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Let's say I have a fish tank with a total volume of 0.25m3 (100x50x50cm) filled with 160l of water. There's no lid and the thing is placed in a room with sea level atmospheric pressure. The whole thing hasn't been moved for a while so the water is absolutely calm.

Now, if I were to lift the tank by 3cm in a seesaw motion in order for the water to start flowing and forming a wave, how would I go about calculating the maximum pressure the water puts on the walls?

Of course I'd wait for a fixed amount of time before lifting the other side of the tank so the wave stays the same in height.

I think I could do the calculations myself but I have no idea where to start or even where to look for a hint (piston pressure? mass inertia? Bernoulli's equation?).

The reason I need this (or at least I think so) is to calculate the wall thickness. It's basically like a wave motion machine, but open and filled with water: Photo

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  • $\begingroup$ A simple, first-order estimate would be to measure max height of the wave at the wall and use the usual pressure vs. depth formula for water. Then add the "slosh" factor by guesstimating the speed and volume of water hitting the wall. (or to be exact, the volume per unit area to get the pressure) $\endgroup$ – Carl Witthoft Mar 21 '18 at 15:37
  • $\begingroup$ Which would be worthless, since you need to compute the wave form and model its impact. Software has a rack-rate of about 75,000 bucks. Training costs up to $1000 per day and you need quite a bit of it. This is a billion dollar industry, and they still have trouble calculating wave impacts on the pipe legs of oil platforms. You can calculate overall momentum in the waves, but max pressure at impact is terribly hard. Just modeling aerates water at fairly low speeds in a pain. The speed of sound can be ridiculously low in bubbly water, and local pressure spikes can be 1000 times average values. $\endgroup$ – Phil Sweet Mar 21 '18 at 15:53
  • $\begingroup$ This is what this area of computaion looks like. I was surprised to find this much detail in a non-paywalled document. rspa.royalsocietypublishing.org/content/470/2172/20140542 $\endgroup$ – Phil Sweet Mar 21 '18 at 15:54
  • $\begingroup$ @PhilSweet are you assuming the OP is wanting to model violent wave impacts based on the link? $\endgroup$ – Solar Mike Mar 21 '18 at 16:07
  • $\begingroup$ I don't know what he's trying to model. It may just be a momentum calculation and he wants the reaction force on the end wall for a soliton. That is doable. MIT has some excellent online material for momentum calculations applied to linear gravity water waves and their interactions with walls and beaches. $\endgroup$ – Phil Sweet Mar 21 '18 at 16:22
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Assuming you time your rocking motion in a way that you get a sinusoidal wave going nicely back and forth you may just add 3*2= 6 centimeter to the height of water. This is without slushing force.

For estimating rough numbers for slushing, I would consider the potential energy of a test rectangular sliver of water standing half the 3 cen at the middle of the tank width as a tide turning into kinetic energy at the wall, without friction breaking into wall of tank. Then you pick another test point along the width and the height and valume corresponding to this point.

After 3-4 trial pints you add the maximum value to the prior value you got for hydrostatic pressure.

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The pressure on the walls of tank consists of a hydrostatic + dynamic component. In all cases of vertical up/down motion, the pressure on the floor of the tank will be the greatest of all the walls, because the hydrostatic pressure is greatest at the floor of the tank and the reaction of the dynamic component is also against the floor. Therefore we will calculate the pressure on the tank's floor and this sets an upper limit for the pressure on the walls.

The hydrostatic component of the pressure is related to the height of the water above the tank. Although sloshing in the tank will vary the height, to a first approximation if we assume that this change is small, the hydrostatic pressure is therefore constant with time and given by (density x gravitational constant x height).

The dynamic component of pressure is proportional to the vertical acceleration of the tank. From F = ma, PxA = rho x V x a P = rho x h x a

Where: P = change in pressure on tank's floor A = unit area on the tank's floor V = volume of water above unit area A = Ah rho = density of water a = acceleration h = height or depth of water

Therefore the total pressure acting on the floor = (rho x g x h) + (rho x h x a)

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The accurate way of solving this problem is through solving the associated free-surface hydrodynamic stability problem and treating the pressure field as having a static plus perturbation components. Depending on your desire for accuracy, of course, this might be too cumbersome and a better way of calculating the wall thickness could be to calculate the wall thickness for the base of the tank (where highest hydrostatic pressure will be experienced) and multiply it by a safety factor instead.

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