CHAPTER XLVIII
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.
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.
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.
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.
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 ÷ √2 |
| = | maximum voltage × maximum current | |
| √2 | ||
| = | 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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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 + (2πfL)2) | |||
| 800 | ||||
| = | = | 6.7 amperes | ||
| √(752 + (2 × 3.1416 × 60 × .25)2) |