Feeder Bonding Jumper

B

bluebuzz81

Member
Location
Greensboro NC
Occupation
Electrical Enginerd
Going over practice questions to prepare for master electrician exam and I have a question,

An individual apartment in a multi-family dwelling is supplied by 255-amp THW copper feeders. The feeder conductors are protected by 225-amp breakers from the main 800 amp building service. The water piping systems in the apartment is metal but, tap is pvc. What size equipment bonding jumper is needed?

Shouldn't the answer be #2 copper?

How I came with this, I went to table 310.16 and figured feeder conductor is 250AWG.
Then I went to table 250.102(C)(1) and got #2 from the table.

Answer shows it should be #4.

HELP!
 
What is a equipment bonding jumper?

For a copper feeder the minimum size for 225 amps is #4/0 not 250 kcmil. The EGC for that feeder would be #4.
 
It should ask for equipment grounding conductor. This is a feeder so no bonding jumper is needed. The equipment grounding conductor is based on 250.122. Since it is over 200 amp breaker you would need #4 as Infinity stated. The wire size is insignificant in this question.
 
bluebuzz81 said:
How I came with this, I went to table 310.16 and figured feeder conductor is 250AWG.
Then I went to table 250.102(C)(1) and got #2 from the table.
Click to expand...
Help us help you here. How did you come up with 250 kcmil?

Also, though this won't likely change any answers, what NEC edition are you using?
 
Seems to me this question is answered by 250.104(A)(2). [I'm using 2017 NEC references].

Per the description, the apartment has a metal water piping system, but it is isolated from the rest of the building by a PVC water pipe connection. 250.104(A)(2) says to size the equipment bonding jumper in accordance with 250.104(D), which says to size the jumper in accordance with Table 250.102(C)(1) based on the largest ungrounded conductor. Table 310.16 does show that "255A copper THW" is 250 kcmil. So the answer per Table 250.102(C)(1) is #2 Cu or 1/0 Al.

Thus I believe the OP is correct, and the answer key is wrong.

Cheers, Wayne
 
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