Industrial Power Systems Salvador Acevedo ELEC 371 Three-phase systems 1 Resistor Inductor Capacitor...

Industrial Power SystemsSalvador Acevedo

ELEC 371 Three-phase systems1

Resistor

Inductor

Capacitor

Passive Elements and Phasor Diagrams

I V

IV

I

V

v RiV = 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

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

AC synchronous generator Single-phase equivalent

AC synchronous motor Single-phase equivalent

AC induction motor (rarely used as generator)



Steady-state Solution

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

jwL+

VS-

I

VS = R I + jwL I

VS = (R + jwL) I

VS = (R + jX) I

VS = Z I

I = VS

Z

IV

I I

max

max

0

Z

VS = R I + jwL I

VS = (R + jwL) I

VS = (R + jX) I

VS = Z I

I = VS

Z

IV

I I

max

max

0

Z

VS

I

From rectangular form to polar form:

I = Ix + Iy Magnitude

Angle or phase

From polar form to rectangular form:

Ix Icos Real part

Iy Isin Reactive or imaginary part

2 2

tan 1 Iy

Ix

From rectangular form to polar form:

I = Ix + Iy Magnitude

Angle or phase

From polar form to rectangular form:

Ix Icos Real part

Iy Isin Reactive or imaginary part

2 2

tan 1 Iy

Ix

vs

i

Iy=I sin

Ix=I cos

I

Rectangular form Polar form

I = Ix + j Iy = Imax



Single-phase Power Definitions

v(t) = Vm sin(wt+ v)volts

i(t) = Im sin (wt+ i) amps

+

-

Load: any R,L,Ccombination

w=2 fw: angular frequency in rad/sec

f: frequency in Hz

v(t) = Vm sin(wt+ v)volts

i(t) = Im sin (wt+ i) amps

+

-

Load: any R,L,Ccombination

w=2 fw: angular frequency in rad/sec

f: frequency in Hz

Instantaneous power

V I

1

2V I

Average Power (or REAL POWER)

P1

2V I V I

Apparent Power

S = V I

Power Factor

pfREAL POWER

APPARENT POWER

For this circuit, the power factor is

pf =V I cos

V I

m m

m m

m m rms rms

rms rms

rms rms

rms rms

p t v t i t wt wt

p t wt

Tp t dt

P

S

v i

v i v i

T

( ) ( ) ( ) sin sin( )

( ) cos( ) cos( )

( ) cos cos

cos

2

1

0



Power Triangle

Ssin =Q

S

P=Scos

Real Power P = S cos = V I cos watts

Reactive Power Q = S sin V I sin vars

Complex Power

S = S P + j Q

S = VI cos + jVIsin

If = -

and assuming a reference = 0

then = -

therefore

S = V Icos(- ) + jIsin(- )

S = V Icos( ) - jIsin( )

S V I *

The magnitude

is called Apparent Power:

S = VI volt - amperes (VA)

v i

v

i

i i

i i



Z =1

jwC= -jX X 90

P VI cos(-90 = 0 watts

Q = VI sin(-90 = -VI = -I X = -V / X

Co

o

o 2L

2L

C

s

)

) var

Purely Capacitive Load

A capacitor absorbs negative Q. It supplies Q.

Power Consumption by Passive Elements

Impedance: Z = R + jX = Z

Z = R = R 0

P = VI cos0 = VI = I R = V / R watts

Q = VI sin 0 = 0 vars

o

o 2 2

o

Resistive Load

A resistor absorbs P

Z = jwL = jX X 90

P VI cos(90 = 0 watts

Q = VI sin(90 = VI = I X = V / X

L Lo

o

o 2L

2L

)

) var s

Purely Inductive Load

An inductor absorbs Q



Creation of the three-phase induction motor

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

adding an extra cable

Savings in magnetic core when constructing Transformers 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 - 2/3) = Vm sin (wt - 120) volts

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

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

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

Va

Vb

Vc

120

120120



Star Connection (Y)

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 ( )

-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

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

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

cos cos cos

cos cos cos

cos

sin

S3 3 V I VALL L



Three-phase Transformers

Use of one three-phase transformer unit

Use of 3 single phase transformers to form a “Transformer Bank



Physical Overview



Three-phase Transformers Connections Y-Y; ; Y- ; -Y

+v1-

+v2-

N1 : N2

a=N1/N2=V1/V2

A single-phase transformer Example:10 KVA, 38.1KV/3.81KV

a' line to line primary voltageline to line secondary voltage

The Bank Transformation Ratio is defined as:

Bank of3 single-phasetransformers

Primaryterminals

Secondaryterminals



Y-Y connection

N

n

+ v1 -

+ v2 -

N1

N2

N1

N2

N1

N2

A B C

a b c

Ratings for Y-Y bankusing 3 single-phase transformers:

3x10KVA = 30 KVA66 KV / 6.6 KV



connection

+ v1 -

+ v2 -

N1

N2

N1

N2

N1

N2

A B C

a b c

Ratings for bank: 30 KVA; 38.1 KV / 3.81 KV



Y- connection

N

+ v1 -

+ v2 -

N1

N2

N1

N2

N1

N2

A B C

a b c

Ratings for Y- bank: 30 KVA; 66 KV / 3.81 KV



-Y connection

n

+ v1 -

+ v2 -

N1

N2

N1

N2

N1

N2

A B C

a b c

Ratings for -Y bank:30 KVA; 38.1 KV / 6.6 KV



Power 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=108

12009. . .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

puactual

basepu

actual

base

puactual

basepu

actual

base

If base voltage and base power are known:

V KV, S MVA

then, base current and base impedance are:

I =S

VI A.

Z =V

IZ

Another way to express base impedance is:

Z =V

I

V

S

V

V

S

Real power base and reactive power base are:

P = S = 100 MW

Q = S = 100 MVAR

base base

basebase

basebase

basebase

basebase

basebase

base

base

base

base

base

base

base base

base base

100 100

100 000 000

100 0001000

100 000

1000100

2

, ,

,

,



Three phase bases

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

why?

V=V

voltage line-to-line = voltage neutral-to-line

unit, per In

:are case phase-three the for

impedance and current base The

S

LL(pu)LN(pu)

3-base

3

2

3

2

3

3

1

3

3

33

3

3

3

base

LLbase

base

LLbase

base

LLbase

base

LLbase

base

base

LNbaseLLbase

base

S

VS

V

Z

V

S

V

S

I

VV

S

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.

0 9

0 9

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

) ( . )08

Three-phase load270 MW100 KVpf=0.8

abc



Transformers in per unit calculations

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 Z base2 = 120/41.667=2.88

+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=(530)/Zbase2 =1.7361 30p.u.

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

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

35 2 05

2 01 4 2

05 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

1 2

1 2 1 2

1 2

, ...

, ... ; , ... ;

, ...

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.

Industrial Power Systems Salvador Acevedo ELEC 371 Three-phase systems 1 Resistor Inductor Capacitor...

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Transcript of Industrial Power Systems Salvador Acevedo ELEC 371 Three-phase systems 1 Resistor Inductor Capacitor...