Ejercicios de DigSilent Series and shunt compensation in Power lines and cables.pdf

7/27/2019 Ejercicios de DigSilent Series and shunt compensation in Power lines and cables.pdf

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-We te(hnis(he universiteit eindhovenCapaciteitsgroep Elektrische EnergietechniekElectrical Power Systenls

Series and shunt compensation inPower lines and cables

door: H. el FarissiEPS.07S368

De faculteit Elektrotechniek van deTechnische Universiteit Eindhovenaanvaardt geen verantwoordelijkheidvoor de inhoud van stage- enafttudeerverslagen

Stagewerk verricht a.1.v.:Dr. ir. J.M.A. Myrzikor december 2006

/ faculteit elektrotechniek


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Contents

1 Introduction1.1 Academic position of the traineeship1.2 The project1.3 Con ten t s . . . . .

2 Problem definition2.1 Introduction .2.2 Voltage drop and phase shift ing .2.3 Simulations . . . . . . . . . . . .

2.3.1 Load demand .2.3.2 Reactive power transport2.3.3 Active power transport

3 STATCOM3.1 Theory.3.2 Control model.3.3 Resu l t s . . . . .

3.3.1 Reactive power transport3.3.2 Active power transport .3.4 STATCOM placed at the end terminal (33kV)

3.4.1 Reactive power transport3.4.2 Active power transport .

4 SSSC4.1 Theory .4.2 Control model. . . . . . . . . . . .4.3 Three phase to dq-transformation .4.4 Results .

4.4.1 Reactive power transport .5 Conclusions and recommendationsA Custom con trol mode lB Voltage source conver terC Load event

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List of Figures2.1 Line model of a simple power system network2.2 Single line diagram, in DIgSILENT . . . . . .2.3 Load sta tic or dynamic. . . . . . . . . . . . .2.4 Problem definition, reactive power load event2.5 Problem definition, vol tage at the terminals2.6 Problem definition, active power flow . . . .2.7 Problem definition, active power load event2.8 Problem definition, reactive power flow . . .2.9 Problem definition, voltage at the terminals3.1 Injection model of the STATCOM3.2 Shunt control frame .3.3 Shunt controller. . . . . . . . . . .3.4 STATCOM layout in DIgSILENT .3.5 STATCOM, Reactive power load event.3.6 STATCOM, active power flow .3.7 STATCOM, voltage at the terminals . .3.8 STATCOM, current and voltage injection3.9 STATCOM, phase between current and voltage3.10 STATCOM, active power load event3.11 STATCOM, reactive power flow .3.12 STATCOM, voltage at the terminals .3.13 Injection model of the STATCOM at terminalEND3.14 STATCOM at terminalEND, reactive power load event.3.15 STATCOM at terminalEND, active power flow . . . .3.16 STATCOM at terminalEND, voltage at the terminals3.17 STATCOM at terminalEND, active power load event.3.18 STATCOM at terminalEND, reactive power flow . ..3.19 STATCOM at terminalEND, voltage at the terminals4.1 Power transfer with and without voltage injection.4.2 SSSC line diagram in DIgSILENT4.3 Series control frame ..4.4 Series current controller4.5 PQ controller . . . . . .4.6 dq-transformation . ..4.7 SSSC, reac tive power load event4.8 SSSC, active power flow . . . . .4.9 SSSC, voltage at the terminals4.10 SSSC, current and voltage injection.A.l common model .

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A.2 common model example .B.l voltage source converter in DlgSILENTB.2 common model .C.l dropdown menu for the load.C.2 Load type menu .

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Chapter 1Introduction1.1 Academic position of the traineeshipThis report is the final report for an internal traineeship at the Eindhoven Universityof Technology (TU/e). This traineeship was carried ou t in the EVT-EPS section ofthe faculty of Electrical Engineering.

The goal of an internal traineeship is that a student learns to carry ou t a relativelysmall research project in a short time. An internal traineeship is carried out in theMaster phase of the students training and is the preparation for the master thesiswhich is the last test for a student. The time the traineeship is supposed to take is240 hours, which corresponds to 8 ECTS (European Credit Transfer System) points.

1.2 The projectAt present, many transmission facilities confront one or more l imit ing network parameters such as, the line reactance, the phase, and the voltage amplitude. In acpower systems, with no electrical storage, the electrical generation and load must bein balance all times.

Pgeneration = Plood + flossesThe replacement of passive and/or electromechanical devices (for example, me

chanically switched capacitors) by active devices allows electrical power transmissioncircuits to operate very close to their parametric l imits. To achieve this capability,FACTS controllers are capable of controlling up to three system parameters; voltagemagnitude, phase angle and transmission impedance.

The main advan tages of FACTS technology over other solutions (for example,electro-mechanical devices and network reinforcement) are [1]

1. Improved steady-state system performance (for example, greater flexibility overthe control of power flows)

2. Improvements in system transient and/or dynamic stability3. FACTS controllers require minimal maintenanceIn this report only two kind of FACTS are discussed:

STATCOM , static compensatorSSSC , static synchronous series compensator5


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These devices have been developed already an d used in some networks. Ou r goalis thus no t t o r ep ea t existing technology, but to implement these devices in th e powersimulation p rog ramme D IgSILE N T an d learn about this. Th e analysis a re done forboth line an d cable networks. for all simulation th e impact of th e FACTS on b ot h t hecable an d line networks are compared.

1.3 ContentsT hi s r ep or t is organised as follows. In chapter 2 th e Power flow in a line is described.This is done to illustrate th e problem of th e voltage drop which exist in th e powersystems. Th e parameters of a line can be controlled such that the power handling ofth e line could be increased. In Chapter 3 th e emphasis is put on th e reactive powercompensation with th e STATCOM. This is done for a simple line diagram to illustrateth e effect clearly.

In th e n ex t c ha pt er 4, th e SSSC is descr ibed an d implemented. This device iscapable of controlling b ot h t he actieve an d reactive power. However du e to lack o f t im ean d th e complexity of th e controller the SSSC is only used as a reactive compensator .Finally some conclusions an d recommendations for future work will be given

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Chapter 2Problem definition2.1 IntroductionThe power flow through an alternating current transmission line is a function of theline impedance, the magnitudes of the sending-end and receiving end voltages, andthe phase angle between these voltages. Essentially, the power is dependent on thevoltage across th e line impedance. In case of a transmission line the impedance can berepresented by an reactance, s ince X > R.

Va L (S

Gridx

Load

Figure 2.1: Line model of a simple power system network

If the phase of the sending end terminal is zero, the active and reactive power flowfor the power transmission circuit as illustrated in Fig 2.1 is

VaVb .Pab = Xsme and VaVb ( )Qab = X l - cose (2.1)

with Va the sending voltage and Vb is the receiving end voltage.When th e reactive power is transported over mainly reactive elements of the power sys-tem, th e reactive power losses may be considerable and these add to the load demand.furthermore this reduces the active power delivery

2.2 Voltage drop and phase shiftingThere is a strong relationship between voltage and reactive power flow. The voltagesin a distribution system and to th e consumers must be maintained within a certainband around the rated equipment vol tage, ideal ly from no load to full load, and under

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varying loading conditions. Sudden load impacts (starting of a large motor),result inshort-time or prolonged vol tage dips. High voltages may occur under light runningload or on sudden load throwing and are of equal considerations, though low voltagesoccur more frequently.A certain balance between the reactive power consuming and generating appara

tuses is required. This must consider losses which may be a considerable percentage ofthe reactive load demand. When the reactive power is transported over mainly reactiveelements of the power system, the reactive power losses may be considerable and theseadd to the load demand (Example 12.5). This reduces th e active power delivery

The European standard NEN-EN 50160 gives the main characteristics of the voltageat the customer's supply terminals in public low voltage and medium voltage electricitydistribution systems under normal operating conditions. The standard gives the limitsor values within which any customer can expect the voltage characteristics to remain.The NEN-EN 50160 (Voltage characteristics of electricity supplied by the public distribution system) gives requirements for low voltage and medium voltage distributionsystems regarding notably:Magnitude of the supply voltageSupply voltage variationsMagnitude of rapid voltageSupply voltage dipsShort interruptions of the supply voltageLong interruptions of the supply voltageAccording to the ranges specified in this standard the voltage range is +6% to

-10% of the nominal voltageIn order to keep th e voltage within the specified ranges it is necessary to use reactive

support. The STATCOM (shunt compensation) and SSSC (series compensation) willbe discussed in the following chapters.

The voltage drop over a certain transmission can be then calculated. Suppose thatthe phases of the sending end is equal to the receiving end voltage.

LB= L6

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Then:

~ Pa+jQa =x=al* (2.2)I (Pa - jQa) (2.3)V*-=---a

~ x=a- jX[ (2.4)

~ V _ "X (Pa - j Qa) with x=a = ~ =VaLO =? (2.5)-=---a J V*-=---a

~ V _ XQa _ j XPa (2.6)a Va VaVb Va - h,V - jbV (2.7)

(2.8)according to equation 2.8 it can see that the d rop of voltage magnitude is caused

mainly from reactive power t ransport . And that the phase shift is caused from activepower transport.

In transmission lines the voltage drop is mainly caused by th e reactance in the lines.For a transmission line (llOkV), X ~ lOR [4]. But in cables the impedance R > X anR must be taken into account . I f R is considered, then it can see that the voltage dropresults from both reactive and active transfer. Equations 2.4 and 2.8 then become

x=a- (R+jX)[

The equations for the voltage drop when R is taken into account wil then be:

(2.9)

2.3 Simulations

and 5V = XPa - RQaVa (2.10)

The effect of a temporary voltage drop resulting from a reactive load demand is ana-lyzed by means of simulations in this section. Fig 2.2 shows the single line diagram inDIgSILENT which is used for the simulation and Table 2.1 shows the correspondingsystem parameters.

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GridTerminalBRON

lineUp

LineDown

TerminalLOAD

TransformerTerminalEND

Load

Figure 2.2: Single line diagram, in DIgSILENT

Table 2.1: System parameters of the circuitTerminalBRON / TerminalLOAD 110 kVTerminalEND 33 kVLength Line/Cable 200 kmR - X (line) 0.02 - 0.2 n/kmR - X (cable) 0.2 - 0.05 n/kmTransformer 50 MVA, 110/33 kV

2.3.1 Load demandThe analysis were done according to RMS/EMT simulations. In the RMS/EMT simulation environment, the network model and associated devices are not represented asconstant steady state impedances, but by the differential equations that govern theirbehavior For RMS/EMT simulations a three-phase load can be modeled as a percentage of sta tic and dynamic load. The static portion is modeled as constant impedancewhereas the dynamic load can be modeled as either a l inear load or a non-linear load.Two-phase, single-phase- and DC loads are generally modeled as constant impedance.In our case we want the load to change in step and within each step th e load is aconstant active or reactive power node. In appendix C it is explained how this couldbe done.

z% %

Dynamicload

Figure 2.3: Load static or dynamic

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2.3.2 Reactive power transportThe load demand is set to change the reactive and active power demand in steps intime. The result is that the voltages a t both terminals change due to changing reactiveand active power. The simulation is done for both lines and cables in order to comparethe two mediums. Furthermore the s imulat ion are also subdivided in a reactive loaddemand and an active load demand. The first s imulat ions are done for a react ive loaddemand and are depicted in Figures 2.4 to 2.6.

The initial demand of th e load is 1 MVar. At the time t = 4 sec it changes to40 MVar. Then at time t = 8 sec the reactive power demand reduces to 20 MVar.Finally th e reactive power demand is set to its nominal reactive power demand of 1MVar at time t = 12 sec. From th e first figure it can be seen that the reactive powerdemand is delivered by the grid. The power grid delivers the reactive power that isdemanded by th e load plus the reactive power for the lines/cables and transformator.From equations 2.1 and 2.10 it is clear that to realize this reactive power support anvoltage drop will be the result at the teminalLOAD and terminalEND respectively (seeFig 2.2).

~ . O O -----------------r-----------------r----------------,-----------------,[MVa 1-'Grid reactive power : :

40.00 1 - --- -- - -- - - - - - - - - f - - : - - - ~ --------- -- ------------------1 Load reactive power 1 1

3 ~ 0 0 ,----------------- ---------------- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~1 1 11 1 I20.00 - - - -- - - -- - - -- - - -- - - - -- - - -- - - - -- - ----------------11 l 1 110.00 1 -------- ---- -- - - - - - -------------__ : ---- -- --- ------------i1 I 1 11 1 10.00 - - ~ ~ ....... ~ ~ .......+- - - ---- -- - - - - - - - - -1-- - - - - - - - - - - - - - - - - j1 1 1 11 1 1 1

16.00s]2.00.00.00_ 1 0 . o o u - - - - ~ - - - - - - - ' - - - - ~ - - - - - - ' - - - - - ~ - - - - - ' - - - - - - - - - - - - l .0.00(a) Lines

50.00 1 - - - - - - - -- - -- - ---- ,- - -- - --- -- --- -- - -'"Grilri'eacnve-pOwer - - - --;- -- - - - - - - - - -- ----11 ~ , - - - - _ . - 1 11 1 137.50 -- ---- - ---------- --Loadreactlvepemer--- ----------------;-----------------i1 1 1

25.00 - - -- -- - - - - - - - - - - - - --- - - - - - - - - - - - - - - - - - - -- - --- ---- - -1--- -------- -- ----1 11 1 1

12.50 :-------------- - -- --------- - -------:---------- -- -- - -- - -- -- -- - -- - - - - - - - j1 1 11 1 1

0.00 1 - - - - ----- ---- - ---:---------------- - -- J1 1 11 1 1

16.00512.00.00.00-12.50 LJ...- ---l ~ _'_ ~ .l.._ . . . . . . . . . ___L

0.00(b) Cables

Figure 2.4: Problem definition, reactive power load event

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1.02 1-- - - - - - - -- - - - - - - - , - - ------ - - - -Llne-voftige-TerminliiBRON -- -- -1- - - - - - - - - - - - - - - - - ,[p.u.] I0 . 9 9 ~ - - - jo 96 -- - - ------------- Une.voJtage-TerminaILOA - J. I II I0.93 ~ - - - - - - - - - - - - - - - - - -- -- ----- -- ----- --- - - - - - - - - - - - - - ~ - - - --- - - - - - - - - - - - j

I I I0.90 -t----------------- ---- -- -- -------- - - - - - - - - - - - - - - - - ~ - - - - --- - - - - - - - - - - jo ~ l - - - - - - - - - - - - - - - - - ~ ~ e y ~ ~ ~ g ! ! ! ~ ~ ~ ~ ~ ~ ~ J J. I I I I II I I I I

18.00s]2.00.00.000 . 8 4 U - - - - - ~ - - - - - ' - - - - - ~ - - - - - ' - - - - - ~ - - - - . l - - - - - ~ - - - - ..0.00(a) Lines

1.02 - ----- ---- - - - - - - - ~ - - - -- - -- - -- - - - - - - , - - ----- - - -- - - - -- - ~ - - - -- - - -- - - -- - - - -I I I I: : Line v o l t ~ g e TerminalBRON :1.00 - -

16.00s]2.00.00.00

0.94

0.98 - - - - - - - - - - - - - - - - - t - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - i : i ~ n : e ~ v : O : l t : a ~ g ~ e ~ f ~ e : r ~ m : r ~ n : a ~ ~ = O : A : : D ~ - - - - - - - - - - - - - - - - - :I I----------------,-----------------,I II II I----------------,-----------------1ine voltage TerminalEND : :

0.92 L ~ ...t:::===::::===:::L ~ .l_ ~ __.:.0.00

0.96

(b) Cables

Figure 2.5: Problem definition, voltage at the terminals

0.30[MW]0.20

0.10

0.00

- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~ - - - ~ - - - - - - - - - ~ - - - ,IIII I_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ JI I I: ~ ~ ~ : :

I I~ - - - - - - - - - - - - - - - - - ----------------- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~I I II II Load active power I I II I I, . . . . . - - - - - - - - - - I - - . _ - - - - - ~ - - - - - - - - - - - - - I - - - - - - - - - - - - o jI II II II I

- o . 1 % . 0 0 u . . . . - - - - " ' - - - - - - - 4 - . o o . . l . - - - - - ~ - - - - 8 - . 0 . l . 0 - - - - ~ - - - - 1 2 - . 0 . l . 0 - - - - [ - S ] ~ - - - - 1 6 - . 0 . l . 0

(a) Lines4.00

[MW]3.00

- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ,III

Grid activepower2.00

1.00

-10%.00 4.00 8.00 12.00 [s] 18.00(b) Cables

Figure 2.6: Problem definition, active power flow

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From the first figure (Fig 2.4) the reactive power support from the grid is shown.This is done for both a line network (F ig 2.4(a)) and a cable network (2.4(b)) . Fromthe figures it can be seen that in case of cable network the reactive support is less thanfor a line network. This is due to the lower reactance for cables than lines. In Figure2.5 the impact of the reactive load demand on th e voltage at the three terminals isshown. For the line network the voltage drop at the terminaEND is more than 10%of i ts nominal voltage. This is more than the tolerable voltage drop specified in theANSI C84.1. In case of a cable network the voltage maintains in the specified voltagerange at all terminals.

The active power which is delivered by the power grid differs also for a cable andline network respectively. For a cable network the active power support from the powergrid is almost 10 times then for a line network. This is because the cable impedanceis a factor 10 larger than the line impedance. This is also a cause for the voltage dropin the cable network.

The same analysis are also done for an active power load demand. This is done inthe next section.2.3.3 Act ive power t ransportNow the initial demand of the load is 1 MW. At the time t = 4 sec it changes to 40MW. Then at time t = 8 sec the active power demand reduces to 20 MW. Finallythe active power demand is set to its nominal active power demand of 1 MW at timet = 12 sec.

1112.50 T - - - - - - - - - -- - - - - - -1

[MWJ : : : : :5 ~ 0 0 ----------------- ---------------- - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - T1 1 11 Grid actille power 1 1

, ----------------- ---------------- ----------------t-----------------t1 Loadact ive power , 1 11 1 1

2 ~ 0 0 ----------------- ---------------- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - T11 1_________________ ~ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ J1 11 1

0.00 ~ .....................!!"!!!"'!. . . . . . . . . . . . . . . . . . . ."""""'-- ---- - -- - - - - - - - - - -------- - - - - - - --"'". . . . . . . . . . . . . . !!"!!!"'!!!"!!!"'! ........ - -1 11 1

-12.58.00 4.00 8.00 12.00 [s ] 18.00(a) Lines

62.50 ----- - - ------ - - ~ - r - - - - - - - - - - - - - - - - -1----- - - - - - - - - -,- -- - - - - - - - - -- - - - -,[MWJ : : : Grid activepowe r : :50.00 , ----------------- - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - i

1 1 137.50 : -------- - -- ------ __ ! : ~ ~ " _ C ! i " " ~ ~ _ _ _ _ - - - - - - - - - - - - - - - - - - ' - - - - - - - - - - - - - - - - - j1 11 1 125.00 ------------- ---- - - ------------- -------- ---------11 1 11 1 112.50 ------- - - - - - - -- - - - - - - -- - - - - - - - - - --1-- - - - - - - - - - - - - - -- - - - - - - - - - - - - - - --- ,1 1 11 1 10.00 ------- -- - ---- --1-----------------1 1 11 1 1

16.00s]2.00.00.00-12.50 LJ...- ~ ___I. ~ ..l..._ ~ ___I. ~ ..J.

0.00(b) Cables

Figure 2.7: Problem definition, active power load event

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I0.00 h - - - - - - - - - - - + - . ~ - _ : _ - - - ~ - _ _ i _Load r eactive p ower II

1250 - - - - - - - - - - - - - - - - - r - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~[MVar) : I :0.00 tr----------------- ----------------- ----------------,-----------------lI I I

~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~ ~ ~ r ~ ~ ~ ~ ~ W ! ~ J-----------------J.50 I I II I II I I5.00 - - - -- - - -- - - -- - - -- - - - -- - - -- - -- - - -- - ----------------,-----------------lI II I~ ~ ----------------- ----------------- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - iI :-1II-2'5%.00 4.00 8.00 12.00 (5) 16.00

(a) Lines

0.00 !"- o!-_

Grid reactive power

8.00[MVar)

6.00

4.00

2.00

- - - - ~ ~ - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ,III I________________ JI II II II I----------------,-----------------lI II II I- - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - 1 I

I I II I- - - - - ~ - - - - - - - - - I - - - - - - - - - - - - iLoad reactive power I II I-20ld.00 4.00 8.00 12.00 [sj 16.00

(b) Cables

Figure 2.8: Problem definition, reactive power flow

Line voltage TerminalEND

1.01

1.000.99

0.98

0.97

- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ,III1 7 - - - - - - - - - J : . : : ; : : . . ; ; ; . : : . . ; ; ; . : : . . ; ; ; . : : . . ; ; ; . : : . . ; ; ; . : : . . ; ; ; . : : . . ; ; ; . ~ - - - - - - - - _ t - - - - - - - - --iLine voltage TerminalBRON I I I________________ .. ~ ~ ~ ~ ~ ; ; ; ; ; ; ~ J

III I__________________________________________________ JI II I_________________ _ ~ ~ ~ ~ ~ ~ ~ I ~ ~ ~ ~ ~ O ~ p J JI II II I

0.91b.OO 4.00 8.00 12.00 (5) 16.00(a) Lines

Line voltage TerminalEND

0.950

0.975

0.925

1.025 - - - - - - - - - - - - - - - - - r - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - , III1.000M----------.....~ - - - - - - - - - - I ~ - - - - - - - - - .....------------iine voltage TerminalBRON I II I- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ~I III I__________________________________________________ J

I II ILine voltage TerminalLOAD I I----------------- ----------------- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~I II II I0.908.00 4.00 8.00 12.00 (5) 16.00

(b) Cables

Figure 2.9: Problem definition, voltage at the terminals

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From the first figure (Fig. 2.7) it can be seen that in case of line network the activepower losses in the lines are almost negligible in comparison to the cable network. Fora cable network the active power support from the grid is more due to the higher resis-tive part of the impedance of the cable than for the line. The reactive power supportfrom the power grid is dep icted in Fig. 2.8. From this figure it can be seen that thereactive support for the line is more due to the higher reactance. However the voltagedrop at the terminalLOAD and terminalEND is larger for the cable network than forthe line network (see Fig. 2.9). This is because the large active power flow whichcauses a higher voltage drop when an higher impedance medium is used.

To compensate these power changes and thus to be able to reduce voltage drops apower support in th e network is needed. The two methods that will be discussed inthe following chapters are the STATCOM and the SSSC. The STATCOM, injects acurrent of variable magnitude at the point of connection. The SSSC injects a voltageof variable magnitude in series with th e transmission line. When these devices operatealone they exchange almost only react ive power. because they don't use a supply.When these devices operate together as an UPFC, i t is capable of exchanging real andreactive power independently. The later will not be discussed in this report.

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Chapter 3STATCOM3.1 TheoryThe STATCOM which is schematically depicted in fig 3.1 consist of a VSC connectedin shunt to the transmission network through a coupling transformer. Therefore it canbe represented by a current source expressed as

(3.1)Where V sh is the output voltage phasor of the VSC, Vi is the voltage phasor of

the connected terminal and Zsh is the shunt transformer impedance. and I; h arethe current components in phase and in quadrature with the voltage at the connectedterminal.

~ I A C B ~-Z-S-h----------IshVdcaStatcomcontroller

Pmr r - - - - - - . . ,VoltageSourceConverter

Figure 3.1: Injection model of the STATCOM

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3.2 Control model

0 ppa-measuremer ataPqmea* 1 0

cosref Pmr1 0 0Phase locked loop Controller VSCElmPII*,ElmPhi* sinref * Pmi - ElmVsc*1 2 1 1Vdc

I3

Vdc Measureme VdcJetl-4taVmea*-"_ref

Figure 3.2: Shunt control frameThe control model frame used for the STATCOM is illustrated in fig 3.2. It con

sist of the actual controller, fig 3.3 and 4 blocks connected with the controller. Thefirst block is the PQ measurement block which measures the Q and supplies it to thecontroller. The next block is the PLL which measures the phase at th e sending endterminal. The output of the PLL is a cosine and sine of the measured phase. Then, tocontrol the DC terminal of the STATCOM the voltage Vdc is needed to be measuredcontinually. That's why the voltage measurement block is needed. The las t block'VSC' represents the shunt converter.

The actual controller illustrated in fig 3.3 exist in two PI-controllers. One for theQ-control and the other one for the DC voltage control at the dc terminal. The Qcontroller controls the amplitude of the AC voltage of the PWM by changing the pulsemodulation index(Pm). The Vdc controller controls the angle difference between theSTATCOM terminal and the grid terminal, dtheta. Because the PWM only supportsthe signals Pm and dtheta, when the voltage angle of the grid terminal is set as theglobal angle, the controller uses the signals Pmr and Pmi:

Pm cos(ere! - de)Pmsin(e re! - de)

with ere! the voltage reference angle of the grid.

17

(3.2)(3.3)


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Syncronizer0 cosref 01 sinref Pmr 01 u

Pmi 12 1

2 Q dQ K(1+1/sT. Pm r--+4 3.. , "\3 Q ref '+ K,T

Q controller

4 Vdc - " dVdc K(1+1/s1 dtheta, ; K1,T1

tVdc refVdc controller

Figure 3.3: Shunt controllerThe voltage at th e AC s ide is then:

KOPmrUDCKOPmiUDC

(3.4)(3.5)

Since th e converter is s up pl ie d b y a dc capacitor, th e voltage across th e capacitorwill fall if th e STATCOM is no t lossless. The dc capacitor voltage ca n be regulatedby replenishing th e losses du e to switching an d in th e coupling tr ansf ormer circuit bydrawing power from th e ac s yst em . T he re fo re phase STATCOM must l ag p ha se g ri dby a small amount such that the dc capacitor voltage is held constant.

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3.3 ResultsThe STATCOM simulation model which is analyzed in DIgSILENT is i llus trated inFig. 3.4. The simulations in this section are all done for the case that the STATCOMis placed at terminalLOAD. (see Fig. 3.4). The other possibility is to place the STATCOM at th e terminalEND. This case is discussed in the next section.

TerminalBRON

External Grid

lineUP

LineDown

TerminalLOADTerminalEND

Transformer 110/33

General LoadStatcom

Shunt terminal

DCterminalShuntlFilter

=-I I--------------------Figure 3.4: STATCOM layout in DIgSILENT

Table 3.1: System parameters of the circuit with STATCOMTerminalBRON / TerminalLOAD 110 kVTerminalEND 33 kVLength Line/Cable 200 kmR - X (line) 0.02 - 0.2 O/kmR - X (cable) 0.2 - 0.05 0/kmLine Transformer 50 MVA, 110/33 kVSTATCOM Transformer 5 MV A, 110/4 kVSTATCOM DC terminal 10 kVSTATCOM shunt capacitor 100 MVar

3.3.1 React ive power t ransportIn order to show the effect of the STATCOM clearly, the same load event is used asin the previous chapter. Thus again the initial demand is 1 MVar. At time t = 4 secthe reactive power demand changes to 40 MVar. Then at time t = 8 sec the reactivepower demand reduces to 20 MVar. Final ly the reactive power demand is se t to it s

19


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nominal reactive power demand of 1 MVar at time t = 12 sec. The simulation resultsare shown in Figures 3.5 to 3.9

- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ,1 1 1 1 11 L ~ ~ ~ ~ ~ ~ ~ ~ n ~ t ~ ~ - - - - J - - - - - - - - - - - - - - - - - J~ . O O 1 1 , . . - --, 1 1

-::-:::::::::::::: ~ ~ : ~ ~ ~ : : : : : : : : : : : : : : : : : : : : ~ : : : : : : : : : : : : : : : : : i,----------------- - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - ----------------- j1 1 10.00 - - - ~ ~ - ~ ~ ~ - - 1 ' " " _J: : Grid reactive power 1 :

16.00_12.50 ........ - - - ~ - - - - . . . . l . . - - - - ~ - - - - - - - l . - - - - ~ - - - - - ' - - - - - - ~ - - - - . . . l .0.00 4.00 8.00 12.00 [8)(a) Lines

62.5050.00[MVar)37.5025.0012.500.00

1 1 1 1 1' L ~ ~ ~ ~ ~ ~ ~ ~ n ~ t ~ ~ J-----------------J: : f ' I . " ' - - - - -1 : :

' ~ : - - : . - - ~ ~ l ~ - ~ " ~ : : l ~ ~ ~ ~ ~ ~ l ~ . L : ~ ~ .: : Grid reactive power , 1 :16.008)2.00.00.00

.12.50 " ' - ~ ....L.. ~ ___l. ~ '_ _ ~ ___.. . l .0.00

(b) Cables

Figure 3.5: STATCOM, Reactive power load event

0.00 t"----------f+f+

0.900.60[MW]0.30

- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ,1 1 1 1Grid active power 1 1 1----------------- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - j1 1 11 1----------------- --------------- ---------------- - - - - - - - - - - - - - - - - - ~1

1~ ~ - - - - _ f l f H j & . ) . . - - - - - f f i ~ . , . . . - - - - - - - i1-0.30

-0.60

_________________ J1 11 1 1__________________________________ L J

1Statcom active power : :16.005)2.00.00.00

-0.90 " '- ~ . l . . ._ ____ ' ' __ ~ ___L. ~ ......L0.00

(a) Lines- - - - - - - - - - - - - - - - - r - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~1 1 1 1_________________ 1 ~ r ~ d _ a ~ " ! ~ ! ' : . _ :--- ----------J ------------J

1 1 I1 1 1________________ ~ - - - - - - - - - - - - - - - - - J1

1. . .- - - - ~ - - - j1_________________ J

1 11 1 1__________________________________L J: Statcomactivepower: : :

0.00 ~ - - - - - - - - --5.00

15.00

-10.00

10.00[MW]5.00

16.005)2.00.00.00-15.00 ........ ~ . l . . ._ L_ ~ ___ ' ~ ___l.

0.00(b) Cables

Figure 3.6: STATCOM, active power flow

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0.85

0.90

0.95

- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~[p.u] I Voltage terminalBRON :1.00 ...------.. .- i ? '.........- .... - - . . ~ , - ....O'===""""-..---lr-..---........I i l r -i

: Voltage terminalLOAD : ::----------------- - - - - - - - - - - - - - - - - - I ~ ~ ~ ~ ~ ~ : : : : ~ - - - - - - - - - - - - - - - - - :ILL I I----------------- ----------------- - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~I I I I II I I; - - - - - - - - - - - - - - - - - t : : : ; ~ ~ ~ m ~ ~ ~ : - - - - - - - - - - - - - - - - 1 - - - - - - - - - - - - - - - - - 1I I I I I

oaq..oo 4.00 8.00 12.00 [5](a) Lines

0.96

0.840.880.92

1.M - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ~ - - - - - - - - - - - - - - - - - ,[p.u] I Voltage terminalBRON I :1.00 1I--. . . . . .- - - -. . . . . .-oAlfld''''''''...... - - - - - - ~ " . . , - - - - - - - - - . . , . . ......__.._ --

Ejercicios de DigSilent Series and shunt compensation in Power lines and cables.pdf

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