Wow! Thank you for that deduction Michael! That is way better than I could have ever done. Inside the cell will be sealed off from air but the outside top is exposed, unless I seal that off too. Ideally the thing will fluoresce and emit a steady glow from the energy generated from the oscillations in the Rochelle and Epsom salts with even a 1.5 or 3 volt battery, 9 volt is the highest I'd like to go. Guess I'll find out...<br>
<br><div class="gmail_quote">On Fri, Jan 20, 2012 at 3:00 PM, Michael <span dir="ltr"><<a href="mailto:krazatchu@hotmail.com">krazatchu@hotmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div bgcolor="#FFFFFF" text="#000000">
This is a interesting challenge..<br>
<br>
The supply droop with resistive load will be a factor of the
batteries internal resistance. <br>
Which is an effect of temperature, chemistry and state of charge.<br>
<br>
Internal resistance can be measured by subjecting the battery to a
known load ...<br>
<a href="http://en.wikipedia.org/wiki/Internal_resistance" target="_blank">http://en.wikipedia.org/wiki/Internal_resistance</a><br>
Then applying (Nominal V - Loaded V) * (Load R) / (Nominal V)<br>
Or you can just figure it's about 0.2 Ohms per Alkaline cell.<br>
A 9V is sometimes made of 6 cells, so about 1.2 Ohms.<br>
<br>
Once you have the internal resistance figured...<br>
The circuit is a 9v voltage source with two series resistances,
internal + load.<br>
The voltage drop over the internal resistance will be 9 - 6 = 3v and
using i = V/R we get the current over the internal resitance...<br>
i = 3/1.2 = 2.5 Amps (there's going to be some serious self heating
which will skew the results)<br>
<br>
As the circuit is series, the current thru the internal resistance
is the same as the current thru the load...<br>
And the final voltage drop over the unknown load is the remainder, 6
volts. Using R = V/i...<br>
R = 6 / 2.5 = 2.4 Ohms ...<br>
<br>
Finally heat is a factor of power dissipation...<br>
Power is V*i = 6 * 2.5 = 15 Watts...<br>
<br>
I just tested this with a Rayovac 9v and a 2.7 ohm load...<br>
The results were a bit off, the internal resitance of the rayovac
must be higher than 1.2 ohms<br>
I got a supply drop to 3.4 volts and a 1.15 amps current. <br>
Using the above calculation, the rayovac has an internal resistance
of 1.73 ohms..<br>
So we could probably guess between 10 to 12 watts of power lost into
load...<br>
<br>
Temperature rise is a factor of thermal conductivity and geometry...<br>
I'd guess that the sugar cell has a pretty low thermal conductivity
and that 10 watts could heat a sugar cell to over 100c given enough
time..<br>
Combustion is going to require an oxidizer so YMMV....<span class="HOEnZb"><font color="#888888"><br>
<br>
Michael <br>
<a href="http://n0m1.com/" target="_blank">http://n0m1.com/</a><br>
<a href="http://krazatchu.ca/" target="_blank">http://krazatchu.ca/</a>
<br></font></span><div><div class="h5">
<br>
<br>
<br>
On 1/20/2012 1:49 PM, Darcy Whyte wrote:
<blockquote type="cite"><br clear="all">
I suppose if you can try it on a couple and see what the current
is, then you can know how many Watts. That would be a start. Then
you just have to figure out how much of the heat is in the load
and how much is in the battery. You can probably do that by
knowing the resistance of the load.<br>
<div>
<div>
<div>
<div><br>
</div>
</div>
</div>
</div>
<br>
<br>
<br>
<div class="gmail_quote">On Fri, Jan 20, 2012 at 1:33 PM, Emily
Daniels <span dir="ltr"><<a href="mailto:emily.daniels@gmail.com" target="_blank">emily.daniels@gmail.com</a>></span>
wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
Hi All,
<div><br>
</div>
<div>Does anyone know how much heat is produced by a 9 volt
battery when the resistance of the material it flows through
reduces it to 6 volts? What would the temperature be inside
the material? Thanks!</div>
<span><font color="#888888">
<div><br>
</div>
<div>Emily</div>
<div>
<div><br>
</div>
-- <br>
Emily Daniels | <a href="http://emilydaniels.com" target="_blank">emilydaniels.com</a>
| @emdaniels | <a href="http://awesomefoundation.org" target="_blank">awesomefoundation.org</a><br>
<br>
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<br></blockquote></div><br><br clear="all"><div><br></div>-- <br>Emily Daniels | <a href="http://emilydaniels.com">emilydaniels.com</a> | @emdaniels | <a href="http://awesomefoundation.org">awesomefoundation.org</a><br><br>