Thought I knew.

[ritelec]

Hello, quick two questions: 1. A lighting circuit draws 16 amps. At 1.25 percent, is 20 amps. Is that circuit 20 amps with the true 16 amp draw at 80 percent to protect the breaker? Or is it a 30 amp amp circuit?
2. voltage drop? 850 feet, one way, 480v 11.6 amp motor. I've figured 6's. I've been told 10's will work. any comments welcome. thank you in advance. rich

 

[charlie b]

I don't understand question #1. I will say that a 20 amp circuit is adequate for this situation. I will also say that adding 25% to a continuous load is not intended to be a way of protecting the breaker. But I am not certain what the real basis is for that extra 25%.

As to question #2, with a #10 I get a 5% voltage drop. A #8 gives you 3.2%. In order to get the VD below 3%, you need a #6. That said, the code does not have a "maximum VD requirement."

 

[bob]

Hello, quick two questions: 1. A lighting circuit draws 16 amps. At 1.25 percent, is 20 amps. Is that circuit 20 amps with the true 16 amp draw at 80 percent to protect the breaker? Or is it a 30 amp amp circuit?

1. Its a 20 amp circuit. It will carry 125% of continuous load(3 hrs or more) +
100% of noncontinuous load. Total should not exceed 20 amps.

2. voltage drop? 850 feet, one way, 480v 11.6 amp motor. I've figured 6's. I've been told 10's will work. any comments welcome. thank you in advance. rich

I assume this is a 3 phase load.
#10 - 21 volt drop 4.2% # 8 - 13 volt drop 2.8% #6 - 8.7 volt drop 1.7%.
Revised voltage drops.

 

[hardworkingstiff]

I assume this is a 3 phase load.
#10 - 29 volt drop 6% # 8 - 18 volt drop 3.8% #6 - 11.5 volt drop 2.4%.
I think I would go with the #6.

Bob,

What formula are you using? I came up with:

#10, 21.2-volt drop, 4.42%
#8, 13.3-volt drop, 2.78%
#6, 8.39-volt drop, 1.75%

I used (SQRT 3)KID/CM (is this correct?)

 

[ritelec]

thank you for the relies...with the lighting being 16 amps... the 25 percent was added because the load was rated "continuous duty" I was and still am not sure if the new number (with the extra 25 percent) is the new real number to work with which would put it over the protecting the breaker at 80 percent of its value ( 20 being 16) or is that 125 percent to rate the circuit but your actual 16 amp load ( 16 true amps) is not more than 16 amps.
AS far as the vd I tend to be in agreement with you.. however.. two "engineers" said the 10's are good. I was useing K(which is 12)X2LXI devided by cm. the engineer said I only needed half the distance. ??

 

[hardworkingstiff]

thank you for the relies...with the lighting being 16 amps... the 25 percent was added because the load was rated "continuous duty" I was and still am not sure if the new number (with the extra 25 percent) is the new real number to work with which would put it over the protecting the breaker at 80 percent of its value ( 20 being 16) or is that 125 percent to rate the circuit but your actual 16 amp load ( 16 true amps) is not more than 16 amps.
AS far as the vd I tend to be in agreement with you.. however.. two "engineers" said the 10's are good. I was useing K(which is 12)X2LXI devided by cm. the engineer said I only needed half the distance. ??

The 20-amp breaker is correct for the scenario you described.

I agree about the "half the distance", ???

 

[infinity]

I used this calculator:

http://www.electrician2.com/calculators/vd_calculatoradv.htm

and came up with a voltage of 461.6 volts at the load with # 10 AWG copper conductors. A 480 volt motor should operate just fine with 461 volts applied to it.

 

[ritelec]

oh no....reading the replies on the vd.....ha ha ha.........oh no....i hope we are wrong... cause if not there are gonna be a few unhappy white collars when the guy i'm installing this for tries to explain why there motors are all bogging down and product isn't moving.oh no.......ha ha...not good.

 

[hardworkingstiff]

oh no....reading the replies on the vd.....ha ha ha.........oh no....i hope we are wrong... cause if not there are gonna be a few unhappy white collars when the guy i'm installing this for tries to explain why there motors are all bogging down and product isn't moving.oh no.......ha ha...not good.

I agree with Trevor, you probably won't have any problems.

 

[bob]

I used this calculator:

http://www.electrician2.com/calculators/vd_calculatoradv.htm

and came up with a voltage of 461.6 volts at the load with # 10 AWG copper conductors. A 480 volt motor should operate just fine with 461 volts applied to it.

Trevor
Did you change the temp to 30C? Using that caculator I came up with
461 volts with #10 also.

 

[bob]

Bob,

What formula are you using? I came up with:

#10, 21.2-volt drop, 4.42%
#8, 13.3-volt drop, 2.78%
#6, 8.39-volt drop, 1.75%

I used (SQRT 3)KID/CM (is this correct?)

I used a software package. I revised the caculations to use 480 volts 3 wire.
The answers were different for some reason. Its close to what you have.
Starting VD using 6 x FLA #10 16% #8 10% and #6 7%

 

[winnie]

I don't understand question #1. I will say that a 20 amp circuit is adequate for this situation. I will also say that adding 25% to a continuous load is not intended to be a way of protecting the breaker. But I am not certain what the real basis is for that extra 25%.

Given that many of the '125%' rules and the '80%' rules permit use of conductors loaded at 100% if they are protected by a _breaker_ that is listed for 100% operation, I am pretty certain that these rules are there for the breaker, not the conductors.

See, for example 215.2(A)(1) and the Exception.

I did a casual look into this (read: I used the internet but didn't go to the library) and didn't come up with the basis of this. My best guess is that this is to protect breaker assemblies from damage or nuisance tripping caused by heat buildup when in enclosures.

-Jon

 

[infinity]

Trevor
Did you change the temp to 30C? Using that caculator I came up with
461 volts with #10 also.

Yes, I usually use the 30 degree temp for calculations when using this calculator.

 

[LarryFine]

A lighting circuit draws 16 amps. At 1.25 percent, is 20 amps. Is that circuit 20 amps with the true 16 amp draw at 80 percent to protect the breaker? Or is it a 30 amp amp circuit?

One of my guys had the same confusion until I explained it to him. 125% and 80% are reciprocals of one another. 125% of 16 is 20. 80% of 20 is 16. The 125% additional capacity used for a continuous load and the limit of 80% of a circuit's rating for a continuous load are two ways of stating the exact same thing.

If you have a 16 amp continuous load, you must multiply it by 125% to arrive at the circuit's required 20-amp capacity. If you have a 20-amp circuit, you may only load it to 16 amps' worth of continuous load. It's merely the exact same rule expressed two different ways.

Which figure to use depends on whether (A) you're starting with a known continuous load and wish to determine the required circuit capacity, or (B) you're starting with an existing circuit and wish to determine the maximum continuous load you may supply from it.

In other words, you don't have to do both for one instance.