67Views3Replies

Author Options:

How to construct a small battery powered immersible thermal element. Answered

Hello,

I am doing a project and part of the project is construct a small battery powered immersible thermal element. The thermal element has to be small and portable, capable of heating 6-8 ounces of water to 95-100 degrees F and hold at that temperature. I plan to use a nine volt battery or a couple of AA batteries. The thermal element has to be small enough to fit in a small 8 ounce bottle. Any suggestions as to how I would go about constructing such a device would be greatly appreciated.

Thank you for your help

Comments

The forums are retiring in 2021 and are now closed for new topics and comments.
0
Jack A Lopez
Jack A Lopez

2 years ago

I suggest you think about this in terms of energy (joules) and power (watts).

Also it will help if you use SI units, instead of this "ounces", "degrees F", nonsense.

For example, let us estimate the amount of heat (in joules) required to increase the temperature of 250 grams of (about 8.8 ounces) of water, from room temp at 25 C (77 F) to 35 C (95 F).

The specific heat of water is about 4.184 joules per gram per degC.

So Q_21 = C*m*(T2-T1) gives

= (4.184 J/g/degC)*(250 g)*(35 degC - 25 degC)

= (4.184 J/g/degC)*(250 g)*(10 degC)

= 10460 J = 10.46 kJ

Next consider the amount of stored electrical energy (again in joules) in a single AA, alkaline battery. I'm going to use the numbers from the Wikipedia article titled, "List of battery sizes", here.

I just multiply current capacity times nominal voltage, like so

U_1AA = (2.7 A*hour)*(1.5 V) = 4.05 W*h

=4.05W*3600s = 14580 J = 14.58 kJ

I notice the amount of heat to ramp up from room temp to warm, Q_21, is roughly the same size as, U_1AA, the total energy in a single AA alkaline cell.

For that reason, I am thinking the number of AA cells to use should be a lot more than just one. I am going to guess about eight of them would be a good number. Also I am going to wire them in series, so the total voltage is 12 VDC ( = 8* 1.5 VDC). This makes the total energy in the batteries, greater than the ramp up heat energy, Q_21, by a factor of about 10.

Next I am going to assume the initial temperature ramp up happens at a leisurely pace of 25 C to 35 C, in 600 seconds (or 10 minutes), and calculate the amount of power (in watts) needed for this, assuming constant power over this time interval, and also ignoring heat lost to the surroundings during this interval.

P = Q_21/(t2-t1) = (14580 J) / (600s) = 24.3 W

Next I assume the heating element can be modeled as a resistor. From Ohm's law, V=I*R, I know the power dissipated by a resistor, is P =V*I = V*V/R = V^2/R. Assuming the battery voltage stays constant, at 12 volts, I can calculate the size of this resistor, in ohms.

R = (V^2)/P = (12V*12V)/(24.3 W) = (144)/(24.3)

= 5.926 ~= 6.0 ohm

Note the current flowing through the heater-resistor is, roughly 2 amperes.

I = V/R = (12V)/(6.0 ohm) = 2.0 A,

So, in summary that is basically my gameplan right there. Use a series stack of 8x AA cells (nominally 8*1.5 = 12 VDC) , with wires and battery holders that can comfortably carry more than 2 A of current, and the heating element itself is a resistor.

I calculated 6 ohms, but I think probably anything in the range from 5 ohms to 10 ohms, would give similar results, in terms of how fast heat will flow out of it, i.e. in watts = joules/second.

By the way, the big unknown, that I have not mentioned yet, is the rate (in watts) at which this, uh, cup of coffee, or whatever it s, is losing heat to its surroundings. Note that if the surroundings are colder than 35 C, the power in those heat losses, whatever their magnitude, has to be supplied by your batteries.

Like ICeng suggests, if you make your container a vacuum insulated (e.g. Thermos (r) brand) bottle, you can make your heat losses very small.

If your container is made out of silver, or copper, with vertical fins on the side... well, you know, that is the opposite of thermal insulation. It would lose heat so fast, the batteries could not keep pace.

A styrofoam(r) cup with a lid, might be a good cheap container, with thermal losses somewhere in between those two extremes.

0
iceng
iceng

2 years ago

With Super thermal insulation like a vacuum enclosed bottle and starting at 99 degrees F there is a possibility it might work.

Wrap the correct Nichrome wire in Kapton https://en.wikipedia.org/wiki/Kapton

insulation in a thin Aluminum tube..

A 9V battery is less energy then AA batteries !