Reactive energy recovery

Could some electronics guru please shed some light on this topic.

What would be a good way to recover reactive electrical energy ? I have searched on Google, several people have claimed it is possible, but lack any details as such information is patent protected.

Any hints/suggestions would be helpful. Thanks in advance.

Reply to
dakupoto
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A Class D amplifier may be a good example. It uses an ultrasonic frequency reactive circuit and it takes some of that stored energy and recovers it to amplify audio frequency signals.

Reply to
omnilobe

What is reactive energy ?

I have heard of reactive (apparent) power measured in [VA] and reactive current measured in [A].

In a reactive circuit there is a phase shift between voltage and current and in order to extract real power from the circuit, the phase shift must be compensated so that voltage and current are in phase, This can be done by using coils or capacitors, depending of the sign of the phase shift.

Reply to
upsidedown

If you mean "reactive power" in the common sense, there is no energy to recover. Actually, there's no power either.

Anything that is "patent protected" is publically available by law. If it's patented, it's openly published by the patent office.

It all sounds bogus.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics
Reply to
John Larkin

Question too vague

NT

Reply to
tabbypurr

Somebody's pulling your leg. You don't have to "recover" it, it comes bac k all by itself.

In circuits carrying alternating current, reactive power is that power us ed to charge the dielectric in a capacitor, or create the magnetic field in an inductor, during one half of the AC cycle. During that time the capacit or or inductor act as a load and power is transferred from the supply to th e load.

For that half cycle, the power draw from the supply times the duration of the cycle could be called the "reactive energy" required to charge the cap /energize the inductor but it's not what I'd call a common phrase.

During the other half of the cycle the capacitor discharges/the magnetic f ield collapses, returning the power from the load back to the supply.

The net flow of power (neglecting losses) is therefore zero over any whol e number of cycles *for ideal capacitors and inductors*. Real caps and coil s (and wires, don't forget them) have some resistance and therefore losses that eat real power.

In other words, all you have to do is wait a little bit to get the "react ive energy" back. It just sloshes back and forth in the wires.

If you're talking about *extracting* that energy/power to do something us eful, you have do something like make the dielectric out of a piezoelectric material that converts the charging energy to sound. Similarly you can put another coil of wire into your inductor's magnetic field and steal some of it while it collapses so you can apply it to a resistive load. In both cas es it doesn't go back to the supply; it becomes real power that the supply has to, er, supply. For pure reactive loads there's no "load" in a sense- c oils and caps just store power for a little bit then give it back.

I hope you weren't chasing down some overunity nonsense. There's no "ther e" there.

Mark L. Fergerson

Reply to
Alien8752

A diode.

Tim

-- Seven Transistor Labs, LLC Electrical Engineering Consultation and Contract Design Website:

formatting link

Reply to
Tim Williams

It's all a filament of his imagination.

It probably is patented. What exactly the patent covers is another question, entirely. Design patents are pretty easy to get.

Reply to
krw

Perfectly put.

Reply to
Bob Engelhardt

ack all by itself.

used to charge the dielectric in a capacitor, or create the magnetic field in an inductor, during one half of the AC cycle. During that time the capac itor or inductor act as a load and power is transferred from the supply to the load.

of the cycle could be called the "reactive energy" required to charge the c ap/energize the inductor but it's not what I'd call a common phrase.

field collapses, returning the power from the load back to the supply.

ole number of cycles *for ideal capacitors and inductors*. Real caps and co ils (and wires, don't forget them) have some resistance and therefore losse s that eat real power.

ctive energy" back. It just sloshes back and forth in the wires.

useful, you have do something like make the dielectric out of a piezoelectr ic material that converts the charging energy to sound. Similarly you can p ut another coil of wire into your inductor's magnetic field and steal some of it while it collapses so you can apply it to a resistive load. In both c ases it doesn't go back to the supply; it becomes real power that the suppl y has to, er, supply. For pure reactive loads there's no "load" in a sense- coils and caps just store power for a little bit then give it back.

ere" there.

Yes, there is no "overunity", but there are ways to get closer to "unity". I have read some papers on coupling the proper cap to an inductive motor. In stead of just drawing pure inductive load in a typical motor. An L/C m otor draws less real energy from the grid. However, you are still paying t he same price for such lousy power factor.

Reply to
edward.ming.lee

IOW, you can save the power company money but your bill won't change.

Reply to
krw

There is some type of scam? going around selling a device to put on motors (freezers, fridges) that is supposed to save you dollars. I get a postcard offering a free meal to come and listen to the spiel. I'm thinking it may be related to that. I have not done it so that's all I know about it. Mikek

Reply to
amdx

It might be possible to build the AC-DC-AC inverter for such. However, the best way is to change it to switched L/C (magnetic/electric) field motor for top efficiency.

Reply to
edward.ming.lee

(snip)

Thank you. I was sort of explaining it to myself as I went along.

Mark L. Fergerson

Reply to
Alien8752

If this is one of those magical "attach this to your power lines and save

90% on your electric bill!!!!" things -- it's bogus.

But, post a link and we'll excoriate it in more detail.

--
Tim Wescott 
Control systems, embedded software and circuit design 
I'm looking for work!  See my website if you're interested 
http://www.wescottdesign.com
Reply to
Tim Wescott

That depends if your electric bill is measured in Watt or VA. However, it appears to be common for household connections to be metered in Watt (accumulated as kWh) and in that case there is no difference.

Industrial connections are measured in VA or have some penalty for bad PF.

Reply to
Rob

e

ck

ve

appears to be common for household connections to be metered in Watt (accu mulated as kWh) and in that case there is no difference.

It depends, this guy explains it best:

"The newer electronic meters read in the voltage and current through A/D co nverters and do the multiplication digitally. They can be programmed to rea d kW-h or KVA-h and which is used depends on your local power company. I th ink most commonly they charge you for kW-h but then they whack you with an extra charge if the power factor drops below a certain value. Since most re sidences have a power factor well above 0.9 they never see the extra charge . "

PS: except for heavy A/C usage.

.

Yes, of course.

Reply to
edward.ming.lee

Interestingly, there is a very useful class of arguments in electrodynamics based on conservation of complex power. My thesis advisor (Gordon Kino) was big on that stuff.

Cheers

Phil Hobbs

Reply to
Phil Hobbs

As often happens, multiple speculations on the OPs motivation, and no more input from the OP.

Mikek

Reply to
amdx

Who needs the OP with the SED crowd around? ;)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

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