# Variation of heat transfer coefficeint with flow rate

1. Jun 20, 2013

### sanka

Hi all,

Doing some calculations on an air-cooled heat exchanger at the moment and could use some help understanding the variation of air-side heat transfer coefficient (htc) with flow rate. It's more of an intuitive problem really as I'm okay with the math, etc.

So I know that as you increase the air flow rate the htc increases due to increased moleular agitation, there is less thermal resistance, etc. But my question is this, why does the RATE of increase of htc reduce as one infinetly increases air flow rate? For example, if I plot htc Vs flow rate, the slope of the line is quite steep at the begining (at lower flow rates) giving a sharp increase in htc, but at some point the slope of this line changes. So, even though the htc still increases with flow rate, the increase is not as great as it was at the lower flow rates.

I have been thinking about it and I think it is due to the onset of fully-developed flow and boundary layers merging in the air-flow channel. The merging of these boundary layers contribute a thermal resistance, thus reducing the increase in htc? Thats just my thoughts anyway.

Any help would be appreciated!
Sanka

2. Jun 20, 2013

Unless you are basing this on an experiment covering all the flow regimes it would be difficult to draw that conclusion. Most forced convection models are empirical, but explicitly assume the flow is already fully developed and either laminar or turbulent but generally for cover transition.

It shouldn't be terribly surprising that the heat transfer rate saturates since all the correlations for flow over a laminar flat plate (or undeveloped pipe flow) tend to have a dependence on $Re_x^{1/2}$, which saturates in the same way. Interestingly, the boundary layer grows with this same parameter for a laminar flow. If you look at most of the data for turbulent boundary layers, you see a similar trend where they depend on $Re_x^{4/5}$, which happens to be the parameter with which a turbulent boundary layer grows. That would imply that the saturation effect you describe is likely related to boundary layer growth.

Once the flow is fully developed, the Nusselt number correlates with $Re_D$ so it no longer correlates to the growth of the boundary layer as you would expect since the boundary layers have merged. At that point it's just a matter of how much fluid you are forcing through the channel. As you increase the mass flow rate you have more thermal mass passing through to which or from which you can move heat, but it has less time in the channel during which it can actually transfer that heat, so it's a trade off at higher flow rates and that can lead to saturation.

3. Jun 27, 2013

### jlefevre76

Another comparison you could make is with internal flow problems (which it sounds like this one may be an internal flow, at least in a manner of speaking?) Remember that flow in a pipe or tube develops until it reaches a constant Nusselt number (and effectively a constant htc). The higher the Reynolds number, the faster it develops. So, as your Reynolds number increases in this case, perhaps it's reaching a "fully developed"* state more quickly, and the Nusselt/htc level out. This assumes laminar flow though, I'm not sure what happens with turbulent flow (I'd have to look it up in my convection heat transfer textbook, and I can't remember where to find it).

If your flow is turbulent, you would at least see a jump or a significant change in the Re vs Nu plot at the point where it transitions to turbulent flow. Whether you've gone that far with your Re numbers or not, only you would know.

* Fully developed thermally

4. Jul 2, 2013

### sanka

Thanks for the reply. I was thinking the exact same thing. Intutively it makes sense, I just never saw it explicitly addressed in any text book or explained in this manner. I know the correlations indicate that the Nu No (essentially heat transfer coefficient) has to level-off at higher Re No's but the correlations are just representing/modelling the physical phemonena. It's nice to understand the actual physics which they are accounting for.