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From calculations to actual design.

  1. Sep 2, 2013 #1
    Hello

    I am trying to put together a system - think of it as a mini oven that reaches temperatures of no higher than about 75-80 degrees C but normally up to about 50 degrees.

    Initially, I'm trying to find out what power would be required to warm a space up to that temperature.

    I started with the equations for some rough calculations
    Heat up 1 m^3 of air to 50 degrees in the time of 10 minutes.

    Started with:

    Q = Cp.m.T

    Q - Amount of heat (KJ)
    Cp - Specific heat (KJ/Kg K)
    m - Mass
    T - Temperature (Kelvin)

    Cp for air is 1.01 (pressure remains constant)
    mass = 1.294 kg (worked it from P = RρT)
    1 Watt = 1 joule per second

    Basically I worked out that it would take a measly 10.89 watts to achieve this.

    Q = Cp.m.T
    Q = 1.01 x 1.294 x 50
    Q = 65.347 KJ

    so to heat up this 1m^3, I'd need:

    Power = 65347 joules / 600 seconds
    Power = 108.91 watts.

    For a start.

    Does this seem feasible to people? Part of me thinks yes... I think of how much heat a 100 watt light bulb emits and that's made to generate more light than heat.

    If it doesn't please let me know where I might be going wrong.
    ----------------------------------------------------------

    The main question of this thread though is:
    What would be the best way to translate that 108 watts into power to heat up that space.
    e.g. a light bulb style filament - only slightly thicker?

    Is there any way of working out what length filament would be ok to use or is that not an issue?

    Many thanks for contributions.
     
  2. jcsd
  3. Sep 2, 2013 #2

    mfb

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    You will most likely need much more power, to heat the walls of the container and to account for heat loss to the environment. Heating air is not an issue usually, heating everything else is.
     
  4. Sep 4, 2013 #3
    Thank you for the reply.

    Thing is... if you heat up the air and continue heating up the air, surely the contents will heat up also wouldn't they?
    But surely... if the power is constant, (using a thermometer and a cut off switch of some kind, to keep the temperature at 50-55 degrees) objects in there will eventually heat up themselves and maintain that heat until the power is eventually turned off.

    What would more power achieve other than reaching the desired temperature more quickly?
    Conventional ovens do just that don't they (of course they can reach 200+ degrees hence more power) but don't they just reach a particular temperature and then maintain it?

    I suppose what I'm trying to establish is whether there is a low power solution to low temperature ovens.

    Any thoughts?
    Or anyone else?
    Many thanks in advance.
     
  5. Sep 4, 2013 #4

    mfb

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    Sure. How is that related to your previous questions?

    Reach it at all.
    Try to cook water in a swimming pool with a small water cooker. It will not take a long time, it will be impossible. Heat will just get conducted away and the temperature in the swimming pool does not change significantly.

    Usually they do that. They need several kW to heat up, and still a lot of power just to maintain the temperature. A lower temperature reduces the required power, but 100W for a surface area of several m^2 looks problematic.
     
  6. Sep 4, 2013 #5
    A few more considerations:
    * Where do you put your heating element?

    * Where do you put your temperature sensor? If it is too close to the heating element it will shut it off before the rest of the space reaches 50 deg. If it's too close to the wall, it will mainly see the temperature of the wall. If it's too far from the heating element, there will be a delay between the time you turn off the heating element and the time the temperature sensor stops increasing in temperature.

    * When you turn off your heating element, it will still be hotter than the air around it so it will continue heating it for a while. How can you avoid overshoot.

    * At what temperature do you turn your heating element on again?
     
  7. Sep 5, 2013 #6
    Thanks for the responses:

    I was responding to where you said that heating air is not usually an issue, heating everything else is.

    I also agree with your swimming pool analogy. I'm just following the maths as to how to heat up a 1 m^3 space. The actual space used will eventually be smaller anyway and I will be using more than 100 watts.
    As for ovens, they are higher power some can be as low as 800 watts if built into microwaves (like ours in the kitchen). The space is smaller, I agree but it's the principle.
    Appreciate your responses though, they are real food for thought.


    Skeptic2
    The things you're asking are some of the questions I'm hoping to have some input on here.
    Thinking of turning the element off at about 52/53 degrees (if looking to maintain roughly 50)
    and back on at about 45-47.

    As for the element.... I don't know where to put it. Is it even the best method to use?
    How thick and how long can the element be given - say - 200 watts?
     
  8. Sep 5, 2013 #7

    meBigGuy

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    You need to consider 1. the thermal conductivity and thermal mass of the walls and 2. their radiation into the ambient, and 3. the minimum ambient temperature. You need enough power to supply what they will radiate at your max temperature (and minimum ambient). If you supply that much power, you will eventually reach that temperature (could be a long wait). To reach it in 10 min you run the calculations you already did with those elements included. I didn't check your calculations.

    skeptic2's points are all important also. The nature of your controller is a significant issue. What you require in that regard is determined partially by how stable and accurate the temperature must be. There are many different controller algorithms that determine the behavior and specifications of your closed loop thermal system.

    You also need to consider how fast you might want to heat a thermal mass that you might place in the already heated oven, and how much the oven temperature should drop during that process.
     
  9. Sep 5, 2013 #8
    You are missing the point - (normal) ovens don't heat things by surrounding them with hot air, they heat things by radiation inwards from the walls. Forget about the air, as the Big Guy says you need to provide heat input to the walls as fast as they can radiate it away.
     
  10. Sep 5, 2013 #9
    Consider the convection currents generated by heating the space. If the space is cubical and the heating element is located bottom center, I'd expect the air to rise from the element to the top of the cube and then to follow the walls back down to the heating element. I would locate the temperature sensor in the center of the cube, if possible, for a quick response from the heating element. In addition, proportional control by adjusting the voltage to the heating element as a function of error between actual and set temperature can increase your performance.

    Furthermore, you can anticipate ahead of time when to turn up or down the heating element by adding the slope (derivative) of the temperature curve to the temperature value. This type of controller is called a PD (Proportional Derivative) controller. By adding the integral of the error temperature voltage you can maintain a very tight control of the temperature. This is called a PID controller. One nice feature of PID controllers is that the integral and differential signals tend to be out of phase with each other which inhibits oscillation.
     
    Last edited: Sep 5, 2013
  11. Sep 7, 2013 #10
    This is really fab. Thanks for the replies. It has made me understand some things I didn't before and think of new things.

    I have taken the point on board that ovens work by radiating heat from the walls.

    Metal walls can be used to radiate the heat.... Metals seem to have a pretty low specific heat - compared to things like air and definitely water so using the same formula I used above... for something like 3 kg of copper (Cp = 0389) I won't be looking at THAT much more power to heat up the copper than I would the air.

    As far as Thermal conductivity goes - meBigGuy - I'd expect to have to consider it for air... rather than the walls (say they are copper) because once the walls have heated up, then (please correct me if I'm wrong) then the heat has to be conducted through the air.
    Although, I'm not sure how to include the thermal conductivity calculation... I know the equation but I'm not sure what the answer means.. if anyone knows.. please let me know - can it include time?

    For this... I believe I can still use the equation I used at the start.. Q = Cp.m.T .. but apply it to the walls rather than the air.. heat up the walls and you heat up the oven (after a while)... Of course... need to keep in mind that they will cool down too so say the insulation is pretty good.

    So if I choose a single square copper plate (density 8290 Kg/m^3) as the base (rather than the wall) of this 'mini oven' (0.4 x 0.45 x 0.003)... keeping 50 degrees and 10 mins... same calculation as above... I'd need power of 145 watts.

    In the end... the idea will be to heat the oven up and keep whatever is inside there for several hours so temperature drops when opening any doors, will not be as important as purely heating up the oven to the desired temperature of about 50 degrees.

    Is what I'm saying making sense?
    Am I thinking along the right lines?
    Do you think what I'm proposing could work - particularly on such low power?

    Any other comments also welcome... e.g. other things to consider... other ways of heating? (e.g. a slightly thicker light bulb style filament) etc

    PS. I understand I still need to think about the closed loop thermal system. I need to look that up before I can start asking questions about that... before I start thinking where to put the thermometer.
     
  12. Sep 7, 2013 #11

    mfb

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    Copper has a low heat capacity per mass, but its density is several thousand times the density of air. Per volume, it needs much more energy to get heated up. In addition, it is a very good heat conductor - if it has contact to cooler objects somewhere, you increase the amount of stuff you have to heat even more.
     
  13. Sep 7, 2013 #12
    Yes totally... I took that into consideration in my calculations.. Feel free to check. It's not impossible that I made a mistake in my numbers.
    The conductivity shouldn't be a massive issue if what it touches is a very bad heat conductor such as wood.
     
  14. Sep 10, 2013 #13
    A 100-watt light bulb can heat things up pretty good. The Easy-Bake Oven used to have 2 of them and it reached temperatures you could bake with. Sure, a lot of it was the size of the enclosure and the insulation used, but that's not a bad feat of engineering in my opinion.

    Really, like mfb says, the problem isn't the heating, it's the heat loss. As you're heating the air, the air is heating the walls of the oven, which is also heating the outside of the enclosure. You have to dump heat into the enclosure faster than it leaves, or else you're never going to get it. If it were me, I'd make it out of sheet metal and put some brick (or something with a lot of thermal mass) or fiberglass insulation around it. You can try to heat things up with light bulbs if you want, but it would be a lot more efficient to use a dedicated heating element. Also, I'd use a thermocouple with a PID controller to monitor and control temperature. As well, PID controllers and thermocouples aren't that expensive, so it's a real option for anyone. It just requires a bit of know-how.

    I guess it really comes down to what your design goals are. Is it optimizing your design for efficiency or actually making something that works? Most people I know with pet projects like this really don't want to try to build anything; they just like the mental exercise.
     
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