Elektrotehniški vestnik 83(3): 122-130, 2016 Original Scientific Paper
A simplified Procedure to determine the Earth-fault Currents in Compensated Networks
Elisabeth Hufnagl, Lothar Fickert, Ernst Schmautzer
Institute of Electrical Power Systems, TU Graz, Inffeldgasse 18, 8010 Graz, +43(0)316/873-7064, lothar.fickert@tugraz.at, www.ifea.tugraz.at
Abstract. The paper proposes a simplified approach for determination of the earth-fault currents to be used in earth-fault compensated systems. By estimating the relative magnitudes of the equipment parameters, the equivalent circuit in symmetrical components at a single-pole fault is significantly reduced enabling the electrical parameters of the impedances to be directly used.
Keywords: earth-fault current, earth-fault calculation, compensated power networks
Učinkovit izračun tokov zemeljskega stika v omrežjih z resonančno ozemljeno nevtralno točko
Članek predstavlja poenostavljen pristop za določanje toka zemeljskega stika, ki je še posebej primeren za omrežja z resonančno ozemljeno nevtralno točko. Zaradi ocenjenih relativnih vrednosti parametrov posameznih elementov omrežja se v primeru enopolnih okvar nadomestno vezje v sistemu simetričnih komponent močno poenostavi. Tak pristop omogoča direktno uporabo znanih parametrov impedanc posameznih elementov omrežja.
1 Introduction
The phase-to-earth-faults, i.e. in the medium-voltage networks constitute the major part of the fault events (see figure 1). Especially in view of the so-called extinction limit and also, if dangerous touch voltages occur at such faults, the determination of the fault current is very important.
□ short circuits ■ earth faults
Jli
110 kV
20 kV
10 kV
Recently, the question of usefulness of operating the earth-fault-compensated networks has been discussed1. As a result of the current state of discussion and regardless of the justified exceptions, such as aged cable networks, this operating regime is often useful. Especially in mixed networks, i.e. those with a large proportion of the overhead lines, operating with no fault-compensated neutral point would lead to a massive increase in the earth-fault current which would affect the substation earthing design and might necessitate upgrading the plant earthing apparatus.
According to the valid standards the plant safety at an earth-fault, the earth-fault current needs to be calculated.
Fig. 1. Statistics for the single- and multi-phase faults, over voltage levels [1]
1 In the case of resonant earthed networks (with Petersen coil) the capacitive earth-fault current is compensated in such a way that at the fault point the earth-fault current as well as the resulting earth potential rise (EPR) are minimized by the arc suppression coil (Petersen coil). In most cases, this induces an automatic termination of the fault arc in overhead systems. Thus the necessity of switching off the faulted line is eliminated. The resonant earthed network can be operated continuously in the event of a single-pole fault and the reliability of supply of this network is not impaired.
Received 12 October 2015 Accepted 4 March 2016
A SIMPLIFIED PROCEDURE TO DETERMINE THE EARTH-FAULT CURRENTS IN COMPENSATED NETWORKS
123
3 Methodology
Based on the theory of symmetrical components, the equivalent circuit for a phase-to-earth-fault is a series circuit of positive-, negative-, and zero-sequence components (see Fig. 3). It is simplified by neglecting the positive- and the negative-sequence system in order to calculate the earth-fault current (ground fault in the earth-fault-compensated network).
Fig. 2. Standing arc in a substation switchgear
2 Task
The earth-fault current is calculated mainly for two reasons:
•	To determine of the residual current at the earth-fault location (i.e. the earth potential rise (EPR) and the associated hazards)
•	To calculate the phasors of the zero-sequence currents and the accompanying residual voltages at various points in an electrical network in order to correctly determine the protection settings concerning their sensitivity and pick-up reliability correctly. This approach is often referred to as "earth-fault engineering."
Usually, the method of symmetrical components is used to accomplish the above. In practice, the use of the network calculation programs for a detailed analysis is recommended.
Experience shows that using the method of symmetrical components requires a considerable amount of theoretical knowledge. Because of the necessary matrix operations and scaling factors, it is often complicated and may lead in many cases to calculation errors.
Fig. 3. Equivalent circuit for the single-pole fault with relevant parameters, [1]
The derivations are based on the theory of symmetrical components. Accordingly, the methodology is explained stepwise.
The accuracy of the simplified calculations is usually ± 10%. The zero-sequence impedances of any neutral star-point transformer and its positive-sequence impedances or of the upstream high-voltage grid -expressed by the short-circuit power Sk ", as well as the angle of the earth return-path factor ("k0"), are neglected.
Comments regarding the simplifications: Numerical analysis of the current flow - represented by the symmetrical component system - shows that the series connection of the positive-, negative-, and zero-sequence components can be interpreted as a series connection of the phase voltage of phase L1 as a driving source voltage with an internal source impedance, with a high-ohmic load. The positive- and negative-sequence system can be considered as a voltage source
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HUFNAGL, FICKERT, SCHMAUTZER
impedance, whereas the zero-sequence system represents the source load.
As to the typical numerical values, the series connection of the positive- and negative-sequence is only a part of the total impedance (positive-, negative-and zero-sequence) and can therefore be neglected.
Comment on the following derivation steps: Starting from a preliminary stage (considering the earth-fault current in a TN-system), the earth-fault current is determined consecutively for the following network types:
a)	earth-fault in a TN-system (preliminary stage)
b)	network with a current-limiting star-point resistor
c)	network with an insulated neutral
d)	network with a Peterson arc-suppression coil: fundamental and harmonic currents
e)	network with a Peterson arc-suppression coil and short-term resistance earthing of the neutral point
At selected locations, the results are demonstrated by simple calculations. In each of the above steps, only the network elements are added which are necessary to understand the methodology and to obtain the required calculation accuracy.
4 Earth-Fault Calculation in a TN
SYSTEM (PRELIMINARY STAGE)
In a first step, an earth-fault in a TN system (= short circuit) between a conductor and PEN conductor is considered (see Fig. 4 and 5)
L§—
Rl
-in
UpfHHl
RL
1
Fig. 5: Physical current flow during earth-fault in a TN system; F ... fault location
According to the illustration the earth-fault current is given by
If, tn = Uph / (Rl+ Rc + Rl) = Uph / (2-Rl + Rc) (1) IF, TN ... earth-fault current at the fault location UPh ... phase-to-earth voltage Rl ... resistance of the phase conductor Rc . contact resistance at the fault location When taking into account the influence of the inductive line impedance, this becomes
If = Uph / (Zl + Rc + Zpen) ~Uph / (2-Zl + Rc)	(2)
ZL. impedance of the phase conductor ZpEN. impedance of the pEN conductor
This relationship can be simplified in approximation to the system of symmetrical components, neglecting the series connection of the positive and negative sequence respectively (see figure 6).
Fig. 4: earth-fault in a TN system: circuit diagram; F fault location
Fig. 6. Simplified representation of a ground fault in TN systems - equivalent circuit, F ... fault location
A SIMPLIFIED PROCEDURE TO DETERMINE THE EARTH-FAULT CURRENTS IN COMPENSATED NETWORKS
125
5 Calculation of the earth-fault
CURRENT IN A RESISTANCE-GROUNDED NETWORK
To answer the core question of the earth-fault calculation "How is the current path of the earth-fault current closed, looking at the return-path of the current loop from the ground into the elsewhere highly galvanically insulated power network?" it is necessary to consider the following current loop (see figure 7): • substation busbar ^ phase conductor ZL ^ line-to-earth transition resistance Rc ^ earth-return-
path impedance Zm ^ star-point resistance RStp substation busbar

Fig. 7. Earth-fault in a resistance-grounded network Electrical diagram; F ... fault location
Jw=-
M
V3
As shown above, this value results also (in accordance with the system of symmetrical components) from the slightly modified representation shown in Fig. 9.
Fig. 9. Earth faults in a resistance grounded network -equivalent circuit; F ... fault location
In practice the fault current is dominated by the neutral point resistance RStp.
6 Calculation of the Earth-fault current in a network with an Insulated neutral
To answer above question for this case the earth-to-conductor capacitances have to be considered (see Fig. 10).
Fig. 8. Physical current flow at an earth-fault in a resistance-grounded network; F ... fault location
For the relationship between line impedance ZL and earth return circuit impedance Zm, one can use the well-known expression of from the power-system protection.
Zm= ko •Zl with ko,cable = 0,8 or ko,OHL = 1
In addition, with a good approximation the circuit impedance ZL can be replaced by the line reactance Xl. As shown in figure 9, the earth-fault current IF is given by (1) where UN is the nominal voltage, RC is the contact resistance at the fault location and RStp is the neutral-point resistance.
(fic+fistp)2+[(1+fco)-*i]2
(1)
Fig. 10. Arc and capacitive earth-to-conductor current paths
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HUFNAGL, FICKERT, SCHMAUTZER
In a network with a galvanically insulated neutral, the path of the earth-fault current returning from the fault location is divided into different sub-current paths (IF,i to IF,n) by their respective single earth-to-conductor capacitances (see Fig. 11).
Fig. 11. Physical current flow in case of an earth fault in a network with an insulted neutral; F ... fault location
By adding these individual earth-to-conductor capacitances to the total capacity Xcap as well as the individual sub-current paths to the total current (IF, CAP), the following schematic diagram is obtained (Fig. 12).
I
©
Fig. 12. Earth fault in a network with an insulated neutral -electrical diagram; F ... fault location
As shown in Figs. 11 and 12, the earth-fault current IF consists of the capacitive current contributions of each network segment as follows:
In a cable network with a cable length lcable, each km of the cable length contributes an earth fault current of about 3 A. 1 km of the overhead line contributes approximately 1/30 of this value. As a result, the capacitive earth-fault current flowing in the network is
(with a neglectable line-to-ground transition resistance, worst-case):
^F = ¡-cable
■ I
cable
+ 1
OHL
OHL
(2)
where: LaHe, lo
1 cable
1 OHL
-	total length of the cables/overhead lines,
-	capacitive current contribution per kilometer of the cable
-	capacitive current contribution per kilometer of the overhead line
The value calculated according to (2) can also be derived (similarly to the system of symmetrical components) from a slightly modified presentation shown in Fig. 13.
X grid.
cap-
Uphii^JC	
	fi:
If	©
	
= ju°	
Fig. 13. Earth fault in a network with an insulated neutral -equivalent circuit; F... fault location
According to Fig. 13, the equivalent total network capacitance X?ll,LC(l|;i for the fault location "F" is the internal impedance of the network.

uPhase
grid,cap lcableicable+lOHLÎoHL
(3)
Remark: If the line-to-ground transition resistance cannot be neglected (Rc > some 100 Q), the earth-fault current is calculated from the complete current loop:
• Transformer ^ line-to-ground transition resistance Rc ^ equivalent total network capacitance Xgrid,cap
h =
UN
Rc+Xgrld,cap
(4)
A SIMPLIFIED PROCEDURE TO DETERMINE THE EARTH-FAULT CURRENTS IN COMPENSATED NETWORKS
127
7 Calculation of the earth-fault
CURRENT IN A RESONANCE-GROUNDED NETWORK - CALCULATION OF THE 50 HZ COMPONENTS
To answer the above core question for this case it is necessary that besides the earth-to-conductor capacitances the newly created current return path through the Petersen arc suppression coil is also considered (see Fig. 14).
Fig. 14. Physical current flow for an earth fault in an earth-fault compensated network; F ... fault location
As in the procedure for a network with an insulated neutral, the following circuit diagram (Fig. 15) is used:
result, the following total earth-fault current is obtained (with the line to ground transition resistance at the fault location):
^F Ipet Igrid,cap,ef
T _ Uphase _ fi
'F = v	\lcable
xPet
Ip W. ■ Igrid,cap,ef
■ I.
cable
+ 1
OHL ■ lOHL
(5)
(6) (7)
IPet... current through the Petersen arc-suppression coil XPet ... reactance of the Petersen arc-suppression coil m ... degree of over-compensation of the network (in pu.)
As in the previous section the above value (7) can be obtained from a slightly modified presentation (similarly as for the system of symmetrical components) given in Fig. 16.
Fig. 16. Earth fault in an earth-fault compensated network; equivalent circuit
The total reactance of the network given in Fig. 16 can be calculated using (8).
y
nri,
uPhase
grid,total	¡p	m.I
uPhase
m^'grid,cap,ef
(8)
Fig. 15. Earth fault in an earth-fault compensated network; F ... fault location
As shown in Figs. 14 and 15, the earth-fault current IF is again obtained by superposing the capacitive current contribution of individual sections of the network (Igrid,cap,ef), but this time it is also superimposed with the inductive coil current IPet of an opposite sign. As a
Remark: The value of Xgnd totai can also be obtained for control purposes from the following relationship
Xgrid,total = Xgrid,cap // XPet
f) // (UPhase/IPet)
The earth-fault current is calculated from the complete current loop according to (4).
)
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HUFNAGL, FICKERT, SCHMAUTZER
• substation busbar ^ line-to-ground transition resistance Rc ^ source reactance of the network Xgrid,total ^ substation busbar
Numerical example 1:
•	The earth fault in an extended 20 kV cable network (nine feeders of 10 km each)
•	The Petersen arc suppression coil is by 3.7 % overcompensated
•	The line-to-ground transition resistance is 100 Q
UPhase = 20.000 V / V3 = 11.547 V
Igrid,cap,ef = Uable^'cable + lOHL •i' OHL
= nine feeders of 10 km 3 A/km + 0 = 270 A
Ipet = (1+m) Igrid,cap,ef = (1 +0,037)^270 A = 280 A
Xgrid,total = UPhase / IF
= Uphase / (m^ Igrid,cap,ef)
= 11.547 V / 10 A =1155Q
If = Uphase / V(Rc2+ Xgrid,total2) = 11.547 / V (1002+11552)
= 9,9 A
network with an insulated neutral. The factor of the line reactances taken into account is f/50Hz.
Fig. 17. Earth fault in an earth-fault compensated network; equivalent circuit
8 Calculation of the earth-fault
CURRENT IN A RESONANCE- GROUNDED NETWORK - CALCULATION OF THE 250-HZ-COMPONENTS
When calculating the 250 Hz conditions, the following is taken into account (see Fig. 17):
1.	The driving voltage is the 250 Hz component of the pre-fault phase-to-earth voltage UPhase,250 = p250 UN/V3, where p250 is the level of the 5th harmonic in the line voltage.
2.	At 250 Hz each capacitive reactance is five times smaller than at 50 Hz.
3.	At 250 Hz the reactance of the Petersen coil is five times higher than at 50 Hz. Therefore, its reactance in relation to the system capacitances can be neglected due to the parallel circuit
4.	The line reactances are five times higher than at 50 Hz and should not be neglected.
In accordance with Fig. 17, the Petersen coil XPet is omitted because of the now negligeably high impedance. This means that at higher frequencies, an earth-fault-compensated network behaves like a
Remark 1: To allow for simplification (from the whole line impedance only the line reactance is taken) at a certain line length, the reactive components are mutually canceled which leads to a resonant condition. In such an idealised case, the current is infinite which is not realistic. As the effective resistances are still present, the harmonic content in the earth-fault current is effectively attenuated. It is only here that an exact calculation is required.
Remark 2: The value of Xgrid,total is obtained for control purposes from the following relationship:
Xgrid,total,250=(Xgrid,cap50/5)//(XPet 5) ~ (-UPhase/Igrid,cap,ef)/5
Remark 3: At 250 Hz the reactance value of the loop from the fault location to the substation busbar and back is five times the value of the 50-Hz reactance. The earth-fault current is calculated from the complete current loop:
• substation busbar ^ line-to-ground transition resistance Rc ^ phase conductor XL 250 ^ network capacity Xgrid,total,250 ^ earth return path impedance Zm250 = k0XL250 ^ substation busbar
Ujv
h.250 = P250 • ,	V3	(9)
lR^+[(l+k0)-XL,250-Xgrid,cap,250]2
A SIMPLIFIED PROCEDURE TO DETERMINE THE EARTH-FAULT CURRENTS IN COMPENSATED NETWORKS
129
Numerical example 2:
•	The earth fault in an extended 20 kV cable network (nine feeders of 10 km each with x' = 0,13 Q/km)
•	The Petersen arc-suppression coil is by 3.7 % overcompensated
•	The line-to-ground transition resistance is 100 Q
•	The fault location is 3 km away from the substation busbar
XL = 3 km • 0.13 Q/km = 0.39 Ohm
ko = 0.7
P250= 2 %
Igrid,cap,ef = lcable i'cable + lOHL ^OHL
= nine feeders • 10 km- 3 A/km + 0 = 270 A
UPhase,250= p250U/V3 = 0,02 -20.000 V / V3 = 231 V
Xgrid,total,250 ~ (-UPhase/Ignd,total) /5
= (11547 V / 270 A) / 5 = 8.6 Q
X
l • 5 = 3 km • 0.13 Q • 5 = 1.95 Q
IF,250 = p250-(UN/V3)W(Rc2+[(1+k0)-XL,250-Xgnd,cap,250.
= 231 / V (1002 + [(1+0.7) 1.95 - 8.6] 2)= 2.3 A
nan ?So]2)
Fig. 18. Physical earth-fault current flow in a network with a resonance- and short-term resistance-grounded neutral point; F ... fault location
9 Calculation of the earth-fault
CURRENT IN A RESONANCE- AND SHORT-TIME RESISTANCE-GROUNDED NETWORK
In a resonance-grounded network, a more or less low-ohmic resistance is in some cases briefly switched parallel to the Petersen coil to improve the earth-fault location by evaluating the resistive component. A resistor parallel to the Petersen coil is taken into account. Hence, another current path is activated through this neutral-point resistor causing resistive component IR. Figs. 19 and 20 show the physical current flow and the derived circuit diagram, respectively.
Fig. 19. Earth fault in a network with a resonance- and short-term resistance-grounded neutral point -Electrical diagram; F ... fault location
According to Fig. 20, the earth-fault current IF is composed of the capacitive current contribution of the individual sections of the network (Igridcap,ef), the inductive coil current IPet and the current through the neutral-point resistance IR.
When it comes to a real network technology, resistive current IR becomes so high that the line impedance to fault point "F" can - in contrast to the previous section -no longer be neglected.
As in the previous sections, the method of symmetrical components is taken as a basis for a simplified presentation.
= x
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HUFNAGL, FICKERT, SCHMAUTZER
Fig. 20. Earth fault in a network with a resonance-g and short-term resistance-grounded neutral point-equivalent circuit diagram; F ... fault location
To answer the core question of the earth fault calculation, the following current loop has to be considered (see Fig. 20). • substation busbar ^ line-to-ground transition resistance ^ phase conductor ZL ^ parallel connection of the source impedance of the network Xgrfd.totai (network capacity // Petersen coil inductance) and the neutral point resistor RStp ^ line-to-ground transition resistance Rc ^ earth return path impedance Zm = k0 ZL ^ star point resistance RStp ^ substation busbar
h =
uPha
(ZL+Rc+Zm+*f,ridXotal+RRStV) Agrid,total +KStp
(10)
10 Conclusion
The paper proposes a procedure to calculate the earth-fault current, for the medium-voltage network with a tolerable imprecision.
Using these attained simplified calculations makes the task of determination of the earth-fault currents considerably easier compared to the other approaches.
References
[1]	Lothar Fickert: Schutz und Versorgungssicherheit elektrischer Energiesysteme. Vorlesungsskript, Institut für elektrische Anlagen, Technische Universität Graz, 2008
[2]	D. Oeding, B.R. Oswald: Elektrische Kraftwerke und Netze, Springer-Verlag, 6. Auflage, Berlin Heidelberg New York, 2004
[3]	G. Herold: Elektrische Energieversorgung III (Drehstrommaschinen - Sternpunktbehandlung -Kurzschlussströme, J. Schlembach Fachverlag, Weil der Stadt, 2002
[4]	ÖVE/ÖNORM EN 60909-3, Kurzschlussströme in Drehstromnetzen - Teil 3: Ströme bei Doppelerdkurzschluss und Teilkurzschlussströme über Erde
Elisabeth Hufnagl received her diploma degree in electrical engineering from Graz University of Technology in 2014. Since then she has been a scientific assistant at the Institute of Electrical Power Systems at the same university.
Lothar Fickert received his Diploma Engineer and PhD degree in 1972 and 1975, respectively. For 25 years he worked in power industry. In 1998 he became full professor and head of the Institute for Electrical Power Systems, Graz University of Technology. His main research activities are protection, power quality and grid design and operation.
UPhase
RStpXgrid,total RStp+xgrid,total
+ Zv (1+ko) + Rc
RStp ... star-point resistor
Xgnd,totai .. source impedance of the network
(11) Ernst Schmautzer is a senior researcher at the Institute of Electrical Power Systems, Graz University of Technology. His main research interests are in efficient use of electric energy, smart grids, smart building installations, low-frequency electromagnetic fields, EMC, EMI, grounding problems and protection measures.

Uphase
grid,total
mlgrid,cap,ef
(12)
m ... degree of the network overcompensation in p.u. ZL ... impedance of the phase conductor, Zm ... earth return path impedance of the "ground" Rc ... line-to-ground transition resistance at the fault location
k0 ... Ratio between the earth return-path impedance and the impedance of the phase conductor