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 Page 1


4.2 Example of [Z
Bus
] matrix building algorithm
The single line diagram of a power system is shown in the Fig. 4.16. The line impedances in pu are
also given. The step-by-step procedure for [
¯
Z
Bus
] matrix formulation is explained as given below:
Figure 4.16: Single Line Diagram of the Power System for the example
Preliminary Step: The graph of the network and a tree is shown in Fig. 4.17. Elements 1,2,4
and 5 are the tree branches while 3, 6 and 7 are the links.
Figure 4.17: Graph and a tree of the network of Fig. 4.17
Step 1: The step-by-step [
¯
Z
Bus
] matrix building algorithm starts with element 1, which is
a tree branch connected between nodes 1 and the reference node 0 and has an impedance of
¯ z
10
=j0.10 pu. This is shown in the accompanying ?gure ,Fig. 4.18.
Page 2


4.2 Example of [Z
Bus
] matrix building algorithm
The single line diagram of a power system is shown in the Fig. 4.16. The line impedances in pu are
also given. The step-by-step procedure for [
¯
Z
Bus
] matrix formulation is explained as given below:
Figure 4.16: Single Line Diagram of the Power System for the example
Preliminary Step: The graph of the network and a tree is shown in Fig. 4.17. Elements 1,2,4
and 5 are the tree branches while 3, 6 and 7 are the links.
Figure 4.17: Graph and a tree of the network of Fig. 4.17
Step 1: The step-by-step [
¯
Z
Bus
] matrix building algorithm starts with element 1, which is
a tree branch connected between nodes 1 and the reference node 0 and has an impedance of
¯ z
10
=j0.10 pu. This is shown in the accompanying ?gure ,Fig. 4.18.
Figure 4.18: Partial network of Step 1
The resulting [
¯
Z
Bus
] matrix is
¯
Z
Bus
= 
(1)
(1) z
10
= 
(1)
(1) j0.10 
Step 2: Next, the element 2 connected between node 2 (q = 2) and the reference node ‘0’ is
selected. This element has an impedance of ¯ z
20
=j0.10 p.u. As this is the addition of a tree branch
it will add a new node ‘2’ to the existing [
¯
Z
Bus
] matrix. This addition is illustrated in Fig. 4.19.
Figure 4.19: Partial network of Step 2
The new bus impedance matrix is given by :
¯
Z
Bus
= 
(1) (2)
(1) j0.10 0
(2) 0 z
20
	= 
(1) (2)
(1) j0.1 0
(2) 0 j0.10
	
Step 3: Element 3 connected between existing nodes, node 1 (p = 1) and node 2 (q = 2),
having an impedance of ¯ z
12
=j0.20 p.u. is added to the partial network, as shown in Fig. 4.20.
Since this is an addition of a link to the network a two step procedure is to be followed. In the
Page 3


4.2 Example of [Z
Bus
] matrix building algorithm
The single line diagram of a power system is shown in the Fig. 4.16. The line impedances in pu are
also given. The step-by-step procedure for [
¯
Z
Bus
] matrix formulation is explained as given below:
Figure 4.16: Single Line Diagram of the Power System for the example
Preliminary Step: The graph of the network and a tree is shown in Fig. 4.17. Elements 1,2,4
and 5 are the tree branches while 3, 6 and 7 are the links.
Figure 4.17: Graph and a tree of the network of Fig. 4.17
Step 1: The step-by-step [
¯
Z
Bus
] matrix building algorithm starts with element 1, which is
a tree branch connected between nodes 1 and the reference node 0 and has an impedance of
¯ z
10
=j0.10 pu. This is shown in the accompanying ?gure ,Fig. 4.18.
Figure 4.18: Partial network of Step 1
The resulting [
¯
Z
Bus
] matrix is
¯
Z
Bus
= 
(1)
(1) z
10
= 
(1)
(1) j0.10 
Step 2: Next, the element 2 connected between node 2 (q = 2) and the reference node ‘0’ is
selected. This element has an impedance of ¯ z
20
=j0.10 p.u. As this is the addition of a tree branch
it will add a new node ‘2’ to the existing [
¯
Z
Bus
] matrix. This addition is illustrated in Fig. 4.19.
Figure 4.19: Partial network of Step 2
The new bus impedance matrix is given by :
¯
Z
Bus
= 
(1) (2)
(1) j0.10 0
(2) 0 z
20
	= 
(1) (2)
(1) j0.1 0
(2) 0 j0.10
	
Step 3: Element 3 connected between existing nodes, node 1 (p = 1) and node 2 (q = 2),
having an impedance of ¯ z
12
=j0.20 p.u. is added to the partial network, as shown in Fig. 4.20.
Since this is an addition of a link to the network a two step procedure is to be followed. In the
Figure 4.20: Partial network of Step 3
?rst step a new row and column is added to the matrix as given below :
¯
Z
(temp)
Bus
=
?
?
?
?
?
?
?
?
(1) (2) (`)
(1) j0.10 0.0 (
¯
Z
12
-
¯
Z
11
)
(2) 0.0 j0.10 (
¯
Z
22
-
¯
Z
21
)
(`) (
¯
Z
21
-
¯
Z
11
) (
¯
Z
22
-
¯
Z
12
)
¯
Z
``
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
(1) (2) (`)
(1) j0.10 0.0 -j0.10
(2) 0.0 j0.10 j0.10
(`) -j0.10 j0.10 j0.40
?
?
?
?
?
?
?
?
where,
¯
Z
``
=
¯
Z
11
+
¯
Z
22
- 2
¯
Z
12
+ ¯ z
20
=j0.10+j0.10- 0.0+j0.20=j0.40p.u.
Next this new row and column is eliminated to restore the size of [
¯
Z
Bus
] matrix as given below:
[
¯
Z
Bus
] = 
j0.10 0.0
0.0 j0.10
	-

-j0.10
j0.10
	-j0.10 j0.10
j0.40
Hence, the impedance matrix after the addition of element 3 is found out to be :
[
¯
Z
Bus
]= 
(1) (2)
(1) j0.075 j0.025
(2) j0.025 j0.075
	
Step 4: The element 4 , which is added next, is connected between an existing node, node 2
(p = 2) and a new node, node 3 (q = 3). The impedance of this element is ¯ z
23
=j0.30 p.u. and
it is a tree branch hence, a new node, node 3 is added to the partial network. This addition, shown
in Fig. 4.21, thus increases the size of [
¯
Z
Bus
] to (3× 3).
Page 4


4.2 Example of [Z
Bus
] matrix building algorithm
The single line diagram of a power system is shown in the Fig. 4.16. The line impedances in pu are
also given. The step-by-step procedure for [
¯
Z
Bus
] matrix formulation is explained as given below:
Figure 4.16: Single Line Diagram of the Power System for the example
Preliminary Step: The graph of the network and a tree is shown in Fig. 4.17. Elements 1,2,4
and 5 are the tree branches while 3, 6 and 7 are the links.
Figure 4.17: Graph and a tree of the network of Fig. 4.17
Step 1: The step-by-step [
¯
Z
Bus
] matrix building algorithm starts with element 1, which is
a tree branch connected between nodes 1 and the reference node 0 and has an impedance of
¯ z
10
=j0.10 pu. This is shown in the accompanying ?gure ,Fig. 4.18.
Figure 4.18: Partial network of Step 1
The resulting [
¯
Z
Bus
] matrix is
¯
Z
Bus
= 
(1)
(1) z
10
= 
(1)
(1) j0.10 
Step 2: Next, the element 2 connected between node 2 (q = 2) and the reference node ‘0’ is
selected. This element has an impedance of ¯ z
20
=j0.10 p.u. As this is the addition of a tree branch
it will add a new node ‘2’ to the existing [
¯
Z
Bus
] matrix. This addition is illustrated in Fig. 4.19.
Figure 4.19: Partial network of Step 2
The new bus impedance matrix is given by :
¯
Z
Bus
= 
(1) (2)
(1) j0.10 0
(2) 0 z
20
	= 
(1) (2)
(1) j0.1 0
(2) 0 j0.10
	
Step 3: Element 3 connected between existing nodes, node 1 (p = 1) and node 2 (q = 2),
having an impedance of ¯ z
12
=j0.20 p.u. is added to the partial network, as shown in Fig. 4.20.
Since this is an addition of a link to the network a two step procedure is to be followed. In the
Figure 4.20: Partial network of Step 3
?rst step a new row and column is added to the matrix as given below :
¯
Z
(temp)
Bus
=
?
?
?
?
?
?
?
?
(1) (2) (`)
(1) j0.10 0.0 (
¯
Z
12
-
¯
Z
11
)
(2) 0.0 j0.10 (
¯
Z
22
-
¯
Z
21
)
(`) (
¯
Z
21
-
¯
Z
11
) (
¯
Z
22
-
¯
Z
12
)
¯
Z
``
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
(1) (2) (`)
(1) j0.10 0.0 -j0.10
(2) 0.0 j0.10 j0.10
(`) -j0.10 j0.10 j0.40
?
?
?
?
?
?
?
?
where,
¯
Z
``
=
¯
Z
11
+
¯
Z
22
- 2
¯
Z
12
+ ¯ z
20
=j0.10+j0.10- 0.0+j0.20=j0.40p.u.
Next this new row and column is eliminated to restore the size of [
¯
Z
Bus
] matrix as given below:
[
¯
Z
Bus
] = 
j0.10 0.0
0.0 j0.10
	-

-j0.10
j0.10
	-j0.10 j0.10
j0.40
Hence, the impedance matrix after the addition of element 3 is found out to be :
[
¯
Z
Bus
]= 
(1) (2)
(1) j0.075 j0.025
(2) j0.025 j0.075
	
Step 4: The element 4 , which is added next, is connected between an existing node, node 2
(p = 2) and a new node, node 3 (q = 3). The impedance of this element is ¯ z
23
=j0.30 p.u. and
it is a tree branch hence, a new node, node 3 is added to the partial network. This addition, shown
in Fig. 4.21, thus increases the size of [
¯
Z
Bus
] to (3× 3).
Figure 4.21: Partial network of Step 4
The new impedance matrix can be calculated as:
¯
Z
Bus
=
?
?
?
?
?
?
?
?
(1) (2) (3)
(1) j0.075 j0.025
¯
Z
12
(2) j0.025 j0.0.075
¯
Z
22
(3)
¯
Z
21
¯
Z
22
¯
Z
22
+ ¯ z
23
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
(1) (2) (3)
(1) j0.075 j0.025 j0.025
(2) j0.025 j0.0.075 j0.075
(3) j0.025 j0.075 j0.375
?
?
?
?
?
?
?
?
Step 5: Element 5 is added next to the existing partial network. This is a tree branch connected
between an existing node, node 3 (p = 3) and a new node, node 4 (q = 4). This is illustrated
in Fig. 4.22.
Since a new node is added to the partial network, the size of [
¯
Z
Bus
] increases to (4× 4). The
impedance of the new element is ¯ z
34
=j0.15 p.u. The new bus impedance matrix is :
¯
Z
Bus
=
?
?
?
?
?
?
?
?
?
?
?
(1) (2) (3) (4)
(1) j0.075 j0.025 j0.025
¯
Z
31
(2) j0.025 j0.075 j0.075
¯
Z
32
(3) j0.025 j0.075 j0.375
¯
Z
33
(4)
¯
Z
13
¯
Z
23
¯
Z
33
¯
Z
33
+ ¯ z
34
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
(1) (2) (3) (4)
(1) j0.075 j0.025 j0.025 j0.025
(2) j0.025 j0.075 j0.075 j0.075
(3) j0.025 j0.075 j0.375 j0.375
(4) j0.025 j0.075 j0.375 j0.525
?
?
?
?
?
?
?
?
?
?
?
Step 6: Next,the element 6 connected between two existing nodes node 1 (p = 1) and node
4 (q = 4) is added to the network, as shown in the Fig. 4.23. The impedance of this element
is ¯ z
23
=j0.25 p.u. As this is a link addition, the two step procedure is used. The bus impedance
Page 5


4.2 Example of [Z
Bus
] matrix building algorithm
The single line diagram of a power system is shown in the Fig. 4.16. The line impedances in pu are
also given. The step-by-step procedure for [
¯
Z
Bus
] matrix formulation is explained as given below:
Figure 4.16: Single Line Diagram of the Power System for the example
Preliminary Step: The graph of the network and a tree is shown in Fig. 4.17. Elements 1,2,4
and 5 are the tree branches while 3, 6 and 7 are the links.
Figure 4.17: Graph and a tree of the network of Fig. 4.17
Step 1: The step-by-step [
¯
Z
Bus
] matrix building algorithm starts with element 1, which is
a tree branch connected between nodes 1 and the reference node 0 and has an impedance of
¯ z
10
=j0.10 pu. This is shown in the accompanying ?gure ,Fig. 4.18.
Figure 4.18: Partial network of Step 1
The resulting [
¯
Z
Bus
] matrix is
¯
Z
Bus
= 
(1)
(1) z
10
= 
(1)
(1) j0.10 
Step 2: Next, the element 2 connected between node 2 (q = 2) and the reference node ‘0’ is
selected. This element has an impedance of ¯ z
20
=j0.10 p.u. As this is the addition of a tree branch
it will add a new node ‘2’ to the existing [
¯
Z
Bus
] matrix. This addition is illustrated in Fig. 4.19.
Figure 4.19: Partial network of Step 2
The new bus impedance matrix is given by :
¯
Z
Bus
= 
(1) (2)
(1) j0.10 0
(2) 0 z
20
	= 
(1) (2)
(1) j0.1 0
(2) 0 j0.10
	
Step 3: Element 3 connected between existing nodes, node 1 (p = 1) and node 2 (q = 2),
having an impedance of ¯ z
12
=j0.20 p.u. is added to the partial network, as shown in Fig. 4.20.
Since this is an addition of a link to the network a two step procedure is to be followed. In the
Figure 4.20: Partial network of Step 3
?rst step a new row and column is added to the matrix as given below :
¯
Z
(temp)
Bus
=
?
?
?
?
?
?
?
?
(1) (2) (`)
(1) j0.10 0.0 (
¯
Z
12
-
¯
Z
11
)
(2) 0.0 j0.10 (
¯
Z
22
-
¯
Z
21
)
(`) (
¯
Z
21
-
¯
Z
11
) (
¯
Z
22
-
¯
Z
12
)
¯
Z
``
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
(1) (2) (`)
(1) j0.10 0.0 -j0.10
(2) 0.0 j0.10 j0.10
(`) -j0.10 j0.10 j0.40
?
?
?
?
?
?
?
?
where,
¯
Z
``
=
¯
Z
11
+
¯
Z
22
- 2
¯
Z
12
+ ¯ z
20
=j0.10+j0.10- 0.0+j0.20=j0.40p.u.
Next this new row and column is eliminated to restore the size of [
¯
Z
Bus
] matrix as given below:
[
¯
Z
Bus
] = 
j0.10 0.0
0.0 j0.10
	-

-j0.10
j0.10
	-j0.10 j0.10
j0.40
Hence, the impedance matrix after the addition of element 3 is found out to be :
[
¯
Z
Bus
]= 
(1) (2)
(1) j0.075 j0.025
(2) j0.025 j0.075
	
Step 4: The element 4 , which is added next, is connected between an existing node, node 2
(p = 2) and a new node, node 3 (q = 3). The impedance of this element is ¯ z
23
=j0.30 p.u. and
it is a tree branch hence, a new node, node 3 is added to the partial network. This addition, shown
in Fig. 4.21, thus increases the size of [
¯
Z
Bus
] to (3× 3).
Figure 4.21: Partial network of Step 4
The new impedance matrix can be calculated as:
¯
Z
Bus
=
?
?
?
?
?
?
?
?
(1) (2) (3)
(1) j0.075 j0.025
¯
Z
12
(2) j0.025 j0.0.075
¯
Z
22
(3)
¯
Z
21
¯
Z
22
¯
Z
22
+ ¯ z
23
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
(1) (2) (3)
(1) j0.075 j0.025 j0.025
(2) j0.025 j0.0.075 j0.075
(3) j0.025 j0.075 j0.375
?
?
?
?
?
?
?
?
Step 5: Element 5 is added next to the existing partial network. This is a tree branch connected
between an existing node, node 3 (p = 3) and a new node, node 4 (q = 4). This is illustrated
in Fig. 4.22.
Since a new node is added to the partial network, the size of [
¯
Z
Bus
] increases to (4× 4). The
impedance of the new element is ¯ z
34
=j0.15 p.u. The new bus impedance matrix is :
¯
Z
Bus
=
?
?
?
?
?
?
?
?
?
?
?
(1) (2) (3) (4)
(1) j0.075 j0.025 j0.025
¯
Z
31
(2) j0.025 j0.075 j0.075
¯
Z
32
(3) j0.025 j0.075 j0.375
¯
Z
33
(4)
¯
Z
13
¯
Z
23
¯
Z
33
¯
Z
33
+ ¯ z
34
?
?
?
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
?
?
?
(1) (2) (3) (4)
(1) j0.075 j0.025 j0.025 j0.025
(2) j0.025 j0.075 j0.075 j0.075
(3) j0.025 j0.075 j0.375 j0.375
(4) j0.025 j0.075 j0.375 j0.525
?
?
?
?
?
?
?
?
?
?
?
Step 6: Next,the element 6 connected between two existing nodes node 1 (p = 1) and node
4 (q = 4) is added to the network, as shown in the Fig. 4.23. The impedance of this element
is ¯ z
23
=j0.25 p.u. As this is a link addition, the two step procedure is used. The bus impedance
Figure 4.22: Partial network of Step 5
Figure 4.23: Partial network of Step 6
matrix is modi?ed by adding a new row and column as given below:
¯
Z
(temp)
Bus
=
?
?
?
?
?
?
?
?
?
?
?
?
?
(1) (2) (3) (4) (`)
(1) j0.075 j0.025 j0.025 j0.025 (
¯
Z
14
-
¯
Z
11
)
(2) j0.025 j0.075 j0.075 j0.075 (
¯
Z
24
-
¯
Z
21
)
(3) j0.025 j0.075 j0.375 j0.375 (
¯
Z
34
-
¯
Z
31
)
(4) j0.025 j0.075 j0.375 j0.525 (
¯
Z
44
-
¯
Z
41
)
(`) (
¯
Z
41
-
¯
Z
11
) (
¯
Z
42
-
¯
Z
12
) (
¯
Z
43
-
¯
Z
13
) (
¯
Z
44
-
¯
Z
14
)
¯
Z
``
?
?
?
?
?
?
?
?
?
?
?
?
?

FAQs on Example of ZBUS Matrix Formulation - Electrical Engineering (EE)

1. What is ZBUS matrix formulation in electrical engineering?

Ans. ZBUS matrix formulation is a method used in electrical engineering to model and analyze power systems. It involves representing the network elements such as generators, transformers, and transmission lines in a matrix form, known as the ZBUS matrix. This matrix helps in solving power flow and fault analysis problems efficiently.

2. How is the ZBUS matrix derived in electrical engineering?

Ans. The ZBUS matrix is derived by gathering the impedance data of all the network elements and arranging them in a matrix form. The diagonal elements of the matrix represent the self-impedances of the buses, while the off-diagonal elements represent the mutual impedances between the buses. The ZBUS matrix is typically obtained by using the bus admittance matrix and converting it to the impedance matrix using the formula ZBUS = inv(YBUS).

3. What are the applications of ZBUS matrix formulation in electrical engineering?

Ans. The ZBUS matrix formulation has several applications in electrical engineering. It is primarily used for power flow analysis, which helps in determining the voltage and power flow in a power system. Additionally, the ZBUS matrix is utilized for fault analysis, where it assists in identifying the location and magnitude of faults in the system. It is also used for contingency analysis, optimal power flow, and stability analysis.

4. How does the ZBUS matrix formulation aid in power flow analysis?

Ans. The ZBUS matrix formulation is instrumental in power flow analysis as it provides a systematic and efficient way to solve complex power system equations. By solving the power flow equations using the ZBUS matrix, engineers can determine the voltage magnitudes and angles at each bus, as well as the power flow in the transmission lines. This information helps in assessing the performance of the power system and identifying potential issues.

5. Can the ZBUS matrix formulation handle large-scale power systems?

Ans. Yes, the ZBUS matrix formulation is capable of handling large-scale power systems. It offers a scalable approach that can accommodate a high number of buses and network elements. However, as the size of the system increases, the computation time and memory requirements for solving the ZBUS matrix equations also increase. Therefore, efficient algorithms and computational techniques are employed to ensure timely and accurate analysis of large-scale power systems.

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Example of ZBUS Matrix Formulation - Electrical Engineering (EE)

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Example of ZBUS Matrix Formulation - Electrical Engineering (EE)

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Additional Information about Example of ZBUS Matrix Formulation for Electrical Engineering (EE) Preparation

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