This preview shows page 1. Sign up to view the full content.
Unformatted text preview: power flow  evaluate normal operating conditions
fault analysis  evaluate abnormal operating conditions threephase singleline to ground and doubleline to ground
linetoline faults unbalanced faults balanced faults specifying ratings for circuit breakers and fuses
protective relay settings
specifying the impedance of transformers and generators Results used for: F a u l t ty p e s : Analysis types Power Systems I Fault Analysis the impedance of the network
the internal impedances of the generators
the resistance of the fault (arc resistance) transmission line impedances
transformer connections and impedances
grounding connections and resistances subtransient period, lasting for the first few cycles
transient period, covering a relatively longer time
steady state period Generator behavior is divided into three periods Network impedances are governed by Magnitude of fault currents depend on: Power Systems I Fault Analysis determine the interrupting capacity of HV circuit breakers
determine the operation timing of the protective relay system for
highvoltage networks determine the interrupting capacity of MV circuit breakers
determine the operation timing of the protective relay system for
mediumvoltage networks
transient stability studies Transient period, XG = Xd’ Subtransient period, XG = Xd" Power Systems I Fault Studies equivalent to the addition of an impedance at the place of the
fault
if the fault impedance is zero, the fault is referred to as a bolted
fault or solid fault the faulted network can be solved conveniently by Thévenin’s
method
network resistances are neglected
generators are modeled as an emf behind the subtransient or
transient reactance
shunt capacitances are neglected
system is considered as having noload First order method A fault represents a structural network change Power Systems I Fault Representation 1
3 2 1 3 j 0.4 j0.2 j0.8 2
j0.4 j0.4 The fault is simulated by switching a fault impedance at
th e f a u l t e d b u s
The change in the network voltages is equivalent to
adding the prefault bus voltage with all other sources
short curcuited Power Systems I Thévenin’s Method 1
j 0.4 j0.2 3 j0.8 Xf = 0.16 2
j0.4 j0.4 = 1 3phase fault with Zf = j0.16 on bus 3 Power Systems I Thévenin’s Method If j 0.4 j0.2 3 j0.8 + V th 2 Xf = 0.16 j0.4 j0.4 1 3 j 0.2 If j0.1 j0.2 j0.4 Xf = 0.16 2 V th Power Systems I j0.2 V3[ 0]
=
Z 33 + Z f j 0.4 + j 0.6
Z 33 = j 0.34 Z 33 ( j 0.4)( j 0.6) + j 0.1
= Z 1s = Z 2 s = ( j 0.4)( j 0.8) =
( j1.6)
( j 0.4)( j 0.4) = j 0.1
Z 3s =
( j1.6) V1[ 0 ] = V2[ 0] = V3[ 0] = 1.0 [
I3 f ] Thévenin’s Method j 0 .2 If 3
j0.16 V th j0.1 j0.24 j0.16 V th Power Systems I If 3 Z33 = j0.34 I [f]
3 V3[ 0]
1.0
=
=
= − j 2. 0
Z 33 + Z f
j 0.34 + j 0.16 Z 33 = j 0.34 Thévenin’s Method use the prefault bus voltages which can be obtained from the
results of a power flow solution
include loads  to preserve linearity, convert loads to constant
impedance model
Thevenin’s theorem allows the changes in the bus voltages to be
obtained
bus voltages are obtained by superposition of the prefault
voltages and the changes in the bus voltages
current in each branch can be solved For more accurate solutions Power Systems I Thévenin’s Method in per unit: SB
X kk Vk[ pre−f ]
=
X kk SCC = I [f ]
k SCC = 3 VL − L ,k I k Measures the electrical strength of the bus
Stated in MVA
Determines the dimension of bus bars and the
interrupting capacity of circuit breakers
Definition:
[ pre − f ] [ f ] Power Systems I Short Circuit Capacity (SCC) 1 3 j 0.4 j0.2 S base 100 MVA
SCC3 =
=
= 294 MVA
Z 33
0.34 S base = 100 MVA Z 33 = j 0.34 Find the SCC for bus #3 Power Systems I Short Circuit Capacity (SCC) j0.8 2
j0.4 j0.4 difficult to apply to large networks operating under
balanced conditions
each generator
represented by a
constant emf behind
a proper reactance
(Xd, Xd+ , or Xd +
)
lines represented by
their equivalent π model
G j i Si Sk k Zf seek a matrix where the diagonal elements represent the source
impedance for the buses
consider the following system Matrix algebra formation Network reduction by Thévenin’s method is not efficient Power Systems I Fault Analysis Using Impedance Matrix Z i −load = The change in the network
voltage caused by the fault
is equivalent to placing a
fault voltage at the faulted
bus with all the other
sources shortcircuited Replace the loads by a
constant impedance
model using the prefault
bus voltages
j Si*−load Vi [pre − f] 2 Place the prefault voltages into a vector Power Systems I i Zi [pre
Vbus −f] Zk k Zf Vk ȺV1[pre −f] Ⱥ
Ⱥ
Ⱥ
Ⱥ
Ⱥ
= ȺVk[pre −f] Ⱥ
Ⱥ
Ⱥ
Ⱥ Ⱥ
ȺV [pre −f] Ⱥ
Ⱥ n
Ⱥ Fault Analysis Using Impedance Matrix [
I kf ] The change in bus voltages
can be calculated from the
network matrix ΔVbus Ⱥ 0 Ⱥ
Ⱥ Ⱥ
Ⱥ
Ⱥ
[Fault]
[Fault]
I bus = Ⱥ− I k Ⱥ
Ⱥ
Ⱥ
Ⱥ Ⱥ
Ⱥ 0 Ⱥ
Ⱥ
Ⱥ I[Fault] = Ybus ΔVbus
bus I bus = Ybus Vbus [f
[ pre
Vbus] = Vbus −f ] + ΔVbus Using superpositioning, the fault voltages
are calculated from the prefault voltages
by adding the change in bus voltages
due to the fault Power Systems I ȺΔV1
Ⱥ
Ⱥ
= ȺΔVk
Ⱥ
Ⱥ
ȺΔV
Ⱥ n Fault Analysis Using Impedance Matrix
Ⱥ
Ⱥ
Ⱥ
Ⱥ
Ⱥ
Ⱥ
Ⱥ
Ⱥ
y1k
ykk
ynk Power Systems I Fault]
ΔVbus = Z bus I[bus −1
Z bus = Ybus −1
ΔVbus = Ybus I[Fault]
bus Ⱥ 0 Ⱥ Ⱥ y11
Ⱥ Ⱥ Ⱥ
Ⱥ
Ⱥ Ⱥ
[f]
Ⱥ− I k Ⱥ = Ⱥ yk 1
Ⱥ
Ⱥ Ⱥ
Ⱥ Ⱥ Ⱥ
Ⱥ 0 Ⱥ Ⱥ yn1
Ⱥ
Ⱥ Ⱥ I[Fault] = Ybus ΔVbus
bus
y1n Ⱥ
Ⱥ
Ⱥ
ykn Ⱥ
Ⱥ
Ⱥ
ynn Ⱥ
Ⱥ
ȺΔV1
Ⱥ
Ⱥ
ȺΔVk
Ⱥ
Ⱥ
ȺΔV
Ⱥ n
Ⱥ
Ⱥ
Ⱥ
Ⱥ
Ⱥ
Ⱥ
Ⱥ
Ⱥ Fault Analysis Using Impedance Matrix Power Systems I [
I kf ] Vk[ pre−f ]
=
Z kk + Z f Ⱥ 0 Ⱥ
Ⱥ Ⱥ
Ⱥ
Ⱥ
[f]
Ⱥ− I k Ⱥ
Ⱥ
Ⱥ
Ⱥ
Ⱥ
Ⱥ 0 Ⱥ
Ⱥ
Ⱥ [
& Vk[ f ] = Z f I kf ] Ⱥ Ⱥ z11 z1k z1n Ⱥ
Ⱥ Ⱥ
Ⱥ
Ⱥ Ⱥ
Ⱥ
Ⱥ = Ⱥ z k1 z kk z kn Ⱥ
Ⱥ Ⱥ
Ⱥ
Ⱥ
Ⱥ Ⱥ
Ⱥ Ⱥ z n1 z nk z nn Ⱥ
Ⱥ
Ⱥ Ⱥ
[f
[ pre
Vbus] = Vbus −f ] + ΔVbus
[f
[ pre
f]
Vbus] = Vbus −f ] + Z bus I[bus [
Vk[ f ] = Vk[ pre−f ] + Z kk I kf ] ȺΔV1
Ⱥ
Ⱥ
ȺΔVk
Ⱥ
Ⱥ
ȺΔV
Ⱥ n Fault Analysis Using Impedance Matrix 1 bus 3 j 0.4 j0.2 3 j0.8 Xf = 0.16 2
j0.4 j0.4 3  p h a s e fa u l t w i t h Z f = j 0 . 1 6 o n Power Systems I Example 1 If j 0.4 j0.2 3 j0.8 V th 2 Xf = 0.16 j0.4 j0.4 j 0.16 j 0.08
j 0.24 j 0.12Ⱥ
j 0.16Ⱥ
Ⱥ
j 0.34Ⱥ
Ⱥ j 2.50 Ⱥ
− j 6.25 j 2.50 Ⱥ
Ⱥ
j 2.50 − j 5.00Ⱥ
Ⱥ j1.25 V3[ pre − f ]
1.0 pu
=
=
= − j 2.0 pu
Z 33 + Z f
j 0.34 + j 0.16 Power Systems I I [f]
3 Ⱥ− j8.75
Ybus = Ⱥ j1.25
Ⱥ
Ⱥ j 2.50
Ⱥ
Ⱥ j 0.16
Z bus = Ⱥ j 0.08
Ⱥ
Ⱥ j 0.12
Ⱥ Example Power Systems I [
V3[ f ] = V3[ pre − f ] − Z 33 I 3 f ] = 1.0 pu − ( j 0.34 )(− j 2.0 ) = 0.32 pu [
V2[ f ] = V2[ pre − f ] − Z 23 I 3 f ] = 1.0 pu − ( j 0.16 )(− j 2.0 ) = 0.68 pu [
V1[ f ] = V1[ pre − f ] − Z13 I 3 f ] = 1.0 pu − ( j 0.12 )(− j 2.0 ) = 0.76 pu Example ...
View Full
Document
 Fall '11
 THOMASBALDWIN

Click to edit the document details