What is heat-transfer coefficient alpha

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The heat-transfer coefficient α is discussed in terms of its units, which can be interpreted as W/(m² K) depending on notation. The formula for heat-transfer resistance includes components such as material thickness (δ) and thermal conductivity (λ). Users express confusion over the overall heat transfer coefficient (k) and its relation to α, clarifying that k represents the overall heat transfer coefficient in W/(m² K). The discussion highlights the importance of understanding the series resistances involved in heat transfer calculations. Ultimately, participants resolve their confusion regarding the terminology and calculations involved.
joaquinjbs
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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!
 
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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
 
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BvU said:
yes and no. Depends on the interpretation of W/m2 K as W/m2 K (no...) or as W/ (m2 K) (yes!)
see wiki

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

joaquinjbs said:
I'm obtaining incoherent results... :(
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) )
 
BvU said:
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) )

I've just checked my problem and I think all it's OK now. Thank you for your time and for responding quickly!
 
The OVERALL HEAT TRANSFER COEFFICIENT in the table is supposed to be the k in your first post.
 
Chestermiller said:
The OVERALL HEAT TRANSFER COEFFICIENT in the table is supposed to be the k in your first post.
I thought the OVERALL HEAT TRANSFER COEFFICIENT was α, and k was the HEAT-TRANSFER RESISTANCE... Thank you!
 
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: !
 
BvU said:
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: !

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.
 
  • #10
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}}##.
 
  • #11
BvU said:
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}}##.
The symbol U is often used for the overall heat transfer coefficient, so that could cause confusion with regard to your equation.
 
  • #12
Sleepiness. Thanks, Chet. ##1/U## it is. I'm used to ##\dot Q = UA\; {\rm LMTD}## and was too hasty. Coffee !
 
  • #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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