AC Circuits Three-Phase Circuits

AC Circuits

Contents

•

What is a Three-Phase Circuit?

•

Balance Three-Phase Voltages

•

Balance Three-Phase Connection

•

Power in a Balanced System

•

Unbalanced Three-Phase Systems

•

Application – Residential Wiring

A simple AC generators

Sinusoidal voltage sources

When a conductor rotates in a constant magnetic field, a sinusoidal wave is generated.

Single-phase system

When the loop is moving perpendicular to the lines of flux, the maximum voltage is induced.

When the conductor is moving parallel with the lines of flux, no voltage is induced.

Single-phase two-wire

Normal household: Single-phase three-wire

neutral

More pole pieces? More coils?

Three-phase

circuit:

A system produced by a generator consisting of three sources having the same amplitude and frequency but out of phase with each other by 120°.

4

Polyphase

Three sources with 120° out of phase Four wired system

Advantages of a three-phase circuit

•

Most of the electric power is generated and distributed

in three-phase:

– Less/more than 3-phase required…taken from 3-phase

•

The instantaneous power in a three-phase system can

be constant (not pulsating):

– uniform power transfer and less vibration of three-phase machine

•

The amount of power, the three-phase system is more

economical that the single-phase.

•

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

system is less than that required for an equivalent

single-phase system.

Balance

Three-Phase Voltages

6

•

A three-phase generator

consists of a rotating magnet

(rotor) surrounded by a stationary winding (stator).

A three-phase generator _{The generated voltages}

It can supply power to both single-phase and three-single-phase load

7

Balance Three-Phase Voltages

•

Two possible configurations:

Three-phase voltage sources: (a) Y-connected ; (b) Δ-connected

Four wires _{Three wires}

Phase voltage Balanced voltages: cn bn an cn bn an V V V V V V      0

8

Balance Three-Phase Voltages

• Balanced phase voltages are equal in magnitude and are out of phase

with each other by 120°.

• The phase sequence is the time order in which the voltages pass through their respective maximum values.

• A balanced load is one in which the phase impedances are equal in magnitude and in phase

3 2 3 4 3 2    j p j p j p p e V e V e V V       bn cn an V V V p V    _{bn cn} an V V V 3 2 3 4 3 2    j p j p j p p e V e V e V V       cn bn an V V V _abc/positive sequence acb/negative sequence abc/positive sequence acb/negative sequence Y 3 2 1 Z Z Z Z    Z_a Z_b Z_c Z_ Y Z Z_  3 Transform:

9

Example

Example 1

Determine the phase sequence

of the set of voltages.

)

110

cos(

200

)

230

cos(

200

)

10

cos(

200



















t

v

t

v

t

v

cn bn an







Solution:

The voltages can be expressed in

phasor form as

We notice that

V

_an

leads

V

_cn

by

120°

and

V

_cn

in turn leads

V

_bn

by

120°

.

Hence, we have an acb sequence.

V 110 200 V V 230 200 V V 10 200 V            cn bn an

10

Balance Three-Phase Connection

Four possible connections

1. Y connection (connected source with a

Y-connected load)

2. Y-Δ connection (Y-connected source with a

Δ-connected load)

3. Δ-Δ connection

4. Δ-Y connection

11

Balance Y-Y system

A balanced Y-Y system is a three-phase system with a balanced y-connected source and a balanced y-connected load.

ca bc ab L cn bn an p p L V V V V V V V V V V where , 3        L l S Y Z Z Z Z    6 3 2 3 2 1 2 3 3 2 3 2 1 1   j j p p Vpe j Vp j Vp e V V                         _{an nb an bn}  ab V V V V V Line voltage:    3 4 3 2 3 2 , j j j e e e ,         a c a Y an Y bn b Y an a I I I Z V Z V I Z V I Line current (KVL):   0, , 0        _{b c n a b c} a I I I I I I I VnN 0

Voltage across the neutral line is zero: the neutral line can be removed (earth acting as the neutral line).

The same as the phase current in Y-Y

12

Balanced Y-Y connection

Example 2

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

shown below:

A 2 . 98 81 . 6 I A 8 . 141 81 . 6 I A 8 . 21 81 . 6 I Ans            c b a

13

Balance Y-



Connection

•

A

balanced Y-Δ system

is a three-phase system with a

balanced y-connected source and a balanced Δ-connected load.

CA BC AB p c b a L p L I I I I I I I I I I where , 3        Line current Phase current

Line voltage = the voltage across the load

ab AB V V  Phase current Δ AB AB Z V I  Line current   6 1 3 4 3 j j a e e       I_AB I_CA I_AB I_AB I_AB I

14

Balance Y-



Connection: Example

Example 3

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

Solution

Using single-phase analysis,

Other line currents are obtained using the abc phase sequence







100

10

V

_an

A

57

.

16

54

.

33

57

.

26

981

.

2

10

100

3

/

Z

V

I

an





















 a

15

Balance



-



Connection

•

A

balanced Δ-Δ

system is a three-phase system with a

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

Line voltage = the phase voltage

ab AB V V  Phase current Δ Δ AB AB Z V Z V I   ab Line current   6 1 3 4 3 j j a e e        AB AB AB CA AB I I I I I I p L I I  3

16

Balance



-



Connection: Example

Example 4

A balanced Δ-connected load having an impedance 20-j15  is connected to a Δ-connected positive-sequence generator having ( ). Calculate the phase currents of the load and the line currents.

Ans:

The phase currents

The line currents

V

0

330

V

_ab







A 87 . 156 2 . 13 I A; 13 . 81 2 . 13 I A; 87 . 36 2 . 13 I_AB    _BC    _AB    A 87 . 126 86 . 22 I A; 13 . 113 86 . 22 I A; 87 . 6 86 . 22 I_a    _b    _c   

17

Balance



-Y Connection

•

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

balanced Y-connected source and a balanced Y-connected load.

18

Balance



-Y Connection: Example

Example 5

A balanced Y-connected load with a phase impedance 40+j25  is

supplied by a balanced, positive-sequence Δ-connected source with a line voltage of 210 V. Calculate the phase currents. Use V_ab as

reference. Answer

The phase currents

A;

58

57

.

2

I

A;

178

57

.

2

I

A;

62

57

.

2

I























CN BN AN

20

Power in a Balanced System

• Instantaneous power (Y-load, Z_Y = Z)

• Comparing the power loss in (a) a single-phase system, and (b) a three-phase system

phase -single , 2 ' ₂ 2 L L loss V P R P  ' ' ₂ , three-phase 2 L L loss V P R P 

• If same power loss is tolerated in both system, three-phase system use only 75% of materials of a single-phase system

      _        _   3 2 cos 2 , 3 2 cos 2 , cos 2V t v V t  v V t  vAN p BN p CN p         _{ }        _{ }    3 2 cos 2 , 3 2 cos 2 , cos 2I t  i I t   i I t   ia p b p c p  cos 3 _{p p} c CN b BN a AN c b a p p v i v i v i V I p

p       For onephase: SP jQV_pI_p*, PV_pI_pcos, QV_pI_psinθ

21

Unbalanced Three-Phase Systems

•

An unbalanced system is due to unbalanced voltage sources

or an unbalanced load.



a b c



n C CN c B BN b A AN a I I I I , Z V I , Z V I , Z V I       

• 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.

22

Unbalanced Three-Phase Systems: Example

Example 6

Determine the total average power, reactive power, and complex power at the source and at the load

Ans At the source: S_s = -(2087 + j834.6) VA P_a = -2087W P_r = -834.6VAR At the load: S_L = (1392 + j1113) VA P_a = 1392W P_r = 1113VAR

23

Application – Residential Wiring

24

Application – Residential Wiring (2

)

Single-phase three-wire residential wiring