applied sciences Article

Application of a High-Power Reversible Converter in a Hybrid Traction Power Supply System Gang Zhang 1,2, *, Jianglin Qian 1,2 and Xinyu Zhang 1,2 1 2

*

School of Electrical Engineering, Beijing Jiaotong University, Beijing 100044, China; [email protected] (J.Q.); [email protected] (X.Z.) Beijing Electrical Engineering Technology Research Center, Beijing 100044, China Correspondence: [email protected]; Tel.: +86-10-5168-7082

Academic Editor: Eric Ka-wai Cheng Received: 24 December 2016; Accepted: 7 March 2017; Published: 14 March 2017

Abstract: A high-power reversible converter can achieve a variety of functions, such as recovering regenerative braking energy, expanding traction power capacity, and improving an alternating current (AC) grid power factor. A new hybrid traction power supply scheme, which consists of a high-power reversible converter and two 12-pulse diode rectifiers, is proposed. A droop control method based on load current feed-forward is adopted to realize the load distribution between the reversible converter and the existing 12-pulse diode rectifiers. The direct current (DC) short-circuit characteristics of the reversible converter is studied, then the relationship between the peak fault current and the circuit parameters is obtained from theoretical calculations and validated by computer simulation. The first two sets of 2 MW reversible converters have been successfully applied in Beijing Metro Line 10, the proposed hybrid application scheme and coordinated control strategy are verified, and 11.15% of average energy-savings is reached. Keywords: urban railway transit; traction power system; regenerative braking; reversible converter; direct current short-circuit

1. Introduction At present, diode rectifiers are still widely used in urban rail transit traction power supply systems to provide energy for the trains [1]. Since the energy can only be transmitted from alternating current (AC) to direct current (DC), the braking resistor must be employed to consume the surplus regenerative braking energy of the trains in order to avoid the abnormal rise of the DC voltage. This scheme will cause a very large waste of energy, increasing the tunnel temperature and the burden of the cooling system. Energy storage based on electric double-layer capacitors (EDLC) and flywheels are widely used in the transportation field [2–4]. EDLC has good performance in terms of power density, charge and discharge time, long lifetime cycles, and lower internal resistance, so it can be used for storing regenerative braking energy of electric vehicles [5–10]. However, in the field of urban rail transit, the maximum regenerative braking power is usually up to several megawatts, so thousands of EDLCs would be needed, which requires a large installation space and cost [11]. Flywheels were firstly installed in Japan in 1988 for storing regenerative energy [12]. The rated energy and power of the flywheels are 25 kWh and 2000 kW, respectively. After that, flywheels were verified in tests at the London underground and Spain railways. However, all of them are just prototypes. Complex mechanisms and high costs limit their wide utilization. Flywheel systems for energy saving may be an important research direction in the future [13–15]. It is a good energy-saving method to invert the regenerative braking energy to the AC grid, which has been already applied in several subway lines [16–21]. However, it requires additional installation Appl. Sci. 2017, 7, 282; doi:10.3390/app7030282

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space, which will increase the cost of substation construction, and the utilization rate of the inverter is not very high. It will be a good idea to introduce high-power reversible converters into the traction power supply system of urban rail transit. Not only will braking energy be recovered, but traction power supply can also be realized by the reversible converter. Thus, the capacity of diode rectifiers can be reduced, or one of the two existing 12-pulse rectifiers may be replaced by a reversible converter for the purpose of saving space. Generally, the DC output characteristic of the reversible converter is constant, which can be realized by a simple voltage closed-loop. However, because it needs to work with the diode rectifier units in parallel, in order to share the traction load, the reversible converter must have a similar voltage droop characteristic as the diode rectifier. Droop control methods are commonly used in DC microgrids [22,23] and other distribution power systems [24–26]. Few studies have paid attention to the DC voltage control of the reversible converter used in traction power supply systems. In the field of urban rail transit, the existing literature is mainly aimed at the steady state and transient short-circuit characteristics of the diode rectifiers [27,28]. A mathematical model of the 12-pulse rectifier is established, and the approximate expression of the short-circuit current is deduced at the outlet. The accuracy of the calculation method was verified by modeling and simulation. However, the DC short-circuit characteristics of the reversible converter are different from the diode rectifier. The DC short-circuit fault of the grid-connected inverter used in wind power generation systems was studied in the literature [29,30], which can be used as an important reference for this paper. This paper aims to introduce the high-power reversible converter into the traction power supply system of urban rail transit, and lay a foundation for further application. The following sections are organized as follows: A hybrid traction power supply system composed of a diode rectifier and reversible converter is proposed in Section 2. The DC output characteristic of the 12-pulse rectifier is introduced and a droop control method based on load current feed-forward is proposed for the reversible converter in Section 3. In order to provide guidance for system protection, the DC short-circuit fault of the reversible converter is studied, the peak short current and rise time are both calculated, which agree with the simulation results in Section 4. The effectiveness of the coordination control strategy is verified by a field test at Beijing Metro Line 10 in Section 5. Finally, conclusions are made in Section 6. 2. Proposed Hybrid Traction Power Supply Scheme At present, the traction power supply equipment installed in the traction substation is a typical 24-pulse rectifier, which is made up of two 12-pulse rectifiers. In this way, AC grid current harmonics and DC voltage ripple can be reduced, and the system redundancy can be increased. For diode rectifiers, DC output voltage is uncontrollable, and the energy can only be transmitted from AC to DC. In order to reuse the braking energy and reduce the DC voltage fluctuation, a high-power reversible converter is introduced into the traction power supply system. Figure 1 shows the proposed hybrid traction power supply scheme. In Figure 1a, an additional reversible converter branch is added, but in Figure 1b, a 12-pulse rectifier is replaced by a high-power reversible converter. It is obvious that no additional installation space is required in Figure 1b. The arrows indicate the direction of the energy flow. No matter what kind of scheme, the reversible converter has a four-quadrant operation capability. When the train is in the traction state, the reversible converter can provide energy for the train, and reduce the DC voltage drop. When the train is in the braking state, the reversible converter can invert the braking energy back to the AC grid, and save energy. The overvoltage caused by regenerative braking can be suppressed effectively. Since the diode rectifier and reversible converter will work in parallel, their work characteristic must be studied, respectively, and then a coordinated control strategy should be adopted.

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3 of 19 rectifier

Transformer

rectifier

Transformer

Udc

Udc

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12-pulse rectifier

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12-pulse rectifier

rectifier

alternating current grid

Transformer

rectifier

Transformer

Appl. Sci. 2017, 7, 282 UABC

Udc

alternating current grid

Udc 3 of 19 converter Transformer 12-pulse rectifier rectifier Transformer

rectifier converter rectifier

Transformer

Udc

Transformer

Udc

alternating current grid

12-pulseconverter rectifier reversible 12-pulse rectifier rectifier

UABC

reversible converter 12-pulse rectifier

UABC

(a)

(b) Transformer

converter

Figure 1.Transformer Proposed hybrid traction power supply

alternating scheme. current (a) grid Hybrid

UABC

converter

scheme I; (b) Hybrid scheme II.

UABC

12-pulse rectifier 3. Coordinated Control Strategy

reversible converter converter

reversible converter

Transformer

rectifier

Transformer

UABC

12-pulse rectifier 12-pulse rectifier

Transformer Transformer

alternating current grid

alternating current grid

Transformer

converter

(a)

(b)

3.1. Direct Current (DC) Output Characteristics of the 12-Pulse Rectifier

Figure 1. Proposed hybrid traction power supply scheme. (a) Hybrid scheme I; (b) Hybrid scheme II.

A 12-pulse rectifier cantraction be equivalent an ideal DC voltage source Uk in I; series with anscheme internalII. Figure 1. Proposed hybrid power to supply scheme. (a) Hybrid scheme (b) Hybrid reversible converter reversible converter resistance Req and a diode D, as shown in Figure 2. 3. Coordinated Control Strategy (a) (b) The DCControl output characteristic curve of a 12-pulse rectifier is shown in Figure 3. The critical 3. Coordinated Strategy Figure 1. Proposed hybrid Itraction power supply scheme. Hybrid scheme I; (b) Hybrid scheme II. current (or Current transition current) is bounded. The is (a) divided 3.1. Direct (DC) OutputdgCharacteristics of thecurve 12-Pulse Rectifierinto two curves: ① is the curve that two 6-pulse(DC) rectifiers work in a push-pull mode; ② is Rectifier the curve that two six-pulse rectifiers 3.1. Direct Current Output Characteristics of the 12-Pulse 3. Coordinated Control Strategy A can be equivalent to an ideal DC voltage source Uk in series with an internal work in 12-pulse parallel. rectifier U doo denotes the no-load voltage of a 6-pulse rectifier, Udio denotes the ideal no-load A 12-pulse rectifier can be equivalent to an ideal DC voltage with an internal resistance and a diode D, as shown Figure 2.the rated k in DC voltageReq of a 12-pulse rectifier, UdNindenotes outputsource voltageUof a series 12-pulse rectifier, 3.1. Direct Current (DC) Output Characteristics of the 12-Pulse Rectifier The DC output characteristic curve of a 12-pulse rectifier is shown in Figure 3. The critical resistance R and a diode D, as shown in Figure 2. eq IdN denotes the rated output voltage of a 12-pulse rectifier. current transition current) Idgequivalent is bounded. Theideal curve dividedsource into two curves: is an theinternal curve A (or 12-pulse rectifier can be to an DCisvoltage Uk in series ① with that two 6-pulse rectifiers work in a push-pull mode; ② is the curve that two six-pulse rectifiers resistance Req and a diode D, as shown in Figure 2. work in parallel. Udoo denotes the no-load of a 6-pulse rectifier, Udio denotes the ideal no-load The DC output characteristic curvevoltage of a 12-pulse rectifier is shown in Figure 3. The critical DC voltage of a 12-pulse rectifier, U dN denotes the rated output voltage of a 12-pulse rectifier, current (or transition current) Idg is bounded. The curve is divided into two curves: ① is the curve Req rectifier. IdN denotes the rated output work voltage 12-pulse that two 6-pulse rectifiers inofa apush-pull mode; ② is the curve that two six-pulse rectifiers work in parallel. Udoo denotes the no-load voltage of a 6-pulse U dc rectifier, Udio denotes the ideal no-load D DC voltage of a 12-pulse rectifier, UdN denotes the rated output voltage of a 12-pulse rectifier, IdN denotes the rated output voltage of a 12-pulse rectifier.

U Reqk

U dc

Figure 2. Equivalent circuit rectifier. Figure 2. Equivalent D circuitofofa 12-pulse a 12-pulse rectifier.

Req Uk The DC output characteristic curve of a 12-pulse rectifierUis shown in Figure 3. The critical current dc 1 is the curve that two (or transition current) Idg is bounded. The curveDis divided into two curves: ① U doo Figure 2. Equivalent circuit of a 12-pulse rectifier. 2② 6-pulse rectifiers work in a push-pull mode; is the curve that two six-pulse rectifiers work in parallel. U dio Uk Udoo denotes the no-load voltage of a 6-pulse rectifier, Udio denotes the ideal no-load DC voltage of a U dN U dc 12-pulse rectifier, UdN denotes the rated output voltage of a 12-pulse rectifier, IdN denotes the rated Figure 2. Equivalent circuit of a 12-pulse rectifier. output voltage of a 12-pulse rectifier.

U dc

① U doo U dio U dc U dN U0doo I① dg

②

I dN

I dc

② U dio FigureU3.dN12-pulse rectified output characteristic curve.

0

I dg

I dN

I dc

Figure 3. 12-pulse rectified output characteristic curve.

0

I dg

I dN

I dc

Figure 3. 12-pulse rectified curve. Figure 3. 12-pulse rectifiedoutput output characteristic characteristic curve.

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1 3.1.1. Curve

The DC voltage and current show typical 12-pulse characteristics. However, at every moment there is actually a single three-phase bridge rectifier at work. Its no-load voltage is Udoo

1 = π/6

Z

π 12 π − 12

√

√ π 6 2 × 2U2L sin = 1.398U2L 2U2L cos θdθ = π 12

(1)

where U2L is the voltage effective value at the secondary side of the transformer. Thus, the mathematical 1 can be described as model of the output curve Udc = Udoo −

3 Xc Idc . π

(2)

1 is as follows: The equivalent resistance for curve

Req1 =

3 Xc . 2π

(3)

2 3.1.2. Curve

When the DC load current is larger than the critical current Idg , the output curve is shown 2 The two six-pulse rectifiers work completely in parallel. The load is shared equally by them. as . 2 extended to vertical axis, is called the ideal The voltage Udio , which is the cross point of curve no-load DC voltage. The voltage of Udio is equal to 1.35U2L . Since the two six-pulse rectifiers work in parallel at the same time, the equivalent resistance for 2 will be reduced to half of the curve , 1 as follows: curve Req2 =

3 Xc 2π

2 is The mathematical description of the curve

Ud = Udio −

3 Xc Id . 2π

(4)

3.2. DC Output Characteristic of the Reversible Converter Typical DC output voltage characteristic of the reversible converter is constant. However, due to the demand of operating in parallel with the 12-pulse diode rectifier, it should have similar voltage droop characteristics. That means DC output voltage will drop with the increase of DC current, as follows: Udc = Uk − Rdp × Idc

(5)

where Uk is the no-load output voltage for reversible converter. Rdp is the equivalent resistance. A voltage droop control method based on output current feed-forward is presented in this paper, as shown in Figure 4. It includes three parts, which are called droop control, DC voltage control, and dq current control, respectively. Droop control is employed to calculate the target voltage Udc *, which decreases with the increase of the DC, complying with the law of Equation (5). The LIM1 will limit the value of the target DC voltage within the allowable range. DC voltage control adopts a classic proportional integral (PI) regulator without static error, so the influence of system parameter variation and external disturbance for the DC voltage can be eliminated, and the actual DC voltage can follow the target well. The output of the DC voltage control loop is the active current given value id *, which will be limited by LIM2, in order to avoid the overload of the reversible converter. The dq current control is based on rotating synchronous coordinates d-q. Thanks to the coordinate transformation, id

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and iq are DC components, and the PI compensator can reduce the error of the fundamental component to zero. the With the of proposed control method above,to thezero. linearity theproposed output curve canmethod be guaranteed reduce the error of the the fundamental fundamental component to zero. Withof the proposed control method above, reduce error component With the control above, and strong system stability can be obtained. the linearity of the output curve can be guaranteed and strong system stability can be obtained. the linearity of the output curve can be guaranteed and strong system stability can be obtained. droop control control droop

dq current current control control loop loop dq

DC voltage voltage control control loop loop DC

Udc*dc* U

Udkdk U

PI PI LIM2 LIM2

LIM1 LIM1

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iiaa,,bb,,cc

abc abc dq dq

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space space vector vector pulse pulse width width modulation modulation

θθ

Eqq E

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pulses pulses

Ua U a Grid U b Grid U b

Figure 4. 4. Voltage Voltage droop droop control Figure 4. Voltage droop control control method method based based on on output output current current feed-forward. feed-forward. Figure method based on output current feed-forward.

3.3. Coordinated Coordinated Control Control Strategy Strategy 3.3. Figure 5a 5a shows shows the the ideal ideal scheme scheme for for output output curves curves of of the the reversible reversible converter and the the 12-pulse 12-pulse Figure converter and diode thethe ideal no-load DC voltage of both 12-pulse diode rectifier the reversible diode rectifier. rectifier.UU U doisis is the ideal no-load DC voltage voltage ofthe both the 12-pulse 12-pulse diode and rectifier and the the dodo diode rectifier. ideal no-load DC of both the diode rectifier and converter. U is the DC voltage target value when the reversible converter works in the inverting reversible converter. U LIM is the DC voltage target value when the reversible converter works in the LIM reversible converter. ULIM is the DC voltage target value when the reversible converter works in the state. A and B are the limit points. With the increase of increase the trainof traction power, the DC voltage inverting state. A and andpower B are are the the power limit points. With the the increase of the train train traction power, the inverting state. A B power limit points. With the traction power, the will drop fast. Above voltage output are completely thecompletely same, and the theysame, shareand the DC voltage voltage will dropthe fast. AboveUthe the voltage UBcurves B,, the the output output curves are are completely the same, and B , the DC will drop fast. Above voltage U curves traction load equally. However, because the value of R will change with the output current, it will they share the traction load equally. However, because the value of R dp will change with the output they share the traction load equally. However, becausedpthe value of Rdp will change with the output increase difficulty of software implementation. 5b shows a simplified which the current, the it will will increase the difficulty difficulty of software softwareFigure implementation. Figure 5b 5b scheme shows in simplified current, it increase the of implementation. Figure shows aa simplified value ofin Rdp is fixed. scheme in which the value value of of R Rdp dp is is fixed. fixed. scheme which the Udcdc U

U dc U dc

U LIM U LIM

U LIM U LIM

Udodo U

U do U do

IIdcdc

IIdcdc

(a) (a)

(b) (b)

Figure 5. 5. Proposed coordinated coordinated control strategy. strategy. (a) (a) Ideal Ideal scheme; scheme; (b) (b) Simplified Simplified scheme. scheme. Figure Figure 5. Proposed Proposed coordinated control control strategy. (a) Ideal scheme; (b) Simplified scheme.

4. Analysis of of the DC DC Short-Circuit Characteristics Characteristics of of the the Reversible Reversible Converter Converter 4. 4. Analysis Analysis of the the DC Short-Circuit Short-Circuit Characteristics of the Reversible Converter Figure 6 shows the the DC short-circuit short-circuit fault of of the reversible reversible converter. R R and LL are are the the equivalent equivalent Figure Figure 66 shows shows the DC DC short-circuit fault fault of the the reversible converter. converter. R and and L are the equivalent resistance and inductance of the DC line, and the value of R and L varies according to the position of resistance and inductance inductance of of the the DC DC line, resistance and line, and and the the value value of of R R and and L L varies varies according according to to the the position position of of the short-circuit point from the converter DC output. C is the DC capacitor of the converter. L r is the the point from from the the converter converter DC DC output. output. C C is is the the DC DC capacitor capacitor of of the the converter. converter. LLr is the the short-circuit short-circuit point r is the AC filter filter inductance. inductance. R Rcc is is the the equivalent equivalent series series resistance resistance of of the the DC DC capacitor capacitor C. C. It It can can be be seen seen that that it it AC AC filter inductance. Rc is the equivalent series resistance of the DC capacitor C. It can be seen that it is is a complex nonlinear circuit when a DC short-circuit fault occurs. The whole period can be divided is a complex nonlinear circuit when a DC short-circuit fault occurs. The whole period can be divided into the the following following three three stages stages [29]. [29]. into

is i1 ia ,b ,c

i2

iL

i3

iC

C

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Lr

R

L

+

v_C

6 of 19

Rc

a complex nonlinear circuit when a DC short-circuit fault occurs. The whole period can be divided Appl. Sci. 2017, 7, 282 6 of 19 into the following three stages [29]. Figure 6. DC short-circuit fault of theisreversible iL converter.

i3 i2 4.1. Stage of Resistance-Inductance-Capacitor i(RLC) Second-Order Response 1 R iC

L

ia ,b ,c fault occurs instantly, the +capacitor C discharges quickly, and In this stage, the DC short-circuit vC C the DC voltage drops quickly from the initial value to zero. The_ DC fault current correspondingly rises rapidly. During this period, (R + Rc), L and C constitute an Resistance-Inductance-Capacitor Lr C and L have initial Rc values when the DC short-circuit fault (RLC) series second-order circuit. The occurs. Due to the presence of the AC inductor Lr, the three-phase AC current cannot be mutated, but flows through the six anti-parallel diodes, so there is an external excitation current to the second-order RLC circuit. As shown in Figure 7, thisofstage can be seen as a full response (including Figure DCshort-circuit short-circuit fault thereversible reversible converter. Figure 6.6.DC fault of the converter. zero-input response and zero-state response) process of the RLC series circuit. Zeroofinput response process: Regardless of the influence of the AC system (without considering 4.1. Stage Stage Resistance-Inductance-Capacitor (RLC) Second-Order Response 4.1. of Resistance-Inductance-Capacitor (RLC) Second-Order Response the is), it is the zero-input response process of the RLC second-order circuit. From Kirchhoff Voltage In(KVL): this stage, stage, the the DC DC short-circuit short-circuitfault faultoccurs occursinstantly, instantly, the the capacitor capacitor CC discharges dischargesquickly, quickly, and and In this Laws the DC voltage value to to zero. TheThe DCDC faultfault current correspondingly rises the voltage drops dropsquickly quicklyfrom fromthe theinitial initial value zero. current correspondingly 2 (6) vc LCand dvan c rapidly. During this period, (R + R(R ),+LdRand constitute Resistance-Inductance-Capacitor (RLC) cLC rises rapidly. During this period, c), C constitute an Resistance-Inductance-Capacitor + ( R + Rc ) C + vc = 0 2 dt dt values series series second-order circuit.circuit. The C The and C L and haveLinitial when. the DCthe short-circuit fault occurs. (RLC) second-order have values initial when DC short-circuit fault Due to the of the AC Lr , the three-phase AC current becannot mutated, but flows occurs. Duepresence to the presence of inductor the AC inductor Lr, the three-phase ACcannot current be mutated, 2 L / C , Equation (6) may have different According to the the six relationship (R +is an Rc)external through thethrough six anti-parallel diodes, soofthere currentexcitation to the second-order but flows anti-parallel diodes, soand there excitation is an external current to RLC the circuit. As shown in FigureAs 7, shown this stage can be 7, seen asstage a fullcan response (including zero-input response ( R + R ) < 2 L / C second-order RLC circuit. in Figure this be seen as a full response (including c solutions, considering R and Rc are small, this paper only analyses . and zero-state response) process of response) the RLC series circuit. zero-input response and zero-state process of the RLC series circuit. Zero input response process: Regardless of the influence of the AC system (without considering is process iof L the RLC second-order circuit. From Kirchhoff Voltage the is), it is the zero-input response Laws (KVL): L R iC 2 (6) d vc+ dv LC + R + Rc ) C c + vc = 0 C 2 vC ( dt _ dt . According to the relationshipRof (R + Rc) and 2 L / C , Equation (6) may have different c solutions, considering R and Rc are small, this paper only analyses

( R + Rc ) < 2 L / C .

Figure 7. Resistance-Inductance-Capacitor (RLC) second-order circuit response equivalent circuit. Figure 7. Resistance-Inductance-Capacitor i (RLC) second-order circuit response equivalent circuit.

is

L

Lof initial R t0, the Zero input response process: of influence the AC system (without Assume that the short circuit Regardless fault occurs atthe moment condition is vc(t0) = considering V0, iL(t0) = I0. iC the ), it is the zero-input response process of the RLC second-order circuit. From Kirchhoff Voltage Theiscircuit is in the oscillation discharge process and, from Equation (6), the zero-input response + Cand current v_C Laws (KVL): of the DC side voltage components after fault are as follows: d2 vc dvc LCV0 ω20 + (6) ( R + Rc )C I 0+ v−δc t = 0. (7) −δt vc = dt e sin ( ωt + β ) −dt e sin ( ωt ) Rcω √ ωCd According to the relationship of (R + Rc ) and 2 L/C, Equation (6) may have different solutions, √ considering R and Rc are small, this paper only analyses ( R + Rc ) < 2 L/C. (8) dv I ω occurs at moment V t , the initial condition is vc (t0 ) = V 0 , Assume that the short circuit fault iL1 = C c = − 0 0 e −δt sin(ωt − β) + 0 e −δ0t sin(ωt ) 7. Resistance-Inductance-Capacitor (RLC) second-order circuit response equivalent circuit. iL (t0 ) Figure = I0 . The circuit is in thedoscillation process from Equation (6), the zero-input ω discharge ωL and, t response components of the DC side voltage and current after fault are as follows: Assume that the short circuit fault occurs at moment t0, the initial condition is vc(t0) = V0, iL(t0) = I0. V0 ω0 −δt process and, from I0 Equation The circuit is in the oscillation (6), vc = discharge e sin(ωt + β) − e−δt sin(ωt (7) ) the zero-input response ω ωC components of the DC side voltage and current after fault ared as follows: vc =

iL1 = C

d vc dt

V0 ω0 −δt I e sin ( ωt + β ) − 0 e −δt sin ( ωt ) ω ωCd

(7)

I 0 ω0 −δt V e sin(ωt − β) + 0 e −δt sin(ωt ) ω ωL

(8)

=−

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dvc I ω V = − 0 0 e−δt sin(ωt − β) + 0 e−δt sin(ωt) iL1 = C dt ω ωL 2 p 1 1 R + R R + R ω c c ω2 = − , δ= , ω0 = δ2 + ω2 = √ , β = arctan . LC 2L 2L δ LC

(8)

From Equation (7), when C discharges to zero, that is, vc = 0, the instant time is t1 = t0 +

γ , γ = arctan[(V0 ω0 Cd sin β)/( I0 − V0 ω0 Cd cos β)]. ω

(9)

Substituting Equation (9) into Equation (8): iL1 (t1 ) = −

I0 ω0 −δ( γ ) V0 −δ( γ ) ω sin(γ − β) + ω sin γ. e e ω ωL

(10)

Generally, when a metal short-circuit fault occurs in the DC line, (( R + Rc )/2L)2 is far smaller than 1/LC, It can be assumed that ω = ω0 , that is, the zero-input response component of the DC inductor current can be reduced to ! r C −δt iL1 = e V0 sin ωt + I0 cos ωt . (11) L Making θ = arctan( VI00

q

L C );

then Equation (11) becomes

iL1 = e

Rc −( R+ 2L ) t

"r

# C 2 2 V0 + I0 sin(ωt + θ) . L

(12)

Zero state response process: As shown in Figure 7, assume that there is no initial value on C and L. The current supplied to the DC side by the AC system is regarded as a constant value is . At this moment, it can be considered that the RLC second-order circuit is in the process of zero-state response, and from Kirchhoff Current Laws (KCL), LC

d2 iL2 di + ( R + Rc )C L2 + iL2 = is . dt dt2

(13)

Equation (13) can be solved the same way as calculating the zero-state response. The zero-state response of the DC inductor current can be obtained as Is ( R+ Rc ) −δt sin(ωt) 2Lω q e Is ( R+ Rc ) C −δt sin(ωt) 2 Le

iL2 = Is − Is e−δt sin(ωt + β) −

= Is − Is e−δt sin(ωt + β) −

(14)

where the δ, ω, β is the same as in Equation (8). The DC inductor current in this stage is the sum of the zero-input response component and the zero-state response component from Equations (12) and (14), that is iL = iL1 + iL2 .

(15)

From Equations (12), (14) and (15), it can be seen that the DC inductor current iL is composed of two parts in this stage. Since the zero-input response component is of a large proportion, when the zero-input response component reaches its peak value, the total DC inductor current also reaches the peak value, and the following conclusions can be obtained: When a DC short-circuit fault occurs, the larger the initial DC voltage V 0 of C and the initial current I0 of L are, the higher the peak value of the DC inductor will be. When the initial values of DC voltage and DC inductor current are constant, the peak current of the DC inductor will increase with the increase of DC capacity C, but decrease with the

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capacitance value is, the more electric field energy can be released by the capacitor C after the decrease of DC line inductance L. This is because, under a certain DC the larger capacitance short-circuit fault, and the more electromagnetic energy stored involtage, the inductor, thethe larger the DC value is, the more electric field energy can be released by the capacitor C after the short-circuit fault, inductor current peak value can reach. The smaller the inductance value is, the greater the current and the more electromagnetic energy in thethe inductor, larger the DC current peak required to store the same energy, andstored the larger currentthe is, the greater theinductor total inductor current value can reach. The smaller the inductance value is, the greater the current required to store the same peak value will be. energy, and the larger the current is, the greater the total inductor current peak value will be. 4.2. Stage of Resistance-Inductance-Diode (RLD) First-Order Response 4.2. Stage of Resistance-Inductance-Diode (RLD) First-Order Response This stage begins with the capacitor DC voltage falling to zero, and continues until the capacitor This stage begins with the capacitor DC voltage falling to zero, and continues until the capacitor is recharged. When vc > 0, the capacitor discharges and the current flows through the line is recharged. When vc > 0, the capacitor discharges and the current flows through the line impedance. impedance. With the capacitor discharge, DC voltage vc will decrease. When vc drops to zero, that is With the capacitor discharge, DC voltage vc will decrease. When vc drops to zero, that is moment t1 , moment t1, the inductor begins to discharge, and the fault current will flow through the anti-parallel the inductor begins to discharge, and the fault current will flow through the anti-parallel diodes of the diodes of the three-phase bridge, assuming that the initial condition is iL(t1) = I0′. The equivalent three-phase bridge, assuming that the initial condition is iL (t1 ) = I0 0 . The equivalent circuit is shown in circuit is shown in Figure 8, the Ron is the total equivalent internal resistance of the six anti-parallel Figure 8, the Ron is the total equivalent internal resistance of the six anti-parallel diodes, and V f is the diodes, and Vf is the their total turn-on voltage drop. their total turn-on voltage drop. _ Vf

is

iL

R

iC

+

C

Ron

L

+

v_C = 0

Rc

Figure 8. Equivalent circuit of Resistance-Inductance-Diode (RLD) first-order circuit response. Figure 8. Equivalent circuit of Resistance-Inductance-Diode (RLD) first-order circuit response.

At this this moment, moment, the circuit becomes a first-order circuit circuit and and the the DC DC inductor inductor current current decreases decreases with with the the exponential exponential law law as as follows: follows: ) )t ′ (R+ Ronon )/L t ) I=00 eI−(( iL (ti)L (= 0e . . − (R+ R

)/ L

t

(16) (16)

This process processisisthethe most dangerous fordiodes, the diodes, a largedischarge inductor This most dangerous timetime for the becausebecause there is there a largeisinductor discharge currentflowing suddenly flowing through the period, diodes and in this period,can and diodesinstantly can be current suddenly through the diodes in this the diodes be the damaged damaged instantly because of the large initial current. Therefore, the diode over-current capability because of the large initial current. Therefore, the diode over-current capability should be considered should be considered in the devices. selection of switching devices. in the selection of switching The time timethat thatcapacitor capacitorCCdischarges discharges zero, value of short-circuit the short-circuit current attime, that The to to zero, andand thethe value of the current at that time, can be simplified as can be simplified as q √ γ C 0 (17) t1 = t0 + ω = t0 + LCarctan VIV (17) L

γ 0 t1 = t0 + q = t0 + LC arctan 0 ωC 2 2 −δt1 − δtI10

iL (t1 ) = iL1 (t1 ) + iL2 (t1 ) = e

where γ = arctan

q

L V0

+ I0 + Is − Is e

C L Is (R+Rc )√C −δt1 √ cos γ − e sin γ 2 L

I (R + Rc ) C − δt1 C 2 Rc 2 V0 − δt1 C cos γ − s e sin γ =iL2 (t√11) = ,eδ−δt1= RV+ 1 )+ 0 + I.0 + I s − I s e I0iL (t1 )L= iL1,(tω LC L 2L 2 L

(18)

(18)

From Equation (17), the time of capacitor C discharging to zero, that is, the maximum over-current Vis0 related Rc C, as well as V 0 and I0 . The relationships among C to the1 value ofR L+ and time of γthe diode, = arctan , δ= . where , ω = 2L and capacitance value are shown in Figure 9 when LC capacitance discharge value, I 0 Ltime, inductance V 0 = From 750 V, Equation and I0 = 1300 It can seen that theCcapacitor discharge timethat will is, increase with the (17),A.the timebeof capacitor discharging to zero, the maximum increase of the total inductance value and capacitance value. over-current time of the diode, is related to the value of L and C, as well as V0 and I0. The relationships among capacitance discharge time, inductance value, and capacitance value are shown in Figure 9 when V0 = 750 V, and I0 = 1300 A. It can be seen that the capacitor discharge time will increase with the increase of the total inductance value and capacitance value. It can be seen from Equation (18) that the inductor peak current is related to the value of L and C, as well as V0, I0 and Is at the fault instant. The relationships among the inductor peak current,

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inductance value, and capacitance value are shown in Figure 10, when V0 = 750 V, I0 = 1300 A, and Is = 1800 A. It can be seen that the inductor peak current increases with the increase of the capacitance value, and decreases with the increase of the total inductance, which is in accordance Appl. Sci. 2017, 7, 282 9 of 19 with the analysis conclusions in Section 4.1. 9 of 19 Capacitor discharge time t1（ms） Capacitor discharge time t1（ms）

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inductance value, and capacitance value are shown in Figure 10, when V0 = 750 V, I0 = 1300 A, and Is = 1800 A. It can be seen that the inductor peak current increases with the increase of the capacitance value, and decreases with the increase of the total inductance, which is in accordance with the analysis conclusions in Section 4.1.

Equivalent inductor L (mH)

DC Capacitor C (mF)

The relationships relationships among among capacitor capacitor discharge discharge time, time, the the equivalent equivalent inductance inductance L, L, and Figure 9. The capacitance C when V V00==750 750V, V,II00==1300 1300A. A.

It can be seen from Equation (18) that the inductor peak current is related to the value of L and C, as well as V 0 , I0 and Is at the fault instant. The relationships among the inductor peak current, inductorvalue L (mH) are shown in Figure 10, when V = 750 V, I = 1300 A, inductance value, and Equivalent capacitance DC Capacitor C (mF) 0 0 and Is = 1800 A. It can be seen that the inductor peak current increases with the increase of the Figure 9. The and relationships capacitor discharge time, the equivalent L, and with capacitance value, decreasesamong with the increase of the total inductance, whichinductance is in accordance capacitance C when V0in = 750 V, I0 =4.1. 1300 A. the analysis conclusions Section

Figure 10. The relationships among inductor peak current, the equivalent inductance L, and capacitance C when V0 = 750 V, I0 = 1300 A, and Is = 1800 A.

The discharge time of the DC capacitor and the inductor peak current are analyzed by the RLC second-order discharge circuit. It can be seen from Equation (16) that the peak current begins to discharge as the initial current in the RLD first-order response. The AC system provides current for the DC side through the diode in the actual first-order discharge process, so the actual DC inductor Figure 10. among inductor peak current, the equivalent inductance L, and current. capacitance current should be the sum of the inductor discharge and system supply As the Figure 10.The Therelationships relationships among inductor peakcurrent current, theAC equivalent inductance L, and C when Vcurrent V,decays IV0 0==1300 A,Ithe and Is = A, 1800 A. 0= capacitance C750 when 750 V, 0 = 1300 and Is = 1800increases, A. DC inductor and AC grid current the capacitor starts charging again and the DC voltage increases at the same time. The The discharge discharge time time of of the the DC DC capacitor capacitor and and the the inductor inductor peak peak current current are analyzed analyzed by the RLC second-order circuit. ItIt can 4.3. Stage of thedischarge Uncontrolled Rectification second-order discharge circuit. can be be seen seen from from Equation Equation (16) (16) that that the the peak peak current current begins begins to discharge as the initial current in the RLD first-order response. The AC system provides current discharge as the initial current The AC system provides current for for When a short-circuit fault occurs, the reversible converter cannot be immediately separated the the DC DC side side through through the diode diode in the actual first-order discharge discharge process, process, so so the actual DC inductor inductor from the AC and DC network by breakers. As a result, the anti-parallel diodes will form a current should should be be the thesum sumofofthe theinductor inductordischarge discharge current and system supply current. As the current and ACAC system supply current. As the DC three-phase bridge rectifier circuit, and the AC system will continue to deliver power to the DC side DC inductor current decays andAC thegrid ACcurrent grid current increases, the capacitor starts charging again inductor current decays and the increases, the capacitor starts charging again and the until the breakers open, as shown in Figure 11. and the DC voltage at the same time. DC voltage increasesincreases at the same time. Stage of of the the Uncontrolled Uncontrolled Rectification 4.3. Stage thethe reversible converter cannot be immediately separated from When aa short-circuit short-circuitfault faultoccurs, occurs, reversible converter cannot be immediately separated the AC and DC network by breakers. As a result, the anti-parallel diodes will form a three-phase from the AC and DC network by breakers. As a result, the anti-parallel diodes will form a bridge rectifier circuit, and circuit, the ACand system willsystem continue deliver to power to power the DCtoside the three-phase bridge rectifier the AC willtocontinue deliver the until DC side breakers open, as shown Figurein 11.Figure 11. until the breakers open, asinshown

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is i1

i2

iL

i3 C

Lr

R

iC

ia ,b ,c

L

+

v_C

Rc

Figure 11. 11. The The stage stage of of uncontrolled uncontrolled rectification. rectification. Figure

The system is equivalent to a three-phase uncontrolled rectifier working in the DC-side short-circuit state. state.The Theshort-circuit short-circuit current be calculated the transient current can can be calculated by theby transient analysisanalysis method. method. Assume Assume that the phasebefore current before short-circuit is as follows: that the phase current short-circuit occurs is occurs as follows:

(

)

i |0| = I m|0| sin ωt + α − ϕ|0| ia|0| a= Im|0| sin ωt + α − ϕ|0|

(19) (19)

I m|0| where thecurrent currentamplitude, amplitude,ω ω is the synchronous angular frequency, the phase α is where Im|0| isisthe is the synchronous angular frequency, α is the phase angle, ϕ angle, is the impedance angle. and ϕ|and is the impedance angle. |0| 0| Three-phase short-circuit current can be obtained: h i τ ia (=ωt I m+sin( I m 0(sin( ) −ImI msin sin( ωtϕ+)α+− ϕI) + sin αϕ− ϕ)0− α−− ϕ ϕ)) ee−−t /t/τ ia = Im sin α− α − ( α m|0| |0| h i ◦ − ϕ) + I sin(α − 120◦ − −t/τ ◦ − ϕ)− t /e ib = Im sin ( ωt + α − 120 sinα (α−− 120 ib = I m sin(ωt + α − 120 − ϕh) m 120 + |0|I m 0 sin(α − 120 −ϕϕ|00| ) −−IImmsin( − ϕ) e i τ ◦ − ϕ) e−t/τ ic = Im sin(ωt + α + 120◦ − ϕ) + Im|0| sin(α + 120◦ − ϕ|0| ) − Im sin(α + 120 −t / τ

(20) (20)

ic = I m sin(ωt + α + 120 − ϕ) + I m 0 sin(α + 120 − ϕ 0 ) − I m sin(α + 120 − ϕ) e where I is the short-circuit current period component amplitude, ϕ is the impedance angle of the where Imm is the short-circuit current period component amplitude, ϕ is the impedance angle of the circuit loop, Lr is the AC inductance, and τ is the time constant of circuit loop. circuit loop, Lr is the AC inductance, and τ is the time constant of circuit loop. √ 2V 2Vn (21) √n ϕ = arctan ω( Lr+ /R, (21) ϕ =[arctan ω ( LLr )+/R L ) ]/,Rτ , = τ =((LLrr + + LL) /)R , I m I≈m ≈ 3ωLr Rx + R3xω+ Lr the root-mean-square root-mean-square voltage voltage of of the the transformer transformer secondary. where Vnn isisthe Equation (21), when the AC inductance Lrr is constant, the short-circuit current period From Equation component amplitude amplitudeIImmisisrelated relatedtoto the line equivalent resistance R. The farther location of component the line equivalent resistance R. The farther the the location of the the short-circuit point is from reversible converter,the thesmaller smallerthe the short-circuit short-circuit current current period short-circuit point is from thethe reversible converter, willbe. be. component Imm will short-circuit current, whose value is greater than Three-phase uncontrolled uncontrolledrectifier rectifiertransfers transfersthethe short-circuit current, whose value is greater zero, zero, to thetoDC Thus,Thus, the short-circuit current on theonDC as follows: than theside. DC side. the short-circuit current theside DCisside is as follows:

i = i + i + i3 = ia ,( > 0) + ib,( > 0) + ic ,( > 0) . is = i1s + i12 +2i3 = ia,(>0) + ib,(>0) + ic,(>0) .

(22) (22)

5. Simulation Simulation and and Experiment Experiment 5. 5.1. DC Short-Circuit Fault Simulation 5.1. DC Short-Circuit Fault Simulation A simulation model was built according to the reversible converter circuit topology in Figure 6, A simulation model was built according to the reversible converter circuit topology in Figure 6, and the specific parameters are shown in Table 1. and the specific parameters are shown in Table 1.

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Table 1. Converter parameters for the simulation. AC: alternating current; DC: direct current. Table 1. Converter parameters for the simulation. AC: alternating current; DC: direct current.

Items Values AC input voltage Vn 450 V Items Values DC output voltage Vd 750 V AC input voltage V n 450 V AC side inductor Lr 300 μH DC output voltage V d 750 V DC Capacitance C 36 mF AC side inductor Lr 300 µH Capacitance parasitic DC Capacitance C resistance Rc 36 0.02 mF mΩ Capacitance Rc of diode Ron 0.02 0.3 mΩmΩ Equivalentparasitic internalresistance Resistance Equivalent internal Resistance of diode Ron 0.3 mΩ Rated power P 1 MW Rated power P 1 MW Switching frequency f s 2 kHz Switching frequency f 2 kHz s

Due to the fact that the location of the short-circuit point is unpredictable, the simulation study Due to out the fact that the location of the short-circuit is unpredictable, the(PSC) simulation is is carried in two situations, which are calledpoint proximal short-circuit and study remote carried out in two situations, which are called proximal short-circuit (PSC) and remote short-circuit short-circuit (RSC). Suppose the resistance of the overhead contact line is r1 = 0.028 Ω/km, the (RSC). Suppose the of the overhead contact is ris Ω/km, of 1 = inductance of that is resistance l1 = 2.6629 mH/km, the resistance of line the rail r2 0.028 = 0.023 Ω/km,the the inductance inductance of that is l = 2.6629 mH/km, the resistance of the rail is r = 0.023 Ω/km, the inductance of the rail is 2 the rail 1is l1 = 1.78 mH/km. Thus, the equivalent line resistance is r = r1 + r2 = 0.051 Ω/km, and the l1 = 1.78 mH/km. Thus, theisequivalent line resistance equivalent line inductance l = l1 + l1 = 4.4429 mH/km.is r = r1 + r2 = 0.051 Ω/km, and the equivalent line inductance is l = l + l = 4.4429 mH/km. 1 1 Related parameters and calculated results of PSC are shown in Table 2. Related parameters and calculated results of PSC are shown in Table 2. Table 2. Related parameters and calculated results for proximal short-circuit (PSC). Table 2. Related parameters and calculated results for proximal short-circuit (PSC).

Items Values Items distance x Values Short-circuit 10 m Line resistance R 0.51 Short-circuit distance x 10 m mΩ Line resistance R 0.51 mΩ mH Line inductance L 0.044 Line inductance L 0.044 mHms Theoretical discharge time of capacitance t1 1.9 Theoretical discharge time of capacitance t1 1.9 ms Theoretical peak current inductance iL(t1) 23,000 A Theoretical peak current inductance iL (t1 ) 23,000 A Theoretical periodcomponent component amplitude of short-circuit current Im 38863886 Theoretical period amplitude of short-circuit current Im A A Simulation results of the PSC are shown in Figures 12 and 13: Simulation results of the PSC are shown in Figures 12 and 13:

Inductor Current（kA）

22800A 20

Inductor Current Capacitor Voltage

600

15

400

10

200

5

0

0.5019 s 0

0.5

0.52

0.54 0.56 0.58 Simulation time（s）

Capacitor Voltage vc（V）

800

25

-200 0.6

Figure 12. 12. DC DC inductor inductor current current and and capacitor capacitor voltage voltage for for proximal proximal short-circuit Figure short-circuit (PSC). (PSC).

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Three Phase AC Current（kA）

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8

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ia

ib

ic

ThreeACPhase AC Current（kA） Three Phase Current（kA）

6

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4 2

0

-2 -4 -6

8 6 8 4 6 2 4 0 2 -2 0 -4 0.45 0.5 -2 -6 0.45 0.5 -4 -6

12 of 19

3780A

ia

ib

ic

ia

ib

ic

12 of 19

3780A

3780A

0.55

0.6

0.55

0.65

0.7

0.75

0.6 0.65 time（s） 0.7 0.75 Simulation

0.8

0.85

0.8

0.85

0.75 for PSC. 0.8 Figure AC current for PSC. Figure13. 13.Three-phase Three-phase AC current for PSC. Simulation time（s）

0.85

Simulation time（s） 0.45Figure 0.5

0.6 0.65 0.7 13.0.55 Three-phase AC current

Related parameters and calculated results for RSC are shown in Table 3. Related parameters and calculated results for RSC RSC inin Table 3. 3. Figure 13. Three-phase AC are current for PSC. Related parameters and calculated results for areshown shown Table

Table 3. Related parameters and calculated results for RSC. Table 3. Related parameters and calculated results for RSC. Related parameters and calculated results for RSC are shown in Table 3. Table 3. Related parameters and calculated results for RSC. Items Items Table 3. Related parameters and calculated results for RSC. Short-circuit Short-circuitdistance distance xx Items Lineresistance resistance Line R Items Short-circuit distance Lineinductor inductor Short-circuit distance Line Lx x Line resistance R of capacitor t1 Theoretical discharge time Line resistance R Theoretical discharge time Line inductor L of capacitor t1 Theoretical peak current of inductance iL(t1) Line inductor L Theoretical current Theoretical peak discharge timeof of inductance capacitor t1 iL(t1) Theoretical period component amplitude ofcapacitor short-circuit current Im Theoretical discharge time of peak current of inductance iL (t1 ) t1 Theoretical Theoretical period component amplitude of short-circuit current Im Theoretical peak current of inductance i L (t 1 ) Theoretical period component amplitude of short-circuit current Im Simulation resultsperiod for RSC are shownamplitude in Figuresof14short-circuit and 15. Theoretical component current Im

Simulation results for RSC are shown in Figures 14 and 15. Simulation results for for RSC areare shown 15. Simulation results RSC shownininFigures Figures 14 14 and and 15.

Values Values 1.7 km 1.7 km Values 86.7 mΩmΩ 86.7 Values 1.7 km 7.553 mHmH 1.7 km 7.553 86.7 mΩ 14.6 ms 86.714.6 mΩms 7.553 mH 2572 mH A 7.553 14.62572 ms A 2550 A 14.6 ms 25722550 A A 2572A A 2550 2550 A

Figure 14. DC inductor current and capacitor voltage for RSC.

ThreeACPhase AC Current（kA） Three Phase Current（kA）

Three Phase AC Current（kA）

4 Figure 14. DC inductor voltage RSC. Figure 14.14. DCDC inductor current and capacitorvoltage Figure inductorcurrent currentand and capacitor capacitor forfor RSC. 3000A ia ib ic

4

24

2

02

0

ib ib

ic ic

3000A 3000A

-20

-2 -4

-2

-4

-4

ia ia

0.45

0.45

0.45

0.5

0.5

0.55 0.6 Simulation time（s） 0.55

0.6

Figure 15. Three-phase AC current for RSC. Simulation time（s） 0.5 0.55 0.6 Simulation time（s） Figure 15. Three-phase AC current for RSC.

Figure 15. Three-phase AC current for RSC. Figure 15. Three-phase AC current for RSC.

0.65

0.65

0.65

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The simulation results and the theoretical calculation results are compared in Table 4. Table 4. The comparison between the simulation results and theoretical calculation results. Short-Circuit Point

Parameters

Calculation Results

Simulation Results

PSC, x = 10 m

t1 iL (t1 ) Im

1.9 ms 23,000 A 3886 A

2 ms 22,800 A 3780 A

RSC, x = 1.7 km

t1 iL (t1 ) Im

14.6 ms 2572 A 2550 A

15.5 ms 2000 A 3000 A

From the comparisons in Table 4, some conclusions can be obtained. When the RSC occurs, the simulation results basically agree with the theoretical calculation results. However, when the RSC occurs, there are greater differences between simulation results and theoretical results. The reason is that the discharge current of the DC capacitor in RSC is smaller than the current that is provided by the AC system, so when the capacitor discharge is finished, the DC inductor current has not reached its peak value, but continues to increase slowly to the steady state short current. It is known from the above comparison that the peak short-circuit current in PSC is larger, and the rise is also faster, so it will do greater harm to semiconductor devices. In order to protect the reversible converter effectively, the PSC should be seriously considered. 5.2. Coordinated Control Experiments Two sets of high-power reversible converters have been successfully applied in the SHILIHE and XIDIAOYUTAI traction substations of Beijing Metro Line 10 (Beijing, China) as a demonstration example. The parameters of the reversible converters are shown in Table 5. Table 5. Parameters of the reversible converter. Parameter

Value

Parameter

Value

AC grid voltage DC rated voltage Converter capacity

10 kV 750 V 2 MW

Transformer capacity IGBT specification Switching frequency

1.6 MVA 2400 A/1700 V 2 kHz

The system main circuit of the traction substation is shown in Figure 16. Uac indicates the voltage of the AC 10 kV grid, iac1 indicates the AC current of the reversible converter, iac2 and iac3 indicate the AC current of each 12-pulse rectifier, respectively, idc1 indicates the DC current of the reversible converter, idc2 and idc3 indicate the DC current of each 12-pulse rectifier, respectively, and Udc indicates the DC bus voltage. The reversible converter has the same DC output characteristic as shown in Figure 5b. Figure 17 shows the voltage and current waveforms on the 10 kV side when the reversible converter and two 12-pulse rectifiers are working in parallel. As shown in Figure 17a, when the train is speeding up, the reversible converter and two 12-pulse rectifiers all work in the rectifying state: they share the train power. As shown in Figure 17b, when the train is slowing down, the reversible converter works in the inverting state, and the two 12-pulse rectifiers are blocked because of the high DC bus voltage. It is obvious that the AC current waveforms of the reversible converter are higher qualified than that of the 12-pulse rectifiers.

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Figure 16.16. Main circuit of the traction substation in Beijing Metro Line 10.10. Figure Main circuit of the traction substation in Beijing Metro Line Figure 16. Main circuit of the traction substation in Beijing Metro Line 10.

(a) (a) Figure 17. Cont.

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(b) Figure 17. AC current waveforms when the reversible converter and two 12-pulse rectifiers work in parallel. (a) train in traction state; (b) train in regenerative braking state.

Figure 18 shows the DC voltage and current waveform when the reversible converter and a 12-pulse rectifier are working in parallel. The meanings of the symbols are shown in Figure 16. As seen in Figure 18, along with the increase of traction power of the train, the DC bus voltage (b) reversible converter and 12-pulse rectifier unit decreases significantly, but the DC current of the increase immediately. However, it when is obvious that the current increase rate of thework reversible Figure Figure 17. 17. AC AC current current waveforms waveforms when the the reversible reversible converter converter and and two two 12-pulse 12-pulse rectifiers rectifiers work in in converter is greater than that of the 12-pulse rectifier unit. The trends basically agree with the DC parallel. state. parallel. (a) (a) train train in in traction traction state; state; (b) (b) train train in in regenerative regenerative braking braking state. output curves shown in Figure 5b. When the train work state is changed to braking from traction, the current 12-pulse rectifiers to zero,waveform but the DC current the reversible converter Figureof18both shows the DC voltagereduce and current when the of reversible converter and a Figure 18 shows to the DC voltage and current waveform when braking the reversible converter and a changes from positive negative, which means that the regenerative energy of the train 12-pulse rectifier are working in parallel. The meanings of the symbols are shown in Figure 16. has 12-pulse rectifier are working in parallel. The meanings of the symbols are shown in Figure 16. been fed backin to Figure the AC18, 10 along kV grid through the reversible converter. As seen with the increase of traction power of the train, the DC bus voltage decreases significantly, but the DC current of the reversible converter and 12-pulse rectifier unit increase immediately. However, it is obvious that the current increase rate of the reversible converter is greater than that of the 12-pulse rectifier unit. The trends basically agree with the DC output curves shown in Figure 5b. When the train work state is changed to braking from traction, the current of both 12-pulse rectifiers reduce to zero, but the DC current of the reversible converter changes from positive to negative, which means that the regenerative braking energy of the train has been fed back to the AC 10 kV grid through the reversible converter.

Figure Figure18. 18.DC DCvoltage voltageand andcurrent currentwaveforms waveformswhen whenthe thereversible reversibleconverter converterand andaa12-pulse 12-pulserectifier rectifier work in parallel. work in parallel.

As seen in Figure 18, along with the increase of traction power of the train, the DC bus voltage decreases significantly, but the DC current of the reversible converter and 12-pulse rectifier unit increase immediately. However, it is obvious that the current increase rate of the reversible converter is greater than that of the 12-pulse rectifier unit. The trends basically agree with the DC output curves Figure 18. DC voltage and current waveforms when the reversible converter and a 12-pulse rectifier work in parallel.

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16 of 19

shown in Figure 5b. When the train work state is changed to braking from traction, the current of both 16 of 19 12-pulse rectifiers reduce to zero, but the DC current of the reversible converter changes from positive to negative, which means that the regenerative braking energy of the train has been fed back to the AC Traction and braking energy statistics of one substation are shown in Table 6. A and B denote 10 kV grid through the reversible converter. traction energy provided by each 12-pulse rectifier, respectively, C denotes the traction energy provided Traction and braking energy statistics of one substation are shown in Table 6. A and B denote by the reversible converter, and D denotes the braking energy recovered by the reversible converter. traction energy provided by each 12-pulse rectifier, respectively, C denotes the traction energy provided by the reversible converter, and D denotes the regenerated braking energy recovered Table 6. Used and energy (MWh). by the reversible converter. Appl. Sci. 2017, 7, x

Reversible Converter 12-Pulse 12-Pulse Total Traction Percentage of Table 6. Used and regenerated energy (MWh). Rectifier 1/A Rectifier 2/B Rectifying/C Inverting/D Energy/A + B + C C/(A + B + C) 2016/6/18 4.43212-Pulse 4.482 4.699 0.974 34.52%of Total 13.613 Traction Reversible Converter Percentage 12-Pulse Date Energy/A +B+C C/(A + B + C) Rectifier 1/A 3.928 Rectifier 2/B Rectifying/C 2016/6/19 3.880 4.474 0.769 12.282 36.43% Inverting/D 2016/6/20 5.186 5.227 5.528 1.498 15.941 34.68% 2016/6/18 4.432 4.482 4.699 0.974 13.613 34.52% 2016/6/21 4.505 5.456 1.593 14.424 37.83% 2016/6/19 4.463 3.880 3.928 4.474 0.769 12.282 36.43% 2016/6/20 4.345 5.186 5.227 5.528 1.498 15.941 34.68% 2016/6/22 4.377 5.183 2.154 13.905 37.27% 2016/6/21 4.476 4.463 4.505 5.456 1.593 14.424 37.83% 2016/6/23 4.514 5.196 2.334 14.186 36.63% 2016/6/22 4.345 4.377 5.183 2.154 13.905 37.27% 2016/6/24 4.874 4.916 5.198 1.942 14.988 34.68% 2016/6/23 4.476 4.514 5.196 2.334 14.186 36.63% 2016/6/25 3.560 4.282 0.656 11.363 37.68% 2016/6/24 3.521 4.874 4.916 5.198 1.942 14.988 34.68% 2016/6/26 3.548 4.623 0.511 11.679 39.58% 2016/6/25 3.508 3.521 3.560 4.282 0.656 11.363 37.68% 2016/6/26 4.390 3.508 3.548 4.623 0.511 11.679 39.58% 2016/6/27 4.393 5.327 1.389 14.110 37.75% 2016/6/27 4.642 4.390 4.393 5.327 1.389 14.110 37.75% 2016/6/28 4.654 5.336 2.335 14.632 36.47% 2016/6/28 4.642 4.654 5.336 2.335 14.632 36.47% 2016/6/29 4.717 4.752 5.290 2.519 14.759 35.84% 2016/6/29 4.717 4.752 5.290 2.519 14.759 35.84% 2016/6/30 4.453 5.041 1.370 13.912 36.23% 2016/6/30 4.418 4.418 4.453 5.041 1.370 13.912 36.23% 57.309 65.633 20.044 179.794 36.50% sum sum 56.852 56.852 57.309 65.633 20.044 179.794 36.50% Date

Figure Figure 19 19 shows shows the the histogram histogram of of the the traction traction energy energy provided provided by by the the reversible reversible converter converter and and each 12-pulse rectifier for clarity. The reversible converter has provided nearly one third of the total each 12-pulse rectifier for clarity. The reversible converter has provided nearly one third of the total traction 36.5%). However, traction power power (the (the average average proportion proportion is is 36.5%). However, it it will will be be found found that that the the traction traction power power provided by the reversible converter is slightly more than that of each 12-pulse rectifier. The reason provided by the reversible converter is slightly more than that of each 12-pulse rectifier. The reason for for is that output curve of the reversible converter is above of the 12-pulse rectifier in thisthis is that thethe DCDC output curve of the reversible converter is above thatthat of the 12-pulse rectifier in the the light load region, as shown in Figure 5b. light load region, as shown in Figure 5b.

Traction energy /MWh

6.000

Traction energy distribution (MWh)

12-pulse rectifier1 12-pulse rectifier2 reversible converter

5.000 4.000 3.000 2.000 1.000 0.000

date Figure19. 19.Traction Tractionenergy energy distribution distribution between between the the reversible reversible converter converter and and each each 12-pulse 12-pulse rectifier. rectifier. Figure

In order to evaluate the energy-saving effect of the reversible converter used in the traction power supply system more clearly and intuitively, the average energy-saving percentage over a long period is defined as η: n

E

(23)

inv _ i

η=

i =1 n

Erec _ i i =1

× 100%

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17 of 19

In order to evaluate the energy-saving effect of the reversible converter used in the traction power supply system more clearly and intuitively, the average energy-saving percentage over a long period is defined as η: n

∑ Einv_i

η=

i =1 n

× 100%

(23)

∑ Erec_i

i =1

where Einv_i denotes the inverting energy of the reversible converter every day, and Einv_i denotes the total rectifying energy of the reversible converter and 12-pulse rectifiers every day. In Table 6, regenerated energy and total traction energy change every day, but according to Equation (23), the average energy-saving percentage η (from 18 June 2016 to 30 June 2016) is calculated as high as 11.15%. 6. Conclusions In this paper, the high-power reversible converter, which can not only be used to recuperate regenerative braking energy, but also provide traction energy, is applied to construct a new hybrid traction power supply system for urban rail transit. A droop control method based on load current feed-forward is proposed to realize the load distribution between the reversible converter and the existing 12-pulse diode rectifiers. It is successfully verified by the field test carried out on Beijing Metro Line 10. The DC short-circuit characteristics of the reversible converter is studied, and then the relationship between the peak fault current and the circuit parameters are derived. Theoretical calculation and computer simulation shows that the peak fault current in RSC is larger, and the time to reach the peak value is also shorter. Thus, it will do greater harm to the reversible converter. The field test data indicates that the average value of energy-savings percentage for one substation is as high as 11.15%. Acknowledgments: This research was supported by National Key Research and Development Program 2016YFB1200504. Author Contributions: Gang Zhang proposed the hybrid traction power supply system and the droop control method based on load current feed-forward. Jianglin Qian studied the DC short-circuit characteristics and carried out the simulation. Xinyu Zhang analyzed the field test data. Conflicts of Interest: The authors declare no conflict of interest.

References 1.

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6.

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Application of a High-Power Reversible Converter in a Hybrid Traction Power Supply System Gang Zhang 1,2, *, Jianglin Qian 1,2 and Xinyu Zhang 1,2 1 2

*

School of Electrical Engineering, Beijing Jiaotong University, Beijing 100044, China; [email protected] (J.Q.); [email protected] (X.Z.) Beijing Electrical Engineering Technology Research Center, Beijing 100044, China Correspondence: [email protected]; Tel.: +86-10-5168-7082

Academic Editor: Eric Ka-wai Cheng Received: 24 December 2016; Accepted: 7 March 2017; Published: 14 March 2017

Abstract: A high-power reversible converter can achieve a variety of functions, such as recovering regenerative braking energy, expanding traction power capacity, and improving an alternating current (AC) grid power factor. A new hybrid traction power supply scheme, which consists of a high-power reversible converter and two 12-pulse diode rectifiers, is proposed. A droop control method based on load current feed-forward is adopted to realize the load distribution between the reversible converter and the existing 12-pulse diode rectifiers. The direct current (DC) short-circuit characteristics of the reversible converter is studied, then the relationship between the peak fault current and the circuit parameters is obtained from theoretical calculations and validated by computer simulation. The first two sets of 2 MW reversible converters have been successfully applied in Beijing Metro Line 10, the proposed hybrid application scheme and coordinated control strategy are verified, and 11.15% of average energy-savings is reached. Keywords: urban railway transit; traction power system; regenerative braking; reversible converter; direct current short-circuit

1. Introduction At present, diode rectifiers are still widely used in urban rail transit traction power supply systems to provide energy for the trains [1]. Since the energy can only be transmitted from alternating current (AC) to direct current (DC), the braking resistor must be employed to consume the surplus regenerative braking energy of the trains in order to avoid the abnormal rise of the DC voltage. This scheme will cause a very large waste of energy, increasing the tunnel temperature and the burden of the cooling system. Energy storage based on electric double-layer capacitors (EDLC) and flywheels are widely used in the transportation field [2–4]. EDLC has good performance in terms of power density, charge and discharge time, long lifetime cycles, and lower internal resistance, so it can be used for storing regenerative braking energy of electric vehicles [5–10]. However, in the field of urban rail transit, the maximum regenerative braking power is usually up to several megawatts, so thousands of EDLCs would be needed, which requires a large installation space and cost [11]. Flywheels were firstly installed in Japan in 1988 for storing regenerative energy [12]. The rated energy and power of the flywheels are 25 kWh and 2000 kW, respectively. After that, flywheels were verified in tests at the London underground and Spain railways. However, all of them are just prototypes. Complex mechanisms and high costs limit their wide utilization. Flywheel systems for energy saving may be an important research direction in the future [13–15]. It is a good energy-saving method to invert the regenerative braking energy to the AC grid, which has been already applied in several subway lines [16–21]. However, it requires additional installation Appl. Sci. 2017, 7, 282; doi:10.3390/app7030282

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space, which will increase the cost of substation construction, and the utilization rate of the inverter is not very high. It will be a good idea to introduce high-power reversible converters into the traction power supply system of urban rail transit. Not only will braking energy be recovered, but traction power supply can also be realized by the reversible converter. Thus, the capacity of diode rectifiers can be reduced, or one of the two existing 12-pulse rectifiers may be replaced by a reversible converter for the purpose of saving space. Generally, the DC output characteristic of the reversible converter is constant, which can be realized by a simple voltage closed-loop. However, because it needs to work with the diode rectifier units in parallel, in order to share the traction load, the reversible converter must have a similar voltage droop characteristic as the diode rectifier. Droop control methods are commonly used in DC microgrids [22,23] and other distribution power systems [24–26]. Few studies have paid attention to the DC voltage control of the reversible converter used in traction power supply systems. In the field of urban rail transit, the existing literature is mainly aimed at the steady state and transient short-circuit characteristics of the diode rectifiers [27,28]. A mathematical model of the 12-pulse rectifier is established, and the approximate expression of the short-circuit current is deduced at the outlet. The accuracy of the calculation method was verified by modeling and simulation. However, the DC short-circuit characteristics of the reversible converter are different from the diode rectifier. The DC short-circuit fault of the grid-connected inverter used in wind power generation systems was studied in the literature [29,30], which can be used as an important reference for this paper. This paper aims to introduce the high-power reversible converter into the traction power supply system of urban rail transit, and lay a foundation for further application. The following sections are organized as follows: A hybrid traction power supply system composed of a diode rectifier and reversible converter is proposed in Section 2. The DC output characteristic of the 12-pulse rectifier is introduced and a droop control method based on load current feed-forward is proposed for the reversible converter in Section 3. In order to provide guidance for system protection, the DC short-circuit fault of the reversible converter is studied, the peak short current and rise time are both calculated, which agree with the simulation results in Section 4. The effectiveness of the coordination control strategy is verified by a field test at Beijing Metro Line 10 in Section 5. Finally, conclusions are made in Section 6. 2. Proposed Hybrid Traction Power Supply Scheme At present, the traction power supply equipment installed in the traction substation is a typical 24-pulse rectifier, which is made up of two 12-pulse rectifiers. In this way, AC grid current harmonics and DC voltage ripple can be reduced, and the system redundancy can be increased. For diode rectifiers, DC output voltage is uncontrollable, and the energy can only be transmitted from AC to DC. In order to reuse the braking energy and reduce the DC voltage fluctuation, a high-power reversible converter is introduced into the traction power supply system. Figure 1 shows the proposed hybrid traction power supply scheme. In Figure 1a, an additional reversible converter branch is added, but in Figure 1b, a 12-pulse rectifier is replaced by a high-power reversible converter. It is obvious that no additional installation space is required in Figure 1b. The arrows indicate the direction of the energy flow. No matter what kind of scheme, the reversible converter has a four-quadrant operation capability. When the train is in the traction state, the reversible converter can provide energy for the train, and reduce the DC voltage drop. When the train is in the braking state, the reversible converter can invert the braking energy back to the AC grid, and save energy. The overvoltage caused by regenerative braking can be suppressed effectively. Since the diode rectifier and reversible converter will work in parallel, their work characteristic must be studied, respectively, and then a coordinated control strategy should be adopted.

Appl. Sci. 2017, 7, 282

3 of 19 rectifier

Transformer

rectifier

Transformer

Udc

Udc

Appl. Sci. 2017, 7, 282 Appl. Sci. 2017, 7, 282

12-pulse rectifier

3 of 19 3 of 19

12-pulse rectifier

rectifier

alternating current grid

Transformer

rectifier

Transformer

Appl. Sci. 2017, 7, 282 UABC

Udc

alternating current grid

Udc 3 of 19 converter Transformer 12-pulse rectifier rectifier Transformer

rectifier converter rectifier

Transformer

Udc

Transformer

Udc

alternating current grid

12-pulseconverter rectifier reversible 12-pulse rectifier rectifier

UABC

reversible converter 12-pulse rectifier

UABC

(a)

(b) Transformer

converter

Figure 1.Transformer Proposed hybrid traction power supply

alternating scheme. current (a) grid Hybrid

UABC

converter

scheme I; (b) Hybrid scheme II.

UABC

12-pulse rectifier 3. Coordinated Control Strategy

reversible converter converter

reversible converter

Transformer

rectifier

Transformer

UABC

12-pulse rectifier 12-pulse rectifier

Transformer Transformer

alternating current grid

alternating current grid

Transformer

converter

(a)

(b)

3.1. Direct Current (DC) Output Characteristics of the 12-Pulse Rectifier

Figure 1. Proposed hybrid traction power supply scheme. (a) Hybrid scheme I; (b) Hybrid scheme II.

A 12-pulse rectifier cantraction be equivalent an ideal DC voltage source Uk in I; series with anscheme internalII. Figure 1. Proposed hybrid power to supply scheme. (a) Hybrid scheme (b) Hybrid reversible converter reversible converter resistance Req and a diode D, as shown in Figure 2. 3. Coordinated Control Strategy (a) (b) The DCControl output characteristic curve of a 12-pulse rectifier is shown in Figure 3. The critical 3. Coordinated Strategy Figure 1. Proposed hybrid Itraction power supply scheme. Hybrid scheme I; (b) Hybrid scheme II. current (or Current transition current) is bounded. The is (a) divided 3.1. Direct (DC) OutputdgCharacteristics of thecurve 12-Pulse Rectifierinto two curves: ① is the curve that two 6-pulse(DC) rectifiers work in a push-pull mode; ② is Rectifier the curve that two six-pulse rectifiers 3.1. Direct Current Output Characteristics of the 12-Pulse 3. Coordinated Control Strategy A can be equivalent to an ideal DC voltage source Uk in series with an internal work in 12-pulse parallel. rectifier U doo denotes the no-load voltage of a 6-pulse rectifier, Udio denotes the ideal no-load A 12-pulse rectifier can be equivalent to an ideal DC voltage with an internal resistance and a diode D, as shown Figure 2.the rated k in DC voltageReq of a 12-pulse rectifier, UdNindenotes outputsource voltageUof a series 12-pulse rectifier, 3.1. Direct Current (DC) Output Characteristics of the 12-Pulse Rectifier The DC output characteristic curve of a 12-pulse rectifier is shown in Figure 3. The critical resistance R and a diode D, as shown in Figure 2. eq IdN denotes the rated output voltage of a 12-pulse rectifier. current transition current) Idgequivalent is bounded. Theideal curve dividedsource into two curves: is an theinternal curve A (or 12-pulse rectifier can be to an DCisvoltage Uk in series ① with that two 6-pulse rectifiers work in a push-pull mode; ② is the curve that two six-pulse rectifiers resistance Req and a diode D, as shown in Figure 2. work in parallel. Udoo denotes the no-load of a 6-pulse rectifier, Udio denotes the ideal no-load The DC output characteristic curvevoltage of a 12-pulse rectifier is shown in Figure 3. The critical DC voltage of a 12-pulse rectifier, U dN denotes the rated output voltage of a 12-pulse rectifier, current (or transition current) Idg is bounded. The curve is divided into two curves: ① is the curve Req rectifier. IdN denotes the rated output work voltage 12-pulse that two 6-pulse rectifiers inofa apush-pull mode; ② is the curve that two six-pulse rectifiers work in parallel. Udoo denotes the no-load voltage of a 6-pulse U dc rectifier, Udio denotes the ideal no-load D DC voltage of a 12-pulse rectifier, UdN denotes the rated output voltage of a 12-pulse rectifier, IdN denotes the rated output voltage of a 12-pulse rectifier.

U Reqk

U dc

Figure 2. Equivalent circuit rectifier. Figure 2. Equivalent D circuitofofa 12-pulse a 12-pulse rectifier.

Req Uk The DC output characteristic curve of a 12-pulse rectifierUis shown in Figure 3. The critical current dc 1 is the curve that two (or transition current) Idg is bounded. The curveDis divided into two curves: ① U doo Figure 2. Equivalent circuit of a 12-pulse rectifier. 2② 6-pulse rectifiers work in a push-pull mode; is the curve that two six-pulse rectifiers work in parallel. U dio Uk Udoo denotes the no-load voltage of a 6-pulse rectifier, Udio denotes the ideal no-load DC voltage of a U dN U dc 12-pulse rectifier, UdN denotes the rated output voltage of a 12-pulse rectifier, IdN denotes the rated Figure 2. Equivalent circuit of a 12-pulse rectifier. output voltage of a 12-pulse rectifier.

U dc

① U doo U dio U dc U dN U0doo I① dg

②

I dN

I dc

② U dio FigureU3.dN12-pulse rectified output characteristic curve.

0

I dg

I dN

I dc

Figure 3. 12-pulse rectified output characteristic curve.

0

I dg

I dN

I dc

Figure 3. 12-pulse rectified curve. Figure 3. 12-pulse rectifiedoutput output characteristic characteristic curve.

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1 3.1.1. Curve

The DC voltage and current show typical 12-pulse characteristics. However, at every moment there is actually a single three-phase bridge rectifier at work. Its no-load voltage is Udoo

1 = π/6

Z

π 12 π − 12

√

√ π 6 2 × 2U2L sin = 1.398U2L 2U2L cos θdθ = π 12

(1)

where U2L is the voltage effective value at the secondary side of the transformer. Thus, the mathematical 1 can be described as model of the output curve Udc = Udoo −

3 Xc Idc . π

(2)

1 is as follows: The equivalent resistance for curve

Req1 =

3 Xc . 2π

(3)

2 3.1.2. Curve

When the DC load current is larger than the critical current Idg , the output curve is shown 2 The two six-pulse rectifiers work completely in parallel. The load is shared equally by them. as . 2 extended to vertical axis, is called the ideal The voltage Udio , which is the cross point of curve no-load DC voltage. The voltage of Udio is equal to 1.35U2L . Since the two six-pulse rectifiers work in parallel at the same time, the equivalent resistance for 2 will be reduced to half of the curve , 1 as follows: curve Req2 =

3 Xc 2π

2 is The mathematical description of the curve

Ud = Udio −

3 Xc Id . 2π

(4)

3.2. DC Output Characteristic of the Reversible Converter Typical DC output voltage characteristic of the reversible converter is constant. However, due to the demand of operating in parallel with the 12-pulse diode rectifier, it should have similar voltage droop characteristics. That means DC output voltage will drop with the increase of DC current, as follows: Udc = Uk − Rdp × Idc

(5)

where Uk is the no-load output voltage for reversible converter. Rdp is the equivalent resistance. A voltage droop control method based on output current feed-forward is presented in this paper, as shown in Figure 4. It includes three parts, which are called droop control, DC voltage control, and dq current control, respectively. Droop control is employed to calculate the target voltage Udc *, which decreases with the increase of the DC, complying with the law of Equation (5). The LIM1 will limit the value of the target DC voltage within the allowable range. DC voltage control adopts a classic proportional integral (PI) regulator without static error, so the influence of system parameter variation and external disturbance for the DC voltage can be eliminated, and the actual DC voltage can follow the target well. The output of the DC voltage control loop is the active current given value id *, which will be limited by LIM2, in order to avoid the overload of the reversible converter. The dq current control is based on rotating synchronous coordinates d-q. Thanks to the coordinate transformation, id

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Appl. Sci. Sci. 2017, 2017, 7, 7, 282 282 Appl.

of 19 19 55 of

and iq are DC components, and the PI compensator can reduce the error of the fundamental component to zero. the With the of proposed control method above,to thezero. linearity theproposed output curve canmethod be guaranteed reduce the error of the the fundamental fundamental component to zero. Withof the proposed control method above, reduce error component With the control above, and strong system stability can be obtained. the linearity of the output curve can be guaranteed and strong system stability can be obtained. the linearity of the output curve can be guaranteed and strong system stability can be obtained. droop control control droop

dq current current control control loop loop dq

DC voltage voltage control control loop loop DC

Udc*dc* U

Udkdk U

PI PI LIM2 LIM2

LIM1 LIM1

Rdp R dp

iiaa,,bb,,cc

abc abc dq dq

θθ

Udcdc U

IIdcdc

Edd E

id*d*

PI PI

idd iqq 00

Udd U

ωL ωL

Uqq U

PI PI

space space vector vector pulse pulse width width modulation modulation

θθ

Eqq E

Edd E

phasephaselocked loop loop locked

pulses pulses

Ua U a Grid U b Grid U b

Figure 4. 4. Voltage Voltage droop droop control Figure 4. Voltage droop control control method method based based on on output output current current feed-forward. feed-forward. Figure method based on output current feed-forward.

3.3. Coordinated Coordinated Control Control Strategy Strategy 3.3. Figure 5a 5a shows shows the the ideal ideal scheme scheme for for output output curves curves of of the the reversible reversible converter and the the 12-pulse 12-pulse Figure converter and diode thethe ideal no-load DC voltage of both 12-pulse diode rectifier the reversible diode rectifier. rectifier.UU U doisis is the ideal no-load DC voltage voltage ofthe both the 12-pulse 12-pulse diode and rectifier and the the dodo diode rectifier. ideal no-load DC of both the diode rectifier and converter. U is the DC voltage target value when the reversible converter works in the inverting reversible converter. U LIM is the DC voltage target value when the reversible converter works in the LIM reversible converter. ULIM is the DC voltage target value when the reversible converter works in the state. A and B are the limit points. With the increase of increase the trainof traction power, the DC voltage inverting state. A and andpower B are are the the power limit points. With the the increase of the train train traction power, the inverting state. A B power limit points. With the traction power, the will drop fast. Above voltage output are completely thecompletely same, and the theysame, shareand the DC voltage voltage will dropthe fast. AboveUthe the voltage UBcurves B,, the the output output curves are are completely the same, and B , the DC will drop fast. Above voltage U curves traction load equally. However, because the value of R will change with the output current, it will they share the traction load equally. However, because the value of R dp will change with the output they share the traction load equally. However, becausedpthe value of Rdp will change with the output increase difficulty of software implementation. 5b shows a simplified which the current, the it will will increase the difficulty difficulty of software softwareFigure implementation. Figure 5b 5b scheme shows in simplified current, it increase the of implementation. Figure shows aa simplified value ofin Rdp is fixed. scheme in which the value value of of R Rdp dp is is fixed. fixed. scheme which the Udcdc U

U dc U dc

U LIM U LIM

U LIM U LIM

Udodo U

U do U do

IIdcdc

IIdcdc

(a) (a)

(b) (b)

Figure 5. 5. Proposed coordinated coordinated control strategy. strategy. (a) (a) Ideal Ideal scheme; scheme; (b) (b) Simplified Simplified scheme. scheme. Figure Figure 5. Proposed Proposed coordinated control control strategy. (a) Ideal scheme; (b) Simplified scheme.

4. Analysis of of the DC DC Short-Circuit Characteristics Characteristics of of the the Reversible Reversible Converter Converter 4. 4. Analysis Analysis of the the DC Short-Circuit Short-Circuit Characteristics of the Reversible Converter Figure 6 shows the the DC short-circuit short-circuit fault of of the reversible reversible converter. R R and LL are are the the equivalent equivalent Figure Figure 66 shows shows the DC DC short-circuit fault fault of the the reversible converter. converter. R and and L are the equivalent resistance and inductance of the DC line, and the value of R and L varies according to the position of resistance and inductance inductance of of the the DC DC line, resistance and line, and and the the value value of of R R and and L L varies varies according according to to the the position position of of the short-circuit point from the converter DC output. C is the DC capacitor of the converter. L r is the the point from from the the converter converter DC DC output. output. C C is is the the DC DC capacitor capacitor of of the the converter. converter. LLr is the the short-circuit short-circuit point r is the AC filter filter inductance. inductance. R Rcc is is the the equivalent equivalent series series resistance resistance of of the the DC DC capacitor capacitor C. C. It It can can be be seen seen that that it it AC AC filter inductance. Rc is the equivalent series resistance of the DC capacitor C. It can be seen that it is is a complex nonlinear circuit when a DC short-circuit fault occurs. The whole period can be divided is a complex nonlinear circuit when a DC short-circuit fault occurs. The whole period can be divided into the the following following three three stages stages [29]. [29]. into

is i1 ia ,b ,c

i2

iL

i3

iC

C

Appl. Sci. 2017, 7, 282

Lr

R

L

+

v_C

6 of 19

Rc

a complex nonlinear circuit when a DC short-circuit fault occurs. The whole period can be divided Appl. Sci. 2017, 7, 282 6 of 19 into the following three stages [29]. Figure 6. DC short-circuit fault of theisreversible iL converter.

i3 i2 4.1. Stage of Resistance-Inductance-Capacitor i(RLC) Second-Order Response 1 R iC

L

ia ,b ,c fault occurs instantly, the +capacitor C discharges quickly, and In this stage, the DC short-circuit vC C the DC voltage drops quickly from the initial value to zero. The_ DC fault current correspondingly rises rapidly. During this period, (R + Rc), L and C constitute an Resistance-Inductance-Capacitor Lr C and L have initial Rc values when the DC short-circuit fault (RLC) series second-order circuit. The occurs. Due to the presence of the AC inductor Lr, the three-phase AC current cannot be mutated, but flows through the six anti-parallel diodes, so there is an external excitation current to the second-order RLC circuit. As shown in Figure 7, thisofstage can be seen as a full response (including Figure DCshort-circuit short-circuit fault thereversible reversible converter. Figure 6.6.DC fault of the converter. zero-input response and zero-state response) process of the RLC series circuit. Zeroofinput response process: Regardless of the influence of the AC system (without considering 4.1. Stage Stage Resistance-Inductance-Capacitor (RLC) Second-Order Response 4.1. of Resistance-Inductance-Capacitor (RLC) Second-Order Response the is), it is the zero-input response process of the RLC second-order circuit. From Kirchhoff Voltage In(KVL): this stage, stage, the the DC DC short-circuit short-circuitfault faultoccurs occursinstantly, instantly, the the capacitor capacitor CC discharges dischargesquickly, quickly, and and In this Laws the DC voltage value to to zero. TheThe DCDC faultfault current correspondingly rises the voltage drops dropsquickly quicklyfrom fromthe theinitial initial value zero. current correspondingly 2 (6) vc LCand dvan c rapidly. During this period, (R + R(R ),+LdRand constitute Resistance-Inductance-Capacitor (RLC) cLC rises rapidly. During this period, c), C constitute an Resistance-Inductance-Capacitor + ( R + Rc ) C + vc = 0 2 dt dt values series series second-order circuit.circuit. The C The and C L and haveLinitial when. the DCthe short-circuit fault occurs. (RLC) second-order have values initial when DC short-circuit fault Due to the of the AC Lr , the three-phase AC current becannot mutated, but flows occurs. Duepresence to the presence of inductor the AC inductor Lr, the three-phase ACcannot current be mutated, 2 L / C , Equation (6) may have different According to the the six relationship (R +is an Rc)external through thethrough six anti-parallel diodes, soofthere currentexcitation to the second-order but flows anti-parallel diodes, soand there excitation is an external current to RLC the circuit. As shown in FigureAs 7, shown this stage can be 7, seen asstage a fullcan response (including zero-input response ( R + R ) < 2 L / C second-order RLC circuit. in Figure this be seen as a full response (including c solutions, considering R and Rc are small, this paper only analyses . and zero-state response) process of response) the RLC series circuit. zero-input response and zero-state process of the RLC series circuit. Zero input response process: Regardless of the influence of the AC system (without considering is process iof L the RLC second-order circuit. From Kirchhoff Voltage the is), it is the zero-input response Laws (KVL): L R iC 2 (6) d vc+ dv LC + R + Rc ) C c + vc = 0 C 2 vC ( dt _ dt . According to the relationshipRof (R + Rc) and 2 L / C , Equation (6) may have different c solutions, considering R and Rc are small, this paper only analyses

( R + Rc ) < 2 L / C .

Figure 7. Resistance-Inductance-Capacitor (RLC) second-order circuit response equivalent circuit. Figure 7. Resistance-Inductance-Capacitor i (RLC) second-order circuit response equivalent circuit.

is

L

Lof initial R t0, the Zero input response process: of influence the AC system (without Assume that the short circuit Regardless fault occurs atthe moment condition is vc(t0) = considering V0, iL(t0) = I0. iC the ), it is the zero-input response process of the RLC second-order circuit. From Kirchhoff Voltage Theiscircuit is in the oscillation discharge process and, from Equation (6), the zero-input response + Cand current v_C Laws (KVL): of the DC side voltage components after fault are as follows: d2 vc dvc LCV0 ω20 + (6) ( R + Rc )C I 0+ v−δc t = 0. (7) −δt vc = dt e sin ( ωt + β ) −dt e sin ( ωt ) Rcω √ ωCd According to the relationship of (R + Rc ) and 2 L/C, Equation (6) may have different solutions, √ considering R and Rc are small, this paper only analyses ( R + Rc ) < 2 L/C. (8) dv I ω occurs at moment V t , the initial condition is vc (t0 ) = V 0 , Assume that the short circuit fault iL1 = C c = − 0 0 e −δt sin(ωt − β) + 0 e −δ0t sin(ωt ) 7. Resistance-Inductance-Capacitor (RLC) second-order circuit response equivalent circuit. iL (t0 ) Figure = I0 . The circuit is in thedoscillation process from Equation (6), the zero-input ω discharge ωL and, t response components of the DC side voltage and current after fault are as follows: Assume that the short circuit fault occurs at moment t0, the initial condition is vc(t0) = V0, iL(t0) = I0. V0 ω0 −δt process and, from I0 Equation The circuit is in the oscillation (6), vc = discharge e sin(ωt + β) − e−δt sin(ωt (7) ) the zero-input response ω ωC components of the DC side voltage and current after fault ared as follows: vc =

iL1 = C

d vc dt

V0 ω0 −δt I e sin ( ωt + β ) − 0 e −δt sin ( ωt ) ω ωCd

(7)

I 0 ω0 −δt V e sin(ωt − β) + 0 e −δt sin(ωt ) ω ωL

(8)

=−

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7 of 19

dvc I ω V = − 0 0 e−δt sin(ωt − β) + 0 e−δt sin(ωt) iL1 = C dt ω ωL 2 p 1 1 R + R R + R ω c c ω2 = − , δ= , ω0 = δ2 + ω2 = √ , β = arctan . LC 2L 2L δ LC

(8)

From Equation (7), when C discharges to zero, that is, vc = 0, the instant time is t1 = t0 +

γ , γ = arctan[(V0 ω0 Cd sin β)/( I0 − V0 ω0 Cd cos β)]. ω

(9)

Substituting Equation (9) into Equation (8): iL1 (t1 ) = −

I0 ω0 −δ( γ ) V0 −δ( γ ) ω sin(γ − β) + ω sin γ. e e ω ωL

(10)

Generally, when a metal short-circuit fault occurs in the DC line, (( R + Rc )/2L)2 is far smaller than 1/LC, It can be assumed that ω = ω0 , that is, the zero-input response component of the DC inductor current can be reduced to ! r C −δt iL1 = e V0 sin ωt + I0 cos ωt . (11) L Making θ = arctan( VI00

q

L C );

then Equation (11) becomes

iL1 = e

Rc −( R+ 2L ) t

"r

# C 2 2 V0 + I0 sin(ωt + θ) . L

(12)

Zero state response process: As shown in Figure 7, assume that there is no initial value on C and L. The current supplied to the DC side by the AC system is regarded as a constant value is . At this moment, it can be considered that the RLC second-order circuit is in the process of zero-state response, and from Kirchhoff Current Laws (KCL), LC

d2 iL2 di + ( R + Rc )C L2 + iL2 = is . dt dt2

(13)

Equation (13) can be solved the same way as calculating the zero-state response. The zero-state response of the DC inductor current can be obtained as Is ( R+ Rc ) −δt sin(ωt) 2Lω q e Is ( R+ Rc ) C −δt sin(ωt) 2 Le

iL2 = Is − Is e−δt sin(ωt + β) −

= Is − Is e−δt sin(ωt + β) −

(14)

where the δ, ω, β is the same as in Equation (8). The DC inductor current in this stage is the sum of the zero-input response component and the zero-state response component from Equations (12) and (14), that is iL = iL1 + iL2 .

(15)

From Equations (12), (14) and (15), it can be seen that the DC inductor current iL is composed of two parts in this stage. Since the zero-input response component is of a large proportion, when the zero-input response component reaches its peak value, the total DC inductor current also reaches the peak value, and the following conclusions can be obtained: When a DC short-circuit fault occurs, the larger the initial DC voltage V 0 of C and the initial current I0 of L are, the higher the peak value of the DC inductor will be. When the initial values of DC voltage and DC inductor current are constant, the peak current of the DC inductor will increase with the increase of DC capacity C, but decrease with the

Appl. Appl. Sci. Sci. 2017, 7, 282

88 of of 19 19

capacitance value is, the more electric field energy can be released by the capacitor C after the decrease of DC line inductance L. This is because, under a certain DC the larger capacitance short-circuit fault, and the more electromagnetic energy stored involtage, the inductor, thethe larger the DC value is, the more electric field energy can be released by the capacitor C after the short-circuit fault, inductor current peak value can reach. The smaller the inductance value is, the greater the current and the more electromagnetic energy in thethe inductor, larger the DC current peak required to store the same energy, andstored the larger currentthe is, the greater theinductor total inductor current value can reach. The smaller the inductance value is, the greater the current required to store the same peak value will be. energy, and the larger the current is, the greater the total inductor current peak value will be. 4.2. Stage of Resistance-Inductance-Diode (RLD) First-Order Response 4.2. Stage of Resistance-Inductance-Diode (RLD) First-Order Response This stage begins with the capacitor DC voltage falling to zero, and continues until the capacitor This stage begins with the capacitor DC voltage falling to zero, and continues until the capacitor is recharged. When vc > 0, the capacitor discharges and the current flows through the line is recharged. When vc > 0, the capacitor discharges and the current flows through the line impedance. impedance. With the capacitor discharge, DC voltage vc will decrease. When vc drops to zero, that is With the capacitor discharge, DC voltage vc will decrease. When vc drops to zero, that is moment t1 , moment t1, the inductor begins to discharge, and the fault current will flow through the anti-parallel the inductor begins to discharge, and the fault current will flow through the anti-parallel diodes of the diodes of the three-phase bridge, assuming that the initial condition is iL(t1) = I0′. The equivalent three-phase bridge, assuming that the initial condition is iL (t1 ) = I0 0 . The equivalent circuit is shown in circuit is shown in Figure 8, the Ron is the total equivalent internal resistance of the six anti-parallel Figure 8, the Ron is the total equivalent internal resistance of the six anti-parallel diodes, and V f is the diodes, and Vf is the their total turn-on voltage drop. their total turn-on voltage drop. _ Vf

is

iL

R

iC

+

C

Ron

L

+

v_C = 0

Rc

Figure 8. Equivalent circuit of Resistance-Inductance-Diode (RLD) first-order circuit response. Figure 8. Equivalent circuit of Resistance-Inductance-Diode (RLD) first-order circuit response.

At this this moment, moment, the circuit becomes a first-order circuit circuit and and the the DC DC inductor inductor current current decreases decreases with with the the exponential exponential law law as as follows: follows: ) )t ′ (R+ Ronon )/L t ) I=00 eI−(( iL (ti)L (= 0e . . − (R+ R

)/ L

t

(16) (16)

This process processisisthethe most dangerous fordiodes, the diodes, a largedischarge inductor This most dangerous timetime for the becausebecause there is there a largeisinductor discharge currentflowing suddenly flowing through the period, diodes and in this period,can and diodesinstantly can be current suddenly through the diodes in this the diodes be the damaged damaged instantly because of the large initial current. Therefore, the diode over-current capability because of the large initial current. Therefore, the diode over-current capability should be considered should be considered in the devices. selection of switching devices. in the selection of switching The time timethat thatcapacitor capacitorCCdischarges discharges zero, value of short-circuit the short-circuit current attime, that The to to zero, andand thethe value of the current at that time, can be simplified as can be simplified as q √ γ C 0 (17) t1 = t0 + ω = t0 + LCarctan VIV (17) L

γ 0 t1 = t0 + q = t0 + LC arctan 0 ωC 2 2 −δt1 − δtI10

iL (t1 ) = iL1 (t1 ) + iL2 (t1 ) = e

where γ = arctan

q

L V0

+ I0 + Is − Is e

C L Is (R+Rc )√C −δt1 √ cos γ − e sin γ 2 L

I (R + Rc ) C − δt1 C 2 Rc 2 V0 − δt1 C cos γ − s e sin γ =iL2 (t√11) = ,eδ−δt1= RV+ 1 )+ 0 + I.0 + I s − I s e I0iL (t1 )L= iL1,(tω LC L 2L 2 L

(18)

(18)

From Equation (17), the time of capacitor C discharging to zero, that is, the maximum over-current Vis0 related Rc C, as well as V 0 and I0 . The relationships among C to the1 value ofR L+ and time of γthe diode, = arctan , δ= . where , ω = 2L and capacitance value are shown in Figure 9 when LC capacitance discharge value, I 0 Ltime, inductance V 0 = From 750 V, Equation and I0 = 1300 It can seen that theCcapacitor discharge timethat will is, increase with the (17),A.the timebeof capacitor discharging to zero, the maximum increase of the total inductance value and capacitance value. over-current time of the diode, is related to the value of L and C, as well as V0 and I0. The relationships among capacitance discharge time, inductance value, and capacitance value are shown in Figure 9 when V0 = 750 V, and I0 = 1300 A. It can be seen that the capacitor discharge time will increase with the increase of the total inductance value and capacitance value. It can be seen from Equation (18) that the inductor peak current is related to the value of L and C, as well as V0, I0 and Is at the fault instant. The relationships among the inductor peak current,

Appl. Sci. 2017, 7, 282

9 of 19

inductance value, and capacitance value are shown in Figure 10, when V0 = 750 V, I0 = 1300 A, and Is = 1800 A. It can be seen that the inductor peak current increases with the increase of the capacitance value, and decreases with the increase of the total inductance, which is in accordance Appl. Sci. 2017, 7, 282 9 of 19 with the analysis conclusions in Section 4.1. 9 of 19 Capacitor discharge time t1（ms） Capacitor discharge time t1（ms）

Appl. Sci. 2017, 7, 282

inductance value, and capacitance value are shown in Figure 10, when V0 = 750 V, I0 = 1300 A, and Is = 1800 A. It can be seen that the inductor peak current increases with the increase of the capacitance value, and decreases with the increase of the total inductance, which is in accordance with the analysis conclusions in Section 4.1.

Equivalent inductor L (mH)

DC Capacitor C (mF)

The relationships relationships among among capacitor capacitor discharge discharge time, time, the the equivalent equivalent inductance inductance L, L, and Figure 9. The capacitance C when V V00==750 750V, V,II00==1300 1300A. A.

It can be seen from Equation (18) that the inductor peak current is related to the value of L and C, as well as V 0 , I0 and Is at the fault instant. The relationships among the inductor peak current, inductorvalue L (mH) are shown in Figure 10, when V = 750 V, I = 1300 A, inductance value, and Equivalent capacitance DC Capacitor C (mF) 0 0 and Is = 1800 A. It can be seen that the inductor peak current increases with the increase of the Figure 9. The and relationships capacitor discharge time, the equivalent L, and with capacitance value, decreasesamong with the increase of the total inductance, whichinductance is in accordance capacitance C when V0in = 750 V, I0 =4.1. 1300 A. the analysis conclusions Section

Figure 10. The relationships among inductor peak current, the equivalent inductance L, and capacitance C when V0 = 750 V, I0 = 1300 A, and Is = 1800 A.

The discharge time of the DC capacitor and the inductor peak current are analyzed by the RLC second-order discharge circuit. It can be seen from Equation (16) that the peak current begins to discharge as the initial current in the RLD first-order response. The AC system provides current for the DC side through the diode in the actual first-order discharge process, so the actual DC inductor Figure 10. among inductor peak current, the equivalent inductance L, and current. capacitance current should be the sum of the inductor discharge and system supply As the Figure 10.The Therelationships relationships among inductor peakcurrent current, theAC equivalent inductance L, and C when Vcurrent V,decays IV0 0==1300 A,Ithe and Is = A, 1800 A. 0= capacitance C750 when 750 V, 0 = 1300 and Is = 1800increases, A. DC inductor and AC grid current the capacitor starts charging again and the DC voltage increases at the same time. The The discharge discharge time time of of the the DC DC capacitor capacitor and and the the inductor inductor peak peak current current are analyzed analyzed by the RLC second-order circuit. ItIt can 4.3. Stage of thedischarge Uncontrolled Rectification second-order discharge circuit. can be be seen seen from from Equation Equation (16) (16) that that the the peak peak current current begins begins to discharge as the initial current in the RLD first-order response. The AC system provides current discharge as the initial current The AC system provides current for for When a short-circuit fault occurs, the reversible converter cannot be immediately separated the the DC DC side side through through the diode diode in the actual first-order discharge discharge process, process, so so the actual DC inductor inductor from the AC and DC network by breakers. As a result, the anti-parallel diodes will form a current should should be be the thesum sumofofthe theinductor inductordischarge discharge current and system supply current. As the current and ACAC system supply current. As the DC three-phase bridge rectifier circuit, and the AC system will continue to deliver power to the DC side DC inductor current decays andAC thegrid ACcurrent grid current increases, the capacitor starts charging again inductor current decays and the increases, the capacitor starts charging again and the until the breakers open, as shown in Figure 11. and the DC voltage at the same time. DC voltage increasesincreases at the same time. Stage of of the the Uncontrolled Uncontrolled Rectification 4.3. Stage thethe reversible converter cannot be immediately separated from When aa short-circuit short-circuitfault faultoccurs, occurs, reversible converter cannot be immediately separated the AC and DC network by breakers. As a result, the anti-parallel diodes will form a three-phase from the AC and DC network by breakers. As a result, the anti-parallel diodes will form a bridge rectifier circuit, and circuit, the ACand system willsystem continue deliver to power to power the DCtoside the three-phase bridge rectifier the AC willtocontinue deliver the until DC side breakers open, as shown Figurein 11.Figure 11. until the breakers open, asinshown

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10 of 19

is i1

i2

iL

i3 C

Lr

R

iC

ia ,b ,c

L

+

v_C

Rc

Figure 11. 11. The The stage stage of of uncontrolled uncontrolled rectification. rectification. Figure

The system is equivalent to a three-phase uncontrolled rectifier working in the DC-side short-circuit state. state.The Theshort-circuit short-circuit current be calculated the transient current can can be calculated by theby transient analysisanalysis method. method. Assume Assume that the phasebefore current before short-circuit is as follows: that the phase current short-circuit occurs is occurs as follows:

(

)

i |0| = I m|0| sin ωt + α − ϕ|0| ia|0| a= Im|0| sin ωt + α − ϕ|0|

(19) (19)

I m|0| where thecurrent currentamplitude, amplitude,ω ω is the synchronous angular frequency, the phase α is where Im|0| isisthe is the synchronous angular frequency, α is the phase angle, ϕ angle, is the impedance angle. and ϕ|and is the impedance angle. |0| 0| Three-phase short-circuit current can be obtained: h i τ ia (=ωt I m+sin( I m 0(sin( ) −ImI msin sin( ωtϕ+)α+− ϕI) + sin αϕ− ϕ)0− α−− ϕ ϕ)) ee−−t /t/τ ia = Im sin α− α − ( α m|0| |0| h i ◦ − ϕ) + I sin(α − 120◦ − −t/τ ◦ − ϕ)− t /e ib = Im sin ( ωt + α − 120 sinα (α−− 120 ib = I m sin(ωt + α − 120 − ϕh) m 120 + |0|I m 0 sin(α − 120 −ϕϕ|00| ) −−IImmsin( − ϕ) e i τ ◦ − ϕ) e−t/τ ic = Im sin(ωt + α + 120◦ − ϕ) + Im|0| sin(α + 120◦ − ϕ|0| ) − Im sin(α + 120 −t / τ

(20) (20)

ic = I m sin(ωt + α + 120 − ϕ) + I m 0 sin(α + 120 − ϕ 0 ) − I m sin(α + 120 − ϕ) e where I is the short-circuit current period component amplitude, ϕ is the impedance angle of the where Imm is the short-circuit current period component amplitude, ϕ is the impedance angle of the circuit loop, Lr is the AC inductance, and τ is the time constant of circuit loop. circuit loop, Lr is the AC inductance, and τ is the time constant of circuit loop. √ 2V 2Vn (21) √n ϕ = arctan ω( Lr+ /R, (21) ϕ =[arctan ω ( LLr )+/R L ) ]/,Rτ , = τ =((LLrr + + LL) /)R , I m I≈m ≈ 3ωLr Rx + R3xω+ Lr the root-mean-square root-mean-square voltage voltage of of the the transformer transformer secondary. where Vnn isisthe Equation (21), when the AC inductance Lrr is constant, the short-circuit current period From Equation component amplitude amplitudeIImmisisrelated relatedtoto the line equivalent resistance R. The farther location of component the line equivalent resistance R. The farther the the location of the the short-circuit point is from reversible converter,the thesmaller smallerthe the short-circuit short-circuit current current period short-circuit point is from thethe reversible converter, willbe. be. component Imm will short-circuit current, whose value is greater than Three-phase uncontrolled uncontrolledrectifier rectifiertransfers transfersthethe short-circuit current, whose value is greater zero, zero, to thetoDC Thus,Thus, the short-circuit current on theonDC as follows: than theside. DC side. the short-circuit current theside DCisside is as follows:

i = i + i + i3 = ia ,( > 0) + ib,( > 0) + ic ,( > 0) . is = i1s + i12 +2i3 = ia,(>0) + ib,(>0) + ic,(>0) .

(22) (22)

5. Simulation Simulation and and Experiment Experiment 5. 5.1. DC Short-Circuit Fault Simulation 5.1. DC Short-Circuit Fault Simulation A simulation model was built according to the reversible converter circuit topology in Figure 6, A simulation model was built according to the reversible converter circuit topology in Figure 6, and the specific parameters are shown in Table 1. and the specific parameters are shown in Table 1.

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11 of 19 11 of 19

Table 1. Converter parameters for the simulation. AC: alternating current; DC: direct current. Table 1. Converter parameters for the simulation. AC: alternating current; DC: direct current.

Items Values AC input voltage Vn 450 V Items Values DC output voltage Vd 750 V AC input voltage V n 450 V AC side inductor Lr 300 μH DC output voltage V d 750 V DC Capacitance C 36 mF AC side inductor Lr 300 µH Capacitance parasitic DC Capacitance C resistance Rc 36 0.02 mF mΩ Capacitance Rc of diode Ron 0.02 0.3 mΩmΩ Equivalentparasitic internalresistance Resistance Equivalent internal Resistance of diode Ron 0.3 mΩ Rated power P 1 MW Rated power P 1 MW Switching frequency f s 2 kHz Switching frequency f 2 kHz s

Due to the fact that the location of the short-circuit point is unpredictable, the simulation study Due to out the fact that the location of the short-circuit is unpredictable, the(PSC) simulation is is carried in two situations, which are calledpoint proximal short-circuit and study remote carried out in two situations, which are called proximal short-circuit (PSC) and remote short-circuit short-circuit (RSC). Suppose the resistance of the overhead contact line is r1 = 0.028 Ω/km, the (RSC). Suppose the of the overhead contact is ris Ω/km, of 1 = inductance of that is resistance l1 = 2.6629 mH/km, the resistance of line the rail r2 0.028 = 0.023 Ω/km,the the inductance inductance of that is l = 2.6629 mH/km, the resistance of the rail is r = 0.023 Ω/km, the inductance of the rail is 2 the rail 1is l1 = 1.78 mH/km. Thus, the equivalent line resistance is r = r1 + r2 = 0.051 Ω/km, and the l1 = 1.78 mH/km. Thus, theisequivalent line resistance equivalent line inductance l = l1 + l1 = 4.4429 mH/km.is r = r1 + r2 = 0.051 Ω/km, and the equivalent line inductance is l = l + l = 4.4429 mH/km. 1 1 Related parameters and calculated results of PSC are shown in Table 2. Related parameters and calculated results of PSC are shown in Table 2. Table 2. Related parameters and calculated results for proximal short-circuit (PSC). Table 2. Related parameters and calculated results for proximal short-circuit (PSC).

Items Values Items distance x Values Short-circuit 10 m Line resistance R 0.51 Short-circuit distance x 10 m mΩ Line resistance R 0.51 mΩ mH Line inductance L 0.044 Line inductance L 0.044 mHms Theoretical discharge time of capacitance t1 1.9 Theoretical discharge time of capacitance t1 1.9 ms Theoretical peak current inductance iL(t1) 23,000 A Theoretical peak current inductance iL (t1 ) 23,000 A Theoretical periodcomponent component amplitude of short-circuit current Im 38863886 Theoretical period amplitude of short-circuit current Im A A Simulation results of the PSC are shown in Figures 12 and 13: Simulation results of the PSC are shown in Figures 12 and 13:

Inductor Current（kA）

22800A 20

Inductor Current Capacitor Voltage

600

15

400

10

200

5

0

0.5019 s 0

0.5

0.52

0.54 0.56 0.58 Simulation time（s）

Capacitor Voltage vc（V）

800

25

-200 0.6

Figure 12. 12. DC DC inductor inductor current current and and capacitor capacitor voltage voltage for for proximal proximal short-circuit Figure short-circuit (PSC). (PSC).

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Three Phase AC Current（kA）

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8

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ia

ib

ic

ThreeACPhase AC Current（kA） Three Phase Current（kA）

6

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4 2

0

-2 -4 -6

8 6 8 4 6 2 4 0 2 -2 0 -4 0.45 0.5 -2 -6 0.45 0.5 -4 -6

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3780A

ia

ib

ic

ia

ib

ic

12 of 19

3780A

3780A

0.55

0.6

0.55

0.65

0.7

0.75

0.6 0.65 time（s） 0.7 0.75 Simulation

0.8

0.85

0.8

0.85

0.75 for PSC. 0.8 Figure AC current for PSC. Figure13. 13.Three-phase Three-phase AC current for PSC. Simulation time（s）

0.85

Simulation time（s） 0.45Figure 0.5

0.6 0.65 0.7 13.0.55 Three-phase AC current

Related parameters and calculated results for RSC are shown in Table 3. Related parameters and calculated results for RSC RSC inin Table 3. 3. Figure 13. Three-phase AC are current for PSC. Related parameters and calculated results for areshown shown Table

Table 3. Related parameters and calculated results for RSC. Table 3. Related parameters and calculated results for RSC. Related parameters and calculated results for RSC are shown in Table 3. Table 3. Related parameters and calculated results for RSC. Items Items Table 3. Related parameters and calculated results for RSC. Short-circuit Short-circuitdistance distance xx Items Lineresistance resistance Line R Items Short-circuit distance Lineinductor inductor Short-circuit distance Line Lx x Line resistance R of capacitor t1 Theoretical discharge time Line resistance R Theoretical discharge time Line inductor L of capacitor t1 Theoretical peak current of inductance iL(t1) Line inductor L Theoretical current Theoretical peak discharge timeof of inductance capacitor t1 iL(t1) Theoretical period component amplitude ofcapacitor short-circuit current Im Theoretical discharge time of peak current of inductance iL (t1 ) t1 Theoretical Theoretical period component amplitude of short-circuit current Im Theoretical peak current of inductance i L (t 1 ) Theoretical period component amplitude of short-circuit current Im Simulation resultsperiod for RSC are shownamplitude in Figuresof14short-circuit and 15. Theoretical component current Im

Simulation results for RSC are shown in Figures 14 and 15. Simulation results for for RSC areare shown 15. Simulation results RSC shownininFigures Figures 14 14 and and 15.

Values Values 1.7 km 1.7 km Values 86.7 mΩmΩ 86.7 Values 1.7 km 7.553 mHmH 1.7 km 7.553 86.7 mΩ 14.6 ms 86.714.6 mΩms 7.553 mH 2572 mH A 7.553 14.62572 ms A 2550 A 14.6 ms 25722550 A A 2572A A 2550 2550 A

Figure 14. DC inductor current and capacitor voltage for RSC.

ThreeACPhase AC Current（kA） Three Phase Current（kA）

Three Phase AC Current（kA）

4 Figure 14. DC inductor voltage RSC. Figure 14.14. DCDC inductor current and capacitorvoltage Figure inductorcurrent currentand and capacitor capacitor forfor RSC. 3000A ia ib ic

4

24

2

02

0

ib ib

ic ic

3000A 3000A

-20

-2 -4

-2

-4

-4

ia ia

0.45

0.45

0.45

0.5

0.5

0.55 0.6 Simulation time（s） 0.55

0.6

Figure 15. Three-phase AC current for RSC. Simulation time（s） 0.5 0.55 0.6 Simulation time（s） Figure 15. Three-phase AC current for RSC.

Figure 15. Three-phase AC current for RSC. Figure 15. Three-phase AC current for RSC.

0.65

0.65

0.65

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The simulation results and the theoretical calculation results are compared in Table 4. Table 4. The comparison between the simulation results and theoretical calculation results. Short-Circuit Point

Parameters

Calculation Results

Simulation Results

PSC, x = 10 m

t1 iL (t1 ) Im

1.9 ms 23,000 A 3886 A

2 ms 22,800 A 3780 A

RSC, x = 1.7 km

t1 iL (t1 ) Im

14.6 ms 2572 A 2550 A

15.5 ms 2000 A 3000 A

From the comparisons in Table 4, some conclusions can be obtained. When the RSC occurs, the simulation results basically agree with the theoretical calculation results. However, when the RSC occurs, there are greater differences between simulation results and theoretical results. The reason is that the discharge current of the DC capacitor in RSC is smaller than the current that is provided by the AC system, so when the capacitor discharge is finished, the DC inductor current has not reached its peak value, but continues to increase slowly to the steady state short current. It is known from the above comparison that the peak short-circuit current in PSC is larger, and the rise is also faster, so it will do greater harm to semiconductor devices. In order to protect the reversible converter effectively, the PSC should be seriously considered. 5.2. Coordinated Control Experiments Two sets of high-power reversible converters have been successfully applied in the SHILIHE and XIDIAOYUTAI traction substations of Beijing Metro Line 10 (Beijing, China) as a demonstration example. The parameters of the reversible converters are shown in Table 5. Table 5. Parameters of the reversible converter. Parameter

Value

Parameter

Value

AC grid voltage DC rated voltage Converter capacity

10 kV 750 V 2 MW

Transformer capacity IGBT specification Switching frequency

1.6 MVA 2400 A/1700 V 2 kHz

The system main circuit of the traction substation is shown in Figure 16. Uac indicates the voltage of the AC 10 kV grid, iac1 indicates the AC current of the reversible converter, iac2 and iac3 indicate the AC current of each 12-pulse rectifier, respectively, idc1 indicates the DC current of the reversible converter, idc2 and idc3 indicate the DC current of each 12-pulse rectifier, respectively, and Udc indicates the DC bus voltage. The reversible converter has the same DC output characteristic as shown in Figure 5b. Figure 17 shows the voltage and current waveforms on the 10 kV side when the reversible converter and two 12-pulse rectifiers are working in parallel. As shown in Figure 17a, when the train is speeding up, the reversible converter and two 12-pulse rectifiers all work in the rectifying state: they share the train power. As shown in Figure 17b, when the train is slowing down, the reversible converter works in the inverting state, and the two 12-pulse rectifiers are blocked because of the high DC bus voltage. It is obvious that the AC current waveforms of the reversible converter are higher qualified than that of the 12-pulse rectifiers.

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Figure 16.16. Main circuit of the traction substation in Beijing Metro Line 10.10. Figure Main circuit of the traction substation in Beijing Metro Line Figure 16. Main circuit of the traction substation in Beijing Metro Line 10.

(a) (a) Figure 17. Cont.

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(b) Figure 17. AC current waveforms when the reversible converter and two 12-pulse rectifiers work in parallel. (a) train in traction state; (b) train in regenerative braking state.

Figure 18 shows the DC voltage and current waveform when the reversible converter and a 12-pulse rectifier are working in parallel. The meanings of the symbols are shown in Figure 16. As seen in Figure 18, along with the increase of traction power of the train, the DC bus voltage (b) reversible converter and 12-pulse rectifier unit decreases significantly, but the DC current of the increase immediately. However, it when is obvious that the current increase rate of thework reversible Figure Figure 17. 17. AC AC current current waveforms waveforms when the the reversible reversible converter converter and and two two 12-pulse 12-pulse rectifiers rectifiers work in in converter is greater than that of the 12-pulse rectifier unit. The trends basically agree with the DC parallel. state. parallel. (a) (a) train train in in traction traction state; state; (b) (b) train train in in regenerative regenerative braking braking state. output curves shown in Figure 5b. When the train work state is changed to braking from traction, the current 12-pulse rectifiers to zero,waveform but the DC current the reversible converter Figureof18both shows the DC voltagereduce and current when the of reversible converter and a Figure 18 shows to the DC voltage and current waveform when braking the reversible converter and a changes from positive negative, which means that the regenerative energy of the train 12-pulse rectifier are working in parallel. The meanings of the symbols are shown in Figure 16. has 12-pulse rectifier are working in parallel. The meanings of the symbols are shown in Figure 16. been fed backin to Figure the AC18, 10 along kV grid through the reversible converter. As seen with the increase of traction power of the train, the DC bus voltage decreases significantly, but the DC current of the reversible converter and 12-pulse rectifier unit increase immediately. However, it is obvious that the current increase rate of the reversible converter is greater than that of the 12-pulse rectifier unit. The trends basically agree with the DC output curves shown in Figure 5b. When the train work state is changed to braking from traction, the current of both 12-pulse rectifiers reduce to zero, but the DC current of the reversible converter changes from positive to negative, which means that the regenerative braking energy of the train has been fed back to the AC 10 kV grid through the reversible converter.

Figure Figure18. 18.DC DCvoltage voltageand andcurrent currentwaveforms waveformswhen whenthe thereversible reversibleconverter converterand andaa12-pulse 12-pulserectifier rectifier work in parallel. work in parallel.

As seen in Figure 18, along with the increase of traction power of the train, the DC bus voltage decreases significantly, but the DC current of the reversible converter and 12-pulse rectifier unit increase immediately. However, it is obvious that the current increase rate of the reversible converter is greater than that of the 12-pulse rectifier unit. The trends basically agree with the DC output curves Figure 18. DC voltage and current waveforms when the reversible converter and a 12-pulse rectifier work in parallel.

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shown in Figure 5b. When the train work state is changed to braking from traction, the current of both 16 of 19 12-pulse rectifiers reduce to zero, but the DC current of the reversible converter changes from positive to negative, which means that the regenerative braking energy of the train has been fed back to the AC Traction and braking energy statistics of one substation are shown in Table 6. A and B denote 10 kV grid through the reversible converter. traction energy provided by each 12-pulse rectifier, respectively, C denotes the traction energy provided Traction and braking energy statistics of one substation are shown in Table 6. A and B denote by the reversible converter, and D denotes the braking energy recovered by the reversible converter. traction energy provided by each 12-pulse rectifier, respectively, C denotes the traction energy provided by the reversible converter, and D denotes the regenerated braking energy recovered Table 6. Used and energy (MWh). by the reversible converter. Appl. Sci. 2017, 7, x

Reversible Converter 12-Pulse 12-Pulse Total Traction Percentage of Table 6. Used and regenerated energy (MWh). Rectifier 1/A Rectifier 2/B Rectifying/C Inverting/D Energy/A + B + C C/(A + B + C) 2016/6/18 4.43212-Pulse 4.482 4.699 0.974 34.52%of Total 13.613 Traction Reversible Converter Percentage 12-Pulse Date Energy/A +B+C C/(A + B + C) Rectifier 1/A 3.928 Rectifier 2/B Rectifying/C 2016/6/19 3.880 4.474 0.769 12.282 36.43% Inverting/D 2016/6/20 5.186 5.227 5.528 1.498 15.941 34.68% 2016/6/18 4.432 4.482 4.699 0.974 13.613 34.52% 2016/6/21 4.505 5.456 1.593 14.424 37.83% 2016/6/19 4.463 3.880 3.928 4.474 0.769 12.282 36.43% 2016/6/20 4.345 5.186 5.227 5.528 1.498 15.941 34.68% 2016/6/22 4.377 5.183 2.154 13.905 37.27% 2016/6/21 4.476 4.463 4.505 5.456 1.593 14.424 37.83% 2016/6/23 4.514 5.196 2.334 14.186 36.63% 2016/6/22 4.345 4.377 5.183 2.154 13.905 37.27% 2016/6/24 4.874 4.916 5.198 1.942 14.988 34.68% 2016/6/23 4.476 4.514 5.196 2.334 14.186 36.63% 2016/6/25 3.560 4.282 0.656 11.363 37.68% 2016/6/24 3.521 4.874 4.916 5.198 1.942 14.988 34.68% 2016/6/26 3.548 4.623 0.511 11.679 39.58% 2016/6/25 3.508 3.521 3.560 4.282 0.656 11.363 37.68% 2016/6/26 4.390 3.508 3.548 4.623 0.511 11.679 39.58% 2016/6/27 4.393 5.327 1.389 14.110 37.75% 2016/6/27 4.642 4.390 4.393 5.327 1.389 14.110 37.75% 2016/6/28 4.654 5.336 2.335 14.632 36.47% 2016/6/28 4.642 4.654 5.336 2.335 14.632 36.47% 2016/6/29 4.717 4.752 5.290 2.519 14.759 35.84% 2016/6/29 4.717 4.752 5.290 2.519 14.759 35.84% 2016/6/30 4.453 5.041 1.370 13.912 36.23% 2016/6/30 4.418 4.418 4.453 5.041 1.370 13.912 36.23% 57.309 65.633 20.044 179.794 36.50% sum sum 56.852 56.852 57.309 65.633 20.044 179.794 36.50% Date

Figure Figure 19 19 shows shows the the histogram histogram of of the the traction traction energy energy provided provided by by the the reversible reversible converter converter and and each 12-pulse rectifier for clarity. The reversible converter has provided nearly one third of the total each 12-pulse rectifier for clarity. The reversible converter has provided nearly one third of the total traction 36.5%). However, traction power power (the (the average average proportion proportion is is 36.5%). However, it it will will be be found found that that the the traction traction power power provided by the reversible converter is slightly more than that of each 12-pulse rectifier. The reason provided by the reversible converter is slightly more than that of each 12-pulse rectifier. The reason for for is that output curve of the reversible converter is above of the 12-pulse rectifier in thisthis is that thethe DCDC output curve of the reversible converter is above thatthat of the 12-pulse rectifier in the the light load region, as shown in Figure 5b. light load region, as shown in Figure 5b.

Traction energy /MWh

6.000

Traction energy distribution (MWh)

12-pulse rectifier1 12-pulse rectifier2 reversible converter

5.000 4.000 3.000 2.000 1.000 0.000

date Figure19. 19.Traction Tractionenergy energy distribution distribution between between the the reversible reversible converter converter and and each each 12-pulse 12-pulse rectifier. rectifier. Figure

In order to evaluate the energy-saving effect of the reversible converter used in the traction power supply system more clearly and intuitively, the average energy-saving percentage over a long period is defined as η: n

E

(23)

inv _ i

η=

i =1 n

Erec _ i i =1

× 100%

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In order to evaluate the energy-saving effect of the reversible converter used in the traction power supply system more clearly and intuitively, the average energy-saving percentage over a long period is defined as η: n

∑ Einv_i

η=

i =1 n

× 100%

(23)

∑ Erec_i

i =1

where Einv_i denotes the inverting energy of the reversible converter every day, and Einv_i denotes the total rectifying energy of the reversible converter and 12-pulse rectifiers every day. In Table 6, regenerated energy and total traction energy change every day, but according to Equation (23), the average energy-saving percentage η (from 18 June 2016 to 30 June 2016) is calculated as high as 11.15%. 6. Conclusions In this paper, the high-power reversible converter, which can not only be used to recuperate regenerative braking energy, but also provide traction energy, is applied to construct a new hybrid traction power supply system for urban rail transit. A droop control method based on load current feed-forward is proposed to realize the load distribution between the reversible converter and the existing 12-pulse diode rectifiers. It is successfully verified by the field test carried out on Beijing Metro Line 10. The DC short-circuit characteristics of the reversible converter is studied, and then the relationship between the peak fault current and the circuit parameters are derived. Theoretical calculation and computer simulation shows that the peak fault current in RSC is larger, and the time to reach the peak value is also shorter. Thus, it will do greater harm to the reversible converter. The field test data indicates that the average value of energy-savings percentage for one substation is as high as 11.15%. Acknowledgments: This research was supported by National Key Research and Development Program 2016YFB1200504. Author Contributions: Gang Zhang proposed the hybrid traction power supply system and the droop control method based on load current feed-forward. Jianglin Qian studied the DC short-circuit characteristics and carried out the simulation. Xinyu Zhang analyzed the field test data. Conflicts of Interest: The authors declare no conflict of interest.

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