- 22 Mar '06 21:35This is taken from http://mathproblems.info:

A large tank has a steadily flowing intake and 10 outlet valves, the latter being all of the same size. With 10 outlets open, it takes two and one half hours to empty the tank; with 6 outlets open it takes five and one half hours to empty the tank. After the tank is empty and with all 10 outlets closed, how long will it take to fill the tank?

There is an answer provided on the website, but I think it's wrong, so I decided to see what you all think about it. Let 'er rip! - 22 Mar '06 22:11 / 1 edit

I think we are supposed to assume the outlets are pressure regulated so the flow rate is the same regardless of percent of fill. BTW, the outflow is like the parallel resistor problem, R total= 1/(1/R1+1/R2+1/R3) etc.*Originally posted by leisurelysloth***Are we to assume that the flow rate through the outlets is not dependent upon the level in the tank?**

in this case 1/20.5 for each outlet, 20.5 hours for one outlet to drain.

Not the solution obviously, just pointing that out.

Its almost like a constant current source charging a capacitor and resistors draining off part of that charge. However at the end of the time the tank is drained, there is still flow. Thats just the other end of the same problem. - 23 Mar '06 00:11 / 2 edits

x = constant inward flow in tanks worth per hour*Originally posted by PBE6***This is taken from http://mathproblems.info:**

A large tank has a steadily flowing intake and 10 outlet valves, the latter being all of the same size. With 10 outlets open, it takes two and one half hours to empty the tank; with 6 outlets open it takes five and one half hours to empty the tank. After the tank is empty and with all 10 outlets closed, how long w ...[text shortened]... ebsite, but I think it's wrong, so I decided to see what you all think about it. Let 'er rip!

y = flow from a single outlet (we assume that outlet flow is a constant not related in any way to how much liquid remains) in tanks worth per hour

Therefore if we assume the tank to be 1 (we did so when we defined our units) we have:

1 + 2.5 (x - 10y) = 0

and

1 + 5.5 (x - 6y) = 0

Rearranging we then have:

2.5x - 25y = -1

6.5x - 33y = -1

Solving we find:

x = 0.14545

y = 0.054545

The units for these are tanks per hour. Therefore we can say that the tank would take 6.875 hours to refill from empty with all valves closed.

EDIT: Sonhouse what the hell are you rambling about? Make some sense for once.

EDIT2: Looking at the webpage (it's problem 163) they took the same approach to me except they defined the tank as holding w gallons and found flows in gallons. Seems like wasted effort to me. They also got the same answer I did. So PBE6, explain why you think it's wrong. - 23 Mar '06 00:16

No.*Originally posted by sonhouse***I think we are supposed to assume the outlets are pressure regulated so the flow rate is the same regardless of percent of fill. BTW, the outflow is like the parallel resistor problem, R total= 1/(1/R1+1/R2+1/R3) etc.**

in this case 1/20.5 for each outlet, 20.5 hours for one outlet to drain.

Not the solution obviously, just pointing that out.

Its almost li ...[text shortened]... the time the tank is drained, there is still flow. Thats just the other end of the same problem. - 23 Mar '06 00:23 / 2 edits

Your solution makes perfect sense if you assume the outflow to be constant and independent of the height in the tank. However, Bernoulli's equation models this situation a little better than that, and gives some interesting restrictions (which are actually quite intuitive when you see them).*Originally posted by XanthosNZ***So PBE6, explain why you think it's wrong.**

**(This derivation is a little long, but I hope it's fairly clear and straightforward):**

Setting up Bernoulli's equation for one outlet, where point 1 is the top of the tank and point 2 is on the free jet at the outlet, we have:

P1/gamma + v1^2/(2g) + h1 = P2/gamma + v2^2/(2g) + h2

where:

P1, P2 are pressures

v1, v2 are flow velocities

h1, h2 are heights

gamma is the specific gravity of water = density * gravity

g is the gravitational constant

We set h2 equal to 0 as a reference height. Now, assuming both water surfaces are exposed to the atmosphere, and the height of the tank is not several miles high so atmospheric pressure is constant, we can cancel P1/gamma and P2/gamma. Also, assuming that the water at the top of the tank is not moving quickly compared to the jet, we can assume v1 = 0. After simplifying, we are left with:

h1 = v2^2/(2g)

Solving for v2 we get:

v2 = (2g*h1)^0.5

By the continuity equation, the flowrate from the outlet is equal to the flow velocity v2 times the outlet area A, so:

Qout = A*v2 = A*(2g*h1)^0.5

Assuming a cylindrical tank, the volume of water in the tank V is the height of the water h1 times the cross-sectional area R, so:

Qout = A*[2g*(V/R)]^0.5 = k*(V^0.5) for simplicity's sake.

Now that we have an expression for the flowrate from the outlet as a function of the volume in the tank, we can set up a differential equation to solve for the volume in the tank as a function of time.

change in volume = flow in - flow out

dV/dt = Qin - n*Qout (...n outlets, all assumed to flow at the same rate)

dV/dt = Qin - n*k*(V^0.5)

This is a seperable differential equation. Seperating both parts we get:

dV/(Qin - n*k*(V^0.5)) = dt

Integrating both sides, we get:

(1/nk)^2 * [2*(V0^0.5 - V^0.5) + 2*Qin*LN[(Qin - nk*V0^0.5)/(Qin - nk*V^0.5)]] = t

The interesting thing here is that if Qin = nk*V^0.5, or V = (Qin/nk)^2, then the LN part of this equation will be undefined. Values of V less than (Qin/nk)^2 give negative values for t, which are clearly not possible. The physical explanation for this is that as the volume in the tank is reduced, there is not enough driving force (hydrostatic pressure in this case) to evacuate the fluid in the tank at a faster rate than it's coming in. In effect, it is not possible to empty the tank so long as water in flowing in. I used a cylindrical tank as the basis for this proof for simplicity, but this principle applies to any shaped tank. - 24 Mar '06 01:19

Nice work. Xanth, I was just showing the similarity to using an electronic analogy, solving with R/C time constant equations and the like. You can have constant current feeding a capacitor and have a set of 10 resistors you can switch off and on, which would be like liquid drains in a tank. One point on that, the R/C time constant equations automatically account for lessoning charge rates when the voltage gets less, just like the lowering water levels reduce the water flow through the drains.*Originally posted by PBE6***Michael Shackleford, the puzzlemaster for the site, added my comments to the solution! Hehe, I'm published now. Too bad the prose was a bit cockeyed:**

http://mathproblems.info/prob163s.htm - 04 Apr '06 04:20

my my, we have been busy haven't we?*Originally posted by PBE6***Your solution makes perfect sense if you assume the outflow to be constant and independent of the height in the tank. However, Bernoulli's equation models this situation a little better than that, and gives some interesting restrictions (which are actually quite intuitive when you see them).**

[b](This derivation is a little long, but I hope it's fairly clear ...[text shortened]... as the basis for this proof for simplicity, but this principle applies to any shaped tank.

Since you've raised some real world problems, I'd like to add a further complication to any attempted solution, just to be an ass. Here's a list of variables not accounted for which would lead to minor errors in the solution:

1. Evaporation rate of water into the air above the tank. Some of the water will evaporate into the air before it can be drained. Evaporation rate will depend on several other variables that haven't been accounted for. Some of these are:

water temperature...ambient air temperature...humidity...elevation above sea level...etc.

One other impact on evaporation rate would be whether or not the intake valve achieves laminar flow of the water molecules...non-laminar flow will impart a degree of excited water molecules thereby enhancing evaporation.

2. A second problem not addressed by the question is that the outlets may create a vortex effect directly above the outlet (similar to flushing a toilet). If each outlet is sufficiently spaced apart, the vortexes will not interfere with each other but with 10 outlets it is likely that as the tank reaches minimal levels, the vortexes will create drag with other vortexes and slow the rate at which each vortex can draw water into the outlets. In other words, the tank will empty at a slower rate due to vortex interference since 2 or 3 vortexes will be partially competing for the same water.

Food for thought.