# Heating air through resistor heat dissipation

I am working on a project and have been delving into electronics. I am still quite new, however, so I would love to hear any info/advice you have!

I would like to heat incoming air through a short pipe to a certain temperature. This would be heating up air from inside (60°-75°) to no more than 80°F so a 20°F increase at most. Power would come from a 12V source, and be regulated by a microcontroller.

I am looking at ways to create this heat, and would love to get some advise. Would a typical (large) carbon resistor do the trick? I have looked at getting some Nichrome wire, but that may be overkill for the small amount of heat I want to dissapate.

Your answers (and more questions) made me realize I am missing some info to calculate this. In particular I will need the air volume flow rate before getting numbers.

I think I will try a few different options and find the best one experimentally. Thank you all for your help!

• How do you propose to control that temperature? Or is this to be "open loop?" – jonk Jan 17 '18 at 21:51
• There are power resistors. Or specially designed for this purpose heaters. – Eugene Sh. Jan 17 '18 at 21:52
• A big factor will be how much heat (not the same as temperature) you will need; how much air is being heated, and whether the pipe will try to absorb it along the way. – gbarry Jan 17 '18 at 21:54
• Incoming air suggests air that is moving. If so, you may need significantly more heat than still air. A hair dryer, with blower fan has a heating element that glows. Non-moving air might use a lightbulb to get to those temperatures. – glen_geek Jan 17 '18 at 21:58
• Work out the mass flow rate of the air (kg/second). You can get this from the volume rate (litres/second) and density (kg/litre). Multiply by the heat capacity times temperature rise (J/kg/K) * kg/s * dT giving J/second, i.e. Watts. This is the power you need. (Add a bit extra for heat losses). Choose a resistor to suit. – Brian Drummond Jan 17 '18 at 22:00

I have done this to keep high voltage from leaking across circuit boards. As part of a POST sequence I used several 12 VDC 100 ohm 5 watt ceramic resistors epoxy mounted across the chassis under the board. They dissipated about 1.44 watts each and locally raised the temperature 10 to 15 degrees F.

The purpose was to burn off any moisture accumulated during the night. The board used a SMPS to generate +2.8 kV for a constant-current supply to test MOV's. The high-voltage would not be turned on until software (LabView) detected that the chassis was at least 90 degrees F.

Since it was a sealed enclosure the radiant heat accumulated mostly on the back plane under the board, so other items such as transformers and the couplings to the PC were not exposed to direct heat.

To prevent overheating fans were set to come on at 120 degrees F by a series of parallel thermo-disc. At 150 degrees F other thermo-disc would cut off the resistor heaters. This avoided a run-away condition were the fans and heaters would cycle often in competition for control of the temperature.

If you only want such a slight increase in temperature you can use 200 ohm resistors or 2 each 100 ohm in series. Now you are down to just .75 watts of heat which adds only 5 degrees locally. It is up to you to decide how 'dense' this array of heater resistors are. If you pack them in less than 2 inches apart the heat will rise quickly all over. Be sure to include thermal cut-off disc of 80 degrees or so.

The resistor formula was 12v/100 ohm = 120 mA. 12V*120 mA = 1.44 watts.

If heating is a normal operation you should not do any calibrations until a set stable temperature is reached.

• Not my downvote but how did you get 1.44W heat out of a 1K resistor with 12V across it? – Brian Drummond Jan 18 '18 at 13:34
• @BrianDrummond. My bad. I should not be doing math at 2:00 am. – Sparky256 Jan 18 '18 at 15:42
• @BrianDrummond. I have corrected my silly mistake and added the simple equation to prof it. – Sparky256 Jan 18 '18 at 20:53
• I do not understand the continued down vote. The math issue has been corrected. If it is locked in place then so be it. – Sparky256 Jan 19 '18 at 0:14
• Neither do I. It wasn't mine. – Brian Drummond Jan 19 '18 at 1:56