Energy saver

[pwdrive]

Can't wait:

http://www.kvarenergyproducts.com/

Lawrence.

[AlanBeckett]

Reduce Amperage my life. Haven't they heard of Current.
Sounds akin to a Perpetual Motion Machine to me.

Alan

[pwdrive]

Folks could order one for every room and a sub meter for the same, if each room originally consumed about the same amount of power that means there will be no electric bill and they will get a medal from the electric company for being absolutely green.

Lawrence.

[pwdrive]

Spot the waffle without reading the comments on this video.

https://www.youtube.com/watch?v=dPFKcUxbNuQ

Lawrence.

[Ed_Dinning]

Hi Gents, domestic meters do not record KVAR, only KW.
This sound like a re-hash of an old NASA patent as a means of reducing KW consumed at the expense of a distorted waveform and yet more noise on the mains.

Ed

[pwdrive]

I had a bill from EDF once (domestic) mentioning KVAR, can't remember what was written in the different columns but it did catch my attention, I checked with previous meter readings and the end game was the same cash wise so no problem.

KVAR has never been mentioned on subsequent bills to date.

I suspect the electric companies already factor in an average/mean KVAR cost for domestic customers, nowt's for free.

EDIT: It might have been KVA and not KVAR that was mentioned on that bill, can't say which for certain but as said it did attract my attention.

Lawrence.

[Skywave]

Most of what has been written is just about correct - albeit a bit 'thin' in places. However, let's think about this extract for half a mo':

Reactive power is now reclaimed and recycled by the Energy Controller, resulting in such power being supplied locally.

A statement, which taken word for word, is very nearly correct. But when the word 'recycled' is used in that statement, recycled to whom and to where? In effect, it is 'recycled' back to where that 'power is being supplied locally'. So, in effect, the load that is being supplied has a improved power factor - so the supplier of your electrical energy can now generate less reactive volt-amps than otherwise would be the case. Hence, it is cheaper for him to generate a given amount of kVA - and with that, a given amount of kW. So, who is it that is making any savings? The generating plant, not the consumer! And the (domestic) consumer pays for kWh - kilowatt hours, not kVA - kilovolt-amps.
When I last looked at it, my electricity meter stated 'kWh', not 'kVA'. So I am changed for watts I use, not the volt-amps.

In a nut-shell, this is no more than elementary power factor correction, achievable with capacitors (for inductive loads). And that is for the benefit of the supplier, not the consumer.

Overall, then, this is a con, aimed at the ignorant and the gullible.

Al.

[Skywave]

So let's just remind ourselves how reactive volt-amps (VAr), volt-amps (VA) and watts (W) are related.

1. Watts are the result of the product of voltage across a load and the current through it - and that current is in-phase with that voltage. Hence, the load is resistive.
2. Volt-amps reactive is the product of voltage across a pure reactance and the current through it - and that current is in quadrature with that voltage.
3. Volt-amps is the product of the voltage across a load (resistive and reactive) and the apparent current through that load. That current will make an angle with the voltage - i.e., it is out-of-phase.

Mathematically, we have:
(Watts)² = (Volt-amps)² - (VAr)² . . . by Pythagoras Theorem.

So, what does all that amount to?

For the size of a given resistive load, if the reactive component is reduced, the amount of volt-amps decreases. (Draw a right-angled triangle, based on the above formula, if you in any doubts about that). And if I am measuring the current through a resistive + reactive load, that current will decrease. But the amount of watts that I am consuming stays the same. And it is watts that I am paying for. If I could ultimately reduce the number of reactive volt-amps to zero, the number of watts would be the same as the number of volt-amps. I then have all the current in my load now in-phase with the voltage across that load - i.e., I have achieved a power factor of unity, and the load is now purely resistive - or looks that way to the supplying source. It is by connecting power-factor correction capacitors across an inductive load (or inductors across a capacitive load, e.g. PSUs in PCs) that I can make my load appear to be pure resistive to that source. And that benefits the supplier, not the consumer.

Al.

[pwdrive]

So if a power correction capacitor is connected in parallel with the motor that must mean the circuit is resonant if the purpose of the capacitor is to cancel out the inductive reactance of the motor, would that present any problem?

Lawrence.

[Skywave]

Yes, the resultant circuit is resonant. Would that cause any problems, you ask. Answer: no.
It's back to basic a.c. theory to establish the validity of those remarks.

A parallel resonant circuit presents a high impedance, so the current flowing in that resultant circuit will be a minimum. For the inductive part of the load, the current in it will lag the applied voltage, but the current in the added capacitor will lead the applied voltage. By choosing the right value (µF) of capacitor, the reactance of the capacitor (Xc) just equals the reactance of the inductive part of the load (XL). Since inductive reactance and capacitive reactance are 180° apart and the magnitude of the reactances are now equal, the net reactance must be zero. In other words, resonance.
Therefore, to calculate the capacitance of the added capacitor, its just a simple matter of using the standard formula for a L/C tuned ct. Obviously, the inductance of the load needs to be known and there is the embedded assumption that that inductance will not change over the a.c. cycle or as the load on the motor (say) varies. If either do change, then a mid-value of inductance should be chosen.

Al.