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What is heat-transfer coefficient alpha

  1. Apr 3, 2016 #1
    Hi everyone!

    I would like to confirm if heat-transfer coefficient α units in SI are W/m2 K. I have to use heat-transfer coefficient α in this formula:

    Heat-transfer resistance: 1/k = δ/λ + 1/α

    Where δ is material thickness and λ is thermal conductivity

    Thank you!
     
    Last edited: Apr 3, 2016
  2. jcsd
  3. Apr 3, 2016 #2

    BvU

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    yes and no. Depends on the interpretation of W/m2 K as W/m2 K (no...) or as W/ (m2 K) (yes!)
    see wiki
     
  4. Apr 3, 2016 #3
    Ok, thank you! And If I have two liquids and between them I have a copper plate. I must use 400 W/(m2 K) ;) as this website says: http://www.engineeringtoolbox.com/overall-heat-transfer-coefficients-d_284.html

    It's true?

    I'm obtaining incoherent results... :(
     
  5. Apr 3, 2016 #4

    BvU

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    I only see water - copper - water 340 - 455 W/(m2 ##\cdot## K) ?

    A bit difficult to help here: my telepathic capabilities are very limited :smile: .
    But you'll get a lot of assistance in the homework forum, provided you use the template properly ....:wink:


    (however, the ##\delta/\lambda## might indicate you need a different interpretation...

    and so does the 400 ... but those are W / (m ##\cdot## K) :nb) )
     
  6. Apr 3, 2016 #5
    I've just checked my problem and I think all it's OK now. Thank you for your time and for responding quickly!
     
  7. Apr 7, 2016 #6
    The OVERALL HEAT TRANSFER COEFFICIENT in the table is supposed to be the k in your first post.
     
  8. Apr 8, 2016 #7
    I thought the OVERALL HEAT TRANSFER COEFFICIENT was α, and k was the HEAT-TRANSFER RESISTANCE... Thank you!
     
  9. Apr 8, 2016 #8

    BvU

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    Thanks, Chet. I probably hinted too carefully in post #4 and forgot to follow up on #5 !
    Should have insisted Joaquin showed his work :rolleyes: !
     
  10. Apr 8, 2016 #9
    My english level is extremely basic, and the hint is a bit difficult to me. :frown:
    A part of my work is about to define all component in the formula at #1. I looked for in my old heat transfer notes but I didn't find anything like that formula.
     
  11. Apr 8, 2016 #10

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    In general there are three resistances in series: ##1\over \alpha_1## on the utility side, ##\delta\over\lambda## from the pipe or plate material and another ##1\over \alpha_2## on the process side.
    For the over-all heat transfer coefficient k ( in W / (m2 ##\cdot## K) ) you have $$ U = {1\over k} = {1\over \alpha_1} + {\delta\over\lambda} + {1\over \alpha_2} $$You don't show your work, so I don't know ##\delta##, but the middle term is often negligible. See the over-all k in water-copper-water of 340 - 455 W/(m2 ##\cdot## K) or the 400 you are supposed to use. A 1 mm plate ##{\displaystyle\delta\over\lambda} = {2.5 \times 10^{-6}} ## would contribute very little to the over-all resistance of ##\approx {2.5 \times 10^{-3}}##.
     
  12. Apr 8, 2016 #11
    The symbol U is often used for the overall heat transfer coefficient, so that could cause confusion with regard to your equation.
     
  13. Apr 8, 2016 #12

    BvU

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    Sleepiness. Thanks, Chet. ##1/U## it is. I'm used to ##\dot Q = UA\; {\rm LMTD}## and was too hasty. Coffee !
     
  14. Apr 8, 2016 #13
    Ok, this post has clarified me all! All my books and notes are in spanish with a little bit different nomenclature so I was a little bit confuse with those terms. At the end, it's just calculate the equivalent resistance...

    Thank you so much to all!
     
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