JPH041869B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a distribution line disconnection detection device that detects a single wire disconnection in a three-phase distribution line, and an object of the present invention is to prevent malfunctions and to reliably detect a single wire disconnection. .

Generally, when detecting a disconnection of one line in a three-phase distribution line, for example, as shown in Fig.
A single wire disconnection detection device for a three-phase circuit described in the specification of No. 82774 detects a single wire disconnection in a distribution line.

In the figure, 1 is a power distribution mother ship, 2 is a power distribution line,
3 is a current transformer, 4 is an auxiliary current transformer, 5 is a fundamental wave filter, 6 is a filter that detects the positive phase component I ₁ , 7 is a filter that detects the negative sequence current I ₂ , 8 is a positive phase change component △ I ₁
9 is a detection circuit that detects the negative phase change ΔI ₂ , 10 is a determination circuit that determines whether each of the changes ΔI ₁ and ΔI ₂ is above a predetermined level; 11 is a setting unit that sets the level; 12 is an arithmetic circuit that calculates △I ₂ /△I ₁ ; 13 is a comparison circuit that compares the calculation output of the arithmetic circuit 12 with a preset operating range; and 14 is its 15 is an output relay, and 16 is a contact point thereof.

Phase currents Ia, Ib, and Ic flowing through the distribution line 2 are each detected by a current transformer 3, which is converted into a voltage by an auxiliary current transformer 4. Then, only the fundamental wave is extracted by the fundamental wave filter 5 and applied to each filter 6 and 7. Here, the positive phase component I ₁ and the negative phase component I ₂ are detected and provided to detection circuits 8 and 9. The detection circuits 8 and 9 detect the difference between the normal phase current and negative phase current and the positive phase component I ₁ and negative phase component I ₂ given from the filters 6 and 7, and detect the difference, that is, the change. The components ΔI ₁ and ΔI ₂ are outputted and given to the determination circuit 10. Here, the level of each variation is determined based on the level set by the settling section 11, and the variation greater than or equal to the settling level is output. This excludes changes based on load fluctuations and the like. Arithmetic circuit 1
2 performs an operation of ΔI ₂ /ΔI ₁ on the variation sent from the determination circuit 10 and provides the result to the comparison circuit 13 . Here, the setting unit 1 compares whether or not the calculation result is included in a certain operating range, and if it is included, outputs an output and operates the output relay 15. This closes the contact 16 and issues a warning or notification that a one-wire disconnection has occurred.

By the way, to explain the operating principle of the one-wire disconnection detection device for three-phase circuits, if the a phase is the reference phase, when the a phase is disconnected, △I ₂ / △I ₁ becomes 1,
When the b and c phases are disconnected, the vector operator is α
If (α=−1/2+j√3/2), then α and α ² will be obtained as shown by the thin lines in FIG. ΔI ₂ /ΔI ₁ when switching a single-phase load and when switching a two-phase load are as shown by the dashed line and thick line in the figure, respectively. Furthermore, since △I ₂ /△I ₁ is 0 when the three-phase load is switched, if △I ₂ /△I ₁ is constantly detected, it is possible to determine whether a one-wire disconnection has occurred or whether a load has been switched. I can do it.

However, for example, when △I ₂ /△I ₁ is 1,
According to the above, an a-phase disconnection has occurred, and according to the symmetric coordinate method, the changes △I ₁ and △I ₂ are respectively △I ₁ = 1/3 (△Ia + α△Ib + α ² △Ic) △I ₂ = 1/3 (△Ia+α ² △Ib + α△Ic) Therefore, if △Ib = △Ic, △I ₁ = △I ₂ and △I ₂ /△I ₁ = 1. but,
△Ib = △Ic may occur due to load fluctuations other than one-wire breakage, and even if one-wire breakage has not actually occurred, the same judgment as one-wire breakage is made, resulting in malfunction.

This invention was made to solve the above-mentioned problems, and in addition to detection based on the △I ₂ / △I ₁ principle, it focuses on a system phenomenon that can distinguish between disconnections and load fluctuations. By combining this with detection of phenomena, malfunctions due to load fluctuations can be prevented, and single wire breakage can be reliably detected.

To briefly describe the above system phenomenon, when a disconnection occurs in a distribution line, the ground capacity of the distribution line changes.
This means that the ground capacity does not change due to changes in the load on the distribution line. Changes in the ground capacity can be detected as changes in the zero-sequence voltage of the distribution line busbar, as will be described later.

The present invention provides an arithmetic circuit that calculates the ratio between a change in positive sequence current and a change in negative sequence current, and a zero-sequence voltage controller that detects a change in the ground capacity of a distribution line due to one wire breakage as a change in zero-sequence voltage. It is equipped with a change detection circuit and a detection circuit that outputs an output when the ratio of the two current changes and the change in the zero-sequence voltage are within preset operating ranges to notify one wire disconnection. The present invention provides a distribution line disconnection detection device characterized by the following.

Next, we will prove that the zero-sequence voltage of the bus changes when a single line disconnection occurs in the distribution line.

In FIG. 3, the same signals as those in FIG. 1 indicate the same signals. 18 indicates a grounding transformer connected to the distribution bus 1, and 19 indicates an impedance with the grounding transformer 18.

27 is the three-phase AC voltage generator to which the distribution bus 1 is connected, 28 is the ground capacity of the distribution line 2 of the first circuit, 29 is the ground capacity of the distribution line 2 of the second circuit, and 30 is the distribution line 2 of the Nth circuit. 31 to 32 are the loads of the first, second, and Nth lines.

Now, in Fig. 3, let the ground capacity of each phase in bus 1 be V〓a, V〓b, V〓c, and the size of the ground capacity 28 to 30 of each line is C ₁ (-C ₁ a + C ₁ b + C ₁ c),
C ₂ (=C ₂ a + C ₂ b + C ₂ c), Cn (=Cna + Cnb + Cnc)
The magnitude of the unbalanced portion of the ground capacity 28 to 30 of each line is C′ ₁ (=C ₁ a+α ² C ₁ b+αc ₁ c), C ₂ ′(=C ₂ a
+α ² C ₂ bα ² C ₂ c), C′n (=Cna+α ² Cnb+αCnc), the zero-sequence voltage V〓o in normal condition is V〓o=Va+Vb+Vc/3=-Y ₂ /Yo+1/Zn・V ₁ ... This is the residual zero-sequence voltage in a healthy state, and the unbalanced components of the ground capacity 28 to 30 are C ₁ ′, C ₂ ′,
This is the zero-sequence voltage generated by C ₃ '. here,
Y〓o is the admittance of the ground capacity of the entire system, and is expressed as Y〓o=Y〓 ₁₀ +Y〓 ₂₀ +...+Y〓no... In the formula, Y〓 ₁₀ = jωC ₁ , Y〓 ₂₀ =
ωC ₂ , ..., Y〓n ₀ =jωCn is the admittance of the ground capacity of each line. Y〓 ₂ is the admittance of the total unbalanced ground capacity of the system, and is expressed as Y〓 ₂ = Y〓 ₁₂ + Y〓 ₂₂ +...+Y〓n ₂ .... In the formula, Y〓 ₁₂ =jωC′ ₁ ,
Y〓 ₂₂ jωC′ ₂ Y〓n ₂ =jωC′n is the admittance of the unbalanced ground capacity of each line. Z〓n is the primary conversion value of the impedance 19, that is, the value when this secondary side impedance 19 is viewed from the primary side in order to analyze the phenomenon on the primary side of the grounding transformer 18, and Let the size be Z〓,
If the turns ratio of the primary coil and secondary coil of each phase of the grounding transformer 18 is n:1, the turns ratio of the primary coil and secondary coil of the entire grounding transformer 18 is n:3, and Z It is expressed as 〓n=n ² /9・Z〓. V〓 ₁ is a positive sequence voltage and does not change due to changes in ground capacity, etc. For 6.6KV distribution line, its size is 6600/√3.

For example, if one line disconnection occurs in the first line, the ground capacity and the unbalanced admittance of the ground capacity of the first line at this time are Y〓 ₁₀ ′ and Y〓 ₁₂ ′, respectively. , the ground capacity of the entire system and the admittance of the unbalanced ground capacity
Y〓 ₀ ′ and Y〓 ₂ ′ are respectively expressed by the following formulas.

Y〓 ₀ ′＝Y〓 ₁₀ ′Y〓 ₂₀ ＋…＋Y〓n ₀ ＝Y〓 ₁₀ ＋Y〓 ₂₀ ＋…＋Y〓n ₀ ＋(Y〓 ₁₀ ′-Y〓 ₁₀ ) ＝Y〓 ₀ ＋△Y〓 ₀ …… Y〓 ₂ ′＝Y〓 ₁₂ ′＋Y〓 ₂₂ ＋…＋Y〓n ₂ ＝Y〓 ₁₂ ＋Y〓 ₂₂ ＋…Y〓n ₂ +(Y〓n ₁₂ ′＋Y〓 ₁₂ ) =Y〓 ₂ ＋ △Y〓 ₂ …… Find the zero-sequence voltage V〓 ₀ ′ at this time according to the formula: V〓 ₀ ′=−Y ₂ +△Y ₂ /Y ₀ +△Y ₀ +1/Zn・V ₁ …… becomes.

Furthermore, by calculating the change in zero-sequence voltage △V〓 due to one wire disconnection from the formula and formula, we get △V〓 ₀ =V〓 ₀ ′−V〓 ₀ = (−Y〓 ₂ +△Y〓 ₂ /Y ₀ +△Y ₀
+1/Zn−Y〓 ₂ /Y ₀ +1/Zn)・V〓 ₁ =△Y〓 ₂ /Y ₀ +△
Y ₀ +1/Zn・V〓 ₁ +△Y〓 ₀ /Y ₀ +△Y ₀ +1/Zn・Y〓 ₂ /Y ₀ +1
/Zn・V〓 ₁ = −△Y〓 ₂ /Y ₀ +△Y ₀ +1/Zn ・V〓 ₁ −△Y〓 ₀ /Y ₀ +△Y ₀ +1/ZN・V ₀ ... It is expressed as In general, the magnitude of the residual zero-sequence voltage V〓 ₀ in the second term on the right side of the equation is sufficiently small compared to the voltage V〓 ₁ , so ignoring the residual zero-sequence voltage V〓 ₀ , the change △V〓 ₀ is It is expressed as △V〓 ₀ −△Y〓 ₂ /Y ₀ +△Y ₀ +1/Zn・V〓 ₁ ... From the above, it was found that the bus zero-sequence voltage changes due to a change in ground capacity due to one wire disconnection.

Next, the change in zero-sequence voltage when one wire is disconnected △V〓 ₀
In order to find the specific value of , consider the case where one line disconnection occurs in the first line of the system shown in Figure 3, and calculate the ground capacity 28 of the first line by marking it with an x in the figure as shown in Figure 4. The capacitance is divided at the A-phase disconnection point, and the size of each capacitance is C _1a1 ,
Let C _1b1 , C _1c1 , C _1a2 , C _1b2 , and C _1c2 . The change in ground capacity admittance of the line due to disconnection △Y〓 ₀ can be calculated using the formula △Y〓 ₀ = Y〓 ₁₀ ′−Y〓 ₁₀ = jw (C _1a1 +C _1b1 +C _1c1 +C _1b2 +C _1c2 − jw (C _1a1 + C _1b1 + C _1c1 + C _1a2 + C _1b2 + C _1c2 ) = -jwC _1a2 ... Similarly, the change in admittance due to the unbalanced ground capacity in the first line due to disconnection △Y〓 ₂ is , using the formula △Y〓 ₂ ′=Y〓 ₁₂ ′−Y〓 ₁₂ =jω(C _1a1 +α ² C _1b1 +αC _1c1 +α ² C _1b2 +αC _1c2 ) −jw(C _1a1 +α ² C _1b1 +αC _1c1 +C _1a2 + α ² C _1b2 + αC _1c2 ) = −jwC _1a2 ... From the formula, △Y〓 ₀ = △Y〓 ₂ .

Here, the ground capacity of each distribution line 2 is calculated as
Let's consider a case where the total length of the three phases is 500 km, and a break occurs in the a phase at a point 1 km from the end, with a setting of 0.01 μF/Km. When the power supply frequency is 60Hz, the admittance of the total ground capacitance Y〓 ₀ is Y〓 ₀ = jω (500 × 0.01) = j1885 [μ] ......, and the change in the admittance of the total ground capacitance △
Y〓 ₀ , △Y〓 ₂ is from the formula, △Y〓 ₀ = △Y〓 ₂ = −jω (1 × 0.01) = −j3.77 [μ]
...... Furthermore, when the impedance 19 consists of only a current limiting resistor, for example, the current limiting resistor is R=
100Ω, and the turns ratio of the grounding transformer 18 is n=
If it is 6600/√3/110/3, the impedance is 1 of 19.
The next conversion value Rn (=Z〓n) is Rn=1/9・(6600/√3/110/3) ² ×100=120[K
Ω], that is, 1/Rn=8.33[μ]... From the formula, the change in electric phase voltage △V〓 ₀ is △V〓 ₀ = −−j3.77／j1885−j3.77＋8.33・V〓 ₁ 3.77
/1885・V〓 ₁ , and due to disconnection, the zero-sequence voltage is about 0.2% of the voltage V〓 ₁ ,
In other words, it will change by about 7.6V.

As explained above, it has been understood that a change in ground capacitance due to a disconnection of one wire can be detected as a change in zero-sequence voltage.

Here, it will be explained that the zero-sequence voltage V〓 ₀ in a healthy state is expressed as in the above equation.

Each phase current on the secondary side of the three-phase AC voltage generator 27 is I〓 _a ,
I〓 _b , I〓 _c , the current flowing in the primary coil of each phase of the grounding transformer 18 is 1/3 I〓 _o , each phase current flowing in the distribution line 2 of the first, second, ... Nth circuit each
I〓 _1a , I〓 _1b , I〓 _1c , I〓 _2a , I〓 _2b , I〓 _2c ,…, I〓 _{o
If a} , I〓 _ob , I〓 _oc , then in the power distribution system shown in Figure 3, I〓 _a I〓 _b I〓 _c + 1/3 I〓 _o 1/3 I〓 _o 1/3 I〓 _o = I〓 _1a I〓 _1b I〓 _1c +I〓 _2a I〓 _2b I〓 _2c +…+I〓 _oa I〓 _ob I〓 _oc ……〓〓 holds, I〓 _a +I〓 _b +I〓 _c +I〓 _o = (I〓 _1a +I〓 _1b +I〓 _1c
) + (I〓 _2a +I〓 _2b +I〓 _2c ) +… + (I〓 _oa +I〓 _ob +I
〓 _oc )...〓〓 becomes.

At this time, the zero-sequence current on the secondary side of the three-phase AC transformer 27 is I〓 ₀ , and the distribution line 2 of the first, second, ... Nth circuit is
The zero-sequence currents at I〓 ₁₀ , I〓 ₂₀ , ..., I〓 _o0
Then, the formula becomes as follows.

3I〓 ₀ +I〓 _o =3I〓 ₁₀ +3I〓 ₂₀ +...3I〓 _o0 ...〓〓 On the other hand, in the grounding transformer 18, the ground voltage of each phase on the primary side V〓 _a , V〓 _b , V〓 The following formula is established using _c , −n/3I〓 _o・Z〓=1/n(V〓 _a +V〓 _b +V〓 _c )=3
/nV〓 ₀ ...〓〓 Here, since Z〓 _o = (n ² /9)・Z〓, I〓 _o = -1/Z _o V〓 ₀ ...〓〓.

In addition, in the distribution line 2 of each line, i=1,
2,..., _o , then 3I〓 _i0 =I〓 _ia +I〓 _ib +I〓 _ic = (jωC _ia , jaC _ib , jaC _i
c )×V〓 _a V〓 _b V〓 _c = (jω(C _ia , C _ib , C _ic ), jω(C _ia , α ² C _ib , αC _i
c ), jω (C _ia , α ² C _ib , α ₂ C _ic )) V〓 ₀ V〓 ₁ V〓 ₂ = (jωC _i , jωC _i ′, jωC _i ″)×V〓 ₀ V〓 ₁ V〓 ₂ =Y〓 _i0 V〓 ₀ +Y〓 _i2 V〓 ₁ +Y〓 _i1 V〓 ₂ ...〓〓.However, Y〓 _i0 =jωC _i ,Y〓 _i2 =jωC _i ′,Y〓 _i
1 =
jωC _i ″, and V〓 ₁ ′ is the positive sequence voltage, and V〓 ₂ is the negative sequence voltage.

Furthermore, since zero-sequence current is not supplied from the △ circuit on the secondary side of the three-phase AC voltage transformer 27, the following conditions are obtained: I〓 _a + I〓 _b + I〓 _c = 3I〓 _p = 0 ...〓〓 .

Furthermore, if the phase-to-phase voltage on the secondary side of the three-phase AC transformer 27 is E〓 _ab , E〓 _bc E〓 _ca , then V〓 _a −V〓 _b V〓 _b −V〓 _c V〓 _c −V〓 _a = 1 -1 0 0 1 -1 -1 0 1V〓 _a V〓 _b V〓 _c ＝E〓 _ab E〓 _bc E〓 _ca ...〓〓, where E〓 _ab ,E〓 _bc ,E〓 Assuming that _ca is balanced, from 〓〓, 0 0 00 1−α ² 00 0 1−αV〓 ₀ V〓 ₁ V〓 ₂ =0 E〓 _ab 0 ∴V〓 ₁ =E〓 _ab /1 −α ² ...〓〓 V〓 ₂ =0 ...〓〓 is obtained.

Then, using the expressions 〓〓, 〓〓, 〓〓, and 〓〓
Rewriting the formula, -1/Z _o V〓0=(Y〓 ₁₀ V〓 ₀ +Y〓 ₁₂ V〓 ₁ ) + (Y〓 ₂₀ V〓 ₀ +Y〓 ₂₂ V〓 ₁ ) +...+(Y〓 _o0 V〓 ₀ +Y〓 _o2 V〓 ₁ ), so -(1/Z _o V ₀ +Y ₁₀ +Y ₂₀ +...+V〓 _o0 )V〓 ₀ = (V〓 ₁₂ +Y〓 ₂₂ +...+Y〓 _o2 )V〓 ₁ − (1/Z _o +Y ₀ )V〓 ₀ = Y〓 ₂ V〓 ₁ V〓 ₀ = −Y〓 ₂ /Y ₀ +1/Z _o・V〓 ₁ , and the formula is derived.

Next, the impedance 19 of the grounding transformer 18
The reason why the linear conversion value Z〓 _o for can be expressed as Z〓 _o = (n ² /9)・Z〓 will be explained.

According to the theory of symmetrical coordinates, the zero-sequence current is a single-phase alternating current, and in the grounding transformer 18, as shown in Figure 3, the zero-sequence current is shunted to three-phase coils on the primary side. , a current flows in the secondary side to cancel the current on the primary side, and the primary side coil can be considered as 3-phase parallel, and the secondary side coil as 3-phase series.

Now, as shown in FIG. 6, the turns ratio of the primary coil and secondary coil in each phase of the grounding transformer 18 is set to n:1, and the magnitude of the impedance 19 is Z〓,
The primary conversion value is Z〓 _o , the current flowing through each coil on the primary side is i _a , i _b , i _c , the current flowing through each coil on the secondary side is i ₀ , and impedance 1
If the voltage across the 9 is v〓 ₀ and the primary voltage is V〓 ₀ , then i _a = i _b = i _c = 1/ni ₀ v〓 ₀ = i ₀・Z〓 V〓 ₀ = n/3v〓 ₀ Since I〓 ₀ = i _a = i _b = i _c = 3/ni ₀ , Z〓 _o = V〓 ₀ /I〓 ₀ = n/3v〓 ₀ /3/ni ₀ = n/3i ₀・
z〓/3/ni ₀ =n ² /9・z〓.

FIG. 5, which shows one embodiment of the present invention, will be described below.

In Figure 5, the same symbols as in Figures 1 and 3 indicate the same or equivalent items.
0 is an auxiliary transformer, 10 is a fundamental wave filter, 22
is a zero-phase voltage change detection circuit that detects a change in zero-phase voltage, and the detected change is input to the comparison circuit 23. Here, a comparison is made to see if it is within a predetermined operating range set by the setting unit 24. 2
5 is an AND gate, and this output is relay 15
to drive. 16 is a contact point of the relay, and the relay 15 and the contact point 16 constitute an output circuit 26. When △I〓 ₂ /△I〓 ₁ and the change in the zero-sequence voltage are both set by the setting sections 14 and 24 and are included in a certain operating range, the comparison circuits 13 and 23
The AND gate 25 from which an output is output is turned on. Then, the relay 15 is driven, the contact 16 is closed, and the occurrence of one wire breakage is reported.

As described in detail above, according to the present invention, malfunctions due to load fluctuations other than one wire disconnection can be prevented by setting the detection of changes in zero-sequence voltage as one of the conditions, and in particular, detection of changes in zero-sequence voltage is Since it is not affected by load current fluctuations, accurate detection is possible, and 1
Wire breakage can be detected reliably.
Therefore, it is possible to provide a highly practical distribution line disconnection detection device.

[Brief explanation of drawings]

Fig. 1 is a block diagram of a conventional one-wire disconnection detection device, Fig. 2 is a vector diagram for explaining the operation of Fig. 1,
Figure 3 and the following drawings show one embodiment of the distribution line disconnection detection device of the present invention, Figure 3 is a wiring diagram of a distribution line system, Figure 4 is an explanatory diagram of the ground capacity of a distribution line, and Figure 5 is an illustration of the ground capacity of a distribution line. The block diagram, FIG. 6, is a wiring diagram of the grounding transformer. 2... Distribution line, 8, 9... Positive phase, negative phase change detection circuit, 12... Arithmetic circuit, 22... Zero phase voltage change detection circuit, 23... Comparison circuit, 25... AND gate, 26...Output circuit.

Claims (1)

[Claims] 1. A change detection circuit that detects each change in the positive sequence current and negative sequence current of a three-phase distribution line, and a change in the negative sequence current that corresponds to the change in the positive sequence current detected by the change detection circuit. an arithmetic circuit that calculates a ratio; a zero-sequence voltage change detection circuit that detects a change in ground capacity of the distribution line due to one wire disconnection as a change in zero-sequence voltage; A distribution line disconnection detection device comprising: an output circuit that outputs an output when both voltage changes are within a preset operating range.