Chapter 7 AC Power Analysis Chapter Objectives: Know the difference between instantaneous power and...

Chapter 7Chapter 7AC Power AnalysisAC Power Analysis

Chapter Objectives: Know the difference between instantaneous power and average

power *Learn the AC version of maximum power transfer theorem Learn about the concepts of effective or rms value Learn about the *complex power, apparent power and power

factor Understand the principle of conservation of AC power Learn about power factor correction

Instantenous AC Power

( ) cos( ) ( ) cos( )

( ) ( )11

cos( ) cos )( (22

2) m m m vm

m v m

v i

i

i

v t V t i t I t

p t v t ti V I V It

Instantenous Power p(t) is the power at any instant of time.

Instantenous AC Power Instantenous Power p(t) is the power at any instant of time.

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

1c( ) ( ) ( )

1cos( )

2os(2 )

2m m v m m v iip t v t i t V I tV I

The instantaneous power is composed of two parts.

• A constant part.

• The part which is a function of time.

Assume a sinusoidal voltage with phase , ( ) cos( )

Assume a sinus

1cos( )

2CONSTANT POWER

oidal current with phase , ( ) cos( )

( ) ( ) ( )1

cos(2 )2

SINUSO A( ) + ID L

v m v

i

m m vv i im

m i

m

v t V t

i t I t

p t v t i Vt

p

I I t

t

V

(frequen POW cyER 2 )

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Trigonometri identityTrigonometri identity

Instantenous and Average Power The instantaneous power p(t) is composed of a constant part (DC) and a time dependent part having frequency 2ω.

Instantenous Power p(t)

12

Average Po we

c )

r

os(m m v iP V I

1co

( ) ( ) ( )

( ) c

1cos(2 )

2

os( ) ( ) cos( )

( s2

) ( )

m v m i

m m vv i im mV I

p t v t i t

v t V t i t I t

I tp t V

Instantenous and Average Power 1 1

1 22 2( ) cos( ) cos(2 ) ( ) ( )m m v i m m v ip t V I V I t p t p t

Average PowerThe average power P is the average of the instantaneous power over one period .

0

1( ) Average

( ) ( ) (

Powe

) Instantaneous Power

( ) cos

r

( ) ( ) cos( )m v

T

m i

p t v t i t

v t V t i

P p t

tT

t

dt

I

1 12 20 0 0

1 1 1( ) cos( ) cos(2 )

T T T

m m v i m m v iP p t dt V I dt V I t dtT T T

12

12

cos( )

1Re cos( )

2

m m v i

m m v i

P V I

P V I

VI

1 12 20 0

12

1 1cos( ) cos(2 )

= co (Integral of a Sinusoidal=0)s( ) 0

T T

m m v i m m v i

m m v i

P V I dt V I t dtT T

V I

Average PowerThe average power P, is the average of the instantaneous power over one period .

12

12

cos( )

1Re cos( )

2

m m v i

m m v i

P V I

P V I

VI

A resistor has (θv-θi)=0º so the average power becomes:

221 1 12 2 2R m m mP V I I R R I

1. P is not time dependent.

2. When θv = θi , it is a purely resistive load case.

3. When θv– θi = ±90o, it is a purely reactive load case.

4. P = 0 means that the circuit absorbs no average power.

Instantenous and Average Power

1cos( )

2385.7

1cos(2 )

2 600cos(20t 10 )

P= 385.7 W is the average power

(

)

low

=

f

W

m mm iv vi mVI Ip V tt

Example 1 Calculate the instantaneous power and average power absorbed by a passive linear network if:

)60 10( sin 15 )(

)20 10( cos80 )(

tti

ttv

Average Power Problem Practice Problem 11.4: Calculate the average power absorbed by each of the five elements in the circuit given.

Average Power Problem

Maximum Average Power Transfer

a) Circuit with a load b) Thevenin Equivalent circuit

Finding the maximum average power which can be transferred from a linear circuit to a Load connected.

• Represent the circuit to the left of the load by its Thevenin equiv.• Load ZL represents any element that is absorbing the power generated by the circuit.• Find the load ZL that will absorb the Maximum Average Power from the circuit to which it is connected.

Maximum Average Power Transfer Condition

• Write the expression for average power associated with ZL: P(ZL). ZTh = RTh + jXTh ZL = RL + jXL

2

2

2 2

2

22 2

2

L

2

L

2

2

Ajust R and X to get maximum P

1 2( ) ( ) 2 ( ) ( )

( )

( ) ( )

( ) ( ) 2 ( )

2 ( )

LThTh Th

LTh L Th Th L L Th L Th L

Th L Th L

L Th L Th L

Th Th L Th L L Th L

L Th L

RVV VI P I R

Z Z R jX R jX R R X X

V R X XP

X R R X X

V R R X X R R RP

R R R

22

2 2

0 0 (

( )

)

L Th L Th Th L Th

Th L

L L

L L L Th Th Th

P PX X R R X X

R

X X

RX

Z R jX R jX Z

Maximum Average Power Transfer Condition

For Maximum average power transfer to a load impedance ZL we must choose ZL as the complex conjugate of the Thevenin impedance ZTh.

2

max 8

L L L Th Th Th

Th

Th

Z R jX R jX Z

VP

R

• Therefore: ZL = RTh - XTh = ZTh will generate the maximum power transfer.• Maximum power Pmax

22

max 2 8ThL L

Th

VI RP

R

*Maximum Average Power Transfer Practice Problem 11.5: Calculate the load impedance for maximum power transfer and the maximum average power.

Maximum Average Power Transfer

Maximum Average Power for Resistive Load

When the load is PURELY RESISTIVE, the condition for maximum power transfer is:

Now the maximum power can not be obtained from the Pmax formula given before.

Maximum power can be calculated by finding the power of RL when XL=0.

2 2 2 20 ( ) L L Th Th L Th Th ThX R R X X R X Z

●

●

RESISTIVE LOAD


Practice Problem 11.6: Calculate the resistive load needed for maximum power transfer and the maximum average power.


Notice the way that the maximum power is calculated using the Thevenin Equivalent circuit.

RL

a) AC a) AC circuitcircuit

2 2

0 0

1 1( ) ( )

T T

eff Rms eff RmsI I i t dt V V v t dtT T

Effective or RMS Value The EFFECTIVE Value or the Root Mean Square (RMS) value of a periodic current is the DC value that delivers the same average power to a resistor as the periodic current.

b) DC circuitb) DC circuit

2 2 2 2

0 0

1( ) ( )

T T

eff Rms

RP i t Rdt i t dt I R I R

T T

Effective or RMS Value of a Sinusoidal The Root Mean Square (RMS) value of a sinusoidal voltage or current is equal

to the maximum value divided by square root of 2.

22 2

0 0

1 1cos (1 cos 2 )

2 2

T Tm m

Rms m

I II I tdt t dt

T T

12 cos( ) cos( )m m v i Rms Rms v iP V I V I

The average power for resistive loads using the (RMS) value is:

22 Rms

R Rms

VP I R

R

Effective or RMS Value Practice Problem 11.7: Find the RMS value of the current waveform. Calculate the average power if the current is applied to a 9 resistor.

4 0 1( )

8 4 1 2

t ti t

t t

A309.23

16rmsI

W48)9(3

162

RIP rms

1 22 2 2 2

0 0 1

1 1(4 ) (8 4 )

2

T

rmsI i dt t dt t dtT

1 22 2 2

0 1

16 (4 4 )2rmsI t dt t t dt

3

2 2 21

1 168 4 2

3 3 3rms

tI t t

2T 4t

8-4t

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An Electical An Electical PowerPower Distribution CenterDistribution Center

Apparent Power and Power Factor


1

2 m m Rms RmsS V I V I

Apparent Power

cos

Power Factor

( )v i

Ppf

S pS

P f

The Apparent Power is the product of the Rms value of voltage and current. It is measured in Volt amperes (VA).

The Power Factor (pf) is the cosine of the phase difference between voltage and current. It is also the cosine of the angle of load impedance. The power factor may also be regarded as the ratio of the real power dissipated to the apparent power of the load.

The Average Power depends on the Rms value of voltage and current and the phase angle between them.

Apparent Power and Power Factor Not all the apparent power is consumed if the circuit is partly reactive.

Purely resistive load (R)

θv– θi = 0, Pf = 1 P/S = 1, all power are consumed

Purely reactive load (L or C)

θv– θi = ±90o,

pf = 0

P = 0, no real power consumption

Resistive and reactive load (R and L/C)

θv– θi > 0

θv– θi < 0

• Lagging - inductive load• Leading - capacitive loadP/S < 1, Part of the apparent

power is consumed

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Power equipment are rated using their appparent power in KVA.

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Apparent Power and Power

Factor

Both have same P

Apparent Powers and pf’s are different

Generator of the second load is overloaded

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Apparent Power and Power Factor

Overloading of the generator of the second load is avoided by applying power factor correction.

Complex Power The COMPLEX Power S contains all the information pertaining to the power absorbed by a given load.

221

2Rms

Rms

VI Z

Z

Rms RmsS VI V I

( )

cos( )

Re{ } Real Powe

sin( )

Im{ } Reactive Power+ r

Rms Rms v i

Rms Rms Rms Rms v i

Rms v Rm

v

s i

i jV I

V I

V I

P j

I

Q

V

j

Rms RmsI

S

S

S

V

The REAL Power is the only useful power delivered to the load.

The REACTIVE Power represents the energy exchange between the source and reactive part of the load. It is being transferred back and forth between the load and the source

The unit of Q is volt-ampere reactive (VAR)

2 2 ( )Rms Rms jXI Z R P QI j S

Re{ }

=Real Power

Im{ }

+Reactive Power

jQP j S S S

2

2= cos( ) Re

= sin( )

{

Im

}

{ }

Rms Rms v i Rm

Rms

s

Rms v i RmsQ

P V I I

V I

R

I X

S

S

Complex Power

Resistive Circuit and Real Power

1si

0 RESISTIVE

n( )sin(2 )2

sin( )sin(2 )

1cos( ) 1 cos(2 )

2cos( ) 1 cos(2 )

( ) sin( ) ( ) sin( )

( ) ( ) ( )

cos(2 )

m m

Rms Rms

Rms R

m m

Rms Rms

Rms Rm

m

ms

m

s

V I t

V I

v t V t i t I t

p V I tt v t

V I t

VV I

t

I

i t

t

p t( ) is always Positive

1cos( ) 1 cos(

1sin( )sin(2 )

2 sin( )sin(2 )

sin(

( ) sin( ) ( ) sin( )

(

90 INDUCTIVE

2 )2

cos(

) (

) 1 cos(2 )

) ( )

2

)

m m

Rms

m m

Rms Rms

Rms Rms

ms

m

L

R

m

V

v t V t i t I t

p t v I t

V I t

V

t V I t

V I t

i t

I t

( ) is equally both positive and negative, power is circula ing tLp t

Inductive Circuit and Reactive Power

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If the average power is zero, and the energy supplied is returned within one cycle, why is a reactive power of any significance?

At every instant of time along the power curve that the curve is above the axis (positive), energy must be supplied to the inductor, even though it will be returned during the negative portion of the cycle. This power requirement during the positive portion of the cycle requires that the generating plant provide this energy during that interval, even though this power is not dissipated but simply “borrowed.”

The increased power demand during these intervals is a cost factor that must that must be passed on to the industrial consumer.

Most larger users of electrical energy pay for the apparent power demand rather than the watts dissipated since the volt-amperes used are sensitive to the reactive power requirement.

The closer the power factor of an industrial consumer is to 1, the more efficient is the plant’s operation since it is limiting its use of “borrowed” power.

Inductive Circuit and Reactive Power

1cos( ) 1 cos(2 )

1sin( )sin(2 )

2 sin( )sin(2 )

si 9

( ) sin( ) ( ) sin( )

(

n(

) ( ) ( )

0 CAPACITIVE

2cos(

) 1 cos(2 )

2

)

m m

Rms

m m

Rms Rms

Rms Rms

Rms

m m

C V I t

V I t

V I t

v t V t i t I t

p t v t V I t

V I t

i t

( ) is equally both positive and negative, power is circulatingCp t

Capacitive Circuit and Reactive Power

Complex Power The COMPLEX Power contains all the information pertaining to the power absorbed by a given load.

2 2

Complex Power

Apparent Power

Real Pow

1

er

Reactive

= ( )2

=

= Re{ } cos( )

=Q Im{ } s Power

Power Fa

in( )

= =cctor os( )

Rms Rms v i

Rms Rms

v i

v i

v i

P jQ V I

S V I P Q

P S

S

P

S

S VI

S

S

S

• Real Power is the actual power dissipated by the load.

• Reactive Power is a measure of the energy exchange between source and reactive part of the load.

Power Triangle

a) Power Triangle b) Impedance Triangle Power Triangle

The COMPLEX Power is represented by the POWER TRIANGLE similar to IMPEDANCE TRIANGLE. Power triangle has four items: P, Q, S and θ.

0 Resistive Loads (Unity )

0 Capacitive Loads (Leading )

0 Inductive Loads (Lagging )

Q Pf

Q Pf

Q Pf

T

100 200 300 600 Watt

0 700 1500 800Var

S 600 800 1000 53.13

T

T

P

Q

j

Power Triangle Finding the total COMPLEX Power of the three loads.

41Eeng 224

1 2 1 2 1 2( ) ( )S P jQ S S P P j Q Q

Power Triangle

Real and Reactive Power Formulation

P is the REAL AVERAGE POWER

Q is the maximum value of the circulating power flowing back and forward

cos sinrms rms rms rmsP V I Q V I

Real and Reactive Power Formulation

( ) cos( ) ( ) cos( )

( ) cos( ) sin( )

=

=Real Power R eactive

sin 2( )

si

1 cos 2(

n

P

)

1 cos 2(

owe

2 )

r

()

m v m i

Rms Rms v i Rms Rms v i

v

v

v

v

v t V t i t I t

p t V I V I

P Q

t

t

t

t

Real and Reactive Powers

CIRCULATING POWER

REAL POWER

47Eeng 224


• Vrms =100 V Irms =1 A Apparent power = Vrms Irms =100 VA• From p(t) curve, check that power flows from the supply into the load for the entire duration of the cycle!• Also, the average power delivered to the load is 100 W. No Reactive power.

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• Vrms =100 V Irms =1 A Apparent power = Vrms Irms =100 VA• From p(t) curve, power flows from the supply into the load for only a part of the cycle! For a portion of the cycle, power actually flows back to the source from the load!• Also, the average power delivered to the load is 50 W! So, the useful power is less than in Case 1! There is reactive power in the circuit.

Power Flowing Back

Practice Problem 11.13: The 60 resistor absorbs 240 Watt of average power. Calculate V and the complex power of each branch. What is the total complex power?

Practice Problem 11.14: Two loads are connected in parallel. Load 1 has 2 kW, pf=0.75 leading and Load 2 has 4 kW, pf=0.95 lagging. Calculate the pf of two loads and the complex power supplied by the source.

LOAD 12 kWPf=0.75Leading

LOAD 24 kWPf=0.95Lagging

Conservation of AC Power The complex, real and reactive power of the sources equal the respective sum of the complex, real and reactive power of the individual loads.

a) Loads in Parallel b) Loads in Series

For parallel connection:

* * * * *1 2 1 2 1 2

1 1 1 1S V I V (I I ) V I V I S S

2 2 2 2

Same results can be obtained for a series connection.

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Complex power is Conserved

Power Factor Correction The design of any power transmission system is very sensitive to the magnitude of

the current in the lines as determined by the applied loads.

Increased currents result in increased power losses (by a squared factor since P = I2R) in the transmission lines due to the resistance of the lines.

Heavier currents also require larger conductors, increasing the amount of copper needed for the system, and they require increased generating capacities by the utility company.

Since the line voltage of a transmission system is fixed, the apparent power is directly related to the current level.

In turn, the smaller the net apparent power, the smaller the current drawn from the supply. Minimum current is therefore drawn from a supply when S = P and QT = 0.

The process of introducing reactive elements to bring the power factor closer to unity is called power-factor correction. Since most loads are inductive, the process normally involves introducing elements with capacitive terminal characteristics having the sole purpose of improving the power factor.

Power Factor Correction

Original Inductive Load Inductive Load with improved power factor correction

Effect of capacitor on total current Power triangle of power factor correction

Increasing the power factor without altering the voltage or current to the load is called Power Factor Correction


Qc = Q1 – Q2

= P (tan θ1 - tan θ2)

= ωCVrms2

P = S1 cos θ1 Q2 = P tan θ2

c 1 22 2rms rms

Q P (tan θ tan θ )C

ωV ω V

Q1 = S1 sin θ1 = P tan θ1

Increasing the power factor without altering the voltage or current to the load is called Power Factor Correction.

Power Factor Correction The process of increasing the power factor without altering the voltage or current to the original load is called power factor correction.

Power factor correction is necessary for economic reasons.

1 2

1 1 1 1 1 1 2 2

Real power stays same

cos sin tan ta

n

P P P

P S Q S P Q P

1 2 1 2(tan tan )CQ Q Q P

1 22 2

(tan tan )C

rms rms

Q PC

V V

2rms

L

VL

Q

• The capacitance value needed to change the pf angle from 1 to 2 .

• Similarly the inductance value needed to change the pf angle from 1 to 2 for a capacitive load.

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Power Factor

Correction

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Power Factor

Correction

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Practice Problem 11.15: Find the value of the capacitance needed to correct a load of 140 kVAR at 0.85 lagging pf to unity pf. The load is supplied by a 110 Volt (rms), 60 Hz line.

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Wattmeter is the instrument for measuring the average Wattmeter is the instrument for measuring the average power. Two coils are used, the high impedance power. Two coils are used, the high impedance Voltage coilVoltage coil and and the low impedance the low impedance Current coilCurrent coil.. Wattmeter measures the average power given by: Wattmeter measures the average power given by:

Applications: Power Measurement

Wattmeter Wattmeter connected to the load


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Problem 11-74

Chapter 7 AC Power Analysis Chapter Objectives: Know the difference between instantaneous power and...

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