The Problem of the Water Tank

Here’s a nice problem I remember hearing in the pub on Friday and figured out this afternoon.

A water tank or sink is open to the air at the top where it can be filled using a tap connected to an infinite reservoir. Water can be drained from the container through an opening at the bottom  by removing a stopper. The effects of viscosity on the outflow from the tank can be neglected.

The time taken for the tank to fill when the tap is fully open and the stopper in place is the  same as the time taken for it to empty from full  when the tap is closed and the stopper is removed.

If the tank is initially empty, the stopper removed and the tap turned full on, how full is the tank when a steady state is reached?

 


21 Responses to “The Problem of the Water Tank”

  1. Bryn Jones Says:

    I voted, and now I think I got it wrong. Pressure is important.

  2. David Thornton Says:

    4/9 full

  3. John Peacock Says:

    There’s not really enough information here. You need to know how the rate of draining (i.e. the velocity in the exit pipe) depends on the depth of water, D. This can be anything between v propto D^1/2 if the pressure purely generates kinetic energy, and v propto D if the drain pipe is long and thin so that viscous drag dictates the exit rate. If you take v propto D^a, then the equilibrium height is D=(1-a)^(1/a) if a<1. Thus any figure between 0 and 1/4 seems possible.

    • telescoper Says:

      I assumed right out that, since no information is given about the size and shape of the drain pipe, this needed to be treated as a problem in non-viscous flow which means it is basically an application of Bernouilli’s theorem. To reinforce this inference I happen to know that the module that it came from didn’t cover viscous flow….

      I agree though that it would have been better to say `neglecting viscosity’ in the question, and have now edited to include that.

      That does, however, remind me that I have another question somewhere about the formation of a hydraulic jump in a sink…

  4. Matt Smith Says:

    After more maths than I expected, I calculated 1/2

  5. If the tap opening is A, the cap opening is B, the water tap speed is v, the steady state is:
    h = v ^2 B^2 / (2 g A^2)
    using
    u A = v B (constant flow at steady state)
    and
    u^2/2 = g h (Bernoulli)
    if it is right, half of the problem is solved

    • When I empty the tank, without tap running; the water has a surface S, and the velocity of the water surface is w, then
      w S = u A (like a syringe)
      where A is the cap surface, then because of the old Bernoulli
      w^2 + 2 g h = u^2
      but
      w = u A / S = u p
      then
      p^2 u^2 + 2 g h = u^2
      (1-p^2) u^2 = 2 g h
      using the reduction of the cylinder tank volume
      h’ = -p u
      so that
      2 (1-p^2) u u’ = – 2 g u’
      (1-p^2) u = – g
      u = -g /(1-p^2)
      the velocity is constant! It is an interesting result.
      h = H – p u t = H – g t /(1-p^2)
      and the emptying time is
      t = H (1-p^2)/g
      the filling time is
      v B t = H S
      t = H S / (B v)
      (1-p^2)/g = S / (B v)
      p^2 = 1- S g /(B v)
      A^2 = S^2 [1-S g/ (B v)]
      A(B) = S [1-S g/ (B v)]^{1/2}
      and this is the constraint on the opening surfaces.
      The rest are calculations, if all is right

      • There are some errors; only for a cylinder tank the Torricelli law can be written exactly:

        2 (1-p^2) u u’ = – 2 g p u

        2 (1-p^2) u’ = – 2 g p

        u = u_0 – g p t/(1-p^2)

        this equation is true for each area of cap surface A, for A -> 0 u=u_0=(2 g h)^{1/2}, so that

        u = (2 g H)^{1/2}-g p t/(1-p^2)

        there is a linear reduction of the velocity, that is null when the tank is empty

        t = (2 g H)^{1/2} (1-p^2)/ (g p)

        there is a difference with the Torricelli solution (differential equation with a constant velocity), a 2 factor, and it seem right.

  6. The key point here is “If the tank is initially empty”. If it was initially full, it would be in a steady state from the beginning… The constant flow of filling depends on the initial condition. So, the question is what is the height of water to get the minimum steady state flow. I guess here that the empty sink is not valid by default because it would be pointless?

  7. You can adjust the input flow to many different steady states…

  8. 1/4

  9. no, I’m wrong it is 3/4. v^2 = H = P

  10. I find 1/4, assuming Torricelli’s law that the rate of flow out of the tank is equal to a*sqrt(2 g h), where a is the area of the hole, h is the height of the water in the tank, and g is the gravitational acceleration. An expression for the full height of the tank in terms of a, g, and Q, the inflow rate, can be obtained from equating the time to fill and to empty as described. An expression for the equilibrium height of the water can be obtained (simply equate inflow and outflow rates). One finds the equilibrium height is 1/4 of the full height. Thanks for the nice question!
    [Integration of 1/sqrt(h) is necessary to obtain the expression for the time to empty the tank, or am I missing a simpler approach?]

  11. I reckon none of these. Won’t it depend on the area of the tank relative to the area of the drain and geometry of the system.

    • telescoper Says:

      It will, but the rate of filling and the rate of emptying also depend on those things, and you’re told something about them.

  12. Finally, I found my last mistake… V(steadystate) = Vmax/2 (at full height),

    and P = H = v^2 => H(steadystate) = Hmax/4

  13. […] all a favour by giving you something interested to do to distract you from the yuletide tedium, The cute problem of the water tank I posted a while ago seemed to provide a diversion for many – although only about 10% of […]

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