EE369POWER SYSTEM ANALYSIS

Lecture 2Complex Power, Reactive Compensation, Three Phase

Tom Overbye and Ross Baldick

1

Reading and Homework

Read Chapters 1 and 2 of the text.HW 1 is Problems 2.2, 2.3, 2.4, 2.5, 2.6, 2.8, 2.11,

2.13, 2.14, 2.16, 2.17, 2.18 and Case Study Questions a., b., c., d. from the text; due Thursday 9/1.

HW 2 is Problems 2.24, 2.25, 2.26, 2.27, 2.28, 2.29, 2.30, 2.32, 2.34, 2.35, 2.36, 2.38, 2.40, 2.41 (need to install PowerWorld); due Thursday 9/8.

2

Review of PhasorsGoal of phasor analysis is to simplify the analysis of constant frequency ac systems:

v(t) = Vmax cos(t + v),i(t) = Imax cos(t + I),

where:•v(t) and i(t) are the instantaneous voltage and current as a function of time t,• is the angular frequency (2πf, with f the frequency in Hertz), • Vmax and Imax are the magnitudes of voltage and current sinusoids,•v and I are angular offsets of the peaks of sinusoids from a reference waveform.

Root Mean Square (RMS) voltage of sinusoid:

2 maxmax

0

1 ( ) , so 2 .2

T VV v t dt V VT

3

Phasor Representationj

( )

Euler's Identity: e cos sin ,

Phasor notation is developed by rewriting using Euler's identity:

( ) 2 cos( ),

( ) 2 Re .

(Note: is the RMS voltage).Given complex phasor (magnitu

V

Vj t

j

v t V t

v t V e

V

de and angle), we can determine sinusoidal waveform (magnitude and phase) and vice versa. 4

Phasor Representation, cont’d

The RMS, cosine-referenced voltage phasor is:

,

( ) Re 2 ,cos sin ,cos sin .

V

V

jVjj t

V V

I I

V V e V

v t V e eV V j VI I j I

• (Note: Some texts use “boldface” type for complex numbers, or “bars on the top”.)

• Also note that the convention in power engineering is that the magnitude of the phasor is the RMS voltage of the waveform:

• contrasts with circuit analysis. 5

Advantages of Phasor Analysis

0

2 2

Resistor ( ) ( )( )Inductor ( )

1 1Capacitor ( ) ( ) (0)C

= Impedance ,= Resistance,= Reactance,

Z = , =arctan .

t

v t Ri t V RIdi tv t L V j LIdt

v t i t dt v V Ij C

Z R jX ZRX

XR XR

Device Time Analysis Phasor

(Note: Z is a complex number but

not a phasor).

6

RL Circuit Example

2 2 1

( ) 2 100cos( 30 ), so 100 30 ,60Hz,4 , 2 3 ,

4 3 5, tan (3/ 4) 36.9 ,100 30 ,5 36.9

20 6.9 Amps,

( ) 20 2 cos( 6.9 ).

v t t VfR X L fL

ZVIZ

i t t

7

Complex Power

max

max

max max

( ) ( ) ( ),( ) = cos( ),(t) = cos( ),

1cos cos [cos( ) cos( )],2

1( ) [cos( )2cos(2 )].

V

I

V I

V I

p t v t i tv t V ti I t

p t V I

t

Instantaneous Power :

8

Complex Power, cont’d

max max

0

max max

1( ) [cos( ) cos(2 )],21 ( ) ,

1 cos( ),2

cos( ),

= = .

V I V I

T

avg

V I

V I

V I

p t V I t

P p t dtT

V I

V I

Instantaneous Power is

Power F

sum

acto

of average

r Angl

and varying te

e

rms :

9

Complex Power, cont’d

max max

max max

max max

max max

1( ) [cos( ) cos(2 )],

21

[cos( ) cos(2 2 ( ))],21

[cos( ) cos(2 2 )cos( )]2

1sin(2 2 )

2

V I V I

V I V V I

V I V V I

V

p t V I t

V I t

V I t

V I t

Re - interpretation of instantaneous Power :

sin( ),V I

10

Instantaneous power into resistive component

Instantaneous power into electric and magnetic fields

Complex Power

*

cos( ) sin( ) ,,

= Real Power (W, kW, MW),= Reactive Power (VAr, kVAr, MVAr),= magnitude of power into electric and magnetic fields, = Complex power (VA, kVA, MVA),

Power Factor

,

(pf

V I V IS V I jP jQ

VPQ

I

S

) = cos ,If current leads voltage then pf is leading,If current lags voltage then pf is lagging.

(Note: S is a complex number but not a phasor.)

11

= average of instantaneous power into resistive load

Complex Power, cont’d

12

|S|

P

Q

2 2S P Q

S P jQ cos( )P PS

pf

1tan QP

2 2

PpfP Q

Power Triangle

Complex Power, cont’d

2

1

Relationships between real, reactive, and complex power:cos ,

sin 1 pf ,

Example: A load draws 100 kW with a leading pf of 0.85.What are (power factor angle), and ?

cos 0.85 31.8 ,10

P S

Q S S

Q S

S

0kW 117.6 kVA,0.85

117.6sin( 31.8 ) 62.0 kVAr.Q

13

negative since leading pf

Load consumes -62 kVAr, i.e. load supplies +62 kVAr capacitive load

Conservation of Power

At every node (bus) in the system:– Sum of real power into node must equal zero,– Sum of reactive power into node must equal zero.

This is a direct consequence of Kirchhoff’s current law, which states that the total current into each node must equal zero.– Conservation of real power and conservation of

reactive power follows since S = VI*.

14

Conservation of Power ExampleEarlier we foundI = 20-6.9 amps

*

* *

2

* *

2

100 30 20 6.9 2000 36.9 VA,36.9 pf = 0.8 lagging,

( ) 4 20 6.9 20 6.9 ,

1600W ( 0),

( ) 3 20 6.9 20 6.9 ,

1200VA r , ( 0).

R R

R R

L L

L L

S V I

S V I RI I

P I R Q

S V I jXI I j

Q I X P

= 1600W + j1200VAr

Power flowing from source to load at bus

15

Power Consumption in Devices

2Resistor Resistor

2Inductor Inductor L

2Capacitor Capacitor C

Capaci

Resistors only consume real power:

,Inductors only "consume" reactive power:

,Capacitors only "generate" reactive power:

1 .C

P I R

Q I X

Q I X XC

Q

2Capacitor

torC

(Note-some define negative.). CXVX

16

Example

*

40000 0 400 0 Amps100 0

40000 0 (5 40) 400 042000 16000 44.9 20.8 kV

44.9k 20.8 400 017.98 20.8 MVA 16.8 6.4 MVA

VI

V jj

S V Ij

First solvebasic circuitI

17

Example, cont’dNow add additionalreactive power loadand re-solve, assumingthat load voltage is maintained at 40 kV.

70.7 0.7 lagging564 45 Amps59.7 13.6 kV33.7 58.6 MVA 17.6 28.8 MVA

LoadZ pfIVS j

18

Need higher source voltage to maintain load voltage magnitude when reactive power load is added to circuit. Current is higher.

59.7 kV

17.6 MW28.8 MVR

40.0 kV

16.0 MW16.0 MVR

17.6 MW 16.0 MW-16.0 MVR 28.8 MVR

Power System NotationPower system components are usually shown as“one-line diagrams.” Previous circuit redrawn.

Arrows areused to show loads

Generators are shown as circles

Transmission lines are shown as a single line

19

Reactive Compensation

44.94 kV

16.8 MW 6.4 MVR

40.0 kV

16.0 MW16.0 MVR

16.8 MW 16.0 MW 0.0 MVR 6.4 MVR

16.0 MVR

Key idea of reactive compensation is to supply reactivepower locally. In the previous example this canbe done by adding a 16 MVAr capacitor at the load.

Compensated circuit is identical to first example with just real power load. Supply voltage magnitude and line current is lower with compensation.

20

Reactive Compensation, cont’d Reactive compensation decreased the line flow

from 564 Amps to 400 Amps. This has advantages: – Lines losses, which are equal to I2 R, decrease,– Lower current allows use of smaller wires, or

alternatively, supply more load over the same wires,– Voltage drop on the line is less.

Reactive compensation is used extensively throughout transmission and distribution systems.

Capacitors can be used to “correct” a load’s power factor to an arbitrary value.

21

Power Factor Correction Example

1

1desired

new cap

cap

Assume we have 100 kVA load with pf=0.8 lagging,and would like to correct the pf to 0.95 lagging

80 60 kVA cos 0.8 36.9

PF of 0.95 requires cos 0.95 18.280 (60 )

60 -ta

80

S j

S j Q

Q

cap

cap

n18.2 60 26.3 kVAr

33.7 kVAr

Q

Q

22

Distribution System Capacitors

23

Balanced 3 Phase () SystemsA balanced 3 phase () system has:

– three voltage sources with equal magnitude, but with an angle shift of 120,

– equal loads on each phase,– equal impedance on the lines connecting the

generators to the loads. Bulk power systems are almost exclusively 3.Single phase is used primarily only in low

voltage, low power settings, such as residential and some commercial.

Single phase transmission used for electric trains in Europe.

24

Balanced 3 -- Zero Neutral Current

* * * *

(1 0 1 1

3Note: means voltage at point with respect to point .

n a b c

n

an a bn b cn c an a

xy

I I I IVIZ

S V I V I V I V IV x y

25

Advantages of 3 Power

Can transmit more power for same amount of wire (twice as much as single phase).

Total torque produced by 3 machines is constant, so less vibration.

Three phase machines use less material for same power rating.

Three phase machines start more easily than single phase machines.

26

Three Phase - Wye ConnectionThere are two ways to connect 3 systems:

– Wye (Y), and– Delta ().

an

bn

cn

Wye Connection VoltagesV VV VV V

27

Wye Connection Line Voltages

Van

Vcn

Vbn

VabVca

Vbc

-Vbn

(1 1 120

3 30

3 90

3 150

ab an bn

bc

ca

V V V V

V

V V

V V

Line to linevoltages arealso balanced.

(α = 0 in this case)

28

Wye Connection, cont’dWe call the voltage across each element of a

wye connected device the “phase” voltage.We call the current through each element of a

wye connected device the “phase” current.Call the voltage across lines the “line-to-line” or

just the “line” voltage.Call the current through lines the “line” current.

6

*3

3 1 30 3

3

j

Line Phase Phase

Line Phase

Phase Phase

V V V eI I

S V I

29

Delta Connection

IcaIc

IabIbc

Ia

Ib

*3

For Delta connection, voltages across elementsequals line voltages

For currents

3

3

a ab ca

ab

b bc ab

c ca bc

Phase Phase

I I I

II I II I I

S V I

30

Three Phase Example

Assume a -connected load, with each leg Z = 10020 is supplied from a 3 13.8 kV (L-L) source

13.8 0 kV13.8 0 kV13.8 0 kV

ab

bc

ca

VVV

13.8 0 kV 138 20 amps

138 140 amps 138 0 amps

ab

bc ca

I

I I

31

Three Phase Example, cont’d

*

138 20 138 0239 50 amps239 170 amps 239 0 amps

3 3 13.8 0 kV 138 amps5.7 MVA5.37 1.95 MVA

pf cos 20 lagging

a ab ca

b c

ab ab

I I I

I I

S V I

j

32

Delta-Wye Transformation

Y

Linephase

To simplify analysis of balanced 3 systems:1) Δ-connected loads can be replaced by

1Y-connected loads with 3

2) Δ-connected sources can be replaced by

Y-connected sources with 3 30

Z Z

VV

33

Delta-Wye Transformation Proof

For the Suppose

side the two sides have identical te

we getrminal behavior

Hence

.

ab ca ab caa

ab ca

a

V V V VIZ Z Z

V VZI

34

+

-

Delta-Wye Transformation, cont’dFor the side we get

( ) ( )(2 )

Since 0Hence 3

3

1Therefore3

ab Y a b ca Y c a

ab ca Y a b c

a b c a b c

ab ca Y a

ab caY

a

Y

YV Z I I V Z I IV V Z I I I

I I I I I IV V Z IV VZ ZI

Z Z

35

Three Phase Transmission Line

36