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Rate of energy transfer: interior/exterior through glass

  1. Apr 13, 2009 #1
    1. The problem statement, all variables and given/known data

    The attached image shows the temperature as a function of the distante to a glass window. The window dimensions are given ( 60cm, 60cm, 0.5cm) and the thermal conductivity of the air and glass are, respectively, [tex]1W/(m.K)[/tex] and [tex]0,025W/(m.K)[/tex]. What is the energy transfer to the exterior through the window?


    2. Relevant equations

    I belive this is the 'rate of energy transfer', it mentions above:
    [tex]\frac{Q}{\Delta T}=\frac{A(T_{2}-T_{1})}{\sum_{i} d_{i} / k_{i}}[/tex]
    This equation is generalized for any number of materials joined together. the [tex]d_i[/tex] are the thickness and [tex]k_i[/tex] the thermal conductivity of each material and [tex]A[/tex] is the area of the surface the exchange is being made.


    3. The attempt at a solution

    So, I tried to solve this problem in all possible ways.. none of them gave me the right answer!
    The one that seems more reasonable is to consider the 'interior' as one material in wich the left extremity is at a high temperature, the exterior also as a material in wich the right extremity is at a low temperature. In the middle is, of course, the glass. So the above equation should look something like this:
    [tex]\frac{Q}{\Delta T}=\frac{3600cm^{2}30K}{(7,75+0,5/0,025+7,75)cm/\frac{W}{m.K}}[/tex]
    Simplifying, using meters we have - if I didn't screwd up- this:
    [tex]\frac{Q}{\Delta T}=31,3W[/tex]

    Wich is NOT the correct answer. The correct answer is [tex]1,75W[/tex]... :uhh:

    Also, this illustration is really troubling me. I don't really undestand the physics behind it. Except the fact that the temperature gets lower. But why is it constant in the glass?
     

    Attached Files:

  2. jcsd
  3. Apr 13, 2009 #2

    Mapes

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    A better equation to use would be Fourier's law of conduction,

    [tex]\dot{Q}=kA\frac{dT}{dx}=kA\frac{\Delta T}{\Delta x}[/itex]

    where [itex]\dot{Q}[/itex] is the energy transfer rate. Note that since the problem assumes steady state, the energy transfer rate is the same everywhere; in other words, you could take the values for k, A, [itex]\Delta T[/itex], and [itex]\Delta x[/itex] from any part of the diagram.

    This equation also explains the different slopes in the temperature diagram and thus could provide an answer to your last question.
     
  4. Apr 13, 2009 #3
    Thanks for the answer, Mapes :]

    I'm going to take a further look into Fourier's law of conduction (tomorrow! :P). But for the values from any part of the diagram I'm not sure. If I take the values in the middle of the glass, for example, it will be zero because there is no change in the temperature!

    PS: I suspect the diagram is wrong..
     
    Last edited: Apr 13, 2009
  5. Apr 13, 2009 #4

    Mapes

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    From the numerical data, it looks like glass conducts heat about 40 times better than air. What would that imply about the slope of the temperature line within the glass?
     
  6. Apr 14, 2009 #5
    That makes sense. The better the conductor, the smaller the slope! But I don't think 'small' means zero, right? Maybe the graphic wasn't supposed to be precise, or the slope can be neglected...

    Anyway, I tried using Fourier's law of conduction in the interior. I calculated the [tex]dT/dx[/tex] geometrically, and found [tex]-20/11\cong -1,818181...[/tex]. Using this result I came up with the final answer of [tex]58,18W[/tex]. Different from my first result and different from the correct answer too.

    You know that feeling, that you know you're doing something reeeally dumb, but doesn't know what it is yet? :bugeye:
     
  7. Apr 14, 2009 #6

    Mapes

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    I don't think it is zero. Once you calculate the rate of energy transfer, you can see what the temperature difference in the glass really is.

    Here you're just working too fast and not checking units. I have no idea where 58.18 W could have come from.
     
  8. Apr 14, 2009 #7
    It was really late when I wrote that! :D Let me elaborate a little more that answer.

    The idea is that it's a steady system, so the rate of energy transfer is the same in the three regions. I'll calculate this quantity in the interior region. First thing we want to know is the [tex]dT/dx[/tex], and for this we want to find [tex]T(x)[/tex].

    Using two points in the diagram, [tex](293K,-8cm)[/tex] and [tex](278K,-0.25cm)[/tex], with [tex]T(x)=a+bx[/tex] we have
    [tex]a=278K+0.25(K/cm)x[/tex]
    [tex]b=\frac{a-293K}{8cm}[/tex]

    Since [tex]dT/dx=b[/tex] we are interested in [tex]b[/tex]. And, solving the above system, we have that [tex]b=-(20/11)K/cm=-1.818181K/cm[/tex].

    Now that we have [tex]dT/dx[/tex], we can calculate the rate of energy transfer using [tex]kAdT/dx[/tex]:
    [tex]1\frac{W}{Km}3200cm^{2}(-1.818181K/cm)[/tex]

    [tex]1\frac{W}{Km}32m(-1.818181K)[/tex]

    [tex]32(-1.818181)W=58.18W[/tex]

    With this, we could find the [tex]Dt/dx[/tex] in the glass region! But I don't know why is this result wrong..
     
  9. Apr 14, 2009 #8

    Mapes

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    Ah, this may be the problem: the thermal conductivity values of glass and air are switched in your original post. The thermal conductivity of glass is much higher than that of air. Does this resolve the contradiction?

    (Also, 60 x 60 = 3600, not 3200.)
     
  10. Apr 14, 2009 #9
    lol
    Yes!! That solves the problem! :) But there are some things still troubling me. Using the differential form you posted I got something like 1,63W. Wich is close enough! But using the equation I posted I have exactly 1,74W! Can I really use the differential form with only one region?

    Anyway, the important part is done now. :) thank you very much Mapes! I'll post a solution in the next post.

    lol :rofl:

    I knew I was doing something dumb... Just didn't know it was two things! :P
     
  11. Apr 14, 2009 #10
    The solution

    The first equation I posted
    [tex]\frac{Q}{\Delta T}=\frac{A(T_{2}-T_{1})}{\sum_{i} d_{i} / k_{i}}[/tex]
    is in fact, according to the Wikipedia, (http://en.wikipedia.org/wiki/Heat_conduction#Fourier.27s_law") the integral form of the Fourier's law of conduction, and integrated over a "simple exponential situation" (and refers to a non existent diagram) Ok.. And then it's generalized for more materials, as I said before.

    [tex]\frac{Q}{\Delta T}=\frac{3600cm^{2}30K}{(7,75/0,025+0,5+7,75/0,025)cm/\frac{W}{m.K}}[/tex]

    (notice it's the same thing I did earlier, only with the thermal conductivity corrected)

    [tex]
    \frac{Q}{\Delta T}=1,74W
    [/tex]

    And that's it! :D
     
    Last edited by a moderator: Apr 24, 2017
  12. Apr 14, 2009 #11

    Mapes

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    OK, I see what you're doing. Note that if you used the correct 7.75 cm instead of 8 cm in your post #9, you would have gotten 1.74 W also, so the approximation by looking at the temperature slope in the air region alone is a valid one.
     
  13. Apr 14, 2009 #12
    It does appears that 8cm in post #7, but also appears 0,25cm, because I was using two points. I could have done it with 7.75cm and 0cm that the linear equation would have the same slope (wich is what matters for the differential form).

    It's possible that 1,63W is more correct, because the integrated form assumes something I'm not sure what it is. What is "simple exponential situation"? I hope this doesn't refer to the [tex]T(x)[/tex]. The one we have is linear :P (but all this would imply the answer of this question is wrong!)

    I'm going to take a look at this Fourier's law in the integral form to see if I can integrate it myself and reach the same equation I used.
     
  14. Apr 14, 2009 #13

    Mapes

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    Last edited by a moderator: Apr 24, 2017
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