Lecture 2

Complex Power, Reactive Compensation, Three Phase

Dr. Youssef A. MobarakDepartment of Electrical Engineering

EE 351

POWER SYSTEM ANALYSIS

2

Announcements

For lectures 2 through 3 please be reading Chapters 1 and 2

3

Review of Phasors

Goal of phasor analysis is to simplify the analysis of constant frequency ac systems

v(t) = Vmax cos(wt + qv)

i(t) = Imax cos(wt + qI)

Root Mean Square (RMS) voltage of sinusoid

2 max

0

1( )

2

T Vv t dt

T

4

Phasor Representation

j

( )

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)

V

V

j t

j

v t V t

v t V e

V

5

Phasor Representation, cont’d

The RMS, cosine-referenced voltage phasor is:

( ) Re 2

cos sin

cos sin

V

V

jV

jj t

V V

I I

V V e V

v t Ve e

V V j V

I I j I

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

6

Advantages of Phasor Analysis

0

2 2

Resistor ( ) ( )

( )Inductor ( )

1 1Capacitor ( ) (0)

C

Z = Impedance

R = Resistance

X = Reactance

XZ = =arctan( )

t

v t Ri t V RI

di tv t L V j LI

dt

i t dt v V Ij C

R jX Z

R XR

Device Time Analysis Phasor

(Note: Z is a complex number but not a phasor)

7

RL Circuit Example

2 2

( ) 2 100cos( 30 )

60Hz

R 4 3

4 3 5 36.9

100 305 36.9

20 6.9 Amps

i(t) 20 2 cos( 6.9 )

V t t

f

X L

Z

VI

Z

t

8

Complex Power

max

max

max max

( ) ( ) ( )

v(t) = cos( )

(t) = cos( )

1cos cos [cos( ) cos( )]

21

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

V

I

V I

V I

p t v t i t

V t

i I t

p t V I

t

Power

9

Complex Power, cont’d

max max

0

max max

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

2

1( )

1cos( )

2cos( )

= =

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

Power Factor

Average

P

Angle

ower

10

Complex Power

*

cos( ) sin( )

P = Real Power (W, kW, MW)

Q = Reactive Power (var, kvar, Mvar)

S = Complex power (VA, kVA, MVA)

Power Factor (pf) = cos

If current leads voltage then pf is leading

If current

V I V I

V I

S V I j

P jQ

lags voltage then pf is lagging

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

11

Complex Power, cont’d

2

1

Relationships between real, reactive and complex power

cos

sin 1

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

-cos 0.85 31.8

1000.

P S

Q S S pf

S

kWS

117.6 kVA85

117.6sin( 31.8 ) 62.0 kVarQ

12

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 power follows since S = VI*

13

Conversation of Power Example

Earlier we found

I = 20-6.9 amps

*

*R

2R R

*L

2L L

100 30 20 6.9 2000 36.9 VA

36.9 pf = 0.8 lagging

S 4 20 6.9 20 6.9

P 1600 (Q 0)

S 3 20 6.9 20 6.9

Q 1200var (P 0)

R

L

S V I

V I

W I R

V I j

I X

14

Power Consumption in Devices

2Resistor Resistor

2Inductor Inductor L

2Capacitor Capacitor C

CapaCapacitor

Resistors only consume real power

P

Inductors only consume reactive power

Q

Capacitors only generate reactive power

1Q

Q

C

I R

I X

I X XC

V

2

citorC

C

(Note-some define X negative)X

15

Example

*

40000 0400 0 Amps

100 0

40000 0 (5 40) 400 0

42000 16000 44.9 20.8 kV

S 44.9k 20.8 400 0

17.98 20.8 MVA 16.8 6.4 MVA

VI

V j

j

V I

j

First solve

basic circuit

16

Example, cont’d

Now add additional

reactive power load

and resolve

70.7 0.7 lagging

564 45 Amps

59.7 13.6 kV

S 33.7 58.6 MVA 17.6 28.8 MVA

LoadZ pf

I

V

j

17

59.7 kV

17.6 MW

28.8 MVR

40.0 kV

16.0 MW

16.0 MVR

17.6 MW 16.0 MW-16.0 MVR 28.8 MVR

Power System Notation

Power system components are usually shown as

“one-line diagrams.” Previous circuit redrawn

C:\Program Files (x86)\PowerWorld\SimulatorGSO17\5th Ed. Book Cases\Chapter2\Problem2_32.pwb

Arrows are

used to

show loadsGenerators are

shown as circles

Transmission lines are shown as a single line

18

Reactive Compensation

44.94 kV

16.8 MW

6.4 MVR

40.0 kV

16.0 MW

16.0 MVR

16.8 MW 16.0 MW 0.0 MVR 6.4 MVR

16.0 MVR

Key idea of reactive compensation is to supply reactive

power locally. In the previous example this can

be done by adding a 16 Mvar capacitor at the load

Compensated circuit is identical to first example with

just real power load

19

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 utility to use small wires, or

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

Reactive compensation is used extensively by utilities

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

20

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 18.2

S 80 (60 Q )

60 - Qta

80

S j

j

cap

cap

n18.2 60 Q 26.3 kvar

Q 33.7 kvar

21

Distribution System Capacitors

22

Balanced 3 Phase () Systems

A 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

23

Balanced 3 -- No Neutral Current

* * * *

(1 0 1 1

3

n a b c

n

an an bn bn cn cn an an

I I I I

VI

Z

S V I V I V I V I

24

Advantages of 3 Power

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

Torque produced by 3 machines is constrant Three phase machines use less material for same

power rating Three phase machines start more easily than single

phase machines

25

Three Phase - Wye Connection

There are two ways to connect 3 systems– Wye (Y)– Delta ()

an

bn

cn

Wye Connection Voltages

V

V

V

V

V

V

26

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 line

voltages are

also balanced

(α = 0 in this case)

27

Wye Connection, cont’d

Define voltage/current across/through device to be phase voltage/current

Define voltage/current across/through lines to be line voltage/current

6

*3

3 1 30 3

3

j

Line Phase Phase

Line Phase

Phase Phase

V V V e

I I

S V I

28

Delta Connection

IcaIc

IabIbc

Ia

Ib

a

b

*3

For the Delta

phase voltages equal

line voltages

For currents

I

3

I

I

3

ab ca

ab

bc ab

c ca bc

Phase Phase

I I

I

I I

I I

S V I

29

Three Phase Example

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

13.8 0

13.8 0

13.8 0

ab

bc

ca

V kV

V kV

V kV

13.8 0138 20

138 140 138 0

ab

bc ca

kVI amps

I amps I amps

30

Three Phase Example, cont’d

*

138 20 138 0

239 50 amps

239 170 amps 239 0 amps

3 3 13.8 0 kV 138 amps

5.7 MVA

5.37 1.95 MVA

pf cos 20 lagging

a ab ca

b c

ab ab

I I I

I I

S V I

j

31

Delta-Wye Transformation

Y

phase

To simplify analysis of balanced 3 systems:

1) Δ-connected loads can be replaced by 1

Y-connected loads with Z3

2) Δ-connected sources can be replaced by

Y-connected sources with V3 30Line

Z

V

32

Delta-Wye Transformation Proof

From the side we get

Hence

ab ca ab caa

ab ca

a

V V V VI

Z Z Z

V VZ

I

33

Delta-Wye Transformation, cont’d

a

From the side we get

( ) ( )

(2 )

Since I 0

Hence 3

3

1Therefore

3

ab Y a b ca Y c a

ab ca Y a b c

b c a b c

ab ca Y a

ab caY

a

Y

Y

V Z I I V Z I I

V V Z I I I

I I I I I

V V Z I

V VZ Z

I

Z Z

34

Three Phase Transmission Line