0. ECE 476 POWER SYSTEM ANALYSIS
Lecture 11
YBus and Power Flow
Professor Tom Overbye
Department of Electrical and Computer Engineering

1. Announcements Homework #4 is due now Homework 5 is due on Oct 4
5.26, 5.27, 5.28, 5.43, 3.4
Homework 5 is due on Oct 4
3.12, 3.19, 27, 60
Oct 2 class (Tuesday) will be in 50 Everitt Lab
First exam is 10/9 in class; closed book, closed notes, one note sheet and calculators allowed
Power plant, substation field trip, 10/11 during class.
Start reading Chapter 6 for lectures 10 to 13

2. Bus Admittance Matrix or Ybus
First step in solving the power flow is to create what is known as the bus admittance matrix, often call the Ybus.
The Ybus gives the relationships between all the bus current injections, I, and all the bus voltages, V, I = Ybus V
The Ybus is developed by applying KCL at each bus in the system to relate the bus current injections, the bus voltages, and the branch impedances and admittances

3. Ybus Example Determine the bus admittance matrix for the network
shown below, assuming the current injection at each
bus i is Ii = IGi - IDi where IGi is the current injection into the
bus from the generator and IDi is the current flowing into the
load

4. Ybus Example, cont’d

5. Ybus Example, cont’d symmetric matrix (i.e., one where Aij = Aji)
For a system with n buses, Ybus is an n by n
symmetric matrix (i.e., one where Aij = Aji)

6. Ybus General Form
The diagonal terms, Yii, are the self admittance terms, equal to the sum of the admittances of all devices incident to bus i.
The off-diagonal terms, Yij, are equal to the negative of the sum of the admittances joining the two buses.
With large systems Ybus is a sparse matrix (that is, most entries are zero)
Shunt terms, such as with the p line model, only
affect the diagonal terms.

7. Modeling Shunts in the Ybus

8. Two Bus System Example

9. Using the Ybus

10. Solving for Bus Currents

11. Solving for Bus Voltages

12. Power Flow Analysis
When analyzing power systems we know neither the complex bus voltages nor the complex current injections
Rather, we know the complex power being consumed by the load, and the power being injected by the generators plus their voltage magnitudes
Therefore we can not directly use the Ybus equations, but rather must use the power balance equations

13. Power Balance Equations

14. Power Balance Equations, cont’d

15. Real Power Balance Equations

16. Power Flow Requires Iterative Solution

17. Gauss Iteration

18. Gauss Iteration Example

19. Stopping Criteria

20. Gauss Power Flow

21. Gauss Two Bus Power Flow Example
A 100 MW, 50 Mvar load is connected to a generator
through a line with z = j0.06 p.u. and line charging of 5 Mvar on each end (100 MVA base). Also, there is a 25 Mvar capacitor at bus 2. If the generator voltage is 1.0 p.u., what is V2?
SLoad = j0.5 p.u.

22. Gauss Two Bus Example, cont’d

23. Gauss Two Bus Example, cont’d

24. Gauss Two Bus Example, cont’d

25. Slack Bus
In previous example we specified S2 and V1 and then solved for S1 and V2.
We can not arbitrarily specify S at all buses because total generation must equal total load + total losses
We also need an angle reference bus.
To solve these problems we define one bus as the "slack" bus. This bus has a fixed voltage magnitude and angle, and a varying real/reactive power injection.

26. Gauss with Many Bus Systems

27. Gauss-Seidel Iteration

28. Three Types of Power Flow Buses
There are three main types of power flow buses
Load (PQ) at which P/Q are fixed; iteration solves for voltage magnitude and angle.
Slack at which the voltage magnitude and angle are fixed; iteration solves for P/Q injections
Generator (PV) at which P and |V| are fixed; iteration solves for voltage angle and Q injection
special coding is needed to include PV buses in the Gauss-Seidel iteration

29. Inclusion of PV Buses in G-S

30. Inclusion of PV Buses, cont'd

31. Two Bus PV Example Consider the same two bus system from the previous
example, except the load is replaced by a generator

32. Two Bus PV Example, cont'd

33. Generator Reactive Power Limits
The reactive power output of generators varies to maintain the terminal voltage; on a real generator this is done by the exciter
To maintain higher voltages requires more reactive power
Generators have reactive power limits, which are dependent upon the generator's MW output
These limits must be considered during the power flow solution.

34. Generator Reactive Limits, cont'd
During power flow once a solution is obtained check to make generator reactive power output is within its limits
If the reactive power is outside of the limits, fix Q at the max or min value, and resolve treating the generator as a PQ bus
this is know as "type-switching"
also need to check if a PQ generator can again regulate
Rule of thumb: to raise system voltage we need to supply more vars

35. Accelerated G-S Convergence

36. Accelerated Convergence, cont’d

37. Gauss-Seidel Advantages
Each iteration is relatively fast (computational order is proportional to number of branches + number of buses in the system
Relatively easy to program

38. Gauss-Seidel Disadvantages
Tends to converge relatively slowly, although this can be improved with acceleration
Has tendency to miss solutions, particularly on large systems
Tends to diverge on cases with negative branch reactances (common with compensated lines)
Need to program using complex numbers