Voltage drop - Load vs Overcurrent Device

[rpg]

Our electricians calculated the conductor size of a single instrument circuit (industrial plant) @ 120 vac, 1210' in length, with a 20 amp breaker (per schedule) to require #3/0 conductor. The designers responded that the actual load of the instrument is 24W and that a #12 conductor is sufficient. NEC 210.18 seems to indicate that conductors should be sized on the overcurrent device installed on the circuit. Could someone please provide more insight on this?

 

[LarryFine]

The #12 would be protected by the 20a breaker, as would anything larger. The general rule is that the breaker's rating cannot exceed that of the conductor.

24w at 120v is only 0.2a,. Even if the load requires up-sizing the conductors, it is proper to keep the same sized breaker.

To add: it is the load current that determines voltage drop, not the circuit's capacity.

 

[rpg]

The #12 would be protected by the 20a breaker, as would anything larger. The general rule is that the breaker's rating cannot exceed that of the conductor.

24w at 120v is only 0.2a,. Even if the load requires up-sizing the conductors, it is proper to keep the same sized breaker.

To add: it is the load current that determines voltage drop, not the circuit's capacity.

Thanks Larry, sounds logical to me. So, if the instrument is replaced in the future with one requiring more current, the conductors may need to be upsized. Correct?

 

[LarryFine]

Thanks Larry, sounds logical to me. So, if the instrument is replaced in the future with one requiring more current, the conductors may need to be upsized. Correct?

Always a possibility, yes. One option later would be to step the voltage up at the source, which reduces the current for a given load (power) by the same ratio, and back down at the load end.

For example, if you had a 10a, 120v load, and stepped the 120v up to 600v (1:5), the current at 600v would only be (5:1) 2a, reducing the voltage drop by the same ratio for a given conductor size.

 

[ptonsparky]

Thanks Larry, sounds logical to me. So, if the instrument is replaced in the future with one requiring more current, the conductors may need to be upsized. Correct?

Can you even imagine an instrument that will require even 50 times the power of now?




Let the future worry about itself. Undersizing the conduit would be more of a problem at 1210’.

 

[winnie]

As a design issue, a dead short at the end of that circuit would probably take several minutes to clear with a normal 20A breaker, and might _never_ clear.

-Jon

 

[rpg]

So are we looking at a potential fire hazard here?

 

[synchro]

To address Jon's concern, adding supplementary overcurrent protection with a fuse significantly smaller than 20A but still large enough to power your instrument load reliably would be helpful. Of course, this additional OCPD should be on the supply side of your long wire run.

 

[topgone]

To address Jon's concern, adding supplementary overcurrent protection with a fuse significantly smaller than 20A but still large enough to power your instrument load reliably would be helpful. Of course, this additional OCPD should be on the supply side of your long wire run.

Correct! We use fuses for protection of instrument circuits.

 

[winnie]

So are we looking at a potential fire hazard here?

A '20A' breaker might carry 30A indefinitely, but 30A is probably not a fire hazard in 12awg wire.

You should simply understand that with 4 ohms of resistance in a 120V circuit short circuits are a different sort of beast.

Jon

 

[designer82]

As a design issue, a dead short at the end of that circuit would probably take several minutes to clear with a normal 20A breaker, and might _never_ clear.

-Jon

Can you explain why that is? didn't know about this

 

[winnie]

All wire has resistance; that is why we have to consider voltage drop. For the circuit in question the expected resistance is nearly 4 ohms because it is so long.

With 120V at the supply, if you short circuit the end of the circuit, 30A will flow through this resistance.

Now you have to look at the 'trip curve' of a circuit breaker. When the handle says '20A' this doesn't mean that the breaker will carry 19.999A and instantly trip at 20.001A.

A perfect circuit breaker follows an inverse time curve, meaning that it takes a certain amount of time to trip at a given overload, and trips more rapidly the greater the overload. On top of this real world breakers have a tolerance range where operation anywhere in a range of times is acceptable.

Well it turns out that for a normal 20A breaker, it is perfectly reasonable for it to carry 30A for minutes or longer...or to trip in 20 seconds.

See for example https://download.schneider-electric...+curves&p_File_Name=730-3.pdf&p_Doc_Ref=730-3 where for this particular 20A breaker a 30A load will trip the breaker between 22 seconds and 125 seconds.

-Jon

 

[hbiss]

If this worries you just use a 15A breaker instead of a 20. Should help some.

-Hal