CN114977091A - Grounding fault isolation method for neutral point arc suppression coil and medium resistance grounding power grid - Google Patents
Grounding fault isolation method for neutral point arc suppression coil and medium resistance grounding power grid Download PDFInfo
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- H—ELECTRICITY
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- H02H3/16—Emergency protective circuit arrangements for automatic disconnection directly responsive to an undesired change from normal electric working condition with or without subsequent reconnection ; integrated protection responsive to fault current to earth, frame or mass
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Abstract
A ground fault isolation method for a power grid with a base neutral point grounded through an arc suppression coil and a middle resistor belongs to the technical field of relay protection of power systems. And the ratio of the active component of the earth fault current to the zero sequence voltage is used as a starting signal for earth fault occurrence and line selection judgment. And finally, determining the protection action time of the ground fault according to the ratio of the ground fault quantity of the monitoring point to the ground fault quantity measured at the monitoring point when the actual ground fault occurs, and realizing the quick isolation of the ground fault of the cable distribution network with a neutral point grounded through an arc suppression coil and a middle resistor. The method has simple steps and good use effect.
Description
Technical Field
The invention relates to a ground fault isolation method for a power grid with a neutral point grounded through an arc suppression coil and a medium resistor, which is suitable for a three-core cable power distribution network with a neutral point grounded through an arc suppression coil and a medium resistor or a small resistor and belongs to the technical field of relay protection of power systems.
Background
Due to the pursuit of people for the beautiful urban environment and the requirement of higher power supply reliability, and due to the improvement of the production technology level of the power cable and the reduction of the application cost of the cable, the urban 10kV power distribution network increasingly adopts the power cable laying power supply mode. Due to the symmetry of the three-core cable core, the three-phase copper foil shielding layers are in contact with each other, and no induced voltage exists on the copper foil layers when the power grid operates normally, the two ends of the cable are grounded. Thus, the earth wire (the copper foil shielding layer and the armor layer of the cable) of the cable connects the grounding electrodes of each substation, the switching station, the ring network station and each user to form a grounding network.
The grounding fault current of the power distribution network is greatly increased due to the use of a large amount of cables, so that the neutral point is largely adopted in a grounding operation mode through the arc suppression coil, and practice proves that the grounding fault current brings great positive effects on the safe and reliable operation of the power distribution network. However, when the neutral point is grounded through the arc suppression coil, the power grid greatly reduces the ground fault current, and simultaneously brings some difficulties to the ground fault line selection, increases the difficulty of ground fault isolation, and in order to improve the accuracy of the ground fault line selection, a mode that the arc suppression coil is grounded with a middle resistor (100-132 omega) (or a small resistor (10-15 omega)) appears, and the accuracy of the ground fault line selection (the transverse selectivity of relay protection) is greatly improved. When the power grid has a ground fault, the arc suppression coil timely compensates the capacitance current of the power grid to the ground, so that the current of the ground fault point is effectively reduced, the instantaneous ground fault arc is quickly extinguished, and the power grid recovers to normal operation; if the earth fault cannot automatically disappear in 3-5 seconds, the fault is a permanent earth fault, QF in the graph 1 is closed, and a resistor (or a small resistor) in a neutral point is merged between a neutral point and an earth electrode of the power grid for 1 second, during the period, the longitudinal selectivity of relay protection of the earth fault of the neutral point arc suppression coil and the medium resistor (or the small resistor) earth power grid is effectively realized by measuring the earth fault current and the active component of the earth current of the power grid and setting a brand new fault quantity and the ratio of the active component of the earth current and the active component of the earth fault current, and the earth fault is quickly and selectively isolated. The invention provides a brand new relay protection method, which effectively draws the fault quantity of each section of the longitudinal feeder line of the fault feeder line through the fault quantity of the ratio of the ground line current to the effective value of the ground fault current, and realizes the longitudinal selectivity of the ground fault relay protection of the power grid with a neutral point grounded through an arc suppression coil and a middle resistor (omega) (or a small resistor). The relay protection mode not only gives full play to the advantages that the neutral point is grounded through the arc suppression coil, the electric arc of the instantaneous earth fault is automatically extinguished, isolation is not needed, and the power supply reliability is improved, but also can quickly isolate the permanent earth fault, greatly reduce the isolation range and the fault searching range of the earth fault, and effectively improve the power quality of the power supply.
Disclosure of Invention
The technical problem is as follows: aiming at the defects of the technology, the method for quickly isolating the grounding fault of the power grid with the neutral point grounded through the arc suppression coil and the middle resistor based on the measurement of the ground wire current belongs to the technical problem of relay protection of a power system, and solves the problem that the longitudinal selective fault isolation cannot be realized when the grounding fault occurs in a power distribution network with the neutral point grounded through the arc suppression coil and the middle resistor or a small resistor.
The technical scheme is as follows: in order to achieve the above object, the ground fault isolation method of the neutral point arc suppression coil and medium resistance grounding power grid of the present invention uses the ratio of the active component and the zero sequence voltage of the grounding fault current as the starting signal for the ground fault generation and line selection judgment, if the ratio of the active component and the zero sequence voltage of the grounding fault current exceeds the setting value, the grounding fault occurs, and the feeder line is determined to be the grounding fault feeder line, and finally, determining the protection action time of the ground fault according to the ratio of the ground fault quantity of the monitoring point to the ground fault quantity measured at the monitoring point when the actual ground fault occurs, so that the ground fault of the cable power distribution network with the neutral point grounded through the arc suppression coil and the middle resistor is quickly isolated.
The method comprises the following specific steps:
a. synchronously sampling zero sequence voltage and ground fault current and ground wire current of each feeder line in real time, and calculating the ratio I of active component to zero sequence voltage of ground fault current of each feeder line KRe /U 0 And judging: if the zero sequence voltage exceeds the setting value U 0qd And I is KRe /U 0 If the frequency is more than 3 omega C, the earth fault occurs, and then the feeder is a fault feeder, U 0 Is zero sequence voltage, I KRe C is the relative ground capacitance of the feeder line for the active component of the earth fault current, and then the following judgment is carried out;
b. calculating active component value I of ground wire current of fault feeder Reli.j I is the number of the power consumer sides from the main power substation, the i-th grounding electrode and j is the grounding wire where the j-th grounding fault radiated by the i-th grounding electrode is located;
c. setting a fault quantity equal to the active component I of the ground wire current Reli+x.j With fault current active component I ReK Of a ratio epsilon i.i+x ＝I Reli+x.j /I ReK X is 0,1,2, x is 0, wherein epsilon i.i The ground fault amount of the section of cable with ground fault detected for the ith level of ground fault protection, x is 1, epsilon i.i+1 The ground fault amount of the next section of cable with ground fault for the ith level of ground fault protection, x is 2, epsilon i.i+2 The ground fault amount of the ground fault of the lower section cable detected by the ith level ground fault protection;
d. determining the actuation time of ground fault protection according to the magnitude of the fault：t i ＝(ε maxi.i /ε i.i+x -1)·Δt i The quick isolation of the grounding fault of the cable distribution network with the neutral point grounded through the arc suppression coil and the middle resistor is realized by utilizing the action time; wherein the i-th stage earth fault protection means the action time of the earth fault protection installed in the substation, switching station or ring network station where the i-th stage earth electrode is located, epsilon i.i+x The amount of ground fault, ε, occurring in the section of cable for which the ith level of ground fault protection is detected maxi.i Maximum earth fault amount, delta t, of earth fault occurring in the cable section for the ith stage earth fault protection i Delay time set for ith level ground fault protection: Δ t i ＝ε maxi.i ·Δt/(ε maxi.i -ε maxi.i+1 ) And delta t is the delay time required by the first-level near backup, 0.1-0.2S is taken, and when epsilon is i.i+1 In the absence of a backup protection, i.e. no backup protection is required, then Δ t is taken i 0, instantaneous action.
As long as an earth fault occurs, the zero sequence voltage of the power grid is increased inevitably, and therefore, the zero sequence voltage exceeds a setting value U 0qd Necessarily meaning that the power grid has a ground fault, the ratio I of the active component of the zero-sequence current to the zero-sequence voltage of each feeder line KRe /U 0 The actual ground electric conduction of the feeder line is about equal to 1% of the capacitance resistance of the ground capacitor for the non-fault feeder line, and the ratio of the dielectric loss of the cable to the non-fault feeder line is about equal to the reciprocal of the sum of the ground resistance of the ground fault point and the ground resistance of the neutral point, and the number is far more than 3 omega C, so that the feeder line is judged to be a fault feeder line.
The grounding fault feeder has 3 sections which are respectively: for a power grid with a neutral point grounded through an arc suppression coil and a medium resistor of 100-132 omega or a small resistor of 10-15 omega, when the active component of grounding fault current flows in a power grid conductor and a ground wire, the fault current is mainly distributed from the fault point to the neutral point in the power grid conductor, and the current is the active component I of the current flowing through the neutral point NRe (ii) a And fault current in the earthThe distribution of the active components, in particular of the active components of the fault current in the earth of the faulty feeder, is composed of 2 parts, the resistance current I of the neutral point NRe Ground wire current distribution and fault current active component I of fault point K under independent action kRE Ground wire current distribution under independent action; zero sequence voltage U 0 The ground wire current distribution and neutral point arc suppression coil compensation current under independent action are reactive components of current and are not discussed here;
A. fault current I at fault point K when active component of neutral point current acts alone kRe Regarded as open circuit, zero sequence voltage U 0 If the zero-sequence current reactive component is ignored, the neutral point current active component I is considered as a short circuit NRe And the active component I of the fault point current kRe Are equal, i.e. have I NRe ＝-I KRe ＝3I 0Re And because the lines of the power distribution network are not very long, the middle resistance or the small resistance of the neutral point is 10-15 omega far larger than the zero sequence impedance of the lines, and the size of the single-phase earth fault current mainly depends on the resistance value of the resistance of the neutral point and the impedance of the earth fault point, no matter K is K 1 Dot or K 2 Dot or K 3 Point generation of single-phase earth fault, I NRe And I kRE The value of (a) is almost constant; i is NRe After the current flows into the grounding grid, the current flows into each grounding electrode along the grounding grid and enters the ground, and the active components of the current in the ground wire of the fault feeder are as follows:
between the first stage main substation and the switching station:
the second section of switching station is connected with the ring network station:
and between the third ring network station and the transformer of the power consumer:
defining: r cei.j Is I NRel(i-1).j Grounding resistance R of injection point ei.j With equivalent resistance of the user-side grounding network, i.e. disconnecting grounding resistance R ei.j Ground wire to power side ground net, I NRel(i-1).j Injecting a wire, observing the grounding grid from the power supply side to the user side, and connecting the grounding resistor R ei.j When the single-port network is regarded as an inlet, the equivalent resistance of the single-port network is R cei.j The resistance can be measured directly to: i.e. disconnect the ground resistance R ei.j Ground wire to power-supply-side grounding grid, i.e. disconnection I NRel(i-1).j Injecting a wire, measuring ground resistance R ei.j The resistance of the upper breaking point to the ground E is R cei.j ；R cei.j Calculated by equation (4):
I cNReli.j is a neutral point current I NRe When the grounding current of the jth feeder line radiated by the ith grounding electrode is in the grounding line, if the longitudinal power supply line exceeds three levels, the calculation can be analogized in turn;
B. active component I of fault current at fault point K kRE When acting alone, neutral point current I N Regarded as open circuit, zero sequence voltage U 0 When the fault is regarded as a short circuit, the fault current of the fault point K enters the ground through the fault point, returns to the grounding grid through each grounding electrode and returns to the fault point through each grounding wire;
if the ground fault occurs at K 1 The active component of the current between the fault point and the power supply is as follows:
defining: r is uei.j Is I uReli.j Ground resistance R of injection point ei.j Equivalent resistance of power-supply-side grounding network, i.e. breaking grounding resistance R ei.j Ground wire of grounding net connected to user side, I kReli.j Injecting wires, observing the grounding net from the side of the user to the power supply, and electrifying the grounding gridResistance R ei.j When the single-port network is regarded as an inlet, the equivalent resistance of the single-port network is R uei.j The resistance can be measured directly to: i.e. disconnect the ground resistance R ei.j Measuring ground resistance R with the ground wire of the user side grounding network ei.j The resistance of the upper breaking point to the ground E is R uei.j ；R uei.j Calculated by equation (6).
The current from the fault point to the load side is:
because the resistance of the grounding electrode of the main substation is required to be less than or equal to 0.5 omega, the number of the cables radiated by the main substation is large, the resistance of the grounding electrode of the switching station is required to be less than or equal to 2 omega, the resistance of the grounding electrode of the ring network station and the power consumer is required to be less than 4 omega, and the number of the radiated cables is small, the equivalent resistance R of the grounding electrode at the power supply side is small uei.j Much smaller than the load side ground resistance R cei.j So that the fault point has a current active component I to the load side ckReli.j Much less than the current real component I from the fault point to the power supply side kReli.j In which I ckReli.j And I cNReli.j Is in the opposite direction, the active component I of the earth current Reli.j Is the algebraic sum of 2 currents, and is substantially the difference of 2, obviously, the current active component from the fault point to the load side in the ground wire is much smaller than the current active component to the power supply side, so the current active component from the fault point to the load side in the ground wire is not discussed below;
if the ground fault occurs at K 2 The current active component between the fault point and the switching station is:
the active component of the current from the switching station to the power supply terminal is：
If the ground fault occurs at K 3 The active component of the current between the fault point and the ring network station is as follows:
the current active component from the ring network station to the switching station is as follows:
the active component of the current between the switching station and the power supply is as follows:
thus, K 1 When the point earth fault occurs, the active component of the earth current from the fault point to the power supply is as follows:
I Rel1.j ＝I cNRel1.j +I k1Rel1.j (13)
K 2 when the point ground fault occurs, the active components of the ground wire current from the fault point to the power supply are as follows in sequence:
I Rel1.j ＝I cNRel1.j +I k2Rel1.j (14)
I Rel2.j ＝I cNRel2.j +I k2Rel2.j (15)
K 3 when the point ground fault occurs, the active components of the ground wire current from the fault point to the power supply are as follows in sequence:
I Rel1.j ＝I cNRel1.j +I k3Rel1.j
(16)
I Rel2.j ＝I cNRel2.j +I k3Rel2.j (17)
I Rel3.j ＝I cNRel3.j +I k3Rel3.j (18)。
has the advantages that: when the power grid has a ground fault, the arc suppression coil compensates the capacitance current of the power grid to the ground in time, so that the current of a ground fault point is effectively reduced, an instantaneous ground fault arc is rapidly extinguished, and the power grid recovers normal operation; if the earth fault cannot automatically disappear in 3-5 seconds, the neutral point is a permanent earth fault, QF in the graph 1 is closed, a middle resistor of 100-132 omega or a small resistor of 10-15 omega is merged between a neutral point and an earth electrode of the power grid for 1 second, and during the period, the longitudinal selectivity of the neutral point arc suppression coil and the relay protection of the earth fault of the medium-resistance or small-resistance earth power grid is effectively realized by measuring the earth fault current and the active component of the earth fault current of the power grid and setting a brand new fault quantity and the ratio of the active component of the earth fault current and the active component of the earth fault current, and the earth fault is quickly and selectively isolated. The relay protection mode not only gives full play to the advantage that the neutral point is grounded through the arc suppression coil, and the electric arc of the instantaneous ground fault is automatically extinguished without isolation, thereby being beneficial to improving the reliability of power supply, but also can quickly isolate the permanent ground fault, greatly reducing the isolation range of the ground fault and the searching range of the fault, and also effectively improving the power quality of the power supply. Moreover, the fault quantity has no dimension which is independent of the impedance of the ground fault point and the type of the ground fault, namely, the single-phase ground fault or the two-phase ground fault, and the fault isolation performance is more excellent.
Description of the drawings:
fig. 1 is a schematic diagram of a zero sequence circuit of a simplified power supply system of the present invention;
fig. 2(a), (b), and (c) are model diagrams obtained by replacing a cable distribution network simulation model diagram by applying an alternative theorem of a circuit;
FIG. 3 is a schematic diagram of a feeder zero sequence current transformer and a ground zero sequence current transformer acquiring a current signal sample;
fig. 4 is a schematic diagram of sampling current signals obtained by the feeder zero sequence current transformer and the ground zero sequence current transformer.
FIG. 5 is a schematic diagram of a cable distribution network simulation model;
Detailed Description
The invention is further described in the following with reference to the accompanying drawings:
the implementation of the invention needs to have a zero sequence current transformer for each feeder line fault current measurement and a current transformer for ground wire current measurement, wherein the feeder line zero sequence current transformer is arranged on the outer side of the high-voltage cable for movement, the ground wire zero sequence current transformer is arranged on the outer side of the ground wire of the high-voltage cable, the feeder line zero sequence current transformer is arranged on the outer side of the high-voltage cable for movement, and 2 current signal sampling methods acquired by the feeder line zero sequence current transformer and the ground wire zero sequence current transformer are shown in figures 4 and 5; 2 current sampling signals are sent to a relay protector for signal conditioning, and then the following calculation and judgment are carried out:
as shown in fig. 1, the method for quickly isolating the ground fault of the power grid with the neutral point grounded through the arc suppression coil and the middle resistor based on the ground current measurement of the invention comprises the following specific steps:
1) synchronously sampling zero sequence voltage and ground fault current and ground wire current of each feeder line in real time, and calculating the ratio I of active component to zero sequence voltage of ground fault current of each feeder line KRe /U 0 And judging: if the zero sequence voltage exceeds the setting value U 0qd And I is KRe /U 0 If the current is more than 3 omega C, the earth fault occurs, and the feeder is the fault feeder, wherein U 0 Is zero sequence voltage, I KRe Is the active component of the ground fault current and C is the capacitance per phase to ground of the present feed line.
As long as an earth fault occurs, the zero sequence voltage of the power grid is increased inevitably, and therefore, the zero sequence voltage exceeds a setting value U 0qd This necessarily means that a ground fault has occurred in the grid. Ratio I of active component to zero sequence voltage of zero sequence current of each feeder line KRe /U 0 The actual ground conductance of the individual feed lines, and the dielectric loss of the cable for non-faulty feed lines, is approximately equal to about 1% of the capacitive reactance to ground. For the grounding fault feeder, the ratio is approximately equal to the reciprocal of the sum of the grounding resistance of the grounding fault point and the grounding resistance of the neutral point, and the number is far more than 3 omega C, so that the feeder is necessarily a fault feeder.
Further, the following judgment is made:
2) calculating the active component value I of the ground wire current of the fault feeder Rel : the ground fault feeder has 3 sections, which are assumed as follows: the zero sequence circuit of the power supply system is simplified as shown in figure 1 by the inlet of a 10/0.4kV transformer from a main power substation to a switching station, a switching station to a ring network station and a ring network station to a power consumer, and figure 1 can be replaced by figures 2(a), (b) and (c) according to the theory that zero sequence current is distributed in the ground wire and the substitution theorem of the circuit, wherein the figures 2(a), (b) and (c) are K respectively 1 At fault, K 2 At fault, K 3 In fig. 1 and fig. 2(a), (b), and (c), i ═ 1 to n are numbers from the grounding electrode of the main substation to the grounding electrode of the user side; j is 1-m, which is the serial number of the ground wire connected to the i-th grounding resistor, and m is the total number of the ground wires connected to the i-th grounding resistor; i is eli.j Is the current of the jth ground connected to the ith ground; r ei.j Is the ith ground resistance; r N A neutral point ground resistor; x N Is the reactance of the arc suppression coil; r is eli.j Is the ground resistance, R 'of the j feeder radiating from the i ground' eli.j +R” eli.j ＝R eli.j 。
For a power grid with a neutral point grounded through an arc suppression coil and a medium resistance (100-132 omega) (or a small resistance of 10-15 omega), if only the active component of the grounding fault current flows in a power grid conductor and a ground wire, the fault current is mainly distributed from the fault point to the neutral point in the power grid conductor, and the current is the active component I of the current flowing through the neutral point NRe (ii) a The distribution of the active components of the fault current in the earth line, in particular in the earth line of the fault feeder, is composed of 2 parts, the resistance current I of the neutral point NRe Ground wire current distribution and fault current active component I of fault point K under independent action kRE Ground wire current distribution under independent action; zero sequence voltage U 0 The ground current distribution and neutral crowbar compensation current under separate action are reactive components of the current and are not discussed here.
A. Fault current at fault point K when the active component of neutral point current acts aloneI kRe Regarded as open circuit, zero sequence voltage U 0 Regarded as a short circuit, neutral point current active component I NRe The distribution in the ground line is shown by the light-colored dashed arrow lines in fig. 2(a), (b), (c). As can be seen from FIGS. 2(a), (b) and (c), neglecting the reactive component of the zero-sequence current, the active component I of the neutral point current NRe And fault point current active component I kRe Are equal, i.e. have I NRe ＝-I KRe ＝3I 0Re Moreover, because the lines of the power distribution network are not very long, the middle resistance (or small resistance 10-15 omega) of the neutral point is far larger than the zero sequence impedance of the lines, and the size of the single-phase earth fault current mainly depends on the resistance value of the resistance of the neutral point and the impedance of the earth fault point, no matter K is K 1 Dot or K 2 Dot or K 3 Point generation of single-phase earth fault, I NRe And I kRE Is almost constant. I is NRe After the current flows into the grounding grid, the current flows into each grounding electrode along the grounding grid and enters the ground, and the active components of the current in the ground wire of the fault feeder are as follows:
between the first stage main substation and the switching station:
the second section of switching station is connected with the ring network station:
and between the third ring network station and the transformer of the power consumer:
here, it is defined that: r cei.j Is shown as I NRel(i-1).j Ground resistance R of injection point ei.j With equivalent resistance of the user-side grounding network, i.e. disconnecting grounding resistance R ei.j Ground wire (I) connected to power supply side grounding grid NRel(i-1).j Injected wire), the grounding grid is observed from the power supply side to the user side, and the grounding resistance R is measured ei.j When the single-port network is regarded as an inlet, the equivalent resistance of the single-port network is R cei.j . The resistance can be measured directly until: i.e. disconnect the ground resistance R ei.j Ground wire to power-supply-side grounding grid, i.e. break I NRel(i-1).j Injecting a wire, measuring ground resistance R ei.j The resistance of the upper breaking point to the ground E is R cei.j ；R cei.j Calculated by equation (4):
I cNReli.j is a neutral point current I NRe The ground current of the j feeder radiated by the i-th grounding electrode when in the ground. If the longitudinal power supply line exceeds three levels, calculation can be analogized in sequence.
B. Active component I of fault current at fault point K kRE When acting alone, neutral point current I N Regarded as open circuit, zero sequence voltage U 0 And when the fault is regarded as a short circuit, the fault current of the fault point K enters the ground through the fault point, returns to the grounding grid through each grounding electrode and returns to the fault point through each grounding wire. The fault current only forms a loop in the grounding grid and flows. The distribution in the ground of the grounding grid is shown by the dark dashed arrow lines in fig. 2(a), (b), and (c).
If the ground fault occurs at K 1 The active component of the current between the fault point and the power supply is as follows:
here, it is defined that: r uei.j Is I uReli.j Ground resistance R of injection point ei.j Equivalent resistance of power-supply-side grounding network, i.e. breaking grounding resistance R ei.j Ground wire (I) of grounding network connected with user side kReli.j Inject line), observe the grounding grid from the user side to the power supply side, and set the grounding pole resistance R ei.j When the single-port network is regarded as an inlet, the equivalent resistance of the single-port network is R uei.j . The resistance can be measured directly to: i.e. disconnect the ground resistance R ei.j Measuring the ground wire of the grounding network connected with the user sideGround resistance R ei.j The resistance of the upper breaking point to the ground E is R uei.j ；R uei.j Calculated by equation (6).
The current from the fault point to the load side is:
because the resistance of the grounding electrode of the main substation is required to be less than or equal to 0.5 omega, the number of cables radiated by the main substation is large, the resistance of the grounding electrode of the switching station is required to be less than or equal to 2 omega, the resistance of the grounding electrode of the ring network station and the power consumer is required to be less than 4 omega, and the number of cables radiated by the switching station and the power consumer is small, the equivalent resistance R of the grounding electrode of the power supply side is small uei.j Much smaller than the load side ground resistance R cei.j Therefore, the current active component I from the fault point to the load side ckReli.j Much smaller than the current active component I from the fault point to the power supply side kReli.j . As can also be seen from FIGS. 2(a), (b), (c), I ckReli.j And I cNReli.j Is in the opposite direction, the active component I of the earth current Reli.j Is the algebraic sum of 2 currents, which is essentially the difference of 2. Obviously, the current real component to the load side of the fault point in the ground line is much smaller than the current real component to the power supply side, and therefore, the current real component to the load side of the fault point in the ground line will not be discussed below.
If the ground fault occurs at K 2 The current active component between the fault point and the switching station is:
the active component of the current between the switching station and the power supply end is as follows:
if grounded soThe barrier occurs at K 3 The active component of the current between the fault point and the ring network station is as follows:
the current active component from the ring network station to the switching station is as follows:
the active component of the current between the switching station and the power supply is as follows:
thus, K 1 When the point is in ground fault, the active component of the ground wire current from the fault point to the power supply is as follows:
I Rel1.j ＝I cNRel1.j +I k1Rel1.j (13)
K 2 when the point ground fault occurs, the active components of the ground wire current from the fault point to the power supply are as follows in sequence:
I Rel1.j ＝I cNRel1.j +I k2Rel1.j (14)
I Rel2.j ＝I cNRel2.j +I k2Rel2.j (15)
K 3 when the point ground fault occurs, the active components of the ground wire current from the fault point to the power supply are as follows in sequence:
I Rel1.j ＝I cNRel1.j +I k3Rel1.j
(16)
I Rel2.j ＝I cNRel2.j +I k3Rel2.j (17)
I Rel3.j ＝I cNRel3.j +I k3Rel3.j (18)
3) setting a fault quantity equal to the active component I of the ground wire current Reli+x.j With fault current active component I KRe Of a ratio epsilon i.i+x ＝I Reli+x.j /I KRe ，x＝0,1,2，x＝0，ε i.i The grounding fault quantity of the grounding fault of the section of cable measured by the ith-level grounding fault protection is x ═ 1, epsilon i.i+1 The grounding fault quantity of the next section of cable subjected to grounding fault detection by the ith stage of grounding fault protection, x is 2, epsilon i.i+2 The value is the amount of ground fault of the next lower cable section which is detected by the ith level of ground fault protection.
Here, epsilon is the failure amount, subscript i is also the number from the power source to the consumer, and k is the failure point. Due to I NRe ＝-I kRe ＝3I 0Re Then, k is 1 When a point ground fault occurs, the fault amount from the fault point to the power supply:
k 2 when the point ground fault occurs, the fault quantity from the fault point to the power supply is as follows in sequence:
k 3 when the point ground fault occurs, the fault quantity from the fault point to the power supply is as follows in sequence:
4) determining earthing by fault magnitudeAction time of the fault protection: t is t i ＝(ε maxi.i /ε i.i+x -1)·Δt i In the formula, the operation time of the i-th ground fault protection (ground fault protection in a substation, a switching station, or a ring main station in which the i-th ground electrode is located) is ε i.i+x Is the ground fault quantity, epsilon, of each section of cable with ground fault detected by the ith level ground fault protection maxi.i Is the maximum earth fault amount, delta t, of the earth fault of the cable section detected by the ith level earth fault protection i The delay time set for the ith-level ground fault protection is generally: Δ t i ＝ε maxi.i+1 ·Δt/(ε maxi.i -ε maxi.i+1 ) And delta t is the delay time required by the common first-level near backup, can be 0.1-0.2S, when epsilon i.i+1 In the absence of a backup protection, i.e. no backup protection is required, then Δ t is taken i When it is 0, it acts instantaneously.
The first embodiment,
As shown in fig. 5, a simplified description of the distribution network of urban 10kV cables is used: if it has the level four earthing pole vertically, be main substation, switching station, looped netowrk cabinet to with the electric consumer transformer respectively, then i ═ 4, earthing pole resistance is respectively: r e1.1 ＝0.5Ω、R e2.j ＝2Ω、 R e3.j ＝4Ω、R e3.j 4 omega, all 9 feeders of main transformer are connected to each switching station, all 6 feeders of each switching station are connected to ring network station, each ring network station has 2 feeders to each user transformer, and all the feeders are 240mm 2 The power supply distance of the three-core armored cable is 1km, and the ground wire resistance R of all the feeder lines eli,j 0.242 Ω. Thus, it can be calculated by equation (4): r ce1.0 ＝0.0483Ω、R ce2.j ＝0.2389Ω、R ce3.j ＝1.386Ω、R ce4.j 4 Ω. Similarly, calculated according to equation (6): r ue1.j ＝0.0537Ω、R ue2.j ＝0.1438Ω、R ue3.j ＝0.3249Ω。
1 st stretch measured the most distal (R) of this stretch " el1.j 0) and nearest end (R " el1.j 0.242 Ω) the minimum and maximum fault amounts when the single-phase ground fault occurs are:
the minimum and maximum fault quantities when the farthest end and the nearest end of the lower section have single-phase earth faults are measured as section I1:
the minimum and maximum fault quantities measured in section I-1 when a single-phase earth fault occurs at the nearest end and the nearest end of the lower section are respectively:
the minimum and maximum fault quantities of the section when the farthest end and the nearest end of the section have single-phase earth faults are measured as the section I-2 are respectively as follows:
the minimum and maximum fault quantities when the farthest end and the nearest end of the lower section have single-phase earth faults are measured as the section I2:
the minimum and maximum fault quantities measured in section I-3 when the single-phase earth fault occurs at the farthest end and the nearest end of the section are respectively:
when Δ t is 0.2S, Δ t is obtained 1 ＝0.163S，Δt 2 ＝0.760S，Δt 3 0. If the fault occurs between the ring network station and the primary side of the transformer of the power consumer, the fault amount detected by the ring network station is the largest and is between 0.8807 and 9337, so that the main protection of the ring network station acts instantaneously to isolate the fault. If the looped network station fails to isolate faults, and the fault amount detected by the switching station is 0.7523-0.7969, the grounding fault protection of the switching station takes 0.2-0.257S of time delay as a backup protection action to isolate the faults. If the substation isolation fault fails again and the fault amount detected by the main transformer is 0.4591-0.4808, the backup protection of the main transformer is delayed for 0.276-0.297S to operate, and the fault is isolated. If the earth fault occurs between the switching station and the ring network station, the fault amount detected by the switching station is the largest, and the main protection action time of the switching station is between 0.7970S and 0.9335S, so that the earth fault is isolated. If the switching station fails to isolate the fault, the main substation detects that the fault amount is 0.4808-0.5472, the action time of the backup protection of the main substation is 0.2-0.266S, and the ground fault is isolated. If connect toThe ground fault is generated between the main substation and the switching station, the fault amount detected by the main substation is 0.5473-0.9999, the ground fault protection action time of the main substation is 0-0.2S, and the fault is isolated.
Through the 4 steps, the longitudinal selectivity of the ground fault of the power distribution network with the neutral point grounded through the small-resistance cable is realized. The invention can be independently implemented by a microcomputer chip and is realized by inserting a microcomputer relay protection device, and the functions of the protection device are increased.
Claims (4)
1. A method for isolating a ground fault of a power grid with a neutral point grounded through an arc suppression coil and a middle resistor is characterized in that: the method comprises the steps of taking the ratio of an active component of a ground fault current to a zero sequence voltage as a starting signal for ground fault generation and line selection judgment, if the ratio of the active component of the ground fault current to the zero sequence voltage exceeds a setting value, generating a ground fault, determining that a feeder line is a ground fault feeder line, calculating the active component value of the ground line current of the fault feeder line, setting the ratio of the active component value of the ground line current to the active component of the ground fault current as a ground fault amount, and finally determining the protection action time of the ground fault according to the ratio of the ground fault amount of a monitoring point to the ground fault amount measured at the monitoring point when the actual ground fault occurs, so that the ground fault of the cable distribution network with a neutral point grounded through an arc suppression coil and a middle resistor is quickly isolated.
2. The method for isolating the ground fault of the power grid with the neutral point grounded through the arc suppression coil and the medium resistance grounded according to claim 1 is characterized by comprising the following specific steps of:
a. synchronously sampling zero sequence voltage and ground fault current and ground wire current of each feeder line in real time, and calculating the ratio I of active component to zero sequence voltage of ground fault current of each feeder line KRe /U 0 And judging: if the zero sequence voltage exceeds the setting value U 0qd And I is KRe /U 0 If the frequency is more than 3 omega C, the earth fault occurs, and then the feeder is a fault feeder, U 0 Is zero sequence voltage, I KRe C is the active component of the earth fault current, and C is the relative position of the feeder lineA ground capacitance, and further making the following judgment;
b. calculating active component value I of ground wire current of fault feeder Reli.j I is the number of the power consumer sides from the main power substation, the i-th grounding electrode and j is the grounding wire where the j-th grounding fault radiated by the i-th grounding electrode is located;
c. setting a fault quantity equal to the active component I of the ground wire current Reli+x.j With fault current active component I ReK Of (a) is i.i+x ＝I Reli+x.j /I ReK X is 0,1,2, x is 0, wherein epsilon i.i The ground fault amount of the section of cable with ground fault detected for the ith level of ground fault protection, x is 1, epsilon i.i+1 The ground fault amount of the next section of cable with ground fault for the ith level of ground fault protection, x is 2, epsilon i.i+2 The ground fault amount of the ground fault of the lower section cable detected by the ith level ground fault protection;
d. determining the action time of the ground fault protection according to the fault magnitude: t is t i ＝(ε maxi.i /ε i.i+x -1)·Δt i The quick isolation of the grounding fault of the cable distribution network with the neutral point grounded through the arc suppression coil and the middle resistor is realized by utilizing the action time; wherein the i-th stage earth fault protection means the action time of the earth fault protection installed in the substation, switching station or ring network station where the i-th stage earth electrode is located, epsilon i.i+x The amount of ground fault, ε, occurring in the section of cable for which the ith level of ground fault protection is detected maxi.i Maximum earth fault amount, delta t, of earth fault occurring in the cable section for the ith stage earth fault protection i Delay time set for ith level ground fault protection: Δ t i ＝ε maxi.i ·Δt/(ε maxi.i -ε maxi.i+1 ) And delta t is the delay time required by the first-level near backup, 0.1-0.2S is taken, and when epsilon is i.i+1 In the absence of a backup protection, i.e. no backup protection is required, then Δ t is taken i 0, instantaneous action.
3. The method of isolating a ground fault in a neutral-point crowbar coil-and-medium-resistance-grounded power grid as claimed in claim 2, wherein: as long as there isWhen the earth fault occurs, the zero sequence voltage of the power grid is increased, so that the zero sequence voltage exceeds the setting value U 0qd Necessarily meaning that the power grid has a ground fault, the ratio I of the active component of the zero-sequence current to the zero-sequence voltage of each feeder line KRe /U 0 The actual ground electric conduction of the feeder line is about equal to 1% of the capacitance resistance of the ground capacitor for the non-fault feeder line, and the ratio of the dielectric loss of the cable to the non-fault feeder line is about equal to the reciprocal of the sum of the ground resistance of the ground fault point and the ground resistance of the neutral point, and the number is far more than 3 omega C, so that the feeder line is judged to be a fault feeder line.
4. The method of isolating a ground fault in a neutral-point crowbar coil-and-medium-resistance-grounded power grid as claimed in claim 2, wherein: the grounding fault feeder has 3 sections which are respectively: for a power grid with a neutral point grounded through an arc suppression coil and a medium resistor of 100-132 omega or a small resistor of 10-15 omega, when the active component of grounding fault current flows in a power grid conductor and a ground wire, the fault current is mainly distributed from the fault point to the neutral point in the power grid conductor, and the current is the active component I of the current flowing through the neutral point NRe (ii) a The distribution of the active components of the fault current in the earth line, in particular in the earth line of the faulty feeder, is composed of 2 parts, the resistance current I of the neutral point NRe Ground wire current distribution and fault current active component I of fault point K under independent action kRE Ground wire current distribution under independent action; zero sequence voltage U 0 The ground wire current distribution and neutral point arc suppression coil compensation current under independent action are reactive components of current and are not discussed here;
A. fault current I at fault point K when active component of neutral point current acts alone kRe Regarded as open circuit, zero sequence voltage U 0 If the zero-sequence current reactive component is ignored, the neutral point current active component I is considered as a short circuit NRe And the active component I of the fault point current kRe Are equal, i.e. haveI NRe ＝-I KRe ＝3I 0Re And because the lines of the power distribution network are not very long, the middle resistance or small resistance of the neutral point is 10-15 omega and is far larger than the zero sequence impedance of the lines, and the size of the single-phase earth fault current mainly depends on the resistance value of the resistance of the neutral point and the impedance of the earth fault point, no matter K is K 1 Dot or K 2 Dot or K 3 Point generation of single-phase earth fault, I NRe And I kRE Is almost constant; I.C. A NRe After the current flows into the grounding grid, the current flows into each grounding electrode along the grounding grid and enters the ground, and the active components of the current in the ground wire of the fault feeder are as follows:
between the first stage main substation and the switching station:
the second section of switching station is connected with the ring network station:
and a third section is connected between the ring network station and the power consumer transformer:
defining: r cei.j Is I NRel(i-1).j Ground resistance R of injection point ei.j With equivalent resistance of the user-side grounding network, i.e. disconnecting grounding resistance R ei.j Ground wire to power side ground net, I NRel(i-1).j Injecting a wire, observing the grounding grid from the power supply side to the user side, and connecting the grounding resistor R ei.j When the single-port network is regarded as an inlet, the equivalent resistance of the single-port network is R cei.j The resistance can be measured directly to: i.e. disconnect the ground resistance R ei.j Ground wire to power-supply-side grounding grid, i.e. disconnection I NRel(i-1).j Injecting a wire, measuring ground resistance R ei.j The resistance of the upper breaking point to the ground E is R cei.j ；R cei.j Calculated by equation (4):
I cNReli.j is a neutral point current I NRe When the grounding current of the jth feeder line radiated by the ith grounding electrode is in the grounding line, if the longitudinal power supply line exceeds three levels, the calculation can be analogized in turn;
B. active component I of fault current at fault point K kRE Neutral point current I when acting alone N Regarded as open circuit, zero sequence voltage U 0 When the fault is regarded as a short circuit, the fault current of the fault point K enters the ground through the fault point, returns to the grounding grid through each grounding electrode and returns to the fault point through each grounding wire;
if the ground fault occurs at K 1 The active component of the current between the fault point and the power supply is as follows:
defining: r uei.j Is I uReli.j Ground resistance R of injection point ei.j Equivalent resistance of power-supply-side grounding network, i.e. breaking grounding resistance R ei.j Ground wire to the grounding net of the user side, I kReli.j Injecting wire, observing the grounding net from the side of user to the power supply, and making the resistance R of grounding electrode ei.j When the single-port network is regarded as an inlet, the equivalent resistance of the single-port network is R uei.j The resistance can be measured directly to: i.e. disconnect the ground resistance R ei.j Measuring ground resistance R with the ground wire of the user side grounding network ei.j The resistance of the upper breaking point to the ground E is R uei.j ；R uei.j Calculated by equation (6).
The current from the fault point to the load side is:
because the resistance of the grounding electrode of the main substation is required to be less than or equal to 0.5 omega, the number of the cables radiated by the main substation is large, the resistance of the grounding electrode of the switching station is required to be less than or equal to 2 omega, the resistance of the grounding electrode of the ring network station and the power consumer is required to be less than 4 omega, and the number of the radiated cables is small, the equivalent resistance R of the grounding electrode at the power supply side is small uei.j Much smaller than the load side ground resistance R cei.j So that the fault point has a current active component I to the load side ckReli.j Much smaller than the current active component I from the fault point to the power supply side kReli.j In which I ckReli.j And I cNReli.j Is in the opposite direction, the active component I of the earth current Reli.j Is the algebraic sum of 2 currents, and is substantially the difference of 2, obviously, the current active component from the fault point to the load side in the ground wire is much smaller than the current active component to the power supply side, so the current active component from the fault point to the load side in the ground wire is not discussed below;
if the ground fault occurs at K 2 The current active component between the fault point and the switching station is:
the active components of the current between the switching station and the power supply end are as follows:
if the ground fault occurs at K 3 The active component of the current between the fault point and the ring network station is as follows:
the current active component from the ring network station to the switching station is as follows:
the active component of the current between the switching station and the power supply is as follows:
thus, K 1 When the point is in ground fault, the active component of the ground wire current from the fault point to the power supply is as follows:
I Rel1.j ＝I cNRel1.j +I k1Rel1.j (13)
K 2 when the point ground fault occurs, the active components of the ground wire current from the fault point to the power supply are as follows in sequence:
I Rel1.j ＝I cNRel1.j +I k2Rel1.j (14)
I Rel2.j ＝I cNRel2.j +I k2Rel2.j (15)
K 3 when the point ground fault occurs, the active components of the ground wire current from the fault point to the power supply are as follows in sequence:
I Rel1.j ＝I cNRel1.j +I k3Rel1.j (16)
I Rel2.j ＝I cNRel2.j +I k3Rel2.j (17)
I Rel3.j ＝I cNRel3.j +I k3Rel3.j (18)。
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| CN117390863B (en) * | 2023-10-18 | 2024-06-04 | 国网四川省电力公司电力科学研究院 | Modeling analysis method and system for single-phase grounding arc of three-element power distribution network under multiple scenes |
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| CN117390863A (en) * | 2023-10-18 | 2024-01-12 | 国网四川省电力公司电力科学研究院 | Modeling analysis method and system for single-phase grounding arc of three-element power distribution network under multiple scenes |
| CN117390863B (en) * | 2023-10-18 | 2024-06-04 | 国网四川省电力公司电力科学研究院 | Modeling analysis method and system for single-phase grounding arc of three-element power distribution network under multiple scenes |
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