Chapter 10 – ACSS Power

Chapter 10 – ACSS Power

Objectives:

•Power concepts: instantaneous power,

average power, reactive power, complex

power, power factor

•Relationships among power concepts – the

power triangle

•Balancing power in AC circuits

•Condition for maximum power transfer to a

load in AC circuits

Chapter 10 – ACSS Power Instantaneous power:

tIV

tIVIV

tItV

titvtp

ivmm

ivmm

ivmm

imvm

2sin)sin(2

2cos)cos(2

)cos(2

......

)cos()][cos([

)()()(

ry)trigonomet of(lots

Notes:

•The first term is a constant – instantaneous

power is not symmetric about the time axis!

•The second and third terms have a

frequency that is twice the frequency of v(t)

and i(t)!

Chapter 10 – ACSS Power Instantaneous power

Chapter 10 – ACSS Power Instantaneous power, rewritten:

)sin(2

)cos(2

2sin2cos)(

ivmm

ivmm

IVQ

IVP

tQtPPtp

where

Notes:

•P is the average (or real) power and has the

units watts [W]

•Q is the reactive power and has the units

volt-ampere-reactive [var]

Tt

tdttp

TP

0

0

)(1


Look at the expressions for P (real power) and Q

(reactive power) for the three different circuit

components – R, L, and C. The expressions

depend on the difference between the phase

angle of the voltage drop across the component

and the phase angle of the current through the

component.

A. Dependent on the source frequency

B. Dependent on the resistance

C. Equal to the voltage phase angle

D. The negative of the voltage phase angle

Chapter 10 – ACSS Power Remember – for a resistor in the presence of

an AC source,

iiv

imvm IRV

R

0

))(0(

IV

The voltage and current for a resistor are in-

phase! Thus, for a resistor,

tPPtp

IVQ

IVIVP iv

mmmmiv

mm

iviv

2cos)(

0)sin(2

;2

)cos(2

0)sin(;1)0cos()cos(

A. The same as the voltage phase angle

B. 90o more than the voltage phase angle

C. 90o less than the voltage phase angle

D. Dependent on the inductance

Chapter 10 – ACSS Power How can you remember the relationship

between the voltage and current phase angles

for inductors and capacitors?

ELI the ICE man!

Huh?

•In the early days of electrical engineering,

voltage was known as “electromotive

force”, and was represented by the symbol

“e” in equations!

•Current is represented by the letter “i”, as

usual.

•L is inductance, C is capacitance.


ELI the ICE man!

ELI – voltage is “ahead of” (i.e. LEADS) current

in an inductor, by 90o.

ICE – voltage is “behind” (i.e. LAGS) current in

a capacitor, by 90o.

Where did the 90o come from? Remember

Ohm’s law for phasors:

iv

imvm

iv

imvm

ICVCj

ILVLj

90

))(901()(

90

))(90(

IV

IV




D. Dependent on the inductance

ELI the ICE man!


The voltage leads the current by 90o in an

inductor (ELI!) Therefore,

tQtp

IVIVQ

IVP mm

ivmm

ivmm

iviv

2sin)(

2)sin(

2;0)cos(

2

1)90sin()sin(;0)90cos()cos(




D. Dependent on the capacitance

ELI the ICE man!


The voltage lags the current by 90o in a

capacitor (ICE!) Therefore,

tQtp

IVIVQ

IVP mm

ivmm

ivmm

iviv

2sin)(

2)sin(

2;0)cos(

2

1)90sin()sin(;0)90cos()cos(


Summary:

•Resistors P > 0, Q = 0

Resistors absorb real power and have

no reactive power

•Inductors P = 0, Q > 0

Inductors absorb reactive power and

have no real power

•Capacitors P = 0, Q < 0

Capacitors generate reactive power

and have no real power

Chapter 10 – ACSS Power Power factor (of a single component or a collection

of components):

This is a term that appears in the definition of

average power.

•When pf = 1, the component is purely resistive.

•When pf = 0, the component is purely reactive.

•To distinguish between inductive and

capacitive reactance, we use the modifiers

“leading” and “lagging”:

•When the pf is leading, the current leads the

voltage; when pf is lagging, the current lags

the voltage.

1 pf 0),cos(pf iv

A. A capacitor

B. An inductor

C. A capacitor and a resistor

D. An inductor and a resistor

ELI the ICE man!

Chapter 10 – ACSS Power Example (AP 10.1)

V )45(100cos v(t)

A )15cos(20)(

t

tti

Find the average power, the reactive power, and

the power factor.

5.0)1545cos()cos(pf

var866)1545sin(2

)100)(20()sin(

2Q

W 500)1545cos(2

)100)(20()cos(

2P

iv

ivmm

ivmm

IV

IV

A. Leading

B. Lagging

C. Can’t tell from the information

V )45(100cos v(t)

A )15cos(20)(

t

tti

A. Generating P and generating Q

B. Generating P and absorbing Q

C. Absorbing P and generating Q

D. Absorbing P and absorbing Q

Chapter 10 – ACSS Power Example (AP 10.1), continued

V )45(100cos v(t)

A )165cos(20)(

t

tti

Find the average power, the reactive power, and

the power factor.

leading 866.0)16545cos()cos(pf

var500)16545sin(2

)100)(20()sin(

2Q

W 866)16545cos(2

)100)(20()cos(

2P

iv

ivmm

ivmm

IV

IV

A. A resistor and a capacitor

B. A resistor and an inductor

C. None of the above

Warning – trick question!

Chapter 10 – ACSS Power Example (AP 10.1), continued

V )45(100cos v(t)

A )165cos(20)(

t

tti

The circuit in the box is generating both

average and reactive power.

•Capacitors generate reactive power,

•Only sources generate average power,

•The pf < 1, so there must also be a resistor.

Thus, the simplest circuit in the box has a

source, a resistor, and a capacitor!

leading 866.0pf var,500QW, 866P

Chapter 10 – ACSS Power Complex power:

Notes:

•|S| is apparent power

•pf is the power factor angle

•Units for both complex power and apparent

power are volt-amperes [VA]

The power triangle brings everything together!

VA |S|QPS pfj

Q

P

|S|

pf pfcospf

A. Purely resistive B. RL C. RC D. Purely capacitive

Q

P

|S|

pf


The complex power in an AC circuit balances!

0Qand0P0S

Example: Find the phasor

voltage and current for

the load, and show that

the power in the circuit

balances.

A 36.87-5A )34(2639

13234

2639

V 3.18-234.36V )13234()0250(3040

2639

jj

j

j

jj

j

LL

L

VI

V






balances.

var325)1](2)5)(265([Q;0P

var50)1](2)5)(45([Q;0P

0Q;W 5.487)1](2)5)(395([P

0Q;W 25)1](2)5)(15([P

var375)87.360sin(]2)5)(250([Q

W 500)87.360cos(]2)5)(250([P

A 36.87-5V, 3.18-234.36

1j26

1j4

139

11

S

S

LL IV






balances.

Component P[W] Q[var] S[VA]

Source 500 375 500j375

1 12.5 0 12.5+j0

j4 0 50 0+j50

39 487.5 0 487.5+j0

j26 0 325 0+j325

Total 0 0 0

Chapter 10 – ACSS Power Review:

•Instantaneous power

•Average (real) power

•Reactive power

•Power factor

(leading, i(t) leads v(t), RC; lagging, i(t) lags

v(t), RL)

1 pf 0),cos(pf iv

[W] 2sin2cos)()()( tQtPPtitvtp

)0( [W] )cos(2

CLivmm PP

IVP

)0( [var] )sin(2

Rivmm Q

IVQ

Chapter 10 – ACSS Power Review, continued:

•Complex power

•Apparent power

•The complex power in an AC circuit balances!

•The power triangle shows how the different

types of power are related:

[VA] || pfSjQPS

[VA] || 22 QPS

0and0or0 QPS

Q

P

|S|

pf pfcospf

A. Purely resistive

B. RL

C. RC

D. Can’t tell from the triangle

Q

P

|S|

pf pfcospf

A. Purely resistive

B. RL

C. RC

D. Purely capacitive

P=|S|

Chapter 10 – ACSS Power An aside, using an example:

Find an expression for the

average power delivered to

the resistor in the circuit

shown here.

R

VdttV

TR

dtR

tv

Tdttp

TP

rmsTt

tvm

Tt

t

Tt

t

222

2

0

0

0

0

0

0

)(cos11

)(1)(

1

The rms value of a periodic function, say f(t)

with period T, is the root of the mean of the

square of that function:

Tt

trms dttf

TF

0

0

)(1 2

Chapter 10 – ACSS Power RMS, continued:

Find the rms value of a cosine waveform.

What use is the rms value of voltage or current in

ac circuits?

2math) of (lots )(cos

1

)cos()(

0

0

22 mTt

tvmrms

vm

VdttV

TV

tVtv

The rms value of any periodic voltage or current delivers the same average power to a resistor as a dc voltage with the same value – rms values allow us to compare the effect of various periodic voltages to the effect of a dc voltage, so are sometimes called effective values!

A. 120 V

B. Greater than 120 V

C. Less than 120 V

Chapter 10 – ACSS Power RMS, continued:

Often, we use rms values in the equations for

real and reactive power:

)sin()sin(22

)sin(2

pfpf22

pf2

ivrmsrmsivmm

ivmm

rmsrmsmmmm

IVIVIV

Q

IVIVIV

P

You must pay attention to the statement of problems involving power calculations – you pick the correct equations for P and Q depending on whether the magnitude of the voltage (or current) is given, or the rms value of the voltage (or current) is given!

Chapter 10 – ACSS Power Example 10.4

A load requires a voltage of 240 Vrms, absorbs

8kW and has a power factor of 0.8 lagging.

Find the complex power of the load and the

load impedance.

We can find the complex power of the load by

drawing the power triangle:

A. 8 kW of power

B. Power factor is 0.8

C. Power factor is lagging

Chapter 10 – ACSS Power Example 10.4, continued




load impedance.

Q

8000 W

|S|

pf

VA 000,10

60008000||

var6000

87.36tan8000

87.368.0cospfcos

2222

-11

QPS

Q

Q

pf

VA 87.36000,10VA 60008000 jS





load impedance.

•How can we find the load impedance?

Start from its definition.

46.361.487.3676.576.567.41

240||

A 67.41)8.0)(240(

8000

pfpf

? ||about but what V 240||

87.36||

||

||

||)(

||

||

rms

jZI

VZ

V

PIIVP

I

Z

L

rms

rmsL

rms

rmsrmsrms

rmsLrmsL

L

Lpf

L

Liv

L

L

L

LL

IV

I

V

I

V

I

V

I

V

A. An inductor and a capacitor

B. A resistor and a capacitor

C. An inductor and a resistor





load impedance.

Q

8000 W

|S|

pf

VA 000,10

60008000||

var6000

87.36tan8000

87.368.0cospfcos

2222

-11

QPS

Q

Q

pf

VA 87.36000,10VA 60008000 jS





load impedance.

•How can we find the load impedance?

Start from its definition.

46.361.487.3676.576.567.41

240||

A 67.41)8.0)(240(

8000

pfpf

? ||about but what V 240||

87.36||

||

||

||)(

||

||

rms

jZI

VZ

V

PIIVP

I

Z

L

rms

rmsL

rms

rmsrmsrms

rmsLrmsL

L

Lpf

L

Liv

L

L

L

LL

IV

I

V

I

V

I

V

I

V

A. An inductor and a capacitor

B. A resistor and a capacitor

C. An inductor and a resistor


Two industrial loads each require 2500 V(rms)

at 60 Hz. Load 1 absorbs 8 kW at 0.8 leading;

load 2 absorbs 20 kVA at 0.6 lagging. The line

that transmits the power from the source to the

load has an impedance of (0.05 + j0.5) . To

begin to analyze the circuit described, draw a

block diagram.

A. In series

B. In parallel


Two loads each require 2500 V(rms) at 60 Hz.

Load 1 absorbs 8 kW at 0.8 leading; load 2

absorbs 20 kVA at 0.6 lagging. The line that

transmits the power from the source to the load

has an impedance of (0.05 + j0.5) . To begin

to analyze the circuit described, draw a block

diagram.


Find the power supplied by the source.

lagging 6.0at kVA 20:

leading 8.0at kW 8:

2

1

L

L

-6 kvar

8 kW

-36.87o

Find the total power required for the loads.

+ 53.13o

12 kW

16 kvar

20 kVA

= 26.57o

20 kW

10 kvar

22.36 kVA

A. True

B. False


Find the power supplied by the source.

VA 57.2636.22 loadS

Find the power “lost” in the line. How, since we

don’t know the source voltage, so can’t find the

voltage drop across the line impedance? Find the

line current, which is the same as the load current.

A(rms) 443.89)89.0)(250(

000,20

pf

pf

load

load

load

loadloadline

loadloadload

V

PII

IVP


VA 000,14400,20

VA 4000400

var4000)1)(443.89)](443.89(5.0[

W 400)1)(443.89)](443.89(05.0[

A(rms) 443.89

jSSS

jjQPS

Q

P

I

loadlinesource

linelineline

line

line

line

The power company bills for the real power it

supplies to you – this includes the real power you

actually use, and the real power lost in the

transmission line.

A. Gives the power company more money

B. Increases the cost of products produced by the industry

C. Warms the feet of the birds who sit on the power lines

D. All of the above


)1)()](([ linelinelineline IIRP

How can we reduce the real power lost in the line?

We can’t do much about the resistance of the

transmission line, and usually it is pretty small. Can

we make the line current smaller?

loadpfload

loadloadline

V

PII

A. The real power required by the load(s)

B. The voltage drop across the load(s)

C. The power factor of the load(s)

loadpfload

loadloadline

V

PII

A. Maximize the load power factor (set it to 1)

B. Minimize the load power factor (set it to 0)

C. Neither (A) or (B)

A. Capacitor in series

with the loads

B. Inductor in series with the loads

C. Capacitor in parallel with the loads

D. Inductor in parallel with the loads

26.57o

20 kW

22.36 kVA

10 kvar


We add a capacitor in parallel with the loads

because

•We don’t want to change the voltage drop

across the loads

•We want to “cancel” the positive (inductive)

reactive power required by the loads

•We want to have no effect on the real power

required by the loads

26.57o

20 kW

22.36 kVA

10 kvar

+ -10 kvar

= 20 kW

1pf )load(total


The capacitor added must generate 10 kvar of

reactive power. Its voltage drop is 250 V(rms).

Calculate the value of capacitance.

F4.424)25.6)](60(2[

125.6

1

25.640

250Z

A(rms) 40)1)(250(

000,10

)1(

)1(

C

CC

I

V

V

QI

IVQ

cap

cap

cap

cap

cap

capcapcap


This process is called power factor correction, and

is usually required by the power companies for

large industrial loads, to conserve energy – there is

a monetary penalty associated with net power

factors less than a pre-set value, like 0.95.

Chapter 10 – ACSS Power Example 10.6, concluded

Did this power factor correction actually reduce the

power lost to the line? By how much?

W) 004 be to(usedW 320)1)(80)](80(05.0[

A(rms)) 443.89 be to(usedA(rms) 80)1)(250(

000,20

pfload

line

load

loadloadline

P

V

PII

Therefore, there was a 20% reduction in the real

power lost to the line. This would be even more

dramatic if the original power factor of the

combined loads was less than 0.89.

Chapter 10 – ACSS Power AP 10.6

In the figure below, Load 1 absorbs 15 kVA at

0.6 lagging and Load 2 absorbs 6 kVA at 0.8

leading. Find the source voltage phasor, Vs.

The plan of attack:

pfpfvi

load

loadload

iloadloadlineline

loadlineS

V

PI

Ijjj

0andpf

)(111 IIV

VVV

Chapter 10 – ACSS Power AP 10.6, continued




If we find the total complex power of the load,

we can find the two unknowns we need – the

total real power of the load, and the total power

factor of the load!





-3.6 kvar

4.8 kW

-36.87o +

53.13o

9 kW

12 kvar

15 kVA

= 31.33o

13.8 kW

8.4 kvar

16,155.5 VA

6 kVA


In the figure below, Load 1 absorbs 15 kVA at 0.6

lagging and Load 2 absorbs 6 kVA at 0.8 leading.

Find the source voltage phasor, Vs.

rms

rms

rms

rms

V9.156.251)0200()67.5878.80(

V 67.5878.80)33.3178.80(11

A33.3178.80

33.3100

A 78.80.33)200)cos(31(

800,13

pf

loadlineS

lineline

loadline

pfi

load

loadload

jj

V

PI

VVV

IV

II

31.33o

13.8 kW

8.4 kvar

16,155.5 VA

A. Iline

B. Iline

C. Ilineline

A. True

B. False


Maximum power transfer: Suppose we attach a load impedance to the Thevenin equivalent of a source circuit. For what value of load impedance will the maximum real power be delivered to the load impedance?

A. True

B. False

Chapter 10 – ACSS Power Maximum power transfer, continued

22

2

2

)()(

)1(

ThLThL

LTh

LLLLLLLZL

XXRR

R

RRPPR

V

IIIIV

To find the resistive and reactive values of load

impedance for maximum power, take the partial

derivative of PL with respect to RL and XL, set the

partial derivatives to zero, and solve for RL and XL.

Chapter 10 – ACSS Power Maximum power transfer, continued

*

ThL ZZ

The text sketches the derivation for the condition

for maximum real power to the load impedance:

Note that these two impedances are complex

numbers; the “star” (*) operator represents the

complex conjugate.

A. (18 – j36)

B. (36 – j18)

C. (18 + j36)

A. 6045

B. 4560

C. 6045

Chapter 10 – ACSS Power AP 10.7

The source current is 3 cos 5000t A(rms). Find

the impedance that should be connected to the

terminals a,b for maximum real power delivered

to that impedance, and find the maximum real

power delivered.


Phasor-transform the circuit:

W 361|10201020|

67.53

2

53.67pf

)1020(

)1020()40||20(184Z

V(rms) 57.2667.53V(rms) )2448()40||20)((03(

*

Th

jjIVP

jZZ

jjj

jj

LLL

ThL

ocV


Objectives:

•Power concepts: instantaneous power,

average power, reactive power, complex

power, power factor

•Relationships among power concepts – the

power triangle

•Balancing power in AC circuits

•Power factor correction

•Condition for maximum power transfer to a

load in AC circuits

Chapter 10 – ACSS Power

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Transcript of Chapter 10 – ACSS Power