Hence,

If L is the inductance fo the arc suppression coil (Peterson coil or ground-fault neutralizer) connected between the neutral and the ground, then

In order to obtain satisfactory cancellation (neutralization) of arcing grounds, the fault current IL flowing through the arc suppression coil should be equal to resultant charging current Icr

Therefore, for balance condition

IL = Icr

Or [From Eqs. (4.161) and (4.162)]

Or

The inductance of the arc suppression coil is calculated from Eq. (4.163).

The method of resonant grounding is usually confined to medium-voltage overhead transmission lines which are connected to the system generating source through interruption due to transient line-to-ground faults which will not be possible with other methods of grounding. In this method of grounding, the tendency of a single phase to ground fault developing into two or three-phase fault is decreased.

Normal time-rating of arc suppression coils used on systems on which permanent ground faults can be located and removed promptly is ten minutes whereas continuous time-rated coils are used on all other systems. In such cases, it is also usual to provide automatic means as shown in Fig. 4.77 to bypass the arc suppression coil after some time. If for any reason more current flows through the arc suppression coil, a circuit breaker closes after a certain time-lag and the ground-fault current flows through the parallel circuit by-passing the arc-suppression coil.

The circuit breaker (C.B.) which is normally open is closed by the trip coil when the relay operates after a predetermined time. Thus the fault current is by-passed through the resistor branch.

(v) Voltage transformer grounding In this method of neutral grounding, the primary of a single phase voltage transformer is connected between the neutral and the ground as shown in Fig. 4.78. A low resistor in series with a relay is connected across the secondary of the voltage transformer. The voltage transformer acts as a very high reactance grounding device and does not assist in mitigating the overvoltage conditions which are associated with ungrounded (isolated) neutral system. A sytem grounded through voltage transformer operates virtually as an ungrounded system. A ground fault on any phase produces a voltage across the relay which causes operation of the protective device.

(vi) Grounding through grounding transformer In cases where the neutral of a power system is not available for grounding or where the transformer or generator is delta connected, an artificial neutral grounding point is created by using a zig-zag transformer called ‘grounding transformer’. This transformer does not have secondary winding and it is a core-type transformer having three limbs built in the same fashion as that of the power transformer. Each limb of the transformer has two identical windings wound differentially (i.e., directions of current in the two windings on each limb are opposite to each other) as shown in Fig. 4.79(a). Since the two identical windings on each limb are wound differentially, under normal conditions, the total flux in each limb is negligibly small and, therefore, the transformer draws very little magnetizing current. The grounding transformer is of short-time rating, usually 10 seconds to 1 minute. Therefore, the size of such transformer is small as compared to the power transformer of the same rating.

• If a zig-zag transformer is not available for grounding, a special small size stardelta transformer can be used without loading the delta side as shown in Fig. 4.79(b). This transformer is also known as grounding transformer and it is a step-down transformer. The star connected is in delta and generally does not supply any load but provides a closed path for triple harmonic currents to circulate in them. Under normal conditions, the current in the transformer is only its own magnetizing current. However, large current may flow in the event of a single line to ground fault condition. Hence, it should be of sufficient rating to withstand the effect of line to ground faults.

Present Practice in Neutral Grounding

(i)Generally, one grounding is prvided at each voltage level. In power system, there are various voltage levels between generation and distribution. At least one grounding is normally provided at each voltage level.

(ii) Grounding is provided at the power source end and not at the load end. If power source is delta connected, grounding is provided by means of grounding transformer rather than grounding at load end.

(iii) Grounding is provided at each major source bus section.

(iv)The generators are normally provided with resistance grounding whereas synchronous motors and synchronous capacitors are provided with reactance grounding.

(v) When several generators are operating in parallel, only one generator neutral is grounded. If more neutral are grounded, disturbance is created by the zero sequence components of the circulating currents.

(vi)When there are one or two power sources, no swithing equipment is used in the grounding circuit.

(vii) When several generators are connected to a common neutral bus, the bus is grounded either directly or through reactance.

(vIII)Effectively (solid) grounding is used for the systems of low voltages up to 600 volts and for systems of high voltages above 33 kV whereas resistance or reactance grounding is used for systems of medium voltages between 3.3 kV and 33 kV.

4.8.3 Equipment Grounding (Safety Grounding)

Equipment grounding means electrically connecting non-current carrying metallic parts (i.e., metallic frame, metallic enclosure, etc.) of the equipment to ground. The object of equipment grounding is to ensure safety against electrical shocks to operating personnel and other human or animal bodies by discharging electrical energy to ground. Equipment grounding also ensures protection against lightning. If insulation fails, there will be a direct contact of the live conductor with the metallic parts (i.e., frame) of the equipment. Any person in contact with the metallic part of the equipment will be subjected to electrical shock which can be fatal. Thus equipment grounding deals with grounding of non-current carrying metallic parts of the equipment to ensure safety of personnel and protection against lightning. Equipment grounding is also called ‘safety grounding”.

• Under fault conditions, the non-current carrying metallic parts of an electrical installation such as frames, enclosures, supports, fencing, etc., may attain high voltages with respect to the ground so that any person or stray animal touching these or approaching these will be subjected to voltage which may result in the flow of a current through the body of the person or the animal of such a value as may prove fatal. The magnitude of the tolerable current flowing through the body is related to duration.

It has been found experimentally that the safe value of the current which a human body can tolerate is given by

for f < 3 secs

And Ib = 9 mA for t > 3 secs

Where Ib is the rms value of the body current in amps and t is the time in seconds

Tolerable Step and Touch Voltages

When a fault occurs, the flow of fault current to ground results in voltage gradient on the surface of the ground in the vicinity of the grounding system. This voltage gradient may affect a person in two ways, viz., step or foot to foot contact and hand to both feet or touch contact.

Step Voltage

It is the voltage between the feet of a person standing on the floor of the substation. With 0.5 m spacing between two feet (i.e., one step), during the flow of ground fault current through the grounding system. Fig. 4.80 shows the circuit for step contact. Rf is the grounding resistance of one foot (is ohms), and Rb is the resistance of body (in ohms).

The grounding resistance of one foot (Rf) may be assumed to be 3ps where ps is the resistivity of the soil near the surface of ground. Rb is assumed to be 1000 ohms.

Therefore, tolerable value of the step voltage is given by [from Fig. 4.80(b)].

Vstep = (Rb + 2Rf) Ib

Substituting the values of Rb, Rf and Ib [from Eqs. (164) and (165)], we get

Vstep = (1000+6ps)0.165/ for t < 3 sec

= (1000 + 6ps)0.009 for t > 3 sec

Touch Voltage

It is the voltage between the fingers of raised hand touching the faulted structure and the feet of the person standing on substation floor. Figure 4.81 shows the circuit for touch contact.

From Fig. 4.81(b), the tolerable value of touch voltage is given by

Vtouch = (Rb + Rf/2) Ib

= (1000 + 1.5 Ps) 0.165/ for t < 3 sec

= (1000 + 1.5Ps) 0.009 for t > 3 sec

The actual values of step voltage and touch voltage should be less than the tolerable values given by Eqs. (4.167) and (4.168) respectively.

Functions of Substation Grounding System

(i) To ensure safety against electrical shocks to operating and maintenance personnel

(ii) To discharge electrical charges to ground during ground faults and lightning

(iii) To provide grounding overhead shielding wires

(iv)To ensure freedom from electromagnetic interference in communication and data processing equipment in the substation.

EXERCISES

1) What do you mean by a fault? Discuss the causes of fault on a power system. What are the types of faults that occur in any power system?

2) What is the necessity of fault analysis? What do you mean by fault level at any point on the power system?

3) What is a per-unit system? What is its significance in power system analysis? What are the advantages of per-unit system? How can the per unit value of impedance on one base be changed to per unit value on a different base?

4) What is an impedance diagram of a power system? What are the steps to be followed for drawing the impedance (reactance) diagram from the one-line diagram of a power system.

5) Distinguish between symmetrical and unsymmetrical faults. What assumptions are made to simplify symmetrical fault calculations?

6)What do you mean by symmetrical short circuit capacity of a power system? How can it be calculated?

7)Discuss the steps to be followed in the calculations for a symmetrical fault.

8) Discuss the criterion for the selection of circuit breaker for a particular location in a power system.

9.A generating station consisting of two generators, one of 20 MVA, 11 kV, 0.20 pu reactance with the unit transformer of 11/132 kV, 20 MVA, 0.08 pu reactance and another of 30 MVA, 11 kV, 0.20 pu reactance with the transformer of 11/132 kV, 30 MVA, 0.10 pu reactance, transmits power over two 132 kV, 3 phase transmission lines each with reactance of 300 ohm per phase.

The 132 kV lines are bussed at both ends. Determine current in the windings of generators when a 3-phase symmetrical fault takes place on the 132 kV bus bars at the receiving end station

10. The power system shown in Fig. 4.82 is initially at no load. Calculate the subtransient fault currents that results when a three-phase symmetrical fault occurs at F, given that the transformer voltage on high voltage side is 66 kV.

11. Figure 4.83 shows a power system whre an infinite bus feeds power to two identical synchronous motors through a step down transformer. The ratings are given on the diatram. Neglecting cable impedances, find the subtransient fault current and subtransient current at breaker B, when 3-phase fault occurs at F.

12. There 25 MVA generators each having a reactance of 0.2 pu are operating in parallel. They feed a transmission line through a 75 MVA transformer having a per unit reactance of 0.25 pu. Find the fault MVA for a fault at the sending end of the line.

13. A 20 MVA, 11 kV, 3-phase generator having a transient reactance of 0.2 pu feeds a 20 MVA, 11kV, 3-phase synchronous motor having a transient reactance of 0.5 pu through a short line. The line has a reactance of 0.30 ohm. A three-phase fault occurs at the motor terminals when the motor is taking an input of 15 MW at unity p.f. The line voltage across motor terminals at the time of fault is 10.5 kV. Find the total current supplied by the generator and motor to the fault.

14. What are the types of reactors used for limiting the symmetrical fault current? Discuss the various locations of the current limiting reactors.

15. There are two generators connected to a common 6.6 kV bus bar and a feeder is taken out through a step-up transformer as shown in Fig. 4.84. Find the ohmic value of the current limiting feeder reactor in order that fault current is limited to four times the full load current, if there is a 3-phase fault just near the raceactor.

16. The section bus bars A and B in Fig. 4.85 are linked by a bus bar reactor rated at 5 MVA with 10% reactance drop. On bus bar A there are two generators each of 10 MVA with 10% reactance and of B two generators, each of 8 MVA with 12% reactance. Find the steady MVA load fed into a dead short circuit between all phases on B with bus bar reactor in circuit.

17. Each of the three generators in a generating station has a short circuit reactance of 10% based on the respective ratings of 75 MVA, 90 MVA and 110 MVA. Each machine is connected to its own sectional bus bar and ech bus bar is connected to a tiebar through a rreactor of 5% reactance based on the rating of the generator connected to it. Calculate the short circuit current in amperes and the MVA fed into the short circuit occurirring between the bus bars of the section to which the 110 MVA generator is connected. The generator voltage is 33 kV.

18. A hydro station has five 80 MVA, 11kV generators each of 10% reactance connected in parallel. The generated voltage is stepped up to 220 kV by three 160 MVA transformers, each of 15% reactance, also connected in parallel. The 220 kV system supplies power to two sub-stations over two feeders having reactances of 12.1 ohms and 24.2 ohms per phase. Calculate for each sub-station separately, the rms value of the symmetrical fault current when a 3-phase short circuit (fault) occurs at the sub-station. Assume that the generators are not loaded prior to the fault.

19. For the system shown in Fig. 4.86 determine the fault current and rupturing capacity of the breaker of feeder C.

20. A small generating station has two generators of 2.5 MVA and 5 MVA and percentage reactances 8 and 6 respectively. The circuit breakers are rated 150 MVA. Due to increase in system load it is intended to extend the system by a supply from the grid via a transformer of 10 MVA rating and 7.5% reactance. If the system voltage is 3.3 kV. Find the reactance necessary to protect the switchgear.

21. There 6.6 kV generators G1, G2 and G3, each of 10% leakage reactance and MVA ratings of 40, 50 and 25 respectively are interconnected electrically, as shown in Fig. 4.87, by a tie bar through current limiting reactors, each of 12% reactance based upon the rating of the machine to which it is connected. A three-phase feeder is supplied from the bus bar of generator G1at a line voltage of 6.6 kV. The feeder has a resistance of 0.06Ω / phase and an inductive reactance of 0.12Ω/ phase. Estimate the maximum MVA that can be fed into a symmetrical short circuit at the far end of the feeder.

22. Discuss the principle of symmetrical components. Derive the necessary equations to convert phase quantities into symmetrical components, and vice versa.

23. Derive the necessary equation to determine the fault current for a single line to ground fault. Discuss the interconnection of sequence networks for this type of fault.

24. Discuss the situation under which a single line-to-ground (L-G) fault can be more severe than a 3-phase symmetrical fault.

25. Show that positive and negative sequence currents are equal in magnitude but out of phase by 1800 in case of a line-to-line fault. Draw a diagram showing the interconnection of positive and negative sequence networks. Why zero-sequence network remains unconnected?

26. Derive equations for sequence currents in case of a double line to ground fault. Draw a diagram showing interconnection of sequence networks for this type of fault.

27. The positive, negative, and zero-sequence reactances of a 20-MVA, 13.2 kV synchronous generator and 0.3 pu, 0.2 pu and 0.1 pu respectively. The generator is solidly grounded and in not loaded. A line-to-ground fault occurs on phase a. Neglecting all the resistance, determine the fault current.

28. The one-line diagram of system is shown in Fig. 4.88. The ratings of the generator and transformers are given on the diagram. Fault occurs at point F. Determine the fault current for

(i) 3-phase symmetrical fault(ii) L-G fault(iii) L-L fault(iv) 2L-G fault.Assume system to be initially on no load.picture29. Six 6.6 kV, 3-phase generators are connected to a common set of bus

bars. Each has positive, negative and zero sequence reactances of 0.90Ω, 0.72Ω and 0.30Ω respectively. A ground fault occurs on one bus bars. Determine the value of the fault current if

(i) All generator neutrals are solidly grounded.(ii) If one generator neutral is grounded through a resistance of 0.2 ohm,

the others being isolated.(iii) If only one generator neutral is solidly grounded, the others being

isolated.

30. Two generators rated 11 kV, 5 MVA having X1 = 0.15, pu X2 = 0.12pu, X0 = 0.15 pu are operating in parallel. A single line-to-ground fault occurs on the bus bar. Calculate the fault current if

(i) Both generator neutrals are solidly grounded,(ii) Only one generator neutral is solidly grounded,(iii) Both neutrals are isolated period 31. A double line to ground fault occurs at F in the system shown in

Fig. 4.89. Draw the sequence networks for the system, and calculate the line current Ib.

32. For a generator the ratio of fault current for line-to-line fault and three-phase faults is 0.866. The positive sequence reactance is 0.15 pu. Calculate negative sequence reactance.

33. Three 6.6 kV, 3-phase, 10 MVA generators are connected to a grid. The positive sequence reactance of each generator is 0.15 pu while the negative and zero-sequence reactances are 75% and 30% of positive sequence reactances respectively. A single line-to-ground fault occurs on the grid bus. Determine the fault current if