Modelling Water Tanks: Solving with Bernoulli Equation

[Alexander350]

I tried using the Bernoulli equation to solve this. I took two points at the surface of the water in both the containers and formed this equation:
gh_{b}=\frac{1}{2}v^2+gh
This is assuming that the velocity of the water in the large tank is approximately zero and using the fact that both the surfaces are at atmospheric pressure. Then, I solved for the velocity and said that this is equal to the rate of change of the height of the water in the narrow cylinder.
\frac{dh}{dt}=\sqrt{2g(h_{b}-h)}
Finally, solving this with the assumption that h starts at 0, I got:
h=\sqrt{2gh_{b}}t-\frac{1}{2}gt^2
But looking at this function, it increases to the height h_{b} and then decreases again. This obviously does not happen; it would just stay at that height forever. So what have I done wrong?

 

[Alexander350]

Integrating the equation correctly might help.

It looks correct to me. Substituting it back into the original equation satisfies it.

 

[DoItForYourself]

Your equation is valid from h:0 to h:h_b. So, if you solve it for h=h_b, you will find your upper limit for t (call it t_b).

From 0 to t_b you can use your solution for h(t).
For t>t_b, dh/dt=0 and so h=h_b.

 

[Chestermiller]

It looks correct to me. Substituting it back into the original equation satisfies it.

Sorry. I misread your equation. You are, of course, correct.

It is interesting to reduce your final equation to dimensionless form as follows:

$$\frac{h}{h_b}=\left(\frac{t}{\tau}\right)-\frac{1}{4}\left(\frac{t}{\tau}\right)^2$$ where the characteristic time $\tau$ is given by $$\tau=\sqrt{\frac{h_b}{2g}}$$
The level in the small tube rises to that in the large tank when $$t=2\tau=\sqrt{\frac{2h_b}{g}}$$