# What size equipment ground for this feeder?



## NolaTigaBait (Oct 19, 2008)

i looked at 250.122 and it says 2/0 copper or 4/0 aluminum...


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## NolaTigaBait (Oct 19, 2008)

250.122(f) im not sure i understand, does it mean you need 3x 2/0?


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## william1978 (Sep 21, 2008)

Take a look at 250.122 (B)


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## NolaTigaBait (Oct 19, 2008)

william1978 said:


> Take a look at 250.122 (B)


you're not increasing the size, just adding more conductors...honestly , i'm not sure of the correct answer


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## redbeard43 (Mar 20, 2009)

whats wrong with the 4/0? Why are there 7 -750s?

3 phase?
Pipe 1 ......................Pipe 2
3 hots..................... 3 hots
1 neutral..................1 neutral
1 ground................. 1 ground

1 phase?
2 hots
1 neutral
1 ground

Somethings not adding up


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## RePhase277 (Feb 5, 2008)

We can assume that the engineer has up sized the conductors for voltage drop, but damn! Seven 750s? Whew! Somebody has got some cheese!

But take it and run with it. If we keep everything proportional, figure what size wire you would need to run seven conductors to achieve 1000 A with. It should turn out to be less than the code minimum of 1/0 for parallel conductors. But size your ground to that, then scale up by the ratio of 750 to that size.


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## nick (Feb 14, 2008)

Well i see alu ? 7 runs of 750 mcm is over kill !!! at 1100 feet there is not that much of a voltage drop at 480 volts 3 phase ? the load you have is not going to be 1000amps whats the load really ?

The ground conductor is per [250.120] or[ 250.4a 5][ or 250.4b 4] take voltage drop into it you increase for that but in your case there is not that much VD on a three phase 480 v at that distance your within the 3% factor on branch circuit .
You would run just one ground in each pipe per breaker size in ampacity. 

[310 .16] =75DEG =385 amps x7 = 2695 amps In my book 3 runs of 750 mcm would be enough with a 4/0 ground . But there maybe something i dont know or missing ? 3 x 385 amps = 1155 thats 750 mcm alu ?http://i611.photobucket.com/albums/tt195/stringking/P1010049_01.jpg 
Look at gear its 7 runs 400 mcm copper at 2000 amps .
In my case we dont need a ground its a service feed 3 hots one neutral each pipe .




Take care ask or RFI the engineer ask if he made a error on his oneline .


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## Jeff000 (Jun 18, 2008)

That seems WAY overkill. 
We have a 1600 amp (but 600v) parallel run that is just 2 750's per phase, 8 total and then a ground. Its ~300 meters.


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## electricista (Jan 11, 2009)

Although I agree with Nick and assuming their is no mistake in requiring 7 - 750 mcm cables then I would say you need 7- 750MCM conduction for EGC in each conduit. Here is why.

As Nick stated, Using aluminum conductors then 7- 750 MCM cables i s equal to 2695 amps- outrageous but not important. What is important is what we would need to accomplish this with 7 parallel feeds. From T. 310.16 we see that 2/0 AL is equal to 155 amps so 7 x 155= 1085. This would satisfy the 1000 amp breaker. 

Now T. 250.122 says that for a 1000 amp OCPD we need a 4/0 AL as the egc. Furthermore art. 250. 122 (F) requires this EGC in each conduit. 

Art. 250.122 (B) says we need to increase the size of the EGC proportionately.

750 MCM= 750000 circulat mils
2/0 = 133100 circular mils

The difference in these conductors is approx. 5.6 times larger than what would have bben required without VD involved. 750000/133100

Now we take the required 4/0 AL EGC and increase it by 5.6
4/0 = 211600
211600 X 5.6 = 1184960 cir. mils
This is equivalent to 1250 mcm

Now we go back to 250.122(A) and it tells us in no case does the egc need to be larger than the circuit conductors supplying the equipment.

So we are back to 750 MCM AL as the EGC

William I am curious how you got 700 MCM. I double checked my math but I am home sick and may have missed something.


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## nick (Feb 14, 2008)

Well the ground does not total up to the paralleled group it is the ground in each conduit alone that must be able to control the fault .
We did some calculating and came up with 4 runs of 750 mcm alu. and you can use the 4/0 ground at 1000amps and at 1100 feet .
cm=1.73 x21.2 x1100x1000/ vd 14.4 == 40343600/14.4 = [2801638.8 ]now divide this by how many paralleled runs you need . 4 runs =750 mcm is fine . this is all you need for that load . How do you come up with 750 mcm explain from what to what is this based on ? take care best to yas


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## electricista (Jan 11, 2009)

nick said:


> Well the ground does not total up to the paralleled group it is the ground in each conduit alone that must be able to control the fault .
> We did some calculating and came up with 4 runs of 750 mcm alu. and you can use the 4/0 ground at 1000amps and at 1100 feet .
> cm=1.73 x21.2 x1100x1000/ vd 14.4 == 40343600/14.4 = [2801638.8 ]now divide this by how many paralleled runs you need . 4 runs =750 mcm is fine . this is all you need for that load . How do you come up with 750 mcm ? We are takling grounding conductor not grounded conductor correct ? take care best to yas


Nick you are changing the question. Given the upsize to 7 pairs of 750MCM from what could be 7 pairs of 2/0 AL you must upsize the EGC proportionately. 



> How do you come up with 750 mcm?


 If that question is for me I believe I showed where that came from. The EGC must be upsized in each conduit based on the upsize of the ungrounded conductors.


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## nick (Feb 14, 2008)

Well were not changing the question were stating the fact that the 7 runs is not needed at that length so in this case its not correct to start with .

Now you are using 250.122 b increased in size ? from what size ? And how do you get 750 mcm were not giving you a hard time we just might learn something can you show me .take care


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## electricista (Jan 11, 2009)

nick said:


> Well were not changing the question were stating the fact that the 7 runs is not needed at that length so in this case its not correct to start with .
> 
> Now you are using 250.122 b increased in size ? from what size ? And how do you get 750 mcm were not giving you a hard time we just might learn something can you show me .take care


We all seem to think the 7 runs are not needed but the specs by the EE call for the 750 MCM's. Whether we agree or not is not pertinent. We were asked to answer the question as stated. I cannot guess what the EE is thinking I can only try and respond to the given info.

That being said you determined to change the number of runs to 4 parallel runs but even that does not enter into the equation.

Let's suppose that the problem states "what size egc would be needed for a run of 750mcm AL conductors for a 125 amp panel.

If we use T. 310.16 we see that a 2/0 AL is all that is needed for this panel. T. 250.122 states that we need a #4 Alum for the 125 amp panel.

The 2/0 has a cir mil of 133100 and the 750 has a cir mil of 750000. So we have upsized our conductors by a factor of 5.6. Now we look at #4 with a cir mil of 41740. Since we have upsized our conductors by 5.6 we must also upsize our EGC by the same proportion. So 41740 X 5.6= 233744 which is equivalent to a 250MCM conductor. We will need to use a 250 MCM conductor for our EGC. art. 250.122(B)

Does this make sense? Do you agree?

Unfortunately the code is not very explicit about where we start from in this upsizing.


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## nick (Feb 14, 2008)

Well yes it does make sense and i know the article i just dont have the same thoughts on how the code can say which feeder is picked .

Meaning the one that needs correcting as any feeder could be used for any load with out voltage drop so to me its a pick a choose process which is the one we pick at that moment . Meaning if i picked a 1000 cm wire on most runs we would not have a vd on most runs or loads .

What were looking for is the reason formula that the length in distance to the voltage dropped is the average between the two conductors of different sizes so at a fault in the line at that lower resistance we better change the ground resistance to match the conductors ampacity ? Is this it 

yes we understand and how to use it. I see more of a one conduit single feeder type VD ?
But i dont agree with it . take care best to ya good points your ok nice job !


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## electricista (Jan 11, 2009)

well this has been a good question. I just looked at the problem from a different stand point and got another answer. 

T. 310.16 will allow 3 parallel runs of 750 MCM to feed a 1000 amp load. Thus if we increase the parallel runs to 7- 750 MCM's we increase the size by 2.3. 

Now if 4/0 AL is given in T. 250.122 to be used for 1000 amp OCPD then we must increase the 4/0 by 2.3.

4/0 has 211600 cir mils. Multiply this by 2.3 we get 486680 or a 500 MCM conductor as the EGC.

Well I am convinced this section needs clarification and new wording. Where do we start from?? UGH


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## nick (Feb 14, 2008)

Well 2/0 alu is what ampacity ? Its really 135 amps not 155 amps . Its funny but most conduits cut or damaged its a fault of all conductors not just one to ground . The load now if shorted or faulted is not just one ground conductor in a parallel group its the phase to phase fault .
Ground faults to me are minimal damage in a electrical short .

But i do see the formula its protecting in the most damaging way possible which is good geeeez what did they do years ago before when no one ran a ground in conduits ? guess they up size the rigid pipe ? 

just joken think about it what makes that breaker trip fast . We all think about new code issues or how to look at each others input and we all learn from talking and others input each day .

take care Electricista


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## william1978 (Sep 21, 2008)

This is how I came up with 700 mcm. They could have used 600 paralleled 3 times which equals 1020 amps, but they used 7- 750 mcm which equals 2695amps which was increased by 2.91666%. A 4/0 ground which is good for a 1000a breaker has 211,600 circular mils X 2.91666% equales 617,165 circular mils which would be rounded up to 700mcm.


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## william1978 (Sep 21, 2008)

I agree that 7--750's is over kill.


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## william1978 (Sep 21, 2008)

I belive that I heard that the EE was fired over this screw up. There is about 50 conduits that have the wrong size grounds in them.


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## nick (Feb 14, 2008)

Well iam not a engineer or a expert in the code .http://i611.photobucket.com/albums/tt195/stringking/P1010019_02.jpg

But i look at a ground wire as a extra point to ground as to just a safe practice but is it really ? 
In a fault condition to ground does voltage drop really meaning anything to time to trip and why ? 
look at a fault condition the voltage is dropped lots of voltage dropped so you dont have normal voltage its at 2volts to ground by example how can a larger wire size help ? 
The current is at peak the voltage is at minimum a leakage to ground is itself a higher resistance to ground when it happens. 
this air gap or point of potential to conduct to ground between is what takes the time to trip the breaker not the resistance alone of the wire per the NEC code . Thats why and how they designed art 250 .122.

Look at wire size to ampacity this is just my thinking and i know iam out of line i always have a thought or a odd point to present but more damage is done by a larger ground wire than a smaller ground wire ? comments 

When we current test a breaker we dont use a smaller wire we dont use the ground wire size we use a conductor 5 times larger than the phase conductor it helps with the smoke coming off the conductors attached to the breaker .
If a ground wire is the only protection we have were in trouble as the connections are by far the worst we have in most panels or equipment and there not able to test these grounds even today no one really cares about a ground wire its just there ? comments take care


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## JohnJ0906 (Jan 22, 2007)

If the EGC is undersized, it will act as a resistor, keeping the fault current lower, perhaps too low to open the breaker/blow the fuse.

It is the resistance of the wire itself that causes voltage drop - it is the resistance of the wire (EGC) that makes it an issue as an acceptable fault path


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## nick (Feb 14, 2008)

JohnJ0906 said:


> If the EGC is undersized, it will act as a resistor, keeping the fault current lower, perhaps too low to open the breaker/blow the fuse.
> 
> It is the resistance of the wire itself that causes voltage drop - it is the resistance of the wire (EGC) that makes it an issue as an acceptable fault path


Well we agree and dont agree to me its the wire and the connection meaning the mechanical make up of lug and wire the bolts and nuts the contact of components parts all parts of circuit , the gap meaning the air gap the spark between parts when it passes from point a to point b that has resistance in time to trip the breaker space between the insulator or dielectric of the air gap by example . 

We all see a short to ground as a good or perfect connection i disagree with that part to me a breaker that trips sees current heat and time if anything can change this it will effect the trip time .yes a breaker will trip slower if current is controlled by other unknow factors like air its a insulator but also a conductor that gap controls voltage and current with out the resistance of wire . We dont see how the nec code is totally correct by a increase in some cases if there is no voltage drop in a parralled run to start with then why do we need to increase the ground wire just because its done in the code book ? take care be safe


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## Pierre Belarge (Feb 3, 2007)

electricista said:


> Although I agree with Nick and assuming their is no mistake in requiring 7 - 750 mcm cables then I would say you need 7- 750MCM conduction for EGC in each conduit. Here is why.
> 
> As Nick stated, Using aluminum conductors then 7- 750 MCM cables i s equal to 2695 amps- outrageous but not important. What is important is what we would need to accomplish this with 7 parallel feeds. From T. 310.16 we see that 2/0 AL is equal to 155 amps so 7 x 155= 1085. This would satisfy the 1000 amp breaker.
> 
> ...


 

I did not check your math, but you do have the procedure down, as per 250.122(B).
Regardless of what we feel or think, if the Engineer is crazy or not, the specs call for it... we have to still install to the code.


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