reading / British practice / Vol D - 1990 (ocr) ELECTRICAL SYSTEM & EQUIPMENT

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ASSOCIATED EQUATIONS

V, 7 V, + V2 = 0 = /2 =

V. = V2 + 1 2 = 0 • =0

V, = V2 = V,

I, + 1 2 + 0

V. = V, = 0 = 0

V, V 2 = V o 0

reactances (direct axis); T', T" are transient and subtransient short circuit time constants; and E, E and E" are the generator voltages behind synchronous, transient and subtransient reactances, respectively.

X„ is generator negative sequence impedance, X4' is generator subtransient reactance (quadrature axis) and R a generator stator resistance.

Induction motor representation

Data for induction motor performance and analysis are frequently in terms of starting and running values of stator and rotor equivalent resistance and reactance. If so, a suitable method of calculating fault current is based on:

X„ X m , X„, = stator reactance, magnetising reactance, rotor reactance at start.

Alternatively, a method similar to that used for synchronous machines may be used.

Evaluation of fault currents and voltages

Once the data for the network to be analysed have been assembled, the nodal admittance matrix Y is formed, as described earlier, and we have:

YV = I

remembering that, in a system with n nodes,

The impedance to ground reference at any node can be found from Y -1 , the inverse of the admittance matrix.

Consider a network with n nodes, the i th node of which the impedance to ground reference is to be found. If a current of I per-unit is injected into the network at node i, then looking at the matrix and two vectors in general terms we have:

Similarly, it can be shown that, with the same unit current injected at node i, Vj = Y 1 .

Thus the impedance to ground reference at node i is given by element Y j 1 , and the fault currents can be determined by knowledge of the element Y', the voltage at the node prior to the fault Vi, and the type of fault.

Further, it can be seen that only one row or column, the 11I, of the 1( -1 matrix is required in the calculation

of V.I. The inverses of the admittance matrices for positive, negative and zero sequence networks can be obtained and the fault current calculated.

If, for example, the fault is a line to ground fault, the three sequence impedances will be connected in series with the value of the fault resistance (if any) inserted. Let the summed network impedance be Z, having real and imaginary parts of R and jX, and the node voltage V be

Then the positive sequence real current, calculated from:

'real = pi real X (R + V imag ) X X]/(R 2 + X 2 )

and the positive sequence imaginary current I iniag from:

Iimag = [Vimag X (R — V,-„i) x X]/(R 2 + X2)

In this example, a single line to ground fault, the

negative and zero sequence currents will equal the poSitive sequence values due to the series connection

between the networks. In general, the fault current in each sequence network can be obtained from the solution of the symmetrical components network interconnected according to Table 2.19.

Once the fault currents Ifm, if(2) and I1(0) have been obtained, th? phase 'a' voltage under fault conditions for all busb It's can be evaluated by the application of the superposition theorem. For a short-circuit fault on busbar i he 'a' phase fault voltages at busbar j

are:

row and P h column of the inverses of the sequence admittance matrices.

branches can be determined. The fault current in a branch is the product of the branch admittance and the voltage difference between the ends of the branch. For branch i — j, the current flowing in phase 'a' is given by:

'rum = Yiicolvti(t) — vrio)]

Ini(2) = Y1 J (2) Pi1i(2) — v1;(2) I

where Yij(I), Yij(2) Yii(0) are the sequence admittances of branch i — j.

For a transformer branch, the sending end current is evaluated. A correction for tap is then applied to positive and negative sequence currents.

The fault voltages for positive, negative and zero sequence networks can then be calculated.

Using Thevenin's Theorem,

Vi(t) = Vi(P) — Y1j ( 1 ) 1i(1)

Generator and motor fault current contributions

These can be calculated from knowledge of the generator and motor equivalent admittances, the faulted network voltages and the generator and motor current contributions fed to the system prior to the fault.

=Yi(I)G Vf(1)

Y(2)6 t't. 2)

=Yi(0)G Vf(0)

where YIG is the value of the generator equivalent admittance. The subject of fault current contribution from loaded synchronous machines is discussed in references fill and [12].

3.2.3 Use of programs

The 'Unit at full load' condition provides a convenient base example for fault analysis of a station electrical system. All plant is assumed to be in service and normal running arrangements adopted. Figure 2.73 shows such a study. The system is for an advanced gas-cooled reactor (AGR) nuclear power station. It provides a convenient starting point or reference for further fault studies having different switching arrangements. Switchgear ratings for this station are:

On the diagram, 35* are 11 kV boards, B6* are 3.3 kV boards and B7* are 415 V boards (* = alphanumeric symbol(s))

Figure 2.73 shows that all fault levels are well within switchgear ratings.

Suppose now that the 132 kV/11 kV/I1 kV station transformer is to be taken out of service. This transformer has three windings; each of the two 11 kV secondary windings feeds an 11 kV station board. It is shown on Fig 2.73 as three separate transformers bounded by 33, J5A1 and J581. The procedure to take the transformer out of service is:

(a)Switch the first unit/station 11 kV interconnector on-load. Check that it picks up load.

(b) Open the associated station transformer 11 kV circuit-breaker.

(c)Switch the second unit/station 11 kV interconnector on-load. Check that it picks up load.

(d) Open the associated station transformer 11 kV circuit - breaker.

(e)Check that there is no load on the station transformer 132 kV circuit-breaker. If so, open it.

Figure 2.74 shows fault levels after operation (a), the first unit/station 11 kV interconnector B5A to B513, closed. The fault levels on the newly interconnected Ii kV boards are now much higher than those in Fig 2.73. Fault levels also rise to a lesser extent at other boards. All fault levels are within switchgear ratings — if this were not so, the switching operation would not be permitted. Figure 2.75 shows fault levels at the next stage in the switching sequence, after the first station transformer 11 kV circuit - breaker (J5A1 to B5A) is opened. Comparing this figure with the base case (Fig 2.73), note a moderate increase in fault level at the 11 kV unit board, B5D. This is due to the additional induction motor contribution, via the unit/station 11 kV interconnector, from the station boards. Figures 2.76 and 2.77 show fault levels for the next two stages of switching. Switching the station transformer out of service will not affect fault levels, hence fault levels after stage (e) are the same as after stage (d).

Because the Ii kV switchgear has different make and break ratings and the studies show make fault levels greater than 750 MVA, the above sequence is repeated to determine break values. Break values are always less than make values because the fault current from synchronous machines (assuming constant excitation) and induction machines decreases with time. The break value is calculated at the shortest time postfault that the switches can be activated to break the fault current. This depends on the speed of the sensing transducers, relays and the circuit-breaker operating time. An example of a break fault level calculation is shown in Fig 2.78. This is a repeat of the study shown in Fig 2.76, with current decrements from contributing machines taken into account. A value of 0.07 s is used here for the post-fault time interval before the circuitbreakers start to open. Fault levels at 11 kV boards are less than the break switchgear rating, 750 MVA.

If the SES is designed to allow 3.3 kV parallel operation, checks must be made to ensure that closing the 3.3 kV interconnector does not raise fault levels to greater than switchgear ratings. Figure 2.79 shows an example of this; starting with the reference Fig 2.73, a 3.3 kV interconnector is added between boards B6AX and B6BX. Because 11 kV board fault levels are all less than 750 MVA, it is not necessary to repeat this study to determine break values.

Figure 2.80 shows a 3.3 kV auxiliary diesel generator connected to board B6AY running in parallel with the main system. The associated 3.3 kV boards have fault levels significantly higher than those in reference Fig 2.73, but are less than the switchgear rating of 250 MVA.

So far, the examples of fault levels on station electrical systems (Figs 2.73 to 2.80) have shown symmetrical RMS values. These are satisfactory for initial investigations and yield much useful information; however, the DC component is not included and, of course, the values of real interest are the actual peak current

that the circuit-breaker has to interrupt, and the maximum prospective current, if closed onto earthed equipment. Thus the symmetrical RMS value of current has to be increased by two factors, the DC offset and the relation between AC peak and RMS values which is

2. This was illustrated in Fig 2.70. The DC offset depends on its initial value, determined by the value of the circuit voltage when the fault occurs; subsequently the rate of d-cay of the DC component depends on the relative value; of resistance and reactance between the current sources and the fault. Figure 2.81 is similar to reference Fig 2.73, but peak asymmetric make fault levels are shown at each board. Figure 2.82 shows RMS asymmetric break fault levels at each board. In both Figs 2.81 and 2.82, maximum initial asymmetry is

assumed.

When a switch interrupts fault current, the process of starting and maintaining arc extinction depends on current zeros occurring when the switch contacts start to open. For current zeros to occur, the peak AC component of current must exceed the DC offset at the post-fault time considered. The X:R ratio of transformers and associated circuits in power station electrical systems is generally low enough to make the DC component decay rapidly; however, problems can occur where components have high X:R ratios. The most likely place to look for this is at the generator circuit-breaker (used in some nuclear stations). Figure

2.83 shows a trace of fault current adjacent to a generator circuit-breaker and illustrates the relationship between DC and AC current components.

Values of fault current against time are also taken at various locations in a supply system and are used to confirm that equipment thermal specifications are adequate. Here, the value of interest is the total energy released before fault clearance, assuming pessimistic design fault clearance time.

Sometimes phase-to-earth fault current can exceed three-phase fault current. This occurs where the zero sequence impedance of a network is less than the positive and negative sequence impedances. High phase- to-earth fault currents are not expected in power station electrical systems at voltages where earth fault current is deliberately restricted by neutral earthing resistors. Checks can be made of prospective phase- to-earth fault current with the neutral earthing resistor in service, or with the neutral earthing resistor shortcircuited (which might happen through error or flashover). Similar checks can be made at generator voltages, where earthing arrangements use a distribution type transformer.

System minimum fault levels are also required for protection setting purposes. These are obtained by reducing grid infeed to its minimum value, disconnecting local generation and disconnecting the induction motors which run only when plant is loaded.

THREE PHASE FAULT LEVEL (MVA) FOR EACH BUSBAR ASYMMETR 0. PEAK VALUE AT T. 10.0 mS NOTE: TRANSFORMER PHASE SHIFTS NOT INCLUDED

JSAI

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860 ACE 255

JSOE

J60E