Meter with two consumer-side connections

[winston1]

It's quite common for repositioned meters (such as those lowered for health and safety reasons) to have only one neutral connection. The other terminal being fitted with a golf tee.

OK. What is a golf tee in this context?

 

[Echo the husky]

http://www.cdg-uk.com/product-secto...ctricity-meter-security-plugs-blanking-plugs/

 

[ban-all-sheds]

... such as those lowered for health and safety reasons ...

Lowered into the flood zone, you mean?

 

[ban-all-sheds]

Rubbish. The meter measures watts

No, it measures volt-amps.

and needs to know current, voltage, and phase angle.

It can measure current without a N connection. It can assume a supply voltage. It has no idea about phase angles.

How can it determine voltage without a neutral connection?

It doesn't.

 

[wundaboy]

Oh dear, didn't realise I'd cause such a kerfuffle! What I meant to express, albeit rather inelegantly, was that current is measured only in the live conductor not the neutral - the neutral terminals are linked with a brass block or a copper braid inside the meter terminal box, as seen in this video of an Ampy meter teardown. Therefore it doesn't make any difference whether the supply neutral passes through the meter or not, though the meter does of course need a neutral connection as a reference point.

 

[Spark123]

But the neutral connection always does pass through the meter.

It doesn't always, as many of us know.

 

[EFLImpudence]

I didn't see anything wrong with wundaboy's original post.

I think the confusion with this has arisen because people are are thinking that "The Neutral" is the installation neutral tails.

It is not necessary for this/these to be attached to the meter at all but obviously "A Neutral" has to be connected to the meter for it to function. This could be any small conductor and a similar small neutral conductor does indeed pass through the meter to the neutral side of its workings.

 

[DetlefSchmitz]

No, it measures volt-amps.



It can measure current without a N connection. It can assume a supply voltage. It has no idea about phase angles.



It doesn't.

This is the second misleading post from BAS that I have read today. This is not a good omen. Domestic meters measure watts, only industrial meter measure VA, because as large users they have to be responsible for keeping their own PF low. Domestic import meters do not assume the voltage, they measure it.

 

[ban-all-sheds]

This is the second misleading post from BAS that I have read today. This is not a good omen.

Sorry - it was a very badly worded post - I failed to use either ?s instead of .s to indicated possible mechanisms, or the word "could".

What I meant to say was that (in theory) a meter could perfectly well just measure current and use a nominal voltage.

Domestic meters measure watts, only industrial meter measure VA, because as large users they have to be responsible for keeping their own PF low.

I think you've got your W vs VA the wrong way round...

 

[ban-all-sheds]

This is the second misleading post from BAS that I have read today.

OOI, which was the other?

 

[winston1]

No, it measures volt-amps.

It can assume a supply voltage. It has no idea about phase angles.

Domestic electricity meters measure watts NOT VA.

Domestic electricity meters do not assume a supply voltage they measure it. They can determine phase angles as well otherwise they could not measure watts.

 

[DetlefSchmitz]

They can determine phase angles as well otherwise they could not measure watts.

Actually I suspect that is not quite true. Phase angles assume pure sine waves. I suspect that the meters calculate a number of samples per mains cycle of instantaneous volt * amp measurements and add them together.This would be simpler and result in a wattage value. The current (and indeed voltage) waveform shape would then not matter. It could derive a notional phase angle from that, but there wouldn't be much point.

 

[ban-all-sheds]

I suspect that the meters calculate a number of samples per mains cycle of instantaneous volt * amp measurements and add them together.This would be simpler and result in a wattage value.

No - it will result in a sum of VA values.

 

[endecotp]

Mechanical meters measure true power. They integrate the instantaneous product of current and voltage.

Quoting https://en.m.wikipedia.org/wiki/Electricity_meter :

The electromechanical induction meter operates by counting the revolutions of a non-magnetic, but electrically conductive, metal disc which is made to rotate at a speed proportional to the power passing through the meter. The number of revolutions is thus proportional to the energy usage.

The disc is acted upon by two sets of coils, which form, in effect, a two phase induction motor. 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, due to the coil's inductive nature, and calibrated using a lag coil.[16] 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, voltage and phase angle (power factor) between them. A permanent magnet acts as an eddy current brake, exerting 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 or rate of energy usage. The disc drives a register mechanism which counts revolutions, much like the odometer in a car, in order to render a measurement of the total energy used.

 

[ban-all-sheds]

The electromechanical induction meter operates by counting the revolutions of a non-magnetic, but electrically conductive, metal disc which is made to rotate at a speed proportional to the power passing through the meter.

Power?

Or the product of voltage and current?