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## peak load summary

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?

2. 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.

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Originally Posted by charlie b
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.
I have the peak load analysis for each month 1 -12 provided by Utility. Sorry Actual Demand load. the bill has a KWH and actual demand load for the 2016 cycle. Which is almost double the KWH. The service is 277480v 3p 4w

KW/.8*1.25%/ 831=KW Does look better?
Last edited by charlie b; 06-13-18 at 12:13 PM. Reason: Repaired the quotation

4. 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.

5. Originally Posted by MBLES
KW/.8*1.25%/ 831=KW Does look better?
That equation will give you an answer in amps, not in KW. But if the utility information really is the peak KW for each month, and you use the highest value for the year, then this will give you a result that you can use as the existing load.

6. Originally Posted by electrofelon
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.
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.

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Originally Posted by charlie b
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.

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.

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Originally Posted by charlie b
That equation will give you an answer in amps, not in KW. But if the utility information really is the peak KW for each month, and you use the highest value for the year, then this will give you a result that you can use as the existing load.

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.

9. Originally Posted by MBLES
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.
Multiply the demand load by 1.25, then add the new load.

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