Module 8 Three Phase Systems

Three-Phase Systems

EE 102 Circuits 2

Module 8

by:Cesar G. Manalo, Jr.

Objectives Define a three-phase system. Describe the wye and delta-type connections in three-phase

systems. Solve voltages and currents in three-phase system. Describe a single-phase three-wire system. Solve voltages and current in single-phase three-wire system.

Introduction The three-phase (3-φ) AC generators (alternators) has 3 sets of

armature windings called phase windings. Each of the 3 windings develop exactly the same sinusoidal voltages

(same magnitude and frequency) called phase voltage but are 120 electrical degrees apart.

N S

120 e deg

120 e deg

3-phase 2-pole alternator

Introduction In most cases, the winding that produces the flux is the one

revolving and the armature winding (that develops phase voltages) is stationary.

The 3 sets of windings can be wired together to form either a delta(Δ) connection or a wye(Y) connection.

a

c

ba’

c’ b’

Eaa’

Ecc’

Ebb’ c

c’bb’

a

a’

N S

Introduction

Y-connected alternator

a

c

ba’

c’ b’

Eaa’

Ecc’ Ebb’ N S

c

c’bb’

a

a’

a

b

c

a

b

c

3-phase 2-pole Y-connected alternator

Introduction


a

c

ba’

c’ b’

Eaa’

Ecc’ Ebb’

a

b

c

Introduction


a

c

ba’

c’ b’

Eaa’

Ecc’ Ebb’

A

B

C

a

c

b

Ea

Ec Eb

A

B

C

n

Y-connected alternator with common point called neutral (n)

Line-to-line Voltage in 3-Phase Y-connected Alternators

In a balanced system, each of the three instantaneous voltages a,b, & c have equal amplitudes but are separated from the other voltages by a phase angle of 120o.

EAN

EBN

ECN

120o

120o

120o

EAN

EBNECN

A

B

C

EAB

EBC

ECAN

3-phase alternator with the phases replaced by 3 single-phase alternatorsWye-connected 3-phase alternator

Phasor diagram of phase voltages

a

c

b

Ea

Ec Eb

A

B

C

N

+

+ +

-

- -


EAN

EBN

ECN

120o

120o

120o

)( BNANNBANAB EEEEE

-EBN

a

bc

A

B

C

EAB

EBC

ECA

N

EAN

ENB

60o

where EAN is the phase voltage of alternator a


EAN

)( BNANNBANAB EEEEE

-EBN

E AB = E AN

+ (-E BN)

ANANAB EEE 73.13

BNBNBC EEE 73.13

CNCNCA EEE 73.13

a

bc

A

B

C

EAB

EBC

ECA

N

EAN

ENB

60o30o

Therefore, the line-to-line voltages of a wye-connected alternator is 1.73 times its phase voltage.


a

bc

A

B

C

EAB

EBC

ECA

NEAN

EBN

ECN

120o

120o

120o

EABECA120o

120o 120o

Therefore, the line-to-line voltages of a wye-connected alternator is 1.73 times its phase voltage, and they are 120o out of phase from each other.

EBC

Phasor diagram of phase and line voltages of a 3-phase Y-connected alternator


ACOUTPUT

• The line-to-line voltage of a 3-phase alternator is 230 V as measured by an AC voltmeter. What is the phase voltage if measured by the same voltmeter if the alternator is connected in wye ? Write the polar form of all line-to-line and phase voltages.

Illustrative Problem 11

Solution:Let: EAB = line-to-line voltage between A & B

EAN = phase voltage of phase AN.

VEE

EE

ABAN

ANAB

9.13273.1

23073.1

73.1

VE

VE

VE

VE

VE

referenceVE

CA

BC

AB

CN

BN

AN

150230

90230

30230

1209.132

1209.132

)(09.132EAN

EBN

ECN

120o

120o

120o

EAB120o

120o 120o

EBC

ECA

Line Current in 3-Phase Y-connected Alternators

a

bc

A

B

C

VAB

VBC

VCA

N Ia

Ib

Ic

120o

120 o

120o

IA

IB

IC

Ia

IbIcIA

IC

IB Phase and line current phasor diagram of a balanced-loaded 3-phase alternator

• The current on a line is equal to the current through the phase to which that line is connected. That is, IA = Ia, IB = Ib, IC = Ic.

• When the external load on a 3-phase alternator is “balanced”, the line and phase currents are all 120o out of phase from each other.

EAN

EBN

ECN

Alternator

Line-to-line Voltage in 3-Phase Δ-connected Alternators

• In a Δ-connected 3-phase alternator, there is no neutral pt.• The phase windings are connected one after the other forming a

delta or closed loop.• The line-to-line voltages equals the phase voltages both in

magnitude and in direction because the phases are connected directly to the lines.

EA

EB

EC

120o

120o

120o

EA

EB

EC

A

B

C

EAB

EBC

ECA

Δ-connected 3-phase alternators Phasor diagram

+

+

+

-

-

-

EAB

EBC

ECA

Line-to-line Current in 3-Phase Δ-connected Alternators

EA

EB

EC

120o

120o

120o

EA

EB

EC

A

B

C

EAB

EBC

ECA

Δ-connected 3-phase alternators Phasor diagram

+

+

+

-

-

-

EAB

EBC

ECA

CCA

BBC

AAB

EEEEEE

IAB

IBCICA

IABIBC

ICA

CACABCC

BCBCABB

ABABCAA

IIII

IIII

IIII

3

3

3

IA

IB

IC

-IAB

IA

IB

IC

• In a Δ-connected balanced 3-phase alternator, the line-currents are 1.73 times the phase currents.

Balanced 3-Phase Loads

a

bc

A

B

C

VAB

VBC

VCAN

IA

IB

IC

Ia

IbIc

• A load is said to be balanced, if all phase impedances are equal.

ZA

ZBZC

3-phase Y-connected Balanced Load, ZA = ZB = ZC

a

bc

A

B

C

VAB

VBC

VCAN

IA

IB

IC

Ia

IbIc

ZC

ZB

ZA

3-phase Δ-connected balanced load, ZA = ZB = ZC

3-phase rectifier

3-phase induction motor


a

bc

A

B

C

230 V

N

IA

IB

IC

Ia

IbIc

Z

Z

Z

230 V 230 V

3-phase Alternator

Balance3-phase load, Z : R = 75Ω, XL = 50 Ω

A balance 3-phase load is connected to a 3-phase alternator with line-to-line voltage of 230 volts. Find:a) Load phase currentsb) Load line currentsc) Phasor diagram of phase and line currents.



a

bc

A

B

C

230 V

N

IA

IB

IC

Ia

IbIc

Z

230 V 230 V

3-phase Alternator

Balance3-phase load, Z : R = 75Ω, XL = 50 ΩZ

Z

IAB

IBC

ICA

Let:VAB = 230 30∠ o

VBC = 230 -90∠ o

VCA = 230 150∠ o

14.90)50()75( 22 Z

69.33

7550tan 1

AIII CABCAB 55.214.90

230

for balance 3-phase load

Since Z is inductive, the load phase currents will lag the line voltages by 33.69o. Hence,

69.355.2)69.3330(55.2ABI

69.12355.2)69.3390(55.2BCI

116.3155.2)69.33150(55.2CAI

Balanced 3-Phase Loads 69.355.269.333055.2ABI

69.12355.2)69.3390(55.2BCI

116.3155.2)69.33150(55.2CAI

33.69o

33.69o

33.6

9o

IAB

IBC

ICA

)( CAABCAABA IIIII

)( ABBCABBCB IIIII

)( BCCABCCAC IIIII

IA

IB

IC

33.69o

86.31o

153.69o AIIII

AB

CBA

41.4)55.2(73.173.1

VABVCA120o

120o 120o

VBC

30o

(equal in RMS value but not in phase)

IA

IB

IC

Z


Z

Z

IAB

IBC

ICA

69.33

Vector calculator

http://www.1728.org/vectors.htm

Balanced 3-Phase Power

a

bc

A

B

C

VAB

VBC

VCAN

IA

IB

IC

Ia

IbIc

ZC

ZBZA

ZA = ZB = ZC

φAB

IAB

IBC

ICA VABVCA120o

120o 120o

VBC

30o

IAB

IBC

ICA

φBC

φCA

CACAC

BCBCB

ABABA

IVSIVSIVS

Apparent power per phase

CACACAC

BCBCBCB

ABABABA

IVQIVQIVQ

sinsinsin

Reactive power per phase

CACACAC

BCBCBCB

ABABABA

IVPIVPIVP

coscoscos

Real power per phase


CACACAC

BCBCBCB

ABABABA

IVQIVQIVQ

sinsinsin

Reactive power per phase

CACACAC

BCBCBCB

ABABABA

IVPIVPIVP

coscoscos

Real power per phase

For balanced delta-connected load:

PCABCAB

LPCABCAB

LPCABCAB

IIIII

VVVVV

3

CACAC

BCBCB

ABABA

IVSIVSIVS

Apparent power per phase

PPPPCBA

PPPPCBA

PPPCBA

IVPPPPIVQQQQIVSSSS

cossin

RMS value only not the phase

Value only not the phase

PLLPL

LPPPT IVIVIVQ sin3sin3

3sin3

PLLPL

LPPPT IVIVIVP cos3cos3

3cos3

LLTTT IVQPS 322

where:VP = phase voltageVL = line voltageIP = phase currentIL = line currentSP = phase apparent powerQP = phase reactive powerPP = phase real power

Let:QT = total reactive power taken by loadPT = total real power taken by loadST = total apparent power taken by load


a

bc

A

B

C

230 V

N

IA

IB

IC

Ia

IbIc

Z

Z

Z

230 V 230 V

3-phase Alternator


Find the per phase apparent, reactive, and true power consumed by the load as shown by the figure below.


AIIII PCABCAB 55.2

69.33P

VAB = VBC =VCA = VP = 230 V

VAIVS

WIVP

VARIVQ

PPP

PPPP

PPPP

586.5)55.2)(230(

0.48869.33cos)55.2)(230(cos

33.32569.33sin)55.2)(230(sin

KVAVAPQS

WIVP

VARIVQ

TTT

PPPT

PPPT

76.15.759,1)24.845()49.563(

99.146369.33cos)55.2)(230(3cos3

99.97569.33sin)55.2)(230(3sin3

2222

Solution:


A 3-phase 74.6 kW delta-connected induction motor is supplied by a 3-phase star-connected alternator generating 1000 V between phases. If the full load efficiency and the power factor of the induction motor are 92 % and 0.85 respectively, calculate;a) Current in each motor phaseb) Current in each alternator phase



A 3-phase, 50 Hz, 415 V star-connected induction motor has an output of 50 kW with an efficiency of 90 % and a power factor of 0.85. calculate the line current. If the motor windings are now connected in delta, what would be the correct voltage of a 3-phase supply suitable for the motor?

Seatwork

Balanced 3-Phase PowerIllustrative Problem 14

Two 3-phase balance loads are connected in parallel to a 400-V line. The first load is delta-connected with a phase impedance of Z = 30+j40 and the second is a star-connected purely resistive load of R = 25 ohms per phase. Find:a) The phasor line currents IA, IB, and IC.b) The apparent, reactive, and real power consumed by the combined load.c) The PF of the combined load.

Z

Z ZR

R

R

A

B

C

IA

IB

IC

N

Phase Sequence

N S

Perspective view of armature

N S

Side view of armature

Phase Sequence

N S

Perspective view of armature

N S

Side view of armature (turned 90o)

Phase Sequence

N S

120 e deg

N S

A B CPerspective view of armature

Side view of armature

Phase Sequence

N S

120 e deg

N S

A B C

Phase Sequence

N S

120 e deg

N S

A B CPHASE SEQUENCE = ABC = BCA = CAB

Va

Vb

Vc

Phasor representation of ABC sequence

Phase Sequence

120 e deg

N S

A C BPHASE SEQUENCE = ACB = CBA = BAC

ARMATURE WINDINGS YELLOW AND RED INTERCHANGED

Va

Vc

Vb

Phasor representation of ACB sequence

Phase Sequence

120 e deg

N S

A C B

A phase sequence defines which of the three phase voltages or line voltages comes in sequence.

There are only two possible phase sequence in three-phase system: phase sequence ABC or phase sequence ACB.

Phase sequence ABC is called positive sequence while phase sequence ACB is called negative sequence.

Phase Sequence

120 e deg

N S

A B C

3-PHASE 4-WIRE

a

bc

A

B

C

VAB

VBC

VCA

N

NVANVBNVCN

AB

CN

3-phase 4-wire alternator with output connected to lines A, B, C, & N

3-phase 4-wire lines

NA

B

C

Phase Sequence

3-PHASE 4-WIREAB

CN

Phase sequence ABC:

120 e deg

N S

A B C

oPAN VV 0

oPBN VV 120

oPCN VV 120

where VP is the line-to-neutral voltage or the phase voltage

Checkpoint 1

AB

CN

Phase sequence ABC:

120 e deg

N S

A B C

oPAN VV 0

oPBN VV 120

oPCN VV 120


Given the phase sequence of phase voltages VAN, VBN, and VCN, below, what is the phase sequence of the line voltages, VAB, VBC, VCA?

Phase Sequence

3-PHASE 4-WIREAB

CN

Phase sequence ABC:

120 e deg

N S

A B C

oPAN VV 0

oPBN VV 120

oPCN VV 120

oPAB VV 303

oPBC VV 903

oPCA VV 1503


Phase Sequence

3-PHASE 4-WIREAB

CN

Phase sequence ACB:o

PAN VV 0

oPBN VV 120

oPCN VV 120

oPAB VV 303

oPBC VV 1503

oPCA VV 903


120 e deg

N S

A C B

Phase Sequence

3-PHASE 3-WIREAB

C

Phase sequence ABC:

120 e deg

N S

A B C

oLAB VV 0

oLBC VV 120

oLCA VV 120

Phase Sequence

3-PHASE 3-WIREAB

C

Phase sequence ACB:

oLAB VV 0

oLBC VV 120

oLCA VV 120

120 e deg

N S

A C B

where VL is the line-to-line voltage

Single-Phase 3-Wire System

A

n

BN

S

N

S

n

AB180 e deg

Bn

An

oPAn VV 0 o

PBn VV 180

)sin(max tvv PAn )180sin(maxo

PBn tvv

VBn

BnAnAnAnBnAnnBAnAB VVVVVVVVV 22

VAB

4-pole alternator with 2 coils A & B

Coils A & B, in series

n

VAn

VBn

VAn VAB

+

+

-

-

Find the currents and total power dissipation as shown in Fig. 1 and Fig. 2 if VAn = - VBn = 100 V.


Single-Phase 3-Wire System

A

n

B

n

Z1 = 30 + j40

Z2 = 30 + j40

IA

IB

In

A

n

B

n

Z1 = 30+j40

Z2 = 30 + j40

IA

IB

In

Z3 = 30+j40 Z3 = 30+j40

+

+

-

-

+

+

-

-

Find the currents and total power dissipation as shown in Fig. 3 VAn = - VBn = 150 V.


Seatwork

A

n

B

n

IA

IB

In

Z

Z = 40 + j50

Z

+

+

-

-

END OF SESSION

Documents

Module 8 Three Phase Systems