Electric Circuits, 9th Edition P40 docx

366 Sinusoidal Steady-State Power Calculations Example 10.1 Calculating Average and Reactive Power a Calculate the average power and the reactive power at the terminals of the network s

366 Sinusoidal Steady-State Power Calculations

Example 10.1 Calculating Average and Reactive Power

a) Calculate the average power and the reactive

power at the terminals of the network shown in

Fig 10.6 if

v = 100 cos (tat + 15") V,

i = 4sin(W - 15°) A

b) State whether the network inside the box is

absorbing or delivering average power

c) State whether the network inside the box is

absorbing or supplying magnetizing vars

'-*•

+

V

Figure 10.6 A A pair of terminals used for calculating power

Solution

a) Because i is expressed in terms of the sine func-tion, the first step in the calculation for P and Q

is to rewrite i as a cosine function:

i = 4cos(o>f - 105°) A

We now calculate P and Q directly from

Eqs 10.10 and 10.11 Thus

P = -(100)(4) cos [15 - (-105)] = -100 W,

Q = -100(4) sin [15 - (-105)] = 173.21 VAR

b) Note from Fig 10.6 the use of the passive sign convention Because of this, the negative value

of -100 W means that the network inside the box is delivering average power to the terminals c) The passive sign convention means that, because

Q is positive, the network inside the box is

absorbing magnetizing vars at its terminals

^ A S S E S S M E N T P R 0 B L E

Objective 1—Understand ac power concepts, their relationships to one another, and how to calcuate them in a circuit

10.1 For each of the following sets of voltage and

current, calculate the real and reactive power

in the line between networks A and B in the

circuit shown In each case, state whether the

power flow is from A to B or vice versa Also

state whether magnetizing vars are being

trans-ferred from A to B or vice versa

a) v = 100 cos (at - 45°) V;

i = 20cos(wr + 15°) A

b) v = 100 cos (cot - 45°) V;

i = 20cos(&rf + 165°) A

c) v = 100 cos (at - 45°) V;

i = 20 cos (w* - 105°) A

d) V = 100 cos at V;

i = 20 cos (art + 120°) A

A

i

—^-+

Answer: ( a ) P

Q

500 W -866.03 VAR

(b) P = -866.03 W

Q = 500 V A R

(c) P = 500 W

Q = 866.03 VAR

(d) P = -500 W

Q = -866.03 V A R

A to B),

B to A);

B to A),

A to B);

A to B),

A t o B ) ;

B t o A ) ,

B t o A )

10.2 Compute the power factor and the reactive

fac-tor for the network inside the box in Fig 10.6, whose voltage and current are described in Example 10.1

Hint: Use -i to calculate the power and

reac-tive factors

Answer: pf = 0.5 leading; rf = -0.866

NOTE: Also try Chapter Problem 10.1,

Appliance Ratings

Average power is used to quantify the power needs of household appliances

The average power rating and estimated annual kilowatt-hour consumption

of some common appliances are presented in Table 10.1 The energy

con-sumption values are obtained by estimating the number of hours annually

that the appliances are in use For example, a coffeemaker has an estimated

annual consumption of 140 kWh and an average power consumption during

operation of 1.2 kW Therefore a coffeemaker is assumed to be in operation

140/1.2, or 116.67, hours per year, or approximately 19 minutes per day

Example 10.2 uses Table 10.1 to determine whether four common

appliances can all be in operation without exceeding the current-carrying

capacity of the household

| Making Power

The branch circuit supplying the

Calculations Involving Household Appliances

: outlets in a typical Solution

home kitchen is wired with #12 conductor and is

protected by either a 20 A fuse or a 20 A circuit

breaker Assume that the following 120 V

appli-ances are in operation at the same time: a

cof-feemaker, egg cooker, frying pan, and toaster Will

the circuit be interrupted by the protective device?

From Table 10.1, the total average power demanded

by the four appliances is

P = 1200 + 516 + 1196 4- 146

The total current in the pro

4058

e f r" 120 '

= 4058 W ective device is 33.82 A

Yes, the protective device will interrupt the circuit TABLE 10.1 Annual Energy Requirements of Electric Household Appliances

Appliance

Food preparation

Coffeemaker

Dishwasher

Egg cooker

Frying pan

Mixer

Oven, microwave (only)

Range, with oven

Toaster

Laundry

Clothes dryer

Washing machine, automatic

Water heater

Quick recovery type

Comfort conditioning

Air conditioner (room)

Dehumidifier

Fan (circulating)

Heater (portable)

Average

Wattage

1,200 1,201

516 1,196

127 1,450 12,200 1,146

4,856

512 2,475 4,474

860

257

88 1,322

NOTE: Assess your understanding of this

Est kWh Consumed Annually 3

140

165

14

100

2

190

596

39

993

103 4,219 4,811

860 b

377

43

176

Appliance Health and beauty Hair dryer Shaver Sunlamp

Home entertainment

Radio Television, color, tube type Solid-state type

Housewares Clock Vacuum cleaner a) Based on normal usage When using such factors as the size of the specific

Average

Wattage

600

15

279

71

240

145

2

630 hese figure appliance, I

Est kWh Consumed Annually 3

25 0.5

16

86

528

320

17

46

; for projections

he geographical area of use, and individual usage should be taken into considera-tion Note that the wattages are not additive, since all units are normally not in operation at the same time

b) Based on 1000 hours of operation pe r year This "igure will vary widely depending on the area and the specific size of the unit See EEI-Pub #76-2, "Air Conditioning Usage Study," for an estimate for your location

Source: Edison Electric Institute

material by trying Chapter Problem J0.2

368 Sinusoidal Steady-State Power Calculations

K, „cos (tot + 6 r ) | R

Figure 10.7 • A sinusoidal voltage applied to the

terminals of a resistor

10.3 The rms Value and Power

Calculations

In introducing the rms value of a sinusoidal voltage (or current) in Section 9.1, we mentioned that it would play an important role in power calculations We can now discuss this role

Assume that a sinusoidal voltage is applied to the terminals of a resis-tor, as shown in Fig 10.7, and that we want to determine the average power delivered to the resistor From Eq 10.12,

h+T

Vl 1 cos 2 (cot + $ v )dt (10.18)

Comparing Eq 10.18 with Eq 9.5 reveals that the average power

deliv-ered to R is simply the rms value of the voltage squared divided by R, or

If the resistor is carrying a sinusoidal current, say, I,,, cos (cot + <£,-), the

average power delivered to the resistor is

The rms value is also referred to as the effective value of the

sinu-soidal voltage (or current) The rms value has an interesting property:

Given an equivalent resistive load, R, and an equivalent time period, T, the rms value of a sinusoidal source delivers the same energy to R as does

a dc source of the same value For example, a dc source of 100 V delivers

the same energy in T seconds that a sinusoidal source of 100 Vnns delivers, assuming equivalent load resistances (see Problem 10.12) Figure 10.8 demonstrates this equivalence Energywise, the effect of the two sources

is identical This has led to the term effective value being used inter-changeably with rms value

The average power given by Eq 10.10 and the reactive power given

by Eq 10.11 can be written in terms of effective values:

V I

P = cos (0 V - 0,-)

V I

= —•=—7= cos (0,, - Bs)

= K-ff'eff cos (fl„ - 0 , ) ; (10.21)

lOOV(rms) R V s = 100V(dc) R

Figure 10.8 A The effective value of v, (100 V rms) delivers the

same power to R as the dc voltage V s (100 V dc)

and, by similar manipulation,

The effective value of the sinusoidal signal in power calculations is so

widely used that voltage and current ratings of circuits and equipment

involved in power utilization are given in terms of rms values For

exam-ple, the voltage rating of residential electric wiring is often 240 V/120 V

service These voltage levels are the rms values of the sinusoidal voltages

supplied by the utility company, which provides power at two voltage

lev-els to accommodate low-voltage appliances (such as televisions) and

higher voltage appliances (such as electric ranges) Appliances such as

electric lamps, irons, and toasters all carry rms ratings on their nameplates

For example, a 120 V, 100 W lamp has a resistance of 1202/100, or 144 ft,

and draws an rms current of 120/144, or 0.833 A The peak value of the

lamp current is 0.833 V2~, or 1.18 A

The phasor transform of a sinusoidal function may also be expressed

in terms of the rms value The magnitude of the rms phasor is equal to the

rms value of the sinusoidal function If a phasor is based on the rms value,

we indicate this by either an explicit statement, a parenthetical "rms"

adja-cent to the phasor quantity, or the subscript "eff," as in Eq 10.21

In Example 10.3, we illustrate the use of rms values for calculating power

Example 10.3 Determining Average Power Delivered to a Resistor by Sinusoidal Voltage

a) A sinusoidal voltage having a maximum

ampli-tude of 625 V is applied to the terminals of a

50 fl resistor Find the average power delivered

to the resistor

b) Repeat (a) by first finding the current in the

resistor

Solution

a) The rms value of the sinusoidal voltage is

625/V2, or approximately 441.94 V From

Eq 10.19, the average power delivered to the

50 Cl resistor is

P = (441.94)2

50 = 3906.25 W

b) The maximum amplitude of the current in the resistor is 625/50, or 12.5 A The rms value of the current is 12.5/V2, or approximately 8.84 A Hence the average power delivered to the resistor is

P = (8.84)250 = 3906.25 W

i / A S S E S S M E N T PROBLEM

Objective 1—Understand ac power concepts, their relationships to one another, and how to calculate them in a drcuit 10.3 The periodic triangular current in Example 9.4,

repeated here, has a peak value of 180 mA

Find the average power that this current

deliv-ers to a 5 kQ, resistor

Answer: 54 W

NOTE: Also try Chapter Problem 10.15

370 Sinusoidal Steady-State Power Calculations

10.4 Complex Power

Complex power •

TABLE 10.2 Three Power Quantities and

Their Units

Quantity

Complex power

Average power

Reactive power

Units volt-amps watts var

|.V| - apparent power

reactive power

P - average power

Figure 10.9 • A power triangle

Before proceeding to the various methods of calculating real and reactive power in circuits operating in the sinusoidal steady state, we need to intro-duce and define complex power Complex power is the complex sum of real power and reactive power, or

As you will see, we can compute the complex power directly from the volt-age and current phasors for a circuit Equation 10.23 can then be used to

compute the average power and the reactive power, because P = !R {S}

and<2 = 3 { S } Dimensionally, complex power is the same as average or reactive power However, to distinguish complex power from either average or reactive power, we use the units volt-amps (VA).Thus we use volt-amps for complex power, watts for average power, and vars for reactive power, as summarized in Table 10.2

Another advantage of using complex power is the geometric

interpre-tation it provides When working with Eq 10.23, think of P, Q, and \S\ as

the sides of a right triangle, as shown in Fig 10.9 It is easy to show that the

angle 6 in the power triangle is the power factor angle 0 I} — 0, For the

right triangle shown in Fig 10.9,

Q

But from the definitions of P and Q (Eqs [10.10] and [10.11 J, respectively),

Therefore, 0 = 0 V - 0,- The geometric relations for a right triangle mean

also that the four power triangle dimensions (the three sides and the power factor angle) can be determined if any two of the four are known The magnitude of complex power is referred to as apparent power Specifically,

Apparent power, like complex power, is measured in volt-amps The apparent power, or volt-amp, requirement of a device designed to convert electric energy to a nonelectric form is more important than the average power requirement Although the average power represents the useful output of the energy-converting device, the apparent power represents the volt-amp capacity required to supply the average power As you can see from the power triangle in Fig 10.9, unless the power factor angle is 0°

(that is, the device is purely resistive, pf = 1, and Q = 0), the volt-amp

capacity required by the device is larger than the average power used by the device As we will see in Example 10.6, it makes sense to operate devices at a power factor close to 1

Many useful appliances (such as refrigerators, fans, air conditioners, fluorescent lighting fixtures, and washing machines) and most industrial loads operate at a lagging power factor The power factor of these loads sometimes is corrected either by adding a capacitor to the device itself or

by connecting capacitors across the line feeding the load; the latter

method is often used for large industrial loads Many of the Chapter

Problems give you a chance to make some calculations that correct a

lag-ging power factor load and improve the operation of a circuit

Example 10.4 uses a power triangle to calculate several quantities

associated with an electrical load

An electrical load operates at 240 V rms The load

absorbs an average power of 8 kW at a lagging

power factor of 0.8

a) Calculate the complex power of the load

b) Calculate the impedance of the load

Solution

a) The power factor is described as lagging, so we

know that the load is inductive and that the

algebraic sign of the reactive power is positive

From the power triangle shown in Fig 10.10,

P = \S\ cos 0,

Q = \S\ sin 0

cos 0 - 0.8, sin 0 = 0.6

Now, because

Therefore

Q =

SkW cos0

10 sine

0.8

6 kVAR,

= lOkVA,

and

5 = 8 + /6 kVA

b) From the computation of the complex power of

the load, we see that P = 8 kW Using Eq 10.21,

= K>ff4ffcos(0t;

= (240)/eff(0.8)

= 8000 W

0i)

Solving for /efr,

/cff = 41.67 A

We already know the angle of the load imped-ance, because it is the power factor angle:

0 = cos-1(0-8) = 36.87°

We also know that 0 is positive because the power factor is lagging, indicating an inductive load We compute the magnitude of the load impedance from its definition as the ratio of the magnitude of the voltage to the magnitude of the current:

\Z\ = lK-a-1

l/effl

240 41.67 5.76

Hence,

Z = 5.76 /36.87° D, = 4.608 + y'3.456 O

Figure 10.10 • A power triangle

10.5 Power Calculations

We are now ready to develop additional equations that can be used to

cal-culate real, reactive, and complex power We begin by combining Eqs 10.10,

10.11, and 10.23 to get

VI VI

S = ~Y~cos (0 V - 6,) + j—^—sm(e v - $i)

V I

r in 1 m [cos ( 0 , , - 0i) + ; sin (0,, - 0,)]

i g g e f f l r t t o \v m I m /{6 n - 9d (10.27)

372 Sinusoidal Steady-State Power Calculations

If we use the effective values of the sinusoidal voltage and current,

Eq 10.27 becomes

S = K f f W ( 0 , ~ 0/) (10.28)

Equations 10.27 and 10.28 are important relationships in power calcula-tions because they show that if the phasor current and voltage are known at

a pair of terminals, the complex power associated with that pair of terminals

is either one half the product of the voltage and the conjugate of the cur-rent, or the product of the rms phasor voltage and the conjugate of the rms phasor current We can show this for the rms phasor voltage and current in Fig 10.11 as follows:

= V ea ei e >-I cii e-! 9 >

Note that left = h&eJ$i follows from Euler's identity and the trigonomet-ric identities cos(—0) = cos(fl) and s i n ( - 0 ) = — sin (#):

I^e* - /cff cos (Si) + //e f fsin (-0,-)

= / e f f cos (0/) - jl ei{ sin (6i)

= Ifo

The same derivation technique could be applied to Eq 10.27 to yield

S = -VI* (10.30)

Both Eqs 10.29 and 10.30 are based on the passive sign convention If the current reference is in the direction of the voltage rise across the termi-nals, we insert a minus sign on the right-hand side of each equation

To illustrate the use of Eq 10.30 in a power calculation, let's use the same circuit that we used in Example 10.1 Expressed in terms of the pha-sor representation of the terminal voltage and current,

V = 100 / 1 5 ° V,

I = 4 / - 1 0 5 ° A

Therefore

S = -(100 /15°)(4 / + 105°) = 200 /120°

•ell

•

-

4-V t ff

Circuit

Figure 10.11 • The phasor voltage and current

associ-ated with a pair of terminals

= - 1 0 0 + /173.21 V A

Once we calculate the complex power, we can read off both the real and

reactive powers, because S = P + jQ Thus

P = -100 W,

Q = 173.21 VAR

The interpretations of the algebraic signs on P and Q are identical to those

given in the solution of Example 10.1

let +

vc f f

—

z

1 Figure 10.12 A The general circuit of Fig 10.11 replaced with an equivalent impedance

Substituting Eq 10.31 into Eq 10.29 yields

S - ZIefrIe[f

= Heff|2Z

= |Ieff|2(fl + IX)

= HeffPi? + /|Icff|2X = P + jQ, (10.32)

from which

In Eq 10.34, X is the reactance of either the equivalent inductance or

equivalent capacitance of the circuit Recall from our earlier discussion of

reactance that it is positive for inductive circuits and negative for

capaci-tive circuits

A second useful variation of Eq 10.29 comes from replacing the

cur-rent with the voltage divided by the impedance:

S = V e f f ( ^ y = ^ = P + jQ- (10.35)

Alternate Forms for Complex Power

Equations 10.29 and 10.30 have several useful variations Here, we use the

rms value form of the equations, because rms values are the most common

type of representation for voltages and currents in power computations

The first variation of Eq 10.29 is to replace the voltage with the

prod-uct of the current times the impedance In other words, we can always

rep-resent the circuit inside the box of Fig 10.11 by an equivalent impedance,

as shown in Fig 10.12 Then,

374 Sinusoidal Steady-State Power Calculations

Note that if Z is a pure resistive element

P =

and if Z is a pure reactive element,

Q =

R '

|V eff | 2

(10.36)

In Eq 10.37, X is positive for an inductor and negative for a capacitor

The following examples demonstrate various power calculations in circuits operating in the sinusoidal steady state

Example 10.5 Calculating Average and Reactive Power

In the circuit shown in Fig 10.13, a load having an

impedance of 39 + /26 O is fed from a voltage

source through a line having an impedance of

1 + /4 O The effective, or rms, value of the source

voltage is 250 V

a) Calculate the load current IL and voltage VL

b) Calculate the average and reactive power

deliv-ered to the load

c) Calculate the average and reactive power

deliv-ered to the line

d) Calculate the average and reactive power

sup-plied by the source

Thus the load is absorbing an average power of

975 W and a reactive power of 650 VAR

in J T Y Y \ » _ /4 a

6 D 250/0°

V(rms)

Line Source

Figure 10.13 • The circuit for Example 10.5

3912

I I

/26 ft; Load

Solution

a) The line and load impedances are in series across

the voltage source, so the load current equals the

voltage divided by the total impedance, or

I, 250 / 0 °

40 + /30 = 4 - /3 = 5 / - 3 6 8 7 ° A (rms)

Because the voltage is given in terms of its

rms value, the current also is rms The load

volt-age is the product of the load current and load

impedance:

VL = (39 + /26)1 L = 234 - / 1 3

= 234.36 / - 3 1 8 ° V (rms)

b) The average and reactive power delivered to the

load can be computed using Eq 10.29 Therefore

S = \Jl = (234 - / 1 3 ) ( 4 + / 3 )

= 975 + /650 VA

c) The average and reactive power delivered to the line are most easily calculated from Eqs 10.33 and 10.34 because the line current is known Thus

P = (5)2(1) = 25 W,

Q = (5)2(4) = 100 VAR

Note that the reactive power associated with the line is positive because the line reactance is inductive

d) One way to calculate the average and reactive power delivered by the source is to add the com-plex power delivered to the line to that delivered

to the load, or

S = 25 + /100 + 975 + /650

= 1000 + / 7 5 0 V A The complex power at the source can also be cal-culated from Eq 10.29:

S s =

-250IL-The minus sign is inserted in Eq 10.29 whenever

the current reference is in the direction of a

volt-age rise Thus

5, = -250(4 + /3) = -(1000 4- /750) VA

The minus sign implies that both average power and magnetizing reactive power are being deliv-ered by the source Note that this result agrees with the previous calculation of 5, as it must, because the source must furnish all the average and reactive power absorbed by the line and load

Example 10.6 Calculating Power in Parallel Loads

The two loads in the circuit shown in Fig 10.14 can

be described as follows: Load 1 absorbs an average

power of 8 kW at a leading power factor of 0.8 Load

2 absorbs 20 kVA at a lagging power factor of 0.6

0.05 n

• V W

-+

/0.50 a

Figure 10.14 • The circuit for Example 10.6

a) Determine the power factor of the two loads in

parallel

b) Determine the apparent power required to

sup-ply the loads, the magnitude of the current, Iv,

and the average power loss in the transmission

line

c) Given that the frequency of the source is 60 Hz,

compute the value of the capacitor that would

correct the power factor to 1 if placed in parallel

with the two loads Recompute the values in (b)

for the load with the corrected power factor

Solution

a) All voltage and current phasors in this problem

are assumed to represent effective values Note

from the circuit diagram in Fig 10.14 that

Iv = l { + I2 The total complex power absorbed

by the two loads is

S = (250)i;

= (250)(1, + I2)8

= (250)1^ + (250)I2

= sx + s2

We can sum the complex powers geometrically,

using the power triangles for each load, as shown

in Fig 10.15 By hypothesis,

8000(.6)

= 8000 - /6000 VA,

S 2 = 20,000(.6) + /20,000(.8)

= 12,000 +/16,000 VA

-36.87°

16kVAR

10 kVAR

Figure 10.15 • (a) The power triangle for load 1 (b) The power triangle for load 2 (c) The sum of the power triangles

It follows that

S = 20,000 + /10,000 VA,

and , 20,000 + /10,000

Therefore

Iv = 80 - /40 = 89.44 / - 2 6 5 7 ° A

Thus the power factor of the combined load is

pf = cos(0 + 26.57°) = 0.8944 lagging

The power factor of the two loads in parallel is lagging because the net reactive power is positive b) The apparent power which must be supplied to these loads is

\S\ = |20 + /10| = 22.36 kVA

The magnitude of the current that supplies this apparent power is

II.sl = 180 - j40| = 89.44 A