1.hvdc basic theory

Introduction to HVDC

Basics of HVDC

Why HVDC ?

HVDC Theory

Basics of HVDC

Principles of HVDC

HVDC Applications

HVDC Long Distance Transmission Systems

Why Back-to-Back Links

HVDC Station

Principles of HVDC

Simplified Block Diagram Characteristics

DC AC AC Id in one direction only System

System 1 System 2

Magnitude of P or Id controlled depending on difference in terminal

Equivalent Circuit voltages (U1, U2)

I R

d P U U Direction of P controlled 1 2 depending on polarity

of terminal voltages (U1, U2)

HVDC Applications

Long Distance

AC AC

System 1 DC line System 2

Sea Cable

AC AC

System 1 DC cable System 2

Back-to-Back

AC AC System 1 System 2


Monopolar

Transmission Terminal Line Terminal B

A

Bipolar

Pole 1

Pole 2

Transmission Terminal Terminal B Line

A


Bipolar System: Operating Modes

Bipolar Monopolar, ground return one DC line pole


Multi Terminal

Terminals in Parallel

Terminals in Series

Back-to-Back Links


Technical Reasons

Different system frequencies (f1 „ f2)

Different system control (Df1 „ Df2)

Exchange of low power compared to the size of the interconnected AC systems

HVDC Station

Bipolar HVDC Terminal


1 AC Switchyard To/ from

Controls, Protection, Monitoring other terminal

2 AC Filters, Capacitor Banks

DC 3 Converter Transformers filter Pole 1

4 Thyristor Valves AC filter

DC Pole 2 filter 5 Smoothing Reactors

and DC Filters

6 DC Switchyard 1 2 3 4 5 6

HVDC Station

Tasks of Equipment

AC Switchyard (1)

Connect the terminal to the AC system

AC Filters, Capacitor Banks (2)

Reactive power supply

Filter harmonic currents

Converter Transformer (3)

Obtain the AC voltage needed for the required DC voltage

Obtain 12-pulse operation (star and delta connection) Allow

for series connection of 6-pulse bridges

HVDC Station

Tasks of Equipment

Thyristor Valves (4)

Convert AC to DC and vice-versa

Connect 6-pulse bridges in series for required DC voltage

Smoothing Reactors and DC Filters (5)

Smoothen the DC current

Avoid resonance with DC line

Limit interference caused by DC side harmonics

DC Switchyard (6)

Achieve required DC-side transmission configuration

Converter Theory - Commutation Groups

Converter Theory - Commutation Groups

Three-Pulse Converter

(Commutation Reactances neglected)

Converter Theory - Voltage Formation

Idealized un-controlled Three-Pulse Converter


π +3 3

U = ⋅u ˆ ⋅ cosω t d ω t di0 ph ∫⋅2p π −

3

3 π U = ⋅ˆ u ⋅ 2 ⋅sin di0 ph 2 ⋅π 3

3 U 3 v U = ⋅ 2 ⋅ ⋅2 ⋅

di0 2 ⋅π 2 3

1 35 U = ⋅U di0 v 2

Idealized un-controlled Three-Pulse Converter


Idealized controlled Three-Pulse Converter


π + +α 3 3

U = ⋅u ⋅ cosω t d ω t di α ph ∫

2 ⋅ π − +α

3

3 π U = ⋅u ⋅2 ⋅sin ⋅cosα diα ph 2 ⋅ 3

1 35 U = ⋅U ⋅cosα diα v 2

U =U ⋅cosα diα di0



-


Inverter Operation β = 180 - α

Converter Theory - Current Commutation

Three-Pulse Converter

(Commutation Reactances considered)


Commutation at Rectifier

cos α - cos (α+u) = 2 dx


Commutation at Inverter

cos γ - cos β = 2 dx

Converter Theory - DC Voltage Calculation

a) Three pulse MP circuit

b) Three phase bridge circuit

c) Twelve pulse group

Converter Theory - Valve Voltages (12-Pulse)

Main Equations - Angle Definitions

180 = α + u + γ

α firing angle

u overlap angle

γ extinction angle

β = u + γ

β advance angle, ‘firing angle of inverter’

δ = α + u

δ ‘extinction angle of inverter’

Main Equations - Udc

Relative Voltage Drop dx for 6-Pulse Bridge:

α +u 1 u dt∫ v 2

α Dx = 6 ⋅T

Dx = 6 ⋅ f ⋅L ⋅ I Tr d

with

1 π U di0 N L = ⋅u ⋅ ⋅Tr k 2 ⋅π ⋅ f 6 I dNu I k d Dx = ⋅ ⋅U di0 N 2 I

dN

Dxdx =

U di0 u I U k d di0 N dx = ⋅ ⋅2 I U dN di0


U =U (cosα − dx− dr )=U (cosα − dx )d dio dio tot I d i =d i d I dx = dx ⋅ dNtot totN u v U U v diou = =v U U

vN dioNU = u ⋅U dio v dioN

i d U = u U cosα − dxd v dioN totN u v

U =U (u cosα − dx ⋅ i ) (1) d dioN v totN d

U =U (cosα − dx ) (2) dN dioN N totN

U u cosα − dx ⋅ i d v totN d = u = (1)/(2) d U cosα − dxdN N totN


U =U (cosα − dx − dr)=U (cosα − dx ) dxtot = dx + drd dio dio tot

At Inverter α>90, i.e. Ud<0:

Ud Rec > 0 Ud Inv < 0

cosα = cos( 180 − β ) = −cos β

U =U (− cos β − dx − dr) − cos β = −cosγ + 2 ⋅dxd dio

U =U (− cosγ + dx − dr)= −U (cosγ − dx + dr)d dio di0

U = −U (cos β + dx − dr)d dio

with different Reference-Arrow System at Inverter:

Ud Rec > 0 Ud Inv > 0

U =U (cosγ − dx + dr)=U (cosγ − dx ) dxtot = dx − drd dio dio tot


-

a) Rectifier with α = const

b) Inverter with β = const

c) Inverter with γ = const

Main Equations - Iv (6-Pulse)

2 Effective Current: I = I

vN dN3

3 3 2 6 Fundamental Current: I = ⋅ I = I = I

vN1 vN dN dNπ π 3 π

Main Equations - STr (6-Pulse)

S = 3 ⋅U ⋅ I TrN vN vN

2 I = I vN dN

3

S = 2 ⋅U ⋅ I TrN vN dN

π U = ⋅U vN dioN

3 2

π S = U ⋅ I TrN dio dNN 3

U dNU =

dioN cosα − dxN totN

π 1 S = ⋅U ⋅ I TrN dN dN3 (cosα − dx )

N totN

Main Equations - XTr

2 U vNX = u TrN K S TrN

S = 2 ⋅U ⋅ I TrN vN dN

U vNX = u

TrN K 2 ⋅ I

dN

π U = ⋅U vN di0 N

3 2

U U π u dio π dioK N N X = = dx ⋅TrN N 3 2 I 3 I

dN dN

U dNU =

dioN cosα − dxN totN

π dx U N dNX =TrN 3 (cosα − dx ) I

N totN dN

Main Equations - Reactive Power (simplified)

P 1 S = 3 ⋅U ⋅ I S 1 1 1 1=

cosϕ

1 = 3 ⋅U ⋅ ⋅ I , mit ü = transformer voltage ratio 1 V 1 ü

1 3 = 3 ⋅U ⋅ ⋅ ⋅ I

1 veffü π

3 2 = 3 ⋅U ⋅ ⋅ I v 1 d π 3

π 3 2 = 3 ⋅ ⋅U ⋅ ⋅ ⋅ I dio d

3 ⋅ 2 π 3

π U 3 2 d = 3 ⋅ ⋅ ⋅ ⋅ ⋅ I d 3 ⋅ 2 cosα −dx π 3 tot

P P d 1 = ≈cosα −dx cosϕ tot

ϕ ≈ arccos(cosα −dx )tot

Q ≈ P ⋅ tan [arccos(cosα −dx )]1 dc tot

Main Equations - Reactive Power

Q = P ⋅ tan Φ 1 dc

u − sin u ⋅cos(2 α + u )tan Φ =

sin u ⋅sin (2 α + u )

For Inverter use γ instead of α

Main Equations - Overlap

u = arccos ( cos α − 2 ⋅ dx ) − α

U d u = Arc cos − dx −α

U dioU d = cos ( α ) − dx

U dio

For Inverter use g instead of a

Main Equations - Reactive Power (Alternatives)

u − sin u ⋅cos(2 α + u )Q = U ⋅ I ⋅1 d d

sin u ⋅sin( 2 α + u )

2 ⋅u + sin (2 α )− sin (2 α + 2 u )Q = U ⋅ I ⋅1 d d cos( 2 α ) − cos(2 α + 2 u )

2 ⋅u + sin (2 α )− sin (2 α + 2 u )Q = U ⋅(cosα − dx )⋅ I ⋅1 di0 tot d

cos( 2 α ) − cos(2 α + 2 u )

cosα + cos( α + u ) 2 ⋅u + sin (2 α )− sin (2 α + 2 u )Q = U ⋅ ⋅ I ⋅1 di0 d

2 2 ⋅ (cosα + cos( α + u )) ⋅ (cosα − cos( α + u ))

2 ⋅u + sin (2 α )− sin (2 α + 2 u )Q = U ⋅ I ⋅1 di0 d

4 ⋅ (cosα − cos( α + u ))

2 ⋅u + sin (2 α )− sin (2 α + 2 u )Q =U ⋅ I ⋅1 di0 d

8 ⋅dxtot For Inverter use γ instead of α

Design Considerations - Valve Short Circuit

L L N Tr

s u 2 φλ u i v k

s L L u N Tr1

L

u = u −u = 2 U sin ω t v 2 1 v

t t 1 2 U v i = u dt = sin ω tdtk v ∫

2 (L + L )∫ 2 L N Tr t α

1 min

ω

Design Considerations - Valve Short Circuit

U t U v v i = [− cosω t ] = [cosα − cosω t ]k α min min 2 ω L 2 ω L ω

U 1 97 U v v Î = [cos5 °+1 ]= ⋅ =k 2 ω L 2 (X + X )

Tr N

1 97 U 1 97 S 1 v Tr= ⋅ = ⋅ ⋅ =

2 2 S 2 U U 2 U Trv v v u +u + k k S S S N Tr N I d =1 97 S Tru +k S N

Design Considerations - Minimum Load

Current discontinuity depend on: - minimum power - control angles - dc-side inductances

Should be avoided, because: -increased stress on snubber circuits -current interruption may cause control instability

Design Considerations - SCR

Short Circuit Ratio and Effective Short Circuit Ratio:

S SC SCR =

P dcN

S - Q Q SC filter filter ESCR = =SCR -

P P dc dc

Thank you

Acknowledgements:-

SAG Training presentation

Direct current Transmission –KimbarkHigh Voltage Direct Current Transmission 2ND Edition-Arrilaga

1.hvdc basic theory

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