Calucating feeder load requirements for a commerical building.

[Oldbull]

I work at the local VA hosiptal. I am having a problem verifying the loading on an installed riser in the building. We have a 600 Amp bussway. Itis the emergency feeder power riser. It is fed by a 600 amptransfer switch which feeds a 500 amp circuit breaker. The combined circuitbreaker rated loads adds up to 970 amps for this riser. These are the combined current rating of disconnects attached to the buss.
The measured loading/constant load for the buss averages arround 175 amps.

Am I correct in saying that the riser is overloaded by 175 amps x 125% =218.75 constant load + 970 amps of attached load = 1188 amps or 297 % .

Or is there some kind of magic calculation I am missing. Those in the "Know" believe that we can keep adding Buss disconnects to the riser and redistrubute the load thus giving us the availability of more power.

 

[david luchini]

I work at the local VA hosiptal. I am having a problem verifying the loading on an installed riser in the building. We have a 600 Amp bussway. Itis the emergency feeder power riser. It is fed by a 600 amptransfer switch which feeds a 500 amp circuit breaker. The combined circuitbreaker rated loads adds up to 970 amps for this riser. These are the combined current rating of disconnects attached to the buss.
The measured loading/constant load for the buss averages arround 175 amps.

Am I correct in saying that the riser is overloaded by 175 amps x 125% =218.75 constant load + 970 amps of attached load = 1188 amps or 297 % .

Or is there some kind of magic calculation I am missing. Those in the "Know" believe that we can keep adding Buss disconnects to the riser and redistrubute the load thus giving us the availability of more power.

The sum of the ratings of the disconnects or c/b's attached to the bus are irrelevant to the "load" on the bus.

If the load measures 175A, then that is the actual load. To calculate the load for NEC purposes, you'd have to know what the actual loads connected to the bus are.

 

[augie47]

IMHO, 517.30(D) would apply and the capacity of the feeder can be calculated in more than one manner including using Art 220 and historical data.
I don't see a requirement to consider the sum of the disconnects attached to the buss.

This would be somewhat akin to adding all the branch breakers in a panel and determining the panel main from that number rather than the calculated load,

 

[Oldbull]

Do these sections cover my calculations.

Do these sections cover my calculations.

I have read what you replied to me. I went to the code book and reviewed what I am using as a basis for the calculations and I still seem to be confused.

In article 215.2 A 1 it says " the minimum feeder-circuit conductor size, before any application of any adjustment or correction factors, shall have an allowable ampacity not less than the noncontinuous load plus 125% of the continuous load".
Then in 220.87 Determining existing loads. It says you can utilize the "actual maximum demand to determine the existing load under all the following conditions" and it goes on to say the max. demand data for a year. with an exception for a 30 day recording on the highest phase.

And the maximum demand at 125% plus the new load

And the feeder has overcurrent protection in accordance with 240.4 and overload protection according to 230.90

So which one do I use?

Article 215 says, The actual load/ continuous load plus 125% and the noncontinuous load (the potential load, Sq. Ft. calculations and sometime use equipment) are used to determine the minimum size feeder.

So my thought is that if the measured load plus 125% and the potential calculated load are added you come up with the minimum size feeder needed for a particular feeder installation. I did use a short cut for my calculations by just adding all the buss disconnect breaker sizes at 80% ( which is the max. allowed loading ) But this should still put me in the ball park as to the loading/ potential loading on the buss duct riser.


Then Article 220.87 says the maximum demand plus 125% and the new load.


What if the new load is undetermined and your just adding expansion space for possible future expansion?

Shouldn't the potential loading (The main breaker size) be factored into the calculations to ensure that an overloading condition could not possibly become a potential problem?

I don't know for sure can someone set me straight? :?

It doesn't seem correct to say that we can just keep adding more and more to the feeder until it starts tripping the breaker. Then someone plugs in a buffer and knocks out the complete system.

 

[GoldDigger]

Well, without going into the whole issue, I will make the statement that the formula does not involve taking the continuous load and adding 125% of that load. You either multiply by 1.25 or you add just 25%.
Then you add the noncontinuous load to that.
The continuous load determination has to be made by analyzing individual loads, not by looking at historical load figures.
You also seem to be mixing and matching sections on branch, feeder and service load determination. Some calculations may refer you to other sections, but in the absence of that you need to stay within the boundaries of the appropriate section.


Tapatalk!

 

[Oldbull]

Well, without going into the whole issue, I will make the statement that the formula does not involve taking the continuous load and adding 125% of that load. You either multiply by 1.25 or you add just 25%.
Then you add the noncontinuous load to that.
The continuous load determination has to be made by analyzing individual loads, not by looking at historical load figures.
You also seem to be mixing and matching sections on branch, feeder and service load determination. Some calculations may refer you to other sections, but in the absence of that you need to stay within the boundaries of the appropriate section.


Tapatalk!

So what is the value of the noncontinus load?

Which method do you use? 80% max loading or some other calucation? You need to figure them too. This where the potental loading problem comes into the equation and causes all the problems.