=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.
[Illustration: FIGS. 2,671 to 2,676.--=The effect of self-induction.=
In a non-inductive circuit, as in fig. 2,672, the whole of the
virtual pressure is available to cause current to flow through the
lamp filament, hence it will glow with maximum brilliancy. If an
inductive coil be inserted in the circuit as in fig. 2,674, the
reverse pressure due to self-induction will oppose the virtual
pressure, hence the effective pressure (which is the difference
between the virtual and reverse pressures), will be reduced and
the current flow through the lamp diminished, thus reducing the
brilliancy of the illumination. The effect may be intensified to such
degree by interposing an iron core in the coil as in fig. 2,676, as
to extinguish the lamp.]
=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.
[Illustration: FIG. 2,677.--Measurement of self induction when
the frequency is known. The apparatus required consists of a high
resistance or electrostatic a.c. voltmeter, d.c. ammeter, and a
non-inductive resistance. Connect the inductive resistance to be
measured as shown, and close switch M, short circuiting the ammeter.
Connect alternator in circuit and measure drop across R and across
X_{_i_}. Disconnect alternator and connect battery in circuit, then
open switch M and vary the continuous current until the drop across R
is the same as with the alternating current, both measurements being
made with the same voltmeter; read ammeter, and measure drop across
X_{_i_}. Call the drop across X_{_i_} with alternating current E, and
with direct current E_{_i_}, and the reading of the ammeter J. Then
____________________
L = √E^{2} + E_{_i_}^{2} ÷ 2π _f_ I. If the resistance X_{_i_}
be known, and the ammeter be suitable for use with alternating
current, the switch and R may be dispensed with.
______________________________
Then L = √E^{2} - X_{_i_}^{2} I_{_i_}^{2} ÷ 2π _f_ I,
where I_{_i_} is the value of the alternating current. The resistance
of the voltmeter should be high enough to render its current
negligible as compared with that through X_{_i_}.]
=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.
[Illustration: FIG. 2,678.--Transposition diagram for two parallel
lines consisting of two wires each.]
[Illustration: FIG. 2,679.--Transposition diagram for three phase,
three wire line, transposing at the vertices of an equilateral
triangle. The line is originally balanced and becomes unbalanced on
transposing, a procedure which should be resorted to only to prevent
_mutual induction_.]
[Illustration: FIG. 2,680.--Transposition diagram of three phase,
three wire line, with center arranged in a straight line.]
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.
[Illustration: FIG. 2,681.--Capacity effect in single phase
transmission line. The effect is the same as would be produced by
shunting across the line at each point an infinitesimal condenser
having a capacity equal to that of an infinitesimal length of
circuit. For the purpose of calculating the charging current, a very
simple and sufficiently accurate method is to determine the current
taken by a condenser having a capacity equal to that of the entire
line when charged to the pressure on the line at the generating end.
The effect of capacity of the line is to reduce the pressure drop,
that is, improve the regulation, and to decrease or increase the
power loss depending on the load and power factor of the receiver.]
[Illustration: FIG. 2,682.--Capacity effect in a three phase
transmission line. It is the same as would be produced by shunting
the line at each point by three infinitesimal condensers connected in
star with the neutral point grounded, the capacity of each condenser
being twice that of a condenser of infinitesimal length formed by
any two of the wires. The effect of capacity on the regulation and
efficiency of the line can be determined with sufficient accuracy
in most cases by considering the line shunted at each end by three
condensers connected in star, the capacity of each condenser being
equal to that formed by any two wires of the line. An approximate
value for the charging current per wire is the current required to
charge a condenser, equal in capacity to that of any two of the
wires, to the pressure at the generating end of the line between any
one wire and the neutral point.]
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.
[Illustration: FIGS. 2,683 to 2,687.--Skin effect and shield effect.
Fig. 2,683, section of conductor illustrating skin effect or tendency
of the alternating current to distribute itself unequally through the
cross section of a conductor as shown by the varied shading, which
represents the current flowing most strongly in the outer portions
of the conductor. For this reason it has been proposed to use hollow
or flat instead of solid round conductors; however, with frequency
not exceeding 100, the skin effect is negligibly small in copper
conductors of the sizes usually employed. In figs. 2,684 and 2,685,
or 2,686 and 2,687, if two adjacent conductors be carrying current
in the same direction, concentration will occur on those parts of
the two conductors remote from one another, and the nearer parts
will have less current, that is to say, they will be =shielded=.
In this case, the induction due to one conductor will exert its
opposing effect to the greatest extent on those parts of the other
conductor nearest to it; this effect decreasing the deeper the latter
is penetrated. After crossing the current axis, the induction will
still decrease in magnitude, but will now aid the current in the
conductor. Hence, the effect of these two conductors on one another
will make the current density more uniform than is the case where
the two conductors adjacent to one another are carrying current in
opposite directions, as in figs. 2,685 and 2,686, therefore, the
resistance and the heating for a given current will be smaller. If
the two return conductors be situated on the line passing through
the center of the conductors just considered, the effect will be to
still further concentrate the current; the distribution symmetry will
be further disturbed, and the resistance of the conductor system
increased. It is therefore difficult to say which of the two cases
considered holds the advantage so far as increasing the resistance is
concerned. The case, however, in which the phases are mixed has much
the smaller reactive drop.]
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.
[Illustration: FIG. 2,688.--Electromagnetic and electrostatic
fields surrounding the conductors of a transmission line. The
electromagnetic field is represented by lines of magnetic force
that surround the conductors in circles, (the solid lines), and
the electrostatic field by (dotted) circles passing from conductor
to conductor across at right angles to the magnetic circles. For
any given size of wire and distance apart of wires there is a
certain voltage at which the critical density or critical gradient
is reached, where the air breaks down and luminosity begins--the
critical voltage where corona manifests itself. At still higher
voltages corona spreads to further distances from the conductor
and a greater volume of air becomes luminous. Incidentally, it
produces noise. Now to produce light requires power and to produce
noise requires power. Air is broken down and is heated in breaking
down, and to heat also requires power; therefore, as soon as corona
forms, power is consumed or dissipated in its formation. When this
phenomenon occurs on the conductors of an alternating current circuit
a change takes place in relation to current and voltage. On the
wires of an alternating current transmission line, at a voltage
below that where corona forms--at a voltage where wires are not
luminous--considerable current, more or less depending on voltage
and length of wire, flows into the circuit as capacity current or
charging current.]
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).
[Illustration: FIGS. 2,689 to 2,692.--Triangles for obtaining
graphically, impedance, impressed pressure, etc., in alternating
current circuits. For a full explanation of this method the reader is
referred to Guide 5, Chapter XLVII on Alternating Current Diagrams. A
thorough study of this chapter is recommended.]
=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).
[Illustration: FIG. 2,693.--Mechanical analogy of power factor, as
exemplified by a locomotive "poling" a car off a siding. The car and
locomotive are shown moving in parallel directions, and the pole
AB, inclined at an angle ϕ. Now, if the length of AB be taken to
represent the pressure exerted on the pole by the locomotive, then
the imaginary lines AC and BC, drawn respectively parallel and at
right angles to the direction of motion will represent respectively
the useful and no energy (wattless) components; that is to say, if
the pressure AB be applied to the car at an angle ϕ, only part of it,
AC, is useful in propelling the car, the other component, BC, being
wasted in tending to push the car off the track at right angles to
the rails, being resisted by the flanges of the outer wheels.]
=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.
[Illustration: FIG. 2,694.--Diagram showing that the apparent current
is more than the active current, the excess depending upon the angle
of phase difference.]
[Illustration: FIG. 2,695.--Diagram showing components of impedance
volts. Compare this diagram with figs. 2,689 and 2,671, and note that
the term "reactance" is the difference between the inductance drop
and the capacity drop if the circuit contain capacity, for instance,
if inductance drop be 10 volts and capacity drop be 7 volts then
reactance 10-7 = 3 volts.]
=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?=
[Illustration: FIGS. 2,696 to 2,698.--Example of wiring showing where
inductance is negligible, and where it must be considered in wire
calculations.]
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
[Illustration: FIGS. 2,699 to 2,703.--Stranded copper cables. For
conductors of large areas and in the smaller sizes where extra
flexibility is required it becomes necessary to employ stranded
cables made by grouping a number of wires together in either
concentric or rope form. The concentric cable as here illustrated is
formed by grouping six wires around a central wire thereby forming
a seven wire cable. The next step is the application in a reverse
direction of another layer of 12 wires and a nineteen wire cable is
produced. This is again increased by a third layer of eighteen wires
for a 37 wire cable and a fourth layer of 24 wires for a 61 wire
cable. Successive layers, each containing 6 more wires than that
preceding, may be applied until the desired capacity is obtained. The
cuts show sectional views of concentric cables each formed from No.
10 B. & S. gauge wires.]
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
[Illustration: FIG. 2,704.--Rope type of stranded copper cable
which is used when a high degree of flexibility is required. The
construction of this cable is the stranding together of seven groups,
each containing seven wires and producing a total of 49 wires.
In cases when a greater carrying capacity is desired than can be
obtained through the use of the 7 × 7 or 49 wire cable, the number of
groups is increased to nineteen thereby making a total of 133 wires
(19 × 7).]
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.