CHAPTER XXXVII
WIRES AND WIRE CALCULATIONS
The wireman who is called upon to plan and install a system of wiring will find it necessary first to have a knowledge of the various kinds of wire so as to select the one best suited for the work, and to be able to make simple calculations in order to determine the proper sizes of wire for the various circuits.
Wires are generally made of circular cross section. The process of manufacture consists in drawing the material through steel dies, when its properties permit this treatment. In the case of some substances, as for instance, tin and lead, difficulties arise in the drawing process, and these are therefore "squirted."
The metals most extensively used for wires are copper and iron; German silver, tin and lead are also employed, but only at points where it is desirable to have a comparatively high resistance in the circuit.
=Copper Wire.=—Copper is used in nearly all cases of wiring because it combines high electrical conductivity with good mechanical qualities and reasonable price. In conductivity it is only surpassed by silver, but the cost of the latter of course prohibits its use for wiring purposes.
Copper wire is used for electric light and power lines, for most telephone and some telegraph lines, and for all cases where low resistance is required at moderate cost.
Hard drawn copper wire is ductile, and has a high tensile strength; these properties allow it to be bent around corners and drawn through tubes without injury.
Pure annealed copper has a specific gravity of 8.89 at 60° Fahr. One cubic inch weighs .32 pound; its melting point is about 2,100° Fahr.
Good hard drawn copper has a tensile strength of about three times its own weight per mile length. Thus, a number 10 B. & S. gauge copper wire, weighing 166 lbs. per mile, will have a breaking strength equal to approximately 3 × 166 = 498 lbs.
=Iron Wire.=—This kind of wire is largely used for telegraph and telephone lines, although it is rapidly being replaced by copper in long lines.
There are three grades of iron wire:
1. =Extra best best (E. B. B.)= which has the highest conductivity and is the nearest to being uniform, in quality, being both tough and pliable;
2. =Best best (B. B.)=, which varies more in quality, is not so tough, and is lower in conductivity. _It is frequently sold as_ =E. B. B.=;
3. =Best (B.)=, which is the poorest grade made, being more brittle, and lowest in conductivity. Iron wire should be well galvanized.
=German Silver Wire.=—German silver is an alloy consisting of 18 to 30% nickel, and the balance about four parts copper to one part zinc. It is very largely used as a resistance material in making resistance coils, and is sold in the form of wire, and strip. The resistance of this wire varies with its composition.
The resistance of the 18% alloy at 25° C. is 18 times that of copper, and of the 30% alloy about 28 times that of copper.
The safe carrying capacity of the wire in spirals in open air for continuous duty is such that the circular mils per ampere varies from about 1,500 in No. 10 wire to about 475 in No. 30. For intermittent duty the capacity is twice as great.
=Standard of Copper Wire Resistance.=—Matthiessen's standard for resistance of copper wire is as follows: _A hard drawn copper wire one meter long, weighing one gramme, has a resistance of .1469 B. A. unit at 32° Fahr._ Relative conducting power: silver, 100; hard or un-annealed copper, 99.95; soft or annealed copper, 102.21.
A committee of the Am. Inst. Electrical Engineers recommends the following form of Matthiessen's standard, taking 8.89 as the specific gravity of pure copper: _A soft copper wire one meter long and one millimeter in diameter has an electrical resistance of .02057 B. A. unit at 0°C._[2] From this the resistance of a soft copper wire one foot long and .001 in. in diameter (mil-foot) is 9.72 B. A. units at 0°C.
[2] NOTE.—The international ohm ÷ B. A. ohm = 1 ÷ .9866. The B. A. ohm ÷ International ohm = 1 ÷ 1.0136. Hence, to reduce British Association ohms to International ohms, divide by 1.0136. or multiply by .9866.
For every degree Fahr., the resistance of copper wire increases .2222%. Thus a piece of copper wire having a resistance of 10 ohms at 32°, would have a resistance of 11.11 ohms at 82°.
Relative Conductivity of Different Metals and Alloys.
(According to Lazare Weiler.)
Pure silver 100 Pure copper 100 Alloy, ½ copper, ½ silver 86.65 Telephonic siliceous bronze 35 Pure zinc 29.9 Brass with 35% zinc 21.5 Swedish iron 16 Pure platinum 10.6 Copper with 10% nickel 10.6 Pure lead 8.88 Pure nickel 7.89 Phosphor-bronze, 10% tin 3.88
=Conductors.=—Copper is used more than any other metal for transmitting electrical energy, and for interior wiring it is used exclusively. Copper conductors should be of the highest commercial conductivity, not less than 97%.
For conductors up to sizes as large as No. 8 B. & S. gauge, single conductors may be used, but for larger sizes the necessary conductivity should be obtained by conductors made up of strands of smaller wires. The size of these strands depend upon the size of the conductors and the conditions under which they are to be used.
Where conductors are very large (as for instance dynamo leads), and where it is essential that they should be as flexible as possible, strands as small as No. 20 or 22 B. & S. gauge may be used.
Conductors for flexible cords, pendants, fixtures, etc., should also consist of very fine strands, so that they may be perfectly pliable and flexible.
The individual strands for instance, for a No. 16 B. & S. gauge flexible cord should be as fine as No. 30.
=Covered Conductors.=—For most conditions of service, wires are protected with an insulating covering. Wires used in interior circuits should have a covering which shall act both as an electrical insulator and as a mechanical protection. In some instances, however, the insulating qualities are of secondary importance.
The various forms of covering now in use commercially for wires are:
1. Rubber; 2. Weather proof; 3. Slow burning; 4. Slow burning weather proof; 5. Armoured.
=Rubber Covered Conductors.=—This class of conductor consists of a tinned copper wire with a rubber covering, protected by an outside braiding of cotton saturated with a preservative compound.
=Ques. What are the advantages of rubber insulation for conductors?=
Ans. It is waterproof, flexible, fairly strong, and has high insulating qualities.
=Ques. What are the disadvantages of rubber insulation?=
Ans. It deteriorates more or less rapidly and is quickly injured by temperatures above 140° Fahr.
=Ques. For what service are rubber covered conductors adapted?=
Ans. For interior wiring.
=Ques. Is pure rubber used?=
Ans. No. The covering should be made from a compound containing from 20 to 35 per cent. of pure rubber.
It would be difficult to place pure rubber on a wire, and moreover a covering made of pure rubber would not be durable and would deteriorate very rapidly, particularly at temperatures above 120° Fahr. Accordingly, it is mixed with other materials, such as French chalk, silicate of magnesia, sulphur, red lead, etc.
=Weather Proof Conductors.=—In this class of conductor, the wire is protected from the weather by a waterproof covering, consisting usually of braided cotton of two or three thicknesses saturated with a moisture resisting insulating compound.
=Ques. Where are weather proof conductors used?=
Ans. In places subject to dampness, such as cellars, tunnels, open sheds, breweries, etc.
=Ques. What are the advantages of weather proof conductors?=
Ans. The insulation is cheap, very durable, and does not deteriorate unless exposed to high temperatures such as will melt the compound.
=Ques. State the disadvantages.=
Ans. The covering is more or less inflammable and is not very efficient as an insulator.
=Ques. What precaution should be taken in using weather proof conductors?=
Ans. On account of the inflammable character of the covering, care should be taken in wiring at points where any considerable number of conductors are brought together, or where there is much woodwork or other combustible material.
=Ques. For what use are weather proof conductors especially adapted?=
Ans. For outside wiring where moisture is certain and where fireproof quality is not necessary.
Obviously conductors of this class should not be used in conduits, nor in fact, in any way except exposed on glass or porcelain insulators.
=Slow Burning Wire.=—This class of conductor is defined as: _one that will not carry fire_. The covering consists of layers of cotton or other thread, all the interstices of which are filled with the fireproofing compound, or of material having equivalent fire resisting and insulating properties. The outer layer is braided and specially designed to withstand abrasion. The thickness of insulation must not be less than that required for slow burning weather proof wire and the outer surface must be finished smooth and hard.—_Underwriters' requirements._
=Ques. Where should slow burning wires be used?=
Ans. In hot dry places, where ordinary insulations would be injured, and where wires are bunched, as on the back of a large switchboard or in a wire tower.
A slow burning covering is considered good enough when the wires are entirely on insulating supports. Its main object is to prevent the copper conductors coming into contact with each other or anything else.
=Ques. What must be done before using weather proof wire?=
Ans. Permission to use the wire must first be obtained from the local Inspection Department.
=Slow Burning Weather Proof Wire.=—The covering of this type wire is a combination of the underwriters and weather proof insulations. The fireproof coating comprises a little more than half of the total covering. When the fireproof coating is placed on the outside, the wire is called "slow burning weather proof."
=Ques. How does slow burning weather proof wire compare with weather proof wire?=
Ans. It is less inflammable and less subject to softening under heat.
=Ques. Where should slow burning weather proof wire be used?=
Ans. In places where the wires are to be run exposed and where moisture resisting quality is desired, also where at the same time it is desirable to avoid an excess of inflammable covering.
=Ques. How should it be installed?=
Ans. It should be set on glass or porcelain insulators.
=Miscellaneous Insulated Conductors=
+————————————-+————————————+————————-+——————————————+——————————+——————————-+ | No. of | Gauge of | No. of | Gauge of | Outside | Weight | | Conductors | Conductors | Armour | Armour Wires | Diameter | per 1,000 | | | (B. & S.) | Wires | B. W. G. | (inch) | Feet | +————————————-+————————————+————————-+——————————————+——————————+——————————-+ | | | | | | | | 1 | 14 | 12 | 8 | ⅞ | 1,150 | | 2 | 14 | 16 | 8 | 1-31/32 | 1,675 | | 3 | 14 | 14 | 6 | 1¼ | 2,400 | | 4 | 14 | 16 | 6 | 1-5/16 | 2,750 | | 5 | 14 | 19 | 6 | 1⅜ | 3,100 | | 6 | 14 | 21 | 6 | 1½ | 3,500 | | 7 | 14 | 21 | 6 | 1½ | 3,600 | | 10 | 14 | 22 | 4 | 1⅞ | 4,600 | | | | | | | | +————————————-+————————————+————————-+——————————————+——————————+——————————-+
NOTE.—The above specifications refer only to river and harbor cables. Ocean cables are of an entirely different character, and consist of "shore end," "intermediate" and "deep sea" types.
=Ques. For what service is slow burning weather proof wire not suited?=
Ans. It is not adapted to outside work.
=Safe Carrying Capacity of Wire.=—All wires will heat when a current of electricity passes through them. The greater the current or the smaller the wire, the greater will be the heating effect. Large wires are heated comparatively more than small wires because the latter have a relatively greater radiating surface.
The temperature of a wire increases approximately as the square of the current, and inversely as the cube of the diameter of the wire.
The elevation in temperature of a wire carrying a current represents so much lost energy.
From these considerations it must be clear that it is important not to overload conductors in order to secure efficient working, and to avoid risk of fire on inside installations.
The Board of Underwriters specifies that the carrying capacity of a conductor is safe when the wire will conduct a certain current without becoming painfully hot.
In the following table of carrying capacity, prepared by the underwriters, a wire is assumed to have a safe carrying capacity when its temperature is not increased by the given current over 30° Fahr. above that of the surrounding air.
SAFE CARRYING CAPACITIES OF WIRES
(Maximum amperes allowed by the Underwriters.)
+————————+——————————————-+————————————+————————————+ | Brown | | Rubber | Other | | and | | insulation | insulations| | Sharpe | Circular mils | —————— | —————— | | Gauge | | Amperes | Amperes | | | | | | +————————+——————————————-+————————————+————————————+ | | | | | | 18 | 1,624 | 3 | 5 | | 16 | 2,583 | 6 | 8 | | 14 | 4,107 | 12 | 16 | | 12 | 6,530 | 17 | 23 | | 10 | 10,380 | 24 | 32 | | 8 | 16,510 | 33 | 46 | | 6 | 26,250 | 46 | 65 | | 5 | 33,100 | 54 | 77 | | 4 | 41,740 | 65 | 92 | | 3 | 52,630 | 76 | 110 | | 2 | 66,370 | 90 | 131 | | 1 | 83,690 | 107 | 156 | | 0 | 105,500 | 127 | 185 | | 00 | 133,100 | 150 | 220 | | 000 | 167,800 | 177 | 262 | | 0000 | 211,600 | 210 | 312 | | | 200,000 | 200 | 300 | | | 300,000 | 270 | 400 | | | 400,000 | 330 | 500 | | | 500,000 | 390 | 590 | | | 600,000 | 450 | 680 | | | 700,000 | 500 | 760 | | | 800,000 | 550 | 840 | | | 900,000 | 600 | 920 | | | 1,000,000 | 650 | 1,000 | | | 1,100,000 | 690 | 1,080 | | | 1,200,000 | 730 | 1,150 | | | 1,300,000 | 770 | 1,220 | | | 1,400,000 | 810 | 1,290 | | | 1,500,000 | 850 | 1,360 | | | 1,600,000 | 890 | 1,430 | | | 1,700,000 | 930 | 1,490 | | | 1,800,000 | 970 | 1,550 | | | 1,900,000 | 1,010 | 1,610 | | | 2,000,000 | 1,050 | 1,670 | | | | | | +————————+——————————————-+————————————+————————————+
The lower limit is specified for rubber covered wires to prevent gradual deterioration of the high insulations by the heat of the wires, but not from fear of igniting the insulation. The question of drop is not taken into consideration in the table on page 731.
The carrying capacity of Nos. 16 and 18 B. & S. gauge wire is given, but no smaller than No. 14 is to be used, except as allowed under rules for fixture wiring.—_Underwriters' Rules._
=Circular Mils.=—The unit of measurement in measuring the cross sectional area of wires is the _circular mil_; it is the area of a circle one mil (.001 in.) in diameter.
_The area of a wire in circular mils is equal to the square of the diameter in mils._
Thus a wire 2 mils in diameter (.002 in.) has a cross sectional area of 2 × 2 = circular mils. Accordingly to obtain the area of a wire in circular mils, _measure its diameter with a micrometer which reads directly in mils or thousandths of an inch, and square the reading_.
The circular mil (abbreviated C.M.) applies to all _round_ conductors, and has a value of .7854 times that of the square mil, that is, 1 circular mil = .7854 square mil. If the diameter be expressed as a fraction of an inch, as for instance ⅓ in., the circular mil area may be found as follows: Reduce the fraction ⅓ to the decimal of an inch, multiply the result by 1,000 to express the diameter in mils, and square the diameter so expressed, thus: ⅓ = 1,000 ÷ 3 = .333. .333 × 1,000 = 333 mils; 333 × 333 = 110,889 circular mils.
The diameter of any wire may be found when its circular mil area is known by extracting the square root of the circular mil area.
=Square Mils.=—For measuring conductors of square or rectangular cross section, such as bus bars, copper ribbon, etc., the square mil is used. A square mil is the area of a square whose sides are one mil (.001 in. long) and is equal to .001 × .001 = .000001 square inch.
EXAMPLE.—A copper ribbon for a field coil measures ⅝ inch by ⅛ inch. What is its area in square mils? What is its area in circular mils?
⅝ = .625 in., or 625 mils; ⅛ = .125 in., or 125 mils.
Area in square mils = 625 × 125 = 78,125.
Area in circular mils= { 78,125 ÷ .7854 } {or 78.125 × 1.2732} = 99,469.
=Mil Foot.=—This unit is used as a basis for computing the resistance of any given wire. A mil foot means _a volume one mil in diameter and one foot long_.
_The resistance of a wire of commercially pure copper one mil in diameter and one foot long is taken as a standard in calculating the resistance of wires, and has been found to be equal to 10.79 ohms at 75° Fahr._
The calculation is made according to the following rule:
_The resistance of a copper wire is equal to its length in feet, multiplied by the resistance of one mil foot (10.79 ohms) and divided by the number of circular mils, or the square of its diameter._
Expressed as a formula:
length of wire in ft. × 10.79 resistance in ohms = —————————————————————————————— (1) circular mils
EXAMPLE. What is the resistance of a copper wire 1,500 feet long and having a transverse area of 10,381 circular mils?
Substituting these values in formula (1)
1,500 × 10.79 resistance = ———————————————————— = 1.559 ohms. 10,381
_The transverse area of a copper wire is found by multiplying the resistance of a mil foot (10.79) by its length in feet and dividing the result by its resistance in ohms._
This is obtained directly from the formula (1) by solving the equation for circular mils, thus:
length of wire in ft. × 10.79 circular mils = ——————————————————————————————- (2) resistance in ohms
EXAMPLE. What is the circular mil area of a wire 1,500 feet long and having a resistance of 1.559 ohms?
Substituting the values in equation (2)
1,500 × 10.79 circular mils = ————————————- = 10,381 1.559
=Lamp Foot.=—This unit facilitates laying out wiring and calculating the drop. A lamp foot is defined as _one 16 candle power lamp at a distance of one foot from the point of supply_. Accordingly the number of lamp feet in any circuit is equal to the number of 16 candle power lamps (or equivalent in other sizes) in the circuit multiplied by the distance in feet from the fuse block to the center of distribution.
When no point is specified, the feet are always measured from the supply point to the center of distribution. When other than 16 c.p. lamps are in the circuit they must be reduced to 16 c.p. lamps. Thus two 8 c.p. lamps would be counted one 16 c.p. lamp, one 32 c.p. lamp would be counted two 16 c.p. lamps, etc.
=Ampere Foot.=—From the foregoing explanation of _lamp foot_, the significance of _ampere foot_ is easily understood—the two terms are in fact self-defining.
An ampere foot may be defined as _the product of one ampere multiplied by one foot_.
The unit ampere foot is used in figuring motor circuits or currents designed to carry a mixed load.
The ampere feet of a main are found by _multiplying the maximum load in amperes by the distance from the fuse block to the electrical center of the load_.
Thus if the center of distribution be 50 feet from the fuse block and the maximum load is 9 amperes, the number of ampere feet is equal to 9 × 50 = 450.
=Electrical Center of Distribution.=—The electrical center of a circuit depends upon the distances between the lamps and the fuse block; also the relative sizes of the lamps.
It may be defined as _the sum of the lamp feet for each section divided by the number of 16 candle power lamps in the circuit_.
If the lamps be of uniform capacity, and placed at equal distances apart, the center of distribution will coincide with the geometrical center of the group of lamps. However, if the lamps vary in size, and be irregularly spaced, the electrical center will not coincide with the geometrical center unless the lamps be symmetrically arranged so as to compensate for the difference in sizes and spacing.
In such cases, as shown in fig. 821, the electrical center can be determined by adding together the lamp feet of the several sections A, B, C, etc., of the main and dividing the result by the 16 c.p. units. Thus the lamp feet of
Section A = 10 lamps × 10 feet= 100 " B = 9 " × 5 " = 45 " C = 7 " × 6 " = 42 " D = 6 " × 4 " = 24 " E = 5 " × 5 " = 25 " F = 4 " × 10 " = 40 " G = 2 " × 5 " = 10 ——- which added together gives a total of 286 lamp feet.
This when divided by the ten 16 c.p. units comprising four 16 c.p. lamps and three 32 c.p. lamps, gives a little over 28½ feet as the distance from the fuse block to the center of distribution, the position of which is shown by the line M N in fig. 821, while that of the geometrical center is shown by the line K L.
When the center of distribution is at a considerable distance from the supply circuit, and it becomes advisable to divide the wiring into two distinct elements—a feeder and one or more mains, the junction of the feeder and the mains should be located at the electrical center of the mains whenever possible. When this is done, it is obvious that the wire size of only one half the main needs to be calculated, as both halves of the main are identical.
=Wire Gauges.=—For the purpose of facilitating the measurement of wire, a number of gauges have been designed by various wire manufacturing concerns. The principal gauges used in the United States are the American or Brown & Sharp's gauge; the English standard or Birmingham gauge; Washburn & Moen's standard gauge; Imperial wire gauge; Stubs' steel wire gauge, and the U. S. Standard wire gauge.
The several gauges are here given with explanation of their use.
=The American Standard or Brown and Sharp's Gauge.=—This gauge is commonly designated as A. W. G. or B. & S., and has been adopted by brass manufacturers and is used mostly in measuring brass, copper, silver, German silver, and gold in both wire and plate.
=Birmingham or Stub's Wire Gauge (B. W. G.).=—Old English Standard and Iron Wire Gauge. _Birmingham or Stubs' Iron Wire Gauge is not the same as Stubs' Steel Wire Gauge_. A table of Stub's Steel Wire Gauge is given on page 741.
=Washburn and Moen's Standard Wire Gauge.=—Commonly designated as W. & M. G. has been adopted by the U. S. Steel Corporation in making their wire.
=New British Standard (N. B. S.).=—British Imperial English Legal Standard and Standard Wire Gauge, and is variously abbreviated by S. W. G. and I. W. G.
=Roebling Gauge.=—Washburn Moen, American Steel & Wire Co.'s Iron Wire Gauge.
=U. S. Standard Wire Gauge.=—This gauge is used for measuring sheet and plate iron, and steel, by the U. S. Government in assessing duties, and in making requisitions for supplies.
=Old English Standard Wire Gauge.=—The old English gauge is the same as the Birmingham or Stubs' standard gauge, commonly designated as B. W. G. It is used chiefly for measuring sheet iron and steel, also soft steel and iron wire.
=London Gauge.=—Old English (_not Old English Standard_).
From the foregoing it is seen that great confusion exists with such a multiplicity of gauges and emphasizes the importance of specifying the gauge and of knowing what gauge to use.
In using the gauges known as Stubs' Gauges, there should be constantly borne in mind the difference between the Stubs' Iron Wire Gauge and the Stubs' Steel Wire Gauge. The Stubs' Iron Wire Gauge is the one commonly known as the English Standard Wire, or Birmingham Gauge and designates the Stubs' _soft_ wire sizes. The Stubs' Steel Wire Gauge is the one that is used in measuring drawn steel wire or drill rods of Stubs' make and is also used by many makers of American drill rods.
=STUBS' STEEL WIRE GAUGE=
——————-+————————-++——————+————————-++——————+————————-++——————+———————- | Size of || | Size of || | Size of || | Size of Letter.| Letter ||No. of| Number ||No. of| Number ||No. of| Number | in || Wire | in || Wire | in || Wire | in |Decimals.||Gauge.|Decimals.||Gauge.|Decimals.||Gauge.|Decimals. ——————-+————————-++——————+————————-++——————+————————-++——————+———————- Z | .413 || 1 | .227 || 28 | .139 || 55 | .050 Y | .404 || 2 | .219 || 29 | .134 || 56 | .045 X | .397 || 3 | .212 || 30 | .127 || 57 | .042 W | .386 || 4 | .207 || 31 | .120 || 58 | .041 V | .377 || 5 | .204 || 32 | .115 || 59 | .040 U | .368 || 6 | .201 || 33 | .112 || 60 | .039 T | .358 || 7 | .199 || 34 | .110 || 61 | .038 S | .348 || 8 | .197 || 35 | .108 || 62 | .037 R | .339 || 9 | .194 || 36 | .106 || 63 | .036 Q | .332 || 10 | .191 || 37 | .103 || 64 | .035 P | .323 || 11 | .188 || 38 | .101 || 65 | .033 O | .316 || 12 | .185 || 39 | .099 || 66 | .032 N | .302 || 13 | .182 || 40 | .097 || 67 | .031 M | .295 || 14 | .180 || 41 | .095 || 68 | .030 L | .290 || 15 | .178 || 42 | .092 || 69 | .029 K | .281 || 16 | .175 || 43 | .088 || 70 | .027 J | .277 || 17 | .172 || 44 | .085 || 71 | .026 I | .272 || 18 | .168 || 45 | .081 || 72 | .024 H | .266 || 19 | .164 || 46 | .079 || 73 | .023 G | .261 || 20 | .161 || 47 | .077 || 74 | .022 F | .257 || 21 | .157 || 48 | .075 || 75 | .020 E | .250 || 22 | .155 || 49 | .072 || 76 | .018 D | .246 || 23 | .153 || 50 | .069 || 77 | .016 C | .242 || 24 | .151 || 51 | .066 || 78 | .015 B | .238 || 25 | .148 || 52 | .063 || 79 | .014 A | .234 || 26 | .146 || 53 | .058 || 80 | .013 | || 27 | .143 || 54 | .055 || | ——————-+————————-++——————+————————-++——————+————————-++——————+———————-
The following table gives the diameters, in decimal parts of an inch, of the various sizes of wire corresponding to the number of gauge numbers of the different standard wire gauges used in the United States.
=TABLE OF VARIOUS WIRE GAUGES=
=In decimal parts of an inch=
————————+————————-+————————-+——————————+——————-+——————-+——————-+——————- Number | Ameri- | Birming-| Wash- | Tren- |G.W. | Old | Brit- of | can, or | ham, or | burn & | ton | Pren- | Eng- | ish Wire | Brown & | Stubs | Moen | Iron | tiss, | lish, | Stan- Gauge | Sharpe | (B.W.G.)| Mfg. Co.,| Co., | Holy- | From | dard | (B.&S.) | | Worces- | Tren- | oke, | Brass | (S.W.- | | | ter, | ton, | Mass. | Mfrs' | G.) | | | Mass. | N.J. | | List | ————————+————————-+————————-+——————————+——————-+——————-+——————-+——————- 0000000 | | | | | | | .500 000000 | | | .460 | | | | .464 00000 | | | .430 | .450 | | | .432 0000 | .46000 | .454 | .393 | .400 | | | .400 000 | .40964 | .425 | .362 | .360 | .3586 | | .372 00 | .36480 | .380 | .331 | .330 | .3282 | | .348 0 | .32486 | .340 | .307 | .305 | .2994 | | .324 1 | .28930 | .300 | .283 | .285 | .2777 | | .300 2 | .25763 | .284 | .263 | .265 | .2591 | | .276 3 | .22942 | .259 | .244 | .245 | .2401 | | .252 4 | .20431 | .238 | .225 | .225 | .2230 | | .232 5 | .18194 | .220 | .207 | .205 | .2047 | | .212 6 | .16202 | .203 | .192 | .190 | .1885 | | .192 7 | .14428 | .180 | .177 | .175 | .1758 | | .176 8 | .12849 | .165 | .162 | .160 | .1605 | | .160 9 | .11443 | .148 | .148 | .145 | .1471 | | .144 10 | .10189 | .134 | .135 | .130 | .1351 | | .128 11 | .090742 | .120 | .120 | .1175 | .1205 | | .116 12 | .080808 | .109 | .105 | .1050 | .1065 | | .104 13 | .071961 | .095 | .0920 | .0925 | .0928 | | .0920 14 | .064084 | .083 | .0800 | .0800 | .0816 | .08300| .0800 15 | .057068 | .072 | .0720 | .0700 | .0726 | .07200| .0720 16 | .050820 | .065 | .0630 | .0610 | .0627 | .06500| .0640 17 | .045257 | .058 | .0540 | .0525 | .0546 | .05800| .0560 18 | .040303 | .049 | .0470 | .0450 | .0478 | .04900| .0480 19 | .035890 | .042 | .0410 | .0400 | .0411 | .04000| .0400 20 | .031961 | .035 | .0350 | .0350 | .0351 | .03500| .0360 21 | .028462 | .032 | .0320 | .0310 | .0321 | .03150| .0320 22 | .025347 | .028 | .0280 | .0280 | .0290 | .02950| .0280 23 | .022571 | .025 | .0250 | .0250 | .0261 | .02700| .0240 24 | .020100 | .022 | .0230 | .0225 | .0231 | .02500| .0220 25 | .017900 | .020 | .0200 | .0200 | .0212 | .02300| .0200 26 | .015940 | .018 | .0180 | .0180 | .0194 | .02050| .0180 27 | .014195 | .016 | .0170 | .0170 | .0182 | .01875| .0164 28 | .012641 | .014 | .0160 | .0160 | .0170 | .01650| .0148 29 | .011257 | .013 | .0150 | .0150 | .0163 | .01550| .0136 30 | .010025 | .012 | .0140 | .0140 | .0156 | .01375| .0124 31 | .008928 | .010 | .0130 | .0130 | .0146 | .01225| .0116 32 | .007950 | .009 | .0120 | .0120 | .0136 | .01125| .0108 33 | .007080 | .008 | .0110 | .0110 | .0130 | .01025| .0100 34 | .006305 | .007 | .0100 | .0100 | .0118 | .00950| .0092 35 | .005615 | .005 | .0095 | .0095 | .0109 | .00900| .0084 36 | .005000 | .004 | .0090 | .0090 | .0100 | .00750| .0076 37 | .004453 | | .0085 | .0085 | .0095 | .00650| .0068 38 | .003965 | | .0080 | .0080 | .0090 | .00575| .0066 39 | .003531 | | .0075 | .0075 | .0083 | .00500| .0052 40 | .003145 | | .0070 | .0070 | .0078 | .00450| .0048 41 | | | | | | | .0044 42 | | | | | | | .0040 ————————+————————-+————————-+——————————+——————-+——————-+——————-+——————-
NOTE.—The sizes of wire are ordinarily expressed by an arbitrary series of numbers. Unfortunately there are several independent numbering methods, so that it is always necessary to specify the method or wire gauge used. The above table gives the numbers and diameters in decimal parts of an inch for the various wire gauges in general use.
=Wiring Terms.=—The various members of a complex wiring installation are designated feeders, sub-feeders, mains, branches, and taps.
A _feeder_ is a stretch of wiring to which no connection is made except at its two ends.
A _sub-feeder_ is of the same class as a feeder, but is distinguished either by being one of two or more connecting links between the end of a single feeder and several distributing mains, or by constituting an extension of a feeder.
A _main_ is a stretch of wiring supplied from one or more feeders or sub-feeders and distributing current to a number of taps, or else to a number of branches.
A _branch_ distributes current among a number of lamps, etc.
A _tap_ almost invariably delivers current to a single lamp or other device.
Reference to fig. 826 will make these definitions clearer. This diagram is intended merely to illustrate the above definitions and does not represent any special plan of wiring.
The simplest possible wiring installation is one in which a single lamp or compact cluster of lamps is located at the end of a main, as shown in figs. 827 and 828. In such cases calculations are almost always unnecessary, for the reason that No. 14 wire, the smallest size allowed by the underwriters, will supply several lamps at a long distance (as interior wiring goes) with a very moderate drop. For example, if the three lamps shown at the end of the main in fig. 828, be of 16 candle power each, and the voltage of the supply circuit be 110 volts, a main of No. 14 wire would supply the lamps at a distance of 135 feet from the fuse block with a drop of only 1 per cent.
When the lamps are strung along the main, however, as in fig. 826, it is sometimes necessary to choose the size of wire with regard to the drop, and in order to do this the main must be measured for either "ampere feet" or "lamp feet."
=Wire Calculations.=—The problem of calculating the size of wire will be presented here in as simple a form as possible, with explanation of the various steps so that any one can understand how the formula is derived.
In determining the size of wire, there are four known factors which enter into the calculation, viz.:
1. Length of circuit in feet; 2. Maximum current in amperes; 3. Drop or volts lost in the circuit, _in % of the impressed voltage_; 4. Heating effect of the current.
The calculation is based on the _mil foot_, which as previously explained, is a foot of copper wire one mil in diameter and whose resistance is equal to 10.79 ohms at 75° Fahr.
The first step is to find an expression for the resistance of the wire which may be later substituted in Ohm's law formula. Accordingly, the resistance of any conductor is equal to _its length in feet multiplied by its resistance per mil foot and the product divided by its area in circular mils_, thus:
length in feet × resistance per mil foot resistance in ohms = ————————————————————————————————————————— circular mils
feet × 10.8 or ohms = ————————————- (1) circular mils
(calling the resistance per mil foot 10.8 instead of 10.79 to facilitate calculation).
LAMP TABLE FOR RUBBER COVERED WIRES
Showing the maximum number of 16 candle power 110 to 240 volt lamps in parallel that may be carried by the various sizes of wire without violating the underwriters' rules.
+——————————+———————+——————————————+————————————————+————————————-+ |Wire size |Amperes|3.1-watt lamps|3.5-watt lamps. |4-watt lamps.| |B. & S. | +——————————————+————————————————+————————————-+ | gauge | | At 110 220 | At 110 220 | 220 230 240 | | | | volts V. | volts V. | V. V. V. | +——————————+———————+——————————————+————————————————+—————————————+ |0000 | 210 | 462 924 | 412 825 | 722 754 787 | | 000 | 177 | 389 778 | 347 695 | 608 636 663 | | 00 | 150 | 330 660 | 294 589 | 515 539 562 | | | | | | | | 0 | 127 | 279 558 | 249 499 | 436 456 476 | | 1 | 107 | 235 470 | 210 420 | 367 384 401 | | 2 | 90 | 197 396 | 176 353 | 309 323 337 | | | | | | | | 3 | 76 | 167 334 | 149 298 | 261 273 285 | | 4 | 65 | 143 286 | 127 255 | 223 233 243 | | 5 | 54 | 118 237 | 106 212 | 185 194 202 | | | | | | | | 6 | 46 | 101 202 | 90 180 | 158 165 172 | | 8 | 33 | 72 145 | 64 129 | 113 118 123 | | 10 | 24 | 52 105 | 47 94 | 82 86 90 | | | | | | | | 12 | 17 | 37 74 | 33 66 | 58 61 63 | | 14 | 12 | 26 52 | 23½ 47 | 41 43 45 | | 16[3] | 6 | 13 .. | 11 .. | 20 21 22 | +——————————+———————+——————————————+————————————————+—————————————+
[3] This size can be used only in the shape of flexible cord.
Now, according to Ohm's law,
volts = amperes × ohms (2)
hence, substituting in (2) the value for the resistance in ohms, as obtained in (1):
feet × 10.8 volts = amperes × ————————————- circular mils
or using the usual symbols
feet × 10.8 E = I × ————————————- (3) circular mils
or expressed in words, formula (3) means that the volts lost or _drop_ between the beginning and end of a circuit is equal to the current flowing through the circuit multiplied by the product of the conductors' length in feet multiplied by the resistance of one mil foot of wire, divided by the area of the conductor in circular mils.
LAMP TABLE FOR WEATHER PROOF WIRES
Showing the maximum number of 16 candle power 120 to 240 volt lamps in parallel that may be carried by various sizes of weather proof wire without violating the underwriters' rules.
+———————+————————+———————————————+———————————————+———————————————-+ |Wire |Amperes |3.1-watt lamps.|3·5-watt lamps.| 4-watt lamps. | |size | +——————————————-+——————————————-+————————————————| |B. & S.| | 110 220 | 110 220 | 220 230 240 | |gauge | | V. V. | V. V. | V. V. V. | +———————+————————+———————+———————+————————+——————+———————————————-+ | 0000 | 312 | 686 | 1372 | 612 | 1225 | 1072 1121 1170 | | 000 | 262 | 576 | 1152 | 514 | 1029 | 900 941 982 | | 00 | 220 | 484 | 968 | 432 | 864 | 756 790 825 | | | | | | | | | | 0 | 185 | 407 | 814 | 363 | 726 | 636 665 693 | | 1 | 156 | 343 | 686 | 306 | 612 | 536 560 585 | | 2 | 131 | 288 | 576 | 257 | 514 | 450 470 491 | | | | | | | | | | 3 | 110 | 242 | 484 | 216 | 432 | 378 395 412 | | 4 | 92 | 202 | 404 | 180 | 361 | 316 330 345 | | 5 | 77 | 169 | 338 | 151 | 302 | 264 276 288 | | | | | | | | | | 6 | 65 | 143 | 286 | 127 | 255 | 223 233 243 | | 8 | 46 | 101 | 202 | 90 | 180 | 158 165 172 | | 10 | 32 | 70 | 140 | 62 | 125 | 110 115 120 | | | | | | | | | | 12 | 23 | 50 | 101 | 45 | 90 | 79 82 86 | | 14 | 16 | 35 | 70 | 31 | 62 | 55 57 60 | +———————+————————+———————+———————+————————+——————+———————————————-+
Now, since the length of the circuit is given as the "run" or distance one way, that is, one half the total length of wire in the circuit, formula (3) must be multiplied by 2 to get the total drop, that is:
feet × 10.8 X 2 I × feet × 21.6 E = I × ——————————————- = ———————————————- (4) circular mils circular mills
Solving the last equation for the unknown quantity, the following equation is obtained for size of wire:
I × feet × 21.6 amperes × feet × 21.6 circular mils = ———————————————- = ————————————————————— (5) E "drop"
The following practical example is given to illustrate the application of the formula just obtained:
EXAMPLE.—What size wire should be used on a 250 volt circuit to transmit a current of 200 amperes a distance of 350 feet to a center of distribution with a loss of three per cent. under full load?
The volts lost or drop is equal to 250 × .03 = 7.5 volts.
=PROPERTIES OF COPPER WIRE=
+————————+————————+——————————+—————————————————————————————+———————————————————————-+ | Number |Diameter| Area in | Weight in pounds | Resistance at 68° Fahr.| |of gauge| in | circular |——————————————————————————————————————————————————————| |B. & S. | mils | mils | 1,000 feet| mile |feet per| 1,000 feet | mile | | | | | | | pound | | | +————————+————————+——————————+———————————+—————————————————+———————————————————————-+ | 0000 | 460 | 211,600 | 640.5 |3,381 | 1.561 | .04893 | .2583 | | 000 | 409.6 | 167,800 | 508 |2,682 | 1.969 | .06170 | .3258 | | 00 | 364.8 | 133,100 | 402.8 |2,127 | 2.482 | .07780 | .4108 | | 0 | 324.9 | 105,500 | 319.5 |1,687 | 3.130 | .09811 | .5180 | | | | | | | | | | | 1 | 289.3 | 83,690 | 253.3 |1,337 | 3.947 | .12370 | .6531 | | 2 | 257.6 | 66,370 | 200.9 |1,062 | 4.977 | .1560 | .8237 | | 3 | 229.4 | 52,630 | 159.3 | 841.1 | 6.276 | .1967 | 1.0386 | | 4 | 204.3 | 41,740 | 126.4 | 667.4 | 7.914 | .2480 | 1.3094 | | | | | | | | | | | 5 | 181.9 | 33,100 | 100.2 | 529.0 | 9.980 | .3128 | 1.6516 | | 6 | 162.0 | 26,250 | 79.46 | 419.5 | 12.580 | .3944 | 2.0824 | | 7 | 144.3 | 20,820 | 63.02 | 332.7 | 15.87 | .4973 | 2.6257 | | 8 | 128.5 | 16,510 | 49.98 | 263.9 | 20.01 | .6271 | 3.3111 | | | | | | | | | | | 9 | 114.4 | 13,090 | 39.63 | 209.2 | 25.23 | .7908 | 4.1754 | | 10 | 101.9 | 10,380 | 31.13 | 166.0 | 31.82 | .9972 | 5.2652 | | 11 | 90.74| 8,234 | 24.93 | 131.6 | 40.12 |1.257 | 6.6370 | | 12 | 80.81| 6,530 | 19.77 | 104.4 | 50.59 |1.586 | 8.374 | | | | | | | | | | | 13 | 71.96| 5,178 | 15.68 | 82.79| 63.79 |2.000 | 10.560 | | 14 | 64.08| 4,107 | 12.43 | 65.63| 80.44 |2.521 | 13.311 | | 15 | 57.07| 3,257 | 9.858 | 52.05|101.4 |3.179 | 16.785 | | 16 | 50.82| 2,583 | 7.818 | 41.28|127.9 |4.009 | 21.168 | | | | | | | | | | | 17 | 45.26| 2,048 | 6.200 | 32.74|161.3 |5.055 | 26.690 | | 18 | 40.30| 1,624 | 4.917 | 25.96|203.4 |6.374 | 33.655 | | 19 | 35.89| 1,288 | 3.899 | 20.59|256.5 |8.038 | 42.440 | | 20 | 31.96| 1,022 | 3.092 | 16.33|323.4 |10.14 | 53.540 | +————————+————————+——————————+———————————+—————————————————+———————————————————————-+
Substituting the given value in formula (5)
350 × 200 × 21.6 circular mils = —————————————————— = 201,600. 7.5
Diameter = 2 √201,600 = 449 circular mils or .449 in.
From the table (on page 731 or on page 742) the nearest (=larger=) size of wire is 0000 B. & S. gauge.[4]
[4] =CAUTION.=—The size thus obtained should be compared with the table of carrying capacity of wires as given on page 731 to see if the wires would have to carry more than the allowable current.
WIRING TABLE FOR LIGHT AND POWER CIRCUITS
===================+======================================================= VOLTS | PERCENTAGE OF LOSS ===================+======================================================= 2000 | 1.7 1.5 1.4 1.2 1.1 1.0 0.75 0.5 1000 | 3.4 2.9 2.7 2.4 2.2 2.0 1.5 1.0 500 | 6.5 5.7 6.2 4.8 4.3 3.9 2.9 2.0 220 | 13.7 12.0 11.0 10.3 9.3 8.3 6.5 4.4 110 | — — 20.0 18.5 17.0 15.4 12.0 8.4 52 | — — — — — — 22.4 16.1 ===================+======================================================= ACTUAL VOLTS LOST =========+=========+======================================================= Carrying | | Capacity | Size | 35 30 27.5 25 22.5 20 15 10 Amperes. | B. & S. | =========+=========+======================================================= 300 | 0000 | 345800 296400 271700 247000 222300 197600 148200 98800 245 | 000 | 274400 235200 215600 196000 176400 156800 117600 78400 215 | 00 | 217525 186450 170912 155375 139837 124300 93225 62150 190 | 0 | 172550 147900 135575 123250 110925 98600 73950 49300 160 | 1 | 136850 117300 107525 97750 87975 78200 58650 39100 135 | 2 | 108500 93000 85250 77500 69750 62000 46500 31000 115 | 3 | 86100 73800 67650 61500 55350 49200 36900 24600 100 | 4 | 68250 58500 53625 48750 43875 39000 29250 19500 90 | 5 | 54250 46500 42625 38750 34875 31000 23250 15500 80 | 6 | 43050 36900 33825 30750 27675 24600 18450 12300 60 | 8 | 26965 23130 21202 19275 17347 15420 11565 7710 40 | 10 | 16975 14550 13337 12125 10912 9700 7275 4850 30 | 12 | 10675 9150 8388 7625 6862 6100 4575 3050 22 | 14 | 6720 5760 5280 4800 4320 3840 2880 1920 5[5] | 16 | 4235 3630 3328 3025 2723 2420 1815 1210 ————————-+————————-+——————————————————————————————————————————————————————-
===================+===================================================== VOLTS | PERCENTAGE OF LOSS ===================+===================================================== 2000 | 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 1000 | 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 500 | 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 220 | 3.9 3.5 3.1 2.7 2.2 1.8 1.4 0.9 0.45 110 | 7.6 6.8 6.0 5.2 4.4 3.5 2.7 1.8 0.9 52 | 14.7 13.3 11.8 10.3 8.8 7.1 5.5 3.7 1.9 ===================+===================================================== ACTUAL VOLTS LOST =========+=========+===================================================== Carrying | | Capacity | Size | 9 8 7 6 5 4 3 2 1 Amperes. | B. & S. | =========+=========+===================================================== 300 | 0000 | 88920 79040 69160 59280 49400 39520 29640 19760 9880 245 | 000 | 70560 62720 54880 47040 39200 31360 23520 15680 7840 215 | 00 | 55935 49720 43505 37290 31075 24860 18645 12430 6215 190 | 0 | 44370 39440 34510 29580 24650 19720 14790 9860 4930 160 | 1 | 35190 31280 27370 23460 19550 15640 11730 7820 3910 135 | 2 | 27900 24800 21700 18600 15500 12400 9300 6200 3100 115 | 3 | 22140 19680 17220 14760 12300 9840 7380 4920 2460 100 | 4 | 17550 15600 13650 11700 9750 7800 5850 3900 1950 90 | 5 | 13950 12400 10850 9300 7750 6200 4650 3100 1550 80 | 6 | 11070 9840 8610 7380 6150 4920 3690 2460 1230 60 | 8 | 6939 6168 5397 4626 3855 3084 2313 1542 771 40 | 10 | 4365 3880 3395 2910 2425 1940 1455 970 486 30 | 12 | 2745 2440 2135 1830 1525 1220 915 610 305 22 | 14 | 1728 1536 1344 1152 960 768 576 384 192 5[5] | 16 | 1089 968 847 726 605 484 363 242 121 ————————-+————————-+————————————————————————————————————————————————————-
[5] NOTE.—In case a larger loss than any given in the table is required, proceed as follows:—Divide the ampere feet by 10 and then refer to column of Actual Volts Lost divided by 10, from which the size of wire is found as before.
RULE.—_Multiply current in amperes by single distance and refer to the nearest corresponding number under column of actual volts lost, to find size of wire._ It should also be noted that the underwriters prohibit the use of wire smaller than No. 14 B. & S. gauge, except as allowed for fixture work and pendant cord.
TABLE OF WIRE EQUIVALENTS
GAUGE NUMBER OF WIRES B. & S. 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384
0000 0 3 6 9 12 15 18 21 24 27 30 33 36 39 000 1 4 7 10 13 16 19 22 25 28 31 34 37 40 _____ 00 2 5 8 11 14 17 20 23 26 29 32 35 38| 0+6 0 3 6 9 12 15 18 21 24 27 30 33 36 39| 1+7 1 4 7 10 13 16 19 22 25 28 31 34 37 40| 2+8 _____| 2 5 8 11 14 17 20 23 26 29 32 35 38|3+9 4+6 3 6 9 12 15 18 21 24 27 30 33 36 39|4+10 5+7 4 7 10 13 16 19 22 25 28 31 34 37 40|5+11 6+8 _____| 5 8 11 14 17 20 23 26 29 32 35 38|6+12 7+9 6 9 12 15 18 21 24 27 30 33 36 39|7+13 8+10 7 10 13 16 19 22 25 28 31 34 37 40|8+14 9+11 _____| 8 11 14 17 20 23 26 29 32 35 38|9+15 10+12 9 12 15 18 21 24 27 30 33 36 39|10+16 11+13 10 13 16 19 22 25 28 31 34 37 40|11+17 12+14 _____| 11 14 17 20 23 26 29 32 35 38|12+18 13+15 12 15 18 21 24 27 30 33 36 39|13+19 14+16 13 16 19 22 25 28 31 34 37 40|14+20 15+17 ____| 14 17 20 23 26 29 32 35 38|15+21 16+18 15 18 21 24 27 30 33 36 39|16+22 17+19 16 19 22 25 28 31 34 37 40|17+23 18+20 ____| 17 20 23 26 29 32 35 38|18+24 19+21 18 21 24 27 30 33 36 39|19+25 20+22 19 22 25 28 31 34 37 40|20+26 21+23 ____| 20 23 26 29 32 35 38|21+27 22+24 21 24 27 30 33 36 39|22+28 23+25 22 25 28 31 34 37 40|23+29 24+26 ____| 23 26 29 32 35 38|24+30 25+27 24 27 30 33 36 39|25+31 26+28 25 28 31 34 37 40|26+32 27+29 ____| 26 29 32 35 38|27+33 28+30 27 30 33 36 39|28+34 29+31 28 31 34 37 40|29+35 30+32 ____| 29 32 35 38|30+36 31+33 30 33 36 39|31+37 32+34 31 34 37 40|32+38 33+35 ____| 32 35 38|33+39 34+36 33 36 39|34+40 35+37 34 37 40|36+39 ___| 35 38|37+39 36 39|38+40 37 40|
=Ques. If the calculated size of wire be larger than any in the table how is the required area obtained?=
Ans. By using two or more smaller wires in parallel, whose combined area is equal to the required area.
To facilitate finding the equivalent sizes the above table of wire equivalents has been prepared.
=Ques. How is the table of wire equivalents used?=
Ans. To use the table, find in the vertical column at left the size of conductor desired; then follow along horizontally until the size of wire that is desired to use for the strands, and the corresponding number at top of column will give the number of strands of that size wire required.
=Ques. What is the significance of the zig-zag line?=
Ans. The figures below this line give the gauge numbers of two wires which will have the same conductivity as the corresponding conductor in left hand column.
TABLE OF CABLE CAPACITIES
=Incandescent Lamps on 660 Watt Circuits.=—The standard incandescent lamp is rated as equivalent to the light given by 16 candles, and may consume, according to type and make, from 50 to 56 watts, or from 3.1 to 3.5 watts per candle power. Therefore, a 660 watt circuit will carry thirteen 16 candle power 49.6 watt lamps, or eleven 56 watt lamps.
The proper size of wire for a 660 watt circuit will depend upon the voltage for which the lamps are made. For example: a 16 candle power lamp which consumes 56 watts on 110 volt circuit will take, 56 ÷ 110 = .5 or ½ ampere of current, while the same lamp, if made for 220 volts, will take only 55 ÷ 220 = .25 or ¼ ampere. Therefore, eleven 16 candle power 56 watt lamps will require a current of 5½ amperes at 110 volts, or 2¾ amperes at 220 volts.
According to the laws of resistance, the resistance of a round wire is inversely proportional to the square of the diameter, and if the circuit be taken at 100 feet, and the allowable percentage of drop at 1 volt, then according to formula, (5) on page 748, the wire required for a circuit carrying eleven 16 candle power 56 watt 110 volt lamps, will have a cross sectional area of,
5.5 × 100 × 21.6 ———————————————— = 11,880 circular mils. 1
while the same number of lamps on a 220 volt circuit will require wire having a cross sectional area of,
2.75 × 100 × 21.6 ————————————————- = 5,940 circular mils. 1
In order to conform to the underwriters' requirements, No. 8. B. & S. gauge, wire must be used for the circuit carrying the 110 volt lamps, while No. 12, B. & S. wire, would be sufficient for the 220 volt circuit.
In the case shown in fig. 829, the branch circuits A and B are identical, each supplying four 16 candle power lamps requiring 3.5 watts per candle power at 110 volts or carrying a load of 4 × 16 × 3.5 = 224 watts, = 224 ÷ 110 = 2 _amperes_.
The distance from the feeder junction or cut out to the electrical center of each branch circuit is 12.5 _feet_. The compact area of distribution permits the reduction of the loss of volts to 1 per cent, or 110 × .01 = 1.1 volts "drop." Then substituting in formula (5) on page 748 the values for _amperes_, _feet_ and _drop_ as obtained above
2 × 25 × 21.6 ————————————- = 981 circular mils, 1.1
or a value far below that of even No. 18 wire, B. & S. gauge (see table on page 731), but the smallest wire allowed by the underwriters for the mains A and B is No. 14, B. & S. gauge.
In calculating the size of wire for the feeders the total load must be considered. This is equal to eight 16 candle power lamps, requiring 3.5 watts per candle power at 110 volts = 8×16×3.5 = 448 watts = 4 amperes.
The distance from the entrance cut out to the feeder cut out is 200 feet. The drop should not be greater than 1.5 per cent. or 110×1.5 = 1.6 volts. Then,
4×200×21.6 —————————— = 10,800 circular mils 1.6
a value which indicates that No. 8 wire, B. & S. gauge, must be used for the feeders in order to keep the drop within the limit of predetermined value.
TABLE FOR TAPS, BRIDGES OR OTHER WIRES AT NEGLIGIBLE DROP
————————————————————————————————————————————————————————————————————————— Wire Nos. |0 |1 |2 |3 |4 |5 |6 | 7 |8 |10 |12 |14|16|18 ————————————————————————————————————————————————————————————————————————— Lamp Feet|52 v.| 300| 260|200|160|130|100| 80| 65| 50| 38| 24|15| 9| 6 |110v.|1,280|1,085|860|680|560|435|345|280|220|160|100|60|40|25 —————————————————————————————————————————————————————————————————————————
NOTE.—In using this table, it is only necessary to calculate the lamp feet of the tap and take the size of wire corresponding to the nearest greater number of lamp feet in the table. The lamp feet specified by this table should not be exceeded by more than 10 per cent. Thus, if a tap measure 108 lamp feet, in 110 volt lamps, No. 12 wire would be used. But if it measure 115 lamp feet, it would be advisable to use No, 10 wire.
=Constant Voltage Arc Lamp Circuits.=—The branch conductor should have a carrying capacity about fifty per cent. greater than the normal current required by the lamp, so as to provide for the heavy current required when the lamp is started. The underwriters prohibit the use of any size wire under No. 12 for parallel connected arc light circuits.
=Constant Current Series Arc Lamp Circuits.=—The wiring for series connected arc lamps should never be concealed nor encased unless requested by the electrical inspector.
For all interior wiring of this class, approved rubber covered wire should be used, and the wire should always be rigidly supported on porcelain or glass insulators which will hold the wires at a distance of at least one inch from the surface wired over. The wires on all circuits up to 750 volts, should be kept at least 4 inches from each other, and 8 inches apart on circuits of over 750 volts. No wires carrying a current having a pressure exceeding 3,500 volts should be carried into or over any building except central stations and sub-stations.
=Wire Calculations for Motors.=—The proper size of wire for a motor may be readily determined by means of the following formula:
H.P. × 746 × D × 21.6 circular mils = ————————————————————- (6) E × L × K
in which
H.P. = horse power of motor; 746 = watts per H.P.; D = length of motor circuit from fuse block to motor;
21.6 = ohms per foot run in circuit where wires are one mil in diameter; E = voltage at the motor; L = drop in percentage of the voltage at the motor; K = efficiency of the motor expressed as a decimal.
The average values for K are about as follows: 1 H.P., .75; 3 H.P., .80; 5 H.P., .80; 10 H.P. and over, 90 per cent.
EXAMPLE.—What is the proper size of wire for a 10 H.P. motor, run at 220 volts, allowable drop 2 per cent. on 200 foot circuit.
Substituting the given values in the formula on page 758:
10 × 746 × 200 × 21.6 Circular mils = ————————————————————- = 36,991. 220 × 4.4 × .9
The nearest larger value to this result, in the table of carrying capacities of copper wire (page 731), is 41,740, corresponding to No. 4 wire, B. & S. gauge.
_In all cases the size of the wire thus formed should be compared with that allowed by the underwriters for full load current of motor, plus 25 per cent. of that current_, and if the size calculated happen to be smaller than the allowable size, it should be increased to the latter, otherwise it will not pass inspection.
TABLE OF AMPERES PER MOTOR
——————————————————————————————————————————————————————————————————————————————- H.P. Per Cent. Eff. Watts Input 50 Volts 100 Volts 220 Volts 500 Volts ¾ 70 800 16 7 4 2 1½ 70 1600 32 15 7 3 3 75 2980 60 27 14 6 5 80 4660 93 42 21 9 7½ 85 6580 132 60 30 13 10 85 8780 176 80 40 18 15 85 13200 264 120 60 26 20 85 17600 352 160 80 35
25 85 21900 438 199 100 44 30 90 24900 498 226 113 50 40 90 33200 664 301 151 66 50 90 41400 828 376 188 83
60 90 49700 994 452 226 99 70 90 58000 1160 527 264 116 80 90 66300 1330 608 302 133 90 90 74600 1490 678 339 149
100 90 82900 1660 755 377 166 120 90 99500 1990 905 453 199 150 90 124000 2480 1130 564 248 ——————————————————————————————————————————————————————————————————————————————-
TABLE OF AMPERES PER DYNAMO
——————————————————————————————————————————————————————————————————————————————- Appx. Appx. K.W. 125 Vs. 250 Vs. 500 Vs. H.P. K.W. 125 Vs. 250 Vs. 500 Vs. H.P. 1. 8 4 2 1.3 30. 240 120 60 40. 2. 16 8 4 2.7 37.5 300 150 75 50. 3. 24 12 6 4.0 40. 320 160 80 53. 5. 40 20 10 6.7 50. 400 200 100 67. 7.5 60 30 15 10. 60. 480 240 120 80. 10. 80 40 20 13. 75. 600 300 150 100. 12.5 100 50 25 17. 100. 800 400 200 134. 15. 120 60 30 20. 125. 1000 500 250 167. 20. 160 80 40 27. 150. 1200 600 300 201. 25. 200 100 50 34. 200. 1600 800 400 268. ——————————————————————————————————————————————————————————————————————————————-
To determine the current required for a motor, as for instance, the 10 H. P. motor under consideration, _multiply the horsepower by 746, and divide the product by the voltage of the motor multiplied by its efficiency as follows_: (10 × 746) ÷ (220 × .90) = 37.6 amperes.
This value increased by 25 per cent. of itself (37.6 × .25 = 9.4 amp.) is equal to 37.6 plus 9.4 = 47 amperes. In the table of carrying capacities of copper wire (page 731), 46 amperes is given as the allowable carrying capacity of No. 6, B. & S. gauge, rubber covered wire; therefore No. 5 wire must be used.
=Calculations for Three Wire Circuit.=—In all cases of interior conduit work, and in most cases of inside open work, the main feeders from a three wire source of supply are installed on the three wire plan, and the sub-feeders and distributing mains, on the two wire plan, except where the application of the method necessitates the use of unwieldy sizes.
In laying out sub-feeders and mains, the total load, under normal operating conditions, should be divided as nearly as possible into two equal parts, and one part connected on each side of the neutral part of the entrance cut out, or the neutral bus bar of the switch board or panel board in an isolated plant, thus making the load on each side of the neutral wire of the feeder as near equal as possible.
Fig. 841 shows a three wire panel board with connection for 12 mains; those shown in solid lines as A, B, C, etc., being connected between the neutral wire and the negative wire of the feeder, and those shown by dotted lines as G, H, I, etc., being connected between the neutral wire and the positive wire of the feeder. The total load consists of ninety-one 16 candle power lamps, which are so distributed that the positive wire of the feeder carries the current for 46 lamps, and the negative wire, 45 lamps, the neutral wire carrying the difference or current for 1 lamp.
The proper size of wire for the mains may be calculated as already explained, but in calculating the outer wires of the three wire feeder, the neutral wire should be disregarded and the outer wires connected as a _two wire circuit carrying the total load of 91 lamps at the over all pressure of 220 volts_.
EXAMPLE.—Ninety-one 16 candle power lamps consuming 3.1 watts per candle power at a pressure of 110 volts, will require a current of
16 × 3.1 × 91 ————————————- = 41 amperes. 110
The distance from the entrance cut out to the main or feeder switch is 200 feet, then for a 2 per cent. drop, or a loss of 110×.02=2.2 volts, the cross sectional area of the wire will be,
41 amperes × 200 feet × 21.6 ———————————————————————————— = 80,509 circular mils. 2.2 volts
The joint resistance of the lamps on a three wire system, however, would be four times greater than on a two wire system; consequently the resistance of the outer wires of the feeder in this case will be four times greater for the same percentage of loss, and the cross sectional area of each of the outer wires will be, 80,509÷4=20,127 circular mils. According to the underwriters' rules, this value compels the use of No. 6 B. & S. gauge wire.
If the _lamp_ voltage, 110 volts, be used, the two wire formula (5) given on page 748 must be modified to,
amperes × feet × 21.6 circular mils = ————————————————————- drop × 4
but if an _over all_ voltage, (that is, the voltage between the outer wires), of 220 volts be used, the two wire formula does not require any modification.
The proper size of wire may also be calculated by means of the formula
drop ———————————————————————- = resistance per foot . . . . (1) 2 × distance × amperes
Example.—If in calculating a three wire feeder, the over all voltage be 220 volts, the drop = 4.4 volts, twice the distance = 400 feet, and the current = 20.5 amperes, then,
4.4 volts ————————————————————————- = .0005365 ohms per foot. 400 feet × 20.5 amperes
In the table of the properties of copper wire which gives the resistance of various sizes of wire, it will be noted that at all of the given temperatures No. 8 wire has a resistance greater than the value just calculated, therefore, No. 6 B. & S. gauge wire should be used for the outer wires of the feeder. In the table the resistance is given per 1,000 feet, hence it is only necessary to move the decimal point to obtain the resistance per foot.
If the calculation be based on the lamp voltage, 110 volts, the formula (1) must be modified to
drop × 4 ———————————————————————— = resistance . . . .(2) 2 × distance × amperes
In this case, drop = 2.2 volts, 2 × distance = 400 feet, and current = 41 amperes, then,
2.2 volts × 4 8.8 ———————————————————— = ———————— = .005364 ohms. 400 feet × 41 amp. 16,400
=Size of the Neutral Wire.=—In three wire circuits, the size of the neutral wire will depend to a great extent upon operating conditions. In the case of installations which occasionally have to be worked as two wire systems, the cross section of the neutral wire should be equal to the combined cross section of the two outer wires.
For interior wiring which must pass inspection, the neutral wire must always be twice the size of one of the outside wires. However, for general distribution, if it be reasonably sure that the system will always be worked three wire and that the drop in the two outer wires does not exceed 1½ per cent., the cross section of the neutral wire may be made smaller than that of one of the outer wires. In such a case the size of the neutral wire may be calculated for a maximum unbalancing of 25 per cent., when the current in one of the outer wires is 75 per cent. of the current in the other outer wire.
For instance, suppose that in a balanced system, the total load on each of the outer wires of a feeder be 211 amperes, and that on account of certain operating conditions, this load has to be divided unequally so as to put 242 amperes on one of the outer wires, and 181 amperes on the other outer wire. In this case the neutral wire will carry 60 amperes, or 25 per cent. of the current carried by the heavier outer wire.
If the drop in the outer wires exceed 1½ per cent., the cross section of the neutral wire will have to be equal to or larger than that of each of the outer wires, otherwise the drop in the neutral wire will exceed ½ volt with an unbalancing of 25 per cent.