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COP 17 Heater | Rosemary Ainslie

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  • Hoppy
    replied
    Originally posted by Aaron View Post
    In Poynt's inductance thread, MH says:



    I just wanted to post that because I'll remember he said that.

    HH is right but now I've had time to think in more depth about the proposed test procedure, lets look more closely at this. Rosemary says: -

    OK this is how I would suggest the test be conducted if there are no storage scope meters available.

    Set the duty cycle and check the temperature of the resistor when it's
    stable.


    I assume this is done using the test battery?


    Then apply the same resistor to a variable power supply and adjust the voltage until the same temperature is found and stable over the same resistor

    Then do a v^2/r analysis to determine the wattage dissipated at the start of the experiment.

    So correct me if I'm wrong, but the 'control' is actually based on the stable temperature of a resistor originally heated by a battery that will be used in future stages of the test and therefore very likely to have a different capacity and condition as the test progresses. If this is the case, then how can this be considered a control?

    Then record the start time to run the experiment until the battery is depleted to say, 11 volts from a 12volt supply or 22 volts from a 24 volt supply.

    Don't you mean total run time?

    I assume to do this stage of the test, the battery used to establish the control dissipation wattage must be re-charged. How does the test procedure ensure that the energy placed into the battery during recharge does not alter the total energy contained in the battery prior to the control wattage being established? I mean, how do you know when to stop charging given that a battery requires considerably more charging energy to replace that taken on discharge?

    Then recharge the batteries and apply a resistor in series with them to draw down the same amperage as recorded at the start of the experimental test.

    The same amperage? Don't you mean battery terminal voltage as this is all I read above as being recorded as a reference point?

    As above, how does the test procedure ensure that the energy placed into the battery during recharge does not alter the total energy contained in the battery, this time prior to the first experiment being run?

    Then recharge the batteries and apply a resistor in series with them to draw down the same amperage as recorded at the start of the experimental test.

    As above, how does the test procedure ensure that the energy placed into the battery during recharge does not alter the total energy contained in the battery, this time prior to when the series resistor was applied?


    Then rerun both tests.

    Does this include doing a new v^2/r calculation?

    If you've got two sets of batteries - run them concurrently until one or other hits that critical voltage level. Then recharge both and swap them, control to experiment and vice versa.

    How does the test procedure ensure that a second set of batteries would have produced the same v^2/r control result as the first set of batteries?

    Please note that of the above questions don't even address the battery conditioning issue which I've now dropped.

    Hoppy
    Last edited by Hoppy; 08-03-2009, 10:24 AM.

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  • witsend
    replied
    TK - I'm hoping I can bend your mind around this problem.

    The battery recharges, power through, voltage first drops - then a spike to, what was it - say 50 volts or thereby? At that same moment the value across the shunt say 0.4volts positive drops to about 1.2volts negative, (aproximate because I couldn't see the actual value) . Then how do you work out the product of the energy available at that moment? In my reckoning it is 1.2/0.25 = plus/minus 4.8 amps. So. I need to be reasonably certain that the actual energy calculated at that moment as v*i = 240 watts BACK TO THE SYSTEM. (Again not shouting. Just emphasis)

    Now. If you do not do the integration simultaneously how is this advantage or gain made evident? And more alarmingly, if you simply do the product of both values and add it to the general loss - then your methodology is wrong.

    If, as you say the LeCroy is doing the math - then I'm afraid I need to see it. I'm still concerned that it is not capable of doing that DC offset. To my primitive way of thinking this means that it cannot gauge zero in order to make a comparison as to where the waveform was found in relation to zero.

    Please explain this. Sorry to impose.

    You are right. Its's the best I can do to explain my concern. I use primative example and analogy. But it serves its purpose.
    Last edited by witsend; 08-03-2009, 08:59 AM.

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  • sucahyo
    replied
    Winding?

    Messy winding heat up faster. Have the best coil winding for heater been discussed before?

    Which is better, a cross 90 degree winding or 180 degree opposite winding? or maybe 45 degree?

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  • Aaron
    replied
    Mh's quote

    In Poynt's inductance thread, MH says:

    Originally posted by MileHigh View Post
    This is why just measuring the temp of the coil-resistor body under controlled conditions is such an accurate way of comparing the running circuit power and the pure-DC-equivalent power. Even if the thermal resistance is likely somewhat non-linerar with respect to temperature, you are still left with the fact that the energy in must equal the energy out.
    I just wanted to post that because I'll remember he said that.
    Last edited by Aaron; 08-03-2009, 07:43 AM.

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  • Aaron
    replied
    gate pot

    Just a quick note on my circuit. It is still running like a champ. However, remember when MH recommended making resistor connection to gate very short?

    Well, that specifically helps to PREVENT oscillation. Look it up!

    I thought it was a good idea because I usually do hyper short connections between everything. I even tried soldering the inductive resistor straight to the battery terminal!

    Anyway, keep the 1k pot or whatever value you use 8 inches or so away from the gate and long curly wires are fine for that!

    The evidence mounts that each recommendation from the skeptics so far has done nothing more than down-tune the circuit from what it is supposed to do.

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  • witsend
    replied
    Sorry guys - yet another note for TK.

    Have read up and am amazed at the sampling ability of the LeCroy. My concern is that you're applying your own computations to that data. Especially as it relates to the DC average.

    I can live with the fact that readings aren't simultaneous. I now understand that point. Thank you.

    Now TK. We will need to see the actual software relating to your computations - if you don't mind. Alternatively just hook the FLUKE 123 probes across the shunt and SHOW US the DC average. The sample range is not as broad but it has the real advantage of giving an IMMEDIATE voltage value. (emphasis is not shouting - just intended for emphasis) within a reasonable level of margin for error.

    The logic applied for the computation of energy delivered is definitely a 'nested if'. If below zero - subtract - from previous'. You get the picture? Again - back to the Fluke. It has the advantage of being able to deduct the one voltage from the other and give the difference. We can then apply our own math. So. My question. Why are you ignoring the use of that little instrument? I cannot understand it.

    And please attend to OC's PM on your youtube. If it's what I suspect - it may be that you have to adjust the conclusions from that sad display of mental arthimetic.

    And could you please address the fact that the 'recharge' may, indeed, be a recharge cycle. The more so as it self-evidently IS. Comments such as 'doesn't mean it is recharging - mind' - could otherwise indicate a preference and a bias.

    It's really hard to keep up with your duplicity TK - the more so as I'm constrained by that stupidity and lack of technical know how which you gratuitiously bring to everyone's attention.

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  • EgmQC
    replied
    I dont realy understand why there a focus on the batterie discharge rate in this experiment. We are not talking about a 10%-20% over unity here , the claim is "COP 17", its 1700% OU. Even if is its not all fine tune , you should be able to see a 50% OU from time to time when you experiment with the circuit, there no mysterious part or complexe assembly to make this realy simplistic circuit.

    A realy simple way to get result is to use the SAME input for all experiment, for ex: 24v 0.5A, so the discharge rate is locked , when you begin so see some excess heat who is REALY unexplainable , well now will be the time to look deeply on the input side, but dont forget the base reference for the heat, its the ambient temperature, both must be monitored, the claim is about excess heat.

    Best whish for you all,
    EgmQC

    EDIT: Forgot to add, apply the same wattage to your resistor than the circuit will use, let it run for 1 hour, nothing connected to it , just the input source, that will show you what is the maximum heat your resistor should have after 1 hour for a X ambient temperature. Or better, use 2 input source, the first one run the circuit and the second input source only run a IDENTICAL resistor(voltage ajusted for same wattage as the first input source ). You will have the best setup to identify excess heat.
    Last edited by EgmQC; 08-03-2009, 01:14 AM.

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  • Hoppy
    replied
    Aaron wrote: -

    "My batteries are thoroughly conditioned. They are very consistent."

    That's good but you clearly don't understand my point about the importance of doing cyclic conditioning directly before or preferably as part of the test procedure itself. Lets drop this now and I'll not bother you any further on this issue.

    Hoppy

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  • Aaron
    replied
    time test

    The battery stabilizes fine after a while running at stabilized temp. I know this because when stable and battery stays at exact voltage (exact up to the hundredth of a volt for many hours, I can turn it off and the battery might climb a few hundredths of a volt meaning it is really running where the battery is at.

    My batteries are thoroughly conditioned. They are very consistent.

    Testing time is testing time of draw and calculating joules is irrelevant. It is testing TIME.

    If Rosemary's circuit with certain settings gets the resistor to say 120F and is stable for hours and hours...that is the temp to use with a control supply.

    Hook up supply to resistor to see exactly the wattage necessary to get to same temp. that watts is what the math says is absolutely necessary to get the resistor to x temp.

    so if the circuit makes that temp and it takes 24 hours to bring the batteries down to 23 volts....that is the time we get with Rosemary's producing equal heat.

    Then whatever control wattage says is necessary, we add that much resistance to the batteries after recharged and let it go, the temp of that resistor is irrelevant since we already know that wattage is what was required to make that heat.

    and if it drops the battery to 23 volts in 12 hours, then Rosemary's produced heat for twice as long compared to what it should have.

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  • Hoppy
    replied
    Originally posted by Aaron View Post
    Exactly!

    The steady DC supply clearly shows what it is "supposed" to take according to basic math in terms of wattage to have the resistor at a certain temp.

    We run your circuit to make equivalent heat for however long on the batt to drop it to x volts.

    Then we charge batts and put resistor + extra resistance to match the wattage that the dc supply shown is necessary. Then we time it until it gets to the x voltage like in the experiment circuit.

    The second will drop the battery much faster but the rate of how much faster is irrelevant. ANY extended run whatsoever is over 1.0 COP.

    I'll post this whole method that anyone can use without a scope.
    Then we charge batts and put resistor + extra resistance to match the wattage that the dc supply shown is necessary. Then we time it until it gets to the x voltage like in the experiment circuit.

    Whenever a battery low termination voltage 'x' (LTP) as a reference, then every time you repeat the same test, assuming you have re-charged the battery with exactly the same energy each cycle (not at all easy because of varying SOC), then your calculated COP based on the energy used to reach the LTP will drop until it stabilises. The test battery must be cycle conditioned to a point where their internal resistance stabilises before a final COP is taken. This is why so many people who use this LTP voltage reference level of testing end up with misleading high COP results. Twice is not enough, unless the batteries have been pre-conditioned adequately to stabilise internal resistance.

    You say: "The second will drop the battery much faster but the rate of how much faster is irrelevant. ANY extended run whatsoever is over 1.0 COP." I agree, the rate of drop is irrelevant but the overall time to reach the 'x' reference voltage certainly is relevant and if I understand you correctly this is what you are measuring in time to calculate your energy in Joules? If this is the case, then your method is flawed unless the batteries are conditioned as explained above.

    Hoppy
    Last edited by Hoppy; 08-02-2009, 08:30 PM.

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  • Harvey
    replied
    Will Be Back Later.

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  • Harvey
    replied
    @ Aaron
    1.
    Consider placing a Schmitt Trigger buffer between the 555 output and the gate. There are some very fast devices like the 74F14 inverter with switching times of less than 12 nanoseconds (yep nano, not micro). You can put two of those gates in series to get the buffer you need, but mind the supply voltage and HEXFET gate requirements. It doesn't look like the IRFPG50 really turns on until it reaches 8V and the 74F14 is a 5V device - look around, you may find a 40Fxx CMOS (higher operating voltage) that has the same characteristics. If not, let me know and I will try and squeeze that in.

    2.
    The Bat-Cap is in the order 10x Farads, like 30F IIRC. The tests I have seen have been using capacitor values a thousand times less, as in 33,000uF. Much more energy storage, and better defined for scientific reasons.

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  • Aaron
    replied
    basic test

    Originally posted by witsend View Post
    Firstly - the COP is determined not by battery draw down but by actual measurement of energy delivered compared to dissipated.
    Exactly!

    The steady DC supply clearly shows what it is "supposed" to take according to basic math in terms of wattage to have the resistor at a certain temp.

    We run your circuit to make equivalent heat for however long on the batt to drop it to x volts.

    Then we charge batts and put resistor + extra resistance to match the wattage that the dc supply shown is necessary. Then we time it until it gets to the x voltage like in the experiment circuit.

    The second will drop the battery much faster but the rate of how much faster is irrelevant. ANY extended run whatsoever is over 1.0 COP.

    I'll post this whole method that anyone can use without a scope.

    Leave a comment:


  • Aaron
    replied
    Time

    Originally posted by Hoppy View Post
    Aaron, Assuming you are saying that the gain will be shown in the battery, exactly how do you intend measuring a gain in your test battery?
    Extended TIME for equivalent watt hours. I'll post this basic test first.

    And I do not recognize ANY authority whatsoever in this matter that I
    need to convince. There is no such thing.
    Last edited by Aaron; 08-02-2009, 07:24 PM.

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  • Hoppy
    replied
    Originally posted by witsend View Post
    Not defensive Hoppy - just trying to appeal to reason. Thanks for letting go here. Just edit the post number. I got it wrong.
    We reason differently Rosemary. Don't worry about the post number.

    EDIT - Agreed.

    Hoppy
    Last edited by Hoppy; 08-02-2009, 07:11 PM.

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