*2.2. Mathematical Model of the 2* × *25 kV System*

Each high speed line is fed with a 2 × 25 kV autotransformer system capable of delivering 2 MW of power for each kilometre of line. This allows "the simultaneous presence of 12 MW trains travelling at 300 km/h (180 mph) 5 min apart with no limits but with margin" [47]. This has been achieved through two 60 MVA feeding transformers in each electrical substation that are located 50 km (30 miles) apart; during normal operation each transformer feeds a section that is 25 km long. The substations are fed through a dedicated power-line at the nominal voltage of 132 kV (in northern Italy) or 150 kV (in southern Italy) capable of delivering 200 MW that connects all the substations of the railway line. Both ends of the power-line are connected to the national grid in very high-voltage (400 kV) nodes through two 250 MVA autotransformers. This permits the system to feed the railway line from a single very high-voltage substation when needed [48].

Paralleling posts are located every 12.5 km on average and are equipped with two 15 MVA autotransformers; they allow for putting the two tracks electrically parallel and connecting them to the autotransformers. During normal operation, only one of the two autotransformers is in use, except when a phase break is present, in which case they are both energised for each phase. This normally happens halfway between two substations. The railway line comprises seven conductors of each track at three different voltage levels: contact wire and messenger wire are fed at +25 kV, the feeder is connected to the −25 kV bus, the two rails, the overhead earth wire and the buried-earth wire are connected to earth, i.e., they have zero volt potential.

The model of a two-track line is made of the following four elements: (a) wires and rails; (b) inductive couplers; (c) the feeder transformer; and (d) autotransformers.

The hypothesis the model is based on is that the current flow through the soil and the buried-earth conductor are negligible; therefore, it is possible to simulate the line as though all current flows through the rails themselves or the overhead earth-wire. Electrical lines are usually modelled with a series of resistances and inductances and with capacitances and conductances in parallel. In most cases, a loaded line can be modelled to exclude the parallel capacitances and conductances, as their effect is negligible. In the case of the high-speed railway line, the power-line is not symmetric on each track even if the two tracks are built symmetrically such that inductance and resistance are represented by a big matrix that, in our case, has dimensions twelve by twelve. The big advantage that the symmetry between the two tracks brings us is that the matrices are symmetric themselves, potentially reducing

the amount of calculation needed. The reason why such big matrices are required is that the distance between the two track centrelines is 5 m, so the tracks cannot be considered independent in terms of magnetic coupling. Such a hypothesis would mean that the two tracks are very far apart, which is not viable, both in terms of environmental impact and economics. If it is assumed that all conductors are parallel and the messenger wire has a constant height, then self and mutual inductances of each conductor can be computed using Neumann formulas, which are:

$$L\_{i\bar{j}} = \frac{\mu\_0}{2\pi} \ln \frac{2 \times l}{D \times e} \tag{3}$$

$$L\_{i\bar{j}} = \frac{\mu\_0}{2\pi} \ln \frac{2 \times l}{D \times e} \tag{4}$$

$$L\_{ij} = \frac{\mu\_0}{2\pi} \ln \frac{2 \times l}{K \times r\_0 \times \varepsilon} \tag{5}$$

where *D* is the distance between the wires, *l* is the wire length, *r*<sup>0</sup> is the wire radius and *K* is a coefficient used to represent how the current is distributed inside the wire. These equations can be simplified if the sum of the currents is taken into account; in this case, the sum of all currents is zero for each track, which means that the inductance equations can be written as following:

$$L\_{ij} = \frac{\mu\_0}{2\pi} \ln \frac{l}{D} \tag{6}$$

$$L\_{ij} = \frac{\mu\_0}{2\pi} \ln \frac{l}{\mathcal{K} \times r\_0} \tag{7}$$

If *l* is equal to 1 and all units are SI, the result is that the inductance per length unit is henry per meter. Exceptions to these formulas are the rails themselves: since steel is a magnetic material and the cross section is high and not entirely used, the preceding formula cannot be used, but the value for the self-inductance is available in the literature either measured or computed through finite elements simulation. In this case, the value 0.359 mH/km has been used. In opposition to the inductances, only the self-resistances of each conductor are meaningful, so the resistances matrix is diagonal. Another attractive feature of this matrix is due to how the line is built. As each track uses the same type of wires, the values can be computed for only one track, because the other track values are the same. The line resistances of each wire are computed with the usual formula:

$$R = \rho \times \frac{l}{S} \tag{8}$$

where *ρ* is the resistivity of the material, *l* is the wire length and *S* is the useful cross section of the wire itself. The rail resistances are again taken from the literature for the same reason it is not possible to compute the self-inductances easily: the value used is 0.116 Ω/km.

All calculations are done using MATLAB software for an Italian high-speed rail line built on an embankment. The wires and rail positions in Table 2 are given with coordinates on a plain whose origin is placed in the middle between the two tracks on rail level.

The line model itself has been built in MATLAB-Simulink using inductances and resistances. The blocks used to make the line model are mutual inductances, resistors and inductors. Each block represents a base line section that is 1.5 km long; this distance has been chosen because inductive couplers are installed every 1.5 km. The inductive couplers are inductors that are installed every 1.5 km and which connect the two running rails of each track together in order to allow the traction current to flow in the earth-wire without short-circuiting the rails themselves.

The inductance is 1.2 mH, which is low enough to let the 50 Hz traction current through and high enough to block the audio-frequency current and allow track circuits to function properly. The inductor is centre-tapped and has been modelled as two separate inductors in series to have the same total inductance and the tap to connect to the earth-wire. The installation requires the centre tap to be connected to the earth-wire and the two edges to be connected to each rail.


**Table 2.** Wires and rail positions on an Italian high-speed line built on an embankment.

The voltage drop along the transmission line is evaluated using a phasorial equation in steady-state conditions as follows:

$$-\left[\frac{dV}{d\mathbf{x}}\right] = \left[Z'\right]\left[I\right] \tag{9}$$

where [*V*] and [*I*] are the phasor vectors of the line-to-ground voltage and of the currents flowing in the conductor, respectively.

It is assumed that the ground is the node to which all the voltages are referred. Because of the ground presence, the resistive factors are introduced in the mutual couplings. The *Z'ii* and *Z'ik* values are expressed using the Carson equations, which are accurate for power systems if used with homogenous ground [14]. The impedance matrix elements are found from the conductor's placement geometry and their characteristics as shown in Figure 2. The self-impedance is then evaluated as follows:

$$Z'\_{\rm ii} = \left(R'\_{\rm i-int} + \Delta R'\_{\rm ii}\right) + j(\omega \frac{\mu\_0}{2\pi} \ln \frac{2h\_{\rm i}}{r\_{\rm i}} + X'\_{\rm i-int} + \Delta X'\_{\rm ii}) \tag{10}$$

**Figure 2.** Geometry of the conductor's placement.

The mutual one is expressed as follows:

$$Z'\_{ik} = Z'\_{ki} = \Delta R'\_{ik} + j(\omega \frac{\mu\_0}{2\pi} \ln \frac{D\_{ik}}{d\_{ik}} + \Delta X'\_{ik}) \tag{11}$$

The conductor internal impedance *R'int* + *jX'int* is also evaluated. The internal reactance is usually combined in a single equation with the external reactance *ω <sup>μ</sup>*<sup>0</sup> <sup>2</sup>*<sup>π</sup>* ln <sup>2</sup>*<sup>h</sup> <sup>r</sup>* , where the radius, *r* is replaced with the minor Geometric Mean Radius (GMR), which is available from the conductor datasheets, in order to take into account the internal magnetic field:

$$
\omega \frac{\mu\_0}{2\pi} \ln \frac{2\hbar}{r} + X'\_{int} = \omega \frac{\mu\_0}{2\pi} \ln \frac{2\hbar}{\text{GMR}} \tag{12}
$$

The internal reactance can be evaluated as a part of the internal impedance. Since for non-magnetic conductors, the internal impedance represents a minor contribution of the total reactance, its accurate estimation is therefore not required. However, the evaluation of the internal resistance *R'int* is more significant due to the increase of its rate with the frequency caused by the skin effect.

In this investigation, as the system is complex and comprises several wires, the traction line has been applied through an integrated parameter model, which can be considered rather precise. The resistance and the inductance can be held practically constant until 1 kHz. As the application of this study is on lower frequencies, therefore, this assumption is suitable. Furthermore, the traction line has been distributed into many multipoles in order to avoid using unnecessary approximations.

Additionally, all the joint connections between the 14 wires founding the system have been considered in each cell. There are actually 7 wires for each way, particularly:


The system is built differently depending on the number of tracks available. Each track has a complete set of wires if the rail is double tracked (Figure 3a), four track railways can be built in two different configurations: it is possible to feed the fast tracks separately, thus requiring two autotransformers at each AT site or feed all the tracks together that means only two feeders and one autotransformer are needed (Figure 3b).

**Figure 3.** (**a**) Two track railway line; (**b**) Four track railway line.

*Energies* **2017**, *10*, 1268

The model details are shown in Figure 4. The feeding substation is fed through a high-voltage power-line connecting the various substations, so each substation has an incoming power-line connected.

**Figure 4.** Electric model of the line including all mutual coupling among live and earthling conductors.

It is also possible to replace the three-phase generator with an AC ideal voltage source with the right voltage and frequency. There may be the need to represent a longer section of the high-speed line fed with multiple substations. In that case, it is possible to add a more detailed representation of the power-lines and the feeding 400 KV node. An appropriate power-line model can be made with the "distributed parameters line" Simulink block; the autotransformer can be replaced by an equivalent transformer and a suitable three-phase voltage source. It is also possible to skip the 400/132 kV autotransformers as long as the three-phase voltage source has the correct short-circuit power.

The feeder transformer is single phase with a centre tap on the secondary winding and it is modelled as a three-winding transformer whose low voltage windings are connected in series. Another important component is the autotransformer which is necessary to connect the 50 kV catenary to feeder transmission line to the 25 kV train feeding system. The real machine is made of two windings wound on each column of the steel core connected together so that both windings share the same magnetic flux. The model parameters are taken from the real machine and are:


The main load of the line are going to be trains. The train model varies depending on the electronic converter used to rectify the AC current to DC. Old trains used diode or thyristor bridges, which make the train absorb highly distorted currents, i.e., the current absorbed is rich in low order harmonics especially the third, fifth, and seventh. This means that the fastest way to model these trains is by using a rectifier bridge feeding an appropriate load. When the use of GTOs and later insulated-gate bipolar transistors (IGBTs) became widespread, rectifier bridges were replaced by four quadrant converters in all newly constructed trains. This allows to convert AC to DC while absorbing a sinusoidal current in phase with the voltage (to be clear there are other harmonics, but their order is multiple of the switching frequency of the converter usually in the range of kilohertz). This means that, unless necessary, the load can be represented by a sinusoidal current injected through a current generator. This is achieved with a controlled generator; the control is very straightforward: the voltage is measured on site where the current will be injected and the signal is scaled with a gain block to the value of the current drawn by the load itself. If High Speed Trains (HST) is used as an example, the current drawn at the maximum power (8.8 MW) is 352 A at 25 KV; the gain block will scale the voltage to the current with the ratio 352/25,000 so that the current is in phase with the voltage and the power drawn is the nominal train power. The model can be used either to calculate the voltage profile along the line or to calculate the waveforms of voltages and currents wherever it is needed. As the line is built with 1.5 km long elementary line pieces, it is possible to simulate the case in which everything works correctly as well as the case with one or more faulty pieces of equipment.
