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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.

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  • \$\begingroup\$ ”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. \$\endgroup\$
    – winny
    Jul 4 at 18:04
  • \$\begingroup\$ @winny are electrics meters incapable of reading apparent power? \$\endgroup\$
    – Jake quin
    Jul 4 at 18:15
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    \$\begingroup\$ 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. \$\endgroup\$
    – winny
    Jul 4 at 18:22
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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.

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  • \$\begingroup\$ 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? \$\endgroup\$
    – Jake quin
    Jul 4 at 18:41
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    \$\begingroup\$ @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. \$\endgroup\$
    – DKNguyen
    Jul 4 at 18:42
  • \$\begingroup\$ 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 \$\endgroup\$
    – Jake quin
    Jul 4 at 18:49
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    \$\begingroup\$ @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. \$\endgroup\$
    – DKNguyen
    Jul 4 at 18:55
  • \$\begingroup\$ @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. \$\endgroup\$ Jul 4 at 18:55
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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.

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  • \$\begingroup\$ 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 ? \$\endgroup\$
    – Jake quin
    Jul 4 at 18:16
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    \$\begingroup\$ @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. \$\endgroup\$ Jul 4 at 18:54
  • \$\begingroup\$ 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: ednasia.com/… \$\endgroup\$
    – DKNguyen
    Jul 4 at 19:37
  • \$\begingroup\$ 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 \$\endgroup\$
    – Jake quin
    Jul 4 at 19:38
  • \$\begingroup\$ @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. \$\endgroup\$
    – Andy aka
    Jul 5 at 8:45

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