What is the 2RID equation?

[97catintenn]

Is it?

(2RID)/cmil?

Does R=resistance from Chp. 9 Table 8?
I=amps
D=does D equal distance one way?

and just clarify that I do divide by cmil at the end

 

[Howard Burger]

Sounds like R is sub. for K

Sounds like R is sub. for K

Bottom line: I don't know. But this sure sounds like the top half of the single-phase formula for voltage drop where the R is in place of K. K for these formulas is derived from table 8 of the NEC: for K multiply the area of a wire in circular mills by the resistance in ohms per 1000 feet as found in Table 8. You're pretty safe remembering K for Cu is 12.9 and Al is 21.2 (or 13 and 21 for simplicity), and yes, the bottom half of the formula is CM.

 

[97catintenn]

It's a short formula to help you remember the calculation. The saying goes, "To Rid yourself of voltage drop" which then you would write 2RID

I just wanted to check on the exact numbers/varibles

 

[Smart $]

Is it?

(2RID)/cmil?

Does R=resistance from Chp. 9 Table 8?
I=amps
D=does D equal distance one way?

and just clarify that I do divide by cmil at the end

Close. You can use CH9-T8, but because you are likely running AC circuits, why not use CH9-T9?

R=resistance/1000ft from table
I=amps
D=one-way distance

...and the formula is Vd=2RID/1000 for single-phase. Replace 2 with 1.732 for 3? circuits.

You can also restructure the formula to backsolve for maximum R, then compare to the table value...

R=1000(V?%)?(2ID)

where "V" is the nominal voltage and "%" is the allowable voltage drop in per cent.

After solving for R, look in the table for the conductor, and its application if using T9, having a lesser R (or Z) value than your Vd-calculated R.

 

[97catintenn]

Great! Thanks for posting.

I have 37kva 230v ciruit wtih 1/0 copper and a 450' run (one way) in PVC.

So, then I would calculate the amps 37va/230v= 160.86amps R from Chp 9 Table 9 0.39ohm (don't forget that you have divide 0.39/1000ft!)

Vd=(2RID)/1000
Vd=(2*.00039ohms*160.86amps*450')/1000

So, my voltage drop comes to 0.056volts.

That doesn't sound right. I'm double checking my answer by taking the 0.39ohm/1000ft*450'*160.86amps and I get 28.23v

Am I totally lost?

 

[david luchini]

Great! Thanks for posting.

I have 37kva 230v ciruit wtih 1/0 copper and a 450' run (one way) in PVC.

So, then I would calculate the amps 37va/230v= 160.86amps R from Chp 9 Table 9 0.39ohm (don't forget that you have divide 0.39/1000ft!)

Vd=(2RID)/1000
Vd=(2*.00039ohms*160.86amps*450')/1000

So, my voltage drop comes to 0.056volts.

That doesn't sound right. I'm double checking my answer by taking the 0.39ohm/1000ft*450'*160.86amps and I get 28.23v

Am I totally lost?

The ohms/1000ft from table 9 is 0.12, not 0.39.

And you divided by 1000 twice. Using 0.39 would give you a 56 Volt drop, not a 0.056 Volt drop.

 

[97catintenn]

Got it. I was confused about it because I saw one reference Vd=2RID and then later saw it as Vd=2RID/1000

So, now it makes sense why guys write it both ways. And you're right 0.39ohm is per 1000 meters...:dunce: