4. In an ас circuit, the peak stored energy in a capacitor is 10 juJ. W hat is the timeaverage stored energy?
5. W hat is the tim e-average value o f v(/) = 1 0 + 5 cos (1 0 0 0 V?
6 . If the pow er factor is 0.75, lagging, by w hat angle does the current lag the voltage?
7. Is the reactive pow er into a capacitor positive or negative?
8 . For a dc voltage, a dc voltm eter m easures 10 V. W hat is the tim e-average voltage?
W hat is the rm s value o f the voltage? For an ac voltage, an ac voltm eter measures
10 V. W hat is the tim e-average value o f the voltage? W hat is the rms value of the
voltage?
Answers. (1)7.07 A; (2) 10 A; (3) true; (4) 5 (xJ; (5)10V ; (6)41.4°; (7) negative; (8 ) 10 V,
10 V, 0 V, 10 V.
POWER AMD ENERGY IN THE FREQUENCY DOMAIN
Introduction.
T he tim e-average (real) pow er in an ac circuit, given in Eq. (5.39), in­
volves the peak (or rms) values o f the voltage and current and the pow er factor, which
is the cosine of the phase angle betw een the voltage and current. All these quantities
can be expressed in the frequency dom ain; indeed, the frequency-dom ain concept of im­
pedance provides the m ost efficient w ay to determ ine the m agnitude and the phase an­
gle o f the current. Thus, we can calculate tim e-average pow er w ithout transforming to
the tim e dom ain. In this section, w e show how to determ ine real, reactive, and apparent
pow er in the frequency dom ain.
Real and Reactive Power from Phasors
Real pow er
To understand
how to calculate
real, reactive, an d
a p p a re n t pow er
in the frequency
dom ain
in-phase current
T he tim e-average pow er into the load is given by Eq. (5.39) as
Vn I n
p — —
c o s 0 = —|V | x
III cos<9 watts
(5.:
in-phase current
Figure 5.10(b) and Eq. (5.39) suggest an interpretation o f the pow er factor. If we asso­
ciate the cos в w ith the m agnitude o f I , the term |I | cos 9 is the projection of the cur­
rent phasor onto the voltage phasor. T his part o f the current is in phase w ith the voltage,
and we conclude that the tim e-average pow er is given by the product o f the voltage and
the current in phase with the voltage. Thus, in the circuit o f Fig. 5.9, repeated in
Fig. 5.10(a),
P = 1 120л/2 x 3.67 7 2 = 440 W
w hich agrees with Eq. (5.47).
out-of-phase
current
II
Reactive power. Equation (5.40) for the reactive pow er can be interpreted as the |
product o f the voltage with the out-of-phase com ponent o f the current, as shown in |
Fig. 5.10(b):
Шят
W hat is the
Z = 10 + j 15.1 £1
nt lag the voltage?
= 18.1Z52.4°a
/15.1
a
120~^2 Z0°
18.1 Z 5 6 .4 £2
e-average voltage?
■oltmeter measures !
e rms value of the
= 6 .6 3 V 2 Z (-5 6 .4 ’) A
= (3 .6 7 -;'5 .5 з /2 A
V = 120V2 Л Г У
) negative; (8) 10 V, i
F i g u r e 5 . 1 0 (a) Solution in the frequency domain, (b) The phasor diagram shows
the “in-phase” and “out-of-phase” components of the current.
n in Eq. (5.39), in-'
Dwer factor, which
J 1 these quantitii
tain concept of im- 1
and the phase an>ut transforming to
ctive, and apparent
out-of-phase current
■
r ■■
where we consider lagging current to produce a positive out-of-phasc com ponent . 13 In
the circuit in Fig. 5.10(a),
Q = A 120л/2 x 5.53 Л
= + 6 6 3 VAR
which agrees with Eq. (5.49). We show ed in Eq. (5.42) the relation betw een reactive
pow er and the stored-energy requirem ent o f the inductor. We explore further the m ean­
ing o f the reactive pow er in the follow ing example.
Real and reactive pow er in the RLC circuit
factor. If we asso>jection of the i
e with the volti
o f the voltage and
g. 5.9, repeated in
Find the real and reactive power into the load in Fig. 5.11 with R = 10 Q L = 40 mH, and
С = 120 |J.F, and interpret these powers as lost or stored energy. The voltage is 120 V (rms)
at 60 Hz.
S o l u t io n :
Figure 5.11 shows the circuit in the frequency domain.
The impedance as seen by the voltage source is
Z = 7 15.1 + 10 || (—/22.1) = 14.0 / 53.8° Q
( 5 .5 7 )
so the current is
interpreted as the :
rrent, as shown :
13Because в is negative for lagging current, Q would be negative for lagging current if strict mathematics were
followed. In a power system, however, lagging current is “norm al” and reactive power is defined to be positive
for lagging, or inductive, current.
5.2 POWER AND ENERGY IN THE FREQUENCY DOMAIN
40 mH
(/15.1 Д)
-
120 ЦБ
' (-722.1 О)
v = 120/2 Z0 °= nofl+jov
Figure 5.11
General M.C load.
|I| = 12Q7 2 / 0 °_ 8 5 5Ay5‘z _ 53.8o
14.0 Z 53.8°
= (5.05 - j 6 . S 9 ) j 2 A
The real power to the load is the product of the rms voltage with the in-phase component
the rms current, P = 120 x 5.05 = 606 watts, and the reactive power to the load is the
uct with the out-of-phase rms current, Q — 120 x 6.89 = 827 VAR. Lagging current
positive reactive power.
To interpret these powers in terms of lost or stored energy, we must complete the phasor
analysis of the circuit. We find the voltage across the ЙПС by multiplying the current by the
parallel impedance:
Y KC = (8.5572 Z -53.8°) x 10|| ( ~ j 2 2 A ) = l i s j l Z -7 8 .1 ° V
(5
The energy lost to the circuit will be in the time-average power to the resistor, which is
R
where
itor is
10
is the rms voltage across the resistor. The peak electric energy stored in the
i
Wep = i CV l = 1(120
v
X
10-6)
X
(77.972 )2 = 0.728 J
where Vc is the peak voltage across the capacitor. The peak magnetic energy stored in the in­
ductor is
Wmp = i H I = 1(0.040) x (8.5572 )2 = 2.92 J
where IL is the peak current through the inductor. These stored energies relate to the reactive
power through Eq. (5.44):
Q = 0)(Wmp - Wep) = 120 к x (2.92 - 0.728) = 827 VAR
which agrees with our earlier result.
■
t*
~ '
-
W h a t if ?
Summary.
What if the inductor and capacitor were exchanged? Find the real and re­
active power for this case.14
The real (time-average) and reactive powers can be determined wholly in
the frequency domain. The real power is the product of the voltage with the current in
phase with the voltage, and the reactive power is the product of the voltage with the
current out of phase with the voltage, with lagging current considered positive. The
real power is the sum of the powers dissipated in the resistances of the load, and the re­
active power is the product of the angular frequency with the difference between the
magnetic and electric peak stored energies within the load.
Complex Power
Definition Of complex power. The complex sum of the real and reactive power
in Eqs. (5 .3 9 ) and (5 .4 0 ) is called the complex power and can be written
tase component of
;Y P
г load is the prodging current gives
>mplete the phasor
the current by the
- P + jQ VA
where S is the complex power in volt-amperes. The complex conjugate of the phasor
current changes the sign o f the imaginary part and thus introduces mathematically the
customary change of sign in the out-of-phase current. The expression for the complex
power in Eq. (5 .6 4 ) also permits us to relax the requirement that the phasor voltage, V ,
be a real quantity. Note that if
where 6V is the phase of the voltage, no longer assumed zero, then
1
=
I
= 1 1 1 Z (6V
-
в)
=> r = HI Z
(-6 V
+
в)
( 5 .6 6 )
Z j
because the complex conjugate changes the sign of the current phase angle. The com ­
plex power, as defined in Eq. (5 .6 4 ), would be
s = \V i*
=
j |v| ze„ x |i|z(-ev+ в)
= * IYIIH
i|V||I| (cos в
+ j sin#)
( 5 .6 7 )
Thus, the phase reference drops out for the complex power defined in Eq. (5.64) and
only the phase difference between voltage and current remains.
|4Р = 2 8 2 watts and Q = - 7 1 1 VARs.
5.2 POWER AND ENERGY IN THE FR
тщщщщ I
Complex power
Find the complex power into the circuit in Fig. (5.9).
power
S o l u t io n :
The phasor voltage is V = \2 0 j2 Z 0° V and the current we found in Fig. 5.10 to be
I =6.63л/2
56.4CA in polar form or I = (3.67 —j5.53)*j2 A in rectangular form. The
complex power is
S = \ VI* i (120^2 + jO) (3.67 V2 + j 5.5372)
(5.68)
= 120 x 3.67 + j 120 x 5.53 =
440
+ ;663
VA
"~ cT "
Thus, the real power is 440 W and the rcactive power is +663 VAR. The positive sign indi­
cates predominant magnetic energy storage. The J l ’s all dropped out, as they always do in
power calculations.
ад
д&й
What if the resistor were increased to 12 O? What would be the complex
power in that case? 15
Apparent power.
The com plex pow er is a com plex num ber yielding information
about the flow of tim e-average pow er and the shuttling o f loaned energy betw een source
and load in an ac circuit. The m agnitude o f the com plex power, 5 = |S | is the apparent
power. The apparent pow er results when you m easure the voltage and current with meters and m ultiply the m easured values w ithout regard for phase . 16
S = |S | — | lYIIH = V t It VA
(5.69)
A pparent pow er is im portant as a m easure o f the operating lim its in electrical equip­
m ent such as transform ers, m otors, and generators. Losses in the w ires are proportional
to the square o f the current in the m achine regardless o f the phase, w hereas losses in the
m agnetic m aterials are roughly proportional to the square o f the operating voltage. Ma­
chine lim its are established by losses. B ecause electrical m achinery is operated with the
voltage m ore or less constant, apparent pow er lim its im ply current lim its. In the previ­
ous exam ple the apparent pow er is 120 x 6.63 = 796 VA.
Summary. The com plex pow er gives concise inform ation about pow er and energy
flow in an ac circuit. T he real part o f the com plex pow er is the tim e-average power in
watts. The real pow er heats resistors, turns m otors, and m akes the electric meter re­
volve. The im aginary part o f the com plex pow er is the reactive pow er in VARs, which
is proportional to the electrical energy lent to the load by the ac source tw ice each cy­
cle. The reactive pow er is considered positive when the load is inductive and negative
§
when the load is capacitive, although the latter is rare in pow er system s. The m agnitude
o f the com plex pow er is the apparent pow er in volt-am peres and indicates the operating
level o f a pow er system.
triangle
n Fig. 5.10 to be
angular form. The
Power triangles. The apparent power, real power, and reactive pow er form a right
triangle. Figure 5.12 show s this p o w e r tria n g le for the previous exam ple. The angle at
the origin is в, the angle o f the im pcdance, + 56.4° in this case. W hen the current lags
the voltage, the pow er triangle is draw-n above the л-axis.
C om plex power
Reactive power
Q = 663 VAR
S= P + j Q
= 4 4 0 + j 663 VA
positive sign indithey always do in
A p p a re n t p o w e r = 7 9 6 VA
6 = + 5 6 .4 •
id be the complex
Figure 5.12 The power triangle
pictures complex, apparent, real, and
reactive power.
Real power
P = 440 W
M ing information
iy between source
S| is the apparent
1 current with me-
Because the various kinds o f pow ers form a triangle, and because the pow er factor
is the cosine o f an angle o f that triangle, there follow a host o f form ulas relating pow er
factor with the various types o f pow ers. We now list several o f these, w hich you can
verify from the definitions and com m on trigonom etric identities.
5 = J p 2 + Q2
P
PF= - = ■
l electrical equip;
:s are proportional
ereas losses in the ■
iting voltage. Ma-: i
operated with the
rnits. In the previ- 4
jow er and energy;
-average power in
electric meter re:r in VARs, which
•ce twice each cy:tive and negative
(5 70 )
P
(5.71)
Q = ± S j 1 - (P F )2 ( + for lagging PF)
( 5.72)
Based on these form ulas, and others that can easily be derived, a variety o f problem s in
ac pow er system s can be analyzed, o f w hich the follow ing is typical.
Determining apparent power, real power, and reactive power from
measurements. Figure 5.8 show s an am m eter (A), voltm eter (V), and w attm eter (P)
m etering an RLC load. We assum e these m eters are ideal and indicate m easured values
of Im, Vm, and Pm. Because the m eters indicate effective values, the m easured apparent
pow er would be the product o f the m eter indications:
»— V
r mIлт
(5.73.);
and Eq. (5.71) w ould give the m easured pow er factor m agnitude, (PF),ri --PJS,,,.11 The
m easured reactive pow er follow s from Eq. (5.70).
17W hether leading or lagging cannot be determ ined from these measurements.
;
5.2
POWER AND ENERGY IN THE FREQUENCY DOMAI
■ иви?
л-
■
■ 4.s>
-ff8r£‘rwteи&и
G« = ± Js 2m
- P 2m = ± J(vaim)2 -P]
Motors
A 230-V motor has a mechanical output power of 3 horsepower (hp). The input current, volt­
age, and power are measured to be 226 V, 15.6 A, and 2920 W, respectively. Calculate the ef­
ficiency, the power factor, and reactive power. Draw a phasor diagram, assuming lagging
current.
S o l u t io n :
The measured apparent power is
Sm = Vm
= 226 x 15.6 = 3530 VA
The measured power factor is
^ = 2_920 = o 828
S„
3530
which corresponds to a phase angle of cos_ l(0.828) = 34.1°, Eq. (5.41), assumed lagging cur­
rent. The reactive power follows from Eq. (5.74):
Q = + J s 2m - P2„, = V(3530)2 - (2920)2 = +1980 VAR
For the phasor diagram in Fig. 5.13, we have used the voltage for the phase reference, and
shown the current lagging. The phase angle of the current follows from the power factor,
9 = cos - 1 (0.828) = 34.1°. The power triangle showing the real, reactive, and apparent powers
would have the same angle but would be drawn above the real axis, similar to Fig. 5.12.
Im
и
V = 2 26V T Z 0"
Re
34.1*
1= 15.61 /Tz-
What if the current were leading? What would change ? 18
The vocabulary and units of power in ac circuits. We now summarize the
four types o f pow er in ac circuits:
18Only the signs o f the phase angle and the reactive power.
..
: si Ш И П
Л
| Щ. *$1
Щ
f if *мЁ!
,
i *:“ '
4
■ж
,
m Sjt s mm
r
, , a >['1Ш
ii
Я
1
Ыш
P = real pow er = tim e-average pow er in watts
Q = reactive pow er in V A R s,19 volt-am peres reactive
s = apparent pow er in VAs, volt-am peres
com plex pow er in VAs
s=
The three units for ac pow er— w atts, volt-am peres reactive, and volt-am peres— all have
the scientific units for power, 3/s. T he related units o f kW, к VAR , and kVA (к = 103)
and MW, MVAR, and MVA (M = 106) are also com m on in the pow er industry. We use
different units to clarify com m unication when speaking o f the various kinds of “pow er”
in ac circuits.
p u t current, voltC a lc u la te the efissu m in g lagging
Reactive Power in Power Systems
Importance Of reactive power: Pow er com panies have to be careful about the
reactive pow er load on their system s. As stated before, the lim its o f larger pow er equip­
ment such as generators and transform ers are described by the apparent power, the Py­
thagorean sum o f the real and reactive pow ers. Thus, if the reactive pow er becom es
large, a piece o f equipm ent may becom e overloaded even though the real pow er is m od­
erate. Also, the reactive energy m ust be transported from generator to user, often over
great distances. Reactive pow er increases line currcnt, and hence increases line losses.
The pow er com pany m ight surcharge industrial custom ers w hose requirem ents for reac­
tive pow er are great to pay for line losses.
ise reference, and
th e p o w er factor,
To understand
how to correct the
power factor of a
i ap p aren t powers
to F ig. 5.12.
Power-factor correction. O ften, the industrial consum er can save m oney by plac­
ing a bank o f capacitors in parallel with an inductive load to store energy locally. In ef­
fect, they receive the stored energy from the pow er com pany only once and then keep
it “ in house” w ith the capacitors. The custom er thus corrects his pow er factor by creat­
ing a resonance betw een the electric energy stored by the added capacitance and the
m agnetic energy used by m otors or other heavy equipm ent.
Power-factor correction: current method. The effect o f adding parallel ca­
pacitance to correct the pow er factor can be understood from the phasor diagram of the
currents. Figure 5 .1 4 (b ) show s the current phasors with the voltage as the phase refer­
ence. The load current, Itomh lags the voltage due to the m agnetic equipm ent in the load.
The in-phase current does useful work, and the out-of-phase current supplies the re-
power-factor
s representing
:nt.
I out-of-phase
F i g u r e 5 . 1 4 (a) The capacitor corrects
the lagging power factor o f the load; (b) the
leading current of the capacitor cancels p an
or all of the lagging out-of-phase current of
the load. The input current is thus reduced.
ja c iio r
led fo r
:r- fa c lo r
v
s u m m a r iz e the
rectio n
19Rhymes with “jars.”
5 .2
POWER AND ENERGY IN THE FREQUENCY DOMAII
■
Ж Я Л fVb
i
" ‘Hi
ШЩШШР
■
h
-
шш
quired m agnetic energy. If the pow er factor is too low, capacitors are added in parallel
with the load to draw leading current, w hich has the effect to im prove the pow er
from PF = cos в to PF' = cos в'. The capacitive current is Ic = a>CVe and the details
am ount to a sim ple problem in trigonom etry, as show n by the follow ing exam ple.
Power factor correction (current method)
A 460-V, 60-H z load uses 12 kW at a lagging pow er factor o f 0.75. W hat capacitance should
be placed in parallel w ith the load to correct the pow er factor to 0.9 lagging?
S o l u t io n :
T he rm s load current is
12,000
P
load ‘
= 34.8 A
V x P F ~ 460 x 0.75
at a lagging phase angle o f
d = c o s” 1(0.75) = 41.4°
(5
W ith the voltage as the phase reference, the current is
I load = 3 4 .8 7 2 Z —4 1 .4 ° = (26.1 - j 2 3 . 0 ) j 2
A
(5
T he parallel capacitor will draw leading current but w ill not change the in-phase component.
F or a pow er factor o f 0.9 lagging ( 0 = —25.9°), the resultant out-of-phase com ponent must
23.0 - l c = 26.1 tan 25.9° = 12.6
(5
Ic = 10.4 A
w here a lagging out-of-phase com ponent is considered positive. T he required capacitance is
therefore, 59.8 ЦР.
W hat if the corrected pow er factor m ust be 0.95 lagg in g ?20
P0Wer-f3Ct0r correction: power method. Pow er capacitors m anufactured for
pow er-factor correction are rated by voltage and reactive pow er .21 T hus, the real and
reactive pow er can be calculatcd directly, as show n by rew orking the previous example
on this basis.
2083.2 |iF.
21 The frequency is assumed to be 60 Hz in the United States.
’
>. V<- с -.V'vW.’ -• .Vi!-».
И'* *
~
ш я р
з ’У!
<
equivalent ci
to establish 1
load to with<