Development of Six-phase Transmission line parameters
E. Afjei, B.Mazloomnezhad
Dept. Electrical & Computer Engineering
Shahid Beheshti University & Saadat Research Center
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 ThreePhase and Six-phase lines. Finally, using a computer program compares the performance of a sixphase 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-ofway has long been a problem for electric
utilities. Transmission of large amounts of
power 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 is also found. It then
compares the performance of a six-phase line
with an equivalent three-phase line. In sixphase 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
by the 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
6o
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 the full transposition arrangement
of conductors.
Substituting in equation 1 and collecting
terms;
(8)

2 x 10 7 
1
1
 I b  I f ln
6 I a ln
6
D
D
D
D
S
12
23
34 D45 D65 D16

 I C  I e  ln

1
1
 I d ln

D13 D24 D35 D46 D15 D26
D14 D25 D36 D14 D25 D36 
Since; I b  I f  I a , I C  I e   I a ,and I d   I a
Then;
(9)

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 :
Where  avr  [  1  2  3  4  5   6 ] / 6 (1)

1
1
1
1
1  2 x 10 7  I a ln
 I b ln
 I C ln
 I d ln
DS
D12
D13
D14


1
1
I e ln
 I f ln

D15
D16 

1
1
1
1
 2  2 x 10 7  I a ln
 I b ln
 I C ln
 I d ln
D
D
D
D
S
23
24
25

(3)
(4)
(5)
1
1 
 I f ln

D24
D34 

1
1
1
1
 5  2 x 10  I a ln
 I b ln
 I C ln
 I d ln
DS
D56
D15
D25



1
1 
 I f ln

D46
D56 
6
2
2
D13 D24 D35 D46 D15 D26 D14 D25 D36
2
(10)
DS 6 D12 D23 D34 D45 D56 D16
 D13 D15 D24 D26 D35 D46 D14 2 D25 2 D36 2


D12 D23 D34 D45 D56 D16





1/ 6

 (11)


Let:
D D D D D D D 2 D 2 D 2
Deq   13 15 24 26 35 46 14 25 36
D12 D23 D34 D45 D56 D16





1/ 6
Equation (11) becomes:
Deq
La  2 x 10 7 ln
H / meter
DS
Series reactance XL of the Line is:
X 1  2 fL
(12)
(13)
(14)
3 Bundle Conductors
(6)
10
1 
 I f ln

D35
D45 
1
1
1
1
 I b ln
I C ln
 I d ln
DS
D16
D26
D36
 I e ln
  2 x10 7 I a ln
1
La  2 x 10 ln 
 DS

7
 6  2 x 10 7  I a ln
Which, reduces to
7
1
1 
 I e ln
 I f ln

D13
D23 

1
1
1
1
 4  2 x 10 7  I a ln
 I b ln
 I C ln
 I d ln
DS
D45
D46
D14

 I e ln

1
1
 I a ln

D13 D24 D35 D46 D15 D26
D14 D25 D36 D14 D25 D36 
(2)
7
 I e ln
 I a ln
or
1
1 
 I e ln
 I f ln

D26
D12 

1
1
1
1
 3  2 x 10  I a ln
 I b ln
 I C ln
 I d ln
DS
D34
D35
D36

2 x 10 7 
1
1
 I a ln
6 I a ln
6
DS
D12 D23 D34 D45 D56 D16

(7)
The equations developed in the last section
are to 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 .
Deq
La  2 x 10 7 ln
H / meter
(15)
DSL
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)
DSL  GMR  d aa
Where, GMR = Geometric Mean Radius of
each conductor
daa = Bundle spacing
For more than 2 bundle in standard
arrangements, the following equations are used
to calculate DSL.
Three conductor bundle:
DSL  3 GMR  d aa
Four conductor bundle :
2
DSL  1.09 4 GMR  d aa
3
(17)
(18)
Vaa '  Va  Va ' 
  

Vbb'   Vb  Vb '   zaa
  
 z
V '   Vc  V '   ba
c
 cc  
  zca
V  V  V  
 dd '    d d '    zda
  
  zea
Vee '   Ve  Ve '  
  
  z fa
V ff '  V f  V f '   z ga
  
 
Vgg '   V gVg ' 
  

zab
z bb
zcb
zdb
zeb
z fb
z gb
zac
zbc
zcc
zdc
zec
z fc
z gc
zad
zbd
zcd
zdd
zed
z fd
z gd
zae
zbe
zce
zde
zee
z fe
z ge
zaf
zbf
zcf
zdf
zef
z ff
z gf
zag   I a 
zbg   I b 
zcg   I c 
 
zdg   I d 
zeg   I e 
 
z fg   I f 
z gg   I g 
(20)
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.
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:
Va  ( z aa  2 z ag  z gg ) I a  ( z ab  z ag  z gb  z gg ) I b
( z ac  z ag  z gc  z gg ) I c  ( z ad  z ag  z gd  z gg ) I d
 ( z ae  z ag  z ge  z gg ) I e  ( z af  z ag  z gf  z gg ) I f
(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 i  j
(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 Capacitance of Six-Phase
lines
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:
Capacitance C is the ratio of change q on
one conductor to the voltage v between that
conductor and another conductor.
q
C
(25)
v
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:
D13
D23
D34
D35
D36 
1 
r
qa ln  qc ln  qb ln  qd ln  qe ln  q f ln 
2 
r
D13
D12
D14
D15
D16 
Vac 
(26)
Position 2:
D
D
D
D24
D 
1 
r
 qc ln
 qb ln 34  qd ln 45  qe ln 46  q f ln 14 
qa ln
2 
r
D24
D23
D25
D26
D12 
Vac 
(27)
position 3:
Vac 
D35
D
D
D
D 
1 
r
 qc ln
 qb ln 45  qd ln 56  qe ln 15  q f ln 25 
qa ln
2 
r
D35
D34
D36
D13
D23 
(28)
position 4:
Vac 
D46
D
D
D
D 
1 
r
 qc ln
 qb ln 56  qd ln 16  qe ln 26  q f ln 36 
qa ln
2 
r
D46
D45
D14
D24
D34 
(29)
position 5:
Vac 
D15
D16
D
D
D 
1 
r
 qd ln 12  qe ln 13  q f ln 14 
qa ln  qc ln  qb ln
2 
r
D15
D56
D25
D35
D45 
(30)
position 6:
Vac 
D26
D
D  (31)
D
D
1 
r
 qc ln
 qb ln 12  qd ln 23  qe ln 24  q f ln 25 
qa ln
2 
r
D26
D16
D36
D46
D56 
Now taking the average of Vac for all six
position and simplifying it knowing that
qa = -qd and qc = -qf
Vac 
1
1



6 
1 
1  D D D D D D D2 D2 D2  6
D16 D12 D23D34 D45D56

qa ln  13 24 35 46 15 26 14 25 36   qc ln r 
2 2
2  

2
r
D34 D45D56 D16 D12 D23

 D13D24 D35D46 D15D26 D14 D25D36  


(32)
the same procedure is done for Vae for each
position and then averaged, the result is
Vae 
1
1



6 
1 
1  D D D D D D D2 D2 D2  6
D12 D23D34 D45 D56 D16

qa ln  15 26 13 24 35 46 14 25 36   qe ln r 
2
2
2

2 
r
D45 D56 D16 D12 D23D34
D13D24 D35 D46 D15 D26 D14 D25 D34  




(33)
Since Vac + Vae =3Van and qc + qe = -qa then;
3Van=Vac+Vae
= q ln 1  D D D D D D D D D  (34)
a
r 3 
2
13
24
35
46
15
26
2
14
D34 D45 D56 D16 D12 D23
Now is obtained by C n 
2
25
2
36
1
2


qa
, hence
Van
F/m
1
1  D D D D D D D2 D2 D2 2
ln( 3 ) 13 24 35 46 15 26 14 25 36 
D34 D45 D56 D16 D12 D23
r 

(35)
Let,
1
 D D D D D D D2 D2 D2  2
D´=  13 24 35 46 15 26 14 25 36 
D34 D45 D56 D16 D12 D23


And defining Deq as,
1
6
( D' )  Deq
Then,
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 shunt capacitance of a
bundled conductor line is:
2
F/m
(39)
Cn 
Deq
ln
Dsc
Where Dsc replaces r, for a two conductor
bundle:
(40)
Dsc  rd aa
Three conductor bundle
2
Dsc  3 rd aa
four conductor bundle
3
Dsc  1.094 rd aa
Where, daa is the bundle spacing.
(41)
(42)
7 Comparison of Three-phase and
Six-phase Equations
6
Cn 
2
F/m
(38)
Deq
ln
r
It is worth mentioning that Deq is the same as
found in obtaining series inductance.
Cn 
(36)
(37)
The following is a comparison of
corresponding series inductance equations for
Three-Phase[4-5] and Six-phase system.
Three-Phase:
Deq
La  2  10 7 ln
H./m (43)
DS
2
F/m
(44)
Cn 
Deq
ln
r
Deq  3 D12 D13 D23
Where,
(45)
Six-phase:
La  2  10 7 ln
2
Deq
ln
r
where,
Deq
DS
Cn 
H/m (46)
F/m
D D D D D D D 2 D 2 D 2
Deq   13 15 24 26 35 46 14 25 36
D12 D23 D34 D45 D56 D16

(47)




1/ 6
(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 sixphase 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, threephase 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