6
\$\begingroup\$

I've searched everywhere for this but can't find the answer...

Its possible to have 3 phase electricity meters that measure both imported and exported kwh with each value being recorded in separate registers.

How does the meter differentiate between imported and exported energy? Doesn't it just see voltage and current going up and down?

\$\endgroup\$
11
\$\begingroup\$

Voltage and current waveforms have an amplitude but they also have a phase relationship: -

enter image description here

The left picture shows voltage and current waveforms in phase i.e. they rise and fall together. The right hand picture illustrates a phase lag on the current.

The red curves for both is power and even though the amplitudes of voltage and current are the same in both pictures, the average power in the right hand picture (what you are billed on) is lower.

Here are more extreme examples: -

enter image description here

The left picture shows current lagging voltage by 90 degrees and the average power is zero (there are equal amounts of power taken from the supply as given back to the supply). The right picture shows a negative average power because the current's phase angle has moved to a point where it is generally inverted i.e. it's positive peaks are closer to the inverted voltage waveform's negative peaks. This is an example where power is flowing out from a "user" and back into the power grid.

How does the meter differentiate between imported and exported energy?

Imported energy is when the current is flowing into the load as typified by the top two diagrams. Average power is positive. Exported power would be a negative value of average power as typified by the lower right hand diagram.

For sine waves, reactive power is the average of VI sin(\$\phi\$). True power is the average of VI cos(\$\phi\$). So, as soon as the phase angle (\$\phi\$) is not zero you get some reactive power and this peaks when \$\phi\$ = 90 degrees.

Pictures taken from here and here.

The above explanation is for a single phase but the same applies to 3 phase systems.

| improve this answer | |
\$\endgroup\$
  • 2
    \$\begingroup\$ Ha! I was waiting for your well-diagrammed answer to this one! \$\endgroup\$ – Neil_UK Aug 14 '17 at 12:53
  • \$\begingroup\$ In examples shown, aren't voltage and current out of phase.. ie isnt this reactive power as opposed to exported active power? \$\endgroup\$ – dazza Aug 14 '17 at 14:13
  • \$\begingroup\$ @dazza when average power is not-zero it is real power (measured in watts) either flowing in or flowing out. This is v x i. Reactive power will be zero when v and i are in phase and will be maximum when v and i are at 90 degreees. See this: electronics-tutorials.ws/articles/reactive2.gif. Because of the cosine/sine relationship reactive power is maximum when real power is zero. At phase angles other than 0 degrees and 90 degrees or multiples thereof, reactive and real power both exist. \$\endgroup\$ – Andy aka Aug 14 '17 at 14:55
  • \$\begingroup\$ @Andyaka OK Im struggling. Before you used voltage leading current as an example of exporting. To me that just looks like a lot of reactive power as current and voltage are out of sync. Our generation equipment runs with a PF of 0.99 to 1.0 which I interpret (perhaps incorrectly) to mean we're not creating or consuming a lot of reactive power to/from the grid, ie current and voltage in phase... so again.. with current and voltage in phase how do you know which way power is going? \$\endgroup\$ – dazza Aug 15 '17 at 15:37
  • \$\begingroup\$ Exporting (as in giving average power back to the incoming supply) is when the current lags voltage between 90 degrees and 270 degrees. This naturally causes average power to be negative. Biggest export is when current is exactly 180 degrees out of phase with voltage. \$\endgroup\$ – Andy aka Aug 15 '17 at 16:35
2
\$\begingroup\$

Found a perfect illustration from an experiment. This trace shows two voltage sources connected via a resistor - one (call it Va) fixed at 170V the other (Vb) increasing from about 75V up to 250V. Both voltages are in-phase. On the left hand side current is in-phase with voltage (lets call it import), When both voltages are 170V there is no net flow and then as Vb increases the current phase shifts to exactly 180 deg (call that export). Voltage is the green line and current is blue.

enter image description here

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ I'm really sorry for the pedantry, but the spelling bugs me here. A "resister" is someone who resists. A "resistor" is the circuit element which follows Ohm's law. Please edit. Otherwise, great illustration, +1. \$\endgroup\$ – Dmitri Aug 16 '17 at 15:39
  • \$\begingroup\$ @Dmitri I can only apologise for my auto-correct and not noticing it! I have now corrected it :) \$\endgroup\$ – dazza Aug 16 '17 at 15:53
  • \$\begingroup\$ ..only thing is Im now trying to get my head around how (if at all) this relates to real, apparent and reactive power! Even then it may depend on whether resistive or inductive load.. Clearly Im not an electrical engineer, but hard to find people who can actually explain all this! \$\endgroup\$ – dazza Aug 16 '17 at 15:56
  • \$\begingroup\$ Oh I get it... reactive power is 90 deg phase shifted. \$\endgroup\$ – dazza Aug 16 '17 at 15:58
  • \$\begingroup\$ For sine waves, reactive power is the average of VI sin(\$\phi\$). True power is the average of VI cos(\$\phi\$). So, as soon as the phase angle (\$\phi\$) is not zero you get some reactive power and this peaks when \$\phi\$ = 90 degrees. \$\endgroup\$ – Andy aka Aug 16 '17 at 16:25
1
\$\begingroup\$

The main thing you're missing is that the phase of the current relative to the voltage is important. Just the RMS voltage and RMS current alone can't tell you which way power is being transferred, if any at all.

The instantaneous power being transferred is the instantaneous voltage times current. Note that both of those are signed quantities. If the voltage and current have the same polarity, then the power is positive. If they differ, it is negative.

The accumulated transferred energy is the time-integral of the above. This is what electric meters measure.

Old meters did the integration by putting the voltage and current on two separate coils that effectively were the rotor and stator of a motor. The mechanical movement averaged the instantaneous power thru inertia. The total energy was accumulated by keeping track of the total number of revolutions of the motor. This was done with gears so that there were a series of dials that each moved 1/10 the speed of the previous. This allowed for reading out the energy value as decimal digits.

Modern meters measure the voltage and current many times per second. These two are multiplied and the result digitally accumulated. Readings are taken 100s to 1000s of times per power line cycle so that the system doesn't get confused by harmonics.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Hi, what exactly do you mean by polarity in this context? \$\endgroup\$ – dazza Aug 14 '17 at 14:24
  • \$\begingroup\$ @daz: Same as the sign. \$\endgroup\$ – Olin Lathrop Aug 14 '17 at 14:39
  • \$\begingroup\$ what sign please? \$\endgroup\$ – dazza Aug 15 '17 at 15:41
0
\$\begingroup\$

Doesn't it just see voltage and current going up and down?

It does see voltage and current going up and down, but it doesn't just see voltage and current going up and down? It also see in what polarity they go up and down.

When current is in phase with the voltage (as when a battery supplies a load), then the meter knows the power is being imported.

When current is out of phase with the voltage (as when a battery is being charged from a supply), then the meter knows the power is being exported.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Would I be right in thinking that during import the current might lag voltage, but during export the voltage would lag current? \$\endgroup\$ – dazza Aug 14 '17 at 12:01
  • \$\begingroup\$ @dazza No, that would be positive or negative VAr, real power is always in phase. \$\endgroup\$ – Neil_UK Aug 14 '17 at 12:51
  • \$\begingroup\$ Hmm, Im not sure about this. Any time current is out of phase with voltage... thats reactive power isnt it? Active power (imported or exported) would have voltage and current in-phase. \$\endgroup\$ – dazza Aug 14 '17 at 14:22
-1
\$\begingroup\$

mostly the export measurements are done in Q3, Q4 coordinates and import measurements are done in Q1,Q2 coordinates the sign of import parameters are negitive while export parameters are possitive

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ It's unlikely that talking about Q1 to Q4 is going to be much help without a picture explaining those concepts. Also the sign that you mention can be the exact opposite of what you state because it is arbitrary - providing the sign is opposite for import compared to export the system can work. For instance most "users" expect to see a positive power on their bill for import i.e. the opposite to what you said. \$\endgroup\$ – Andy aka Aug 14 '17 at 12:28
  • \$\begingroup\$ import refers to the power which a substation intakes from solar panels or wind energy while export refers the power given by substation to household which our meters gives the amount of power consumed regarding my point \$\endgroup\$ – sunil chowdary chaganti Aug 16 '17 at 5:10

Not the answer you're looking for? Browse other questions tagged or ask your own question.