et201 unit 5.ppt - ~ my edu blog~ • three-phase electricity supply is a common way of transmission...

UNIT 5UNIT 5UNIT 5UNIT 5

THREE PHASETHREE PHASESYSTEMSYSTEMPREPARED BY:PREPARED BY:

NABILAH BINTI MAZALANNABILAH BINTI MAZALAN

OBJECTIVES• Understand three phase system• Explain the basic principles of a three phase

system• Understand three system configurations• Advantage and application of three phase

compare to single phasecompare to single phase• Explain three phase e.m.f generation• Identify connection and vector phase diagram

for delta and star system• Determine the phase voltage, phase current,

line voltage, line current• Define balance load in three phase system• Calculate power for three phase system

INTRODUCTION• Three-phase electricity supply is a common way of

transmission of alternating current electricity supply.• It is a multi-phase systems, and is the most

commonly used in electrical distribution grid systemin the world. It is also used as a source of power forlarge electric motors and other high load.

• In general, three-phase system is the most• In general, three-phase system is the mosteconomical system as it uses less conductormaterial to transmit electricity from the systemsingle-phase or two phases are equal at the samevoltage.

• In the three-phase system, three conductors carrythree alternating current that reached peak valuesat different times, with intervals between the phasesof 1 / 3 cycle.

THREE PHASE SYSTEM

• A three phase ac power systems consistsof three phase generators, transmissionlines and load

• A three phase generator consists of threesingle phase generators, with voltagessingle phase generators, with voltagesequal in magnitude but differing in phaseangle from the others by 120

GENERATION OF 3-PHASE

• A 3-phase generator consists of 3 single phasegenerators, with voltages equal in magnitude butdiffering in phase angle from the others by 120

• Rotating at a constant speed in a uniformmagnetic field.magnetic field.

• Each voltage source is called phase.

• Usually, three phases are labeled as– Red (R-Red)

– Yellow (Y-Yellow)

– Blue (B-Blue).

GENERATION OF 3-PHASE

Flux densityB [T]

l

d

L

A

M

N

Current induces in the coil as thecoil moves in the magnetic field

Current produced at terminal

Generator for single phase

NoteInduction motor cannot start byitself. This problem is solved byintroducing three phase system

GENERATION OF 3-PHASE

Instead of using one coil only , three coils are used arranged inone axis with orientation of 120o each other. The coils are R-R1, Y-Y1 and B-B1. The phases are measured in this sequenceR-Y-B. I.e Y lags R by 120o , B lags Y by 120o.

GENERATION OF 3-PHASE

The three winding can be represented by the above circuit. Inthis case we have six wires. The emf are represented by eR ,eY, eB.

te sinEmR

Finish R

L1eR

)120-sin(EmYte

)240-sin(EmBte

LoadFinish

Finish

start

start

start

R1

Y

Y1

B

B1

L2

L3

eY

eB

GENERATION OF 3-PHASE

vector diagram for three-phase system

GENERATION OF 3-PHASE

ADVANTAGES OF 3 PHASESYSTEM

1. 3-phase power has a constant magnitude butsingle phase power is a pulsating one

2. A 3-phase system can be set up a rotatingmagnetic field in stationary windings. This is notpossible in single phase systempossible in single phase system

3. For the same rating, 3-phase machine are smallerin size and have better operating characteristicsthan single phase machine

4. 3-phase induction motors are self startingwhereas single phase induction motors are notself starting

ADVANTAGES OF 3 PHASESYSTEM

5. 3-phase motors have better power factor andefficiency over single phase motor.

6. Generation, transmission and utilization ofpower is more economical in three phasepower is more economical in three phasesystem compared to single phase system.

7. There is saving of conductor material whenthe same power is to be transmitted over agiven distance by a 3-phase systemcompared to a single phase system

PHASE SEQUENCE

• Definition :-

The order in which the voltages in thethree phases reach their maximumvaluesvalues

PHASE & LINE ELEMENT

• The voltage between any one line and naturalis called the phase voltage (VPH)

• The current flowing in a phase is known asthe current phase (IPH)the current phase (IPH)

• The voltage between any two lines or phasesis called the line voltage (VL)

• The current flowing in a line is called the linecurrent (IL)

DELTA CONNECTION

a) Physical connection diagram

b) Conventional connectiondiagram

DELTA CONNECTION

DELTA CONNECTION

• IR, IY and IB are called line current• I1, I2 and I3 are called phase current

)(III 3131R II

From Kirchoff current law we haveI3

R

Y

IR

IY

I

I1

VL

VP

)(III 1212Y II

)(III 2323B II

In phasor diagram

BIB

I2

DELTA CONNECTIONSince the loads are balanced, the magnitude of currents areequaled but 120o out of phase. i.e I1 =I2=I3 ,=IP Therefore:-

I1 = VP30;

I2 = VP-90;

I3 = VP150;

IR = IL30;

IY = IL-90;

IB = IL150;

PR III )3(30cos2 1

PY III )3(30cos2 2

PB III 330cos2 3

Thus IR=IY=IB = IL

Where IP is a phase current and IL is a line current

3 PB L

DELTA CONNECTION

Voltage

PL VV

PL II 3

Current

STAR CONNECTION

a) Physical connection diagram

b) Conventional connectiondiagram

STAR CONNECTION

)( VVVVV

STAR CONNECTION

R

Y

B

N

VRY

VYB

IR

IY

R

Y

B

N

V

VRYVBR

VYB

IR

IY

R

VYVB

•VRY, VYB and VBR are called line voltage•VR, VY and VB are called phase voltage

From Kirchoff voltage law we have

)( YRYRRY VVVVV BIB

BIB

)( BYBYYB VVVVV

)( RBRBBR VVVVV

In phasor diagram

For balanced load VR , VY andVB are equaled but out of phase

VR = VP30;

VY = VP-90;

VB = VP150;

VRY = VL30;

VYB = VL-90;

VBR = VL150;

STAR CONNECTION

Po

RRY VVV 330cos2

therefore

Po

YYB VVV 330cos2

Po

BBR VVV 330cos2

PL VV 3

Voltage

STAR CONNECTION

PL II

Current

POWER IN 3-PHASESYSTEM

• DC power system

• AC power system

POWER IN 3-PHASESYSTEM

• Power for 3-phase system (refer to phaseelement) cos3 PP IVP

• Power for 3-phase system (refer to lineelement)

cos3 LL IVP

SUMMARYCharacteristic Star Connection Delta Connection

Symbol or

Voltage

Current

BalanceBalanceCondition

Power in 1

Power in 3

- Phase element

- Line element

EXAMPLE 1

Three similar resistors are connectedin star acroos 400V, 3 phase lines. Theline current is 5A. Calculate the valueof each resistor. To what value shouldof each resistor. To what value shouldthe line voltage be changed to obtainthe same line current with theresistors are in delta connected?

SOLUTION 1

• Line voltage, VL = 400V– Phase voltage, Vph = VL / = 400 / =230.94V

• Line current, IL = 5A

3 3

• Line current, IL = 5A– Phase current Iph = 5A (in star IL = Iph)

• Zph = Rph = Vph / Iph = 230.94 / 5 =46.19

• If the same resistors are connected indelta– VL = Vph and IL = Iph

– Vph = Iph Zph = Iph Rph = IL / x Rph =

3

3

EXAMPLE 2

Three identical impedances areconnected in delta to a three phase 400Vsupply. The line current is 34.65A and thetotal power taken from the supply is14.4kW. Calculate the resistance and14.4kW. Calculate the resistance andreactance values of each impedance.

SOLUTION 2

– Line Voltage, VL = 400V

• Phase Voltage, Vph = VL = 400V

– Line Current, IL = 34.65A

• Phase Current, IPH = IL / 3 = 34.65 / 3 = • Phase Current, IPH = IL / 3 = 34.65 / 3 =20A

– Total power, P = 14.4kW

• But P = 3 Iph Rph

Rph = 14.4k / 3 Iph = 14.4k / 3(20) = 12

– Zph = Vph / Iph = 400 / 20 = 20

• XL = Zph - Rph = 20 - 12 = 16

2

2 2

2 2 2 2

EXAMPLE 3

Three similar coils are connected instar taken at a total power of 1.5kW ata p.f of 0.2 lagging from a 3-phase400V 50Hz supply. Calculate theresistance and inductance of eachresistance and inductance of eachphase.

SOLUTION 3

• P = 3 VL IL cos θ

• IL = P / 3 VL cos θ = 1.5k / ( 3 x 400x 0.2)

= 10.83A

√

√ √

• IL = Iph = 10.83A

• Vph = VL / 3 = 400 / 3 = 230.94V

• Zph = Vph / Iph = 230.94 / 10.83 =21.32Ω

• Cos θ = Rph / Zph

Rph = Zph Cos θ = 21.32 (0.2) =4.264Ω

√ √

SOLUTION 3

• XL = Zph2 - Rph2 = 21.322 – 4.2642

= 20.89Ω

XL = 2πfL

√ √

XL = 2πfL

L = XL / 2πf = 20.89 / (2π x 50) =66mH

EXAMPLE 4

Find :-

a) Phase Impedance, ZPH

b) Phase Current, IPH

c) Line Current, Ic) Line Current, IL

d) Phase Voltage, VPH

e) Single phase power, P1

f) Three phase power (phase element),P3

g) Three phase power (line element), P3

SOLUTION 4

EXAMPLE 5

Find :-Find :-

a) Phase Impedance, ZPH

b) Phase Current, IPH

c) Line Current, IL

d) Three phase power (line element), P3

SOLUTION 5

EXAMPLE 6

Three coil each having a resistance of20Ω and inductive reactance of 15Ωare connected to a 400V, 3 phase 50Hzsupply. Calculate:-

a)The line currenta)The line current

b)Power factor

c)Power drawn from the supply

SOLUTION 6a) The line current

– Zph = Rph2 + Xph2 = 202 + 152 = 25Ω

– Vph = VL / √3 = 400 / √3 = 230.94V

– Iph = Vph / Zph = 230.94 / 25 = 9.24A

– IL = Iph = 9.24A

√ √

– IL = Iph = 9.24A

b) Power factor

- cos θ = Rph / Zph = 20 / 25 = 0.8

c) Power

- P = √3 VL IL cos θ = √3(400)(9.24)(0.8)

= 5.121kW

THE END