EIEN15 Electric Power Systems Lecture 1 Olof … · natural or surge impedance loading (SIL) – P....

Voltage

Olof Samuelsson

1 EIEN15 Electric Power Systems L5

Outline

• Affecting bus voltage • Voltage along a line

– Transmission, Zline=0+jX (inductive lines) – Distribution, Zline=R+j0 (resistive lines)

• Power transfer across X – both voltages fixed • Power transfer across X – one voltage fixed • Voltage stability • Voltage during faults


Vth

Zth ZL V

IL

V1

Z

V2

I12

Affecting bus voltage +

• Q injection increases V – Shunt capacitors – Synchronous generators

• IC through X(TH) negative voltage drop!

VTH

VL

XTH

IC

jXC=-j/(ωC) ~

VTH=VL+jXTHIC

IC=VL/(jXC)=jωCVL

VTH=VL+jXTHjωCVL

VTH =(1-XTHωC)VL

VL>VTH VTH

VL

IC

jXTHIC


jXTHIC

Affecting bus voltage –

• V decreased by Q load – Shunt reactors – Synchronous generators

VTH

VL

XTH

IL

jXL=jωL ~

VTH=VL+jXTHIL

IL=VL/(jXL)=–jVL/(ωL)

VTH=VL–jXTH jVL/XL

VTH =(1+XTH/XL)VL

VL<VTH

VL

VTH

IL jXLIL

4 EIEN15 Electric Power Systems L5 L5 Example 1

jXLIL

Shunt capacitor raises V – how much?

• Raising V by α p.u. (∂V=α) requires Q= -SSCα

– Draw negative Q shunt capacitor does that – Capacitor Mvar rating α SSC

• SSC reflects network strength (stiffness) 5 EIEN15 Electric Power Systems L5

XV

XEVQ

2−=Assume E=V=1, δ=0:

VSQ SC∂−=∂

Example: ZTH and SSC in network a. At 3 kV busbar

» X3kV=(X300MVA+X6MVA)//X2MVA

b. After transformers » X3kV +(X0,5MVA//X0,6MVA)

Strengthen: Decrease XSC, increase SSC

» Add parallel Z (transformer/line)

» Add source/parallel Z (generator)

Weaken: Increase XSC, decrease SSC » Add series Z (transformer/line)

a)

b)

3 kV


Off-nominal turns ratio

• Vbase1/Vbase2≠n1/n2

• Tap changer SGO Example 3.12 PW V2 variable in steps of 1-2% V1/V2≠Vnom1/Vnom2

• Phase shifting transformer δ1≠δ2

• Parallel operation Example: 130/10kV and 130/12kV n1≠n2

• How about per unit now?

V1∠δ1

22 δ∠V

n1 n2


Voltages and currents

V1 V2 Zeq c:1

V1 V2

= c Vbase1 Vbase2

c complex

I1 I2

V1 p.u. = cV2 p.u.

0=S1+S2=V1I 1*+V2I 2*

V1/V2=(–I2/I 1)*=c

I2= – c*I 1

p.u. removes nominal turns ratio, leaves factor c

* is complex conjugate


Bus admittance matrix

−=+=

12

121

* IcIIZcVV eq

V1 V2 Zeq c:1 V1 V2

= c Vbase1 Vbase2

c complex I1 I2

( )( )

−−=−=

eq

eq

ZcVVcIZcVVI

/*/

212

211

−

−=

2

12

2

1

* VV

YcYc

cYY

II

eqeq

eqeqeqeq ZY /1=


π-circuit of tap changing transformer

• Try c:1=0.9:1 (1-c)Yeq=0.1Yeq, (c2-c)Yeq= - 0.09Yeq

• Like shunt-X on left side and shunt-C on right side! 10 EIEN15 Electric Power Systems L5

V1 V2

cYeq I1 I2

Can only be drawn for real c <=> symmetric Ybus

(1-c)Yeq (c2-c)Yeq

Check 1: Try c=1 Check 2: Set up Ybus

−

−=

−+−

−−+22

11cc

cY

ccccccc

Y eqeq

X

XI

Voltage at midpoint of a line

Assume V1 and V2 fixed Assume line only series X Add bus along line with no generation/load V3 smaller than V1 and V2

V2 V1

jX1I jX2I

V3 V1 V2 V3 X2 X1

I


Voltage profile of a line

• Move bus 3 Plot V3 as f(x) V(x)=Voltage profile!

V3 V2 V1

V1 V3 V2

x

V1 V2

V(x)

x


Voltage support

• Line draws Q – At line ends I and V not in phase, Q injection from both ends – At midpoint: I and V in phase ↔ only P transfer

• Q infeed at bus 1 and 2 = Voltage supported at line ends

V1 V2

V(x)

x


XI

V2 V1

jX1I jX2I

V3

I

Line model with C but no R (X>>R=0)

• In reality C and L are distributed along line

• Assume V1 fixed, but not V2

• Series L draws reactive power QL=ωLI2 – decreases V along line

• Line charging C generates reactive power QC=ωCV2 – increases V along line


V C

L I

∆x

V1 V2

Two extremes • QL << QC

I low = light load – V increases along line

• QL >> QC

I high = heavy load – V decreases

Fixed V1

V(x)

x

Light

Heavy


Balance • Load P0 so that QL = QC

– No Q transfer no reactive V drop same V along line – P0 natural or surge impedance loading (SIL) – P0=V2/R0, R0=sqrt(L/C)=Zcharacteristic

• In practice: – Series resistance gives voltage drop – Line loaded to more than SIL extra C needed

V1 V(x)

x

Light

Heavy

SIL

Exercise 5.1 Also in PW


Lab1: Distributed Generation at MV or LV

• Before: only load = voltage always falls away from transformer – Vset>Vnom at transformer avoids undervoltage @ remote load

• DG (wind power or photovoltaics) = negative load raises V – Vset<Vnom at transformer would avoid overvoltage @ DG

• Smart Grid Solution: Reduce Vset at low load to permit more DG


Ingmar Leisse, PhD 2013

DG

Power through reactance important case

• Line transfer

• Power through transformer

• Power from generator

Xline

Xeq

Xd


Power through series Z - line transfer 1


V1=V1∠θ1

I1 I2

S12 S21 V2=V2∠θ2

Z=Z∠α V1,V2 constant

• How much power enters line at bus 1?

( ) ( )

( )αθθα

αθθαθθ

α

θθθ

+−

−−−−−−

−=

=−=

=

−=

=

−

==

21

2111

211

212

1

212

1

*

211

*21

1*

1112

jj

jj

j

jjj

eZVVe

ZV

eZVVe

ZV

ZeeVeVeV

ZVVVIVS

Line transfer 2 – How much is S12?

θ1-θ2

P12

V1,V2 constant

Pmax=V1V2/X


ikkiki

ik VVYZ

VV=

( )

( )αθθα

αθθα

+−−==

+−−==

2121

21

1212

2121

21

1212

sinsinIm

coscosRe

ZVV

ZVSQ

ZVV

ZVSP

( ) ( )2121

21

122121

12 cos sin θθθθ −−=−=XVV

XVQ

XVVP

Transmission line: Z=R+jX ⇒ Z=X and α=90°

Also use sin(90°+θ)=cosθ and cos(90°+θ)= - sinθ

P flows to lower angle (if V1=V2)

Q flows to lower V (if θ1=θ2)

Note: PMU

Line transfer 3 – What arrives at bus2?

• Exchange bus numbers and change sign:

• Out from bus 2 comes:

• Can be applied to Thévenin, generator, line, transformer


Q21 ≠ −Q12

L5 Example 2

( ) ( )2121

21

122121

12 cos sin θθθθ −−=−=XVV

XVQ

XVVP

( ) ( )1221

22

211221

1221 cos sin θθθθ −+−=−−−==−XVV

XVQ

XVVPP

( ) ( )X

VXVVQ

XVVP

22

1221

2121 cos sin −−=−= θθθθ

Generator feeding large system: Voltage support at both ends

P+jQ

X E∠δ V∠0

P = EVX

sinδ

Q = EVX

cosδ − V 2

X


E and V fixed

( )

( )X

VXVVQ

XVVP

22

1221

2121

cos

sin

−−=

−=

θθ

θθ

General: Here:

This symbol denotes an infinite(ly strong) bus, where nothing can change the voltage

Large system+load: Voltage fixed only at sending end

P+jQ X

E∠0 V∠–δ

XV

XEVQ

XEVP

2

cos

sin

−=

=

δ

δ


Only E fixed

( )

( )X

VXVVQ

XVVP

22

1221

2121

cos

sin

−−=

−=

θθ

θθ

General: Here:

Let Q=0 and vary P with R

• Vary R and plot V and I • Plot V(P)=VI • Two P for each V!


P X=1

E∠0 =1∠0 V∠- δ

R=5 to 0

P/SSC

V/E

1

0.5

0.7

Reducing R increases P

R>X, P∝1/R R=X

R=0

R=∞

R<X, P∝R

Reducing R reduces P

0123450

0.5

0.71

1

𝑉(𝑅) =𝑅

𝑅 + 𝑗𝑗𝐸

𝐼(𝑅) =𝐸

𝑅 + 𝑗𝑗

𝑃 𝑅 = 𝑉 𝑅 𝐼(𝑅)

R

V(P) – PV curve or ”nose curve”

Use SSC=E2/X and v=V/E:

( ) 2

422

0X

VVEVPQ −=⇒=

( ) 42 vvSvP SC −=

Use trigonometric identity, eliminate δ:

( )22

2

22

,

+−=X

VQXVEQVP


P/SSC

V/E

1

0.5

0.7

Q=0

P X

E∠0 V∠- δ

R

Maximum P from ”nose curve”

dP/dv=0:

Pmax=SSC/2 at

for R=X, δ=45° 2/1=v

P/SSC

V/E

1

0.5

0.7

Q=0

26 EIEN15 Electric Power Systems L5 PW: PV

Q to bus with no V support Q X

E∠0 V∠0

With SSC=E2/X and v=V/E

( ) ( )X

VEVVQP −=⇒=⇒= 00 δ

Xload

( ) ( )2vvSvQ SC −= 27

EIEN15 Electric Power Systems L5

Q/SSC

V/E

1

0.25

0.5

P=0

Max Q:

dQ/dv=0: Qmax=SSC/4 at v=1/2 for Xload=X

Voltage instability scenario • Lost line reduces V P also reduced

• Thermostats assume P~1/R

• If P too small, R is reduced (more parallel heating elements):

• Upper side R>x: P increase … equilibrium

• Lower side R<x: P decrease… R reduced … no equilibrium

• Lower side is voltage unstable

• For R<x also tap changer behavior is reversed

• For low V, induction motors stall (stop with P=0 and high Q)

• All contribute to reducing V voltage collapse 28 EIEN15 Electric Power Systems L5

1

P~1/R P~R

P~1/R

P~R

Voltage collapses 1983, 2003 P

X E∠0 V∠- δ

Lines+generators lost, remaining lines more loaded V low more Q transfer + load P recovers V lower and I higher distance protection: “a fault!”, trips line other lines even more loaded and tripped

North Sweden South Sweden


blackout

seconds

Voltage near Stockholm Southernmost Sweden disconnected (blackout)

3-phase short-circuit: Voltages

• Example: Voltages during fault near Malmö 27 March 2007 • Voltage during fault V = Prefault voltage VF + ∆V • Short-circuit = Current injection –IF at fault bus • First use ZBUS to find IF • Then use ZBUS and IF to find ∆V


Two-bus example Zero ohm fault at bus 1

Fault current: I1=–IF=–VF/Z11

Vector of injections: [–IF 0]T

Voltage changes: ∆V=ZBUS[–IF 0]

ZBUS =Z11 Z12Z21 Z22

V1V2

= ZBUS

I1I2

Z12=Z21

Z22–Z21 Z11–Z12

V1 I1

V2 I2

/1

0 here

//

11122

1

2

1

1112

1111

12

11

2

1

FF

F

FF

F

VZZVV

VVVV

VZZZZ

IZIZ

VV

−

∆+∆+

=

−−

=

−−

=

∆∆


Generalize example

Fault at bus n

Fault current: In=–IF=–VF/Znn

Vector of injections: [0 …–IF…0]T

Voltage change at bus k:

∆Vk = −Zkn IF = −Zkn / ZnnVF

Vk = VF + ∆Vk = 1 − Zkn / Znn( )VF

Visualization of ZBUS for general system?


Visualizing general ZBUS Two-bus system

Z12 Z22 Z11

V1 I1

V2 I2

N-bus system

Z12 Z22 Z11

V1

I1

V2

I2

Z2N ZNN

VN

Z1N

Rake equivalent in textbook


SLG fault at bus n: Voltage at bus k?

• Sequence impedances and currents known • Prefault voltage known • One equation for each sequence network (0, 1, 2)

• Transform back to phase domain


−

=

−

−

−

−

−

−

−

−

−

2

1

0

2

1

0

2

1

0

000000

0

0

n

n

n

kn

kn

kn

F

k

k

k

III

ZZ

ZV

VVV

Conclusions

• P through X flows towards … • Q through X flows towards … • Injecting reactive power … voltage • Two ways to draw Q; using a … or a … • p.u. change in voltage requires p.u. Q injection times … • … = Surge Impedance Loading gives flat voltage profile • Wind power and PV makes the voltage …crease • Nose curve is …=f(…), where the … part is stable • Blackouts in Sweden 1983 + 2003 were caused by … • Voltage during fault is calculated using …


Conclusions

• P through X flows towards lower phase angle • Q through X flows towards lower voltage magnitude • Injecting reactive power raises voltage • Two ways to draw Q; using a reactor or a synch generator • p.u. change in voltage requires p.u. Q injection times -SSC

• SIL = Surge Impedance Loading gives flat voltage profile • Wind power and PV makes the voltage increase • Nose curve is V=f(P), where the upper part is stable • Blackouts in Sweden 1983 + 2003 were caused by voltage instability • Voltage during fault is calculated using ZBUS


EIEN15 Electric Power Systems Lecture 1 Olof … · natural or surge impedance loading (SIL) – P....

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Transcript of EIEN15 Electric Power Systems Lecture 1 Olof … · natural or surge impedance loading (SIL) – P....