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Heating element temperature - correct calculation?

  1. Oct 4, 2011 #1
    Hi everyone. I am not sure if this is the best place for this question but here it goes.

    I am looking to build a small heating element to position in the base of a tank of water. I want to avoid any boiling of the water so want to keep the temperature of the wire at around 90 degrees C. The calculation I have used is as follows:

    Assume the heating element is a circular horizontal flat disc so the length scale is L=A/P where A is the surface area and P is the perimeter (as detailed in Fundamentals of Heat Transfer). (The design doesn't have to be a circular disc although it would be preferred).

    Calculate the averages Nusselt number as: Nu=0.54Ra^(1/4) which yields 13.62 assuming an ambient temperature of the water of 20 degrees C and all properties in the calculation taken at the average temperature of the setup (90+20)/2=55 degrees C.

    Then calculate an average heat transfer coefficient from h=(Nu kappa)/L, where the conductivity of nichrome was taken as 11.3.

    Assuming that the majority of heat loss will be due to convection away from the surface I then calculated the required power from Q=hA(90-20).

    Does this calculation seem reasonable? I anticipate I am doing something hideously wrong as keep getting power ratings of around 300W and I was expecting it to be much less for such a modest temperature. My naivety may be in the picking of the conductivity of the element: is there a better way to get an approximate rating of this value?

    Regards

    Smithy
     
  2. jcsd
  3. Oct 4, 2011 #2
    For a back-of-the-envelope approach, your numbers look reasonable at a glance.

    Also important for these heaters is the power density: power / surface area.

    See what value you get in watt / in^2.
     
  4. Oct 4, 2011 #3

    AlephZero

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    Compare the surface area of your heating element with an electric kettle, to see if your 300W seems reasonable. A typcial kettle element would have a 2 to 3 kW heater, and you don't get any local boilng until the water gets quite hot.
     
  5. Oct 6, 2011 #4
    Well doing the above calculation I seem to get a convection coefficient value of 6 x 10^4 which is surely way to high? That is doing the calculation exactly as listed above.

    I have heard of power density before, and presumably I would obtain this by just rearranging the above equation to Q/A=h(Ts-Tinf) where Q/A will be the power density? What does this actually show that is beneficial though? (I should point out I am not suggesting it doesn't show anything, rather I am genuinely interested in why it is important compared to just the power).

    The only trouble with the kettle scenario is I would be leaving the heating element on. I would like to know what the final temperature of the wire will be, or rather the maximal temperature if that makes sense?

    I basically need a setup where I can have a heating element of 400-500 W power but limited in temperature to 90 degrees C or less. Is this achievable? From my calculations above it would seem like it may be if I make the correct choice of wire and what not.

    Any tips would be much appreciated as has the help given thus far.
     
  6. Oct 6, 2011 #5
    I should add that the box will have openings so heat will be able to escape and not build up in the interior of the tank. So the only crucial thing is that local boiling close to the heating element is avoided.
     
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