Equivalent AWG formula

[psala]

Is there a formula(or chart somewhere) that will give the equivalant AWG of two wires run in parallel and jointed at each end.

For eaxmple; if you had two 50' lengths of 16-awg solid copper wire run in parallel and joined at each end, what would the equivilant AWG be?

I assume the combined AWG would be somewhere around 14 or 12.

 

[MR. S]

Only 1/0 or larger can be run in parallel

 

[raider1]

Take a look at Chapter 9 Table 8. This Table will give you the area in circular mils of each size conductor. You would then add the circuilar mil areas together and find the equivalent AWG.

For example lets take the 16 AWG's in parallel from your example. a 16 AWG conductor has a circular mil area of 2580. If we double that we have 5160 circular mils. The circular mil area of a 14 AWG conductor is 4110 and a 12 AWG conductor is 6530. So the combine circular mil area of 2 16 AWG conductors is between a 14 and a 12 AWG.

Chris

 

[iwire]

if you had two 50' lengths of 16-awg solid copper wire run in parallel and joined at each end, what would the equivalent AWG be?

There is no 'equivalent' AWG.

The ability of a material to carry current diminishes as the conductor size increases.

If you want to parallel conductors you simply add up the capacity of the size your using.

16 AWG has an ampacity of 8 amps per table 402.5.

If 16 AWG was allowed to be paralleled in order to increase ampacity two in parallel would have an ampacity of 16 amps, four in parallel would be 32 amps. But you also have to consider adjustment factors for temperature and the number of current carrying conductors run together.

 

[winnie]

It actually depends upon what you mean by ' equivalent AWG'. If you mean 'same amount of copper', then you get one answer. If you mean 'same current carrying capacity' then you get another answer.

As MR. S notes, for purposes of building wiring and increasing ampacity, you can only parallel 1/0 or larger conductors. When doing so, you are permitted to add up the amp rating of the individual conductors, derating as necessary for number of current carrying conductors. The _equation_ that you use would be some derivative of the Neher Mcgrath formula, which takes into account wire resistance, ambient thermal conditions, and insulation temperature rating to determine the conductor ampacity.

The 'same amount of copper' question comes up all the time when building electric motors. For electric motors it is quite common to run very small conductors and conductors of different size in parallel in order to get easy winding and best slot fill. For these motors, wire of every imaginable AWG is available, including fractional AWG. Sitting on the shelf I have a spool of 21.5 AWG magnet wire with class 220C insulation.

The approximate rule for wire area is that 3 gauge numbers is equal to a doubling of area. The _exact_ number is 19.5 * ln(2)/ln(92) = 2.9891645.....

The diameter of a wire in mils is given by the equation 5 * 92^((36-n)/39) where n is the gauge number. From this you can derive a formula for the area of a wire in circular mils; the equation is 25 * 92^((36-n)/19.5) where n is the gauge number.

From these equations you can make a spreadsheet that shows the equivalence of different wire sizes.

-Jon