[Lab] Odd Question

[Emily Daniels]

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...

On Fri, Jan 20, 2012 at 3:00 PM, Michael <krazatchu at hotmail.com> wrote:

>  This is a interesting challenge..
>
> The supply droop with resistive load will be a factor of the batteries
> internal resistance.
> Which is an effect of temperature, chemistry and state of charge.
>
> Internal resistance can be measured by subjecting the battery to a known
> load ...
> http://en.wikipedia.org/wiki/Internal_resistance
> Then applying (Nominal V - Loaded V) * (Load R) / (Nominal V)
> Or you can just figure it's about 0.2 Ohms per Alkaline cell.
> A 9V is sometimes made of 6 cells, so about 1.2 Ohms.
>
> Once you have the internal resistance figured...
> The circuit is a 9v voltage source with two series resistances, internal +
> load.
> 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...
> i = 3/1.2 = 2.5 Amps (there's going to be some serious self heating which
> will skew the results)
>
> As the circuit is series, the current thru the internal resistance is the
> same as the current thru the load...
> And the final voltage drop over the unknown load is the remainder, 6
> volts. Using R = V/i...
> R = 6 / 2.5 = 2.4 Ohms ...
>
> Finally heat is a factor of power dissipation...
> Power is V*i = 6 * 2.5 = 15 Watts...
>
> I just tested this with a Rayovac 9v and a 2.7 ohm load...
> The results were a bit off, the internal resitance of the rayovac must be
> higher than 1.2 ohms
> I got a supply drop to 3.4 volts and a 1.15 amps current.
> Using the above calculation, the rayovac has an internal resistance of
> 1.73 ohms..
> So we could probably guess between 10 to 12 watts of power lost into
> load...
>
> Temperature rise is a factor of thermal conductivity and geometry...
> 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..
> Combustion is going to require an oxidizer so YMMV....
>
> Michael
> http://n0m1.com/
> http://krazatchu.ca/
>
>
>
> On 1/20/2012 1:49 PM, Darcy Whyte wrote:
>
>
> 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.
>
>
>
>
> On Fri, Jan 20, 2012 at 1:33 PM, Emily Daniels <emily.daniels at gmail.com>wrote:
>
>> Hi All,
>>
>>  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!
>>
>>  Emily
>>
>>  --
>> Emily Daniels | emilydaniels.com | @emdaniels | awesomefoundation.org
>>
>>
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-- 
Emily Daniels | emilydaniels.com | @emdaniels | awesomefoundation.org
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