An overview of High Voltage Direct Current Transmission system

7/29/2019 An overview of High Voltage Direct Current Transmission system

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Department of Electrical

Engineering ECE529: Session 44

An Overview of HVDC Transmission Systems

ECE 529

Spring 2009

HVDC Transmission Systems 1/24


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Steady-State HVDC Converter Representation

Steady state equivalent circuit

Inverter

R

dc

V Vdcr dci

dc acf ( ,I , |V |)

dcI

Rectifier

Have fast, direct control over (firing delay angle) Vdc = Vdo cos (firing delay angle) where Vdo = const |VLL| Some control of|Vac| with tap changing transformer DC current indirectly controlled by changing



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Basic Six-Pulse Converter

Based on line commutated, current source converter

Thyristors used as devices Converter with stiff current source on dc side Stiff voltage source on ac side (turns off thyristors)

Basic 6-pulse bridge:

ane (t)

Smoothing Reactor

+-V

IL s dc

dc

1 3 5

4 6 2

cX

A

+

-

-- -

+

+C

B

e (t)e (t)

cn

bn

Transformer Inductance



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Basic Six-Pulse Converter

Initially assume: 1) Ideal ac sources, 2) ideal switches,3) Xc = 0, and 4) Ls

source)

AC side of converter has an ideal voltage source, dc side of converterhas an ideal current source

Apply Kirchhoffs Current Law:

i1 + i3 + i5 = Idc (one switch always closed)i2 + i4 + i6 = Idc

Apply Kirchhoffs Voltage Law:ean + ebn + ecn = 0 (balanced 3 phase set)

Since Xc = 0, only one switch in (1,3,5) can be closed with a switch in(2,4,6)



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Basic Six-Pulse Converter (cont.)

Allowable combinations:

1 with (2 or 6) (4 shorts dc bus)3 with (2 or 4)

5 with (4 or 6)

2 with (1 or 5)

4 with (1 or 3)6 with (3 or 5)

Need to determine a switching sequence

Start from assumption of positive phase sequence

Typical current waveforms:ia | |ib | |ic

| |HVDC Transmission Systems 5/24


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Possible sequences:Top three switches: 1-3-5-1 or 1-5-3-1

Bottom three switches: 4-6-2-4 or 4-2-6-4

Assume: Vdc = V+dc - VdcSwitch # V+dc Switch # V

dc

1 ean(t) 4 ean(t)3 ebn(t) 6 ebn(t)5 ecn(t) 2 ecn(t)



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Positive sequence ( = 0, 1-3-5-1 and 4-6-2-4)

0.0 6.67 13.33 20.00 26.66 33.33

-10

.00

-6.0

0

-2.0

0

2.0

0

6.0

0

10.0

0

Time (mS)

Vo

ltage

(V)

Negative sequence ( = 0, 1-5-3-1 and 4-2-6-4)

0.0 6.67 13.33 20.00 26.66 33.33

-10.0

0

-6.0

0

-2.0

0

2.0

0

6.0

0

10.0

0

Time (mS)

Vo

ltage

(V)



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Phase currents:

1

dc

3 55

6 2 4

I

I

dc

Look at the line voltages:

Switch Combination Vdc=V+

dc - V

dc

1 - 6 eab= ean - ebn = Vdc1 - 2 eac= ean - ecn3 - 2 ebc= ebn - ecn3 - 4 eba= ebn - ean5 - 4 eca= ecn - ean5 - 6 ecb= ecn - ebn

If = 0, then Vdc =

3

2

|VLL

|= 1.35

|VLL

|We define this as Vdo



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Controlled Firing of Thyristors

Now add a firing delay () for the thyristors. Same delay for all 6switches

0.0 6.67 13.33 20.00 26.66 33.33

-10.0

0

-6.

00

-2.0

0

2.0

0

6.0

0

10.0

0

Time (mS)

Vo

ltage

(V)



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Controlled Firing of Thyristors

Vdc = 36+

6 +

2|VLL|cos()d = 3

2

|VLL|sin() |

6 +

6 +

Then Vdc = 32 |VLL|cos Define Vdo = 3

2

|VLL|

Therefore Vdc = Vdocos

= 0 diode bridge Vdc = Vdo0 < 90 rectifier Vdc > 0 = 90 P =0 Vdc = 090 <

180

inverter Vdc < 0

Current does not reverse



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Commutation Overlap

Now add source inductance (Lc = 0)

i

+

-dcrV

IL s dc

1 3 5

4 6 2

cL+

-

Vdc



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Current Transfer Between Switches

Current does not fall to zero immediately in ac side inductance

Temporarily create line to line short

e (t)cn

cX

ane (t) e (t)bn

cX cX

3

L s

L s

1

2

dcI

0

3

1



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Current Transfer Between Switches (cont.)

What happens if gets to big (i.e. 180)?dcI

1

0

1

This is called a commutation failure

Thyristor 3 fails to turn on and thyristor 1 fails to turn off

This is more common ifLc is large, which is the case looking into aweaker ac system Normally corrects during next interval, although often have a second

failure when thyristor 5 turns on, double commutation failure



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Output Voltage During Commutation

Switch 1 contribution: V+dc1 = ean - Lc di1dt Switch 3 contribution: V

+dc3 = ebn - Lc

di3dt

During overlap we see the average between V+dc1&V+dc3So V+dc =

V+dc1+V+

dc3

2= ean+ebn

2- Lc

2

di1dt

+ di3dt

i1 + i3 = Idc, so di1+di3dt = 0But since its a linear network:

di1+di3dt

=

di1dt

+ di3dt

= 0

So: Vdc = V+dc ecn = ean+ebn2 ecn =eac+ebc

2



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Average DC Voltage with Overlap

Recall: Vdo = 3

2

|VLL| = 3

6

|V| = 3

3

|Em|where Em is peak line to neutral voltage

Then we find:

Vdc =3

+

3

2Emcosd+

+3

+

3|Em|cos(

6)d

Leading to: Vdc = 332 Em[cos+ cos(+)]

Or Vdc =Vdo

2 [cos+ cos(+)]



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Average DC Current

Start out with Lc = 0 and = 0 for nowFundamental Current Component

120

0 30 150 180

210 330 360

Firing delay simply adds a phase shift to the current (always lagging),

and cos = cos

a

Ean

I

3090

Ean

aI

Fundamental Componenti1pk =

2

2

2

iacos()d =2

Idc

3

3

cos()d =2

3

Idc

|I1RMS

|=

6

Idc



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Average DC Current

Then i1(t) = 2

3

Idccos(t)

Also: P = 3I1RMSVPcos = VdcIdcSo: 3I1RMSVPcos =

3

6

VPcosIdc

So: |Ia1RMS| =

6

Idc as expected

During overlap: Idc = Ic =

3Em2Lc =

eLL2Xc

i3(t) = Ic(cos cost) with t +where t = + at the end of the commutation interval

So average current is: I

dc= I

c(cos

cos(+))

Also: Ic =

3Em2Lc

=

32

|Vp|Xc

= |VLL|2Xc



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Average DC Circuit Equations

We have the following equations:

Vdc =Vdo

2 [cos+ cos(+)]Idc = Ic(cos cos(+))Vdo =

3

2

|VLL|

Ic =|VLL|

2Xc= V

do

6Xc

Substitute for the cos(+) in the Vdc equation

Then Vdc = VdocosV

do2IcIdc

Where Vdo2Ic

= Vdo2Vdo6Xc

= 3Xc = Rc (called the commutating resistance)

So Vdc = Vdocos

IdcRc



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Average DC circuit

Rc represents a current dependent voltage drop due to overlap

Rc does not represent any energy dissipation! So using Vdc = VdocosIdcRc we get:

Rc

Rc

V cosdo

line

V cos

do

RECTIFIER INVERTER

R



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Inverter Operation

+ + = Covers positive half cycle of voltage

is defined as the extinction angle o is minimum extinction angle for proper turn-offTypical values: 15 20

So +

180

o gives limits for control settings

Replace with 180 in averaged equations* note: cos(180 ) = cos()



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Inverter Operation (cont.)

Generate equations in terms on instead of

Vdc =Vdo

2[cos+ cos(+)]

Vdc =Vdo

2[cos(180 ) + cos(180 )]

Vdc = Vdo2

[cos(+) + cos()]

Idc = Ic(cos cos(+))Idc = Ic(cos cos(+))

Sign reversal in voltage equation expected for inverter



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Effect of Overlap on Power Transfer

Pac = 3I1RMSVpcos

Pdc = Idc Vdo cos+cos(+)2 = Idc 3

6Vp

cos+cos(+)2

I1RMS = Idc

6

Then cos = cos+cos(+)2

= VdcVdo

Note: overlap equations change if > 60, covered inKimbark, Direct Current Transmission: Volume I. Wiley, 1971.



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Transformer LoadingDC System

x (p.u.)

AC system

Xc = Lc where X 1220%

ZB =V

BIB

Xc = ZBX = VBIB X =3V2B

3VBIBX, this is

V2LLBMVA3B

Xc =V2BL

MVAB3X IB = I1RMS =

6

IdcB

So MVAB3 = 3VBIB = 3VB

6

IdcB= VRdoIdcB



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Transformer Loading (cont.)

Need to use true RMS (transformers sees all harmonic components)

IRMS = 12

3o I2dcd = Idc2

3

So MVAB3 = 3IRMSV = 3VRdoIdcB

Then Xc =

V2BLMVAB3

X with Vdo = 3

2

VBL

Then ZB = Vdo6IdcB Then from Vdc = VdocosIdc 3Xc we get

Vdc = Vdocos

Idc

3

(ZBX)

Vdc = VdocosVdoX2 IdcIdcB Leading to a per unit expression: Vdc

Vdo= cos X

2Idc


An overview of High Voltage Direct Current Transmission system

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