A Novel Power Flow Algorithm for Traction Power Supply Systems Based on the Thévenin Equivalent

energies

Article A Novel Power Flow Algorithm for Traction Power Supply Systems Based on the Thévenin Equivalent

Junqi Zhang ID , Mingli Wu * and Qiujiang Liu

School of Electrical Engineering, Beijing Jiaotong University, Beijing 100044, China; [email protected] (J.Z.); [email protected] (Q.L.) * Correspondence: [email protected]; Tel.: +86-136-2135-9126

Received: 12 December 2017; Accepted: 31 December 2017; Published: 4 January 2018

Abstract: With the rapid development of high-speed and heavy-haul railways throughout China, modern large power locomotives and electric multiple units (EMUs) have been applied in main railway lines. The high power requirements have brought about the problem of insufﬁcient power supply capacity (PSC) of traction power supply systems (TPSSs). Thus, a convenient method of PSC assessment is meaningful and urgently needed. In this paper, a novel algorithm is proposed based on the Thévenin equivalent in order to calculate the PSC. In this algorithm, node voltage equations are converted into port characteristic equations, and the Newton-Raphson method is exploited to solve them. Based on this algorithm, the PSC of a typical high-speed railway is calculated through the repeated power ﬂow (RPF). Subsequently, the effects of an optimized organization of train operations are analyzed. Compared to conventional algorithms, the proposed one has the advantages of fast convergence and an easy approach to multiple solutions and PV curves, which show vivid and visual information to TPSS designers and operators. A numerical analysis and case studies validate the effectiveness and feasibility of the proposed method, which can help to optimize the organization of train operations and design lines and enhance the reliability and safety of TPSSs.

Keywords: electric railway; power ﬂow; power supply capacity (PSC); Thévenin equivalent; repeated power ﬂow (RPF); traction power supply system (TPSS)

1. Introduction Electric railways, as an environmentally friendly and efficient means of producing passenger and freight services, have been selected by many countries [1]. Modern high-speed railway lines are being designed and built throughout China, Japan, and some European countries, and there is also an incremental interest in high-speed railway services in Southeast Asia and the United States [2]. China has the greatest amount of high-speed railways in service in the world, over 22,000 km by the end of 2016. With the rapid development of high-speed and heavy-haul railways, modern high power locomotives and electric multiple units (EMUs) have been designed and applied in main railway lines [3–5]. Their high power consumption has brought about a problem of insufficient power supply capacity (PSC) [6] of the single-phase 25 kV or 2 × 25 kV alternating current (AC) traction power supply systems (TPSSs) that are widely adopted to feed trains [7]. If a TPSS does not have enough PSC, locomotives and EMUs would not be able to operate normally. For the sake of the safe and efficient operation of electric railways, it is essential to calculate and assess the PSCs of TPSSs; hence, a technical scheme is urgently required. A power flow algorithm is expected to converge at the power limit for application to the repeated power flow (RPF), which repeatedly solves power flow equations at a succession of points along a specified power change pattern [8–13]. TPSSs are special distribution systems and have their own features, e.g., more conductors and earth return involved, that are different from three-phase public grids. A number of power flow

Energies 2018, 11, 126; doi:10.3390/en11010126 www.mdpi.com/journal/energies Energies 2018, 11, 126 2 of 17 algorithms have been proposed for TPSSs in the past decades, as listed in Table1. Though the algorithms of [3,14–22] are applicable, the Multiple conductor Nodal Fixed-point Algorithm (MNFA), involving the multiple conductor model, a nodal analysis, and fixed-point iteration [23–30], has acquired the widest use because of its accuracy and convenience for programming. However, the MNFA cannot offer convincing evidence of the power limit for PSC assessment. In a practical implementation, if the MNFA fails to converge, the PSC is considered exceeded. Hence, the maximum power which makes the MNFA converge is treated as the power limit. Instead, the divergence may result from the inability to converge on existent solutions. Therefore, the divergence of the MNFA is not a rigorous proof of PSC insufficiency. Though the slope of the tangent line of a power–voltage curve (PV curve) can be an auxiliary criterion [31], it is still not cogent and explicit enough. On the other hand, the continuation power flow (CPF) [31–35] overcomes such disadvantage through tracing the solution curve, and is capable of multiple solutions. The multiple solutions can form PV curves and show vivid and visual information to power system planners and operators. However, the CPF has been rarely used in TPSSs, and its implementation is more complicated than conventional power flow algorithms. Programmers need knowledge of the numerical continuation and the techniques of parameterization, prediction, correction, and step length control. Recently, a novel power flow method called holomorphic embedding [36] has drawn researchers’ attention, owing to its ability to give the right solution or prove the nonexistence of solutions. Its mathematical foundation is complex analysis instead of iterative methods, and advanced mathematical theory is used. Some application cases have been reported [37–40].

Table 1. Features of traction power supply system (TPSS) power ﬂow algorithms.

Algorithm Model Accuracy Applicability to Feeding Scheme Convergence Speed [14] Accurate All Linear [15–17] Accurate All Quadratic [3,18–21] Accurate AT 1 feeding Linear [22] Accurate AT feeding Quadratic MNFA 2 More accurate All Linear PA More accurate All Quadratic 1 AT: auto transformer. 2 MNFA: Multiple conductor Nodal Fixed-point Algorithm.

In this paper, a novel power ﬂow algorithm (named Port Algorithm, PA) is proposed for the PSC calculation of TPSSs based on the Thévenin equivalent. The main features of the PA are included in Table1 with comparison to the previous algorithms. The main contributions of this work are as follows:

1. The Thévenin equivalent of a TPSS feeding section is introduced, which concentrates efforts on train nodes instead of all nodes. Fewer variables need consideration than in the previous algorithms. It can help with not only power ﬂow analysis, but also the TPSS harmonic impedance calculation in resonance analysis. 2. Node voltage equations are converted into port characteristic equations according to the Thévenin equivalent. The Newton-Raphson method is exploited to solve those equations, which has faster convergence than the MNFA. Besides, it has an easier approach to multiple solutions than the CPF. 3. The RPF based on the PA is utilized to calculate the PSC of a typical China high-speed railway TPSS. Some practical recommendations are proposed to optimize the organization of train operations. The minimum intervals of adjacent trains are estimated.

The rest of this paper is organized as follows. The PA is described in Section2, including the Thévenin equivalent and port characteristic equations solving. Its properties are veriﬁed by numerical results compared with the MNFA in Section3. In Section4, the RPF procedure is given based on the PA, and case studies are conducted to verify the proposed method. The conclusions are summarized in Section5. Energies 2018, 11, 126 3 of 16

Energies2. Algorithm2018, 11, 126 Description 3 of 17

EnergiesFigure 2018, 11 1a, 126 presents the overall process of the previous algorithms. It usually includes3 of 16 two steps: 2. Algorithmformulating Description node voltage or mesh current equations according to input data, and then solving the 2. Algorithm Description equationsFigure1a presents through the numerical overall methods. process of Distinct the previous from those, algorithms. the PA It usuallyinserts an includes additional two step: steps:converting formulatingFigure 1athe presents node node voltage the voltage overall equations or process mesh toof current portthe previous characteristic equations algorithms. accordingequations It usually towith includes input the help data, two of steps: and the Thévenin then solvingequivalentformulating the equations ( nodesee Figure voltage through 1b or). numerical meshThis stepcurrent methods.concentrates equations Distinct according efforts from on to train those,input nodesdata, the PAand instead inserts then solvingof an all additional nodes. the Fewer equations through numerical methods. Distinct from those, the PA inserts an additional step: step:variables converting need the consideration node voltage than equations in the previous to port algorithms. characteristic In this equations section, the with Thévenin the help equivalent of converting the node voltage equations to port characteristic equations with the help of the Thévenin the Thandévenin port equivalent characteristic (see equations Figure1 b). are Thisdescribed step concentrates first, then the efforts equation on train-solving nodes implemen insteadtation of is provided.equivalent (see Figure 1b). This step concentrates efforts on train nodes instead of all nodes. Fewer all nodes.variables Fewer need variablesconsideration need than consideration in the previous thanalgorithms. in the In previousthis section, algorithms. the Thévenin In equivalent this section, the Thandévenin port equivalent characteristic and equations port characteristic are described equations first, then are the described equation ﬁrst,-solving then implemen the equation-solvingtation is implementationprovided. isFo provided.rmulate node voltage / Formulate node voltage mesh current equations equations Formulate node voltage / Formulate node voltage mesh current equations equations Formulate port characteristic equations Formulate port characteristic equations

Solve equations Solve equations Solve equations Solve equations (a) (b) (a) (b) Figure 1. Overall steps of algorithms. (a) Previous algorithms; (b) Port Algorithm (PA). FigureFigure 1. Overall 1. Overall steps steps of of algorithms. algorithms. ((aa)) Previous algorithms algorithms;; (b () bPort) Port Algorithm Algorithm (PA) (PA).. 2.1. Thévenin Equivalent of Feeding Section 2.1. Thévenin Equivalent of Feeding Section 2.1. ThéveninFigure Equivalent 2 shows of an Feeding independent Section feeding section which is a basic unit of a TPSS, and its Thévenin Figure 2 shows an independent feeding section which is a basic unit of a TPSS, and its Thévenin equivalent network, where Ei, Ii, and Vi stand for the open-circuit voltage, current, and voltage of port Figureequivalent2 shows network, an independent where Ei, Ii, and feeding Vi stand section for the whichopen-circuit is a basic voltage, unit current of a TPSS,, and voltage and its of Th portévenin i, respectively. Z is the n × n impedance matrix of the equivalent network. The port characteristic equivalenti, respectively. network, Zwhere is the nE i×, Ini, impedance and Vi stand matrix for theof the open-circuit equivalent voltage, network. current, The port and characteristic voltage of port i, respectively.equationsequations areareZ is the n × n impedance matrix of the equivalent network. The port characteristic equations are n n n EVZIEVZI (1) (1) E =iiikkV iiikk+  ZI (1) i i k∑1 kik1 k k=1 where Zik represents the element in row i and column k of Z. The values of i and k are 1, 2, …, n, and wherewhereZ represents Zik represents the element the element in row in irowand i columnand columnk of Z k. of The Z. values The values of i and of ik andare 1,k are 2, ... 1, ,2,n …, and, n, and willik remain the same for the rest of this paper. will remainwill remain the same the same for the for rest the ofrest this of paper.this paper. − + − ++ I1 + E1 I1 V1 TrainE 11 V1 −Train 1 Traction − + STubrastcattiioonn + − I2 − + E2 + Substation V2 n×n TI2rain 2 Z E2 V2

· − · Zn×n Train 2 Contact wire ·

· − − + · Contact wire · + In Train 1 Train 2 … Train n En − + Vn Train n + In Train 1 TraiRna 2il … Train n En − Vn Tra in n (a) (b) Rail − Figure 2. Thévenin equivalen(a) t of feeding section. (a) Original feeding section; (b) Thévenin(b) equivalen t network. Figure Figure 2. Th 2. éThéveninvenin equivalent equivalen oft of feedingfeeding section section.. (a) ( aOriginal) Original feeding feeding section section;; (b) Thévenin (b) Thé veninequivalent equivalentnetwork network..

Energies 2018, 11, 126 4 of 17

The critical task of the equivalence is to identify Ei and Zik. There are two equations derived from (1). E = V | (2) i i I1=I2=···=In=0

Vi Zik = − (3) Ik Ei = 0 I1 = I2 = ··· = Ik−1 = Ik+1 = ··· = In = 0

These two equations are employed to identify Ei and Zik as follows.

• Identiﬁcation of Ei

1. Set all train currents to 0 A. 2. Solve node voltage equations.

3. Ei equals the voltage of train i.

• Identiﬁcation of Zik

1. Set the current supplied by the substation to 0 A. 2. Set the current of train k to −1 A, and that of the others to 0 A. 3. Solve node voltage equations.

4. Zik equals the voltage of train i.

2.2. Port Characteristic Equations Solving

If the complex power of train i is Pi + jQi, its voltage will be

Pi + jQi Vi = ∗ (4) Ii

Substituting (4) in (1) yields n Pi + jQi Ei = ∗ + ∑ Zik Ik (5) Ii k=1 The real form of (5) is  − n  Pi IiRe Qi IiIm + (R I − X I ) − E = 0  I2 +I2 ∑ ik kRe ik kIm iRe iRe iIm k=1 (6) + n  Pi IiIm Qi IiRe + (R I + X I ) − E = 0  I2 +I2 ∑ ik kIm ik kRe iIm iRe iIm k=1

The subscript “Re” and “Im” denote the real and imaginary part of a complex number, respectively. Rik and Xik are the real and imaginary part of Zik, respectively. The Newton-Raphson method is exploited to solve (6) as follows:

1. Set the initial value of the train current vector

h iT I = I1Re I1Im I2Re I2Im ··· InRe InIm (7)

Alternatively, set the initial values of the train voltages ﬁrst, and then calculate the train currents. 2. Calculate the residual vector

h iT b = b1Re b1Im b2Re b2Im ··· bnRe bnIm (8) Energies 2018, 11, 126 5 of 17

where P I − Q I n = i iRe i iIm + ( − ) − biRe 2 2 ∑ Rik IkRe Xik IkIm EiRe (9) IiRe + IiIm k=1 P I + Q I n = i iIm i iRe + ( + ) − biIm 2 2 ∑ Rik IkIm Xik IkRe EiIm (10) IiRe + IiIm k=1 3. Calculate the Jacobian matrix   J11 J12 ··· J1n    J21 J22 ··· J2n  J =   (11)  . . .. .   . . . .  Jn1 Jn2 ··· Jnn

where " # Rik −Xik Jik = , i 6= k (12) Xik Rik   R + Pi −X − Qi ii I2 +I2 ii I2 +I2 J =  iRe iIm iRe iIm  ii + Qi + Pi Xii 2 + 2 Rii 2 + 2 IiRe IiIm IiRe IiIm  2IiRe(Pi IiRe−Qi IiIm) 2IiIm(Pi IiRe−Qi IiIm)  (13) 2 2 (I2 +I2 ) (I2 +I2 ) − iRe iIm iRe iIm   2IiRe(Pi IiIm−Qi IiRe) 2IiIm(Pi IiIm−Qi IiRe)  2 + 2 2 2 + 2 2 (IiRe IiIm) (IiRe IiIm) This step will not take much time since most elements are constant. 4. Solve the corrective equation J∆I = b (14)

to obtain the corrective vector ∆I. 5. If the norm of ∆I is smaller than a given value e, ﬁnish the calculation successfully. Otherwise, subtract ∆I from I and go to Step 6. 6. If the number of iterations reaches a given value N, ﬁnish the calculation unsuccessfully. Otherwise, go to Step 2.

After they are determined, the values of the train currents are substituted into the node voltage equations to identify the node voltages. Afterwards, the power ﬂow will be available easily. The ﬂowchart of the PA is given in Figure3. Energies 2018, 11, 126 6 of 17 Energies 2018, 11, 126 6 of 16

Start Calculate residual vector b Input data

Calculate Jacobian matrix J I = I  I Calculate nodal admittance matrix of section Solve corrective equation JI = b Conduct Thévenin equivalent No No ||I|| < e ? Iteration = N ? Set initial current vector I Yes Yes Solve node voltage equations with I substituted

Calculate voltage, current or power throughout network

Finish

Figure 3.3. Flowchart of PA.PA.

3. Numerical Results The convergence speed and ability to ﬁndfind multiple solutions of the proposed PA are analyzedanalyzed through a numerical study with a comparison to the MNFA. All of thethe calculationscalculations are conducted on a desktop computer with an IntelIntel CoreCore i5-3470i5-3470 CPUCPU @@ 3.203.20 GHz,GHz, 3.603.60 GHzGHz andand 88 GBGB memory.memory.

3.1.3.1. Test SystemSystem Auto transformertransformer (AT)(AT) feeding feeding tends tends to to be be utilized utilized in in high-speed high-speed and and heavy-haul heavy-haul railways railways due due to itsto its larger larger PSC PSC than than direct direct feeding. feeding. Therefore, Therefore, realistic realistic parameters parameters of anof an AT AT feeding feeding section section are are listed listed in Tablein Table2 and 2 and adopted adopted for thefor the calculations. calculations.

Table 2.2. Parameters of thethe testtest system.system. Parameter Value ParameterSection Value Track Section 2 TrackLength 30 km 2 LengthPublic grid 30 km System capacity Public grid 5000 MVA System capacityVoltage 220 5000 kV MVA VoltageTraction transformer 220 kV Connection Tractionsingle transformer phase with secondary mid-point drawn out ConnectionRated capacity single phase with secondary40 MVA mid-point drawn out RatedRated capacity voltage 220 kV/ 402 × MVA27.5 kV RatedI voltagempedance 220 kV/28.4% × 27.5 kV ImpedanceAT 8.4% Location 15AT km and 30 km away from the substation LeakageLocation impedance 15 km and 30(0.1 km + awayj 0.45) fromΩ the substation Leakage impedanceTrains (0.1 + j 0.45) Ω Locations as depicted in Figure 4 Active power the same value for all trains Power factor 0.97 lagging Conductors as depicted in Figure 5 Rail to earth resistance 100 Ωkm

Energies 2018, 11, 126 7 of 17

Table 2. Cont.

Parameter Value Trains Locations as depicted in Figure4 Active power the same value for all trains Power factor 0.97 lagging Conductors as depicted in Figure5 Energies 2018Rail, 11 to, 126 earth resistance 100 Ωkm 7 of 16 Energies 2018, 11, 126 7 of 16 Earth resistivity 100 Ωm Initial values of the trainEarth resistivity voltages 100 Ωm 40 kV Initial valuesEarth of resistivity the train voltages 10040 ΩkVm Convergence criterion k∆Ik2 < 0.01 InitialConvergence values of the criteriontrain voltages ||∆I40||2 kV< 0.01 Convergence criterion ||∆I||2 < 0.01 Traction STurbascttaitoino n Substation 25 km 25 km Up track coUnpta tcrta wcki re contact wire Train 1 Train 2 Train 1 Train 2 Up track Upr tarialck rail Up track AT1 AT2 posUitpiv ter afceked er AT1 AT2 positive feeder 25 km 25 km Down track cDoonwtanc tt rwacirke contact wire Train 3 Train 4 Train 3 Train 4 Down track Dowrna itlrack rail Down track pDosoitwivne tfreaecdke r 30 km positive feeder Section start 30 km Section end Section start Section end

Figure 4. Outline of test system. FigureFigure 4. 4Outline. Outline of test test system system.. Positive feeder Positive feeder PoLsiGtivJ3e0 f0e/e5d0er Catenary Catenary PoLsiGtivJ3e0 f0e/e5d0er Catenary Catenary (L4G3J0300, 08/55000) THJ120 THJ120 (L9G30J03,0 805/500) (4300, 8500) (T0H, 7J91020) (5T00H0J, 1729000) (9300, 8500) (0, 7900) (5000, 7900) Protective wire Protective wire ProLtGecJt1i5v0e/ w25ire ProLtGecJt1i5v0e/ w25ire (L3G5J0105, 08/02050) Contact wire Contact wire (L8G50J01,5 800/2050) (3500, 8000) CoCnutMacgt 1w5i0re CoCnutMacgt 1w5i0re (8500, 8000) C(u0,M 6g80105)0 (C50u0M0,g 6185000) (0, 6800) (Horizontal, Height) (5000, 6800) (HorizUonnitta: lm, Hmeight) Up track Unit: mm Down track Up track Down track Rail Rail Rail Rail RPa6i0l RPa6i0l RPa6i0l RPa6i0l (-75P5,6 10000) (755P,6 10000) (424P56, 01000) (575P56, 01000) (-755, 1000) (755, 1000) (4245, 1000) (5755, 1000)

Earth wire Origin Earth wire Origin EarTthJ 5w0ire (0, 0) EarTthJ 5w0ire TJ50 (0, 0) TJ50 (3900, 1000) (8900, 1000) (3900, 1000) (8900, 1000) Figure 5. Conductors in feeding section. Figure 5. Conductors in feeding section. Figure 5. Conductors in feeding section. 3.2. Algorithm Properties 3.2. Algorithm Properties 3.2. Algorithm3.2.1. Properties Convergence Speed 3.2.1. Convergence Speed 3.2.1. ConvergenceThe train Speed voltages calculated by the MNFA and PA are listed in Tables 3 and 4, with the active powerThe of train each voltagestrain set calculatedto 10 MW andby the 20 MNFAMW, respectively and PA are. The listed two in sets Table ofs results 3 and 4,are with different the active, and power of each train set to 10 MW and 20 MW, respectively. The two sets of results are different, and The trainthe heavier voltages the loads calculated are, the larger by the the MNFAdifferences and are. PA This are is because listed the in convergence Tables3 and speeds4, with of th thee active the heavier the loads are, the larger the differences are. This is because the convergence speeds of the power of eachtwo algorithmstrain set to contrast 10 MW sharply, and 20 asMW, illustrated respectively. in Figure The6. The two fixed sets-point of results iteration are used different, in the and the twoMNFA algorithms has linear contrast convergence, sharply, while as illustrated the Newton in - FigRaphsonure 6. methodThe fixed used-point in the iteration PA has used quadratic in the heavier theMNFAconvergence loads has are, linear. Hence, the convergence, larger the PA the requires differences while a lowerthe Newton number are. This-Raphson of iteration is because methods to reach used the the convergencein convergencethe PA has quadratic speedscriterion. of the two algorithmsconvergence contrast sharply,. Hence, the as PA illustrated requires a lower in Figure number6. Theof iteration ﬁxed-points to reach iteration the convergence used incriterion the MNFA. has

Energies 2018, 11, 126 8 of 17 linear convergence, while the Newton-Raphson method used in the PA has quadratic convergence. Hence, the PA requires a lower number of iterations to reach the convergence criterion.

Table 3. Train voltages calculated by MNFA and PA under light loads (Unit: V). Energies 2018, 11, 126 8 of 16 Train MNFA PA Difference Energies 2018, 11Table, 126 3. Train voltages calculated by MNFA and PA under light loads (Unit: V). 8 of 16 1 26,162.81 − j 3049.25 26,162.79 − j 3049.25 0.02 2Table Train3. Train 24,639.93 voltages−MNFAj calculated4664.22 by MNFA 24,639.89 andPA −PAj 4664.22under lightDifference loads (Unit: 0.04 V). 31 25,585.1826,162.8−1j −3666.87 j 3049.25 26 25,585.15,162.79 − j− 3049.2j 3666.875 0.02 0.03 Train MNFA PA Difference 42 24,784.1624,639.93− j −4552.74 j 4664.22 24 24,784.12,639.89 − j− 4664.22j 4552.74 0.04 0.04 3 1 2526,585.1,162.88 1− −j 3666.8j 3049.27 5 2526,585.1,162.75 9− −j 3666.8j 3049.27 5 0.030.02 4 2 2424,784.16,639.93 − −j 4552.7j 4664.224 2424,784.1,639.892 − −j 4552.7j 4664.224 0.040.04 Table 4. Train voltages3 25,585.1 calculated8 − j 3666.8 by MNFA7 25,585.1 and5PA − j 3666.8 under7 heavy0.03 loads (Unit: V). Table 4. Train4 voltages24,784.16 calculated − j 4552.7 by 4MNFA 24,784.1 and 2PA − j under4552.7 heavy4 loads0.04 (Unit: V). Train MNFA PA Difference Table 4Train. Train voltagesMNFA calculated by MNFA andPA PA under heavyDifference loads (Unit: V). 11 22,791.54 22,791.5− j46299.11 − j 6299.11 22 22,791.49,791.49 − −j 6299.12j 6299.12 0.05 + j 0.01 0.05 + j 0.01 2Train2 18,037.97 18,037.9− jMNFA79100.00 − j 9100.00 18 18,037.88,037.88 −PA −j 9099.99 j 9099.99 0.09Difference − j 0.01 0.09 − j 0.01 33 21,108.93 1 2122,108.93,791.5− j 7411.10 4− −j 7411.j 6299.1110 21 21,108.8722,108.87,791.4 9− −j 7411.10j j6299.127411.10 0.050.06 + j 0.01 0.06 44 18,459.19 2 1818,459.1,037.9− j98933.77 7− −j 8933.7j 9100.007 18 18,459.1018,459.10,037.8 8− −j 8933.76j j9099.998933.76 0.090.09 − −j 0.01j 0.090.01 − j 0.01 3 21,108.93 − j 7411.10 21,108.87 − j 7411.10 0.06 3 4 18 ,PA459.1 9 − j MNFA8933.77 18,459.103 − j 8933.76 0.09 − j 0.01 10 10 PA MNFA Convergence criterion Convergence criterion 0 3 10 PA MNFA 3 10 10 PA MNFA Convergence criterion 10-1 -3 Convergence criterion 10100 10-1

(A) -6 (A) -5

2 -3 2

|| 1010 || 10

I ||

|| -9

(A) -6 (A) -5

2 10 2

10 || 10-9

I 10 || || -12 1010-9 -9 -1310 -15-12 10 1010 0 5 10 15 20 25 30 0 20 40 60 80 100 -13 10-15 Iteration step 10 Iteration step 0 5 10 15 20 25 30 0 20 40 60 80 100 (a) (b) Iteration step Iteration step Figure 6. Convergence speeds of two algorithms. (a) Light loads; (b()b Heavy) loads. Figure 6. Convergence(a) speeds of two algorithms. (a) Light loads; (b) Heavy loads. Figure 7 presentsFigure 6 . theConvergence calculation speeds speeds of two of thealgorithms two algorithms.. (a) Light loads When; (b )the Heavy active loads. power of each Figuretrain ranges7 presents from 0 the to 20 calculation MW, the MNFA speeds needs of the as much two algorithms. time as the PA When to complete the active the calculation, power of each Figure 7 presents the calculation speeds of the two algorithms. When the active power of each trainroughly ranges from 85 ms 0, tothough 20 MW, the the number MNFAs of needs iteration as are much larger. time However, as the PA as to a completereflection of the the calculation, slow train ranges from 0 to 20 MW, the MNFA needs as much time as the PA to complete the calculation, roughlyconvergence, 85 ms, though the number the of numbers iterations of and iteration the calculation are larger. time of However, the MNFA as will a reﬂectionrise to a great of level the slow roughly 85 ms, though the numbers of iteration are larger. However, as a reflection of the slow if 20 MW is exceeded. In contrast, those of the PA are impacted slightly. convergence,convergence, the number the number of iterations of iteration ands and the the calculation calculation time time of of the the MNFAMNFA willwill rise to to a a great great level level if 20 MWif is23000 exceeded. MW is exceeded In contrast,. In contrast, those PA those of the ofMNFA PAthe arePA are impacted200 impacted slightly. slightly. PA MNFA

250300 PA MNFA 180200 PA MNFA

200250 160180

150200 140160

100150 120140

Number of iteration Calculation time (ms) 50100 100120

Number of iteration 050 Calculation time (ms) 80100 0 5 10 15 20 25 0 5 10 15 20 25

0 Active power of each train (MW) 80 Active power of each train (MW) 0 5 10 15 20 25 0 5 10 15 20 25 (a) (b) Active power of each train (MW) Active power of each train (MW) Figure 7. Calculation(a speed) of the two algorithms. (a) Number of iteration(sb; ()b ) Calculation time.

Figure 7. Calculation speed of the two algorithms. (a) Number of iterations; (b) Calculation time. FigureMost 7. calculationsCalculation are speed concentrated of the two algorithms.around the power (a) Number limit ofduring iterations; the RPF (b), Calculationso the substitution time. of the PA for the MNFA will save considerable time in a PSC assessment. Most calculations are concentrated around the power limit during the RPF, so the substitution of the PA for the MNFA will save considerable time in a PSC assessment.

Energies 2018, 11, 126 9 of 17

Most calculations are concentrated around the power limit during the RPF, so the substitution of the PA for the MNFA will save considerable time in a PSC assessment.

3.2.2. Ability to Find Multiple Solutions If the initial values of the train voltages are set to 1 kV and the other conditions remain the same, the train voltages calculated by the PA will be lower, as listed in Tables5 and6. Heavier loads result in smaller differences between the high and low voltage solutions. As to the MNFA, adequate initial values have not been found which bring convergence on the low voltage solutions.

Energies 2018Table, 11, 126 5. Multiple solutions of train voltages under light loads (Unit: V). 9 of 16

3.2.2. Ability to Find Multiple Solutions Train High Voltage Solutions Low Voltage Solutions Difference If the initial values of the train voltages are set to 1 kV and the other conditions remain the same, t1he train voltages 26,162.79 calculated− j by3049.25 the PA will be low 15,959.64er, as listed− j in4789.04 Tables 5 and 6.10,203.15 Heavier loads + j result1739.79 in2 smaller differences 24,639.89 between− j 4664.22 the high and low 1206.80 voltage solutions− j 3075.46. As to the MNFA, 23,433.09 adequate− j initial1588.76 values3 have not 25,585.15 been found− jwhich3666.87 bring convergence 12,878.89 on the− lowj 5093.25 voltage solutions. 12,706.26 + j 1426.38 4 24,784.12 − j 4552.74 2848.02 − j 3879.99 21,936.10 − j 672.75 Table 5. Multiple solutions of train voltages under light loads (Unit: V).

TableTrain 6. Multiple High Voltage solutions Solutions of train Low voltages Voltage Solutions under heavyDifference loads (Unit: V). 1 26,162.79 − j 3049.25 15,959.64 − j 4789.04 10,203.15 + j 1739.79 2 24,639.89 − j 4664.22 1206.80 − j 3075.46 23,433.09 − j 1588.76 Train High3 Voltage25,585.1 Solutions5 − j 3666.87 Low12,878.89 Voltage − j 5093.2 Solutions5 12,706.26 + j 1426.38 Difference 4 24,784.12 − j 4552.74 2848.02 − j 3879.99 21,936.10 − j 672.75 1 22,791.49 − j 6299.12 16,952.05 − j 7108.26 5839.44 + j 809.14 2 18,037.88Table 6. −Multiplej 9099.99 solutions of train 6071.67 voltages −underj 8086.42 heavy loads (Unit: 11,966.21 V). − j 1013.57 3 21,108.87Train High− Voltagej 7411.10 Solutions Low 13,760.17 Voltage Solutions− j 7954.93 Difference 7348.70 + j 543.83 4 18,459.101 22−,791.4j 8933.769 − j 6299.12 16 6989.39,952.05 − j− 7108.26j 8192.97 5839.44 + j 11,469.71809.14 − j 740.79 2 18,037.88 − j 9099.99 6071.67 − j 8086.42 11,966.21 − j 1013.57 3 21,108.87 − j 7411.10 13,760.17 − j 7954.93 7348.70+j 543.83 PV curves are formed4 by the18,459.10 multiple − j 8933.76 solutions 6989.39 found − j 8192.9 under7 successively11,469.71 − j 740.79 varying power, as shown in Figure8. The PV curves are continuous in the neighborhood of the power limit, which proves that PV curves are formed by the multiple solutions found under successively varying power, as the PA did notshown encounter in Figure 8 numerical. The PV curves difficulty. are continuous Moreover, in the neighborhood most low of voltage the power solutions limit, which are available only throughproves the samethat the setPA did of not initial encounter values, numerical simpler difficulty than. Moreover, the CPF. most As low a voltage result, solutions the programming are is simplified significantly.available only through the same set of initial values, simpler than the CPF. As a result, the programming is simplified significantly.

30 High voltage solutions 30 High voltage solutions Low voltage solutions Low voltage solutions 25 25

20 20

15 15

10 10

5 5

Train voltage magnitude (kV) Train voltage magnitude (kV) 0 0 0 5 10 15 20 25 0 5 10 15 20 25 Active power of train (MW) Active power of train (MW) (a) (b) 30 High voltage solutions 30 High voltage solutions Low voltage solutions Low voltage solutions 25 25

20 20

15 15

10 10

5 5

Train voltage magnitude (kV) Train voltage magnitude (kV) 0 0 0 5 10 15 20 25 0 5 10 15 20 25 Active power of train (MW) Active power of train (MW) (c) (d)

Figure 8. Power-voltage (PV) curves of trains. (a) Train 1; (b) Train 2; (c) Train 3; (d) Train 4. Figure 8. Power-voltage (PV) curves of trains. (a) Train 1; (b) Train 2; (c) Train 3; (d) Train 4. Energies 2018, 11, 126 10 of 17

Energies 2018, 11, 126 10 of 16 The MNFA is merely capable of the upper side of PV curves. Under these circumstances, the slope of the tangentThe MNFA line can is merely be an capable auxiliary of the criterion upper side of theof PV power curves limit.. Under It the isse concluded circumstances, from the Figure 8 that theslope slope of the is at tangent negative line can infinity be an whenauxiliary the criterion power of limit the power is reached. limit. It Nonetheless,is concluded from negative Figure infinity is not available8 that the slope in the is at numerical negative infini calculations.ty when the power Though limit a is value reached. can Nonetheless, be selected negative to represent infinity infinity, this approximationis not available isin notthe numerical universal. calculations. Slopes of Though different a value trains can be are selected not equivalent, to represent asinfinity, presented in this approximation is not universal. Slopes of different trains are not equivalent, as presented in Figure9. The selected value may not be appropriate for other trains or systems. Figure 9. The selected value may not be appropriate for other trains or systems.

-40 -40

-30 -30

-20 -20 Slope Slope

-10 -10

0 0 0 5 10 15 20 25 0 5 10 15 20 25 Active power of train (MW) Active power of train (MW) (a) (b) -40 -60

-50 -30 -40

-20 -30 Slope Slope -20 -10 -10

0 0 0 5 10 15 20 25 0 5 10 15 20 25 Active power of train (MW) Active power of train (MW) (c) (d)

Figure 9. Slopes of tangent lines of PV curves. (a) Train 1; (b) Train 2; (c) Train 3; (d) Train 4. Figure 9. Slopes of tangent lines of PV curves. (a) Train 1; (b) Train 2; (c) Train 3; (d) Train 4. In summary, the PA can offer a more convincing proof of the power limit than the MNFA and Inan summary, easier implementation the PA can than offer the a CPF. more The convincing PA is more prooffeasible of to thecalculate power the limitPSC. than the MNFA and an easier implementation than the CPF. The PA is more feasible to calculate the PSC. 4. Application to PSC Calculation

4. Application4.1. RPF Procedure to PSC Calculation 4.1. RPF ProcedureThe lower side of a PV curve related to low voltage solutions may not have practical meaning, but it provides good verification of the power limit. It is known that the high and low voltage Thesolutions lower approach side of aw PVhile curvethe power related is increasing to low, and voltage coincide solutions when the may power not limit have (4 ×practical 22.602 MW meaning, but it provides= 90.408 MW good for verification the test system of) the is reached power [35] limit.. If Itthe is power known continues that the increasing, high and the lowre will voltage be no solutions approachsolution whiles. theTherefore, power the is increasing, power limit and found coincide by the PA when can therepresent power the limit PSC (4. The× 22.602 RPF is MWperform = 90.408ed MW based on the PA as follows: for the test system) is reached [35]. If the power continues increasing, there will be no solutions. Therefore, the power1. limitSet the found complex by thepower PA of can each represent train to 0 the. PSC. The RPF is performed based on the PA as follows: 2. Choose a load change pattern and step length, namely how much the complex power of each 1. Set thetrain complex increases power or decreases. of each train to 0. 2. Choose3. Increase a load the change power according pattern andto the step chosen length, pattern namely and step how length much. the complex power of each train4. increasesPerform the or PA. decreases. If the calculation converges, go to Step 3. Otherwise, go to Step 5. 3. Increase the power according to the chosen pattern and step length. 4. Perform the PA. If the calculation converges, go to Step 3. Otherwise, go to Step 5. 5. If the step length is smaller than a given value ε, ﬁnish the calculation. Otherwise, go to Step 6. 6. Decrease the step length, for instance, by half. Energies 2018, 11, 126 11 of 16 Energies 2018, 11, 126 11 of 16 Energies 2018, 11, 126 11 of 17 5. If the step length is smaller than a given value ε, finish the calculation. Otherwise, go to Step 6. 5. If the step length is smaller than a given value ε, finish the calculation. Otherwise, go to Step 6. 6. Decrease the step length, for instance, by half. 6. Decrease the step length, for instance, by half. 7. DecreaseDecrease the power according to the pattern and new step length,length, and go to Step 4. 7. Decrease the power according to the pattern and new step length, and go to Step 4. Afterwards, the total active power of the trains is treated as the PSC. The process of power Afterwards, the total active power of the trains is treated as the PSC. The process of power adjustmentadjustment is illustrated in Figure 1010,, andand thethe ﬂowchartflowchart isis givengiven inin FigureFigure 1111.. adjustment is illustrated in Figure 10, and the flowchart is given in Figure 11.

2 30 10 2 30 10

1 10 1 25 10 25 100 100 20 20 10-1 10-1

15 Step length (MW) 15 10-2 Step length (MW) 10-2

Active power of each train (MW) -3

10Active power of each train (MW) 10 10 -3 0 5 10 15 20 10 0 5 10 15 20 Iteration step Iteration step Figure 10. Process of power adjustment. FigureFigure 10. 10Process. Process of of power power adjustment. adjustment.

Start Start

Initialize train power Initialize train power

Choose power change Choose power change pattern and step length pattern and step length

Increase train power Increase train power

Decrease train power Perform PA Decrease train power Perform PA

Yes Decrease step length Converge ? Yes Decrease step length Converge ? No No No No Step < e ? Step < e ? Yes Yes Finish Finish

Figure 11. Flowchart of repeated power flow (RPF). FigureFigure 11. 11Flowchart. Flowchart of of repeated repeated power power ﬂow flow (RPF). (RPF). Although plenty of calculations were performed with the PA successfully, it is still not Although plenty of calculations were performed with the PA successfully, it is still not guaranteedAlthough that plenty it can of always calculations converge were near performed the power with limit. the PA In successfully,the case of a itfailure is still inferred not guaranteed from a guaranteed that it can always converge near the power limit. In the case of a failure inferred from a thatdiscontinuous it can always PV converge curve, the near last the solution power limit.on the In curve the case can of be a failureused as inferred the initial from value a discontinuous of the next discontinuous PV curve, the last solution on the curve can be used as the initial value of the next PVcalculation, curve, the which last solution is a kind on of the discrete curve continuation can be used as method the initial [41]. value of the next calculation, which is calculation, which is a kind of discrete continuation method [41]. a kind of discrete continuation method [41]. 4.2. Case Studies 4.2.4.2 Case. Case Studies Studies The PSC can be enhanced through improving TPSS parameters and the organization of train The PSC can be enhanced through improving TPSS parameters and the organization of train operationThe PSCs. Enhancing can be enhanced the PSC through through improving improving TPSS TPS parametersS parameters and includes the organization connecting of train to a operations. Enhancing the PSC through improving TPSS parameters includes connecting to a operations.stronger public Enhancing grid, employing the PSC through AT feeding improving, or installing TPSS parameters power factor includes correctors. connecting Enhancing to a strongerthe PSC stronger public grid, employing AT feeding, or installing power factor correctors. Enhancing the PSC publicthrough grid, the employingorganization AT of feeding, train operations or installing mainly power affects factor the correctors. number and Enhancing locations the of PSC trains. through Both through the organization of train operations mainly affects the number and locations of trains. Both

Energies 2018, 11, 126 12 of 17 Energies 20182018,, 1111,, 126126 1212 ofof 1616 theof them organization are effective, of train but operations the former mainly is expensive affects and the time number--consuming. and locations For that of trains. reason, Both the ofpriority them areisis givengiven effective, toto thethe but latterlatter the former inin orderorder is expensivetoto realizerealize the andthe fullfull time-consuming. potentialpotential ofof aa Forpresentpresent that reason,TPSS.TPSS. ItsIts the effectseffects priority areare is analyzedanalyzed given to thebelow. latter in order to realize the full potential of a present TPSS. Its effects are analyzed below.

4.2.1.4..2.1..1. CaseCase 11 SupposeSuppose there there areare nn trainstrainstrains inin in thethe the testtest test system,system, system, twotwo two of of them them are are,,, respectivelyrespectively respectively,,, at atthe the section section end end on onthethe the upup and upand and downdown down tracktrack tracks,s,s, andand andthethe distancedistance the distance betweenbetween between adjacentadjacent adjacent trainstrains trains d isis thethed issame,same, the same,asas depicteddepicted as depicted inin FigureFigure in limit P Figure12. Other 12 .conditions Other conditions are identical are identical toto Section to Section3.1. The 3.1 power. The limit power Plimit limit and maximumlimit and maximum active power active of max P d n limit powereach train of each Pmax train are influencedmax are inﬂuenced by d as presented by as presentedin Figure 13. in FigureWhen 13n remains.remains When thetheremains same,same, thePlimit same, risesrises P d n limit P withlimit drises increasing.increasing. with increasing. WhenWhen n becomes When smaller,becomes Plimit smaller, will falllimit steeplywill first, fall steeply and then ﬁrst, rise and more then gently rise more than max gentlybefore. than ItIt has before. a minimum It has a of minimum 55.0 MW, of and 55.0 MW,a maximum and a maximum of 95.5 MW, of 95.5 var MW,ying varyingacutely.acutely. As for P Asmax for,, itit Prisesrisesmax, slightlyslightly it rises slightly with the with increment the increment of d,, andand of itsitsd, maximum andmaximum its maximum isis 27.527.5 MW.MW. is 27.5 MW.

Trraccttiion Subssttattiion

d d Up ttrracck cconttacctt wiirree ··· Up ttrracck rraiill

Up ttrracck AT1 AT2 possiittiivee ffeeeedeerr d d Down ttrracck cconttacctt wiirree ··· Down ttrracck rraiill

Down ttrracck positive feeder 30 km positive feeder 30 km Section end Section end FigureFigure 12.12.. LocationLocation ofof successivesuccessive trainstrains inin CaseCase 1.1..

100 14 100 Plimit Pmax 14 Plimit Pmax 12 80 10 60

60 8 n 6 n 40 6

4 Active power (MW) Active power (MW) 20 2

0 0 0 5 10 15 20 25 30 35 d (km) d (km) Figure 13. Influence of d on Plimit and Pmax in Case 1. Figure 13. Influence of d on Plimit and Pmax in Case 1. Figure 13. Inﬂuence of d on Plimit and Pmax in Case 1.

Energies 2018, 11, 126 13 of 17 Energies 2018, 11, 126 13 of 16 Energies 2018, 11, 126 13 of 16

4.2.2. Case 2 44.2.2..2.2. Case Case 2 2 If there are, respectively, two trains at the section start on the up track and the section end on IfIf there there are are,, respectively respectively,, two two trains trains at at the the section section start start on on the the up up track track and and the the section section end end on on thethe downdown track,track, as depicted in in Figure Figure 14,14, the the results results will will be be distinct. distinct. The The minimum minimum of of PlimitP hashas an the down track, as depicted in Figure 14, the results will be distinct. The minimum of Plimit limithas an increment to 89.7 MW, and the maximum is still 95.5 MW, as given in Figure 15a. The variation incrementan increment to 89.7 to89.7 MW, MW, and and the the maximum maximum is still is still 95.5 95.5 MW, MW, as as given given in in Figure Figure 15a 15a.. The The variation variation becomesbecomes notnot asas severesevere asas inin CaseCase 1.1. OnOn thethe otherother hand,hand, whenwhenn nremains remains the the same, same,P Pmax almostalmost keepskeeps becomes not as severe as in Case 1. On the other hand, when n remains the same, Pmaxmax almost keeps constantconstant despite the change of of dd.. Accordingly, Accordingly, PPmax is isapproximate approximatelyly inversely inversely proportional proportional to ton, asn, constant despite the change of d. Accordingly, Pmaxmax is approximately inversely proportional to n, as shown in Figure 15b. shownas shown in Figure in Figure 15b .15 b.

Traction Traction Substation Substation

d d d d Up track Up track contact wire contact wire ··· ··· Up track Up track rail rail Up track Up track AT1 AT2 positive feeder AT1 AT2 positive feeder d d d d Down track Down track contact wire contact wire ··· ··· Down track Down track rail rail Down track Down track positive feeder 30 km positive feeder 30 km Section start Section end Section start Section end Figure 14. Location of successive trains in Case 2. FigureFigure 14 14.. LocationLocation of of successive successive trains trains in in C Casease 2 2.. P P 50 100 P limit P max 14 50 100 limit max 14 12 80 12 40 80 40 10 10 30 60 8 30 60 8

n (MW) Pmax = 92.22 / n

40 6 max 20 2 P 40 6 max 20 2R = 0.9999

P R = 0.9999 4

Active power (MW) 4 20 Active power (MW) 10 20 2 10 2 0 0 0 0 0 0 5 10 15 20 25 30 35 0 0 2 4 6 8 10 12 14 0 5 10 15 20 25 30 35 0 2 4 6 8 10 12 14 d (km) n d (km) n (a) (b) (a) (b)

Figure 15. Influence of d on Plimit and Pmax in Case 2. (a) Plimit and Pmax; (b) Curve fitting of the relation FigureFigure 15 15.. InfluenceInﬂuence of ofd ond on PlimitP andand PmaxP in Casein Case2. (a) 2.Plimit (a )andP Pmaxand; (b)P Curve;(b fitting) Curve of ﬁttingthe relation of the between Pmax and n. limit max limit max between Pmax and n. relation between Pmax and n. 4.2.3. Discussion 4.2.3. Discussion 4.2.3. Discussion The two cases above indicate how the organization of train operations affects the PSC. Plimit and The two cases above indicate how the organization of train operations affects the PSC. Plimit and Pmax Thein Case two 1 cases are smaller above indicatethan in C howase the2 overall organization under the of trainsame operations number of affects trains. the The PSC. reasonPlimit is andthat Pmax in Case 1 are smaller than in Case 2 overall under the same number of trains. The reason is that Pthemax trainsin Case are 1 closer are smaller to the thansection in Caseend, so 2 overall that larger under conductor the same impedance number of participates trains. The reason in the power is that the trains are closer to the section end, so that larger conductor impedance participates in the power thetransmission, trains are closerand the to theTPSS section supplies end, less so thatpower. larger Thus, conductor it is recommended impedance participatesthat trains on in the the up power and transmission, and the TPSS supplies less power. Thus, it is recommended that trains on the up and down tracks not be concentrated near the section end, especially in weak TPSSs. down tracks not be concentrated near the section end, especially in weak TPSSs.

Energies 2018, 11, 126 14 of 17

transmission,Energies 2018, 11, 126 and the TPSS supplies less power. Thus, it is recommended that trains on the up14 andof 16 down tracks not be concentrated near the section end, especially in weak TPSSs.

In addition, addition,P limitPlimitexceeds exceeds the the rated rated capacity capacity of the traction of the tractiontransformer transformer considerably. considerably. The transformer The doestransformer not match does the not PSC. match If the thePSC PSC. is expected If the toPSC be isfully expected utilized, to the betransformer fully utilized, capacity the transformer needs to be enlargedcapacity needs to 95.5 to MW/0.97 be enlarged≈ 100 to MVA95.5 MW at least./0.97 ≈ 100 MVA at least. China EMUs in serviceservice can be classiﬁedclassified into three levels according to the data of active power consumption, asas givengiven in in Figure Figure 16 16.. Their Their minimum minimum distance distance and and time time intervals intervals corresponding corresponding to Case to 1C andase 1 Case and 2Case are presented2 are presented in Table in Table7, assuming 7, assuming that the that speeds the speeds are 300 are km/h. 300 km/h. It is impliedIt is implied that that the optimizedthe optimized organization organization of train of train operations operation is effectives is effective for the for third the third level level of EMUs, of EMUs, reducing reducing the time the intervaltime interval by 1.4 by min. 1.4 Ifmin. the If opening the opening hours hours are 16 are h (6:00 16 h to (6:00 22:00) to 22:00) a day, 100a day, pairs 100 of pairs trains of more trains will more be ablewill be to passable to the pass feeding the feeding section, section, and the and carrying the carrying capacity capacity will be enhancedwill be enhanced greatly, greatly, supposing supposing that the signalsthat the and signals station and facilities station facilities are able toare cooperate able to cooperate with the with power the supply. power supply.

6 12

5 10

4 8

3 6

2 4 Active power (MW) Active power (MW) 1 2

0 0 0 30 60 90 120 150 180 0 30 60 90 120 Time (s) Time (s) (a) (b) 25

Active power (MW) 5

0 0 30 60 90 120 150 180 210 240 Time (s) (c)

Figure 16.16. ClassiﬁcationClassification of of active active power power consumption consumption of Chinaof China electric electric multiple multiple units units (EMUs). (EMUs (a) Level). (a) 1,Level 5 MW; 1, 5 ( bMW) Level; (b) 2,Level 10 MW; 2, 10 ( cMW;) Level (c) 3,Level 20 MW. 3, 20 MW.

Table 7.7. Typical intervalsintervals ofof ChinaChina EMUs.EMUs. Level Active Power Consumption Minimum Interval in Case 1 Minimum Interval in Case 2 Level1 Active Power5 MW Consumption Minimum<5 km/ Interval<1 min in Case 1 Minimum<5 km/<1 Interval min in Case 2 2 110 MW 5 MW 7.5 <5 km/ km/<11.5 min 7.5 <5 km/1.5 km/<1 minmin 3 220 10MW MW 22 7.5 km/ km/1.54.4 min min 7.515 km/1.5km/3 min min 3 20 MW 22 km/4.4 min 15 km/3 min

Energies 2018, 11, 126 15 of 17

5. Conclusions The PA is proposed for TPSSs based on the Thévenin equivalent. Port characteristic equations, converted from nodal voltage equations, are solved by the Newton-Raphson method. Owing to its quadratic convergence, the calculation time is shorter than the MNFA near the power limit. What is more, an easier approach to multiple solutions than the CPF is provided. The low voltage solutions can be found effortlessly only through another set of initial values, instead of knowledge of the numerical continuation and a complicated programming implementation. PV curves formed by multiple solutions are capable of providing vivid and visual information to TPSS planners and operators. With the help of the RPF based on the PA, the PSC is available conveniently. The organization of train operations has significant effects on the PSC. It is recommended that the trains on the up and down tracks not be concentrated near the section end, especially in weak TPSSs. This optimization can help to shorten the interval of adjacent trains, and is beneficial to sufficient PSC utilization and enhancement of the carrying capacity.

Acknowledgments: This work was supported by the Fundamental Research Funds for the Central Universities (under Grant 2016YJS145) and the National Key Research and Development Program of China (under Grant 2017YFB1200802). Author Contributions: Junqi Zhang and Mingli Wu conceived the algorithm. Junqi Zhang and Qiujiang Liu performed the calculations. Junqi Zhang analyzed the results and wrote the paper. Conﬂicts of Interest: The authors declare no conﬂicts of interest.

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