Calculating the short circuit current and input impedance ...

1Lecture 23 Power Engineering - Egill Benedikt Hreinsson

Calculating the short circuit current and input impedance of each node in

the power system (by matrix methods)


System Equations Summary

1

The conventional form of the load flow equations

−

= ⋅= ⋅

=

bus bus bus

bus bus bus

bus bus

I Y VV Z I

Z Y

• The power system is defined by the following matrix equations:


System Equations Summary (2)

1

2

3

n

VVV

V

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

busV

• Vbus is the vector of complex voltage phasors on each bus.

• The elements of Vbusare complex numbers


System Equations Summary (2)

1

2

3

n

III

I

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

busI

• Ibus is the vector of injected currents into each bus.

• The elements of Ibus are also complex numbers

• In normal operation, each injection can be – from a generator (+) – or into a load (-)

• In abnormal circumstances and operation an injection can feed into a short circuit


A Power System Model for a Short Circuit Study

111 12 1

221 22 1

3 3

1 2

0' ' '

0' ' '

' ' '0

n

n

n n nnn

Vz z z

Vz z z

V I

z z zV

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥= ⋅⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

• Let us review a previous equation describing the system.

• There is 1 “injection” of current at bus #3 .

• This is now the fault current but it could as well be a current from a generator

• The resulting bus voltage changes are on the left hand side

• Multiply the 3rd row by the column vector and we get.

• The z’33 element in the modified Zbusmatrix is the input impedance at node #3

3 31 32 33 3

34 3

' 0 ' 0 '' 0 ... ' 0n

V z z z Iz z= ⋅ + ⋅ + ⋅

+ ⋅ + + ⋅

3 33 3'V z I= ⋅

3 33

33 ,3' in

V VI

z z= =


We get 2 equivalent circuits:

I3,fault

V3,0ΔV1

ΔV2I3,fault

V3,0

z’33= zin,3

The power system

By the previous matrix calculations the power system has been reduced to a single Theveninequivalent circuit

fZ

fZ


Calculation of fault currents• The general equation for calculating the

fault current from the previous simple circuit is therefore:

• The most important element is the diagonal element of the modified impedance matrix, z´33

• If we have a solid fault zf = 0 and the fault current takes its maximum or:– V3,0 is the pre-fault voltage at node 3– I3,fault is the short circuit current– z´33 is the input impedance at node 3

3,03,

33'faultf

VI

z z=

+

3,03,

33'fault

VI

z=

I3,faultV3,0

fZz’33


A Short Circuit Calculation Algorithm (A review)

1. Define the power system and its operational conditions2. Calculate the pre-fault power flow and calculate pre-

fault voltages and currents3. Calculate the internal impedance (input impedance) and

pre-fault voltage (Thevenin equivalent circuit)4. Calculate the fault currents (i.e. system wide changes of

currents)5. Calculate the post fault currents as the sum of the pre-

fault currents and the fault currents (by the superposition principle)


Fault Current Calculations (A review)

• Pre-fault currents I0 (load currents) are usually in phase or near in phase with the voltage

• The fault currents If are primarily inductive, i.e. out of phase with the voltage.

• Short circuit currents >> load currents• Power system components are reactive• Load components are resistive• Pre-fault load currents can often be

ignored

V

I0

If

Total currentI0+If


Zbus building algorithm

Building the matrix “from scratch” in many steps adding one element at a time


Building the Zbus matrix directly

• It is often difficult to invert the Ybus matrix by traditional matrix inversion methods

• The Ybus matrix is symmetric and usually large and sparse

• The Zbus matrix is symmetric and full (and usually large)

• We often need to change the configuration of the underlying power system, (for instance in contingencies where a single link is added/deleted))

– We need to remove a line or add a transmission line or a link• A direct step-by-step matrix building

method is advantageous.


Zbus building algorithm• The Zbus matrix can be built

step by step be a direct method rather than calculating Ybus

-1

• Consider a general n-bus power system

• We add one link at a time• In each step we transform

the matrix as follows

12

34

k

n

JThe earth is a special node = “reference node”

, ,→bus old bus newZ Z


Zbus building algorithm

• As an example take a 3-bus system with 5 impedances (links)

• We build the full matrix in 5 steps and add 1 impedance at a time

• The Zbus matrix is defined as follows

z1 z2z3

z4 z5

1 2

3

= ⋅bus bus busV Z I1 11 1 12 2 13 3v z i z i z i= + +

2 21 1 22 2 23 3v z i z i z i= + +

3 31 1 32 2 33 3v z i z i z i= + += -1bus busZ Y

1

23

J


1st step

5 steps in Zbus building algorithm

• Zbus is going to be a 3x3 matrix built in 5 steps:

12

3

J

2nd step

12

3

J

3rd step

12 3

J

4th step 5th step

12 3

J

2 3

J

1x1 2x2

3x3 3x33x3


4 different types of steps1. Add a new node and from it a new

branch to ground2. Add a new node and from it a new

branch to an existing node3. Add a new branch between 2

existing nodes4. Add a new branch from an

existing node to ground


Numerical network example in building Zbus

z3 z4z5

z1 z2

1 3

22

3

J

1

1z 2z

4z3z

5z

• Consider also the following numerical example

• Assume we have 5 impedances and 3 buses

• Again, we build the matrix in 5 steps

• The branches in the numerical example have impedance values:

1 0.15z j= 2 0.075z j= 3 0.1z j= 4 0.1z j= 5 0.1z j=


1st step

J

1x1

The 1st step in the Zbus building algorithm• We add branch and inject a

current at any node #1 and get a voltage there

1z1 11 1 12 2 13 3v z i z i z i= + +

2 21 1 22 2 23 3v z i z i z i= + +

3 31 1 32 2 33 3v z i z i z i= + +

1i 1v 2v3v

• By comparing the above equations we see that the Zbus matrix after this step is a 1x1 (single element):

[ ]1z=busZ

1 1 1v z i=


1st step

J

1x1

The 1st step in the Zbus building algorithm

1z

1i 1v 2v3v[ ] [ ]

1 2 3

4 5

1

0.15 0.075 0.10.1 0.1

0.15

In our numerical example: , , ,

and Therefore

z j z j z jz j z j

z j

= = == =

= =busZ


2nd step

J 2x2

The 2nd step in the Zbus building algorithm

• We now add a branch to the reference node from a node (#2) not previously connected

1z1 1 1 20v z i i= + ⋅

2 1 2 20v i z i= ⋅ +

1i 1v 2v3v

• The Zbus will now be a 2x2:

2

00 z

⎡ ⎤= ⎢ ⎥⎣ ⎦

bus(old)bus(new)

ZZ

2z

[ ]1z=bus(old)Z

1 1 1

2 2 2

00

v z iv z i⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

busV

2i


The 2nd step in the Zbus building algorithm (2)

• Assume Zbus,old is a matrix of dimension (k x k)

• We want to add a node (# k +1) that has not previously been added to our set of nodes.

• From this node we add the new impedance, znew to the reference node

• We transform Zbus,old to Zbus,new by adding a row and a column to the Zbus,old (# k+1)

• Add zeros to all elements of the new row and column except the new diagonal element which is the new impedance, znew

• The dimensions of the matrix are increased by 1 and are now (k+1) times (k+1)

,

0

00 0 newz

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

bus newZ ,sbu oldZ

column # k+1

row # k+1

Dimension of Zbus(old) = k


2nd step

J 2x2

The 2nd step in the Zbus building algorithm

• In our numerical example we get in step #2:

1z

1i 1v 2v3v

2z

[ ] [ ]

1 1 1

2 2 2

1

2

00

0.15

0 0.15 00 0 0.075

v z iv z i

z j

jz j

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦= =

⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦

bus

bus(old)

bus(old)bus(new)

V

Z

ZZ

2i


3rd step

J 2x2

The 3rd step in the Zbus building algorithm• We now add a branch to a

previously connected node(#2) from a new node (#3) , not previously connected

• We have the following equations:

1z

3 2 4 3v v z i= +

1i 1v 2v3v

12

22

12 22 2 4

zz

z z z z

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥+⎣ ⎦

bus,newZ

2z11 1

2 32 2

00

iv zi iv z⎡ ⎤⎡ ⎤ ⎡ ⎤

= ⎢ ⎥⎢ ⎥ ⎢ ⎥ +⎣ ⎦ ⎣ ⎦ ⎣ ⎦

1 1 1

2 2 2 2

3 2 2 4 3

0 000

v z iv z z iv z z z i

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥+⎣ ⎦⎣ ⎦ ⎣ ⎦

bus,oldZ

2i3i4z

23Lecture 23 Power Engineering - Egill Benedikt HreinssonThe 3rd step in the Zbus building algorithm (2)

• Assume Zbus,old is a matrix of dimension (k x k)

• We want to add a node (# k+1) that has notpreviously been added to our set of nodes.

• From this node we add the new impedance, znew to a previously connected node (# j)

• We transform Zbus,old to Zbus,new by adding a row (number k+1) and a column (number k+1) to the Zbus,old

• Duplicate row/column #j in Zbus,old as the new row/column. The new diagonal element is found by adding the new impedance znew to the diagonal element zjj of Zbus,old

• The dimensions of the matrix are increased by 1

1

2

1 2

j

j

kj

j j jk jj new

zz

zz z z z z

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥+⎣ ⎦

bus,newZ bus,oldZ

column # k+1

row # k+1

Dimension of Zbus(old) = k


3rd step

J 2x2

The 3rd step in the Zbus building algorithm

1z

1i 1v 2v3v

2z

1 2 3

4 5

1 1 1

2 2 2 2

3 2 2 4 3

1

2

3

0.15 0.075 0.10.1 0.1

0 000

0.15 0 00 0.075 0.0750 0.075 0.175

In our numerical example: , , ,

and

Therefore:

z j z j z jz j z j

v z iv z z iv z z z i

j ij j ij j i

= = == =

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥

+⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

2i3i4z


The 3rd step in the Zbus building algorithm

0.15 0 00 0.075 0.0750 0.075 0.175

Therefore, in our numerical example, after the 3rd step:

jj jj j

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

busZ


4th step

J 2x2

The 4th step in the Zbus building algorithm (1)

• We now add a branch between 2 previously connected nodes (#1)and (#2)

• This new impedance, z3 = znew in general causes a circular current, iL to flow.

• The circular current is equivalent to an extra injected current of + iL at node #1 and of of -iLat node #2

• For the 3 x 3 matrix in the figure, this is equivalent to the following adjustments of the voltage/current equations

1z

1i 1v 2v3v

2z

2i

3i4zLi

1 11 12 13 1

2 21 22 23 2

3 31 32 33 3

L

L

v z z z i iv z z z i iv z z z i

+⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

newz

27Lecture 23 Power Engineering - Egill Benedikt HreinssonThe 4th step in the Zbus building algorithm (2)

• In general, assume Zbus,old is a matrix of dimension (k x k)

• Assume that a new element, znew is added between node #m and node #n in a system that already has knodes. (In the example, m = 1 and n = 2)

1,1 1,2 1, 1 1,1

2,1 2,2 2, 1 2,2

,1 ,2 , 1 ,

,1 ,2 , 1 ,

1,1 1,2 1, 1 1,1

,1 ,2 , 1 ,

k k

k k

m m m k m km

n n n k n kn

k k k k k kk

k k k k k kk

z z z zv iz z z zv

z z z zvz z z zv

z z z zvz z z zv

−

−

−

−

− − − − −−

−

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ = ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

1

2

1

m L

n L

k

k

i

i ii i

ii−

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

+⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

bus,oldZ

• We can expand the equations as shown on the next slide

• The equations assuming a circular current of magnitude iLare as follows:


1,1 1,2 1,1 1 1 1

2,1 2,2 2,2 2

,1 ,2 ,

k m n

kL

k k k kk k km kn

z z zv i z zz z zv i

i

z z zv i z z

−⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + ⋅⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦

• We rewrite the former equation• We define a new vector, a which

represents the difference between columns # m and #n in the Zbus,oldmatrix

• Therefore the former equation will be as follows:

• The voltage drop across the new branch, znew will be:

Li= ⋅ + ⋅bus bus,old busV Z I a

1 1m n

km kn

z z

z z

−⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥−⎣ ⎦

a

0m n new Lv v z i− + ⋅ =

n m new Lv v z i− = ⋅


[ ]1 2

1 2

0

, , , ( )

, ,

(1)We expand and in the expression: using lines #m and #n from the matrix equation in the last slide. This expansion gives:

and

m n m n new L

m m m mk mm mn L

n n n

v v v v z i

v z z z z z i

v z z

− + ⋅ =

= ⋅ + − ⋅

=busI

[ ]

[ ]1 1

, ( )

, , (2 ) 0

0 (2 )

(1)

(2)

We substitute these into the equation which leads to:

This is in compact notation: where

nk nm nn L

n m nk mk mn mm nn new L

mn mm nn new L

z z z i

z z z z z z z z i

z z z z i

⋅ + − ⋅

− − + − − − =

= + − − −

bus

bus

bus

I

I

bI

[ ]1 1 , , .(3)

(2) (3)

the k-

dimensional row matrix is befined by: =

The loop current can be solved from equation and substituted into eq. n m nk mkz z z z− −b b


1( 2 )

( 2 )

( 2 )

The result is:

which can be written:

or This leads to:

T

new mm nn mn

new mm nn mn

new mm nn mn

z z z z

z z z z

z z z z

= ++ + −

⎧ ⎫= +⎨ ⎬+ + −⎩ ⎭

= ⋅

= ++ + −

bus bus,old bus

bus bus,old bus

bus bus,new bus

bus,new bus,old

V Z abI

abV Z I

V Z I

abZ Z

his formula shows how to calculate a new version of the matrixbusZ


Example: Zbus building

3

11 22

12

0.15 0 00.10 0 0.075 0.075

0 0.075 0.175

0.15 0.0752 2 0

In our numerical example: and

is the old -bus matrix from the 3rd step.Therefore and and .

new

mm nn

mn

z z j j

Zz z j z z j

z z

⎡ ⎤⎢ ⎥= = = ⎢ ⎥⎢ ⎥⎣ ⎦

= = = == =

bus,oldZ

[ ] [ ]

2 0.32500.15 0 0.15

0 0.075 0.0750 0.075 0.075

0 0.15 0.075 0 0.075 0 0.15 0.075 0.075

Therefore

We get The matrix will be:

new mm nn mnz z z z j

j j

j j

+ + − =

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − = −⎢ ⎥ ⎢ ⎥

− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= − − − =

a b

b


Example: Zbus building (2)

[ ]0.150.075 0.15 0.075 0.075

( 2 ) 0.32500.075

0.0692 0.0346 0.03460.0346 0.0173 0.01730.0346 0.0173 0.0173

0.0808 0.03

By adding this matrix to we get

new mm nn mn

jz z z z

j

j

⎡ ⎤⎢ ⎥= −⎢ ⎥+ + −−⎢ ⎥⎣ ⎦

−⎡ ⎤⎢ ⎥= − −⎢ ⎥

− −⎢ ⎥⎣ ⎦

=

bus,old

bus,new

ab

Z

Z46 0.0346

0.0346 0.0577 0.05770.0346 0.0577 0.1577

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦


4th step

J 2x2


• We finally in step 5 add a branch (z5) between 2 previously connected nodes

• This connects nodes m = 1 and n = 3 and is the same kind of step as step #4

• The new impedance is z5 = znew

1z

1i 1v2v

3v2z

2i

3i4z3z

5z

5 11 33

13

0.1 0.0808 0.15772 2 0.0692

2 0.2693

and and and Therefore

new mm nn

mn

new mm nn mn

z z j z z j z z jz z j

z z z z j

= = = = = == =

+ + − =



[ ][ ]

0.0808 0.0346 0.04620.0346 0.0577 0.02310.0346 0.1577 0.1231

0.0346 0.0808 0.0577 0.0346 0.1577 0.0346

0.0462 0.0231 0.1231

We get The matrix will be:

the incremental matr

j j

j

j

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − = −⎢ ⎥ ⎢ ⎥

− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= − − −

= −

a b

b

b

[ ]0.04620.0231 0.0462 0.0231 0.1231

0.26930.1231

0.0079 0.0040 0.02110.0040 0.0020 0.01060.0211 0.0106 0.0563

ix will be

j

j

⎡ ⎤⎢ ⎥= −⎢ ⎥−⎢ ⎥⎣ ⎦

−⎡ ⎤⎢ ⎥= − −⎢ ⎥

− −⎢ ⎥⎣ ⎦



0.0729 0.0386 0.05570.0386 0.0557 0.04710.0557 0.0471 0.1014

By adding this matrix to we finally getthe final version of the matrix:

This complets our numerical example.

j⎡ ⎤⎢ ⎥= = ⎢ ⎥⎢ ⎥⎣ ⎦

bus,old

bus

bus,new bus

ZZ

Z Z

36Lecture 23 Power Engineering - Egill Benedikt HreinssonTransient synchronous machine short circuit currents

phase “a”

phase “b”

phase “c”

• Subtransient , xd’’ and

transient reactances, xd

’’ are smaller than the steady state synchronous reactances, xd

’’

• A DC component is present in all 3 phases

• The exact shape of the short circuit current will depend on the phase instant of the short circuit and is therefore different in the 3 phases

''dx '

dx dx

sub-transient-áhrif

transient-áhrif

DC compononent


Energy control centers


Energy Control Centers• Energy Control Center (ECC):

– SCADA, EMS, operational personnel– “Heart” (eyes & hands, brains) of the power system

• Supervisory control & data acquisition (SCADA):– Supervisory control: remote control of field devices– Data acquisition: monitoring of field conditions– SCADA components:

• Master Station: System “Nerve Center” located in ECC• Remote terminal units: Gathers data at substations; sends to Master Station• Communications: Links Master Station with Field Devices

• Energy management system (EMS)– Topology processor & network configurator– State estimator and power flow model development– Automatic generation control (AGC), Optimal power flow (OPF)– Security assessment and alarm processing


Early Power System Control (in 1919)


23-Nov-11

Energy Control Centers

Source Measured Success: Constructing PerformanceMetrics for Energy Managementby John Van Gorp, Power Measurement Corp.Í ritinu“Web BasedEnergy Information andControl Systems:Case Studies and Applications”

CRC Press 2005


23-Nov-11

Landsvirkjun’s Dispatch Centre

Landsvirkjun‘s Dispatch Centre in Reykjavík was commissioned in 1989. Its role is coordinating operation of the electricity system. Its chief task is to ensure conditions that allow the system to handle variable loads at all times, thereby safeguarding operational security and efficiency. It monitors the entire power system and controls both production of electricity and its transmission nation-wide.

In order to fullfill its role, the Dispatch Centre must have comprehensive hands-on data about the electricity system and therefore needs to be in constant, reliable contact with all its units. The Dispatch Centre is linked to power plants all over Iceland by means of microwave radio and optical fibre cables. These carry an average of 600 status point indications per minute from 35 remote terminals to its control computer, which gives real-time information about each and every part of the system. It sends warnings of any deviations to the two dispatchers who are on duty at any time, and with a complete overview of the electricity system they are able to respond accordingly and prescribe the correct action to be taken, via the remote control system.

Source: http://www.lv.is


SubstationRemote terminal unit

SCADA Master Station

Com

mun

icat

ion

link

Energy control center with EMS

EMS alarm displayEMS 1-line diagram


More energy control centers


The control of distributed generation

http://www.pserc.org/cgi-pserc/getbig/generalinf/presentati/presentati/Microgrid_Tutorial.pdf

Micro-Grid Operation and Control


References

• O.I. Elgerd, Electric Energy Systems Theory, An Introduction.McGrawHill, 1983

• Electric Energy Systems; Analysis and Operation, Ed by A Gomez Exposito, A J Conejo, C Canizares, CRC press, 2009

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