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

 

[Latexman]

Calculate the initial flow rate with water at the 10 cm height, Q_{10 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 Q_average = Q_{10 cm}/2. Lastly, calculate time to empty as Volume/Q_average.

Good luck,
Latexman

 

[Grizzman]

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.

 

[Ashereng]

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
Have you read FAQ731-376 to make the best use of Eng-Tips Forums?

 

[khardy]

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.

 

[Latexman]

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

Good luck,
Latexman

 

[Grizzman]

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

 

[Latexman]

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

 

[RWF7437]

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

 

[khardy]

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?

 

[Latexman]

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

Good luck,
Latexman

 

[rcooper]

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.

 

[IRstuff]

You might do a websearch for "waterclock." There's supposedly a couple of articles on the math model for such.

TTFN

FAQ731-376

 

[Grizzman]

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

 

[khardy]

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.

 

[IRstuff]

The shape of the waterclock is important to maintaining accurate time:

http://gershwin.ens.fr/vdaniel/Doc-...thodes-Maths/white/deqn/a1/wclock/wclock.html

TTFN

FAQ731-376

 

[Grizzman]

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

 

[Latexman]

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

 

[Grizzman]

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

 

[IRstuff]

I Dogpiled waterclock equation

TTFN

FAQ731-376