L3-BEKG2433-Three_Phase_Part_1.pdf

7/24/2019 L3-BEKG2433-Three_Phase_Part_1.pdf

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CHAPTER 3

Three Phase System

(Part 1)

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CHAPTER 3: THREE-PHASE

SYSTEM

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1) Introduction

2) 3-phase generation3) 3-phase circuit

4) Definitions: phase & line voltages, phase & line currents

5) V & I in star-connected system

6) V & I in delta-connected system

This part will cover the subtopic as below:


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Three-phase electric power is a type of polyphasesystem, and is the most common method used byelectric power distribution grids worldwide to distributepower.

Three circuit conductors carry three alternating currents(of the same frequency) which reach their instantaneouspeak values at different times.

Taking one conductor as the reference, the other two are

delayed in time by one-third (120phase shift) and two-thirds of one cycle (240phase shift).

1) Introduction

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Electric power supply system in a Malaysia comprises of:

generatingunits that produce electricity

high voltage transmission lines that transport electricityover long distances

distributionlines that deliver the electricity to consumers

loadsthat receive the useful power

Electrical Power System

Generation TransmissionDistribution

Loads

11kV20kV (50Hz)

Figure 3: An example of powersystem in Malaysia


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Certain text books use A,B,C instead of R,Y,B

Each phase is separated by 120phase-shiftERR= V sin EYY= V sin (- 120) EBB= V sin (- 240)

or = V sin (+ 120)

3-Phase (3) Generation

Time

Please refer this website : http://www.wisc-online.com/objects/ViewObject.aspx?ID=IAU14008

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http://www.wisc-online.com/objects/ViewObject.aspx?ID=IAU14008http://www.wisc-online.com/objects/ViewObject.aspx?ID=IAU14008http://www.wisc-online.com/objects/ViewObject.aspx?ID=IAU14008http://www.wisc-online.com/objects/ViewObject.aspx?ID=IAU14008


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Phase Sequence

The order of voltage waveform sequences in a polyphasesystem is calledphase rotationorphase sequence.

If we're using a polyphase voltage source to powerresistive loads, phase rotation will make no difference atall. Whether R-Y-B or R-B-Y, the voltage and currentmagnitudes will all be the same.

There are some applications of three-phase power, thatdepend on having phase rotation being one way or theother.

3motor is one of them

Sequence:

Positive Sequence Negative Sequence

R-Y-B @ ABC R-B-Y @ ACB

sin1RR

e

01201

sinYYe

01201 sinBBe

sin1RR

e

01201

sinYYe

01201 sinBBe

A

B

C

A

B

C

10


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Example 1: (Assume the system is balance and sequence is R-Y-B)

a. If VRR = 277V0, what are VYYand VBB

b. If VYY= 347V-120, what are VRR and VBB

c. If VBB= 120V150, what are VRR and VYY

Answer:VYY= 277V-120 VBB= 277V120

Answer:VRR= 347V0 VBB = 347V120

Answer:VRR= 120V30 VYY= 120V-90

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3) 3-Phase Circuit

2 Types:

Star or wye (Y)

connection

Delta () or meshconnection


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Connections Between Generator-Load

1. Star to star

2. Delta to delta

3. Star to delta

4. Delta to star

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Star-star (or Y-Y) Connection

n

With neutral conductor, thesystem is known as 4-wirestar-connected system or 3-phase 4-wire system

In a balanced load, theresultant current in neutralconductor is zero, IR + IY + IB= 0.

Neutral conductor

IR

IY

IB

IN

Neutral points

Without neutral conductor, thesystem is known as three-wirestar-connected system or 3-phase 3-wire system


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Line Currents = Phase Currents : Ir, Iy, Ib

Configuration

Voltages at the GeneratorPhase VoltagesERN, EYN, EBNLine Voltages (Line to Line)ERY, EYB, EBR

Voltages at the LoadPhase VoltagesVRn, VYn, VBnLine Voltages (Line to Line)VRY, VYB, VBR

Ib

Iy

Ir

Definitions (Star) :

Line Voltages

ERY, EYB, EBRLine voltages at the generatorVry, Vyb, VbrLine voltages at the loadPhase Voltages

ERN, EYN, EBNPhase voltages at the generator

Vrn, Vyn, VbnPhase voltages at the loadLine Currents = Phase CurrentsIr, Iy, Ib

4) DEFINITIONS

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5) V & I in Star-Connected System

VYB = VYn- VBn

=3VYn-90

VBR= VBn- VRn

=3VBn150

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oP h a s eP h a s eL in e VV 303

oLine

Line

Ph as e

VV 30

3

0

0

0

303

303

303

BnBnBR

YnYnYB

RnRnRY

VV

VV

VV

Phasor Diagram:

ILine = IPhase

|VLine| = 3 |VPhase|

Only for a balanced star

connection. Otherwise,

derive from KVL.

Summary: V & I in Star-Connected System


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Example (Balanced Star-Connected System)

For the following circuit, suppose ERN= 120V0, Zrn= Zyn= Zbn= (12j9)

a) Determine the phase voltages at the load.b) Determine the line voltages at the load.

c) Show all voltages on a phasor diagram.

d) Compute IR, then determine IYand IBby inspection.

e) Verify answer in d by direct computation.

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Solution:

a.VRn = ERn. Thus,0

0120 VVrn

.

Since the system is balanced,

o

bn

o

yn VV&VV 120120120120

b. Since the system is balanced,

030208303 V)(VV o

rnry

0015020890208 VV&VV bryb

c. The phasors are shown below:


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0

0

0

138308873615

120120..

.

A

V

Z

VI

yn

yn

y

0

0

0

8715608873615

120120..

.

A

V

Z

VI

bn

bnb

e. The currents by direct computation

0

0

00

873608873615

0120

912

0120.A.

.jZ

VI

rn

rnr

d. The currents:

000138308120873608 .A..A.Iy

000 8715608120873608 .A..A.Ib

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Line Currents

Ir, Iy, IbCurrents in the line conductors

Phase Currents

Iry, Iyb, IbrCurrents through load

Line Voltages = Phase Voltages

Vry, Vyb, Vbr

Delta Configuration


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6) V & I in Delta-Connected System

The relationship between line and phase current of a delta circuit,consider:

For Voltage, we can see that VLine= VPhase

For current, using KCL, at node r, we get

Ir + Ibr= Iry Ir= Iry - Ibr

r

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Substuting..

Ir = Iry- Ibr= I0- I120

= 1.723 I-30

=3 I-30

=3 Iry-30

Iy = Iyb- Iry

=3 Iyb-150

Ib = Ibr- Iyb

=3 Ibr90

Now assume a magnitude I for each phase current and take Iryasreference. The sequence is R-Y-B and the system is balance.

o

b r

o

yb

o

r y IIIIII 120&120,0


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0

0

303

303

LineLine

Phase

PhasePhaseLine

II

II

0

0

0

903

1503

303

brbrb

ybyby

ryryr

II

II

II

Phasor Diagram:

VLine = VPhase

|ILine| = 3 |IPhase|

Summary : V & I in Delta-Connected System

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Example 1 (Balanced Delta-Connected System)

For the circuit in the previous slide page, suppose Vry= 240V15,Zry= Zyb= Zbr= 10 + j 3

a) Determine the phase currents.

b) Determine the line currents.

c) Sketch the phasor diagram.


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Example 3

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Example 4

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Determine :

a) Phase current b) Line current

Solution :

13.239.13

129

30208A

jZ

VI

ab

abab

13.1439.13

120withIlagIo

abbc

AIbc

VAB= 208V30

87.969.13

120withIleadI oabca

AIca

13.5324)13.2330()9.13(3

303

A

II aba

b)a)

13.17324

120withIlagI oab

AIb

87.6624

120withIleadIo

ac

AIc

Example 5

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Solution :

22.01170120

1.41.3

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dividervoltageuse

Vj

jVan

ircuitequivalent1Find

EV ABab

c

0120E:referenceasE

1203

208E

generatoratvoltagephase

AN V

V

AN

AN

Line voltage at generator is 208 volts.

Determine line voltage at loads.

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3/)129(3/Z

j

jZY

22.30203303 VVV anab

Example 6

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L3-BEKG2433-Three_Phase_Part_1.pdf