|Jun24-03, 02:16 PM||#1|
I'm doing a research about water flowing from a barrel through a small hole. I am trying to proove that there is an exponential relationship between the height level of the water and the time. This basically implies that
In order to do that , I have to prove first that the rate of change of volume remaining in the barrel is proportional to the pressure.
i.e.: dV/dt = -kp. This is the bit were I got quite stuck. If someone could help me with some cool tecky ideas or some adequate sites, I would be gratefull.
And it's good fun to.
|Jun24-03, 03:25 PM||#2|
here's something on the topic, but it doesn't support your theory. It says there is a definite time after which the barrel is empty.
|Jun24-03, 07:56 PM||#3|
doesn't take into account. One is that water has surface tension, so that the water in the stream syphons out the water in the barrel. Also the atmospheric pressure forces out the water. However, even if you removed all atmospheric pressure the surface tension would insure that the barrel would empty. It would be very annoying in practice if things worked the way you described because every container would have to be pressurized so that it could be emptied. Thank goodness you're wrong this time!
|Jun24-03, 10:12 PM||#4|
Good try but...
Why would you think that the rate of water level change does not depend upon the size of the hole in the barrel?
|Jun25-03, 05:55 AM||#5|
Good try but...
Why would you think that the rate of water level change does not depend upon the size of the hole in the barrel? "
I didnt say it doesnt depent on the size of the hole. And, yes , for starters I am treating this as a model.
Could anyone tell me how to prove that the rate of change of volume in the barrel is proportional to the pressure on the base?
thanks for the site arcnets.
|Jun25-03, 10:11 AM||#6|
|Jun25-03, 10:25 AM||#7|
I'm no cooper, but I'll give it a try.
Use Bernoulli's Law and Archimedes principle kid.
Patm + 1/2 &rho v2 = const.
A1 * v1 = A2 * v2
At some point you should be able to derive the expression:
1/2 &rho * (v12 - v22) = &rho g * (h2 - h1)
where the subscript 1 represents the point at the bottom of the barrel and the subscript 2 represents the point at the top.
Apply Archimedes principle, and you should be able to reduce the equation even further:
1/2 &rho v12 * (1 - (A1/A2)2) = &rho g h,
where h = h2 - h1.
Take the derivative with respect to time of both sides:
&rho v1 * (dv1/dt) * (1 - (A1/A2)2) = &rho g * dh/dt.
The only component of acceleration associated with velocity v1 should be g downward. So, this express reduces further:
dh/dt = v1 * (1 - (A1/A2)2).
If you want to show an exponential relationship (i.e. dh/dt = -k * h), you must find v1 in terms of h.
v1 = Sqrt[2gh + v2].
Since you said the hole is small, I will take that to mean that the radius of the outlet hole is much smaller than the radius of the barrel (i.e. R1 << R2. In this case we can approximate the previously stated expression as:
v1 = Sqrt[2gh],
and the ratio of areas can go to zero:
A1/A2 = 0.
So, here is your relationship kid:
dh/dt = Sqrt[2gh].
Use it wisely.
P.S. It doesn't look like the solution you wanted, but maybe I made a mistake somewhere... or maybe that's just another one of life's little tragedies....
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