based on 12 month load summary provided by utility I have 12 KWH readings on for each month. Do i calculate the
highest KWH/.8pf*1.25%/831=? kw to determine what my load will be?
based on 12 month load summary provided by utility I have 12 KWH readings on for each month. Do i calculate the
highest KWH/.8pf*1.25%/831=? kw to determine what my load will be?
No. You throw all that data away and look for another way to determine the existing load.
Riddle me this: If I told you how many miles I drove each month during 2017, can you tell me whether I was ever at risk at receiving a speeding ticket?
KWH is a measure of the total energy used - analogous to miles driven. KW is a measure of how fast you are using energy - analogous to driving speed.
In order to determine the existing load, you need to know the maximum measured KW, and KWH cannot possibly help you find that number. Ask the utility whether they can give you the peak KW for each month, and you will have what you need.
Charles E. Beck, P.E., Seattle
Comments based on 2017 NEC unless otherwise noted.
Last edited by charlie b; 06-13-18 at 12:13 PM. Reason: Repaired the quotation
Demand in KW is what you need from the utility. 220.87 doesn't say you need to figure power factor or imbalance so it would just be highest kw times 1.25, done that's it. Personally I would give it a little more headroom though.
Ethan Brush - East West Electric. NY, WA. MA
"You can't generalize"
Charles E. Beck, P.E., Seattle
Comments based on 2017 NEC unless otherwise noted.
It doesn't explicitly say that, but I believe it means that. Look at 220.87(2). It says to take the maximum demand (without clearly saying KW or KVA) at 125%, add the new load, and check to make sure you would not exceed the ampacity of the conductors. But you can't compare load to ampacity without taking power factor into account.
Charles E. Beck, P.E., Seattle
Comments based on 2017 NEC unless otherwise noted.
I really do appreciate the answers but I was at work so my question vague and incomplete. I always like to verse engineer a problem to see how the answer is determined. I got my answer with your first response because i was using the KWH and not the column further down on page. that said actual demand on utility summary As far as the equation using the 80% power factor and the load calc. was given to me by engineer on drawing and I wanted to ask why if there was a reason he added the PF to calculation but like thought its not needed per 220.87. Thanks again.
without starting new thread. here is my question
I have 182kw existing 600a 277480 3ph. I need to add 31.44 kw or 131 amps single phase. we are installing a single phase 480/120240 xfmr. when applying 220.87. Would i add the 31.44kw to 182kw = 213,440kw then do .8*1.25/831= 401a ? Or am I missing something.
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