An alternative method for sizing conductors is to use tables provided for that purpose, such as the examples found in Appendixes XIV and XV. These tables have several important limitations. They apply only to wires with insulation types of R, T, TW, RH, RHW, and THW, and they can only be used with a 2% voltage drop and 120 or 240 V. For any other type of conductor, insulation, voltage drop, or voltage, a different table must be used.
Problem: Determine the size of wire needed to supply 120-V electricity to the pump house in Figure 27.1.
Solution: The total length of wire is two times the sum of the distance from the building to the first pole (A), the distance between the two poles, and the distance from the last pole to the pump (B). First determine lengths A and B by using Pythagorean's theorem:
Distance A is:
DistanceA = ^25.02 + 20.02 = V625 + 400 = 32.015... or 32.0 ft
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and distance B is:
a = Vb2 + c2 = V 15.02 + 22.02 = V225 + 484 = 26.627... or 26.6 ft
The total length (LT) of wire is:
Then using Appendix XIV, select the correct size of conductor. Twelve amps and 316 ft are not shown in the table; use the next larger values (15 amps and 350 ft). Then, from Appendix XIV, the required size of conductor is No. 3. Remember to compare this value with the value to the left of the vertical line and use the larger of the two sizes.
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