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Suppose I have a single phase system and I have a sensor attached to to measure both voltage and current. The sensor samples for 50 kHz and the data is saved to a file.

When we then graph out the file we get something like this:

enter image description here

In the beer analogy, power factor is equal to true power multiplied by apparent power. Since the units are in kilowatts I suppose we have to multiply the voltage reading with the current reading.

enter image description here

What is then the result when we multiply the current and voltage data? Is it true power? Apparent power?

I doubt that it is apparent since I have read that establishments are billed for power factor. If the result is the true power, where is the reactive power on the graph? If the result of the multiplication is already the apparent power, then is the meter incapable of reading reactive power? I don’t think so since there are digital meters and they should have information like the graph above.

JRE
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Jake quin
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  • _”Since the units are in kilowatts”_ No they are not as you can see in your beer analogy picture. Apparent power is measured in (k)VA. Most utility companies do not bill individuals for apparent power consumption, only real power. – winny Jul 04 '21 at 18:04
  • @winny are electrics meters incapable of reading apparent power? – Jake quin Jul 04 '21 at 18:15
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    Opposite. The are most likely able to measure it but in most places in the world, apartments and houses (individuals) does not pay for apparent power. Your utility company may differ. Call and ask. – winny Jul 04 '21 at 18:22

2 Answers2

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I find the most illustrative way is to get an out of phase V and I graph, and multiply them together to get instantaneous power graph.

enter image description here Modified from: https://learn.openenergymonitor.org/electricity-monitoring/ac-power-theory/introduction

In the power graph, everything is real power flowing in one direction or the other at any point in time. You can see sometimes positive, sometimes negative.

Here is the illustrative part:

Add up (integrate) all the positive parts. That's all energy flowing from source to load. Dividing that by your time interval T gives you the power going from source to load.

Add up (integrate) all the negative parts. That's all energy flowing from load to source. Dividing that by your time interval T gives you the power going from load to source.

It's all real power flowing in the end, but not always is it being dissipated to do work. The difference between the two tells us the real power actually being supplied or generated.

You can see over time some power just flows to the load and back again, doing no work but still consuming ampacity. This is the reactive power. It's the power flowing from source to load which has an equal amount of power flowing from load to source with which to cancel out.

If I have 1000W of real power flowing to the load over time interval T, and 300W of real power flowing back to the source over T, only 700W of real power is dissipated/consumed by the load to do work. The 300W flowing back and forth is the reactive power.

If you average the power graph over some time interval, to get a single average power number, that is the same as the real power (or net power) delivered or generated.

Apparent power is the maximum real power delivered we would get if we fiddled with the phase alignments and is just used as a reference number (the theoretical best). Power factor is a measure of how close we are to that number with our power delivery.

DKNguyen
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  • Thank you i find your explination quite helpful, i do still have some questions though. so the negative power is actually power that is "returned" to the grid am i correct? – Jake quin Jul 04 '21 at 18:41
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    @Jakequin It is the power flowing in the opposite direction of what you defined as positive. By convention, positive usually means source to load, so yes. – DKNguyen Jul 04 '21 at 18:42
  • Thats makes sense that you are going to be billed by only the positive side of the graph( real power). Then why is having a bad power factor affects the bill of large industrial buisnesses, why are the electric company charging them since those power on the negative graph is essientally returned to the grid – Jake quin Jul 04 '21 at 18:49
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    @Jakequin You can see that real current is always flowing through the wires in one direction or the other, and real current requires real wire to carry. And current produces heat in wires regardless of the direction it is flowing in. It consumes the ampacity of the utility's lines for useless roundabout power flow which could be used for other purposes. – DKNguyen Jul 04 '21 at 18:55
  • @Jakequin Large industrial consumers are not billed in the same way as residential customers. There's no reason to think that they use the same kind of meters that small business and residential customers use. – Elliot Alderson Jul 04 '21 at 18:55
  • Like returning things bought via mail order. Does it take resources to return the item even though the seller is getting the item back? You bet. Does it waste the seller's time preparing the package in the first place for delivery and then processing it again when it's returned? You bet. The seller could have been processing other orders that the buyers would actually keep and that the seller would actually make money from. You'd go out of business if you spent too much time, staff, and money sending out orders only to have them come back unless you charged for the return. Restocking fees. – DKNguyen Jul 04 '21 at 19:01
  • i see so portion of it is lost on the wires and not all of it get will make it to other customers. although small it will add up eventually – Jake quin Jul 04 '21 at 19:02
  • @Jakequin A portion is lost on the wires but I imagine the bigger concern is that it eats up capacity in the lines which prevents those lines from delivering power to other customers. – DKNguyen Jul 04 '21 at 19:02
  • Thank you very much i have pretty much grasp the concept now, also here we have "line losses charge" on our bill you could say this is how electric company saying to pay up for power disspated to those lines – Jake quin Jul 04 '21 at 19:05
  • _the bigger concern is that it eats up capacity in the lines which prevents those lines from delivering power to other customers._ oh ? how so ? is line capacity the same as the maximum current a wire can handle? – Jake quin Jul 04 '21 at 19:06
  • @Jakequin Well, if 10A is coming back to me on the line, I very well can't use that 10A capacity to send power out to another customer. And I'm paying for more a bigger line to support that customer. – DKNguyen Jul 04 '21 at 19:07
  • Oh so all of it is going to be dissipated into heat? – Jake quin Jul 04 '21 at 19:10
  • No. I own a one-way toll road. Am I going to charge you for driving back and forth on it even though you are going nowhere? Of course. Forget heat dissipated. You are taking up space on my road that other drivers could be paying me to use. Or I need to make a bigger road to accommodate you and them at the same time which also costs money. Either way, I pass the cost to you. – DKNguyen Jul 04 '21 at 19:12
  • does line capacity means the maximum current rating of the wire/transformers. if go by your analogy that seems to be the case – Jake quin Jul 04 '21 at 19:18
  • @Jakequin Yes, by capacity I am referring to ampacity in this case. – DKNguyen Jul 04 '21 at 19:19
2

What is then the result when we multiply the current and voltage data? Is it true power?

After averaging your waveforms of \$v\cdot i\$, the result is both true power and apparent power because, the voltage and current waveforms appear to be exactly in-phase. Apparent power is when you use an RMS reading voltmeter and an RMS reading ammeter and multiply the two RMS values together. Only when power equals unity does true power equal apparent power (like the example in your question).

I doubt that it is apparent since I have read that establishments are billed for power factor.

Most households are billed for true power and that is quite simply the average value of the composite waveform of \$v\cdot i\$. Establishments are not billed for power factor because it is meaningless to do so. Some establishments are billed for their use of reactive power because it contributes to larger \$i^2\cdot R\$ losses in the distribution system when delivering a certain level of true power to the customer.

In the beer analogy, power factor is equal to true power multiplied by apparent power.

No, power factor is true power divided by apparent power.

Additionally, the beer glass analogy has to be treated with some caution because the level of foam plus the level of beer aren't linearly related to their equivalent powers. You need to use the power triangle to understand that and not some beer-driven (and quite possibly drunken) linear addition: -

enter image description here

Image from here.

Pythagoras is the trick and not a linear sum of quantities in a beer glass (despite it being a very, very nice drink when watching certain UEFA Euro quarter finals on the TV and usually at all other times except breakfast).

From comments: -

what do electric companies bill you for

Normally true power but, certain companies will bill higher usage customers an extra charge based on poor power factor performance.

dont digital meters read current and voltage simultaneously

They measure current and voltage simultaneously. If they didn't the measurement would have a small phase angle error and the reading could be wrong.

what is that number on the meter

What meter, what setting, what number?

Footnote about power waveforms based on \$v\cdot i\$ and different phase angles: -

enter image description here

Picture taken from this answer and this answer and this answer and this answer and this answer and this answer and this answer and this answer to name a few.

I suggest you read them to get a wider perspective.

From an other answer's comment: -

makes sense that you are going to be billed by only the positive side of the graph( real power).

You are not billed on positive half of the power waveform, you are billed on the average level of the power waveform.

Andy aka
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  • Im still having trouble grasping the concept, let me first ask what do electric companies bill you for ? (aside from those extra charges) is it true or aparent? dont digital meters read current and voltage simultaneously>? what is that number on the meter ? – Jake quin Jul 04 '21 at 18:16
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    @Jakequin As winny commented earlier, in most locales you will be billed for only the real (true) power that you use. Your meter likely displays the real power. Whether or not digital meters read current and voltage simultaneously is not relevant to the question of how you are billed or what the meter displays. – Elliot Alderson Jul 04 '21 at 18:54
  • Technically you don't need to sample voltage and current simultaneously, but ultimately you do need data points for both at the same instant a la signal processing voodoo to "interpolate" the points via point stuffing: https://www.ednasia.com/sample-multiple-channels-simultaneously-with-a-single-adc/ – DKNguyen Jul 04 '21 at 19:37
  • Im sorry i got added to the list person you have to re-explain this stuff hehe. _You are not billed on positive half of the power waveform, you are billed on the average level of the power waveform._ what happens when the avverage power goes negative as the graph on the lower right shows – Jake quin Jul 04 '21 at 19:38
  • @Jakequin that's when a customer is over-producing power and shipping it back on to the power grid. Many utilities allow this and you can get paid (as a customer rather than be charged) for delivering surplus power to the grid. Of course the price for buying electricity may be higher than what you can sell it for but, that's a different thing to the engineering. – Andy aka Jul 05 '21 at 08:45
  • @Andyaka digital meters aside, how will analog electric meters react to this ? from what i know about analog meters they use a magnetic field to turn the disk, so current is the only factor (*) at play, Is that why if you want to sell back to the grid you need another meter? – Jake quin Jul 05 '21 at 09:18
  • @Jakequin I don't know where you get your info from but, even old fashioned analogue rotating-disc power meters would turn in reverse for negative power. Anyway, you have accepted an answer so we should terminate this. – Andy aka Jul 05 '21 at 09:29