How specific is the requirement for parallel conductors to be the same length or is to be taken to be taken literally. 500 KCMIL has a resistance of .000027 ohms per foot. At this level assuming 240 volts the conductors would have to be 50 feet different in length to make a difference of only 1 amp between them. Making up a main with lots of parallel conductors is sloppy if they have to be exactly the same length.
parallel conductors
- Thread starter wyboy
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- Chapel Hill, NC
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- Retired Electrical Contractor
The code is very specific but in reality there wouldn't be much difference in long runs with an inch or so difference. The longer the run the less likelihood of a problem.
If you had 2 inch difference over a 100 feet is not much compared to the same difference in 5'. I think you try the best you can and that's about all you can do. I am sure many installs are not perfect.
If you had 2 inch difference over a 100 feet is not much compared to the same difference in 5'. I think you try the best you can and that's about all you can do. I am sure many installs are not perfect.
fmtjfw
Senior Member
- Location
- Fairmont, WV, USA
Let's try an example
Let's try an example
Make the following assumptions:
700 amps total for two parallel 500CM conductors
conductor A 100 feet
conductor B 120 feet
resistance of 100 ft is .00270
resistance of 120 ft is .00324
total resistance of two paralleled conductors is 1/((1/.00270)+(1/.00324)) which is .00147 ohms
the voltage drop is .00147*700 which is 1.0309 volts
the current in conductor A is 1.0309/.00270 or 381.82 amps
the current in conductor B is 1.0309/.00324 or 318.18 amps
an imbalance of 60+ amps
if you use your example of 50 extra feet you get an imbalance of 420 versus 280 amps 140 amps
Let's try an example
Make the following assumptions:
700 amps total for two parallel 500CM conductors
conductor A 100 feet
conductor B 120 feet
resistance of 100 ft is .00270
resistance of 120 ft is .00324
total resistance of two paralleled conductors is 1/((1/.00270)+(1/.00324)) which is .00147 ohms
the voltage drop is .00147*700 which is 1.0309 volts
the current in conductor A is 1.0309/.00270 or 381.82 amps
the current in conductor B is 1.0309/.00324 or 318.18 amps
an imbalance of 60+ amps
if you use your example of 50 extra feet you get an imbalance of 420 versus 280 amps 140 amps
iceworm
Curmudgeon still using printed IEEE Color Books
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- North of the 65 parallel
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- EE (Field - as little design as possible)
Not exactly. It is an algebra problem. And yes, I know what the code says.
For two parallel conductors, 1% different in length, they share with about .5% difference in current.
For 10% difference in length, they share with about 5% difference in current.
So, how close do you have to have the length? Well I say close enough in length the difference in current does not push the higher current (shorter) one above its ampacity. So:
1.how much headroom - calculated load to conductor ampacity (percent)?
2. what is the percent difference in length?
Without going into the calcs
500kcmil Cu, 2 parallel, loaded to 750A. Ampacity for parallel 500 is 2 x 380 = 760A. As designed, these cables are loaded at 98.7% of NEC ampacity.
Example 1:
100 foot pull. After termination there is a 1 foot difference in the cut-off pieces - 1% difference in length. The conductors will share at 50.25% and 49.75% (not exactly, but close enough for this example). So, the short one will have 377A and the long one will have 373A. The short one is loaded to 99.2% of ampacity. Meets Code.
Example 2:
Pull is between two adjacent swirchboards and is only 10 feet long. Same as before, cut off the same length, pull in, terminate, the cutoff pieces for phase A are again 1 foot shorter. The difference in length is 10%, so the conductors will share at 52.5% and 47.5% (again not exactly, but close enough). The short one will have 394A. The long one 356A. Oops
Morals of the story:
1. Keep the Ampacity headroom (percent) more than the difference in length (percent) and you are at least 2X to the good.
2. Probably the only time it will matter is if the each phase is in a separate conduit. Makes the terminations look nice, but one has to make sure the conduits are sufficiently close in length.
So sayeth the worm
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dnbob
Senior Member
- Location
- Rochester, MN
iceworm,
Would you please share your formula for this problem. It would be a useful tool for helping my guys understand the importance of equal lengths.
Thank you!
Would you please share your formula for this problem. It would be a useful tool for helping my guys understand the importance of equal lengths.
Thank you!
- Location
- Illinois
- Occupation
- retired electrician
I don't understand this comment. If you have all of one phase in a single (hopefully non-ferrous) conduit, there is no reason that the conductors in this raceway be the same length as the conductors of the other phases in other conduits.
I've been trying to wrap my head around iceworm's percentage basis, and I just can't come to terms with it because the percent difference in length vs. ampacity headroom percent will vary with the number of parallel conductors.
Let's say you have a 500A load with 2 parallel conductors rated 300A each. Headroom 20% [(2*300-500)/500]. For one conductor to reach its ampacity of 300A under nominal load, the second will be at 200A. Because length is inversely proportional to current, your length difference percentage from nominal is (300-250)/250 and (250-200)/250, both equalling 20% deviation.
Now let's say you have a 1250A load with 5 parallel conductors rated 300A each. Again, nominal load per conductor 250A. Headroom 20% [(5*300-1250)/1250]. For one conductor to reach its ampacity of 300A under nominal load, the other four will average 237.5A [i.e. (1250 - 300)/4]. Here we have again (300A-250)/250 = 20% for the shortest conductor's deviation from nominal. However that only allows (250-237.5)/250 = 5% deviation on average from nominal for the other four!
Actually, the determination of percentage of length deviation is a bit more complex than that. For the last example with five conductors, the max'd out conductor only has to be 17% short while the other four are 5% long... so if the short conductor were actually 20% short, it would exceed its ampacity.
FWIW, here's a link to another discussion a few years back:
http://forums.mikeholt.com/showthread.php?t=96501I
One of my posts in that thread has a link to an Excel file you can download. In it, you can enter the total load, the number conductors, and the length of each to see how the current changes on each conductor with differing lengths.
iceworm
Curmudgeon still using printed IEEE Color Books
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- North of the 65 parallel
- Occupation
- EE (Field - as little design as possible)
Now that you mention it - I don't either. :dunce: Of course, if all of one phase is in one conduit, then they are pretty well the same length. Each phase would be different, but so what.
thanks
ice
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
The one time I did parallel conductors 35+ years ago we laid them out on the ground and cut them all the same length. If we cut a foot off at the end to terminate then we did the same on all the conductors.
I suspect you can do the same when you have phases in different runs of non ferrous conduit but in many cases that would not be an easy task.
I suspect you can do the same when you have phases in different runs of non ferrous conduit but in many cases that would not be an easy task.
iceworm
Curmudgeon still using printed IEEE Color Books
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- North of the 65 parallel
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- EE (Field - as little design as possible)
Yes. But consider it a model instead of a formula.
View attachment img100.pdf
You get to do the 10% difference :blink:
ice
- Location
- Massachusetts
Unfortunately we are supposed to take the NEC literally, luckily it seems most inspectors see the issue with that.
I have always had a problem with how that section is written as it is impossible to comply with. In my opinion there are no two conductors in the world that 'are the same length'
The NEC should add a +/- percentage to that section.
- Location
- Illinois
- Occupation
- retired electrician
Maybe the code should remove the length wording and replace with something like this" The conductors shall be installed such that the maximum permitted ampacity of any single conductor of the parallel set will not be exceeded at the calculated load"
- Location
- Massachusetts
While I agree that is the most direct and technically correct way to go I think the calculations required for that are beyond the reach of many, including inspectors.
iceworm
Curmudgeon still using printed IEEE Color Books
- Location
- North of the 65 parallel
- Occupation
- EE (Field - as little design as possible)
That's true. That's why I limited the model to two parallel. The purpose of the model is to demonstrate two things:
1. As long as the design is not tight against the ampacity limits it likely is fine. A 1% difference in length puts one at .25% below design current and one at .25% above design current.
2. The effect is multipled in short runs. A one foot difference in a 10 foot run is 10% - giving one at 52.4% of the current and the other at 47.6%
Thats true. Let's take 5 parallel, varing inlength from 90% of nominal length to 1.10% of nominal length (Or pick any other numbers you want)
1/Rt = 1/.90 + 1/.95 + 1/1.00 + 1/1.05 + 1/1.10 => Rt = .1990
At 1 per unit current, Vd = .1990
Ishort = Vd/.90 = .2211 => 22% of what ever you picked for 1 pu current
Ilong = Vd/1.10 = .1809 => 18%
So, for a deviation of 20% long to short,
Ishort is 2% above design current
Ilong is 2% below design current
It doesn't have to be. Your are looking at: Given the ampacity headroom, how far off can the conductor lengths be. My model is looking at: Given the conductor lengths, how does the current share.
I looked at it. The main difference is I used percent of design length and percent of design current. Mine is no better - perhaps just easier to see the concepts.
ice
iceworm
Curmudgeon still using printed IEEE Color Books
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- North of the 65 parallel
- Occupation
- EE (Field - as little design as possible)
That is a good choice. It certainly covers the practical aspect of making the cables the same length.
I give electricians a lot more credit than that. Anybody that can do a load calc, calculate conduit fill derate, or calculate a temperature derate, can handle this.
Yes, I have been lucky in that respect. Generally I have good luck explaining why what I want to do meets the intent and is usually way above minimum spec. Some of that may be from the fact we rarely bump up against code minimums.iwire said:
And, I don't recall an inspector ever asking me to show/proove that parallel cables were the same length.
ice
We sometimes install parallel cables.
There two man reasons for doing this.
As a rule, two smaller cables are easier to handle that one larger cable.
And the two smaller cables usually result in less copper being used.
I've never known sharing to be a problem. The cables take the same route between the items being connected.
Differences in length of run are small so differences in the impedance of the run also small.
And, even with the small differences, there is another equalising mechanism.
The shorter of the two cables will have lower cold resistance and take more current. In doing so, it will get a bit warmer than the other and that will go at least some way in mitigating the disparity between the currents.
There two man reasons for doing this.
As a rule, two smaller cables are easier to handle that one larger cable.
And the two smaller cables usually result in less copper being used.
I've never known sharing to be a problem. The cables take the same route between the items being connected.
Differences in length of run are small so differences in the impedance of the run also small.
And, even with the small differences, there is another equalising mechanism.
The shorter of the two cables will have lower cold resistance and take more current. In doing so, it will get a bit warmer than the other and that will go at least some way in mitigating the disparity between the currents.
- Location
- Massachusetts
Do you think we do it for different reasons here? :huh:
Anywhere from 2 to 10 sets are common.
Hmmm... you are distributing the lengths equally. Nothing wrong with that, but real work isn't always that pretty. Consider the same 20% short to long but make one 90% and the other four 110%. Ishort in this scenario is over 17% above design.
I am looking at it from multiple perspectives. That includes both yours and the one you think I have.
If you want to base it on percentages, a rule of thumb might be the length deviation from average (in percent) should be no more than the ampacity headroom (in percent) divided by the number of parallel conductors.
iceworm
Curmudgeon still using printed IEEE Color Books
- Location
- North of the 65 parallel
- Occupation
- EE (Field - as little design as possible)
Yes. Per unit (or percentage) is an excellent design tool for modeling the real world.
I have no interest in a rule-of-thumb. My goal is to present a model to assist in understanding the issues of current sharing in paralleled cables.
Once one understands the model, the arithmetic to solve is simple - probably take less time than finding a spreadsheet and entering the data :roll:
ice
That's a matter of opinion. I find that per unit and percentages can often be erroneously applied.
That's not exactly what I glean from this statement of yours...
I have no problem with your intent
Perhaps, but as a classroom tool, a spreadsheet is much, much easier to enter different lengths to show how the current on each conductor changes.
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