Power System Analysis 3/24/2023

Lecture On Lecture’s Outline

Algorithm for the • Rules

Formation of the Bus Admittance Matrix • Problems

Course Teacher
Dr. Muhammad Mohsin Aman
Associate Professor
Department of Electrical Engineering,
NED University of Engineering and Technology.

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About Z-Bus Matrix

• Zbus is symmetric when Ybus is symmetric

• Zbus is a full matrix
• In Ybus some of the elements are zero, but in Zbus, zero
elements of Ybus becomes non-zero elements
About Bus Impedance Matrix • Zbus matrix can be found in two ways:
1. Inverse of Ybus matrix
2. Directly found from reactance diagram

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Algorithm for formation of Z-Bus Matrix

• Let Zs is the self admittance

• Adding Zs from a new to a reference bus
• Adding Zs from a new to an old bus
• Adding Zs from an old to a reference bus
Bus Impedance Matrix And Graph
• Adding Zs in between two old buses.
Theory

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Algorithm for Formation of the Z-Bus Matrix Algorithm for Formation of the Z-Bus Matrix
• Using this algorithm we shall build the Bus Impedance Matrix • Some definitions from graph theory are introduced below
Zbus starting with a single element and adding elements one by related to Fig. The system shown in (a) above has a graph
one till all the elements of the system are accounted for in the shown in (b).
Impedance Matrix.

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Algorithm for Formation of the Z-Bus Matrix Algorithm for Formation of the Z-Bus Matrix
• Each bus becomes a node or a corner in the graph. • A tree of a connected graph is a
Branch
• The datum, or ground bus is the node number 0 in the graph. connected subgraph connecting
• Elements in the power system become edges or elements of all the nodes without forming a Link

the graph. For example, the line from bus 1 to bus 2 whose loop (0-1)→(0-2) →(1-3).
impedance is j0.8 ohm becomes the element 4 in the graph. • The elements of a tree are
called branches.
• A graph contains multiple trees.
• The number of branches in a
tree is always one less than the
number of nodes, i.e.
𝑏 =𝑛−1 𝑛=4
𝑏 =4−1 =3
• where n is the number of nodes
and b the number of branches
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in the graph. 10

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Algorithm for Formation of the Z-Bus Matrix Algorithm for Formation of the Z-Bus Matrix
• Once a tree of a graph is determined, the remaining branches • A loop that contains one link is called a basic loop.
are called links. • The number of basic loops is unique: it is equal to the number
• In figure (c) above one such tree is shown using solid lines and of links which is also equal to the number of independent loop
the links are shown with dashed lines. equations the system has.
• The collection of links is called a cotree. • A cut set is a minimal set of elements which when cut will
• If e is the total number of elements in a graph, then the number divide the graph into two connected subgraphs.
of links l in a cotree is • A fundamental cut set is a cut set which contains only one
𝑙 = 𝑒−𝑏 = 𝑒−𝑛+1 branch.
𝑙 =5−3=2 • The number of fundamental cut sets is unique: it is equal to the
𝑙 =5−4+1=2 number of branches which is also equal to the number of
independent node equations the system has.

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Power System Analysis 3/24/2023

Algorithm for Formation of the Z-Bus Matrix

• The bus impedance matrix can be built up starting with a
single element and the process is to add an element at a time
till all the elements of the system are included.
• Suppose we have a partial network as shown on the right with
a total of m buses (in addition to the ground or reference bus).
Algorithm for Formation of the Bus
Impedance Matrix

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Algorithm for Formation of the Z-Bus Matrix Algorithm for Formation of the Z-Bus Matrix
• The corresponding equation for this partial network is • The added element may be a branch or a link described as
𝑉𝑏𝑢𝑠 = 𝑍𝑏𝑢𝑠 𝐼𝑏𝑢𝑠 follows.
• Here the size of m Zbus is m x m .
• It is assumed that the network has n buses where n > m. • Addition of a Branch
• Next we add branches or elements, one by one, till all the • Addition of a Link
elements of the network are accounted for.

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Algorithm for Formation of the Z-Bus Matrix

• The added element may be a branch or a link described as
follows.

• Addition of a Branch
• Addition of a Link Algorithm for Formation of the Bus
Impedance Matrix
Rules

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Power System Analysis 3/24/2023

Rule 1: Rules:
Rule 1: Addition of a Tree Branch to the Reference Rule 2: Addition of a Tree Branch from a New Bus to an Old
Bus
• Start with the branches connected to the reference node. • Addition of a branch zpq between a new node q and an existing
• Addition of a branch zq0 between node q and the reference node p will increase the size of the old bus impedance matrix
node 0 changes the old bus impedance matrix by adding a row by one, from (m x m) to (m +1) x (m + 1) . The change is
and column as shown below: shown below:

• Where i the position of row or column.

• Note that the red zeros are vectors of size m (same size as the • where Zpi is identical to row p of old Zbus and Zip is identical to
old bus impedance matrix). column p of old Zbus.
• Note that by symmetry Zpi = (Zip)T.
• Also, Zpp is the diagonal element of the old bus impedance
19 matrix in the pth position. 20

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Rules: Rule 3 (continue):

Rule 3: Addition of a Co-tree Link Between Two Existing • The element Zll is given by
Nodes 𝑍𝑙𝑙 = 𝑧𝑝𝑞 + 𝑍𝑝𝑝 + 𝑍𝑞𝑞 − 2𝑍𝑝𝑞
• When a link is added between existing nodes p and q, the size • The new row and column are eliminated using the equation
of the old bus impedance matrix is increased by one as follows

• If the added link is between bus p and the reference bus 0, the
above results are changed as follows:
• where (Zqi – Zpi) = (ΔZ)T is the difference between row q and
row p of the old bus impedance matrix
• and Ziq - Zip = ΔZ is the difference between column q and
column p of the old bus impedance matrix.
• where -Zpi = (-Zip)T is the negative of the pth row (column) of
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the old bus impedance matrix and Zll = zpq + Zpp . 22

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Problem 1
• Construct the bus impedance matrix for the network of Fig.
shown again below:

Algorithm for Formation of the Bus

Impedance Matrix
Problem No. 1

• And find out the bus impedance matrix

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Power System Analysis 3/24/2023

Solution Solution
• Note the proper tree is chosen in (c) above. We start with the • Note the proper tree is chosen in (c) above. We start with the
branches connected to the reference node. branches connected to the reference node.

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Solution Solution
• First we include branch 1, z10 = j0.2 between nodes 1 and 0 • Next, we add branch 2, z20 = j0.4 between nodes 2 and 0, we
(ref. node). now have;
• According to rule 1 we have;
(1)
𝑍𝑏𝑢𝑠 = Z11 = z10 = j0.2
(1)
𝑍𝑏𝑢𝑠 = Z11 = z10 = j0.2

• Note that the off-diagonal elements of the bus admittance

matrix are zero. This is because there is no connection between
these buses other than to the reference (till now).
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Solution Solution
• There are no more elements to ground, so we go to the next • There are no more elements to ground, so we go to the next
element, branch 3. element, branch 3.
• We are adding a new node (3) and a new branch (3) to an • We are adding a new node (3) and a new branch (3) to an
existing node (1). existing node (1).
Z11 + z13 = j0.2+j0.4
• We use rule 2: • We use rule 2:
𝑍11 𝑍12 𝑍11 𝑗0.2 0 𝑗0.2
(3)
𝑍𝑏𝑢𝑠 = 𝑍21 𝑍22 𝑍21 = 0 𝑗0.4 0
𝑍11 𝑍12 𝑍11 + 𝑧13 𝑗0.2 0 𝑗0.2 + 𝑗0.4

• This completes all the tree branches. • This will complete all the tree branches.

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Power System Analysis 3/24/2023

Solution Solution
• Add branch 3. i.e. a new node (3) and a new branch (3) to an • Now we go to the co-tree and add the links.
existing node (1). • We add link (4), z12 = j0.8 between existing nodes 1 and 2. We
𝑗0.2 0 𝑗0.2 have;
𝑍11 𝑍12 𝑍11
(3)
𝑍𝑏𝑢𝑠 = 𝑍21 𝑍22 𝑍21 = 0 𝑗0.4 0
𝑍11 𝑍12 𝑍11 + 𝑧13 𝑗0.2 0 𝑗0.2 + 𝑗0.4
Z11 + z13 = j0.2+j0.4

𝑍11 𝑍12 𝑍13 𝑗0.2 0 𝑗0.2

(3)
𝑍𝑏𝑢𝑠 = 𝑍21 𝑍22 𝑍23 = 0 𝑗0.4 0
𝑍31 𝑍32 𝑍33 𝑗0.2 0 𝑗0.6

• This completes all the tree branches.

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Solution Solution
• We add link (4), z12 = j0.8 between existing nodes 1 and 2. • Here 𝑗0.2 0 𝑗0.2 −𝑗0.2
(4) 0 𝑗0.4 0 𝑗0.4
𝑍11 𝑍12 𝑍13 𝑍12 − 𝑍11 𝑍𝑏𝑢𝑠 =
𝑗0.2 0 𝑗0.6 −𝑗0.2
(4) 𝑍21 𝑍22 𝑍23 𝑍22 − 𝑍21 −𝑗0.2 𝑗0.4 −𝑗0.2 𝑍44
𝑍𝑏𝑢𝑠 =
𝑍31 𝑍32 𝑍33 𝑍32 − 𝑍31
𝑍21 − 𝑍11 𝑍22 − 𝑍12 𝑍23 − 𝑍13 𝑍44
𝑍𝑙𝑙 = 𝑧𝑝𝑞 + 𝑍𝑝𝑝 + 𝑍𝑞𝑞 − 2𝑍𝑝𝑞
𝑗0.2 0 𝑗0.2 0 − 𝑗0.2
(4) 0 𝑗0.4 0 𝑗0.4 − 0 𝑍44 = 𝑧12 + 𝑍11 + 𝑍22 − 2𝑍12
𝑍𝑏𝑢𝑠 =
𝑗0.2 0 𝑗0.6 0 − 𝑗0.2 𝑍44 = 𝑗0.8 + 𝑗0.2 + 𝑗0.4 − 2 𝑗0 = 𝑗1.4
0 − 𝑗0.2 𝑗0.4 − 0 0 − 𝑗0.2 𝑍44
𝑗0.2 0 𝑗0.2 −𝑗0.2
𝑗0.2 0 𝑗0.2 −𝑗0.2 (4) 0 𝑗0.4 0 𝑗0.4
0 𝑗0.4 0 𝑗0.4 𝑍𝑏𝑢𝑠 =
𝑍𝑏𝑢𝑠
(4)
= 𝑗0.2 0 𝑗0.6 −𝑗0.2
𝑗0.2 0 𝑗0.6 −𝑗0.2 −𝑗0.2 𝑗0.4 −𝑗0.2 𝑗1.4
−𝑗0.2 𝑗0.4 −𝑗0.2 𝑍44 33 34

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Solution Solution
• Now we need to eliminate row and column. • Now we need to eliminate row and column.

𝑗0.2 0 𝑗0.2 −𝑗0.2

(4) 0 𝑗0.4 0 𝑗0.4
𝑍𝑏𝑢𝑠 =
𝑗0.2 0 𝑗0.6 −𝑗0.2
𝑗0.2 0 𝑗0.2 −𝑗0.2
−𝑗0.2 𝑗0.4 −𝑗0.2 𝑗1.4
(4) 0 𝑗0.4 0 𝑗0.4
𝑍𝑏𝑢𝑠 =
𝑗0.2 0 𝑗0.6 −𝑗0.2 −𝑗0.2
−𝑗0.2 𝑗0.4 −𝑗0.2 𝑗1.4 ∆𝑍∆𝑍𝑇 1
= 𝑗0.4 −𝑗0.2 𝑗0.4 −𝑗0.2
𝑍44 𝑗1.4 −𝑗0.2

−𝑗0.2 𝑗0.0286 −𝑗0.0571 𝑗0.0286

∆𝑍∆𝑍𝑇
∆𝑍 = 𝑍𝑖𝑞 − 𝑍𝑖𝑝 = 𝑗0.4 = −𝑗0.0571 𝑗0.1143 −𝑗0.0571
−𝑗0.2 𝑍44
𝑗0.0286 −𝑗0.0571 𝑗0.0286
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Solution Solution
• From Eq. (9), the new bus impedance matrix is: • Repeating the same process by adding link 5, z23 = j0.4
between nodes 2 and 3 we have:

𝑍11 𝑍12 𝑍13 𝑍13 − 𝑍12

(5) 𝑍21 𝑍22 𝑍23 𝑍23 − 𝑍22
𝑍𝑏𝑢𝑠 =
𝑍31 𝑍32 𝑍33 𝑍33 − 𝑍32
𝑍31 − 𝑍21 𝑍32 − 𝑍22 𝑍33 − 𝑍23 𝑍44

𝑗0.17143 𝑗0.05714 𝑗0.17143 𝑗0.11429

(5) 𝑗0.05714 𝑗0.28571 𝑗0.05714 −𝑗0.22857
𝑍𝑏𝑢𝑠 =
𝑗0.17143 𝑗0.05714 𝑗0.57143 𝑗0.51429
𝑗0.11429 −𝑗0.22857 𝑗0.51429 𝑍44
𝑍11 𝑍12 𝑍13 𝑗0.17143 𝑗0.05714 𝑗0.17143
(4)
𝑍𝑏𝑢𝑠 = 𝑍21 𝑍22 𝑍23 = 𝑗0.05714 𝑗0.28571 𝑗0.05714
𝑍31 𝑍32 𝑍33 𝑗0.17143 𝑗0.05714 𝑗0.05713
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Solution Solution
• where • From Eqn.
𝑍44 = 𝑧23 + 𝑍22 + 𝑍33 − 2𝑍23
𝑍44 = 𝑗0.4 + 𝑗0.28571 + 𝑗0.57143 − 2 𝑗0.05714 = 𝑗1.14
• The new bus impedance matrix is

𝑗0.11429
∆𝑍∆𝑍𝑇 1
= −𝑗0.22857 𝑗0.11429 −𝑗0.22857 𝑗0.51429
𝑍44 𝑗1.4 𝑗0.51429

𝑗0.01143 −𝑗0.02286 𝑗0.05143

∆𝑍∆𝑍𝑇
= −𝑗0.02286 𝑗0.04571 −𝑗0.10286
𝑍44
𝑗0.05143 −𝑗0.10286 𝑗0.23143

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Problem
• The bus impedance matrix for the network shown below is
found to be

Algorithm for Formation of the Bus

Impedance Matrix
Problem No. 3

• The line between buses 1 and 3 with impedance Z13 = j0.56 is

removed by the simultaneous opening of breakers at both ends
of the line.
• Determine the new bus impedance matrix.
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Power System Analysis 3/24/2023

Solution Solution
• One way of solving this problem is to add a line between the • From Previous eqn. of line removal
buses 1 and 3 whose impedance is the negative of Z13 to be
removed, i.e. we add - j0.56 between nodes 1 and 3.
• We proceed using the equation for adding a bus:

• Therefore, we get:

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Solution Solution
• We know that • Therefore the new bus impedance matrix is:

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Problem 3
• Find out the bus impedance matrix for the given tree:

Algorithm for Formation of the Bus

Impedance Matrix
Problem No. 3

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Problem 4
• Find out the bus impedance matrix for the given tree:

Algorithm for Formation of the Bus

Impedance Matrix
Problem No. 4

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Answer
• Bus admittance matrix is:

Algorithm for Formation of the Bus

Impedance Matrix
• Bus impedance matrix Problem No. 5

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Problem

Consider a three bus system with line impedance Zs=5

ohms. Form the Zbus matrix.

• Assume (1) as a reference bus (RB).

RB: Bus 1 Bus 2

Bus 3 53

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