JPH0424936B2 - - Google Patents
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- Publication number
- JPH0424936B2 JPH0424936B2 JP3806184A JP3806184A JPH0424936B2 JP H0424936 B2 JPH0424936 B2 JP H0424936B2 JP 3806184 A JP3806184 A JP 3806184A JP 3806184 A JP3806184 A JP 3806184A JP H0424936 B2 JPH0424936 B2 JP H0424936B2
- Authority
- JP
- Japan
- Prior art keywords
- voltage
- power
- reactive power
- active power
- calculation unit
- Prior art date
- Legal status (The legal status is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.)
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Description
【発明の詳細な説明】
〔発明の技術分野〕
この発明は電力系統を保護する保護継電装置に
関するものである。DETAILED DESCRIPTION OF THE INVENTION [Technical Field of the Invention] The present invention relates to a protective relay device for protecting an electric power system.
電力系統に系統事故が発生し、事故解除後系統
に動揺が生じた時、系統の動揺が安定動揺か、又
は脱調にいたるかを検出する脱調保護継電方式に
ついて以下説明する。第１図は電力系統を模擬し
た模擬系統図で、１ａはａ端の電源、１ｂはｂ端
の電源を示す。２は電源１ａ，１ｂ間の送電線、
３ａは電源１ａの母線、４ａは保護継電器６ａ
（以下リレーと略称する）へ電流を導入する変成
器、５ａは母線３ａからリレー６ａへ電圧を導入
する変成器を示す。
A description will be given below of a step-out protection relay system that detects whether the system fluctuation is a stable fluctuation or a step-out when a system fault occurs in the power system and the system oscillates after the fault is cleared. FIG. 1 is a simulated power system diagram simulating a power system, where 1a shows the power source at the a end, and 1b shows the power source at the b end. 2 is a power transmission line between power sources 1a and 1b;
3a is the bus bar of the power supply 1a, 4a is the protective relay 6a
A transformer 5a introduces a current to a relay (hereinafter referred to as a relay), and a transformer 5a introduces a voltage from a bus 3a to a relay 6a.
電源１ａ，１ｂの電圧V〓a・a〓b（＝V〓ae-j〓）間
の
相差角θに対してａ端の電気量電圧V〓a、電流I〓eよ
り求まる電力W〓＝Pa＋jQe＝V〓a・I〓a（I〓aはI〓aの
共役
複素数）の有効電力Pa、無効電力Qaは第２図に
示すように、有効電力Paは90°において極大値と
なり、無効電力Qaはθが180°において極大値とな
り、図中のＡ点（θ＝90°）において等しくなる。 Power W〓 determined from the voltage V〓 a of the a end and the current I〓 e for the phase difference angle θ between the voltage V〓 a・a〓 b (=V〓 a e -j 〓) of the power supplies 1a and 1b =P a +jQ e =V〓 a・I〓 a (I〓 a is the conjugate complex number of I〓 a ) The active power P a and the reactive power Q a are as shown in Fig. 2, and the active power P a is 90 The reactive power Q a reaches a maximum value when θ is 180°, and becomes equal at point A (θ=90°) in the figure.
有効電力Pa、及び無効電力Qaは次の様にして
求められる。送電線のインピーダンスをZ〓＝Zej
≒jZ（≒90°）とすれば、電流I〓a＝（V〓a−V〓b）
／Z〓
＝V〓a（１−e-j〓）Z〓より次式が得られる。 The active power P a and the reactive power Q a are obtained as follows. The impedance of the transmission line is Z = Ze j
If ≒jZ (≒90°), current I〓 a = (V〓 a −V〓 b )
/Z〓
=V〓 a (1-e -j〓 )Z〓, the following formula is obtained.
W〓＝V〓a・I〓a＝（V〓2 a／Ｚ）｛sinθ＋ｊ（１−cos
θ）｝＝
Pa＋jQa
また横軸を有効電力Pa、縦軸を無効電力Qaと
した時のPQ軌跡を第３図に示す。第３図に示す
ようにPa，Qaの軌跡はP2 a＋（Qa−V〓2／a／Ｚ）2＝（V
〓2／a／Ｚ）2
の円の軌跡であり、図中のＡ点がθ＝90°を示す。
一般に電力系統では、その同期化力の大小も関係
するが、相差角θが90°以上開くと脱調したと判
断してよい。但し、この時の有効電力Qaは増加、
即ちｄ｜Qa｜／dt＞０である。W〓=V〓 a・I〓 a = (V〓 2 a /Z) {sinθ+j(1−cos
θ)}=
P a +jQ a Fig. 3 shows the PQ locus when the horizontal axis is the active power P a and the vertical axis is the reactive power Q a . As shown in Figure 3, the trajectories of P a and Q a are P 2 a + (Q a −V〓 2 / a /Z) 2 = (V
〓 2 / a / Z) 2 , and point A in the figure indicates θ = 90°.
In general, in power systems, the magnitude of the synchronization force is also relevant, but if the phase difference angle θ is 90° or more, it can be determined that synchronization has occurred. However, at this time, the active power Q a increases,
That is, d|Q a |/dt>0.
このように電力系統の脱調を判定するために
は、電源端の電圧・電流より有効電力Ｐ、無効電
力Ｑを求め、そのPQ軌跡がθ＝90°の点を越えた
かどうか判定すれば、良い。第４図に示すように
リレー６ａは電圧電流よりPaQa演算を行う演算
部７ａ、有効電流Pa、無効電力Qaよりθ＝90°を
示すＡ点を越えたかどうかの判定及び脱調出力を
出力する判定部８ａから構成されている。 In this way, in order to determine a loss of synchronization in the power system, calculate the active power P and reactive power Q from the voltage and current at the power supply end, and determine whether the PQ trajectory exceeds the point of θ = 90°. good. As shown in Fig. 4, the relay 6a has a calculation unit 7a that calculates P a Q a from the voltage and current, and determines whether or not the point A indicating θ=90° has been exceeded based on the active current P a and the reactive power Q a. It is composed of a determination section 8a that outputs a modulation output.
次に、リレーの設置点は第５図に示すように電
力系統の電源端のみでなく、ｃ点ｄ点の様に系統
の変電所等に設置される。ｃ点設置リレーへ変成
器４ｃ，５ｃを通して導入される電圧電流V〓c，I〓c
の関係は、電源V〓a，V〓b、ａ端とｃ端間の送電線
のインピーダンスをZ〓ac，cd間をZ〓cd，db間をZ〓db
とす
れば
V〓c＝Z〓cd＋Z〓db／Zac＋Zcd＋Z〓dbV〓a＋Z〓ac／Za
c＋Zcd＋ZdbV〓b
＝Z〓acV〓b／Zac＋Zcd＋Zdb（１＋Z〓cd＋Z〓db／Zac
∠θ）
I〓c＝V〓a−V〓b／Zac＋Zcd＋Zdb＝−V〓b／Zac＋Zcd
＋Zdb（１−１∠−
θ）
但しV〓a＝V〓b∠θ、V〓b基準となる。この電圧V〓O、
電流I〓cより有効電力Pc、無効電力QcのPQ軌跡は
円形にならず第６図のように楕円となる。特に脱
調中心点（Z〓cd＋Z〓db／Zac＝１）の時は
Pc＝Z〓ac・V〓b 2／｜Zac＋Zcd＋Zdb｜2sinθ Qp
＝０
となり、有効電力のみの変化しか表われない。こ
のため、PQ軌跡は有効電力Ｐ軸上を移動するこ
とになり、リレー設置点の電圧、電流によるPQ
軌跡では脱調の判定が出来ない場合が生じる。こ
れを改良するためにｃ点にて計測した電圧V〓e及
びI〓cより下記の如く、電源電圧を求めることによ
つて有効電力Ｐと無効電力Ｑの軌跡は円状にする
ことが出来、脱調判定が可能になる。 Next, the relays are installed not only at the power supply end of the power system as shown in FIG. 5, but also at substations of the system, such as points c and d. Voltage and current V〓 c , I〓 c introduced to the relay installed at point c through transformers 4c and 5c
The relationship between the power supplies V〓 a , V〓 b , the impedance of the transmission line between the a end and the c end is Z〓 ac , the distance between CD is Z〓 cd , and the distance between db is Z〓 db
Then, V〓 c = Z〓 cd +Z〓 db /Z ac +Z cd +Z〓 db V〓 a +Z〓 ac /Z a
c +Z cd +Z db V〓 b =Z〓 ac V〓 b /Z ac +Z cd +Z db (1+Z〓 cd +Z〓 db /Z ac
∠θ) I〓 c =V〓 a −V〓 b /Z ac +Z cd +Z db = −V〓 b /Z ac +Z cd
+Z db (1-1∠-θ) However, V〓 a =V〓 b ∠θ, V〓 b is the standard. This voltage V〓 O ,
From the current I〓 c , the PQ locus of the active power P c and the reactive power Q c does not become circular but becomes an ellipse as shown in FIG. 6. Especially at the center point of step-out (Z〓 cd +Z〓 db /Z ac = 1), P c =Z〓 ac・V〓 b 2 /｜Z ac +Z cd +Z db ｜ 2 sinθ Q p
= 0, and only the change in active power appears. Therefore, the PQ locus will move on the active power P axis, and the PQ trajectory will be affected by the voltage and current at the relay installation point.
There may be cases where it is not possible to determine step-out based on the trajectory. To improve this, the locus of active power P and reactive power Q can be made circular by finding the power supply voltage from the voltages V〓 e and I〓 c measured at point c, as shown below. , it becomes possible to judge out-of-step.
即ち第７図において背後電源電圧は第７図イに
おいてV〓a＝V〓c＋Z〓acI〓c第７図ロにおいてV〓b＝V
〓c−
（Z〓cd＋Z〓db）I〓cによつて求めることが出来るここ
で
電圧V〓a，V〓bは定格電圧、は送電線のインピー
ダンス角を示し、脱調中心点がｃ点よりｂ端側の
時はイ図より、またｃ点よりａ端側に脱調中心点
の時はロ図により電源を求める。求められた背後
電源をV〓BCとすれば電力W〓＝V〓BCI〓＝P〓BC＋jQBC
より
円のPQ軌跡が得られる。 That is, in Fig. 7, the background power supply voltage is V〓 a = V〓 c + Z〓 ac I〓 c in Fig. 7 B, V〓 b = V in Fig. 7 B.
〓 c −
(Z〓 cd + Z〓 db ) I〓 c Here, the voltage V〓 a , V〓 b is the rated voltage, and represents the impedance angle of the transmission line, and the center point of the step out is from point c to b If it is on the edge side, find the power source from figure A, and if the center point of step-out is on the side of point A from point C, find the power source from figure B. If the obtained back power source is V〓 BC , then the power W〓=V〓 BC I〓=P〓 BC +jQ BC
A more circular PQ locus can be obtained.
上述のように系統の各リレー設置点において変
成器より導入した電圧電流から電源端の電圧を求
め、電源端における有効電力Ｐ、無効電力Ｑを求
め、そのPQ軌跡から電源の相差角θが90°を越え
たかどうかの判定をすることで脱調を検出するリ
レーについて説明した。 As mentioned above, the voltage at the power supply end is determined from the voltage and current introduced from the transformer at each relay installation point in the system, the active power P and reactive power Q at the power supply end are determined, and from the PQ trajectory, the phase difference angle θ of the power supply is 90 We have explained a relay that detects step-out by determining whether or not the temperature exceeds °.
上記電力系統の各点に設置されるリレーは保護
すべき系統間に設置されることになる。例えば第
５図のcd間を保護するためにｃ点とｄ点に各々
リレーが設置され、脱調中心点が区間cd内に存
在するときは内部判定、区間外に存在するときは
外部判定と出力する。このためリレー６ｃ、リレ
ー６ｄ相互間で情報の受授が必要となる。即ち第
８図に示す例では第８図イは脱調中心点がcd間
にある場合の各々リレー６ｃ、リレー６ｄのPQ
軌跡、第８図ロが脱調中心点が外部にある場合の
各々のPQ軌跡である。このようにリレー６ｃ、
リレー６ｄは各々相手端の情報を得て区間中に脱
調中心点があるかどうかを判定することが出来
る。相手端の情報がなければ脱調したかどうかの
判定は行なえるが、脱調中心点がどの位置にある
か判別出来ないことになる。 The relays installed at each point of the power system are installed between the systems to be protected. For example, relays are installed at points c and d to protect between CDs in Figure 5, and when the center point of the step-out is within the interval CD, it is determined to be internal, and when it is outside the interval, it is determined to be external. Output. Therefore, it is necessary to exchange information between the relays 6c and 6d. That is, in the example shown in FIG. 8, FIG.
Figure 8 (b) shows each PQ trajectory when the center point of step-out is outside. In this way, relay 6c,
Each relay 6d can obtain information about the other end and determine whether there is a center point of synchronization within the section. Without information about the other end, it is possible to determine whether or not the synchronization has occurred, but it is not possible to determine where the center of synchronization is.
本発明は以上の点を鑑み、みなされたもので、
相手端のリレーのPQ軌跡を自端の電圧電流から
求めることによつて相手端からの情報を不必要に
する保護継電装置を提供することを目的とする。
The present invention was conceived in view of the above points,
It is an object of the present invention to provide a protective relay device that makes information from the other end unnecessary by determining the PQ locus of the relay at the other end from the voltage and current of the other end.
以下この発明の一実施例を図について説明す
る。保護継電装置の構成を示す第９図、第１０図
において、設置点ｃにおけるリレー６ｃは変成器
４ｃ，５ｃを通して電圧V〓c、電流I〓cを導入する。
７ｃは自端の有効電力Pc、無効電力Qcを求める
第１演算部、１０ｃは相手端の有効電力Pdと無
効電力Qdを求める第２演算部８ｃは自端のPQ軌
跡、相手端のPQ軌跡より各々相差角が90°を越え
たかどうかを判定とPQ軌跡の方向による脱調中
心点内外判定をするの内外判定部を示す。
An embodiment of the present invention will be described below with reference to the drawings. In FIGS. 9 and 10 showing the configuration of the protective relay device, a relay 6c at an installation point c introduces a voltage V〓c and a current I〓c through transformers 4c and 5c.
7c is a first calculation unit that calculates the active power P c and reactive power Q c of the own end, and 10c is a second calculation unit 8c that calculates the active power P d and reactive power Q d of the other end. The inside/outside determination part is shown for determining whether each phase difference angle exceeds 90° from the PQ locus at the end and for determining whether the out-of-sync center point is inside or outside based on the direction of the PQ locus.
次に動作について説明する。第１演算部７ｃは
自端の計測電気信号V〓c及びI〓cより背後の電源電圧
V〓CBを求める。電圧V〓CBの求め方は既に説明した
第７図のイ又はロのベクトル関係より求めること
が出来る。この電圧V〓CBと電流I〓cとから電力W〓c＝
V〓CB・I〓c＝Pc＋jQcを求める。有効電力Pcと無効電
力Qcの軌跡は電源端のPQ軌跡となり円である。 Next, the operation will be explained. The first calculation unit 7c calculates the power supply voltage behind the measured electric signals V〓 c and I〓 c at its own end.
Find V〓 CB . The voltage V〓 CB can be determined from the vector relationship A or B in Fig. 7, which has already been explained. From this voltage V〓 CB and current I〓 c , power W〓 c =
Find V〓 CB・I〓 c = P c + jQ c . The locus of active power P c and reactive power Q c is the PQ locus at the power source end, which is a circle.
第２演算部１０ｃは第１０図に示すように入力
信号I〓cから電圧補正量−Z〓bI〓cを作成する補正量作
成部１１c、電圧V〓cに該電圧補正量を加算してV〓d
＝V〓c−Z〓bI〓cを得る補正部１２ｃ、V〓dとI〓cから
背後
の電源電圧V〓dBを求め、この電圧V〓dBと電流I〓cとか
ら電力W〓d＝V〓dB・I〓c＝Pd＋jQcを求める第３演算
部１３ｃとから構成されている。電圧V〓dBの求め
方は電圧V〓CBと同様に求めることが出来る。内外
判定部８ｃは自端の背後Pc，Qc軌跡及び相手の
背後Pd，Qd軌跡より各々の軌跡で相差角が90°を
越えたかどうか、即ち脱調の判定と各々の軌跡の
方向から脱調中心点がリレー設置点聞、即ち保護
すべき系統上に存在するかの内外部判定を行い判
定の出力の出力信号９ｃを出力する。 As shown in FIG. 10 , the second calculation unit 10c creates a voltage correction amount −Z〓 b I〓 c from the input signal I〓 c , and adds the voltage correction amount to the voltage V〓 c . Then V〓 d
The correction unit 12c obtains =V〓 c −Z〓 b I〓 c , calculates the back power supply voltage V〓 dB from V〓 d and I〓 c , and calculates the power W〓 from this voltage V〓 dB and current I〓 c . d =V〓 dB ·I〓 c =P d +jQ c . The voltage V〓 dB can be determined in the same way as the voltage V〓 CB . The inside/outside determination unit 8c determines whether the phase difference angle exceeds 90° from the own end's rear P c , Q c trajectories and the opponent's rear P d , Q d trajectories, that is, whether the phase difference exceeds 90°, that is, it determines whether or not there is synchronization. Based on the direction, it is determined whether the center point of synchronization is between the relay installation point, that is, on the system to be protected, and an output signal 9c is output as the output of the determination.
以上のように、この発明によれば自端にて相手
端の電圧を求め、更に相手端の有効電力無効電力
も同時に演算するようにして構成したので、相手
端からの情報を必要としないため装置が安価にで
き、また、制度の高い保護継電装置が得られる効
果がある。
As described above, according to the present invention, since the voltage of the other end is determined at the own end and the active and reactive power of the other end are calculated simultaneously, information from the other end is not required. This has the effect of making the device inexpensive and providing a highly accurate protective relay device.
第１図は電力系統の模擬系統図、第２図はａ端
ｂ端の相差に相差角に対する有効電力、無効電力
図、第３図はPQ軌跡図、第４図は保護継電器の
構成図、第５図電力系統の模擬系統図で系統の各
点のリレー設置点を示す図、第６図が電源端以外
の各点のPQ軌跡を示す図、第７図はリレー設置
点から電源端電圧を求めるためのベクトル図、第
８図は自端、相手端のPQ軌跡が区間内、区間外
を説明する図、第９図、第１０図はこの発明の一
実施例による保護継電装置を示す構成図である。
１……電源、２……送電線、３……母線、４，
５……変成器、６……リレー、７……第１演算
部、８……内外判定部、９……出力、１０……第
２演算部、１１……補正量作成部、１２……補正
部、１３……第３演算部、尚添字ａ，ｂ，ｃ，ｄ
は各リレー設置点、又は相当部分を示し、また図
中同一符号は同一又は相当部分を示す。
Fig. 1 is a simulated power system diagram, Fig. 2 is a diagram of active power and reactive power with respect to the phase difference angle between ends A and B, Fig. 3 is a PQ trajectory diagram, and Fig. 4 is a configuration diagram of a protective relay. Figure 5 is a simulated power system diagram showing the relay installation points at each point in the system, Figure 6 is a diagram showing the PQ trajectory at each point other than the power supply end, and Figure 7 is the power supply end voltage from the relay installation point. FIG. 8 is a diagram explaining whether the PQ trajectory of the own end and the opposite end is within the interval or outside the interval, and FIGS. 9 and 10 are diagrams showing a protective relay device according to an embodiment of the present invention. FIG. 1...Power supply, 2...Power line, 3...Bus bar, 4,
5...Transformer, 6...Relay, 7...First calculation section, 8...Inside/outside determination section, 9...Output, 10...Second calculation section, 11...Correction amount creation section, 12... Correction unit, 13...Third calculation unit, subscripts a, b, c, d
indicates each relay installation point or a corresponding part, and the same reference numerals in the figure indicate the same or a corresponding part.
Claims (1)
にり有効電力と無効電力を演算する第１演算部
と、前記自端より導入した電圧、電流信号により
相手端の有効電力と無効電力を演算する第２演算
部と、前記第１演算部の有効電力および無効電力
の軌跡と上記第２演算部の有効電力および無効電
力の軌跡より各々相差角が90度を越えたか否かを
判定し、該軌跡方向により脱調中心点の内外判定
信号を送出する内外判定部を設け、前記第２演算
部は、前記電流信号により電圧補正量を作成する
補正量作成部と、この補正量作成部からの補正量
を自端電圧に加算して相手端電圧を作成する補正
部と、該相手端電圧と電流信号により、相手端の
有効電力、無効電力を算出して、上記内外部判定
部に入力する第３演算部とからなることを特徴と
する保護継電装置。1. A first calculation unit that calculates active power and reactive power based on the voltage and current signals introduced from the own end of the power system, and a first calculation unit that calculates the active power and reactive power of the other end based on the voltage and current signals introduced from the own end. a second calculation unit that determines whether the phase difference angle exceeds 90 degrees from the trajectory of the active power and reactive power of the first calculation unit and the trajectory of the active power and reactive power of the second calculation unit, An inside/outside determining section is provided which sends an inside/outside determination signal of the center point of synchronization based on the trajectory direction, and the second calculation section includes a correction amount creation section that creates a voltage correction amount based on the current signal, and a correction amount creation section that outputs a voltage correction amount from the current signal. a correction unit that adds the correction amount to the voltage at its own end to create a voltage at the other end, and calculates active power and reactive power at the other end based on the voltage at the other end and the current signal, and inputs the calculated active power and reactive power to the internal/external determination unit. A protective relay device comprising: a third calculation section that performs the following operations.
Priority Applications (1)
| Application Number | Priority Date | Filing Date | Title |
|---|---|---|---|
| JP3806184A JPS60183934A (en) | 1984-02-28 | 1984-02-28 | Protective relaying system |
Applications Claiming Priority (1)
| Application Number | Priority Date | Filing Date | Title |
|---|---|---|---|
| JP3806184A JPS60183934A (en) | 1984-02-28 | 1984-02-28 | Protective relaying system |
Publications (2)
| Publication Number | Publication Date |
|---|---|
| JPS60183934A JPS60183934A (en) | 1985-09-19 |
| JPH0424936B2 true JPH0424936B2 (en) | 1992-04-28 |
Family
ID=12514979
Family Applications (1)
| Application Number | Title | Priority Date | Filing Date |
|---|---|---|---|
| JP3806184A Granted JPS60183934A (en) | 1984-02-28 | 1984-02-28 | Protective relaying system |
Country Status (1)
| Country | Link |
|---|---|
| JP (1) | JPS60183934A (en) |
-
1984
- 1984-02-28 JP JP3806184A patent/JPS60183934A/en active Granted
Also Published As
| Publication number | Publication date |
|---|---|
| JPS60183934A (en) | 1985-09-19 |
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