120/208VAC panel main breaker size

[RJC2008]

Hello,

I have a 120/208 Vac panel that is fully populated and due to the nature of this panel installation there is not a need to provide additional expansion. The load is a mix of 120 and 208 lighting loads and receptacle loads (no motors etc.)
The total load on the panel is 26,688VA.
My question is: what is the minimum size main breaker that I can use to protect the panel.
I think it should be (26688VA/(208*1.73))*1.25= 92amps. So I should use a 100amp breaker.

Is this correct??

 

[rbulsara]

Match the rating of the panelboard.

 

[RJC2008]

Normally I would size the main breaker based on the panel rating. But, this panel does not need provisions for future expansion and all circuits are used I would like to have the smallest feeder possible and also the smallest circuit breaker possible.

 

[davidbeach]

this panel does not need provisions for future expansion

Famous last words.

Like rbulsara said, match the panelboard rating.

 

[RJC2008]

I understand the concern and normally this would be a non-issue as I would size the breaker and feeder for the panel but this is truly special case. I need to know the absolute minimum sizes for both the breaker and feeder.

The problem is that I am unsure of the calculation because the loads are both 208 and 120.

 

[davidbeach]

If you truly need the absolute minimum size (which I don't believe, but oh well), you need to do a phase by phase load analysis, for the 208V loads half the VA goes on each phase, and then size based on the phase with the highest loading. It will be larger than your 92A unless you have perfect balance between phases.

This type of sizing always leads to regrets and bad mouthing of the person that was penny wise and pound foolish to install it in the first place.

 

[waross]

I have heard those words from customers when installing long feeders. A few years later they started to complain bitterly that the main breaker was blowing after they just added a little load that was about 30% of the existing load. I don't undersize feeders anymore. I can't count the times that I have had to scramble to find capacity because of undersized feeders.
BUT, If you insist;
Step one.> Start with your 208 volt loading. It is a waste of time to try to calculate the exact current in each phase. If you do, so what? You can't get a breaker with one pole 80 amps the second pole 100amps and the third pole 93 amps.
Assume that the 208 volt loads are balanced but use the heaviest phase to phase load for all three phases.
Example; Phase "A" to "B" 37 Amps, Phase "B" to "C" 30 Amps, and Phase "C" to "A" 40 Amps.
Assume that all three phases are loaded to 40 Amps for your calculations.
Step 2.> 40 Amps plus 40 Amps = 80 Amps x .866 = 69 Amps.
This (69 Amps) is your phase current from the 208 volt loads and may now be added directly to the 120 volt loads.
Add the heaviest single phase load. This is your absolute minimum loading. Check your code to see if the feeders must be over sized and check the maximum temperature rating that may be used when determining the feeder ampacity.
Now I expect to hear about a year or so from now;
"Help! Our panel worked well for a year but a component in one of the loads failed and the replacement draws more current, the main breaker..............."

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[RJC2008]

Thanks for the response.
In step 2, I am not clear on why you are adding 40 and 40?
and where does the factor .866 come from?
Could I not just take the sum of the VAs for the heaviest loaded phase and divide it by 120?

 

[waross]

If the heaviest loaded phase is 40 amps, assume the other phases are loaded at 40 amps. This is line to line connected loads. Because of phase angle differences the two 40 amp loads, do not sum to 80 amps but to .866 times 80 amps. This conversion also adjusts the phase angle to be in phase with the line to neutral loads.

Add the heaviest single phase load.
This would be better if I had written "Add the line to neutral current from the heaviest loaded phase."

How about if I start over.
Start with the balanced three phase loads.
Add 1.73 of the unbalanced current from the heaviest loaded line to line connection. (.866 is 1/2 of 1.73)
Add the line to neutral current from the phase with the heaviest line to neutral current.

OR
Consider 13 identical loads each drawing 14 amps single phase at 208 volts.
We can divide these loads into a balanced load with 4 on each phase and 1 temporarily left over.
Each group of 4 will draw 4 x 14 = 56 amps.
But when we combine the 56 amps from A-B to the 56 amps from B-C, we get 1.73 X 56 amps = 97 amps rather than 112 amps.
The resultant current is now at the correct phase angle to be added directly to the line to neutral currents.
Now take the left over load and assume that there are 2 other identical loads and that it is balanced. Multiply by 1.73 to correct the phase angle and add this current to the others. With only one odd load the resultant current is a little high but with two odd loads this will be the current on the highest loaded phase.
The reason that you can't just add the VA is the phase angle difference between line to line loads and line to neutral loads.
Note, with only one

Bill
--------------------
"Why not the best?"
Jimmy Carter