2. 2.1 BUS ADMITTANCE MATRIX
• Network equations can be formulated systematically in a variety of
forms.
• The node-voltage method is the most suitable form for many power
system analyses and is commonly used.
• In order to obtain the node-voltage equations, the bus admittance
matrix needs to be formulated.
• Consider the simple power system shown in Figure 1 where
impedances are expressed in per unit on a common MVA base.
• Resistances are neglected for simplicity.
3. 2.1 BUS ADMITTANCE MATRIX
• Since the nodal solution is based on Kirchhoff’s current law,
impedances are converted to admittance:
𝑦𝑖𝑗 =
1
𝑧𝑖𝑗
=
1
𝑟𝑖𝑗 + 𝑗𝑥𝑖𝑗
• The circuit has been redrawn in Figure 2 in terms of admittances
and transformation to current sources.
4. 2.1 BUS ADMITTANCE MATRIX
Figure 1 – The impedance diagram of
a simple system.
Figure 2 – The admittance diagram for system of Figure 1.
5. 2.1 BUS ADMITTANCE MATRIX
• Applying KCL to nodes 1 to 4:
• Rearranging these equations
yields:
Figure 2 – The admittance diagram for system of Figure 1.
6. 2.1 BUS ADMITTANCE MATRIX
• We introduce the following
admittances:
Figure 2 – The admittance diagram for system of Figure 1.
7. 2.1 BUS ADMITTANCE MATRIX
• The node equations reduce to:
• Since there is no connections
between bus 1 and 4:
𝑌14 = 𝑌41 = 0
• Similarly for bus 2 and 4:
𝑌24 = 𝑌42 = 0
Figure 2 – The admittance diagram for system of Figure 1.
8. 2.1 BUS ADMITTANCE MATRIX
• Extending the above relation to an 𝑛 bus system, the node-voltage
equation in matrix form is:
9. 2.1 BUS ADMITTANCE MATRIX
Terms Comments
𝐼𝑏𝑢𝑠 • The vector of the injected bus current.
• Positive when flowing towards the bus.
• Negative of flowing away from the bus.
𝑉𝑏𝑢𝑠 • Voltage measured from the reference node.
𝑌𝑏𝑢𝑠 • The bus admittance matrix
• The diagonal element (self-admittance) of each node is the
sum of admittances connected to it.
• The off-diagonal element (mutual admittance) is equal to the
negative of the admittance between the nodes.
10. 2.1 BUS ADMITTANCE MATRIX
• The self-admittance element is given by:
𝑌𝑖𝑖 =
𝑗=0
𝑛
𝑦𝑖𝑗 𝑗 ≠ 𝑖
• The mutual-admittance element is given by:
𝑌𝑖𝑗 = 𝑌
𝑗𝑖 = −𝑦𝑖𝑗
• The 𝑛 bus voltages can be solved using the following equation:
𝑉𝑏𝑢𝑠 = 𝑌𝑏𝑢𝑠
−1
𝐼𝑏𝑢𝑠
11. 2.1 BUS ADMITTANCE MATRIX
• The bus admittance matrix for the network in Figure 2 is given as
follows:
12. 2.1 BUS ADMITTANCE MATRIX
Tutorial 1
A power system network is shown in Figure 3. The generators at buses
1 and 2 are represented by their equivalent current sources with their
reactances in per unit on a 100-MVA base. The lines are represented by
𝜋 model where series reactances and shunt reactances are also
expressed in per unit on a 100 MVA base. The loads at buses 3 and 4
are expressed in MW and MVar. Assuming a voltage magnitude of 1.0
per unit at buses 3 and 4, convert the loads to per unit impedances.
Convert network impedances to admittances and obtain the bus
admittance matrix by inspection.
13. 2.1 BUS ADMITTANCE MATRIX
Tutorial 1
Figure 3 – The impedance diagram for Tutorial 1
15. 2.1 BUS ADMITTANCE MATRIX
Tutorial 2
A power system network is shown in Figure 4. The values marked are
impedances in per unit on a base of 100 MVA. Determine the bus
admittance matrix by inspection.
Figure 4 - The impedance diagram for Tutorial 2
17. 2.2 NONLINEAR ALGEBRAIC
EQUATIONS’ SOLUTION TECHNIQUES
The most common techniques used for the iterative solution of
nonlinear algebraic equations are as follows:
• Gauss-Seidel,
• Newton-Raphson,
• Quasi-Newton.
18. 2.2 NONLINEAR ALGEBRAIC
EQUATIONS’ SOLUTION TECHNIQUES
Gauss-Seidel method
• Also known as the method of successive displacements.
• Consider the solution of the nonlinear equation given by:
• The above function is rearranged as:
• If 𝑥(𝑘) is an initial estimate of the variable 𝑥, the following iterative
sequence is formed:
19. 2.2 NONLINEAR ALGEBRAIC
EQUATIONS’ SOLUTION TECHNIQUES
Gauss-Seidel method
• A solution is obtained when the difference between the absolute
value of the successive iteration is less than a specified accuracy
such that:
• 𝜖 is the desired accuracy.
• In some cases, an acceleration factor 𝛼 can be used to improve the
rate of convergence using the following formula:
20. 2.2 NONLINEAR ALGEBRAIC
EQUATIONS’ SOLUTION TECHNIQUES
Example 1
Use the Gauss-Seidel method to find a root of the following equation:
Solution to Example 1:
• Solving for 𝑥, the above expression is written as:
21. 2.2 NONLINEAR ALGEBRAIC
EQUATIONS’ SOLUTION TECHNIQUES
Solution to Example 1:
• Apply the Gauss-Seidel algorithm and use an initial estimate of:
• The first iteration is:
• The second iteration is:
• The subsequent iterations result in 2.8966, 3.3376, 3.1398. 3.9568.
3.9988 and 4.0000.
22. 2.2 NONLINEAR ALGEBRAIC
EQUATIONS’ SOLUTION TECHNIQUES
Solution to Example 1:
• Apply the Gauss-Seidel algorithm and use an initial estimate of:
• The first iteration is:
• The second iteration is:
• The subsequent iterations result in 2.8966, 3.3376, 3.1398. 3.9568.
3.9988 and 4.0000.
23. 2.2 NONLINEAR ALGEBRAIC
EQUATIONS’ SOLUTION TECHNIQUES
Solution to Example 1:
Figure 5 - Graphical illustration of the Gauss-Seidel method
24. 2.2 NONLINEAR ALGEBRAIC
EQUATIONS’ SOLUTION TECHNIQUES
Example 2:
Find a root of the equation in Example 1, using the Gauss-Seidel
method with an acceleration factor of α = 1.25.
Solution to Example 2
• The first iteration is:
25. 2.2 NONLINEAR ALGEBRAIC
EQUATIONS’ SOLUTION TECHNIQUES
Solution to Example 2:
• The second iteration is:
• The subsequent iterations result in 3.0801, 3.1831. 3.7238. 4.0084.
3.9978 and 4.0005.
26. 2.2 NONLINEAR ALGEBRAIC
EQUATIONS’ SOLUTION TECHNIQUES
Solution to Example 2:
Figure 6 - Graphical illustration of the Gauss-Seidel method
using acceleration factor
27. 2.2 NONLINEAR ALGEBRAIC
EQUATIONS’ SOLUTION TECHNIQUES
Consider the system of 𝑛 equations in 𝑛 variables:
Solving for one variable from each equation, the above functions are
rearranged and written as:
28. 2.2 NONLINEAR ALGEBRAIC
EQUATIONS’ SOLUTION TECHNIQUES
Tutorial 3
Use Gauss-Seidel method to find the solution of the following
equations:
with the following initial estimates:
a) 𝑥1
(0)
= 1, 𝑥2
(0)
= 1 (Answer: 𝑥1
(2)
= 5, 𝑥2
(2)
= 1 )
b) 𝑥1
(0)
= 1, 𝑥2
(0)
= 2 (Answer: 𝑥1
(10)
= 2.0006, 𝑥2
(10)
= 3.9994)
Continue the iterations until ∆𝑥1
(𝑘)
and ∆𝑥2
(𝑘)
are less than 0.001.
