# What is heat-transfer coefficient alpha

1. Apr 3, 2016

### joaquinjbs

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. Apr 3, 2016

### BvU

yes and no. Depends on the interpretation of W/m2 K as W/m2 K (no...) or as W/ (m2 K) (yes!)
see wiki

3. Apr 3, 2016

### joaquinjbs

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... :(

4. Apr 3, 2016

### BvU

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

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

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

and so does the 400 ... but those are W / (m $\cdot$ K) )

5. Apr 3, 2016

### joaquinjbs

I've just checked my problem and I think all it's OK now. Thank you for your time and for responding quickly!

6. Apr 7, 2016

### Staff: Mentor

The OVERALL HEAT TRANSFER COEFFICIENT in the table is supposed to be the k in your first post.

7. Apr 8, 2016

### joaquinjbs

I thought the OVERALL HEAT TRANSFER COEFFICIENT was α, and k was the HEAT-TRANSFER RESISTANCE... Thank you!

8. Apr 8, 2016

### BvU

Thanks, Chet. I probably hinted too carefully in post #4 and forgot to follow up on #5 !
Should have insisted Joaquin showed his work !

9. Apr 8, 2016

### joaquinjbs

My english level is extremely basic, and the hint is a bit difficult to me.
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. Apr 8, 2016

### BvU

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. Apr 8, 2016

### Staff: Mentor

The symbol U is often used for the overall heat transfer coefficient, so that could cause confusion with regard to your equation.

12. Apr 8, 2016

### BvU

Sleepiness. Thanks, Chet. $1/U$ it is. I'm used to $\dot Q = UA\; {\rm LMTD}$ and was too hasty. Coffee !

13. Apr 8, 2016

### joaquinjbs

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!