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simple drainage question

simple drainage question

simple drainage question

(OP)
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

RE: simple drainage question

Calculate the initial flow rate with water at the 10 cm height, Q10 cm.  It's a "no-brainer" that the final flow rate = 0 with water at the 0 cm height.  Then, calculate the average flow rate as Qaverage = Q10 cm/2.  Lastly, calculate time to empty as Volume/Qaverage.

Good luck,
Latexman

RE: simple drainage question

(OP)
That's assuming the change in velocity is linear. But isn't it exponential? (V^2=h/(2*g))
I know in this example the numbers will be almost identical, but on a larger scale, the decrease in velocity due to the decrease in head would be noticeable.
Here are the equations I'm working with:
h/(2*g)=v^2
initial volume= Ac(area of the container) times intial height
Q=V*Ah(area of the hole)
intial volume-Q*t=0

combining these three equations yields (I believe) this equation:
t(h)=(initial volume-Ac*h)/( (2*g*h)^1/2*Ah)

then you'd intergrate this from hi(initial height) to h=0

Now I no longer possess the ability to solve that integration so I plugged it into mathcad and the answer I got was .952*length*time
I plugged in cm and sec and the answer was 95.2 cm*seconds.
This makes sense in that the integration would add another h(length) to the equation, but doesn't make sense in that I'm solving for time. So I basically divided by the intial height and got a time of 9.52. This only makes sense in that , w/o a pressure drop, the time is 7.14 seconds and since the drop in pressure (or loss of head)will reduce the velocity, it should take a bit longer to drain. Though an additional 2 seconds seems like a lot.

RE: simple drainage question

This sounds like a homework question. Is it?

"Do not worry about your problems with mathematics, I assure you mine are far greater."   
Albert Einstein
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RE: simple drainage question

The easy to do it is have a look at Lindeburg's "Mechanical Engineering Reference Manual." Look in the fluids section under "time to empty a tank" where you will find the following equiation:

t=(2*Ac*Sqrt(h))/(Cd*Ah*Sqrt(2g))

Cd is the coefficient of discharge.  I used 0.62 for a sharp edged hole.  Other nomenclature is per your example.  

I got 24.5-sec.

RE: simple drainage question

What is the velocity head loss (K value) you assigned to the 1 cm^2 hole?

Good luck,
Latexman

RE: simple drainage question

(OP)
Ashereng:
No this is not a homework problem. I am trying to develop a drainage timer. You know set volume through set diameter hole yields a fairly constant time to delay or trigger an action.

Khardy:
Thanks for the input. Being as I'm a structural engineer, I don't have a mechanical engineers reference manual handy. But I figured it had to have been worked out at some point.
24.5 seconds eh?!? That's probably right, it just feels too long when I draw a 10cm x 10cm square with an approx 1/2" dia. hole in it. it just doesn't seem like it will take that long. But then, Fluids isn't my primary field of engineering.

Latexman:
what do you mean by velocity head loss? wouldn't head loss (which reads to me as decrease in head) result in deceleration as opposed to velocity?

Thanks for all the replies!
G

RE: simple drainage question

I got 24 seconds using the method I outlined, except I used C = 0.595 for a Beta ~ 0.1 instead of using a velocity head loss approach.

Good luck,
Latexman

RE: simple drainage question

Since no one else has suggested it; try:
1. Build 1 box 10cm x 10cm x 10cm with plywood, screws and kinda waterproof glue
2. Drill Hole of desired size or sizes in bottom
3. Fill with water
4. Using Atomic Clock, calibrated to at least 7 GPS satelites, measure the time required for box to empty.
Repeat.

good luck

RE: simple drainage question

Well, it actually wouldn't be that hard to test.  The answer is independant of the shape of the container so you could use any container with a 100-cm^2 cross sectional area.  Maybe you can rummage throught the trash and find a 11-cm diameter tin can?  

RE: simple drainage question

I was thinking a rectangular, paper milk or juice carton might work nicely.

Good luck,
Latexman

RE: simple drainage question

The answer is not independant of the shape of the container.  Consider a shape with 90% of the volume in the bottom 10% of the container.  This will take longer to drain than a container with 90% of the volume in the top 10% of the container.

RE: simple drainage question

(OP)
Wow you guys are really passionate about what I thought was a simple question. That's good! More information than I was looking for, but good that you love your discipline so much.
I really need the equation as a starting point, so I can determine the size hole required and size volume required to furnish a give time delay. Exact timing is not critical, but within a 10-15 seconds would be nice. Once I've built the model, I can bore or plug the hole as required to fine tune the drainage time.
Thanks so much for everybody's replies!
G

RE: simple drainage question

The answer is not independant of the shape of the container.

Yes, you are correct.  I should have said independant of the shape so long as the cross section is constant.

RE: simple drainage question

(OP)
Ah! very interesting read. Way more than I need, but still pertinent. The waterclock is requiring a constant drop, thus the exponential shape of the container to account for the diminshing pressure as the water level decreases.
I am working on a 15 minute timer with only the overall time being important. Since the decrease in height doesn't need to be constant, I don't think a specific shape is required.
When I googled water clock, I did not find this reference.
Thanks!
G

RE: simple drainage question

Grizzman,

If you go to the water clock link above, scroll to the bottom, and click "Last", you'll see the development of Torricelli's Law.  Torricelli's Law is a mathematical relationship between the flow rate of fluid from a draining tank and the height of fluid in the tank.  That's very pertinent to what you are doing.

Good luck,
Latexman

RE: simple drainage question

(OP)
Yeah, I noticed that. Just haven't had the opportunity to read it and let it really sink in yet.
Thanks!
G

RE: simple drainage question

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.

RE: simple drainage question

(OP)
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.

RE: simple drainage question

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.      

RE: simple drainage question

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

RE: simple drainage question

(OP)
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.

RE: simple drainage question

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.

RE: simple drainage question

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

RE: simple drainage question

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)?

RE: simple drainage question

(OP)
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.

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