A cylindrical tank with its axis horizontal has a nozzle at the bottom. The tank is full at first, and it is found that it takes 3 minutes for half the water to flow out. If the radius of the tank is 2 ft. and its length is 8 ft. find the depth of the water after t minutes. Find also the time t for the tank to empty.

Hi,

You surely know the water speed is dependent on the water pressure, and so, the water depth in the tank. Are you looking for the complete physical solution or to a more simpler one, assuming a constant water outer flux? Your last question is trivial in the second case, so I don't have many hopes you are going to choose the easy way.

JPA

****
My candidate solution to the first question, is case I didn't error, is
``x(t)=4 Sech(aSech(1/sqrt(2)) (t/3))^2 = 4 Sech(0.293791 t)^2``
where x is measured in ft, t in minutes and Sech(a)=1/cosh(a) is the hyperbolic secant function. The tank will take an infinite time to be empty.

SOILER ALERT: Demonstration ahead

Demonstration
The solution to this isn't easy, so I'll take the liberty to simplify the problem: I'll add a second hole to the top of the tank so water may flow unrestricted in the bottom nozzle without fighting the air bubbles trying to enter, and have equal atmospheric pressure in and out off the tank.

The water flow isn't constant, it varies with water pressure and so, it changes with the water depth too. My common sense says that the escaping water must have a kinetic energy equal to the potential energy of the water in the tank layers which are substituted by air, more precisely:

``1/2 dm v^2 = dm g x <=> v = sqrt(2 g x)``

This is not exactly true, as part of the potential energy must be used to move the water inside the tank, move the air outside it and lost in heat against friction, but I have the feeling it is accurate enough. Now this speed is related to the flow or water volume variation rate:

``d/dtV = v A``

which can be written in terms of x, substituting v:

``d/dtV = v A<=>(dV)/dx (dx)/dt =  A sqrt(2 g x) <=> (dx)/dt =  A sqrt(2 g x)//((dV)/dx) ``

..where A is the sectional area of the nozzle.  The volume is given by ``V(x)=L R^2 (Pi - acos(x/R-1))`` and its derivative in order to x is ``(dV)/dx=(LR^2)/sqrt((2R - x)x)`` where R is the radius, and L, its length. Substituting `dV//dx`, this leads us to this horrible differential equation:

``(dx)/dt = A/(LR^2) sqrt(2g(2R - x)) x``

The solution to it is

 ``x(t)=2 R Sech((A sqrt(g))/(LR^(3/2)) t + sqrt(R/2) C)^2``

There are two unknows here, A and C, but fortunately, we know that at time zero, x is 2R, and at time T, x is R. The only valid solution to these equations is

``C=0``

and

``A=(LR^(3/2))/T sqrt(g) aSech(1/sqrt(2))``

Substituting, we have

``x(t)=2 R Sech(aSech(1/sqrt(2)) t/T)^2``

 which is the formula we wanted. About the time when x is zero, there is none: Sech or hyperbolic secant (Sech(x)=1/cosh(x)=2/(exp(x)+exp(-x))) is a function without zeros. That means the tank will never be empty. This makes sense as the water being removed approaches the bottom, will have less potential energy to translate in out-speed... and in the "bottom", it will have none. Of course, at this time, an whole bunch of other not accounted phenomenas will be dominating the process. An interesting detail, x(t) only depends on the tank radius... apparently not in its length or total volume. But apparently only, since the tank volume is codified in the time T, which being bigger or smaller relates us the relationship between the tank length and the sectional area of the nozzle. Another interest fact is, we can write the expression for x(t) in the form:

``x/R=2 Sech(aSech(1/sqrt(2)) (t/T))^2``

 The proportion between x and R evolves always in the same way in relation to the proportion t/T.

Very interesting problem,... I couldn't resisted to present my reasonings from the start. Cheers,

Joao Pedro Afonso
PS.: The math expressions are AsciiMathMl compliant so it is enough to copy and paste them to a tiddle wiki or a page with that module, to see the math much better (I hope). Some expressions in ascii text will appear to be wrong because of that, but it is worthy. You may get the AsciiMathMl.js in this address: http://www1.chapman.edu/~jipsen/mathml/asciimath.html

Joao,
Is this similar logic to that where the hare is chasing the tortoise, and it can be proven that it will never be caught, when intuitively one knows that it will be ?
Similarly, a tank with a hole in it (for all intents and purpose in my proposed situation) will empty itself despite all calculations to the contrary.

Bill.

Nope, my intuition runs precisely in the opposite direction. A tank in your conditions will never empty itself completely until the end. You don't even need to look to the obvious fact that you will find the real tanks always wet (meaning they are not empty), either because of surface tension effects or little traps in the walls. Instead, think on the very end, when the water runs so scarce that a single molecule may run a lot of time without collisions. And without them, you don't have the pressure gradients to give the lead to the nozzle. That molecule may take an eternity to find it. However long time it takes, that will be the time you need to empty the tank.

But you have some points here. For one, I'm not sure how this extreme behaviors are reflected on hydrodynamic equations. I have some training in physics but unfortunately, strangely, fluid dynamics was not part of it. Nor did I tried to consult anything... I'm not counting to be right, but I'm curious to see to where my own feet takes me. Second point, I forgot to include in the equations, the atmospheric pressure. If I opened a hole in the top of the tank, then I have work realized by the air equal to PdV, which should also translate in more kinetic energy. I have to see how this works but I have the feeling it might not change my infinite time result: because the tank is an horizontal cylinder, the area were that pressure is applied, dwindles to zero in the end, and the solutions might reflect this.

Last point, I appear to imply that the molecules in the nozzle frontier would be motionless if there was not the possibility of some volume changes in the tank to convert potential energy in kinetic one,... when everyone knows that they would start to free fall immediately. I don't know how to easily answer to this. My reasoning is based on the fact that water is incompressible, so, it should be already in movement in the nozzle mouth, if we want other to replace the one which is exiting. To be in motion, there must energy sources to account. Once calculated, we have the speed and the flow. After exit, it accelerates by gravity forces but that is accounted by the thinning of the water "tube" and eventually by its disintegration in small drops... no scheme like this inside the tank.

I'll write back when I see the effects of atmosphere.

Cheers,
João Pedro Afonso

Hi,

Not a shred of good luck here. When I introduce the atmosphere pressure, the differential equations appear to turns unmanageable (not a clue how to deal with them). My fault, I was the one to introduce it. Not that I'm going to be lucky again if I close the hole: there is something called vapor pressure which given enough time and temperature, will fill the hole being created. I'm going to do what good physics do when they don't know how to handle with a factor: ignore it, hopping it will go way :-).

(Oh, it is never so simple. We must have an idea of the factor impact and compare it with other factors, to see where the ranges are in which the last ones will be much more important than the first. Water vapor at low temperatures, for example, is a lot weaker than the standard atmosphere and perhaps may be discarded)

But my second conclusion stands still: the time derivative of the depth when it approaches zero, is zero too. That means the water flow will go to zero when the depth approaches zero, and the zero will never be reached.

I'm seeing by your remarks, this is not the solution you were expecting. I'll start reading to see where I was wrong. Perhaps the Bernoulli equation has the answers.

Cheers,
Joao Pedro Afonso

Joao,
Wihout detracting in any way from your impeccable logic, I have in front of me an old technical handbook edited by three Heads of Mathematics at 3 English colleges which gives a finite time for the emptying of the tank. I think this demonstates the difference in "mind-set" between practical puzzle composers and theoretical purists.  
Bill

We have a water reservoir of an arbitrary shape that has
a small hole its bottom.  We wish to determine the time
to empty the reservoir.

In an article in this thread, João has argued that
fluid is ejected from the hole with a velocity of v =
sqrt(2gh), where g is the acceleration of gravity and
h is the height of the fluid in the reservoir.  This is
known as Torricelli's Law.

Torricelli's Law gives a reasonably good approximation
for low viscosity fluids, like water, but is not so
good for high viscosity fluids, like honey.  It is common
to replace Torricelli's Law by v = k sqrt(2gh) to allow
for viscosity effects, where k < 1 is a constant.

Let h be the height and V(h) be the volume of the
fluid in the reservoir.  By the fundamental theorem of
calculus we have dV/dh = A(h), where A(h) is the area of
the exposed part of fluid at the top of the reservoir.
Let "a" be the area of the hole in the bottom of the
reservoir.  The conservation of volume is expressed as:

  dV/dt = - av.

But dV/dt = (dV/dh)(dh/dt), therefore:

  A(h) dh/dt = - a k sqrt(2gh).

By solving this differential equation, we obtain h as a
function of t.

Special case: Let the reservoir be a cylinder of length
L and radius R, placed so that its axis is horizontal,
and initially filled with fluid, thus h(0)=2R.

It is easy to see that A(h) = 2 L sqrt(2hR-h^2).
Therefore

    dh/dt = - a k sqrt(2gh) / [2 L sqrt(2hR-h^2)].

We solve this with h(0)=2R to get:

   h(t) = 2R - (1/(2 L)) * (9 g L k^2 a^2 t^2)^(2/3) .

Let us say that the reservoir reaches the half-empty
level at time t=T, that is h(T)=R.  From this we
calculate the value of k:

   k = 2 sqrt(2 g R) L R / (3 g a T ) .

To compute the time T* when the reservoir is completely
empty, we solve h(T*) = 0.  We obtain:

  T* = 2 sqrt(2) T.

--
Rouben Rostamian

Correct.

  Very elegant deduction, Rouben. And thanks for enlight us with the names of its different parts, it helps to keep track of possible references and helpers, to the problem (at least to me). I, for instance, should have remembered that dV/dx is a sectional area, it would have help me a lot and spare me from the mistake I made. That's what we gain for doing things through brute force.

It is interesting to note that this approach to access viscous fluids is the same as to have a smaller hole in the tank than the one it is really there. Since the hole area is unknown, our ignorance of the type of the flow isn't going to be lifted with this problem, but I'm wondering what would have happened if h was very high. I keep think that in that case, k will evolve to be smaller... but I'm probably fretting for nothing: to much speed and the fluid will turn turbulent, and every of this equations will turn in dust :-)

  Arghhhh... to think I restart to do these puzzles to regain confidence on myself. I was becoming very troubled because every time I did the calculations for dV/dx, I obtained LR^2/sqrt((R-x)x), which was obviously wrong when we know dV/dx is the top surface area of the fluid. This morning, it got me: I forgot a term in V. My V should have be

  V = LR^2(Pi-arccos(x/R - 1))+L(x-R)sqrt(R^2-(x-R)^2)

I'll refrain from name myself what I deserve to be called  now :-(

This would had take me to the right track, but it chills me to thinks what kind of expressions I would had to work. Yours is a more simple path, Rouben.

This is a most interesting problem. What worries me is that I was able to semi-justify to my happiness, the result I got. Usually the kind of physical reasonings I used, would had serve me to do reality check on my results and avoid the mistake I made, but that here didn't happened. At least, my physics was sound... I expected until your demonstration that I was forgetting some obvious physical effect. In the end, it was only an error of geometrical nature.

There's something which still puzzles me. To have an infinite discharge time, we only need a tank area proportional to a power of x, with an exponent smaller than 1/2. A tank with walls with a curved V format would had do the trick. Then, we would had had

A(h) dh/dt = -K sqrt(2gh) <=>
Ch^k dh/dt = -K sqrt(2g) h^1/2 <=>
dh/dt = -K/C sqrt(2g) h^(1/2-k), k<1/2

and that means dh(0)/dt=0. So, in this case, we would had a tank taking really an infinite time to be empty... but that shape is not intuitive at all. The answer resides possibly in the impossibility of building a hole of finite area in the zero depth.

The problem of a hole in the top of the tank is still unsolved to me.

Cheers,
João Pedro Afonso

Hello João, in the previous message you wrote:

That argument is not a correct.
dh/dt=0 when h=0 does not imply that
the emptying time is infinite.  To see
that, rearrange the equation as

   h^(k-1/2) dh = -K/C sqrt(2g) dt

then integrate on both sides to get:

 (1/(k+1/2)) h^(k+1/2)  = -K/C sqrt(2g) t + B,

where the integration constant B may be determined
by applying the initial condition h(0)=H (but that
is not relevant to the rest of the argument).

To get the emptying time, T, set h to zero in the
above expression.  We see that:

   T = B C / (K sqrt(2g)).

--
Rouben Rostamian

  Your argument is sharp as a razor knife. Thanks for your insight. Funny how Ontadian remarks about Zenon and the Turtle ended gaining relevance to me.

  I could had retreat to my original (wrong) volume formula to save face, but there is no hope there: it describes a tank with an infinite width at the base, no wonder it will take an infinite time.

  Do you know of any simple way of taking into account the atmosphere work in the problem, if we open an hole in the top of the tank?

Cheers,
Joao Pedro Afonso