1.hvdc Basic Theory 3h1t2m

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 DC

AC

System

System 1

Characteristics

AC System 2

Magnitude of P or Id controlled depending on difference in terminal voltages (U1, U2)

Equivalent Circuit I U 1

Id in one direction only

d

R P

U 2

Direction of P controlled depending on polarity of terminal voltages (U1, U2)

HVDC Applications Long Distance

AC System 1

AC DC line

System 2

Sea Cable

AC System 1

Back-to-Back

AC

AC

System 1

System 2

AC DC cable

System 2

HVDC Long Distance Transmission Systems

Monopolar

Terminal A

Transmission Line

Terminal B

Bipolar Pole 1

Pole 2 Terminal A

Transmission Line

Terminal B

HVDC Long Distance Transmission Systems Bipolar System: Operating Modes

Bipolar

Monopolar, ground return one DC line pole

HVDC Long Distance Transmission Systems

Multi Terminal Terminals in Parallel

Terminals in Series

Back-to-Back Links

AC

AC System 1

System 2

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 AC

AC

System 1

System 2

To/ from other terminal

Controls, Protection, Monitoring

1 AC Switchyard 2 AC Filters, Capacitor Banks 3 Converter Transformers

DC filter

Pole 1

4 Thyristor Valves

AC filter DC filter

1

2

3

4

5

Pole 2

6

5 Smoothing Reactors and DC Filters 6 DC Switchyard

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 - Commutation Groups

Three-Pulse Converter (Commutation Reactances neglected)

Converter Theory - Voltage Formation

Idealized un-controlled Three-Pulse Converter

Converter Theory - Voltage Formation + π3

3 U di0 = ⋅ ⋅uˆ ph ⋅ ∫ cosωt dωt 2p −π 3

3 π U di0 = ⋅ˆ phu ⋅ 2⋅sin 2 ⋅π 3 Uv 3 3 U di0 = ⋅ 2⋅ ⋅2⋅ 2 ⋅π 2 3 1 35 U di0 = ⋅U v 2 Idealized un-controlled Three-Pulse Converter

Converter Theory - Voltage Formation

Idealized controlled Three-Pulse Converter

Converter Theory - Voltage Formation

+ π3 +α

3 U di α= ⋅u ph ⋅ ∫ cosωt dωt 2 − π3 +α ⋅ 3 π U diα = ⋅u ph ⋅2⋅sin ⋅ cosα 2⋅ 3 1 35 U diα = ⋅U v ⋅ cosα 2 U diα = U di 0 ⋅ cosα Idealized controlled Three-Pulse Converter

Converter Theory - Voltage Formation

-

Idealized controlled Three-Pulse Converter Inverter Operation β = 180 - α

Converter Theory - Current Commutation

Three-Pulse Converter (Commutation Reactances considered)

Converter Theory - Current Commutation

Commutation at Rectifier cos α - cos (α+u) = 2 dx

Converter Theory - Current Commutation

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:

1 α +u uv dt ∫ 2 α Dx = 6 ⋅ T Dx = 6 ⋅ f ⋅ LTr ⋅ I d with 1 π U di0 N LTr = ⋅u k ⋅ ⋅ 2 ⋅π ⋅ f 6 I dN Dx =

uk I d ⋅ ⋅U di0 N 2 I dN

Dx dx = U di0 dx =

uk I d U di0 N ⋅ ⋅ 2 I dN U di0

Main Equations - Udc U d = U dio (cosα − dx − dr )= U dio (cosα − dxtot ) dxtot = dxtotN ⋅ U dio = uv ⋅U dioN

id uv

id =

Id I dN

Uv U dio uv = = U vN U dioN

 id  U d = uvU dioN cosα − dxtotN  uv   U d = U dioN (uv cosα − dxtotN ⋅id )

(1)

U dN = U dioN (cosα N − dxtotN )

(2)

Ud uv cosα − dxtotN ⋅id = ud = U dN cosα N − dxtotN

(1)/(2)

Main Equations - Udc U d = U dio (cosα − dx − dr ) = U dio (cosα − dxtot )

dxtot = dx + dr

At Inverter α>90, i.e. Ud<0:

Ud Rec > 0

Ud Inv < 0

cosα = cos(180 − β ) = − cos β U d = U dio (− cos β − dx − dr )

− cos β = − cosγ + 2 ⋅ dx

U d = U dio (− cosγ + dx − dr ) = −U di0 (cosγ − dx + dr ) U d = −U dio (cos β + dx − dr ) with different Reference-Arrow System at Inverter:

Ud Rec > 0

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

Ud Inv > 0

dxtot = dx − dr

Main Equations - Udc

-

a) Rectifier with α = const b) Inverter with

β = const

c) Inverter with

γ = const

Main Equations - Iv (6-Pulse)

Effective Current:

I vN =

Fundamental Current:

I vN1 =

2 I dN 3 3 3 2 6 I dN = I ⋅ I vN = π π 3 π dN

Main Equations - STr (6-Pulse)

STrN = 3 ⋅U vN ⋅ I vN I vN =

2 I dN 3

STrN = 2 ⋅U vN ⋅ I dN U vN STrN

π = U dioN ⋅ I dN 3 U dio

STrN

π = ⋅U dioN 3 2

N

U dN = cosα N − dxtotN

1 π = ⋅U dN ⋅ I dN 3 (cosα N − dxtotN )

Main Equations - XTr X TrN

2 U vN = uK STrN

STrN = 2 ⋅U vN ⋅ I dN X TrN = u K

U vN 2 ⋅ I dN U vN

X TrN

π u K U dio N = 3 2 I dN

π = ⋅ U di 0 N 3 2 U dio π N = dxN ⋅ 3 I dN

U dio = N X TrN

U dN cos α N − dxtotN

dxN U dN π = 3 (cos α N − dxtotN ) I dN

Main Equations - Reactive Power (simplified) S1 =

3 ⋅U1 ⋅ I 1

S1=

P1 cosϕ

1 3 ⋅U1 ⋅ ⋅ IV 1 , mit ü = transformer voltage ratio ü 1 3 = 3 ⋅U1 ⋅ ⋅ ⋅ I veff ü π 3 2 3 ⋅U ⋅ ⋅ Id = v1 π 3 3 2 π ⋅U dio ⋅ ⋅ ⋅ Id = 3⋅ π 3 3⋅ 2 Ud 3 2 π ⋅ ⋅ ⋅ ⋅ Id = 3⋅ 3 ⋅ 2 cosα − dxtot π 3 Pd P ≈ 1 = cosα − dxtot cosϕ ϕ ≈ arccos(cosα − dxtot ) Q1 ≈ Pdc ⋅ tan[arccos(cosα − dxtot )] =

Main Equations - Reactive Power

Q1 =

Pdc ⋅ tan Φ

u − sin u ⋅ cos(2α + u ) tan Φ = sin u ⋅ sin (2α + u ) For Inverter use γ instead of α

Main Equations - Overlap u = arccos ( cos α − 2 ⋅ dx ) − α

u = Arc cos Ud U

Ud − dx −α U dio

= cos ( α ) − dx

dio

For Inverter use g instead of a

Main Equations - Reactive Power (Alternatives) Q1 =

U d ⋅ Id ⋅

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

Q1 =

U d ⋅ Id ⋅

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

Q1 =

U di 0 ⋅ (cosα − dxtot )⋅ I d ⋅

Q1 =

U di 0 ⋅

Q1 =

U di 0 ⋅ I d ⋅

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

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

Q1 = U di 0 ⋅ I d ⋅

2 ⋅ u + sin (2α )− sin (2α + 2u ) 4 ⋅ (cosα − cos(α + u ))

2 ⋅ u + sin (2α )− sin (2α + 2u ) 8 ⋅ dxtot

For Inverter use γ instead of α

Design Considerations - Valve Short Circuit LTr

LN

s u2 s u1

φλuv

ik

LN

LTr

L uv = u 2 − u1 = 2U v sin ωt t 2U v 1 ik = uv dt = ) ∫ 2(LN + LTr t 2L 1

t

∫ sin ωtdt α min ω

Design Considerations - Valve Short Circuit

ik =

Uv U [− cosωt ]tαmin = v [cosα min − cosωt ] 2ωL 2ωL ω

Uv Uv 1 97 [cos5° +1] = ⋅ = 2ωL 2 (XTr + X N ) 1 97 Uv 1 97 STr 1 = ⋅ = ⋅ ⋅ = 2 2 S U 2  Uv Uv  2 v u + Tr u + k k S  SN  Tr S N  Îk =

=1 97

Id uk +

STr SN

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:

SSC SCR = PdcN ESCR =

SSC - Q filter Pdc

=SCR -

Q filter Pdc

Thank you

Acknowledgements:SAG Training presentation Direct current Transmission –Kimbark High Voltage Direct Current Transmission 2ND Edition-Arrilaga