Chapter 12 Three-Phase Circuits. Power Factor Correction A balanced three-phase circuit Medium...

Chapter 12

Three-Phase Circuits

Power Factor Correction

A balanced three-phase circuit

Medium Voltage Metal Enclosed Capacitor Banks

Three-Phase Voltages

(b) Balanced three-phase voltages

(a) The three windings on a cylindrical drum used to obtain three-phase voltages

2 cos

2 cos( 120 )

2 cos( 240 )

aa

bb

cc

v V t

v V t

v V t

Three-Phase Voltages

Generator with six terminals

Three-Phase Balanced Voltages

0

120

240 120

aa

bb

cc

V

V

V V

V

V

V

Phase sequence or phase rotation is abcPositive Phase Sequence

0

120

240 120

a

c

b

V

V

V V

V

V

V

Phase sequence or phase rotation is acbNegative Phase Sequence

neutral terminal

(a) Y-connected sources (b) -connected sources

Two Common Methods of Connection

phase voltage

Phase and Line Voltages

The line-to-line voltage Vab of the Y-connected source

0 120

( 0.5 0.866)

3 30

ab a b

p p

p p

p

V V

V V j

V

V V V

Similarly

3 90

3 210

bc p

ca p

V

V

V

V

The Y-to-Y Circuit

A four-wire Y-to-Y circuit

The Y-to-Y Circuit (cont.)

, , anda b caA bB cC

A B C

V V V

I I IZ Z Z

nN aA bB cC I I I I

The average power delivered by the three-phase sourceto the three-phase load

A B CP P P P When ZA = ZB = ZC the load is said to be balanced

0 120 120, , and

, ( 120 ), and ( 120 )

a b caA bB cC

A B C

aA bB cC

V V V

Z Z Z

V V V

Z Z Z

V V VI I I

Z Z Z

I I I

Four - wire

0nN aA bB cC I I I I

The Y-to-Y Circuit(cont.)

2

cos( ) cos( ) cos( )

3 cos( )

A B CP P P P

V V VV V VZ Z Z

V

Z

The average power delivered to the load is

There is no current in the wire connecting the neutral nodeof the source to the neutral node of the load.


A three-wire Y-to-Y circuit

Three - wire We need to solve for VNn


0

0 120 120

a Nn b Nn c Nn

A B C

Nn Nn Nn

A B C

V V V

V V V V V V

Z Z Z

V V V

Z Z Z

Solve for VNn

( 120 ) 120 0A C A B B CNn

A C A B B C

V V V

Z Z Z Z Z Z

VZ Z Z Z Z Z

, , anda Nn b Nn c NnaA bB cC

A B C

V V V V V VI I I

Z Z Z

When the circuit is balanced i.e. ZA = ZB = ZC


( 120 ) 120 0

0

Nn

V V V

ZZ ZZ ZZV

ZZ ZZ ZZ

The average power delivered to the load is

2

3 cos( )

A B CP P P P

V

Z

A three-wire Y-to-Y circuit with line impedances


Transmission lines


The analysis of balanced Y-Y circuits is simpler than theanalysis of unbalanced Y-Y circuits.

VNn = 0. It is not necessary to solve for VNn . The line currents have equal magnitudes and

differ in phase by 120 degree. Equal power is absorbed by each impedance.

Per-phase equivalent circuit

Example 12.4-1 S = ?rms

rms

rms

110 0 V

110 120 V

110 120 V

a

b

c

V

V

V

50 80

50

100 25

A

B

C

j

j

j

Z

Z

Z

Unbalanced 4-wire

rms

rms

rms

110 01.16 58 A

50 80

110 1202.2 150 A

50

110 1201.07 106 A

100 25

aaA

A

bbB

B

ccC

C

j

j

j

VI

Z

VI

Z

VI

Z

Example 12.4-1 (cont.)*

*

*

68 109 VA

242 VA

114 28 VA

A aA a

B bB b

C cC c

j

j

j

S I V

S I V

S I V

The total complex power delivered to the three-phase loadis

182 379 VAA B C j S S S S

Example 12.4-2 S = ?

rms

rms

rms

110 0 V

110 120 V

110 120 V

a

b

c

V

V

V

50 80

50 80

50 80

A

B

C

j

j

j

Z

Z

Z

Balanced 4-wire

rms

110 01.16 58 A

50 80a

aAA j

V

IZ

* 68 109 VAA aA a j S I V


3 204 326 VAA j S S

Alsorms rms1.16 177 A , 1.16 62 AbB cC I I

68 109 VAB Cj S S

Example 12.4-3 S = ?

rms

rms

rms

110 0 V

110 120 V

110 120 V

a

b

c

V

V

V

50 80

50

100 25

A

B

C

j

j

j

Z

Z

ZUnbalanced 3-wire

Determine VNn

rms

(110 120 ) 110 120 110 0

56 151 V

A C A B B CNn

A C A B B C

Z Z Z Z Z ZV

Z Z Z Z Z Z

, , anda Nn b Nn c NnaA bB cC

A B C

V V V V V VI I I

Z Z Z

Example 12.4-3 (cont.)

1.71 48 , 2.45 3 , and 1.19 79aA bB cC I I I

* *

* *

* *

( ) 146 234 VA

( ) 94 VA

( ) 141 35 VA

A aA a aA aA A

B bB b bB bB B

C cC c cC cC C

j

j

j

S I V I I Z

S I V I I Z

S I V I I Z


287 364 VAA B C j S S S S

Example 12.4-4 S = ? Balanced 3-wire

rms

rms

rms

110 0 V

110 120 V

110 120 V

a

b

c

V

V

V

50 80

50 80

50 80

A

B

C

j

j

j

Z

Z

Z

rms

110 01.16 58 A

50 80a

aAA j

V

IZ

* 68 109 VAA aA a j S I V


3 204 326 VAA j S S

Example 12.4-5 PLoad = ? PLine = ? PSource = ?

Balanced 3-wire

per-phase equivalent circuit

100 0( ) 1.894 18.7 A

50 (377)(0.045)a

aAA j

V

IZ

The phase voltage at the load is

( ) (40 (377)(0.04)) ( ) 81 2 VAN aAj V I


The power delivered by the source is

cos( )2

(100)(1.894)cos(18.7 ) 89.7 W

2

m ma V I

V IP

The power delivered to the load is2 2(1.894)

Re( ) 40 71.7 W2 2m

A A

IP Z

The power lost in the line is2 2(1.894)

Re( ) 10 17.9 W2 2m

aA Line

IP Z

Line loss 20%

The -Connected Source and Load

rms

rms

rms

120 0 V

120.1 121 V

120.2 121 V

ab

bc

ca

V

V

V

( )3.75 A

1ab bc ca

V V V

I

Total resistance around the loop

Circulatingcurrent Unacceptable

Therefore the sources connection is seldom used in practice.

The -Y and Y- Transformation

AZ BZ

CZ

1Z

3Z

2Z

1 3

1 2 3

2 3

1 2 3

1 2

1 2 3

A

B

C

Z Z

Z Z Z

Z Z

Z Z Z

Z Z

Z Z

Z

ZZ

Z

1

2

3

A B B C A C

B

A B B C A C

A

A B B C A C

C

Z Z Z Z Z Z

Z

Z Z Z Z Z Z

Z

Z

Z

ZZ Z Z Z Z Z

Z

Example

The Y- Circuits

aA AB CA

bB BC AB

cC CA BC

I I I

I I I

I I I

where

3

1

2

ABAB

BCBC

CACA

VI

Z

VI

Z

VI

Z

The Y- Circuits (cont.)

cos sin cos( 120 ) sin( 120 )

3 ( 30 )

aA AB CA

I j I j

I

I I I

or

3 3aA L pI I I I

Example 12.6-1 IP = ? IL = ?

rms

rms

rms

22030 V

3220

150 V3

22090 V

3

a

b

c

V

V

V

The -connected load is balanced with 10 50 Z

rms

rms

rms

220 0 V

220 120 V

220 240 V

AB a b

BC b c

CA c a

V V V

V V V

V V V

rms

rms

rms

22 50 A

22 70 A

22 190 A

ABAB

BCBC

CACA

VI

Z

VI

Z

VI

Z

The line currents are

22 3 20 , 22 3 100 , 22 3 220aA AB CA bB cC I I I I I

The Balanced Three-Phase Circuits

3Y

ZZ

per-phase equivalent circuit

Y-to- circuit equivalent Y-to-Y circuit

Example 12.7-1 IP = ?

rms

rms

rms

110 0 V

110 120 V

110 120 V

a

b

c

V

V

V

10 5

75 225L j

j

Z

Z25 75

3Y j Z

Z

rms1.26 66 AaaA

L Y

V

IZ Z


rms rms1.26 186 A and 1.26 54 AbB cC I I

rms

rms

rms

99.6 5 V

99.6 115 V

99.6 125 V

AN aA Y

BN

CN

V I Z

V

V

The voltages in the per-phase equivalent circuit are

rms

rms

rms

172 35 V

172 85 V

172 155 V

AB AN BN

BC BN CN

CA CN AN

V V V

V V V

V V V

The line-to-line voltages arerms

rms

rms

0.727 36 A

0.727 156 A

0.727 84 A

ABAB

BCBC

CACA

VI

Z

VI

Z

VI

Z

Instantaneous and Average Power in BTP Circuits

One advantage of three-phase power is the smooth flow ofenergy to the load.

2 2 2

2

2 2

2

( )

1 cos2 1 cos2( 120 ) 1 cos2( 240 )2

3cos2 cos(2 240 )cos(2 480 )

2 20

3

2

ab bc cav v vp t

R R R

Vt t t

R

V Vt t t

R R

V

R

2

cos , cos( 120 ),

and cos( 240 )

(1 cos2 )cos

2

ab bc

ca

v V t v V t

v V t

tt

The instantaneous power


The total average power delivered to the balanced Y-connected load is

33cos()

3cos()

3cos()3

3cos()

YAPLAVAI

PL

LL

LL

PPVI

VI

VI

VI

,aA L AI AN P AVI V I V Phase A


The total average power delivered to the balanced -connected load is

33cos()

3(3)cos()3

3cos()

ABABAB

LP

PL

PPVI

IV

VI

Example 12.8-1 P = ?

rms

rms

rms

110 0 V

110 120 V

110 120 V

a

b

c

V

V

V

10 5

75 225L j

j

Z

Z

rms1.26 66 AaaA

L Y

V

IZ Z

rms99.6 5 VAN aA Y V I Z

3(99.6)(1.26)cos(5 ( 66 )) 122.6 WP

Two-Wattmeter Power Measurement

cc = current coilvc = voltage coil

W1 read

1 1cosAB AP V I

W2 read

2 2cosCB CP V I

For balanced load with abc phase sequence

1 230 and 30a a

is the angle between phase current and phase voltage of phase a a

Two-Wattmeter Power Measurement(cont.)

1 2

2 cos cos30

3 cos

L L

L L

P P P

V I

V I

To determine the power factor angle

1 2 2cos cos30L LP P V I

1 2 ( 2sin sin 30 )L LP P V I

1 2

1 2

2cos cos30 3

( 2sin sin 30 ) tanL L

L L

P P V I

P P V I

11 2 1 2

1 2 1 2

tan 3 or tan 3P P P P

P P P P

Example 12.9-1 P = ?

10 45 Z

line-to-line voltage = 220Vrms

The phase voltage220

303

A V

The line current220 30

12.7 7510 3 45

AA

V

IZ

and 12.7 195B I

1 1

2 2

cos 2698 W

cos 723 WAC A

BC B

P V I

P V I

1 2 3421 WP P P

Electrodynamic Wattmeter

Digital Power Meter VAR Meter

pf Meter

Summary

Three-Phase voltages

The Y-to-Y Circuits

The -Connected Source and Load

The Y-to- Circuits

Balanced Three-Phase Circuits

Instantaneous and Average Power in Bal. 3 Load

Two-Wattmeter Power Measurement

Chapter 12 Three-Phase Circuits. Power Factor Correction A balanced three-phase circuit Medium...

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Transcript of Chapter 12 Three-Phase Circuits. Power Factor Correction A balanced three-phase circuit Medium...