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