CHAPTER LXV
=ALTERNATING CURRENT WIRING=
In the case of alternating current, because of its peculiar behaviour, there are several effects which must be considered in making wiring calculations, which do not enter into the problem with direct current.
Accordingly, in determining the size of wires, allowance must be made for
1. Self-induction; 2. Mutual-induction; 3. Power factor; 4. Skin effect; 5. Corona effect; 6. Frequency; 7. Resistance.
Most of these items have already been explained at such length, that only a brief summary of facts need be added, to point out their connection and importance with alternating current wiring.
=Induction.=--The effect of induction, whether self-induction or mutual induction, is to set up a back pressure of _spurious resistance_, which must be considered, as it sometimes materially affects the calculation of circuits even in interior wiring.
_Self-induction is the effect produced by the action of the electric current upon itself during variations in strength._
=Ques. What conditions besides variations of current strength governs the amount of self-induction in a circuit?=
Ans. The shape of the circuit, and the character of the surrounding medium.
If the circuit be straight, there will be little self-induction, but if coiled, the effect will become pronounced. If the surrounding medium be air, the self-induction is small, but if it be iron, the self-induction is considerable.
=Ques. With respect to self-induction, what method should be followed in wiring?=
Ans. When iron conduits are used, the wires of each circuit should not be installed in separate conduits, because such arrangement will cause excessive self-induction.
The importance of this may be seen from the experience of one contractor, who installed feeders and mains in separate iron pipes. When the current was turned on, it was found that the self-induction was so great as to reduce the pressure to such an extent that the lamps, instead of giving full candle power, were barely red. This necessitated the removal of the feeders and main and re-installing them, so that those of the same circuit were in the same pipe.
=Ques. What is mutual induction?=
Ans. Mutual induction is the effect of one alternating current circuit upon another.
=Ques. How is it caused?=
Ans. It is due to the magnetic field surrounding a conductor cutting adjacent conductors and inducing back pressures therein.
This effect as a rule in ordinary installations is negligible.
=Transpositions.=--The effect of mutual induction between two circuits is proportional to the inter-linkage of the magnetic fluxes of the two lines. This in turn depends upon the proximity of the lines and upon the general relative arrangement of the conductors.
The inductive effect of one line upon another is equal to the algebraic sum of the fluxes due to the different conductors of the first line, considered separately, which link the secondary line.
The effect of mutual induction is to induce surges in the line where a difference of frequency exists between the two currents, and to induce high electrostatic charges in lines carrying little or no current, such as telephone lines.
INDUCTANCE PER MILE OF THREE PHASE CIRCUIT
---------+-------+----------+------------ | Diam. | Distance | Self Size | in | _d_ in | Inductance B. & S. | inch. | inches. | L henrys. ---------+-------+----------+------------ 0000 | .46 | 12 | .00234 | | 18 | .00256 | | 24 | .00270 | | 48 | .00312 | | | 000 | .41 | 12 | .00241 | | 18 | .00262 | | 24 | .00277 | | 48 | .00318 | | | 00 | .365 | 12 | .00248 | | 18 | .00269 | | 24 | .00285 | | 48 | .00330 | | | 0 | .325 | 12 | .00254 | | 18 | .00276 | | 24 | .00293 | | 48 | .00331 | | | 1 | .289 | 12 | .00260 | | 18 | .00281 | | 24 | .00308 | | 48 | .00338 | | | 2 | .258 | 12 | .00267 | | 18 | .00288 | | 24 | .00304 | | 48 | .00314 | | | 3 | .229 | 12 | .00274 | | 18 | .00294 | | 24 | .00310 | | 48 | .00351 | | | 4 | .204 | 12 | .00280 | | 18 | .00300 | | 24 | .00315 | | 48 | .00358 | | | 5 | .182 | 12 | .00286 | | 18 | .00307 | | 24 | .00323 | | 48 | .00356 | | | 6 | .162 | 12 | .00291 | | 18 | .00313 | | 24 | .00329 | | 48 | .00369 | | | 7 | .144 | 12 | .00298 | | 18 | .00310 | | 24 | .00336 | | 48 | .00377 | | | 8 | .128 | 12 | .00303 | | 18 | .00325 | | 24 | .00341 | | 48 | .00384 | | | 9 | .114 | 12 | .00310 | | 18 | .00332 | | 24 | .00348 | | 48 | .00389 | | | 10 | .102 | 12 | .00318 | | 18 | .00340 | | 24 | .00355 | | 48 | .00396 ---------+-------+----------+------------
This effect may be nullified by separating the lines and by transposing the wires of one of the lines so that the effect produced in one section is opposed by that in another. Of two parallel lines consisting of two wires each, one may be transposed to neutralize the mutual inductance.
Fig. 2,678 shows this method. The length L' should be an even factor of L so that to every section of the line transposed there corresponds an opposing section.
The self inductance of lines is readily calculated from the following formula:
L = .000558 {2.303 log (2A ÷ _d_) + .25} per mile of circuit
where
L = inductance of a loop of a three phase circuit in henrys. _Note._--The inductance of a complete single phase circuit = L × 2 ÷ √3. A = distance between wires; _d_ = diameter of wire.
CAPACITY IN MICRO-FARADS PER MILE OF CIRCUIT FOR THREE PHASE SYSTEM
---------+-------+----------+-------------- | Diam. | Distance | Capacity Size | in | A in | C in B. & S. | inch. | inches. | micro-farads ---------+-------+----------+-------------- 0000 | .46 | 12 | .0226 | | 18 | .0204 | | 24 | .01922 | | 48 | .01474 | | | 000 | .41 | 12 | .0218 | | 18 | .01992 | | 24 | .01876 | | 48 | .01638 | | | 00 | .365 | 12 | .0124 | | 18 | .01946 | | 24 | .01832 | | 48 | .01604 | | | 0 | .325 | 12 | .02078 | | 18 | .01898 | | 24 | .01642 | | 48 | .01570 | | | 1 | .289 | 12 | .02022 | | 18 | .01952 | | 24 | .01748 | | 48 | .0154 | | | 2 | .258 | 12 | .01972 | | 18 | .01818 | | 24 | .01710 | | 48 | .01510 | | | 3 | .229 | 12 | .01938 | | 18 | .01766 | | 24 | .01672 | | 48 | .01480 | | | 4 | .204 | 12 | .01874 | | 18 | .01726 | | 24 | .01636 | | 48 | .01452 | | | 5 | .182 | 12 | .01830 | | 18 | .01690 | | 24 | .01602 | | 48 | .01426 | | | 6 | .162 | 12 | .01788 | | 18 | .01654 | | 24 | .01560 | | 48 | .0140 | | | 7 | .144 | 12 | .01746 | | 18 | .01618 | | 24 | .01538 | | 48 | .01374 | | | 8 | .128 | 12 | .01708 | | 18 | .01586 | | 24 | .01508 | | 48 | .01350 | | | 9 | .114 | 12 | .01660 | | 18 | .01552 | | 24 | .01478 | | 48 | .01326 | | | 10 | .102 | 12 | .01636 | | 18 | .01522 | | 24 | .01452 | | 48 | .01304 ---------+-------+----------+--------------
=Capacity.=--In any given system of electrical conductors, a pressure difference between two of them corresponds to the presence of a quantity of electricity on each. With the same charges, the difference of pressure may be varied by varying the geometrical arrangement and magnitudes and also by introducing various dielectrics. The constant connecting the charge and the resulting pressure is called the capacity of the system.
All circuits have a certain capacity, because each conductor acts like the plate of a condenser, and the insulating medium, acts as the dielectric. The capacity depends upon the insulation.
For a given grade of insulation, the capacity is proportional to the surface of the conductors, and universally to the distance between them.
A three phase three wire transmission line spaced at the corners of an equilateral triangle as regards capacity acts precisely as though the neutral line were situated at the center of the triangle.
The capacity of circuits is readily calculated by applying the following formulae:
C = 38.83 sc 10^{-3} / log (D ÷ d) per mile, insulated cable with lead sheath; C = 38.83 × 10^{-3} / log (4h ÷ d) per mile, single conductor with earth return; C = 19.42 × 10^{-3} / log (2A ÷ d) per mile of parallel conductors forming metallic circuit;
in which
C = Capacity in micro-farads; for a metallic circuit, C = capacity between wires;
sc = Specific inductive capacity of insulating material; = 1 for air, and 2.25 to 3.7 for rubber;
D = Inside diameter of lead sheath; d = Diameter of conductor; h = Distance of conductors above ground; A = Distance between wires.
=Frequency.=--The number of cycles per second, or the frequency, has a direct effect upon the inductance reactance in an alternating current circuit, as is plainly seen from the formula.
X_{i} = 2π_f_L
In the case of a transmission line alone; the lower frequencies are the more desirable, in that they tend to reduce the inductance drop and charging current. The inductance drop is proportional to the frequency.
The natural period of a line, with distributed inductance and capacity, is approximately given by __ P = 7,900 / √LC
where L is the total inductance in millihenrys, and C, the total capacity in micro-farads. Accordingly some lower odd harmonic of the impressed frequency may be present which corresponds with the natural period of the line. When this obtains, oscillations of dangerous magnitude may occur. Such coincidences are less likely with the lower harmonics than with the higher.
=Skin Effect.=--The tendency of alternating current to confine itself to the _outer_ portions of a conductor, instead of passing uniformly through the cross section, is called _skin effect_. The effect is proportional to the size of the conductor and the frequency.
=Ques. What effect has "skin effect" on the current?=
Ans. It is equivalent to an increase of ohmic resistance and therefore opposes the current.
If the conductor be large, or the frequency high, the central portion of the conductor carries little if any current, hence the resistance is therefore greater for alternating current than for direct current.
=Ques. For what condition may "skin effect" be neglected?=
Ans. For frequencies of 60 or less, with conductors having a diameter not greater than 0000 B. & S. gauge.
=Ques. How is the "skin effect" calculated for a given wire?=
Ans. Its area in circular mils multiplied by the frequency, gives the ratio of the wire's ohmic resistance to its combined resistance.
That is to say, the factor thus obtained multiplied by the resistance of the wire to direct current will give its combined resistance or resistance to alternating current.
The following table gives these ratio factors for large conductors.
RATIO FACTOR FOR COMBINED RESISTANCE
+---------------+---------+---------------+---------+ | Cir. mils. | Ratio | Cir. mils. | Ratio | | × frequency | factor | × frequency | factor | +---------------+---------+---------------+---------+ | 10,000,000 | 1.00 | 70,000,000 | 1.13 | | 20,000,000 | 1.01 | 80,000,000 | 1.17 | | 30,000,000 | 1.03 | 90,000,000 | 1.20 | | 40,000,000 | 1.05 | 100,000,000 | 1.25 | | 50,000,000 | 1.08 | 125,000,000 | 1.34 | | 60,000,000 | 1.10 | 150,000,000 | 1.43 | +---------------+---------+---------------+---------+
=Corona Effect.=--When two wires, having a great difference of pressure are placed near each other, a certain phenomenon occurs, which is called _corona effect_. When the spacing or distance between the wires is small and the difference of pressure in the wires very great, a continuous passage of energy takes place through the dielectric or atmosphere, the amount of this energy may be an appreciable percentage of the power transmitted. Therefore in laying out high pressure transmission lines, this effect must be considered in the spacing of the wires.
=Ques. How does the corona effect manifest itself?=
Ans. It is visible at night as a bluish luminous envelope and audible as a hissing sound.
=Ques. What is the critical voltage?=
Ans. The voltage at which the corona effect loss takes place.
=Ques. Upon what does the critical voltage depend?=
Ans. Upon the radius of the wires, the spacing, and the atmospheric conditions.
The critical voltage increases with both the diameter of the wires, and the spacing.
The losses due to corona effect increase very rapidly with increasing pressure beyond the critical voltage.
The magnitude of the losses as well as the critical voltage is affected, by atmospheric conditions, hence they probably vary with the particular locality, and the season of the year. Therefore, for a given locality, a voltage which is normally below the critical point, may at times be above it, depending upon changes in the weather.
Such elements as smoke, fog, moisture, or other particles (dust, snow, etc.) floating in the air, increase the losses; rain, however, apparently has no appreciable effect upon the losses. It follows then that in the design of a transmission line, the atmospheric conditions of the particular locality through which the line passes should be considered.
=Ques. How should live wires be spaced?=
Ans. They should be so spaced as to lessen the tendency to leakage and to prevent the wires swinging together or against towers. The spacing should be only sufficient for safety, since increased spacing increases the self-induction of the line, and while it lessens the capacity, it does so only in a slight degree.
The following spacing is in accordance with average practice.
SPACING FOR VARIOUS VOLTAGES
+---------+----------+ | Volts | Spacing | +---------+----------+ | 5,000 | 28 ins. | | 15,000 | 40 ins. | | 30,000 | 48 ins. | | 45,000 | 60 ins. | | 60,000 | 60 ins. | | 75,000 | 84 ins. | | 90,000 | 96 ins. | | 105,000 | 108 ins. | | 120,000 | 120 ins. | +---------+----------+
=Resistance of Wires.=--For quick calculation the following method of obtaining the resistance (approximately) of wires will be found convenient:
1,000 feet No. 10 B. & S. wire, which is about .1 inch in diameter (.1019), has a resistance of one ohm, at a temperature of 68° F. and weighs 31.4 pounds. A wire three sizes larger, that is No. 7, has twice the cross section and therefore one-half the resistance. A wire three sizes smaller than No. 10, that is No. 13, has one-half the cross section and therefore twice the resistance.
Thus, starting with No. 10, any number three sizes larger will double the cross sectional area and any wire three sizes smaller will halve the cross sectional area of the preceding wire. This is true to the extreme limits of the table, so that the area, weight and resistance of any wire may be at once calculated to a close approximation from this rule, intermediate sizes being obtained by interpolation.
For alternating current, the combined resistance, that is, the total resistance, including skin effect, is obtained by multiplying the resistance, as found above by the "ratio factor" (see table page 1,894).
=Impedance.=--_The total opposition to the flow of electricity in an alternating current circuit_, or the impedance may be resolved into two components representing the ohmic resistance and the spurious resistance; these components have a phase difference of 90°, and they may be represented graphically by the two legs of a right angle triangle, of which the hypothenuse represents the impedance.
Similarly, the volts lost or "drop" in an alternating circuit may be resolved into two components representing respectively
1. The loss due to resistance.
2. The loss due to reactance.
These components have a phase difference of 90° and are represented graphically similar to the impedance components. This has been explained at considerable length in Chapter XLVII (Guide V).
=Power Factor.=--When the current falls out of step with the pressure, as on inductive loads, the power factor becomes less than unity, and the effect is to increase the current required for a given load. Accordingly, this must be considered in calculating the size of the wires. As has been explained, the current flowing in an alternating current circuit, as measured by an ammeter, can be resolved into two components, representing respectively the _active component_ and the _wattless component_ or idle current. These are graphically represented by the two legs of a right triangle, of which the hypothenuse represents the current measured by the ammeter.
This _apparent_ current, as is evident from the triangle, exceeds the _active_ current and lags behind the pressure by an amount represented by the angle ϕ between the hypothenuse and leg representing the energy current as shown in fig. 2,694.
=Ques. What determines the heating of the wires on alternating current circuits with inductive loads?=
Ans. The apparent current, as represented by the hypothenuse of the triangle in fig. 2,694.
=Ques. How is the apparent current obtained?=
Ans. Divide the true watts by the product of the power factor multiplied by the voltage.
Example.--A certain circuit supplies 20 kw. to motors at 220 volts and .8 power factor. What is the apparent current?
true watts 20,000 Apparent Current = -------------------- = -------- = 113.6 amperes power factor × volts .8 × 220
=Ques. What else, besides power factor, should be considered in making wire calculations for motor circuits?=
Ans. The efficiency of the motor, and the heavy starting current.
The product of the efficiency of the motor multiplied by the power factor gives the _apparent efficiency_, which governs the size of the wires, apparatus, etc., necessary to feed the motors.
Allowance should be made for the heavy starting current required for some motors to avoid undue drop.
TABLE OF APPROXIMATE AMPERES PER TERMINAL FOR INDUCTION MOTORS
Column headings: A-110 volts B-220 volts C-440 volts D-550 volts
+-----+----------------+------------------+------------------------+ | | | Two phase | Three phase | |Horse| Single phase | four wire | three wire | |power+----+-----+-----+-----+-----+------+-----+-----+------+-----+ | | A | B | C | A | B | C | A | B | C | D | +-----+----+-----+-----+-----+-----+------+-----+-----+------+-----+ | .5| 6.6| 3.4 | 1.8| 3.3 | 1.7 | .9 | 3.7 | 1.8 | 1 | | | 1 | 14 | 7 | 3.5| 6.4 | 3.2 | 1.6 | 7.4 | 3.7 | 1.9 | | | 2 | 24 | 12 | 6 | 11 | 5.7 | 2.9 | 13 | 6.6 | 3.3 | 2.5 | | 3 | 34 | 17 | 8.5| 16 | 8.1 | 4.1 | 19 | 9.3 | 4.7 | 3.5 | | 4 | 52 | 26 | 13 | 26 | 13 | 6.5 | 30 | 15 | 7.5 | 6 | | 5 | 74 | 37 | 18.5| 38 | 19 | 9.5 | 44 | 22 | 11 | 9 | | 10 | 94 | 47 | 23.5| 44 | 22 | 11 | 50 | 25 | 12.5 | 11 | | 15 | | | | 66 | 33 | 16.5 | 76 | 38 | 19 | 16 | | 20 | | | | 88 | 44 | 22 | 102 | 51 | 25.5 | 22 | | 25 | | | | 111 | 55 | 28 | 129 | 64 | 32 | 25 | | 30 | | | | 134 | 67 | 33.5 | 154 | 77 | 38.5 | 32 | | 40 | | | | 178 | 89 | 44.5 | 204 | 107 | 53.5 | 44 | | 50 | | | | 204 | 102 | 51 | 236 | 118 | 59 | 52 | | 75 | | | | 308 | 154 | 77 | 356 | 178 | 89 | 77 | | 100 | | | | 408 | 204 | 102 | 472 | 236 | 118 | 100 | +-----+----+-----+-----+-----+-----+------+-----+-----+------+-----+
=Ques. What are the usual power factors encountered on commercial circuits?=
Ans. A mixed load of incandescent lamps and induction motors will have a power factor of from .8 to .85; induction motors above .8 to .85; incandescent and Nernst lamps .98; arc lamps, .85.
=Wire Calculations.=--In the calculation of alternating current circuits, the two chief factors which make the computation different from that for direct current circuits, is _induction_ and _power factor_. The first depends on the frequency, and physical condition of the circuit, and the second upon the character of the load.
=Ques. Under what conditions may inductance be neglected?=
Ans. In cases where the wires of a circuit are not spaced over an inch apart, or in conduit work, where both wires are in the same conduit.
Under these conditions the calculation is the same as for direct current after making proper allowance for power factor.
=Ques. Under what conditions must induction be considered?=
Ans. On exposed circuits with wires separated several inches, particularly in the case of large wires.
=Sizes of Wire.=--The size of wire for any alternating circuit may be determined by slightly modifying the formula used in direct current work, and which, as derived in Guide No. 4, page 748, is
amperes × feet × 21.6 circular mils = --------------------- (1) drop
The quantity 21.6, is twice the resistance (10.8) of a foot of copper wire one mil in diameter (_mil foot_). This resistance (10.8) is multiplied by 2, giving the quantity 21.6, because the length of a circuit, or feet in the formula, is given as the "run" or distance one way, that is, one-half the total length of wire in the circuit, must be multiplied by 2 to get the total drop, viz.:
amperes × feet × 10.8 × 2 amperes × feet × 21.6 circular mils = ------------------------- = --------------------- drop drop
It is sometimes however convenient to make the calculation in terms of watts. Formula (1) may be modified for such calculation.
In modifying the formula, the "drop" should be expressed in percentage instead of actual volts lost, that is, instead of the difference in pressure between the beginning and the end of the circuit.
In any circuit the loss in percentage, or
drop % loss = ------------------ × 100 impressed pressure
from which
% loss × impressed pressure drop = --------------------------- (2) 100
Substituting equation (2) in equation (1)
amperes × feet × 21.6 circular mils = ----------------------------- % loss × imp. pressure ------------------------ 100
amperes × feet × 2,160 = -------------------------- (3) % loss × imp. pressure
Equation (3) is modified for calculation in terms of watts as follows: The power in watts is equal to the _applied voltage_ multiplied by the current, that is to say, the power is equal to the _volts at the consumer's end of the circuit_ multiplied by the current, or simply
watts = volts × amperes
from which
watts amperes = ----- (4) volts
Substituting this value for the current in equation (3) and remembering that the pressure taken is the volts at the consumer's end of the line
watts ----- × feet × 2,160 volts circular mils = --------------------- % loss × volts
watts × feet × 2,160 = -------------------- (5) % loss × volts^{2}
This formula (5) applies to a direct current two wire circuit, and to adapt it to any alternating current circuit it is only necessary to use the letter M instead of the number 2,160, thus
watts × feet × M circular mils = ------------------ (6) % loss × volts^{2}
in which M is a coefficient which has various values according to the kind of circuit and value of the power factor. These values are given in the following table:
=VALUES OF M=
--------+----------------------------------------------------------- | POWER FACTOR SYSTEM +-----+-----+-----+-----+-----+-----+-----+-----+-----+----- | 1.00| .98| .95| .90| .85| .80| .75| .70| .65| .60 --------+-----+-----+-----+-----+-----+-----+-----+-----+-----+----- Single |2,160|2,249|2,400|2,660|3,000|3,380|3,840|4,400|5,112|6,000 phase | | | | | | | | | | --------+-----+-----+-----+-----+-----+-----+-----+-----+-----+----- Two |1,080|1,125|1,200|1,330|1,500|1,690|1,920|2,200|2,556|3,000 phase | | | | | | | | | | (4 wire)| | | | | | | | | | --------+-----+-----+-----+-----+-----+-----+-----+-----+-----+----- Three |1,080|1,125|1,200|1,330|1,500|1,690|1,920|2,200|2,556|3,000 phase | | | | | | | | | | (3 wire)| | | | | | | | | | --------+-----+-----+-----+-----+-----+-----+-----+-----+-----+-----
NOTE.--The above table is calculated as follows: For =single phase= M = 2,160 ÷ power factor^{2} × 100; for =two phase= four wire, or three phase three wire, M = ½ (2,160 ÷ power factor^{2})× 100. Thus the value of M for a single phase line with power factor .95 = 2,160 ÷ .95^{2} × 100 = 2,400.
It must be evident that when 2,160 is taken as the value of M, formula (6) applies to a two wire direct current circuit and also to a single phase alternating current circuit when the power factor is unity.
In the table the value of M for any particular power factor is found by dividing 2,160 by the square of that power factor for single phase and twice the square of the power factor for two phase and three phase.
=Ques. For a given load and voltage how do the wires of a single and two phase system compare in size and weight, the power factor being the same in each case?=
Ans. Since the two phase system is virtually two single phase systems, the four wires of the two phase systems are half the size of the two wires of the single phase system, and accordingly, the weight is the same for either system.
=VALUES OF T=
+-------------------+--------------------------------------------+ | | POWER FACTOR | | SYSTEM +--------+--------+--------+--------+--------+ | | 1.00 | .98 | .90 | .80 | .70 | +-------------------+--------+--------+--------+--------+--------+ | Single phase | 1.00 | .98 | .90 | .80 | .70 | +-------------------+--------+--------+--------+--------+--------+ | Two phase, 4 wire | 2.00 | 1.96 | 1.80 | 1.60 | 1.40 | +-------------------+--------+--------+--------+--------+--------+ |Three phase, 3 wire| 1.73 | 1.70 | 1.55 | 1.38 | 1.21 | +-------------------+--------+--------+--------+--------+--------+
NOTE.--This table is for finding the value of the current in line, using the formula I = W ÷ (E × T), in which I = current in line; E = voltage between main conductors at receiving or consumers' end; W = watts. For instance, what is the current in a two phase line transmitting 1,000 watts at 550 volts, power factor .80? I = 1,000 ÷ (550 × 1.60) = 1.13.
=Ques. Since there is no saving in copper in using two phases, what advantage has the two phase system over the one phase system?=
Ans. It is more desirable on power circuits, because two phase motors are self-starting.
That is to say, the rotating magnetic field that can be produced by a two phase current, permits an induction motor to start without being equipped with any special phase splitting devices which are necessary on single phase motors, because the oscillating field produced by a single phase current does not produce any torque on a squirrel cage armature at rest.
=Ques. For equal working conditions, what is the comparison between the single, two and three phase system as to size and weight of wires?=
Ans. Each wire of the three phase system is half the size of one of the wires of the single phase system, hence the weight of copper required for the three phase system is 75% of that required for the single phase system. Since in the two phase system half of the load is carried by each phase, each wire of the three phase system is the same size as one of the wires of the two phase system, hence, the copper required by the three phase system is 75% of that required by the two phase system.
=MISCELLANEOUS FORMULÆ FOR COPPER WIRES=
Diameter squared = circular mils Circular mils × .7854 = square mils .000003027 × circular mils = pounds per foot .003027 × circular mils = pounds per 1,000 feet .0159847 × circular mils = pounds per mile .003879 × square mils = pounds per 1,000 feet .33033 ÷ circular mils = feet per pound .0000002924 × circular mils = pounds per ohm .342 ÷ circular mils = ohms per pound .096585 × circular mils = feet per ohm 10.353568 ÷ circular mils = ohms per foot
Breaking weight of wire ÷ area = breaking weight per square inch.
Breaking weight per square inch × area = breaking weight of wire.
The weight of copper wire is 1-1/7 times the weight of iron wire of same diameter.
EXAMPLE.--What size wires must be used on a single phase circuit 2,000 feet in length to supply 30 kw. at 220 volts with loss of 4%, the power factor being .9?
The formula for circular mils is
watts × feet × M circular mils = ------------------ (1) % loss × volts^{2}
Substituting the given values and the proper value of M from the table, in (1)
30,000 × 2,000 × 2,660 circular mils = ---------------------- = 82,438 4 × 220^{2}
Referring to the accompanying table of the properties of copper wire, the nearest _larger_ size wire is No. 1 B. & S. gauge having an area of 83,690 circular mils.
=TABLE OF THE PROPERTIES OF COPPER WIRE=
Giving weights, length and resistances of wires of Matthiessen's Standard Conductivity for both B. & S. G. (Brown & Sharpe Gauge) and B. W. G. (Birmingham Wire Gauge) from Transactions October 1903, of the American Institute of Electrical Engineers.
__________________________________________________________________ | | | Gauges. To the nearest fourth | | Length.|Resistance. significant digit. | | | _____________________________________| |________|_________ | | | | Weight. | | | | Diameter. | Area. | Lbs. | Feet | Ohms per | | | | per | per lb.|1,000 ft. ______|_______|___________|__________| 1,000 |________|__________ | | | | feet. | | | | | Circular | | | B.& S.| B.W.G.| Inches. | mils. | | | @ 68° F. ______|______ |___________|__________|_________|________|_________ | | | | | | 0000 | | 0.460 | 211,600 | 640.5 | 1.561 | .04893 | 0000 | 0.454 | 206,100 | 623.9 | 1.603 | .05023 | 000 | 0.425 | 180,600 | 546.8 | 1.829 | .05732 | | | | | | 000 | | 0.4096 | 167,800 | 508.0 | 1.969 | .06170 | 00 | 0.380 | 144,400 | 437.1 | 2.288 | .07170 00 | | 0.3648 | 133,100 | 402.8 | 2.482 | .07780 | | | | | | | 0 | 0.340 | 115,600 | 349.9 | 2.858 | .08957 0 | | 0.3249 | 105,500 | 319.5 | 3.130 | .09811 | 1 | 0.3000 | 90,000 | 272.4 | 3.671 | .1150 | | | | | | 1 | | 0.2893 | 83,690 | 253.3 | 3.947 | .1237 | 2 | 0.2840 | 80,660 | 244.1 | 4.096 | .1284 | 3 | 0.2590 | 67,080 | 203.1 | 4.925 | .1543 | | | | | | 2 | | 0.2576 | 66,370 | 200.9 | 4.977 | .1560 | 4 | 0.2380 | 56,640 | 171.5 | 5.832 | .1828 3 | | 0.2294 | 52,630 | 159.3 | 6.276 | .1967 | | | | | | | 5 | 0.2200 | 48,400 | 146.5 | 6.826 | .2139 4 | | 0.2043 | 41,740 | 126.4 | 7.914 | .2480 | 6 | 0.2030 | 41,210 | 124.7 | 8.017 | .2513 | | | | | | 5 | | 0.1819 | 33,100 |100.2 | 9.98 | .3128 | 7 | 0.1800 | 32,400 | 98.08 | 10.20 | .3196 | 8 | 0.1650 | 27,230 | 82.41 | 12.13 | .3803 | | | | | | 6 | | 0.1620 | 26,250 | 79.46 | 12.58 | .3944 | 9 | 0.1480 | 21,900 | 66.30 | 15.08 | .4727 7 | | 0.1443 | 20,820 | 63.02 | 15.87 | .4973 | | | | | | | 10 | 0.1340 | 17,960 | 54.35 | 18.40 | .5766 8 | | 0.1285 | 16,510 | 49.98 | 20.01 | .6271 | 11 | 0.1200 | 14,400 | 43.59 | 22.94 | .7190 | | | | | | 9 | | 0.1144 | 13,090 | 39.63 | 25.23 | .7908 | 12 | 0.1090 | 11,880 | 35.96 | 27.81 | .8715 10 | | 0.1019 | 10,380 | 31.43 | 31.82 | .9972 | | | | | | | 13 | 0.0950 | 9,025 | 27.32 | 36.60 | 1.147 11 | | 0.09074 | 8,234 | 24.93 | 40.12 | 1.257 | 14 | 0.08300 | 6,889 | 20.85 | 47.95 | 1.503 | | | | | | 12 | | 0.08081 | 6,530 | 19.77 | 50.59 | 1.586 | 15 | 0.07200 | 5,184 | 15.69 | 63.73 | 1.997 13 | | 0.07196 | 5,178 | 15.68 | 63.79 | 1.999 | | | | | | | 16 | 0.06500 | 4,225 | 12.79 | 78.19 | 2.451 14 | | 0.06408 | 4,107 | 12.43 | 80.44 | 2.521 | 17 | 0.0580 | 3,364 | 10.18 | 98.23 | 3.078 | | | | | | 15 | | 0.05707 | 3,257 | 9.858 | 101.4 | 3.179 16 | | 0.05082 | 2,583 | 7.818 | 127.9 | 4.009 | 18 | 0.04900 | 2,401 | 7.268 | 137.6 | 4.312 | | | | | | 17 | | 0.045260 | 2,048 | 6.200 | 161.3 | 5.055 | 19 | 0.042000 | 1,764 | 5.340 | 187.3 | 5.870 18 | | 0.040300 | 1,624 | 4.917 | 203.4 | 6.374 | | | | | | 19 | | 0.035890 | 1,288 | 3.899 | 256.5 | 8.038 | 20 | 0.035000 | 1,225 | 3.708 | 269.7 | 8.452 | 21 | 0.032000 | 1,024 | 3.100 | 322.6 | 10.11 | | | | | | 20 | | 0.031960 | 1,022 | 3.092 | 323.4 | 10.14 21 | | 0.028460 | 810.1 | 2.452 | 407.8 | 12.78 | 22 | 0.028000 | 784.0 | 2.373 | 421.4 | 13.21 22 | | 0.025350 | 642.4 | 1.945 | 514.2 | 16.12 | 23 | 0.025000 | 625.0 | 1.892 | 528.6 | 16.57 23 | | 0.022570 | 509.5 | 1.542 | 648.4 | 20.32 | | | | | | | 24 | 0.022000 | 484.0 | 1.465 | 682.6 | 21.39 24 | | 0.020100 | 404.0 | 1.223 | 817.6 | 25.63 | 25 | 0.020000 | 400.0 | 1.211 | 825.9 | 25.88 | | | | | | | 26 | 0.018000 | 324.0 | .9808 | 1,020 | 31.96 25 | | 0.017900 | 320.4 | .9699 | 1,031 | 32.31 | 27 | 0.016000 | 256.0 | .7749 | 1,290 | 40.45 | | | | | | 26 | | 0.015940 | 254.1 | .7692 | 1,300 | 40.75 27 | | 0.014200 | 201.5 | .6100 | 1,639 | 51.38 | 28 | 0.014000 | 196.0 | .5933 | 1,685 | 52.83 | | | | | | | 29 | 0.013000 | 169.0 | .5116 | 1,955 | 61.27 28 | | 0.012640 | 159.8 | .4837 | 2,067 | 64.79 | 30 | 0.012000 | 144.0 | .4359 | 2,294 | 71.90 | | | | | | 29 | | 0.011260 | 126.7 | .3836 | 2,607 | 81.70 30 | | 0.010030 | 100.5 | .3042 | 3,287 | 103.0 | 31 | 0.010000 | 100.0 | .3027 | 3,304 | 103.5 | | | | | | | 32 | 0.009000 | 81.0 | .2452 | 4,078 | 127.8 31 | | 0.008928 | 79.70 | .2413 | 4,145 | 129.9 | 33 | 0.008000 | 64.0 | .1937 | 5,162 | 161.8 | | | | | | 32 | | 0.007950 | 63.21 | .1913 | 5,227 | 163.8 33 | | 0.007080 | 50.13 | .1517 | 6,591 | 206.6 | 34 | 0.007000 | 49.0 | .1483 | 6,742 | 211.3 | | | | | | 34 | | 0.006305 | 39.75 | .1203 | 8,311 | 260.5 35 | | 0.005615 | 31.52 | .09543 |10,480 | 328.4 36 | 35 | 0.005000 | 25.0 | .07568 |13,210 | 414.2 | | | | | | 37 | | 0.004453 | 19.83 | .06001 |16,660 | 522.2 | 36 | 0.004000 | 16. | .04843 |20,650 | 647.1 38 | | 0.003965 | 15.72 | .04759 |21,010 | 658.5 | | | | | | 39 | | 0.003531 | 12.47 | .03774 |26,500 | 830.4 40 | | 0.003145 | 9.888 | .02993 |33,410 |1047. --------------------------------------------------------------------
=Drop.=--In order to determine the drop or volts lost in the line, the following formula may be used
drop = ((% loss × volts) / 100) × S (1)
in which the % loss is a percentage of the applied power, that is, the power delivered to the consumer and not a percentage of the power at the alternator. "Volts" is the pressure at the consumer's end of the circuit.
=VALUE OF "S" FOR 60 CYCLES=
-----------------+------------------------+------------------------+ | .98 power factor | .90 power factor | --------+--------+------------------------+------------------------+ Size of | Area | Spacing of | Spacing of | wire | in | conductors | conductors | B.&S. |circular| | | gauge | mils. | | | | +----+----+----+----+----+----+----+----+----+----+ | | 1" | 3" | 6" | 12"| 24"| 1" | 3" | 6" | 12"| 24"| --------+--------+----+----+----+----+----+----+----+----+----+----+ 500,000 |500,000 |1.21|1.45|1.61|1.77|1.92|1.32|1.80|2.11|2.44|2.75| 300,000 |300,000 |1.15|1.29|1.38|1.48|1.57|1.19|1.47|1.66|1.84|2.02| 0,000 |211,600 |1.12|1.22|1.28|1.34|1.41|1.13|1.33|1.45|1.58|1.63| 000 |167,800 |1.09|1.18|1.22|1.28|1.29|1.08|1.23|1.33|1.44|1.53| 00 |133,100 |1.07|1.14|1.18|1.21|1.25|1.03|1.16|1.24|1.32|1.40| 0 |105,500 |1.05|1.10|1.14|1.17|1.20|1.00|1.09|1.16|1.22|1.28| 1 | 83,690 |1.04|1.08|1.10|1.13|1.15|1.00|1.05|1.09|1.14|1.19| 2 | 66,370 |1.02|1.05|1.08|1.10|1.12|1.00|1.00|1.04|1.08|1.12| 3 | 52,630 |1.02|1.04|1.06|1.07|1.09|1.00|1.00|1.00|1.03|1.06| | | | | | | | | | | | | 4 | 41,740}|1.00|1.02|1.03|1.04|1.07|1.00|1.00|1.00|1.00|1.00| 5 | 33,100}| | | | | | | | | | | | | | | | | | | | | | | 6 | 26,250}|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| 7 | 20,820}| | | | | | | | | | | | | | | | | | | | | | | 8 | 16,510}| | | | | | | | | | | 9 | 13,090}|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| 10 | 10,380}| | | | | | | | | | | --------+--------+----+----+----+----+----+----+----+----+----+----+ --------+--------+----+----+----+----+----+----+----+----+----+----+ Size | | .80 power factor | .70 power factor | of | Area +------------------------+------------------------+ wire | in | Spacing of | Spacing of | B.&S. |circular| conductors | conductors | gauge | mils. +----+----+----+----+----+----+----+----+----+----+ | | 1" | 3" | 6" | 12"| 24"| 1" | 3" | 6" | 12"| 24"| --------+--------+----+----+----+----+----+----+----+----+----+----+ 500,000 | 500,000|1.27|1.89|2.25|2.64|3.03|1.14|1.72|2.12|2.53|2.92| 300,000 | 300,000|1.11|1.46|1.68|1.90|2.12|1.00|1.33|1.56|1.78|2.01| 0,000 | 211,600|1.03|1.27|1.43|1.58|1.75|1.00|1.14|1.29|1.45|1.69| 000 | 167,800|1.00|1.16|1.28|1.41|1.53|1.00|1.02|1.15|1.28|1.50| 00 | 133,100|1.00|1.07|1.15|1.22|1.00|1.00|1.00|1.03|1.13|1.21| 0 | 105,500|1.00|1.00|1.07|1.15|1.00|1.00|1.00|1.00|1.01|1.09| 1 | 83,690}|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| 2 | 66,370}|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| 3 | 52,630}|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| | | | | | | | | | | | | 4 | 41,740}| | | | | | | | | | | 5 | 33,100}|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| | | | | | | | | | | | | 6 | 26,250}| | | | | | | | | | | 7 | 20,820}| | | | | | | | | | | | | | | | | | | | | | | 8 | 16,510}| | | | | | | | | | | 9 | 13,090}|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| 10 | 10,380}| | | | | | | | | | | --------+--------+----+----+----+----+----+----+----+----+----+----+ The coefficient S has various values as given in the accompanying tables. As will be seen from the table, the value of S to be used depends upon the size of wire, spacing, power factor and frequency.
These values are accurate enough for all practical purposes, and may be used for distances of 20 miles or less and for voltages up to 25,000.
The capacity effect on very long high voltage lines, makes this method of determining the drop somewhat inaccurate beyond the limits above mentioned.
=VALUE OF "S" FOR 25 CYCLES=
---------+--------+------------------------+------------------------+ Size | | .98 power factor | .90 power factor | of | Area +------------------------+------------------------+ wire | in | Spacing of | Spacing of | B.&S. |circular| conductors | conductors | gauge | mils. +----+----+----+----+----+----+----+----+----+----+ | | 1" | 2" | 6" | 12"| 24"| 1" | 3" | 6" | 12"| 24"| ---------+--------+----+----+----+----+----+----+----+----+----+----+ 500,000 |500,000 |1.01|1.17|1.23|1.29|1.36|1.02|1.22|1.35|1.43|1.61| 300,000 |300,000 |1.04|1.10|1.13|1.18|1.21|1.00|1.08|1.16|1.25|1.31| 0,000 |211,600 |1.03|1.07|1.09|1.11|1.14|1.00|1.02|1.07|1.13|1.15| 000 |167,800 |1.00|1.05|1.06|1.09|1.10|1.00|1.00|1.02|1.07|1.11| 00 |133,100 |1.00|1.03|1.05|1.06|1.08|1.00|1.00|1.00|1.02|1.05| 0 |105,500 |1.00|1.01|1.02|1.03|1.04|1.00|1.00|1.00|1.00|1.00| | | | | | | | | | | | | 1 | 83,690}| | | | | | | | | | | 2 | 66,370}|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| 3 | 52,630}| | | | | | | | | | | | | | | | | | | | | | | 4 | 41,740}| | | | | | | | | | | 5 | 33,100}|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| 6 | 26,250}| | | | | | | | | | | | | | | | | | | | | | | 7 | 20,820}| | | | | | | | | | | 8 | 16,510}| | | | | | | | | | | 9 | 13,090}|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| 10 | 10,380}| | | | | | | | | | | ---------+--------+----+----+----+----+----+----+----+----+----+----+ Size | | .80 power factor | .70 power factor | of | Area +------------------------+------------------------+ wire | in | Spacing of | Spacing of | B.&S. |circular| conductors | conductors | gauge | mils. +----+----+----+----+----+----+----+----+----+----+ | | 1" | 3" | 6" | 12"| 24"| 1" | 3" | 6" | 12"| 24"| ---------+--------+----+----+----+----+----+----+----+----+----+----+ 500,000 | 500,000|1.00|1.15|1.30|1.47|1.62|1.00|1.00|1.16|1.33|1.49| 300,000 | 300,000|1.00|1.00|1.09|1.16|1.25|1.00|1.00|1.00|1.02|1.12| 0,000 | 211,600|1.00|1.00|1.00|1.03|1.10|1.00|1.00|1.00|1.00|1.00| 000 | 167,800|1.00|1.00|1.00|1.00|1.01|1.00|1.00|1.00|1.00|1.00| 00 | 133,100|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| 0 | 105,500|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| | | | | | | | | | | | | 1 | 83,690}| | | | | | | | | | | 2 | 66,370}|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| 3 | 52,630}| | | | | | | | | | | | | | | | | | | | | | | 4 | 41,740}| | | | | | | | | | | 5 | 33,100}|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| 6 | 26,250}| | | | | | | | | | | | | | | | | | | | | | | 7 | 20,820}| | | | | | | | | | | 8 | 16,510}| | | | | | | | | | | 9 | 13,090}|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00|1.00| 10 | 10,380}| | | | | | | | | | | ---------+--------+----+----+----+----+----+----+----+----+----+----+
EXAMPLE.--A circuit supplying current at 440 volts, 60 frequency, with 5% loss and .8 power factor is composed of No. 2 B. & S. gauge wires spaced one foot apart. What is the drop in the line?
According to the formula
% loss × volts drop = -------------- × S 100
Substituting the given values, and value of S as obtained from the table for frequency 60
5 × 440 drop = ------- × 1 = 22 volts 100
=Current=.--As has been stated, the effect of power factor less than unity, is to increase the current; hence, in inductive circuit calculations, the first step is to determine the current flowing in a circuit. This is done as follows:
apparent load current = ------------- (1) volts
and
watts apparent load = ------------ (2) power factor
Substituting (2) in (1)
watts ------------ power factor watts current = ------------ = --------------------- (3) volts power factor × volts
EXAMPLE.--A 50 horse power 440 volt motor has a full load efficiency of .9 and power factor of .8. How much current is required?
Since the brake horse power of the motor is given, it is necessary to obtain the electrical horse power, thus
brake horse power 50 E.H.P. = ----------------- = -- = 55.5 efficiency .9
which in watts is
55.5 × 746 = 41,403
which is the actual load, and from which
actual load 41,403 apparent load = ------------ = ------ = 51,754 power factor .8
The current therefore at 440 volts is
apparent load 51,754 ------------- = ------ = 117.6 amperes volts 440
EXAMPLE.--A 50 horse power single phase 440 volt motor, having a full load efficiency of .92 and power factor of .8, is to be operated at a distance of 1,000 feet from the alternator. The wires are to be spaced 6 inches apart and the frequency is 60, and % loss 5. Determine: =A=, _electrical horse power_; =B=, _watts_; =C=, _apparent load_; =D=, _current_; =E=, _size of wires_; =F=, _drop_; =G=, _voltage at the alternator_.
=A=. _Electrical horse power_
brake horse power 50 E. H. P. = ----------------- × --- = 54.3 efficiency .92
or,
54.3 × 746 = 40,508 watts
=TABLE OF WIRE EQUIVALENTS= (Brown and Sharpe gauge) ------+--------+--------+--------+---------+---------+--------- 0000 | 2 # 0 | 4 # 3 | 8 # 6 | 16 # 9 | 32 # 12 | 64 # 15 000 | 2 " 1 | 4 " 4 | 8 " 7 | 16 " 10 | 32 " 13 | 64 " 16 00 | 2 " 2 | 4 " 5 | 8 " 8 | 16 " 11 | 32 " 14 | 64 " 17 0 | 2 " 3 | 4 " 6 | 8 " 9 | 16 " 12 | 32 " 15 | 64 " 18 1 | 2 " 4 | 4 " 7 | 8 " 10 | 16 " 13 | 32 " 16 | 64 " 19 2 | 2 " 5 | 4 " 8 | 8 " 11 | 16 " 14 | 32 " 17 | 64 " 20 3 | 2 " 6 | 4 " 9 | 8 " 12 | 16 " 15 | 32 " 18 | 64 " 21 4 | 2 " 7 | 4 " 10 | 8 " 13 | 16 " 16 | 32 " 19 | 64 " 22 5 | 2 " 8 | 4 " 11 | 8 " 14 | 16 " 17 | 32 " 20 | 64 " 23 6 | 2 " 9 | 4 " 12 | 8 " 15 | 16 " 18 | 32 " 21 | 64 " 24 7 | 2 " 10 | 4 " 13 | 8 " 16 | 16 " 19 | 32 " 22 | 64 " 25 8 | 2 " 11 | 4 " 14 | 8 " 17 | 16 " 20 | 32 " 23 | 64 " 26 9 | 2 " 12 | 4 " 15 | 8 " 18 | 16 " 21 | 32 " 24 | 64 " 27 10 | 2 " 13 | 4 " 16 | 8 " 19 | 16 " 22 | 32 " 25 | 64 " 28 11 | 2 " 14 | 4 " 17 | 8 " 20 | 16 " 23 | 32 " 26 | 64 " 29 12 | 2 " 15 | 4 " 18 | 8 " 21 | 16 " 24 | 32 " 27 | 64 " 30 13 | 2 " 16 | 4 " 19 | 8 " 22 | 16 " 25 | 32 " 28 | 14 | 2 " 17 | 4 " 20 | 8 " 23 | 16 " 26 | 32 " 29 | 15 | 2 " 18 | 4 " 21 | 8 " 24 | 16 " 27 | 32 " 30 | 16 | 2 " 19 | 4 " 22 | 8 " 25 | 16 " 28 | | 17 | 2 " 20 | 4 " 23 | 8 " 26 | 16 " 29 | | 18 | 2 " 21 | 4 " 24 | 8 " 27 | 16 " 30 | | 19 | 2 " 22 | 4 " 25 | 8 " 28 | | | 20 | 2 " 23 | 4 " 26 | 8 " 29 | | | 21 | 2 " 24 | 4 " 27 | 8 " 30 | | | ------+--------+--------+--------+---------+---------+---------
=B.= _Watts_
watts = E.H.P. × 746 = 54.3 × 746 = 40,508
=C.= _Apparent load_
apparent load or kva = (actual load or watts ÷ power factor)
= 40,508 ÷ .8 = 50,635
=D.= _Current_
current = (apparent load or kva ÷ volts) = 50,635 ÷ 440 = 115 amperes
=E.= _Size of wires_
cir. mils = (watts × feet × M) ÷ (% loss × volts^{2}) = (40,508 × 1,000 × 3,380) ÷ (5 × 440^{2}) = 141,443
From table page 1,907, nearest size _larger_ wire is No. 00 B.&S. gauge.
=F.= _Drop_
drop = ((% loss × volts) ÷ 100) × S = ((5 × 440) ÷ 100) × 1.17 = 25.74 volts
NOTE.--Values of S are given on page 1910.
=G.= _Voltage at alternator_
alternator pressure = (volts at motor + drop) = 440 + 25.74 = 465.7 volts.