PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2024-05-17, 21:13:03
News: Registration with the OUR forum is by admin approval.

Pages: 1 2 3 4 [5]
Author Topic: Reliable Measurements and Simulations: Power In and Out and Efficiency n  (Read 51367 times)

Group: Administrator
Hero Member
*****

Posts: 3210
It's not as complicated as it may seem...
Hi Professor, and welcome back from your trip.

You are absolutely correct, in that 50% of the energy will be lost in such a direct cap-to-cap transfer. The "lost" energy is dissipated in the wiring between the two capacitors, and it matters not how low that resistance be, it will always amount to 50% loss.

When we insert an inductor in between the two capacitors, things begin to change and we are able to transfer the energy with less loss. The efficiency of this method depends on the inductor. The higher the inductance and the lower its DC resistance, the better. In other words, we want to use an inductor with as high a Q-factor (L/R ratio) as possible.

In my opinion, using two caps in this case as you have suggested as a means to determine circuit efficiency, may be more complicated and less accurate.

I'll let others weigh in on the merit of this approach however.

You may be interested in reading the attached document I put together some time ago, on this very subject.

.99


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
Group: Moderator
Hero Member
*****

Posts: 2995
Hi Professor, and welcome back from your trip.

You are absolutely correct, in that 50% of the energy will be lost in such a direct cap-to-cap transfer. The "lost" energy is dissipated in the wiring between the two capacitors, and it matters not how low that resistance be, it will always amount to 50% loss.

When we insert an inductor in between the two capacitors, things begin to change and we are able to transfer the energy with less loss. The efficiency of this method depends on the inductor. The higher the inductance and the lower its DC resistance, the better. In other words, we want to use an inductor with as high a Q-factor (L/R ratio) as possible.

In my opinion, using two caps in this case as you have suggested as a means to determine circuit efficiency, may be more complicated and less accurate.

I'll let others weigh in on the merit of this approach however.


You may be interested in reading the attached document I put together some time ago, on this very subject.

.99

 Thank you -- I have looked through your document -- very informative, especially (to me) the part that begins thus:  
"In an attempt get obtain meaningful work output from the setup, the coil is replaced with a transformer. The goal is to shuttle the initial energy back and forth between C1 and C2 with as little loss as possible, and to tap the energy flowing through the series inductance in order to make it do some work for us."

I agree with your conclusion that using capacitors in an attempt to measure n = Eout/Ein is complicated and difficult overall, and probably not a good approach.  In the simple experiment I did, charge was conserved as expected -- but energy was NOT conserved in the circuit -- it must have been lost due to sparking or resistance (or other factors -- energy did not evaporate).

However, there is a way using caps that may work:  we use a Cap with VERY large C for the input and such a cap on the "output leg" of the circuit also, then running the circuit for a relatively short time so that the voltage drop in the input cap is small.
  I'm not certain how to wire this into the circuit, frankly, because of difficulties imposed by the necessity of having a complete circuit -- and charge MUST be conserved!   

Now, in your use of two DMM's, .99, to measure Pin and Pout, I also have a problem -- because the "high-resistance" voltmeters affect the circuit, as seen in subtle changes in the brightness of the LED [which I observe sometimes] when a DMM is connected at certain places in the circuit.  How to get around that problem?

General conclusion to this point:  accurately measuring Pin and Pout is not trivial, not at all.  I suspect that many in the "free energy" field are making mistakes in their measurements, mistakes I would like to learn to avoid -- to get at reliable measurements (my goal at this time).
   
Group: Moderator
Hero Member
*****

Posts: 2995
  Here is another method that I tried; see what you think of this, fellow researchers.

  With a standard JT, when I hook up a battery to run it, the voltage across the battery drops.  This is expected for a circuit drawing power from the battery.

  Another circuit B I've developed is different.  I measure the battery voltage as it stands alone, actually two AA batteries in series, 2.65 volts.  (not fully charged in this case)
Then I hook up the battery to circuit B and the voltage "of the battery" goes UP... to about 2.72 volts and stays up for a long time, at least an hour.

NOTE:  Just two caps in the circuit, 151 pF each, and no other batteries.

  Further, after an hour of running the LED brightly, the battery voltage is the same as when I started, within measurement error of 0.01 volt.    When I run the JT, the battery voltage after the run is significantly less than before the run -- but not with circuit B.

How do you explain these observations?
   

Group: Administrator
Hero Member
*****

Posts: 3210
It's not as complicated as it may seem...
Could you please show what circuit B looks like?

Thanks,

.99


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   

Group: Administrator
Hero Member
*****

Posts: 3210
It's not as complicated as it may seem...
Thank you -- I have looked through your document -- very informative, especially (to me) the part that begins thus:  
"In an attempt get obtain meaningful work output from the setup, the coil is replaced with a transformer. The goal is to shuttle the initial energy back and forth between C1 and C2 with as little loss as possible, and to tap the energy flowing through the series inductance in order to make it do some work for us."

I agree with your conclusion that using capacitors in an attempt to measure n = Eout/Ein is complicated and difficult overall, and not a good approach.  In the simple experiment I did, charge was conserved as expected -- but energy was NOT conserved in the circuit -- it must have been lost due to sparking or resistance (or other factors -- energy did not evaporate).

Now, in your use of two DMM's, .99, to measure Pin and Pout, I also have a problem -- because the "high-resistance" voltmeters affect the circuit, as seen in changes in the brightness of the LED [which I observe sometimes] when a DMM is connected at certain places in the circuit.  How to get around that problem?

General conclusion to this point:  accurately measuring Pin and Pout is not trivial, not at all.  I suspect that many in the "free energy" field are making mistakes in their measurements, mistakes I would like to learn to avoid -- to get at reliable measurements (my goal at this time).

Yes, the lost energy in a cap-to-cap transfer will get dissipated in the connecting wire.

First and foremost, understand that I have not been attempting to measure Pout using the DMM method, nor is it advised. It will not work for Pout.

For the Pin measurement, does the meter still affect the intensity of the LED's?

.99


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
Group: Moderator
Hero Member
*****

Posts: 2995
Yes, the lost energy in a cap-to-cap transfer will get dissipated in the connecting wire.

First and foremost, understand that I have not been attempting to measure Pout using the DMM method, nor is it advised. It will not work for Pout.

For the Pin measurement, does the meter still affect the intensity of the LED's?

.99

Ah, I see the difference, .99.  The LED brightness in circuit B changes significantly when I look at output using the DMM (either of the two I'm using), but this is not the case when I do the Pin measurement.  

  About circuit B, I would like to discuss it generally at this time rather giving the specific circuit.  I'd like to do more measurements of different types myself first, I hope you all understand.

About the battery not dropping in voltage -- so I ran this overnight, for over NINE HOURS, with the LED lit albeit dimly in this case for the input battery (rechargeable, but never charged -- purchased yesterday) was just 1.26V at the start.  The LED (resistance not known; how to measure??) was in series with 6.1 ohms in three resistors, so there was clearly significant resistance on the output leg of circuit B.

I just measured the voltage of this battery this morning -- 1.26 volts, measured on two different DMM's, same as at the start last night.   To me, this is itself a significant result...  although I'm not willing to claim anything yet.  I want to do measurements using different methods, you see.  I will say that the Tek 3032 basically agrees -- circuit B draws VERY little mean input power.
   
Group: Moderator
Hero Member
*****

Posts: 2995
  I would like to get to the bottom of this, to understand the results from circuit B finally.

Attached are two schematics, tested on the way to my development of circuit B (which differs significantly from both of these).  Please focus with me for a moment on measuring the input power -- see how the scope probes are connected, the same in both circuits and in circuit B.

Now, the Mean input power -- given the way the probes are connected -- will it show up positive or negative on the scope?  (MEAN Pin)  

Let's consider that first, then I'm willing to show the input power waveforms for this strange circuit B.

   

Group: Administrator
Hero Member
*****

Posts: 3210
It's not as complicated as it may seem...
IF we assume that the circuit is UU and that the net power is FROM the battery, then your MEAN Pin should come out negative. Voltage across the CSR will be net negative wrt GND, and hence V*I will be negative also.

Remember also that we have to ADD the power in the battery to the power in the CSR to get the actual Pin.

.99


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   

Group: Administrator
Hero Member
*****

Posts: 3210
It's not as complicated as it may seem...
As an alternative to obtain the Pin, you can use the scope math to perform (V1-V2), then take the MEAN of that. This will give you the average battery voltage. Then you can multiply that by the MEAN voltage across the CSR to get average battery power.

so in summary, you have:

MEAN[(V1-V2)] * MEAN[V2] = Pin

.99


---------------------------
"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
Group: Moderator
Hero Member
*****

Posts: 2995
  Now, I have been using oscilloscopes, but I'm looking for an accurate alternative for measuring power.  I've been concerned about RF pick-up in the probes, for example.
Hence I've pushed for the used of the cap/time method for measuring Pinput -- direct, straightforward. 

    For Poutput, I'm proposing the use of a simple calorimeter.
  Here's the latest from discussion with my colleague to get a ~5-10% measurement of Poutput.  (After tests at this level are made on several devices, one can use the expensive calorimeter at the university for better precision.)   

This design one may be able to use at home.  I welcome comments!!  especially on the temperature measurement -- is there something better than a Seebeck device, on the block of aluminum?   


From Prof H:
Quote
A block of aluminum (around 4" diameter and 1" thick) to act as a constant temperature heat sink. This does not have to be temperature controlled unless you want to operate away from room temperature.
Stick a 2" square BiTe thermoelectric (Seebeck device or thermopile) on the block with heat sink compound or the right type of silicone glue. Connect the wires from the [Seebeck or thermopile] device to a microvoltmeter and a computer.

Place the [DUT] device on the Seebeck device so it makes good thermal contact, maybe use heat sink compound. [We should think about this part... resistors, LED's etc. need to dump their heat "through" the thermopile, for the measurement of Poutput to be accurate.  In time, and with good insulation, the heat will go through, but it may be very slow unless thermal contact is good... comments?] 

 Run the wires around the aluminum block and tape them down...
Cover the whole thing with a Styrofoam box.
To run the experiment:
Wait until the signal from the voltmeter is stable, record some baseline for a few minutes, trigger the device, and continue recording until the signal returns to baseline.
Integrate the signal above the baseline to get the total heat.
Calibrate the system with an electrical resistor in place of your device.


And note that this cal'r is not a Faraday cage... the DUT sits on top of the thermopile, which sits on the heat sink (Al cylinder). 
Comments welcomed.


   
Pages: 1 2 3 4 [5]
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2024-05-17, 21:13:03