Exploring the Inductive Resistor Heater

[gmeast]

Greg,

I never said that the delta T's were used with regard to drawdown (except for the initial BH run).

Based on your numbers, up to the point where you do compare drawdowns, you demonstrate an efficiency of about 2.5% OU, or just under OU if your calculated driver contributions are added in.

This is determined by comparing the power required from the DC supply to produce same deltaT as the BH, which is used as "output power", to the power calculated from either the rheostat test or the measured SH3 voltage and the average Vbatt, which is used as "input power".

Am I correct so far?

The drawdown numbers from the rheostat test are then used to determine the time required for Vbatt to cross the BH run's end voltage, and the length of time at which that occurs is then used to calculate the watt hours consumed by the BH.

Is this correct?

PW

You are simply wrong, and I'm done here.

[picowatt]

You are simply wrong, and I'm done here.

Greg,

What part is wrong?

I have looked at the slide show several times.  If I am wrong, please tell me where am I wrong.

Surely you would not want others to have the same misunderstanding that I am when looking at your slide show ...

PW

[gmeast]

Greg,

What part is wrong?

I have looked at the slide show several times.  If I am wrong, please tell me where am I wrong.

Surely you would not want others to have the same misunderstanding that I am when looking at your slide show ...

PW

But others don't.

[picowatt]

Greg,

Again, this is what I get from the slide show:

(1).  You run the BH (burst heater) and note the deltaT, SH3 voltage, and also log the battery discharge curve.

(2).  To determine output power, you use the DC supply connected to Rload to produce the same deltaT at Rload as in (1) above and note the supply I and V.  Using the supply I and V you calculate output power, which I believe was determined to be 3.16 watts (from memory, I don't have the slide video open right now)

(3).  To determine input power, you use a rheostat in series with SH3 set to produce the same SH3 voltage as in (1) above (which I believe was 5.6mv) and calculate input power based on the total resistance used and the Vbatt average voltage.  You also log the battery disharge curve over time while performing this rheostat test.

As alternate input power verification, you also use the current calculated from the SH3 voltage measured in (1) above multiplied by the average Vbatt voltage and note that the two methods agree closely.  I believe that fom the input power tests and calculations you arrived at figures that were in close agreement, being 3.09 versus 3.099 watts or something like that, depending on the method used to determine input power.

At this point, by your your own meaurements and calculations, you demonstrate 3.16 watts of output using only 3.09 watts of input, which is just over unity by 2.5% or so.  If your calculated driver contribution is added to the input power, then the figures shift slightly to 3.16 watts out for 3.20 watts in, which is just slightly under OU.

Am I correctly following your slide show up to this point?

PW

[gmeast]

Greg,

Again, tis is what I get from the slide show:

1.  You run the BH and note the deltaT, SH3 voltage, and log the battery discharge curve.

2.  To determine output power you use the DC supply connected to Rload to produce the same deltaT at Rload as in (1) above and note the supply I and V.  Using the supply I and V you calculate output power, which I believe was determined to be 3.16 watts (from memory, I don't have the slide video open right now)

3.  To determine input power, you use a rheostat in series with SH3 set to produce the same SH3 voltage as in (1) above (which I believe was 5.6mv) and calculate input power based on the total resistance used and the Vbatt average voltage.  You also log the battery disharge curve over time while performing this rheostat test.

As alternate input power verification, you also use the SH3 voltage from (1) above multiplied by the average Vbatt voltage and note that the two methods agree closely.  I believe that from the input power tests/calculations you arrived at 3.09 versus 3.099 watts or something like that, depending on the method used to determine input power.

At this point, by your your own meaurements and calculations, you demonstrate 3.16 watts of output using only 3.09 watts of input, which is just over unity by 2.5% or so.  If your calculated driver contribution is added to the input power, then the figures shift slightly to 3.16 watts out for 3.20 watts in, which is just slightly under OU.

Am I correctly following your slide show up to this point?

PW

This is where you have gone wrong. You cannot use REAL-TIME measures of Power to determine EFFICIENCIES of these things. You MUST use 'Energy'. The energy consumed from the batteries is 25.28 Watt-Hours heating RL (3.16W X 8-Hours). That heating over 8-hours drew the battery down .4V and at 8-hours the battery voltage was 27.44V.  For the 1st draw-down, the same Starting Voltage after battery re-charge and stabilization, the Rheostat load drew the batteries down to the SAME 27.44V in 6.38 hours. The Rheostat load was 3.1Watts for ONLY 6.38 Hours for an Energy of only 19.78 Watt-Hours. The 5.6mV SH3 was simply a reference for adjusting the  load rheostats. I could have used anything. I could have used 10mV and the batteries would have drawn down quicker, but the energy would still have been around 19.78 Watt-Hours. I could have used 3mV and the draw-down would have lasted longer than 8-hours, but still would have been around 19.78 Watt-Hours ... at the point where the batteries hit 27.44V.


The 2nd Rheostat load test simply used the ratio of the Energies from the first two tests to adjust the rheostats such that the starting and ending voltages were the same as the 1st test (the circuit test) ... 27.84V to 27.44V.  19.78Wh / 25.28Wh = 0.78  So 5.6mV (SH3) X 0.78 = 4.4mV for the new SH3 voltage drop and I adjusted the rheostats to produce that load. The 2nd draw-down test at 4.56mV(avg) (SH3) resulted in identical starting and ending battery voltages as the circuit test ... 27.84V to 27.445V. Then I measured the load rheostat resistance and calculated the power which was 2.52Watts .... THIS IS THE INPUT POWER.  I then applied 2.52Watts DIRECTLY to RL and it produced a significantly lower Delta-T, and this simply proved everything out.


You are like everyone else that has assumed you can simply use poynty-head's PIN POUT nonsense crap for determining efficiency. YOU CANNOT USE REAL-TIME MEASURES OF POWER FOR THIS STUFF. YOU MUST USE MEASURES OF ENERGY!


The reason I know you haven't done any more than skim my presentation is that ALL OF WHAT I SAID ABOVE IS IN THAT SLIDE SHOW.


Take off your BLINDERS you guys.


Regards,


Greg