carson

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 alongearth. 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.
IConductors

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 
di , 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 isfrequency 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= 60Hz, 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 dropequation 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

orVdrop = 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= 0.11ohm/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...