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The Rayleigh number is generally described as distinguishing between the convective and coductive regime of heat flow in a fluid. I'm looking at a specific geometry. and it's not clear to me precisely what

  1. the characteristic length scale would be, and

  2. what the critical Rayleigh number for the problem is.

The geometry consists of a large reservior of fluid at a temperature $T_1$. In one of the walls, there is a slit of thickness $d$ and length $L$, and the end of the slit is cooled to a temperature $T_0<T_1$, as indicated in the sketch: enter image description here

Gravity is acting downwards (indicated by the green arrow). The box is very long in the third direction (into the screen). The boundary colours indicate the higher temperature (red) , the lower temperature (blue) and insulated boundaries (black). We can assume $L/d\gg1$ if necessary (at least $L/d>5$, more likely $>10$).

The reservoir is much larger than the slit, and I'm really only interested in the flow in the slit (and the heat flow at the cooled tip), in a stationary situation. I assume that, depending on the parameters, there will be a "conduction" regime without fluid flow, and heat flow by diffusion, and a "convection" regime, where the higher density of the cooled fluid at the tip leads to a steady flow into the (top of the) slit, cooling at the tip and flow out of the (bottom of the) slit. Intuitively, convection would be favoured by a large temperature difference (or large coefficient of thermal expansion), small $L$ and large $d$.

As far as I understand, the Rayleigh number depends on the difference in density (which in turn depends on thermal expandsion and temperature difference), viscosity $\eta$ and diffusivity $\alpha$, $$\text{Ra}\propto \frac{\Delta \rho g}{\eta \alpha}\cdot \left(\text{characteristic length scale}\right)^3\,,$$ and it contains the "characteristic scale" and possiby a numerical suppression factor.

The characteristic scale presumably is a function of $L$ and $d$ and encodes the fact that the viscous fluid needs to move along a long narrow slit.

So my questions are:

  1. Is my general intuition correct?
  2. If so, is there a formulation of the Reynolds number for this situation?
  3. What would be the critical value of the Rayleigh number here?
  4. Additionally, what happens if the slit is filled with a porous medium of permability $k$?
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  • $\begingroup$ If the fluid is incompressible, then there is no natural convection. $\endgroup$ Jun 2, 2021 at 14:29
  • $\begingroup$ @ChetMiller I meant to say that we can neglect elastic compression, but of course thermal expansion is nonzero. But I guess that assumtpion is not necessary anyway, so I removed it $\endgroup$
    – Toffomat
    Jun 2, 2021 at 15:01
  • $\begingroup$ We can compute several runs for the range of Rayleigh numbers to get pictures. Is this question about it? $\endgroup$ Jun 4, 2021 at 14:56
  • $\begingroup$ @AlexTrounev I'm not sure I understand your comment. Do you mean you would do simulations with different Rayleigh numbers? (FWIW, I'm planning to do that, and I would like to have an analytical cross-check for the results.) $\endgroup$
    – Toffomat
    Jun 7, 2021 at 6:58
  • $\begingroup$ Where is "downwards" in your picture? Please, add vector $\vec {g}$ . Why there is no any length given for the large cavity (reservoir of fluid)? Actually the Rayleigh number depends on the temperature difference. Therefore for nonzero temperature difference we have convection in a slit and in reservoir as well. As a length scale we can take $d$ for initial nonstationary convection, $L$ for the middle stage, and length of reservoir for the final stage. $\endgroup$ Jun 7, 2021 at 10:38

1 Answer 1

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Let suppose that for initial time slit is filed with a high temperature fluid. Then we can compute convection in a slit and in a reservoir as well. As length scale we take $d$. A system of equations describing free convection is given by

$\nabla .\vec u=0, \frac{d\vec u}{dt}=Pr\nabla ^2\vec u -RaPrT\frac {\vec g}{g}$

$\frac {dT}{dt}=\nabla^2T$

$ d/dt =\partial/\partial t +(\vec u .\nabla )$

$\vec u$ is velocity field vector, $T$ - temperature, $\vec g $ - gravity vector, $Pr$ - the Prandtl number, $Ra$ - the Rayleigh number.

I have used my code published on this page. In the animation below shown the distribution of the flow velocity at $Ra=10^3, Pr=0.72$. We can see that after $t>7$ (in the time scale $\tau=d^2/\nu$) flow expands in reservoir and therefore large vortex is formed. Figure 1

Temperature distribution is shown below Figure 2

This picture shows how flow in the slit and big vortex in the reservoir look like at $t=10$

Figure 3

At $t>10$ the next stage of convection coming when the vortex in the reservoir became stronger and its center goes down, like in this picture computed for $t=15$ (for this stage effectively we have $Ra=10^6$ since it computed with $L$ and we take $L/d=10$)

Figure 4

In a case of water at 30C the Prandtl number is about $Pr=5.534$. For $Ra=10$ it takes more time to compute final state. In pictures below the temperature, module of velocity and stream lines are shown at $t=40$

Figure 5

Figure 6

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  • $\begingroup$ Cool, thank you! Stupid question: What would happen for, say, $Ra=10$? $\endgroup$
    – Toffomat
    Jun 8, 2021 at 8:16
  • $\begingroup$ @Toffomat We need also Prandtl number $Pr$ for computation. Is reservoir filed up with water or air? $\endgroup$ Jun 8, 2021 at 9:17
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    $\begingroup$ @Toffomat See update to my answer. $\endgroup$ Jun 8, 2021 at 11:18
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    $\begingroup$ @Toffomat This is not stationary flow at $t=40$. I have finished computation for $Ra=10, Pr=5.534$. The picture is the same as above, but maximum velocity in reservoir is about 0.4 and in the slit - about 0.073. Your question probably sounds like this: can we neglect by convection in the slit and compute temperature from equation $\nabla^2 T=0$? Yes we can, difference in temperature in the slit-reservoir connection with and without convection is not much, $T=0.8894$ and $T=0.8801$ consequently. $\endgroup$ Jun 9, 2021 at 11:15
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    $\begingroup$ @Toffomat Yes, we can use linear model in a slit for $Ra\le 10$. $\endgroup$ Jun 9, 2021 at 12:57

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