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Grid potential rise (GPR) gets worse as the fault current and earth impedance increases. One way of limiting the GPR in high fault current substations is by adding more copper in the earth mat, thus reducing the earth impedance. This however, is an expensive approach. An overstated fault current can result in an uneconomical substation earth mat design. It is therefore very important to know the maximum current that will flow to earth via the grid for various possible earth-fault locations.For the design and specification of power system equipment, it is necessary to know what the maximum system fault currents will be at the specific location. It is normal practice to compile databases of maximum earth fault levels, using specialised power flow software. However the maximum earth current required in order to calculate the GPR in a substation applies strictly to the current that flows into the earth itself.

In most cases a lower level of resistance is ef fect ive due to the presence of parallel earth paths through shield wires of overhead lines, tower footing resistances, etc. These parallel paths cause the fault current to divide into multiple paths to complete the return path to the source. In such cases, the calculated level of fault current may well be considerably greater than the actual current that will flow through the earth grid. Ignoring the parallel paths and only considering these high fault current levels could result in an over design of the substation earth grid. In this paper, different transformer configurations and fault locations are analysed. Each analysis is approached by drawing the phase-sequence network based on the single line diagram of the faulted circuit. The actual phase currents and grid current are then calculated from the per unit values and shown on the three-phase circuits. Finally the rise in grid potential and the step and touch potentials may be calculated using the maximum grid current values.

Configurations considered

The following transformer configurations within substations are considered:

Dyn, HV/MV transformer

YNd, HV/MV transformer

YNa(d), auto-transformer with delta connected tertiary windings.

Earth current considerations for earth mat design in HV substations

by the late Alan Ware. Edited by Jaco Mostert, Hatch Africa, and Johanette van der Merwe, EON Consulting

The design of substation earth grids is involved with limiting grid potential rise (GPR), which is a result of earth fault currents that flow to earth through an earth grid impedance. Overstated substation fault current can potentially result in an uneconomical substation earth mat design. This paper illustrates that it is necessary to get a clear understanding of the specific substation configuration in order to design a safe and economical earth grid.

The following abbreviations are used in the diagrams:

Rg : Equivalent earth resistance of all parallel paths

T: Transformer

H: High voltage winding

M: Medium voltage winding

L: Low voltage winding (tertiary)

Configuration 1: Dyn, HV/MV transformer

The simplest case is that of a single line feeding a substation containing a Dyn, HV/MV transformer where the MV neutral is earthed via a resistor (Fig. 1a). In the case of an HV earth fault within the substation, fault current flows down the line from the remote source. All of the current returns to the source neutral via Rg . If the earth fault level is high, coupled with a significant value of Rg , the GPR may be excessive.

The phase sequence networks for HV earth faults are straight forward (Fig. 1b). IF flows through Rg , which is series-connected in the fault path.

GPR, Eg , is the product of Rg and the current in it. In this case the current is 3 · I0 , therefore the zero sequence voltage, Eg , is:

Eg = 3I0·Rg (p.u.) (1)

Eg = I0·Rg0 (p.u.) where Rg0 = 3Rg (2)

Eg = I0·Rg0 · Ebase kV

It is clear that I0 is equal to I0S . The zero sequence voltage over Rg is given by:

Eg = I0·Rg0 3·I0S·Rg (p.u.)

For this scenario, the earth fault current is equal to the total fault current.

In the case of an MV earth fault outside the substation, the fault current is limited by the MV earthing impedance ZN in series

Fig. 1a: Single line diagram: HV line terminated on Dyn transformer (HV fault shown).

Fig. 1b: Actual phase currents in the three phase network.(HV fault).

Fig. 1c: Phase sequence network: HV line terminated on Dyn transformer, earth fault on HV

side of transformer.

Fig. 2a: Single line diagram: HV line terminated on YNd transformer (HV fault shown).

Fig. 2b: Actual phase currents in the three phase network.(HV fault).

Fig. 2c: Phase sequence network: HV line terminated on YNd transformer,

earth fault on HV side of transformer.

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with Rg . (A similar phase sequence network can be derived for this scenario). A low ZNO

in series with high earth resistivity, or Rg0 , may result in increased problems with step and touch voltages. In such cases special attention may be necessary.

Configuration 2: YNd, HV/MV transformer

Now consider a substation containing a YNd, HV/MV transformer where the HV neutral is solidly earthed (see Fig. 2a), with an HV earth fault applied.

Fig. 2b shows the paths for actual phase currents. The total earth fault current is 3 · IY + IX (p.u.), while the GPR is given by IX · Rg (p.u.).

In order to calculate the GPR, it is necessary to consider the phase-sequence networks (Fig. 2c).

Rg0 is connected such that only I0S flows through it. Rg0 must neither be connected between F0 and N1 , nor between Z0T and F0 . On the rare occasion that the source earth resistance, RS , is significantly high, its value can be added to Rg , then the total resistance = Rg + RS.

Similar to Eqn. 2, GPR is the product of Rg and the current flowing in it. In this case the current is 3·I0S , therefore the zero sequence voltage, Eg , is:

Eg = 3I0S·Rg (p.u.) (3)

Eg = I0S·Rg0 (p.u.) where Rg0 = 3·Rg (4)

The GPR is determined by the earth current, I0S , which is less than the total fault current, I0. (I0S = I0 – I0T)

Configuration 3: YNa(d), Auto-transformer with delta connected tertiary windings

Now consider a step-down substation with an autotransformer of which the tertiary winding is delta connected (Fig. 3a).

Earth fault on primary side

When the earth fault is on the primary side of the autotransformer the current distribution is very similar to that for an YNd transformer. The three phase networks are shown in Fig. 3b.

The phase sequence networks are shown in Fig. 3c. Rg must be connected in the path of I0S , definitely not in the path of I0 , which is the total zero-sequence component of the fault current.

Similar to Eqn. 4, GPR is Eg = I0S·Rg0 (p.u.) and again the earth current is less than the total fault current.

Earth fault on secondary side

When an ear th fau l t occurs on the secondary side of an auto-transformer, calculations are rather more complicated. There are two distinct paths for zero-sequence current in the t ransformer windings (I0S and I0T in Fig. 4b). Each path and the division of current must achieve magnet ic balance in the core, i .e. ampere-turn balance in the windings.

The currents in the main windings are made up of all three of the phase-sequence components. For the secondary earth-fault, a component of zero-sequence current f lows in the tert iar y winding, matched by components in the main (series and common) windings. Additional components of zero sequence current flow also in the series and common windings, to oppose one another and achieve ampere-turn balance without the aid of the tertiary current.

The star point of the common windings is not representable in the phase-sequence networks. The per uni t values of the currents in the common windings are not manifest in the analysis of these networks, because the currents in the HV windings are combined directly with the currents in the common windings so there is no convenient base. To obtain actual currents the procedure in the example may be followed.

The phase-sequence networks should now be analysed to obtain per unit values. To calculate the current distribution in the windings, Kirchoff's law may be used. In many cases, where the source fault level is very high, the source current, IX , will be greater than 3·IY . But the reduction in GPR is still substantial when 3 · IY is correctly eliminated from the calculation.

Example

The earth current (4 p.u.) in Fig. 4c is therefore less than the total fault current (4,9 p.u.).

Conclusions

I t is clear f rom the analysis that the earth current is often less than the total fault current as may be obtained using fault calculation software. It is therefore necessary to get a clear understanding of the specific substation configuration in order to design a safe and economical earth grid.

References

The books listed refer to various aspects in this paper such as sequence component models and earth currents. [1] CHW Lackey: Fault Calculations, Oliver & Boyd,

Edinburgh, 1st edition, pp. 139 – 147, 1962. (Power transformer sequence impedances). Areva: Network Protection & Automation Guide, Areva T&D, Paris, 1st edition, pp. 57 – 60, July 2002. (Transformer zero sequence equivalent circuits).

[2] J L Blackburn: Symmetrical Components for Power Systems Engineering, Marcel Dekker Inc., New York, pp. 236 – 243, 1993. (Auto-transformer fault calculation).

Contact Jaco Mostert, Hatch, Tel 021 949-7836, jmostert@hatch.co.za

Fig. 3a: Single line diagram: Auto-transformer with delta connected tertiary windings.

Fig. 3b: Three-phase currents for earth fault on primary side of auto-transformer.

Fig. 3c: Phase sequence networks: Auto-transformer, earth fault on HV side.

Fig. 4a: Three-phase currents for earth fault on secondary side of auto-transformer.

Fig. 4b illustrates a 2:1 (e.g. 132:66 kV) step-down autotransformer with a fault on the MV side of the transformer. Feasible per unit

values illustrate the principle.

Fig. 4c: Feasible three-phase currents for earth fault on 66 kV side of 132/66 kV auto-transformer.