BALANCED THREE-PHASE AC CIRCUIT

BALANCED THREE

BALANCED THREE

-

-

PHASE

PHASE

AC CIRCUIT

AC CIRCUIT

• Balanced Three-Phase Voltage Sources Delta Connection

Star Connection

• Balanced 3-phase Load Delta Connection Star Connection

Introduction

Balanced Three Phase Voltages

Three-phase voltage sources

a) wye-connected source b) delta-connected source If the voltage source have the same amplitude and

frequency ω and are out of phase with each other by 120o_{, the voltage are said to be balanced.}

0

V

V

V

_an

+

_bn

+

_cn

=

cn bn an

V

V

V

=

=

Balanced phase voltages are equal in magnitude and

Balanced Three Phase Voltages

0 p 0 p cn 0 p bn 0 p an

120

V

240

V

V

120

V

V

0

V

V

+

∠

=

−

∠

=

−

∠

=

∠

=

abc sequence or positive sequence:

acb sequence or negative sequence:

0 p 0 p bn 0 p cn 0 p an 120 V 240 V V 120 V V 0 V V + ∠ = − ∠ = − ∠ = ∠ = p

V _{is the effective or rms value}

Va Vb Vc

ωt V Vm

Balanced Three Phase Loads

Two possible three-phase load configurations:

a) a Star or Y-connected load b) a delta-connected load For a balanced wye connected load:

Y 3

2

1

Z

Z

Z

Z

=

=

=

Z

a

=

Z

b

=

Z

c

=

Z

∆ For a balanced delta connected load:

Y

Z

3

Z

_∆

=

∆ = Z 3 1 Z_Y

Example 1

V

V_an = 200∠10° V_bn = 200∠− 230°V

Determine the phase sequence of the set of voltages

van = √2 200 cos(ωt + 10◦)

vbn = √2 200 cos(ωt − 230◦), vcn = √2 200 cos(ωt − 110◦)

Solution:

The voltages can be expressed in phasor form as

We notice that Van leads Vcn by 120◦ and Vcn in turn leads Vbn by

120◦.

Hence, we have an acb sequence.

V V_cn = 200∠−110° V V_an =110∠150° V V_bn = 110∠30°

Given that , find Van and Vcn, assuming a

positive (abc) sequence.

Answer: _{V V}

Balanced Y-Y Connection

A balanced Y-Y system is a three phase system with a balanced Y connected source and balanced Y connected load.

L s Y L s

Z

Z

Z

Z

Z

Z

Z

+

+

=

=

=

=

l l Source impedance Line impedance Load impedance L Y

Z

Z

=

Balanced Wye-Wye Connection

L s Y Y L s

Z

Z

Z

Z

Z

Z

Z

Z

+

+

=

=

=

=

=

l l Source impedance Line impedance Load impedance

Total impedance per phase

L Y

Z

Balanced Y-Y Connection

0 p ca 0 p bc 0 p ab

210

V

3

V

90

V

3

V

30

V

3

V

−

∠

=

−

∠

=

∠

=

Line to line voltages or line voltages given that phase a voltage is reference can be shown to be:

p L p L

I

I

V

V

=

=

3

ca bc ab L

V

V

V

V

=

=

=

cn bn an p

V

V

V

V

=

=

=

120o Vcn 30o Vbn Vab Van

Balanced Y-Y Connection

Given the phase voltages, the line current can be calculated as:

Applying KVL to each phase:

Y an a Z V I = 0 a Y 0 an Y bn b I 120 Z 120 V Z V I = = ∠ − = ∠ − 0 a Y 0 an Y cn c I 240 Z 240 V Z V I = = ∠− = ∠− 0 I I I I_a + _b + _c = − _n =

0

I

Z

V

_nN

=

_{n n}

=

Y an a Z V I =

Thus, the per-phase equivalent circuit can be expressed as:

• A balanced positive-sequence Y-connected 60 Hz three-phase source has phase voltage Va=1000V. Each phase of the load consists of a 0.1-H

inductance in series with a 50-ΩΩΩΩ resistance.

• Find the line currents, the line voltages, the power and the reactive power

delivered to the load. Draw a phasor diagram showing line voltages, phase

voltages and the line currents. Assuming that the phase angle of V_an is zero.

0 0 37 37 62 . 62 7 . 37 50 = ∴ ∠ = + = + = θ ωL j j R Z 0 0 0 15.97 37 15.97 157 , 15.97 83 an aA bB cC V I Z I I = = ∠ − ∴ = ∠ − = ∠ & & & &

Y-Y configuration Example:1

0 0 0 0

3 30 1732 30 1732 90 , 1732 150

ab an bc ca

Example 2

1- Calculate the line

currents in the three wire Y-Y system of figure below.

2- A Y-connected balanced three-phase generator with an impedance of 0.4+j0.3 Ω per phase is connected to a Y-connected balanced load with an impedance of 24 + j19 Ω per phase. The line joining the generator and the load has an impedance of 0.6 + j0.7 Ω per phase.

Assuming a positive sequence for the source voltages and that

Find: (a) the line voltages (b) the line currents

V

V

_an

=

120

∠

30

0

Balanced Y-Delta Connection

CA 0 p ca BC 0 p bc AB 0 p ab

V

210

V

3

V

V

90

V

3

V

V

30

V

3

V

=

−

∠

=

=

−

∠

=

=

∠

=

Line voltages:

°

∠

=

−

=

°

−

∠

=

−

=

°

−

∠

=

−

=

90

3

150

3

30

3

AB BC CA c AB AB BC b AB CA AB a

I

I

I

I

I

I

I

I

I

I

I

I

Line currents: ∆ ∆ ∆ = = = Z V I Z V I Z V I_AB AB ; _BC BC ; _CA CA Phase currents:

A balanced Y- _∆ system consists of balanced Y connected source feeding a balanced _∆ connected load.

Balanced Y-Delta Connection

3 Z Z_Y = ∆ ) 240 1 1 ( I I I I 240 I I 0 AB CA AB a 0 AB CA − ∠ − = − = − ∠ = 0 AB a I 3 30 I = ∠− 3 I I_L = _p c b a L

I

I

I

I

=

=

=

I

_p

=

I

_AB

=

I

_BC

=

I

_CA

Magnitude line currents:

A single phase equivalent circuit

3 / ∆ = = Z V Z V I an Y an a

Y-Delta configuration: Example 3

1- A balanced abc sequence Y-connected source with

is connected to a ∆-connected balanced load (8+j4)Ω per phase. Calculate the phase and line currents.

V

10

100

V

_an

=

∠

0 V V_AB =180∠− 200

2-One line voltage of a balanced Y-connected source is

If the source is connected to a ∆ _{-connected load of , find}

the phase and line currents. Assume the abc sequence.

Ω ∠ 0

40 20

Balanced Delta-Delta

Connection

A balanced _{∆ - ∆ system is one in which both balanced} source and balanced load are _{∆ connected.}

Balanced Delta-Delta Connection

∆ ∆ ∆ = = = Z V I Z V I Z V I CA CA BC BC AB AB CA ca BC bc AB ab V V V V V V = = = Line voltages: Line currents: 3 I I_L = _p

Magnitude line currents:

3 Z Z_Y = ∆ Total impedance: Phase currents: ° ∠ = − = ° − ∠ = − = ° − ∠ = − = 90 3 150 3 30 3 AB BC CA c AB AB BC b AB CA AB a I I I I I I I I I I I I

A balanced _{∆ - ∆ system is the one in which both balanced source and balanced} load are _{∆ connected.}

A delta-connected source supplies a delta-connected load through wires having

impedances of Z_line=0.3+j0.4Ω, the load impedance are

Z∆=30+j6 Ω, the balanced

source ab voltage is

Vab=1000<30

Find the line current, the line volatge at the load, the current in each phase of the load, the power delivered to the load, and dissipated in the line.

1- A balanced ∆ connected load having an impedance of 20-j15 Ω is connected to a ∆ connected, positive sequence generator having

Calculate the phase currents of the load and the line currents. V 0 330 V_ab = ∠ 0 A I_a = 22.5∠350

2- A positive-sequence, balanced -connected source supplies a balanced ∆-connected load. If the impedance per phase of the load is 18+j12 Ω and , find IAB and VAB.

Balanced Delta-Y Connection

Replace _{∆ connected source to equivalent Y connected source.}

Phase voltages: 0 p cn 0 p bn 0 p an 90 3 V V 150 3 V V 30 3 V V + ∠ = − ∠ = − ∠ =

A single phase equivalent circuit

Y p Y an a Z V Z V I 0 30 3 ∠ − = =

Y-Delta configuration: Example 5

1-A balanced Y connected load with a phase resistance of 40 Ω and a reactance of 25 Ω is supplies by a balanced, positive

sequence ∆ connected source with a line voltage of 210 V. Calculate the phase currents. Use V_ab as reference.

V

V

_ab

=

240

∠

15

0

2-In a balanced -Y circuit,

POWER IN A BALANCED SYSTEM

) 120 cos( 2 ), 120 cos( 2 , cos 2 = − 0 = + 0 = V t v V t v V t v_{AN p} ω _{BN p} ω _{CN p} ω

For Y connected load, the phase voltage:

θ ∠ = Z Z_Y

If Phase current lag phase voltage by θ.

) 120 t cos( I 2 i ) 120 t cos( I 2 i ) t cos( I 2 i 0 p c 0 p b p a + θ − ω = − θ − ω = θ − ω =

POWER IN A BALANCED SYSTEM

Total instantaneous power:

c CN b BN a AN c b a

p

p

v

i

v

i

v

i

p

p

=

+

+

=

+

+

θ

=

3

V

I

cos

p

_{p p}

Average power per phase:

Apparent power per phase:

Reactive power per phase:

Complex power per phase:

θ

=

V

I

sin

Q

_{p p p}

θ

=

V

I

cos

P

_{p p p} * p p p p p

P

jQ

V

I

S

=

+

=

p p p

V

I

S

=

POWER IN A BALANCED SYSTEM

Total average power:

θ

=

θ

=

=

3

P

3

V

I

cos

3

V

I

cos

P

_{p p p L L}

Total reactive power:

θ

=

θ

=

=

3

Q

3

V

I

sin

3

V

I

sin

Q

_{p p p L L}

Total complex power:

* p 2 p p 2 p * p p p

Z

V

3

Z

I

3

I

V

3

S

3

S

=

=

=

=

θ

∠

=

+

=

P

jQ

3

V

_L

I

_L

S

Power: Example 6

1-A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the line voltage is 220 V and the line current is 18.2 A. Determine the power factor of the motor.

2- Calculate the line current required for a 30-kW three-phase motor having a power factor of 0.85 lagging if it is connected to a balanced source with a line voltage of 440 V.