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Other things to consider that may effect combustion...<br>
100 C is likely the vapor flash point within a narrow air:fuel
ratio.<br>
Submersion in vegetable oil would probably keep the oxygen out
but...<br>
<br>
A mixture will have entirely different properties depending on the
components.<br>
Other reactions besides oxygen + fuel could occur or be catalyzed by
heat + electricity...<br>
For example chlorine is a strong oxidizer, a candle will readily
burn in it.<br>
<br>
Have fun! <br>
<br>
Michael <br>
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On 1/20/2012 3:15 PM, Emily Daniels wrote:
<blockquote
cite="mid:CAH=qAzZnDeUPSWq+c_FYKFaRfjNKo7x4UOz0MGBJK1W+unRyQw@mail.gmail.com"
type="cite">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 moz-do-not-send="true"
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 moz-do-not-send="true"
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>
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target="_blank">http://n0m1.com/</a><br>
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target="_blank">http://krazatchu.ca/</a> <br>
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<div class="h5"> <br>
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<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>
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<div class="gmail_quote">On Fri, Jan 20, 2012 at 1:33
PM, Emily Daniels <span dir="ltr"><<a
moz-do-not-send="true"
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>
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-- <br>
Emily Daniels | <a moz-do-not-send="true"
href="http://emilydaniels.com"
target="_blank">emilydaniels.com</a> |
@emdaniels | <a moz-do-not-send="true"
href="http://awesomefoundation.org"
target="_blank">awesomefoundation.org</a><br>
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