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Small battery powered heating element for Biology experiment?

Hello everyone!

I would like some help designing a small heating element. It will be battery powered (and voltage hopefully 3 or 4 C or D batteries).
It should fit on the top of a quart mason jar, as I will be heating the milk inside to about 90 degrees Fahrenheit for a period of one week. I would hope that battery changing is not necessary, but if so, any help is wanted. I have soldering skills, and would prefer to make as much as possible by hand. I also have lots of spare parts and wires, as my dad is an HVAC Chiller Mechanic.

A coil of wire attached to batteries slid through holes in the lid might work?

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iceng1 year ago

You will need to use a comparator http://www.mouser.com/ProductDetail/Texas-Instrume... IC https://en.wikipedia.org/wiki/Comparator that turns the heater on and off by comparing a set-point to a thermistor in the wg...

http://www.amwei.com/views.asp?hw_id=66

When the milk = white-goo = wg is below the set-point it closes a MOSFET transistor and heats a coil of wire in the wg as soon as the thermistor is above the set-point the power is turned off and when the wg cools the cycle repeats.

BTW I do not believe the batteries will last a week even with a 9 volt battery to run the logic.

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Thermistors for temperature circuit.bmp
EdmondD3 (author)  iceng1 year ago

or i will use Iceng's diagram and connect to the wall power.

EdmondD3 (author)  iceng1 year ago

I will use a wall wort. I probably will construct something along these lines;

http://www.electroschematics.com/8998/arduino-temperature-controlled-relay/

I will just splice in a heating coil in place of the fan. This will be the only design step I may need help with. How to build a coil attachable to the schematic. Insulation, voltage, etc...

Why battery powered ?

EdmondD3 (author)  steveastrouk1 year ago

For the purpose of portability and scale. Small, complicated things are fun!

Why on earth would you want to carry 2 pints of rancid milk with you ?

EdmondD3 (author)  steveastrouk1 year ago

It's for an experiment to test the pH of milk undergoing decay in an incubated form. There is a need for warmth to speed it along. I am looking to find the colonies in the milk and count their existence. My teacher told me toooooo!

That doesn't answer the question. Why carry it around ?

EdmondD3 (author)  steveastrouk1 year ago

Not to carry around, but I guess I could use a power supply. Then I might have a breaker trip or blown fuse. I'd rather keep it away from the wall.

Lets say it will take as little as 1W to do this, if you use a vacuum flask as the container, a wall wart is going to have more than enough power to do it, but a battery would need to hold 1 x 3600 x 24 x 7 Joules = 605,000 Joules. Lets say you have three D cells in series, they hold roughly 25000 Joules

EdmondD3 (author) 1 year ago

I will try a heat loss experiment, figure the loss through an insulated container and report back. In terms of deltaT. Then comes the question of building. Hopefully as cheap as possible with in home materials( i.e. copper wires, batteries). It does not have to remain exactly 90 degrees, but that would be an average temp. I may be able to scrounge some thermostat parts from my dads' scrap. If an Arduino is necessary, I may have to pay. Though, science projects for high school are not big on my spend list:)

Use a vacuum flask to minimise your losses, and report back.

So you want a small volume, about one quart or one liter, in volume, where the temperature stays almost constant, at 90F or 33C.

I am imagining part of your design is an insulated box, for to keep in the heat of the warm stuff inside the box. Moreover, I am imagining the temperature outside the box is always colder than 90F or 33C, and the heat moves through the insulated box in the usual way, from a the hotter inside of the box, to the colder outside.

Because heat is always being lost, maintaining the temperature inside the box will require some way of replacing that heat.

A question you should be interested in is: How fast is heat being lost?

Because the answer to that question tells you how fast heat must be added, by way of your resistive heating element, or whatever.

By the way, I don't know how fast your insulated box loses heat, because I'm not, like, clairvoyant, you know. However I do have some general expectations for the problem, the most important of which is I am expecting heat flow to be directly proportional to temperature difference between the inside and outside of the box, and I think I can explain how you can measure the rate of heat loss for an experimental set up, and thus get an estimate for the heating rate (in watts) needed to keep the inside of the box warm.

I mean, at this point I don't even have a wild guess for how much average heat power is required. So I could not guess if it even possible to supply that power with batteries.

Anyway, you can do some experiments to find out how fast an insulated box loses heat. For example, you build your box, made to fit your liter sized glass jar sample container. Then fill the sample container with hot water, like 40 C or so, then using some thermometers, preferably the kind on long wires, you watch the temperatures inside and outside the box, and make a graph of the temperature difference (also called deltaT), as a function of time.

The graph you are expecting is an exponential decay.

https://en.wikipedia.org/wiki/Exponential_decay

The temperature drop is fast at first, while deltaT is large, and slow later, as deltaT approaches zero.

The rate at which the liter of liquid water is losing heat is almost directly proportional to the rate of temperature loss. A drop in temperature of 1 degree C (or K) corresponds to a loss of about 4200 J of heat.

https://en.wikipedia.org/wiki/Properties_of_water#...

(see table titled "Constant pressure heat capacity")

So, for example, if one kilogram of water is losing 1 degree C per minute, that means the heat loss is deltaQ/deltat = (4200 J)/(60s) = 70 J/s = 70 W. So now you have heat loss in units of watts, which are conveniently the same units you work with designing electric heating elements, e.g. power dissipated (as heat) by a resistor:

P = V*I = I^2*R = V^2/R

(P is in watts, V is in volts, I in amperes, R is in ohms)

Note that I am kind of expecting heat loss to be proportional to temperature difference. So if the loss is 70 W, at deltaT = 33 - 23 = 10 C, (outside the box is 23C) the loss will be twice as much at twice the deltaT, i.e. 140 W when deltaT=33-13 = 20 C.

Some kind of closed loop control, also called a "thermostat" may be desirable; i.e. something that watches the temperature, and turns on heat power when it starts to drift too low from the target temperature, and conversely turns off heating when the temperature is just right, or too high.

If you are comfortable with building things with microcontrollers, e.g. Arduino, then this kind of temperature control should not be too hard.

But like I said previously, I think the most important part of this problem is getting a good estimate of how much heating, time averaged, in watts, is actually required for keeping your sample warm.

EdmondD3 (author) 1 year ago

The heating element will be submerged in the milk

EdmondD3 (author) 1 year ago

It will be on and heating for a whole week.