THE POWER FACTOR
The determination of the power in a direct current circuit is a simple
matter since it is only necessary to multiply together the volts and
amperes to obtain the output in watts. In the case of alternating
current circuits, this holds true only when the current is in phase
with the pressure--a condition rarely found in practice.
When the current is not in phase with the pressure, the product of
volts and amperes as indicated by the voltmeter and ammeter must be
multiplied by a coefficient called the _power factor_ in order to
obtain the _true watts_, or actual power available.
There are several ways of defining the power factor, any of which
requires some explanation. The power factor may be defined as: _The
number of watts indicated by a wattmeter, divided by the apparent
watts_, the latter being the _watts as measured by a voltmeter and
ammeter_.
The power factor may be expressed as being equal to
_true power_ _true watts_ _true watts_
---------------- = ---------------- = -------------------
_apparent power_ _apparent watts_ _volts_ × _amperes_
~Ques. What are the true watts?~
Ans. The watts as measured by a wattmeter.
~Ques. What are the apparent watts?~
Ans. The watts obtained by multiplying together the simultaneous
voltmeter and ammeter readings.
~Ques. What is usually meant by power factor?~
Ans. The multiplier used with the apparent watts to determine how much
of the power supplied is available.
[Illustration: FIG. 1,343.--Marine analogy of power factor. A ferry
boat in crossing a river to a slip C would head for some point B
up stream from C to allow for the effect of the tide. Under such
conditions the actual motion (referred to the water) would be from A
to B, and the apparent motion, from A to C. Accordingly, the energy
expended in propelling the boat from A to B in still water, will
propel it from A to C when the tide is running in the direction of
the arrow. The effect of the tide is the same as that of inductance
or capacity in an alternating circuit, that is, it puts the applied
force or thrust (impressed volts) out of phase with the motion of the
boat (amperes), this phase difference being indicated by the angle
BAC or φ. Now, _work_ (watts) is the product of two factors, pressure
(volts) and distance (amperes); accordingly, the apparent work done in
propelling the boat from A to C is the product of the _thrust of the
paddle wheels multiplied by_ AC, which in analogy corresponds to the
product of voltmeter and ammeter readings at the alternator, called
"kva." Actually, however, the power is only applied from A to B, the
boat being carried sidewise by the tide, as it crosses, a distance BC
which represents no energy expended by the paddle wheels. In analogy,
the actual power, expended in propelling boat from A to B corresponds
to the wattmeter reading in an alternating current power circuit. To
obtain the actual work done on the boat, the product of its apparent
motion × thrust must be multiplied by a coefficient or _power factor_
because the thrust is applied at an angle to the apparent motion,
the power factor being equal to the cosine of this angle, (φ) or
AB ÷ AC. Similarly, when there is phase difference between pressure
and alternating current, the voltmeter and ammeter readings must be
multiplied by the power factor or cos φ to give the output of an
alternator available for external work, the excess power indicated
by ammeter and voltmeter readings, performing no external work, but
causing objectionable heating of the alternator.]
~Ques. Upon what does the power factor depend?~
Ans. Upon the relative amounts of resistance inductance and capacity
contained in the circuit.
~Ques. How does the power factor vary in value?~
Ans. It varies from one to zero.
The power factor, as will be shown later, is equal to _the
cosine of the angle of phase difference_; its range then is
from one to zero because these are the limiting values of the
cosine of an angle (neglecting the + or-sign).
~Ques. What is the effect of lag or lead of the current on the power
factor?~
Ans. It causes it to become less than one.
[Illustration: FIG. 1,344.--Method of drawing the power curve from the
pressure and current curves. As shown, the same scale is used for all
curves. This as a rule, makes the power curve inconveniently high,
hence it is usually drawn to smaller scale as in fig. 1,345.]
~How to Obtain the Power Curve.~--Since under any phase condition, the
power at any instant is equal to the product of the pressure multiplied
by the current at that instant, a curve may be easily plotted from the
pressure and current curves, giving the instantaneous values of the
power through a complete cycle.
In fig. 1,344, from the zero line of the current and pressure
curves, draw any ordinate as at F cutting the current curve at
G and the pressure curve at G'. The values for current and
pressure at this point are from the scale, 2 amperes and 3.7
volts. Since watts = amperes × volts, the ordinate FG is to be
multiplied by ordinate FG' that is,
2 × 3.7 = 7.4.
Project up through F the ordinate FG" = 7.4, and this will
give one point on the power curve.
Similarly at another point, say M, where the current and
pressure are maximum
MS × MS' = MS", that is
3 × 5 = 15
giving S" another point on the curve. Obtaining several points
in this way the power curve is then drawn through them as shown.
[Illustration: FIG. 1,345.--Usual method of drawing power curve from
the pressure and current curves. A smaller scale is employed for the
power curve in order to reduce its height.]
~Ques. Why is the power curve positive in the second half of the period
when there are negative values of current and pressure?~
Ans. Because the product of two negative quantities is positive.
~Ques. Does fig. 1,344 represent the usual way of drawing a power
curve?~
Ans. Since ordinates of the power curve are products of the current and
pressure ordinates, they will be of inconvenient length if drawn to
the same scale; it is therefore customary to use a different scale for
the power ordinates, as in fig. 1,345.
The illustration is lettered identical with fig. 1,344, with
which it should be compared.
~Synchronism of Current and Pressure; Power Factor Unity.~--The current
and pressure would be in phase as represented in fig. 1,346 were it
possible to have a circuit containing resistance only. In actual
practice all circuits contain at least a small amount of reactance.
[Illustration: FIG. 1,346.--Synchronism of current and pressure. Power
curve showing that the power factor is unity. This is indicated by the
fact that the power curve does not project below the base or zero line.]
A circuit supplying nothing but incandescent lamps comes very nearly
being all resistance, and may be so considered in the discussion here.
Fig. 1,347 illustrates a circuit containing only resistance. In such a
circuit the pressure and current (as shown in fig. 1,346) pass through
zero and through their maximum values together.
Multiplying instantaneous values of volts and amperes will give the
power curve, as before explained, whose average value is half-way
between the zero line and the maximum of the curve; that part of the
power curve above the line of average power WW, exactly filling the
open space below the line WW. That is,
average power = maximum power ÷ √2̅
= maximum voltage × maximum current / √2̅
= virtual voltage × virtual current.
This latter is simply the product of the voltmeter and ammeter readings
which gives the watts just the same as in direct current.
[Illustration: FIG. 1,347.--Diagram of circuit containing only
resistance; in such a circuit the power factor is unity.]
~Ques. What should be noticed about the power curve?~
Ans. Its position with respect to the zero line; it lies wholly
above the zero line which denotes that all the power delivered to
the circuit except that dissipated by friction is useful, that is,
the power factor is unity. Hence, _to keep the power factor as near
unity as possible is one of the chief problems in alternating current
distribution_.
~Ques. Can the power factor be less than unity if the current and
pressure be in phase?~
Ans. Yes, if the waves of current and voltage be distorted as in fig.
1,348.
~Effect of Lag and Lead.~--In an alternating circuit the amount of
power supplied depends on the phase relationship of the current and
pressure. As just explained, when there is synchronism of current and
pressure, that is, when they are in phase (as in fig. 1,346) the power
factor is unity, assuming no distortion of current and pressure waves.
In all other cases the power factor is less than unity that is, _the
effect of lag or lead is to make the power factor less than unity_.
[Illustration: FIG. 1,348.--Case of synchronism of current and pressure
with power factor less than unity. Suppose the waves of current and
voltage to be in phase, but distorted in form, and not symmetrical, so
that they do not run uniformly together, as shown in the figure. Then
the real power factor may not be unity, although indicated as such by
the power factor meter. However, the switchboard instruments are made
to show the angle of lag as the power factor, because the error due to
wave distortion is generally too small to be considered.]
[Illustration: FIG. 1,349.--Effect of lag on the power factor. When
the current lags behind the pressure the power factor becomes less
than unity. It will be seen that the power curve projects below the
zero line giving the shaded area which represents negative power which
must be subtracted from the + areas above the zero line to get the net
power. In the figure the line WW' is drawn at a height corresponding
to the average power, and HN at a height corresponding to the average
power that would be developed if the current were in phase with the
pressure. The power factor then is represented by M ÷ S, and by
inspection of the figure it is seen that this is less than unity.]
[Illustration: FIG. 1,350.--Effect of lead on the power factor. When
the current is in advance of the pressure the power factor becomes
less than unity. The curve, as shown, projects below the zero line,
giving the shaded area which represents negative power which must
be subtracted from the + areas above the zero line to get the net
power. As in fig. 1,349, the line WW' at a height M represents the
average power, and HN the average power for synchronism of current and
pressure. The power factor then is M ÷ S which is less than unity.]
The effect of lag on the power factor may be illustrated by fig. 1,349,
in which the angle between the pressure and current, or the angle of
lag is taken as 40°, corresponding to a power factor of .766. Plotting
the power curve from the products of instantaneous volts and amperes
taken at various points, the power curve is obtained, a portion of
which lies below the horizontal line. The significance of this is that
at certain times, the current is flowing in the opposite direction to
that in which the impressed pressure would send it. During this part
of the period conditions are reversed, and the power (indicated by the
shaded area), instead of being supplied by the source to the circuit,
is being supplied by the circuit to the source.
[Illustration: FIG. 1,351.--Steam engine analogy of power factor. The
figure represents an indicator card of an engine in which the steam
distribution is such that the steam is expanded below the back pressure
line, that is below the pressure of the exhaust. This results in
_negative work_ which must be overcome by the _momentum_ or _kinetic
energy_ previously stored in the fly wheel, and which is represented on
the diagram by the shaded loop S. If the exhaust valve had opened at G,
the amount of work done during the revolution would be represented by
the area M, but continuing the expansion below the back pressure line,
the work done is M - S. This latter case as compared with the first
when expansion does not continue below the back pressure line gives an
efficiency (power factor) of (M - S) ÷ M, the shaded area representing
so much loss.]
This condition is exactly analogous to the case of a steam
engine, expanding the steam below the back or exhaust pressure,
a condition sometimes caused by the action of the governor in
considerably reducing the cut off for very light load. An
indicator diagram of such steam distribution is shown in fig.
1,351. This gives a negative loop in the diagram indicated by
the shaded section.
It must be evident that the average pressure of the shaded loop
portion of the diagram must be subtracted from that of the
other portion, because during the expansion below the exhaust
pressure line, the back pressure is in excess of the forward
pressure exerted on the piston by the expanding steam, and the
engine would accordingly reverse its motion, _were it not for
the energy previously stored up in the fly wheel_ in the form
of _momentum_, which keeps the engine moving during this period
of back thrust. Evidently the shaded area must be subtracted
from the positive area to obtain the net work done during the
stroke. Hence following the analogy as far as possible if M
work (watts) be done during each revolution (cycle) when steam
does not expand below back pressure (when current and pressure
are in phase), and S negative work (negative watts) be done
when steam expands below back pressure (when there is lag), the
efficiency (power factor) is (M - S) ÷ M.
[Illustration: FIG. 1,352.--Power curve illustrating the so-called
wattless current in which case the power factor is zero. By noting that
the curve projects equally on each side of the zero line, the + power
areas equal the negative power areas, hence the summation of these
areas for the period is zero, that is, the two + areas minus the two
shaded areas equal zero. It should be noted that the line of average
power WW', which is visible in the other figures, here coincides
with the zero line, and the average power then is zero, since the
positive part above the zero line is equal to and offsets the negative
(shaded) part below the line. This is the case of "wattless" current
and (considering a circuit with resistance so small that it may be
considered as zero) shows plainly the possibility of having full load
current and voltage on a circuit yet delivering no power, the current
simply surging to and fro without an actual transfer of power.]
~"Wattless Current;" Power Factor Zero.~--When the power factor is
zero, it means that the phase difference between the current and the
pressure is 90°.
The term _wattless current_, as understood, does not indicate an
absence of electrical energy in the circuit; its elements are there,
but not in an available form for external work. The false power due
to the so called wattless current pulsates in and out of the circuit
without accomplishing any useful work.
An example of wattless current, showing that the power factor
is zero is illustrated in fig. 1,353. Here the angle of lag is
90°, that is, the current is 90° behind the pressure.
The power curve is constructed from the current and pressure
curves, and, as shown in the diagram, it lies as much below the
zero line as above, that is, the two plus power areas which
occur during each period are equal to the two negative (shaded)
power areas, showing that the circuit returns as much energy
as is sent out. Hence, the total work done during each period
is zero, indicating that although a current be flowing, this
current is not capable of doing external work.
[Illustration: FIG. 1,353.--Example of wattless current showing that
the power factor is zero when the phase difference between current and
pressure is 90°. For zero power factor the current may _lead_ 90° as in
fig. 1,352, or _lag_ 90° as here shown. Since the shaded or negative
areas = the plus areas, the average power (indicated by WW' which
coincides with the zero line) is zero, that is the circuit is carrying
current under pressure yet delivering no power, hence, the power factor
is zero.]
~Ques. Is the condition as just described met with in practice?~
Ans. No.
~Ques. Why not?~
Ans. The condition just described involves that the circuit have no
resistance, all the load being reactance, but it is impossible to have
a circuit without some resistance, though the resistance may be made
very small in comparison to the reactance so that a close approach to
wattless current is possible.
~Ques. Give some examples where the phase difference is very nearly
90°.~
Ans. If an alternator supplies current to a circuit having a very small
resistance and very large inductance, the current would lag nearly 90°
behind the pressure. The primary current of a transformer working with
its secondary on an open circuit is a practical example of a current
which represents very little energy.
[Illustration: FIG. 1,354.--Performance curves of General Electric
single phase repulsion induction motor.]
~Ques. When the phase difference between the current and pressure is
90°, why is the current called "wattless"?~
Ans. Because the product of such a current multiplied by the pressure
does not represent any watts _expended_.
A man lifting a weight, and then allowing it to descend the
same distance to its initial position, as shown in figs. 1,355
to 1,357, presents a mechanical analogy of wattless current.
Let the movement of the weight represent the current and the
weight the pressure. Then calling the weight 10 pounds (volts),
and the distance two feet (amperes). The work done by the man
(alternator) on the weight in lifting it is
10 pounds × 2 feet = 20 foot pounds (1)
(10 volts × 2 amperes = 20 watts.)
The work done on the man by the weight in forcing his hand down
as his muscles relax is
10 pounds × 2 feet = 20 foot pounds (2)
(10 volts × 2 amperes = 20 watts.)
From (1) and (2) it is seen that the _work done by the man
on the weight is equal to the work done by the weight on the
man_, hence no useful work has been accomplished; that is, the
potential energy of the weight which it originally possessed
has not been increased.
[Illustration: FIGS. 1,355 to 1,357.--Mechanical analogy of wattless
current. If a man lift a weight any distance, as from the position of
fig. 1,355 to position of fig. 1,356, he does a certain amount of work
on the weight giving it potential energy. When he lowers it to its
original position, as in fig. 1,357, the weight loses the potential
energy previously acquired, that is, it is given back to the man, the
"system" (man and weight) having returned to its original condition as
in fig. 1,355. During such a cycle, the work done by the man on the
weight is equal to the work done by the weight on the man and no useful
external work has been accomplished.]
~Why the Power Factor is equal to Cos φ.~--In the preceding figures
showing power curves for various phase relations between current and
pressure, the curves show the instantaneous values of the fluctuating
power, but what is of more importance, is to determine the average
power developed.
When the current is in phase with the pressure, it is a simple matter,
because the power or
_watts = amperes_ × _volts_
that is, the product of the ammeter and voltmeter readings will give
the power. However, the condition of synchronism of current and
pressure hardly ever exists in practice, there being more or less phase
difference.
[Illustration: FIG. 1,358.--Method of obtaining the _active component_
of the current; diagram illustrating why the power factor is equal to
cos φ. If AB and AC be respectively the given current and pressure,
or readings of the ammeter and voltmeter, and φ the angle of phase
difference between current and pressure, then drawing from B, BD
perpendicular to AC will give AD the active component. Now, true
power = AC × AD, but AD = AB cos φ, hence true power = AC × AB cos
φ. Again, apparent power = AC × AB, and since true power = apparent
power × power factor, the power factor = cos φ.]
When the current is not in phase with the pressure, it is considered as
made up of two components at right angles to each other.
1. _The active component_, in phase with the pressure;
2. _The wattless component_, at right angles to the pressure.
With phase difference between current and pressure the product of
ammeter and voltmeter readings do not give the true power, and in
order to obtain the latter, the _active component_ of the current in
phase with the pressure must be considered, that is,
_true power_ = _volts_ × _active amperes_ (1)
The active component of the current is easily obtained
graphically as in fig. 1,358.
With any convenient scale draw AB equal to the current as given
or read on the ammeter, and AC, equal to the pressure, making
the angle φ between AB and AC equal to the phase difference
between the current and pressure.
From B, draw the line BD perpendicular to AC, then BD will be
the wattless component, and AD (measured with the same scale as
was used for AB) the active component of the current, or that
component in phase with the pressure.
Hence from equation (1)
true power = AC × AD (2)
Now in the right triangle ABD
AD
-- = cos φ
AB
from which
AD = AB cos φ (3)
Substituting this value of AD in equation (2) gives
true power = AC × AB cos φ (4)
Now the power factor may be defined as: _that quantity by which the
apparent watts must be multiplied in order to give the true power_.
That is
_true power = apparent watts_ × _power factor_ (5)
Comparing equations (4) and (5), AC × AB in (4) is equal to the
apparent watts, hence, the power factor in (5) is equal to cos φ. That
is, _the power factor is numerically equal to the cosine of the angle
of phase difference between current and pressure_.
EXAMPLE I.--An alternator supplies a current of 200 amperes at
a pressure of 1,000 volts. If the phase difference between the
current and pressure be 30°, what is the true power developed?
In fig. 1,359, draw AB to scale, equal to 200 amperes, and
draw AC of indefinite length making an angle of 30° with AB.
From B, draw BD perpendicular to AC which gives AD, the active
component, and which measured with the same scale as was used
in laying off AB, measures 173.2 amperes. The true power
developed then is
true watts = 173.2 × 1,000 = 173.2 kw.
The true power may be calculated thus:
From the table cos 30° = .866, hence
true watts = 200 × 1,000 × .866 = 173.2 kw.
[Illustration: FIG. 1,359.--Diagram for obtaining the active component
of the current in a circuit having a current of 200 amperes and angle
of lag of 30°.]
EXAMPLE II.--If in an alternating current circuit, the
voltmeter and ammeter readings be 110 and 20 and the angle of
lag 45°, what is the apparent power and true power?
The apparent power is simply the product of the current and
pressure readings or
apparent power = 20 × 110 = 2,200 watts
The true power is the product of the apparent power multiplied
by the cosine of the angle of lag. Cos 45° = .707, hence
true power = 2,200 × .707 = 1,555.4 watts.
~Ques. Does the power factor apply to capacity reactance in the same
way as to inductance reactance?~
Ans. Yes. The angles of lag and of lead, are from the practical
standpoint, treated as if they lay in the first quadrant of the circle.
Even the negative sign of the tangent φ when it occurs is simply
used to determine whether the angle be one of lag or of lead, but in
finding the value of the angle from a table it is treated as a positive
quantity.
[Illustration: FIG. 1,360.--Diagram for obtaining the power factor for
example II. With convenient scale, lay off AB = 20 amperes. From A draw
AC at 45° to AB, and from B, draw BD perpendicular to AC. Then, the
power factor which is equal to cosine of angle of lag, = AD ÷ AB = (by
measurement) 14.15 ÷ 20 = .707.]
~Ques. In introducing capacity into a circuit to increase the power
factor what should be considered?~
Ans. The cost and upkeep of the added apparatus as well as the power
lost in same.
~Ques. How is power lost in a condenser?~
Ans. The loss is principally due to a phenomenon known as _dielectric
hysteresis_, which is somewhat analogous to magnetic hysteresis. The
rapidly alternating charges in a condenser placed in an alternating
circuit may be said to cause alternating polarization of the
dielectric, and consequent heating and loss of energy.
~Ques. When is inductance introduced into a circuit to increase the
power factor?~
Ans. When the phase difference is due to an excess of capacity.
EXAMPLE.--A circuit having a resistance of 3 ohms, and a
resultant reactance of 4 ohms, is connected to a 100 volt line.
What is: 1, the impedance, 2, the current, 3, the apparent
power, 4, the angle of lag, 5, the power factor, and 6, the
true power?
1. _The impedance of the circuit._
Z = √(3² + 4²) = 5 ohms.
2. _The current._
current = volts ÷ impedance = 100 ÷ 5 = 20 amperes.
3. _The apparent power._
apparent power = volts × amperes = 100 × 20 = 2,000 watts.
4. _The tangent of the angle of lag._
tan φ = reactance ÷ resistance = 4 ÷ 3 = 1.33. From table
of natural tangents (page 451) φ = 53°.
5. _The power factor._
The power factor is equal to the cosine of the angle of
lag, that is, power factor = cos 53° = .602 (from table).
6. _The true power._
The true power is equal to the apparent watts multiplied by
the power factor, or
true power = volts × amperes × cos φ
= 100 × 20 × .602 = 1,204 watts.
~Ques. Prove that the power factor is unity when there is no resultant
reactance in a circuit.~
Ans. When there is no reactance, tan φ which is equal to
reactance ÷ resistance becomes 0 ÷ R = 0. The angle φ (the phase
difference angle) whose tangent is 0 is the angle of 0 degrees. Hence,
the power factor which is equal to cos φ = cos 0° = 1.
[Illustration: FIGS. 1,361 to 1,365.--Diagrams illustrating why the
power factor is unity or one when there is no resultant reactance
in the circuit, that is, when the circuit is resonant or has only
resistance. The power factor is equal to the cosine of the angle of
lag (or lead). In the figures this angle is BAC or φ and the value of
the _natural cosine_ AC gives the power factor. By inspection of the
figures, it is evident that decreasing the reactance decreases the
angle φ and increases cos φ or the power factor. The circular arc in
each figure being at unity distance from the center A, the power factor
with decreasing reactance evidently approaches unity as its limit, this
limit being shown in fig. 1,365 where the reactance B'C' = 0.]
~Ques. What is the usual value of the power factor in practice?~
Ans. Slightly less than one.
~Ques. Why is it desirable to keep the power factor near unity?~
Ans. Because with a low power factor, while the alternator may be
carrying its full load and operating at a moderate temperature, the
consumer is paying only for the actual watts which are sent over the
line to him.
[Illustration: FIG. 1,366.--Diagram illustrating power factor test,
when on non-inductive and inductive circuits. The instruments are
connected as shown and by means of the double throw switch can
be put on either the non-inductive or inductive circuit. First
turn switch to left so that current passes through the lamps;
for illustration, the following readings are assumed: ammeter
10, voltmeter 110, and wattmeter 1,100. The power factor then is
wattmeter reading ÷ volts × amperes = 1,100 actual watts ÷ 1,100
apparent watts = 1, that is, on non-inductive circuit the power
factor is unity. Now throwing the switch to the right connecting
instruments with the inductive circuits, then for illustration
the following readings may be assumed: ammeter 8, voltmeter 110,
and wattmeter 684. Now, as before, power factor = wattmeter
reading ÷ volts × amperes = 684 ÷ (8 × 110) = 684 ÷ 880 = .78.]
For instance, if a large alternator supplying 1,000 kilowatts
at 6,600 volts in a town where a number of induction motors are
used on the line be operating with a power factor of say .625
during a great portion of the time, the switchboard instruments
connected to the alternator will give the following readings:
Voltmeter 6,600 volts; ammeter 242.4 amperes; power factor
meter .625.
The apparent watts would equal 1,600,000 watts or 1,600
kilowatts, which, if multiplied by the power factor .625 would
give 100,000,000 watts or 1,000 kilowatts which is the actual
watts supplied. The alternator and line must carry 242.4
amperes instead of 151 amperes and the
difference 242.4 - 151 = 91.4 amperes represents a _wattless
current_ flowing in the circuit which causes useless heating of
the alternator.
The mechanical power which is required to drive the alternator
is equivalent to the actual watts produced, since that portion
of the current which lags, is out of phase with the pressure
and therefore requires no energy.
~Ques. How are alternators rated by manufacturers in order to avoid
disputes?~
Ans. They usually rate their alternators as producing so many kilovolt
amperes instead of kilowatts.
[Illustration: FIG. 1,367.--Ayrton and Sumpner method of alternating
current power measurement. Three voltmeters are required, and
accordingly the method is sometimes called the three voltmeter method.
It is a good method where the voltage can be regulated to suit the
load. In the figure, let the non-inductive resistance R be placed in
series with the load AB. Measure the following voltages: V across the
terminals of R, V₁ across the load AB, and V₂ across both, that is from
A to C. Then, true watts = (V₂² - V₁² - V²) ÷ 2R. The best conditions
are when V = V₁, and, if R = ½ ohm, then W = V₂² - V₁² - V².]
~Ques. What is a kilovolt ampere (kva)?~
Ans. A unit of apparent power in an alternating current circuit which
is equal to one kilowatt when the power factor is equal to one.
The machine mentioned on page 1,120 would be designed to carry
151 amperes without overheating and also carry slight overloads
for short periods. It would be rated as 6.6 kilo volts and 151
amperes which would equal approximately 1,000 kilowatts when
the power factor is 1 or unity, and it should operate without
undue heating. Now the lower the power factor becomes, the
greater the heating trouble will be in trying to produce the
1,000 actual kilowatts.
[Illustration: FIG. 1,368--Curves illustrating _power factor_. In a
circuit having no capacity or inductance, the power is given by the
product of the respective readings of the voltmeter and ammeter, as in
the case of a direct current. In the case of a circuit having capacity
or inductance, this product is higher than the true value as found by
a wattmeter, and is known as the _apparent watts_. The ratio _true
watts_ ÷ _apparent watts_ is known as the power factor. The current
flowing in an inductive circuit, such as the primary of a transformer,
is really made up of two components, as already explained, one of which
(the load or active component), is in phase with the pressure, while
the other the magnetizing component, is at right angles to it, that
is, it attains its crest value when the other is at zero, and vice
versa. To illustrate, take a complete cycle divided into 360 degrees
and lay out on it the current required to correspond to a given load
on the secondary of a transformer, say a crest value of 100 amperes,
and at right angles to this lay out the current required for exciting
the magnetic circuit of the transformer, giving A, merely for purposes
of illustration, a crest value of 25 amperes. Combining these curves,
the dotted curve in the figure is obtained and which represents the
resultant current that would be indicated by an ammeter placed in the
primary circuit of the transformer. It will be noted that this current
attains its maximum at a point 14° 2´ later than the load current,
giving the angle of lag. Multiplying the apparent watts by the cosine
of the angle of lag gives the true watts. Now assuming the diagram to
show the full load condition of the transformer, the angle of lag being
14° 2´, the power factor at full load is .97 (.97 being the value of
the natural cosine of 14° 2´ as obtained from table, such as on page
451). With no external load on the transformer, the load component of
the current is that necessary to make up the core losses. For instance,
at 5 amperes, while the magnetizing current remains as before at 25
amperes, the angle of lag becomes 78° 41´ and the power factor .196. It
is thus seen that in transformers, induction motors, etc., the power
factor is a function of the load.]
~Ques. How can the power factor be kept high?~
Ans. By carefully designing the motors and other apparatus and even
making changes in the field current of motors which are already
installed.
~Ques. How is the power factor determined in station operation?~
Ans. Not by calculation, but by reading a meter which forms one of the
switchboard instruments.
[Illustration: FIG. 1,368.--Fleming's combined voltmeter and ammeter
method of measuring power in alternating current circuits. It is
quite accurate and enables instruments in use to be checked. In the
figure, R is a non-inductive resistance connected in shunt to the
inductive load. The voltmeter V measures the pressure across the
resistance XY. A and A₁ are ammeters connected as shown. Then, true
watts = (A₁² - A² - (V/R)²) × R / 2. If the volt meter V take an
appreciable amount of current, it may be tested as follows: disconnect
R and V at Y, and see that A and A₁ are alike; then connect R and V at
Y again, and disconnect the load. A₁ will equal current taken by R and
V in parallel.]
~Ques. When is the power factor meter of importance in station
operation, and why?~
Ans. When rotary converters are used on alternating current lines for
supplying direct currents and the sub-station operators are kept busy
adjusting the field rheostat of the rotary to maintain a high power
factor and prevent overheating of the alternators during the time of
day when there is the maximum demand for current or the peak of the
load.
EXAMPLE.--An alternator delivers current at 800 volts pressure
at a frequency of 60, to a circuit of which the resistance is
75 ohms and .25 henry.
Determine: _a_, the value of the current, _b_, angle of lag,
_c_, apparent watts, _d_, power factor, _e_, true power.
_a. Value of current_
pressure E
current = --------- = -----------------
impedance √(R² + (2π_f_L)²)
800
= --------------------------------- = 6.7 amperes
√(75² + (2 × 3.1416 × 60 × .25)²)
_b. The angle of lag_
reactance 2π_f_L 2 × 3.1416 × 60 × .25
tan φ = ---------- = ------ = --------------------- = 1.25
resistance R 75
φ = angle of lag = 51° 15´ (from table, page 451).
_c. The apparent power_
apparent power = volts × amperes = 800 × 6.7 = 5,360 watts
= 5.36 kva.
_d. The power factor_
power factor = cosine of the angle of lag
= cos 51° 15´ = .626.
_e. The true power_
true power = apparent power × power factor
= 5,360 × .626 = 3,355 watts.
[Illustration: FIG. 1,369.--Wattmeter method of three phase power
measurement. Two wattmeters are required in unbalanced systems as shown
in the illustration. The total power transmitted is then the algebraic
sum of the readings of the two wattmeters. If the power factor be
greater than .5, the power is the arithmetical sum, and if it be less
than .5, the power is the arithmetical difference of the readings.]