Quick coordination Q

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greenspark1

Senior Member
Location
New England
Hi all,
I'm working in an existing building with a 800A 208/120V main service panel with a 200A QGA3220 breaker feeding a Square D NQ panel just 10' away. The 200A breaker is rated for 65kAIC but the panel is only 14kAIC. In this situation doesn't the downstream panel (and breakers) need to match the AIC rating of the upstream breaker? How did this pass a previous inspection?
 

ron

Senior Member
The available fault current might be <14kA, did you calculate it?. There might be a series rating with the upstream main to make the resulting short circuit current rating adequate for whatever the available short circuit current is.

Essentially it doesn't have to match the other equipment.
 

greenspark1

Senior Member
Location
New England
No, not in my scope to calculate or really worry about it. We're supposed to just add branch circuits to the NQ general panel but it jumped out at me. 14kA breakers in a MLO panel fed by a 65kA breaker in an I line panel. The I line panel is the service panel so no upstream breakers.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
If the upstream breaker is not required to have that high an AIC, then there is no problem with the downstream breaker having a lower AIC as long as its AIC exceeds the available fault current.
If the available fault current at the downstream breaker is greater than 14kA, it may still be OK if the two breakers are listed together for series rating.

Tapatalk!
 

Pharon

Senior Member
Location
MA
If that 800A panel is fed by a 225kVA transformer, the fault current on the secondary will be limited to roughly the available full load current divided by the impedance, or 625A/0.0525 (assuming typical dry type) = 11,905A.

Not sure if this is what you have, but it's a typical rule of thumb to go by.
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
If that 800A panel is fed by a 225kVA transformer, the fault current on the secondary will be limited to roughly the available full load current divided by the impedance, or 625A/0.0525 (assuming typical dry type) = 11,905A.

Not sure if this is what you have, but it's a typical rule of thumb to go by.

This is not a good rule for utility supplied transformers.

Most utility transformer banks (<500kVA) rarely have more than 3%Z, 'pole top' banks may be <2%.
 
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