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Chapter 9. Steady-state power analysis

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Contents

1. Instantaneous Power: For the special case of steady state sinusoidal signals

2. Average Power : Power absorbed or supplied during one cycle

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

4. Effective or RMS Values : For the case of sinusoidal signals

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

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

7. Complex Power : Measure of power using phasors

8. Single Phase Three-Wire Circuits : Typical distribution method for households and small

loads

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1. Instantaneous power

)()()( : Power ousInstantane tittp

)cos()(

)cos()(

:Statesteady In

iM

vM

tIti

tVtv

)cos()cos()( ivMM ttIVtp

)2cos()cos(2

)( ivivMM t

IVtp

0 iv 902

or

Net power transfer is positive!

Net power transfer is zero!

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tVtv M cos)(

Power vs. voltage

)2cos()cos(2

)( ivivMM t

IVtp

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2. Average power

TT

22

)2cos()cos(2

)( ivivMM tIV

tp

If voltage and current are in phase MMiv IVP2

1

If voltage and current are in quadrature 090 Piv

Purely inductive or capacitive

Tt

t

dttpT

P0

0

)(1

Tt 00t)cos(

2 ivMM IV

P

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Example 9.2Find the average power absorbed by impedance

)(1553.34522

6010

22

6010A

jI

WP 5.12)45cos(3.35

15,60,53.3,10 ivMM IV qq

[V]1506.71553.32

A][1553.3

IRV

I

[V]10506.79021553.3

A][1553.3

LjIV

I

0P

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Example 9.3Determine the average power absorbed by each resistor, the total average power absorbed and the average powersupplied by the source

)(4534

45121 AI

)(57.7136.537.265

4512

12

45122 A

jI

)(36.522

1 22 WP

57.7136.545321 III

)(10.6215.8 AI

)cos(2 ivMM IV

P

][74.46)10.6245cos(15.8122

1supplied WP

WP 183122

14

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83. Effective or rms Values

)(ti

R

2

20

0

0

0

)(1

)()(1

period with periodic iscurrent If

eff

Tt

t

Tt

t

RI

dttiT

RdttvtiT

P

T

)(tP

P

RIPR

tIti

M

M

2

2

1,

)cos()(

IV

2

then ))(( DC iscurrent If

dcdc

dc

RIP

Iti

22

1 22 MeffMeff

dcav

IIII

PP

Tt

t

eff dttiT

I0

0

)(1 2

Definition is valid for ANY periodicsignal with period T

root mean square

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)cos(2

1 :case sinusoidalFor ivMMav IVP

)cos( iveffeffav IVP

Compute rms value of the voltage waveform

tv 2

4T

Tt

t

rms dttT

V0

0

)(1 2

dttVrms 2

0

2)2(4

1)(

3

8

3

12

0

3 Vt

Example

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4. Complex power

)cos( ivrmsrms IVP ; Average power

irmsrmsVrmsrms IV IV ,

}Re{)cos()(

}Re{)cos()(i

v

jMiM

jMvM

eItIti

eVtVtv

*rmsrmsIVS

}Re{}Re{ *rmsrmsP IVS

; Complex power

; Average power

)( iVrmsrmsirmsVrms IVIVS

jQPS

)sin( ivrmsrms IVQ

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LR

CrmsSV ,

Example

CLjR

VI rmss

rms

1

,

C

LjR

CLR

VV

CLjR

VVVIS rmssrmssrmssrmss

1

11 2

2

*,,

*,,*

2

,2

2 1}Re{ rmssV

CLR

RSP

Evaluate the average power consumed by the following circuit.

rmsI

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125. Maximum average power transfer

THTHTH jXRZ

LLL jXRZ

THL

OCL

OCTHL

LL

ZZ

VI

VZZ

ZV

22

2

)()(

||

2

1

THLTHL

LOCL XXRR

RVP

}Re{2

1}Re{ *VIS P

LTHL

OC

THL

OCOC

THL

LLLL Z

ZZ

V

ZZ

VV

ZZ

Z2*

2

1

2

1

2

1

*IVS

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22

2

)()(

||

2

1

THLTHL

LOCL XXRR

RVP

THL

THL

L

L

L

L

RR

XX

R

PX

P

0

0

*TH

optL ZZ

TH

OCL R

VP

4

||

2

1 2max

Maximum average power transfer condition

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93.1647.164.908.6

57.2694.8

16

48

37

1652

37

)16)(48(

16

48)12(||4

j

j

jjj

j

jjZTH

43.041.193.1647.1* jZL

TH

OCL R

VP

4

||

2

1 2max*

THoptL ZZ

)(45.241.14

26.5

2

1 2max WPL

Example

64.926.5

64.908.6

03204

16

24

jVOC

Find ZL for maximum average power transfer. Compute the maxi-mum average power supplied to the load.

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6. Power factor

Instantaneous power in an ac circuit. Positive p represents power to the load; negative p represents power returned from the load.

Power to a purely resistive load. The peak value of p is VmIm.

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Power to a purely capacitive load. Aver-age power is zero.

Power to a purely inductive load. Energy stored during each quarter cycle is returned during the next quarter-cycle. Average power is zero.

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)cos()cos(2

1ivrmsrmsivMM IVIVP

power) (apparent :apparent rmsrms IVP

zivP

Ppf cos)cos(

apparent

pfIVP rmsrms

Power factor

☆ The reference phase angle is that of voltage.A leading or lagging phase angle is the state that the phase angle of current is leading or lagging behind that of voltage.( 전압의 위상을 기준으로 하여 전류의 위상이 lead 인지 lag 인지 결정됨 .)

V

e)(capacitiv

leadscurrent

090 z

)(inductive

lagscurrent

900 z

)( IVZ I

VZZ

Power factor Phase difference(θv - θi )

Q (Complex power)

leading - - Capacitive

in phase 0 0 Resistive

lagging + + Inductive

*)( VIjQPSS iv

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Transmission line

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Electric company전기 회사

House가정

R1

1k

+-

R1

1k

계량기Power meter

cos rmsrmsav IVP

Z

RIP rmsLoss2

전선에 의한 손실을 줄이기 위해서 가능한 전류를 적게 흘려야 한다 .To save power losses due to resistance of transmission lines, current should be kept as small as possible.

Transmission line loss

220V

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R1

1k

+-

R1

1k

AIL 1220

220

220W220V

1.0

1.0

WRIPLoss 2.02 2

R1

1k

+-

R1

1k

AIL 2110

220

220W110V

1.0

1.0

WRIPLoss 8.02 2

Example

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An industrial load consumes 88 kW at a pf of 0.707 lagging from a 480-V rms line. The transmission line resis -tance from the power company’s transformer to the plant is 0.08. Let us determine the power that must be sup-plied by the power company (a) under present conditions and (b) if the pf is somehow changed to 0.90 lagging. (It is economically advantageous to have a power factor as close to one as possible.)

Example 9.10

45707.0cos zz

pfIVP rmsrms

][3.259707.0480

1088 3

rmsrms

Lrms A

pfV

PI

kWIPP rmsLS 38.933.25908.0000,8808.0 22

(a)

(b)

][7.2039.0480

1088 3

rmsrms

Lrms A

pfV

PI

9.0pf

kWIPP rmsLS 32.917.20308.0000,8808.0 22

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23ExampleA load operates at 20 kW, 0.8 pf lagging. The load voltage is at 60 Hz. The impedance of the line is 0.09 + j0.3 . We wish to determine the voltage and power factor at the in-put to the line.

pfSpfIVP rmsrms

*)( VIjQPSS iv

)cos( ivrmsrmsIVP

)sin( ivrmsrms IVQ

kjkVIS 1520*

][87.3664.1132.689.90220

1520

0220

1520**

rmsAjkjkkjk

V

SI

][86.454.2491.2164.248220)2.689.90()3.009.0( rmsS VjjjV

73.41iv

lagging75.0)cos( ivpf

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247. Power factor correction

Simple approach to power factor correction

LI

LVIndustrial load withlagging pf LS

LP

LQ

LLL jQP S

IV

Electricalsource

2** )( VCjCVjVVIQ

LS

LP

LQ

IV

CQ

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87.36)8.0(cos)( 1IV

500,37000,50)75.01(50)tan(1* jjkjPjQPVIS IV

434,16000,50)329.01(50tan1 jjkjPS newnew

2.18)95.0(cos 1new

2066,21500,37434,16 VCjjjjQ

FC 1155

lagging 0.95 factor topower theincrease torequiredcapacitor theDetermine

.60Hzf Example

lagging8.0

0220,50

pf

VkW rmsL

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268. Single-phase three wire circuits

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R1

1k

+-

R1

1k

1kWW102 RIP rmsLoss 100V(rms)

05.0R

A10rmsI

05.0R

Disadvantage of single phase two wire system ( 단상 2선식의 단점 )

1. If the load increases to 2kW, Ploss becomes 40W. 2. If each wire can sustain heat only to 5W power loss, the cross-section area

should increase 4 times, whereby cost for copper wires increase 4 times.

100V(rms)

R1

1k

+- 2kWW402 RIP rmsLoss 100V(rms)

05.0R

A20rmsI

05.0R

100V(rms)

R1

1k

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R1

1k

+-

1kW100V(rms)

05.0R

05.0R+- 100V(rms) 1kW

R1

1k

R1

1k

A10rmsI

100V(rms)

100V(rms)

1. Although the power load increases to 2kW, Ploss remains 10W.

2. With 2 times power load, only one copper wire is added.

A0rmsI

Single-phase 3 wire circuits

A10rmsI

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30Example 9.27A light-duty commercial single-phase three-wire 60-Hz circuit serves lighting, heating, andmotor loads, as shown below. Lighting and heating loads are essentially pure resistanceand, hence, unity power factor (pf), whereas motor loads have lagging pf.

9.367.419.36240

10kVAIm

07.41120

5kII HL

1.124.1199.367.4107.4107.41mHLa IIII

04.8307.4107.41HLn III

9.367.41mb II

lagging9778.01.12cos

][314)1.124.119)(0120(*

pf

kVAjIVS aana

lagging8.0

][34)9.367.41)(0120(*

pf

kVAjIVS abnb

kWIIIP

R

nbalosses

lines

147.1||||||05.0

05.0222

Unbalanced connection

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4.1806.799.367.4107.41mLa III

4.1806.79ab II

0nI

lagging9487.04.18cos

][39)4.1806.79)(0120(*

pf

kVAjIVS aana

lagging9487.04.18cos

][39)4.1806.79)(0120(*

pf

kVAjIVS bbnb

kWIIP

R

balosses

lines

625.0||||05.0

05.022

)/$08.0(@366/$

522.0

kWhyear

kWPsaved

Balanced connection

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이행선

Safety considerations

Average effect of 60Hz current from hand tohand and passing the heart

Required voltage depends on contact, personand other factors

Typical residential circuit with ground andneutral

Ground conductor is not needed fornormal operation

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Grounding is needed to protect from lightning.

Grounding for safety considerations

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기기 고장에 의한 short

기기 고장에 의한 short

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Increased safety due to grounding

When switched on the tool case is energized

without the ground connector theuser can be exposed to the fullsupply voltage!

Conducting due to wet floor

If case is grounded then the supply is shorted and the fuse acts to openthe circuit

More detailed numbers in a related case study

Example

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Ground prong removed

R(dry skin) 15kOhmR(wet skin) 150OhmR(limb) 100OhmR(trunk) 200Ohm

Suggested resistancesfor human body

150150

Wet skin

400

1

mAIbody 171701

120

Can cause ventricular fibrillation