Calculating heat transfer coefficient from experimental data

  • Thread starter Jakob81
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  • #1
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I have a graph which shows the rate of cooling of a tent from about 35 deg C to 15 deg C, it looks like this:

http://students.bath.ac.uk/en0jma/graph.gif [Broken]

How do I work out the convective heat transfer coefficient of the tent from this data?

Thanks for any help
 
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Answers and Replies

  • #2
FredGarvin
Science Advisor
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Is this free convection or forced convection? What is your experimental set up?
 
  • #3
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this is free convection.

The tent was heated up, and then allowed to cool naturally. The graph is the results of the thermometer readings taken from within the tent.

Ive been working out the reduction in energy of the air between two temperatures, say 34 and 20, and then dividing by the time taken to reduce heat, this gives X watts of heat (Q)

i know I should be using the equation

Q = hA(delta T)

to try and work out h, but im not sure if im using the right values for delta T, already know A, the area of the tent walls.

I want to find h, the convective heat transfer coefficient by re-arranging the above to,

h = Q / A(delta T)
 
  • #4
russ_watters
Mentor
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You are doing it the way I would. In the beginning, when the slope of your graph is near linear, is probably the best place to calculate a Q and a delta-T to plug into the Q=hA(delta-T) equation - say, the first 100 seconds. Otherwise, you are averaging the slope (and Q) over a non-linear period using a linear equation.

To check, plug your numbers back into the equation you've built and see if Excel generates the same graph as your experimental data.
 
  • #5
FredGarvin
Science Advisor
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My 2 cents worth here...

I would pretty much do it the same way, but if I were going to use this number in any kind of calculation, I'd extend the time frame of interest to a point that appears to become asymtotic to [tex]T_{\infty}[/tex] and curve fit a straight line there. That way you will get a bit more conservative average heat transfer coefficient that is a bit more representative over the broader temperature range. That is my opinion though.

If you're not going to do that then I'd do it in the linear portion like Russ mentioned.
 
  • #6
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Cheers, thats exactly what I've done, and it seems to work! Thanks very much for the help. :smile:
 
  • #7
Hi, i think am having the same issue as the guy above, except i'm not sure how to calculate the Q value needed to find heat transfer coefficient. Thanks.
 

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