Carson's Equations
1. What do they model?
a. The earth modifies the magnetic field intensity from conductor
b. In single-phase or unbalanced three phase case some current returns along ground.
x
x
X
X
X
x
x
2. Format for Carson's Equations
Carson considered two overhead conductors above uniform earth. A current I is
injected into the first conductor and returns along earth. If we imagine an ideal short at
the end of both conductors then the ratio of sending end voltage of the first conductor to
current is the self impedance of conductor 1 and the ratio of sending end voltage of
conductor 2 to current in conductor 1 is the(reciprocal) mutual impedance of the pair.
In normal operation these voltages are the physical voltage drops acrross the wires.
I
Conductors
i
I
Dik
hi
φik
hk
+
Vi
-
Zik
+
Vik
-
I
Images
Earth
Return
I
x
Zii
k
Zkk
3. Carson's equations are usually given in the following form. For a specific frequency jf:=

 Di , k 
 ohm
Zi , k := Ri , k + ∆Ri , k + j ⋅ ω ⋅µ 0 ⋅ln
 + ∆Xi , k 
d

 i,k 
 m
−1
Rik is the usual conductor ac resistance when i=k, 0 0therwise. ∆Ri,k and ∆Xi,k are called carson's
corrections for earth return and are given in the form of infinite series
Truncated to two terms the formulas are

1
2
−
⋅a ⋅cosφi , k 
 8 6 ⋅π

∆Ri , k := µ 0 ⋅ω ⋅
∆Xi , k :=
a :=
µ 0 ⋅ω 
  1.851382   + 2 ⋅a ⋅cosφ 
⋅0.5 ⋅ ln

i , k
a
π 
 
  6 ⋅π



5 ⋅1000 ⋅µ 0 ⋅ Di , k ⋅
f 

ρ
f is frequency in Hz, ω=2πf and ρ is the earth bulk resistivity in meters
Distances are in meters
A common for of Carson's formula utilizes only the first term in the correction and is written
Zi , k := Ri , k + µ 0 ⋅
ω
ω    1  

+ j ⋅ µ 0 ⋅
 ⋅ ln 
⋅
8
 2 ⋅π    di , k  
 
 
  ohm
 
f
 1000 ⋅ 5 ⋅µ 0 ⋅
  m
ρ  

1.85138
For ρ=100 ohm-m, f= 60 Hz, di,k in feet and Zi,k converted to ohms per mile
  1  + 7.934 


  di , k 

Zi , k := Ri , k + 0.0953 + j ⋅0.1213 ⋅ ln
ohm
mi
Is EE493, 494 The formula is sometimes also written as
Zi , k := Ri , k + µ 0 ⋅
ω
ω   De 

+ j ⋅ µ 0 ⋅
 ⋅ln

8
 2 ⋅π   di , k 
ohm
mi
De := 658.5 ⋅
ρ
f
m
This, in analogy with the single-phase line, can be interpreted as saying that earth is a return 'image'
conductor with resistance µ0ω/8 and Distance De meters.
This is where the Glover Sarma text starts on P 177
4. Applying Carson's equations to overhead Lines
Carson's equations allow us to describe the inductive part of the voltage current relationship on a line
in terms of self and mutual impedances as shown below. The voltage drop equation for a line with three
phase conductors and 2 shield wires is shown below
 Vs1g − Vs1'g   Zs1s1 Zs1s2
 Vs2g − Vs2'g   Zs2s1 Zs2s2

 
 Vag − Va'g  :=  Zas1 Zas2
 Vbg − Vb'g   Zbs1 Zbs2

 
 Vbg − Vb'g   Zcs1 Zcs2
Zs1a Zs1b Zs1c   Is1 
  Is2 
 
Zaa Zab Zac  ⋅ Ia 
  Ib 
Zba Zbb Zbc
 
Zca Zcb Zcc   Ic 
Zs2a Zs2b Zs2c
or Vdrop = Zcond I cond
Is1
Vs1’g
Is2
Vs1g
Vs2g
Vs2’g
Ia
Vag
Va’g
Ib
Vbg
Vb’g
Ic
Vcg
Vc’g
Example.
We will consider a segment of a three phase 345 kV transmission line. The segment called AIP1 is
described in the ASPEN Oneliner printout in Appendix B.
The parameters used for conductors are:
phase gmr=.038' shield gmr= 0.012'
phase resistance= 0.11 ohm/mi 50 deg C
shield resistance= 8 ohm/mi 50 deg C
The average conductor height is taken as height at attachment- (2/3) sag
Many programs are available to calculate the carson formula based impedances. I used
the Alternative Transients Program and modeled the AIP1 structure with Corncrake conductor. The two bundled
were modeled as a single conductor with Ds= sqrt(gmr*d). ATP uses 20 terms in the Carson series.
The impedance matrix is shown below in ohm/mi
The resistances and inductances in the above matrix give us the primitive parameters needed
to model phenomen such as lightning strikes to shield wire, when appropriate frequency
dependencies are taken into account.
6. Phase Domain model
In many applications( power flow, short circuit), although shield wires may carry currents, we are
interested in currents and voltages along the phase wires only. Since shield wires are grounded( but may
be segmented) it is possible to 'eliminate' them since
Vs1g=0 Vs2g=0 Vs1'g= 0 and Vs2'g = 0
Thus Vs1g-Vs1'g= 0 and Vs2g-Vs2'g=0.
Thus Vs1g-Vs1'g= 0 and Vs2g-Vs2'g=0.
This does not imply that there is no voltage drop along the shield.
The performance equation can now be partitioned as shown below
0 = Zss Is + Zsp Ip
Vdp = ZspTIs+Zpp Ip
Where Is is the vector of shield currents, Ip is the vector of phase current and
Vdp is the voltage drop across phases. The equation can besolved to yield
Vdp=Zp Ip Zp= Zpp- ZpsZss-1ZpsT
Zp is the impedance matrix that relates voltage drops across phases to line currents
Example 2. For our line Zp is shown below in ohm/mi
7. Sequence Models and Impedances
Phase domain models are essential when unbalanced operation is of interest. In balanced studies
and , as a good approximation, in fault studies sequence models are useful. Recall that the symmetrical
component transformation ( Chapter 3) relates phase domain quantities to sequence domain quantities as
summarized below
Vp = A Vs Ip = A Is
Vp = Zp Ip
Vs = Zs Is
Vs = Zs Is
Zs = A-1 Zp A
The sequence Impedance matrix is shown below in ohm/mi
−0.02507 − j ⋅0.01003 
0.02073 − j ⋅0.016
0.5447 + j ⋅2.1133


Zs :=  −0.02507 − j ⋅0.01003 0.06543 + j ⋅0.58565 −0.04838 + j ⋅0.028286 
 0.02073 − j⋅0.016 0.04868 + j⋅0.02775 0.06543 + j⋅0.58565 


Notice it is nearly diagonal. If the line were transposed and the phase domain impedance matrix were
perfectly symmetrical the sequence domain matrix would indeed be diagonal.
When we model components using sequence impedances, we ignore the off-diagonal terms or
sequence-coupling.We then say that the zero positive and negative sequence impedances are
Z0 := 0.5447 + j ⋅2.1133 ohms
mile
Z1 := 0.06543 + j ⋅0.58565ohms
mile
Z2 := 0.06543 + j ⋅0.58565ohms
mile
Now, this is indeed disappointing-- after all this work the positive and negative sequence impedances
are exactly what we calculated without Carson's equations, as illustrated below!!
Ra :=
0.11
2
gmr := 0.038
Ra = 0.055
feet
ohms
phase
Bundle
d := 1.5
Ds :=
gmr⋅d
Ds = 0.239
Deq :=
3
Deq = 29.608
23.5 ⋅23.5 ⋅47
 Deq 
Xa := 0.1213 ⋅ln

 Ds 
60 Hz ac, 50 deg c
Xa = 0.585
ohm
mi
feet
Of course we do not have a pat formula for zero sequence impedance. But here's one out of Hermann
Dommell's EMTP Theory book-- units are ohm/km
Ro= Rac' +3*π*ω*0.0001/2
Xo= 0.0006*ω* ln ( (658.87*(ρ/f)0.5)/Deq0)
Rac' is the equivalent resistance of the zero sequence current path and equals the equivalent resistance of
one phase plus three times the equivalent resistance of parallel connected shield wires.
Deq0 is an equivalent GMR of the line where we imagine all conductors including shields to form one
equivalent conductor!! units are meters !ouch!
Special note: If you compare the results of impedance calculations in this writeup( using EMTP), with
those at the bottom of the Aspen printouts and those in Appendix A you will notice small differences.
These arise for two reasons:
1. I used data from conductor tables in Glover and sarma and some other notes i have. Aspen results
were given to me and presumably more accurate conductor data were used. The data for the same size
conductor can differe slightly due to stranding, etc.
2. EMTP used 20 terms in the series, Aspen used ??, Appendix A uses 1 term
Mutual COupling(magnetic)
Often, at getaways from substations, lines may run for some distance along a common right of way.
These lines are then mutually coupled and modeled as such. For two lines with two shield wires each we
build a 10x10 matrix and then eliminate shields to obtain a 6x6 matrix. For transposed line, or as
anapproximation for untransposed lines we may want to transform to the sequence domain. It turns out
that there is very little coupling expect in the zero sequence circuit.
This is illustrated in the phase and sequence matrices displayed below for A pair of lines out
and 115 kV line out of Luna station. Data are in appendix B, structure AIP5, and calculations are
detailed in Example 2 in Appendix A
Notice that there are two lines in a common right of way They are separated by about 87 feet.
One is a 345 kV line similar to what we just looked at. The other is a 115 kV line with one
conductor per phase.
Note that the primitve impedance matrix is now a 10x10 matrix; each bundled phase conductors
in the 345 kV line are replaced by a single conductor with equivalent gmr and numbered 1-3.
The 115 kV phase conductors are numbered 4-6. The 345 kV shields are numbered 7-8 and the
115 kV shields are numbered 9-10.
The computation of the phase impedance matrix is as before.The submatrix Zpp is now 6x6, Zss is 4x4
and Zps is 6x4
Real part of Zp
 0.235

 0.17
 0.167

 0.168
 0.169

 0.17
0.17 0.167 0.168 0.169 0.17 

0.236 0.168 0.166 0.167 0.169 
0.168 0.229 0.162 0.164 0.165 

0.166 0.162 0.366 0.181 0.18 
0.167 0.164 0.181 0.368 0.181 

0.169 0.165 0.18 0.181 0.367 
Imaginary Part of Zp
 1.075

 0.513
 0.431

 0.319
 0.334

 0.352
0.513 0.431 0.319 0.334 0.352 

1.075 0.515 0.297 0.309 0.323 
0.515 1.079 0.28 0.291 0.302 

0.297 0.28 1.322 0.572 0.479 
0.309 0.291 0.572 1.321 0.542 

0.323 0.302 0.479 0.542 1.32 
Again we often use sequence models. But now we must define the symmetrical component
transformation as
A
0
0
A
A2=
Where A is the conventional transformation
Then Zs= A2-1 Zp A2 yields a 6x6 sequence matrix shown below
Real Part
−0.021
0.024
 0.57

0.065
0.049
 0.024

−0.048
0.065
 −0.021
 0.5
−0.013
0.02

−4
−3
 −0.015 −5.338 × 10 −2.327 × 10

−3
−4
5.479 × 10
 0.011 2.384 × 10
0.5
0.02
−4
5.479 × 10
−3
−0.013 2.384 × 10
0.729
−0.028
0.026
0.186
−0.028
−0.054


−2.327 × 10 
−4 
−5.338 × 10 

0.026

0.054


0.186

−0.015
0.011
−3
Imaginary part
−0.019
−0.021
0.936
−0.018
−0.012
2.049


−3
−3
 −0.018
−1.453 × 10 
−2.639 × 10
0.031
0.028
0.59


−3
−3
 −0.012
−2.741 × 10 
−1.449 × 10
0.038
0.59
0.028


−3
−3
−5.355 × 10 
−4.371 × 10
2.383
0.031
0.038
 0.936


−3
−3
−3
0.011
0.79
 −0.019 −2.741 × 10 −1.453 × 10 −5.355 × 10

 −0.021 −1.449 × 10− 3 −2.639 × 10− 3 −4.371 × 10− 3

0.79
0.012


The thing to notice is again that the off diagonals are near zero except for the zero-sequence mutual
coupling between the lines which has a value of 0.5+j 0.936 ohm per mile. It is important to model
such coupling in short circut studies if sequence models are used.
An interesting interpretation/ application is this. Suppose the 115 kV line is deenergized . A single
pahase fault occurs on the 345 kV with a zero-sequence fault current of 1000 A. Then roughly 1 kV per
mile of voltage will be induced in the 115 kV line! SUch electromagnetic coupling is usually not of
concern. Later we will see capacitive coupling is potentially more hazardous.
Appendix A
Line Impedance calculation using one term approximation to carson's equations
Examample 1 AIP1A with corncrake conductor
Conductors := 5
Coordinates Conductors are numbered 1,2,3; shield wires 4 and 5
x1 := 0
x4 := 10
x2 := 23.5
x5 := 37
x3 := 47
y1 := 42
y4 := 74
y2 := 42
y5 := 72
y3 := 42
gmr := .038
Phases are Corncrake
Ds1 :=
gmr⋅1.5 Ds2 :=
gmr⋅1.5 Ds3 :=
shield is 3/8 in steel
Conductor to Conductor distance
feet
gmr⋅1.5
gmrsteel := 0.012
feet
feet
Ds4 := gmrsteel
D ( m , n) :=
Ds5 := gmrsteel
feet
Dsm if m = n
( xm − xn) 2 + ( ym − yn) 2
otherwise
distance matrix
m := 1 , 2 .. 5
dm , n := D ( m , n)
n := 1 , 2 .. 5
47
33.526
 0.239 23.5
 23.5 0.239 23.5 34.731

d =  47
23.5 0.239 48.918
 33.526 34.731 48.918 0.012

 47.634 32.898 31.623 27.074
47.634 


31.623 

27.074

0.012 
32.898
Primitive impedance matrix
Resistances
Phase conductor 60Hz, 50 dec
steel
m := 1 , 2 .. 3
R4 , 4 := 8.
Rm , m := .055
R5 , 5 := 8.
ohm
mi
Carson formula
i := 1 , 2 .. 5
k := 1 , 2 .. 5
∆Ri , k := 0.0953
  1  + 7.934 


  di , k 

Xi , k := 0.1213 ⋅ ln
Zi , k := Ri , k + ∆Ri , k + i ⋅Xi , k
 0.15 + 1.136i
 0.095 + 0.579i

Z =  0.095 + 0.495i
 0.095 + 0.536i

 0.095 + 0.494i
0.095 + 0.579i 0.095 + 0.495i 0.095 + 0.536i 0.095 + 0.494i 


0.095 + 0.579i 0.15 + 1.136i 0.095 + 0.491i 0.095 + 0.543i 

0.095 + 0.532i 0.095 + 0.491i 8.095 + 1.499i 0.095 + 0.562i

0.095 + 0.539i 0.095 + 0.543i 0.095 + 0.562i 8.095 + 1.499i 
0.15 + 1.136i 0.095 + 0.579i 0.095 + 0.532i 0.095 + 0.539i
Phase Impedance Matrux
Zpp := submatrix( Z , 1 , 3 , 1 , 3)
Zps := submatrix( Z , 4 , 5 , 1 , 3)
Zss := submatrix( Z , 4 , 5 , 4 , 5)
T
−1
Zp := Zpp − Zps ⋅Zss
⋅Zps
 0.204 + 1.099i 0.151 + 0.541i 0.149 + 0.458i 
Zp =  0.151 + 0.541i 0.208 + 1.097i 0.151 + 0.541i 
 0.149 + 0.458i 0.151 + 0.541i 0.204 + 1.099i 


Sequence Impedances
Symmetrical Component Tranformation
1
1
1



i⋅ 120deg
( i⋅ − 120deg )  1 
A :=  1 e
e
 ⋅ 
3
 ( i⋅ − 120deg)
i⋅ 120deg 
e
1 e

Zs := A
−1
0.333
0.333
0.333


A =  0.333 −0.167 + 0.289i −0.167 − 0.289i 
 0.333 −0.167 − 0.289i −0.167 + 0.289i 


⋅Zp⋅A
 0.505 + 2.125i −0.025 − 0.012i 0.022 − 0.015i 
Zs =  0.022 − 0.015i 0.055 + 0.585i 0.049 + 0.028i 
 −0.025 − 0.012i −0.048 + 0.028i 0.055 + 0.585i 


Line Impedance calculation using one term approximation to carson's equations
Example 2
AIP5 2 lines in proximity
Conductors := 10
Circuit 1 (345 kV)Coordinates
Conductors are numbered 1,2,3; shield wires 7 and 8
x1 := 0
x7 := 10
x2 := 23.5
x8 := 37
x3 := 47
y1 := 42
y7 := 74
y2 := 42
y8 := 72
y3 := 42
gmr := .035
Phases are Corncrake
Ds1 :=
gmr⋅1.5 Ds2 :=
shield is 3/8 in steel
gmr⋅1.5 Ds3 :=
feet
feet
gmr⋅1.5
gmrsteel := 0.012
feet
Ds7 := gmrsteel
Ds8 := gmrsteel
feet
Circuit 2 (115 kV)Coordinates Conductors are numbered 4,5,6; shield wires 9 and 10
x4 := −115.5
x9 := −109.5 x5 := −101.5
y4 := 32.7 y9 := 43.75
y5 := 32.7
Ds5 := gmr2
shield is 3/8 in steel
y10 := 72
Conductor to Conductor distance
feet
feet
Ds6 := gmr2
gmrsteel := 0.012
x6 := −87.5
y6 := 43.75
gmr2 := .029
Phases are Hawk
Ds4 := gmr2
x10 := −93.5
feet
Ds9 := gmrsteel
D ( m , n) :=
Ds10 := gmrsteel feet
Dsm if m = n
( xm − xn) 2 + ( ym − yn) 2
otherwise
distance matrix
m := 1 , 2 .. 10
n := 1 , 2 .. 10
dm , n := D ( m , n)
Primitive impedance matrix
Resistances
Phase conductor 60Hz, 50 dec
m := 1 , 2 .. 3
Rm , m := .055
ohm
mi
m := 4 , 5 .. 6
Rm , m := .1859
ohm
mi
steel
m := 7 , 8 .. 10
i := 1 , 2 .. 10
Rm , m := 8
k := 1 , 2 .. 10
∆Ri , k := 0.0953
  1  + 7.934 


  di , k 

Xi , k := 0.1213 ⋅ ln
Zi , k := Ri , k + ∆Ri , k + i ⋅Xi , k
Phase Impedance Matrux
Zpp := submatrix( Z , 1 , 6 , 1 , 6)
Zps := submatrix( Z , 7 , 10 , 1 , 6)
Zss := submatrix( Z , 7 , 10 , 7 , 10)
T
−1
Zp := Zpp − Zps ⋅Zss
⋅Zps
ohm
mi
 0.225

 0.17
 0.167
Re( Zp) = 
 0.168
 0.169

 0.17
 1.075

 0.513
 0.431
Im ( Zp) = 
 0.319
 0.334

 0.352
0.17 0.167 0.168 0.169 0.17 

0.226 0.168 0.166 0.167 0.169 
0.168 0.219 0.162 0.164 0.165 

0.166 0.162 0.366 0.181 0.18 
0.167 0.164 0.181 0.368 0.181 

0.169 0.165 0.18 0.181 0.367 
0.513 0.431 0.319 0.334 0.352 

1.075 0.515 0.297 0.309 0.323 
0.515 1.079 0.28 0.291 0.302 

0.297 0.28 1.322 0.572 0.479 
0.309 0.291 0.572 1.321 0.542 

0.323 0.302 0.479 0.542 1.32 
Sequence Impedances
Symmetrical Component Tranformation
1
1

1


i⋅ 120deg
( i⋅ − 120deg )  1 
A :=  1 e
e
 ⋅ 
3
 ( i⋅ − 120deg)
i⋅ 120deg 
e
1 e

0.333
0.333
0.333


A =  0.333 −0.167 + 0.289i −0.167 − 0.289i 
 0.333 −0.167 − 0.289i −0.167 + 0.289i 


A
 0 0 0 
ZeroM :=  0 0 0 
0 0 0 


AU := augment( A , ZeroM)
A2 := stack ( AU , AL)
−1
Zs := A2
⋅Zp⋅A2
−1
1
1
1


=  1 −0.5 − 0.866i −0.5 + 0.866i 
 1 −0.5 + 0.866i −0.5 − 0.866i 


AL := augment( ZeroM , A)
0.024
−0.021
 0.56

0.049
0.055
 0.024

0.055
−0.048
−0.021
Re( Zs) = 
 0.5
0.02
−0.013

−3
−4
 −0.015 −5.338 × 10 −2.327 × 10

−4
−3
5.479 × 10
 0.011 2.384 × 10
0.5
0.02
−4
5.479 × 10
−3
−0.013 2.384 × 10
0.729
−0.028
0.026
0.186
−0.028
−0.054


−3
−2.327 × 10 
−4 
−5.338 × 10 

0.026

0.054


0.186

−0.015
0.011
−0.019
−0.021
0.936
−0.018
−0.012
2.049


−
3
−
3
 −0.018
−1.453 × 10 
−2.639 × 10
0.031
0.028
0.59


−3
−3
 −0.012
−2.741 × 10 
−1.449 × 10
0.038
0.59
0.028
Im ( Zs) = 

−3
−3
−5.355 × 10 
−4.371 × 10
2.383
0.031
0.038
 0.936


−3
−3
−3
0.011
0.79
 −0.019 −2.741 × 10 −1.453 × 10 −5.355 × 10

 −0.021 −1.449 × 10− 3 −2.639 × 10− 3 −4.371 × 10− 3

0.79
0.012

