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.

~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.

~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.

~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.

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.

~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_.

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.

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.

~"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.

~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.

~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.

~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.

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.

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.

~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.

~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.

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.

~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.

~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.

~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.