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
2 EIEN15 Electric Power Systems L5
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
3 EIEN15 Electric Power Systems L5
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
6 EIEN15 Electric Power Systems L5
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
7 EIEN15 Electric Power Systems L5
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
8 EIEN15 Electric Power Systems L5
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=
9 EIEN15 Electric Power Systems L5
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
11 EIEN15 Electric Power Systems L5
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
12 EIEN15 Electric Power Systems L5
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
13 EIEN15 Electric Power Systems L5
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
14 EIEN15 Electric Power Systems L5
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
15 EIEN15 Electric Power Systems L5
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
16 EIEN15 Electric Power Systems L5
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
17 EIEN15 Electric Power Systems L5
Ingmar Leisse, PhD 2013
DG
Power through reactance important case
• Line transfer
• Power through transformer
• Power from generator
Xline
Xeq
Xd
18 EIEN15 Electric Power Systems L5
Power through series Z - line transfer 1
19 EIEN15 Electric Power Systems L5
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
20 EIEN15 Electric Power Systems L5
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
21 EIEN15 Electric Power Systems L5
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
22 EIEN15 Electric Power Systems L5
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
23 EIEN15 Electric Power Systems L5
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!
24 EIEN15 Electric Power Systems L5
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
25 EIEN15 Electric Power Systems L5
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
29 EIEN15 Electric Power Systems L5
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
30 EIEN15 Electric Power Systems L5
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
−
∆+∆+
=
−−
=
−−
=
∆∆
31 EIEN15 Electric Power Systems L5
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?
32 EIEN15 Electric Power Systems L5
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
33 EIEN15 Electric Power Systems L5
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
34 EIEN15 Electric Power Systems L5
−
=
−
−
−
−
−
−
−
−
−
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 …
35 EIEN15 Electric Power Systems L5
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
36 EIEN15 Electric Power Systems L5
Top Related