450kcmil cable

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Besoeker3

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Location
UK
Occupation
Retired Electrical Engineer
I think it's about 240 mm2. We'd usually use double 120 mm2 flexibles- it's just easier to work with.
 

augie47

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Location
Tennessee
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State Electrical Inspector (Retired)
When I do the math, a 400 kcmil would suffice as the EGC with phase conductors increased to 400.

Disregard: I misread and thought you have (6) sets of 400.
 
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Rock86

Senior Member
Location
new york
Occupation
Electrical Engineer / Electrician
Very well but you still dont get my point but this should illustrate i know the equation:

1. Lets say you take 3/0 awg AL at 65C then 130A.

1200/130= 9.2 which means minimum 10 sets of 3/0AL req.

3/0 is approx 163kcmil. 163x10=1630kcmil

Using 6 sets of 600kcmil so thats 3600kcmil

3600/1630*250= 552kcmil
Why did you multiply by 250?

Your math looks close but I wouldn't say it is right.

What I did was... (All conductors based on Aluminum Values)
1200A - 310.15b16 -> 6 sets of 250mcm can handle a 1230A at 75C. So...

250mcm * 6 = 1500mcm
600mcm * 6 = 3600mcm
250.122(B) wire shall be increased in size proptional to the cmil area of the ungrounded conductors.
3600mcm / 1500mcm = 2.4
table 250.122 tells us that a 1200A load needs a 250mcm EGC there fore...
250mcm * 2.4 = 600mcm Aluminum EGC
 

Rock86

Senior Member
Location
new york
Occupation
Electrical Engineer / Electrician
Why did you multiply by 250?

Your math looks close but I wouldn't say it is right.

What I did was... (All conductors based on Aluminum Values)
1200A - 310.15b16 -> 6 sets of 250mcm can handle a 1230A at 75C. So...

250mcm * 6 = 1500mcm
600mcm * 6 = 3600mcm
250.122(B) wire shall be increased in size proptional to the cmil area of the ungrounded conductors.
3600mcm / 1500mcm = 2.4
table 250.122 tells us that a 1200A load needs a 250mcm EGC there fore...
250mcm * 2.4 = 600mcm Aluminum EGC
I had another look at your work... and it was fine. Using the 10 #3/0 through me off. but your numbers were close enough.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
Very well but you still dont get my point but this should illustrate i know the equation:

1. Lets say you take 3/0 awg AL at 65C then 130A.

1200/130= 9.2 which means minimum 10 sets of 3/0AL req.

3/0 is approx 163kcmil. 163x10=1630kcmil

Using 6 sets of 600kcmil so thats 3600kcmil

3600/1630*250= 552kcmil
You said there are six sets of conductors. The increase in the size of the EGC will be based on that. The 1200 amp breaker would require each of the six sets to carry 200 amps. 200 amps would require a 250kcmil aluminum circuit conductor. You have 600 kcmil so the EGC will have to be increased by a factor of 600/250. The required size of an aluminum EGC for a 1200 amp breaker is 250 kcmil. That will require 600 kcmil aluminum EGC in each of the six raceways.
 
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