Effect of faults near to and far from short-circuit sources

The effect of electrical distance between the short-circuit source and the fault point on the initial magnitude of the short-circuit current, i.e. at the instant of fault is easily understood. The closer the fault location to the short-circuit source is, the smaller is the impedance between the source and the fault location, and the higher is the initial magnitude of short-circuit current. However, what is the effect on the magnitudes of the ac and dc currents afterwards, e.g. a few cycles later? To answer this question we need to understand, besides the effect on the current magnitude, the effect of the external impedance on the ac and dc time constants that govern the rates of decay of the ac and dc currents. This is an important aspect in practical short-circuit analysis. From Equations (6.4c) and (6.5), we have the following.

Three-phase short-circuit fault

     (6.6)

The limiting values of the network subtransient time constant T″_ac are determined by setting X_e = 0 and X_e → ∞ in Equation (6.6). Thus, the minimum value of T″_ac is equal to T″_d of the generator whereas the maximum value of T″_ac is equal to

For most synchronous machines, to 1.5 X″_d giving to 1.5T″_d. However, because the difference between X′_d and X′_d is much smaller than that between X′_d and X_d, the effect of external reactance on the initial magnitude of the subtransient component, as can be seen from Equation (6.4a), is much greater than the effect on the transient current component. In other words, the magnitude of the initial subtransient current component drops in a much greater proportion than the transient component. Following a similar approach for the network transient time constant T′_ac given in Equation (6.4c), we have

     (6.7)

For most synchronous machines, X_d can be up to 8X′_d thus T′_ac ≈ T′_d … 8T′_d.

Similarly, the effect of the external reactance is to cause a proportionately much greater reduction in the transient current component than in the steady state current component. The network dc time constant T_dc given in Equation (6.4c) can be rewritten as

     (6.8)

where T_a(3φ) is the generator’s armature time constant.

In practical power systems, generator X/R ratios X″_d/R_a are significantly greater than the X″_d/R_a ratios of external network elements X_e/R_e of transformers, lines, cables, etc., or any combination of these. Therefore,

giving

Thus, the effect of fault through an external impedance is to reduce the dc time constant and hence increase the rate of decay of the dc short-circuit current. The dc time constant is used to characterise the dc current component for circuit-breaker testing under IEC 62271–100:2001/2003 high voltage switchgear standard. In addition, it is usual power system practice to characterise the network dc current component by the network de X/R ratio given by

     (6.9)

Figure 6.8 shows the variation of the ac component of short-circuit current for a near to generator three-phase short-circuit fault and a far from generator similar fault. The latter is separated from the generator terminal by a large external reactance. The important observation is that the ac current decay is very pronounced in the near to generator case but is negligible in the far from generator case. Conversely, the dc current decay is now greater than that for a generator alone.

Figure 6.8 Illustration of near-to and far-from generator short-circuit faults

Network ac time constant

We consider the network subtransient ac time constant only since similar analysis also applies to the transient time constant.

In the case of a three-phase short circuit at point F, the total subtransient current component at F, using Equation (6.4a) is the sum of the subtransient current components supplied by each individual generator. Thus,

     (6.10a)

where

or

and

     (6.10b)

The expression for X_Eq in Equations (6.10a) and (6.10b) is a parallel equivalent and it ensures that the initial current magnitude, i.e. at t = 0, of the single equivalent exponential in Equation (6.10a) is equal to that obtained from the sum of the individual exponentials.

In the general case, sources have different impedances and time constants and each current component decays at a different rate. Therefore, the magnitudes of the n subtransient current components for t > 0 will be different. The same observation would be made even if the machines were identical, i.e. have similar parameters, but their external impedances were different. Consequently, a single network ac time constant at the fault location for all the ac sources that enables the calculation of the total current magnitude for t > 0 does not exist. Mathematically speaking, the sum of several exponentials of different time constants cannot be made equal to a single equivalent exponential with a single equivalent time constant. However, in this radial network topology this is not required because each source is radially connected to the fault point and the accurate equation for the sum of n exponential terms can be used. It can be shown that similar conclusions apply for single-phase short-circuit faults.

Network dc time constant

In the case of a three-phase short circuit at point F in Figure 6.9, the total dc current component at point F, using Equation (6.4b), is the sum of the dc current components supplied by each individual machine. Thus

     (6.11a)

where

or

and

     (6.11b)

Equation (6.11a) shows that at t = 0, the magnitude of the total dc current is equal to the sum of the magnitudes of the individual dc current components. However, for t > 0, and because each current component decays at a different rate, a single equivalent exponential component that gives an equivalent rate of decay or time constant for the sum of the individual exponentials does not exist. This is best illustrated in the following example.

Example 6.1

Consider three parallel sources or n = 3 in Figure 6.9 having significantly different time constants as follows: T_dc(1) = 50 ms, T_dc(2) = 150 ms and T_dc(3) = 15 ms. Using Equation (6.11b), the corresponding initial dc current magnitudes supplied at t = 0 by the three sources are equal to I_dc(1) = 0.6 pu, I_dc(2) = 0.3 pu and I_dc(3) = 0.1 pu. These initial current magnitudes and their associated dc time constants can represent a variety of sources and external impedances such as synchronous machines, induction machines and equivalent infeeds from transmission or distribution networks. Figure 6.10 curve (a) shows the total dc fault current at point F supplied by the three parallel sources and calculated using Equation (6.11a), i.e. from the accurate sum of three exponential terms. Figure 6.10 also shows the total dc current derived from a single time constant as calculated by two different approximate methods. The first, curve (b), is calculated from the single X/R ratio of the equivalent impedance ‘seen’ at the fault point, Z_Thév. = R + jX, also known as IEC 60909 Standard Method B (this is discussed in Chapter 7). The second, curve (c), is calculated from the ratio of the equivalent reactance at the fault point, X_Thév., (with all resistances set to zero) to the equivalent resistance at the fault point, R_Thév., (with all reactances set to zero), as per IEEE C37.010 Standard (this is discussed in Chapter 7). It can be shown that the single equivalent time constant at the fault point using the IEC method is equal to 48.6 ms (X/R = 15.2 at 50 Hz). Using the IEEE method, the single equivalent time constant at the fault point is equal to 76.3 ms (X/R = 24.0 at 50 Hz). The equations for the dc currents calculated by the three methods are given by where t is in ms. Figure 6.10 shows that the IEC Method B single equivalent time constant underestimates the actual dc current for t ≥ 10 ms and this underestimate increases with time. Conversely, and because the IEEE R_Thév. is significantly smaller than the equivalent resistance calculated using the IEC approach, the IEEE time constant is 1.57 times the IEC time constant. Figure 6.10 shows that the IEEE approach, in this example, significantly overestimates the actual dc current over the period up to 140 ms, then the overestimate disappears. It should be noted that the degree of underestimate of IEC Method B and overestimate of IEEE Method varies with the initial magnitude of each current source and the relative time constants between them. Larger and smaller underestimates and overestimates can be found in practice and the reader is encouraged to analyse the effect of different combinations.

Figure 6.10 Illustration of IEC and IEEE approaches for the calculation of dc short-circuit current variation with time for Figure 6.9

Network ac time constant

Figure 6.11(a) shows a simple meshed network that consists of three independent short-circuit current sources, each is connected by an external impedance to a common node N. The three sources are then connected to the short-circuit fault location F via a common impedance. To illustrate the problem, each source is assumed to be a synchronous machine whose time-dependent reactance is given by

Figure 6.11 Simple meshed network for the analysis of ac short-circuit current decrement (variation with time): (a) network configuration and (b) equivalent circuit

     (6.12)

where i = 1, 2 or 3.

The equivalent circuit is shown in Figure 6.11(b) where each machine is replaced by a constant voltage source E_o behind its time-dependent PPS reactance, given by Equation (6.12), and the armature resistance. The equivalent time-dependent

PPS system impedance at the fault point F is given by

     (6.13)

From Equation (6.13), it is interesting to note that not only the equivalent reactance X_F at the fault point is time dependent but also the equivalent resistance R_F. In many practical calculations, we can assume that X_m(i)(t) + X_(i)(t) » (R_ai + R_i) for i = 1, 2 and 3, and X_c » R_c. Therefore, from Equation (6.13), we have

     (6.14)

In general, the magnitude of the time-dependent ac component of fault current at the fault point F is given by

     (6.15a)

or

     (6.15b)

The substitution of Equation (6.12) in Equation (6.13) or (6.14) shows that the fault current magnitude given by Equation (6.15) would still exhibit subtransient and transient decay. However, the corresponding time constants are very complicated functions of the three individual machine time constants and no general analytical formula can be derived. The accuracy of various methods of calculating the time variation of the ac short-circuit current component is illustrated using an example.

Example 6.2

Consider Figure 6.11. The three machines are generators having the following parameters: X_d = 2.04 pu, X′_d = 0.275 pu, X′_d = 0.19 pu, T′_d = 40 ms, T′_d = 1100 ms, X₁ = X₂ = X₃ = 0.19 pu, X_c = 0.2 pu.

Curve (a) of Figure 6.12 shows the accurate time variation of the magnitude of the rms short-circuit current obtained from Equation (6.15b) with E_o = 1 pu. Curve (b) shows the current waveform obtained from the Thévenin’s equivalent subtransient, transient and steady state impedances calculated at the fault point F as X″_F = 0.3267 pu, X′_F = 0.355 pu and X_F = 0.9434 pu, respectively. The subtransient and transient time constants that include the effect of each machine’s external impedance up to node N but exclude the effect of the common impedance Z_c are calculated as T″_de(i) = 47.3 ms and T′_de(i) = 1701.5 ms. X″_F, X′_F and X_F are calculated usingX″_d(i), X′_d(i) and X_d(i), respectively, for each machine. The current obtained by using the Thévenin’s equivalent subtransient, transient and steady state impedances is given by,

Figure 6.12 Illustration of different approaches for the calculation of ac short-circuit current variation with time for a meshed network

     (6.16)

As expected, curve (b) Equation (6.16) underestimates the current magnitude for t > 0 and this underestimate increases with time. The reason is that the common impedance Z_c between node N and the fault point F will also cause some increase in the effective time constants of each source but Equation (6.16) fails to capture this effect. Curve (c) of Figure 6.12 shows the current variation if maximum known values of practical synchronous machine transient and subtransient time constants are used in Equation (6.16). These are 2000 and 70 ms, respectively. The result shows a small overestimate of up to around 90 ms, then an increasing underestimate thereafter. If the time range of interest is up to around 120 ms, as would be the case in high voltage and extra high voltage systems, then an approximation can be made in Equation (6.16) by setting Also, using an average value for synchronous machine subtransient time constants of around 40 ms, we can write

     (6.17)

Equation (6.17) is plotted as curve (d) in Figure 6.12 where the current falls to the minimum transient value then remains constant. Equation (6.17) gives an increasing overestimate with time for t ≥ 50 ms.

The above analysis can easily be extended to single-phase short-circuit faults using the machine time-dependent PPS reactance equations derived in Chapter 5.

There is no international consensus on how to calculate the time variation of the ac component of short-circuit current in practical meshed networks, when fixed impedance analysis techniques are used. Different empirical approaches are used, and this will be covered in Chapter 7.

Network dc time constant

We will continue to use the meshed network topology shown in Figure 6.11(a), but to simplify the analysis, we now use constant voltage sources instead of synchronous machines. The voltage source equation is given by When a short-circuit fault occurs at point F, the time domain solution of the short-circuit currents supplied by each source and the total fault current can be obtained by replacing all reactances with corresponding inductances and writing the differential equations relating the currents and voltages in this network. We encourage the reader to carry out this simple but time consuming exercise in Kirchoff’s voltage and current laws to obtain the solutions of the four currents. It can be shown that the instantaneous short-circuit current supplied by each voltage source and the total short-circuit current at the fault point F are given by

     (6.18a)

     (6.18b)

where

and

All the constant multipliers and the three dc time constants are complicated functions of the network resistances and inductances.

Equation (6.18b) shows that the dc component of the total short-circuit current consists of three components having different time constants T_dc(1), T_dc(2) and T_dc(3). The following example compares the accurate solution against the methods of characterising the dc current by a single decaying exponential having a time constant derived from the IEC 60909 Method B and the IEEE C37.010 approaches.

Example 6.3

In Figure 6.11(a), the X/R ratios of the four network branches are as follows: X₁/R₁ = 15.7, X₂/R₂ = 47.0, X₃/R₃ = 4.7, X_c/R_c = 10. Figure 6.13 shows that, as in the case of parallel independent sources, the equivalent X/R ratio at the fault point or single exponential method underestimates the true magnitude of the dc current component. However, the underestimate produced using the IEC Method B is generally smaller in meshed network than that produced in a network having parallel independent sources, due to the common impedance branch that is shared by the three sources in the meshed network. The configuration of parallel independent short-circuit sources generally produces the worst case underestimate. An overestimate is observed using the IEEE C37.010 Method. The errors in the dc currents calculated using the IEC and IEEE approaches on one hand and the accurate solution reduce with increasing Z_c that is increasing distance between the machines and the fault location. In other words, as Z_c increases, the IEC and IEEE curves move towards the curve that represents the accurate solution. Although we have used simple parallel and meshed network configurations to present the problem of dc current estimation, the conclusions regarding the errors observed in such network configurations are generally applicable in other networks.

Figure 6.13 Illustration of IEC and IEEE approaches for the calculation of dc short-circuit current variation with time for Figure 6.11(a)

Thévenin’s and Norton’s PPS equivalent of models of machines

The topology of power system networks and the data of all network elements, as well as the initial network loading and generation operating conditions, allow the formation of a nodal equivalent network of admittances and injected current sources. Short-circuit current sources such as generators and, where required, motors can be represented at the time instant of interest as constant voltage sources behind the appropriate machine fixed impedance, e.g. subtransient or transient impedance or any another value, as appropriate. Figure 14.6(a) and (b) shows equivalent models of a loaded synchronous machine. The latter shows a nodal or Norton’s equivalent which consists of a current source in parallel with the machine admittance.

From Figure 6.14(a), the apparent power output of a machine connected to node (i) is given by

Figure 6.14 Machine equivalent PPS model: (a) Thévenin’s equivalent and (b) Norton’s equivalent

     (6.19a)

The short-circuit current supplied by the machine for a fault at node i is given by

     (6.19b)

Using E_m(i) from Equation (6.19a) in Equation (6.19b), we obtain

     (6.19c)

PPS admittance and impedance matrix equations

Figure 6.15 shows a simple power system network that will be used to illustrate the process of formulating the sequences nodal matrices.

Figure 6.15 Simple power system network

The formation of the admittance matrix starts by forming the positive sequence nodal model shown in Figure 6.16, where each network element is replaced by its PPS model.

Figure 6.16 PPS network model of Figure 6.15

Figure 6.17 shows the nodal equivalent model with all shunt elements at every node collected into an equivalent shunt branch. The result is only one shunt branch at each node and a series branch interconnecting two nodes. The nodal admittance matrix equation of the resultant nodal equivalent circuit is given by

Figure 6.17 PPS network model of Figure 6.16 with lumped shunt branches at each node

     (6.20a)

or in concise matrix form

     (6.20b)

where

     (6.20c)

The admittance matrix of Equation (6.20) is in general symmetric, and even for small power systems, it is quite sparse, i.e. it contains only a few non-zero elements, each representing an admittance element connecting two nodes. For example, for a medium size system of 4000 nodes and 3000 series branches, the number of non-zero elements is 4000 + 2 × 3000 = 10 000. This is only 0.0625% of the total number of elements of the full matrix (4000 × 4000 = 16000000).

The use of the nodal admittance matrix in calculating short-circuit currents is inefficient, as it requires a new solution for each fault location and fault type. Therefore, the nodal impedance matrix given in Equation (6.21) below is used. It is important to note that the current vector represents a set of injected nodal current sources. Thus, from Equation (6.20b),

     (6.21a)

or

     (6.21b)

The nodal impedance matrix is generally full, i.e. it contains elements in every position unless there are disconnected parts of the network, as would be the case in the ZPS network as we discuss later. However, in practice, it is inefficient to obtain this matrix by direct inversion of the admittance matrix for medium and large size systems. Direct inversion of large matrices is avoided by the use of various computer-based numerical calculation techniques, such as the method of successive forward elimination and backward substitution (also known as Gaussian elimination). Other similar methods are also used, such as Kron reduction, triangular factorisation, LU decomposition and others. Generally, the sparse admittance matrix can be retained and the impedance matrix is calculated or built directly from the factorised admittance matrix.

Referring back to Equation (6.21b), it is important to appreciate the meaning of the diagonal and off-diagonal elements of the nodal impedance matrix. The diagonal terms , with all other current sources set to zero, is termed the driving point impedance or the Thévenin’s impedance as ‘seen’ from node k. This impedance is equal to the voltage produced at node k divided by the current injected at node k. The off-diagonal terms are, for example, again with all other current sources set to zero is termed the transfer impedance between nodes 1 and k. It is equal to the voltage produced at node 1 divided by the current injected at node k.

General analysis of three-phase short-circuit faults

We will present a general analysis technique suitable for the calculation of not only the short-circuit current at the fault location but also the currents and voltages throughout the network. As explained in Section 6.3.2, the application of the fault will cause current and voltage changes in the network. The calculation of these quantities in a large-scale system of N nodes makes use of the efficiently calculated nodal bus impedance matrix of the system. Since the three-phase short circuit is a balanced fault, the system nodal admittance and nodal impedance matrices are in effect PPS admittance/impedance matrices since all the network elements are replaced by their PPS models. The actual currents and voltages in the network due to the fault are obtained by the superposition of the initial currents and voltages and the Thévenin’s changes. Thus,

or

     (6.22a)

     (6.22b)

Three-phase short-circuit fault at one location

Denote the faulted node as k, the fault impedance as Z_F, the connection fault current as I_F, the Thévenin’s impedance at node k as Z_kk and the prefault voltage at node k as V_k(o). The calculation of the current and voltage changes at the fault location and elsewhere is illustrated using Figure 6.18 and the nodal bus impedance matrix of Equation (6.21b).

Figure 6.18 Calculation of current and voltage changes using Thévenin’s equivalent circuit: (a) representation of the entire network and (b) equivalent at the faulted node k

Since we are considering a fault at one location only, namely node k, there is only one injected node current I_k = −I_F and all other injected currents are zero. Therefore, from Equation (6.21b), the voltage changes are given by

     (6.23a)

From Equation (6.23a), the voltage changes caused by the fault are given by

     (6.23b)

The actual voltage during the fault at the faulted node k is given by

     (6.24)

From Equation (6.23b) row k, we can write and using Equation (6.24), we obtain

     (6.25)

Having calculated the fault current, the nodal voltage changes throughout the network are calculated using Equation (6.23b). The changes in branch currents are calculated from the nodal voltage changes using

     (6.26)

where y_ij is the retained admittance of branch ij.

Denoting the actual nodal voltages during the fault as V₁, V_k … V_N, then from Equation (6.23b), we can write

     (6.27)

Substituting Equation (6.25) in Equations (6.24) or (6.27), the actual nodal voltages can be expressed as

     (6.28)

and

     (6.29)

For a solid short-circuit fault, Z_F = 0. The actual branch currents are calculated using

     (6.30)

The short-circuit current contribution of each machine is calculated from the actual nodal voltages where each machine is connected and using Equation (6.19c)

     (6.31)

or, alternatively, using Equation (6.19b) in Equation (6.31)

     (6.32)

Simultaneous three-phase short-circuit faults at two different locations

In Figure 6.19(a), denote the faulted nodes as j and k, the corresponding fault impedances as Z_j(F) and Z_k(F) and the connection fault currents as I_j(F) and I_k(F). The Thévenin’s impedances at nodes j and k are Z_jj and Z_kk, the transfer impedances between nodes j and k are Z_jk and Z_kj and the prefault voltages at nodes j and k are V_j(o) and V_k(o).

Figure 6.19 Calculation of current and voltage changes for two simultaneous short-circuit faults using Thévenin’s equivalent circuits: (a) representation of the entire network for two simultaneous short-circuit faults and (b) equivalent between faulted nodes j and k

From Equation (6.21b), we can write

     (6.33)

Since the nodal impedance matrix is symmetric, Z_jk = Z_kj. Therefore, the voltage changes in Equation (6.33) caused by the faults are given by

     (6.34)

The actual voltages during the faults at the faulted nodes j and k are given by

     (6.35a)

     (6.35b)

From Equation (6.34) rows j and k, we can write

     (6.36a)

and

     (6.36b)

Substituting Equations (6.35a) and (6.35b) in Equations (6.36a) and (6.36b), respectively, and solving for the fault currents, we obtain

     (6.37)

It is instructive to visualise the Thévenin’s equivalent circuit seen from the two faulted nodes j and k. Adding and subtracting Z_jkI_j(F) to Equation (6.36a) and Z_jkI_k(F) to Equation (6.36b), we obtain

     (6.38a)

and

     (6.38b)

Equations (6.35) and (6.38) can be represented by the equivalent circuit shown in Figure 6.19(b). As expected, the Thévenin’s impedance from nodes j and k to earth, or to the reference node, are Z_jj and Z_kk, respectively. The equivalent Thévenin’s impedance between the faulted nodes j and k is given by

     (6.39)

Having calculated the fault currents using Equation (6.37), the nodal voltage changes are calculated using Equation (6.34). The actual voltages are then calculated using Equation (6.22a). The actual branch and machine currents are calculated using Equations (6.30) and (6.32), respectively.

Three-phase short-circuit fault between two nodes

In Figure 6.20(a), consider a fault between nodes j and k. The fault impedance is Z_F, the connection fault currents are I_j(F) and I_k(F), the Thévenin’s impedances at nodes j and k are Z_jj and Z_kk, the transfer impedance between nodes j and k is Z_jk = Z_kj, and the prefault voltages at nodes j and k are V_j(o) and V_k(o). Consider the fault current to flow out of node j and into node k hence the injected currents are I_j(F) = −I_(F) and I_k(F) = I_(F).

Figure 6.20 Representation of a three-phase short-circuit fault between two nodes: (a) Thévenin’s representation of the entire network and (b) Thévenin’s equivalent between nodes j and k for a short-circuit fault between them

From Equation (6.21b), we can write

     (6.40)

From Equation (6.40), the voltage changes caused by the fault are given by

     (6.41)

The actual voltage during the fault between the faulted nodes j and k is given by

     (6.42)

From Equation (6.41) rows j and k, we can write

     (6.43a)

and

     (6.43b)

From Equations (6.42) and (6.43), the fault current is given by

     (6.44)

Equations (6.42) and (6.44) are represented by the Thévenin’s equivalent circuit shown in Figure 6.20(b). Having calculated the fault current, the nodal voltage changes are calculated using Equation (6.41). The actual voltages are then calculated using Equation (6.22a). The actual branch and machine currents are calculated using Equations (6.30) and (6.32), respectively.

Nodal sequence impedance matrices

In the case of a three-phase short circuit, only the PPS nodal impedance matrix is required, as presented in Section 6.6.2. For a two-phase unbalanced short circuit, the NPS nodal impedance matrix is also required. This can be efficiently derived from the NPS nodal admittance matrix, as we discussed for the PPS impedance matrix. From the early days of digital computer analysis of large power systems, an assumption has generally been made that the NPS and PPS impedance matrices are identical in order to remove the need to form, store and calculate the NPS nodal matrix. This is based on the assumption that the PPS and NPS models and parameters of all passive network elements are identical and that those for machines do not differ significantly. However, this assumption is only approximately valid when calculating the subtransient short-circuit current using machine subtransient impedances. It is not valid when calculating the short-circuit current some time after the instant of fault, e.g. the transient current using machine transient impedances, because these differ significantly from the machine NPS impedances. On the whole, with the huge and cost-effective increases in computing storage and speed, particularly since the early 1990s, this assumption is no longer needed.

In the case of a single-phase to earth and two-phase to earth faults, the nodal ZPS impedance matrix is also required in addition to the PPS and NPS impedance matrices. This matrix is radically different from the PPS and NPS impedance matrices both in connectivity due to e.g. transformer windings and impedance values of most network elements. Figure 6.21 shows the ZPS network model of Figure 6.15.

Figure 6.21 ZPS network model of Figure 6.15

General fault impedance/admittance matrix for unbalanced short-circuit faults

It is useful to derive a single method of analysis that covers all unbalanced short-circuit faults, namely single-phase to earth, two-phase and two-phase to earth. Such a method needs to allow for both solid short circuits and those through fault impedances. Figure 6.22 shows a general fault connection node with four impedances or admittances, the values of which can be appropriately chosen to represent any type of unbalanced short circuit.

Figure 6.22 General fault impedance connection matrix

It can be shown that the phase voltages and currents at the fault location are related by the following equation:

     (6.45a)

and the fault impedance matrix is defined as

     (6.45b)

For example, to represent a solid single-phase to earth fault, we set Z_R = 0, Z_E = 0, Z_Y = ∞ and Z_B = ∞. Similarly, we represent a solid two-phase (Y-B) short circuit by setting Z_Y = 0, Z_B = 0, Z_R = ∞ and Z_E = ∞. The sequence fault impedance matrix is calculated using The result is given by

     (6.46)

However, as we have already seen, some of the impedances will be infinite under certain fault conditions and as a result will become undefined. This problem

is solved by inverting to obtain the sequence fault admittance matrix as follows:

     (6.47)

where

General mathematical analysis

Using Thévenin’s theorem, and Equation (6.22b) the current and voltage changes in the PPS, NPS and ZPS networks, dropping the delta sign for currents for convenience, are given by

     (6.51a)

     (6.51b)

     (6.51c)

where and are the PPS, NPS and ZPS nodal impedance matrices, respectively, of the N node network. Prior to the short-circuit fault, the NPS and ZPS currents between all nodes, and the NPS and ZPS voltages at all nodes are zero because the three-phase system is assumed to be balanced. Where the system loading or operating condition is to be taken into account, the PPS voltages and currents are obtained from an initial loadflow study. Therefore, using the superposition principle, the actual nodal voltages during the fault are given by

     (6.52a)

     (6.52b)

     (6.52c)

where

     (6.53a)

     (6.53b)

The short-circuit fault is applied at node k. Thus, all injected sequence currents in the three sequence networks are zero except the sequence currents injected at the faulted node k. Equations (6.53a) and (6.53b) become

     (6.54a)

     (6.54b)

From Equations (6.52) to (6.54), we can extract the sequence voltage components of row k as follows:

     (6.55a)

or

     (6.55b)

where

     (6.55c)

The PPS, NPS and ZPS networks are interconnected at the fault location in a manner that depends on the fault type, as described in Chapter 2. For the general analysis of any type of unbalanced fault, in fact even balanced faults if desired, we need to describe the conditions at the fault location using a general method. One method is to use of a fault admittance matrix that is dependent on the fault type, as described in the previous section. Thus, the sequence fault currents can be written as

     (6.56)

Substituting Equation (6.56) into Equation (6.55b), we obtain

     (6.57)

where U is the identity matrix. The sequence fault currents at the faulted node k are now obtained by substituting Equation (6.57) back into Equation (6.56), giving

     (6.58)

The sequence voltages at the faulted node k, , are calculated using Equation (6.57). The PPS, NPS and ZPS voltages throughout the PPS, NPS and ZPS networks can be calculated by substituting Equation (6.54) in Equation (6.52). Therefore, for a fault at node k, the sequence voltages at any node i are given by

     (6.59)

The actual sequence currents flowing in the branches of the PPS, NPS and ZPS networks are calculated as follows:

     (6.60)

The sequence current contributions of each machine are calculated as follows:

     (6.61)

Having calculated the sequence components of every nodal voltage and branch current, the corresponding phase voltages and currents are calculated using V^RYB = HV^PNZ and I^RYB = HI^PNZ, where H is the transformation matrix given in Chapter 2. The reader should remember to modify the phase angles of the sequence currents and voltages in those parts of the network which are separated from the faulted node by star-delta transformers, as discussed in Chapter 4.

One-phase open-circuit fault

The connection of the PPS, NPS and ZPS sequence networks was covered in Chapter 2, and shown in Figures 2.10 and 2.11 for one-phase and two-phase open-circuit faults, respectively. In these figures, the PPS, NPS and ZPS impedances are the Thévenin’s impedances seen looking from the open-circuit fault points F and F′. We first illustrate how these impedances are calculated making use of the nodal impedance matrix and the Thévenin’s impedance between two nodes using Figure 6.23.

Figure 6.23 Open-circuit fault on a circuit connected between nodes j and k

In Figure 6.23, consider an open-circuit fault on a circuit connected between two nodes j and k and having sequence impedances Z^P, Z^N and Z^Z. In Equation (6.39), we derived the equivalent Thévenin’s impedance between two nodes j and k and showed it in Figure 6.19(b). This Thévenin’s impedance includes the circuit that is physically connected between nodes j and k. An open-circuit fault can be represented by removing the circuit between nodes j and k completely, i.e. all three phases, and then reinserting the appropriate sequence impedances Z^P, Z^N and Z^Z of the circuit. The disconnection of the circuit can be simulated by connecting −Z^P, −Z^N and −Z^Z between nodes j and k in the PPS, NPS and ZPS Thévenin’s equivalents. Therefore, the equivalent Thévenin’s impedance between the open-circuit points F and F′ is calculated as illustrated in Figure 6.24 and is given by

Figure 6.24 Calculation of (a) PPS, (b) NPS and (c) ZPS Thévenin’s impedances between open-circuit points F and F′

     (6.62a)

where, using Equation (6.39),

     (6.62b)

The prefault current flowing between F and F′ at node j before the open-circuit fault occurs is a PPS current, and is calculated from the initial loadflow study using

     (6.63a)

Alternatively, an approximate value may be calculated as follows:

     (6.63b)

Therefore, Equation (6.62) and the prefault current can be used in the equations derived in Chapter 2 for one-phase open circuit and two-phase open circuits to calculate the sequence currents and voltages at the fault location FF′.

To calculate the sequence voltage changes throughout the PPS, NPS and ZPS sequence networks using their nodal impedance matrices, we need to calculate the equivalent sequence currents to be injected at nodes j and k in each sequence network. These are easily calculated from Figure 6.24 by replacing each sequence voltage V^x in series with the circuit impedance Z^x (x = P, N or Z) with a current source equal to V^x/Z^x in parallel with Z^x, the latter cancelling out when combined with −Z^x. This current source V^x/Z^x flows from node k to node j and can be considered as two separate current sources. The first is equal to V^x/Z^x and is injected into node j, and the second is equal to −V^x/Z^x and is injected into node k. Figure 6.25 illustrates the sequence current sources injected into each sequence network. In summary, the sequence current sources injected into nodes j and k are given by

Figure 6.25 Representation of open-circuit faults by injection of PPS, NPS and ZPS current sources into respective sequence networks

     (6.64)

From Equations (6.51) and using x = P, N or Z, we have

     (6.65)

Using Equation (6.64), we obtain

     (6.66)

The actual NPS and ZPS sequence voltages are equal to the changes calculated using Equation (6.66), as shown in Equations (6.52b) and (6.52c). The actual PPS voltages are calculated by adding the prefault voltages obtained from the initial loadflow study to the PPS voltage changes calculated using Equation (6.66), as shown in Equation (6.52a). The actual sequence currents flowing in the branches of the PPS, NPS and ZPS networks are calculated using Equation (6.60).

Synchronous machines

To maintain generality, we assume that, the synchronous machine NPS impedance is not equal to the PPS impedance. From the three independent sequence networks of the machine, shown in Figure 6.26, the following sequence equation can be written

Figure 6.26 Independent machine sequence circuits

     (6.67a)

or

     (6.67b)

Before the short-circuit fault occurs, the machine is assumed to produce a balanced set of three-phase voltages having a PPS component only. However, the machine may be assumed to be connected to either a balanced or an unbalanced network. If the network is balanced, then only PPS currents flow, but if the network is unbalanced, then NPS and ZPS currents will also flow. Multiplying Equation (6.67b) by the sequence-to-phase transformation matrix H, we obtain

     (6.68a)

where

     (6.68b)

and

     (6.68c)

From Equation (6.67), the machine sequence impedance matrix is given by

     (6.69a)

After some algebra, Equation (6.68c) results in the following machine phase impedance matrix

     (6.69b)

where

     (6.69c)

The machine phase impedance matrix shows equal self-phase impedances for the three phases, full interphase but unequal mutual coupling between the three phases, and is non-symmetric, i.e. Z₁₂ ≠ Z₂₁. In cases where the machine’s NPS impedance is assumed to be equal to the PPS impedance, i.e. Z^N = Z^P, then Equation (6.69b) simplifies to the following familiar symmetric phase impedance matrix

     (6.70a)

where

     (6.70b)

Since the machine’s R^P and R^Z are usually taken to be equal, Z_M is mainly reactive, i.e. Z_M ≈ jX_M, and it is worth noting that X_M is negative because X^Z < X^P.

From Equations (6.68a), (6.68b), (6.69b) and (6.70a), the three-phase Thévenin’s equivalent circuit of the machine is as shown in Figure 6.27(a).

Figure 6.27 Three-phase (a) Thévenin’s and (b) Norton’s equivalent circuit models of machines

From Equation (6.68a), under a three-phase short-circuit fault at the machine terminals, the short-circuit current vector is given by . Substituting for E^RYB from Equation (6.68a) we obtain

     (6.70c)

Equation (6.70c) represents the machine’s three-phase Norton’s equivalent circuit model shown in Figure 6.27(b).

Induction machines

A three-phase induction machine model can be derived in a similar way to a synchronous machine model, with one exception. Viewed from its terminals, an induction machine appears as an open circuit in the ZPS network where the stator winding is connected in either delta or star with an isolated neutral as is usually the case. Thus, the machine’s ZPS impedance is infinite. Using the machine’s sequence admittances Y^P = 1/Z^P, Y^N = 1/Z^N and Y^Z = 0, as well as Equation (6.69a), the machine’s sequence admittance matrix is given by

     (6.71a)

Using Equation (6.68c), the induction machine phase admittance matrix is given by

     (6.71b)

In the case where Y^P ≈ Y^N, Equation (6.71b) reduces to

     (6.71c)

General

Where the magnitude of the dc component of the short-circuit fault current is estimated from the X/R ratio at the fault location, the X/R ratio can be calculated using Thévenin’s impedance matrix at the fault location in accordance with IEC 60909 Method B, IEC 60909 Method C or IEEE C37.010 Method. These will be discussed in Chapter 7.

In general, the Thévenin’s phase impedance matrix representing the mutually coupled three-phase network is bilateral and the elements of are reciprocal, i.e. Z_RY = Z_YR except in special cases. This may be due to the generally unequal PPS and NPS impedances of machines and the non-symmetric admittance matrices of quadrature boosters and phase shifting transformers.

The effective X/R ratio at the fault location is dependent on the short-circuit fault type. To illustrate how the X/R ratio can be calculated, we use Figure 6.28 where the following equation can be written

     (6.83)

We will consider one-phase to earth and three-phase to earth short-circuit faults but the method is general and can be extended to cover other fault types.

One-phase to earth short-circuit fault

For a solid one-phase to earth short-circuit fault on phase R at node K, we have and Therefore, from the first row of Equation (6.83), we have

     (6.84)

Similar equations can be written where the faulted phase is either phase Y or phase B. Therefore, in general, the X/R ratio of the phase fault current is calculated from the self-impedance of the faulted phase, or

     (6.85)

Alternatively, the X/R ratio can be calculated by transforming Equation (6.83) to the sequence reference frame using V^PNZ = H⁻¹V^RYB, I^PNZ = H⁻¹I^RYB and The result is given by

     (6.86)

As we discussed in Chapter 2, the boundary conditions I^R = I^F, I^Y = 0, I^B = 0 and translate to and Using these sequence boundary conditions and Equation (6.86), the symmetrical component equivalent circuit is shown in Figure 6.29.

Figure 6.29 Calculation of X/R ratio of single-phase short-circuit fault current in an unbalanced three-phase network

Now, summing the three rows of Equation (6.86), or using Figure 6.29, and using , the following equation can be written

     (6.87)

Therefore, the equivalent X/R ratio of the phase fault current I^R for a single-phase short-circuit fault on phase R is given by

     (6.88)

It is quite interesting to note that one third of the sum of the nine elements of the sequence impedance matrix for a one-phase fault on phase R is equal to the self-phase impedance of phase R that is Equations (6.84) and (6.87) are identical. Thus,

     (6.89)

Equation (6.89) can be obtained using Equations (2.25) and (2.26a) from Chapter 2 even though Equation (2.26b) was derived assuming a bilateral network.

Three-phase to earth short-circuit fault

For a three-phase to earth short-circuit fault, we have However, unlike the balanced network analysis presented in Chapter 2, because the network is now unbalanced, . Therefore from Equation (6.83) we have

     (6.90)

or

     (6.91)

Therefore, using E^Y = h²E^R, E^B = hE^R and expanding Equation (6.91), the effective impedance seen by phase fault current R is given by

     (6.92a)

and the X/R ratio of phase R fault current is given by

     (6.92b)

The effective impedance seen by phase fault current Y is given by

     (6.93a)

and the X/R ratio of phase Y fault current is given by

     (6.93b)

The effective impedance seen by phase fault current B is given by

     (6.94a)

and the X/R ratio of phase B fault current is given by

     (6.94b)

It is important to note that the three-phase fault currents are generally unbalanced both in magnitude and phase and that the three X/R ratios are generally different. The differences being dependent on the degree of network unbalance.

In the author’s experience, and depending on the type of study being carried out, it is often more advantageous in three-phase unbalanced analysis to use actual physical units, i.e. volts, amps and ohms, instead of per-unit quantities.