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Why is the following formula used when we define heat transfer due to convection on a flat plate?:

$$ q=h(T_s-T_\infty)$$

Where $T_s$ is the temperature of surface at a particular location and $T_\infty$ is the temperature of fluid flowing with free stream velocity. Why are we using $T_\infty$ here and not any other temperature of the fluid flowing over the plate?

I read in a book that this expression is derived from newtons law of cooling i.e $dT/dt=-k(T_s-T_\infty)$, where $T_\infty$ is the temperature of surroundings?

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    $\begingroup$ Hi and welcome to Physics Stack Exchange. Please note that I have edited your question to clean it up a little and I have formatted the formulas with MathJax. We prefer that MathJax be used for formulas, as it helps readability/searchability. $\endgroup$ – Time4Tea Oct 12 '18 at 14:07
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    $\begingroup$ For the OP and any editors, it would help if the nomenclature stayed consistent. Is lower case "t" time or temperature? $\endgroup$ – David White Oct 12 '18 at 16:09
  • $\begingroup$ @DavidWhite that's a good point - I hadn't spotted that. I can change it now? $\endgroup$ – Time4Tea Oct 12 '18 at 17:26
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    $\begingroup$ On lhs it's time . $\endgroup$ – shivam_maestro Oct 12 '18 at 17:28
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One reason why $T_\infty$ is used, rather than the temperature at any other point in the fluid is because that is how the coefficient $h$ is defined. $h$ is defined as the amount of heat that is transferred per unit difference in temperature between the plate and the free stream fluid. It would be possible to use a different temperature than $T_\infty$, if you were to use an alternative definition for $h$, or a different formula for the heat transfer.

Edit: to answer the specific question as to why it makes more sense logically to choose the free-stream $T_\infty$ as the 'datum' temperature, rather than any other temperature of the fluid:

From a thermal point-of-view, the fluid flowing over the plate can be broken down into two regions, the thermal boundary layer and the free-stream, as shown in the following image (taken from Wikipedia):

Thermal Boundary Layer Image

The thermal boundary layer is the region close to the plate that is affected by the heat transfer to/from the plate and the free-stream region is everywhere else that is outside the thermal boundary layer. So, by 'any other temperature of the fluid', I assume you must mean 'some temperature within the thermal boundary layer', because everywhere else is at the same free stream temperature ($T_\infty$).

The problem with using a temperature inside the thermal boundary layer as a baseline is that the thermal boundary layer is governed by some fairly complex physics and depends on many factors, such as flow velocity, fluid properties and also it varies with tangential distance along the plate ($x$ in the above diagram). So, to use a temperature in the thermal b/l as a baseline, you would need to account for all this complex stuff that is going on (which is going to be a hassle).

On the other hand, out in the free-stream, the temperature everywhere is a constant, at $T_\infty$, because the fluid in this region is not affected by the heat transfer with the plate. This is because thermal diffusion in a fluid is quite a slow process (as long as we're not talking about some exotic, highly-conductive metallic treacle ..), so the heat diffusing out from the wall hasn't had time to get to this region yet. So, out here, we've got a constant temperature that doesn't depend on any heat transfer physics and is essentially imposed on the situation as an independent boundary condition. Therefore, it is the ideal choice to use as a temperature baseline, because it is independent, constant and stable, compared to what is happening inside the thermal boundary layer.

(and if that isn't enough to convince you, it is also easier to measure, because it extends over a larger spatial domain and has a very low (essentially zero) thermal gradient)

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  • $\begingroup$ so you are saying that thermal boundary layer thickness increases with time even in steady state conditions? $\endgroup$ – shivam_maestro Oct 13 '18 at 2:30
  • $\begingroup$ I think your answer is right because my teacher didn't gave any explanation about this temperature T infinity so that means it is a convention . $\endgroup$ – shivam_maestro Oct 13 '18 at 2:43
  • $\begingroup$ @shivamsavages in a steady flow, the thermal boundary layer would not grow with time, but it would grow with distance along the plate. $\endgroup$ – Time4Tea Oct 13 '18 at 12:26
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The way I think about $t_\infty$ is that we could pick it by going to extremes: $t_\infty$ should either be the temperature of the fluid right above the plate or it should be the temperature of the fluid far away from the plate. I will argue that it is way more useful to choose the temperature of the fluid far away from the plate.

Assume the plate velocity = zero. If the fluid is not slipping, the velocity of the fluid just barely above the plate is the same as the velocity of the plate: zero. Therefore, theoretically since the fluid above the plate is not moving, the temperature of the fluid at that point will eventually reach the temperature of the plate, thus $t_\infty$ = $t_s$. If $t_\infty$ = $t_s$ then we would have no net heat transfer at all and the equation would be useless. So it is better to pick the temperature of the fluid far away from the plate.

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  • $\begingroup$ Thanks for your vote; however, I'm not sure I agree with the reasoning in your last paragraph. For any heat transfer to occur, there has to be a temperature gradient. Even through the solid plate, there has to be a temperature gradient for heat to transfer. So, I don't think that the fluid just above the plate being at zero/low velocity makes it more likely that the temperature there would be equal to $t_s$. It will be much closer to the temperature of the plate than $t_\infty$, simply due to its proximity, but that isn't because of the low velocity. $\endgroup$ – Time4Tea Oct 12 '18 at 15:15
  • $\begingroup$ I think the main reason that the temperature far from the plate is used is more to do with ease of measurement. Farther from the plate (i.e. outside the boundary layer), the temperature gradient will be lower and so the temperature more consistent to measure. Putting a probe into the thermal boundary layer may be more likely to 'disturb' the system that is being measured. Also, the temperature difference closer to the plate will be lower, which will amplify the effects of any measurement error. $\endgroup$ – Time4Tea Oct 12 '18 at 15:18
  • $\begingroup$ @Time4Tea I agree 95%. However I can't wrap my head around "the fluid just above the plate... will be much closer to [$t_s$] than $t_\infty$" . The closer you get to the plate, the closer the fluid temperature is to $t_s$ because temperature is continuous. It makes no sense logically, but the model I was taught in school was that the fluid temperature at the plate was equal to that of the plate. But there is definitely heat transfer happening... so how to avoid a logical paradox? A: When applying Newton's Law of cooling we should not use the temperature right near the plate. $\endgroup$ – pentane Oct 12 '18 at 17:13
  • $\begingroup$ As to difficulty of measurement: what about IR cameras? They can measure temperatures very accurately without disturbing the flow. $\endgroup$ – pentane Oct 12 '18 at 17:15
  • $\begingroup$ Where the fluid contacts the plate, the temperature will be equal to that of the plate ($t_s$) - as you say, temperature has to be continuous. However, there will still be a gradient of temperature (if there is any heat flux). I.e. if you move an infinitessimal distance away from the plate, there will be a small difference in temperature. Think of it as a continuous curve of varying temperature, that is continuous at the point where the plate and fluid meet. $\endgroup$ – Time4Tea Oct 12 '18 at 17:19

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