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### Topic: How much heat will I need and how should I get it? (Read 2980 times)previous topic - next topic

#### masterwigglestin

##### Jan 23, 2013, 07:15 pm
Hello I am making a cell incubator which will be made from a styrofoam box and controlled by my Arduino. The inside of the box will have about 1 cubic foot of space and I am thinking about heating it with aluminum clad power resistors attached to a CPU heat sink. I don't know how much heat I am going to need or how many resistors create it to maintain 37 C. Can someone direct me to tutorials or tell me what kind of power I will need and what kind of transformer to get power from an American wall power socket? I have a general idea but I am knew to heating and to anything AC power. I am also very open to other types of heating elements if anyone knows of a better option.

#### majenko

#1
##### Jan 23, 2013, 09:20 pm
The amount of heat you generate doesn't directly affect the temperature you run at (as long as it is above the target temperature) - it more effects the speed at which you reach that temperature.

You need a feedback loop consisting of a temperature sensor which can report to the Arduino what the actual temperature is, and if it is below your target temperature then turn the heating element on.  If it is above it, turn it off.

The more powerful your heating element, the quicker you will reach the target temperature, and the less time it will spend below your target temperature.

Anything that generates heat can be used.  I don't know how big this cell incubator needs to be, so I can't recommend what is a good heat source.  Yes, resistors are fine.  There is simple maths that can be used to calculate the temperature they will run at.

The amount of power they will "consume" - V²/R, so if you run at 12V, and have 10? resistors, 12²/10 = 144/10 = 14.4W.

#### KeithRB

#2
##### Jan 23, 2013, 09:28 pm
You haven't given us enough information. If the box is insulated well enough *any* amount of heat will be OK. If you want it to heat up fast, 50-100W should do it. You should look into PID control to avoid overshoot since you don't have any way to *cool* the box.

You could probably get a few crock-pots from a thrift shop and scrounge the elements from those.

#### masterwigglestin

#3
##### Jan 23, 2013, 09:35 pm
I have the feedback loop taken care of. My question is more of how much would be a minimum number of watts to maintain the temp and increase it fairly quickly when it drops and on how I can figure out how much power I need / what kind of transformer would be needed to power the elements. I guess I would just like recommendations on what I should use.

#### KeithRB

#4
##### Jan 23, 2013, 09:40 pm
If you have the box, put in a 25, 50 and 100 W lightbulb and see how long it takes to heat up.  CFL's won't work well for this. 8^)

#### MarkT

#5
##### Jan 24, 2013, 01:27 am
If you want good insulation from heat radiation losses, perhaps line the box or the polystyrene cavity with aluminium foil.
Conduction is not the only way to lose heat.  Expanded polystyrene may well allow heat radiation through (far infra-red).
Maybe someone knows?
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### Chagrin

#6
##### Jan 24, 2013, 05:23 am
My plywood (egg) incubator was ~2 cubic feet with 1" of expanded polystyrene foam. I needed ~50W to keep 100F. I used two lightbulbs for heat with a total of 150W, but with the on/off time I'm guesstimating 50W.

It wasn't perfectly sealed because eggs still require some air turnover (oxygen) when incubating.

#### michinyon

#7
##### Jan 24, 2013, 05:31 am
Its is not too hard to calculate the surface area of your box,   the  R value of the insulation you propose to put on the outside,   the external air temperature and draft,   and calculate the heat flow through the box.

The problem is,  this will be a useless result for your purpose,  because most of the heat loss will come when you open and close
the box yourself,  not the heat which it loses when it is closed up and just sitting there.

Experience suggests that a low powered incandescent light bulb is not a bad solution to this problem.

There do exist heating resistors,   they used to be used in electrical supply cabinets to maintain a reasonble
temperature.    It looks like a fish tank heater but without the glass test-tube enclosure,   and designed
to work in air not water.

#### retrolefty

#8
##### Jan 24, 2013, 05:45 am
Just mount a standard AC lamp threaded fixture. You can always increase or decrease the bulb wattage size used based on actual performance of your controller's tuning values and the heat up time goal you desire. I wouldn't worry trying to guess or calculate the best lamp wattage to use from the start.

Lefty

#### dc42

#9
##### Jan 24, 2013, 03:58 pmLast Edit: Jan 24, 2013, 04:01 pm by dc42 Reason: 1
Assuming the box is thermally insulated apart from perhaps one Pespex side so that you can see inside, I suggest you aim for about 25W heating power.

If you use mains-voltage heating elements (e.g. incandescent lamps or possibly a heated plant propagator), then the low voltage power supply requirements (for the Arduino) will be minimal, however you will need an SSR or opto triac to control the heater from the Arduino.

If you use low-voltage heating elements, then you will need a power supply capable of powering them and the Arduino, such as a laptop power supply. Power resistors are quite pricey, so it may be less expensive to use one or two TIP100 darlingtons along with small signal NPN transistors to turn them into constant current devices. You would need power transistors or mosfets to drive the resistors anyway. [I normally disparage the use of darlingtons because their high saturation voltage causes them to run hot, but that's not a disadvantage in this application.]
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#### cjdelphi

#10
##### Jan 24, 2013, 04:14 pm
If you want to stick to low voltage, buy a couple of 12v automotive bulbs and control them via a mosfet / power transistor (mounting it inside for extra heat?)

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