Chapter 12: Three-Phase Circuits - Amazon...

EEEB123 Chapter 12: Three-Phase Circuits Dr. C.S.Tan

Chapter 12: Three-Phase Circuits 12.1 Introduction 12.2 Balanced Three-Phase Voltages 12.3 Balanced Wye-Wye connection 12.4 Balanced Wye-Delta Connection 12.7 Power in a Balanced System 12.1 INTRODUCTION A single-phase ac power system consists of a generator connected through a pair of wires (i.e., a transmission line) to a load as shown in figure below:

Circuits or systems in which the ac sources operate at the same frequency but different phases are known as polyphase. The following figures below shows a two-phase three-wire system and a three-phase four-wire system.

As distinct from a single-phase system, a two-phase system is produced by a generator consisting of two coils placed perpendicular to each other so that the voltage generated lags the other by 90º. Similarly, a three-phase system is produced by a generator consisting of three sources having a same amplitude and frequency but out of phase with each other by 120º. Since the three-phase system is by far the most prevalent and most economical polyphase system, hence all the discussion in this chapter will be mainly on three-phase systems.


Three-phase systems are important for several reasons: • Nearly all electric power is generated and distributed in three-phase at the operating

frequency of 60Hz or ω=377rad/s (i.e., US) or 50Hz or ω=314rad/s (i.e., East Japan). • When single-phase or two-phase inputs are required, they can be taken from the three-

phase system rather than generated independently. If more phases like 48 phases are required for industry usage, three-phase supplied can be manipulated to provide 48 phases

• The instantaneous power in a three-phase system can be constant (i.e., not pulsating).This results in uniform power transmission and less vibration of three-phase machines.

• For the same amount of power, the three-phase system is more economical than the single-phase system.

• The amount of wire required for a three-phase system is less than that required for an equivalent single-phase system.

12.2 BALANCED THREE_PHASE VOLTAGES This section will begin with a discussion of balanced three-phase voltages. Three-phase voltages are often produced with a three-phase ac generator (or alternator) whose cross-sectional view is shown below.

The generator basically consists of a rotating magnet called the rotor surrounded by a stationary winding called the stator. Three separate windings or coils with terminal a-a’, b-b’ and c-c’ are physically placed 120º apart around the stator. As the rotor rotates, its magnetic field ‘cuts’ the flux from the three coils and induces voltages in the coils. Since the coils are placed 120º apart, the induced voltages in the coils are equal in magnitude but out of phase by 120º shown below.


Since each coil can be regarded as a single-phase generator by itself, the three-phase generator can supply power to both single-phase and three-phase loads. A 3-phase voltage system is equivalent to three single-phase systems. The voltage sources (generator connection) can be connected in wye-connected shown in (a) or delta-connected as in (b)

The voltages Van, Vbn and Vcn are between lines a, b and c and the neutral line, n respectively. Van, Vbn and Vcm are called phase voltages. It is called balanced if they are same in magnitude and frequency and are out of phase from each other by 120º such that

0an bn cnV V V+ + = ….. (12.1)

an bn cnV V V= = ….. (12.2)

There are two possible combinations; positive sequence and negative sequence Positive/ abc Sequence (Balanced) Negative / acb Sequence (Balanced)

Let 0an pV V °= ∠ vector diagram for phase abc phase sequence gives

Assuming anV as the reference hence

0 an p p pV V and V V θ° °= ∠ = ∠

Where pV =amplitude of phase voltage in rms

Thus; 0

120

( 120 ) ( 240 )

240

( 240 ) ( 120 )

an p p

bn an

p p

cn an

p p

V V V

V V

V V

V V

V V

θ

θ θ

θ θ

° °

°

° ° ° °

°

° ° ° °

= ∠ = ∠

= ∠−

= ∠ − = ∠ +

= ∠−

= ∠ − = ∠ +

….(12.3)

This sequence is produced when the rotor rotates in the counterclockwise direction.

Let 0an pV V °= ∠ vector diagram for phase acb phase sequence gives

Assuming anV as the reference hence

0 an p p pV V and V V θ° °= ∠ = ∠

Where pV =amplitude of phase voltage in rms

Thus; 0

240

( 240 ) ( 120 )

120

( 120 ) ( 240 )

an p p

bn an

p p

cn an

p p

V V V

V V

V V

V V

V V

θ

θ θ

θ θ

° °

°

° ° ° °

°

° ° ° °

= ∠ = ∠

= ∠−

= ∠ − = ∠ +

= ∠−

= ∠ − = ∠ +

…. (12.4)

This sequence is produced when the rotor rotates in the clockwise direction.


Mathematically, both phase sequences in Eq. (12.3) and Eq. (12.4) satisfy Eq. (12.1). For example, from Eq. (12.4)

( )( )

( 240 ) ( 120 )

1 0 1 240 1 120

1 ( 0.5 0.866) ( 0.5 0.866) 0

an bn cn p p p

p

p

V V V V V V

V

V j j

θ θ θ

θ

θ

° ° ° ° °

° ° ° °

°

+ + = ∠ + ∠ − + ∠ −

= ∠ ∠ + ∠− + ∠−

= ∠ + − + + − − =

Like the generator connections, loads can also be connected in Y or ∆. Balanced load is where the phase impedances are equal in magnitude and are in phase

Y-connected Load (Balanced) ∆-connected Load (Balanced) Three-phase load configuration:

For balanced Y-connected load:

1 2 3 YZ Z Z Z= = = ….. (12.5) Where YZ is the Total load impedance per phase

Three-phase load configuration:

For balanced Y-connected load:

a b cZ Z Z Z Δ= = = ….. (12.6) Where ZΔ is the Total load impedance per phase Recall from earlier chapter 9, Eq. (9.69) that

13 3

Y YZ Z or Z ZΔ Δ= = …..(12.7)

Hence, Y-connected load can be transformed into a ∆-connected load or vice versa using Eq. (12.7)

Since both sources and loads can be connected in either Y or ∆, there are 4 possible connections:

• Y-Y connection (i.e., Y-connected source with Y-connected load) • Y-∆ connection (i.e., Y-connected source with ∆-connected load) • ∆-∆ connection (i.e., ∆-connected source with ∆-connected load) • ∆-Y connection (i.e., ∆-connected source with Y-connected load)


12.3 BALANCED WYE-WYE CONNECTION The balanced four-wire Y-Y system shown below has a Y-connected source on the left and Y-connected load on the right. Assume a balanced load (i.e., impedances are equal).

YZ = Total load impedance per phase, Ω/phase sZ = Source impedance per phase, Ω/phase lZ = Line impedance per phase, Ω/phase LZ = Load impedance per phase, Ω/phase nZ = impedance of the neutral line per phase, Ω/phase

Thus in general Y s l LZ Z Z Z= + + ….. (12.8)

s lZ and Z are often very small compared with LZ therefore, one can assume that Y LZ Z= if no source or line impedance is given. By lumping the impedances together, the Y-Y system shown above can be simplified as below

Voltages: Assuming the source rotates in positive sequence, the phase voltages (i.e., line-to-neutral voltages) are (from Eq. (12.3)) p pwhere V V θ °= ∠

0 ,

120 ( 120 ) ( 240 )

120 ( 240 ) ( 120 )

an p an p

an p bn p p

an p cn p p

V V V V

V V or V V V

V V V V V

θ

θ θ

θ θ

° °

° ° ° ° °

° ° ° ° °

= ∠ = ∠

= ∠− = ∠ − = ∠ +

= ∠+ = ∠ − = ∠ +

….. (12.9)


From the phasor diagram, the line voltages (i.e., line to line voltages) , ab bc caV V and V are related to the phase voltages. For example,

( )

( )( )

( )( ) ( )( )( )

( )( )

120

(1 0 1 120 ) 1 ( 0.5 0.866)

1.5 0.866 3 30

3 30

3 30 ( )

ab an nb an bn p p

p p

p p

p

an

V V V V V V V

V V j

V j V

V

V proved

θ θ

θ θ

θ θ

θ

°°

° ° ° °

° ° °

°

°

= + = − = ∠ − ∠ −

= ∠ ∠ − ∠− = ∠ − − −

= ∠ + = ∠ ∠

= ∠ +

= ∠

Similarly, we can get bc caV and V as follows:

( ) ( )( )( ) ( )( )( ) ( )( )

3 30 3 30 3 ( 30 )

3 30 ( 120 ) 3 30 3 ( 90 )

3 30 ( 120 ) 3 30 3 ( 150 )

ab an p p

bc bn p p

ca cn p p

V V V V

V V V V

V V V V

θ θ

θ θ

θ θ

° ° ° ° °

° ° ° ° ° °

° ° ° ° ° °

= ∠ = ∠ ∠ = ∠ +

= ∠ = ∠ − ∠ = ∠ −

= ∠ = ∠ + ∠ = ∠ +

….. (12.10)

Note: The magnitude of the line voltages LV is 3 times the magnitude of the phase voltages pV

and the line voltages lead their corresponding phase voltage by 30° (See phasor diagram below)

3L pV V=

Where an bn cnp

ab bc caL

V V V V and

V V V V

= = =

= = =

From Phasor diagram (Positive Sequence)

a b a n n b

b c b n n c

c a c n n a

V V V

V V V

V V V

= +

= +

= +

….. (12.11)

Phase Voltages:

120 240

( 120 ) ( 240 )

240 120

( 240 ) ( 120 )

an p

bn an an

p p

cn an an

p p

V V

V V V

V V

V V V

V V

θ

θ θ

θ θ

°

° °

° ° ° °

° °

° ° ° °

= ∠

= ∠− = ∠+

= ∠ − = ∠ +

= ∠− = ∠+

= ∠ − = ∠ +

….. (12.12)


Currents: Balanced Y-Y connection:

Line Current, aI = Phase Current, ABI . Line Current = Current in each line Phase Current= Current in each phase

of the source or load

Applying KVL to each phase, we obtain the line currents as:

120 120

120 120

ana

Y

bn anb a

Y Y

cn anc a

Y Y

VIZV VI IZ ZV VI IZ Z

°°

°°

=

∠−= = = ∠−

∠+= = = ∠+

….. (12.12)

And (See phasor diagram on the left)

( ) 0

0

0

a b c

n a b c

n nnN

I I I

I I I I

or V Z I

+ + =

= − + + =

= =

….. (12.13)

This implies that the voltage across the neutral wire is equal to zero. Therefore the neutral line can be removed without affecting the system (balanced Y-Y). Since it is a balanced Y-Y system, the system can be represented or analyzed using a single phase equivalent circuit with or without neutral line (see below)

. The single-phase analysis yields the line current, aI as

an

aY

VIZ

= ….. (12.14)


12.4 BALANCED WYE-DELTA CONNECTION

Balanced Y-∆ connection is a 3-phase system with balanced Y-connected sources supplying ∆-connected load. From the figure above, there is no neutral connection from source to load in this type of connection. Balanced Y-∆ connection:

Phasor Diagram: Relationship between

phase and line voltages

Note: Using KVL, the phase currents can

be determined. For example, applying KVL around loop aABbna

to find phase currents ABI

Assuming positive sequence, the phase voltages (source) are (i.e., Eq. (12.9))

,

( 120 ) ( 240 )

( 240 ) ( 120 )

an p

bn p p

cn p p

V V

V V V

V V V

θ

θ θ

θ θ

°

° ° ° °

° ° ° °

= ∠

= ∠ − = ∠ +

= ∠ − = ∠ +

As shown and derive in Section 12.3, the line voltages across the source are (i.e., Eq. (12.10))

3 ( 30 ) 3 ( 30 )

3 ( 90 ) 120

3 ( 150 ) 120

ab abp p

bc bc abp

ca ca abp

V V V V

V V or V V

V V V V

θ θ

θ

θ

° ° ° °

° ° °

° ° °

= ∠ + = ∠ +

= ∠ − = ∠−

= ∠ + = ∠+

Looking at the balanced Y-∆ connection, it shows that the line voltages (at the source) are equal to the voltage across the load impedances. Assuming positive sequence,

3 30

3 30 3 90

3 30 3 150

ab AB an

bc BC bn an

ca CA cn an

V V V

V V V V

V V V V

°

° °

° °

= = ∠

= = ∠ = ∠−

= = ∠ = ∠+

….. (12.15)

From these voltages (at the load), we can obtain the phase currents as

, , AB BC CAAB BC CA

V V VI I IZ Z Z

= = =

….. (12.16)

Thus, taking ABV as the reference

0

120 120

120 120

ABAB

BC ABBC AB

CA ABCA AB

VIZ

V VI IZ Z

V VI IZ Z

°

°°

°°

∠=

∠−= = = ∠−

∠+= = = ∠+

….. (12.17)


Balanced ∆ connection at the load

Note:

, , ,

, , ,

L a b c

p AB BC CA

I line current I I I

I phase current I I I

=

=

Phasor Diagram: Relationship between

phase and line currents

To obtain the line currents, , , a b cI I and I . Applying KCL at nodes A, B, and C we obtain

:

:

:

a AB CA

b BC AB

c CA BC

At node A I I I

At node B I I I

At nodeC I I I

= −

= −

= −

….. (12.18)

Thus

(1 1 240 )

(1 0.5 0.866)

3 30

a AB CA AB

AB

AB

I I I I

I j

I

°

°

= − = − ∠−

= + −

= ∠−

Similarly, , b cI and I , gives

3 30

3 30

b BC

c CA

I I

I I

°

°

= ∠−

= ∠−

Generally,

3L pI I= ….. (12.19)

a b cL

AB BC CAp

I I I I and

I I I I

= = =

= = =

Note: The magnitude of the line current LI is 3 times the magnitude of the phase current pIand the line currents lag their corresponding phase currents by 30° (See phasor diagram beside)

Another way to analyze the Y-∆ circuit is by transforming the ∆-connected load to an equivalent Y-connected load. Using the transformation formula in Eq. (12.7), we will have a Y-Y system. Hence, the three-phase Y-∆ system can be replaced by a single-phase equivalent circuit as shown below.

where

3Y

ZZ = ….. (12.20)

This will allows us to calculate the line currents. The phase currents are then obtained using Eq. (12.19) with the fact that each of the phase currents leads the corresponding line current by 30º.


12.7 POWER IN A BALANCED SYSTEM The total instantaneous power in the load is the sum of instantaneous powers in the three phases: that is

( )( ) ( )( )( )( )

2 cos 2 cos( ) 2 cos( 120 ) 2 cos( 120 )

2 cos( 120 ) 2 cos( 120 )

a b c

AN a BN b CN c

p p p p

p p

p p p pv i v i v i

V t I t V t I t

V t I t

ω ω θ ω ω θ

ω ω θ

° °

° °

= + += + +

= − + − − −

+ + − +

Applying trigonometric identity, gives (If you are interested to derive please referred to pp.520) 3 cosp pp V I θ= ….. (12.21)

p

p

where V phase voltage

I phase current

angle of the load impedanceangle between the phase voltage and the phase current

θ

=

=

==

From Eq. (12.21), one can notice that the total instantaneous power in a balanced three-phase system is constant. In other words, it does not change with time as one can see in the instantaneous power of each phase. This result is true for both Y and ∆-connected loads. Since the total instantaneous power is independent of time, the average power per phase pP either for ∆-connected load or Y-connected load is / 3p , yields / 3 cosp p pP p V I θ= = ….. (12.22)

And the average reactive power per phase is sinp p pQ V I θ= ….. (12.23)

The average apparent power per phase is p p pS V I= ….. (12.24) Total real power in a balanced 3-phase system is: 3 cosa b c p pP P P P V I θ= + + = ….. (12.25)

Total reactive power in a balanced 3-phase system is: 3 sina b c p pQ Q Q Q V I θ= + + = ….. (12.26) Total apparent/complex power in a balanced 3-phase system is:

22 3

3 3 pp pa b c p p

p

VS S S S V I I Z

Z

∗

∗= + + = = = ….. (12.27)

p pwhere Z Z load impedance per phaseφ°= ∠ =

Alternatively, we may write Eq. (12.27) as S P jQ= +


Power Balance Y-Y Connection

Assuming abc sequence

3 30

3 30

°

°

= ∠

= ∠−

L p

L p

V V

I I

power absorbed by the load is

1

2

φ∗ ∗

= =

=

p p pp L

L p

S V I I Z I

Z I

23 13 3φ φ= = L LS S I Z ..... (12.28)

Or the above 3-phase can be represented as per phase diagram since they are balanced

Where =p anV V

From the single-phase one-line diagram above, power absorbed by the load is

2

1

3 13

φ

φ φ

=

=

L LS Z I

S S

2 2

3 3 3φ = =L L L aS Z I Z I

Where =L aI I

Power Balanced for Y-∆ Connection

Assuming abc sequence

3 30

3 30

°

°

= ∠

= ∠−

L p

L p

V V

I I

power absorbed by the load is

21

221

33

φ∗ ∗

= = =

= =

p p pp L L p

pL L L

S V I I Z I Z I

IZ Z I

23 13φ φ= = L LS S Z I ..... (12.29)

Or the above 3-phase can be represented as per phase diagram since they are balanced

Where =p anV V

From the single-phase one-line diagram above, power absorbed by the load is

21

3 1

33

φ

φ φ

=

=

LL

ZS I

S S

2 2

3φ = =L L a LS I Z I Z

Where =L aI I

Chapter 12: Three-Phase Circuits - Amazon...

Documents

Transcript of Chapter 12: Three-Phase Circuits - Amazon...