signals and systems chapter 12

8/19/2019 signals and systems chapter 12

1/34

1

Chapter 12

Three-Phase Circuit

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.


2/34


3/34

3

Structure of ower System

: 10~20kV

: 154, 345, 765kV

: 154, 66kV : 380V

: 120/220V: 6.6, 11.4, 22.9kV

Energy

Source

Fossil

Fuel

Hydro

Power

Power Plant

Ther

mal

Mecha

nical

Electri

cal

Prime Mover Generator

Transf

ormati

on

Transf

ormer

Power

Transmission

Line

Transf

ormati

on

High

Voltage

Transmis

sion

Netwo

rk

Trans

format

ion

Distribu

tion

Station

Small

Consu

mption

Transf

ormer

Transformer

for Distribution

Ho

me

Small

Consu

mer

Fac

toryLarge

Consu

mer

L

O

A

D

Motor

Heater

Light

and Etc.

Atom

Fuel

Ther

mal

Mecha

nical

Electri

cal

Mecha

nical

Electri

cal

Substation Substation

Power

Distribution

Line


4/34

4

12.1 What is a Three-Phase Circuit?

a

b

A

B

V pÐf Z L

Single Phase vs. Three Phase

N

S

a

n

b

n

c n

N

S

a

b

magneto

a

b

A

B

V pÐ0°

Z L

c C

V pÐ-120°

n N

V pÐ+120°

Z L

Z L


5/34

5

• It is a system produced by a generator consisting ofthree sources having the same amplitude andfrequency but out of phase with each other by 120°.


Three sources

with 120° out

of phaseFour wired

system

neutral line


6/34

6

Advantages:

1. Most of the electric power is generated anddistributed in three-phase.

2. The instantaneous power in a three-phase systemcan be constant.

3. The amount of power of the three-phase systemis more economical than that of the single-phase.

4. In fact, the amount of wire required for a three-

phase system is less than that required for anequivalent single-phase system.



7/34

7

12.2 Balanced Three-Phase Voltages

• Balanced three-phase voltages are defined as the voltagesequal in magnitude and are out of phase with each other by120 o.

Sinusoidal form of the balanced three-phase voltages


8/34

8


• Two possible configurations:

Three-phase voltage sources:

(a) Y-connected configuration; (b) Δ-connected configuration


9/34

cnbnan

p pcn

pbn

pan

leadsturninwhichleads

V V

V

V

V,VV

V120V240V

V120V

V0V

°Ð°-Ð

°-Ð°Ð

9

12.2 Balanced Three-Phase VoltagesThree-phase voltage configuration has two possible combinations

with different phase sequence.

In both cases,

abc or positive sequence(clockwise direction) acb or negative sequence(counterclockwise direction)

bncnan

p pbn

pcn

pan

leadsturninwhichleads

V V

V V

V,VV

V120V240V

V120VV0V

°Ð°-Ð

°-Ð°Ð

effective or rms

0an bn cn

an bn cn

v v v

v v v

+ +


10/34

10


• Balanced phase voltages are equal inmagnitude and are out of phase with each otherby 120°.

• The phase sequence is the time order in whichthe voltages pass through their respectivemaximum values.

Graph in the time domain for the positive sequence


11/34

11


Example 1

Determine the phase sequence of the

set of voltages.

)110cos(200)230cos(200

)10cos(200

°- °-

°+

t vt v

t v

cn

bn

an


12/34

12


Solution:

The voltages can be expressed in phasor formas

We notice that Van leads Vcn by 120° and Vcn inturn leads Vbn by 120°.

Hence, we have an acb (negative) sequence.

V110200V

V230200V

V10200V

°-Ð

°-Ð

°Ð

cn

bn

an


13/34

13

12.3 Balanced Three-Phase Connection

Balanced Load Configuration

Z1

Z2

Z3

a

b

c

n

Za

ZbZc

a

b

c

Y connected Load D connected Load

- A balanced load is one that in which the phase impedance are equal in

magnitude and phase.

1 2 3, =Y a b c Z Z Z Z Z Z Z Z D

- Y-connected Load can be interchangeable with D connected load each other.

13 , ,3

Y Y Z Z Z Z D D


14/34

Delta-Y Conversion

Similarly

321

213 )(

R R R

R R R

R R cb ++

+

+ 321321 )(

R R R

R R R

R R ac ++

+

+ Rbc = Rca =(2) (3)

From (1), (2), (3)

R R R

R R R

R R R

R R R

R R R

R R

R

c

b

a

++

++

++

321

13

321

32

321

21

++

++

++

R

R R R R R R R

R

R R R R R R R

R

R R R R R R R

a

accbba

c

accbba

b

accbba

3

2

1

Delta to Y

Y to Delta

Delta to Y

conversion

Y to Delta

conversion

(4)

(5)

(6)14


15/34

15


•Four possible connections

1. Y-Y connection (Y-connected source with a

Y-connected load)

2. Y-Δ connection (Y-connected source with aΔ-connected load)

3. Δ-Δ connection

4. Δ-Y connection


16/34

16

Three Line (Wire) System

, , L ab bc cav v v v

- Line Current

, , L a b c I I I I - Line Voltage

A

B

C

a

c

b

Ia

Ib

Ic

Three Phase

Signal Source

Three Phase

Load

wrt the

interconnection

Ib

Ic

ZY

ZY ZY

A

B C

Ia

N

I AN

IBN ICN

Ib

Ic

A

B C

ZD ZD

ZDIBC

I AB

ICA

Ia

1

2

, ,, ,

L AN AN CN

L AB BC CA

v v v vv v v v

- Phase Current

1

2

, ,

, ,

P AN BN CN

P AB BC CA

I I I I

I I I I

- Phase Voltage

Y-connected

Δ -connected

Y-connected

Δ -connected

In Y-connected system

phase current = line current

phase voltage ≠ line voltage

In Δ -connected system

phase current ≠ line current

phase voltage = line voltage

Need to find phase (line) current & phase (line) voltage


17/34


18/34

18


The line voltages lead their corresponding phase voltage by 30o.

°-Ð-

°-Ð-

°Ð

++

°-Ð-°Ð-

°Ð

°-Ð

°Ð

2103VVV

903VVV

similarly,

3032

3

2

11

1200VVV

V,V,V

voltageslineorvoltagesline-to-lineThe

V120V

V120V

V0V

pancnca

pcnbnbc

p p

p pbnanab

cabcab

pcn

pbn

pan

V

V

V jV

V V

areand

V

V

V

Phasor diagram illustrating the relationship between line and phase voltages in

a balanced Y-Y system.

Van

Vbn

Vcn Vab

Vbc

Vca

Line voltage:abc sequence


19/34

19


The voltage across the neutral line is zero, so the neutral line can be removed

without affecting the system.

0IZV

0III-I

0III

240I240VV

I

120I

120VV

I

VI

++

++

-Ð-Ð

-Ð

-Ð

nnnN

cccn

ccc

o

a

Y

o

an

Y

cnc

o

aY

o

an

Y

bn

b

Y

ana

or

Z Z

Z Z

Z

Applying KVL to each phase,

we obtain line currentsby KVL

- Only three wires are needed in the balanced three phase system

- Find line currents, which are phase currents in Y-configuration

Line current:

abc sequence


20/34

20


An alternative way of analyzing a balanced Y-Y system is to do so on

a “per phase” basis.

From Ia, we use the phase sequence to obtain other line currents. Thus, as long

as the system is balanced Y-Y connected, we need only analyze one phase.

Y

ana

Z

VI


21/34

21


Example 2

Calculate the line currents in the three-wire Y-Y

system shown below:

A

A

A

j j j Z

Z

ooo

ac

oo

ab

o

o

o

a

o

Y

Y

ana

2.9881.68.26181.6240II

8.14181.6120II

8.2181.68.21155.16

0110I

8.21155.1661581025

VI

Ð-Ð-Ð

-Ð-Ð

-ÐÐÐ

Ð+++-

Balanced Y-Y system ? OK. Then,line impedance

load impedance


22/34


23/34

23


°-Ð°-Ð

°-Ð-+

°-Ð-°Ð

°-Ð

°-Ð°-Ð°Ð

303II,303II,

303I866.05.01I

240101II-II

240II

240II,120II,0II

I-III-III-II ,,,

CAc BC b

AB AB

ABCA ABa

ABCA

ABCA AB BC m AB

BC CAc AB BC bCA ABa

similarly

j

Since

The line currents are obtained from the phase currents.

CA BC AB p

cba L

p L

I

I

I I

III

III

where,3

Phasor diagram illustrating the

relationship between phase

and line currents in delta

connection.

An alternative way of analyzing the Y-∆ circuit is to

transform the ∆ connected load to an equivalent Y

connected load. Using Y-∆ transformation formula

3

ZZ DY

phase current

line current

Each line current lags the

corresponding phase

current by 30o.

abcsequence


24/34

24

12.3 Balanced Three-Phase ConnectionExample 3

A balanced abc -sequence Y-connected source with( ) is connected to a Δ-connected load (8+j4) per phase. Calculate the phase and line currents.

SolutionUsing single-phase analysis,

Other line currents are obtained using the abc phasesequence

°Ð 10100Van

A57.1654.33

57.26981.2

10100

3/Z

VI an °-Ð

°Ð

°Ð

D

a

A43.10354.33120II

A57.13654.33120II

°Ð°+Ð

°-Ð°-Ð

ac

ab

°-Ð 303II ABa


25/34

25

• A balanced Δ- Δ system is a three-phase system with a

balanced Δ -connected source and a balanced Δ -connectedload.


V120V

V120V

V0V

°Ð

°-Ð

°Ð

pca

pbc

pab

V

V

V

Assuming positive sequence,

In this configuration, line voltages are same asphase voltages.

If there is no line impedance, the phase voltages

of the delta-connected source are equal to the

voltages across the impedances;

CAca Bbc ABab VV,VV,VV C

Hence, the phase currents are

DD

DD

DD

Z Z

Z Z

Z Z

caCACA

bc BC BC

ab AB A

VVI

,VV

I

,VV

I B


26/34

26

• A balanced Δ- Δ system is a three-phase system with a balanced Δ -

connected source and a balanced Δ -connected load.


The line currents are obtained from the phase currents by

applying KCL. (same w/ balanced Y-Δ system)

°-Ð°-Ð

°-Ð-+

°-Ð-°Ð°-Ð

303II,303II,

303I866.05.01I

240101II-II240II

I-III-III-II ,,,

CAc BC b

AB AB

ABCA ABa

ABCA


similarly

j

Since

CA BC AB p

cba L

p L

I

I

I I

III

III

where,3


corresponding phase current

by 30o.

An alternative way of analyzing the Δ- Δ circuit is to convert both the source and the load to their Y equivalents.

3ZZ DY see the next section.



and line currents.

Each line current lags thecorresponding phase

current by 30o.


27/34

27

• A balanced Δ- Δ system is a three-phase system with a balanced Δ -

connected source and a balanced Δ -connected load.


The line currents are obtained from the phase currents by

applying KCL. (same w/ balanced Y-Δ system)

°-Ð°-Ð

°-Ð-+

°-Ð-°Ð°-Ð

303II,303II,

303I866.05.01I

240101II-II240II

I-III-III-II ,,,

CAc BC b

AB AB

ABCA ABa

ABCA


similarly

j

Since

CA BC AB p

cba L

p L

I

I

I I

III

III

where,3


corresponding phase current

by 30o.

An alternative way of analyzing the Δ- Δ circuit is to convert both the source and the load to their Y equivalents.

3ZZ DY see the next section.



and line currents.

Each line current lags thecorresponding phase

current by 30o.


28/34

• A balanced Δ-Y system is a three-phase system with abalanced Δ -connected source and a balanced Y-

connected load.


Assuming positive sequence,

V120VV,120VV,0V °Ð°-Ð°Ð pca pbc pab V V V

These are also the line voltages and phase voltages.

We can obtain the line currents in many ways.

Method1.

Apply KVL to loop aANBba, writing

Y

o

pba

o pabbaY

bY aY ab

Z V I I V I I Z

or I Z I Z

00V

0V-

Ð-Ð-

-+

Since, ,120 sequenceabcin I I oab -Ð

oaa

o

a ba

303I2

3 j

2

11I12011III Ð

++-Ð--

o

ac

o

ab

Y

o

p

a I I I I Z

V I 120,120,

303/Ð-Ð

-Ð

2 3 l d h h


29/34

Balanced Δ-Y system

12.3 Balanced Three-Phase ConnectionMethod2.

To replace the delta-connected source with its equivalent Y-

connected source.

☞ Line voltages of a Y-connected source lead their correspondingphase voltage by 30o. Therefore, we obtain each phase voltage of

the equivalent Y-connected source by and shifting its phase -30o.3

°+Ð°-Ð°-Ð 903

V,1503

V,303

V p

cn

p

bn

p

an

V V V

single phase equivalent circuit

Y

p

a Z

V 303/I

°-Ð

°-Ð 303

V p

an

V

V0V °Ð pab V

Recalling

V120VV,120VV,0V °Ð°-Ð°Ð pca pbc pab V V V positive

sequence

Same result

from the KVL

method


30/34

30

12.4 Power in a Balanced System

• Let’s examine the instantaneous power observed by the load.

☞ In the time domain, for a Y-connected load, the phase voltages are

o pCN o p BN p AN t V vt V vt V v 120cos2,120cos2,cos2 +- necessary because V p has been defined as the rms value of the phase voltage.

, Ð Z Z If Y

o pco pb pa t I it I it I i 120cos2,120cos2,cos2 +----

☞ The total instantaneous power in the load is the sum of the instantaneous powers in the three phases

oooo

p p

cCN b BN a AN cba

t t t t t t I V

iviviv p p p p

120cos120cos120cos120coscoscos2 +-++---+-

++++

☞ Applying the trigonometric identity B A B A B A -++ coscos2

1coscos

12 4 P i B l d S t


31/34

31

12.4 Power in a Balanced System

• Let’s examine the instantaneous power observed by the load.

cos3cos2

12coscos3

)2(

240sinsin240coscos240sinsin240coscoscoscos3

2402cos2402cos2coscos3

p p p p

oooo

p p

oo

p p

I V I V

t where

I V

t t t I V p

-++

-

-++++

+-+--+-+

Total instantaneous power in a balanced three-phase system is constant !

The average power per phase P p for either Y-connected or ∆-connected load is p/3 or cos p p p I V P

and the reactive power per phase is sin p p p I V Q

The apparent power per phase is p p p I V S , the complex power per phase is*IVS p p p p p jQ P +

phase voltage and phase current, respectively.The total average/reactive power:

)3:,3:(

sin33sin3,cos3cos33

p L p L p L p L

L L p p p L L p p pcba

V V but I I load connected V V but I I load connected Y

I V Q I V Q I V I V P P P P P

-D-

++

The complex power: Ð+ L L p p p p p jQ P IV3Z3II3VS3S2*


32/34

32

12.5 Unbalanced Three-Phase Systems

• An unbalanced system is due to unbalanced voltage sources or an

unbalanced load.

0)III(I

,Z

VI,

Z

VI ,

Z

VI

++-

cban

C

CN c

B

BN b

A

AN a

• To calculate power in an unbalanced three-phase system requires that we find the

power in each phase.

• The total power is not simply three times the power in one phase but the sum of

the powers in the three phases.

☞ Unbalanced three-phase Y-connected load

(1) The source voltages are not equal in magnitude and/or differ in phase by angles are unequal, or (2)

load impedances are unequal.

Neutral line can not be removed.


33/34

33

12.5 Unbalanced Three-Phase SystemsExample 7

Consider the ∆- ∆ system shown below. Take Z1=8+6j ,Z2=4.2-2.2j , Z3=10+0j . (a) Find the phase current I AB, IBC,

and ICA. (b) Calculate the line currents IaA, IbB, and IcC.


34/34

34

Problems

12.8

12.10

12.13

12.18

12.21

12.25

12.31

12.40

12.42

12.54

signals and systems chapter 12

Documents