Line currents Ia, Ib and Ic of A phase, B phase and C phase
The changes in these vector values ΔIa, ΔIb when the vector values change and become Ia′, Ib′, and Ic′, respectively.
and ΔIc are shown. Figure 4A, Figure 5A,
6A and 7A are vector diagrams showing vector value changes ΔIa, ΔIb, and ΔIc in FIGS. 3A to 3D, respectively. Note that the vectors indicating the changes in the line current of each phase ΔIa, ΔIb, and ΔIc are the ratio of their absolute values, that is, |ΔIa|:|
For convenience, ΔIb|:|ΔIc| is set to 2:1:1 for Figure 4A, and 0:1:1 and √3: for Figures 5A, 6A, and 7A, respectively.
This is when the ratio is 1:1 and 1:1:1. (1) A phase 1 wire disconnection (i) In equation (10), multiplying ΔIc by a is
Since the purpose is to rotate ΔIc counterclockwise by 120 degrees, the vectors ΔIa and ΔIa shown in Figure 4A are
The relationship between ΔIb and ΔIc is as shown in FIG. 4B.
As is clear from this figure, |ΔIb− of equation (10)
Since ΔIc·a| is √3 and |ΔIa| is 2, STa is √3/2. (ii) Calculate STb at this time as follows. If ΔIa in equation (11) is rotated by 120° as in (i), the relationship between vectors ΔIa, ΔIb, and ΔIc shown in Figure 4A becomes as shown in Figure 4C, |ΔIc−ΔIa・a | is √3 and |ΔIb| is 1, so STb is √
It becomes 3. (ii) Furthermore, calculate STc at this time as follows. If ΔIb in equation (12) is rotated by 120° as in (i), we get
The relationship between ΔIa, ΔIb and ΔIc in Figure 4A is
As shown in Figure D, |ΔIa−ΔIb·a| is √3, and |ΔIc| is 1, so STc is √3. (2) BC phase single-phase load release If you do the same thing as described above, Fig. 5A will become as shown in Fig. 5B, Fig. 5C, and Fig. 5D, respectively, and since ΔIa is 0, STa becomes 1/0=∞. Further, both STb and STc are 1/1=1. (3) AB/CA phase V load release If you do the same thing as described above, Figure 6A will become like Figure 6B, Figure 6C, and Figure 6D, respectively, and STa will be 1/√3 , STb and STc are both 1/1=1. (4) 3-phase load release If you do the same thing as described above, Fig. 7A will become as shown in Fig. 7B, Fig. 7C, and Fig. 7D, respectively, |ΔIb−ΔIc・a|, | Since ΔIc−ΔIa・a| and |ΔIa−ΔIb・a| are both 0, STa=
STb=STc=0. Here, Table 1 shows the values of STa, STb, and STc when one wire of the A phase is disconnected and when various loads are released.

[Table] This table shows the calculated values of STa, STb, and STc when one wire of A phase is disconnected and when the load is opened under a three-phase balanced load.These values differ between these two states. It shows that it is on.
Therefore, the value for determining a disconnection is, for example, 1.5.
and set STa, STb and STc as the adiabatic conditions.
If any two of these are greater than 1.5 (however, the value of ∞ is not taken), it is considered a wire breakage, then as can be seen from Table 1, a single wire breakage in a three-phase balanced load is detected. be able to. Additionally, STa, STb and
For example, the sum of the three values of STc is 4.0 (however,
Assume that the value of ∞ is not taken. ) may be used as a value to determine that the wire is disconnected. However, in the case of an unbalanced load, the values shown in Table 1 will change depending on the degree of unbalance.
In particular, the situation may overlap when one wire of the A phase is disconnected and when the AB/CA phase V load is released. Therefore, in order to distinguish between a single wire break and a V load release, it is necessary to set a certain range of unbalance, and under that condition, to distinguish between a single wire break in the A phase and a V load release in the AB and CA phases. Time STa,
By calculating STb and STc, the value at which it is determined that the wire is disconnected must be determined. The calculation of this value will be explained below. (a) A phase 1 wire disconnection Set Y = ica/iab (>0), substitute this into equations (4) to (6), and then convert these equations into equations (10) to (12). The equations obtained by substituting are as follows. STb=√1+ ² −2(−−120°
……(14) STc=1/Y√1+ ² −2(−−120°) ……(15) Also, (b) AB/CA phase V load released As shown in Figure 2C, each line current Ia', Ib' and Ic' after V load is released is Ia'=0 Ib'=ibce ^j 〓・a ² Ic'= −ibce ^j 〓・a ² , and the changes in each current ΔIa, ΔIb and ΔIc
is as follows. ΔIa=−iabe ^j 〓+icae ^j 〓・a ...(17) ΔIb=iabe ^j 〓 ...(18) ΔIc=−icae ^j 〓・a ...(19) Here, Y=ica/iab(>0 ), then substitute these into equations (17) to (19), and then substitute these equations into equations (10) to (12) to obtain the following equations. For STa, STb, and STc obtained in this way, |α−τ
｜When Y and A phase 1 wire is disconnected in the range of ≦30° and V
Draw a graph of the relationship between STa, STb, and STc when the load is released (not shown). In this graph, a value A for determining a wire breakage is set in advance from the values of STa, STb, and STc when Y=1, and the following three wire breakage conditions are considered based on the value A. P: When any two of STa, STb, and STc are greater than A, the wire is broken. Q: When at least one of STa, STb, and STc is greater than A, the wire is broken. However, ∞
Assume that the value of is not taken. R: When ST is greater than A, the wire is broken. However, the value of ∞ shall not be taken. Therefore, in order to find the value of A,
By variously changing |α−τ|≦30°, 0.1≦Y
In order to satisfy each of the disconnection conditions P, Q, and R in the range of ≦2, the range in which each disconnection can be detected and the range in which V load release can be detected are tabulated from the above-mentioned graph (not shown) (not shown). For example, if a disconnection or load release can be detected, mark it with a circle, and if it cannot be detected, mark it with an x. From this table, it is possible to know, for each of the disconnection conditions P, Q, and R, A that makes the disconnection detection range as wide as possible and minimizes the malfunction when the V load is released. Next, based on each A, the range in which disconnection can be detected and the range in which V load release can be detected are determined from the table (not shown) mentioned above.
By creating a table with Y on the horizontal axis and (α-τ) on the vertical axis, it is possible to confirm the detection of A phase 1 wire disconnection in a certain Y and (α-τ) range. can. In this way, for the disconnection condition P, A=
1.15, and the conditions for Y and (α−τ) at this time are
0.4≦Y≦2.0, −30°≦α−τ≦+10°. Also, regarding the disconnection condition Q, A=1.63, and the condition of Y and (α-τ) at this time is 0.1≦Y≦
2.0, −30°≦α−τ≦+10°. Furthermore, regarding the disconnection condition R, A=3.56, and the conditions for Y and (α-τ) are 0.5≦Y≦2.0, -10°≦α
−τ≦+30°. Therefore, if any one of the wire breakage conditions P, Q, and R is satisfied, it is possible to distinguish between a single wire break and a load release. This is clear from the fact that a single wire break in the A phase can be detected in a three-phase balanced load from the values shown in Table 1. Although the above explanation has been about detecting a disconnection in the A phase, it is clear that the same method as described above can also be used to detect a disconnection in the B phase or C phase. be. As described above, according to the wire breakage detection method of the present invention, by setting values in advance according to each breakage condition, it is possible to detect not only a three-phase balanced load but also an unbalanced load under any of the conditions. Even if it is,
It is advantageous that load changes such as three-phase load opening, V load opening, or single-phase load opening, and one-wire breakage can be detected at a substation.

[Brief explanation of drawings]

Figure 1 is a diagram showing the lines where load L ₁ and load L ₂ are connected to substation T _1. Figure 2 A to D are equivalent circuits when one wire of A phase is disconnected, and when a single phase load of BC phase is opened. Figures 3A to 3D show the equivalent circuit when the AB/CA phase V load is opened, and the equivalent circuit when the 3-phase load is opened.
When one phase wire is disconnected, when the BC phase single phase load is opened, AB/CA
Vector diagrams of line currents in the substation when phase V load is released and phase 3 load is released, Figures 4A, 5A, 6A, and 7A are respectively Figure 3A
Vector diagrams showing changes in vector values ΔIa, ΔIb, and ΔIc in D to D, FIG. 4 B to D, FIG. 5 B to D, FIG. 6 B to D, and FIG. 7 B to D are respectively shown in FIG. A is a vector diagram after the vectors shown in FIG. 5A, FIG. 6A, and FIG. 7A are rotated counterclockwise by 120 degrees.

Claims (1)

[Claims] 1. Changes in each vector value when the line currents Ia, Ib and Ic of the three-phase electric line change to Ia', Ib' and Ic, respectively ΔIa = Ia' - Ia, ΔIb = Ib' From −Ib and ΔIc=Ic′−Ic, the disconnection index STa, STb, and STc at the time of A-phase, B-phase, and C-phase disconnection STa=|ΔIb−ΔIc・a/ΔIa| STb=|ΔIc−ΔIa・a/ΔIb A method for detecting a break in an electric line by determining |STc=|ΔIa−ΔIb・a/ΔIc| and determining that the wire is broken by detecting that the values of STa, STb, and STc have a predetermined relationship. 2 The predetermined relationship is STa, STb and
The method for detecting disconnection in an electric line according to claim 1, wherein any two of the values of STc are larger than 1.15. 3 The predetermined relationship is STa, STb and
At least one of the STc values is 1.63 (but
Assume that the value of ∞ is not taken. ) The method for detecting disconnection of an electric line according to claim 1, wherein the wire breakage detection method is larger than . 4. Claim 1, wherein the predetermined relationship is when the value of STa + STb + STc is larger than 3.56 (however, the value of ∞ is not taken).
Disconnection detection method of electric line as described in section.