3 phase service load

[Ponchik]

I am doing a small commercial location that will have several 3 phase machines. (drill presses, table saws, lathes, welders....)

I like to do a load calculation before I submit my bid.

The shop said 75% of the time all machines will be running at the same time, however, they will turn ON for about 30 to 60 minutes then turn OFF. Another words they are not all running at the same time for 8 hours

Not counting the lighting and general use office receptacles, where do I begin in doing the calculation?

Thank you

 

[GoldDigger]

To start, I would not consider any of the fixed loads to be continuous loads. But the continuous/non-continuous distinction does not apply to motor loads anyway.
Next I would not be able to use any kind of diversity factor in the calculation, since there is no interlock to keep all of them from running at the same time, and in fact you are told that they will often all be running at the same time. I would not worry about starting multiple machines simultaneously though.

 

[Ponchik]

To start, I would not consider any of the fixed loads to be continuous loads. But the continuous/non-continuous distinction does not apply to motor loads anyway.
Next I would not be able to use any kind of diversity factor in the calculation, since there is no interlock to keep all of them from running at the same time, and in fact you are told that they will often all be running at the same time. I would not worry about starting multiple machines simultaneously though.

Thanks GD.

Do I just add up the KVA ratings of the equipment and then size the service according to that?

>>example<<

machine #1: 34KVA
#2: 16
#3: 78
#4: 45
#5: 25

total 198KVA

198kva/240/1.732 = 476A

600A service is required

 

[kwired]

Thanks GD.

Do I just add up the KVA ratings of the equipment and then size the service according to that?

>>example<<

machine #1: 34KVA
#2: 16
#3: 78
#4: 45
#5: 25

total 198KVA

198kva/240/1.732 = 476A

600A service is required

Those are just a little more then your average "drill presses, table saws, lathes, welders...."

Smallest one you have listed is around 15 HP.

Even if all were running at the same time I doubt they all run at full load capacity at same time or for very long if they do.

Still wouldn't know how to calculate such load in advance, but I bet you would be surprised if you could look at usage data maybe a year after it has been in operation.

Without further details though I'd probably still go with at least 400 amps service and designed to somewhat easily add more capacity if needed. They may add more machines down the road as well and then you are still somewhat started at adding more capacity.

I bet the POCO only gives you a 75 or 112.5 kVA max transformer

 

[Ponchik]

Those are just a little more then your average "drill presses, table saws, lathes, welders...."

Smallest one you have listed is around 15 HP.

Even if all were running at the same time I doubt they all run at full load capacity at same time or for very long if they do.

Still wouldn't know how to calculate such load in advance, but I bet you would be surprised if you could look at usage data maybe a year after it has been in operation.

Without further details though I'd probably still go with at least 400 amps service and designed to somewhat easily add more capacity if needed. They may add more machines down the road as well and then you are still somewhat started at adding more capacity.

I bet the POCO only gives you a 75 or 112.5 kVA max transformer

Thanks Kwired.

Those numbers that i listed are just examples not real machines.

I just want to know what the calculation process and procedure is for 3phase small (600-1000A) service would be?

 

[kwired]

Thanks Kwired.

Those numbers that i listed are just examples not real machines.

I just want to know what the calculation process and procedure is for 3phase small (600-1000A) service would be?

Well one thing you need to consider with those types of machines you listed to get a realistic load on a service or feeder is how often are they loaded to full capacity? Sure you want your branch circuits to handle the full load because they may reach that at times. But a heavy duty table saw that has say a 2 or even 3 hp motor on it maybe only gets a demand of 3 hp when ripping a 4 inch thick board with higher moisture content, or worse yet trying to do that with a dull blade. The rest of the time it maybe only cuts 2 inch or less stock and isn't loaded all that hard in relation to it's capacity.

Same with a welder. How often does a person weld with the settings at maximum amps? Some so called portable welders maybe the exception, especially those handy portable 120 volt welders that don't have that high of capacity to start with, but they also may have a pretty low duty cycle, and will often cut out once you have pushed them too hard until they have had a chance to cool again.

 

[Sahib]

Calculation of actual load from connected load boils down to taking damand factors of individual loads into account.

 

[Smart $]

Calculation of actual load from connected load boils down to taking damand factors of individual loads into account.

There's essentially no demand factor applicable to motor operated machines. While agree with what kwired points out, the NEC just does not allow it to be calculated that way.

 

[iwire]

There's essentially no demand factor applicable to motor operated machines. While agree with what kwired points out, the NEC just does not allow it to be calculated that way.

I agree.

At the same time perhaps an AHJ would allow some sort of interlocking and the application of non-coincident loads?

 

[Sahib]

There's essentially no demand factor applicable to motor operated machines.

Better to take demand factor of motor loads as one for making calculation comprehensive.

 

[Sahib]

Smart$: If OP's loads are segregated into 2 or more feeders, is then application of demand factor to feeders possible?

 

[Sahib]

Smart$: If OP's loads are segregated into 2 or more feeders, is then application of demand factor to feeders possible?

Yes that may be only code way to reduce load demand (and feeder size) IMO...........

 

[Smart $]

Smart$: If OP's loads are segregated into 2 or more feeders, is then application of demand factor to feeders possible?

No.

 

[kwired]

Better to take demand factor of motor loads as one for making calculation comprehensive.

So what do you base demand factor on?

Many of the allowable demand factors that are in art 220 are based on studies of typical load patterns in specific equipment or specific applications. OP's loads don't really fit any of the situations in art 220.

The only other thing you can go on is historical load data for a similar operation, and in some cases you may need to do some convincing to get an AHJ to accept that. It is hard to use historical data from OP's new installation since it has no history yet. One can use historical load data on an existing installation to determine if there is enough capacity for adding more load though.

 

[iwire]

Smart$: If OP's loads are segregated into 2 or more feeders, is then application of demand factor to feeders possible?

Yes that may be only code way to reduce load demand (and feeder size) IMO...........

The demand does not decrease if you break it up into multiple feeders. In fact the smaller the feeder the less likely there will be diversity. (If the NEC even had demand tables for this application)

 

[Ponchik]

So from what I understand from your replies, basically add up all of the name plate KVA and then size my service base on that.

 

[electrofelon]

Calculation of actual load from connected load boils down to taking damand factors of individual loads into account.

Per the NEC, it really depends on what the load is as far as how close to "actual" it will be. For example, if I have a building full of grow lights, in that case the nameplate is used and is pretty much what the load will draw (If it is continuous of course there will be a 25% fudge on top). I think lights are one of the things that will result in closest to actual load.

In general, the nameplate of something gives a rather high unrealistic value, as they are are usually calculated at lowest acceptable input voltage, and other "worst case scenario" situations.

Motors of course we use the NEC tables which are quite high, and also very few motors are run at FLC.

just a few thoughts.

 

[Fulthrotl]

I am doing a small commercial location that will have several 3 phase machines. (drill presses, table saws, lathes, welders....)

I like to do a load calculation before I submit my bid.

The shop said 75% of the time all machines will be running at the same time, however, they will turn ON for about 30 to 60 minutes then turn OFF. Another words they are not all running at the same time for 8 hours

Not counting the lighting and general use office receptacles, where do I begin in doing the calculation?

Thank you

125% of the largest load, plus the sums of the smaller loads?

my experience with machine shops, and it's a fair bit, is that they will
pack a marshmallow into a piggy bank, and load that thing to the limits
of the physical space, and electrical capacity.

so, if you assume everything will be running full out all the time, then
you'll have some wiggle room when they add four more machines in march.

and they *will* add four more machines in march.

one thing to consider is that a lot of CNC stuff seldom is loaded to max,
but it has a higher minimum draw than a lot of old school machine tools.

 

[Sahib]

The demand does not decrease if you break it up into multiple feeders. In fact the smaller the feeder the less likely there will be diversity. (If the NEC even had demand tables for this application)

Are you considering 430.24 or not?

 

[Smart $]

Are you considering 430.24 or not?

430.24 is what it is.

125% largest motor
100% other motors
125% non-motor continuous
100% non-motor noncontinuous