0
$\begingroup$

enter image description here

I have heated water in a tank that is undisturbed (no inlet/outlet, no stirring). Due to the cooling of the sides of the tank, it will over time naturally stratify into a top temperature layer and a bottom temperature layer. I want to prevent this stratification.

If I add a thermally conductive material inside the tank, say a vertical copper rod with copper sheets at both ends, to act as a heat exchange between the top and bottom layers, how well can it reduce the top/bottom temperature difference?

I assume this will lead to an equilibrium between the push for stratification (tank sides cooling down), and push for equalization (heat exchanger copper rod). The more efficient heat exchanger, and the lower cooling down speed of sides, the lower the temperature differential. Or am I wrong?

Improve this question
$\endgroup$
3
  • $\begingroup$ does your tank have an insulated lid on top, or is it open to the air? $\endgroup$
    – niels nielsen
    Jan 1, 2018 at 7:28
  • $\begingroup$ Couldn't you just provide some insulation to all the surfaces and sides of the tank? That would have cost less than copper. Anyways, if the lid of your tank is closed and the lid is insulated, it would work. $\endgroup$
    – Wrichik Basu
    Jan 1, 2018 at 7:43
  • $\begingroup$ The tank is insulated on all sides, to make the water cool down at a slower rate. But I don't think it matters much to the general principle at work though. Insulation or not, over time any tank with hot water will become stratified. Unless it has some mechanism to prevent stratification, which is what I'm trying to create with the copper rod. $\endgroup$
    – Björn Morén
    Jan 2, 2018 at 9:28

4 Answers 4

Reset to default
1
$\begingroup$

As heat escapes from the sides of the tank, it creates cooler heavier fluid which runs down the edges and pools at the bottom. If this process were perfectly efficient (it isn't, as there is certainly some mixing between the cool sinking water and the warmer water further in), then the heat flux you'd be asking your copper rod to carry would be equal to the heat lost through the sides of the tank. You can estimate that rate, F, if you know the mass of water in the tank and the characteristic rate at which it cools (ok, it doesn't cool uniformly, just take some kind of average). The heat flux carried by the copper bar can be estimated as F = C * A * DT / L where C is the thermal conductivity of copper (401 W/m/deg), A is the cross sectional area of the rod (m^2), DT is the temperature difference between the two ends (deg C [or K]), and L is the length of the rod (height of the tank) (m). If you have a desired maximum temperature difference DT, you can find the size of rod you need to carry the heat by plugging in to the above equation and solving for A.

My guess is that you will need a darn fat piece of copper. Probably more effective to stick in a small submersible pump with intake and outlet near the top and bottom of the tank.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks Ben. Seemed like a simple problem at first, but I realize that it's hard to build an accurate theoretical model of this. Just have to try it in real life. Have looked around for similar solutions, but failed to find any. Found solutions where they stir the water though, just as you mention, so maybe that is a hint of the only practical way. $\endgroup$
    – Björn Morén
    Jan 4, 2018 at 7:41
1
$\begingroup$

I know that there would be several variables, thickness, length of rod, DT, etc; but wouldn't it work much better if the rod were insulated, which would carry more heat to the bottom of the tank to facilitate conduction? Also, wouldn't it work better if the plate at the bottom of the tank were to one side, or possibly just have a copper rod in the shape of a "J" with the lower part of the "J" uninsulated and in a vertical tube, say ~ 1/3 to 2/3 of barrel height in length,the tube preferably of low conductivity, i.e. insulated - these to facilitate convection.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Yes, good point about the insulated rod. $\endgroup$
    – Björn Morén
    Sep 1, 2020 at 11:40
0
$\begingroup$

Your experiment will work if and only if the tank has an insulated lid that is closed. Otherwise, even the upper part of the copper rod will be at same temperature as the lower one, and your aim will not be fulfilled.

Keeping in mind the insulated lid, your experiment will theoretically work.

Improve this answer
$\endgroup$
0
$\begingroup$

Suppose all the walls are well insulated. Initially you have hot water at uniform temperature inside the container. Cooling happens at the free surface of water which is exposed to cooler ambient air. Cooling is due to both convection currents set up in air above water's surface as well as due to evaporation. As the top layer of water cools it sinks down and hot water from deep inside wells up to the surface. In other words, convection currents are set up inside hot water too. During convection if you were to measure temperature, then the average temperature of water would increase with depth. Convection progressively reduces this unstable temperature gradient.

When temperature difference between top and bottom of water body falls below a critical value (check out Rayleigh Benard convection and the associated Rayleigh number) convection stops and now heat transfer happens only due to conduction within the body of water. I presume you are referring to this state of stratification. Providing a highly conducting path for heat will help quickly equalize temperature in this case. Having multiple fins on the copper plate would do a better job.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.