Example D5(a)

[woodworkingsparky]

Hello, I am designing a 20 unit apartment building and have run into some confusion while computing the service size(120/208v, three phase).

I have figured 20 , 12kw ranges at 29,000 va per table 220.19. In the example this load is divided by two in order to figure a per phase demand. The per phase demand is then multiplied by three to arrive at an equivalent three phase load. Why is this done? Is not a load a load when we are talking in VA?

In the above example, I computed my equivalent three phase load to be 43,500 va. If a load is truly a load why is this step needed.

Where I'm going with this is that I also have roughly 53,000 va of air conditioning load - single phase. Do I also need to convert this to an equivalent three phase load, if so, why?

Thank you in advance for your time.

 

[charlie b]

Re: Example D5(a)

Originally posted by woodworkingsparky:I have figured 20 , 12kw ranges at 29,000 va per table 220.19.

The code allows you to size your service to include 20 ranges times 12 KVA each, or 240 KVA, if you like. But that is more than is needed, and you would be underbid for the job. The code also allows you to size your service to include 35 KVA for this many ranges, looking at Column C of Table 220.19. Example D(5)(a) gives an approach that allows you to size your service to include 14 ranges, as follows: 7 ranges will be fed between Phase A and Phase B, 7 ranges will be fed between Phase B and Phase C, and 6 ranges will be fed between Phase A and Phase C. So any single phase will supply power to a max of 14 ranges. That is why the Example selects the value of 29 KVA from Table 220.19.

In the example this load is divided by two in order to figure a per phase demand. The per phase demand is then multiplied by three to arrive at an equivalent three phase load. Why is this done? Is not a load a load when we are talking in VA?

A load is truly a load. But you have more than one load, so you must add them up. The total load on three phases is the sum of the loads on each of the three phases. The 29 KVA calculated above represents the load on two phases. Presuming things are reasonably balanced, the total load (i.e., on three phases) is one and a half times the load on two phases. 1.5 times 29 is 43.5.

Where I'm going with this is that I also have roughly 53,000 va of air conditioning load - single phase. Do I also need to convert this to an equivalent three phase load, if so, why?

No you don?t. The difference is that the demand factor table has different entries for different amounts of ranges. Example D(5)(a) performed a calculation, using demand factors, to arrive at a total range load. You don?t need to perform such a calculation, because there are no demand factors for air conditioning. The value of 53 KVA already represents the total load. In order to follow the process used in the example, but without the demand factor table, you could divide the 53 KVA by 3, to get the single phase load, then double it to get the load on two phases, and finally divide by 2 and multiply by 3, to arrive at the 3-phase load. But that would bring you back to 53 KVA.

NOW MY QUESTION:

Why not just use 35 KVA, directly from the table, instead of the larger value of 43.5 KVA that Example D(5)(a) derives?

 

[woodworkingsparky]

Re: Example D5(a)

quote:

NOW MY QUESTION:

Why not just use 35 KVA, directly from the table, instead of the larger value of 43.5 KVA that Example D(5)(a) derives?

Very good point Charlie, thanks for your input. Your response was very clear and well thought out.

In response to your question, yes that would be a better way to go. I sometimes get the blinders on and need someone to re-direct me a little.

Thanks again!