Need desperate help on fluid dynamics

[siddharth]

yes, but that formula oversimplifies the problem since it ignores the fact that there is no input so the flux rate ot the hole drops as volume drops, and volume drop rate is not consatnt and relies of the past flux rates and how much water was previously lost.

Er.. what? I don't understand. Could you elaborate?

 

[gtabmx]

SOLVED! I will explain when I get home. Thanks everyone!

 

[vanesch]

If you use Bernoulli's law in this case, we simply have:

v^2/2 + h g = constant
(the density of the fluid drops out).
If we are at the surface, v = 0, and h = d, so constant = g d
If we are at the hole, h = 0 and hence v = sqrt(2 d g)

So the velocity of the fluid that squirts out of a hole at depth d from the surface of the fluid, equals v = sqrt(2 d g)

This is independent of the area of the hole or even the density of the fluid.

However, the fluid flux will be equal to this velocity times the area of the hole A:
phi = v.A = A sqrt(2 d g).

Now, in a time dt, a fluid flux phi will correspond to a volume dV = phi.dt of water that has escaped. This will then lower the surface of the water with an amount - d d (silly notation, meaning that depth d has diminished by d d). If the section of your bottle equals S, we obtain that when a volume dV has been taken away, we have that d d . S = - dV

So: d d. S = - dV = - phi.dt = - A.sqrt(2 d g).dt

which gives you a differential equation for the function d(t):

(hum, I did a bit too much here, you should work this out yourself...)

I almost gave the solution in my quoted post (actually I did, and then I quickly erased it for it is against the policy of PF). Given all the work and the solution, I think it is fair now to go to the final solution.

Now that we got this far, once we have the differential equation:

d d. S = - A. sqrt(2 d g).dt

with:
d the fluid level (in meters, say) above the hole, S the cross section of the bottle (horizontal area in m^2), A the cross section of the hole (also m^2), g the gravitational acceleration (9.81 m/s^2), we can write this as:

d d / sqrt(d) = - A/S sqrt(2 g) dt

and integrate, which gives us:

sqrt(d) + cte = - A/S sqrt(g/2) t

at t = 0, we have d = d0 (the initial fluid level), so cte = - sqrt(d0)

sqrt(d) = sqrt(d0) - A/S sqrt(g/2) t

Or:

d(t) = (sqrt(d0) - A/S sqrt(g/2) t)^2

which gives you the evolution of the fluid level above the hole d(t) as a function of t...

It has the form of a parabola, and I hope I didn't make any silly mistake.