# Behaviour of water erupting from hydrothermal vent

Imagine a hydrothermal vent far underwater that is connected to the surface by a tube, so that everything that erupts from the vent is contained by the tube. I'm trying to calculate some realistic numbers on the behaviour of the flow of heated fluids from the vent towards the surface. In particular, I'm trying to find out how far the fluids will rise before cooling to the temperature of the surrounding water.

I've done some calculations assuming natural convection and no turbulence. I'm curious however of whether these assumptions are realistic in the given circumstances. From what I have read the ocean is "chaotic" enough that turbulence has a big impact, although I am unsure of whether the fact that the water is contained by a tube reduces the effect of turbulence. Also, I'm unsure of how to handle the initial velocity of the water when erupting from the vent. Does the initial velocity lead to a forced convection scenario, or can the effects from the initial velocity likely be ignored without too big an issue?

• A sort of natural model related for this is either the flow of magma through vents, or heat rising from a fire, through a chimney although obviously viscosity is on a different scale in both cases. I suggest these as sources of information, that almost certainly have been researched. And I am guessing you know about the N-S equations. (Apologies if you do). Possibly having a resources recommended tag might help.
– user146020
Commented Mar 7, 2017 at 22:00
• Yes. I took some fluid dynamics as part of my degree, but not--in my opinion--enough to confidently calculate realistic estimates for dynamic behaviour of fluids in real-world scenarios. Thanks for the suggestions! Commented Mar 7, 2017 at 22:05
• Unless your tube is quite wide it's existence will affect (perhaps inhibit) the formation of convection cells. Commented Mar 7, 2017 at 22:15
• Complicated question! Is the vent at the bottom of a large depression, and thus isolated from prevailing currents for the ostensible entirety of its path? Commented Mar 7, 2017 at 22:29
• In my head the tube is made out of some sort of non-permeable cloth or soft plastic, so definitely not completely rigid. You're right, I guess the tube is a major complicating factor. Commented Mar 7, 2017 at 22:36

There are many complicating factors in your stated problem, and nothing short of experiments (or numerical simulations) would satisfactorily answer your question. We can only make crude estimates based on simplified models. Even then because of presence of turbulence, predictions due to a simple model would perhaps be in serious error. Here is my attempt at simplification, and I am sure anyone could raise varied objections to the argument presented below.

Let us assume that the tube in which the hot fluid from the vent rises, is rigid and thermally conducting with thermal diffusivity, $\alpha_{tube}$. We shall assume that the fluid flowing in the tube remains in one phase. Let us assume that flow inside the tube is turbulent.

You must be aware that in turbulent flows mixing is very efficient, and results in enhanced transport of heat (among other things). As far as average heat transport is concerned, this enhancement may be modeled using the concept of eddy diffusivity, and mean heat transport due to molecular diffusion may be neglected. Mixing is achieved by energy containing eddies, which are typically the large eddies in the flow. For turbulent flow inside the tube, large eddies are of the size of tube diameter, $d$. If $u_{rms}$ is the scale for velocity fluctuations for turbulent flow inside the tube, then eddy diffusivity for flow inside the tube may be taken in the first approximation as $\alpha_{fluid}\sim u_{rms}\times d$. To calculate $u_{rms}$ this might be of some help. A similar procedure for the ocean outside the tube which performs the role of heat sink, gives for the eddy diffusivity of the turbulent ocean, $\alpha_{ocean}\sim u_{rms,ocean}\times L_{ocean}$, where $L_{ocean}$ may be taken as the integral length scale of the turbulent ocean (see here). As a first approximation, all quantities involved in the calculation are taken as constant. We shall neglect any mean flow of the ocean itself in the vicinity of the tube.

Now we may model the problem as follows: if a hot fluid of thermal diffusivity $\alpha_{fluid}$ and temperature $T_0$ at its source, is flowing with mean speed $U$ inside a thermally conducting tube, immersed in a static infinite fluid of thermal conductivity $\alpha_{ocean}$ and whose temperature far away from the tube is $T_{ocean}<T_0$, what distance does the fluid inside the tube has to travel, beginning from its source, before its mean temperature becomes equal to that of the ocean, $T_{ocean}?$

Given the crudity of the model there is no point in solving the above problem exactly. From the link I have given previously, $L_{ocean}\sim 100~ m$ and $u_{rms,ocean}\sim 1~m/s$ , which means $\alpha_{ocean}\sim 100~ m^2/s$. I don't know what $\alpha_{fluid}$ will be for fluid inside the tube, because while size of large eddies is much reduced (limited to tube diameter) the velocity fluctuation scale will perhaps be higher than that for the turbulent ocean because of the added effect of buoyancy. However I seriously doubt whether this can compensate enough, and perhaps it is a good guess that $\alpha_{fluid}\ll \alpha_{ocean}$. Normal tube material will have a much lower thermal diffusivity still (see here), i.e. $\alpha_{tube}\ll \alpha_{ocean}$. Therefore we may approximate the ocean as a heat bath, and the outer wall of the tube in contact with the ocean may be assumed to be at temperature $T_{ocean}$ all along its length.

Now temperature along the inner wall of the tube will vary from $T_0$ to $T_{ocean}$ over a distance $H$. We shall assume that the flow is statistically steady. If you take a control volume enclosing the fluid over this distance, then the difference between the heat flux entering and leaving this control volume is $\rho C_p (T_0-T_{ocean})\times \frac{\pi}{4}d^2\times U$, which is also equal to the heat lost through the tube wall to the ocean over this length. A scale for mean speed $U$ is perhaps $\sqrt{g\beta (T_0-T_{ocean})d}$, where $\beta$ is the thermal expansion coefficient of the fluid inside the tube. If we take the average temperature at the inner wall of the tube to be $\frac{T_0+T_{ocean}}{2}$, then the average temperature gradient (across tube wall) along the length $H$ of the tube is $\frac{T_0-T_{ocean}}{2t}$, where $t$ is the tube wall thickness. Therefore in statistically steady state the following balance must hold approximately: \begin{align} k_{tube}\times \pi d H \times \frac{T_0-T_{ocean}}{2t} & \approx\rho_{fluid} C_{p,fluid} (T_0-T_{ocean})\times \frac{\pi}{4}d^2\times U \\ H & \approx \frac{\rho_{fluid} C_{p,fluid}dUt}{2k_{tube}} \end{align} where $k_{tube}$ is the thermal conductivity of the tube. However this estimate must be taken with extreme caution, and verified by experiments. Surprisingly, according to this estimate $H$ does not depend on $T_0-T_{ocean}$ explicitly, but only through $U$.

• @Aqwis You are welcome. Needless to say, do not trust this estimate unless you verify it against some experiments or simulations.
– Deep
Commented Mar 13, 2017 at 5:32