9/12/2006 ELE A6 Three-phase Circuits 1

Lecture NotesELE A6

Ramadan El-Shatshat

Three Phase circuits

9/12/2006 ELE A6 Three-phase Circuits 2

Three-phase Circuits

9/12/2006 ELE A6 Three-phase Circuits 3

Advantages of Three-phase Circuits

• Smooth flow of power (instantaneous power is constant).

• Constant torque (reduced vibrations).• The power delivery capacity tripled

(increased by 200%) by increasing the number of conductors from 2 to 3 (increased by 50%).

9/12/2006 ELE A6 Three-phase Circuits 4

Three-phase CircuitsWye-Connected System

• The neutral point is grounded

• The three-phase voltages have equal magnitude.

• The phase-shift between the voltages is 120 degrees.

V 0 V =°∠=anV

bnV 120 V °−∠=

cnV 240 V °−∠=

9/12/2006 ELE A6 Three-phase Circuits 5

Three-phase CircuitsWye-Connected System

• Line-to-line voltages are the difference of the phase voltages

30V 3 ∠== bnanab V -VV

90- V 3 ∠== cnbnbc V -VV

150V 3 ∠== ancnca V -VV

9/12/2006 ELE A6 Three-phase Circuits 6

Three-phase CircuitsWye-Connected System

• Phasor diagram is used to visualize the system voltages

• Wye system has two type of voltages: Line-to-neutral, and line-to-line.

• The line-to-neutral voltages are shifted with 120o

• The line-to-line voltage leads the line to neutral voltage with 30o

• The line-to-line voltage is timesthe line-to-neutral voltage

3

9/12/2006 ELE A6 Three-phase Circuits 7

Three-phase CircuitsWye-Connected Loaded System

• The load is a balanced load and each one = Z• Each phase voltage drives current through the

load.• The phase current expressions are:

IVz

IVz

IVza

anb

bnc

cn= = =, ,

9/12/2006 ELE A6 Three-phase Circuits 8

Three-phase CircuitsWye-Connected Loaded System

• Since the load is balanced (Za = Zb = Zc) then: Neutral current = 0

• This case single phase equivalent circuit can be used (phase a, for instance, only)

• Phase b and c are eliminatedIo

a

Van

aZa

Ia

a

n

9/12/2006 ELE A6 Three-phase Circuits 9

Wye-Connected System with balanced load

• A single-phase equivalent circuit is used

• Only phase a is drawn, because the magnitude of currents and voltages are the same in each phase. Only the phase angles are different (-120o

phase shift)

• The supply voltage is the line to neutral voltage.

• The single phase loads are connected to neutral or ground.

Three-phase Circuits

lnV Load

9/12/2006 ELE A6 Three-phase Circuits 10

Three-phase CircuitsBalanced Delta-Connected System

• The system has only one voltage : the line-to-line voltage ( )

• The system has two currents :– line current– phase current

• The phase currents are:

LLV

IVzaab=

9/12/2006 ELE A6 Three-phase Circuits 11

Three-phase CircuitsDelta-Connected SystemThe line currents are:

• In a balanced case the line currents are:

caaba III −=

abbcb III −=

bccac III −=

I Iline phase= ∠ −3 30

9/12/2006 ELE A6 Three-phase Circuits 12

Three-phase CircuitDelta-Connected System

• The phasor diagram is used to visualize the system currents

• The system has two type of currents: line and phase currents.

• The delta system has only line-to-line voltages, that are shifted by

• The phase currents lead the line currents by

• The line current is times the phase current and shifted by 30 degree.

°30

3

9/12/2006 ELE A6 Three-phase Circuits 13

Three-phase Circuit

• Circuit conversions– A delta circuit can be converted to an equivalent wye

circuit. The equation for phase a is (will not be used for this course):

– Conversion equation for a balanced system is:

ab

ab

cabc

caa ZZZ

ZZ Z++

=

3

abZ Za =

9/12/2006 ELE A6 Three-phase Circuits 14

Three-phase CircuitPower Calculation• The three phase power is equal the sum of the phase powers

• If the load is balanced:

• Wye system:

• Delta system:

cba PPP P ++=

( )φ cosIV 3P 3 P phasephasephase ==

LNLLLphaseLNphase V 3V IIV V ===

( ) ( )φφ cosIV 3cos IV 3 P LLLphasephase ==

phaseLLphaseLine VV I 3I ==

( ) ( )φφ cosIV 3cos IV 3 P == LLLphasephase

9/12/2006 ELE A6 Three-phase Circuits 15

Three-phase CircuitPower measurement

• In a four-wire system (3 phases and a neutral) the real power is measured using three single-phase watt-meters.

• In a three-wire system (three phases without neutral) the power is measured using only two single-phase watt-meters.- The watt-meters are supplied by the line

current and the line-to-line voltage.

9/12/2006 ELE A6 Three-phase Circuits 16

Three-phase Circuit

Power measurement

• The total power is the algebraic sum of the two watt-meters reading.

Load Watt meter 1

Wattmeter 2

a

b

c

)cos(3

)cos()cos(

21

2

1

θ

θθθθ

LLT

Icvcbccb

Iavabaab

IVWWP

IVWIVW

=+=

−=−=

9/12/2006 ELE A6 Three-phase Circuits 17

Example 1

A 345 kV, three phase transmission line delivers 500 MVA, 0.866 power factor lagging, to a three phase load connected to its receiving end terminals. Assume the load is Y connected and the voltage at the receiving end is 345 kV, find:

• The load impedance per phase.

• The line and phase currents.

• The total real and reactive power.

Three-phase Circuits

9/12/2006 ELE A6 Three-phase Circuits 18

MVAR 9.249)sin(3

MW 433)cos(3 )(

307.836 )(

11920630238307.836)866.0(cos7.836

7.8363453

5003

I ,03

345

)(

1

==

==

−∠==

+=∠=

−∠=−∠=

===∠=

=

−

θ

θ

φ

φ

φ

φφ

φ

φφ

LL

LL

L

L

IVQ

IVPc

IIb

jZI

AkV

MVAV

SVkVV

IV

Za

9/12/2006 ELE A6 Three-phase Circuits 19

Example 2

Repeat example 2 assuming the load is Delta connected.

Three-phase Circuits

9/12/2006 ELE A6 Three-phase Circuits 20

MVAR 9.249)sin(3

MW 433)cos(3 )(

607.836303 )(

3573.61830714

301.483)866.0(cos1.483

1.483345*3

5003

I

,0345 )(

1

==

==

−∠=−∠=

+=∠==

−∠=−∠=

===

∠==

−

θ

θ

φ

φ

φφ

φ

φ

φ

LL

LL

L

L

L

IVQ

IVPc

IIb

jIV

Z

I

AkV

MVAVS

VkVVVa

9/12/2006 ELE A6 Three-phase Circuits 21

Power factor correction

• Power factor (p.f.) correction is the process of making P.f. = 1.

• In order to correct the power factor in any system, a reactive (either inductive or capacitive) will be added to the load.

• If the load is inductive, then a capacitance is added.– Note: Correcting the P.f. WILL NOT affect the active

power, why?

9/12/2006 ELE A6 Three-phase Circuits 22

Example 3

A 3-phase load draws 120 kW at a power factor of 0.85 lagging from a440 V bus. In parallel with this load, a three phase capacitor bank that is rated 50 kVAR is inserted, find:

• The line current without the capacitor bank.

• The line current with the capacitor bank.

• The P.F. without the capacitor bank.

• The P.F. with the capacitor bank.

Three-phase Circuits

9/12/2006 ELE A6 Three-phase Circuits 23

98.0)49.11(ccapacitor)(with .F. )(

0.85capacitor) (no .F. )(

49.116.160

906.65

6.65440310*50

3

1)sin(loadcapacitor pure have weSince

)sin(3 )(

,8.312.185)85.0(cos25.185

25.185)85.0(4403

10*120I

)cos(3 )(

)()(

)(

3

)(

)(

1

3

==

=

−∠=+=

∠=

==

=

=

=

−∠=−∠=

==

=

−

osPd

Pc

III

I

AI

IVQ

IVQb

I

A

IVPa

capacitorLLnewL

capacitorL

capacitorL

capacitorLLcapacitor

LLcapacitor

L

L

LL

θ

θ

θ

9/12/2006 ELE A6 Three-phase Circuits 24

Home Work

9/12/2006 ELE A6 Three-phase Circuits 25

Home Work