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I want to look at the thermal heat plume of a human in still air in a room. So I have a tank of water with a heating coil/cylinder in it. How do I calculate what the heat output of my heating coil will need to be in order to mimic that of a human's ($100W$) in air?

enter image description here

This is based on dimensional similarity between water:air, Reynolds' number and Reyleigh number. The Rayleigh number, which governs buoyancy is given by: $\dfrac{g\beta}{\nu \alpha \kappa}q x^4$, where:

$\alpha=$ thermal diffusivity $\beta=$ thermal expansion coefficient $\kappa=$ thermal conductivity $\nu=$ kinematic viscosity $x$= distance from the heated surface

But I don't really understand if the Rayleigh number is the right value to be attempting to maintain similar between both scenarios. Any help would be much appreciated.

EDIT:

So using the Q* formula to get similarity between the two media

$\dfrac{Q_{air}}{\left(\rho C_pT_{\infty}g^{1/2}x\right)_{air}}=\dfrac{Q_{water}}{\left(\rho C_pT_{\infty}g^{1/2}x\right)_{water}}$

and rearranging to get Q_water:

$Q_{water}=Q_{air}\dfrac{\left(\rho C_pT_{\infty}g^{1/2}x\right)_{water}}{\left(\rho C_pT_{\infty}g^{1/2}x\right)_{air}}$

Then substituting in:

AIR: $\rho=1.225kg/m^3$, $C_p=1.005kJ/kg\,K$, $T_{\infty}=21^{\circ}C$, $x=0.1m$, $Q_{air}=100W$.

WATER: $\rho=1000kg/m^3$, $C_p=4.19kJ/kg\,K$, $T_{\infty}=8^{\circ}C$, $x=0.1m$, $Q_{air}=?W$.

I get $Q_{water}\simeq 1.2\times10^4W$... But this seems much too high. Can this be right?

HCAI
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1 Answers1

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Thermal plumes have been studied extensively for fire safety applications. Often you know the heat release rate $Q$ but little more. A dimensionless group called $Q^*$ (pronounced "Q star") is used instead of more common parameters like the Reynolds number and Rayleigh number. This parameter can be thought of as the strength of the heat source at a particular distance. It correlates well for thermal plumes. You can derive this group by non-dimensionalizing the Navier-Stokes equations and setting dimensionless groups equal to 1 to define the characteristic length and velocity. For more information, check out Gunnar Heskestad's paper on this dimensionless group.

In the fire modeling case, generally people ignore Prandtl number similarity and some other things, so they say the dimensionless temperature and velocity distributions are only functions of $Q^*$.

The most relevant parameters are:

$$T^* \equiv \frac{T - T_\infty}{T_\infty}$$

$$Q^* \equiv \frac{Q}{\rho c_p T_\infty (g x)^{1/2} x^2}$$

To be more explicit, if you know the temperature ($T$) as a function of height ($x$) above the hot object, you can find $T^*$ as a function of $Q^*$. $Q^*$ is like a dimensionless spatial coordinate.

Strictly speaking, your setup is not going to be exactly similar because your coil and a human are not geometrically similar (and the heat flux distribution on the coil is probably not similar either). In your photo, I assume the human would be lying down if any reasonable geometric similarity is desired. The far-field should be okay, and I'll assume this is what interests you [2].

It's also not exactly clear what quantity you are interested in. I assumed you want to get the temperature distribution in the plume, say, at a height $x_1$ above in reality which would be $x_2$ in your model. Correct me if this is wrong.

Also, while I don't do experiments, I imagined your heating coil has an output of $W$, not heat flux. Let me know if I'm mistaken and I'll change my answer.

Ignoring the other parameters may or may not be valid in your case (it seems to be okay for fire safety [1]), so I'll do the analysis assuming it's not. You can skip the remainder if you want to assume the two mentioned parameters are all you need.

You can get the number of required groups from the Buckingham $\pi$ theorem.

The relevant parameters I've identified are $T$ (temperature at height $x$), $x$, $Q$, $g$, $\alpha$, $\beta$, $\nu$, $T_\infty$, $\rho$, and $c_p$. The Buckingham $\pi$ theorem suggests there will be 6 dimensionless groups here. (Assuming that I am not missing a parameter. I also need to check that the dimensional matrix is not rank deficient. For more details about dimensional analysis, I recommend reading Dimensional Analysis and Theory of Models by Henry Langhaar.)

So, the first 5 dimensionless groups are:

$$T^* \equiv \frac{T - T_\infty}{T_\infty}$$

$$Q^* \equiv \frac{Q}{\rho c_p T_\infty (g x)^{1/2} x^2}$$

$$Pr \equiv \frac{\nu}{\alpha}$$

$$Gr_x \equiv \frac{g \beta (T - T_\infty) x^3}{\nu^2}$$

$$\rho^* \equiv \beta (T - T_\infty)$$

This fifth group is inspired by the Boussinesq approximation. In that approximation, the density difference is modeled as a temperature difference. Similarity in this parameter ensures that your density field is similar.

For the remaining group, I needed to get a little creative. Similarity does not require this group takes any particular form, but it's best to stick with parameters with known physical meanings (or parameters which can be derived from governing equations, which usually have physical meanings). I can't think of anything good off hand, but the following works:

$$\Pi_6 \equiv \frac{g x}{c_p(T - T_\infty)}$$

You need to match all of these for similarity. It should be clear that matching all of these will be a challenge. As I said, it appears to be common practice in fire safety to ignore everything but $T^*$ and $Q^*$. I don't know if this is because the other parameters don't matter, or if it's just for convenience. Sorry if this is not the answer you expected, but as with many things in engineering, the answer is not easy.

[1] I remembered later that the non-dimensionalization of the Navier-Stokes equations suggests that $Q^*$ is the only parameter in the solution. So perhaps $T^*$ and $Q^*$ are all you need, and the Buckingham $\pi$ approach just gives you superfluous parameters. I don't recall all the details of the non-dimensionalization, but if there is interest I'm sure I could reproduce it.

[2] The theoretical argument that supports the use of $Q^*$ assumes the heat source is a point source. So it's really only correct far away, because the temperature goes to infinity at the point source in the model. This is because $Q^*$ goes to infinity at $x = 0$, as you can see from its definition. If you are developing a correlation, say $T^* = a (Q^*)^b$ where $a$ and $b$ are coefficients, you can get around this by defining a "virtual origin", which will allow you to develop a correlation without singularities. Basically, instead of using $x$ you define use instead $x_\text{virtual} = x + x_\text{origin}$. That is, $Q^*$ is now written:

$$Q^* \equiv \frac{Q}{\rho c_p T_\infty (g [x + x_\text{origin}])^{1/2} (x + x_\text{origin})^2}$$

You pick $x_\text{origin}$ such that your correlation fits better. It's another parameter in the correlation. If you know the surface temperature, you can pick $x_\text{origin}$ such that the surface temperature is what the correlation returns at $x = 0$.

Also, as the argument supporting the use of $Q^*$ really makes the far-field assumption from the beginning, it's not clear that simply using a virtual origin is enough to make a correlation valid in the near-field (even if you have geometric similarity). I can't say whether or not the other factors I've identified factor in or not.

Ben Trettel
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    +1 for bringing in an example from a different discipline and highlighting assumptions – pandita Feb 08 '15 at 22:26
  • Hi Ben, welcome to engineering.SE! This is an excellent answer, great job! – Chris Mueller Feb 09 '15 at 02:00
  • Thank you very much indeed! This is exactly what I was hoping for and more. I do have one query though and that's relating to the value to choose for x when comparing models as it's a local variable rather a global quantity. Does it mean I should try to keep Q* the same at all points in both domains? I am actually interested in the plume very close to the cylinder... How does this change the analysis or prediction? Again many thanks for such a great answer! – HCAI Feb 10 '15 at 14:22
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    @HCAI: I added a brief part noting that $Q^*$ takes the place of the spatial coordinate, so it's impossible to make $Q^*$ constant (that would mean there's only one spatial location). I also added a footnote about using $Q^*$ in the near-field. Let me know if you have any other questions. Also, I want to highlight that fire safety experimentalists use a technique called "salt-water modeling" that you might find relevant to this; try some Google searches for more information. – Ben Trettel Feb 10 '15 at 22:20
  • @BenTrettel I worked out that the Q_water needs to be about 12000W but this seems enormous. Can this be right or have I completely misunderstood the formulae? (I've added my working in the question) – HCAI Feb 27 '15 at 09:36
  • @HCAI I believe the trick to this is that the $Q^*$ computation used assumes a medium (gas) with volumetric expansivity $\gamma=T^{-1}$. That's way off for water. Related: Is there a reason that you want to use water rather than air for the scaled experiments? – Dan Feb 27 '15 at 15:09
  • @Dan Oh that's a bit unexpected!... Is there a workaround? If not then I'm a bit stuck eek! I wanted to use water (or other viscous liquid) because it's easier to use PIV, PTV and LDA visualisation techniques. – HCAI Feb 27 '15 at 19:45
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    I just checked, and the derivation of $Q^*$ uses the Boussinesq approximation, which does not assume that $\beta$ takes the form for an ideal gas. As far as I know, this sort of similarity is used in experiments with salt water (again, I'd suggest looking for papers detailing this). – Ben Trettel Feb 28 '15 at 01:57
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    As for the numbers, you seem to have a few math errors. First, the power on $x$ is $2.5$, not $1$, though this will not change your result as you are not scaling anything in space. I get that $Q_\text{water} \approx 3 \cdot 10^5~\text{W}$, which is still large. It seems that you need to reduce the size of your model to get a reasonable number, e.g., using $x_\text{water} = 1~\text{cm}$ seems reasonable (about 1 kW). – Ben Trettel Feb 28 '15 at 01:59
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    @HCAI: Forgot to include you so that you get a notification of my response. – Ben Trettel Mar 01 '15 at 17:55
  • @BenTrettel Thank you, I see the mistake. The dilema that I have is that I'm actually interested in the near-field temperature. Do you think that the thermal plume/boundary layer will actually behave in a similar way between both scenarios? My other option is to ignore similarity of heat input and focus on plume speed above the heater by Reynolds' number comparison. What do you think? – HCAI Mar 02 '15 at 17:48
  • @HCAI: There certainly will be similarities. The difficulty is that without geometric similarity it's not clear how to compare locations, or even if a comparison is valid. I don't know how to answer either question right now. As for focusing on plume speed above the heater, I'm not quite sure what you mean. Could you explain in more detail? – Ben Trettel Mar 03 '15 at 02:47