ece 330 power circuits and … · mechanical energy into electric energy through a three- ......

ECE 330 POWER CIRCUITS AND

ELECTROMECHANICS

LECTURE 1

COURSE INTRODUCTION

AND REVIEW OF PHASORS

Acknowledgment-These handouts and lecture notes given in class are based on material from Prof. Peter

Sauer’s ECE 330 lecture notes. Some slides are taken from Ali Bazi’s presentations.

Disclaimer- These handouts only provide highlights and should not be used to replace the course textbook.

1/16/2018

WHY POWER CIRCUITS AND ELECTROMECHANICS?

Electromechanics combines electrical and mechanical

processes and plays key roles at several levels:

At generation level: synchronous machines convert

mechanical energy to electric energy.

Source: slideshare.net Source: water.usgs.gov

1/16/2018 2


At transmission-distribution level: Circuit breakers trip the

faults in affected areas.

Source: abb.com Source: engineeringtutorial.com

1/16/2018 3

I

Ngxlp μ

μ

Xo

Core mean length: lc

Core area: Ac


At the level of loads: Electric energy is consumed by

mechanical loads through motors, mostly induction (three-

phase and single-phase)

Source: pumpsandsystem.com Source: electrical-know-how.com

1/16/2018

4

WHY POWER CIRCUITS AND ELECTROMECHANICS ?

In advanced technology: renewable energy, electric cars,

robots,….. etc. contain electromechanic systems.

Wind Turbine Electric Vehicle Robot

Source: energy.gov Source: exchangeev.aaa.co m Source: yaskawa.co.jp

1/16/2018 5

Goals:

To impart basics of three-phase power circuits, transformers,

electromechanical systems, and rotating machines.

Objectives:

To build cognitive skills, such as analytical thinking and

problem solving. The course integrates fundamentals of

advanced math and science to enhance the ability to design a

system and meet desired needs.

1/16/2018

6

POWER CIRCUITS AND ELECTROMECHANICS

COURSE OUTLINE

Review of phasors, complex power

Three-phase circuits, three-phase power, wye-delta conversion

Magnetic circuits, self and mutual inductance, Maxwell’s equations

Ideal transformers, practical transformers, equivalent circuits

Electromechanical systems, energy, co-energy, energy cycles,

computation of forces and torques.

Dynamic equations, numerical integration of electromechanical

systems

Equilibrium points, linearization, stability

Synchronous machines

Induction machines

Other machines

1/16/2018 7

STRUCTURE OF POWER SYSTEMS

• Power system’s main components: generation, transmission,

distribution, and loads.

Source: image.slidesharecdn.com

1/16/2018 8


Generating System:

The generating system consists of a fuel source such as

coal, water, natural gas or nuclear power.

Hydropower accounts for about 8%, and nuclear power

20%, renewable energy 14% of electric energy production

in the US.

The turbine acting as the prime-mover converts

mechanical energy into electric energy through a three-

phase synchronous machine.

The three-phase voltages of the generators range from

13.8 kV to 24 kV.

1/16/2018 9


Transmission System:

A three-phase transformer at each generator steps up the

voltage to a high value ranging from 138 kV to 765 kV.

The transmission lines carry power over these lines to load

centers and, where appropriate, step it down to lower

voltages up to 34 kV at the bulk-power substations.

Some industrial customers are supplied from these

substations. This is known as the transmission—sub-

transmission system.

1/16/2018 10


Distribution System:

Transformers step down the voltage to a range of 2.4 kV to

69 kV.

Power is carried by main feeders to specific areas where

there are lateral feeders to step it down further to customer

levels, such as 208, 240, or 600 volts.

The distribution transformers serve anywhere from 1 to 10

customers.

11 1/16/2018


The classes of entities in the electricity market are:

Generator companies (GENCOs), also called

independent power producers (IPPs).

Transmission companies TRANSCOs,

Their primary responsibility is to transport power

from generators to customers and make the

transmission system available to all.

12 1/16/2018


Distribution companies DISCOs,

owning and operating the local distribution network.

Independent system operator (ISO)

The ISO is charged with ensuring the reliability and

security of the entire system.

13 1/16/2018

DYNAMICS OF POWER SYSTEMS AND COMPONENTS

The power system is a dynamic one, it is described by a

set of vector of differential equations

The time scales in the response of this equation range

from milliseconds, in the case of electromagnetic

transients, to a few seconds in the control of frequency,

and a few hours in the case of boiler dynamics.

Therefore, we analyze such equations for time-domain

response, steady-state sinusoidal behavior, equilibrium

points, stability, etc. 1/16/2018 14

( )x f x u


15 1/16/2018

Power system dynamic structure

v, i

Fuel

Source

Furnace

& Boiler Turbine

Generator

Fuel Steam

Torque

Electrical

Energy

Mechanical

Energy

Supply

Control

Pressure

Control

Speed

Control

Voltage

Control

Network &

Network

Control

Load

P,Q

Load

Control

Energy

Control

Center


Diagram does not show all the complex dynamic

interaction between components and their controls.

Electrical side contains mechanical dynamics(TCUL)

Mechanical side contains components with electrical

dynamics (electrical valves)

After the model is derived we put it in the state-space

form.

1/16/2018 16

REVIEW OF PHASORS

• Phasors represent quantities with magnitude and angle

with respect to a reference and are commonly used in

energy systems.

• Example:

Euler’s expansion:

Then, v (t ) can be written as

where is the RMS phasor with

cosine reference. 1/16/2018 17

( ) cos ( )m vv t V t

cos( ) sin( )je j

( )( ) Re{ } Re{ } Re{ 2 }v vj t jj t j t

m mv t V e V e e Ve

2

vmrms v

jVV e V

REVIEW OF PHASORS

• Recall: RMS (root-mean square) where v is a

function of time, t0 is the initial time, and T is the

period of v.

• How do phasors apply in electric circuits?

• Example:

Find

1/16/2018 18

2 2 21 1( ) cos ( )

2

o o

o o

t T t T

mrms m v

t t

VV v t dt V t dt

T T

R L

+

v(t)

-

i(t)

( ) sin( ) Amv t for i t I t

REVIEW OF PHASORS

1/16/2018 19

R L

+

v(t)

-

i(t)

• Time-domain approach:

For R = 2Ω, L = 1/377H, ω = 377 rad/s, and Im = 10A,

2

2 11 2 10( ) 2 10 377 10 cos 377 tan

1377377 10

377

v t t

2 2 1

( )( ) ( )

( ) sin( ) cos( )

( ) cos tan

m m

mm m

m

di tv t Ri t L

dt

v t RI t L I t

RIv t RI L I t

L I

( ) 22.36cos(377 63.4 )v t t V

REVIEW OF PHASORS

20

• Frequency-domain approach (phasors)

• In frequency-domain (Ohm’s law)where Z can

be a series-parallel combination of R, XC , and/or XL.

Time-domain Frequency-domain

( )

( ) ( )

1 1 ( )

L L

C C

v t V R R

i t I L j L jX X L

jC jX X

j C C C

V Z I

1/16/2018

REVIEW OF PHASORS

1/16/2018 21

R jωL

+

V

-

I

• Frequency-domain approach:

2 2 1

10( ) where 90 A (RMS)

2

1 102 377 90

377 2

1 102 1 tan 90

2 2

10 526.56 90

2

o

o

o

o o

V R j L I I

V j

V

V

15.81 63.4 V(RMS)oV

TWO-TERMINAL NETWORK

• A two-terminal electrical network has voltage at its

terminals and current flowing in and out of its

terminals.

• The instantaneous power is

• For and

we get

1/16/2018 22

( ) ( ) ( ).p t v t i t

+

v(t)

-

i(t)

Load

Notation

Source

Notation

+

v(t)

-

i(t)

( ) cos( ) Am ii t I t ( ) cos( ) Vm vv t V t

( ) cos( )cos( )m m v ip t V I t t


The first term is time-independent, while the second

term is a sinusoid at double frequency.

1/16/2018 23

( ) cos( )+ cos(2 ) W2 2

m m m mv i v i

V I V Ip t t

1cos cos [cos( )+cos( )]

2


• The average power is thus

,

• This is called the active or real power and its unit is

watts (W).

• The power factor is the cosine of the phase angle

between v (t) and i(t). 1/16/2018 24

cos( ).2

m min v i

V IP

0

1( ) ( )

T

P P t d tT

2

T

POWER FACTOR

• The power factor (P.F.) is thus

• The power factor can be:

– Lagging:

– Leading:

– Unity:

Therefore, ,

and the highest real power

exists when P.F.=1. Source: grupovision.com

1/16/2018 25

. . cos( ).v iP F

0 90o o

v i

90 0o o

v i

0v i

0 . . 1P F

APPARENT POWER AND REACTIVE POWER

• The apparent power is

• The apparent power unit is volt-amps (VA).

• The reactive power is

• The reactive power unit is volt-amps-reactive

(VARs).

1/16/2018 26

sin( ).2

m min v i

V IQ

2

m mV IS

COMPLEX POWER

• The instantaneous power is

• The time varying component

1/16/2018 27

( ) cos( )+ cos(2 ) W2 2

m m m mv i v i

V I V Ip t t

cos (2 ) cos[ (2 2 ) ( )]2 2

cos (2 2 )cos ( ) sin (2 2 )sin ( )2 2

m m m mv i i v i

m m m mi v i i v i

V I V It t

V I V It t

COMPLEX POWER

• Define

• The real power can be written as

1/16/2018 28

sin( ), (Reactive power)2

m min v i

V IQ

( ) cos(2 2 ) sin(2 )

(1 cos(2 )) sin(2 2 )

in in i in i

in i in i

p t P P t Q t

P t Q t

cos( ) V cos( )2

m min v i rms rms v i

V IP I

PHASOR REPRESENTATION

• The reactive power can be written as

• The voltages and currents can be written as phasors:

Source: Tonex.com

1/16/2018 29

Re{ } Re{ }2

v i v ij j j jm min rms rms

V IP e e V e I e

Im{ } Im{ }2

v i v ij j j jm min rms rms

V IQ e e V e I e

and .v ij j

rms rmsV e V I e I

Re( ) V cos( )in rms rms v iP V I I

Im( ) V sin( )in rms rms v iQ V I I

Complex Power

• Define the complex power as

• Then can be written as

• The quantity is the complex conjugate of

• can also be written as

• Note that

1/16/2018 30

in inS P jQ

* S V I

*I .I

( )v iS S

S

S

2 2

2

m min in

V IS P Q

S

inP

inQ

( )v i

ALTERNATE FORMS OF COMPLEX POWER

• If the load is connected across the source

By Ohm’s law: , but

Then can be written as Also,

and

• Thus, Q > 0 when is inductive,

and Q < 0 when is capacitive,

• and are not phasors but complex quantities.

1/16/2018 31

,Z R jX

2 2S I R jI X

S

S

V Z I

Z

2 2 and ,P I R Q I X Z

Z

. . cos( ( )).P F angle Z

X L

1X

C

V

*S V I

EXAMPLE: LC FILTER AND R LOAD

• The circuit shown is commonly used as an LC filter

to supply a load, which is resistive in this case.

• Find the current, real, reactive, and complex powers,

and the P.F. for

1/16/2018 32

2

/ /

( )

jZ j L R

C

L j RLC RZ

RC j

R

L

+

v(t)

-

i(t)C

( ) 2 cos(377 )rmsv t V t

EXAMPLE: LC FILTER AND R LOAD

• Let

1/16/2018 33

*

0.0197 39.41 0.0152 0.0125

120 06091.4 39.41 A

0.0197 39.41

( ) 6091.4 2 cos(377 39.41 )

731 39.41 kVA

731cos( 39.41 ) 564.8kW

731sin( 39.41 ) 464.1kVAR

P.F.= cos( 39.41 ) 0.77

o

oo

o

o

o

o

in

o

in

o

Z j

VI

Z

i t t

S V I

P

Q

3 leading ( 39.41 )o

v i

R

L

+

v(t)

-

i(t)C

120 , 1mH, 6.8mF, and R 10 .rmsV V L C

ece 330 power circuits and … · mechanical energy into electric energy through a three- ......

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