11.2. Example Far-Field Analysis

This example considers the electric conduction problem with infinite boundaries described as problem # 476 in "Worked problems in Applied Mathematics," by N.N. Lebedev, I.P. Skalskaya, Y.S. Ufland, Dover Publications, Inc., NY, p. 226 (1979).

11.2.1. Problem Description

A DC current (I) enters ground of conductivity σ through a plate in the form of a disk of radius a. Find the distribution of electric current under the plate and calculate the resistance of the plate.

Figure 11.7: Grounding Plate with Current Loading

Grounding Plate with Current Loading

The geometric and electrical parameters are:

Radius a = 1 cm
Applied current I = 5 mA
Soil electrical conductivity σ = 0.5 e-2 S/m

Axisymmetric PLANE230 triangular elements model a rectangular area of width 2a and depth a below the circular electrode. A single layer of axisymmetric INFIN110 quadrilateral elements represents the exterior semi-infinite domain (IFE). The ratio of the IFE domain dimensions to the FE domain dimensions is equal to 1.8 (that is, the length of the infinite elements is approximately equal to the depth of the FE domain). The electrode is defined by coupling the VOLT degrees of freedom of the ground surface. The current load is applied as a concentrated nodal load.

11.2.2. Results

The electric current results are shown in the following figure.

Figure 11.8: Electric Current Distribution

Electric Current Distribution

The calculated resistance is 4943 Ohms. The expected result from the above reference is:

R = 1/(4σa) = 5000 Ohms

11.2.3. Command Listing

You can perform this electric conduction analysis using the Mechanical APDL commands shown below. Text prefaced by an exclamation point (!) is a comment.

/title, Current distribution of circular electrode
! Problem parameters
I=5e-3               ! current, A
sig=0.5e-2           ! conductivity of soil, S/m
a=1e-2               ! disk radius, m
w=2*a                ! finite region width
d=a                  ! finite region depth
r=1.8                ! ratio infinite to finite

/prep7
et,1,PLANE230,,,1    ! axisymmetric 8-node electric quad
et,2,INFIN110,3,1,1  ! axisymmetric 8-node infinite electric quad
mp,rsvx,1,1/sig      ! resistivity input

rect,,w,,-d          ! enveloped area
k,5,,r*(-d)
k,6,r*w,r*(-d)
k,7,r*w
a,1,5,6,2
a,2,6,7,3

lesize,3,,,16         ! specify line divisions
lesize,1,,,16
lesize,6,,,16

lesize,4,,,8
lesize,2,,,8
lesize,8,,,8

lesize,9,,,1
lesize,5,,,1

msha,1,2d            ! mesh finite region with triangles
type,1
amesh,1

mshk,1               ! mesh infinite region with quads
msha,0,2d

type,2
amesh,2,3           
nummrg,node

lsel,s,line,,6       ! flag exterior surface of INFIN elements
lsel,a,line,,8
sfl,all,INF
lsel,all

nsel,s,loc,y,0       ! define electrode
nsel,r,loc,x,0,a
cp,1,volt,all        ! couple VOLT dof
n_load=ndnext(0)     ! master node
nsel,all
fini

/solu
antype,static
f,n_load,amps,I      ! apply total current
solve
fini

/post1
plesol,jc,sum        ! plot electric current vector sum
/com
/com  Calculated resistance R = %volt(n_load)/I%  Ohm
/com
/com  **************************************************************************
/com   Expected results:  R = %1/(4*sig*a)% Ohm
/com  **************************************************************************
finish