Paralled Conductors different size allowed?

[Johny98]

Hi,

can I parallel two 500 MCM and one 2/O per phase in a 3 phase system to get 1000 Amp? What would be the min. ground size and how should the phases be run in three conduits , 4". I would appreciate the reference to NEC for it too.

Thanks

Johny

 

[iwire]

Hi,

can I parallel two 500 MCM and one 2/O per phase in a 3 phase system to get 1000 Amp?

No, see 310.4

What would be the min. ground size

See Table 250.122

and how should the phases be run in three conduits , 4".

See 310.4

 

[Johny98]

where does it say you can't and if not what is the reason behind it?

thx

 

[bob]

John
If you look at table 9, the resistance of the 2/0 is close to 10 times the resistance of the 2 500 kcm in parallel. All most all of the load would flow through the 2 500 and almost none through the 2/0. That is why the conductors need to be the same size.

 

[Johny98]

Bob,

thanks, I am thinking that it is like to resistors parallel with 500s lower resistance lowering 2/0 resistance and therefore 2/0 carriyng more current. right?

 

[480sparky]

where does it say you can't and if not what is the reason behind it?

Thx

310.4(b)(3).

 

[bob]

Bob,

thanks, I am thinking that it is like to resistors parallel with 500s lower resistance lowering 2/0 resistance and therefore 2/0 carriyng more current. right?

No that is not correct. The resistance of each conductor is fixed. The fact that you have 2 #500 in parallel with the 2/0 does not lower the 2/0 resistance. You must consider the combination of the 3 conductors. The load will see 2 conductors of low resistance and 1 conductor with high resistance. Which route do you think it will take?

 

[Johny98]

Thanks Bob.lower resistance carries more current.

 

[steve66]

Look at the math. Assume 1000 amps, and 100 meters of feeder.

From table #9, the resistance of 2/0 is 0.3170 ohms per KM. That's 0.03170 ohms for our 100 meter feeder.

For 500 KCM, its 0.0845 ohms per KM. That's 0.00845 ohms for our 100 meter feeder.

For (2) sets of 500 KCM, and one set of 2/0, the total resistance of our feeder is:

[ (.03170 ^ -1 ) + (.00845 ^ -1) + (.00845 ^ -1)] ^ -1 = .003728 ohms.



The voltage dropped across our feeder at 1000 amps of load current is:

1000 amps * .003728 ohms = 3.728 volts.

Since that voltage is across each feeder, we can use it to find the current through each feeder. For the 2/0, the current is:

3.728 / .03170 = 118 amps.

For the 500's, the current is:

3.728/ .00845 = 441 amps (for each set of 500's).

Check our answer by adding the three currents and making sure we get our 1000 amp totol current:

118 + 441 + 441 = 1000 amps.

The smaller 2/0 wire does get the smaller current, but it is a lot smaller current than we would want.

The 500 KCM's each get 441 amps, but they are only rated at 380 amps. So they would be overloaded by 60 amps each. A significant amount.

Steve