It frequently happens that a very large wire is necessary for a feeder, and the resistance is not given in the wire tables, so that the formulas already considered will not apply. In this case, the area of the wire in circular mils (Art. 8) is calculated by the following formula, for 110-volt circuits:

A = 10.8 N F/E,

(9.) in which N, F, and E have the same significance as in the preceding formulas.

Example

The feeders from the switchboard to the distributing center in a certain installation are 160 feet in length, and carry the current for six hundred 16-candlepower 110-volt lamps, a drop of 2.5 volts being allowed. What is the size of the wire ?

Solution

The number of lamps is 600 = N; the distance is 160 feet = F; and the drop is 2.5 volts = E. Then, by formula 9, the area of the wire in circular mils

A = 10.8 x 600 x 160 = 414,720 circ. mils. Ans.

2.5

09. To check the results obtained from formula 9, it is necessary to find the safe carrying capacity of the wire. For this purpose reference may be made to the accompanying table, which gives the carrying capacity of concealed cables, the difference between successive sizes being 100,000 circular mils. Any intermediate size will have proportional capacity.

Table 2. Current Capacity of Cables

Area in Circular Mils.

Current in Amperes.

Area in Circular Mils.

Current in Amperes.

200,000

200

1,100,000

673

300,000

272

1,200,000

715

400,000

336

1,300,000

756

500,000

393

1,400,000

796

600,000

445

1,500,000

835

700,000

494

1,600,000

873

800,000

541

1,700,000

910

900,000

586

1,800,000

946

1,000,000

630

1,900,000

981

2,000,000

1,015

In the example of Art. 68, the current required by 600 lamps of 16 candlepower and 110 volts is 600 x .5 = 300 amperes. Referring to Table 2, it is seen that a wire of the calculated area will carry over 336 amperes; consequently, the size computed may be used.

70. In some cases, the loss in E. M. F. in a circuit is given as a certain rate per cent. It is obvious, on consideration, that this is not to be taken as referring to the voltage at the lamps, for the drop in potential is between the point of supply and the lamps, and the initial voltage is therefore higher than at the lamp terminals. In order, then, to ascertain the actual drop in the feeders or conducting wires, the initial E. M. F. must be determined and the lamp E. M. F. subtracted therefrom.

Rule

To find the drop in potential in a wiring system when the percentage of loss is given and the E. M. F. at the lamp terminals is known, multiply the lamp E. M. F. by 100 and divide by 100 minus the rate per cent. of loss. The quotient will give the initial E. M. F., from which is subtracted the E. M. F. at the lamp terminals, the remainder being the actual drop in volts.

This may also be expressed in algebraic form:

V' = 100 V ;

100 - p

(10.)

E = V '-V,

(11.) where V is the initial voltage, V the voltage at the lamp terminals, E the drop in volts, and p the rate per cent. of loss.

Example

What will be the actual drop in voltage on a 110-volt circuit when a 3 per cent. loss is allowed for ?

Solution

The E. M. F. at the lamp terminals is 110 volts = V, and the per cent. loss is 3 = p. Then, by formula 10, the initial voltage V' =100x100 = 113.4, 100-3 and by formula 11, the drop in voltage

E = 113.4 - 110 = 3.4 volts. Ans.