Lecture 23
Power Engineering - Egill Benedikt Hreinsson
Calculating the short circuit current
and input impedance of each node in
the power system (by matrix methods)
1
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
System Equations Summary
• The power system is defined by the
following matrix equations:
The conventional form of the load flow equations
I bus = Ybus ⋅ Vbus
Vbus = Z bus ⋅ I bus
Z bus = Ybus
−1
2
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
System Equations Summary (2)
• Vbus is the vector of
complex voltage
phasors on each bus.
• The elements of Vbus
are complex numbers
⎡V1 ⎤
⎢V ⎥
⎢ 2⎥
Vbus = ⎢V3 ⎥
⎢ ⎥
#
⎢ ⎥
⎢⎣Vn ⎥⎦
3
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
4
System Equations Summary (2)
• 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
I bus
⎡ I1 ⎤
⎢I ⎥
⎢ 2⎥
= ⎢ I3 ⎥
⎢ ⎥
#
⎢ ⎥
⎢⎣ I n ⎥⎦
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
A Power System Model for a Short Circuit Study
• 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 Zbus
matrix is the input impedance
at node #3
⎡V1 ⎤
⎡0⎤
⎢V ⎥ ⎡ z '11 z '12 " z '1n ⎤ ⎢ 0 ⎥
⎥ ⎢ ⎥
⎢ 2 ⎥ ⎢z '
z
z
'
'
"
22
1n ⎥
⎢V3 ⎥ = ⎢ 21
⋅ ⎢ I3 ⎥
# % # ⎥ ⎢ ⎥
⎢ ⎥ ⎢ #
⎥ ⎢#⎥
⎢ # ⎥ ⎢z '
z 'n 2 " z 'nn ⎦
⎢⎣Vn ⎥⎦ ⎣ n1
⎢⎣ 0 ⎥⎦
V3 = z '31 ⋅ 0 + z '32 ⋅ 0 + z '33 ⋅ I 3
+ z '34 ⋅ 0 + ... + z '3n ⋅ 0
V3 = z '33 ⋅ I 3
V3
V3
I3 =
=
z '33 zin ,3
5
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
6
We get 2 equivalent circuits:
I3,fault
ΔV1
V3,0
Zf
By the previous matrix
calculations the power
system has been
reduced to a single
Thevenin
equivalent
circuit
ΔV2
V3,0
z’33= zin,3
The power system
I3,fault
Zf
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
7
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
V3,0
I3,fault
z’33
Zf
I 3, fault =
V3,0
z f + z '33
I 3, fault =
V3,0
z '33
Lecture 23
Power Engineering - Egill Benedikt Hreinsson
A Short Circuit Calculation Algorithm (A review)
1. Define the power system and its operational conditions
2. Calculate the pre-fault power flow and calculate prefault voltages and currents
3. 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 prefault currents and the fault currents (by the
superposition principle)
8
Lecture 23
Power Engineering - Egill Benedikt Hreinsson
9
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 current
I0+If
Lecture 23
Power Engineering - Egill Benedikt Hreinsson
Zbus building algorithm
Building the matrix “from scratch” in many steps adding one element
at a time
10
Lecture 23
Power Engineering - Egill Benedikt Hreinsson
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.
11
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
12
Zbus building algorithm
• The Zbus matrix can be built
step by step be a direct
1
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
Z bus,old → Z bus,new
2
k
4
3
n
J
The earth is a special node = “reference node”
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
13
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
1
3
z1
z3
z2
1
2
z4
2
3
z5
v1 = z11i1 + z12i2 + z13i3
Vbus = Z bus ⋅ I bus
v2 = z21i1 + z22i2 + z23i3
Z bus = Ybus
v3 = z31i1 + z32i2 + z33i3
-1
J
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
14
5 steps in Zbus building algorithm
1st step
• Zbus is going to be a 3x3
matrix built in 5 steps:
3rd step
2
3
2
1
4th step
2
3
2nd step
3
J
3x3
J
J
1x1
2x2
5th step
2
J
J
3x3
3
1
1
1
2
3x3
3
Lecture 23
Power Engineering - Egill Benedikt Hreinsson
4 different types of steps
1. Add a new node and from it a new
branch to ground
2. Add a new node and from it a new
branch to an existing node
3. Add a new branch between 2
existing nodes
4. Add a new branch from an
existing node to ground
15
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
16
Numerical network example in building Zbus
• Consider also the following
z3
numerical example
2
• Assume we have 5
impedances and 3 buses 1
• Again, we build the matrix in
5 steps
• The branches in the
numerical example have z1
impedance values:
J
2
z4
z3
3
z5
z5
z2
z4
3
1
z1
z2
z1 = j 0.15 z2 = j 0.075 z3 = j 0.1 z4 = j 0.1 z5 = j 0.1
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
The 1st step in the Zbus building algorithm
• We add branch and inject a
current at any node #1 and
get a voltage there v = z i
1
11
1st step
i1
v1 = z11i1 + z12i2 + z13i3
v2 = z21i1 + z22i2 + z23i3
v3 = z31i1 + z32i2 + z33i3
• By comparing the above equations we
see that the Zbus matrix after this step is
a 1x1 (single element):
v2
v1
v3
z1
J
1x1
Z bus = [ z1 ]
17
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
The 1st step in the Zbus building algorithm
In our numerical example:
z1 = j 0.15 , z2 = j 0.075 , z3 = j 0.1 ,
z4 = j 0.1 and z5 = j 0.1
Therefore Z bus = [ z1 ] = [ j 0.15]
1st step
i1
v2
v1
z1
J
1x1
v3
18
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
19
The 2nd step in the Zbus building algorithm
• We now add a branch to the
reference node from a node
(#2) not previously
connected
v1 = z1i1 + 0 ⋅ i2
v2 = 0 ⋅ i1 + z2 i2
2nd step
i1
v1
Vbus
Z bus(old) = [ z1 ]
J
0 ⎤ ⎡ i1 ⎤
⎥
⎢
⎥
z2 ⎦ ⎣i2 ⎦
Z bus(new)
v3
z2
z1
• The Zbus will now be a 2x2:
⎡ v1 ⎤ ⎡ z1
=⎢ ⎥=⎢
⎣ v2 ⎦ ⎣ 0
i2 v2
2x2
⎡ Z bus(old)
=⎢
⎣ 0
0⎤
⎥
z2 ⎦
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
20
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)
Dimension of Zbus(old) = k
Z bus,new
row # k+1
column #
k+1
0 ⎤
⎡
⎢
⎥
#
Z
⎥
= ⎢ bus ,old
⎢
0 ⎥
⎢
⎥
⎣0 " 0 znew ⎦
Lecture 23
The 2nd step in the Zbus building
algorithm
Power Engineering - Egill Benedikt Hreinsson
• In our numerical example we get in
step #2:
2nd step
v1
i1
⎡ v1 ⎤ ⎡ z1 0 ⎤ ⎡ i1 ⎤
Vbus = ⎢ ⎥ = ⎢
⎥
⎢
⎥
⎣ v2 ⎦ ⎣ 0 z2 ⎦ ⎣i2 ⎦
Z bus(old) = [ z1 ] = [ j 0.15]
Z bus(new)
⎡ Z bus(old)
=⎢
⎣ 0
0 ⎤ ⎡ j 0.15
=⎢
⎥
z2 ⎦ ⎣ 0
i2 v2
z2
z1
J
⎤
⎥
j 0.075⎦
0
v3
2x2
21
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
22
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:
⎡ v1 ⎤ ⎡ z1
⎢v ⎥ = ⎢ 0
⎣ 2⎦ ⎣
v3 = v2 + z4i3
⎡ v1 ⎤ ⎡ z1
⎢v ⎥ = ⎢ 0
⎢ 2⎥ ⎢
⎢⎣ v3 ⎥⎦ ⎢⎣ 0
0
z2
z2
3rd step
i1
0 ⎤ ⎡ i1 ⎤
⎢
⎥
⎥
z2 ⎦ ⎣i2 + i3 ⎦
i2
v1
v2
z4
z2
z1
J
0 ⎤ ⎡ i1 ⎤
⎢
⎥
⎥
z2
i2
⎥⎢ ⎥
z2 + z4 ⎥⎦ ⎢⎣ i3 ⎥⎦
Zbus,new
v3
⎡Z
bus,old
⎢
=
⎢
⎢⎣ z12 z22
2x2
z12 ⎤
z22 ⎥
⎥
z2 + z4 ⎥⎦
i3
Lecture 23
The 3rd step in the Zbus building algorithm
(2)
Power Engineering - Egill Benedikt Hreinsson
• 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 a previously connected
node (# j)
Dimension of Zbus(old) = k
• 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
⎢
Z bus,old
found by adding the new impedance znew to Z
⎢
bus,new =
the diagonal element zjj of Zbus,old
⎢
⎢
• The dimensions of the matrix are increased by
⎢ z j1 z j 2 " z jk
row # k+1
1
⎣
23
column # k+1
⎤
z2 j ⎥
⎥
⎥
#
⎥
zkj ⎥
z jj + znew ⎥⎦
z1 j
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
24
The 3rd step in the Zbus building algorithm
In our numerical example:
z1 = j 0.15 , z2 = j 0.075 , z3 = j 0.1 ,
z4 = j 0.1 and z5 = j 0.1
0 ⎤ ⎡ i1 ⎤
⎡ v1 ⎤ ⎡ z1 0
Therefore: ⎢ v2 ⎥ = ⎢ 0 z2
z2 ⎥ ⎢i2 ⎥
⎢ ⎥ ⎢
⎥⎢ ⎥
⎢⎣ v3 ⎥⎦ ⎢⎣ 0 z2 z2 + z4 ⎥⎦ ⎢⎣ i3 ⎥⎦
0
0 ⎤ ⎡ i1 ⎤
⎡ j 0.15
=⎢ 0
j 0.075 j 0.075⎥ ⎢i2 ⎥
⎢
⎥⎢ ⎥
j 0.075 j 0.175⎥⎦ ⎢⎣ i3 ⎥⎦
⎢⎣ 0
3rd step
i1
i2
v1
v2
v3
z4
z2
z1
J
2x2
i3
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
The 3rd step in the Zbus building algorithm
Therefore, in our numerical example,
after the 3rd step:
Z bus
⎡ j 0.15
⎢
0
=
⎢
⎢⎣ 0
0
j 0.075
j 0.075
0 ⎤
⎥
j 0.075
⎥
j 0.175⎥⎦
25
Lecture 23
The 4th step in the Zbus building algorithm
(1)
Power Engineering - Egill Benedikt Hreinsson
• We now add a branch between 2
4th step
previously connected nodes (#1)
and (#2)
• This new impedance, z3 = znew in general causes
1
a circular current, iL to flow.
• The circular current is equivalent to an extra
injected current of + iL at node #1 and of of -iL
at node #2
• For the 3 x 3 matrix in the figure, this is
equivalent to the following adjustments of the
voltage/current equations
i
⎡ v1 ⎤ ⎡ z11
⎢v ⎥ = ⎢ z
⎢ 2 ⎥ ⎢ 21
⎢⎣ v3 ⎥⎦ ⎢⎣ z31
z12
z22
z32
z13 ⎤ ⎡ i1 + iL ⎤
z23 ⎥⎥ ⎢⎢i2 − iL ⎥⎥
z33 ⎥⎦ ⎢⎣ i3 ⎥⎦
i2
znew v2 v
3
v1
i
z
3
4
i
L
z2
z1
J
2x2
26
Lecture 23
The 4th step in the Zbus building algorithm
(2)
Power Engineering - Egill Benedikt Hreinsson
• 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 k
nodes. (In the example, m = 1 and n = 2)
⎡ v1 ⎤ ⎡ z1,1
• The equations ⎢ v ⎥ ⎢ z2,1
2 ⎥
⎢
⎢
assuming a
⎢ # ⎥ ⎢ #
circular
⎢
⎥ ⎢
current of
⎢ vm ⎥ = ⎢ zm ,1
magnitude iL ⎢ v ⎥ ⎢ z
n ,1
n
⎥ ⎢
are as follows: ⎢
⎢ # ⎥ ⎢ #
⎢v ⎥ ⎢ z
⎢ k −1 ⎥ ⎢ k −1,1
⎢⎣ vk ⎥⎦ ⎢⎣ zk ,1
z1,2
" "
• We can expand the
equations as
shown on the next
slide
"
z2,2
z1, k −1
z2, k −1
#
%
zm , k −1
zn , k −1
#
zk −1,2
zk ,2 " "
zk −1, k −1
" zk , k −1
zm ,2
zn ,2
Z bus,old
%
z1, k ⎤ ⎡ i1 ⎤
z2, k ⎥⎥ ⎢ i2 ⎥
⎢
⎥
# ⎥⎢ # ⎥
⎥⎢
⎥
zm , k ⎥ ⎢im + iL ⎥
zn , k ⎥ ⎢ in − iL ⎥
⎥⎢
⎥
# ⎥⎢ # ⎥
zk −1, k ⎥ ⎢ ik −1 ⎥
⎥⎢
⎥
zk , k ⎥⎦ ⎢⎣ ik ⎥⎦
27
Lecture 23
The 4th step in the Zbus building algorithm
(3)
• We rewrite the former equation
• We define a new vector, a which
represents the difference between
columns # m and #n in the Zbus,old
matrix
• Therefore the former equation will be as
follows:
• The voltage drop across the new
branch, znew will be:
Power Engineering - Egill Benedikt Hreinsson
⎡ v1 ⎤ ⎡ z1,1
⎢v ⎥ ⎢ z
⎢ 2 ⎥ = ⎢ 2,1
⎢#⎥ ⎢ #
⎢ ⎥ ⎢
⎣ vk ⎦ ⎣ zk ,1
z1,2 " z1,k ⎤ ⎡ i1 ⎤ ⎡ z1m − z1n ⎤
⎥
z2,2
z2,k ⎥ ⎢i2 ⎥ ⎢ #
⎥⎢ ⎥+⎢
⎥ ⋅ iL
% # ⎥⎢#⎥ ⎢ #
⎥
⎥⎢ ⎥ ⎢
⎥
zk ,2 " zk ,k ⎦ ⎣ik ⎦ ⎣ zkm − zkn ⎦
Vbus = Z bus,old ⋅ I bus + a ⋅ iL
vn − vm = znew ⋅ iL
vm − vn + znew ⋅ iL = 0
⎡ z1m − z1n ⎤
⎢ #
⎥
⎥
a=⎢
⎢ #
⎥
⎢
⎥
−
z
z
⎣ km kn ⎦
28
Lecture 23
The 4th step in the Zbus building algorithm
(4)
Power Engineering - Egill Benedikt Hreinsson
We expand vm and vn in the expression: vm − vn + znew ⋅ iL = 0
29
(1)
using lines #m and #n from the matrix equation in the last slide.
This expansion gives:
vm = [ zm1 , zm 2
vn = [ z n 1 , z n 2
, " , zmk ] ⋅ I bus + ( zmm − zmn ) ⋅ iL and
, " , znk ] ⋅ I bus + ( znm − znn ) ⋅ iL
We substitute these into the equation (1)
which leads to: [ zn1 − zm1 , " , znk − zmk ] I bus + (2 zmn − zmm − znn − znew )iL = 0
This is in compact notation: 0 = bI bus + (2 zmn − zmm − znn − znew )iL
(2)
where the k-
dimensional row matrix b is befined by: b= [ zn1 − zm1 , " , znk − zmk ]
(3)
The loop current can be solved from equation (2) and substituted into eq. (3)
.
Lecture 23
The 4th step in the Zbus building algorithm
(5)
Power Engineering - Egill Benedikt Hreinsson
The result is: Vbus = Z bus,old +
which can be written: Vbus
( znew + zmm
1
abI bus
+ znn − 2 zmn )
⎧
⎫
ab
= ⎨ Z bus,old +
⎬ I bus
( znew + zmm + znn − 2 zmn ) ⎭
⎩
or Vbus = Z bus,new ⋅ I bus
This leads to:
Z bus,new = Z bus,old +
( znew
30
ab
+ zmm + znn − 2 zmn )
This formula shows how to calculate a new version of the Z bus matrix
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
31
Example: Zbus building
In our numerical example: znew = z3 = j 0.10 and Z bus,old
0
0 ⎤
⎡0.15
= j⎢ 0
0.075 0.075⎥
⎢
⎥
0.075 0.175⎥⎦
⎢⎣ 0
is the old Z -bus matrix from the 3rd step.
Therefore zmm = z11 = j 0.15 and znn = z22 = j 0.075
and 2 zmn = 2 z12 = 0 . Therefore znew + zmm + znn − 2 zmn = j 0.3250
⎡ 0.15 − 0 ⎤
We get a = j ⎢0 − 0.075⎥ =
⎢
⎥
⎢⎣0 − 0.075⎥⎦
b = j [ 0 − 0.15 0.075 − 0
⎡ 0.15 ⎤
j ⎢ −0.075⎥ The b matrix will be:
⎢
⎥
⎢⎣ −0.075⎥⎦
0.075 − 0] = j [0.15 0.075 0.075]
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
Example: Zbus building (2)
⎡ 0.15 ⎤
ab
j ⎢
=
−0.075⎥ [ 0.15 0.075 0.075]
⎥
( znew + zmm + znn − 2 zmn ) 0.3250 ⎢
⎢⎣ −0.075⎥⎦
⎡ −0.0692 0.0346 0.0346 ⎤
= j ⎢ 0.0346 −0.0173 −0.0173⎥
⎢
⎥
⎢⎣ 0.0346 −0.0173 −0.0173⎥⎦
By adding this matrix to Z bus,old we get
Z bus,new
⎡ 0.0808 0.0346 0.0346⎤
= j ⎢0.0346 0.0577 0.0577 ⎥
⎢
⎥
⎢⎣0.0346 0.0577 0.1577 ⎥⎦
32
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
33
The 5th step in the Zbus building algorithm (1)
• 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
i2
4th step
i1
v1
z3
J
znew = z5 = j 0.1 and zmm = z11 = j 0.0808 and znn = z33 = j 0.1577
Therefore znew + zmm + znn − 2 zmn = j 0.2693
z4
z5 v3
z2
z1
and 2 zmn = 2 z13 = j 0.0692
v2
2x2
i3
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
The 5th step in the Zbus building algorithm (2)
⎡ 0.0808 − 0.0346 ⎤
⎡ 0.0462 ⎤
We get a = j ⎢ 0.0346 − 0.0577 ⎥ = j ⎢ −0.0231⎥ The b matrix will be:
⎢
⎥
⎢
⎥
⎢⎣ 0.0346 − 0.1577 ⎥⎦
⎢⎣ −0.1231⎥⎦
b = j [ 0.0346 − 0.0808 0.0577 − 0.0346 0.1577 − 0.0346]
b = j [ −0.0462 0.0231 0.1231]
the incremental matrix will be
⎡ 0.0462 ⎤
j ⎢
=
−0.0231⎥ [ 0.0462 0.0231 0.1231]
⎥
0.2693 ⎢
⎢⎣ −0.1231⎥⎦
0.0211 ⎤
⎡ −0.0079 0.0040
= j ⎢ 0.0040 −0.0020 −0.0106⎥
⎢
⎥
⎢⎣ 0.0211 −0.0106 −0.0563⎥⎦
34
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
The 5th step in the Zbus building algorithm (3)
By adding this matrix to Z bus,old we finally get
the final version of the Z bus matrix:
Z bus,new = Z bus
⎡0.0729 0.0386 0.0557 ⎤
⎢
⎥
= j 0.0386 0.0557 0.0471
⎢
⎥
⎢⎣0.0557 0.0471 0.1014 ⎥⎦
This complets our numerical example.
35
Lecture 23
Transient synchronous machine short
circuit currents
• Subtransient , xd’’ and
transient reactances,
xd’’ are smaller than the
DC compononent
steady state
synchronous
sub-transient-áhrif
reactances, xd’’
• A DC component is
present in all 3 phases
• The exact shape of the
phase “b”
short circuit current will
depend on the phase
instant of the short
circuit and is therefore
phase “c”
different in the 3 phases
Power Engineering - Egill Benedikt Hreinsson
phase “a”
xd''
xd'
xd
transient-áhrif
36
Lecture 23
Power Engineering - Egill Benedikt Hreinsson
Energy control centers
37
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
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
38
Lecture 23
Power Engineering - Egill Benedikt Hreinsson
Early Power System Control (in 1919)
39
Lecture 23
Power Engineering - Egill Benedikt Hreinsson
Energy Control Centers
Source Measured
Success: Constructing
Performance
Metrics for Energy
Management
by John Van Gorp, Power
Measurement Corp.
Í ritinu
“Web Based
Energy Information and
Control Systems:
Case Studies and
Applications”
CRC Press 2005
23-Nov-11
40
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
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
23-Nov-11
41
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
Remote terminal unit
Communication link
Substation
SCADA Master Station
EMS 1-line diagram
Energy control center with EMS
EMS alarm display
42
Lecture 23
More energy control
centers
Power Engineering - Egill Benedikt Hreinsson
43
Power Engineering - Egill Benedikt Hreinsson
Lecture 23
The control of distributed generation
Micro-Grid
Operation and
Control
http://www.pserc.or
g/cgipserc/getbig/gener
alinf/presentati/pre
sentati/Microgrid_T
utorial.pdf
44
Lecture 23
Power Engineering - Egill Benedikt Hreinsson
45
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