Three phase-circuits

Three-phase CircuitsWorkshop on Basic Electrical Engineering held at

VVCE, Mysuru, on 30-April-2016

R S Ananda Murthy

Associate ProfessorDepartment of Electrical & Electronics Engineering,

Sri Jayachamarajendra College of Engineering,Mysore 570 006

R S Ananda Murthy Three-phase Circuits

Learning Outcomes

After completing this lecture the student should be able to –State the advantages of three-phase supply.State the meaning of phase sequence, balanced supply,balanced load, and balanced system.Derive the relationship between line and phase values ofvoltages and currents in a balanced system.Derive equation for the power consumed by a three-phasebalanced load.Show that power in a three-phase three-wire system canbe measured using two wattmeters.Find the power factor of a balanced three-phase load usingtwo wattmeter readings.


Advantages of Three-phase Supply

The amount of conducting material required to transfer agiven amount of power is minimum in a three-phasesystem.The instantaneous power in a three-phase system neverfalls to zero resulting in smoother and better operatingcharacteristics of the load.Three-phase supply is required by three-phase inductionmotors which are widely used in industry because of theirruggedness, longer life, higher torque, low initial andmaintenance costs.


Advantages of Three-phase Supply

Domestic as well as industrial and commercial power canbe supplied from the same three-phase distribution system.Three-phase system has better voltage regulation.For a given size of the machine, the power generated by athree-phase alternator is higher.


Generation of Three-phase Supply

NS

A1

A2

C2

Stator

C1 B1

B2

vA = Vm sinωt =⇒ VA = |Vph|∠0◦

vB = Vm sin(ωt−120◦) =⇒ VB = |Vph|∠−120◦

vC = Vm sin(ωt−240◦) =⇒ VB = |Vph|∠−240◦


Meaning of Phase Sequence

ABC Sequence ACB Sequence

Phase sequence can be changed by reversing the direction ofrotation of rotor of the alternator.


Meaning of Balanced and Unbalanced Supply

(a) (b) (c) (d)

If |VA|= |VB|= |VC |= |Vph| and if the phase differencebetween VA and VB, VB and VC , VC and VA is equal to120◦ as shown in (a) then, the supply is said to bebalanced or symmetrical.Phasor diagrams (b), (c), and (d) represent unbalancedsupply. Can you explain why?


Meaning of Balanced/Unbalanced Load and System

If the three impedances, which may be Y or ∆ connectedare equal, then, the three-phase load is said to bebalanced.If load and supply are both balanced, then three-phasesystem is said to be balanced.Under normal working conditions, a three-phase systemcan be taken to be balanced.


Relation between Line and Phase Voltages

ABC Sequence

If supply is balanced, then, line voltage magnitudes will be|VAB|= |VBC |= |VCA|=

√3|Vph|= |V |.

When phase sequence is ABC, VAB leads VAN by 30◦, VBCleads VBN by 30◦, and VCA leads VCN by 30◦.


Relation between Line and Phase Voltages

If supply is balanced, then, line voltage magnitudes will be|VAB|= |VBC |= |VCA|=

√3|Vph|= |V |.

When phase sequence is ACB, VAB lags VAN by 30◦, VBClags VBN by 30◦, and VCA lags VCN by 30◦.


Line and Phase Currents in Three-phase Circuits

Star Point

LineCurrent

Current flowing througheach impedance iscalled phase current

Current flowing througheach suppy line iscalled line current

Load can beY or deltaconnectedas shownabove


Relation between Line and Phase Currents

In Y-connected load, the line current is equal to the phasecurrent.From the phasor diagram given in the previous slide, it isclear that

|IA|= 2|IAB|cos30◦ =√

3 · |IAB|=√

3 · |V ||Z |

= |IB|= |IC |= |I|

i.e., in a balanced ∆-connected three-phase load, the linecurrent is

√3×|Iph| where |Iph|= |V |/|Z |.


Zero Neutral Shift Voltage in Balanced System

A

C B

Balanced Supply Balanced Load

Neutral ShiftVoltage

It can be shown that in a balanced system the neutral shiftvoltage is zero so that VAN = VAN ′ , VBN = VBN ′ , andVCN = VCN ′ .


Power in Balanced System

A

B

C

BalancedThree-phase

Supply

Total power consumed by the load is

P = 3Pph = 3|Vph| · |Iph|cosφ =√

3 ·√

3|Vph| · |Iph|cosφ

But√

3|Vph|= |V | and |Iph|= |I| in a Y-connected load and∆-connected load can always be replaced by equivalent Y. So,

P =√

3 · |V | · |I| ·cosφ


Two Wattmeter Method to Measure 3-phase Power

M L

M L

COM

COM

V

V

Y or DeltaConnected

Three-phaseBalanced

Load

A

BC

N

Balanced Supply

Wattmeter has current coil with terminals marked as M and L,and voltage coil terminals marked as COM and V.



The reading of wattmeter W1 is given by

P1 = |VAB|× |IA|×cos(φ + 30◦) = |V ||I|cos(30◦+ φ) (1)

where |V | and |I| are the line voltage and current respectively.The reading of wattmeter W2 is

P2 = |VCB|× |IC |×cos(30◦−φ) = |V ||I|cos(30◦−φ) (2)

So, the sum of the two wattmeter readings is

P1 + P2 = |V | · |I| · [cos(30◦+ φ) + cos(30◦−φ)]

= 2|V | · |I| ·cos30◦ cosφ

=√

3|V | · |I| ·cosφ (3)

which is nothing but the total three-phase active power.


Finding Power Factor from Two Wattmeter Readings

We can also write

P1−P2 = |V | · |I| · [cos(30◦+ φ)−cos(30◦−φ)]

= −2|V | · |I| ·sin30◦ sinφ

= −|V | · |I| ·sinφ (4)

Dividing Eq.(4) by Eq. (3) we get

P1−P2

P1 + P2=− tanφ√

3=⇒ φ = tan−1

[√3(P2−P1)

P1 + P2

](5)

from this the load power factor cosφ of the load can be found.But the above equation can be applied only to balanced load.


License

This work is licensed under aCreative Commons Attribution 4.0 International License.


Three phase-circuits

Engineering

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