The Power of Iron

Wiring, circuits, batteries etc.

The Power of Iron

Postby cj6s on Sat Jun 28, 2008 12:18 pm

I'm developing an iron fuel cell/battery that has proven in experiments (done at the local Department of Science and Technology [DOST-CAR] lab with staff witnesses) that seems to fly in the face of what an <<Answers.com>> site claims to be the amount of energy stored per kg (or liter) of iron. The data was 7.4 megajoules per kg (from a table comparing it to such data as MJ/kg or /liter of diesel, coal, and other oxidizeable substances. (To be precise, the data, 7.4MJ, was from iron burned to Fe2O3.) Now, if I use my average watt output per gram per cell I get something like 21MJ/kg. So either wikianswers is wrong, both my testers are wrong, the nrg value is different (because I am not producing Fe2O3), or I don't know how to take readings for continuous amp output.
Assuming it's the latter that's true--how does one use a tester, set on DC amps to get a reading for how many amps are coming out of a cell per hour? This has been a bane on my r&d since the beginning. Someone told me (an electrician) to take the spike reading and use that to calculate watt hours using the normal V reading over time equation. If so, then I can get a maximum of 108MJ/kg which seems impossible because that would put the price of iron energy at about US$ 0.02/kwh... and this seems too good to be true... The problem is that when I leave my tester connected to my test cell the ampere reading drops to zero after less than a minute--but if I just keep touching the tester lead to my negative wire, I get consistent spikes of up to 36mA on my best cells. I've touched and released for a period of minutes, waiting a few seconds between readings and keep getting good spikes but no steady (just a fast-dropping) current.
If I power a device like a motor or bulb, I still won't get an accurate reading because both can work on lower wattages than their specs (yes, slower or dimmer, but how do I know by how much?) Could I use the voltage setting and a resistor to get a steady amp output reading? (BTW I have no way of calculating electrode surface since it may not be a smooth LxW calculation, due to a bumpy surface. What is of most importance though is total watt hours produced per kg fuel consumed, and without a true ampere reading I cant get this...) Any and all help would be greatly appreciated......[If it does require a setup w/ a resistor could you send a diagram to my email, please? I accept attachments...thanks!]
ceej

Remember: Electricity to power vehicles still needs to come from somewhere...EV's are not the solution, alt nrg is.
cj6s
 
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Re: The Power of Iron

Postby windgat on Sun Jun 29, 2008 1:09 pm

Sounds like what you need is a watt/hour meter, or an Amp/hour meter. Or take frequent amp readings and integrate. Remember that for non constant current a digital meter can get confused and report an incorrect reading - I think that's how a few over-unity myths have started.

Also, an amp meter is (ideally) zero resistance, which means you need to connect a realistic load in series with the amp meter. Otherwise you may be getting some current, but at practically zero volts, so no power. US$ 0.02 per Kw/h is indeed too good to be true! Using the spike amps reading to calculate power will give a nonsense result - far too high.

Can I ask - have you compared the energy cost of producing iron to the energy you are getting out of it?
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Re: The Power of Iron

Postby cj6s on Mon Jun 30, 2008 11:25 am

Sounds like what you need is a watt/hour meter, or an Amp/hour meter.[I can try and look but have never seen Wh or Ah meters, except the ones for AC home power consumption, and I don't think they sell private individuals those.] Or take frequent amp readings and integrate. [The thing is what is a reading? I have taken readings every few seconds for about a three second duration and still seem to het readings that are a little too high. In desperation what I tried to do is use the actual number of electrons that are theoretically released by a kg of iron, so, the maximum, without accounting for losses, and maybe since my math is terrible I came up with a value that is less than half than what internet sources say is the available energy from full oxidation of a kilo of iron into Fe2O3!?!?!? The internet source from Answers.com stated iron gives 7.4 megajoules, which is about 2.06 kW/h. If I can extract even 75% of this figure in my battery I beat fuel prices in many countries, the thing is I can't even get a good reading to calculate my efficiency...]Remember that for non constant current a digital meter can get confused and report an incorrect reading - I think that's how a few over-unity myths have started.[Yeah, I agree with you there.]

Also, an amp meter is (ideally) zero resistance, which means you need to connect a realistic load in series with the amp meter. [What value resistor would you recommend? And when connected in series do I take the voltage reading as well ir connect the resistor across the terminals and read the V in parallel across the resistor with my meter? BTW, another weird thing is that when I measure voltage without a load, it starts low, about 0.3V and takes many minutes to climb to over 0.9?!?!?!? But it stays there...]Otherwise you may be getting some current, but at practically zero volts, so no power. US$ 0.02 per Kw/h is indeed too good to be true! Using the spike amps reading to calculate power will give a nonsense result - far too high.[How did you get your meter to stay still at an amp reading foe the pictures on your lead acid batt page? Do you have a hold button? The thing is my amps just keep going down to zero, quite quickly. But there is an electrochemical reaction going on for sure and iron isn't inert enough to give off less than 0.1mA. I mean, it rusts readily enough when no one is paying attention but seems to hide its secrets when I bring a meter near it.]

Can I ask - have you compared the energy cost of producing iron to the energy you are getting out of it? [Of course. One of my main objectives is to prove that iron nrg is cheaper than petroleum derived fuel here. And all I lack now is a proper amp reading. I have already discovered that rusty scrap is better than pure, shiny iron (because the "bad" readings are way lower for new iron) so I use the current price of scrap metal which is US$0.48/kg. So, yes, I measure the nrg efficiancy of iron by way of the cost method.] I just need a few bleedin' amp readings! Can I use the V = IR equation to figure out current if I know V and R (which would be the resistance across the terminals???)...?
ceej

Remember: Electricity to power vehicles still needs to come from somewhere...EV's are not the solution, alt nrg is.
cj6s
 
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Re: The Power of Iron

Postby windgat on Thu Jul 03, 2008 10:24 pm

This is how:
First guess the wattage of the battery (or wattage at which you wish to test). Eg. 2watts.
Measure the open circuit volts of the (charged) battery. Say its 8V.
Now calculate current: P = VI, so I = 0.25A
Now use V=IR, so R = 8 / 0.25 = 32 ohm

Put the resistor in series with a ampmeter across the battery. Put a voltmeter in parallel to the resistor.

Watch and note the volts simultaneously with the amps every say 10 sec.

If the volts and amps drop quickly, the resistor is too small for the test. Adjust the resistor (make it bigger, and repeat).
Finally calc the power your battery is giving by P = VI, and plot against time. To be useful, a battery needs to maintain most of its voltage over the expected useful life of the battery.

Hope that helps!

PS: Forget about the 'spike' readings.
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Re: The Power of Iron

Postby cj6s on Sat Jul 05, 2008 2:24 pm

I can make a guess using the theoretical power of the battery but that's it. I can calculate the total number of theoretical electrons released per gram that I use but the thing is I still won't have a rate of flow to calculate wattage...Is there any way to make an amp-hour meter using resistors and V=IR? I mean if the cell wattage can't be guessed? I have a light bulb that is 2.3V and 0.27 amps. My single cell won't light it at all but can it be used as a resistor of 8.52 Ohms? When the cell is in series with a tester measuring amps I get 0.01A with agitation of the cell and a parallel tester on the bulb gives me 0.01V (Open circuit varies from 0.3V to 0.95V) For example, half the time if I leave the tester connected to the cell open circuit, it will read 0.3 to 0.45 to start, and go up over an hour to 0.7 or 0.91V. Then, overnight, still connected it bottoms at 0.45 or 0.55. ... It all really confuses me. The chenmistry I got grips on but the freshman physics eludes me, man.
ceej

Remember: Electricity to power vehicles still needs to come from somewhere...EV's are not the solution, alt nrg is.
cj6s
 
Posts: 7
Joined: Thu Jun 26, 2008 10:36 pm

Re: The Power of Iron

Postby windgat on Sun Jul 06, 2008 1:20 pm

So then guess the power and go from there!

An amp hour meter needs sophisticated digital circuitry, so I think forget about making one from simple components.

If your cell won't light the bulb it means the cell is far lower power than the bulb. That bulb looks like about 0.6 watts, so maybe you should start at around 0.1 watts. Or maybe less.

My guess is that your cell is very, very very... low power (probably due to the speed of the chemical reaction and the surface area reacting) and so is capable of delivering only a milliamp or less. Which means it would need to be 1000 times bigger to start being useful (i.e. being able to deliver 1watt or more). Would that be practical?

PS: Putting an amp meter in 'series' with the cell is a bit of a useless exercise - all you are doing there is shorting out the cell. And if the cell ever starts to work well, you will blow the amp meter.
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