three phase circuit. single phase two wire objectives explain the differences between single- phase,...

THREE PHASE THREE PHASE CIRCUITCIRCUIT

SINGLE PHASE TWO WIRESINGLE PHASE TWO WIRE

pV

ObjectivesObjectivesExplain the differences between

single-phase, two-phase and three-phase.

Compute and define the Balanced Three-Phase voltages.

Determine the phase and line voltages/currents for Three-Phase systems.

SINGLE PHASE SYSTEMSINGLE PHASE SYSTEMA generator connected through a pair

of wire to a load – Single Phase Two Wire.

Vp is the magnitude of the source voltage, and is the phase.

SINLGE PHASE THREE WIRESINLGE PHASE THREE WIRE

pV

pV

SINGLE PHASE SYSTEMSINGLE PHASE SYSTEMMost common in practice: two

identical sources connected to two loads by two outer wires and the neutral: Single Phase Three Wire.

Terminal voltages have same magnitude and the same phase.

POLYPHASE SYSTEMPOLYPHASE SYSTEM

Circuit or system in which AC sources operate at the same frequency but different phases are known as polyphase.

TWO PHASE SYSTEM THREE TWO PHASE SYSTEM THREE WIREWIRE

pV

90pV

POLYPHASE SYSTEMPOLYPHASE SYSTEMTwo Phase System:

◦A generator consists of two coils placed perpendicular to each other

◦The voltage generated by one lags the other by 90.

POLYPHASE SYSTEMPOLYPHASE SYSTEM Three Phase System:

◦A generator consists of three coils placed 120 apart.

◦The voltage generated are equal in magnitude but, out of phase by 120.

Three phase is the most economical polyphase system.

THREE PHASE FOUR THREE PHASE FOUR WIREWIRE

IMPORTANCE OF THREE PHASE IMPORTANCE OF THREE PHASE SYSTEMSYSTEM

All electric power is generated and distributed in three phase.◦One phase, two phase, or more than

three phase input can be taken from three phase system rather than generated independently.

◦Melting purposes need 48 phases supply.


Uniform power transmission and less vibration of three phase machines.◦The instantaneous power in a 3 system

can be constant (not pulsating).◦ High power motors prefer a steady torque

especially one created by a rotating magnetic field.


Three phase system is more economical than the single phase.◦The amount of wire required for a three

phase system is less than required for an equivalent single phase system.

◦Conductor: Copper, Aluminum, etc

THREE PHASE THREE PHASE GENERATIONGENERATION

FARADAYS LAWFARADAYS LAW Three things must be present in

order to produce electrical current:a) Magnetic fieldb) Conductorc) Relative motion

Conductor cuts lines of magnetic flux, a voltage is induced in the conductor

Direction and Speed are important

GENERATING A SINGLE PHASE

Motion is parallel to the flux.

No voltage is induced.

N

S

x

N

S

Motion is 45 to flux. Induced voltage is 0.707 of maximum.



x

N

S

Motion is perpendicular to flux. Induced voltage is maximum.


Motion is 45 to flux.

x

N

S

Induced voltage is 0.707 of maximum.


N

S

Motion is parallel to flux. No voltage is induced.


x

N

S

Notice current in the conductor has reversed.




N

S

x

Motion is perpendicular to flux.

Induced voltage is maximum.


N

S

x




Motion is parallel to flux. N

S

No voltage is induced.Ready to produce another cycle.

THREE PHASE GENERATORTHREE PHASE GENERATOR

GENERATOR WORKGENERATOR WORKThe generator consists of a rotating

magnet (rotor) surrounded by a stationary winding (stator).

Three separate windings or coils with terminals a-a’, b-b’, and c-c’ are physically placed 120 apart around the stator.

As the rotor rotates, its magnetic field cuts the flux from the three coils and induces voltages in the coils.

The induced voltage have equal magnitude but out of phase by 120.

GENERATION OF THREE-PHASE AC

N

xx

S

THREE-PHASE WAVEFORM

Phase 2 lags phase 1 by 120 Phase 2 leads phase 3 by 120Phase 3 lags phase 1 by 240 Phase 1 leads phase 3 by 240

Phase 1 Phase 2 Phase 3

120 120 120

240120 120 120

240

Phase 1Phase 2 Phase 3

GENERATION OF 3 VOLTAGES

Phase 1 is ready to go positive.Phase 2 is going more negative.Phase 3 is going less positive.

N

xx

S

THREE PHASE THREE PHASE QUANTITIESQUANTITIES

BALANCED 3BALANCED 3 VOLTAGESVOLTAGESBalanced three phase voltages:

◦same magnitude (VM )

◦120 phase shift

120cos240cos)(

120cos)(

cos)(

tVtVtv

tVtv

tVtv

MMcn

Mbn

Man

BALANCED 3BALANCED 3 CURRENTS CURRENTSBalanced three phase currents:

◦same magnitude (IM )

◦120 phase shift

240cos)(

120cos)(

cos)(

tIti

tIti

tIti

Mc

Mb

Ma

PHASE SEQUENCEPHASE SEQUENCE

120cos)(

120cos)(

cos)(

tVtv

tVtv

tVtv

Mcn

Mbn

Man

120

120

0

Mcn

Mbn

Man

VV

VV

VV

120

120

0

Mcn

Mbn

Man

VV

VV

VV

POSITIVESEQUENCE

NEGATIVESEQUENCE

PHASE SEQUENCEPHASE SEQUENCE

EXAMPLE # 1EXAMPLE # 1Determine the phase sequence of

the set voltages:

110cos200

230cos200

10cos200

tv

tv

tv

cn

bn

an

BALANCED VOLTAGE AND BALANCED VOLTAGE AND LOAD LOAD

Balanced Phase Voltage: all phase voltages are equal in magnitude and are out of phase with each other by 120.

Balanced Load: the phase impedances are equal in magnitude and in phase.


POWER◦The instantaneous power is constant

)cos(3

cos2

3

)()()()(

rmsrms

MM

cba

IV

IV

tptptptp


Three Phase Power,

SSSSS 3 CBAT

THREE PHASE QUANTITIESTHREE PHASE QUANTITIES

QUANTITY SYMBOL

Phase current I

Line current IL

Phase voltage V

Line voltage VL

PHASE VOLTAGES and LINE PHASE VOLTAGES and LINE VOLTAGESVOLTAGES

Phase voltage is measured between the neutral and any line: line to neutral voltage

Line voltage is measured between any two of the three lines: line to line voltage.

PHASE CURRENTS and PHASE CURRENTS and LINE CURRENTSLINE CURRENTS

Line current (IL) is the current in each line of the source or load.

Phase current (I) is the current in each phase of the source or load.

THREE PHASE THREE PHASE CONNECTIONCONNECTION

SOURCE-LOAD CONNECTIONSOURCE-LOAD CONNECTION

SOURCE LOAD CONNECTION

Wye Wye Y-Y

Wye Delta Y-

Delta Delta -

Delta Wye -Y

SOURCE-LOAD CONNECTIONSOURCE-LOAD CONNECTIONCommon connection of source: WYE

◦Delta connected sources: the circulating current may result in the delta mesh if the three phase voltages are slightly unbalanced.

Common connection of load: DELTA◦Wye connected load: neutral line may

not be accessible, load can not be added or removed easily.

WYE CONNECTIONWYE CONNECTION

WYE CONNECTED WYE CONNECTED GENERATORGENERATOR

WYE CONNECTED LOADWYE CONNECTED LOAD

ZY

ZY

ZY

a

c

b

nLoad

ZY

a

b

c

Load

n

OR

BALANCED Y-Y CONNECTIONBALANCED Y-Y CONNECTION

PHASE CURRENTS AND PHASE CURRENTS AND LINE CURRENTSLINE CURRENTS

In Y-Y system:

φL II

PHASE VOLTAGES, VPHASE VOLTAGES, V

Phase voltage is measured between the neutral and any line: line to neutral voltage

PHASE VOLTAGES, VPHASE VOLTAGES, V

an M

bn M

cn M

V V 0 volt

V V 120 volt

V V 120 volt

LINE VOLTAGES, VLINE VOLTAGES, VLL

Line voltage is measured between any two of the three lines: line to line voltage.

n

a

b

c

Vab

Vbc

Vca

Vbn

Vcn

Van

Ia

Ib

Ic

Vab

Vbc

Vca

LINE VOLTAGES, VLINE VOLTAGES, VLL

ancnca

cnbnbc

bnanab

VVV

VVV

VVV

150V3V

90V3V

30V3V

Mca

Mbc

Mab

ab M

bc M

ca M

V 3 V 30 volt

V 3 V 90 volt

V 3 V 150 volt

an M

bn M

cn M

V V 0 volt

V V 120 volt

V V 120 volt

PHASE VOLTAGE (V)

LINE VOLTAGE

(VL)

PHASE DIAGRAM OF VPHASE DIAGRAM OF VL L AND VAND V

30°

120°

Vca Vab

Vbc

Vbn

Van

Vcn

-Vbn

PROPERTIES OF PHASE PROPERTIES OF PHASE VOLTAGEVOLTAGE

All phase voltages have the same magnitude,

Out of phase with each other by 120

an bn cnV V V V = =

PROPERTIES OF LINE PROPERTIES OF LINE VOLTAGEVOLTAGEAll line voltages have the same

magnitude,


ab bc caV V V VL = =

RELATIONSHIP BETWEEN RELATIONSHIP BETWEEN VV and V and VLL

30VVL

1. Magnitude

2. Phase

- VL LEAD their corresponding V by 30

LV 3 V

EXAMPLE 1 EXAMPLE 1 Calculate the line currents

DELTA CONNECTIONDELTA CONNECTION

DELTA CONNECTED DELTA CONNECTED SOURCESSOURCES

DELTA CONNECTED DELTA CONNECTED LOADLOAD

OR

BALANCED BALANCED - - CONNECTION CONNECTION

PHASE VOLTAGE AND PHASE VOLTAGE AND LINE VOLTAGELINE VOLTAGE

In - system, line voltages equal to phase voltages:

φL VV

PHASE VOLTAGE, VPHASE VOLTAGE, VPhase voltages are equal to the

voltages across the load impedances.

PHASE CURRENTS, IPHASE CURRENTS, I

Δ

CACA

Δ

BCBC

Δ

ABAB Z

VI,

Z

VI,

Z

VI

The phase currents are obtained:

LINE CURRENTS, ILINE CURRENTS, ILLThe line currents are obtained from the

phase currents by applying KCL at nodes A,B, and C.

LINE CURRENTS, ILINE CURRENTS, ILL

BCCAc

ABBCb

CAABa

III

III

III

120I I

120I I

30I 3I

ac

ab

ABa

PHASE CURRENTS (I)

LINE CURRENTS (IL)

Δ

CACA

Δ

BCBC

Δ

ABAB

Z

VI

Z

VI

Z

VI

120I I

120I I

30I 3I

ac

ab

ABa

PHASE DIAGRAM OF IPHASE DIAGRAM OF IL L AND IAND I

PROPERTIES OF PHASE PROPERTIES OF PHASE CURRENTCURRENT

All phase currents have the same magnitude,


Δ

φCABCABφ Z

VIIII

PROPERTIES OF LINE PROPERTIES OF LINE CURRENTCURRENT

All line currents have the same magnitude,


cbaL IIII

RELATIONSHIP BETWEEN IRELATIONSHIP BETWEEN I and Iand ILL

1. Magnitude

2. Phase

- IL LAG their corresponding I by 30

IIL 3

30IIL

EXAMPLE EXAMPLE A balanced delta connected load having an impedance 20-j15 is connected to a delta connected, positive sequence generator having Vab = 3300 V. Calculate the phase currents of the load and the line currents.

Given QuantitiesGiven Quantities

0330V

87.3625 j1520Z

ab

Δ

Phase CurrentsPhase Currents

A87.15613.2120II

A13.83-13.2120II

A36.8713.236.8725

0330

Z

VI

ABCA

ABBC

Δ

ABAB

A87.12686.22120II

A13.311-86.22120II

87.686.22

A30336.8713.2

303II

ac

ab

ABa

Line CurrentsLine Currents

BALANCED WYE-BALANCED WYE-DELTASYSTEMDELTASYSTEM

THREE PHASE POWER THREE PHASE POWER MEASUREMENTMEASUREMENT

Unbalanced loadUnbalanced load

In a three-phase four-wire system the line voltage is 400V and non-inductive loads of 5 kW, 8 kW and 10 kW are connected between the three conductors and the neutral. Calculate: (a) the current in each phase

(b) the current in the neutral conductor.

VV

V LP 230

3

400

3

AV

PI

P

BB 7.21

230

105 3

AV

PI

P

YY 8.34

230

108 3

AV

PI

P

RR 5.43

230

104

Voltage to neutral

Current in 8kW resistor



IR

IYIB

IYH

IYV

IBH

IBV

A.III oBYH 3.117.218.34866030cos30cos

AIIII oBYRV 0.13)7.218.34(5.05.4360cos60cos

INV

INH

IN

A.III NVNHN 2.170.13311 2222

Resolve the current components into horizontal and vertical components.

R

B

Y

400V

400V

400V

IR

IB

IY

I3

I2

I1R 1=100

R 2=20 C=30 F

X 2=60

A delta –connected load is arranged as in Figure below. The supply voltage is 400V at 50Hz. Calculate:

(a)The phase currents;(b)The line currents.

AR

VI RY 4

100

400

11

I1 is in phase with VRY since there is only resistor in the branch

(a)

22

22Y XRZ 22 6020

2

21Y R

Xtan

20

60tan 1

C

BR3 X

VI 90)1030502/(1

4006

A77.3

AZ

VI

Y

YB 32.66020

400222

o90

In branch between YB , there are two components , R2 and X2

In the branch RB , only capacitor in it , so the XC is -90 out of phase.

'3471

V RY

IR

I3

I2

V YBV BR

I1

71 o34'90 o

30 o

-I3

30o

(b) 31R III

2331

21

2 cos2 IIIIIR

3.5677.330cos77.30.420.4 222 oRI

AIR 5.7

I1

-I 1I2

IY

120 o

71 o 34'

60 o

12Y III

2121

22

2 cos2 IIIIIY

5.1050.4'3411cos32.60.4232.6 222 oYI

AIY 3.10

= 71o 34’ -60o= 11o 34’

=30o

2223

23

2 cos2 IIIIIB

5.1877.3'26138cos32.677.3232.6 222 oBI

AIB 3.4

= 180-30o-11o 34’ = 138o 34’

23B III

-I 2

I2

I3

I2

71 o 34'90 o

30 o11 o 34'

Power in three phasePower in three phase

Active power per phase = IPVP x power factor Total active power= 3VPIP x power factor

cos3 PP IVP

If IL and VL are rms values for line current and line voltage respectively. Then for delta () connection: VP = VL and IP = IL/3. therefore:

cos3 LLIVP

For star connection () : VP = VL/3 and IP = IL. therefore:

cos3 LLIVP

A three-phase motor operating off a 400V system is developing 20kW at an efficiency of 0.87 p.u and a power factor of 0.82. Calculate:

(a)The line current;(b)The phase current if the windings are delta-connected.

(a) Sincewattsinpowerinput

wattsinpoweroutputEfficiency

fpVI

wattsinpoweroutput

LL .3

82.04003

10002087.0

LI

Acurrentline

currentPhase 1.233

0.40

3

And line current =IL=40.0A

(b) For a delta-connected winding

Three identical coils, each having a resistance of 20 and an inductance of 0.5 H connected in (a) star and (b) delta to a three phase supply of 400 V; 50 Hz. Calculate the current and the total power absorbed by both method of connections.

20R P 1575.0502X P

22PPPPP XRjXRZ

8315820

157tan15720 122

First of all calculating the impedance of the coils

P

P1

R

Xtanwhere

1264.083coscos

20

400V400V

400V 0.5H

0.5H 0.5H20

20

Star connection

V400VV LP A38.4158

400

Z

VI

P

PP

cos3 LLIVP W3831264.038.44003

A balanced three phase load connected in star, each phase consists of resistance of 100 paralleled with a capacitance of 31.8 F. The load is connected to a three phase supply of 415 V; 50 Hz.Calculate: (a) the line current;

(b) the power absorbed;(c) total kVA;(d) power factor .

415

V2403

415

3

VV L

P

PPP X

1

R

1Y Cj

1X P

Admittance of the load

where

CjR

1

P

6108.31502j100

1 S)01.0j01.0(

)01.001.0(240 jYVII PPPL 4539.34.24.2 j

PPVA IVP 454.8144539.3240

57645cos4.814 PAP

kW728.15763PA

Line current

Volt-ampere per phase

Active power per phase

Total active power

kW728.1j5763jPR

Reactive power per phase 576j45sin4.814jPPR

Total reactive power

kVA44.24.8143 Total volt-ampere

(b)

(c)

(d) Power Factor = cos = cos 45 = 0.707 (leading)

A three phase star-connected system having a phase voltage of 230V and loads consist of non reactive resistance of 4 , 5 and 6 respectively.Calculate:(a) the current in each phase conductor

(b) the current in neutral conductor and (c) total power absorbed.

A5.574

230I4

A465

230I5

A3.386

230I6

X-component = 46 cos 30 + 38.3 cos 30 - 57.5 = 15.5 A

Y-component = 46 sin 30 - 38.3 sin 30 = 3.9 A

16A3.915.5I 22N Therefore

(b)

(c) kW61.323.38465.57230P

57.5 A

46 A

38.3 A

three phase circuit. single phase two wire objectives explain the differences between single- phase,...

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