Fluid mechanics,interesting problem

[pcm]

Homework Statement

An open-topped vertical tube of diameter D is filled with water up to a height h.The narrow bottom-end of the tube ,of diameter d, is closed by stop .
when the stop is removed ,the water starts flowing out through the bottom orifice with approximate speed v=√(2gh). However ,this speed is reached by the liquid only after a certain time t.obtain an estimate of the order of magnitude of t.Ignore viscous effects.

Homework Equations

equation of continuity(mass conservation), A.v=constant
bernoulli's equation, P+0.5ρv²+ρgh =constant
the total energy,potential +kinetic ,of water is conserved.

The Attempt at a Solution

i assumed that i) water level falls with "constant" acceleration a.
then in a small time dt, i found the mass of water that came out of the hole,and decrease in water level.this gives me the loss of potential energy.i know velocity at the end of that small interval,i used it to find kinetic energy.using energy conservation ,i got a=g.
i don't know how to proceed now. also i am not sure about my first assumption.and once the liquid starts falling out at v=√(2gh),will this acceleration magically disappear? does the fact that a=g,implies normal force from the bottom surface is 0 during the acceleration period?

 

[BruceW]

I'm not sure about your assumption either. But I'm not certain what the answer is meant to be. Are you allowed to look up information for this question? I know what my guess would be for the answer, but it would require me to look up one fact online.

Anyway, I'd start by thinking about the equation v=√(2gh) where does this equation come from? and keeping this in mind, why do you think it would require a short amount of time before it started to be true?

 

[pcm]

no, i am not allowed to use other information.i am required only to give a rough estimate, taking reasonable values for the parameters,if required.numerical prefactor need not be accurate.

i get v=√2gh ,by applying bernoulli's theorem at 1)a point just inside the water from top.
2)apoint just outside hole.
P₁=P₂=atmospheric pressure
v₁can be neglected at first,in comparison to v₂.
therefore, after cancelling pressures and rearranging,i get, v₂=√2gh.
now ,D²v₁=d²v₂,
this gives v₁=(d/D)²√2gh.
it would require a short amount of time before it started to be true,because velocity cannot change instantaneously.

this is how i got a=g:-
(a is the acceleration of water in the tube,acceleration of water coming out would be a(D/d)²)
mass of water that came out in time dt after opening the stop=0.5a(dt)²*0.25πD²*ρ_water
height descended =0.5a(dt)²
velocity at time dt=a(dt).
now using energy conservation ,
mass*height descended*g=0.5*mass*(velocity)².
from there i get a=g.

 

[BruceW]

I think you get a=g because the assumption you're effectively making is that the water is in free-fall. This is good, it gives us the parameter g. You need to get a timescale. And you have the parameter g and the height h. So from these two parameters, what powers should they be raised to, to give a parameter of time?

Also, the reason I was asking about the equation v=√2gh was because this works by pressure being transmitted through the water, forcing water to come out at that speed. So my guess was that you could use the parameter of the speed of a pressure wave in water. But, you said that you can't look up this value, so I think it's best to ignore this idea.

 

[asteriatic]

I would first commend the student for their efforts in attempting to solve this problem. Fluid mechanics is indeed an interesting and challenging subject, and this problem presents an opportunity to apply various principles and equations to arrive at a solution.

To address the student's concerns, I would suggest considering the following points:

1. The assumption of constant acceleration may not be accurate, as the velocity of the water will change as it flows out of the tube. Instead, it may be more appropriate to consider the change in velocity over small time intervals, using the principles of calculus.

2. The acceleration of the water will indeed decrease as it reaches the end of the tube and exits through the orifice at a constant velocity. This is due to the fact that the pressure at the bottom of the tube will decrease as the water level falls, resulting in a decrease in the net force on the water.

3. The normal force from the bottom surface will not be zero during the acceleration period. In fact, it will be equal to the weight of the water above it, which will decrease as the water level falls. This can be seen by considering the forces acting on a small volume of water at the bottom of the tube.

To estimate the time taken for the water to reach a velocity of √(2gh), one could use the equation of continuity (A.v=constant) to relate the change in velocity to the change in cross-sectional area of the tube. Then, using Bernoulli's equation and the principle of energy conservation, one could find the time taken for the water to reach this velocity. However, it should be noted that this is only an approximation, as it ignores viscous effects and assumes a constant acceleration. A more accurate solution would require considering the dynamics of the flow using principles such as the Navier-Stokes equations.