EE369POWER SYSTEM ANALYSIS

Lecture 3Three Phase, Power System Operation

Tom Overbye and Ross Baldick

1

Reading and HomeworkFor lecture 3 read Chapters 1 and 2For lectures 4 through 6 read Chapter 4– we will not be covering sections 4.7, 4.11, and

4.12 in detail,– We will return to chapter 3 later.

HW 2 is Problems 2.26, 2.27, 2.28, 2.29, 2.30, 2.32, 2.33, 2.35, 2.37, 2.39, 2.40 (need to install PowerWorld); due Thursday 9/11.

HW 3 is Problems 2.42, 2.44, 2.45, 2.47, 2.49, 2.50, 2.51, 2.52; due Thursday 9/18. 2

Per Phase Analysis

Per phase analysis allows analysis of balanced 3 systems with the same effort as for a single phase system.

Balanced 3 Theorem: For a balanced 3 system with:– All loads and sources Y connected,– No mutual Inductance between phases.

3

Per Phase Analysis, cont’d

Then– All neutrals are at the same potential,– All phases are COMPLETELY decoupled,– All system values are the same “sequence” as

sources. That is, peaks of phases occur in the same order. The sequence order we’ve been using (phase b lags phase a and phase c lags phase b) is known as “positive” sequence; in EE368L we’ll discuss “negative” and “zero” sequence systems.

4

Per Phase Analysis ProcedureTo do per phase analysis 1. Convert all load/sources to equivalent Y’s.2. Solve phase “a” independent of the other

phases3. Total system power S = 3 Va Ia

*

4. If desired, phase “b” and “c” values can be determined by inspection (i.e., ±120° degree phase shifts)

5. If necessary, go back to original circuit to determine line-line values or internal values.

5

Per Phase ExampleAssume a 3, Y-connected generator with Van = 10 volts supplies a -connected load with Z = -j through a transmission line with impedance of j0.1 per phase. The load is also connected to a -connected generator with Va’’b’’ = 10 through a second transmission line which also has an impedance of j0.1 per phase.Find1. The load voltage Va’b’

2. The total power supplied by each generator, SY and S

6

Per Phase Example, cont’d

First convert the delta load and source to equivalent

Y values and draw just the "a" phase circuit

7

+-

Per Phase Example, cont’d

a' a' a'

To solve the circuit, write the KCL equation at a'

1(V 1 0)( 10 ) V (3 ) (V j

3j j

8

Per Phase Example, cont’d

a' a' a'

a'

a' b'

c' a'b'

To solve the circuit, write the KCL equation at a'

1(V 1 0)( 10 ) V (3 ) (V j

310

(10 60 ) V (10 3 10 )3

V 0.9 volts V 0.9 volts

V 0.9 volts V 1.56

j j

j j j j

volts

9

Per Phase Example, cont’d

• What is real power into load?• Is this a reasonable dispatch of generators?• What is causing real power flow from Y-connected

generator to -connected generator?

10

** '

ygen

*

'' '''

S 3 5.1 3.5 VA0.1

3 5.1 4.7 VA0.1

a aa a a

a agen a

V VV I V j

j

V VS V j

j

Power System Operations Overview

Goal is to provide an intuitive feel for power system operation

Emphasis will be on the impact of the transmission system

Introduce basic power flow concepts through small system examples

11

Power System BasicsAll power systems have three major

components: Generation, Load and Transmission/Distribution.

Generation: Creates electric power.Load: Consumes electric power.Transmission/Distribution: Moves electric power

from generation to load. – Lines/transformers operating at voltages above 100

kV are usually called the transmission system. The transmission system is usually networked.

– Lines/transformers operating at voltages below 100 kV are usually called the distribution system. The distribution system is usually radial except in urban areas. 12

Small PowerWorld Simulator Case

Bus 2 Bus 1

Bus 3Home Area

204 MW

102 MVR

150 MW

150 MW 37 MVR

116 MVR

102 MW 51 MVR

1.00 PU

-20 MW 4 MVR

20 MW -4 MVR

-34 MW 10 MVR

34 MW-10 MVR

14 MW -4 MVR

-14 MW

4 MVR

1.00 PU

1.00 PU

106 MW 0 MVR

100 MWAGC ONAVR ON

AGC ONAVR ON

Load withgreenarrows indicatingamountof MWflow

Usedto controloutput ofgenerator

Direction of arrow on line is used to Indicate direction of real power (MW) flow

Note realand reactivepower balance ateach bus13

Closed circuit breakeris shown as red box

Pie chart and numbers show real and reactive power flow

Voltageshown innormalized“per unit”values

Power Balance Constraints

Power flow refers to how the power is moving through the system.

At all times in the simulation the total power flowing into any bus MUST be zero!

This is due to Kirchhoff’s current law. It can not be repealed or modified!

Power is lost in the transmission system:If losses are small, the sending and receiving end power

may appear the same when shown to two significant figures.

14

Basic Power Control

Opening a circuit breaker causes the power flow to (nearly) instantaneously change.

No other way to directly control power flow in an AC transmission line.

By changing generation (or, in principle, by changing load) we can indirectly change this flow.

15

Transmission Line LimitsPower flow in transmission line is limited by

heating considerations.Losses (I2 R) can heat up the line, causing it to sag.Each line has a limit:

Simulator does not allow you to continually exceed this limit.

Many transmission owners use winter/summer limits.Some transmission owners, eg Oncor, are moving to

“dynamic” ratings that consider temperature etc.

16

Overloaded Transmission Line

17