Rosemary Ainslie COP>17 Circuit / A First Application on a Hot Water Cylinder

[SkyWatcher123]

Hi folks, these videos by gotoluc still have merit in my opinion and I think is similar in effect to Rosemary's circuit. All I can say is that I think gotoluc was convinced he had nothing special, when there is something special going on here.
http://www.youtube.com/profile?user=gotoluc#p/u/70/WsmPyUzZtgQ
http://www.youtube.com/profile?user=gotoluc#p/u/66/xvE7IGCra14
peace love light
Tyson

[Rosemary Ainslie]

Hi folks, these videos by gotoluc still have merit in my opinion and I think is similar in effect to Rosemary's circuit. All I can say is that I think gotoluc was convinced he had nothing special, when there is something special going on here.
http://www.youtube.com/profile?user=gotoluc#p/u/70/WsmPyUzZtgQ
http://www.youtube.com/profile?user=gotoluc#p/u/66/xvE7IGCra14
peace love light

Hello Tyson.  I love goto's work.  But that's work that was done over a year ago and we're not discussing apples and apples in this test.  He has not got inductive resistors - or not highly inductive ones as we require - and there's no self oscillating resonance - which is very much required.  But I do NOT want to detract from gotoluc's work.  I'm a very big fan of his.  Just not sure that this reference is in any way appropriate.  Not sure why you favour this over those videos that show this very clearly. ??

Regards,
Rosemary

[Rosemary Ainslie]

Hi Guys,

I need to point out a quandary that I've been battling with.  I was rather hoping that someone would point it out.  Another question.  In that battery confirguration (post 75).  Take away the resistor and assume that during the 'on' period of any cycle the discharge battery is 24 volt to the recharge battery 12.  No impedance in the circuit other than a nominal voltage drop across the diodes.  All things being equal then the 24V source will discharge plus minus 2 amps = vi 48 watts.

Here's the kicker.  2 amps is supplied to a 12 volt battery.  Two amps * 12 volts (recharging battery) = vi 24 watts.  Net loss to the system is therefore 24 watts and in terms of conventional measurement protocols there's no useful point to this circuit or this excercise.  Every time that the 24 volt supply discharges it recharges the supply with 24 watts - it marginally heats up two diodes - and the remaining plus/minus 24 watts are lost?  Where?  It's not in heat at the supply batteries - because discharging batteries are NOT known to heat up.  The only impedance/resistance in the circuit is the recharge battery and IT only has 12 volts.  The assumption is that this will restrict the supply current to plus/minus 2 amps.  SO.  Where does that surplus wattage get wasted?  Technically it never reaches the recharge battery.  And the only thing connecting the two of them are some circuit wires. 

Hopefully someone can enlighten me.  At this stage I'm wondering if this circuit is simply pointing to a measurement paradox.  Either that or - as is more likely - I've made a blunder.  The one thing I would point out is that the 24 watts is not in the wires.  Those batteries are effectively in parallel.  And the maximum current flow discharge from the source batteries would be 2 amps. 

regards,
Rosemary

[gyulasun]

Hi Rosemary,

Perhaps the first thing to examine is the 2A current (true RMS, average, peak) value and how dependable the measurement was.

It may start from the switch's duty cycle that eventually controls the current value able to flow due to the 12V voltage difference between the batteries. Pulsed current measurements are not easy you surely know.

And in case it turns out the true current is maybe less than the 2A value, then it can only warm up slowly the inside body volumes of the batteries, making it hardly noticeable, especially in case the measurements lasted for only say 10-15 minutes. (I guess this time duration of course.)

rgds,  Gyula

PS:  Sorry for 'chiming' in.  I am convinced that your switched heating circuit with the inductive resistor can only be evaluated by calorofic measurements (Sandy nicely described this above.)

So this means an enclosure of the inductive resistor into an isolated oil volume, (into a thermos bottle filled with vegetable oil you use for cooking) and using a thermometer inside to note the temperature changes etc then compare the input DC energy consumption to the calculated heat energy.

This was already suggested by Dr Stiffler too last year, Peter Lindemann agreed with him then, but still nobody did it from the energetic forum decent team, instead they dealt with day long scope measurements and you know what.
Yes, I know that measurements, evaluating the waveforms are also important but where is a down to earth proof, a practical 'gadget' that utilizes the extra heat coming from somewhere?
I mean I put on the table two heat sources, I run both simultenously, I measure both input power, suppose they are about the same and then I find one of the heat sources gives out a definitely more, well sensible heat 'stream' than the other.  (Of course, the the heatsource performing better than the other would include your circuit.)
Sorry if such tests have already been done, I surely have missed them.

[fritz]

Take away the resistor and assume that during the 'on' period of any cycle the discharge battery is 24 volt to the recharge battery 12.  No impedance in the circuit other than a nominal voltage drop across the diodes.  All things being equal then the 24V source will discharge plus minus 2 amps = vi 48 watts.

If we assume 3 identical 12V batteries  20Ah capacity, 25mOhm internal Resistance, identical charge condition (1+1 == 24V, 1 == 12V), I would expect the following:
Battery Voltage each 12.6V, 3times internal resistance = 75mOhm, Voltage loss diode+switch 2V-> voltage difference==10.6V, resistance 75mOhm
Well, that would give a theoretical value of 140Amps. The major impact will be the wiring and contacts involved (which is missing but would play the dominant role). I would expect something around 30 Amps.
The problem with 140Amps diodes is that they are not the fastest ones.

I would expect the transfered charge as u(diff)(10.6)*i(==u(diff)/(internal resistances+wiring)) *t(on)
->@ 320W (@30Amps) - adding some losses, heating up - we transfer maybe @250W

The diodes would heat up (@30Amps, 1V drop) with 30W each.

So I would assume that you blow any diodes and switches using 20Ah lead-acid with fast diodes at the first pulse.

Here's the kicker.  2 amps is supplied to a 12 volt battery.  Two amps * 12 volts (recharging battery) = vi 24 watts.  Net loss to the system is therefore 24 watts and in terms of conventional measurement protocols there's no useful point to this circuit or this excercise.  Every time that the 24 volt supply discharges it recharges the supply with 24 watts - it marginally heats up two diodes - and the remaining plus/minus 24 watts are lost?  Where?  It's not in heat at the supply batteries - because discharging batteries are NOT known to heat up.  The only impedance/resistance in the circuit is the recharge battery and IT only has 12 volts.  The assumption is that this will restrict the supply current to plus/minus 2 amps.  SO.  Where does that surplus wattage get wasted?  Technically it never reaches the recharge battery.  And the only thing connecting the two of them are some circuit wires. 

Do you use batteries with 2Amps short circuit current ?
I would expect that the heating up of batteries is the same on charging and discharging - only dependent from the current. On top of that you would have extra heating if the battery is full and you try to overcharge it - than heating up will increase dramatically. Same if you charge with too much current.

rgds, fritz.

BTW: You would achieve the maximum transfer if the internal resistance of the battery which is derived from the charge condition is identical. If a full battery charges an empty one - the current would be limited by the internal resistance of the empty one which would be definitely higher. If you charge the full battery with the empty one - again the high internal resistance of the empty one will limit the maximum current.