Given a load current of 1 AMP, and using 18 AWG wire, how much voltage drop can we expect at the load end for a 350 foot run of paired wire?

Using the chart, we match the row for 18 AWG and the column for 1 AMP and determine that voltage drop per 100 feet is 1.27 Volts. By dividing the paired wire length by 100, we get the factor by which we need to multiply voltage drop per 100 feet to determine total voltage drop. Therefore, 350 feet divided by 100 equals 3.5. Multiply 3.5 by 1.27 volts drop per 100 feet to get your total voltage drop. Thus the total voltage drop is 3.5 times 1.27, or 4.445 voltage drop for 350 feet.


 

EXAMPLE TWO:

Given a camera load of 2 AMPs, that is 400 feet from the power source, which wire gauge should be selected to keep voltage drop at the camera to less than 3 volts?

To use the chart, we need to determine what the maximum voltage drop per 100 feet is. We calculate that 100 feet is 1/4 of 400 feet, thus the voltage drop allowed for 100 feet is 1/4 times 3 volts (which is the equivalent of 0.75 volts per 100 feet):

voltage drop per 100 feet = 3/4 = .75 volts per 100 feet.

So, knowing that we can not allow anything greater than a voltage drop of .75 volts per 100 feet, we can now look at the chart and select the wire gauges that will give us lower voltage drops per 100 feet at a 2 AMP load current. In this case, wire gauges of 10 (.40 V), 11 (.50 V), and 12 AWG (.64) will all suffice, with 13 AWG (.80) being a possibility.

Thus, in order to keep voltage drop at the camera to less than 3 volts given a camera load of 2 AMPs and a 400 foot paired wire run, we need to use a wire gauge in the range of 10-13 AWG.

 

Voltage Drop Per 100 FT Run of Paired Wire


 

FORMULA METHODS:

These handy equations can be used to determine voltage drop per 100 feet or wire gauge as an alternative to the chart, even for values that are not on the chart. To arrive at total voltage drop, always divide paired wire run length by 100, and then multiply that number by voltage drop per 100 Feet:

1. To determine voltage drop per 100 feet given load current and wire gauge:

VD = Voltage drop per 100 feet (Volts)
IL = Current load (AMPs)
AWG = Wire gauge

2. To determine wire gauge necessary given paired wire length, load current, and desired voltage drop per 100 feet:


With these useful tools, voltage drop problems can be avoided before installation, saving time, money and ensuring a correctly working system.

1. 24VAC solenoids generally require a minimum of 20 VAC for reliable operation. The corresponding allowable voltage drop is therefore 24- 20 = 4VAC. The wire sizing chart is based on a 4 volt drop allowance. With a SWELL used on the valve, a 6 volt drop is permitted.
2. Solenoid inrush current is used to calculate wire size. When multiple valves are energized simultaneously on the same line, add the solenoid inrush currents. If inrush is given in volt-amps (VA), divide by 24. DO NOT USE THE WATT
   RATING. VA is not the same as watts for AC solenoids. Inrush current is listed along the top of the chart. Up to 5 SWELLS are also listed across the top with their inrush currents pre-calculated.
3. The chart vertical axis shows the distance from the controller to the valve and back in thousands of feet.
4. Even though wire sizes on the chart are listed from 18 to 2 gauge, the smallest direct burial size recommended is 14 UF solid copper. Sizes larger than 2 are omitted and normally not used because of cost.
5. For a single valve or multiple valves at the same distance from the controller, use the chart as follows:

• Add the total inrush currents of all the solenoids and locate the closest value to this figure along the top of the chart.
  If SWELLS are used, up to five are listed along the top.
• Select the closest round trip distance along the vertical axis.
• The box at the intersection indicates the recommended gauge.
• Compare wire and energy savings if SWELLS are used.

EXAMPLE 1:
What gauge wire is needed to operate two solenoids, each with an inrush current of 1 .0 amps, located 2000 feet away
from the controller? What if SWELLS (one per valve) are used?
 

• Two solenoids add up to 2.0 amps inrush.
• The distance loop is 2000 feet X 2 = 4000 feet.
• Chart shows that 6 gauge wire is required, shown as circle A on chart.
• With two SWELLS (one per valve), the chart shows that 18 gauge could be used at point circle B. For direct burial, it is
  however recommended that 14 gauge be use.

   6.  For multiple valves on a line at various distances from controller:

• Add the inrush currents as before.
• Add distances from controller to each valve and divide by number of valves to get average distance. Multiply aver-
  age by 2 to get loop distance.
• Use chart as before to determine gauge.
• Compare wire and energy savings with SWELLS.

EXAMPLE 2:
There are 4 valves along one line. The first is 2000 feet from controller; second is 3000 feet away from controller; third is 3300 beet away; and the last 3700 feet away. Each valve solenoid has an inrush of .5 amps. What size wire should be
specified conventionally or with the SWELLS?
 

• Total inrush current is .5 + .5 + .5 + .5 = 2.0 amps.
• Adding distances: 2000 + 3000 + 3300 + 3700 = 12,000 feet-;- 4 = 3000 feet average X 2 = 6000 feet loop.
• Chart shows size 4 wire as designed by circle C
• With 4 SWELLS, chart shows 14 gauge at circle D

SWELL BENEFITS:
In example 1 above, the estimated wire cost savings after subtracting for the SWELL cost is about 50%. In example 2, the saving is about 70%. Additional benefits in using the SWELL are as follows:
 

• Holding current draw is reduced by about 80% in example 1 and over 90% in example
• This saves energy.
• Allows operation of a valve at significantly lower AC voltages and much greater distances.
• Extends solenoid life because solenoid coil cannot overheat.
• Cuts installation cost because 14 gauge is easier to handle than heavier wires.
• Reduces dealer inventory of wire to one size (14 gauge) for most applications.