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Perpetual motion transformer

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FlapJack

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Can someone explain these specs to me, it looks like the latest and greatest oil furnace ignition transformer put's out more power than it uses.


Background

I have been looking at ways to make my oil fired furnace more efficient. One of the things that caught my eye was the new oil ignition transformer that has a 35 VA input compared to my existing 250 VA input transformer.

The only problem is i do not understand how you get more power out than power in.


My current ignition transformer says
120v 60hz
250VA PRI
10KV 23mA SEC
Allanson 21-436
cat no 2721
type no 628
secondary mid tap grounded

Allanson #: 2721-628G
https://www.patriot-supply.com/products/showitem.cfm/ALLANSON_2721_628G

10KV x 23mA = 230 VA - This matches nicely with the 250 VA input rating.


Replacement - Beckett electronic ignition transformer
120vac 60hz 35va
output 6kv rms 20kv peak at 35mA rms
Beckett model L39-149
https://www.beckettcorp.com/product2/productdetail.asp?detailid=14

6KV x 35mA = 210 VA - compared to their 35 VA claim

Beckett is a large and respected company that has been in the oil fired furnace business for more than 60 years. So is this a typo, or am i not seeing something.
 

BradtheRad

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Perhaps the output is for a few seconds at most. Is there a hint of internal components? Which store 35VA for a minute, then release it as 210VA in a narrow space of time?
 

SunnySkyguy

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1) No such thing exists as a Perpetual Motion transformer, power supply or anything over unity in energy ( except in fairy tales and bad measurements by Test Engineering standards)
2) Nowhere is it stated the output is Sinusoidal or even constant RMS voltage under load.
In fact the Peak to RMS ( open circuit) voltage is a clue it is not a sinusoidal transformer like the former unit.
3) If the we assume 100% efficiency (90% is realistic)
... can you estimate the voltage out after the arc begins? (Hint: VA)

All arcs have a negative resistance slope characteristic but net positive resistance (like Triac) but the secondary coil could have very large resistance.

There is a bonus point for the right answer

You will not save much energy as the controller disables it until flame ignition is needed. It is just a cost Reduction design.
 

FlapJack

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https://en.wikipedia.org/wiki/High_voltage
The dielectric breakdown strength of dry air, at Standard Temperature and Pressure (STP), between spherical electrodes is approximately 33 kV/cm. This is only as a rough guide, since the actual breakdown voltage is highly dependent upon the electrode shape and size.


per Beckett drawing electrode spacing is 5/32 = .156" = 3.97 mm = .397 cm.

33 kV/cm x .397 cm = 13 KV min required to jump the gap, unless they are pointed which they are. But the original 10 KV transformer has been the standard for 60 years so it has to be a good reference value.

Also remember that in a oil fired furnace the ignition spark is on continuously with oil flow. If the ignition spark stops for just a fraction of a second the flame stops also. So i do not believe that a capacitor could be charged up and it would just get a spark every 1/4 of a second or so.

- - - Updated - - -

Thinking about BradTheRad and SunnySkyGuy's post's, The answer is probably in wave shape and duty cycle.

Assuming a switching power supply, a rectangular wave is plausible and to get the power down you would have to cut back on the duty cycle. So maybe a 10 or 20% duty square wave?

Start with this assuming 100% duty.

6KV x 35mA = 210 VA - compared to their 35 VA claim.
At 90% efficiency 233 VA

So 50% duty would be 233 VA / 2 = 116 VA

25% duty would be 233 VA / 4 = 58 VA

12.5% duty would be 233 VA / 8 = 29 VA

So

35 VA / 233 VA = .150 = 15% duty


Does this sound realistic.
 

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  • Electrode adjustment USE 5-16 NOT 7-16 for my furnace Beckett AF burner 1-16 ahead of nozzle.JPG
    Electrode adjustment USE 5-16 NOT 7-16 for my furnace Beckett AF burner 1-16 ahead of nozzle.JPG
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schmitt trigger

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Indeed your explanation sounds realistic.

If the pulse repetition rate is fast enough, it won't allow the flame to extinguish before the next pulse appears. As such, the duty cycle can be made low enough to have a low AVERAGE current consumption.
 

FlapJack

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It must be right, i came up with 15% duty and your meter is pointing to 15.
 

dick_freebird

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You have to distinguish "envelope ratings" from an actual
delivered output power and input power case. Worst times
worst on paper has nothing to do with real times real on
the bench.
 

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