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Electric Power measurement Current Transformers and reactive power

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w0z
#1
Jan11-10, 04:18 PM
P: 4
Apparently my modern electronic home electric meter uses a 'current transformer' to measure the current in order to calculate the wattage I am using.
My question: is the flux generated in the current transformer also proportional to the reactive part of the current as well as the in-phase current or is the reactive power somehow ignored?
Accepted wisdom seems to that power companies do not charge homeowners* for reactive (apparent) power, but only for real power, so how do the modern electronic meters discriminate between the two.
(*I understand that the power companies reserve the right to adjust their charges for factories and industrial users etc. who have persistently poor power factors, because they still have to generate the reactive power)

Thanks, w0z
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russ_watters
#2
Jan11-10, 05:46 PM
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Electric meters do not have to sense both and discriminate - the meters measure wattage by combining that amperage measured with the CT with the voltage. And if you make no allowance for the voltage and amperage being out of phase, you'll only get the active power.
w0z
#3
Jan11-10, 06:19 PM
P: 4
Quote Quote by russ_watters View Post
Electric meters do not have to sense both and discriminate - the meters measure wattage by combining that amperage measured with the CT with the voltage. And if you make no allowance for the voltage and amperage being out of phase, you'll only get the active power.
Thank you for the reply but I dont think it really answered my question, and I'm sorry if I didn't make my question clear. What I am asking is whether the current transformer generates flux from the part of the current that is out of phase with the voltage, and if it doesn't then why doesn't it OR if it does then surely the phase of the voltage relative to the current (at the instant of measurement of course) must be taken into account when calculating the wattage IF it is to indicate real power only.

Wikipedia on electric meters is somewhat vague on this question, but does clearly say that solid state meters can measure PF and apparent power, not that they necessarily do. In any event my question is one on the physical level of how a current transformer operates. By the way for the sake of simplicity I am assuming that no clipping of the sinewave occurs.

Snoogans
#4
Jan11-10, 06:55 PM
P: 36
Electric Power measurement Current Transformers and reactive power

A current transformer will basically operate according to amperes law, that is a current is induced proportional to the current in the cable under measurement. I think the part that may be confusing you is that CT meters are also provided a Voltage reference from the same conductor which allows power factor to be calculated.

It is likely to be a law or something similar that the electricy suppliers only charge for real power. It is generally required that power factor be above a certain value to protect the suppliers and the cost of reactive power is likely to be built into the cost anyway. This would have originated from whole current kilowatt-hour meters that could not measure reactive power that were widely used in the past.
w0z
#5
Jan11-10, 07:47 PM
P: 4
[QUOTE=Snoogans;2527164] I think the part that may be confusing you is that CT meters are also provided a Voltage reference from the same conductor which allows power factor to be calculated.

Thanks, that was what I said in the last post in a roundabout way. I don't think I'm confused, I think I'm either ignorant or stupid. On the other hand it is possible that I might just be too confused to know if I'm confused or not!
So am I correct in thinking that at a physical level in terms of flux generation, the CT will produce an output which is proportional to all the current flowing through it, both real and reactive, and it's only by taking into account the phase of the voltage that the real (i.e. non reactive) power is calculated? I'm sorry to say it's a long time since I did my studies and they didn't include current transformer theory!
regards
w0z
russ_watters
#6
Jan11-10, 09:56 PM
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Quote Quote by w0z View Post
Thank you for the reply but I dont think it really answered my question, and I'm sorry if I didn't make my question clear. What I am asking is whether the current transformer generates flux from the part of the current that is out of phase with the voltage, and if it doesn't then why doesn't it OR if it does then surely the phase of the voltage relative to the current (at the instant of measurement of course) must be taken into account when calculating the wattage IF it is to indicate real power only.
The current transformer measures the total current, not just the in or out of phase part.
So am I correct in thinking that at a physical level in terms of flux generation, the CT will produce an output which is proportional to all the current flowing through it, both real and reactive...
Correct.
...and it's only by taking into account the phase of the voltage that the real (i.e. non reactive) power is calculated?
Well, no. It doesn't need to take into account the phase of the voltage specifically. When you mutiply the instantaneous amperage and voltage, you get real power exactly because they are out of phase with each other.
russ_watters
#7
Jan11-10, 10:06 PM
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From the wiki:
The electromechanical induction meter operates by counting the revolutions of an aluminium disc which is made to rotate at a speed proportional to the power. The number of revolutions is thus proportional to the energy usage. It consumes a small amount of power, typically around 2 watts.

The metallic disc is acted upon by two coils. One coil is connected in such a way that it produces a magnetic flux in proportion to the voltage and the other produces a magnetic flux in proportion to the current. The field of the voltage coil is delayed by 90 degrees using a lag coil.[5] This produces eddy currents in the disc and the effect is such that a force is exerted on the disc in proportion to the product of the instantaneous current and voltage. A permanent magnet exerts an opposing force proportional to the speed of rotation of the disc. The equilibrium between these two opposing forces results in the disc rotating at a speed proportional to the power being used. [emphasis added]
http://en.wikipedia.org/wiki/Electri...hanical_meters
sophiecentaur
#8
Jan12-10, 05:15 AM
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If your meter measured the (average) value of the Current and multiplied it by the (average) value of the Volts, you would get an answer which corresponds to the so-called VA. This is useful to know but doesn't tell you the Power you are consuming.
Multiplying the instantaneous values of V and I and then averaging, will tell you the Energy consumed. Electronic meters can do either, quite easily, using the values of actual I (from a current transformer) and V .
I think it's an historical thing. The rotating disc meter actually does this instantaneous multiplication sum which produces the Energy consumed. The regulations and the billing has been done based on this information, which sounds like just what you want. But it ignores the problems (and costs) associated with poor Power Factors. There are requirements for large equipment users to keep their Power Factor reasonable in order to limit the extra losses which are associated with V and I being out of phase. I believe you have to pay on the basis of Max VA, in some circumstances, because this has implications on the spec of the supply equipment.
w0z
#9
Jan15-10, 06:29 PM
P: 4
...and it's only by taking into account the phase of the voltage that the real (i.e. non reactive) power is calculated?

Well, no.
It doesn't need to take into account the phase of the voltage specifically. When you mutiply the instantaneous amperage and voltage, you get real power exactly because they are out of phase with each other.


Don't you mean "well yes"? Insofar as it clearly does take into account the instantaneous phase of the volts relative to the current (or vice versa), however I think it's just a matter of how you or I see it.

Thank you to all for the replies.


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