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Heat loss question.

  1. Feb 8, 2010 #1
    Hello, I'm engineering a heat-pump for my senior project. My professor is out of town and these questions have been on my mind.

    I'm trying to calculate ACTUAL heat loss through the tent walls and ceiling.

    I'm using a tent 10ft x 5ft x 6ft.
    I'm trying to heat it with approx. 750 btus/hr

    There is one person inside

    I'm running calcs for heat transfer through thin nylon (k=.25 in SI) and am coming up with some incredible numbers.

    I then realized this is more of a natural and forced convection problem.... I know a temperature gradient forms due to lack of any real air movement and therefore wont transfer the approx 95000 watts/hr the nylon will transfer.

    How do I PROPERLY approach this problem?
  2. jcsd
  3. Feb 8, 2010 #2
    I'm assuming that you are not a Mechanical Engineering Student, otherwise you would have had Heat Transfer by the time you reach your senior year... indiependent of that, the basic approach you will need to take is the following:
    - Break your problem up into logical chuncks. If you have a classical (pyramid shaped) tent, you will need to consider three surfaces: the inclined side, the flat side/end and the base.
    - Analyze each chunk in terms of heat transfer on both/all sides of the material. For the inclined surface, you can use textbook values for the Nusselt Number to figure out your h on both sides. The same goes for the ends. The base is a bit different since you will have soil below it so only calculate h on one side, the other side will just be conduction into the ground.
    - Finally, add all the numbers together. You should have five surfaces (three unique) that conduct heat.
    - You may want to check your assumptions too. Do you need to calculate Radiation losses? Does the Nylon fabric really have an impact or could you just calculate the convective losses (my guess).

    This should be fairly straight forward if you have taken a Heat Transfer Class (and covered Natural Convection).
  4. Feb 8, 2010 #3
    Well I do love being called a liar right off the bat.

    I'm mechanical engineering with a specialization in manufacturing, meaning while everyone else was taking heat transfer i was taking advanced cad/cam, welding, metal casting (our school has a foundry)

    I know I need to "break it into chunks"

    What I don't know are simply non-calc formulas for calculating a thermal gradient through air.

    The high numbers for heat transfer are assuming the heat is in constant contact with the material, which it isnt due to a partial "blanket" forming within the air
  5. Feb 8, 2010 #4
    He didnt call you a liar. He assumed that you were not a Mechanical Engineer is all.

    Be respectful.

  6. Feb 8, 2010 #5
    Whoa... I wasn't calling you a liar, just made the assumption that you are not a ME. After all, that isn't stated anywhere. For all I know, you are a biology major doing some work on a green house... but please accept my appologies if I came across as anything but helpful.

    The basic thing is, there are no "simple" non-calc tables or formulae (what is a non-calc forumula anyway...). You will need to calculate the convection transfer coefficient ("h" for those who did take Heat transfer). The way to go about it is by finding the Nusselt number (Nu). For simple cases, there are simple correlations for Nu that you can use. This is the key point here. But you can't just "look up" the value for Q in a natrual convection.
    Engineering isn't that simple... I am sure you know that.

    If this sounds too complicated, I suggest you get the help from someone who did take Heat Transfer or at least read through their text book about Natural convection, Nusselt numbers and the like.

    Good luck.
  7. Feb 8, 2010 #6
    Sorry, I wrote the question when I went to bed, and read your response when I woke up... needless to say I was very groggy.

    I'm just trying to not have to differentiate through the layers of air, I know most equations are calc-based but as far as using calc, you dont have to :)

    I'll ask the "smarter" kid from the Applied Heat Transfer class today and see what he says, thanks.

  8. Feb 8, 2010 #7


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    If this is my doing this problem, I first realize that I'm probably going to have to live with a decent amount of error. Having said that, I find the heat convection coefficients (use either natural assuming no wind or even with wind). Then, I'll ratio those convections by the exposed area (as said before the sides will be slightly different than the sloped sides). This gives me one surface area and an equivalent heat transfer coefficient.

    From there I assume the interior is a lumped capacitance body, that is, uniform temperature. That means that at steady steady, you have a simple heat in = heat out. Heat out will be a function of interior temperature, exterior temperature and coefficient, heat in is given.
  9. Feb 8, 2010 #8
    Thanks minger! I'll start looking at the lumped capacitance equations.
  10. Feb 8, 2010 #9


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    Well, typically when we look at lumped capacitance (or it's sister semi-infinite solid) models, we are looking at unsteady response. The lumped capacitance model simply assumes that bodies internal heat transfer is much greater than its external heat transfer such that the body can be assumed to be uniform temperature throughout.

    In your situation, you would therefore need not look at the actual lumped equations, rather just take that general approach and write a heat balance. Since the inside of your tent is just a uniform temp body, you can essentially ignore it. Heat in is from your heater, and heat out can be determined from two thermal resistances. First the thickness of the tent material as conduction, and then convection to the outside.

    I'm basically just saying ignore gradients inside the tent.
  11. Feb 8, 2010 #10
    Well when I run the conduction I get something like 96000 watts.

    Does that sound right? Does this balance out somewhere else down the road?
  12. Feb 9, 2010 #11


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    Maybe you can show the calculation you used to get that number. I'm not even real sure what you mean by "that's the conduction".
  13. Feb 9, 2010 #12
    No problem,

    I was using Fouriers Law

    k = .25 for nylon
    A= 60sqft for a wall
    dT= 35deg F
    s = .003"

    This equation yields some insanely high transfer rates.... I'm assuming that is MAXIMUM thermal transmitting capacity of the material if heat were constantly applied to one side of it.... but internal gradients form which is why I was asking about the whole lump sum deal...

    According to the heat losses, I shouldn't ever be able to heat the tent.
  14. Feb 9, 2010 #13


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    I've found one of your problem(s)

  15. Feb 9, 2010 #14
    Are you talking the s value?

    If I change it to feet, the value is 12x higher :(
  16. Feb 9, 2010 #15


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    Write out your equation using units. Don't guess. What do you get?
  17. Feb 9, 2010 #16


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    HINT: You alluded to it in your first post...
  18. Feb 9, 2010 #17
    Well I reran the equation with the value of 1.7 BTUs/ft^2 deg F and got

    14280000 BTUs/sqft
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