CN104965151A - Fault distance detecting method based on voltage fault component of fault point - Google Patents

Abstract

A fault location method based on the sudden change in voltage at the fault point. This method first calculates the voltage at the fault point and the sudden change in voltage at the fault point, and then lists the faults according to the principle of the same phase between the voltage at the fault point and the sudden change in voltage at the fault point. The distance is an unknown quadratic equation, and the equation is solved to obtain the fault distance. This design is not only suitable for single-phase and phase-to-phase faults of the line, but also has strong practicability, high calculation accuracy and high efficiency.

Description

A Fault Location Method Based on Voltage Change at Fault Point

technical field

The invention belongs to the technical field of electric power system relay protection, and in particular relates to a single-end ranging method for a short-circuit fault of a power transmission line.

Background technique

Transmission line is the basic equipment for power generation and transmission in the power system, and occupies a very important position in the power system. When the transmission line is faulty, if it cannot be cut off in time or by mistake, it will have a great impact on the main system, and it will easily cause major accidents such as grid splitting and transformer tripping. High-voltage/ultra-high voltage/ultra-high voltage line protection undertakes the important task of power transmission and is an important hub of the power system. They are extremely expensive to manufacture, and once they are damaged due to malfunctions, they will be difficult to repair and take a long time, causing huge direct and indirect losses to the national economy.

The single-end fault location method of transmission lines only uses the electrical quantity at one end of the transmission line for fault location, does not require communication and data synchronization equipment, has low operating costs and stable algorithms, and has been widely used in transmission lines. Single-ended fault location methods for transmission lines are mainly divided into traveling wave method and impedance method. The traveling wave method utilizes the transmission properties of the fault transient traveling wave to perform single-ended fault location. It has high precision and is not affected by the operation mode and transition resistance. However, it requires a high sampling rate, requires a special wave recording device, and has high application costs. . The impedance method uses the voltage and current after the fault to calculate the fault loop impedance, and performs single-ended fault location according to the characteristic that the line length is proportional to the impedance. It is simple and reliable, but the fault location accuracy is seriously affected by factors such as transition resistance and load current. , especially when the transition resistance is large, due to the influence of the additional fault distance caused by the transition resistance, the measurement result of the fault distance will seriously deviate from the real fault distance, and even the failure of fault distance measurement will occur.

Chinese patent: The application publication number is CN104090212A, and the invention patent published on October 8, 2014 discloses a line-phase fault location method without pre-fault voltage memory. The method first measures the two fault phases at the transmission line protection installation. The voltage mutation, the two fault phase current mutations and the normal phase voltage, use the normal phase voltage to calculate the fault phase-to-phase voltage when the transmission line is in normal operation, and then use the amplitude of the voltage mutation at the phase-to-phase short-circuit fault point after the line fault (that is, equal to the transmission line normal During operation, the fault phase-to-phase voltage amplitude) is the largest and monotonically decreases from the phase-to-phase short-circuit fault point to both ends of the transmission line, which realizes the single-end location of the line-phase short-circuit fault. Although this method does not need to memorize the voltage before the line fault, and the relay protection hardware does not need to set the voltage holding circuit before the line fault, it is only applicable to the phase-to-phase fault of the line, and the scope of application is narrow, and the fixed step is substituted into the comparison to solve the problem. If the step size is large, the calculation accuracy is not enough, and if the step size is small, the calculation time will be too long.

Contents of the invention

The purpose of the present invention is to overcome the problems of narrow application range, low calculation accuracy and efficiency in the prior art, and provide a fault distance measurement method based on fault point voltage mutation with wide application range and high calculation accuracy and efficiency.

For realizing above object, technical scheme of the present invention is as follows:

A method for fault distance measurement based on the sudden change in voltage at a fault point, comprising the following steps in sequence:

Step 1: The protection device samples the voltage and current waveform of the line transformer where the protection device is installed to obtain the instantaneous value of voltage and current;

Step 2: If it is a single-phase fault, use the Fourier algorithm to obtain the fault phase current at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Protect the phase voltage at the installation The phasor form of the zero-sequence current I ₀ in the case of a single-phase ground fault; if it is a phase-to-phase fault, the Fourier algorithm is used to obtain the fault phase-to-phase voltage at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Fault phase-to-phase current at the protective installation in, Indicates the three phases a, b, and c of the line;

Step 3: Calculate the fault point voltage U' according to formula 1:

U'＝U-IZ _l x Formula 1

In the above formula,

Z _l is the positive sequence impedance per kilometer, x is the fault distance from the protection installation to the short-circuit fault point;

If it is a single-phase fault, then k is zero sequence compensation coefficient;

In the case of a phase-to-phase fault, the

Step 4: First take the N-cycle sampling point before the fault and the M-cycle sampling point after the fault, and then use the method described in step 2 to obtain the fault phase/phase-to-phase voltage U ₁ at the protection installation before the fault, and the fault phase/phase voltage at the protection installation before the fault. Phase-to-phase current I ₁ , fault phase/phase-to-phase voltage U ₂ at the post-fault protection installation, fault phase/phase-to-phase current I ₂ at the post-fault protection installation, and then use the method described in step 3 to obtain the post-fault fault point voltage U' ₂ , the fault point voltage U' ₁ before the fault, and finally put the obtained U' ₁ and U' ₂ into Equation 2 to obtain the fault point voltage mutation value ΔU′:

ΔU'＝U' ₂ -U' ₁ Formula 2

Among them, M and M are both positive integers;

Step 5: Put U' obtained in Step 3 and ΔU' obtained in Step 4 into Equation 3, and combine Equation 4 to obtain the quadratic equation of the fault distance x from the protection installation to the short-circuit fault point, and then calculate its value :

$\frac{\mathrm{Re}({\mathrm{\Δ\; U}}^{\′}.C)}{\mathrm{Im}({\mathrm{\Δ\; U}}^{\′}.C)}=\frac{\mathrm{Re}\left(u^{\′}\right)}{\mathrm{Im}\left(u^{\′}\right)}$ Formula 3

C＝1.0∠(180°-(ArgZ _l +δ)) Formula 4

In the above formula,

Arg(Z _l ) is the positive sequence impedance angle of the line, δ is the angle at which the system impedance angle is greater than the positive sequence impedance angle of the line, Im(·) is the imaginary part of the variable, Re(·) is the real part of the variable, and C is the amplitude is a vector of 1s.

In the step 4, M is a positive integer greater than or equal to 2 and less than the number of continuous fault cycles, and N is 1 or 2.

In the step 5, if it is a single-phase fault, then δ is -8° to 0°; if it is an interphase fault, then δ is 0° to +4°.

Compared with prior art, the beneficial effect of the present invention is:

1, a kind of fault location method based on fault point voltage sudden change of the present invention utilizes the phase relation that exists between these two vectors of fault point voltage and fault point voltage sudden change, has listed to protect installation place to short-circuit fault point A quadratic equation with unknown distance to fault. Since this equation has nothing to do with load current and ground resistance, it is not affected by load current and high-resistance ground fault. The calculation accuracy and efficiency are high. Therefore, the present invention not only has a wide application range, but also has high calculation accuracy and efficiency.

Description of drawings

Fig. 1 is a schematic diagram of a line grounding short circuit fault according to the present invention.

In the figure, for the fault phase current at the protection installation Protect the phase voltage at the installation Fault point voltage U', fault point current I _f , fault distance x from the protection installation to the short-circuit fault point.

Detailed ways

The present invention will be described in further detail below in combination with specific embodiments.

A method for fault distance measurement based on the sudden change in voltage at a fault point, comprising the following steps in sequence:

Step 1: The protection device samples the voltage and current waveform of the line transformer where the protection device is installed to obtain the instantaneous value of voltage and current;

Step 2: If it is a single-phase fault, use the Fourier algorithm to obtain the fault phase current at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Protect the phase voltage at the installation The phasor form of the zero-sequence current I ₀ in the case of a single-phase ground fault; if it is a phase-to-phase fault, the Fourier algorithm is used to obtain the fault phase-to-phase voltage at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Fault phase-to-phase current at the protective installation in, Indicates the three phases a, b, and c of the line;

Step 3: Calculate the fault point voltage U' according to formula 1:

U'＝U-IZ _l x Formula 1

In the above formula,

Z _l is the positive sequence impedance per kilometer, x is the fault distance from the protection installation to the short-circuit fault point;

If it is a single-phase fault, then k is zero sequence compensation coefficient;

In the case of a phase-to-phase fault, the

Step 4: First take the N-cycle sampling point before the fault and the M-cycle sampling point after the fault, and then use the method described in step 2 to obtain the fault phase/phase-to-phase voltage U ₁ at the protection installation before the fault, and the fault phase/phase voltage at the protection installation before the fault. Phase-to-phase current I ₁ , fault phase/phase-to-phase voltage U ₂ at the post-fault protection installation, fault phase/phase-to-phase current I ₂ at the post-fault protection installation, and then use the method described in step 3 to obtain the post-fault fault point voltage U' ₂ , the fault point voltage U' ₁ before the fault, and finally put the obtained U' ₁ and U' ₂ into Equation 2 to obtain the fault point voltage mutation value ΔU′:

ΔU'＝U' ₂ -U' ₁ Formula 2

Among them, M and M are both positive integers;

Step 5: Put U' obtained in Step 3 and ΔU' obtained in Step 4 into Equation 3, and combine Equation 4 to obtain the quadratic equation of the fault distance x from the protection installation to the short-circuit fault point, and then calculate its value :

$\frac{\mathrm{Re}({\mathrm{\Δ\; U}}^{\′}.C)}{\mathrm{Im}({\mathrm{\Δ\; U}}^{\′}.C)}=\frac{\mathrm{Re}\left(u^{\′}\right)}{\mathrm{Im}\left(u^{\′}\right)}$ Formula 3

C＝1.0∠(180°-(ArgZ _l +δ)) Formula 4

In the above formula,

Arg(Z _l ) is the positive sequence impedance angle of the line, δ is the angle at which the system impedance angle is greater than the positive sequence impedance angle of the line, Im(·) is the imaginary part of the variable, Re(·) is the real part of the variable, and C is the amplitude is a vector of 1s.

In the step 4, M is a positive integer greater than or equal to 2 and less than the number of continuous fault cycles, and N is 1 or 2.

In the step 5, if it is a single-phase fault, then δ is -8° to 0°; if it is an interphase fault, then δ is 0° to +4°.

Principle of the present invention is described as follows:

The method of the invention is applied to high-voltage, ultra-high-voltage, and ultra-high-voltage lines of the power system, and based on the principle that the fault point voltage mutation amount and the fault point voltage are in phase, it can better eliminate the influence of load current and grounding resistance on grounding distance protection. The specific calculation process is as follows:

Since the voltage mutation at the fault point ΔU'=-I _f Z _Σ (I _f is the current at the fault point, Z _Σ is the total impedance seen from the fault point, that is, the equivalent impedance of the system), and both single-phase faults and phase-to-phase faults belong to resistance grounding, that is, the fault point current I _f is in the same phase as the fault point voltage U', so U' is ahead of ΔU'(180°-Arg(Z _Σ )) degrees, so that Arg(Z _Σ )=Arg(Z _l )+δ , where Arg(Z _l ) is the line positive sequence impedance angle, δ is the angle at which the system impedance angle is greater than the line positive sequence impedance angle;

Let C=1.0∠(180°-(ArgZ _l +δ)), then Arg(ΔU′.C)=Arg(U’), that is, the angles of the two phasors ΔU′.C and U’ are equal, so Their real part divided by the imaginary part is equal, and the following equation can be listed:

$\frac{\mathrm{Re}({\mathrm{\Δ\; U}}^{\′}.C)}{\mathrm{Im}({\mathrm{\Δ\; U}}^{\′}.C)}=\frac{\mathrm{Re}\left(u^{\′}\right)}{\mathrm{Im}\left(u^{\′}\right)}$ Formula 3

and

ΔU'=U' ₂ -U' ₁ ＝U ₂ -I ₂ Z _i x-(U ₁ -I ₁ Z _i x)=(U ₂ -U ₁ )-(I ₂ -I ₁ )Z _i x= ΔU-ΔIZ _l x,

U'＝U-IZ _l x

Among them, ΔU is the voltage mutation before and after the fault, and ΔI is the current mutation before and after the fault;

So formula 3 can be changed to:

$\frac{\mathrm{<span\; class="notranslate">\; Re</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; \&Delta;\; U</span>}<span\; class="notranslate">\; .</span>\mathrm{<span\; class="notranslate">\; C</span>}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; x\&Delta;IZ</span>}}_{\mathrm{<span\; class="notranslate">\; l</span>}}<span\; class="notranslate">\; .</span>\mathrm{<span\; class="notranslate">\; C</span>}<span\; class="notranslate">\; )</span>}{\mathrm{<span\; class="notranslate">\; Im</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; \&Delta;\; U</span>}<span\; class="notranslate">\; .</span>\mathrm{<span\; class="notranslate">\; C</span>}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; x\&Delta;IZ</span>}}_{\mathrm{<span\; class="notranslate">\; l</span>}}<span\; class="notranslate">\; .</span>\mathrm{<span\; class="notranslate">\; C</span>}<span\; class="notranslate">\; )</span>}<span\; class="notranslate">\; =</span>\frac{\mathrm{<span\; class="notranslate">\; Re</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; u</span>}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; wxya</span>}}_{\mathrm{<span\; class="notranslate">\; l</span>}}<span\; class="notranslate">\; )</span>}{\mathrm{<span\; class="notranslate">\; Im</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; u</span>}<span\; class="notranslate">\; -</span>{\mathrm{<span\; class="notranslate">\; wxya</span>}}_{\mathrm{<span\; class="notranslate">\; l</span>}}<span\; class="notranslate">\; )</span>}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; )</span>$

Multiply (1) left and right to get (2):

Re(ΔU.C)Im(U)-x.Re(ΔU.C)Im(IZ _l )-xRe(ΔIZ _l C)Im(U)+x ² Re(ΔIZ _l C)Im(IZ _l )＝

Im(ΔU.C)Re(U)-x.Im(ΔU.C)Re(IZ _l )-xIm(ΔIZ _l C)Re(U)+x ² Im(ΔIZ _l C)Re(IZ _l )

(2)

After rearranging the equations, we get (3):

x ² [Re(ΔIZ _l C)Im(IZ _l )-Im(ΔIZ _l C)Re(IZ _l )]+

x[Im(ΔU.C)Re(IZ _l )+Im(ΔIZ _l C)Re(U)-Re(ΔU.C)Im(IZ _l )-Re(ΔIZ _l C)Im(U)]+ ( 3)

Re(ΔU.C)Im(U)-Im(ΔU.C)Re(U)＝0

The above formula is a quadratic equation with x as the unknown. One of the reasonable solutions of this quadratic equation is the fault distance from the protection installation to the short-circuit fault point.

M: In order to avoid the decay time of the DC component, M should be greater than or equal to 2.

Example 1:

In the following, according to this method, fault location is carried out on a line B-phase ground fault (fault location 350 kilometers, grounding resistance 100 ohms), see Figure 1;

This method carries out the following steps in sequence:

Step 1: The protection device samples the voltage and current waveform of the line transformer where the protection device is installed to obtain the instantaneous value of voltage and current;

Step 2: Using the Fourier algorithm, calculate the fault phase current at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Protect the phase voltage at the installation The phasor form of the zero-sequence current I ₀ in case of a single-phase-to-ground fault, where, Indicates phase b of the line;

Step 3: Calculate the fault point voltage U' according to formula 1:

U'＝U-IZ _l x Formula 1

In the above formula,

k is the zero-sequence compensation coefficient, Z _l is the positive-sequence impedance per kilometer, and x is the fault distance from the protection installation to the short-circuit fault point;

Step 4: Take the sampling point of the 1 cycle before the fault and the sampling point of the 2 cycles after the fault, and then use the method described in step 2 to obtain the fault phase voltage U ₁ of the protection installation before the fault and the fault phase current I ₁ of the protection installation before the fault , the fault phase voltage U ₂ at the post-fault protection installation, the fault phase current I ₂ at the post-fault protection installation, and then use the method described in step 3 to obtain the post-fault fault point voltage U' ₂ and the pre-fault fault point voltage U' ₁ , and finally put the obtained U' ₁ and U' ₂ into formula 2 to obtain the voltage mutation value ΔU' at the fault point;

ΔU'＝U' ₂ -U' ₁ Formula 2

Step 5: Put the U' obtained in Step 3 and ΔU' obtained in Step 4 into Equation 3, and combine Equation 4 to obtain the quadratic equation of the fault distance x from the protection installation to the short-circuit fault point, and then calculate its value :

$\frac{\mathrm{Re}({\mathrm{\&Delta;\; U}}^{\&prime;}.C)}{\mathrm{Im}({\mathrm{\&Delta;\; U}}^{\&prime;}.C)}=\frac{\mathrm{Re}\left(u^{\&prime;}\right)}{\mathrm{Im}\left(u^{\&prime;}\right)}$ Formula 3

C＝1.0∠(180°-(ArgZ _l +δ)) Formula 4

In the above formula,

Arg(Z _l ) is the positive sequence impedance angle of the line, δ is the angle at which the system impedance angle is greater than the positive sequence impedance angle of the line, Im(·) is the imaginary part of the variable, Re(·) is the real part of the variable, and C is the amplitude is a vector of 1;

Since the system impedance angle is unknown, it is similar to the line impedance angle and slightly larger than the line impedance angle. In the case of a single-phase fault, there is zero-sequence current, and the line zero-sequence impedance angle is smaller than the line positive-sequence impedance angle. , it is desirable to take δ as -6°.

Solving x ₁ =-11, x ₂ =348. Take x=x ₂ =348, that is, the fault location is 348 kilometers away from the protection installation, which is very close to the actual fault location of 350 kilometers.

Example 2:

Carry out fault distance measurement to a line AB phase-to-ground fault (fault location 350 kilometers, grounding resistance 50 ohms) according to this method as follows;

This method carries out the following steps in sequence:

Step 1: The protection device samples the voltage and current waveform of the line transformer where the protection device is installed to obtain the instantaneous value of voltage and current;

Step 2: Using the Fourier algorithm, calculate the fault phase-to-phase voltage at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Fault phase-to-phase current at the protective installation in, Indicates the a and b phases of the line;

Step 3: Calculate the fault point voltage U' according to formula 1:

U'＝U-IZ _l x Formula 1

In the above formula,

Z _l is the positive sequence impedance per kilometer, x is the fault distance from the protection installation to the short-circuit fault point;

Step 4: First take the sampling points of the 1 cycle before the fault and the sampling points of the 2 cycles after the fault, and then use the method described in step 2 to obtain the fault phase-to-phase voltage U ₁ at the protection installation before the fault and the fault phase-to-phase current I ₁ at the protection installation before the fault , the fault phase-to-phase voltage U ₂ at the post-fault protection installation, the fault phase-to-phase current I ₂ at the post-fault protection installation, and then use the method described in step 3 to obtain the post-fault fault point voltage U' ₂ and the pre-fault fault point voltage U' ₁ , and finally put the obtained U' ₁ and U' ₂ into formula 2 to obtain the voltage mutation value ΔU' at the fault point;

ΔU'＝U' ₂ -U' ₁ Formula 2

Step 5: Put the U' obtained in Step 3 and ΔU' obtained in Step 4 into Equation 3, and combine Equation 4 to obtain the quadratic equation of the fault distance x from the protection installation to the short-circuit fault point, and then calculate its value :

$\frac{\mathrm{Re}({\mathrm{\&Delta;\; U}}^{\&prime;}.C)}{\mathrm{Im}({\mathrm{\&Delta;\; U}}^{\&prime;}.C)}=\frac{\mathrm{Re}\left(u^{\&prime;}\right)}{\mathrm{Im}\left(u^{\&prime;}\right)}$ Formula 3

C＝1.0∠(180°-(ArgZ _l +δ)) Formula 4

In the above formula,

Arg(Z _l ) is the positive sequence impedance angle of the line, δ is the angle at which the system impedance angle is greater than the positive sequence impedance angle of the line, Im(·) is the imaginary part of the variable, Re(·) is the real part of the variable, and C is the amplitude is a vector of 1;

In the phase-to-phase fault, since there is no zero-sequence current, it is desirable to take δ as +2° at this time.

Solving x ₁ =-19, x ₂ =348. Take x=x ₂ =349, that is, the fault location is 349 kilometers away from the protection installation, which is very close to the actual fault location of 350 kilometers.

The above examples show that the method of the present invention has high practicability and calculation accuracy.

Claims (3)

1. A fault location method based on fault point voltage mutation amount, is characterized in that: The method sequentially includes the following steps: Step 1: The protection device samples the voltage and current waveform of the line transformer where the protection device is installed to obtain the instantaneous value of voltage and current; Step 2: If it is a single-phase fault, use the Fourier algorithm to obtain the fault phase current at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Protect the phase voltage at the installation The phasor form of the zero-sequence current I ₀ in the case of a single-phase ground fault; if it is a phase-to-phase fault, the Fourier algorithm is used to obtain the fault phase-to-phase voltage at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Fault phase-to-phase current at the protective installation in, Indicates the three phases a, b, and c of the line; Step 3: Calculate the fault point voltage U' according to formula 1: U'＝U-IZ _l x Formula 1 In the above formula, Z _l is the positive sequence impedance per kilometer, x is the fault distance from the protection installation to the short-circuit fault point; If it is a single-phase fault, then k is zero sequence compensation coefficient; In the case of a phase-to-phase fault, the Step 4: First take the N-cycle sampling point before the fault and the M-cycle sampling point after the fault, and then use the method described in step 2 to obtain the fault phase/phase-to-phase voltage U ₁ at the protection installation before the fault, and the fault phase/phase voltage at the protection installation before the fault. Phase-to-phase current I ₁ , fault phase/phase-to-phase voltage U ₂ at the post-fault protection installation, fault phase/phase-to-phase current I ₂ at the post-fault protection installation, and then use the method described in step 3 to obtain the post-fault fault point voltage U' ₂ , the fault point voltage U' ₁ before the fault, and finally put the obtained U' ₁ and U' ₂ into Equation 2 to obtain the fault point voltage mutation value ΔU′: ΔU'＝U' ₂ -U' ₁ Formula 2 Among them, M and M are both positive integers; Step 5: Put the U' obtained in Step 3 and ΔU' obtained in Step 4 into Equation 3, and combine Equation 4 to obtain the quadratic equation of the fault distance x from the protection installation to the short-circuit fault point, and then calculate its value : $\frac{\mathrm{Re}({\mathrm{\&Delta;\; U}}^{\&prime;}.C)}{\mathrm{Im}({\mathrm{\&Delta;\; U}}^{\&prime;}.C)}=\frac{\mathrm{Re}\left(u^{\&prime;}\right)}{\mathrm{Im}\left(u^{\&prime;}\right)}$ Formula 3 C＝1.0∠(180°-(ArgZ _l +δ)) Formula 4 In the above formula, Arg(Z _l ) is the positive sequence impedance angle of the line, δ is the angle at which the system impedance angle is greater than the positive sequence impedance angle of the line, Im(·) is the imaginary part of the variable, Re(·) is the real part of the variable, and C is the amplitude is a vector of 1s. 2. a kind of fault location method based on fault point voltage mutation amount according to claim 1, is characterized in that: In the step 4, M is a positive integer greater than or equal to 2 and less than the number of continuous fault cycles, and N is 1 or 2. 3. according to claim 1 or 2 described a kind of fault location method based on fault point voltage mutation amount, it is characterized in that: In the step 5, if it is a single-phase fault, then δ is -8° to 0°; if it is an interphase fault, then δ is 0° to +4°.