Chap 4 Three Phase Circuits Karady-Holbert FINAL 2 Slides 97

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EEE 360Chapter 4:

Three-Phase Circuits




Practically all electrical energy generation andtransmission systems use a three-phase circuit.

The three-phase energy is transported through

three or four conductors to large customers. Only the small household and light commercial

loads are supplied by a single phase.

The major advantage of the three-phase systemis the efficiency of power transmission.




Another advantage is generation of constanttorque, which reduces vibration, for rotating

machines. This is particularly important for

industries with large motors.

A three-phase transmission line carries three

times the power of a single-phase line while

requiring practically the same right-of-way.

Polyphase systems (six-phase and twelve-phase) have been tried but proved to be un-

economical for power transmission but are used

for large rectifiers.




Three-Phase QuantitiesBasic Definitions



Basic Definitions

In this chapter only the balanced th ree-phasesystems are presented.

A three-phase system has three sinusoidal

voltages sources.

In a balanced system, each voltage source has

the same magnitude (V M ) and frequency (ω),

and each is 120° out-of-phase with the other two

240240cos)(

120120cos)(

0cos)(

M M cn

M M bn

M M an

V t V t v

V t V t v

V t V t v

cn

bn

an

V

V

V



Basic Definitions

When Vbn lags Van by 120° and Vcn lags Vbn by120°, then the system is said to have a positive

phase sequence, or also called an abc phase

sequence.

The phase sequence describes the order in

which the phase voltages reach their maximum

(peak) values with respect to time.

If Vcn and Vbn lag Van by 120° and 240°,respectively, the system has a negative (or acb)

phase sequence.

This class presents only the positive sequence

system.



Basic Definitions

A balanced three-phase circuit is one in whichthe loads are such that the currents produced by

the voltages are also balanced.

The balanced load currents are described by the

equations below

where the voltage of each phase leads its

corresponding current by an angle of θ

240240cos)(120120cos)(

cos)(

M M c

M M b

M M a

I t I t i I t I t i

I t I t i

c

b

a

II

I



Basic Definitions

The sum of the balanced voltages and the sumof the balanced currents equal zero

The instantaneous power in each phase is the

product of voltage and current. The total

instantaneous power is

Hence, the instantaneous three-phase power is

constant over time.

0)()()(

0)()()(

t i t i t i

t v t v t v

c ba

cnbnan

)cos(3cos2

3)()()()( rmsrmsM M

c baT I V I V t pt pt pt p



Basic Definitions

The sinusoidal varying voltages and currents canbe represented by complex phasors.

Vcn

Vbn

Van

Ia

Re

Im

120°

120°

θ

Ib

Ic

120°

120°

Vbn

Vcn

Van + Vbn + Vcn = 0

Ic

Ib

Ia + Ib + Ic = 0

Figure 4.1 Phasor diagram for a

balanced three-phase circuit

with a positive phase sequence

Figure 4.1 presents the

phasor diagram of athree-phase balanced

system.

The phase shift is –120°

The sums of the voltage

vectors and current

vectors are zero.



Basic Definitions

Similarly the total three-phase complex poweris three times the complex power of any of the

phases: ST = S A + SB + SC = 3 S1

Figure 4.2 shows the sinusoidal voltage and

currents as well as the instantaneous power

variation in time.

The frequency of the voltage and current is 60

Hz, but the power of each phase varies at afrequency of 120 Hz.

The sum of the three instantaneous powers is

constant power .



Basic Definitions

Figure 4.2 Time-varying voltage, current and power

in a balanced three-phase circuit.

0 5 10 15 20 25 30 35 40-500

0

500

Voltage(kV) Three-Phase Balanced System at 60 Hz, and V Leads I by 45°

0 5 10 15 20 25 30 35 40-1

0

1

Current(kA

)

0 5 10 15 20 25 30 35 40-100

0100

200

300

Time (ms)

Power(

MW)

Va

V b

Vc

Ia

I b

Ic

Pa

P b

Pc

Ptotal




Three-Phase QuantitiesDelta –Wye Load Connections



Delta-Wye Connections

A balanced three-phase load can be connectedin a delta ( Δ) or in a wye (Y).

Figure 4.3 shows wye-connected loads.

Z Y

Z Y

Z Y

a

c

b

n Load

Z Y

Z Y Z Y

a

b

c

Load

n

Figure 4.3 Wye-connected loads.




Figure 4.4 shows delta-connected loads; the seriesand parallel impedance combination techniques

can not be used for a delta connected load.

Z

Z

Z

a

c

b

Load

Z

Z

Z

a

b

c

Load

Figure 4.4 Delta-connected loads.




The analysis and simplification of a delta-connectedsystem requires the transformation of the delta-

connected impedances to wye-connected impedances.

Figure 4.5 shows the concept of delta –wye impedance

transformation.

Zc

Z b Z a

a b

c

Z 2 Z 3

Z1

a b

c

Y-

Figure 4.5 Delta –

wye impedance transformation




The equations for the delta to wye transformationare

321

32

321

31

321

21

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

c

b

a

These relations can

be read as the

impedance next to aparticular Y node is

the product of the two

delta impedances

connected to that

node divided by the

sum of the three delta

impedances.




The reverse transformation (Y to Δ) can beperformed by the equations below

For balanced load, when the impedances are

equal

a

ac c bba

b

ac c bba

c

ac c bba

Z

ZZZZZZZ

Z

ZZZZZZZ

Z

ZZZZZZZ

3

2

1

YΔ

ZZ 3




Three-Phase QuantitiesWye Connected Generator



Wye Connected Generator

Figure 4.6 Three-phase wye-connected generator

n

a

Vab

Ia

Van

VbnVcn

bIb

Icc

Vbc

Vca

Figure 4.6 exhibits a three-phase generator connectedin a wye (Y) configuration.

The generator is represented by three voltage sources.

The magnitudes of the ac

source voltages are the

same.

The phase shift between the

voltages is 120º.

This system has three phase

conductors (a, b and c) and

a grounded neutral (n).




The voltage of phase a (Van) is selected as thereference with a phase angle of .

The phasor voltage expressions are

where V P is the phase voltage magnitude between

the phase conductors and neutral, and is called thel ine-to-neutral vo ltage .

An important property of this balanced voltage set is

240

120

P

P

P

V

V

V

cn

bn

an

V

V

V

0cnbnan

VVV

V P is the rms valueof the voltage




Figure 4.6 Three-phase wye-connected generator

Figure 4.6 shows the three line-to-neutral voltages aswell as the three l ine-to- l ine voltages (Vab, Vbc, Vca).

The line-to-line voltage is calculated using KVL.

Figure 4.6, the KVL relationfor the n→a→b→n loop is

The line-to-line voltage from

this equation is

0bnaban

VVV

bnanab VVV

n

a

Vab

Ia

Van

VbnVcn

bIb

Icc

Vbc

Vca




We substitute for Vbn in terms of Van to define thel ine-to -l ine vo ltage

The line voltage (Vab) leads the corresponding

phase voltage (Van) by 30 degrees.

The line-to-line voltage magnitude is times

larger than the line-to-neutral (phase) voltage.

303

1

120120

an

an

anan

bnanab

V

V

VV

VVV

j e




The relations for the three-phase line-to-linevoltages and the line-to-neutral voltages are

The abc phase sequence exhibits the ±120°

phase shift between the three line voltages.

2103303

903303

303

ancnca

anbnbc

anab

VVV

VVV

VV




Figure 4.7 Comparison of the relative magnitudes of and the

phase shift between the line, vab(t), and phase, van(t), voltages

t 0 0. 1m s 30m s

0 10 20 30200

100

0

100

200

van t( )

vab t( )

t

m s

Figure 4.7 shows the relative magnitudes of theline and phase voltages and the phase shift

between them. Figure shows that:

• The amplitude of the

line-to-line voltageis larger than the

line-to-neutral

voltage.

• The line-to-neutralvoltage is 30°

behind the line-to-

line voltage.




Figure 4.8 Phasor diagram of a

wye-connected balanced system.

30°120°

Vca

Vab

Vbc

Vbn Van

Vcn

–Vbn

Re

Im The Figure 4.8 phasor diagram of the voltages

shows that:

The phase shift between

the line-to-line and line-to-neutral voltages is 30°

(e.g., Van lags Vab by 30°)

The difference of the two

line-to-neutral voltages is

the line-to-line voltage

(e.g., Vab = Van – Vbn)




Three-Phase QuantitiesWye Connected Loads



Wye Connected Loads A wye-connected generator can be loaded by three

impedances connected in a wye configuration, as seen inFigure 4.9.

Most high-voltage transmission lines use a three-wire

grounded system.

When the neutral point at both the supply and the load aregrounded, the earth interconnects the two neutral points.

Van

Vbn Vcn

Za

Z c Z b

Van

Vbn Vcn

Za

Z c Z b

(b) Four-wire system(a) Three-wire system

Figure 4.9 Three and four wire wye-connected systems



Wye Connected Loads

The earth acts a conductor and transforms the three-wire

system to a practical four-wire system.

In the distribution level, a true four-wire system is

frequently used. The neutral wire is insulated, although it

is grounded at the transformer sites.

At low voltage the larger industrial complexes use a three-

wire 460 V system for larger motor loads.

The smaller loads are supplied by a 208 V four-wire

system, where

The lighting and small loads are connected between

the phases and the neutral wire at 120 V.

The larger loads are connected between the phases

and supplied by 208 V.



Wye Connected Loads

Figure 4.10 shows a wye-connected generator loaded bythree impedances connected in a wye.

Three phase conductors (a, b, c) connect the load

impedances (Za, Zb, Zc) to the sources, and the neutral

conductor connects the neutral point of the sources to

the neutral point of the load.

Figure 4.10 Three-phase wye-connected generator with an impedance load

Van

Vcn

Vbn

n

a

b

c

Ic

Ia

Ib

Vab

VbcVca

I0

Za

Zb

Zc

The neutral point of the

sources is grounded to assure

that the potential of the phase

conductors to the groundremains constant.

This is an important safety

consideration.




Three-Phase QuantitiesWye Connected Four-wire

System



Wye Connected Loads

For the four-wire circuit, the source voltages areconnected directly to the load impedances and the

neutral conductor provides a return path for each phase

current.

Each source line-to-neutral voltage drives a current

through the corresponding impedance.


Figure 4.10 reveals that the

current in each phase can be

calculated by dividing theappropriate line-to-neutral

voltage by the load impedance

in that phase.

Van

Vcn

Vbn

n

a

b

c

Ic

Ia

Ib

Vab

Vbc

Vca

I0

Za

Zb

Zc



Wye Connected Loads

With a balanced load, the phases do not affect oneanother; it is sufficient to calculate the current in phase a

using Ohm’s Law in the circuit of Figure 4.10


Since Za = Zb = Zc = Z Y, the line

currents in phases b and c areidentical in magnitude to Ia, but

lagging by 120° and 240°,

respectively.

Using this fact and applying KCLat the neutral node yields

a

an

a

Z

VI

0cba0

IIII

Van

Vcn

Vbn

n

a

b

c

Ic

Ia

Ib

Vab

VbcVca

I0

Za

Zb

Zc



Wye Connected Loads

The complex power in the load of phase a is theproduct of the voltage and the conjugate of the

current:

The total three-phase complex power is threetimes the single-phase power

*

aana IVS

aT SS 3

LoadP3ph_load / 3

Van

pf Load

Ia

This implies that athree-phase circuit

can be represented by

an equivalent single-

phase circuit.

Figure 4.11 Single-phase

equivalent of a three-phasesystem with a balanced load




Three-Phase QuantitiesWye Connected Three-wire

System



Wye Connected 3-wire System

Three wye-connected sources supply three impedancesalso connected in a wye, as shown in Figure 4.12.

The neutral point (g) of the generator is grounded. But the

supply and load neutral points are not connected together.

This eliminates the direct return path.

Za

Zb

Zc Vngg n

a a

b bc c

Va

Vb Vc

Ia

IbIc

Figure 4.12 Three-wire wye-connected system

The unequal line

currents generate a

voltage difference (Vng)

between the neutralpoint of the generator

and the neutral point (n)

of the load impedances.




The system contains three lines, which areg→a→n, g→b→n and g→c→n.

The currents are calculated from the loop

voltage equations (KVL).

As an example in loop

g→a→n→g, the

difference of the line-

to-neutral voltage and

the voltage betweenthe neutral points (Vng)

drives the current

though impedance Za.

Za

Zb

Zc Vngg n

a a

b bc c

Va

Vb Vc

Ia

IbIc





The loop equations produced line currents are

c

ngc

c

b

ngb

b

a

nga

aZ

VVI

Z

VVI

Z

VVI

Za

Zb

Zc

Vng

g n

a a

b bc c

Va

Vb Vc

Ia

IbIc





The sum of the currents in node point n is zero.

The node point equation (KCL) for node n is

The voltage

difference

between the

neutral points

is calculated

from this

equation.

0

c

ngc

b

ngb

a

nga

Z

VV

Z

VV

Z

VV

Za

Zb

Zc Vngg n

a a

b bc c

Va

Vb Vc

Ia

IbIc





After algebraic manipulation, the voltagedifference between the neutral points is

The currents are calculated by substituting theobtained voltage difference in the earlier current

equations:

cba

c

c

b

b

a

a

ng

ZZZ

Z

V

Z

V

Z

V

V111

c

ngc

c

b

ngb

b

a

nga

a

Z

VVI

Z

VVI

Z

VVI




For balanced load the equations are simplified by

substituting Za = Zb = Zc = Z Y in the voltage

difference equation

But the sum of the three phase voltages is zero,

which implies that in case of a balanced load the

voltage difference between the neutral points is

also zero.

33

1

111

cba

Y

cba

Y

cba

c

c

b

b

a

a

ng

VVV

Z

VVVZ

ZZZ

ZV

ZV

ZV

V




Three-Phase QuantitiesNumerical Example for Wye

Connected Balanced System



Balanced Wye System Example

Example: Calculate the currents in a balanced three-phase system, when the wye connected generator

supplies the wye-connected load of 3 kW with a

power factor of 0.8 lagging, and the generator line-to-

line voltage is 480 V. The balanced three-phase Y-Y system is represented

by a single-phase equivalent circuit.

The circuit is energized by the line-to-neutral voltage.

LoadP 3ph_load / 3

Van

pf load

Ia This circuit carries

one-third of the totalthree-phase power.



The system data are

The line voltage is 480 V, so the line-to-neutral is

The single-phase load power is one-third of the

three-phase load power:

. pf P load load ph lagging80W3000 _ 3

W10003

3

1

ph_load

ph

P P


. V12773

V480

anV



The single-phase complex load power is

The current in phase a is

36.87 A4.51

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V1277V·A7501000

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. j

*

an

1pha

VSI


LoadVan

Ia

V·A7501000)8.0(cosexp80W1000

)(cosexp

1

11

j j .

pf j pf

P load

load

ph

1phS Lagging pf



With a balanced positive sequence system, thecurrents in phases b and c are simply lagging

phase a by 120° and 240°, respectively

13.83 A514

87276 A514)2401)(87.36 A514(

87156 A514)1201)(87.36 A514(

240

120

.

...e

...e

j

j

ac

ab

II

II





Three-Phase QuantitiesNumerical Example for Wye

Connected Unbalanced System



Unbalanced Wye System Example

Calculate the currents in an unbalanced four-wire Y-Y system.

The loads are

13

712

510

c

b

a

Z

Z

Z

j

j

Van

Vcn

Vbn

n

a

b

c

Ic

Ia

Ib

Vab

Vbc

Vca

I0

Za

Zb

Zc



The line-to-neutral (phase) voltage is V P = 120 V

The line-to-neutral phasor voltages are

024V120

012V120

0V120

cn

bn

an

V

V

V


Van

Vcn

Vbn

n

a

b

cIc

Ia

Ib

Vab

Vbc

Vca

I0

Za

Zb

Zc



Using Ohm’s Law, the phase a current is

626 A7310

A8469

)510(

V120

..

. j .

j a

an

aZ

VI


Van

Vcn

Vbn

n

a

b

c

Ic

Ia

Ib

Vab

Vbc

Vca

I0

Za

Zb

Zc



Similarly, the phase b current is

The phase c current is

A63880390789 A6388

3.303.891

120V120

712

120V120

. j...

j

b

bnb

Z

VI

A99476154

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13

240V201

. j.

.

c

cnc

Z

V

I


Van

Vcn

Vbn

n

a

b

c

Ic

Ia

Ib

Vab

Vbc

Vca

I0

Za

Zb

Zc



Using KCL, the neutral conductor current is

As expected, the

neutral current isnot zero for the

unbalanced four-

wire system.

347 A4087

A44450245

)99476154()63880390()8469(

..

. j.

. j.. j.. j.

cba0 IIII


Van

Vcn

Vbn

n

a

b

c

Ic

Ia

Ib

Vab

Vbc

Vca

I0

Za

Zb

Zc




Three-Phase QuantitiesDelta Connected Generators



Delta Connected Generators

Figure 4.13 depicts a three-phase generatorconnected in a delta ( Δ).

The voltage between terminals ab, bc and ca

are the line-to-line voltages.

240

120

L

L

L

V

V

V

ca

bc

ab

V

V

V

Figure 4.13 Delta-connected generators

The delta source

line voltages areVab

a

bc

Vca

Vbc

Vab

a

b

c Vbc

Vca




For a balanced three-phase system, themagnitudes of these line voltages are equal, and

the phase shift between them is 120º.

The voltage between phase ab is typically

selected as the reference with a zero phaseangle (ψ = 0).

The delta-connected generator can be

converted to an equivalent wye-connectedsupply consisting of three voltage sources, as

illustrated in Figure 4.13.




The voltages of the equivalent wye-connectedsources are the line-to-neutral voltages, which

are calculated from the line-to-line voltages

Vab

a

bc

Vca

Vbc

Vab

a

b

cVbc

Vca

Vab

a

b c

Vca

Vbn

Van

a

b

n

Vbc

Vcn

c

Figure 4.13 Delta-connected

generators and the equivalent

wye-connected system.




The equations for the calculation of the equivalentline-to-neutral voltages from the line-to-line voltages

are

The practical use of this conversion is that we can

assume either a wye or a delta configuration for the

supply depending on the load.

In the case of a delta load, the delta supply

connection is more advantageous.

In the case of a wye load, the assumption of a wye-

connected source simplifies the calculation.

303030

333

j j j eee ca

cn

bc

bn

ab

an

VV

VV

VV




Three-Phase QuantitiesDelta Connected Loads



Balanced Delta Load

Figure 4.14 diagrams a delta-connected load in whichthe line-to-line (source) voltages are directly applied to

the load if the line impedance is negligible.

Figure 4.14 Delta connected load.

The calculation of only the

load current ab issufficient in the case of a

balanced load.

The currents in phases bc

and ca will have the sameamplitude and relative

phase angle as Iab, but

will be shifted by ±120º.Ic

Ib

Ia a

b c

Vca

b

c

a

Vab

Vbc Ibc

Iab

Ica

Zbc

Z a b

Z c a



Delta Connected Loads

The current through impedance Zab is computed usingOhm’s Law:

To determine the line currentIa, we employ KCL at node a:

Substituting the Ohm’s Lawexpressions for the phase

currents yields

ab

ab

abZ

VI

caaba III

ca

ca

ab

ab

aZ

V

Z

VI

Ic

Ib

Ia a

b c

Vca

b

c

a

Vab

Vbc Ibc

Iab

Ica

Zbc

Z a b Z c

a




For balanced load and voltage, the equation issimplified

The line current is times larger than the delta

load (phase) current.

303

1

24011

240

240

ab

ab

Δ

ab

Δ

ab

Δ

ab

a

I

I

Z

V

Z

V

Z

VI

j e




The voltage Vab and current Iab for Zab obey the

passive sign convention, and the complex power

is thus

The total three-phase complex power is three

times the single-phase power

*

ababab IVS

C




The phasor diagram shows

the generated voltages and

currents.

The chart visualizes that in

a delta system there are

two currents (phase and

line) and one voltage (the

line-to-line voltage).

The line current is times

the phase current and the

phase shift is –30º.

Ic

Vca

Ica

Ibc

Vbc

Vab Re

Im

Ib

Ia

Iab

–Ica

30°

f

Figure 4.15 Phasor diagram of a

delta-connected balanced system.




Three-Phase QuantitiesExample for Delta Connected

Balanced Load

Delta Connected Balanced




Load Example

Calculate the line and phase currents in abalanced delta connected three-phase system,

when the sources are loaded by a three-phase

load of 3 kW with a power factor of 0.8 lagging

Vab

a

bc

Vca

Vbc

Vab

a

b

cVbc

Vca

a

b c

Ica

Ia

Ic

Ib

Iab

Ibc





Load Example

The system data are

Phase ab line-to-line voltage is the reference

voltage with zero phase shift: Vab = 208 V

Phase ab carries one-third of the total three-

phase load and the line-to-line voltage supplies

this load

V208,8.0W,30003 _ Lload phload V pf P

W10003

W3000

3

3 _

1

phload

ph

P P





Load Example

The complex power for each phase is

VA7501000

)8.0(cosexp8.0

W1000

)(cosexp

1

11

j

j

pf j pf

P load

load

ph

1phS





Load Example

The load current in the delta ab branch is

For balanced conditions, the other delta branchcurrents are

8736 A016 A6384

V208

VA7501000... j.

j*

ab

1ph

abV

SI

1383 A016

240

87156 A016

120

..

..

abca

abbc

II

IIVab

a

b

c

Vbc

Vca

a

bc

Ica

Ia

Ic

Ib

Iab

Ibc





Load Example

The line currents are

Was there an easier way to find Ib and Ic?

53.13 A41.10 A327.8245.6

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j

j j ... j.

. j.. j.

... j.. j.

. j.

bccac

abbcb

caaba

III

III

IIIVca

a

bc

Ica

Ia

Ic

Ib

Iab

Ibc




Three-Phase QuantitiesExample for Delta Connected

Unbalanced Load

Delta Connected Unbalanced



e ta Co ected U ba a ced

Load Example

A delta connected generator, with a 208 V line-to-line voltage supplies unbalanced delta

connected loads.

Calculate the phase and line currents, and thesupply complex and

active powers.

The figure

shows theequivalent

circuit.

Vab

a

bc

Vca

Vbc

Vab

a

b

c

Vbc

Vca

a

bc

Ica

Ia

Ic

IbIab

Ibc




Load Example

The load impedances are

Zab = 23 Ω Zbc = 22 + j 15 Ω

Zca = 25 – j 22 Ω

The reference line voltage is Vab = 208 V

Vab

a

bc

Vca

Vbc

Vab

a

b

c

Vbc

Vca

a

bc

Ica

Ia

Ic

Ib

Iab

Ibc




Load Example

The three-phase equivalent circuit shows thateach line-to-line voltage is directly applied to the

corresponding impedance

This implies that the current through eachimpedance can be calculated by dividing the

appropriate

line-to-line

voltage by thecorresponding

impedance

(Ohm’s Law).

Vab

a

bc

Vca

Vbc

Vab

a

b

c

Vbc

Vca

a

bc

Ica

Ia

Ic

Ib

Iab

Ibc




Load Example

The line-to-line voltages drive a different currentthrough each impedance.

The line-to-line voltages are

V1.180104

240V208

V1.180104

120V208

V208

j

j

ca

bc

ab

V

V

V




Load Example

The phase currents through the impedances arecalculated using Ohm’s Law.

161.3 A246.6 A.9981918.5

2225

V1.180104

3.154 A812.7

A389.3038.7

1522

V1.180104

A044.9 23

V208

j

j

j

j

j

j

ca

ca

ca

bc

bc

bc

ab

ab

ab

Z

VI

Z

VI

Z

VI

Vab

a

b

c

Vbc

Vca

a

bc

Ica

Ia

Ic

Ib

Iab

Ibc




Load Example

The currents in the conductors connecting thegenerator to the loads are the line currents.

The line currents are calculated with the node

point equations (KCL).

Vca

a

bc

Ica

Ia

Ic

Ib

Iab

Ibc

6.7 A09.15 A.00296.14

).9981918.5(044.9

j

j

caaba III




Load Example

The other two line currents are

3.78 A50.5

A39.512.1

)042.9038.7().9981918.5(

168.1 A43.16

A.39308.16

044.9)389.3038.7(

j

j j

j

j

bccac

abbcb

III

III

Vca

a

bc

Ica

Ia

Ic

Ib

Iab

Ibc




Load Example

The complex powers of the phases are

VA858975

3.41VA1299)3.161 A246.6)(120V208(

VA9151343

34.3VA1625 )3.154 A812.7)(120V208(

VA1881) A044.9)(V208(

*

*

j

j

*

cacaca

*

bcbcbc

*

ababab

IVS

IVS

IVS




Load Example

The three-phase (supply) complex power is thesum of the complex power of each phase

The total apparent power is 4199 VA

The total real power is 4199 W

8.0VA4199

VA754199

)858975()9151343(1881

j

j j

cabcab3ph SSSS




Three-Phase PowerMeasurement

Three-Phase Power



Three Phase Power

Measurement

Power measurements generally require

wattmeters.

An analog wattmeter has two inputs: current

and voltage. The current input is connected in series with

the load.

The voltage input is in parallel with the supplyor load.

The wattmeter provides a measurement of the

average power (P ).

Three-Phase Power



Measurement A voltmeter and an ammeter are used in

conjunction with an analog wattmeter todetermine the rms values of the voltage andcurrent (V rms and I rms).

Digital power meters sample the voltage andcurrent waveforms, i.e., v (t ) and i (t ).

A microprocessor calculates the rms values ofthe voltage and current.

The instantaneous voltage and current readingsare multiplied to calculate the instantaneouspower, p(t ).

The average value of p(t ) gives the real power.

Three-Phase Power



Measurement

The measured data permit the calculation of thepower factor ( pf ) and the apparent power (|S|).

The magnitude of the complex power, that is, the

apparent power, is

The power factor is

rmsrms I V

*IVS

S

P pf




Three-Phase PowerMeasurement

Four and Three Wire Systems

Four Wire System



Four-Wire System

Figure 4.21 Measurement connections

for a four-wire, three-phase system.

A

A

V

W

W

Phase a

Phase b

Phase c

Van

Ia

IcV

A

V

W

Ib

Vbn

Vcn

Neutral (n)

P a

P b

P c

In a four-wire system, the loads can beconnected between the phase conductors, or

between the phase conductors and the neutral.

For unbalanced

loading, themeasurement

requires three

wattmeters to

measure each

phase separately

Figure 4.21 shows the

connection diagram

Four Wire System



Four-Wire System

The measurement system provides a current, avoltage, and a power reading for each of the

three phases

The sum of the

measured threepowers gives the

total three-phase

power

c baT P P P P

Figure 4.21 Measurement connections

for a four-wire, three-phase system.

A

A

V

W

W

Phase a

Phase b

Phase c

Van

Ia

IcV

A

V

W

Ib

Vbn

Vcn

Neutral (n)

P a

P b

P c

Three Wire System



Three-Wire System

Figure 4.22 Measurement

connections for a three-wire,

three-phase system.

In a three-wire system, the loads are connectedbetween the phase conductors.

The sum of the three line currents is zero, which

permits the measurement of the three-phase

power with just two wattmeters.

Figure 4.22 sketches the

connection diagram for a

three-wire system.

A

A

V

V

W

W

Phase a

Phase b

Phase c

Vab = Va – Vb

Vcb = Vc – Vb

Ia

Ic

P ab

P cb

Three Wire System



Three-Wire System

The current of phase a and the voltage betweenphases a and b feed one of the wattmeters.

The current of phase c and the voltage between

phases c and b supply the other wattmeter.

The wattmeter multiplies

the voltage and current

values, and determines

the average powervalue.

Figure 4.22 Measurement

connections for a three-wire,

three-phase system.

A

A

V

V

W

W

Phase a

Phase b

Phase c

Vab = Va – Vb

Vcb = Vc – Vb

Ia

Ic

P ab

P cb

Three-Wire System



Three-Wire System

The readings of the wattmeters can beexpressed in complex notation.

The power is the real value of the voltage timesthe current conjugate.

The reading of the wattmeter that is connectedto current Ia and voltage Vab is

The reading of the wattmeter that is connectedto current Ic and voltage Vcb is

*

ab

*

aa

*

aba

*

aab IVIVIVVIV ReReReabP

*

cb

*

cc

*

cbc

*

ccb IVIVIVVIV ReReRecbP

Three-Wire System



Three-Wire System

We shall find that the sum of the two wattmeterreadings is the total three-phase real power:

The substitution of the two power values results

in

Simplification of the expression yields

cbabT P P P

*

cb

*

cc

*

ab

*

aa IVIVIVIV ReReT P

*

c

*

ab

*

cc

*

aa IIVIVIV Re

T P

Three-Wire System



Three-Wire System

The sum of the three line currents is zero suchthat

Substituting the current formula into the power

equation gives

The real value of the components can be

separated individually, which results in:

*

c

*

a

*

bcab IIIIII or

*

bb

*

cc

*

aa IVIVIV Re

T P

cbabc ba

T

P P P P P

P

*

cc

*

bb

*

aa IVIVIV ReReRe




Three-Phase PowerExample for Three-Phase

Transmission

Three-Phase Transmission Example




Figure 4.23 Generator supplies a load

through a short transmission line.

Example: A three-phase generator supplies a loadthrough a transmission line.

Figure 4.23 shows the equivalent circuit.

Calculate the generator terminal and excitation

voltages together with the voltage regulation.

The load data are

active power: 40 MW

load voltage: 22 kV

power factor: 0.8

(lagging)

Egen P load /3

Transmission LineGenerator Load

X g Zline

VgenIload




The transmission line impedance is

The generator rating is

The latter value ( x g ) is the per unit value of the

generator impedance per phase, which must be

converted from percent to ohms via

915.042.0 j line

Z

125%kV22MVA50 g g g x V S

12.1VA1050

V)(220001.25)(

6

22

g

g

g g S

V x X





Calculate the load current using the power perphase and the line-to-neutral voltage since the

given load voltage is the line voltage

36.9kA31.1

A7871050

)8.0(acosexp

)8.0(3

V000,22

3)W1040(

)(acosexp

3

3

6

j

j

pf j

pf V

P load

load load

load loadI


Eq. (3.38)




The generator excitation voltage is the sum ofthe load line-to-neutral voltage and the voltage

drop on the line and generator impedances.

The calculation

results in:

7.29kV9.26V335,13386,23

)1.12915.042.0)( A7871050(3

V000,22

)(

3

j

j j j

j X g lineloadload

gen ZIV

E

Three Phase Transmission Example

Egen P load /3


X g Zline

Vgen

Iload




The generator terminal voltage is the sum of theload line-to-neutral voltage and the voltage drop

on the line and generator impedances.

The calculation

gives:

kV0.243

2.6kV88.13kV63.086.13

)915.042.0)(9.36kA31.1(3kV22

3

gengen_ll

lineloadload

gen

VV

ZIV

V

j

j


Egen P load /3


X g Zline

Vgen

Iload

Generator line-to-line voltage




Compute the voltage regulation. The generatorregulates the voltage such that in no-load the

load voltage will be the rated generator voltage.

The voltage regulation is

9%%)100(kV22

kV22kV24Reg

load

loadgen_ll

V

VV


Egen P load /3


X g Zline

Vgen

Iload

Why isn’t the

no-load voltage

l t E ?

Chap 4 Three Phase Circuits Karady-Holbert FINAL 2 Slides 97

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Transcript of Chap 4 Three Phase Circuits Karady-Holbert FINAL 2 Slides 97