simple drainage question 1

[Grizzman]

I have a 10 x 10 x 10 (cm^3) container filled with water and open at the top. If I open a hole in the bottom with an area of 1cm^2, how long will it take to drain? Please consider ideal, but include the decrease in head.
Also, could you post the equation for it?
thanks,
I've gotten answers of both 4.76 sec and 9.52 sec. w/o pressure drop, I believe the answer is 7.14 sec, but I'm not real comfortable with the solution as the Integration is solved through mathcad and I don't have the resultant equation.
Hopefully I'm just making this difficult and there's an easy solution somewhere.
Thanks,
G

 

[mjpetrag]

Set up a Bernoulli eq

P1 + .5*(rho)* (v1^2) + (rho)*g*h1 = P2 + .5*(rho)*(v2^2) + (rho)*g*h2

Assume P1, V1, P2, and h2 to be zero... the equation becomes

sqrt(2*rho*g*h1) = v2

Now you have V2, from the hole area you can find your initial flow rate, and your final flow rate is 0

I think that's right.

 

[Grizzman]

yes I came to that conclusion. However since h1 diminishes with time, v2 becomes a function of time. you have to integrate to find your solution and you end up with h/(h)^0.5 in the integration which I do not remember how to solve. That's what compelled me to ask the question in the first place. If you look at khardy's first response you'll see a lot of similar expressions in the equation he provided.
I never thought it was a new concept. I just didn't have the solved solution available to me.

 

[rconner]

Not claiming experise in draining cubes or tanks, but if one were uncomfortable with “integration” (or unsure of more complicated formulae of unknown derivation) but trusted an available simple relationship for coefficient of discharge, I wonder if one could perhaps approximate the time to empty the cube as follows:

1. Put your finger over the hole at the bottom of the cube and fill the cube with water. Assume the velocity exiting the bottom of the hole in the cube when you release your finger is equal to the relationship VL=(2gHL/Cd). In other words, the (driving) head above the hole is entirely converted to velocity (only) when a volume of water exits the hole. I suspect the coefficient of discharge would be arguably dictated largely by the geometry of the walls and the hole edges etc., but lets just say it is Cd =~0.6 for a relatively thin wall container with a drilled hole in the bottom.
2. Divide the cube into several thin vertical intervals or layers of water, lets say 10 layers at 1 cm each high. Calculate the individual volume of water in each layer, and also calculate the initial and final velocities (VL ) per level HL (as each lower volume drains due to surface head above it. Calculate a simple average velocity between initial and final velocities for draining a given layer of water (I realize this value isn’t exactly right, but I would think it would be very close for a small vertical layer or interval of changing head). In other words, if the initial velocity due to a full container/10 cm of head were 1.81m/sec and that due to 9 cm falling water level head were 1.72m/sec, one could perhaps reasonably assume the first lower layer of water in a full container might drain at a velocity of say 1.76 m/sec etc.
3. The time to evacuate each succeeding lower layer of water could then of course be determined as T=(Volume of Layer/Area of Drain)/(Average Velocity for the Specific Layer), when these values are in comparable units, and the ten values tabulated for adding together. It is of course correct that the time to evacuate each layer is not linear, and will increase as the water level falls.

I did my arguably sort of Cro-Magnon “integration” exercise and (if I did the math correctly) got ~8.6 seconds total time to drain this cube. I thus do not necessarily distrust at least the higher of the initially provided “answers”. Also, (as I suspect most tanks may most often be fitted and drained with short lengths of small spigot pipes and also probably other minor loss valves etc., and also there could be quite an argument about how long it takes for the very last water to seep, drip, or evaporate etc. from the very bottom of the container of whatever surface condition) I won’t necessarily argue either with the even longer answers or formulae. I like better the idea of making a container and timing it, so the observer/experimenter can explain exactly what he finds/means with regard to the latter. Everyone have a good weekend.

 

[RWF7437]

Galileo used a water clock, but you don't have to. Is there some reason you want to ?

 

[Grizzman]

rconner
you solution still requires integration and is dependent on the continuous change in volume.
Also, the problem with just timing a container, is its very trial and error. with an equation, you at least start in the ball park of what you want.plus you can vary your parameters to adjust your results.
RWF7437
I am trying to produce a cycle timer that is not electrically dependent.

 

[BarryEng]

Time to empty tanks by means of an orifice: -

t = (2 At / C Ao SQRT (2 g)) (h1^^0.5 - h2^^0.5)
constant cross section with no inflow.

t = integral h1 to h2 (-At dh)/ (Qout - Qin)
inflow < outflow, constant cross section.

That felt good typing this out when I have no idea how to do the integral without assistance.

I will now come clean - go to Schaum's Fluid Mechanics & Hydraulics SI (metric) edition. Chapter 9. Measurement of the flow of fluids.

Time to empty tanks by means of an orifice - see problem 38.

Tank cross section not constant - see problem 41.

Time to empty tanks using weirs - see problem 43.

Time to establish flow in a pipe - see problem 45.

How about that for a complete answer using only one source.

 

[hydtools]

Mark's handbook has a solution for unsteady tank flow. The example problem was for a tank with flow in and flow out. One solution was for the time for the level to change some value.

Reducing the integrated result of the example in Mark's to this problem yielded:

t2 - t1 = 2*A*sqrt(h) / (C*a*sqrt(2*g))

For this problem I got time to empty to be 23.4 sec.

A = tank area
h = initial water height
C = orifice coefficient (0.61)
a = orifice area

You could rearrange the equation to determine the tank size and shape for the time interval you want. Or the orifice size.

Ted

 

[rconner]

Just saw that I wrote the discharge equation I used (to calculate the decreasing velocities/increasing time values to drain the successive layers out the area of the hole) down wrong -- I should have written VL=(2gHL/Cd)^0.5. [I used this equation to obtain the agggregate 8.6 sec total time to discharge all successive thin vertical layers .

Grizzman,
If one makes the discrete vertical lenses of water short/numerous enough as I explained, would not the "continuous change of volume" be accounted for (with increasing time values per layer as the cube drains)?

 

[Grizzman]

rconner,
indeed if you made the numerous layers and calculated each successive layer, you are accounting for the volume change. You're basically using the foundation principles of calculus without the integration.
and if this were a singular calculation I would do that or just build it and time it.
I actually chose those dimensions for ease of calculation. I was looking for the general equation so I could design a container and hole to yield a specific time of discharge.