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Re: Gap Losses
At 04:26 AM 9/27/96 +0000, you wrote:
>> Subject: Re: Gap Losses
>> >Subject: Gap Losses
>
>From hullr-at-whitlock-dot-comThu Sep 26 21:46:35 1996
>Date: Thu, 26 Sep 1996 10:11:00 -0700
>From: Richard Hull <hullr-at-whitlock-dot-com>
>To: tesla-at-pupman-dot-com
>Subject: Re: Gap Losses
>
>Tesla List wrote:
>>
>> >From couturejh-at-worldnet.att-dot-netWed Sep 25 22:10:41 1996
>> Date: Wed, 25 Sep 1996 22:08:02 +0000
>> From: "John H. Couture" <couturejh-at-worldnet.att-dot-net>
>> To: tesla-at-pupman-dot-com
>> Subject: Re: Gap Losses
>>
>> At 04:25 AM 9/25/96 +0000, you wrote:
>> >From rwall-at-ix-dot-netcom-dot-comTue Sep 24 22:21:55 1996
>> >Date: Tue, 24 Sep 1996 04:18:51 -0700
>> >From: Richard Wayne Wall <rwall-at-ix-dot-netcom-dot-com>
>> >To: tesla-at-pupman-dot-com
>> >Subject: Gap Losses
>> >
>> >9/24/96
>> >
>> >Has anyone ever experimentally measured gap losses? Where and in what
>> >form is this energy lost from the primary circuit? Is this energy
>> >merely heat, light and other forms of EM energy? Can it be measured
>> >and analysed or is it merely assumed to be the difference between
>> >energy input into the tank circuit minus energy coupled to the
>> >secondary coil?
>> >
>> >
>> >RWW
>> >
>>
>> Richard -
>>
>> If you could measure this loss it is interesting to speculate on what you
>> would find.
>> In the Tesla Coil Notebook a computer printout is shown for a typical 1000
>> watt input coil. The overall efficiency is 31 percent and the RMS voltage is
>> 9020 volts. The maximum losses are, therefore, (1000)x(1-.31) = 690 watts.
>> Assuming half of the losses are in the spark gap, the gap losses are 690 x
>> .5 = 345 watts. If the current thru the spark gap is measured it should be
>> about 345/9020 = .038 amps or 38 ma RMS. Is there anyone who would like to
>> try this? I would be willing to send them a JHCTES computer printout on
>> their coil to give them the necessary information for this test.
>>
>> What are your comments? Do you have a better method of solving this problem?
>>
>> Jack Couture
>
>
>Jack, I think you are talking about averaged or RMS current here. In a
>1000 watt coil, depending on the capacitor,input voltage, and gap
>construction, the current in the gap's arc channel should be as high as
>200 amps in a super system and as low as 50 amps in a poor one. These
>are instantaneous values based on the capacitor discharge. It is a very
>complex waveform of voltage and current across the gap. It is not
>predictable or caluculable to a precise degree.
>
> A really crude and imprecise value would to divide the peak input
>voltage(12.75KV) by the surge impedance (SQRT L/C) of the tank circuit.
>Let's assume .02ufd and a 40uh primary. Zsurge=~45ohms... Thus Igap=
>285amps peak! Peak impulse power = 3.6 megawatts! This will never
>happen! The curent will always be far less due to a number of causitive
>agents. (Gap impedance, distributed circuit losses, etc) As a matter of
>fact for 99% of the time, the current will be zero (0). A 1000 watt coil
>might expect 100 peak amps in the tank and a peak power of about a
>megawatt.
>
>Richard Hull, TCBOR
>
Richard -
You are correct. I was talking about RMS current. This is not AVG current. I
agree the peak current could be 200 amps or greater. In fact for this coil
the JHCTES printout shows a theoretical peak primary current of 703 amps.
This gives a theoretical peak power of 6.3 megawatts or an output of about
6.3 x .31 = 1.9 megawatts.
Spark gap losses, however, are in the form of energy, not in power. Coilers
should be cautious when talking about power loss. When wattage loss is
mentioned it should be understood that it is RMS wattage x time = energy. If
the time interval is very short a small amount of energy loss can produce a
spectacular effect as in the operating spark gap or in the spark from the
secondary terminal. When a Tesla coil is rated at 1000 watts input it is
assumed that it is consuming 1000 watts RMS per second. The losses in the
spark gap, etc. must be less than this amount.
Tesla Coil Energy vs Power -
1 - OUTPUT ENERGY = input energy minus losses. If there is no energy there
are no losses.
2 - OUTPUT POWER can be much greater than input power when time intervals
are short. Losses are not involved. In the Tesla Coil Construction Guide a 5
million volt Tesla coil is described that shows how energy, power, and
losses are calculated. This coil was built in the 1930's by three
scientists. It had an input of 3KW RMS and an output transient of 1700 KW, a
power gain of over 500. The overall efficiency was estimated at about 25
percent. A method to accurately measure the output voltage is also shown.
The spark gap losses can be measured in several ways including;
1 - Messy peak currents, waveforms, differential equations, etc.
2 - Heat and other radiation, differential equations,etc.
3 - Measure voltage across the gap, differential equations,etc.
4 - The easy way, simple math, requires estimates
The first two methods are very difficult tests and it may be impossible to
get accurate results. The third method can only be useful if the gap
resistance is known. The resistance varies from almost zero to infinity in
an unknown way over time. I estimated the fourth method would give about 690
watts RMS of loss which is 69% of the input of a 1000 watt RMS, 9000 volt
RMS Tesla coil. This small amount of loss can give impressive sparks because
of the short time intervals of Tesla coil operation.
You can test your coil by the 'easy way' test. This test does not measure
the Tesla coil input in the usual way which involves losses that are not
part of the Tesla coil circuit. To test for the true TC wattage input, the
power transformer secondary (not primary) voltage and current must be
measured. This eliminates the losses in the variacs, etc.
The 'easy way' test is to insert a standard AC meter in the power
transformer secondary HV WIRE to the Tesla primary circuit. Do not insert
ammeter in the GROUND wire which could have large peak currents in it. Also,
the ammeter must be well insulated from ground. Both the secondary RMS
current and RMS voltage must be measured. This current times the voltage is
the true RMS wattage input to your coil. This wattage multiplied by 69% is
the approximate total RMS wattage loss of your coil. This wattage loss
multiplied by 50% will give an RMS wattage that is approximately the spark
gap loss. Note that the loss percentages are estimated efficiencies and
depend on the size of the coil. A graph showing Percent Efficiency vs Watts
per foot of spark is shown in the Tesla Coil Notebook.
The 'easy way' test would have to be with a complete Tesla coil operating
with a continuous stream of sparks from the secondary terminal (length not
important). The load at the ammeter could then be considered at almost 100%
POWER FACTOR. The advantage of this test, of course, is that many coilers
have the equipment to try it but very few coilers, if any, will be able to
do the other three tests.
I must admit the above is based on estimates made by me and all comments are
welcome.
I would be especially interested in comments from R. Hull and R. Wall.
Jack Couture
JHC Engineering
San Diego, CA