ECE 333 Renewable Energy Systems

Lecture 3:Basic Circuits, Complex Power

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

overbye@illinois.edu

Announcements

• Be reading Chapters 1 and 2 from the book• Be reading Chapter 3 from the book• Homework 1 is 1.1, 1.11, 2.6, 2.8, 2.14. It will be

covered by the first in-class quiz on Thursday Jan 29• As mentioned in lecture 2, your two lowest

quiz/homework scores will be dropped

2

Engineering Insight: Modeling

• Engineers use models to represent the systems we study

• Guiding motto: “All models are wrong but some are useful” George Box, 1979

• The engineering challenge, which can be quite difficult sometimes, is to know the limits of the underlying models.

3

• Ideal Voltage Source

• Ideal Current Source

Basic Electric Circuits

sv

+

-

i

+

-

si

Load

Load

si

i

i

v

sv

v

4

Example – Power to Incandescent Lamp

• Find R if the lamp draws 60W at 12 V

• Find the current, I• What is P if vs doubles and R stays the same? 240W

12sv V+

-

i

Load

125

2.4v

i AR

2vP v i

R

60P W

2 2122.4

60v

RP

5

Equivalent Resistance for Resistors in Series and Parallel

• Resistors in series – voltage divides, current is the same

v

+

-

1R

2R

NR

i

1 2EQ NR R R R

+

-

v

i

nodevoltages

6

Equivalent Resistance for Resistors in Series and Parallel

• Resistors in parallel – current divides, voltage is the same

1 2

11 1 1

...EQ

N

R

R R R

Simplification for 2 resistors

1 2

1 2EQ

R RR

R R

+

-

i

i

1R 2R NRv

branch currents

v

7

Voltage and Current Dividers

i1R

2R

+

-

v+

-outv

1 2EQ

v vi

R R R

2outv i R

2

1 2out

Rv v

R R

i

1R2R

v+

-2i1i 2

2

vi

R

1 2

1 2EQ

R Rv i R i

R R

12

1 2

Ri i

R R

Voltage Divider

Current Divider

8

Wire Resistance

• For dc systems wire resistance is key; for high voltage ac often the inductance (reactance) or capacitance (susceptance) are limiting

• Resistance causes 1) losses (i2R) and 2) voltage drop (vi)• Need to consider wire resistance in both directions

9

AC: Phase Angles

• Angles need to be measured with respect to a reference - depends on where we define t=0

• When comparing signals, we define t=0 once and measure every other signal with respect to that reference

• Choice of reference is arbitrary – the relative phase shift is what matters

• Relative phase shift between signals is independent of where we define t=0

10

Example: Phase Angle Reference

• Pick the bottom wave as the reference

1 sin4

v V t

2 sin 0v V t

1 sin 0v V t

2 sin4

v V t

1 2 4

1 2 4

• Or pick the top as the reference- it does not matter!

11

Important Properties: RMS

• RMS = root of the mean of the square• RMS for a periodic waveform

• RMS for a sinusoid (derive this for homework)

21( )

o

o

t T

RMSt

V v tT

let ( ) cos( )pv t V t

2p

RMS

VV

T period

12

In 333 we are mostly only concerned withsinusoidals

Important Properties:Instantaneous Power

• Instantaneous power into a load

p(t)= ( ) ( )v t i t( )v t

+

-

( )i t

( )= cos( )

( )= cos( )

p V

p I

v t V t

i t I t

( )= ( ) ( )p t v t i t

( )= cos cos 22p p

V I V I

V Ip t t

“Load sign convention” with current and power into load

positive

1cos cos cos cos

2

Identity

13

Important Properties: Average Power

• Average power is found from

• Find the average power into the load (derive this for homework)

( )= cos cos 22p p

V I V I

V Ip t t

1( )

o

o

t T

t

P p t dtT

T period

P= cos2p p

V I

V I P= cosRMS RMS V IV I or

14

Important Properties:Real Power

• P is called the Real Power

• cos(θV-θI) is called the Power Factor (pf)

• We’ll review phasors and then come back to these definitions…

P= cosRMS RMS V IV I

P= Re{VI*}

15

Review of Phasors

• Phasors are used in electrical engineering (power systems) to represent sinusoids of the same frequency

• A quick derivation…

2 f ( ) cos( )pA t A t

1cos( )

2jx jxx e e

cos( )2

j t j tAA t e e

Identity

Ap denotes the peak value of A(t)

16

Review of Phasors

• Use Euler’s Identity

• Written in phasor notation as

cos sinjxe x j x Identity

( ) cos( )

( ) Re

p

j t jp

A t A t

A t A e e

cos Re jxx e

or jRMS RMSA A e A A Tilde denotes a phasor

or jA A e A A Other, simplified notation

Regardless of what notation you use, it helps to be consistent.

Note, a convention- the amplitude used here is the RMS value, not the peak value as used in some other classes!

17

Why Phasors?

• Simplifies calculations– Turns derivatives and integrals into algebraic equations

– Makes it easier to solve AC circuits

dA j A

dt

R( )

R i (t)= Rv tR

L( )

L (t)= Ldi tv L

dt

C( )

C (t)= Cdv ti C

dt

=V

RI

=Lj IV Vj L

I

I=Cj V 1VI j C

LjX j L

1cjX j

C

18

Why Phasors: RLC Circuit

R j L

1j C

R L

C

( ) cosv t V t V V

1( ) ( ) ( )

div t Ri t L i t dt

dt C

I( )i t

1V RI j LI I

j C

Solve for the current- which circuit do you prefer?

+

-

+

-

19

RLC Circuit Example

( ) 2 100cos 30v t t

3LX L

2 f

60Hzf

2 24 3 5Z 1 3

tan 36.94Z

100 30

20 6.95 36.9

VI

Z

( ) 2 20cos( 6.9 )i t t 20

Complex Power

VV= RMSV

II= RMSI

Asterisk denotes complex conjugate

*

*

VI

VI cos sin

RMS RMS V I

RMS RMS V I RMS RMS V I

V I

V I jV I

S

Apparent

power

P

Real Power

Q

Reactive Power S = P+jQ

SQ

P(θV-θI)

Power triangle

21

Apparent, Real, Reactive Power

• P = real power (W, kW, MW)• Q = reactive power (VAr, kVAr, MVAr)• S = apparent power (VA, kVA, MVA)• Power factor angle• Power factor

*

*

*

VI

VI

VI cos sin

RMS RMS V I

RMS RMS V I RMS RMS V I

S P jQ

V I

V I jV I

V I cos( )pf

22

Apparent, Real, Reactive Power

• Remember ELI the ICE man

ELI ICEInductive loads

I lags V (or E)

Capacitive loads

I leads V (or E)

S Q

P(θV-θI) P

QS(θV-θI)

Q and θ positive Q and θ negative

(producing Q)

“Load sign convention” – current and power into load are assumed positive

23

Apparent, Real, Reactive Power

• Relationships between P, Q, and S can be derived from the power triangle just introduced

• Example: A load draws 100 kW with leading pf of 0.85. What are the power factor angle, Q, and S?

cos

sin

P S

Q S

-1cos 0.85 31.8

100 kW117.6 kVA

0.85Q=117.6 kVA sin( 31.8 ) 62.0 kVAr

S

24

Conservation of Power

• Kirchhoff’s voltage and current laws (KVL and KCL)– Sum of voltage drops around a loop must be zero– Sum of currents into a node must be zero

• Conservation of power follows – Sum of real power into every node must equal zero– Sum of reactive power into every node must equal zero

25

Conservation of Power Example

* 100 30 20 6.9 2000 36.9S VI 36.9 0.8 laggingpf

Resistor, consumed power

Inductor, consumed power

* 4 20 6.9 20 6.9 1600 R R RS V I 2 1600 WR RP I R

* 3 20 6.9 20 6.9 1200 L L LS V I j j 2 1200 VArL L LQ I X

0 VArRQ

0 WRP 26

Power Consumption in Devices

• Resistors only consume real power

• Inductors only consume reactive power

• Capacitors only produce reactive power

2R RP I R

2L L LQ I X LX L

2C C CQ I X

1CX

C

2R

RV

PR

2L

LL

VQ

X

2C

CC

VQ

X

27

Example

40000 0400 0 Amps

100 0I

40000 0 5 40 400 0

42000 16000 44.9 20.8 kV

V j

V j

* 44.9 20.8 400 0

17.98 20.8 MVA 16.8 6.4 MVA

S VI

S j

Solve for the total power delivered by the source

28

Reactive Power Compensation

• Reactive compensation is used extensively by utilities

• Capacitors are used to correct the power factor• This allows reactive power to be supplied locally• Supplying reactive power locally leads to decreased

line current, which results in– Decrease line losses– Ability to use smaller wires– Less voltage drop across the line

29

Power Factor Correction Example

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

S 80 60 kVAj 1cos (0.8) 36.9 We have:

We want: 1cos (0.95) 18.2desired

S Qdes.=?

P=8018.2

.tan(18.2 ) desQP

. tan(18.2 )*40 26.3 kVArdesQ This requires a capacitance of:

60 26.3 33.7 kVArcapQ Q=60 Q=-33.7

PQdes=26.3

P

30

Distribution System Capacitors for Power Factor Correction

31