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?
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In the Dark 
November 26, 2017 at 8:10 pm
I voted, and now I think I got it wrong. Pressure is important.
November 26, 2017 at 10:57 pm
4/9 full
November 27, 2017 at 12:55 pm
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.
November 27, 2017 at 1:52 pm
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…
November 27, 2017 at 1:22 pm
After more maths than I expected, I calculated 1/2
November 27, 2017 at 7:03 pm
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
November 28, 2017 at 11:01 am
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
November 29, 2017 at 10:09 pm
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.
November 27, 2017 at 8:30 pm
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?
November 27, 2017 at 8:32 pm
Not really. If it were full at the start it would still drain faster than it was filled…. for a time.
November 27, 2017 at 8:32 pm
You can adjust the input flow to many different steady states…
November 27, 2017 at 8:33 pm
No you can’t. The tap is fully open.
November 27, 2017 at 8:36 pm
1/4
November 27, 2017 at 8:52 pm
no, I’m wrong it is 3/4. v^2 = H = P
November 29, 2017 at 11:12 am
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?]
November 29, 2017 at 6:49 pm
According to my calculation, you are right. To be sure, I verified with numerical calculations.
November 29, 2017 at 7:04 pm
I made another mistake. My calculations gives me a steady state v of 1/4 the max, so the height should be 1/16???
November 29, 2017 at 6:00 pm
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.
November 29, 2017 at 6:51 pm
It will, but the rate of filling and the rate of emptying also depend on those things, and you’re told something about them.
November 30, 2017 at 7:09 am
Finally, I found my last mistake… V(steadystate) = Vmax/2 (at full height),
and P = H = v^2 => H(steadystate) = Hmax/4
December 19, 2017 at 4:06 pm
[…] 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 […]