Neutral & Dirt Current Computation in Distribution Systems

The Computation of Neutral and Dirt
Currents and Power Losses
W. H. Kersting, Life Fellow IEEE
Abstract - Line segments in distribution systems are
inherently unsymmetrical. That is, the spacing distances
between phases are not equal and the lines are not
transposed; as is typically done on high voltage
transmission lines. The non-symmetry results in unequal
self and mutual impedances. Couple this with line
currents that are typically unbalanced and a potential for
severe voltage unbalances can occur along with
additional line power losses. For a four wire grounded
wye line, the unbalanced operation leads to currents
flowing in the neutral and dirt (ground). This paper will
develop a method for the computation of the neutral and
dirt currents and the power losses that are a result of
these currents.
Index Terms – Carson’s Equations, distribution
systems, dirt current, line segment impedance, neutral
current, power loss.
I. Introduction
Distribution feeders are inherently unbalanced.
That is loads are generally unbalanced and lines are
not transposed.
The analysis of the operating
conditions for a distribution feeder must model the line
segments as accurately has possible. The application
of the modified Carson’s equations allows for a very
accurate model of the line segment and does provide a
means for computing the current flowing in the neutral
conductor and “dirt”. The term “dirt” is used in this
paper to distinguish it from “ground” current which
usually is the sum of the phase currents without any
regard to the return path for the “ground” current.
II. Line Segment Model
A. The Modified Carson’s Equations
The analysis of the unsymmetrical line with
unbalanced currents starts with a model of the line
where the impedances have been calculated using the
modified Carson’s equations. In order to apply these
equations, it is important to understand how they are
developed.
The development starts with the application of the
general flux linkage equation [1], [2]. This application
leads to the general equations for the self and mutual
impedances of conductors.
zii = ri + j 0.12134 ⋅ ln
zij = j 0.12134 ⋅ ln
1
GMRi
Ω/mile
(1)
1
Ω/mile
Dij
(2)
where:
zii = self impedance of conductor i
zij = mutual impedance between i and j
ri = resistance of the conductor I
GMRi = Geometric Means Radius of the conductor
Dij = spacing between conductors i and j ft.
These equations can be applied to a one mile long
line consisting of two conductors that are grounded at
the remote end as shown in Figure 1.
z ii
+
Vig
Ii
+
Vjg
-
z jj
z ij
Ij
ground
-
I d z dd
z jd
z id
Fig. 1 Two Conductor Line
Figure 1 shows the currents flowing in the lines (Ii
and Ij) and the current flowing in dirt (Id). Using these
currents Kirchhoff’s Voltage Law (KVL) is used to
write the equation for the voltage between conductor i
and ground.
Vig = z ii ⋅ Ii + zij ⋅ I j + zid ⋅ I d
−( z dd ⋅ I d + z di ⋅ Ii + z dj ⋅ I j )
Collect terms in Equation 3
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(3)
(
)
(
)
Vig = z ii − z dn ⋅ Ii + z ij − z dj ⋅ I j
(
)
$z
+ z id − z dd ⋅ I d
(4)
A key to understanding how the dirt current can be
computed later is to understand how Kirchhoff’s
Current Law (KCL) is now applied to this simple
circuit. Apply KCL at the remote end of the lines
where they are connected to ground.
Ii + I j + I d = 0
(5)
I d = − Ii − I j
Note here that the current flowing in dirt must be
equal to the negative sum of the actual line currents.
Substitute Equations 1 and 2 into Equations 3 and 4
and collect terms:
Vig = ( z ii + z dd − z di − zid ) ⋅ Ii
+( zij + z dd − z dj − z id ) ⋅ I j
(6)
Equation 6 is of the general form:
Vig = z$ ii ⋅ Ii + $zij ⋅ I j
(7)
⎛
ji = rd + j 0.12134 ⎜ ln
⎜
⎝
Ddj ⋅ Did
1
+ ln
D ji
GMRd
(8)
Substitute Equations 1 and 2 into Equations 8 and
after some algebra manipulations results in:
⎛
z$ ii = rd + ri + j 0.12134 ⋅ ⎜ ln
⎜
⎝
Did ⋅ Ddj
1
+ ln
GMRi
GMRd
⎞
⎟
⎟
⎠
(9)
$z
⎛
ij = rd + j 0.12134 ⎜ ln
⎜
⎝
Ddj ⋅ Did
1
+ ln
Dij
GMRd
⎞
⎟
⎟
⎠
(10)
A similar procedure is followed for conductor j
which will lead to:
V jg = z$ ji ⋅ Ii + z$ jj ⋅ I j
(11)
where:
⎛
z$ jj = rd + r j + j 0.12134 ⋅ ⎜ ln
⎜
⎝
⎛
$
ii = zii + z dd − z di − zid
$z
ij = zij + z dd − z dj − zid
Did ⋅ Ddj
1
+ ln
GMR j
GMRd
⎞
⎟
⎟
⎠
(12)
(13)
Equations 9,10,12 and 13 are interesting in that
they give a method for computing the self and mutual
impedances of conductors taking into account the
effect of the current flowing in dirt. Unfortunately
these four equations require knowledge of the
Geometric Mean Radius (GMR) of dirt and the
distance from the conductors to dirt. This presents a
problem since the GMR of dirt is not found in tables
and the distance from a conductor to the path of the
current flowing in dirt is unknown.
Fortunately John Carson in 1926 wrote a classic
paper [3] that helps define the unknowns in the
equations. In his paper Carson postulates currents
injected into conductors that are tied together and
grounded at the end of the lines (Figure 1). Carson
then develops two equations that will determine the
self and mutual impedances of the conductors taking
into account the dirt return path. His equations are
quite long and difficult to evaluate. However, it is
possible to develop what will be referred to as the
“modified Carson’s Equations”. The development of
the modified equations can be found in Reference [1].
The end result of the modified Carson’s equations is:
where:
$z
⎞
⎟
⎟
⎠
zii = ri + 0.09530 + j 0.12134 ⎜ ln
⎝
$z
⎛
ij = 0.09530 + j 0.12134 ⎜ ln
⎜
⎝
⎞
1
+ 7.93402 ⎟
GMRi
⎠
(14)
⎞
1
+ 7.93402 ⎟
⎟
Dij
⎠
(15)
Note that Equations 14 and 15 are in exactly the
same form as Equations 10 and 11. In fact, a
comparison of the two sets of equations gives
numerical values for the unknowns. Equation 14 gives
the self impedance of the phase conductor and
Equation 15 gives the mutual impedance between
conductors i and j. In these two equations the
equivalent resistance of dirt is shown to be 0.0953
ohms/mile and the equivalent mutual inductive
reactance between a conductor and dirt is given by the
7.93402 term in the two equations. Recall that the first
step in the development of these equations was to
recognize that the dirt current has to be the negative
sum of the conductor currents (Equation 5).
With all of the terms for the computation of self
and mutual impedance known, a new equivalent circuit
of the line segment can be developed. Removing the
short circuit to ground in Figure 1 and applying
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Equations 7 and 11 results in the equivalent
“primitive” circuit of Figure 2.
z ii
+
Ii
Vig
z jj
+
Ij
Vjg
-
-
+
z ij
+
V'ig
-
-
V'jg
ground
It is important to understand that the impedance of
dirt has now been “folded” into the values of the
“primitive” impedances in Figure 2 and that the
negative sum of the conductor currents will be equal to
the current flowing in dirt as shown in Figure 1.
B. Phase Impedance Matrix for Overhead Lines
A four wire grounded wye three-phase line is
shown in figure 3.
zaa
+
Ia
+
Ib
Vbg
+
Ic
z nn
Vcg
+
-
-
V'ag
V'bg
+
z cn
V'cg
V'ng
-
-
-
=
⎡V
⎢
⎢V
⎢
⎢V
⎢
V
⎣
⎡ z$
aa
⎢
⎢ z$ ba
⎥+⎢
$
⎥ ⎢ z ca
⎥ ⎢$
⎦ ⎢
⎣ z na
z$ ab
z$ ac
z$ an ⎤
$z
bb
$z
bc
z$ bn ⎥
'cg
'ng
z$
cb
z$
cc
z$ nc
⎡Ia ⎤
⎢ ⎥
I
⎢ b⎥
⋅
⎥
$z
⎢I ⎥
cn ⎥ ⎢ c ⎥
⎥
I
z$ nn ⎥⎦ ⎣ n ⎦
⎥
(16)
Partition between the third and fourth rows and
columns. In partitioned form Equation 16 becomes:
[
]
⎡ Vabc ⎤
⎢
⎥
⎢ ⎡Vng ⎤ ⎥
⎦⎦
⎣⎣
[V 'abc ]⎤
⎡
=⎢
⎢ ⎡V
⎣⎣
⎡ ⎡ $z ⎤
⎣ ij ⎦
⎥+⎢
'ng ⎤⎦ ⎥ ⎢ ⎡ $z nj ⎤
⎦ ⎢⎣ ⎣
⎦
−1
⋅ ⎡ $z nj ⎤ ⋅ [ I abc ]
⎣
(20)
⎦
Before going on note that when the line currents are
known, then the current flowing in the neutral
conductor can be determined using Equation 20.
Because of this, the “neutral current transformation
vector” is defined as:
[ znt ] = − ⎡⎣ z$ nn ⎤⎦
−1
⋅ ⎡ $z nj ⎤
⎣
(21)
⎦
(22)
The primitive impedance matrix can be reduced to
the phase impedance matrix by first substituting
Equation 20 into Equation. 18:
'ag ⎤
⎥
'bg ⎥
z$ nb
[ I n ] = − ⎡⎣ z$ nn ⎤⎦
-
In Figure 3 all of the self and mutual impedances
can be calculated using the modified Carson’s
equations (Eqs. 14 and 15). The general voltage
equations in matrix form for this line are determined
by applying KVL and results in:
⎤
⎥
⎥
⎥
⎥
⎥
⎦
(19)
I n = ⎣⎡ znt ⎦⎤ ⋅ ⎣⎡ I abc ⎦⎤
+
Fig. 3 Grounded Wye Three-Phase Line
⎡Vag
⎢
⎢Vbg
⎢
⎢Vcg
⎢
V
⎣ ng
[0] = [0] + ⎡⎣ $z nj ⎤⎦ ⋅ [ I abc ] + ⎡⎣ z$ nn ⎤⎦ ⋅ [ I n ]
When the phase currents have been computed, the
current flowing in the neutral (in the same direction as
the phase currents) is given by:
+
zbn z an
In
Vng
-
-
zbc
z cc
(18)
+
z ab z ac
z bb
[Vabc ] = [V 'abc ] + ⎡⎣ z$ ij ⎤⎦ ⋅ [ I abc ] + ⎡⎣ $zin ⎤⎦ ⋅ [ I n ]
Solve Equation 19 for [In]:
Fig. 2 Equivalent Primitive Circuit
Vag
Because the neutral is grounded the voltages Vng
and V’ng are equal to zero. Substituting those values
into Equation 17 and expanding results in:
[Vabc ] = [V 'abc ]
⎛
+ ⎜ ⎡ $z abc ⎤ − ⎡ $z in ⎤ ⋅ ⎡ z$ nn ⎤
⎝
⎣
⎦
⎣
⎦ ⎣
⎦
−1
⎞
⋅ ⎡ $z nj ⎤ ⎟ ⋅ [ I abc ]
⎣
⎦
⎠
(23)
(24)
[Vabc ] = [V 'abc ] + [ zabc ] ⋅ [ I abc ]
where:
⎡⎣ zabc ⎤⎦
= ⎡ $zij ⎤ − ⎡ $zin ⎤ ⋅ ⎡ $z nn ⎤
⎣
⎦
⎣
⎦ ⎣
⎦
−1
⋅ ⎡ $z nj ⎤ (25)
⎣
⎦
This process of eliminating the row and column of the
primitive impedance matrix is known as the “Kron
Reduction [4] and results in Figure 4.
⎡ $z in ⎤ ⎤
⎣
⎦⎥ ⎡
⋅⎢
$
⎡ z nn ⎤ ⎥ ⎣
⎣
⎦ ⎥⎦
[ I abc ]⎤
[ I n ] ⎥⎦
(17)
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5.
Node n
Ia
Zaa
Ib
Zbb Zab Zca
Ic
Zcc Zbc
+
Vag
n
+
Vbg
-
-
Fig. 4
n
+
Vcgn
Node m
+
Vagm
+
Vbg
m
+
Vcg
- m -
-
=
n
⎡V
ag ⎤
⎢
⎥
⎢Vbg ⎥
⎢
⎥
Vcg ⎥
⎢
⎣
⎦
m
⎡ Z aa
⎢
+ ⎢ Zba
⎢ Z ca
⎣
Z ab
Zbb
Z cb
Z ac ⎤ ⎡ I a ⎤
⎥ ⎢
⎥
Zbc ⎥ ⋅ ⎢ Ib ⎥
Zcc ⎥⎦ ⎢⎣ Ic ⎥⎦
(27)
-
The real power loss of a line segment that has been
represented by the phase impedance matrix must be
computed for each phase as:
Equivalent Three-Phase Line
⎤
⎥
⎥
⎥
⎥
⎦
I d = − ( I a + Ib + Ic + I n )
D. Computation of Power Loss
Writing KVL for the circuit of Figure 4 gives:
⎡V
ag
⎢
⎢Vbg
⎢
V
⎢
⎣ cg
The current flowing in ground is computing
by:
Plossa,b,c = Pina,b, c − Pouta,b, c
(26)
where: Zij = zij ⋅ length
The impedance matrix of Equation 26 is defined as
the “phase impedance matrix”. The process of
developing the phase impedance matrix for the four
wire grounded wye line can be extended to both twophase and single-phase lines. In the two phase line
applying the modified Carson’s equations will result in
a 3x3 matrix. The Kron reduction will reduce this to a
2x2 matrix. The phase impedance matrix will still be a
3x3 matrix with the row and column corresponding to
the missing phase set to zero and the elements of the
Kron reduced matrix filling in the remaining elements.
For a single-phase line modified Carson’s equations
gives a 2x2 matrix and Kron reduction reduces this to a
single element. The phase impedance matrix then has
the rows and columns of the missing phases set to zero
and the single element will be the diagonal term of the
phase present.
C.
Computation of neutral and dirt current
calculations
The steps to take in computing the neutral and dirt
currents are as follows:
1. Compute the “primitive impedance” matrix
using the modified Carson’s equations
2. Kron reduce the “primitive impedance” matrix
down to the “phase impedance matrix”. In this
process
define
the
“neutral
current
transformation matrix” of Equation 21.
3. For a given load condition determine the phase
currents flowing in the line. This will usually
require a three-phase power flow program
4. With the line currents known, apply Equation
22 to compute the neutral current
(28)
It is not unusual to have one of the phase power
losses to be negative in a line carrying unbalanced
currents. Typically the lightly loaded phase will
display a negative power loss. Even with what appears
to be a problem, the total three-phase power loss is
correctly computed as the sum of the three individual
phase power losses even if one of them is negative.
The total three-phase power loss of the line is then
given by:
Plosstotal = Plossa + Plossb + Plossc
(29)
By computing the power losses in this way, the
effect of neutral and dirt power losses are included.
However, it is critical that it is understood that the
power loss per phase can not be computed as the phase
resistance times the current squared. That process will
give only the power loss in the phase conductors and
will ignore the power loss in the neutral and dirt.
In order to determine the power losses in the
neutral and dirt, the neutral and dirt currents must be
computed. When that is done the appropriate power
losses are computed as:
Plossneutral = Rneutral ⋅ I n
2
Plossdirt = 0.0953 ⋅ length ⋅ I d
2
(30)
When this has been done, now the power losses in
the individual phases can be computed by:
Pxlossa = Ra ⋅ I a
2
Pxlossb = Rb ⋅ Ib
2
Pxlossc = Rc ⋅ Ic
2
(31)
The total three phase power loss using this
approach is given by:
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Plosstotal = Plossa + Plossb + Plossc
+ Plossn + Plossd
(32)
The total power loss computed by Equation 32 will
be exactly the same as that given by Equations 28 and
29.
III. Example
Consider a three-phase overhead distribution line to
be constructed as shown in Figure 5.
2.5'
a
Assume that the line is three-miles long and
operating at 12.47 kV and serves an unbalanced threephase grounded wye connected load of:
Phase-a = 1500 kVA at 0.88 lagging power factor
Phase-b = 1000 kVA at 0.95 lagging power factor
Phase-c = 2000 kVA at 0.8 lagging power factor
In order to avoid an iterative solution, the loads are
treated as constant impedance computed assuming
nominal voltage.
The load impedance matrix
becomes:
4.5'
c
b
⎡⎣ Z
3.0'
load
⎤⎦
0
⎡30.409 + j16.413
⎢
0
49.242 + j16.185
=⎢
⎢
⎣
0
0
0
0
20.733 +
⎤
⎥
⎥
j15.550 ⎦⎥
4.0'
Kirchhoff’s voltage law for the circuit is:
⎡⎣VLGabc ⎤⎦
source
n
=
( ⎡⎣ Z abc ⎤⎦ + ⎡⎣ Zload ⎤⎦ ) ⋅ ⎡⎣ Iabc ⎤⎦
25.0'
where:
⎣⎡VLN abc ⎦⎤ source
=
⎡ 7199.6/ 0 ⎤
⎢
7199.6/ − 120 ⎥⎥
⎢
⎢⎣ 7199.6/120 ⎥⎦
V
Solving for the line currents gives:
Fig. 5 Three-Phase Distribution Line Conductor
Spacings
For this line the phase conductors are 556,550 26/7
ACSR and the neutral conductor is 4/0 ACSR.
Applying the modified Carson’s equations to this line
results in the following “primitive” impedance matrix
in ohms/mile:
⎡.2812 +
⎢
.0953 +
⎢
⎣⎡ zp ⎤⎦ = ⎢
.0953 +
⎢
⎣⎢.0953 +
j1.383 .0953 + j.7266 .0953 + j.8515 .0953 + j.7524 ⎤
⎥
j.7266 .2812 + j1.383 .0953 + j.7802 .0953 + j.7674 ⎥
j.8515 .0953 + j.7802 .2812 + j1.383 .0953 + j.7865 ⎥
⎥
j.7524 .0953 + j.7674 .0953 + j.7865 .6873 + j1.546 ⎦⎥
Partitioning this matrix between the third and
fourth rows gives the “neutral current transformation
vector” of Equation 21 as:
⎡⎣ I abc ⎤⎦
=
⎡ 202.84/ − 32.0 ⎤
⎢
137.94/ − 139.1⎥⎥
⎢
⎢⎣ 255.95/ 81.7 ⎥⎦
A
The current flowing in the neutral conductor is given
by (Equation 21):
I n = ⎣⎡ znt ⎦⎤ ⋅ ⎣⎡ I abc ⎦⎤ = 54.62/ − 131.0
A
The current flowing in “dirt” is computed to be:
I d = − ( I a + Ib + Ic + I n ) = 70.3/ − 168.3
A
The line-to-ground voltages at the load are computed
to be:
znt = ⎡⎣ −.4292 − j.1291 −.4373 − j.1327 −.4476 − j.1373⎤⎦
Performing the Kron reduction gives the phase
impedance matrix as:
⎡⎣ z
abc
⎤⎦
⎡.3375 + j1.0478 .1535 + j.3849
⎢
= ⎢ .1535 + j.3849 .3414 + j1.0348
⎢ .1559 +
⎣
j.5017
.1580 + j.4236
.1559 + j.5017 ⎤
⎥
.1580 + j.4236 ⎥
⎡⎣VLGabc ⎤⎦
load
=
⎡ 7009.3/ 3.6 ⎤
⎢
7150.1/ − 120.9 ⎥⎥
⎢
⎣⎢ 6633.3/118.6 ⎦⎥
V
The input complex power for the line is:
.3465 + j1.0179 ⎥⎦
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⎡
⎣ Sin ⎤
⎦
=
⎡ 1238.6 + j 773.7 ⎤
⎢
938.5 + j 325.0 ⎥⎥
⎢
⎢⎣1446.1 + j1142.0⎥⎦
The total three-phase real power loss can be computed
as the sum of the five losses shown above:
kW + jkvar
Pxlosstotal = 22.9460 + 10.6122 + 36.5342
The load complex powers are:
⎡ 1251.1 + j 675.3 ⎤
⎢
=
937.0 + j 308.0 ⎥⎥
S
⎡ out ⎦
⎤
⎣
⎢
⎢⎣1358.2 +
+5.2981 + 1.4137 = 76.8041
kW + jkvar
j1018.7 ⎥⎦
The real power loss is computed as:
⎡
⎣ Ploss ⎤
⎦
= Re ⎡⎣ Sin ⎤⎦ − Re ⎡⎣ Sout ⎤⎦ =
⎡ −12.5654 ⎤
⎢
1.4666 ⎥⎥
⎢
⎢⎣ 87.9030 ⎥⎦
kW
Computing real power loss this way would seem to
give some wrong answers. However, remember that
power loss by definition is input power minus output
power. The negative power loss in phase a is a result
of including the effects of the neutral and ground
resistances and currents in the three-phase model of
Figure 4.
Although the individual phase power losses must
be taken with a grain of salt, the correct total threephase real power loss is the sum of the phase real
power losses:
The total three-phase real power loss is computed
to be exactly the same using the first method of power
in minus power out. If only the total three-phase loss
is desired, the direct method is the first method. Using
method one sometimes will result in a negative power
loss on one phase, but that is OK since this model has
incorporated into the phase loss the effects of the
neutral and dirt losses. If there is interest in the losses
in the neutral and dirt, then the second method must be
used.
IV. Conclusions
This paper has developed and demonstrated how
the neutral and “dirt” currents and associated real
power losses can be computed. The only “trick” is that
the impedances of the line must be computed using the
modified Carson'’ equations followed by the Kron
reduction.
V. References
Plosstotal = Plossa + Plossb + Plossc = 76.8041 kW
1.
Another way of computing the total three-phase
power loss is to compute the real power losses using
I 2 ⋅ R for each phase conductor, neutral conductor and
2.
3.
“dirt”.
Pxlossa = I a
2
⋅ Ra = 202.842 ⋅ 0.5577 = 22.9460
Pxlossb = Ib
2
⋅ Rb = 127.942 ⋅ 05577 = 10.6122
Pxlossc = Ic
2
⋅ Rc = 255.952 ⋅ 0.5577 = 36.5342 kW
Pxlossn = I n
2
⋅ Rn = 54.622 ⋅1.7766 = 5.2981
Pxlossd = I d
2
⋅ Rd = 70.32 ⋅ 3 ⋅ .0953 = 1.3137
kW
4.
Kersting, W.H., Distribution System Modeling and
Analysis, CRC Press, Boca Raton, Florida, 2002.
Glover, J.D. and Sarma M., Power System
Analysis and Design, 2nd edition, PWS-Kent
Publishing, Boston, 1994.
Carson, John R., “Wave propagation in overhead
wires with ground return”, Bell System Technical
Journal, Vol. 5, New York, 1926.
Kron, G., “Tensorial analysis of integrated
transmission systems”, Part 1, the six basic
reference frames, AIEE Trans., Vol 71, 1952.
W. H. Kersting (SM’64, F’89) was born in Santa Fe, NM. He
received the BSEE degree from New Mexico State University, Las
Cruces, and the MSEE degree from Illinois Institute of Technology.
He joined the faculty at New Mexico State University in 1962 and
served as Professor of Electrical Engineering and Director of the
Electric Utility Management Program until his retirement in 2002.
He is currently a consultant for Milsoft Utility Solutions. He is also
a partner in WH Power Consultants.
Note that now all phases have positive power
losses, which makes more sense. The phase power
losses are quite a bit different from those computed
previously. Note also that the power losses in the
neutral conductor and “dirt” have indeed been
computed.
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