Download - Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege

[email protected] • ENGR-43_Lec-09-2_Complex_Power.ppt1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 43

Chp 9 [5-7]Chp 9 [5-7]Complex Complex PowerPower



Outline – AC SS Power cont.Outline – AC SS Power cont. Effective or RMS Values

• Heating Value for Sinusoidal Signals

Power Factor• A Measure Of The Angle Between Current

And Voltage Phasors within a Load

Power Factor Correction• Improve Power Transfer To a Load By

“Aligning” Phasors

Single Phase Three-Wire Circuits• Typical HouseHold Power Distribution



Outline – AC Steady State PowerOutline – AC Steady State Power

Instantaneous Power Concept• For The Special Case Of Steady State

Sinusoidal Signals

Average Power Concept• Power Absorbed Or Supplied During in

Integer Number of Complete Cycles

Maximum Average Power Transfer• When The Circuit Is In Sinusoidal Steady

State



Power FactorPower Factor Consider A Complex

Current Thru a Complex Impedance Load

The Current and Load-Voltage Phasors (Vectors) Can Be Plotted on the Complex Plane

By Ohm & Euler

in the Electrical Power Industry θZ is the Power Factor Angle, or Simply the Phase Angle

LZiMI

vMV

iv

VIz

ivz

izv

or

IZV

ZIV



Power Factor contPower Factor cont The Phase Angle Can

Be Positive or Negative Depending on the Nature of the Load

Typical Industrial Case is the INDUCTIVE Load• Large Electric Motors are

Essentially Inductors

Now Recall The General Power Eqn

Measuring the Load with an AC DMM yields• Vrms

• Irms

Zrmsrmsivrmsrms

ZMMivMM

IVIVP

IVIVP

cos)cos(

cos2

1)cos(

2

1

0MVV

e)(capacitiv

leadscurrent 090 z

)(inductive

lagscurrent 900 z

ZZ

ZVI

0

V is the BaseLine



Power Factor cont.2Power Factor cont.2 The Product of the

DMM Measurements is the APPARENT Power

The Apparent Power is NOT the Actual Power, and is thus NOT stated in Watts.• Apparent Power

Units = VA or kVA

Now Define the Power Factor for the Load

Some Load Types

rmsrmsapparent IVP pfIVP

P

Ppf

rmsrmsactual

zivapparent

actual

and

cos)cos(

inductive pure

inductiveor lagging

resistive

capacitiveor leading

capacitive pure

90

900

0

10

01

09010

900

z

z

z

pf

pf

pf



pf – Why do We Care?pf – Why do We Care? Consider this case

• Vrms = 460 V

• Irms = 200A• pf = 1.5%

Then• Papparent = 92kVA

• Pactual =1.4 kW

This Load requires The Same Power as a Hair Dryer

However, Despite the low power levels, The WIRES and CIRCUIT BREAKERS that feed this small Load must be Sized for 200A!• The Wires would be

nearly an INCH in Diameter



Example Example Power Factor Power Factor The Local Power

Company Services this Large Industrial Load

Find Irms by Pwr Factor

Then the I2R Loses in the 100 mΩ line

Improving the pf to 94%

1.0

kW100080V4

Power company

rmsrms

rmsrms

VpfPI

pfIVP

22

22 1

pfV

RPRIP

rms

linelinermslosses

234.4

)(707.0

1

480

1.010)707.0(

22

10

kW

WpfPlosses

13.134.4

)(94.0

1

480

1.010)94.0(

22

10

kW

WpfPlosses

kWkWPsaved 77.334.487.0

I lags V



Example - Power Factor contExample - Power Factor cont For This Ckt The

Effect of the Power Factor on Line Losses

1.0

kW100080V4



Complex PowerComplex Power Consider a general Ckt

with an Impedance Ld Mathematically

For this Situation Define the Complex Power for the Load:

Converting to Rectangular Notation

*rmsrmsIVS

Ziv

ivrmsrms

irmsvrms

irmsvrms

IV

IV

IV

:recall

*

S

S

S

)sin()cos( ivrmsrmsivrmsrms IVjIVS

PActive Power

Q

Reactive Power



Complex Power contComplex Power cont Thus S in Shorthand Alternatively,

Reconsider the General Sinusoidal Circuit

S & Q are NOT Actual Power, and Thus all Terms are given Non-Watt Units• S→ Volt-Amps (VA)

• Q → Volt-Amps, Reactive (VAR)

P is Actual Power and hence has Units of W

First: U vs. Urms

jQP S

rmsrms

Mrms

M

U

UU

U

U

U

2and



Complex Power cont.2Complex Power cont.2 Now in the General Ckt

By Ohm’s Law

In the Last Expression Equate the REAL and Imaginary Parts

And Again by Ohm

iviv

rmsrms

ivrms

rms

irms

vrms

rmsrms

jZZ

j

soZIV

I

V

I

V

sincos

ImRe

and

ZZ

Z

Z

IVZ

Z

Z

iv

iv

Z

Z

Imsin

Recos

ZVI

Z

V

Z

V

rmsrms

irms

iv

vrmsrms

rmsrms

So

I

ZVI



Complex Power cont.3Complex Power cont.3 And by Complex Power

Definition

Using the Previous Results for P

Similarly for Q

ivrmsrms

ivrmsrms

IVQ

IVP

jQP

sinIm

cosRe

then

S

S

S

RIIP

IZ

V

ZIVP

rmsrms

rmsrms

rmsrms

22 Re

Re

ReRe

Z

Z

ZS

XIIQ rmsrms22 Im Z

So Finally the Alternative Expression for S

ZS 2rmsI



Complex Power TriangleComplex Power Triangle The Expressions for S

Plotting S in the Complex Plane

From The Complex Power “Triangle” Observe

ZS

S2rmsI

jQP

P

Qiv tan

Note also That Complex Power is CONSERVED

kkrmsktot I ZSS 2,



Example - Complex PowerExample - Complex Power For the Circuit At Right

• Zline =0.09 Ω + j0.3 Ω

• Pload = 20 kW

• Vload = 2200°

• pf = 80%, lagging

• f = 60 Hz → ω = 377s-1

Lagging pf → Inductive

From the Actual Power

Thusinductive

capacitive

pfSP iv )cos(||Re SS

kVAkW

pf

PSL 25

8.0

20

And Q from Pwr Triangle

kVAR15222 QPSQ L



Example - Complex Power contExample - Complex Power cont Then SL

Recall the S Mathematical Definition

Alternatively

87.36251520 kVAkVAjLS

*LLL IVS

Note also that [U*]* = U In the S Definition,

Isolating the Load Current and then Conjugating Both Sides

)(86.3664.113

0220

87.3625**

A

V

kVA

L

L

LL

I

V

SI

)(18.6891.90

220

000,15000,20*

Aj

j

L

L

I

I

Lagging



Example - Complex Pwr cont.2Example - Complex Pwr cont.2 Now Determine VS

Then VS Then The Phase Angle

To find the Src Power Factor, Draw the I & V Phasor Diagram

)(220)18.6891.90)(3.009.0(

0220)3.009.0(

Vjj

j

S

LS

LlineS

V

IV

VVV

86.453.249

14.2163.248

rmsS

S

V

j

V

V

86.4

86.36

SV

LI

7464.0

72.41coscos

also and

Load Inductive I Leads V

72.4186.3686.4

pf

pf iv

iv



Example - Complex Power kVARExample - Complex Power kVAR For the Circuit At

Right, Determine• Real And Reactive

Power losses in the Ln

• Real And Reactive Power at the Source

Lagging pf → Inductive

From the Actual Power

Thusinductive

capacitive

kVAkW

pf

PSL 62.47

84.0

40

And by S Definition

laggingpf

kW

84.0

40

1.0 25.0j

pfSP iv )cos(||Re SS

rmsL

LL A

V

SI )(45.216* VIS



Example - Complex kVAR cont.Example - Complex kVAR cont. Also from the S

Relation

Now the PowerFactor Angle Then for Line Loses

Quantitatively

laggingpf

kW

84.0

40

1.0 25.0j

)(839,25|||| 22 VARPLL SQ

86.3284.0acosiv

pf = cos(θv − θi); hence

2

** )(

Lline

LLlinelinelineline

IZ

IIZIVS

VAj

j

line

line

117134685

)45.216)(25.01.0( 2

S

S

I Lagging V



Example - Complex kVAR cont.2Example - Complex kVAR cont.2 Find Power Supplied

by Conservation of Complex Power

Then to Summarize the Answer• Pline = 4.685 kW

• Qline = 11.713 kVAR

• PS = 44.685 kW

• QS = 37.552 kVAR

laggingpf

kW

84.0

40

1.0 25.0j

kVA

kVAj

j

jjSup

04.4037.58

552.37685.44

839.25713.1140685.4

839.2540713.11685.4S

LoadlineSupplied SSS

In this Case



Power Factor CorrectionPower Factor Correction As Noted Earlier, Most

Industrial Electrical Power Loads are Inductive• The Inductive

Component is Typically Associated with Motors

The Motor-Related Lagging Power Factor Can Result in Large Line Losses

The Line-Losses can Be Reduced by Power Factor Correction

To Arrive at the Power Factor Correction Strategy Consider A Schematic of a typical Industrial Load



Power Factor Correction cont.Power Factor Correction cont. Prior to The Addition of

the Capacitor

For The Capacitive Load

)cos(

||

oldold

oldoldoldoldold

pf

SjQPS

22

2

C

Im

90

LL

CCC

LC

LLLC

L

CVC

CVIQ

CVI

CVCj

Z

VZ

VI

After Addition of the Capacitor

)cos(

||

newnew

newnew

Coldold

Coldnew

pf

jQjQP

S

SSS



Power Factor Correction cont.2Power Factor Correction cont.2 Find θnew

• Cap is a Purely REACTIVE Load

The Vector Plot Below Shows Power Factor Correction Strategy

old

new

old

Coldnewtan

P

Q

P

QQ

Use Trig ID to find QC to give desired θnew

2tan1

1cos

1cos

1

new2

old

Cold

P

QQ

QL

QC

QL-QC

P

Qnew



Trig ID DigressionTrig ID Digression

Start with the ID

2tan1

1cos

Solve for tan

old

Coldnewtan

P

QQ

Recall tanθnew

1cos

1

new2

old

Cold

P

QQ

Or

new

new2

new2

new2

new2

new2

new2

old

Cold

cos

cos1

cos

cos1

cos

cos

cos

1

P

QQ

But: cosθnew = pfnew

new

new

pf

pf 2

new

1tan

1cos

1tan

22

Substituting



Example – pf CorrectionExample – pf Correction Kayak Centrifugal

Injection-Molding Power Analysis• Improve Power Factor

to 95%

Find Sold

Now Qold

Adding A Cap DoesNOT Change P• Use Trig ID to Find Tan(new)

And by S Relation

laggingpf

VkW rmsL

8.0

0220,50

Roto-molding

process

pfSSP iv )cos(Re S

kVAkW

pf

PSold 5.62

80.0

50

)(5.37|||| 22 kVARPoldold SQ

329.01

tan95.0cos2

new

newnewnewnew pf

pf

P

Q

kVARPQP

Qnew

new 43.16329.0



Example – pf Correction contExample – pf Correction cont Then the Needed QC

Recall The Expression for QC

laggingpf

VkW rmsL

8.0

0220,50

Roto-molding

process

kVAQ

QQQ

C

newoldC

07.21

43.165.37

CVQ LL 2CC |||| IV

Then C from QC

FFC

V

QC

L

C

1155)(001155.0

)220()602(

1007.212

3

2



WhiteBoardWhiteBoard Work Work

Let’s Work (w/ 2-changes)

Problem 9.81• Determine at the input

SOURCE – Voltage & Current

– Complex Power

– Δθ & pf

19V220 rms

60 kVA

Download - Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege

Top Related