Bus Impedance Matrix in Faults Calculations

ELCT 751 Matrix Methods for Short Circuits 1

Lecture 8: Matrix Methods for Short-Circuit Studies

Bus Impedance MatrixNetwork Changes

Short-Circuit CalculationsConsideration of Large

Systems


Introduction For systems with more than a few

generators, use matrix methods forshort-circuit studiesNodal admittance matrix is easy to

calculateNodal impedance matrix is more

difficult to calculate but gives short-circuit currents easily


Example-j 4.0

-j 5.0 -j 3.0

-j 0.5

-j 0.1I1

I2

Admittances in per unit

1 2

3


ExampleNodal analysis:

I1 = -j 9.0 V1 + j 4.0 V2 + j 5.0 V3I2 = j 4.0 V1 j 7.1 V2 + j 3.0 V30 = j 5.0 V1 + j 3.0 V2 j 8.5 V3or I = Ybus Vwhere Ybus is the nodal (bus)

admittance matrix


Bus Impedance MatrixSolve for V in terms of I:

Then V = Ybus-1 I = Zbus Iwhere Zbus is the nodal impedance matrix or bus impedance matrix

The bus impedance matrix can be computed by inverting the bus admittance matrix or by direct formation


For example, line/transf data:

4

3

3

2

0.00 - j10.00.00 + j0.1034

0.00 - j2.500.00 + j0.4023

0.00 - j4.000.00 + j0.2512

0.00 - j5.000.00 + j0.2011

Admittance [per unit]

Impedance [per unit]

Bus Numbers

Branch Number


and shunt admittances:

0.00 j0.604

0.00 j0.003

0.00 j0.802

0.00 j0.801

Admittance [per unit]

Bus Number


Ybus =

-j 10.6j 10.0j 0.00j 0.00

j 10.0-j 16.5j 2.50j 4.00

j 0.00j 2.50-j 8.30j 5.00

j 0.00j 4.00j 5.00-j 9.80


Zbus =

j 0.5762j 0.5108j 0.4035j 0.4143

j 0.5108j 0.5415j 0.4277j 0.4392

j 0.4035j 0.4277j 0.5117j 0.4357

j 0.4143j 0.4392j 0.4357j 0.5036


Short-Circuit Calculations with Zbus

Fault

Rest of the system

1

2

3

4


Short-Circuit Calculations with Zbus

Fault simulation

Rest of thesystem

1

23

4

1.0

0

1.0


Z43

Z33

Z23

Z13

0Z34Z32Z31V30

-If

0

V4

-1

V1=

Z44Z42Z41

Z24Z22Z21

Z14Z12Z11


If = 1/Z22

V1 = - Z12 /Z22 V1 = 1 - Z12 /Z22

Vi is voltage of node i to 0 node,Vi = 1+Vi is voltage of node i to neutral


Previous Example

Zbus =j

0.57620.51080.40350.4143

0.51080.54150.42770.4392

0.40350.42770.51170.4357

0.41430.43920.43570.5036


ExampleFault at bus k

Current into fault (at k):

I k= 1/Zkk

Voltage at bus j:

Vj = 1 - Zjk /Zkk


0.11350.0

0.1643

0.1278

V3

0.00.0566

0.2116

0.1772

V4

0.29980.2809-j 1.73540.21020.1889-j 1.8473

0.00.1487-j 1.9542

0.13490.0-j 1.9861

V2V1Ikk

Fault at bus k is three-phase short circuit


Adding a Branch to Zbus To modify an existing Zbus to add a

new branch. Four cases:1) Radial branch connecting a new bus

to the reference node2) Radial branch connecting a new bus

to an existing system bus3) Branch from an existing system bus

to the reference node4) Branch between two existing buses


Case 1 is straightforwardAdd a new axis (row and column

n+1) to the Z matrix for the new bus having the branch impedance z in the diagonal and zeros off-diagonal

n+1n

1...n+11 ... n

Znew =z0 ... 0

0...0

Z


Case 2 a new branch from old bus p to new bus q (= n+1)A current Iq injected into q has the

same effect on the existing system as if it were injected at p

Copy axis p to a new axis q andZqq = z + Zpp


n+1

n1...

n+11 ... n

Znew =Zpp+ zZp1 ... Zpn

Z1p...Znp

Z

where 1 p n


Case 3 a new branch from old bus p to the reference node (loop-closing)Add a new axis by the algorithm of

Case 2, creating a fictitious new busShort the new bus to reference (it now

has zero voltage)Use Kron reduction to eliminate the

new axis of the matrix


Case 4 a new branch z between existing buses p and q (loop closing)Create a new axis (row and column)

to represent the new loop createdDiagonal:

Zn+1,n+1 = Zpp+Zqq-Zpq-Zqp+z

Off diagonal: Zn+1, j = Zpj-ZqjZi, n+1 = Zip-Ziq


Case 4 (continued):Use Kron reduction to eliminate the

new axis representing the loopZijnew = Zij (Zi, n+1 Zn+1,j )/ Z n+1,n+1

Notice that this will change almost every element in the existing part of the matrix


Zbus Building Method The Zbus modification outline

above can also be used to create Zbus, by starting with any Case 1 branch to get the 1 1 matrix for the starting point:Z = z

Then proceed to add other branches in any convenient order


Removing a Branch

Any branch (of impedance z) may be removed during a study by adding a new branch in parallel with impedance of zZeq = -z2/(z-z) = -z/0 an open

circuit


Axis Discarding Method Large-scale systems have full Zbus

matrices that can be difficult to store and use Instead we can store the sparse LU

factors (as discussed previously)Or we can discard unneeded axes by

simply striking them off (if their current injections are zero)


Axis Discarding

If I1= 0 and we dont need V1 then:V1 = Z11 I1 + Z12 I2 + Z13 I3V2 = Z21 I1 + Z22 I2 + Z23 I3V3 = Z31 I1 + Z32 I2 + Z33 I3

becomes:V2 = Z22 I2 + Z23 I3V3 = Z32 I2 + Z33 I3


Axis discarding for short-circuit calculation: Divide system into study system and

external systemStart building Z matrix in external

systemAfter all branches are added to external

node, its axis is discardedAll study system nodes are kept

Axis Discarding

ELCT 751 Matrix Methods for Short Circuits 290.0379076

0.2000084

0.0599083

0.0414073

0.0820053

0.0420043

0.1160052

0.1737071

0.1763031

0.1983021

0.0185060

0.0175010

XRji

Branch Z [pu]Bus Numbers

Example: Set up Z matrix, discarding all axes except 1, 2 and 5


Solution: Add 0-6: Z = 6j 0.0185

Add 6-7: Z = 7j 0.0564j 0.0185

j 0.0185 6j 0.0185

Discard 6: Z = 7j 0.0564

Add 7-3: Z = 3j 0.0978j 0.0564

j 0.0564 7j 0.0564


Add 7-1:

Z = j 0.0564j 0.0978j 0.0564

3j 0.0564j 0.05641j 0.2301j 0.0564

j 0.0564 7j 0.0564

Discard 7: Z = 1j 0.2301j 0.0564

j 0.0564 3j 0.0978


Add 3-1:Z =

-j 0.1737j 0. 2301j 0.0564

1-j 0.1737j 0.0564Lj 0.3914j 0.0414

j 0.0414 3j 0.0978

Kron reduction on L:

1j 0.15301j 0.07477j 0.07477 3j 0.09342

Continue this process, adding 3-4, 3-5, 3-8

Z =


0.093420.093420.135420.07477

40.093420.093420.07477

0.093420.17542

0.07477

50.093420.0747780.153320.07477

0.07477 10.15301

Discard 3:

Z= j


Add 4-8 creating fictitious loop axis L. Note that 4 and 8 are to be immediately discarded, so there is no need to compute those elements in the Kron reduction

Result: Z = j50.175420.07477

0.07477 10.15301


Add 1-0 perform Kron Reduction. Then add 2-5 and 1-2 and perform another Kron Reduction:

0.068560.102800.01004

50.068560.0100420.120900.01208

0.01208 10.01556Z = j


0.068560.102800.01004

50.068560.0100420.120900.01208

0.01208 10.01556Z = j

Fault at bus 2:If = 1/j0.1209 = -j 8.27 puV5 = 1 - j0.06856/j0.1209 = 0.433 puI52 = (V5-V2)/z52 = (0.433-0.00)/j0.116

= -j 3.73 pu


Bus-Tie Circuit Breakers A normally-open circuit breaker may

connect two buses when closedStudy might require closing and/or

opening of bus-tie circuit breakerNeed to represent a zero-impedance

branch that can be openedAdd z in series with z between buses


Bus-Tie Circuit Breakers1

2

NC

NC

NO

0 2

1Zg1

Zg2

z-z N+1

One-linediagram

Circuitrepresentation

Fictitious new bus

Bus Impedance Matrix in Faults Calculations

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