Chapter 11 Balanced Three-Phase Circuits

Chapter 11

Balanced Three-Phase

Circuits

11.1-2 Three-Phase Systems

11.3 Analysis of the Y-Y Circuit 11.4 Analysis of the Y- Circuit

11.5 Power Calculations in Balanced Three-Phase Circuits

11.6 Measuring Average Power in Three- Phase Circuits

Overview

 An electric power distribution system looks like:

where the power transmission uses “balanced three-phase” configuration.

Why three-phase?

 Three-phase generators can be driven by

constant force or torque (to be discussed).

 Industrial applications, such as high-power motors, welding equipments, have constant

power output if they are three-phase systems (to be discussed).

Key points

 What is a three-phase circuit (source, line, load)?

 Why a balanced three-phase circuit can be analyzed by an equivalent one-phase circuit?

 How to get all the unknowns (e.g. line voltage of the load) by the result of one-phase circuit

analysis?

 Why the total instantaneous power of a

Section 11.1, 11.2

Three-Phase Systems

1. Three-phase sources 2. Three-phase systems

One-phase voltage sources

 One-phase ac generator: static magnets, one rotating coil, single output voltage v(t)=V_m cos



t.

Three-phase voltage sources

 Three static coils, rotating magnets,

three output voltages

Ideal Y- and -connected voltage sources

Real Y- and -connected voltage sources

 Internal impedance of a generator is usually

Balanced three-phase voltages

 Three sinusoidal voltages of the same

amplitude, frequency, but differing by 120

phase difference with one another.

 There are two possible sequences:

1. abc (positive) sequence: v_b (t) lags v_a (t) by 120. 2. acb (negative) sequence: v_b (t) leads v_a (t) by

abc sequence  v_b (t) lags v_a (t) by 120 or T/3. . 120 , 120 , 0           _{m b m c m} a V V V V V V 

Three-phase systems

 Source-load can be connected in four configurations: Y-Y, Y-, -Y, -

 _{It’s sufficient to analyze Y-Y, while the others}

can be treated by -Y and Y- transformations.

(Y or ) (Y or )

Section 11.3

Analysis of the Y-Y Circuit

1. Equivalent one-phase circuit for

balanced Y-Y circuit

General Y-Y circuit model

Ref.

The only essential node.

Unknowns to be solved

Phase voltage

 Line (line-to-line) voltage: voltage across any pair of lines. Line current Phase current  Phase (line-to- neutral) voltage: voltage across a single phase. Line voltage

Solution to general three-phase circuit

 No matter it’s balanced or imbalanced three- phase circuit, KCL leads to one equation:

Impedance of neutral line.

Total

impedance along line aA.

Total impedance along line bB. Total impedance along line cC. ), 1 ( , 1 1 1 0 0  C c gc N n c B b gb N n b A a ga N n a N cC bB aA Z Z Z Z Z Z Z Z Z Z                    V V V V V V V I I I I

which is sufficient to solve V_N (thus the entire circuit).

 

 

Solution to “balanced” three-phase circuit

 For balanced three-phase circuits,

1. {V_a'n , V_b'n , V_c'n } have equal magnitude and 120

relative phases;

2. {Z_ga = Z_gb = Z_gc }, {Z_1a = Z_1b = Z_1c }, {Z_A = Z_B = Z_C };

 total impedance along any line is the same

Z_ga + Z_1a + Z_A =… = Z_. , 0 Z Z Z Z N n c N n b N n a N V V V V V V V _{ }  _{ }  _{ }   Eq. (1) becomes:

Meaning of the solution

 V_N = 0 means no voltage difference between

nodes n and N in the presence of Z₀ .  Neutral

line is both short (v = 0) and open (i = 0).

 The three-phase circuit can be separated into 3 one-phase circuits (open), while each of them has a short between nodes n and N.

Equivalent one-phase circuit

 Directly giving the line current & phase voltages: I_nn = 0 I_aA Line current Phase voltage of load





, 1 Z Z Z Z_{ga a A} N n a aA      V  V I Phase voltage of source , A aA AN I Z V  V_an  I_aA



Z_1a  Z_A



.

, 120      aA n b bB Z I V I 

The 3 line and phase currents in abc sequence

aA

I

I

cC

I

 Given the other 2 line currents are:I_aA  V_an Z_ ,

which still

follow the abc sequence relation. , 120     aA n c cC Z I V I 

The phase & line voltages of the load in abc seq. Phase voltage Line voltage





, 30 3 120            AN AN AN BN AN AB V V V V V V . 120 , 120 ,          _{ } B _{AN CN AN} n b BN A n a AN Z Z Z Z V V V V V V V   (abc sequence)



 





120



, 90 3 120 120                  AN AN AN BC V V V V V V V

The phase & line voltages of the load in acb seq. Phase voltage Line voltage







 







. 150 3 120 , 90 3 120 120 , 30 3 120                                 AN AN AN CA AN AN AN BC AN AN AN BN AN AB V V V V V V V V V V V V V V

 Line voltages are 3 times bigger, leading (abc) or lagging (acb) the phase voltages by 30.

(acb

Example 11.1 (1)

Phase voltages

Z_ga Z_1a

Z_A

 Q: What are the line currents, phase and line voltages of the load and source, respectively?

= 40 + j30 .

Example 11.1 (2)

 The 3 line currents (of both load & source) are:

 The 3 phase voltages of the load are:











2.4 83.13



A. 120 , A 87 . 156 4 . 2 120 , A 87 . 36 4 . 2 30 40 0 120 1                             aA cC aA bB A a ga n a aA j Z Z Z I I I I V I



















115.22 118.81



V. 120 , V 19 . 121 22 . 115 120 . V 19 . 1 22 . 115 28 39 87 . 36 4 . 2                           AN BN A aA AN Z j V V V V I V

Example 11.1 (3)

 The 3 line voltages of the load are:























199.58 148.81



V. 120 , V 19 . 91 58 . 199 120 , V 81 . 28 58 . 199 19 . 1 22 . 115 30 3 30 3                              AB CA AB BC AN AB V V V V V V

Example 11.1 (4)

 The three line voltages of the source are:

 The 3 phase voltages of the source are:



















118.9 119.68



V. 120 , V 32 . 120 9 . 118 120 , V 32 . 0 9 . 118 5 0 2 0 87 . 36 4 . 2 120                             _ an cn an bn ga aA n a an Z . j . V V V V I V V























205.94 90.32



V, 120 , V 68 . 29 94 . 205 32 . 0 9 . 118 30 3 30 3                            ab bc an ab V V V V

Section 11.4

Load in  configuration

Phase current Line current

Line voltage = Phase voltage

-Y transformation for balanced 3-phase load

 The impedance of each leg in Y-configuration (Z_Y ) is one-third of that in -configuration (Z_):

. , , 3 2 1 c b a b a c b a a c c b a c b Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z         

Equivalent one-phase circuit

 The 1-phase equivalent circuit in Y-Y config. continues to work if Z_A is replaced by Z_/3:

Line current

directly giving the line current: ,

1a A ga n a aA Z Z Z    V  I

and line-to-neutral voltage: 

Line-to-neutral voltage 

Phase voltage  Line voltage

The 3 phase currents of the load in abc seq.

 Can be solved by 3 node equations once the 3 line currents I_aA , I_bB , I_cC are known:

Line current Phase current

.

,

,

_{bB BC AB cC CA BC} CA AB aA

I

I

I

I

I

I

I

I

I













Phase current (abc sequence)

Section 11.5

Power Calculations in

Balanced Three-Phase

Circuits

1. Complex powers of one-phase and

the entire Y-Load



































.

,

,

3

,

cos

A L aA L AN A

Z

I

V

I

I

V

V

I

V

P

       





I

V

Average power of balanced Y-Load

 The average power delivered to Z_A is:

(rms value)

Complex power of a balanced Y-Load

 The reactive powers of one phase and the entire Y-Load are:















.

sin

3

sin

3

,

sin

       







L L tot

V

I

V

I

Q

I

V

Q

 The complex powers of one phase and the entire Y-Load are:























.

3

3

3

;

*                 j L L j tot j

e

I

V

e

I

V

S

S

e

I

V

jQ

P

S

V

I

One-phase instantaneous powers

 The instantaneous power of load Z_A is:

).

cos(

cos

)

(

)

(

)

(

t



v

t

i

t



V

I



t



t





_

p

_{A AN aA m m}  The instantaneous powers of Z_A , Z_C are:











120



cos ) ( , 120 cos 120 cos ) ( ) ( ) (           









t I V t p t t I V t i t v t p m m bB BN B (abc sequence)

Total instantaneous power

  

2

2

cos

3

cos

.

5

.

1

cos

5

.

1

)

(

)

(

)

(

)

(

      







I

V

I

V

I

V

t

p

t

p

t

p

t

p

_{tot A B C m m}













 The instantaneous power of the entire Y-Load is a constant independent of time!

 The torque developed at the shaft of a 3-phase motor is constant,  less vibration in

machinery powered by 3-phase motors.

 The torque required to empower a 3-phase generator is constant,  need steady input.

Example 11.5 (1)

 Q: What are the complex powers provided by

the source and dissipated by the line of a-phase?

 The equivalent one-phase circuit in Y-Y configuration is:

Z_1a

Example 11.5 (2)

 The line current of a-phase can be calculated by the complex power is:







577

.

35

36

.

87



A.

,

3

600

10

120

160

,

3 * * 















aA aA

j

S

I

I

I

V

_{ } 

 The a-phase voltage of the source is:











357

.

51

1

.

57



V.

025

.

0

005

.

0

87

.

36

35

.

577

3

600

1  



















j

Z

_a aA AN an

V

I

V

Example 11.5 (3)

 The complex power provided by the source of a- phase is:

 The complex power dissipated by the line of a- phase is:









206

.

41

38

.

44



kVA.

87

.

36

35

.

577

57

.

1

51

.

357

*   













_{an aA} an

S

V

I



 





577



.

35

0

.

005

0

.

025

2 1 2







Z

j

S

_aA

I

_{aA a}

Key points

 What is a three-phase circuit (source, line, load)?

 Why a balanced three-phase circuit can be analyzed by an equivalent one-phase circuit?

 How to get all the unknowns (e.g. line voltage of the load) by the result of one-phase circuit

analysis?

 Why the total instantaneous power of a