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Re: SSTC does 10 foot sparks
Original poster: Sean Taylor <sstaylor-at-uiuc.edu>
John,
I'm sure there will be many people replying to this, so I'll keep it as
short as possible :-)
>I agree you cannot "get more average power out than what is coming in".
>However, it is very possible to get much more pulse power out than average
>power in.
I completely agree, this is the whole thing with capacitor discharges, you
can charge them at a low rate, and discharge much faster to create high
peak currents. This concept is used all over the place in technology.
With Steve's TC it appears that the average power in is about
>4800
>watts and the pulse power out is about 300 KW giving a power gain of 62.5 .
The concept of "power gain" is a VERY misleading one. Power is DEFINED as
energy transfer per unit time, so by definition when comparing two powers,
unless otherwise stated, you are comparing a total power transfer. The
difference is when you consider peak power, which is the instantaneous RATE
of energy transfer.
>With a potential power output of 300 KW it is obvious that a very
>long spark would be possible depending on the TC design. That is why using
>power instead of energy units is not good for rating Tesla coils. It can
>exagerate the output possibilities of a coil so you have to be very specific
>about the input conditions.. If you use energy units you will not have this
>type of problem. The energy output units will always be less than the input
>units.
Not necesarily true. Energy output at a chosen time (perhaps between
bangs) will be much less, about 0, than the input energy. Also, saying
input or output energy entails energy transfer, implying a rate, not just a
quantity of energy. With any sort of energy storage device, energy in and
out can be very different from each other.
>"Peak power out will be larger than peak power in" is another example of the
>confusion caused by using power with Tesla coils.
I wouldn't call this a confusion so much as a difference of measurement
techniques.
"Peak power out"
>cannot be
>larger than "peak power in" unless there is a time difference between the
>two
>stated powers.
Peak power out and peak power in can be very different, and either one can
be greater than the other. I think what you mean by "unless there is a
time difference . . . " is that the total time that the power is measured
over is different for the input and output. You can have a single spike of
power at say 10W, and measure for however long you want, and still only
have a 10W peak, the time doesn't matter. I think you are confusing
integrating over the two times (yielding energy) rather than recording the
peak power transfer.
This means bringing in time into the process which gets
>you
>into an energy process.
Not really, depending on how you use the time. Dividing by time will give
an average power transferred per time, multiplying/integrating will give
you an amount of energy transferred.
It would be preferable to say that "Peak power
>out
>will be larger than average power in". This still requires more
>explanation. The time period involved in the output vs the time period
>involved in the input. And we are back again into energy out vs energy in.
A peak power is an instantaneous event, there is no measurement over time
for the peak. It happens, and it's over with, there is no amount of time
that matters. The time that energy is being transferred overall may be
(and will be) different between the input and output, but this is not a
concern for peak power measurements, and is the whole essence of power
storage devices/pulse discharges. It's why a TC works!
>Note that when using average power that you are adding time to the power
>units which brings you into the energy unit solution. This has caused great
>confusion for coilers in the past. Average power is actually energy because
>you have to use time to find the average power.
Again, see above, just because you use time doesn't mean you get energy.
There is a big difference between average power and energy. Average power
is calcualted from W/sec, over a specified period of time yielding Watts
again. Energy is just a specific quantity of energy, no time involved
whatsoever.
In other words when you
>connect a wattmeter to the input of a TC you are measuring many parameters
>depending on how you want to use them. For example the wattmeter gives you
>at the TC input
>
> 1. wattage
> 2. average wattage
> 3. peak wattage
> 4. instantaneous wattage
> 5. volt amps
> 6. RMS wattage ??
Strictly speaking, wattmeter doesn't give you all these things, it gives
you one: average "wattage", or power. Some, with storage functions, will
give you peak power, but this can be the peak over 1 cycle, or the peak
instantaneous power. In an AC circuit, you have instantaneous power, which
is defined as instantaneous current times instantaneous voltage, but is not
very meaningful in terms of what is actually going on because both I and V
are going positive and negative continuously. This is where average power
comes in - the average over one AC cycle.
Because of non-resistive loads, the power transfer can be going in to or
out of the "load", meaning the instantaneous power is positive sometimes,
negative other times, so an average "power" is used to represent what work
is actually being done - also know as the real power, measured in Watts.
The RMS current and RMS voltage, considered without and phase difference is
the "apparent power" - Volt-Amps, and often most devices are rated to a
certain VA because the wire has to handle a certain amount of current, and
it doesn't care if it's in phase with voltage or not, there is still that
amount of current to be passes. The imaginary power, measured in VA
reactive, is just the part of the current that is purely reactive,
imaginary, or 90 degrees out of phase with the voltage that does absolutely
no work whatsoever, and can't because the average power is zero - half the
time energy is flowing into the load, the other half out of the load, the
effective energy transfer is zero, and power is zero.
>Correctly using all of these parameters can be very confusing. You can avoid
>all of the above confusion by properly using energy units to rate Tesla
>coils. If the wattmeter is used as an energy meter you have to do some calcs
>and you end up with different numbers compared to using it as a power meter.
>For example a 100 watt wattmeter will give you 50, 100, 200, etc, watt
>seconds when used as an energy meter if the times are 1/2, 1, 2, etc,
>seconds.
So how is this less confusing than using power? I can run my 1" TC for
days on end and claim that "consumed" more than 30 MJ. Then I'll go run my
15" 10 kVA pig coil for under an hour, and it'll "consume" the same amount
of energy. So what's the point? I can also tell you that one coil has a
bang energy of 2 J, and another 10 J. If the breakrate of the first is 600
bps, and the second is 120 bps, they "consume" the same amount of energy
per time, or use the same power. I can also tell you that the National
Ignition Facility at LLNL consumes over 2 MJ in one shot, much less than
one second, while running my small TC will take over 5.5 hours to process
the same amount of energy. So, how do you propose we use energy to compare
TCs? I'm not seeing how it would work.
>There is a much more to comparing power vs energy and I find that in some of
>my past posts I have used the words incorrectly. Coilers are correct when
>they say that power and energy can muddy the waters.
I think trying to compare energy and power is utterly useless. I think we
can all agree that when we talk about power input, we are talking about
average power, or just a rate of energy transfer into our coils. Steve's
less than 4800 W input is the average power going into his coil, and also
must leave at the same rate, whether it be in the form of heat, light, or
electricity. However, instead of entering at a (relatively) constant rate
as happens on the 60 Hz line (since 60 Hz is slow compared to RF), the
power is leaving in large pulses that happen as often as he dictates by the
breakrate of the coil, and while these peak powers occur at a lower duty
cycle than the input power has, there are much larger peak powers (maximum
of instantaneous power) present on the output.
Okay, so that wasn't as short as I expected, but I hope that clears up some
nomenclature questions for everbody (and maybe for myself, as I'll probably
be corrected on some things I wrote).
Sean Taylor
Urbana, IL