Chapter1 - 3 phase system 17feb13 - Universiti Malaysia...

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EKT103 ELECTRICAL ENGINEERING Chapter 1 Chapter 1 Three Three - - Phase System Phase System 1

Transcript of Chapter1 - 3 phase system 17feb13 - Universiti Malaysia...

EKT103 ELECTRICAL ENGINEERING

Chapter 1Chapter 1

ThreeThree--Phase SystemPhase System

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COURSE OUTCOME (CO)

CO1: Ability to define and explain the concept of single-phase and three-phase system.

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• A sinusoid is a signal that has the form of the sine or cosine function.

• A general expression for the sinusoid,

where

Vm = the amplitude of the sinusoidω = the angular frequency in radians/s

Ф = the phase

Revision

)sin()( φω += tVtv m

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Revision

A periodic function is one that satisfies v(t) = v(t + nT), for all t and for all integers n.

ω

π2=T

HzT

f1

= fπω 2=

• Only two sinusoidal values with the same frequency can be compared by their amplitude and phase difference.

• If phase difference is zero, they are in phase; if phase difference is not zero, they are out of phase.

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Revision

Example 1

Given a sinusoid, , calculate its amplitude, phase, angular frequency, period, and frequency.

)604sin(5 ot −π

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Revision

Example 1

Given a sinusoid, , calculate its amplitude, phase, angular frequency, period, and frequency.

Solution:

Amplitude = 5, phase = –60o, angular frequency = 4π rad/s, Period = 0.5 s, frequency = 2 Hz.

)604sin(5 ot −π

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Revision

Example 2

Find the phase angle between and , does i1 lead or lag i2?

)25377sin(41

oti +−=

)40377cos(52

oti −=

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Revision

Example 2

Find the phase angle between and , does i1 lead or lag i2?

)25377sin(41

oti +−=

)40377cos(52

oti −=

Solution:

Since sin(ωt+90o) = cos ωt

therefore, i1 leads i2 155o.

)50377sin(5)9040377sin(52

oootti +=+−=

)205377sin(4)25180377sin(4)25377sin(41

oooottti +=++=+−=

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Revision

Impedance transformation

YZ3Z =∆ ∆= Z3

1ZY

SingleSingle--Phase CircuitPhase Circuit

Three wired system

• same magnitude

• same phase

A single phase circuit consists of a generator connected through a pair of wires

to a load

Two wire system

Aa

TwoTwo--Phase CircuitPhase Circuit

Three wired

system

Second source

with 90° out

of phase

Three wired system

• same magnitude

• different phase

• It is 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°.

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What is a ThreeWhat is a Three--Phase Circuit?Phase Circuit?

Three sources

with 120° out

of phaseFour wired

system

• A three-phase generator consists of a rotating magnet (rotor) surrounded by a stationary winding (stator).

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Balance ThreeBalance Three--Phase Voltages Phase Voltages

A three-phase generator The generated voltages

• Two possible configurations:

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Balance ThreeBalance Three--Phase Voltages Phase Voltages

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

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Phase sequences

a) abc or positive sequence b) acb or negative sequence

Balance ThreeBalance Three--Phase Voltages Phase Voltages

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If the voltage source have the same amplitude and frequency ω and are out of phase with each

other by 120o, the voltage are said to be balanced.

0VVV cnbnan =++

cnbnan VVV ==

Balanced phase voltages are equal in magnitude and out of phase with each other by

120o

Balance ThreeBalance Three--Phase Voltages Phase Voltages

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0

p

0

pcn

0

pbn

0

pan

120V240VV

120VV

0VV

+∠=−∠=

−∠=

∠=

abc sequence or positive sequence:

acb sequence or negative sequence:

0

p

0

pbn

0

pcn

0

pan

120V240VV

120VV

0VV

+∠=−∠=

−∠=

∠=

pV is the

effective or

rms value

Balance ThreeBalance Three--Phase Voltages Phase Voltages

Example 1

Determine the phase sequence of the set of

voltages.

)110cos(200

)230cos(200

)10cos(200

°−=

°−=

°+=

tv

tv

tv

cn

bn

an

ω

ω

ω

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Balance ThreeBalance Three--Phase VoltagesPhase Voltages

Solution:

The voltages can be expressed in phasor form as

We notice that Van leads Vcn by 120°and Vcn in turn

leads Vbn by 120°.

Hence, we have an acb sequence.

V 110200V

V 230200V

V 10200V

°−∠=

°−∠=

°∠=

cn

bn

an

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Balance ThreeBalance Three--Phase VoltagesPhase Voltages

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Two possible three-phase load configurations:

a) a wye-connected load b) a delta-connected load

Balance ThreeBalance Three--Phase Voltages Phase Voltages

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

For a balanced wye connected load:

Y321 ZZZZ ===

∆=== ZZZZ cba

For a balanced delta connected load:

YZ3Z =∆ ∆= Z3

1ZY

Balance ThreeBalance Three--Phase Voltages Phase Voltages

• Four possible connections

1. Y-Y connection (Y-connected source with

a Y-connected load)

2. Y-∆ connection (Y-connected source with

a ∆-connected load)

3. ∆-∆ connection

4. ∆-Y connection

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Balance ThreeBalance Three--Phase Connection Phase Connection

Balance YBalance Y--Y Connection Y Connection

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

load.

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=

=

=

=

Y

L

s

Z

Z

Z

Z

l

Source impedance

Line impedance

Load impedance

Total impedance per phase

LY ZZ =

Balance YBalance Y--Y Connection Y Connection

LsY ZZZZ ++=l

Since all impedance are in series,

Thus

Balance YBalance Y--Y Connection Y Connection

Y

ana

Z

VI =

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Applying KVL to each phase:

Y

ana

Z

VI =

0

a

Y

0

an

Y

bnb 120I

Z

120V

Z

VI −∠=

−∠==

0

a

Y

0

an

Y

cnc 240I

Z

240V

Z

VI −∠=

−∠==

0IIII ncba =−=++

0IZV nnnN ==

Balance YBalance Y--Y Connection Y Connection

Balance YBalance Y--Y Connection Y Connection

0

pca

0

pbc

0

pab

210V3V

90V3V

30V3V

−∠=

−∠=

∠=

Line to line voltages or line voltages:

Magnitude of line voltages:

pL V3V =

cabcabL VVVV ===

cnbnanp VVVV ===

Example 2

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

shown below:

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Balance YBalance Y--Y Connection Y Connection

Example 2

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

shown below:

A 2.9881.6I

A 8.14181.6I

A 8.2181.6I

Ans

°∠=

°−∠=

°−∠=

c

b

a

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Balance YBalance Y--Y Connection Y Connection

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BalancBalance e Y-∆ CConnection onnection

• A balanced Y-∆ system is a three-phase system with a balanced y-connected source and a balanced ∆-connected

load.

BalancBalance e Y-∆ CConnection onnection

A single phase equivalent circuit

3

ZZY

∆=

3/∆

==Z

V

Z

VI an

Y

an

a

BalancBalance e Y-∆ CConnection onnection

A single phase equivalent circuit

CA

0

pca

BC

0

pbc

AB

0

pab

V210V3V

V90V3V

V30V3V

=−∠=

=−∠=

=∠=

Line voltages:

BalancBalance e Y-∆ CConnection onnection A single-phase equivalent circuit of a balanced Y-∆∆∆∆ circuit

°∠=−=

°−∠=−=

°−∠=−=

903

1503

303

ABBCCAc

ABABBCb

ABCAABa

IIII

IIII

IIII

Line currents:

=

=

=

Z

VI

Z

VI

Z

VI

CA

CA

BC

BC

AB

ABPhase currents:

BalancBalance e Y-∆ CConnection onnection A single-phase equivalent circuit of a balanced Y-∆∆∆∆ circuit

)24011(IIII

240II

0

ABCAABa

0

ABCA

−∠−=−=

−∠=

0

ABa 303II −∠=

pL II 3=

cbaL IIII ===

CABCABp IIII ===

Magnitude line currents:

Example 3

A balanced abc-sequence Y-connected source with ( )

is connected to a ∆-connected load (8+j4)Ω per phase. Calculate the phase and line currents.

Solution

Using single-phase analysis,

Other line currents are obtained using the abc phase sequence

°∠= 10100Van

A 57.1654.3357.26981.2

10100

3/Z

VI an °−∠=

°∠

°∠==

a

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Balance Balance Y-∆ ConnectionConnection

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Balance Balance ∆-∆ CConnectiononnection

• A balanced ∆-∆ system is a three-phase system with a balanced ∆ -connected source and a balanced ∆ -connected

load.

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=

=

=

Z

VI

Z

VI

Z

VI

CA

CA

BC

BC

AB

AB

CAca

BCbc

ABab

VV

VV

VV

=

=

=

Line voltages: Line currents:

3II pL =

Magnitude line currents:

3

ZZY

∆=

Total impedance:

Phase currents:

°∠=−=

°−∠=−=

°−∠=−=

903

1503

303

ABBCCAc

ABABBCb

ABCAABa

IIII

IIII

IIII

Balance Balance ∆-∆ CConnectiononnection

Example 4

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

Ans:

The phase currents

The line currents

V 0330Vab °∠=

A 87.1562.13IA; 13.812.13IA; 87.362.13I °∠=°−∠=°∠= ABBCAB

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A 87.12686.22IA; 13.11386.22IA; 87.686.22I °∠=°−∠=°∠= cba

Balance Balance ∆-∆ CConnectiononnection

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Balance Balance ∆-Y ConnectionConnection

• A balanced ∆-Y system is a three-phase system with a balanced y-connected source and a balanced y-connected

load.

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Balance Balance ∆-Y ConnectionConnection

Applying KVL to loop aANBba:

Y

p

baZ

VII

00∠=−

From:

0

ab 120II −∠=

0

aba 303III ∠=−

Y

p

aZ

V

I

0303

−∠

=

Line currents:

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Replace ∆ connected source to equivalent Y

connected source.Phase voltages:

0p

cn

0p

bn

0p

an

903

VV

1503

VV

303

VV

+∠=

−∠=

−∠=

Balance Balance ∆-Y ConnectionConnection

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A single phase equivalent circuit

Y

p

Y

an

aZ

V

Z

VI

0303

−∠

==

Balance Balance ∆-Y ConnectionConnection

Example 5

A balanced Y-connected load with a phase impedance 40+j25 Ωis supplied by a balanced, positive-sequence ∆-connected source with a line voltage of 210V. Calculate the phase currents. Use Vab as reference.

Answer

The phase currents

A; 5857.2I

A; 17857.2I

A; 6257.2I

°∠=

°−∠=

°−∠=

CN

BN

AN

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Balance Balance ∆-Y ConnectionConnection

phase-single ,2'2

2

L

Lloss

V

PRP =

( ) phase- three,'3

'3''3'2

2

2

22

L

L

L

LLloss

V

PR

V

PRRIP ===

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Power in a Balanced SystemPower in a Balanced System

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

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

( ) phase-single ,222

22

L

LLloss

V

PRRIP ==

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)120cos(2

)120cos(2

cos2

0

0

+=

−=

=

tVv

tVv

tVv

pCN

pBN

pAN

ω

ω

ω

For Y connected load, the phase voltage:

Power in a Balanced SystemPower in a Balanced System

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Phase current lag phase voltage by θ.

θ∠= ZZY

If

)120cos(2

)120cos(2

)cos(2

0

0

+−=

−−=

−=

θω

θω

θω

tIi

tIi

tIi

pc

pb

pa

The phase current:

Power in a Balanced SystemPower in a Balanced System

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Total instantaneous power:

cCNbBNaANcbaivivivpppp ++=++=

θ= cosIV3ppp

Average power per phase:

Apparent power per phase:

Reactive power per phase:

Complex power per phase:

θ= sinIVQppp

θ= cosIVPppp

*

pppppIVjQPS =+=ppp

IVS =

Power in a Balanced SystemPower in a Balanced System

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Total average power:

θ=θ== cosIV3cosIV3P3PLLppp

Total reactive power:

θ=θ== sinIV3sinIV3Q3QLLppp

Total complex power:

*

p

2

p

p

2

p

*

pppZ

V3ZI3IV3S3S ====

θ∠=+=LL

IV3jQPS

Power in a Balanced SystemPower in a Balanced System

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Power loss in two wires:

2

L

2

L2

LlossV

PR2RI2P ==

Power loss in three wires:

2

2

2

22

333

L

L

L

LLloss

V

PR

V

PRRIP === PL : power absorbed by the load

IL : magnitude of line current

VL : line voltage

R : line resistance

Power in a Balanced SystemPower in a Balanced System

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Example 6

A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the

line voltage is 220 V and the line current is 18.2 A.

Determine the power factor of the motor.

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Example 6

A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the

line voltage is 220 V and the line current is 18.2 A.

Determine the power factor of the motor.

The apparent power is

VAIVS LL 13.6935)2.18)(220(33 ===

The real power is WSP 5600cos == θ

The power factor is8075.0cos ===

S

Ppf θ

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Exercise 6

Calculate the line current required for a 30-kW three-phase motor having a power factor of 0.85

lagging if it is connected to a balanced source

with a line voltage of 440 V.

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Exercise 6

Calculate the line current required for a 30-kW three-phase motor having a power factor of 0.85

lagging if it is connected to a balanced source

with a line voltage of 440 V.

Answer : 31.46=LI