Masters G.M. Renewable and Efficient Electric Power Systems

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58 FUNDAMENTALS OF ELECTRIC POWER

=√2

cos w

=√2w

cos(w

+ p/2)

−p/2

Figure 2.6 Current through a capacitor leads the voltage applied to it.

which says that the rms current I is given by

I = ωCV (2.22)

and the phase angle between current and voltage is

θ = π/2 (2.23)

Rearranging (2.22) gives

V =



ωC



I(2.24)

Equation (2.24) is beginning to look like an ac version of Ohm’s law for capaci-

tors. It should be used with caution, however, since it does not capture the notion

that current and voltage are 90

◦

out of phase with each other.

Also of interest is the average power dissipated by a capacitor subjected to

a sinusoidal voltage. Since instantaneous power is the product of voltage and

current, we can write

p = vi =

√

2V cos ωt ·

√

2I cos



ωt +



(2.25)

Using the trigonometric identity cos A · cos B =

[cos(A + B) + cos(A − B)]

gives

p = 2VI ·



cos



ωt + ωt +



+ cos



ωt −



ωt +



(2.26)

Since cos(−π/2) = 0, this simpliﬁes to

p = VI cos



2ωt +



(2.27)

IDEALIZED COMPONENTS SUBJECTED TO SINUSOIDAL VOLTAGES 59

Since the average value of a sinusoid is zero, (2.27) tells us that the average

power dissipated by a capacitor is zero.

avg capacitor

= 0 (2.28)

Some of the time the capacitor is absorbing power (charging) and some of the

time it is delivering power (discharging), but the average power is zero.

Example 2.4 Current in a Capacitor. A 120-V, 60-Hz ac source sends cur-

rent to a 10-microfarad capacitor. Find the rms current ﬂowing and write an

equation for the current as a function of time.

Solution. The rms value of current is given by (2.22) as

I = ωCV = 2π60 ·10 × 10

−6

· 120 = 0.452 A

The phase angle is θ = π /2, so from (2.21) the complete expression for current is

i =

√

2I cos(ωt + θ) =

√

2 · 0.452 cos



2π60t +



= 0.639 cos



377t +



2.2.3 Idealized Inductors

A sinusoidal voltage applied across an inductor is shown in Fig. 2.7.

We want to ﬁnd the current through the inductor. Starting with the fundamental

relationship for inductors,

v = L

(2.29)

and then solving for current:

i =



di =



dt =



vdt (2.30)

=√2

cos w

−

Figure 2.7 A sinusoidal voltage across an ideal inductor.

60 FUNDAMENTALS OF ELECTRIC POWER

and inserting the equation for applied voltage

i =



√

2V cos ωt dt =

√



cos ωt dt =

√

ωL

sin ωt (2.31)

Applying the trigonometric relationship sin ωt = cos(ωt − π/2) gives

i =



ωL



√

2V cos



ωt −



√

2I cos(ωt + θ) (2.32)

Equation (2.32) tells us that (1) the current through the inductor has the same

frequency ω as the applied voltage, (2) the current lags behind the voltage by

an angle θ =−π /2, and (3) the rms value of current is

I =



ωL



V(2.33)

Rearranging (2.33) gives us something that looks like an ac version of Ohm’s

law for inductors:

V = (ωL)I (2.34)

So, for an inductor, you have to supply some voltage before current ﬂows;

for a capacitor, you need to supply current before voltage builds up. One way to

remember which is which, is with the memory aid

“ELI the ICE man”

That is, for an inductor (L), voltage (E, as in emf) comes before current (I),

while for a capacitor C, current (I ) comes before voltage (E).

Finally, let us take a look at the power dissipated by an inductor:

p = vi =

√

2V cos ωt ·

√

2I cos



ωt −



(2.35)

Using the trigonometric identity for the product of two cosines, gives instanta-

neous power through the inductor

p = 2VI ·



cos



ωt + ωt −



+ cos



ωt −



ωt −



= VI cos



2ωt −



(2.36)

The average value of (2.36) is zero.

avg inductor

= 0 (2.37)

That is, an inductor is analogous to a capacitor in that it absorbs energy while

current is increasing, storing that energy in its magnetic ﬁeld, then it returns that

POWER FACTOR 61

energy when the current drops and the magnetic ﬁeld collapses. The net power

dissipated when an inductor is subjected to an ac voltage is zero.

2.3 POWER FACTOR

Those rather tedious derivations for the impact of ac voltages applied to ide-

alized resistors, capacitors, and inductors has led to three simple but important

conclusions. One is that the currents ﬂowing through any of these components

will have the same ac frequency as the source of the voltage that drives the cur-

rent. Another is that there can be a phase shift between current and voltage. And

ﬁnally, resistive elements are the only components that dissipate any net energy.

Let us put these ideas together to analyze the generalized black box of Fig. 2.8.

The black box contains any number of idealized resistors, capacitors, and

inductors, wired up any which way. The voltage source driving this box of

components has rms voltage V , and we will arbitrarily assign it a phase angle

of θ = 0.

v =

√

2V cos ωt (2.38)

Since the current delivered to the black box has the same frequency as the voltage

source that drives it, we can write the following generalized current response as

i =

√

2I cos(ωt + θ) (2.39)

The instantaneous power supplied by the voltage source, and dissipated by the

circuit in the box, is

p = vi =

√

2V cos ωt ·

√

2I cos(ωt + θ) = 2VI [cos ωt · cos(ωt + θ)] (2.40)

Once again, applying the identity cos A · cos B =

[cos(A + B) + cos(A − B)]

gives

p = 2VI



[cos(ωt + ωt + θ) + cos(ωt − ωt − θ)]



(2.41)

p = VI cos(2ωt + θ)+ VI cos(−θ) (2.42)

−

Figure 2.8 A black box of ideal resistors, capacitors, and inductors.

62 FUNDAMENTALS OF ELECTRIC POWER

The average value of the ﬁrst term in (2.42) is zero, and using cos x = cos(−x)

lets us write that the average power dissipated in the black box is given by

avg

= VI cos θ = VI × PF (2.43)

Equation (2.43) is an important result. It says that the average power dissipated

in the box is the product of the rms voltage supplied times the rms current

delivered times the cosine of the angle between the voltage and current. The

quantity cos θ is called the power factor (PF):

Power factor = PF = cos θ(2.44)

The power expressed by (2.43) tells us the rate at which real work can be

done in the black box. That black box, for example, might be a motor, in which

case (2.43) gives us power to the motor in watts.

Why is power factor important? With an “ordinary” watt-hour meter on the

premises, a utility customer pays only for watts of real power used within their fac-

tory, business, or home. The utility, on the other hand, has to cover the i

R resistive

power losses in the transmission and distribution wires that bring that power to the

customer. When a customer has voltage and current way out of phase—that is, the

power factor is “poor”—the utility loses more i

R power on its side of the meter

than occurs when a customer has a “good” power factor (PF ≈ 1.0).

Example 2.5 Good Versus Poor Power Factor. A utility supplies 12,000 V

(12 kV) to a customer who needs 600 kW of real power. Compare the line losses

for the utility when the customer’s load has a power factor of 0.5 versus a power

factor of 1.0.

Solution. To ﬁnd the current drawn when the power factor is 0.5, we can start

with (2.43):

P = VI · PF

600 kW = 12 kV · I(A) · 0.5

I =

600

12 × 0.5

= 100 A

When the power factor is improved to 1.0, (2.43) now looks like

600 kW = 12 kV · I(A) · 1.0

so the current needed will be I =

600

= 50 A

THE POWER TRIANGLE AND POWER FACTOR CORRECTION 63

When the power factor in the plant is improved from 0.5 to 1.0, the amount

of current needed to do the same work in the factory is cut in half. The utility

line losses are proportional to current squared, so line losses for this customer

have been cut to one-fourth of their original value.

2.4 THE POWER TRIANGLE AND POWER FACTOR CORRECTION

Equation (2.43) sets up an important concept, called the power triangle.The

actual power consumed by a circuit is the rate at which real work can be done

(in watts). Because voltage V and current I may not be in phase, their product

does not, in general, equal real power. Figure 2.9 sets up a power triangle in

which the hypotenuse is the product of rms volts times rms amps. This leg is

called the apparent power, S, and it has units of volt-amps (VA). Those volt-amps

are resolved into the horizontal component P = VI cos θ,whichisreal power in

watts. The vertical side of the triangle, Q = VI sin θ, is called reactive power and

has units of VAR (which stands for volt-amps-reactive). Reactive VAR power is

incapable of doing any work: It corresponds to voltage 90

◦

out of phase with

current so any work absorbed in one half of the cycle is returned, unchanged, in

the other half.

Apparent power,

volt-amps

Real power,

(watts)

sin q

Volt-Amps-Reactive (VAR)

Reactive

power,

(VAR)

cos q

Figure 2.9 Showing apparent power S (volt-amps) resolved into reactive power (VAR)

and real power P (watts).

Example 2.6 Power Triangle for a Motor. A 230-V induction motor draws

25 A of current while delivering 3700 W of power to its shaft. Draw its power

triangle.

Solution

Real power P = 3700 W = 3.70 kW

Apparent power S = 25 A × 230 V = 5750 volt-amps = 5.75 kVA

64 FUNDAMENTALS OF ELECTRIC POWER

Power factor PF =

Real power

Apparent power

3700 W

5750 VA

= 0.6435

Phase angle θ = cos

−1

(0.6435) = 50

◦

Reactive power Q = S sin θ = 5750 sin 50

◦

= 4400 VAR = 4.40 kVAR

The power triangle is therefore

q = 50°

Real power

= 3.70 kW

Reactive power

= 4.40 kVAR

Apparent power

= 5.75 kVA

Utilities are very concerned about customers who draw a lot of reactive

power—that is, customers with poor power factors. As suggested in Example 2.5,

reactive power increases line losses for the utility, but doesn’t result in any more

kilowatt-hours (kWh) of energy sales to the customer. To discourage large cus-

tomers from having a poor power factor, utilities will charge a penalty based on

how low the power factor is, or they will charge not only for kWh of energy but

also for kVAR of reactive power.

Many large customers have loads that are dominated by electric motors, which

are highly inductive. It has been estimated that lagging power factor, mostly

caused by induction motors, is responsible for as much as one-ﬁfth of all grid

losses in the United States, equivalent to about 1.5% of total national power

generation and costs on the order of $2 billion per year. Another reason for

concern about power factor is that transformers (on both sides of the meter) are

rated in kVA, not watts, since it is heating caused by current ﬂow that causes

them to fail. By correcting power factor, a transformer can deliver more real

power to the loads. This can be especially important if loads have increased to

the point where the existing transformers can no longer handle the load without

overheating and potentially burning out. Power factor correction can sometimes

avoid the need for additional transformer capacity.

The question is, How can the power factor be brought closer to a perfect 1.0?

The typical approach is fairly intuitive; that is, if the load is highly inductive,

which most are, then try to offset that by adding capacitors as is suggested in

Fig. 2.10. The idea is for the capacitor to provide the current that the inductance

needs rather than having that come from the transformer. The capacitor, in turn,

THE POWER TRIANGLE AND POWER FACTOR CORRECTION 65

(a) Original circuit (b) With PF correcting capacitor

Load,

lagging PF

Fully loaded

transformer

(PF < 1)

correcting

capacitor

(PF = 1)

Load,

lagging PF

Transformer

with extra

capacity

Figure 2.10 Correcting power factor for an inductive load by adding a parallel capacitor.

gets its current from the inductance. That is, the two reactive elements, capacitor

and inductance, oscillate, sending current back and forth to each other.

Capacitors used for power factor compensation are rated by the volt-amps-

reactive (VAR) that they supply at the system’s voltage. When rated in these

units, sizing a power-factor correcting capacitor is quite straightforward and is

based on the kVAR of a capacitor offsetting some or all of the kVAR in the

power triangle.

Example 2.7 Avoiding a New Transformer by Improving the Power Factor.

A factory with a nearly fully loaded transformer delivers 600 kVA at a power

factor of 0.75. Anticipated growth in power demand in the near future is 20%.

How many kVAR of capacitance should be added to accommodate this growth

so they don’t have to purchase a larger transformer?

Solution. At PF = 0.75, the real power delivered at present is 0.75 × 600 kVA =

450 kW. And the phase angle is θ = cos

−1

(0.75) = 41.4

◦

. If demand grows by

20%, then an additional 90 kW of real power will need to be supplied. At that

point, if nothing is done, the new power triangle would show

Real power P = 450 + 90 = 540 kW

Apparent power S = 540 kW/0.75 = 720 kVA (too big for this transformer)

Reactive power Q = VI sin θ = 720 kVA sin(41.4

◦

) = 476 kVAR

For this transformer to still supply only 600 kVA, the power factor will have to

be improved to at least

PF = 540 kW/600 kVA = 0.90

66 FUNDAMENTALS OF ELECTRIC POWER

The phase angle now will be θ = cos

−1

(0.90) = 25.8

◦

. The reactive power will

need to be reduced to

Q = 600 kVA sin 25.8

◦

= 261 kVAR

The difference in reactive power between the 476 kVAR needed without power

factor correction and the desired 261 kVAR must be provided by the capaci-

tor. Hence

PF correcting capacitor = 476 − 261 = 215 kVAR

The power triangles before and after PF correction are shown below:

q = 41.4°

= 450 kW

= 397 kVAR

= 600 kVA

= 540 kW

261 kVAR

q = 25.8°

(a) Before correction (b) With 215-kVAR capacitor

′= 476 kVAR

90 kW

Anticipated

growth to 720 kVA

Without

PF correction

215-kVAR

Capacitor

While rating capacitors in terms of the VAR they provide is common in power

systems, there are times when the actual value of capacitance is needed. From

(2.24), we have a relationship between current through a c apacitor and voltage

across it:

V =



ωC



I(2.24)

The power through a capacitor is all reactive, so that

VA R = VI = V(ωCV) = ωCV

(2.45)

That is, the conversion from VARs to farads for a capacitor is given by

C(farads) =

VA R s

ωV

(2.46)

Notice, by the way, that the VAR rating of a capacitor depends on the square of

the voltage. For example, a 100-VAR capacitor at 120 V would be a 400-VAR

THREE-WIRE, SINGLE-PHASE RESIDENTIAL WIRING 67

reactance at 240 V. That is, the VAR rating itself is meaningless without knowing

the voltage at which the capacitor will be used.

Example 2.8 Power-Factor-Correcting Capacitor for a Motor What size

capacitor would be needed to correct the power factor of the 230-V, 60-Hz, 5-hp

motor in Example 2.6?

Solution. The capacitor must provide 4.40 kVAR of capacitive reactance to cor-

rect for the motor’s 4.40 kVAR of inductive reactance. Since this is a 230-V

motor operating at 60 Hz, (2.46) indicates that the capacitor should be

C =

VA R s

ωV

4400

2π × 60 × (230)

= 0.000221 F = 221 µF

2.5 THREE-WIRE, SINGLE-PHASE RESIDENTIAL WIRING

The wall receptacle at home provides single-phase, 60-Hz power at a nominal

voltage of about 120 V (actual voltages are usually in the range of 110–120 V).

Such voltages are sufﬁcient for typical, low-power applications such as light-

ing, electronic equipment, toasters, and refrigerators. For appliances that requires

higher power, such as an electric clothes dryer or an electric space heater, special

outlets in your home provide power at a nominal 240 V. Running high-power

equipment on 240 V rather than 120 V cuts current in half, which cuts i

R heat-

ing of wires to one-fourth. That allows easy-to-work-with, 12-ga. wire to be

used in a household, for both 120-V and 240-V applications. So, how is that

240 V provided?

Somewhere nearby, usually on a power pole or in a pad-mounted rectan-

gular box, there is a transformer that steps down the voltage from the utility

distribution system at typically 4.16 kV (though sometimes as high as 34.5 kV)

to the 120 V/240 V household voltage. Figure 2.11 shows the basic three-wire,

single-phase service drop to a home, including the transformer, electric meter,

and circuit breaker panel box.

As shown in Fig. 2.11, by grounding the center tap of the secondary side of

the transformer (the neutral, white wire), the top and bottom ends of the windings

are at the equivalent of + 120 V and − 120 V. The voltage difference between

the two “hot” sides of the circuit (the red and black wires) is 240 V. Notice the

inherent safety advantages of this conﬁguration: At no point in the home’s wiring

system is the voltage more than 120 V higher than ground.

The ± 120-V lines are 120 V (rms) with a 180

◦

phase angle between them. In

fact, it would be reasonable to say this is a two-phase system (but nobody does).