I am trying to make a thermostatic water tank using an arduino. The tank will hold about a liter of water, maybe a little bit more.
However, I am kind of stuck when it comes to a solution to heat the water. I tried to use a relay to turn on a circuit that consisted of a battery leading to a thin wire that passed through the water. This was kind of a long shot, and unsurprisingly, it didn't work. The wire got hot, but simply did not heat up the water very effectively.
Does anyone have a suggestion for a heating solution I could use? The relay I have is a small hobby one and so I am not really comfortable with pushing much through it, and generally, I would welcome any solution that doesn't require me to deal with high voltages.
I would like it to get to at least 40 C, and as long as it takes less than about an hour to get to there, it would be fine.
Sorry for not including this stuff earlier.
To raise the temperature of 1000 grams of water from room temperature (22C ?) to 40C (18 C) would take 18000 gram calories or about 21 Watt hours or 21 Watts for one hour. You might try something like this wrapped around the container.
Look up the specific heat capacity of water on Wikipedia.
Then calculate how much energy you need to heat up 1L+ of water 20+ degrees C in <=3600 sec.
I calculated almost 30watt to only get there in a 100% insulated vessel.
I would say you need at least 50watt to compensate for the heat loss, depending on room temp.
Leo..
sr_6:
Thanks for the suggestion about the glow plug Idahowalker. That might work. Do you think it might be overkill for 1-1.5 liters of water?
One concern I have is when the water reaches boiling, will the container keep pressure to prevent air bubbles forming and can the container hold the pressure created. My next concern is about what to use, and the mounting/placement of a temperature probe.
For a temperature probe, I'd use a diode and measure the reverse current, you can look it up on the internet. A diode temperature probe can be covered with epoxy resin. Next the hole made for the probe must be sealed.
So, I was figuring:
(1 liter of water = 1 kg of mass * 2.2 = 2.2 lbs) and (18°C = 32.4°F) so 2.2 * 32.4 = 71.28 BTU = 20.89 Wh.
What is the correct ISO expression for that amount of heat?
Idahowalker:
One concern I have is when the water reaches boiling, will the container keep pressure to prevent air bubbles forming and can the container hold the pressure created. My next concern is about what to use, and the mounting/placement of a temperature probe.
For a temperature probe, I'd use a diode and measure the reverse current, you can look it up on the internet. A diode temperature probe can be covered with epoxy resin. Next the hole made for the probe must be sealed.
All in all quite feasible.
Water doesn't boil at 40C.
Use a DS1820 sealed sensor, easy to interface and use.
Amazon Prime has a 1.5 Quart (1.4 liters) "Slow Cooker" for $14 or a 2 Quart (1.89 liters) one for $15.07 if you don't mind the Captain America shield motif. :). Both have glass lids.
Just bypass the bimetallic thermostat with a relay module. I expect there is a thermal fuse to disable power if the temperature gets too high (above 110°C or so). Keep that part.
