Ask Me Help Desk - Maximum allowable length of electrical wire

Maximum allowable length of electrical wire

Can you tell me the max allowable length (in feet) of a 12/2 electrical wire which will carry a maximum of 20 amps at 240 V?

It may or may not be allowed to.
Loads that are continuous must be multiplied by 1.25
Loads that are not are just added.
Continuous loads are loads that are expected to be on more than 3 hours.
Thus the maximum continuous load is 80% of 20 A.

Can you re-phrase your question?

Homework?

I bet he wants to know the maximum length allowable on 12-2 not to exceed 80%, with a maximum 5% total Voltage drop. Don't have formula handy.If no one answers in a day, I will dig up.
Mikeholt.com and under free stuff, they have a voltage drop calculator. My Microsoft is having problems with .xls extensions, give it a try

I'll rephrase my question: Can I run 100 feet of 12/2 copper wire from a 20 amp 240 V breaker to feed three 1000 W baseboard heaters? Thanks

Quote:

Originally Posted by howardl00

I'll rephrase my question: Can I run 100 feet of 12/2 copper wire from a 20 amp 240 V breaker to feed three 1000 W baseboard heaters? Thanks

You should be fine.

Thanks!:)

The magic number is 119 feet for a 3% drop. 3*1000/240*1.25 Amps

Kiss, if I may, and with all due respect, I would like to ask that you explain:

Quote:

Originally Posted by KeepItSimpleStupid

The magic number is 119 feet for a 3% drop. 3*1000/240*1.25 Amps

The result of completing your formula is 10 amps. How is that related to 119 feet?

The 1.25 multiplier is used to determine the minimum size branch circuit conductor and the minimum size circuit breaker.

Why is the 1.25 relevant for voltage drop?

The circuit load of 3000 watts at 240 volts is 12.5 amps that must be delivered on #12 solid copper wire 100 feet.

This whole voltage drop issue, in general, is so confusing to most, and your answer makes it even more confusing to me. Not trying to trip you, if your answer is confusing to me, I am sure others will be also.

The most accurate basic voltage drop formula, used for DC circuits or AC circuits with a Power factor of 1, as published by ANSI/IEEE Std 141, and recommended by NEC is:

Vd=2 x L x R x I

Formula Variables

Vd = Voltage drop
R = DC Resistance from NEC® Chapter 9 Table 8
L = Distance
I = Current in amperes (A)

Circuit Data

.00193 Ohms per foot for solid copper #12 wire (1.93 per 1000 Feet)
100 Feet
12.5 amps

This formula does not consider any reactance, and since the question mentions resistance load as electric heat, this formula will work fine for this example. There are other formulas that will take reactance (Inductance, capacitance, Power factor) into consideration, along with ambient temperature.

The maximum recommended voltage drop for a branch circuit is 3%, and the voltage mentioned is 240 volts, so the max Vd allowed is 7.2 volts.

So, using the data given by Howard, and using the Vd formula above, I arrive at 4.825 volts dropped, well below the max allowed Vd of 7.2 volts.

By the way, the answer to the first question:

Quote:

Originally Posted by howardl00

Can you tell me the max allowable length (in feet) of a 12/2 electrical wire which will carry a maximum of 20 amps at 240 V?

using the above formula and the data given happens to be 93.26425 feet to arrive at the maximum of 7.2 volts dropped.

Of course, since the question is by a layperson that is not aware of the derating and safety factors involved, the question unknowingly is flawed for his purpose. The actual data given makes the question relevant.

And since the circuit load is 12.5 amps, and the circuit breaker must be rated at least 125% of the branch circuit load (increase 12.5 amps by 25%), a 20 amp circuit breaker is perfect, and no additional load is allowed on this circuit.

Tk:

Since this load is continuous, i.e. expected to be on for more than 3 hours. 12.5 * 125% is 15.625 Amps.

Vd of 3% was based on the derated current, not actual current. I suppose that's wrong?

Quote:

Originally Posted by KeepItSimpleStupid

tk:

Since this load is continuous, i.e. expected to be on for more than 3 hours. 12.5 * 125% is 15.625 Amps.

This is correct, electric heat is to be considered as a continuous load, and to size the wire and circuit breaker, multiply the actual load of 12.5 amps by 1.25, as you did, so the wire must be rated at least to carry 15.625 amps, #12 THHN or #12 Romex, and the breaker needs to be the next higher standard size, a 20 amp breaker.

Quote:

Originally Posted by KeepItSimpleStupid

Vd of 3% was based on the derated current, not actual current. I suppose that's wrong?

Algebraically, you can see that is the result, derating the actual current.
Actual load is never derated, that I am aware of, for any purpose.

I am sure you get this now, so for the benefit of others:

For voltage drop especially, the actual calculated load is what will be needed at the very end of the circuit, as the actual measured current will be very close or dead on as per the nameplate load.

Each wire, the feed and the return, is in series with the load. Each wire , having a resistance of .00193 ohms per foot times 100 feet is a total of .193 ohms each, and a total of .386 ohms in the circuit along with the electric heat, or which ever load is on a circuit.

Using Ohms Law, W=I2R, or Amps Squared times Resistance will equal Watts, the wire alone will act as a load of consuming 60 watts, and dissipate this as heat, called Power Loss, or I Squared R Loss.

(This is one, probably the most important, reason for the 125 % factor increase to size conductors.)

To see how this compares:

Again using Ohms Law, E=Square Root of PR, or Volts equals the square root of watts times ohms, E= SqRt of 60 Watts times .386 Ohms = 4.8124 Volts.

Appears that this is very close to the result of the voltage drop calculation I did earlier arriving at 4.825 Vd.

If you use Vd=2 x L x R x I in a spreadsheet, and plug in different loads in amperages, and leave the wire and distance the same, you can then see that the Volts Dropped will increase as the load amps increase, or directly proportional. This is why knowing the actual load in watts or amps is crucial to doing proper voltage drop calculations.

Thank you all! I'll keep your data for reference when my work is inspected - it will make the process a lot easier.:) :) :)

Quote:

Originally Posted by howardl00

Thank you all! I'll keep your data for reference when my work is inspected - it will make the process a lot easier.:) :) :)

Like I said, you should be fine! :) :D ;)

Dumb answer on my part. I should have known better. :o