Quick FE by resonance

[wizkycho]

Hi all !

Lets say that R1 (and R2) represents electric heater of say 48 Ohm (1kW at 220V mains).
Is it possible to make resonant circuit by adding Capacitor and Inducor and make complete resistance bigger, drawing
less current and have high current through heater.

I tried to calculate this, but two methods gives me different results.

How to calculate this properly ? For resonant freq. 50 or 60 hz what would be the values of L and C ?
Which has better chance to lower payment bill for heating 1 or 2 ?

Wiz

[z.monkey]

http://en.wikipedia.org/wiki/LC_circuit

[wizkycho]

http://en.wikipedia.org/wiki/LC_circuit

Yes it would be nice if so, there is an significant R (50Ohms) (not just small parasitic) and will certanly prolong Tau of charging capacitor so the res. frequency is changed.

Anyone sees any posibility with this circuit or simmilar?
Electrical Heaters have pretty much constant resistance and is stable value through wide temperature range. (depends on fillament)
Scince resistance is stable it should be easy to put heaters in resonance. (would be of course nice without additional trafo)
And we spend so much money and energy on heating.
Can someone develop some formula or experiment (someone who has bipolar caps of enough capacitance) ?


Wiz

[gyulasun]

Hello Wiz,

The principle you would like to use is similar to the rotoverter from Ark Research (Hector). I would choose your circuit setup shown as 1.

Here is an online calculator for the LC values: http://www.cvs1.uklinux.net/cgi-bin/calculators/tuned_circuit.cgi

First write 5 for the inductance and choose 1 for its multiplier on the right, this will be 5 Henry then.
Leave empty the capacitor window.
Write 50 for frequency and also choose 1 for its multiplier, this will be 50 Hz then.
Write 48 for the resistance and also choose 1 for its multiplier, this will be 48 Ohms.  Then press Calculate Now.

I got 2.02... uF for C and 32.7.. for Q. Now if you play with the inductance numbers by increasing or decreasing, you will see how Q will change (the higher the L value, the bigger Q you get of course).

Now there is a matching problem to the mains you have to solve somehow because if you connect directly your circuit shown as 1 to the 220V AC mains, then the calculated Q will be attenuated to around 1 so you end up with a low Q resonant circuit and every reactive current multiplication due to the high Q will be lost. The mains has a very low inner impedance by default.

A possible solution is to use a very hefty, at least 1.5-2kW transformator like a welding transformator. And you need unfortunately two transformers. The first connects directly to the mains and steps down it to a low voltage which feeds the welding transformer low voltage coil (usually it is its original secondary coil). And you use the welding transformer original primary coil as your L in your circuit shown as 1.  Use motor run capacitors for the C with 500-600V rating at least.
The best is to meause the welding or any high power 1-2kW mains transformer primary coil inductance and choose the C accordingly.  The first transformer you connect to the mains should be rated for approximatly Q times less power (so for 1kW heater and with a resonant Q of say 10 the first transformer should be rated for at least 100W).

I have not built such circuit but in theory this should work if we accept that reactive currents inside a parallel resonant circuit dissipate power in heat form in a resistor just like a normal current does.

(A good thing with the heater is that it does not change its resistance once it is heated up, hence the resonant Q value is more or less constant, once it is tuned properly, so I mean the load is constant, unlike to a rotoverter where the electric motor or generator tuned to resonance would need a continuous retuning of the C value whenever the load on the motor or generator shaft changes.)

rgds,  Gyula

[wizkycho]

Now there is a matching problem to the mains you have to solve somehow because if you connect directly your circuit shown as 1 to the 220V AC mains, then the calculated Q will be attenuated to around 1 so you end up with a low Q resonant circuit and every reactive current multiplication due to the high Q will be lost. The mains has a very low inner impedance by default.

I don't quite follow this...I know that efficiency is highest when source and load impendances mach. but somehow I don't see this situation
within mains AC as source.

But wouldn't the nature of simple LC resonator render less amperage draw from mains (at res freq.)? and according slow down the speed of aluminium plate counter ?

Wiz