Development of Six-phase Transmission line parameters
E. Afjei, B.Mazloomnezhad
Dept. Electrical & Computer Engineering
ShahidBeheshtiUniversity& SaadatResearchCenter
Tehran-Iran
Abstract: - In this paper, the electrical parameters namely, series reactance by two different methods and shunt admittance formulas are developed for steady state analysis of six-phase transmission lines. It then, compares the corresponding series inductance and the shunt admittance equations for Three-Phase and Six-phase lines. Finally, using a computer program compares the performance of a six-phase line with an equivalent three phase line and present the results.The results show that six-phase transmission line is an alternative to three-phase line and has an advantage of having smaller line clearances compared to equivalent three phase line.
Key- words: Six-phase transmission line, six-phase line parameters, line parameter
1 Introduction
The efficient use of transmission right-of-way has long been a problem for electric utilities. Transmission of large amounts ofpower over six-phase transmission lines, rather than standard three-phase lines, may offer some advantages that will allow more compact lines to be built, or existing double circuit lines to be upgraded without additional line clearances. Six-phase transmission lines may also be an alternative to ultra high voltage lines, which are considered to be objectionable by environmentalists[1-3]. In this paper calculations of electrical parameters for steady state analysis of transmission lines are considered. Formulas for the series reactance using two different methods namely the summation of the flux linkages due to all conductors and also the utilization of the Carson’s equation are developed. The equation for the shunt admittance isalso found. It then compares the performance of a six-phase line with an equivalent three-phase line. In six-phase systems each phase is 60º out of phase with each of its adjacent phases. Fig. 1 shows the phasor diagram for a set of six-phase voltages.
Fig.1 Six-phase phasor diagram
2 Flux Linkage and Inductance, Any Spacing
The equivalent inductance of each phase is the flux linkages of that phase divided bythe current going through it. This section will derive the equations for series inductance and reactance for a six phase transmission line with non-regular spacing as shown in fig. 2
6 o o 1
5 o o 2
4 o o 3
Fig. 2 General Configuration
In order to find the inductance of a Line with for any general configuration, full transposition of the conductors is assumed. Fig. 3 shows thefull transposition arrangement of conductors.
Fig. 3 Transposition cycle
The approach to finding the flux Linkage of one conductor is to find the Linkage of a conductor for each of its positions in the transposition cycle, and then average these flux Linkages. That is find avr :
(1) (2)
(3)
(4)(5) (6)(7)
Substituting in equation 1 and collecting terms;
(8)Since;,,and
Then;
(9)
Which, reduces to
(10)
or
(11)
Let:
(12)
Equation (11) becomes:
(13)
Series reactance XL of the Line is:
(14)
3 Bundle Conductors
The equations developed in the last section areto determine the series inductance of a transmission line with one conductor per phase. If bundled conductors are used, the effect is to alter the effective radius of the phase conductor. The DS term in Equation 13 becomes an equivalent radius, DSL . (15)
The distance used in determining Deq , are the distances measured from the center of the bundle. The equation for the equivalent DSL of a two conductor bundle is;
(16)
Where, GMR = Geometric Mean Radius of each conductor
daa = Bundle spacing
For more than 2 bundlein standard arrangements, the following equations are used to calculate DSL.
Three conductor bundle: (17)
Four conductor bundle :
(18)
4- Sequence impedances utilizing Carson’s equations
This approach, incorporate the application of the Carson’s equations to find the series impedance matrix of the line by considering current flow through the earth and then, obtaining the sequence impedances. The configuration of the circuits is shown in fig. 4.
Fig. 4 Six-phase line with earth return
Since all wires are grounded at the remote point a´,b´,c´,d´,e´, and f ´, one can recognize that
Ig = -( Ia + Ib + Ic + Id + Ie + If ) (19)
Now, the voltage drop equation in the direction of current flow can be written:
(20)
It can be seen from fig. 4 that
Va´-Vg´= 0 , Vb´-Vg´= 0 , Vc´-Vg´= 0 ,
Vd´ -Vg´= 0 , Ve´ - Vg´= 0 , Vf´ - Vg´= 0
Subtracting the first equation ,Vaa´ from the last equation , Vgg´ and also substitute for Ig results in:
(21)
Substituting for all the zs, one in general can write:
Zii = (ra+ rd)+j .0683 ln De/Ds (22)
Zij = rd+j .0683 ln De/Deq for (23)
Now the sequence impedance can easily be obtained by:
Zseq= [A-1][Zabcdef][A] (24)
where, A is a 6×6 transformation matrix.
5 Shunt Capacitanceof Six-Phase lines
Capacitance C is the ratio of change q on one conductor to the voltage v between that conductor and another conductor.
(25)
Considering the conductor configuration in Fig. 2 and also assuming complete transposition for each phase. Here Vac for each position is found and then the average is obtained.
Position 1:
(26)
Position 2:
(27)
position 3:
(28)
position 4:
(29)
position 5:
(30)
position 6:
(31)
Now taking the average of Vac for all six position and simplifying it knowing that
qa = -qd and qc = -qf
(32)
the same procedure is done for Vae for each position and then averaged, the result is
(33)
Since Vac + Vae =3Van and qc + qe = -qa then;
3Van=Vac+Vae = (34)
Now is obtained by , hence
F/m (35)
Let,
D´= (36)
And defining Deq as,
(37)
Then,
F/m (38)
It is worth mentioning that Deq is the same as found in obtaining series inductance.
6 Bundled Conductors
A similar method is used in calculating the effect of bundled conductors on capacitance as with calculating the effect on inductance. In both cases a new effective radius is found and used in the denominator of the natural logarithm (ln) argument. One significant difference is that the outside radius of a strand conductor is used instead of the GMR. The equation for shuntcapacitance of a bundled conductor line is:
F/m (39)
Where Dsc replaces r, for a two conductor bundle:
(40)
Three conductor bundle
(41)
four conductor bundle
(42)
Where, daa is the bundle spacing.
7Comparison of Three-phase and Six-phase Equations
The following is a comparison of corresponding series inductance equations for Three-Phase[4-5] and Six-phase system.
Three-Phase: H./m (43)
F/m (44)
Where, (45)
Six-phase:
H/m (46)
F/m (47)
where, (48)
8 Comparative study
In order to find impedances of three-phase and six-phase fully transposed lines by the two methods, the following data are used:
Conductors:ACSR795000 Circular mils
Length of the line: 100 miles
Line configuration: regular hexagon for six-phase and flat type for three-phase
Phase spacing: 10 feet between adjacent phase conductors (six-phase) and 13 feet, 13 feet, 26 feet (three-phase)
Earth resistivity, ρ:100 Ω-m
Using equation (44 ) for three-phase and (47) for six-phase yields
Za=12.88+j 61.46 Ω (three-phase)
Za=12.88+j 69.04 Ω (six-phase)
Using equations (22) and (23) three-phase and six-phase with different De results in:
Zii= 20.82 + j 113.42 Ω (three-phase)
Zij= 7.94 + j 51.948 Ω (three-phase)
Zii= 20.82 + j 113.42 Ω (six-phase)
Zij = 7.94 + j 44.38 Ω (six-phase)
Employing equation (24 ) results in;
Z0 = 36.68 + j 217.3 Ω (three-phase)
Z1=Z2= 12.86 + j 61.45 Ω (three-phase)
Z0 = 60.52 +j 335.3 Ω (six-phase)
Z1=Z2= Z3=Z4=Z5=12.88 + j69.02 Ω for (six-phase)
The sequence impedances (except zero sequence) calculated for fully transposed line has same values of (12.88 + j69.02) for both methods. The zero sequence has much higher values, almost 3 times larger than positive sequence for the three-phase line
(ie. 36.68 + j 217.3 ) and about five times higher for the six-phase line
(ie. 60.52 +j 335.3).
A Program is written to determine the relative performance of two different transmission lines, namely, a 230KV, three-phase line and a 138KV six-phase line. The 230KV line configuration is flat, with phase spacing of 10 feet, 10 feet, and 20 feet. The 138KV line configuration is a regular hexagon, with phase spacings of 10 feet between adjacent phases Conductor. The length of the line is 100 miles and the load is 150 MVA with a power factor of .8 lagging for the six-phase and 75MVA with the same power factor for the three-phase. The conductors are ACSR 795000 Circular mils. The comparison shows the six-phase line can carry twice as much power as the three-phase and the loss on the six-phase line is almost twice as the three-phase since it transfers twice as much power. The efficiencies of lines are about 98.6%.
9 Conclusion
This paper has shown that six-phase transmission line is an alternative to three-phase line. It is possible to convert an existing double circuit lines to six-phase operation. The main advantage of six-phase lines is that the line clearances can be smaller than for an equivalent three-phase. This is because the line to neutral voltage is increased, but the line to adjacent line voltage is not.
References:
1-W. C. Guyker, W. H. Booth, M. A. Jassen, S. S. Venkata, E. K. Stanek, N. B. Bhatt,” 38KV, six-phase transmission system Feasibility”, American power conference, Chicago, Illinois, April 25,1978
2-N. B. Bhatt , S. S. Venkata, W. C. Guyker, W. H. Booth,” six-phase power transmission systems: Fault analysis”, IEEE transaction on power apparatus and systems, vol. pas-96 no. 3, pp 758-767, may/june 1997.
3-W. C. Guyker, W. H. Booth, M. A. Jassen, S. S. Venkata, E. K. Stanek, N. B. Bhatt,” 38KV, six-phase transmission system Feasibility: A overview”, Presented at the Pennsylvania electrical association’s planning committee meeting, Valley Forge, Pennsylvania, may -, 1979.
4-P. Anderson, Analysis of faulted power systems, Iowa state university press, Iowa,1981
5-J. Duncan Glover, M. S. Sarma, Power system analysis and design, Brooks/Cole, California,2002