[Stoves] Hand Crank Dynamo for Fanned Stove?

[Robert Penn Taylor]

Paul, David, Crispin, and the other half-dozen folks who've weighed in 
on this topic:

David G. LeVine wrote:
> At 06:49 PM 12/4/2007, you wrote:
> I just looked at three "muffin" fans (computer fans.)  They were "5V, 
> 0.55W" or 0.11 Amperes, "12 Volt, 0.08 Amperes" and "12 Volt, 0.12 
> Amperes"  Now, a typical LED runs about 3.5 Volts for white and can 
> be in the 20 mA to 1,500 mA range.  The 3 LED flashlight is PROBABLY 
> running either no more than 10.5 Volts OR has a complex 
> driver,  

More likely the LEDs are wired in parallel, and the battery provides 3.5 
Volts.

After typing that last sentence, I dug out my 3-LED handcrank flashlight 
and took it apart. Here's the story: the LEDs are wired in parallel. The 
battery is 3 button-style NiMHs wired in series, the total stack 
nominally providing 3.6V. The dynamo provides ~5V (in case it isn't 
obvious, the battery "charger" -- the dynamo in this case -- must 
provide a higher voltage than the open voltage of the battery or no 
charging takes place.)

Now, the batteries in this flashlight are rated at 80mAh capacity, so 
3.6V * 80mAh * 3600s/h = 1037 Joules. When you're looking at battery 
_capacity_, that's the important metric: Joules. Not amp-hours, not 
volts, not amps, not Watts, but Joules. This tells us how much energy 
can be stored by the battery. It comes out at some set voltage (in 
theory, but reality is a little different), but we can easily transform 
that voltage to a higher or lower voltage, so it's really Joules that we 
care about.

Paul said his fan requires 4 Watts. The question now becomes how long 
will it takes to dissipate 1037 Joules at 4 watts (Joules/second)? Well, 
that's 1037/4 = 260 second = ~4.3 minutes. That's if the battery in this 
flashlight is fully charged. The other question about whether 3.6V will 
run the fan is less important because we can always transform the 
voltage with a DC-DC transformer (which, it's true, will eat away some 
of the energy coming through.) The 0.55W muffin fan David mentions would 
give 1037J/0.55W = ~30 minutes.

The next question is then: How many turns must the handle on the dynamo 
be cranked in order to fully charge the battery? Unfortunately I don't 
have a multimeter here to hook up to the dynamo to find that out.


On to the pulse-width modulation unit that Paul is using on this thing: 
There have been a couple posts about whether the PWM is providing 4 
Watts, 2 Watts, etc. Hopefully the following will help clarify the matter.

For starters, Crispin mentioned that
1 Volt*amp*second = 1 Watt
which is not quite right and should instead read:
1 Volt*amp*second = 1 Joule
         or
1 Volt*amp = 1 Watt

With that settled, it seems there may be some confusion on what the PWM 
does. The idea behind a PWM unit is that it does not provide power to 
the load continuously. It pulses on and off. If the fan is set up to 
draw "4 Watts" (that is, it assumes it will get 0.3A at 12V), then when 
the PWM is sending power to the fan, it's sending 4 Watts. But when the 
PWM is not sending power to the fan, it isn't sending anything. If the 
PWM is set to reduce the fan speed by one-half, then it turns on and off 
very rapidly, but it is on for exactly the same amount of time that it's 
off. If you look at the total power dissipation over one second, it's 2 
Watts since
((0.5s * 4W) + (0.5s * 0W))/1s = 2W

Hope that helps somewhat in this discussion.

-Penn Taylor