Chapter 10curt.nelson/engr228/lecture/chap10.pdf · Engr228 -Chapter 10, Nilsson 11e 1 Chapter 10 Sinusoidal Steady-State Power Engr228 Circuit Analysis Dr Curtis Nelson Chapter 10

Engr228 - Chapter 10, Nilsson 11e 1

Chapter 10Sinusoidal Steady-State

Power

Engr228

Circuit Analysis

Dr Curtis Nelson

Chapter 10 Objectives

• Understand the following sinusoidal steady-state power concepts:• Instantaneous power;• Average power;• Root Mean Squared (RMS) and effective values;• Reactive power;• Complex power;• Power factor;• Power factor correction;• Power generation;• Maximum real power delivery to a load.


Instantaneous Power

RvRiviP2

2 ===

• Instantaneous power is the power measured at any given instant in time.

• In DC circuits, power is measured in watts as:

• In AC circuits, voltage and current are time-varying, soinstantaneous power is time-varying. Power is still measured in watts as:

RvRiviP2

2 ===

Instantaneous Sinusoidal Steady-State Power

)t(i)t(v)t(p ´=

)cos()( vm tVtv qw +=

• In an AC circuit, voltage and current are expressed in general form as:

)cos()( im tIti qw +=

• Instantaneous power is then:

)cos()cos()( ivmm ttIVtp qwqw ++=


Instantaneous Power - Continued

• Using the trig identities:

it can be shown that instantaneous power is:

)cos()cos()( ivmm ttIVtp qwqw ++=

• Note that the instantaneous power contains a constant term as well as a component that varies with time at twice the input frequency.

cos $ cos % = 0.5cos $ − % + cos($ + %)cos $ + % = cos($)cos(%) − sin($)sin(%)

cos(2t+5V − 5I) =cos(5V − 5I)cos(2t) − sin(5V − 5I)sin(2t)

( ) cos( ) cos( )cos(2 ) sin( )sin(2 )2 2 2 m m mv v vm m mi i i

I I Ip t t V tV Vq q q q wq wq= - + - - -

Instantaneous Power Plot( ) cos( ) cos( )cos(2 ) sin( )sin(2 )2 2 2 m m mv v vm m m

i i iI I Ip t t V tV Vq q q q wq wq= - + - - -


Instantaneous vs. Average Power

The equation above can be simplified as follows:

p(t) = P + Pcos(2t) – Qsin(2t)

where

" = $%&%' cos(,- − ,/) (Average Power)

Q = $%&%' sin(,- − ,/) (Reactive Power)



Instantaneous vs. Average Power

• Average power is sometimes called real power because it describes the power in a circuit that is transformed from electric to nonelectric energy, such as heat. Average power is also the average value of the instantaneous power over one period:


Average Power - Continued

The last two terms in the equation above will integrate to zero since the average value of sin and cosine signals over one period is zero. Therefore:

)cos(2 ivmm

AVGIVP qq -=



Example - Power Calculations

Calculate the instantaneous and average power if:

Pinst (t) = 459.6 + 600cos(20t + 80º)WPavg = 459.6W

v(t) = 80cos(10t + 20º)Vi(t) = 15cos(10t + 60º)A


Example - Average Power

Calculate the average power absorbed by the resistor and inductor and find the average power supplied by the voltage source.

PR = 9.6W absorbedPL = 0WPSource = 9.6W delivered

The Power Scene for Resistors

• Since the voltage and current are in phase across a resistor, the instantaneous power becomes:

p(t) = P + Pcos(2t)

• Integrating over one period shows again that average power in a resistor is:

" = $%&%'

• Note that average power in a resistor is always positive meaning that power cannot be extracted from a purely resistive network.




The Power Scene for Inductors

• For inductors, !" − !$ = +90º or the current lags the voltage by 90º. The instantaneous power then reduces to:

p = −Qsin2t

• Note that Q is always a positive quantity for inductors. The average power absorbed by an inductor is zero, meaning energy is not transformed from electric to non-electric form. Instead, the instantaneous power is continually exchanged between the circuit and the source at a frequency of 2. When p is positive, energy is stored in the magnetic field of the inductor. When p is negative, energy is transferred back to the source.



The Power Scene for Capacitors

• For capacitors, !" − !$ = −90º or the current leads the voltage by 90º. The instantaneous power then reduces to (just like inductors):

p = −Qsin2t

• Note that Q is always a negative quantity for capacitors. The average power absorbed by a capacitor is also zero, meaning energy is not transformed from electric to non-electric form. Instead, the instantaneous power is continually exchanged between the circuit and the source at a frequency of 2. When p is positive, energy is stored in the electric field of the capacitor.




Power - Units

• Power in resistors is called average, or real power. Power in capacitors and inductors is called reactive power, recognizing that their impedances are purely reactive. To distinguish between these different types of power, we use different units – average power is measured in watts (W) and reactive power is measured in volt-amp-reactive (VAR).

Example 10.1 – Nilsson 11th

1. Calculate the average and reactive power if:

2. State whether the circuit is absorbing or delivering average and reactive powers.

Pavg = -100WQreac = 173.21VARDelivering average power and absorbing reactive (magnetizing vars) power

v(t) = 100cos(t + 15º)Vi(t) = 4sin(t - 15º)A


Effective or RMS Values

• Sometimes using average AC power values can be confusing. For instance, the average DC power absorbed by a resistor is P = VM IM while the average AC power is P = VM IM /2. By introducing a new quantity called effective value, the formulas for the average power absorbed by a resistor can be made the same for dc, sinusoidal, or any general periodic waveform.

• The effective value of a periodic voltage is the DC voltage that delivers the same average power to a resistor as the periodic AC voltage.

Effective or RMS Value

dtxT1

XT

0

2rms ò=

For any periodic function x(t), the effective or root mean squared (rms) value, is given by:


Effective or RMS Value

dttcosIT1

IT

0

22mrms ò w=

2I

dt)t2cos1(21

TI

I mT

0

2m

rms =w+= ò

• For example, the effective value of I = Imcost is

• Average power can then be written in two ways:

)cos( ivrmsrmsAVG IVP qq -=

)cos(2 ivmm

AVGIVP qq -=

Effective Values of Common Waveforms


a) Find the rms value of the periodic voltage shown in the figure below.

b) If the voltage is applied across a 40Ω resistor, find the average power dissipated by the resistor.

Vrms = 28.28VRMS

P = 20W

Textbook Problem 10.13 - Nilsson 11th

• θV - θi is known as the Power Factor Angle.• cos(θV - θi) is known as the Power Factor (PF)

• For a purely resistive load, PF=1;• For a purely reactive load, PF=0;• For any practical circuit, 0 ≤ PF ≤ 1.

• We must distinguish between positive and negative arguments for the power factor since cos(x) = cos(-x).

The Angle θV - θi


Leading and Lagging Power Factor

• pf is lagging if the current lags the voltage (θV - θI is positive like in an inductive load);

• pf is leading if the current leads the voltage (θV - θI isnegative like in a capacitive load).

Power Factor - Example

• Calculate the power factor of each circuit below if the applied voltage is V = 10cos(2500 t) V

pfind = 0.707 laggingpfcap = 0.371 leading


Complex Power and Impedance Triangles

Q

P

SX

R

IZI

Power triangle Impedance triangle

qq

ZR

SPpf === qcos

The Power Factor is also defined as:

Complex Power - Triangle

Complex power contains both a real part (average power) and an imaginary part (reactive power) that together are represented in a power triangle. In equation form:

S = P + jQ


Complex Power - Units

Complex Power – The Equations

PAVG = VRMSIRMS cos(θV - θi) watts

PREACTIVE = VRMSIRMS sin(θV - θi) volt-amps-reactive

PAPPARENT = VRMSIRMS volt-amps


Apparent Power and Power Factor

Power factor can also be defined as:

!" = $%&'$(& )*+&'$))$'&,- )*+&' =

!./012345234

Example - Power Factor

Calculate the power factor of the circuit below as seen by the source. What is the average power supplied by the source?

pf = 0.936 (lagging)PAVG = 118W


Example Problem

Find the average power delivered to each of the two loads, the apparent power supplied by the source, and the power factor of the combined loads.

Pavg2 = 288 WPavg1 = 144 WPapp = S = 720 VAPF = 0.6 (lagging)

For ig = 50 + j0 mA (rms), find the average and reactive power for the current source and state whether it is absorbing or delivering this power.

PAVG = -50.0 mW (delivering)PREAC = 25.0 mVAR (absorbing)



Find the average power, the reactive power, and the apparent power absorbed by the load if vg = 150cos250t V.

PAVG = 180 WPREAC = 90 VARPAPP = 180 + j90 VA


Alternative Forms for Complex Power

S = P + jQ= VrmsIrms∠"V - "I= Vrms ∠"V Irms∠-"I= VrmsI*rms


Complex Power - Summary

Real (Average) power:

)cos(S)SRe(P iv q-q== Watts

Reactive power:)sin(S)SIm(Q iv q-q== VAR

Apparent power:22

rmsrms QPIVSS +=== VA

• Q = 0 for resistive loads (unity power factor);• Q < 0 for capacitive loads (leading power factor);• Q > 0 for inductive loads (lagging power factor).

Complex power:

RMSRMS IVjQPS *=+= VA

Example - Apparent Power and Power Factor

• A series connected RC load draws a current of

when the applied voltage is

• Find the apparent power and the power factor of the loadand determine the element values that form the load.

Atti )10100cos(4)( °+= p

Vttv )20100cos(120)( °-= p

S = 240 Vpf = 0.866 (leading)C = 212.2µFR = 26.0Ω


Power Factor Correction• The process of increasing the power factor without

altering the voltage or current to the original load is known as power factor correction.

• Most loads are inductive. The power factor of the load can be improved (to make it closer to unity) by installing a capacitor in parallel with an inductive load.

a) Original inductive load b) Inductive load with improved power factor

Power Factor Correction Capacitors


Power Factor Correction Capacitors

Power Factor Correction by Power Triangle

The triangle below shows that the reactive power (Q1) in a large inductive load can be reduced by an amount (QC) by adding a capacitor in parallel to the inductive load.

Q1 = Ptan θ1

Q2 = Ptan θ2

QC = Q1 – Q2

= P(tan θ1 – tan θ2)


Power Factor Correction by Power Triangle

2rms

212rms

C

V)tan(tanP

VQC

wq-q

=w

=

2rms

C

2rms

C CVXVQ w==

• The value of required parallel capacitance is then:

• But QC is also the amount of reactive power the capacitor must provide:

Example - PF Correction

Find the value of parallel capacitance needed to correct a load of 140kVAR at 0.85 lagging pf to unity pf. Assume that the load is supplied by a 110VRMS, 60Hz line.

C = 30,691µF

2rms

212rms

C

V)tan(tanP

VQC

wq-q

=w

=


A group of small appliances on a 60 Hz system requires 20 kVA at 0.85 pf lagging when operated at 125 Vrms. The impedance of the feeder supplying the appliances is 0.01+ j0.08 Ω. Find:

a) The rms value of the voltage at the source end of the feeder.

b) The average power loss in the feeder.

c) What size capacitor at the load end of the feeder is needed to improve the load power factor to unity?

VS = 133.48 VRMS

PREAC = 256 WC = 1188.7 µF


Hydro-Power Generation

Solar Power

Nuclear Power Generation

Fossil Fuel Generation

Wave Power Generation

Gravity Light Project

Gravity Light Project – Version 2

Power Generation

https://www.youtube.com/watch?v=ABv631t1OKI

https://www.youtube.com/watch?v=xKxrkht7CpY

https://www.youtube.com/watch?v=MGj_aJz7cTs

https://www.youtube.com/watch?v=SeXG8K5_UvU

https://www.youtube.com/watch?v=iJUM4pCi71Q

https://www.youtube.com/watch?v=iJUM4pCi71Q

https://www.youtube.com/watch?v=GisQYoYw05k


Maximum Power Transfer

An independent voltage source in series with an impedance Zthdelivers a maximum average power to a load impedance ZL when ZL is the conjugate of Zth.

ZL = Zth*

Maximum Average Power Transfer

Th

RMSTH

RVP4

,2

max =

• When ZL = Z*TH, the maximum power delivered to the load

is:

• If the load is purely resistive, then the value of RL that maximizes the power is the magnitude of the Thevenin impedance.

• Note that the 4 factor in the denominator above becomes 8 if the voltage is expressed as an average or peak value.


Example: Maximum Average Power

Determine the load impedance ZL that maximizes the average power drawn from the circuit below. Calculate the maximum average power.

W-== 4667.49333.2* jZZ ThL

WP 3675.2max =

Chapter 10 Summary

• Understand the following sinusoidal steady-state power concepts:• Instantaneous power;• Average power;• Root Mean Squared (RMS) and effective values;• Reactive power;• Complex power;• Power factor;• Power factor correction;• Power generation;• Maximum real power delivery to a load.

Documents

Chapter 10curt.nelson/engr228/lecture/chap10.pdf · Engr228 -Chapter 10, Nilsson 11e 1 Chapter 10 Sinusoidal Steady-State Power Engr228 Circuit Analysis Dr Curtis Nelson Chapter 10