interpreting Utility meter data and calculating amps

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Could someone take a look at the attached image and tell me your interpretation of column "T"? I assume that is the peak KW recorded from all the 15 minute reads during that billing period. Then, how does one convert this to amps? Is there sufficient information from one of the VAR fields? Thanks.
 

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texie

Senior Member
Location
Fort Collins, Colorado
Occupation
Electrician, Contractor, Inspector
120/208 3 phase

This looks like a typical POCO metering document. Your highlighted column labeled KW Usage is POCO parlance for demand that will be stated on the customers bill as KW Demand. The highest number I see is 148.32 KW so the current would be 148,320/208/1.732 or 411 amps. Of course this assumes a perfectly balanced load but gives you an idea of the current on each phase.

Remember, KWH is energy and KW is power.
 
This looks like a typical POCO metering document. Your highlighted column labeled KW Usage is POCO parlance for demand that will be stated on the customers bill as KW Demand. The highest number I see is 148.32 KW so the current would be 148,320/208/1.732 or 411 amps. Of course this assumes a perfectly balanced load but gives you an idea of the current on each phase.

Remember, KWH is energy and KW is power.

But the current will be higher due to power factor. I am not sure how to interpret/calculate this from the VARS. ?
 

texie

Senior Member
Location
Fort Collins, Colorado
Occupation
Electrician, Contractor, Inspector
But the current will be higher due to power factor. I am not sure how to interpret/calculate this from the VARS. ?

Yes, that's true. I think I now understand what you getting at in your original question. I was just assuming a unity PF as I did not pay any attention to the other columns. While the heading of the next 2 columns to the right are cut off, I think the first to the right is KW Current Reactive and the next one is KW Last Reactive. Looks like .8 or so is the lowest number so that would be 411 X 1.25 for a more accurate line current.
 

texie

Senior Member
Location
Fort Collins, Colorado
Occupation
Electrician, Contractor, Inspector
Ok well that makes sense and was my first thought, however those columns read "KW curr read" and "KW last read". Of course those numbers seem like they must be PF but the title doesn make much sense.

Point noted. After a closer look at the numbers in those columns, it looks like they are just the actual meter reading X the multiplier in Column W. I'm not well versed on what basis a POCO calculates PF. Sounds like we need one of the POCO metering guys to chime in that are better versed on this than me.
 
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mivey

Senior Member
You can sorta get the info but not exactly. The average pf for each month can be calculated by: kWh/sqrt(kWh^2+kvarh^2). For June 2015 this would be 56640 / sqrt( 56640^2 + 21280^2) = 0.936. You could calculate this each month to determine how the average pf varies with the seasons.

You have the peak demand for the month (146.56 kW) but not the pf during the peak period. You could probably make some assumptions but you need some info about the load as there may be some motors on during the peak.

Another useful part of the puzzle will be the load factor. For June 2015 (30 DOS or days of service) this is approximately 56640 / 146.56 / (30*24) = 0.537. If the load factor were very high, you might assume the peak kW is closer to the average kW. Not the case here.

However, for amp calculation example purposes, let's assume 90% pf. With a peak kW of 146.56 we get the peak kVA as 146.56 / 0.9 = 162.84 kVA. At 208Y/120 this yields 162.84 / sqrt(3) / 0.208 = 452 amps. I would feel comfortable saying the pf for this load on peak does not drop below around 85% at most and that would give you 479 amps.

Looking at several months you get:

Code:
  kW        kWh      days     kvarh       LF       pf       kVA       A @208V    kVA@85%pf  A @85%pf
146.56     56,640     30     21,280     53.7%    93.6%     156.56      434.57      172.42      478.60 
132.32     48,000     28     17,120     54.0%    94.2%     140.48      389.95      155.67      432.10 
137.76     50,560     31     17,440     49.3%    94.5%     145.73      404.49      162.07      449.86 
134.08     49,600     29     16,800     53.2%    94.7%     141.56      392.94      157.74      437.85 
132.32     50,720     30     17,120     53.2%    94.7%     139.65      387.64      155.67      432.10 
125.12     57,440     35     19,200     54.7%    94.8%     131.92      366.19      147.20      408.59 
125.92     56,160     33     19,520     56.3%    94.5%     133.31      370.03      148.14      411.20 
133.92     50,240     29     17,600     53.9%    94.4%     141.90      393.87      157.55      437.32 
128.64     51,680     29     20,960     57.7%    92.7%     138.82      385.32      151.34      420.08 
146.88     60,960     33     25,280     52.4%    92.4%     159.01      441.36      172.80      479.64 
146.08     54,720     29     22,880     53.8%    92.3%     158.34      439.50      171.86      477.03 
141.76     54,880     30     22,240     53.8%    92.7%     152.96      424.57      166.78      462.93 
132.16     56,000     32     22,400     55.2%    92.8%     142.34      395.10      155.48      431.58 
148.32     53,280     29     19,840     51.6%    93.7%     158.27      439.31      174.49      484.35 
                                    
148.32     750,880     427     279,680     49.4%    93.7%     158.27      439.33      174.49      484.35
 
You can sorta get the info but not exactly. The average pf for each month can be calculated by: kWh/sqrt(kWh^2+kvarh^2). For June 2015 this would be 56640 / sqrt( 56640^2 + 21280^2) = 0.936. You could calculate this each month to determine how the average pf varies with the seasons.

You have the peak demand for the month (146.56 kW) but not the pf during the peak period. You could probably make some assumptions but you need some info about the load as there may be some motors on during the peak.

Another useful part of the puzzle will be the load factor. For June 2015 (30 DOS or days of service) this is approximately 56640 / 146.56 / (30*24) = 0.537. If the load factor were very high, you might assume the peak kW is closer to the average kW. Not the case here.

However, for amp calculation example purposes, let's assume 90% pf. With a peak kW of 146.56 we get the peak kVA as 146.56 / 0.9 = 162.84 kVA. At 208Y/120 this yields 162.84 / sqrt(3) / 0.208 = 452 amps. I would feel comfortable saying the pf for this load on peak does not drop below around 85% at most and that would give you 479 amps.

Looking at several months you get:

Code:
  kW        kWh      days     kvarh       LF       pf       kVA       A @208V    kVA@85%pf  A @85%pf
146.56     56,640     30     21,280     53.7%    93.6%     156.56      434.57      172.42      478.60 
132.32     48,000     28     17,120     54.0%    94.2%     140.48      389.95      155.67      432.10 
137.76     50,560     31     17,440     49.3%    94.5%     145.73      404.49      162.07      449.86 
134.08     49,600     29     16,800     53.2%    94.7%     141.56      392.94      157.74      437.85 
132.32     50,720     30     17,120     53.2%    94.7%     139.65      387.64      155.67      432.10 
125.12     57,440     35     19,200     54.7%    94.8%     131.92      366.19      147.20      408.59 
125.92     56,160     33     19,520     56.3%    94.5%     133.31      370.03      148.14      411.20 
133.92     50,240     29     17,600     53.9%    94.4%     141.90      393.87      157.55      437.32 
128.64     51,680     29     20,960     57.7%    92.7%     138.82      385.32      151.34      420.08 
146.88     60,960     33     25,280     52.4%    92.4%     159.01      441.36      172.80      479.64 
146.08     54,720     29     22,880     53.8%    92.3%     158.34      439.50      171.86      477.03 
141.76     54,880     30     22,240     53.8%    92.7%     152.96      424.57      166.78      462.93 
132.16     56,000     32     22,400     55.2%    92.8%     142.34      395.10      155.48      431.58 
148.32     53,280     29     19,840     51.6%    93.7%     158.27      439.31      174.49      484.35 
                                    
148.32     750,880     427     279,680     49.4%    93.7%     158.27      439.33      174.49      484.35

Wow great thanks! The load is pretty much all motors - refrigeration, and AC to pull out the heat - so I thought the Pf would be worse. There is 40 KW of resistance heat (3 water tanks and 3 autoclaves) which would be about 112 amps when all on. I am sizing the new service equipment off of this data so I want to corrrect it to amps reasonably accurately, although I am certainly am not going to cut it too close. I am proposing an 800 amp service which I think is reasonable considering phase imbalance, system additions, etc.
 

Electric-Light

Senior Member
That's the DEMAND. Its the calculated by highest kWh used per 15 minutes or kWh per 30 minutes, depending on your rate. The 15 or 30 minute "window" is always moving forward.

If you're on 30 minute system and you use 10kW for 15 minutes, you're 5kW, because it averages out to 2.5kWh consumed over 30 minutes. The maximum captured sticks for the billing period.

If you're on a 15 minute rate, then that's considered 10kW demand.

What does it mean?

If you left a 100W light bulb on 24/7 for a month, you use 72kWh.

If you used no power at for the billing cycle, except on the last day, you used ran a 288kW load for 15 minutes. (72kWh consumed over 15 minutes)
Your energy bill is still 72kWh, but it will be about $2,000 for demand. (at $7/kW)
 
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Electric-Light

Senior Member
AC and refrigeration will cycle. Can you see the kWh used per hour, 30min or 15 minutes from the utility's site?

Picture mountains sitting on a bookshelf. You can save money if you can shrink the shelf height.
If the peaks are steep, you can usually clip off the top. Lower the temp a bit before the peak and set back the temp enough during the peak (as determined by looking at YOUR own usage, not just the rate period...), then program it back to resume after you're past the time frame.

Raising the overall load factor will be difficult since that's weather dependent. You can't precool for 3PM at 8AM.. but you try overshooting at 2PM, set back until. 4PM to clip the summit at 3PM. If the demand continues for hours, you may not be able to do much.
 
AC and refrigeration will cycle. Can you see the kWh used per hour, 30min or 15 minutes from the utility's site?

Picture mountains sitting on a bookshelf. You can save money if you can shrink the shelf height.
If the peaks are steep, you can usually clip off the top. Lower the temp a bit before the peak and set back the temp enough during the peak (as determined by looking at YOUR own usage, not just the rate period...), then program it back to resume after you're past the time frame.

Raising the overall load factor will be difficult since that's weather dependent. You can't precool for 3PM at 8AM.. but you try overshooting at 2PM, set back until. 4PM to clip the summit at 3PM. If the demand continues for hours, you may not be able to do much.

Ok. I am not really sure what you are trying to say though.....?

So their figures are not the actual peak KW of all the instantaneous measurements the meter did, but the highest 15 minute average?
 

Electric-Light

Senior Member
Ok. I am not really sure what you are trying to say though.....?

So their figures are not the actual peak KW of all the instantaneous measurements the meter did, but the highest 15 minute average?

Correct. It's reset every billing cycle or per your rate schedule.
You need to check with the PoCo or the statement what the demand average period is.
It could be 15min, 30 min or possibly something else.
 

kwired

Electron manager
Location
NE Nebraska
Could someone take a look at the attached image and tell me your interpretation of column "T"? I assume that is the peak KW recorded from all the 15 minute reads during that billing period. Then, how does one convert this to amps? Is there sufficient information from one of the VAR fields? Thanks.
I haven't figured out exactly what some of those VAR fields actually are, none seem to be a current VAR to go with the current kW, if there were one you could easily figure out the corresponding KVA for that period.

You must also remember that just because kW peaks at a certain point doesn't necessarily mean power factor is in a similar peak. Some instances it could peak at same (high or low), and other instances peak PF is somewhere else in relation to kW peaks.

Lightly loaded motors usually will have lower power factor then fully loaded motor - but at same time are drawing less kW, so peak kVA is still likely going to coincide with peak kW most of the time.
 

mivey

Senior Member
Ok. I am not really sure what you are trying to say though.....?

So their figures are not the actual peak KW of all the instantaneous measurements the meter did, but the highest 15 minute average?
There are no instantaneous readings. Even the one shown on the display is not instantaneous but is a reasonably close representation if the load is reasonably steady. The kW readings are chunks of energy divided by a time interval, usually 15 or 30 minutes. These are usually calculated using a sliding window.

Take a 30-minute demand for example. A typical measurement would be a watt value calculated every five minutes by taking the energy (Wh) consumed that five minutes and dividing by the time (Wh / (5/60 h) or Wh * 12/h). This is averaged along with the previous 5 watt values to get the latest sliding 30-minute watt demand.
 

mivey

Senior Member
I haven't figured out exactly what some of those VAR fields actually are, none seem to be a current VAR to go with the current kW, if there were one you could easily figure out the corresponding KVA for that period.
I explained this in my post. They go with the monthly kWh and I did calculate the monthly kVAh in my post #9 (the denominator in the pf formula). The kvar fields include readings for kvarh just like readings for kWh. You have a current reading, prior reading and multiplier. The kvarh for the month comes from (current - prior) * multiplier.
 

kwired

Electron manager
Location
NE Nebraska
I explained this in my post. They go with the monthly kWh and I did calculate the monthly kVAh in my post #9 (the denominator in the pf formula). The kvar fields include readings for kvarh just like readings for kWh. You have a current reading, prior reading and multiplier. The kvarh for the month comes from (current - prior) * multiplier.
Sorry, for some reason the posts after about #6 or 7 I had not seen when I posted, not sure if I somehow missed them or if there was an error loading the page or if I am just losing it. :huh:
 
Ok great. I am not too concerned about the power factor as I am giving plenty of leeway in the service sizing, just wanted an approximate (the PF is not as bad as I would have thought), and had some academic curiosity as well.

So a more broad academic question is what is the "best" interval to measure peak power? If the interval is too long you could/will be getting readings lower than peak, but too short and you could start picking up inrush values and such. Does something like a fluke 1735 logger have adjustable parameters for power calculation interval?
 

mivey

Senior Member
So a more broad academic question is what is the "best" interval to measure peak power? If the interval is too long you could/will be getting readings lower than peak, but too short and you could start picking up inrush values and such.
That will depend on the thermal constant of the equipment & conductor. But let's talk in broad general terms.

From a heating standpoint there is normally not much value in going to less than 2 minutes for sizing measurements as you are near momentary times (then you need to make PQ measurements). You might consider a 2-5 minute interval for a severe bump load (a sawmill blade comes to mind). This gets closer to the classic thermal demand where you had an exponential response (a fast response rise to the 90% point). You could even opt for a thermal or induction type demand if equipment is over-heating and the normal 15-30 minute demand looks within equipment capacity.

Most of our loads do not tend to be severe and the thermal capacity of the equipment tends to be enough that we can measure in 15-30 minute intervals. I think 15 minutes is fine for most things: not too short, not too long.

Does something like a fluke 1735 logger have adjustable parameters for power calculation interval?
It should have. Most PQ meters I have used do. Just measure in small enough intervals and convert to what you need. I like 1-minute intervals for PQ stuff unless it is for noise or start-up analysis in which I will use 1-cycle intervals or even wave-form captures.
 
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