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Re: Tube Coil Wattage?
From: Alfred A. Skrocki[SMTP:alfred.skrocki-at-cybernetworking-dot-com]
Sent: Friday, November 14, 1997 5:16 PM
To: Tesla List
Subject: Re: Tube Coil Wattage?
On Friday, November 14, 1997 12:13 PM Andrew Chin
[SMTP:chinny-at-ozemail-dot-com.au] wrote;
> Wouldn't the power of the coil be equal to the power drawn from the
> mains?
Never, there are far too many looses; the spark gap being probably
the bigest.
> Granted, a Tesla coil is far more efficeint than a conventional
> transformer (>100%), so maybe you could say that the power of the coil
> is at LEAST the power drawn from the mains?
Huh, there ain't no such thing as a free lunch sport. You will NEVER get
out more than you put in, in fact you will never get out even AS MUCH as
you put in. There are ALWAYS going to be losses; resistive, radiated, heat,
light, ect. all of these take power from the system. BTW Tesla coils are
far less efficient than conventional transformers, conventional
transformers can approach 95 percent efficiency I seriously doubt that
any Tesla coil ever got over 50 percent if that. At the voltages and
frequency Tesla coils are operated at losses are inherently high, also
the losses are minimized in conentional transformers by their closed
magnetic circuit while Tesla coils being air core are open magnetic
circuits so the magnetic losses are going to be high as well.
> If a Tesla coil was a conventional transformer, then power calculations
> would be so much easier. All you would need to do is use the power
> formula, Power=current x voltage. A transformer will change current and
> voltage, but the input power will always equal the output power. (For a
> 100% efficient transformer of course)
The reason it is easy to calculate the power in or out on a 'standard'
transformer is because you are dealing with a continous sine wave, in
a Tesla coil you are dealing with a dampened sign wave at best and the
problem is compounded by the difficulty in measuring the output voltage
of a tesla coil. To calculate the power you would have to know the exact
wave shape of the output and calculate the area under the curves formed,
a real calculous nightmare.
Sincerely
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Alfred A. Skrocki
Alfred.Skrocki-at-CyberNetworking-dot-com
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