ECEN 460 Power System Operation and Control - PowerPoint Presentation

Slide1

ECEN 460Power System Operation and Control

Lecture 10: The power balance equations

Adam BirchfieldDept. of Electrical and Computer EngineeringTexas A&M Universityabirchfield@tamu.edu

Material gratefully adapted with permission from slides by Prof. Tom

Overbye

.

AnnouncementsPlease read Chapter 6

Homework 2 and 3 solutions will be postedQuizzes most ThursdaysLab 4 Sept. 28, Oct. 1, and Oct. 2.No lab Oct 5-9 due to examExam 1 will be Tuesday October 9Closed-book, closed-notes, regular calculator and one 8.5”x11” notesheet are allowed

Power flow analysis

We now have the necessary models to start to develop the power system analysis toolsThe most common power system analysis tool is the power flow (also known sometimes as the load flow)Power flow is a steady-state analysis toolPower flow determines how

the current and power flows in a networkThe solution variables that specify the system state” are the bus complex voltages

Once we’ve solved for these we can get the rest

Bus admittance matrix or Y

busFirst step in solving the power flow is to create what is known as the bus admittance matrix, often call the Ybus. The Y

bus gives the relationships between all the bus current injections, I, and all the bus voltages, V,

I

=

Y

bus

VThe 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

Ybus example

Determine the bus admittance matrix for the network

shown below, assuming the current injection at each

bus

i

is I

i

=

I

Gi

-

IDi where IGi is the current injection into thebus from the generator and IDi is the current flowing into the load

Ybus example, cont’d

Ybus example, cont’d

For a system with n buses,

Y

bus

is an n by n

symmetric matrix (i.e., one where

A

ij

=

A

ji

)

Modeling shunts in the Ybus

Two bus system example

Using the Ybus

Solving for bus currents

Solving for bus voltages

Ybus in PowerWorld

To see the Ybus in PowerWorld, select Case Information, Solution Details, YbusFor large systems most of the Ybus elements are zero, giving what is known as a sparse matrix

Ybus

for Lab 3 Three Bus System

Generator models

Engineering models depend upon applicationGenerators are usually synchronous machinesFor generators we will use two different models:a short term model treating the generator as a constant voltage source behind a possibly time-varying reactance (used in Lab and earlier in class)a steady-state model, treating the generator as a constant power source operating at a fixed voltage; this model will be used for power flow and economic analysis

Load models

Ultimate goal is to supply loads with electricity at constant frequency and voltageElectrical characteristics of individual loads matter, but usually they can only be estimatedactual loads are constantly changing, consisting of a large number of individual devicesonly limited network observability of load characteristicsAggregate models are typically used for analysisTwo common modelsconstant power: Si

= Pi + jQiconstant impedance: Si

=

|

V

|

2 / Zi

Power flow analysis

When analyzing power systems we know neither the complex bus voltages nor the complex current injectionsRather, we know the complex power being consumed by the load, and the power being injected by the generators plus their voltage magnitudesTherefore we can not directly use the Ybus equations, but rather must use the power balance equations

Linear versus nonlinear systems

A function H is linear if H(a1m1

+ a2m2) =

a

1

H

(

m1) + a2H(m2)

That is

1) the output is proportional to the input

2) the principle of superposition holds

Linear Example:

y = H(x) = c x y = c(x1+x2) = cx1 + c x2Nonlinear Example:

y

=

H

(

x

) = c

x

2

y

= c(

x

1

+

x

2

)

2

≠ (c

x

1

)

2

+ (c

x

2

)

2

Linear power system elements

Nonlinear power system elements

Constant power loads and generator injections are nonlinear and hence systems with these elements can not be analyzed by superposition

Nonlinear problems can be very difficult to solve,

and usually require an iterative approach