WIND ENERGY SYSTEMS - Cd3wd

Chapter 9—Wind Power Plants 9–42
the other adjacent sizes. For wires that are two gauge sizes apart, like 6 gauge and 4 gauge,
the ratio is 41.74/26.25 = 1.590 = 1.261 2 . A span of 10 gauge numbers yields a change of area
of (1.261) 10 = 10.166 or approximately a factor of 10. The resistance is inversely proportional
to area, so if the resistance of a 10 gauge copper wire is about 1 Ω per 1000 ft, then the
resistance of a 20 gauge wire will be about 10 Ω per 1000 ft.
The selection of a value of 1.261 for the ratio between areas of adjacent gauge sizes means
that the potential benefit of having diameters expressed in integers has been lost, except for
the 250 and 1000 kcmil conductors. It could have been done in a manner similar to resistor
sizes, where each standard value is about 10 % larger than the previous size, but rounded off
to two digits. But it is too late to make such a choice now, so we must learn to live with the
values given in the table.
There are two different constraints which must be considered in selecting a wire size. First,
the I 2 R loss in the wire must not result in a temperature rise in the wire that will damage the
insulation. Second, the voltage drop in the wire must not excessively lower the load voltage.
We will illustrate the second constraint with a short example.
Example
The new well at your country home is 400 ft from your house. The 120-V single-phase induction
motor driving the pump draws 30 A of starting current and 10 A of running current. The starting
current flows only for a few seconds at most, so is not a factor in determining the wire temperature.
You are told that 12 gauge copper wire is rated for 20 A service in residential wiring. This is twice the
running current for this situation, which seems like an ample safely margin for thermal considerations.
But you want to check the voltage drop also. The current has to flow to the well and back to the
house, so the total length of conductor is 2(400) = 800 ft. The resistance from Table C.2 is
R = 800 1.588 = 1.270 Ω
1000
The voltage drop while the starting current is flowing is
V drop = IR = 30(1.27) = 38.1 V
The voltage available to the pump motor is only 120 - 38.1 = 81.9 V during the starting interval.
Single-phase motors do not have very good starting characteristics, and may not start under load if
the terminal voltage is less than about 90% of rated, or 108 V for the 120 V system. The proposed
wire size is too small.
We see that selecting wire sizes based on thermal considerations alone may easily lead to
situations where performance is poor due to voltage drop. In general, we have two choices
in such situations. We can increase the voltage rating or we can increase the wire size (or
both). A 240 V pump motor costs about the same as a 120 V motor, but draws only half the
current for the same power. Insulation on most 12 gauge wire is rated at 600 V, so that is not
Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001