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I need some help finding the best way to make a lightweight heater. Lots more info in details!

So for one of my classes we work with a random company to solve some sort of problem that they have.  It’s an engineering class so the problems are all solved with physical things that we have to design; I just thought I should be clear that we weren’t solving financial problems or anything like that.  Anyways, my project has to do with keeping water in a tray (48mm x 60mm) at a certain temperature for a period of a couple of hours.  The water needs to be held constant at any temperature between 20 and 40 degrees Celsius.  My first thought was to use NiChrome wire to heat it but I don’t know how well that would work at lower temperatures so then I thought to use etched pcb as a heater but I’m not sure if I could get enough resistance from so little space.  The other restriction I’m facing is that you must be able to see clear through the center of the tray and whatever I’m using to heat it (lets just say a hole 1cm x 2cm must be left in the center) which reduces the resistance I could get from pcb even further given there is less area to run traces.  The heating apparatus must be as light as possible and I feel like electronic heating is my best bet since it is the easiest to control.  If you have any suggestions or think that my ideas are dumb or anything just let me know.  I’m happy to get any input here.  THANKS!

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frollard4 years ago
Steveastrouk has it...you need to do the math.

Work backwards:

This system needs a few things;
-heating element -temperature sensor/thermostat to control heating element
Distilled Water between 0 and 100C (at sea level) takes 1 joule heat energy in 1 mL (1cc) to increase 1 degree Centigrade.
Handily enough 1 joule per second is 1 watt.

Planning phase: So, you need to figure out how much energy your water loses to its surroundings, which will require you to know the ambient temperature. 40C water will transfer more energy to 20C air than 20C water, which will arguably transfer net zero energy. Once you are at the correct temperature, this 'leakage' between your thermal mass and the surroundings is the amount of energy you need to input to maintain temperature.
THAT is the amount of energy required (in joules per second or watts)

THAT is the amount of energy your heater needs to put out to maintain...

...and from that range (ish) you can decide on how much resistance you need for a given voltage power supply to generate that much heat in the water.
If you want to heat faster...you can do that, but your bare minimum wattage heater will be the heater that can maintain at the temperature you need to maintain at (or lower) in the ambient air you need to be in.

http://circuitcalculator.com/wordpress/2006/01/24/trace-resistance-calculator/
camtheman1283 (author)  frollard4 years ago
Thank you so much for the specific answer. Honestly this is stuff I could have figured out myself with enough research but hearing it from another person really solidifies the information. The link is great too. Is there any chance you could help me find the connective heat transfer coefficient for still water to still air. I cant find a good resource to accurately calculate the energy transfer to air to so I can find the energy needed to maintain the heat in the water. Also, after all your help I don't want to correct you but I'm pretty sure the specific heat capacity of water is 4.18 J/g°C. Too many damn Chem classes haha. Thanks again!
Right, crossed calorie with joule in my head. but the rest of the math still applies if you convert for joules :) my bad!

As for transfer water to air, I don't know - somewhere there must be a listing of the thermal resistance of the junction. For a 20 degree C difference, it's not zero, but its pretty small. You could calculate it by putting water in the container and measuring its temperature over time to see how fast it cools (which can be thought of in watts) - that would include the water/container/air/table and the water/air junctions.
camtheman1283 (author)  frollard4 years ago
Ah I love physics. It all just makes so much sense. That's awesome though I'm pretty sure that's about all I need to figure this out. You're the best. Thanks!
You're doing an engineering class. DO THE MATH. Either of the ideas you suggest are viable. PCB is entirely viable.
camtheman1283 (author)  steveastrouk4 years ago
I would do the math if I knew what math to do haha. I don't know the resistance on the PCB trace or how the current relates to the resistance and heat that it produces. I don't have the resources to experiment with any of this to test it out. If you or anyone else could help me find this kind of information thats really where I'm stuck.
What you need is very, very basic physics. You must have done high school physics to get to engineering classes.
camtheman1283 (author)  steveastrouk4 years ago
I learned quite a bit in my high school physics class but the equation for calculating the voltage drop over a certain length of copper trace wasn't one of them. The AP tests aren't quite that rigorous haha. Thanks for the help though
Really ? It was part of ours. Its trivially simple maths, which is why I am surprised. Resistance is proportional to length and inversely proportional to crosssectional area. The constant is the resistivity of copper. Haha.
onrust4 years ago
How could you NOT be solving a financial problem for this company.
camtheman1283 (author)  onrust4 years ago
I'm not sure what you mean.