Home >> Cyclopedia-of-architecture-carpentry-and-building-a-general-reference-v-09 >> Ab Cd to Skylight Work >> Cm X

Cm X

conductors, lamps, current, loss, conductor, three-wire, system and voltage

CM X V D— 4 Formulæ are frequently given for calculating sizes of conductors, etc., where the load, instead of being given in amperes, is stated in lamps or in horse-power. It is usually advisable, however, to reduce the load to amperes, as the efficiency of lamps and motors is a variable quantity, and the current varies correspondingly.

It is sometimes convenient, however, to make the calculation in terms of watts. It will readily be seen that we can obtain a formula expressed in watts from Formula 1. To do this, it is advisable to express the loss in volts in percentage, instead of actual volts lost. It must be remembered that, in the above formulae, V represents the volts lost in the circuit, or, in other words, the difference in potential between the beginning and the end of the circuit, and is not the applied E. M. F. The loss in percentage, in any circuit, is equal to the actual loss expressed in volts, divided by the line voltage, multiplied by 100; or, P — E X 100.

From this equation, we have: V= PE 100 If, for example, the calculation is to be made on a loss of 5 per cent, with an applied voltage of 250, using this last equation, we would have: V = 5 = 12.5 volts. Substituting the equation V= 100 in Formula 1, we have: This equation, it should be remembered, is expressed in terms of applied voltage. Now, since the power in watts is equal to the applied voltage multiplied by the current (W = EC), it follows that By substituting this value of C in the equation even above (C M= D X 2,160` , the formula is expressed in terms of watts instead PE J of current, thus: in which W = Power in watts transmitted; D = Length of the circuit (one way)—that is, the length of one conductor; P = Figure representing the percentage loss; E*= Applied voltage.

All the above formulae are for calculations of two-wire circuits. In making calculations for three-wire circuits, it is usual to make the calculation on the basis of the two outside conductors; and in three wire calculations, the above formulae can be used with a slight modifi cation, as will be shown.

In a three-wire circuit, it is usually assumed in making the cal culation, that the load is equally balanced on the two sides of the neutral conductor; and, as the potential across the outside conductors is double that of the corresponding potential across a two-wire circuit, it is evident that for the same size of conductor the total loss in volts could be doubled without increasing the percentage of loss in lamps.

Furthermore, as the load on one side of the neutral conductor, when the system is balanced, is .virtually in series with the load on the third side, the current in amperes is usually one-half the sum of the current required by all the lamps. If C be still taken as the total current in amperes (that is, the sum of the current required by all of the lamps) in Formula I, we shall have to divide this current by 2, to use the formula for calculating the two outside conductors for a three-wire system. Furthermore, we shall have to multiply the voltage lost in the lamps by 2, to obtain the voltage lost in the two out side conductors, for the reason that the potential of the outside con. ductors is double the potential required by the lamps themselves. In other words, Formula I will become: in which C = Sum of current required by all of the lamps on both sides of the neutral conductor; D = Length of is, of any one of the three conductors; V = Loss allowed in the lamps, i. e., one-half the total loss in the two outside conductors.

In the same manner, all of the other formulae may be adapted for making calculations for three-wire systems. Of course the calcula tion of a three-wire system could be made as if it were a two-wire system, by taking one-half the total number of lamps supplied, at one-half the voltage between the outside conductors.

It is understood, of course, that the size of the conductor in Formula 6 is the size of each of the two outs de ones; but, inasmuch as the Rules of the National Electric Code require that for interior wiring the neutral conductor shall be at least equal in size to the outside conductors, it is not necessary to calculate the size of the neutral conductor. It must be remembered, however, that, in a three-wire system where the neutral conductor is made equal in capacity to the combined size of the two outside conductors, and where the two outside conductors are joined together, we have virtually a two-wire system arranged so that it can be converted into a three-wire system later. In this case the calculation is exactly the same as in the case of the two-wire circuits, except that one of the two conductors is split into two smaller wires of the same capacity. This is frequently done where isolated plants are installed, and where the generators are wound for 125 volts and it may be desired at times to take current from an outside three-wire 125-250-volt system.