batteriesheat
How should I go about creating a heater from batteries (AA/AAA/9V/C/D any will do) that last as long as possible (the longer lasting the better)?
I have seen this post.
But that only lasts 45 mins and doesn't seem very efficient. Is there any way that I could make the heater last long? Maybe by making it not heat up as hot?
I need to have something heat up to \$60^{\circ}\$F deg (for example if a pipe is at \$30^{\circ}\$F deg, I want to heat it up to \$60^{\circ}\$F).
This is due primarily to internal corrosion of the lead frames and the electrodes, short circuits cause the sulfation of the lead. This causes sulfation, that makes the PbSO4 crystals merge into ever larger networks. This reduces the electrochemical active surface of the PbSO4. Due to the now smaller surface, the PbS04 cannot be "come off" that good , as it could before. So it takes very long until a sufficiently high concentration of Pb2+ is present. In addition, the electrical conductivity of the sulfate is lower than that of lead. The resulting increased Internal resistance of the load cell leads to increased voltage drop. Give that process enough time and your at the point where you wont have any cell voltage at all.
I hope this helped you.
P.S.: i didnt know a good word for "ablösen" in english, since unsolder is not the one i was looking for and peel of is not the right term. but i think youll still get it.
Constant LED brightness means constant current, and constant current means a current limiter.
simulate this circuit – Schematic created using CircuitLab
Select R2 as (VD2 + VD3 - VbeQ1)/ID1. Don't be afraid to use a current slightly lower than the spec (i.e. 3 to 6mA).
Best Answer
My gut feeling is that using that kind of battery to create any more than a hand warmer is not very practical. To get an estimate if the idea of a battery powered heater is doable you have to look at the energy needed:
When heating things the energy is used to do two things:
raising the temperature: to heat 1kg of water from 30°F to 60°F you need the following energy: (4180 J/K) * 16.7K = 70KJ (heat capacity times temperature difference in Kelvin). For reference: An alkaline AA battery has about 9360J (source). You can substitute your own data for e.g. your pipe (water filled?) combination to get the result for your application.
replacing the heat that is lost due to bad insulation: If e.g. your pipe is warmer than the ambient temperature it will get colder. Here you can also calculate the energy that is lost this way: P = (Temperature difference) / (thermal resistance) with P being the heat lost measured in W (= J/s)
I would start with plugging in the numbers for part 1 and then add an estimate for part 2 ("if I use insulation X, how much energy would I need to have the temperature be constant for time X?"). Then compare the energy needed with the amount stored in the type of battery you want to use.
I think you can assume pretty much 100% efficiency, but better have the batteries be warm or they will not deliver their full power.