PROBLEMS. 489 



the lamps, when the group of lamps forms a nearly reentrant row and when the lamps 

 are always either all on or all off. 



On the other hand, the result of problem 158 shows that the return loop as speci- 

 fied in problem 156,15 not at all suited to the case in which part of the lamps in the 

 group are turned off, the effect being to cause a very considerable rise of voltage at 

 the remaining lamp (or lamps). 



160. A group of 50 lamps each taking i.o ampere is to be 

 installed in a church at a distance of one mile from a lighting 

 station. It is understood that whenever any of the lamps are in 

 use all are in use, so that the drop in the feeders which supply 

 the lamps may have any value that economy demands. The 

 lamps are to be operated for 300 hours each year. The cost of 

 power at the station is 3. 5 cents per kilowatt-hour, the cost of cop- 

 per is 1 6 cents per pound, the annual charge on the cost of the 

 wire is 10 per cent, (interest 6 per cent., depreciation 3 per cent., 

 and taxes I per cent.), and the station voltage is 125 volts. 

 Find : (a) The size of the feeders to give a balance between loss 

 of power and cost of copper, and (&) the voltage at the lamps. 

 Ans. (a) 76,370 circular mils ; (b) 50.45 volts. 



Note i. The only objection to the application of the economic principle of the 

 balance between loss of power and cost of copper to a case like the one here con- 

 sidered is that the voltage at the lamps may be very different from the voltage which 

 prevails in the other parts of the lighting system, so that the station management 

 would have to be careful to supply special lamps suited to the special voltage. It is 

 evident that it is expensive at best to supply the fifty lamps at a distance of a mile, 

 for, under the conditions of problem 160, it requires $391.40 worth of copper with a 

 loss of $39.14 worth of power each year, and to transmit the required power, 2.522 

 kilowatts (or 22.93 amperes with no volts at the lamps), with 15 volts drop would 

 take $892 worth of copper with a loss of $3.61 worth of power each year. 



Note 2. It is instructive to solve problem 160 graphically as follows : Assume, 

 say, 50, OCX), 60,000, 70,000, 80,000, 90,000, and 100,000 circular mils. Use these 

 sectional areas as abscissas of two curves, A and B ; the ordinates of curve A repre- 

 senting the values in dollars of the power lost each year, and the ordinates of curve 

 B, representing the annual charge in dollars on the total cost of the copper. Then 

 plot a third curve, C, of which each ordinate is the sum of the corresponding ordi- 

 nates of curves A and B, and the abscissa corresponding to the minimum ordinate of 

 this curve, C, is the required sectional area. 



161. A consumer pays 10 cents per kilowatt-hour not only for 

 the energy he uses in his lamps but also for the energy that is 

 lost in the wires that lead from the watt-hour meter to his lamps. 





