So far circuit analysis has been performed on single-

Three phase systems

Three phase systems

Introduction

Introduction

 So far circuit analysis has been performed on

single-phase circuits i e there has been one circuit with a phase circuits, i.e. there has been one circuit with a number of different voltage and current sources which were not synchronised in any purposeful way.

 In a three phase system three voltages are generated  In a three-phase system, three voltages are generated

with specific relationship in amplitude, frequency and phase. The amplitude and frequency of the voltages are the same but each voltage is displaced from the other by the same but each voltage is displaced from the other by 120.

 Three phase systems are used to deliver power from a

power station to a local substation before being power station to a local substation before being

distributed to home and factories. In the UK, the voltage on the transmission system is 250kV-440kV and at

Merits

e ts

of t

o t

hree phase systems

ee p ase syste s

 With same sizes，3-phase generators and

electrical motors have high output power, simple g p p p in construction and better capacity than single phase systems.

 Transmission of electricity in three phases  Transmission of electricity in three phases

economically.

 Single phase instantaneous power is time-variant

h l l f h h

while total instantaneous power of three phase is time-invariant.



Instantaneous expression

Instantaneous expression

)

cos(

2

)

(

t

U

t

u





B + A + C +

)

120

cos(

2

)

(

o B

t



U

t



u



)

cos(

2

)

(

A

t

U

t

u





+ – u_B + – u_A + – u_C u_A u_{B uC} Y X _Z o C

( )

2

cos(

120 )

u

t



U



t



u_A u_{B uC}  t 0

Phasor expression

Phasor expression



U



C o B o A

120

0













U

U

U

U

A 

U

120° 120° o C







240





120



U

U

U

 A

U

U



120° 120 B

U

Phase Sequence: it refers to the order in which three-phase voltage generated

C 

generated

A B

Positive phase sequence：

Y connection of source

Y connection of source

A  A + Three-wire systems N B A 

U

CA 

U

A

I

– X Y Z A 

U

Four-wire systems B B 

I

B C Z B 

U

C 

U

BC 

U

y

 The neutral point is known as the star point.

C B C 

I

C . . .  Phase voltage  Line voltage

 Phase current = line current

, , A B C U U U I I I   . . . , , A B B C C A U U U

 Phase current line current

C B

A I I

Relation between phase and line voltages

Relation between phase and line voltages



U

o

0







U

U

 CN

U

BN U  AB 

U

30o 30o CA 

U

o B A

120

0













U

U

U

U

N A 

U



U

30o o C







240





120



U

U

U

 BN

U

B C 

U

o

3

30

AB A B

U



U



U



U



o

3

90

U

_BC



U

_B

U

_C



3

U



90

U



U



U



U

 

o

3

150

CA C A

U



U



U



U



The phase relation of

three-phase voltages and currents must be clear.

Summary for symmetric Y connection

Summary for symmetric Y connection



Line current is equal to phase current



Phase voltage is symmetric

，

and line voltage is

also symmetric

also symmetric



Line voltage is times the phase voltage

3



Line voltage is times the phase voltage

3

3

l p

U



U



Line voltage phase leads phase voltage by 30

o

CN CA BN C B AN B A      







U

U

U

U

U

U

_{AB BC CA}

Delta connection of source

A



I

A

Delta connection of source

 A A Z C 

U

A 

U

B A 

U

U

_CA B 

I



I

B X + – _B C Y B 

U

BC 

U

C

I

C B

U

U

BC

 Line voltage = phase voltage  Ph lt U. U. U.

？

 Phase voltage  Line voltage

 Line current ≠ phase current

, , A B C U U U C B A

I

I

I



,



,



. . . , , A B B C C A U U U

？

Relation between phase and line currents

Relation between phase and line currents

A 

I



I

a Z ca 

I

a o ab





0



I

I



I

ab

I

b Z b o ca o bc

120

120











 

I

I

I

I

  B

I

_Z bc 

I

c ca

I



120

I

ca 

I

C 

I

C

I

c ab 

I

30 30o ca ab A   





I

I

I



3

ab





30

o 

I

ab 

I



I



A 

I

B 

I

30 o   

I

I

I

ab bc B   





I

I

I

ca ab A ab o bc

30

3







I

 o  bc

I

I

ca bc ca C



I



I

I

o ca

30

3







I

Summary for delta connection

Summary for delta connection



Line voltage is identical to phase voltage



Phase current is symmetric and line current is

also symmetric

also symmetric



Line current is times the phase current

3



Line current is times the phase current

3

3

l p

I



I



Line current lags phase current by 30

o ca bc ab

I

I

I



,



,



Reference direction

ab

,

bc

,

ca

Reference direction

Balanced Y-Y connection

Balanced Y Y connection

+ A 

I

Z₂

Y - Y

+ _ _{N n} Z₁ A 

U

- ₂ + B 

U

B 

I

 -Z₃ C 

U

Example: Given in symmetric 3 phase source line

C



I

Example: Given in symmetric 3-phase source, line voltage is 38030V，symmetric load Z＝10030.

Voltage at the neural point

o A





0



U

U

o B A

120







 

U

U

U

C



120° o C







240





120



U

U

U



The character of balanced

A 

U

120° 120° 120°

The character of balanced three-phase voltage:

0

B

U



120°





0

0

C B A C B A













  

U

U

U

u

u

u

B A

U

U







0

C B A



U



U

U

Balanced Y-Y connection

Balanced Y Y connection

+ Z  A 

I

V

U

Given



_AB



380



30

o _ + N n Z Z Z A 

U

C 

U



-V

U

So



_AN



220



0

o + + Z B 

U

I

B C 

I

+ AN

U



A Z

I



o o o

220 0

2.2

30

100 30

AN A

U

I

A

Z









 







-N n o o o

2.2 ( 30

120 )

2.2

150

B

I





 





 

A

o o o o

2 2 ( 150

120

360 )

2 2 90

I



_C

2.2 ( 150



o

120

o



360 )

o

2.2 90



o

A

I



 









A

Balanced delta-delta connection

Balanced delta delta connection

A  I A a ZZ Z  B c b ab  I ca  I A C 

U

U

_AB B  I C  I B C c bc b  I + – BC 

U

Example: Given in symmetric 3-phase circuit, line voltage 380 0V，Symmetric load Z＝10030

Balanced delta-delta connection

Balanced delta delta connection

Choose A phase to compute

V

U

If



_AB



380



0

o o o o

380 0

3.8

30

100 30

AB ab

U

I

A

Z









 







+

U



ab Z

I



A a -B b AB

U

Z o o o

3 3.8 ( 30

30 ) 6.58

60

A

I





  





 

A

o

6.58

180

B

I





 

A

o o

6 58 ( 180 120 360)

6 58 60

I



_C



6.58 ( 180 120 360)

 







6.58 60



A

I



 









A

Balanced Y-delta connection

Balanced Y delta connection

A 

I

+ A a _ _N Z Z Z A 

U

B 

U

C 

U

b c ab 

I

ca 

I

Y -

∆

+ + B B 

I

C 

I

B C b c bc 

I

A  I A A C 

U

U

_AB Z_n

∆

-

Y

B  I  I B C _{– +} 

U

A C

U

_AB n Z Z

∆

-

Y

C I + BC

U

Y

-

∆

connected systems

co

ected syste s

 There are a number of methods for dealing with such

systems One method is to change the sources form star systems. One method is to change the sources form star connected to delta connected. But remember to add in the associated phase difference, this as the phase and line voltage relationships as derived for the star

line voltage relationships as derived for the star connected system. A A A C 

U

U

_AB + _ _N A 

U

B C _{– +} BC 

U

+ _ _ + B 

U

C 

U

B C

Y

-

∆

connected systems

co

ected syste s

Star Voltages Delta Voltages

o A





0



U

U

U

_AB



U

_A



U

_B



3

U



30

o o B A

120







 

U

U

o

3

90

BC B C

U



U



U



U

 



The calculation proceeds as normal

o C



U





240



U



120

U

 o

3

150

CA C A

U



U



U



U



∆

-

Y

connected systems

co

ected syste s



Similarly change the delta connected

sources to star. Then proceed as usual.

Delta Voltages Star Voltages

o

3

0

AB

U



U



Delta Voltages _{Star Voltages}

o

30

A

U

  

U

AB o

3

120

BC

U



U

 

A o

150

B

U

  

U

o

3

120

CA

U



U



U

_C

 

U

90

o

Unbalanced three-phase systems

Unbalanced three phase systems



So far, the analysis has been on balanced

three-phase systems.



For unbalanced three phase systems the load



For unbalanced three-phase systems, the load

in each of the phases is not the same.



The calculations are similar but they need to be

done for each phase of the load.



The analysis of unbalanced loads and the

displacement of middle point

displacement of middle point.

Unbalanced Y-Y 4 wire connection

+ Z_a AN 

U

I



A



Unbalanced Y Y 4 wire connection

a b c

Z



Z



Z

_ _{N N'}a Z_b Z_c AN

U

CN 

U

BN 

U

A N N

I



_ B

I



0

a b c n

I



Z_c BN C

I



B

I

CN C b BN B AN A

Z

U

I

Z

U

I

Z

U

I



















c b a

Z

Z

Z

0















N A B C AN BN CN N

Z

U

Z

U

Z

U

I

I

I

I















c b a

Z

Z

Z

Unbalanced delta connected load

Unbalanced delta connected load

+ A _Zb 

I

A a _ _N Zc Zb Za A 

U

B 

U

C 

U

b c ab 

I

ca 

I

+ + B B 

I

C 

I

B C b c bc 

I

 The specified line voltage will be across each phase.  Take three phases and put them into 3 independent  Take three phases and put them into 3 independent

single phase systems with voltage 120 apart in phase.

/

b AB b

I

_ab



U

_AB

/

Z

_b

I



U

/

Z

I



I



I

I

U

Z

A ab ca

I



I

I

/

ca CA a

I



U

Z

Unbalanced 3 wire Y connected load

Unbalanced 3 wire Y connected load

+ Z 

U

A _ _{N N'} Z_a Z_b Z AN

U

CN 

U



U

_{B N B Zc}

U

_B C

 This is a difficult case. Without the neutral conductor, the

voltage at n’ will vary.

 The currents in the load sum to zero at n’ and then

Three-phase power

Three-phase total power

：

Three phase power

C

B

A

P

P

P

P







p

p

When Loads are balanced

：



cos

3

U

p

I

p

P





cos

3

U

l

I

l

P



3 phase power

3 phase power



The power per phase = V

_ph

I

_ph

cos

φ



Thus TOTAL power (in a balanced) system

 = 3V_phI_phcos φ



Delta

Star



V

_L

= V

_ph

, I

_L

=

√

3I

_ph

V

_L

=

√

3V

_ph

, I

_L

= I

_ph



V

_L

V

_ph

, I

_L

√

3I

_ph

V

_L

√

3V

_ph

, I

_L

I

_ph



P

_T

=

√

3V

_L

I

_L

cos

φ

P

_T

=

√

3V

_L

I

_L

cos

φ



Apparent power:

* P jQ S   V I P = Real Power (W, kW, MW)

Q = Reactive Power (var, kvar, Mvar) S Complex power (VA kVA MVA)

jQ

Measurement of single phase wattmeter

*

i

Current loop Voltage loop

g

p

*

i

1

i

₂

u

loads

In AC circuit





kI

U



cos



In AC circuit





kI

1

U



cos



is the angle between



i

₁

and

u

3-phase four-wire connection

Loads is Y connection and use three power meters to measure.

The total power is equal to the sum of three wattcmeters

3 phase four wire connection

1

W

*

*

A

The total power is equal to the sum of three wattcmeters.

At this

1

W

W

*

*

A

B

time each voltage loop

2

W

*

W

*

*

C

B

loop measure the phase loads

3

W

*

C

N

voltage of each phase

A three-wire system with only two

wattmeter

A

i

*

wattmeter

A

i

u

A

A

u

A

Z

1

W

*

B

i

u

AC

B

u

C

A

u

A

B

Z

2

W

*

*

B

u

C

i

u

BC

Z

C

The total power is the algebraic sum of the

meter readings

C

Principal of measure power using the

two-wattmeter method

i

u

i

u

i

u

p







wattmeter method

B

A

C

i

i

i









A

B

B

 

C

C



A

u

u

i

u

u

i

i

u

i

u

i

u

p











C

A

B



A

C

 

B

B

C



A

u

i

u

i

u

u

i

u

u

i













BC

B

AC

A

u

i

u

i





Notes

：

three-phase total power is equal to

Notes

：

three phase total power is equal to

the sum of two meter readings.

Just a single meter reading has no meaning

Just a single meter reading has no meaning.