Load Calculation Question

[cooltc2004]

Hello,

I'm trying to size my main breaker in a control cabinet I'm making, and need just a little bit of guidance. Incoming power is going to be 240V 3PH. Each leg is going into and coming out of a circuit breaker and than going into a distribution block, where I pull single phase circuits from (L1+L2, L2+L3, L1+L3) to power single phase components (25kW worth). When running load calculations, am I taking into account the 3 phase power in the equation (25kW/sqrt(3)*240V) or not?

 

[bob]

Hello,
When running load calculations, am I taking into account the 3 phase power in the equation (25kW/sqrt(3)*240V) or not?

the equation should be 25kW/(sqrt(3)*0.240V)
you are taking into account all load and assuming that the single phase load is balanced.

 

[cooltc2004]

Just to verify, even though the load is single phase, since it is ultimately connecting to 3 phase power, you use the 3 phase equation (assuming a balanced load)?

Also, that was poor use of units on my part. I am using 25000 in my calculations.

 

[cooltc2004]

Ok, so I am constantly going over calculations and would like to cover this one more time. I've attached a piece of my excel document that will be used for reference.

I have a control cabinet that contains 1-12 circuits. In this example, I have 240VAC 3PH power coming into a main breaker (defined in column G based off calculations). Each circuit contains two sets of 50A breakers (column D), which are rated for 40A continuous (hence 80A total in column C). I'm calculating max amp draw for the main breaker, and I'm calculating it based off max draw on a single leg of 3 phase power (column E).

My question is this: Am I looking at the max amp draw for the main breaker correctly?

Please excuse me if this seems rudimentary; I am a Mechanical Engineer by degree, thrown into an Electrical Engineering position and told to make it work So far so good, just 3 phase power calculations are not my forte and I want to become knowledgeable on the subject.

 

[Smart $]

... Each circuit contains two sets of 50A breakers (column D), which are rated for 40A continuous (hence 80A total in column C). ...

This part is confusing. If you have two sets of 50A breakers, you have two circuits, not one. Or are you saying 40A on one leg and 40A on another leg is 80A?

 

[cooltc2004]

This part is confusing. If you have two sets of 50A breakers, you have two circuits, not one. Or are you saying 40A on one leg and 40A on another leg is 80A?

I apologize, let me clarify. We classify circuits as breakers contained within one GFI ring and what the customer would need for one area of what they are powering. So while it is two circuits per ring as properly defined, in the industry I am in it circuit = area for power required.

 

[Smart $]

I apologize, let me clarify. We classify circuits as breakers contained within one GFI ring and what the customer would need for one area of what they are powering. So while it is two circuits per ring as properly defined, in the industry I am in it circuit = area for power required.

Okay.

Just to add some clarity, what is the kW rating of this "one" circuit?

 

[cooltc2004]

Okay.

Just to add some clarity, what is the kW rating of this "one" circuit?

In the 240VAC 3PH application (cabinets are available in 240/480VAC and 1/3 Phase) it is

Each circuit is single phase power, so 19.2kW by calculation.

 

[Smart $]

In the 240VAC 3PH application (cabinets are available in 240/480VAC and 1/3 Phase) it is

Each circuit is single phase power, so 19.2kW by calculation.

And you have two Code branch circuits of 9.6kW, 40A @ 240V each, correct?

If yes, your spreadsheet determination appears to be in order.

 

[cooltc2004]

Yes I do.

Thank you for the help!

 

[cooltc2004]

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