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
Zii
k
hi Dik
+ Zik Zkk
hk +
Vi
φik -
Vik
-

Images
Earth
Return x
I
3. Carson's equations are usually given in the following form. For a specific frequency jf:= −1

  Di , k   ohm
Zi , k := Ri , k + ∆Ri , k + j ⋅ ω ⋅µ 0 ⋅ln  + ∆Xi , k 
  i,k 
d  m

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 
∆Ri , k := µ 0 ⋅ω ⋅ − ⋅a ⋅cosφi , k 
 8 6 ⋅π 

µ 0 ⋅ω    1.851382   + 2 ⋅a ⋅cosφ 
∆Xi , k := ⋅0.5 ⋅ ln  i , k
π    a   6 ⋅π 
 f 
a := 5 ⋅1000 ⋅µ 0 ⋅ Di , k ⋅ 
 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

ω  ω    1   1.85138   ohm
Zi , k := Ri , k + µ 0 ⋅ + j ⋅ µ 0 ⋅  ⋅ ln  ⋅
8  2 ⋅π    di , k   f
 
   1000 ⋅ 5 ⋅µ 0 ⋅   m
   ρ  

For ρ=100 ohm-m, f= 60 Hz, di,k in feet and Zi,k converted to ohms per mile

ohm
Zi , k := Ri , k + 0.0953 + j ⋅0.1213 ⋅ ln
  1  + 7.934 
  mi
  di , k  
Is EE493, 494 The formula is sometimes also written as

ohm
ω  ω   De  ρ
Zi , k := Ri , k + µ 0 ⋅ + j ⋅ µ 0 ⋅  ⋅ln  mi De := 658.5 ⋅ m
8  2 ⋅π   di , k  f

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 Zs1a Zs1b Zs1c   Is1 

 Vs2g − Vs2'g   Zs2s1 Zs2s2   Is2 
Zs2a Zs2b Zs2c
    
 Vag − Va'g  :=  Zas1 Zas2 Zaa Zab Zac  ⋅ Ia 
 Vbg − Vb'g   Zbs1 Zbs2 Zba Zbb Zbc
  Ib 
    
 Vbg − Vb'g   Zcs1 Zcs2 Zca Zcb Zcc   Ic 

or Vdrop = Zcond I cond

Is1

Vs1g Is2 Vs1’g

Vs2g Vs2’g
Ia

Vag Ib Va’g

Vbg Ic Vb’g

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


0.5447 + j ⋅2.1133 0.02073 − j ⋅0.016 −0.02507 − j ⋅0.01003 

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!!

0.11 ohms
Ra := Ra = 0.055 60 Hz ac, 50 deg c
2 phase
gmr := 0.038 feet Bundle d := 1.5 feet

Ds := gmr⋅d Ds = 0.239
3
Deq := 23.5 ⋅23.5 ⋅47 Deq = 29.608
 Deq 
Xa := 0.1213 ⋅ln
ohm
 Xa = 0.585
 Ds  mi
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 0.236 0.168 0.166 0.167 0.169 
 0.167 0.168 0.229 0.162 0.164 0.165 
 
 0.168 0.166 0.162 0.366 0.181 0.18 
 0.169 0.167 0.164 0.181 0.368 0.181 
 
 0.17 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 1.075 0.515 0.297 0.309 0.323 
 0.431 0.515 1.079 0.28 0.291 0.302 
 
 0.319 0.297 0.28 1.322 0.572 0.479 
 0.334 0.309 0.291 0.572 1.321 0.542 
 
 0.352 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
A2=
0 A

Where A is the conventional transformation

Then Zs= A2-1 Zp A2 yields a 6x6 sequence matrix shown below

Real Part
 0.57 −0.021 0.024 0.5 0.011 −0.015 
 −4 −3

 0.024 0.065 0.049 0.02 5.479 × 10 −2.327 × 10 
 −3 −4 
 −0.021 −0.048 0.065 −0.013 2.384 × 10 −5.338 × 10 
 0.5 −0.013 0.02 0.729 −0.028 0.026 
 −4 −3

 −0.015 −5.338 × 10 −2.327 × 10 0.026 0.186 0.054 
 −3 −4 
 0.011 2.384 × 10 5.479 × 10 −0.028 −0.054 0.186 
 Imaginary part

 
2.049 −0.012 −0.018 0.936 −0.021 −0.019
 −0.018 0.59 0.028 0.031 −2.639 × 10
−3
−1.453 × 10 
−3
 
−3 −3
 −0.012 0.028 0.59 0.038 −1.449 × 10 −2.741 × 10 
 −3 −3

 0.936 0.038 0.031 2.383 −4.371 × 10 −5.355 × 10 
 −3 −3 −3 
 −0.019 −2.741 × 10 −1.453 × 10 −5.355 × 10 0.79 0.011 
 −0.021 −1.449 × 10− 3 −2.639 × 10− 3 −4.371 × 10− 3 
 0.012 0.79 

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 feet
Phases are Corncrake gmr := .038 feet

Ds1 := gmr⋅1.5 Ds2 := gmr⋅1.5 Ds3 := gmr⋅1.5 feet

shield is 3/8 in steel gmrsteel := 0.012 Ds4 := gmrsteel Ds5 := gmrsteel feet

Conductor to Conductor distance D ( m , n) := Dsm if m = n

( xm − xn) 2 + ( ym − yn) 2 otherwise

distance matrix

m := 1 , 2 .. 5 n := 1 , 2 .. 5  0.239 23.5 47 33.526 47.634 

 23.5 0.239 23.5 34.731 32.898

 
dm , n := D ( m , n) d =  47 23.5 0.239 48.918 31.623 
 33.526 34.731 48.918 0.012 27.074

 
 47.634 32.898 31.623 27.074 0.012 

Primitive impedance matrix

Resistances

ohm
Phase conductor 60Hz, 50 dec m := 1 , 2 .. 3 Rm , m := .055
mi
R4 , 4 := 8. R5 , 5 := 8.
steel
Carson formula
i := 1 , 2 .. 5 k := 1 , 2 .. 5

∆Ri , k := 0.0953

  1  + 7.934 
Xi , k := 0.1213 ⋅ ln  
  di , k  

Zi , k := Ri , k + ∆Ri , k + i ⋅Xi , k

 0.15 + 1.136i 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 + 0.579i 0.095 + 0.532i 0.095 + 0.539i

 
Z =  0.095 + 0.495i 0.095 + 0.579i 0.15 + 1.136i 0.095 + 0.491i 0.095 + 0.543i 
 0.095 + 0.536i 0.095 + 0.532i 0.095 + 0.491i 8.095 + 1.499i 0.095 + 0.562i

 
 0.095 + 0.494i 0.095 + 0.539i 0.095 + 0.543i 0.095 + 0.562i 8.095 + 1.499i 

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 0.333 0.333 0.333
 i⋅ 120deg

( i⋅ − 120deg )  1 
A :=  1 e e  ⋅  A =  0.333 −0.167 + 0.289i −0.167 − 0.289i 
 ( i⋅ − 120deg) 3
i⋅ 120deg   0.333 −0.167 − 0.289i −0.167 + 0.289i 
 
1 e e 

−1
Zs := A ⋅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 feet
Phases are Corncrake gmr := .035 feet

Ds1 := gmr⋅1.5 Ds2 := gmr⋅1.5 Ds3 := gmr⋅1.5 feet

shield is 3/8 in steel gmrsteel := 0.012 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 x10 := −93.5 x6 := −87.5

y4 := 32.7 y9 := 43.75 y5 := 32.7 y10 := 72 y6 := 43.75 feet

Phases are Hawk gmr2 := .029 feet

Ds4 := gmr2 Ds5 := gmr2 Ds6 := gmr2 feet

shield is 3/8 in steel gmrsteel := 0.012 Ds9 := gmrsteel Ds10 := gmrsteel feet

D ( m , n) := Dsm if m = n
Conductor to Conductor distance
( 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

ohm
m := 1 , 2 .. 3 Rm , m := .055
mi

ohm
m := 4 , 5 .. 6 Rm , m := .1859
mi

steel
ohm
m := 7 , 8 .. 10 Rm , m := 8
mi

i := 1 , 2 .. 10 k := 1 , 2 .. 10

∆Ri , k := 0.0953

  1  + 7.934 
Xi , k := 0.1213 ⋅ ln  
  di , k  
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
  0.225 0.17 0.167 0.168 0.169 0.17 
 
 0.17 0.226 0.168 0.166 0.167 0.169 
 0.167 0.168 0.219 0.162 0.164 0.165 
Re( Zp) =  
 0.168 0.166 0.162 0.366 0.181 0.18 
 0.169 0.167 0.164 0.181 0.368 0.181 
 
 0.17 0.169 0.165 0.18 0.181 0.367 

 1.075 0.513 0.431 0.319 0.334 0.352 

 
 0.513 1.075 0.515 0.297 0.309 0.323 
 0.431 0.515 1.079 0.28 0.291 0.302 
Im ( Zp) =  
 0.319 0.297 0.28 1.322 0.572 0.479 
 0.334 0.309 0.291 0.572 1.321 0.542 
 
 0.352 0.323 0.302 0.479 0.542 1.32 

Sequence Impedances

Symmetrical Component Tranformation

1 
 
1 1 0.333 0.333 0.333
 i⋅ 120deg

( i⋅ − 120deg )  1 
A :=  1 e e  ⋅  A =  0.333 −0.167 + 0.289i −0.167 − 0.289i 
 ( i⋅ − 120deg) 3
i⋅ 120deg   0.333 −0.167 − 0.289i −0.167 + 0.289i 
 
1 e e 

 
1 1 1
−1
A =  1 −0.5 − 0.866i −0.5 + 0.866i 
 1 −0.5 + 0.866i −0.5 − 0.866i 
 
 0 0 0 
ZeroM :=  0 0 0 
0 0 0 
 

AU := augment( A , ZeroM) AL := augment( ZeroM , A)

A2 := stack ( AU , AL)

−1
Zs := A2 ⋅Zp⋅A2
  0.56 −0.021 0.024 0.5 0.011 −0.015 
 −4 −3

 0.024 0.055 0.049 0.02 5.479 × 10 −2.327 × 10 
 −3 −4 
−0.021 −0.048 −0.013 2.384 × 10 −5.338 × 10 
Re( Zs) = 
0.055
 0.5 −0.013 0.02 0.729 −0.028 0.026 
 −4 −3

 −0.015 −5.338 × 10 −2.327 × 10 0.026 0.186 0.054 
 −3 −4 
 0.011 2.384 × 10 5.479 × 10 −0.028 −0.054 0.186 

 
2.049 −0.012 −0.018 0.936 −0.021 −0.019
 −0.018 0.59 0.028 0.031 −2.639 × 10
− 3
−1.453 × 10 
− 3
 
−3 −3
 −0.012 0.028 0.59 0.038 −1.449 × 10 −2.741 × 10 
Im ( Zs) =  −3 −3

 0.936 0.038 0.031 2.383 −4.371 × 10 −5.355 × 10 
 −3 −3 −3 
 −0.019 −2.741 × 10 −1.453 × 10 −5.355 × 10 0.79 0.011 
 −0.021 −1.449 × 10− 3 −2.639 × 10− 3 −4.371 × 10− 3 
 0.012 0.79 