Lecture Notes 2

7/17/2019 Lecture Notes 2

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

Lecture Notes

ELE A6

Ramadan El-Shatshat

Three Phase circuits




Three-phase Circuits

http://en.wikipedia.org/wiki/Image:Transmission_Towers.jpg




Advantages of Three-phase

Circuits

• Smooth flow of power (instantaneous poweris 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%).




Three-phase CircuitsWye-Connected System

• The neutral point is grounded

• The three-phase voltages haveequal magnitude.

• The phase-shift between thevoltages is 120 degrees.

V0V=anV

bnV 120V°

cnV 240V°





Wye-Connected System

• Line-to-line voltages are the

difference of the phase voltages

30V3 ∠== bnanab V-VV

90-V3 ∠== cnbnbc V-VV

150V3 ∠== ancnca V-VV




Three-phase CircuitsWye-Connected System

• Phasor diagram is used to visualizethe 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 times

the line-to-neutral voltage

3




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:

I

V

z I

V

z I

V

za

an

b

bn

c

cn

= = =, ,




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 eliminated

Io

a

Van

aZa

Ia

a

n




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.


lnV

Load




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

I V

za

ab=





Delta-Connected System

The line currents are:

• In a balanced case the line

currents are:

caaba III

abbcb III

bcac III

I I line phase= ∠ −3 30




Three-phase CircuitDelta-Connected System

• The phasor diagram is used tovisualize 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




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 =




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 PPPP ++=

( )φ cosIV3P3P phasephasephase ==

LNLLLphaseLNphase V3V IIVV ===

( ) ( )φ φ cosIV3cosIV3P LLLphasephase ==

phaseLLphaseLine VV I3I ==

( ) ( )φ φ cosIV3cosIV3P == LLLphasephase




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 withoutneutral) the power is measured using only twosingle-phase watt-meters.

- The watt-meters are supplied by the linecurrent and the line-to-line voltage.




Three-phase Circuit

Power measurement

• The total power is

the algebraic sum ofthe two watt-meters

reading.

Load Watt meter 1

Wattmeter 2

a

b

c

)cos(3

)cos(

)cos(

21

2

1

θ

θ θ

θ θ

L LT

Icvcbccb

Iavabaab

I V W W P

I V W

I V W

=+=

−=

−=




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.





MVAR 9.249)sin(3

MW433)cos(3)(

307.836)(

11920630238

307.836)866.0(cos7.836

7.8363453

500

3I,0

3

345

)(

1

==

==

−∠==

+=∠=

−∠=−∠=

===∠=

=

−

θ

θ

φ

φ

φ

φ φ

φ

φ

φ

L L

L L

L

L

I V Q

I V Pc

I I b

j Z

I

AkV

MVA

V

S V

kV V

I

V Z a




Example 2

Repeat example 2 assuming the load is Delta connected.





MVAR 9.249)sin(3

MW433)cos(3)(

607.836303)(

3573.61830714

301.483)866.0(cos1.483

1.483345*3

500

3I

,0345 )(

1

==

==

−∠=−∠=

+=∠==

−∠=−∠=

===

∠==

−

θ

θ

φ

φ

φ

φ

φ

φ

φ

L L

L L

L

L

L

I V Q

I V Pc

I I b

j I

V Z

I

AkV

MVA

V

S

V kV V V a




Power factor correction

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

• In order to correct the power factor in any system,a reactive (either inductive or capacitive) will beadded 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?




Example 3

A 3-phase load draws 120 kW at a power factor of 0.85 lagging from a

440 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.





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

0.85capacitor)(no.F.)(

49.116.160

906.65

6.654403

10*50

3

1)sin(

load capacitor purehaveweSince

)sin(3)(

,8.312.185)85.0(cos25.185

25.185

)85.0(4403

10*120I

)cos(3 )(

)()(

)(

3

)(

)(

1

3

==

=

−∠=+=

∠=

==

=

=

=

−∠=−∠=

==

=

−

osPd

Pc

I I I

I

A I

I V Q

I V Qb

I

A

I V Pa

capacitor L Lnew L

capacitor L

capacitor L

capacitor L Lcapacitor

L Lcapacitor

L

L

L L

θ

θ

θ




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