Question

Hole in a bottle question.

Answer 1

Find a function h(t) which describes the height of water level with respect to time. I tried using Bernoulli equation but couldn't make reasonable result out of it.

Answer 2

I'm not exactly sure this is right, it's been a while since I've done fluid mechanics, but this is probably how I would solve this.

So you have bernoulli's equation and it should simplify to this I think (I use variables here because I'm too lazy to put in the actual value):

\[\frac{1}{2} v_{1}^{2} + gh(t) = \frac{1}{2}v_{2}^{2}\]

Then you want to invoke flow continuity so you get this:
\[A_{1}v_{1} = A_{2}v_{2}\]

where A1 is the area at the top, A2 is the area of the hole, v1 is the velocity at the top, and v2 is the velocity at the bottom.

Now we define v1 to be this:
\[v_{1} = \frac{dh(t)}{dt}\]

since the rate of change of the height is the same as the velocity of the water moving at the top.

So now you just do some algebraic substitutions, and then you get this:

\[(\frac{dh(t)}{dt})^{2} + 2gh(t) = (\frac{A_{1}}{A_{2}})^{2}(\frac{dh(t)}{dt})^{2}\]

which is a differential equation that I don't feel like solving at the moment.

Hope this help, and takes this all with a grain of salt.

Answer 3

Which term in the Bernoulli Equation describes the pressure of the water?

Answer 4

pressure head

Answer 5

Could you explain more?

Answer 6

I ignored the pressure term in the equation since I think we can make the assumption that the pressure between the top of the bottle and opening is negligible for the bottle.

I don't think the problem is solvable if this assumption doesn't hold true, since there's no way to find the pressure difference between the top and the opening without it being given to you.

Answer 7

@4n1m0s1ty

In the first equation stating Bernoulli's law for a tank with a hole in the bottom there is no flow of water in the tank at the top therefore \[v _{1} =0\]

The change in height per unit time does not constitute a flow in the tank.

Answer 8

pressure at the top of the tank and at the exit flow section have same atmospheric pressure so what in bernouli eqn it cancels each other

Answer 9

one more thing : velocity in the tank is not zero. its negligible when compared to exit velocity

Answer 10

@gleem

I think you are right. Since the area is much larger than the inlet, it makes sense that v1 would be negligible compared to v2, so v1 is essentially 0.

Answer 11

@THIRURAJMECH

I was answering thomas5267 question about "which term describes the pressure of the water?". The fact that the top and exit have the same pressure was what I was pointing out to him. Which is why it explains the missing pressure term from the simplified Bernoulli equation in my answer.

Answer 12

So I guess the new equation would be this instead:

\[2gh(t) = (\frac{A_{1}}{A_{2}})^{2}(\frac{dh(t)}{dt})^{2}\]

since v1 is ~0