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February 23

Liquid flow

Consider liquid emanating from a corner of a flat square tray with a containing wall at the edges. If the outflow and depth are constant the liquid will spread on a quarter-circular front with the radius increasing at a rate proportional to the square root of the elapsed time, until it has reached the two adjacent corners. What will happen then, I.e. What shape will the front take and how will it move with time until the whole surface is covered?→31.54.246.96 (talk) 17:49, 23 February 2015 (UTC)[reply]

Have a look at wave propagation, maybe check out this video [1]. Also consider doing some empirical tests. That is probably more fun than setting up a thin-film navier stokes equation and trying to solve it ;) SemanticMantis (talk) 18:04, 23 February 2015 (UTC)[reply]

The viscosity of the liquid (usually temperature dependent) will matter, as will it's surface tension and other properties. Other factors are the rate the liquid is added and the tray material and shape (whether there are rounded corners and edges). A thicker fluid poured slowly will tend to spread as you described, while a thin liquid poured quickly will splash around, exhibiting turbulent flow instead of the laminar flow you described. In any case, there will be waves generated off collisions with the walls and corners, although those waves may be quite small for certain conditions. Note that laminar flow is quite predictable, while the actual path taken by the liquid under turbulent flow each time is chaotic, and thus unpredictable, other than as probabilities.

Also, under the laminar flow you described, you may find that the liquid expands more slowly adjacent to the walls, so you don't get a perfect quarter circle. StuRat (talk) 20:30, 23 February 2015 (UTC)[reply]

I think the previous respondents have over-complicated the issue. This looks like a homework question to me. Observe that the questioner explicitly notes that the depth of the liquid is constant. This simplifies the problem to simply requiring that the increase in area covered is a constant. So from the point at which the two corners are reached the liquid will simply continue to expand along a front that is a steadily narrowing arc of an increasing circle. If the square is of unit side, the circle will reach maximum radius of sqrt(2), with the front moving faster the closer it gets to the final corner of the square. RomanSpa (talk) 21:04, 23 February 2015 (UTC)[reply]

It's not homework, just something that I was trying to get an answer to out of interest. I'm long past the age of taking tests and exams, nor have I ever worked with the complications referred to in the first two responses. Taking the much simpler case referred to the third response, is it possible to get an explicit relation between radius of front and time after reaching the corners adjacent to the source? My analysis led to t being the integral of r(π/4-arccos(1/r)), ignoring a scaling factor, which I couldn't evaluate but which seemed unlikely to lead to r as a function of t.→31.54.246.96 (talk) 23:12, 23 February 2015 (UTC)[reply]

Another way of describing it would be viscous fluid being injected between two sheets of glass, with the air freely escaping somehow. The density, surface tension and dynamic effects other than viscosity are all to be neglected. The only facts needed are the geometry and the fact that the viscosity is uniform and dominates such that all other effects (inertia etc.) can be neglected. You would have a scalar field describing the pressure at every point, the flow vector field, and a function giving the boundary's shape. This will produce something along the lines of a partial differential equation, with a nonlinearity at the boundary. Intuitively, the boundary touching the square will grow faster than the central part once the second pair of edges is reached. I expect that numerical modelling would deal with this most easily. —Quondum 01:18, 24 February 2015 (UTC)[reply]

Random walk

I have a random walk variable, call it x, each step moving up by 1 with a probability of p and -1 with probability of 1-p. I also have another variable y that is defined as y_k=max{x_1,x_2...x_k}-x_k. How do I get the expected value of y and does it depend on k? Thx 93.136.7.126 (talk) 19:08, 23 February 2015 (UTC)[reply]

Well it definitely depends on k. Consider k=1, then y_k=0. But in general y_k need not equal zero. SemanticMantis (talk) 20:08, 23 February 2015 (UTC)[reply]

That's true, how stupid that I didn't think of that :) I've ran some Monte Carlo sims since; y_k should obviously diverge to infinity if p<0.5, as random walk then tends to negative infinity, and probably also for p=0.5. It appears to converge quickly (no apparent tendency at k=10^3 ~ 10^5) to n/(p-0.5), with n≈0.21. That's a good enough result for me, but I'd still be interested in the math to calculate it exactly, if anyone knows how. 78.0.233.239 (talk) 02:42, 24 February 2015 (UTC)[reply]

Yes I think your reasoning is correct for p not equal to 0.5. As for the analytic calculation, I'm not optimistic that there is an easy way to get this. It can be tricky to get distributions for maxima of fixed lists, and growing lists will be much harder. Anyway, here's some decent refs on simple random walks [2] [3] [4] [5], they may have tidbits that help you work it out. SemanticMantis (talk) 15:07, 24 February 2015 (UTC)[reply]