This document discusses system earthing in industrial and commercial electrical systems. It covers: 1. Reasons for earthing systems, including carrying fault currents safely and preventing electric shock. 2. Factors that influence earth resistance, such as soil moisture content and electrode depth. 3. Thresholds for body current that can cause adverse health effects like fibrillation. 4. Concepts of touch and step voltages caused by potential differences in earthed systems. 5. Calculation of maximum earth fault currents and factors like ground resistivity and grid geometry that influence potential rises. 6. Types of earth electrodes and equations for calculating resistance and potential gradients around hemispherical and spherical electrodes

1. ECNG 3015
Industrial and Commercial Electrical
Systems
Lecturer
Prof Chandrabhan Sharma
#7
System Earthing

2. SYSTEM EARTHING
Reason for earthing:
- To provide means to carry electric currents to earth
under normal and fault conditions without exceeding
any operating and equipment limits or adversely
affecting continuity of service.
- To ensure that a person in the vicinity of the
grounded facilities is not exposed to the danger
of critical electric shock.

3. Earth resistivity:
- measured either Ω.m
- affected by moisture content of the soil
→ since conduction in soil is electrolytic in nature
Soil resistivity falls sharply when moisture content falls below 22 % by
weight.
Electrodes should be buried deep enough
- to ensure that the moisture content of the
surrounding soil does not vary from season to
season.
- and have permanent contact with moist soil

4. The Human factor
- normal, healthy person can feel 1 mA
- 10-25 mA can cause loss of muscular control
- 100 mA can result in ventricular fibrillation
(death), burns etc.
Note:
For practical grounding studies, the threshold of ventricular fibrillation is
the major concern, i.e. most grounding mats are designed to limit the current
through the average human body to below threshold.

5. Studies have shown that 99.5 % of healthy persons can tolerate a
current through the heart without going into V.F. given by
0.116
Ib ....................................(1)
T
where I b maximumbody currentin A
T - is thedurat ionof current in seconds
Note:
Large currents can be accommodated once ‘T’ is kept below the value given by
equation 1.

6. STEP AND TOUCH VOLTAGES
The body can tolerate more current flowing from one leg to the
other than it can from one hand to the legs.
TOUCH VOLTAGE: (refer to the equivalent circuit below.)
Ib
Rg
Ig earth rod
resistivity of the soil must be
very low to ensure that most
of the current flows to ground.

7. The circuit below shows the hazards due to touch voltages.
The body current Ib is driven by the potential between A and B.
The worst possible touch potential (mesh potential) occurs at or
near to the centre of a grid mesh.

8. STEP VOLTAGE: (refer to the equivalent circuit below.)
Here, current flows from one foot to another due to the potential
difference between A and B.
Hence the step voltage is the voltage which a man would receive
across the body by taking a 1m step in a radial direction from the earth
electrode.

9. In both cases;
Rb = 1000 Ω (IEEE Std 80 -2000)
Rf = 3 s where s – soil resistivity and
treating the foot as a circular plate
electrode
Actually, the body itself has a total measured resistance of 2300 Ω hand to
hand and 1100 Ω hand to foot.
From the equivalent circuit for touch voltage, the effective
resistance under consideration for hand to foot would be:
Rf but R b 1000 and Rf 3
R effective,Touch Rb s
2
R effective,Touch 1000 1.5 s

10. And for foot to foot currents:
R effective,Step R b 2R f again, R b 1000 and Rf 3 s
R effective,Step 1000 6 s
Applying the above two equations to equation (1), we have:
(0.116)(10 1.5 s ) 116 0.17
00 s
E Touch max I b (1000 1.5 s )
T T
(0.116)(10 6 s )
00 116 0.7 s
and EStep max I b (1000 6 s )
T T
The grid voltage depends upon three basic factors:
1. Ground resistivity
2. Available fault current
3. Grid geometry

11. 1. Ground Resistivity:
Type of ground Resistivity (Ωm)
Wet organic soil 10
Moist soil 102
Dry soil 103
Bed rock 104
Eelctricfield (V/l ) RA
Volume Resistivity m
Current Density(I/A) l
Where
R resistance
A cross- sectionarea and
l length

12. 2. Available fault current:
The maximum ground fault current (S.L.G) is given by :
3V
I "
3Rg (R1 R 2 R 0 ) j(X1 X 2 X0 )
Where
V phase voltage
R g grid resistanceto earth
R1 , R 2 , R 0 positive,negativeand zero seq. resistance
"
X1 positiveseq. subtransient reactance
and X 2 , X 0 negativeand zero seq. reactance

13. To account for the initial d.c. offset of the fault current, I is
multiplied by an appropriate correction factor known as the
decrement factor.
Shock Cycles (60 Hz) Decrement factor
Duration (s) Fault duration
0.008 1/2 1.65
0.1 6 1.25
0.25 15 1.10
5 30 1.00
Shock and Fault Duration

14. A more accurate value of the decrement factor (D) is given by
-2 T
1 1 X X
R
D T . 1 e
T R
Where
T durationof fault (s) (obtained from relay setting)
system frequency in radians/s
X totalsystem reactance
R totalsystem reactance

15. 3. Grid Geometry:
The potential rise of points protected by a grounding mat depends
on such factors as:
a. grid burial depth
b. length and diameter of conductor
c. space between each conductor
d. distribution of current throughout the grid
e. proximity of the fault electrode and the system
grounding electrodes to the grid conductors

16. EARTH ELECTRODES:
1. Hemisphere electrode at the earth surface
For a hemisphere of radius ‘r’ from which a total current ‘I’ spreads
out in a radial direction into the earth:
I
Current density (i ) at a distance ( x)
2πx 2

17. The voltage drop at distance x across a hemispherical shell
of thickness dx is given by:
I
Voltage drop (dv) i dx 2
dx (V)
2πx
dv I
Potential gradient e x (V/m)
dx 2πx 2
Vx = voltage drop from ‘r’ to ‘x’
x I I 1 1
i.e. Vx 2
dx Vx
r 2πx 2π r x
Voltage at t heelectrodesurface w.r.t remoteearthpot ential
I
(x ) E0
2πr

18. I
and at distance x Ex
2π x
E0
The resistance R at the electrode surface is R
I 2π r
Question:
For a hemispherical electrode, r = 0.5 m, = 10 .m and
I = 100 A. Find the resistance at the electrode surface and the
potential gradient at distance 0.6m from the electrode surface.
Soln :
a. T heresistanceat t heelect rodesurface is R
2π r
10
3.18
2π (0.5)

19. I
b. P otentialgradient,e x 2
,
2πx
(where x is thehorizontal distancefrom thecentreof theh/sphere)
10 x 100
131.5 V/m
2π(0.5 0.6) 2
If 50 m
th e x
en 657.6V/m E step is fatal

20. 2. Spherical electrode
Hemisphere electrode at the earth surface with insulated lead wire
buried at depth, ‘h’ having radius ‘r’.
Case 1.
‘h’ very large:
The area of the sphere is twice that of the hemisphere so all the
previous equations would be halved.
I
Potential gradient, e x
4πx 2
and potential thesurface of thesphereand at distance x is
at
I I
E0 and E x respectively
4πr 4πx

21. E0
AND the resistance R at the electrode surface is R
I 4π r
Case 2.
‘h’ finite – h>> r
The non-uniformity due to the earths surface may be eliminated by
considering identical and equal current flowing from an image
electrode a distance ‘h’ above the earth surface.
See diagram on next page.

23. T hepot ent ial point P1 is due t o bot h elect rodes:
at
I 1 1
i.e. E P1
4π x1 x'
For a point on the surface of the electrode: x1 = r , x’ = 2h
I 1 1
potential thesurface E 0
at
4π r 2h
I r
E0 1
4π r 2h
E0 r
But R 1
I 4π r 2h

24. For a point P2 on t heEart hSurface :
I 2
i.e. E P2
4π x2 h2
Potential
gradient at thesurface:
dE P2 I 2x
e P2 3
dx 4π
(x2 2 2
h )
and when x 0 thepotential theearthsurface verticall aboveis :
at y
I 2
EV
4π h

25. thepotential
between th surface of thesphereand
e
theearthsurface verticall aboveit is (T ouchVoltage):
y
I 1 3
E0 - EV
4π r 2h
Touch voltage increases with increase in ‘h’
Similarly,it can be shown thatwhen x 0.6 themaximum voltageis :
I 2
e max 2
i.e. (Step Voltage)
4π (1.6h)
Step voltage decreases with depth of electrode.

26. 3. Rod electrode
A rod or pipe in which the length of the rod (l ) >> the diameter
of the rod (d) can be broken up into ‘n’ spherical electrodes.
For therod theearthresistanceis :
4l
R ln
2π l d
d
If theradius of thesphere then the
2
l
diameteris d and thenumber of spheresn
d

27. I
also ex
2π x(l 2 x2 )
2x
and E 0 E x 0.366 i log (Step Voltage)
d
where i current/le
ngth
x horizontal
distanceon theearth' surface from therod
s

28. EARTH MAT DESIGN
Knowing the tolerable step and touch voltage limits, we can now
design and construct the ground system for a substation using a
step by step procedure as follows:
1. Determine the soil characteristics (resistivity).
2. Determine the maximum possible fault current (SLG).
3. Prepare preliminary design of grounding system.
4. Calculate the resistance of the designed grounding system in (3).
5.Calculate the maximum grid potential rise.
6.Calculate the step voltages at periphery of the mat.

29. 7. Calculate the internal step and touch voltages.
8. Refine the design so as to satisfy 5, 6 and 7 for safety.
9. Construct the mat as per design (make sure measured values
compares favourably with the calculated values)
10. Review 5, 6, 7, 8 based on actual measurement and modify (
additional electrodes, screens, barriers etc.).

30. Step 3 – Preliminary grounding design
Having obtained (1) and (2), we can now do a grounding design.
Assumption – the grounding system will take the form of a grid of
horizontally buried conductors
3.1 Determination of conductor size on the grid:
This can be estimated by the equation below;
Tm Ta
log10 1
234 Ta
I A
33s

31. where
I currentin Amps
A coppercross- sectionarea (circular mills or mm2 )
s timein secondsduring which I is applied
Tm maximumallowable temp. C in
Ta ambient te in C
mp.
1974circular mills 1 mm2

32. 3.2 Determination of conductor length required for gradient
control
In this part of the analysis, Touch voltages are used for the
basis of analysis
Also, Touch voltages from a grounded structure to the center
of a rectangle of grid mesh are used instead of touch voltages
at a horizontal distance of 1 m.
Laurent developed the following approximate equations
Estep = 0.1 0.5 i
Etouch = 0.6 0.8 i
Emesh = i

33. Where
Estep = step voltage over a horizontal distance of 1 m
Etouch = touch voltage over a horizontal distance of 1 m
from the grid conductor
Emesh = potential difference in volts, from grid
conductor to ground surface at center of a grid
mesh
= soil resistivity (Ωm)
i = current in amp (per m of buried conductor),
flowing into the ground

34. In order to allow for the non-uniformity in flow of ground current
per unit length of buried conductor we write:
I
E mesh k mki
L
where
km coefficien which ta intoaccount th physicalparameters
t kes e
of thegrid D, d, h
ki an irregularity correctionfactor
averageresistivit ( m)
y
I max.totalr.m.scurrentflowing between ground grid and earth
[adjust for future growth]
L totallengthof buried conductor(m)

35. For safe touch vol
tages we must have:
116 0.17 s I
E mesh k mki
t L
k mki I t
From which L
116 0.17 s
where
s resistivit immediatel beneath th foot
y y e
t max.durationof shock (s)

36. Step 4 – Calculation of grounding system resistance
R
4r L
where
R resist anceof t hegrounding syst em
r t heradius in m of a circular plat ehavingt hesame
area as t hatoccupiedby t hegrid
derived from t hecircular plat eelect rodeformula
4
eliminat eerrorsdue t o t hefact t hatt hegrid is not a
L
t ue circular plat e
r

37. Step 5 – Maximum Grid Potential Rise : E
E IR
where I max.s/c currentin thegrid
R resistanceof thegrid
116 0.17 s
if E E touch then the
preliminar design is o.k.
y
t
Step 6 – Step voltage at the periphery
for homogeneous soil = s
I 116 0.7 s
E stepmax kski
L t
similar toE mesh but k m ks

38. Step 7 –
If the length of the conductor calculated before is used (L), then
the internal step and touch voltages should be within the limits.
The values can be decreased by increasing the length of
conductor.
Aside:
Actually: if d = diameter of conductor
D = spacing
n = no. of parallel conductors
h = depth of burial
1 D2 1 3 5 7 2n 1
km ln ln ..........
2π 16hd π 4 6 8 2n 2
where t henumber of t ermsin ln is (n - 2)

39. 1 1 1 1 1
and k s .........
π 2h D h 2D 3D
where thenumber of termsin thebrackets n
Step 8 –
If calculations based on preliminary design indicate dangerous
potential can exist, the following remedies can be taken:
a. Decrease R by either
- Increasing area of grid, or
- Increase L by the use of earth rods
- Spacing the grid closer which increases L
b. Addition of a relatively high resistance surface layer (crushed
stone) to increase the resistance in series with the body. Use
approximately 4’ thickness of ½’ washed gravel.

41. Example:
An electric utility plans to install a new 132 kV substation to feed an
industrial load. The proposed area of the substation is a rectangle of
dimension 42 m x 100 m. The soil in the area is non-homogenous with s =
3000 Ωm and = 1316 Ωm. The system has the following impedances X” =
40.5, X2 = 41 and X0 = 41 Ω respectively. The protection is so designed that
the clearing time is 0.4 s. Given km = 0.568 and ks = 0.814 and an irregularity
factor ki = 2.0. Estimate, taking account of a decrement factor of D of 1.3
and future growth of fault level of 125%:
a. The minimum length of copper conductor required to keep touch and
step voltages within safe limits.
b. The resistance of the earth mat using the estimated copper in (a).
c. Give a proposed layout of the grid calculate the actual length of
conductor used (Maximum grid spacing = 6 m and length of earth rods
= 3 m).
d. For the layout in (c), check whether the maximum step voltage
developed is within tolerable limits.

42. Solution:
Estimated fault current for existing system is:
3VLN 3 132 x 103
Ifault (SLG fault) "
x 1260A
j(X1 X 2 X0 ) 181.5 3
Design fault current 1260 x D x growth factor
1260 x 1.3 x 1.25
2047.5A
k mki I t
(a) Recall, t heconduct orlengt h L
116 0.17 s
0.568x 2.0 x 1316 x 2047.5x 0.4
116 (0.17 x 3000)
3092.5m

43. (b) Resistanceof earthmat is given by R
4r L
1 1
1316
4r L
Now πr 2 42 x 100 r 36.6m
1 1
R 1316 9.4
4(36.6) 3092.5
(c) Layout
Let the
spacingon the m side be 4 m and on the42 m side be 6m
100
100 42
T hen the
lengthof conductorfor grid 1 42 1 100 1892m
4 6

44. 3092 5 1892
.
Min.no.of 3 m earthrods required 400
3
Eitherreduce thespacingor bury 440rods (10%)
Let the of earthrods be 440
no.
totallengthof conductoris now 440 x 3 1320m
Hence totallengthof copper to used 1892 1320
be
3212m
I 0.814x 2.0 x 1316 x 2047.5
(d) Max.E step kski
L 3212
1365.7V

45. 116 0.7 s 116 (0.7 x 3000)
T olerableE step
t 0.4
3503.8V
Since Estepmax Esteptolerable then the
design is good
Aside
T hemax.grid potential
rise E IR
2047.5x 9.4
19.276kV
To avoid this being transferred, it is necessary to isolate the
substation circuits from the supply circuits i.e. substation ground
not the common ground, therefore all m/c and other equipment
neutrals should be grounded at their site.

46. Earth mat
Cu