6 ac power 2 book 1/6. power in...power in an ac circuit training development transmission &...

T R A I N I N G

M A N U A L0 2 / 0 7 / 0 5

Power in anAC Circuit

Training Development Centre

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PP O W E R I N A N A CC I R C U I T

OBJECTIVE:Given a calculator and formula sheet,you will be able to calculate the powerfactor of an AC circuit.

WHY?Maintaining a good power factor allows morecapacity from equipment, and understandingpower factor is a necessary basis formetering.

LEARNING OBJECTIVES:1. Define true power.

2. Define reactive power.

3. Define apparent power.

4. Define power factor.

5. Draw and label a power triangle.

Preface

Start Here

Intro to Hotsticks

Alternating Current

Street Lights

Single Phase Transformation

Three Phase Systems

Class 1 Rubber Gloves

DCRACF

RMSRINRIR

PACCXC

RXCSRX

PRX

SLIISL

MSLSLP

TOP

TRRPTR

TRLTWC

Standard Protection

Code

TPSYCO

DCOCTP

Copyright © 1998 by the Training and Development Centre, SaskPower. All rights reserved. 0 2 / 0 7 / 0 5

IN THIS MODULE:1. True Power

2. Reactive Power

3. Apparent Power

4. Power Factor

5. Power Triangle

RESOURCES:• Calculator

ESTIMATED TIME:1.5 hours

PREREQUISITES:• “Resistance and Inductive Reactance in

an AC Circuit” module

LEARNING STEPS:1. Cover the module.

2. Complete the Review Questions.

3. Clarify any questions or concerns youmay have.

Copyright © 1998 by the Training and Development Centre, SaskPower. All rights reserved.0 2 / 0 7 / 0 5

IntroductionIn a purely resistive circuit, our sine waves would look like this:

Our power wave would like the above example due to the productof voltage and current, which are “in phase.” Due to the nature ofan AC circuit, we are only concerned with the average power ofthe circuit, which is simple:

ERMS x IRMS = Poweraverage

Average power means one half of the power peak value, which wearrive at by using effective voltage and current values.

(E).707 x .707(I) = .5(P) or one half peak power

This power value can be referred to as:

• True power

• Apparent power

• Reactive power

4 T R A I N I N G M A N U A L .

0 2 / 0 7 / 0 5Copyright © 1998 by the Training and Development Centre, SaskPower. All rights reserved.

True Power• Symbol - TP

• Unit of measurement is watts

• Definition: The rate at which volts and amps are convertedto heat, light or mechanical energy in the purely resistiveportion of the circuit.

TP is the product of E x I when voltage and current are in phase.True power is the power used by the customer as recorded on ourwatt meters.

The power produced when both E and I values are either positiveor negative is referred to as true power and is measured in watts, aproduct which is always positive.

When E and I values are both positive, TP is positive, and wheneverE and I are both negative, TP is positive.

True power = Voltage drop at the resistor X Current flow throughthe resistor.

TP (watts) in an ACcircuit is always

positive

TP E x I

TP I x R

TP ER

R R

R

=

=

=

2

2

P O W E R I N A N A C C I R C U I T 5 .

0 2 / 0 7 / 0 5 Copyright © 1998 by the Training and Development Centre, SaskPower. All rights reserved.

Reactive Power• Symbol - RP

• Unit of measurement is vars (volt amps reactive)

• Definition: Reactive power is the product of E and I whenthey are out of phase with each other by 90 degrees.

Reactive power is the wasted power of the circuit.

RPL = EL x IL

Reactive power of the inductor = voltage drop at inductormultiplied by the current flow through the inductor. RPL ismeasured in vars.



Apparent Power• Symbol - AP

• Unit of measurement is VA (volt amps)

• Definition: Apparent power is the product of the Eapplied andIT readings of a circuit. Apparent power is the vectorial sumof true power and reactive power.

The answer will be in volt amps (VA).

AP E x I or AP TP RPapp T= = +2 2



Power FactorDue to the fact that true power and apparent power are dissimilar,there is a waste in our system which we measure in vars. We canexpress the efficiency or inefficiency of an electrical system bysimply dividing the larger number of apparent power (VA) into thetrue power (watts) of the circuit. The result would be the efficiencyof the system expressed as a percentage (%).

We refer to this relationship as power factor.

Example 1:

The cos-1 (inverse) of 0.894 will give the angle theta; therefore, thecurrent lags the voltage by 26.6 degrees. This is referred to as thephase angle.

As calculated, this particular system is operating at close to 90%efficiency and is very acceptable. Since regular revenue metersonly measure watts, some utilities have had to attach an additionalmeasuring device called VA demand on the meters to measure theproduct of circuit volts and amps. By making the customer pay forthe wasted power (vars), as well as the true power (watts) , there issome incentive to upgrade the efficiency of their systems.

p f TPAP

x

p f WVA

x

p f

. .

. .

. . .

=

=

=

100

20002236

100

89 4%



Example 2:

The following example illustrates the procedure to solve variablesin an AC series circuit.

Given the following, find Z, IT, ER, EL, Eapp, angle theta, TP, RP,AP, p.f.:

We can find:

I EZ

V amps

E I x R A x voltsE I x X A x voltsE volts

adjhyp

EE

lag

AP E x I V x A VA

TP E x I V x A wattsRP E xI V x A s

p f TPAP

WVA

T

R

L L

app

R

app

app T

R T

L T

= = =

= = == = =

=

= = = = °

= = =

= = == = =

= = =

110233

4 7

4 7 20 944 7 12 56 4

109 6

94109 6

30 9

109 6 4 7 515

94 4 7 441856 4 4 7 265

4418515

86%

..

.. .

.

cos.

.

. .

. .. . var

. . .

ΩΩ

Ω

θ

We found Z by vectorally adding R and XL.

Z R XL= + =2 2 23 3. Ω



Power Factor Analogy

Using a glass of root beer will help illustrate the advantage ofmaintaining a good power factor. The TP is the root beer, the RP isthe foam, and the AP is the root beer and the foam.

Always draw a vector to aid in calculating variables.



Frequently Asked QuestionsWhat is considered to be a good power factor?

• A power factor of 90% is considered optimum.

How does poor power factor affect me?

Poor power factor affects the average power consumer infour ways:

• Increased power costs — power companies chargemore for low power factor.

• Reduced system capacity — larger currents are drawnthrough the plant distribution system for a givenamount of useful work. Thus, you need a largersystem similar to the stronger tow rope for the waterskier.

• Increased power losses — larger currents meanincreased power losses.

• Poorer voltage regulation — larger currents also meangreater voltage drop.

Why do some utilities charge more for low power factor?

• Low power factor at the customer’s facilities has thesame effect on the electric system as it has in yourplant. Poor power factor means that a largergeneration and transmission system must be built,more power is lost in transmission, and moreequipment must be installed to maintain good linevoltage.

• TP is the power produced by the volts and amps in phase.

• RP is the power produced by the volts and amps out of phase by 90 degrees.

• AP is the total power required to supply the load and the waste. Apparent power is the size of transformer required to supply the actual load and the wasted power.



How can I correct poor power factor?

• There are three methods normally considered —synchronous motors, synchronous condensers, andcapacitors. The most economical installation in mostcases is the capacitor.

How does a capacitor help correct power factor?

• Capacitors supply leading reactive current whichoffsets the lagging reactive current required by motorsand other inductive devices and thus reduces the totalcurrent flow from the electric system.

Where do I install a capacitor?

• The most benefit is obtained by installing the capacitoras close to the load as possible. In the case of anelectric motor, the capacitor should be connected to themotor terminals. In this way, the reactive currentflows from the capacitor to the motor directly, leavingmore feeder capacity available for other loads.

How much do capacitors cost?

• The cost of capacitors varies according to the voltage,size and type. Large, higher voltage indoorinstallations are less costly than small, lower voltageoutdoor installations. For example, a 240 volt,2.5kvar, outdoor, weather-proof capacitor may cost inthe neighbourhood of $50/kvar whereas a 480 volt,10kvar indoor capacitor costs in the neighbourhood of$13/kvar.

How long will it take to pay for the capacitor installation throughreduced power costs?

• This will depend on the installation, of course, butnormally between one and three years.

Is there any relationship between the wise and efficient use ofenergy and power factor?

• Yes! Power factor correction at the load decreases thecurrent flow and thus reduces the power losses in theelectrical system.



Power TriangleThis triangle is a visual aid which will enable you to see andcalculate the three kinds of power in an electrical system or circuit.This power triangle is a right angle triangle; therefore,Pythagorean's Theorem may be used to solve for unknown values(a2 + b2 = c2).

c = hypotenuse

The construction of the triangle goes as such:

• The adjacent or the base of the triangle is always the watts ortrue power (TP) of the circuit.

• The hypotenuse or diagonal of the triangle is always theapparent power of the circuit since it is always of the highestvalue.

• The opposite or vertical of the triangle is always the reactivepower, or wasted power, of the circuit.

Current (IT) is the constant and, in this case, is lagging theapplied voltage by the angle theta.



SummaryTo summarize this module, you have learned:

• AC power and values.

• Power factor and phase angle.

• Power triangle and variables.

Now...

• Complete the Review Questions.

• Clarify any questions or concerns you may have.



REVIEW QUESTIONS

Power in an AC Circuit

Define true power.

1. True power is measured in:

(a) waste.

(b) vars.

(c) volt amps (VA).

(d) watts.

2. TP is calculated in an AC series circuit by:

(a) Eapp x IT

(b) ER x IT

(c) EL2 x ER

2

(d) none of the above

3. If R = 10 ohms and IT = 24A, the TP of the circuit is:

This section is a review for your Power Line Apprenticeshipwritten exams. Complete the questions, and verify your answerswith the Review Question Solutions located at the end of thismodule.



Define reactive power.

4. Reactive power is measured in:

(a) waste.

(b) vars.

(c) volt amps (VA).

(d) watts.

5. RP is calculated in an AC series circuit by:

(a) Eapp x IT

(b) ER x IT

(c) EL x IT

(d) EL2 x ER

2

6. If XL = 40 ohms and RPL = 4000 vars, EL and IL are:

(a) IL = 10A, EL = 400V

(b) IL = 400V, EL = 10A

(c) IL = 400A, EL = 10V

(d) IL = 256.8A, EL = 600V

Define apparent power.

7. Apparent power is measured in:

(a) waste.

(b) vars.

(c) volt amps (VA).

(d) watts.



8. AP is calculated in an AC circuit by:

(a) Eapp x IT

(b) ER x IT

(c) Eapp x IR

(d) EL2 x ER

2

9. Calculate the apparent power of the circuit, given RPL = 4kvars and TP = 5760W.

Define power factor.

10. Power factor is:

(a) the efficiency of the system expressed as a percentage.

(b) cos-1(p.f.) (power factor as a whole, not a percent).

(c) the relationship of usable power to actual power required.

(d) all of the above.

11. Power factor can be found by:

(a) Eapp x IT

(b) E x R x 100

(c) EL2 = ER

2 x 100

(d) cos∠θ x 100



12. If Eapp = 240V, ER = 176V, and IT = 13.6A:

(a) What type of circuit is it?

(b) What is the AP of the circuit?

(c) What is the TP of the circuit?

(d) What is the p.f. of the circuit?

(e) What is the angle theta of the circuit?

Draw and label a power triangle.

13. Given an AC series circuit with a 15 ohm resistor, a 30 ohm coiland a current flow of 17 amps, find the following and draw avector.

Z _______________

EL _______________

ER _______________

Eapp _______________

TP _______________

RPL _______________

AP _______________

p.f. _______________

_______________∠θ



14. Using the information from Question #13, draw and label a sinewave.



1. d

2. b

3. Method #1:

Method #2:

4. b

5. c

6. a

7. c

8. a

9.AP TP RP

AP VA or kVA

= +

= +=

2 2

2 25760 40007012 7 7 012. .

TP I x R

A xTP W

T=

==

2

224 105760

Ω

E I x RA x

E V

TP E x IV x A

TP W

R

R

R

===

===

24 10240

240 245760

Ω


REVIEW QUESTION SOLUTIONS



10. d

11. d

12a. series

12b. 3264VA

12c. 2393.6 watts

12d. 73.3%

12e. 42.8 degrees

13. Z = 33.5 ohmsEL = 510 voltsER = 255 voltsEapp = 570.2 voltsTP = 4335 wattsRPL = 8670 varsAP = 9693.4VAp.f. = 44.7%∠ = °θ 62 7.


REVIEW QUESTION SOLUTIONS (CONTINUED)Power in an AC Circuit


14.


REVIEW QUESTION SOLUTIONS



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