Basic Concept Operation and Control of HVDC Transmission System

Basic Co Concept, Op Operation an and Control of HVDC Transmission System 13.00-16.00 hrs. July 29, 2008 Room 2003, T.102, EGAT Head Office

Nitus Voraphonpiput, Ph.D. Engineer Level 8 Technical Technic al Analysis Analysis – Foreign Power Power purchase purchase Agreement Agreement Branch Power Purchase Agreement Division Electricity Generating Authority of Thailand

Objective 

Note:

Introducing operation, and control of the High Voltage Direct Current Transmission System.

This presentation continues from the morning session. Basic mathematics and electrical engineering knowledge will be useful for attendee. 2

Contents  

1. HVAC vs. HVDC 2. HVDC Principle Q&A for 15 minutes Coffee break 10 minutes



3. Control of DC Transmission Q&A for 15 minutes

3

1. HVAC vs. HVDC 

Why use DC transmission?

This question is often asked. One response is that losses are lower, lower, but is it true? Reference [2] has been explained using Insulation ratio and Power capacity in order to proof this statement.

4


Insulation ratio of HVAC and HVDC (Ref. 1-2) A given insulation length for an overhead line, the ratio of continuous working withstand voltage factor (k ( k ) is expressed as, (note 1 ≤ k ≤ 2 ) k =

DC withstand voltage AC withstand voltage(rms)

= 1.0

A line has to be insulated for over-voltages expected during faults, switching operations, etc. Normally AC transmission line is insulated against over-voltages of more than 4 times the normal effective (rms ) voltage. 5

1. HVAC vs. HVDC This insulation requirement can be met by insulation corresponding to an AC voltage of 2.5-3.0 times the normal rated voltage. k 1

=

AC Insulation level Rated AC Voltage(rms) (phase - ground)

= 2 .5

For suita suitable ble conve converte rterr control control the the cor corresp respond onding ing HVDC transmission ratio is expressed as k 2

=

DC Insulation level Rated DC Voltage (pole - ground)

= 1.7 6

1. HVAC vs. HVDC Insulation ratio for a DC pole-ground voltage (V ( V d ) and AC phase-ground (V (V p ) is expressed as insulation ratio ( K ) =

insulation length required for each AC phase insulation length required for each DC Pole AC insulation level

=

AC withstand voltage(rms) DC insulation level

=

k

k 1 V P k 2 V d

DC withstand voltage

It can be seen that the actual ratio of insulation levels is a function of AC/DC voltage. Next, determine AC/DC 7 voltage.

1. HVAC vs. HVDC Determine AC/DC voltage Assumed resistances (R (R ) of the lines are equal in both cases (HVDC and HVAC). AC Loss = 3 x R x I L2 and DC Loss = 2 x R x I d 2 Let losses in both cases are equal, so that, I d =

3 2

I L

The power of a HVAC system and a bipolar HVDC system are as: AC Power

= 3 V P I L cos φ

DC Power

= 2 V d I d 8

1. HVAC vs. HVDC At the same power transfer, AC Power DC Power

So that, V p = 2

=

1

3 cos φ

3V P I L cos φ 2 V d I d

=

3 V P cos φ 2

V d

=1

V d

Thus, insulation ratio (K) can be written as K = k

k 1

2

1

k 2

3 cos φ

≈

1.2 cos φ

It can be seen that HVDC requires insulation ratio at least 20% less that the HVAC which essentially reflects the cost. 9


Power Capacity Compared a double circuit HVAC line (6 lines) and double circuit DC line of Bipolar HVDC.

Power transmitted by HVAC (P (P ac ) and HVDC (P (P dc ) are Pac

= 6V P I L cos φ

Pdc

= 6V d I d

On the basic of equal current and insulation, I d = I L, K =1: =1:

Pdc

⎛ k 1 ⎞ k 1 ⎜ ⎟ = 6⎜ k V P I L ⎟ = k k 2 ⎝ k 2 ⎠

Pac

cos φ

=

1.47 cos φ

Pac 10


For the same values of k , k 1 and k 2 as above and pf is assumed to 1.0, the power transmitted by overh ove rhead ead lin lines es can can be in incre crease ased d to 147 147%. %. Th The e percentage line losses, which is inversion of the power transmit, are reduced to 68%.



In addition, for underground or submarine cables, power transmitted by HVAC cable can be increase 294 % and line loss reduced to 34%. Note: for cable k equals at least two. 11

1. HVAC vs. HVDC From referen reference ce [3], losses losses are lower is not correct. correct. “The level of losses is designed into a transmission system and is regulated by the size of conductor selected . . DC and AC conductors , either as overhead transmission lines or submarine cables can have lower losses but at higher expense since the larger cross -sectional area will generally result in lower losses but cost more .” The reasons that HVDC have been used are: 1. An overhead d.c. transmission line with its towers can be designed to be less costly per unit of length. 2. It is not not practical practical to consi consider der AC cable systems systems exceedin exceeding g 50 km (due to VAR charging of the cable). 3. Some a.c. electric power systems are not synchronized to neighboring networks even though their physical distances 12 between them is quite small.

2. HVDC Principle The HVDC The HVDC val valve ve com compri prise ses s the the th thyr yris isto tors rs ac acti ting ng as cont co ntro rolle lled d sw swit itch ch.. In th the e ‘O ‘OFF FF’’ st stat ate, e, th the e th thyr yris isto torr bl bloc ocks ks the current to flow, as long as the reverse or forward breakdown voltages is not exceeded.  It chang changes es to ‘ON ‘ON’’ sta state te if it is forwa forward rd biase biased d (VAK > 0) and has has small small positive positive ‘Gate ‘Gate’’ volta voltage ge applied applied between between the Gate and the Cathode. 

Anode (A) Gate (G) Cathode (K) 13

2. HVDC Principle 



Thyristor switches between conducting state (ON) and nonconducting (OFF) state in response to control signal (firing) as its characteristic. The Gate voltage need not to be presen pre sentt when when the thyri thyristo storr is alread already y in ON state.

14

2. HVDC Principle Anode (A) R d = V AK / I A

i A

R d V T

Cathode (K) Anode (A) R r

i r

Cathode (K)

R r = V AK / I A

V T

P loss-ON state = V T .I A(avg.) + R d .I A2 (rms) P loss-OFF state = R r .I r 2 (rms) 15


ON-OFF state - ON state continues continues until until current current drops to zero, even even reverse reverse bias appears across the thyristor. - The cri crittical time to cl clear charg rge e ca carri rier ers s in the semi-conductor is referred as the turn-off time t off . If forward bias appears to soon, t t < < t off , thyri thyristo storr can not OFF OFF.. V AK > 0 and V G >0

OFF

< t off V AK > 0 and t t <

ON I A < 0

t > t > t off

OFF

16

2. HVDC Principle

ON State

OFF State 17

2. HVDC Principle Single Phase Bridge Rectifier Th1

I d Th3

Ld

I s V s

V d Rd = 10 Th4

US

Th2

= 220 V α = 30

o

18

2. HVDC Principle V s

I s

α = 30 ° °

Voltage waveform of inductor (Ld ), V Ld = Vd – Rd I d

V d Voltage waveform of resistor (R (R d ), V Rd = R d I d

Th 3

Th 1

Th 3

Th 4

Th 2

Th 4

I d 19

2. HVDC Principle Harmonics in the voltage and current waveform.

I s

150 Hz

DC

250 Hz

350 Hz

100 Hz

V d

200 Hz 300 Hz

DC

I d 100 Hz 50 Hz

20


Even DC side does not have reactive power (Q), the reactive power still presents on the AC side. The reactive power occurrence is caused by the delay angle ( ) (or called firing angle) of the current waveform. P = |V S | |I S | cos

V s

Q = |V S | |I S | sin VS

I s

30°

° 360 ° 20 ms

α = 30 ° °

time

IS Phasor Pha sor of funda fundamen mental tal component 21

2. HVDC Principle 50 Hz

I s

150 Hz

250 Hz

350 Hz

100 Hz

V d

200 Hz 300 Hz

I d

Product of V d and I d is (active) power (P (P ). ). 100 Hz

Product of phasor V S and phasor I S is not the apparent power (S (S ) . It represents the active power (P (P ) and reactive power (Q (Q ). ). There are harmonic distortion power, which is a new term caused by the higher harmonics (more than 50 Hz). It D (distortion is represented by D (distortion power). Finally, S 2 = P 2 + Q 2 becames S 2 = P 2 + Q 2 + D 2 . 22

2. HVDC Principle Ith1

I d

Lk

V s

I s

V d Ith2

The inductance Lk represents reactance on AC side (called commutating reactance). Due to nature of an inductor, The Ld inductor current can not change suddenly. Thus, during turn-off of the Th1 (and Th2) and turn-on Rd of the Th3 (and Th4), both are in conducting state for a short time (overlap time). This phenomena occurs during commutation of the thyristors.

Increasing I d Ith2 Ith1 is overlap angle

commutation Increasing Lk

It can be seen that if current is high, overlap angel is increased. In addition, if inductance is high, overlap angle is also increased. 23

2. HVDC Principle I s

Inductor current can not suddenly be changed, thus there is a slope.

V s α = 30 ° ° µ

cos φ ≈

cos α + cos(α + µ ) 2

V d Th 3

Th 1

Th 3

Th 4

Th 2

Th 4

I d 24




The impact of the overlap angle (µ ) is the reduction of the average dc voltage (V (V d ). It decreases the harmonic content of the ac current (I ( I s ) and power factor of the AC side. V d

V d

V d

= V do −

X K

2

π

V T

V d

Idea Id eall ca case se V do

D R

R d I d

X K I d D X

= 2 π f LK I d

Voltage drop due to commutating reactance is represented as D X

X K

I d Overall voltage drop V T and D R are very less compared to D X . Thus, there are usually neglected.

25

2. HVDC Principle Natural commutation

3-pulse converter V A

V d

Th1

Ld R d

I L B V B

→∞

I L

Th2

t C V C

Th3

Ld V d α α R d d

V A = √ 2 VP sin ω t t V B = √ 2 VP sin ω t-120 t-120 ° ° V C = √ 2 VP sin ω t+120 t+120 ° °

α = 0

V d

o

α = 60

o

α = 90

o

α = 120

= 1.17V P cos α = V d 0 cos α

o

26

2. HVDC Principle V d

V d

V d 0

V d 0

= cos α Rectifier mode can be performed when firing angle is less than 90 degrees.

1.0

Positive average voltage

Rectifier 0.5

α = 60

o

Average voltage is zero when the firing angle is 90 degrees. 45

Negative average voltage

o

90

o

-0.5

Inverter -1.0

135

o

180

o

α Inverter mode can be performed when firing angle is more than 90 degrees. 27

2. HVDC Principle

V d =60 ° α =60

=30 ° α =30 °

I d 28

2. HVDC Principle V A, I A 120°

V B , I B

V C , I C Th 1

Th 2 Th 3 Th 1

Th 2 Th 3

I d 29

2. HVDC Principle Reversing phase sequence

V A, I A α=30°

V d

α=120°

Positive voltage Negative voltage

Inverter mode can be performed as long as the DC current continues flow.

I d 30

2. HVDC Principle VA

V d α

Lk

D X Vk

VB

Lk Id t

VA

α

α Vk

IC

IA

IB

IA

IC

IB t

V d VB

= V d 0 cos α − D X

D X

=

3 2π

ω Lk I d

31

2. HVDC Principle The commutating reactance (X (X k ) results in decreasing of DC voltage, but it increases DC voltage in inverter mode. It can also be seen that the overlap time will γ increase when DC current is high and this can cause commutation failure in inverter mode.

V d

D X Vk

γ t

180 ° °

° 180 °

α

α IA

IA

IB

V d

= V d 0 cos α + D X

° Note: α + µ < 180 °

γ ) = 180 - α - µ The extinction angle ( γ

IB

D X

=

3 2π

ω Lk I d

32


V d+

6-pulse converter =0 ° α =0 °

V d = V = V d+ - V - V d- V d+ V d V d+

V d-

-V d-

V d- =0 ° α =0 °

The 6-pulse bridge consists of two 3-pulse bridges (positive and negative) connected in parallel.

33


6-pulse bridge HVDC Smoothing reactor

power

Smoothing reactor

power

I d V dr

Reactive power

DC line

power I d

V di Reactive power

DC line

The HVDC comprises two converters connected in anti-parallel through smoothing reactors and DC lines. One converter is operated in rectifier mode to transmit power from the AC network to the other side whereas the other side converter is operated in inverter mode to receive power into the (other 34 side) AC network.

2. HVDC Principle Rectifier Operation of the 6-pulse bridge converter

° and µ = 25 ° ° Assume α = 15 ° cos φ ≈

≈

cos α + cos(α + ) 2 cos 15

φ ≈ 30

o

+ cos(15 + 25 o

o

)

2 o

φ I.sin φ

φ I.cos φ

= 0.866

V

30°

I The converter operates in rectifier mode. It transmits active power while consumes reactive power. 35

2. HVDC Principle Inverter operation of the 6-pulse bridge converter

° and µ = 25 ° ° Assume α = 135 ° cos α + cos(α + ) cos φ ≈ 2 cos135 + cos(135 + 25 ) ≈ = 0.823 2 φ ≈ 145 φ I.cos φ V o

o

o

o

145°

I

φ I.sin φ The converter operates in inverter mode. It receives active power while consumes reactive power.

36


For convenience, the converter operated in inverter mode is often referred to extinction angle ( ). Thus direct voltage in inverter mode (V (V di ) are expressed as

V d

= V d 0 cos α + D X , α > 90

D X



=

3 2π

ω Lk I d

o

= V d 0 cos γ − DX γ = π − α − µ

V d

Actually, inverter is commonly controlled at constant extinction angle to prevent commutation failure. Therefore, it is not only for convenience, but also for converter control purpose. It is important to note that voltage drop caused by commutating reactance (Dx) is now negative. 37


Voltage vs. current (VI) characteristics at steady state

V d

V d Slope is D X

V d 0 1.0

V d 0 1.0

α = 0 ° ° g n i s a e r c n I

Rectifier

1.0

g n i s a e r c n I

Rectifier

I d I dN

is the control variable for rectifier and is the control variable for inverter.

1.0

Inverter

-1.0

α = 0 ° °

Inverter

α max < 180 °

-1.0

I d I dN

g n i s a e r c n I

γ = 0 ° ° 38

2. HVDC Principle 12-pulse bridge HVDC



Y

Y

V dr

Y

V drY

I d

I d

V di

V diY

Y

Y

Y

The 12-pulse converter is required to improve harmonic current current on AC AC sides. It comprises two 6-pulse converters connected in series. Harmonic current on AC sides are odd orders starting from 11 th, 13th …. whereas even orders present on the DC side (12 th, 14th…). To achieve 12-pulse, phase displacement of 30 generated by Star (Y) and Delta ( ) connection of 39the transformers are employed.

2. HVDC Principle V d

V dY

V d

Rectifier operation of the 12-pulse bridge converter

° Assume α = 15 ° ° and µ = 25 °

I AY I A

V d

Y

I A I A

I A

Y

V d Y

V dY

I AY

40

2. HVDC Principle ½ R d power

power

I d

Y

power Y

Y

V dr

V di

Y

Y

Y

½ R d Reactive power

Reactive power

α min < α α min = 5 ° ° - 7 ° ° To ensure all thyr th yris isto torr va valv lves es ar are e enough forward bias to turn on.

γ min < γ

voltage

decreasing

γ min = 15 ° ° - 17 ° °

V dr V di I d

To keep reactive power requirement on inverter side as low as possible.

current

Voltage drop caused by line resistance (R ( R d ) is taken into account and the VI characteristic presents operating point of the HVDC system. 41


Detail Configuration of the HVDC

42


Alternatives for the implementation of a HVDC power tran tr ansm smis issi sion on sys syste tem m

a) Earth Return

ii) Bipolar Configuration

b) Metallic Return

i) Mono-polar Configuration

iii) Homo-polar Configuration

43


Alternatives for the implementation of a HVDC power tran tr ansm smis issi sion on sy syst stem em (c (con onti tinu nued ed))

44

3. Control of the th e DC Tra ran nsmi miss ssio ion n 





Can we use manual control for the rectifier (vary ) and the inverter (vary )? If we can not do that, which side should be controlled (rectifier or inverter) or control them both? What is/are the control purpose(s)? 45

3. Control of the DC Transmission 

Typical control strategies used in a HVDC system consists of:        

Firing Control {Rectifier} Current Control (CC) {Inverter} Constant Extinction Angle (CEA) Control {Inverter} Current Margin Control (CM) {Inverter} Voltage Control (VC) Voltage Dependent Current Limit (VDCL) Tap change Controls (TCC) Power Reversal 46


Firing Control

Function of the firing control is to convert the firing angle order ( *) demanded fed into the valve group control system. There might be voltage distortions due to non-characteristic harmonics, faults and other transient disturbances such as frequency variation. Thus, phase-locked loop (PLL) based firing system is generally applied. θ v A v B v C

Phase Detector

v error

PI Controller

K

u A u B u C

(1 + Ts ) Ts

v o

Voltage Controlled Oscillator

comparator

…

sin(.)

…

…

comparator

sin(.) sin(.)

comparator

-

⅔

π *

Gate firing 47

3. Control of the DC Transmission Firing Control (Continued)



u A

v A

0

time v error 0

time

2 π π

* 0

α

Firing pulse of phase A

time 48


Curr Cu rren entt Co Cont ntro roll (C (CC) C)

The firing angle is controlled with a feedback control system as shown in figure. The dc voltage of the converter increases (by decrease *) or decreases (by increase *) to adjust the dc current to its set-point (Id*). K

(1 + Ts )

I d

Ts Y

V dr v A, v B , v C

id*

- +

id

αmax PI

αmin

*

Firing Control

Y

Y

6 6

Current measurement 49


Constant Extinction Angle Control (CEA)

The firing angle of the inverter is controlled at minimum angle ( min) to reduce reactive power requirement. This can be achieved by using Gamma control ( -control). Current measurement Y

V di Y

Y

v A, v B , v C

Valve voltage

6 6

Firing Control

*

αmin measurement

αmax PI

-

*

+ 50

3. Control of the DC Transmission VI Characteristic of the CC and the CEA voltage

voltage

V dr

α*

X

V di

γ* = γmin

V dr

V di

γ* = γmin

α*=α min

AC voltage decreasing

I d

current

I d

current

If AC voltage on rectifier side decreases, CC decreases * down to The intersection (X) is the operating ( I d ), but min to increase DC current (I point of the DC transmission line. there is no operating point (X). This problem can be solved using CMC. 51 VI Characteristic


Current Margin Control (CMC)

A better way is to use the inverter to control current less than of the rectifier by an amount of current margin ( I d ) when the rectifier can not perform CC.

Y

V di Y

Y

v A, v B , v C

Current measurement

6

*

- Co Con ntr tro ol

Firing Control

*

m n o u i t m c i e n l i e M s

αmin

id = 0.1 to 0.15

αmax

+

id*

PI

-

+

id 52

3. Control of the DC Transmission VI Characteristic of CC, CEA and CMC voltage

voltage

V dr

α*

CEA X

V di

γ* = γmin CMC

I d I d

CC

current

Combined characteristics of CC, CEA and CMC This method can maintain stable operation when AC voltage of both sides are fluctuated.

V dr α*=α min AC voltage decreasing

V di

γ* = γmin

X

I d I d

current

If AC voltage on rectifier side decreases, CC decreases * down to (I d ), but min to increase DC current (I there is no operating point (X). This problem can be solved by CMC. 53

3. Control of the DC Transmission What will happen if AC network of the t he inverter side is too weak! voltage

voltage

V dr

More Weak

α*

X

Weak AC

V dr

CEA γ* = γmin

α* X

V di CMC

I d I d

CEA

γ* = γmin current

In this range the intersection is poorly to define and both current controllers will hunt between the operating points.

CMC

I d I d

V di VC γ* > γmin current

This problem can be solved by adjust VI characteristic of the inverter to voltage control (VC) in order to avoid hunting between two 54 controllers.


Voltage Control (VC)

it is very effective when the inverter is connected to a weak AC network. The normal operating point X corresponds to a value of higher than the minimum. Thus, the inverter (rectifier as well) consumes more reactive power compared to inverter with CEA. Y

V di Y

Y

v A, v B , v C 6 6 Voltage measurement

Firing Control

*

m n o u i t m c i e n l i e M s

m n o u i t m i c e x l a e M s

CMC

*

- Co Cont ntro roll αmax

vdi* -

PI

+

αmin

vdi 55


Voltage Dependent Current Limit (VDCL)

Commutation failures can occur during an AC fault on the inverter side. It results in continue conduction of a valve beyond its 120 conduction interval. The CC will regulate the DC current to its rated value, but in the worst case, two inverter valves may form DC short circuit and continue conducting for a long time, which can cause valve damage. To prevent this problem, DC current must be reduced. One possible to detect the AC side fault is the lowering of the DC voltage. This voltage is typically chosen at 40% of the rated voltage. I d

56



The VDCL is a limitation imposed by the ability of the AC system to sustain the DC power flow when the AC voltage at the rectifier bus is reduced due to some disturbance as well. The VDCL characteristics is presented below. voltage

voltage

VC

VC

V dr α*

V dr α* X

X

VDCL 0.4

I d I d-min

CMC

V di

I d

CMC VDCL

V di I d

0.4

VDCL

I dmax

current

I d I d-min

VDCL

I dmax

VDCL I d

current 57



id* vd

v

i

i

s M e i l n e i c m t i u o n m

V d

CC

v

1 1 + Ts

VDCL

vd

Voltage measurement

58


Tap Change Control (TCC)

When voltage of the AC system of the rectifier and/or of the inverter is fluctuated, transformer taps (both side) can adjust to keep the DC voltage within desired limits or suitable operating point.. Genera point Generally, lly, the the tap will will be changed changed when the the firing firing angle angle of the rectifier/inverter still reach its more than 10-15 minutes to avoid interaction of other controls. Example: if the firing angle ( ) of the rectifier reaches minimum limit ( min) for long time. It means that the AC voltage of the converter is not appropriate. Thus, AC voltage of the converter must be reduced by tap changing of the converter transformer to free the firing angle of the rectifier. 59


Power Reversal

The VI characteristic of power reversion is presented below (VDCL and VC are not included). The station 1 (rectifier) increases firing angle ( ) into the inverter region region and the station 2 (inverter) decreases its firing angle ( ) into rectifier region. This can be performed without altering the direction of current flow. voltage

voltage

V1dr α*

γ* = γmin

X

V2 di I d

I d

current

current

V1di V2 dr α*

X

γ* = γmin 60

3. Control of the DC Transmission I d

Y

Y

V dr Y

M a x .

CC CAE VC TCC

M i n .

V di

Y

α*

Y

C F o i n r i t n r g o l

VDCL min

V d , I d , ,

l g o r n t i r i n o F C

Master Control i d* p*/vd po

VDCL

p*

Vd* p Modulation Signal

Power order

α*

Y

. x a M

. n i M

id CC

min

CAE

Vd*

VC TCC 61


CIGRE’s HVDC benchm CIGRE’s benchmark ark was was simulate simulated d on ATP-EM ATP-EMTP TP with with the typical HVDC control schemes, which the CC mode was employed at rectifier and VC mode was applied at inverter. All simulation results are presented in normalized values. val ues.

Rectifier Current Control

Start Up HVDC

Inverter Voltage Control 62


The HVDC started at 0.1 sec. The firing angle of rectifier started at 90 while the extinction angle of inverter started at 90 .

Firing Angle (α) of Rectifier

Firing Angle (α) of Inverter Extinction angle (γ) is also shown

Start Up HVDC

63


The HVDC started to reverse power flow direction at 0.5 sec. Firing angle of the rectifier increased (with a ramp rate) into inverter zone while firing angle of the inverter decreased (with a ramp rate) into rectifier zone.

Firing Angle (α) of Rectifier and Inverter

Power Reversal

DC Current 64


The power flow direction of the HVDC reversed at 0.9 sec.

Power Reversal

65


VDCL performance during 1-phase fault at AC network of the rectifier station. V

a

V

b

V

c

1 –phase Fault at AC network of the rectifier station

66

3. Control of the DC Transmission p. u .

I REF I d Vdi

Id

Degree

Alpha _ r ( α r ) Alpha_i ( α i ) α

i

I REF

Vdi

Fault at AC network of rectifier station

α

r

67


VDCL performance during 1-phase fault at AC network of the inverter station. V

a

V

b

V

c

1-phase Fault at AC network of the inverter station

68

3. Control of the DC Transmission p. u.

I REF I d Vdi Id

Degree

Alpha _ r ( α r ) Alpha_i ( α i ) α

Vdi

i

I REF α

Fault at AC network of inverter station

r

69




Modulation signal is employed when a power system has a special requirement such as frequency control, power oscillation damping, etc. For example, the addition frequency control loop is included into HVDC control system to stabilize frequency of the AC system.

70

3. Control of the DC Transmission

Modulation Function of EGAT-TNB HVDC

71

3. Control of the DC Transmission

Power Swing Damping (PSD) Function of EGAT-TNB HVDC 72

Thank you very much for your attention

References 1. Ani Gole, “HVDC Transmission Lecture Note”, University of Manitoba, Manitoba, 2000. nd

2. Jos Arrilaga, “High Voltage Direct Current Current Transmission”, Transmission”, 2

, IEE-Press, 1998.

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ศน วรพนพิพัฒน , วิทวัส 8. กฤตยา สมสัย , นิทัศน

ผองญาติ องญาติ, “การจําลองระบบส าลองระบบสงไฟฟาแรงสู าแรงสูง กระแสตรงโดย ATP-EMTP”, สัมมนาวิ มมนาวิชาการระบบส ชาการระบบสง กฟผ. 2548. 74

Basic Concept Operation and Control of HVDC Transmission System

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