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Topic Title: Energy kWhr versus amps comparison Topic Summary: Trying to compare smart meter data with amps drawn Created On: 06 February 2014 11:20 AM Status: Read Only 
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06 February 2014 11:20 AM


A quick sanity check question.
If I have a three phase HH smart meter which tells me I drew 100kWhr from the mains for a HH period, how can I make a sensible estimate of average current drawn at that time? My assumption/guess/ramblings: 1 Assume that 100kWhr was an averaged over that 30mins and therefore also for 1hr 100kWhr. 2  Assume voltage is standard 400V (although never seen our voltage dip below 425V 3  100kWhr means that over 1hr we averaged 100kW load 4  100kW with a unity power factor gives 100kVA. 5  100kVA /400V / sqrt3 = 144A drawn. In fact: we have a TP 400A fused incomer where current reading can go up 250+ amps but the HH data indicates a maximum of 80kWhr at peak over day. These peaks of 250+ amps are farly flat over the 30min/hour duration (seen on logger equipment) If current stays at 250A, * 400V * sqrt3 = 173kVA at unity would give 173kWhr. Factor of 2 out. So does a HH data read of approx 70kWhr incorrect and should be interpreted as 70kWhalfhours? 



06 February 2014 01:07 PM


Assumption 1 doesn't make any sense to me. If you drew 100 kWh in 30 minutes, then at the same rate, you would draw 200 kWh in 1 hour.
By analogy, if your water meter says you used 1 cubic metre in 30 minutes, and you carried on using water at the same rate, you would have used 2 cubic metres after 1 hour.  S P Barker BSc PhD MIET 



06 February 2014 03:19 PM


Smart/Large industrial tariff meters record kWhr over half hourly periods and also accumilated the kW just like home domestic meters which record kWhr used.
Unless I am totally wrong, I assumed that kWhr is a unit with like velocity, Miles per hour. You can go 30miles in one half hour or a 60 miles in one hour, giving an average speed of 60mph, although your instantaneous speed could vary over the distance. Your water flow example was not a good analogy. I am trying match the kWhr consumption data given to me by the electricity supply to the amps logged over the same/similar period. The data of HH (half hourly) kWhr produces an inversed bath tub type graph with old spikes a break/lunch times and a base load overnight 



06 February 2014 03:34 PM


pedants' corner:
miles per hour is not velocity, it is speed. velocity has a direction. further, kwh is power multiplied by time: i.e. If you double the time (at a steady power consumption) you double your energy usage. whilst your speed is distance divided by time: i.e If you double the time for a given journey, you halve your speed. Simply, kwh is merely the power you use, multiplied by the length of time you use it for. Power is defined as: Energy / Time and kwh is defined as: Power * Time therefore kwh is Energy / time * time which is just energy The units for kwh are therefore the units of energy: joules, and one kwh is 3600 kJ From the number 3600 it follows that you can see that one joule is one wattsecond (there are 3600 seconds on one hour) ecto's analogy is correct: if you used 1 kwh for 1 hour and again for another hour, you have doubled your energy use (to 7200 kJ); use the same energy again for another hour and you have used 10.8 MJ and so on. Edited: 06 February 2014 at 04:18 PM by Zuiko 



06 February 2014 03:40 PM


Think of it this way
In one day I have 100mm of rain  it is a steady drizzle and all my gutters, downpipes and below ground drainage work fine  so 100mm/day doesn'tcause a problem Tomorrow, I also expect to see 100mm of rain  however, it arrives and dumps 100mm of water in a half hour, then it turns sunny  however my gutters overflow, my downpipes surcharge and the car park floods  but it was still 100mm/day. You can't directly deduce power from energy  ie you can't derive instantaneous Amps from accumulative KWh. Regards OMS  Failure is always an option 



06 February 2014 03:45 PM


That's right OMS,
If need be, you can calcualte an average current throughout the day  but if the current varies a lot, this average may be useless which is why maximum maybe more helpful a lot of the time. (if you lived in an area of erratic rainfall, you would want your gutter to be able to handle the maximum amount of expected rain at a given time, not the average  something the nondredgers on the somerset levels have learned at terrible cost) 



06 February 2014 03:50 PM


Zuiko  I am afaid you have a typo in your response!
Your post should read: :: :: Power is defined as: Energy / Time and kWh is defined as: Power * Time therefore kWh is: Energy / Time * Time which is just Energy. The units for kWh are therefore the units of Energy: Joules, and one kWh is 3600 kJ From the number 3600 you can see that one Joule is one Wattsecond. :: :: OMS  I agree, but for reference, you can deduce average Amps from cumulative kWh. Regards Henry Kafeman  Henry Kafeman HDK Solutions Ltd. Investigate, Analyse, Resolve Tel: +44 (0)1908 866921 



06 February 2014 04:19 PM


well spotted,
my proof reading is terrible! (btw, oms said you can't deduce instantaneous amps, which is correct; of course you can deduce average (mean) amps, which may or may not be useful) Edited: 06 February 2014 at 05:07 PM by Zuiko 



06 February 2014 05:15 PM


I am trying to match a near continuous current measure from a logger against the kWHr data recorded by my energy suppliers for a particular halfhourly period. The graphs produced on the Energy suppliers viewer shows vertical axis of kWhr. The graph for the day is an inverted bathtub.
So... where graph is level as say 70kWhr over a 30minute period then the average consumption is 140kW over an hour equivalent period . The next subsequent 30minute period has a 65kWhr indicating I was drawing an average 130kW. Right or wrong? As initial posting just a sanity check. As for direction, joules etc just keeping terminology in simple relant laymans terms. 



06 February 2014 05:20 PM


It's best to use engineers' terms rather than laymans' terms! It keeps definitions and therefore calculations accurate. (this is, after all, the IET and we are not laymen)




06 February 2014 05:44 PM


Indeed, but unless you have a pretty steady demand, the variance of average to peak to minimum might well be significant  so much so as to be unusable for any practical purpose  my rainfall analogy would be even more of a problem for cable sizing Regards OMS  Failure is always an option 



06 February 2014 05:49 PM


exactly! We may (fingers crossed) turn out to have a drier than average summer and autumn, which will in turn result in a mean annual rainfall below average  this means nothing for the people under several feet of water right now!
Depending on the application, you may well need to design for maximum demand (or even a safety factor of several times the maximum demand)  as oms rightly points out in cable (and guttering!) sizing. 



06 February 2014 11:55 PM


rjcomber
Yes, an average 130 kW for 30 minutes is 65 kWh. You say you want to match your "near continuous current measure from a logger against the kWh data recorded by my energy suppliers". Assuming the logger is recording at regular intervals and you can match the timestamps with those from the energy supplier readings. Then if you can get all the data into a spreadsheet, it is fairly straightforward to do a numerical integration multiplying the logger current readings (I assume you are measuring all three phases?) by the observed voltage and time between them and summing the result. The overal accuracy will depend on the accuracy of the individual logger measurements, the interval between each logger measurement and the actual instantaneous waveforms of the current (and voltage) concerned. If that is really want you want, then if you send me all the data directly in a suitable form, then for a consideration of my time and effort involved I could analyse the data in a spreadsheet(s) and send you back the results, graphs, etc. If you send me full details of the equipment concnerned, logger settings, etc. then I could give you an estimate of the accuracy that could be expected as well. I have some experience of doing just this from captured data from digital Oscilloscopes although my focus has been at about a 0.1 second to 3 second timescales rather than a 30 second one.... Also I have been working on single phase systems. But yours is a three phase system which complicates things somewhat? One final set of questions  Why do you want to try to match the two?  Are you trying to validate the suppliers meter?  If so could you send details of the meter and why you want to do so? It sounds from your earlier post that you are not technical (at least not in this field), so can I just say that there is the further complication of "Real Power" (kW) versus "Apparent Power" (kVA) and "Power Factor" to consider for other fellow Engineers..... BUT actually going back to your original post. 250A * 425V * sqrt 3 = 184 kW so for half an hour that would give you 92 kWh. You say your meter readings are about 80 kWh, so that would give you a Power Factor of about 0.87 assuming your current measurements are for the "Apparent Current" and the smart meter is reading "Real Power"..... So there is no problem?  Right?  Henry Kafeman HDK Solutions Ltd. Investigate, Analyse, Resolve Tel: +44 (0)1908 866921 



09 February 2014 01:30 PM


Henry. Thank you for your answer in last paragraph of last post. Just what wanted to get a sanity check.
As for not being technical, I do not trumpted my nominals and happy to doubt myself and bounce ideas etc. Thanks to all that responded. 



15 February 2014 01:33 PM


hi
i need some help regarding my factory. the problem is that on one machinery i have 20 hp 960 rpm motor which is taking 17 ampere and on the same other machinery 20 hp 1440 rpm motor is taking only 12 ampere. i could not get any explanation anywhere and i need to save electricity bill of my factory. without consultation if i change other motors of 960 rpm to 1440 rpm it would be huge waste of money. so i thought i should take some help from technical experts. plz help me regarding this. 


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