Resistor i v = Ri + v - V = R I v i I Vsacevedo/eece458/threeph.pdf · VS-I VS = R I + jwL I VS = (R + jwL) I VS = (R + jX) I VS = Z I I = VS Z I V I I max max = ∠ ∠ = ∠−

© Salvador Acevedo, 2000

EECE 458/571Three-phase systems1

Resistor

Inductor

Capacitor

Passive Elements and Phasor Diagrams

I V

IV

I

V

v Ri=V = R I

Ri

+ v -

v Ldi

dt=

V = jwL I

i

+ v -

L

i Cdv

dt=

I = jwC V

Ci

+ v -

v

i

v i

vi



+V2-

+V1-

I1 I2

Z

V2 V1

I1

I2

Assuming a RL loadconnected to secondaryand ideal source to primary

Ideal Transformer

av

v

i

i

a

= = =

= = =

1

2

2

1

N1

N2V1

V2

I2

I1

N1

N2

i1 i2+v1-

+v2-

N1:N2

Transformer feeding load:

V2 = V1/a

I2 = V2/Z

I1= I2/a



Two Winding Transformer Model

à The linear equivalent model of a real transformer consists of an ideal transformer and some passive elements

i1 i2

+

v1

-

+

v2

-N1:N2



AC Generators and Motors

à AC synchronous generator« Single-phase equivalent

à AC synchronous motor« Single-phase equivalent

à AC induction motor (rarely used as generator)



Steady-state Solution

In sinusoidal steady-state a circuit may be solved using phasorsR

jwL+

VS-

I

V S = R I + j w L I

V S = ( R + j w L ) I

V S = ( R + j X ) I

V S = Z I

I = V S

Z

IV

I I

m a x

m a x

=∠

∠

= ∠ −

0

Z θ

θ

VS

I

θ

F r o m r e c t a n g u l a r f o r m t o p o l a r f o r m :

I = I x + I y M a g n i t u d e

A n g l e o r p h a s e

F r o m p o l a r f o r m t o r e c t a n g u l a r f o r m :

I x I c o s R e a l p a r t

I y I s i n R e a c t i v e o r i m a g i n a r y p a r t

2 2

θ

θθ

=

==

−t a n 1 I y

I x

θ

vs

i

0 π 2π

Iy=I sinθ

Ix=I cosθ

θ

I

Rectangular form Polar form

I = Ix + j Iy = Imax ∠ θ



Single-phase Power Definitions

v(t) = Vm sin(wt+θv)volts

i(t) = Im sin (wt+θi) amps

+

-

Load: any R,L,Ccombination

w=2πfw: angular frequency in rad/sec

f: frequency in Hz

[ ][ ]

I n s t a n t a n e o u s p o w e r

V I

1

2V I

A v e r a g e P o w e r ( o r R E A L P O W E R )

P1

2V I V I

A p p a r e n t P o w e r

S = V I

P o w e r F a c t o r

p fR E A L P O W E R

A P P A R E N T P O W E R

F o r t h i s c i r c u i t , t h e p o w e r f a c t o r i s

p f =V I c o s

V I

m m

m m

m m r m s r m s

r m s r m s

r m s r m s

r m s r m s

p t v t i t w t w t

p t w t

Tp t d t

P

S

v i

v i v i

T

( ) ( ) ( ) s i n s i n ( )

( ) c o s ( ) c o s ( )

( ) c o s c o s

c o s

= = + +

= − − + +

= = =

= =

=

∫

θ θ

θ θ θ θ

θ θ

θθ

2

1

0



Power Triangle

Ssin θ=Q

S

P=Scosθ

θ

Real Power P = S cos θ = V I cos θ watts

Reactive Power Q = S sin θ = V I sin θ vars

[ ][ ]

C o m p l e x P o w e r

S = S P + j Q

S = V I c o s + j V I s i n

I f = -

a n d a s s u m i n g a r e f e r e n c e = 0

t h e n = -

t h e r e f o r e

S = V I c o s ( - ) + j I s i n ( - )

S = V I c o s ( ) - j I s i n ( )

S V I *

T h e m a g n i t u d e

i s c a l l e d A p p a r e n t P o w e r :

S = V I v o l t - a m p e r e s ( V A )

v i

v

i

i i

i i

∠ =

=

θ

θ θ

θ θ θθ

θ θ

θ θ

θ θ



Z =1

j w C= - j X X 9 0

P V I c o s ( - 9 0 = 0 w a t t s

Q = V I s i n ( - 9 0 = - V I = - I X = - V / X

Co

o

o 2L

2L

C

s

= ∠ −

= )

) v a r

Purely Capacitive Load

A capacitor absorbs negative Q. It supplies Q.

Power Consumption by Passive Elements

Impedance: Z = R + jX = Z ∠θ Ω

Z = R = R 0

P = V I c o s 0 = V I = I R = V / R w a t t s

Q = V I s i n 0 = 0 v a r s

o

o 2 2

o

∠

Resistive Load

A resistor absorbs P

Z = j w L = j X X 9 0

P V I c o s ( 9 0 = 0 w a t t s

Q = V I s i n ( 9 0 = V I = I X = V / X

L Lo

o

o 2L

2L

= ∠

= )

) v a r s

Purely Inductive Load

An inductor absorbs Q



à Creation of the three-phase induction motor

à Efficient transmission of electric powerð 3 times the power than a single-phase circuit by

adding an extra cable

à Savings in magnetic core when constructingð Transformers

ð Generators

Advantages of Three-phase Systems

Three-phaseinduction motor

Single-phaseinduction motor

Starting torque yes noneeds auxiliarystarting circuitry

Steady state torque Constant Oscillating causingvibration

Single-phaseLoad

+v-

i

Three-phaseLoad

va

vb

vc

ia

ib

ic

p = vi p = va ia + vb ib + vc ic



Three-phase Voltages

va vb vc

va(t) = Vm sin wt volts

vb(t) = Vm sin (wt - 2π/3) = Vm sin (wt - 120°) volts

vc(t) = Vm sin (wt - 4π/3) = Vm sin (wt - 240°) voltsor

vb(t) = Vm sin (wt + 2π/3) = Vm sin (wt + 120°) volts

w = 2 π f w: angular frequency in rad/sec f : frequency in Hertz

Va

Vb

Vc

120°

120°120°



Star Connection (Y)

à Y-connected Voltage Source

a

bc

n

+Van

-

Vbn

Vcn+

-

+

-

Line - to - neutral voltages Van, Vbn, Vcn.

(phase voltages for Y - connection)

same magnitude: V

V Van Vbn Vcn

Line - to - line voltages Vab, Vbc, Vca

same magnitude: V

Vab = Van - Vbn

V = 3 V

P

P

LL

LL P

= = =



Delta Connection (∆∆ )

à ∆-connected Voltage Source

a

b

c

+Vab

-

Vbc

Vca

+

-

+

-

Line - to - line voltages Vab, Vbc, Vca.

(phase voltages for - connection)

same magnitude: V V

Phase currents Iab, Ibc, Ica.

same magnitude: I

Line currents Ia, Ib, Ic.

same magnitude: I

I = 3 I

LL P

P

L

L P

∆=



Y-connected Load

a

bc

n

+Van

-

Vbn

Vcn+

-

+

-

ia

Ia

ia

Ib

ia

Ic

Za

ZbZcn'

Balanced case: Za = Zb = Zc = Z

Ia + Ib + Ic = 0

Ib = Ia -120

Ic = Ia - 240

∠ °∠ °



∆∆-connected Load

a

bc

n

+Van

-

Vbn

Vcn+

-

+

-

ia

Ia

ia

Ib

ia

Ic

Zbc

ZabZca



Y-∆∆ Equivalence

Za

ZbZcn'

Zbc

ZabZca

Balanced case:

Za = Zb = Zc = Zy

Z = 3Zy

Zab = Zbc = Zca = Z = 3Zy

∆

∆



Power in Three-phase Circuits

( ) ( )( ) ( )( ) ( )

( ) ( ) ( ) ( )

Three-phase voltages and currents:

The three-phase instantaneous power is:

v V wt i I wt

v V wt i I wt

v V wt i I wt

p t p v i v i v i

p V Iwt wt wt wt

wt

a m v a m i

b m v b m v

c m v c m i

a a b b c c

m m

v i v v

= + = +

= + − ° = + − °

= + − ° = + − °

= = + +

=+ + + + − ° + − °

+ +

sin sin

sin sin

sin sin

( )

sin sin sin sin

sin

θ θ

θ θ

θ θ

θ θ θ θ

θ

φ

φ

120 120

240 240

120 120

3

3 ( ) ( )

( )

v i

m m v i

P P

Pm

Pm

v i

wt

p V I

P p

P V I

VV

II

− ° + − °

= −

=

=

= = = −

240 240

3

2 2

332

3 3

3

sin

cos

cos

θ

θ θ

θ

θ θ θ

φ

φ φ

φ

This expression can easily be reduced to:

Since the instantaneous power does not change with the time,

its average value equals its intantaneous value:

where:



Three-phase Power

In a Y-connection

In a -connection

Regardless of the connection (for balanced systems),

the average power (real power) is :

Similarly, reactive power and apparent power expressions are:

Q vars3

V V I I

P V IV

I V I

V V I I

P V I VI

V I

P V I watts

V I

LL P L P

P PLL

L LL L

LL P L P

P P LLL

LL L

LL L

LL L

= =

= =

=

= =

= =

=

=

=

3

3 33

3

3

3 33

3

3

3

3

3

3

φ

φ

φ

φ

θ θ θ

θ θ θ

θ

θ

cos cos cos

cos cos cos

cos

sin

∆

S3φ = 3V I VALL L



àPower lines operate at kilovolts (KV)

and kilowatts (KW) or megawatts (MW)

To represent a voltage as a percent of a reference value, we first define this BASE VALUE.

Example:

Base voltage: Vbase = 120 KV

** The percent value and the per unit value help the analyzer visualize how close the operating conditions are to their nominal values.

Per unit modelling

Circuit voltage Percent ofbase value

Per unit value

108 KV 90% 0.9

120 KV 100% 1.0

126 KV 105% 1.05

60 KV 50% 0.5

per unit quantity = actual quantitybase quantity

Voltage_1=108120

= 0 9. . .p u



Defining bases

4 quantities are needed to model a network in per unit system:

V: voltage VBASE

I: current IBASE

S: power SBASE

Z: impedance ZBASE

Given two bases, the other two quantities are easily determined.

VV

VI

I

I

SS

SZ

Z

Z

p ua c t u a l

b a s ep u

a c t u a l

b a s e

p ua c t u a l

b a s ep u

a c t u a l

b a s e

= =

= =

( )

I f b a s e v o l t a g e a n d b a s e p o w e r a r e k n o w n :

V K V , S M V A

t h e n , b a s e c u r r e n t a n d b a s e i m p e d a n c e a r e :

I =S

VI A .

Z =V

IZ

A n o t h e r w a y t o e x p r e s s b a s e i m p e d a n c e i s :

Z =V

I

V

S

V

V

S

R e a l p o w e r b a s e a n d r e a c t i v e p o w e r b a s e a r e :

P = S = 1 0 0 M W

Q = S = 1 0 0 M V A R

b a s e b a s e

b a s eb a s e

b a s eb a s e

b a s eb a s e

b a s eb a s e

b a s eb a s e

b a s e

b a s e

b a s e

b a s e

b a s e

b a s e

b a s e b a s e

b a s e b a s e

= =

= =

= =

=

=

1 0 0 1 0 0

1 0 0 0 0 0 0 0 0

1 0 0 0 0 01 0 0 0

1 0 0 0 0 0

1 0 0 01 0 0

2

, ,

,

,Ω



Three phase bases

In three-phase systems it is common to have data for the three-phase power and the line-to-line voltage.

( )

S

T h e b a s e c u r r e n t a n d i m p e d a n c e

f o r t h e t h r e e - p h a s e c a s e a r e :

I n p e r u n i t ,

l i n e - t o - n e u t r a l v o l t a g e = l i n e - t o - l i n e v o l t a g e

V = V

w h y ?

b a s e - 3

L N ( p u ) L L ( p u )

Φ Φ

Φ

Φ

Φ Φ

=

=

=

=

=

=

−

− −

−

−

−

−

−

−

−

−

3

3

3

33

3

3

1

3

3

3

2

3

S

V V

I

S

VS

V

Z

V

S

V

S

b a s e

b a s e L L b a s e L N

b a s e

b a s e

b a s e L L

b a s e

b a s e L L

b a s e

b a s e L L

b a s e

b a s e L L

b a s e

With p.u. calculations, three-phase values of voltage, current and power can be used without undue anxiety about the result being a factor of √3 incorrect !!!

With p.u. calculations, three-phase values of voltage, current and power can be used without undue anxiety about the result being a factor of √3 incorrect !!!



Example

The following data apply to a three-phase case:

Sbase=300 MVA (three-phase power)

Vbase=100 KV (line-to-line voltage)

+V=1 p.u.-

Single-phase equivalent:

I=1.125 p.u.

Using the per unit method:

P =270

300 p.u.

V =1.0 p.u.

P = V I pf

then

I =P

V pf

This current is 12.5% higher than its base value!

To check: 1.125xIbase = (1.125)300,000

3 100 x 1732 A.

=

= =

= =

09

09

10 081125

1125 19485

.

.

( . )( . ). . .

. .

p u

Normally, we'd say:

P = 3 V I cos = 3 V I pf

I =P

3 V pf

270x10

3 (100x101948.5 A.

L L L L

L

6

3

θ

= =) ( . )0 8

Three-phase load270 MW100 KVpf=0.8

abc



Transformers in per unit calculations

à With an ideal transformer

2400 V. 4.33 + j 2.5 ohms

2400:120 V5 KVA

+V1-

+V2-

High Voltage Bases Low Voltage BasesSbase1 = 5 KVA Sbase2 = 5KVAVbase1= 2400 V Vbase2 = 120 VIbase1 = 5000/2400=2.083 A I base2 = 5000/120=41.667 AZbase1= 2400/2.083=1152 Ω Z base2 = 120/41.667=2.88 Ω

+1.0-

+1.0-

From the circuit: V1=2400 V.V2=V1/a=V1/20=120 V.

In per unit: V1=1.0 p.u.V2=1.0 p.u.

The load in per unit is: Z=(5∠ 30°)/Zbase2 =1.7361 ∠ 30° p.u.

The current in the circuit is: I=(1.0 ∠0°)/ (1.7361 ∠ 30°) =0.576 ∠-30° p.u.

The current in amperes is:Primary: I1=0.576 x Ibase1= 1.2 A.Secondary: I2=0.576 x Ibase2= 24 A.



One line diagrams

à A one line diagram is a simplified representation of a multiphase-phase circuit.

GENERATOR

TRANSFORMERTRANSFORMER

GENERATOR

Transmission line

Transmission line

Transmission line

LOAD



Nodal Analysis

Suppose the following diagram represents the single-phase equivalent of a three-phase system

V1= 1 p.u.

z1=j1 p.u.

z13=j2 p.u.

z12=j0.5 p.u. z23=j0.5 p.u.

z2=10 p.u.

z3=j2 p.u.

V3= -j1 p.u.

I1= -j1 p.u.

y1=-j1 p.u.

y13=-j0.5 p.u.

y12=-j2 p.u. y23=-j2 p.u.

y2=0.1 p.u. y3=-j0.5 p.u.

I3= -0.5 p.u.

1 2 3

Finding Norton equivalents and representing impedances as admittances:

I1=y1 V1 + y12(V1-V2) + y13(V1-V3)0 = y12 (V2-V1) + y2 V2 + y23(V2-V3)I3=y13(V3-V1) + y23(V3-V2) + y3 V3

In matrix form:

y + y + y - y - y

- y y + y + y - y

- y - y y + y + y

V1

V2

V3

I1

0

I3

V1

V2

V3

- j1

0

- 0.5

1 12 13 12 13

12 12 2 23 23

13 23 13 23 3

=

−−

−

=

j j j

j j j

j j j

3 5 2 0 5

2 0 1 4 2

0 5 2 3

. .

.

.

solving

V1

V2

V3

=∠ − °∠ − °∠ − °

0 77 24

0 73 35

0 71 44

.

.

.

. .p u



General form of the nodal analysis

The system of equations is repeated here to find a general solution technique:

y +y +y -y -y

-y y +y +y -y

-y -y y +y +y

V1

V2

V3

I1

0

I3

or

Y Y Y

Y Y Y

Y Y Y

V1

V2

V3

J1

J2

J3

In general:

1 12 13 12 13

12 12 2 23 23

13 23 13 23 3

11 12 13

21 22 23

31 32 33

=

=

Y = y

Y =-y

J = I

ii ijj=1

N

ij ij

i i (from current sources flowing into the node)

∑

∑

=

= = ≠

=

i N

i N j N i j

i N

12

12 12

12

, ...

, ... ; , ... ;

, ...

Once the voltages are found, currents and powers are easily evaluated from the circuit. We have solved one of the phases of the three-phase system (e.g. phase ‘a’). Quantities for the other two phases are shifted 120 and 240 degrees under balanced conditions.

Actual quantities can be found by multiplying the per unit values by their corresponding bases.

Documents

Resistor i v = Ri + v - V = R I v i I Vsacevedo/eece458/threeph.pdf · VS-I VS = R I + jwL I VS = (R + jwL) I VS = (R + jX) I VS = Z I I = VS Z I V I I max max = ∠ ∠ = ∠−