208V three phase load calculation

[mull982]

I have someone that is requiring a 208Y/120 40A load, with both three phase and single phase loads? My question is, that even though some of the legs may have single phase loads do I still use the 208V as my voltage multiplier to calculate kVA. I am trying to size a transformer due to the fact that I will have to step down from 480V to the required 208/120. The size of the transformer I am figuring I need is 208V * 40A * 1.73 = 14.39kVA or a 15kVA transformer. I can then use this 15kVA to figure out my 480V current draw by taking 15kVA/.48/1.73= 18.06A. Am I correct in my calculations?

Does the same apply if I have single phase 240/120V. If I have a 240V service feeding both 204V and 120V single phase loads, do I use the 240V as my voltage in figuring out kVA load calculations?

 

[mo2004]

i think you go with 208V

 

[mo2004]

i hope this will help

Wye voltage calculations



for a 150 kVA, 480 to 208Y/120V, 3? transformer


ILine = 150.000 VA/(208V x 1.732) = 416A, or IPhase = 50,000 VA/120 =
Remember, in a wye system, ILine = IPhase.

 

[kingpb]

Not sure if I understand the loads you have. Regardless, a simple way is to look at the loads (in KVA) in a schedule format. Total up each leg, take the largest leg, and multiply by three (3 phase). That will be the minimum size transformer you will need.

 

[mull982]

Not sure if I understand the loads you have. Regardless, a simple way is to look at the loads (in KVA) in a schedule format. Total up each leg, take the largest leg, and multiply by three (3 phase). That will be the minimum size transformer you will need.

Not sure if I understand the loads you have. Regardless, a simple way is to look at the loads (in KVA) in a schedule format. Total up each leg, take the largest leg, and multiply by three (3 phase). That will be the minimum size transformer you will need.

Why would you multiply by three? For instance if the largest leg added up to be be 10A then you are saying that I would need a transformer capable of supplying 30A? Wouldn't this 10A be the maximum that could occur on any of the legs, thus sizing the transformer for supplying 10A.

In my particular situation I was told by a cabinet vendor that he has a cabinet that has a FLA of 40A at 208V/120V. The cabinet is used for controlling railroad crossing gate arms and the lights and instrumentation associated with it. Based off of his 40A load, I am calculating that I will need about a 15kVA transformer with a 208/120V secondary. Since the primary side will be 480V I am calculating that I will have an 18A current draw on my primary feeders. The additional questions that I had were:

1) when sizing this step-down xfmr should I use a 15kVA (calculated form voltage time current load) or shoud I use a larger one with some more room on it? Is anything additional factored in when selecting a transformer?

2) The circuit breaker I on the primary feeders I was planning to use was going to be a 25A circuit breaker. (18A * 1.25 = 22.5 or 25A) Does this sound like a logical choice?

3) The total circuit length feeding the transformer in the cabinet will be 1200 ft, so to account for the voltage drop I will need a 3/c #3. Do I still only need my breaker to be sized for the 18A load, and not the additional capacity on the upsized feeder cable?

Thanks for the help?

 

[bob]

Why would you multiply by three?
Thanks for the help?

Take a look at King's post again.

Regardless, a simple way is to look at the loads (in KVA) in a schedule format.Total up each leg, take the largest leg, and multiply by three (3 phase). That will be the minimum size transformer you will need.

He said use the KVA, not amps.

 

[mull982]

Take a look at King's post again.


He said use the KVA, not amps.

That is my mistake. I see how that works now using the kVA of each leg. With a three phase load if you were adding up a leg, for that particular phase of the three phase load would you use the L-L, or L-N voltage rating.

 

[kingpb]

It depends on whether it is a 2 pole load, or a single pole load.

A two pole load would be L-L voltage, and a single pole load is L-N voltage.

The reason to use KVA is because KVA = KVA = KVA, you won't usually mess it up. But if you switch to Amps in your panel schedule it can cause you problems with 2-pole loads on a 3-phase panel because there is a tendency to split the load, then divide by the L-N voltage, which will put you off by a factor of 1.155. See following example.

View attachment 1048

So, by leaving the load in KVA, you can split it equally between poles and sum up the KVA without an error.

 

[mull982]

kingpb: Thanks for the example!

A two pole load would be L-L voltage, and a single pole load is L-N voltage.
.

I understand this now, if i have a load across two or three poles I use the L-L voltage when calculating the kvA for that phase, and if I have a single phase load on that phase I use the L-N voltage for calculating the kVA for that phase. In other words if you had a single phase load on one of the legs in your example that was a 5A single phase load then you would take 120*5=600VA. You would then add this 600VA to your 1000VA load (using 208V) for a total kVA of 1600VA.?

In the exapmple that you posted, does this mean that there would be 4.808A on each leg?

You mentioned in a previous post that you could add up the kVA in each load individually and then multiply it by (3) for a three phase load to come up with your kVA. How would that same method be done for the two legs used in your example?

If I am not given the individual loads but am just told that the overall three phase load is 40A is it safe to just take 40 * 208 * 1.73 = 14.3kVA? Do you have a three phase example that is similar?

Thanks for the help

 

[kingpb]

kingpb: Thanks for the example!



I understand this now, if i have a load across two or three poles I use the L-L voltage when calculating the kvA for that phase, and if I have a single phase load on that phase I use the L-N voltage for calculating the kVA for that phase. In other words if you had a single phase load on one of the legs in your example that was a 5A single phase load then you would take 120*5=600VA. You would then add this 600VA to your 1000VA load (using 208V) for a total kVA of 1600VA.?

In the exapmple that you posted, does this mean that there would be 4.808A on each leg?

You mentioned in a previous post that you could add up the kVA in each load individually and then multiply it by (3) for a three phase load to come up with your kVA. How would that same method be done for the two legs used in your example?

If I am not given the individual loads but am just told that the overall three phase load is 40A is it safe to just take 40 * 208 * 1.73 = 14.3kVA? Do you have a three phase example that is similar?

Thanks for the help

Don't forget the sqrt of 3 if you have a 3 pole load.

The sum of the single phase KVA loads is the total 3 phase KVA load. Think of it as using either (3) single phase transformers or (1) 3 phase transformer. So yes, the 1600VA would be the total connected VA.

Yes, in the example the 4.8 A is on each leg.


Summing each leg KVA, taking the largest and multiplying by 3 is useful for sizing a transformer ahead of the panel, and for selecting the main breaker or fuses.

The beauty of using KVA instead of amps, is that you can split the KVA between poles, e,g, in the example 1000VA is the total load and each pole would have 500VA. If you combined it with the 600VA single pole load, you would have 1100VA on one leg, and 500VA on the other, and techincally 0 VA on the third. the total connected VA is 1600VA.

But, since you have to have the capability to feed the 1100VA on leg one, you take 1100VA x 3 = 3300VA, and 3300VA/(208*1.73) = 9.2A. But now say you have 100VA 2 pole, and 600VA single pole. You could have 500VA on leg 1, 500VA on leg 2, and 600VA on leg 3. in this case your largest leg is the 600VA, and therefore your breaker would only be 3 x 600VA = 1800VA, 1800VA/(208*1.73) = 5A (plus 125% factor)

If you are simply told 40A, 3pole total load then you can determine the total VA as you described.