Analysis on the fault characteristics of three-phase short-circuit for half-wavelength AC transmission lines

1 Introduction

Half-wavelength alternating current transmission(HWACT) is an ultra-long distance three-phase AC power transmission technology with an electrical distance near to half-wavelength at the system power frequency. In 50Hz system, the distance is 3000km, while it is 2500km in 60Hz system. The advantages of HWACT are infinite theoretical maximum transmission capacity and no additional reactive power compensation. Hence it is suitable to serve as the long-distance backbone frame in the future Global Energy Interconnection.

The concept of HWACT is initially proposed by the experts from Soviet Union in 1940s [1]. After some following exploratory research in this field [2-4], the principal transmission characteristics and some correlative key technical problems, such as corona loss and overvoltage [5, 6], have been solved out. But the research on HWACT has not been continuously and deeply carried out due to the limited demand and lower technical level at that time.

In the new century, the ultra-long distance, large capacity transmission has attracted global attention because of the serious energy shortage problem. Many researchers in different countries focus on HWACT once again. In Brazil, HWACT is used as the backup scheme to transmit large scale hydropower from Amazon River basin to the load centers [7, 8]. And in China, the scheme of HWACT has also been proposed to transmit the coal and hydro energy to the east coast load centers [9-16]. In particular,the State Grid Corporation of China proposed the idea of building Global Energy Interconnection in July 2014. The UHV power grid serves as the backbone to mainly deliver clean energy and achieve a globally interconnected smart grid. Long distance UHV transmission lines are needed for the long distance power transmission or exchange.Therefore, the study on HWACT to satisfy the requirement of large capacity long distance point-to-point power transmission has broader significance and practical value.

The fault characteristics of an HWACT line can be different from those of a conventional transmission line.It is very important for the construction and operation of HWACT to analyze its fault features and corresponding protection technology. [13-15] all deal with the analysis of fault for HWACT line, but the studies are limited to overvoltage along the line. Reference [12] gives a simple analysis of the fault characteristics for HWACT line, but not further.

When a symmetrical short-circuit fault occurs, the three-phase system can be regarded as a single-phase system in order to simplify the fault analysis. The symmetrical fault analysis is the basis of the asymmetrical fault analysis. Therefore, the steady-state voltage and current characteristics of the bus bar and fault point and the steady-state overvoltage distribution along the line will be analyzed in this paper when a three-phase symmetrical short-circuit fault occurs on an HWACT line.

2 HWACT system model

The steady-state voltage and current characteristics will be analyzed when the symmetrical faults occur on an HWACT line in the cases of different fault distances.

Fig. 1 Model of an HWACT system

The analysis of fault characteristics of an HWACT line will be done by using the HWACT system model shown in Fig. 1. The line mn is a 1000kV HWACT line, and the rest parts of the system are equivalent to two voltage sources with impedances. The phase angle difference between two equivalent voltage sources of the bus m and bus n is larger than 180°. The power frequency is 50Hz. The line parameters in sequence quantities are shown as Table 1.

Table 1 Parameters of HWACT Line

Parameters Positive sequence Zero sequence Resistance(Ω/km) 0.00805 0.205 Inductance (mH/km) 0.825 2.375 Capacitance(nF/km) 14.0 9.03

According to the line parameters, the length of line mn can be calculated as follows, where λ is the wavelength with the power frequency f=50Hz. L and C are the line positive sequence inductance and capacitance per unit length respectively.

When a three-phase short-circuit fault occurs at the point f on the HWACT line, the fault characteristics will be analyzed without considering the fault resistance.

3 Fault characteristics of bus bar

Only single phase should be considered for a symmetrical fault. And when the fault resistance is 0, the right part of the fault point has no influence on the voltage and current of bus m. According to the hyperbolic function solution of the uniform transmission line equations, the voltage and current of the bus m and bus n are related to the voltage and current of the fault point f,

where the parameters Zc and γ are the surge impedance and propagation constant of the HWACT line. x is the distance from the fault point f to the bus m. and are the voltage and current of the bus m respectively.represents the voltage of the fault point f and its value is 0. is the current which flows into the fault point f from the bus m.

The boundary condition at the bus m is as follows,

whereis the equivalent voltage source out of the bus m and Zm is the equivalent impedance.

Analyzing (2) (3), the voltage of the bus m can be obtained as follows.

For simplicity, the resistances in the system will be neglected. Therefore,

where Lm is the inductance in Zm. Evidently, k is real.

The value of k will be discussed while the fault distance x changes from 0 to λ/2. With x increasing, the cotangent decreases from +∞ to -∞ and k increases. When x=0,the cotangent equals to +∞ and k=0. When x=λ/4, the cotangent equals to 0 and k=1. When the denominator in(5) decreases to 0, an extreme point appears and here k increases to +∞. And after that, k increases from -∞. When x=λ, the cotangent equals to -∞ and k=0. The extreme point is:

The extreme point exists in the interval (λ/4, λ/2)and it is related to the line parameters and the equivalent impedance out of the analyzed bus.

Fig. 2 Voltage RMS of bus m calculated with the change of the fault distance

The voltage RMS (Root Mean Square) of bus m is calculated with the change of the fault distance mf, as shown in Fig. 2. And it can be also regarded as |k| with the change of x. From Fig. 2, we can see that the voltage RMS of bus m increases from 0 to +∞ (extreme point) and then decreases to 0 with the fault distance mf increasing.

For the current of the bus m, according to (3),

Evidently, k’ is pure imaginary. When x=λ/4, k’=0. And the extreme point of k’ is the same as that of k.

The current RMS of bus m is calculated with the change of the fault distance mf, as shown in Fig. 3. And it can be also regarded as |k’| with the change of x. From Fig. 3, we can see that the current RMS of bus m firstly decreases with the fault distance mf increasing. It decreases to 0 at the middle point of the line. After that, it increases to +∞ (extreme point) and then decreases.

In the cases of longer transmission lines, when the symmetrical faults occur, there is a half-wavelength periodic extension phenomenon for the voltage and current RMS of bus m with the change of the fault distance mf. The extreme points are:

Fig. 3 Current RMS of bus m calculated with the change of the fault distance

When a symmetrical fault occurs on a full-wavelength line, the voltage RMS of bus m is calculated with the change of the fault distance mf, as shown in Fig. 4.We can see that the voltage RMS of bus m displays a periodic change and the change period is the halfwavelength.

Fig. 4 Voltage RMS of bus m calculated with the change of the fault distance when a symmetrical fault occurs on a fullwavelength line

4 Fault characteristics of fault point

The current of the fault point is the sum of and,which are respectively the currents that flow into the fault point f from the bus m and bus n. That is

where Am and An are the impedance coefficients.

The currents that flow into the fault point f from the bus m can be obtained according to (2) (3),

If the resistances in the system are neglected,

In the same way,

When either of the sines in Am and An is 0, the current of the fault point will be infinite. The extreme points are:

The two extreme points exist in the interval (0, λ/4)and interval (λ/4, λ/2) respectively. The extreme point in the interval (0, λ/4) is related to the line parameters and the equivalent impedance of the bus n. And the one in the interval (λ/4, λ/2) is related to the line parameters and the equivalent impedance of the bus m.

Fig. 5 Current RMS of fault point calculated with the change of the fault distance

The current RMS of fault point is calculated with the change of the fault distance mf, as shown in Fig. 5. We can see that the current RMS of fault point firstly increases to infinite (first extreme point) and then decreases (not to 0) with the fault distance mf increasing. Subsequently, it increases to infinite again (second extreme point) and then decreases. In particular, if the equivalent impedances of the bus m and bus n are equal, the current RMS of fault point will be symmetric with respect to λ/4.

In the cases of longer transmission lines, when the symmetrical faults occur, the impedance coefficient Am remains the same while An changes. If the line length is l,

Here, the extreme points are:

Fig. 6 Current RMS of fault point calculated with the change of the fault distance when a symmetrical fault occurs on a line with the length of 1.25λ

When a symmetrical fault occurs on a line with the length of 1.25λ, the current RMS of fault point is calculated with the change of the fault distance mf, as shown in Fig. 6.We can see that the current RMS of fault point displays a periodic change and the change period is the halfwavelength.

5 Overvoltage distribution along the line

According to (2), when a three-phase short-circuit fault occurs on the HWACT line at point f, the overvoltage distribution along the line is as follows,

where x is the distance from the fault point f to the bus m. represents the voltage of the point from which the distance is y to the bus m. If the resistances in the system are neglected,

If the fault point is determined, andare constant values. By setting the fault point as a reference point, the voltage RMS along the line at bus m side varies as sine with the distance between fault point and measuring point increasing. The peak value of the sine is related toSimilarly, the voltage RMS along the line at bus n side also varies as sine and the peak value is related to.

Fig. 7 Voltage RMS along the line calculated for different fault distances

The voltage RMS along the line is calculated for different fault distances, as shown in Fig. 7. Here, the curves of different colors mean different distances between fault point and bus m.

In the cases of longer transmission lines, when a symmetrical fault occurs, by setting the fault point as a reference point, the voltage RMS along the line at bus m side varies as sinusoidal positive half cycle periodically with the distance between fault point and measuring point increasing. So does the voltage RMS along the line at bus n side. However, if the equivalent impedances of the bus m and bus n are not equal, the sinusoidal peaks of the two sides are different.

When a symmetrical fault occurs on a full-wavelength line, the voltage RMS along the line is calculated for different fault distances, as shown in Fig. 8. It shows the same result to the analysis above.

Fig. 8 Voltage RMS along the line calculated for different fault distances when a symmetrical fault occurs on a fullwavelength line

6 Fault characteristics considering fault resistance

When there is a fault resistance, the voltage at the fault point f is no longer 0,

After ignoring the resistors in the equivalent impedances, Am and An are sinusoidal pure imaginary.

Fig. 9 Voltage RMS of fault point calculated for different fault distances

The voltage RMS of fault point is calculated for different fault distances under different fault resistances, as shown in Fig. 9. The curves of different colors in the figure represent different fault resistances. It can be seen from the figure that when the fault resistance is 0, the voltage of fault point is 0. And when the fault resistance is infinite,the voltage curve of fault point is the same with the voltage distribution curve along the line. When the fault resistance exists, the voltage of fault point is fixed at two pointsAnd it will not be influenced by the fault resistance.

7 Conclusions

The fault characteristics of the HWACT line are quite different from those of the conventional transmission line. In this paper, the steady-state voltage and current characteristics of the bus bar and fault point and the steadystate overvoltage distribution along the line are analyzed when a three-phase symmetrical short-circuit fault occurs on an HWACT line. The conclusions are summarized as follows.

1) When a three-phase symmetrical short-circuit fault occurs in the HWACT line, if there is no fault resistance,the steady-state voltage and current RMS at the bus have an extreme point between the line middle point and the opposite bus bar. If the resistances in the system are neglected, the extreme value is infinite. And the current at the middle point is zero.

2) The steady-state current RMS at the fault point has two extreme points, which locate at the two sides of the line middle point. When the equivalent impedances on both sides are equal, the variation of the RMS current at the fault point is symmetrical about the line middle point.

3) By setting the fault point as a reference point, the voltage RMS along the line varies as sine. The peak values of the sine on both sides are related to the corresponding equivalent impedances.

4) For longer distance transmission system, the threephase short-circuit fault characteristics will display a periodic change and the change period is the half-wavelength.

Acknowledgements

This work was supported by National Key Research and Development Program of China (2016YFB0900100).