What is heat-transfer coefficient alpha

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    Alpha Coefficient
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Discussion Overview

The discussion revolves around the heat-transfer coefficient α, its units in the SI system, and its application in heat-transfer resistance calculations. Participants explore the interpretation of these units and the relationship between different coefficients in heat transfer scenarios, particularly involving liquids and solid materials.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • Some participants confirm that the units of the heat-transfer coefficient α can be expressed as W/(m² K), but there is ambiguity regarding its interpretation.
  • Others argue that the expression W/m² K can be understood differently, leading to confusion about its application in formulas.
  • One participant mentions using a specific value of 400 W/(m² K) for a copper plate between two liquids, referencing an external website for support.
  • Another participant provides values for overall heat transfer coefficients in a water-copper-water system, suggesting a range of 340 to 455 W/(m² K).
  • There is discussion about the overall heat transfer coefficient k and its relationship to α, with some participants expressing confusion over the definitions and roles of these coefficients in calculations.
  • One participant highlights the importance of showing work in calculations to clarify misunderstandings and assumptions.
  • Another participant notes that the middle term in the heat transfer resistance equation is often negligible, which may affect the overall resistance calculations.
  • Confusion arises from different nomenclature in various languages, impacting participants' understanding of the terms used in the discussion.

Areas of Agreement / Disagreement

Participants express differing interpretations of the heat-transfer coefficient α and its units, leading to unresolved disagreements about its application and the relationships between various coefficients in heat transfer equations.

Contextual Notes

Participants mention the need for clarity regarding the definitions of terms and the potential for confusion due to different nomenclature in various languages. There are also indications of missing assumptions and the importance of showing work in calculations to avoid misunderstandings.

Who May Find This Useful

This discussion may be useful for students and professionals interested in heat transfer principles, particularly those dealing with coefficients in thermal resistance calculations and seeking clarification on terminology and unit interpretations.

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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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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