ECE 3183 – Chapter 5 – Part D CHAPTER 5 - D1 ECE 3183 – EE Systems Chapter 5 – Part D AC Power, Power Factor

ECE 3183 – Chapter 5 – Part D CHAPTER 5 - D1

ECE 3183 – EE Systems

Chapter 5 – Part D

AC Power, Power Factor


EFFECTIVE OR RMS VALUES

)(ti

R Rtitp )()( 2

power ousInstantane

The effective value is the equivalent DCvalue that supplies the same average power



2

)(

dcdc

dc

RIP

Iti

then )( DC iscurrent If

dcaveff PPI :



Tt

t

Tt

tav dtti

TRdttp

TP

T

0

0

0

0

)(1

)(1 2

period with periodic iscurrent If

Tt

teff dtti

TI

0

0

)(1 22

Tt

teff dtti

TI

0

0

)(1 2

Definition is valid for ANY periodicsignal with period T



If the current is sinusoidal the average power is

RIP Mav2

2

1

22

2

1Meff II


LEARNING EXAMPLECompute the rms value of the voltage waveform

3T


LEARNING EXAMPLECompute the rms value of the voltage waveform


)(ti

R

LEARNING EXAMPLECompute the rms value of the voltage waveform and use it todetermine the average power supplied to the resistor

2R Ampstti 20000cos5)(


)(ti

R

LEARNING EXAMPLECompute the rms value of the voltage waveform and use it todetermine the average power supplied to the resistor

)(4 sT

2R


LEARNING EXTENSIONCompute rms value of the voltage waveform

tv 2

4T


)(ti

R

LEARNING EXTENSION

4R

Compute the rms value for the current waveforms and usethem to determine average power supplied to the resistor

6T

Tt

trms dttx

TX

0

0

)(1 2

RIP rmsav2

8T


)(ti

R

LEARNING EXTENSIONCompute the rms value for the current waveforms and usethem to determine average power supplied to the resistor

6T

Tt

trms dttx

TX

0

0

)(1 2

RIP rmsav2

4

2

6

4

2

0

2 41646

1dtdtdtIrms 8

6

8328

)(3248 WP

8T

816168

1 6

4

2

0

2

dtdtIrms )(32 WP

4R


STEADY-STATE POWER ANALYSISLEARNING GOALS

Instantaneous Power For the special case of steady state sinusoidal signals

Average Power Power absorbed or supplied during one cycle

Maximum Average Power Transfer When the circuit is in sinusoidal steady state



Effective or RMS Values For the case of sinusoidal signals

Power Factor A measure of the angle between current and voltage phasors

Power Factor Correction How to improve power transfer to a load by “aligning” phasors



Complex Power Measure of power using phasors

Single Phase Three-Wire Circuits Typical distribution method for households and small loads


INSTANTANEOUS POWER

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

Impedance to

SuppliedPower

ousInstantane

Voltsttv 60cos4)(

302Z

FIND i(t) and p(t)

)cos()cos()()()( ivmm ttIVtitvtp We know that:

)cos()cos(2

1)cos()cos( 212121

where

)2cos()cos(2

)( ivivmm tIV

tp

yielding:


INSTANTANEOUS POWER LEARNING EXAMPLE

)(),(

302

),60cos(4)(

tpti

Z

ttv

:Find

:Assume

)(302302

604A

Z

VI

))(30cos(2)( Atti

30,2

60,4

iM

vM

I

V

)902cos(430cos4)( ttp


INSTANTANEOUS POWER


AVERAGE POWER

For sinusoidal (and other periodic signals) we compute averages over one period

2T

)2cos()cos(2

)( ivivMM t

IVtp

)cos(2 iv

MM IVP DOES IT MATTER WHO LEADS?

Tt

t

dttpT

P0

0

)(1


AVERAGE POWER

If voltage and current are in phase

MMiv IVP2

1

If voltage and current are in quadrature

090 Piv

Purelyresistive

Purely inductive orcapacitive

(cos(0)=1)

(cos(90)=0)


AVERAGE POWER

LEARNING EXAMPLE

Find the average power absorbedby the impedances

)(1553.34522

6010

22

6010A

jI

_____,_____

_____,_____,

iv

MM IV

RV


AVERAGE POWER

LEARNING EXAMPLE

Find the average power absorbedby the impedances

RV

)(1506.7601022

2V

jVR

W48.1253.306.72

1 AVP

Since an inductor does not absorb power, one can use voltages and currents for the resistive part only.


LEARNING EXAMPLEDetermine the average power absorbed by each resistor, the total average power absorbed and the average powersupplied by the source


LEARNING EXTENSIONFind the AVERAGE power absorbed by each PASSIVE component and the total power supplied by the source

1I2I

3010243

241 j

jI

)(62.4014.6301095.1528.7

57.2647.41 AI

)(14.632

1

2

1 223 WRIP M

62.4014.630102I

)(05.1412.4

95.1528.7

30303010

243

32

A

jI

)(12.442

1 24 WP

)(02 WPj

Power supplied by source

Method 1. absorbedsupplied PP

WPPP 50.9043 supplied

Method 2: )cos(2

1ivMM IVP

sV

62.4042.183 1IVs

)3062.40cos(1042.182

1 P


LEARNING EXAMPLEDetermine average power absorbed or supplied by each element

)(3062

30122 AI

)(36622

1

2

1 222 WRIP M

01 jP

To determine power absorbed/suppliedby sources we need the currents I1, I3

)(19.3643.7

39.466639.10

1

0630123

A

jj

j

jI

07.728.11

)(39.12.1139.46320.5321 AjjjIII

)(54

)07.730cos(28.11122

13012

W

P

)1836(

)cos(2

1ivMM IVP

Power Average

WP 18)19.360cos(43.762

106

Passive sign convention

R

VRIP M

M

22

2

1

2

1

resistorsFor


LEARNING EXTENSIONDetermine average power absorbed/supplied by eachelement

V

1I

2I



V

1I

925.062.42

85.177.1424

2

0241 j

jVI

)(32.1171.41 AI

2I

VoltsjV 85.177.14



V

1I

2I

VoltsjV 85.177.14

j

j

j

j

j

VI

2

85.177.1412

2

0122

)(73.12367.1)(385.1925.02

77.285.12 AAj

jI



V

1I )(32.1171.41 AI 2I

VoltsjV 85.177.14

)(73.12367.12 AI

)(18.2271.422

1 22 WP

)(67.274

88.14

2

1 2

4 WP

)(565.5)73.1230cos(67.1122

1012 WP

)(42.55)32.110cos(71.4242

1024 WP

R

VRIP M

M

22

2

1

2

1

resistorsFor

)(42.55

)(565.567.2718.22

WP

WP

supplied

absorbed

:Check


THE POWER FACTOR

LZiMI

vMV

iv

VI

izv

IZVZIV

z

)cos()cos(2

1ivrmsrmsivMM IVIVP

rmsrms IVP apparent

zivP

Ppf cos)cos(

apparent

V

e)(capacitiv

leadscurrent 090 z

)(inductive

lagscurrent 900 zpfIVP rmsrms


THE POWER FACTOR

inductive purely

inductive or lagging

90

900

0

10

resistive01

capacitiveor leading09010

capacitive purely900

z

z

z

pf

pf

pf


LEARNING EXAMPLEFind the power supplied by the power company.Determine how it changes if the power factor is changed to 0.9

Power company



Power company

pfIVP rmsrms

rmsV )(480

rmsA)(453.259

Current lags the voltage

45707.0cos zz

To solve for z, take the arccos of p.f.

z = cos-1(0.707) = 45°. Which one to choose?

Since the load is lagging (inductive), we know 0 < z < 90°

So pick z = + 45°.



Power company

pfIVP rmsrms

rmsAW

Irms )(3.259707.0480

)(1088 3

rmsV )(480

rmsA)(453.259


)(378.93)(000,88

378.508.03.259 22

kWWPP

kWRIP

lossesS

rmslosses

If pf=0.9

kWRIP

rmsAI

rmslosses

rms

32.3

)(7.2039.0480

000,88

2

Losses can be reduced by 2kW!

45707.0cos zz



Power company

pfIVP rmsrms

rmsAW

Irms )(3.259707.0480

)(1088 3

rmsV )(480

rmsA)(453.259

45707.0cos zz


93.378(kW)88,000(W)PP

5.378kW0.08259.3RIP

lossesS

22rmslosses

480453.25908.0

08.0

LrmsS VIVrms

)(7.14957.147.494

480)4.1834.183(08.0

Vj

jVrmsS


LEARNING EXTENSION

1.0

707.0,)(480,100

line

LL

R

pfrmsVVkWP

Determine the power savings if the power factor can be increased to 0.94

pfIVP rmsrms

kW

WpfPlosses

34.42

)(707.0

1

480

1.010)707.0(

22

10

kW

WpfPlosses

34.413.1

)(94.0

1

480

1.010)94.0(

22

10

kWkWPsaved 77.334.487.0

22

22 1

pfV

RPRIP

pfV

PI

rms

linelinermslosses

rmsrms


COMPLEX POWER

**rmsrms IV

2

1IVS

PowerComplex of Definition

PowerReactive

PowerReal

PowerApparent

KVARimaginaryjQ

KWrealP

KVAcomplexS

jQPS


COMPLEX POWER

jQPS

IVjIVS

IVS

IVS

IVS

IVrmsrmsIVrmsrms

IVrmsrms

IrmsVrms

rmsrms

sincos

*

*

P=Active Power Q=Reactive Power


COMPLEX POWER

Let a load be ZL = R + jX.

Then Q is:•Positive if the load is inductive (X>0)•Negative if the load is capacitive (X<0)•Zero if the load is resistive (X=0)

inductive

capacitive

Documents

ECE 3183 – Chapter 5 – Part D CHAPTER 5 - D1 ECE 3183 – EE Systems Chapter 5 – Part D AC Power, Power Factor