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Suppose there is a large reservoir of water, and a small horizontal pipe at one of the walls of the pipe. I have done some calculations using Bernoulli's theorem to show that the velocity in the pipe when the end is opened is approximately $$v(t)=2k\tanh\left(\frac{kt}{L}\right)$$ for some positive constant $k$. (It is approximate since we assume the height of the reservoir stays the same, and the water near the top is not affected).

Question: Calculate the timescale for the flow to reach its maximum speed.

I don't understand how to do this, since $\tanh$ doesn't ever reach its maximum. However since they say "timescale" rather than "time", perhaps this means something slightly different? If someone could help me to understand this, and give some idea on how to do this, it would be appreciated.

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If your liquid is stationary, and you suddenly open the pipe, it will take some time for the liquid to accelerate to "something like steady state". You know the initial pressure difference (when there is no flow, the full pressure difference due to the column of water is experienced), so you know the acceleration of the water at the exit. As it accelerates, the pressure difference decreases and the rate of acceleration slows down.

It is a reasonable "time scale" to say "final velocity divided by initial acceleration is of the right order"; I think if you do this a bit more carefully you get at least another factor 2x.

It's a start.

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