Time it takes for a water tank to empty

[Flipmeister]

Homework Statement

A cylindrical tank of diameter 2R contains water to a depth d. A small hole of diameter 2r is opened in the bottom of the tank. r<<R, so the tank drains slowly. Find an expression for the time it takes to drain the tank completely.

Homework Equations

p_1+\frac{1}{2}ρv_1^2+ρgh_1=p_2+\frac{1}{2}ρv_2^2+ρgh_2\\<br /> Q=vA=\frac{\delta V}{\delta t}

The Attempt at a Solution

I believe, since $p_1$ and $p_2$ are the same, that Bernoulli's equation becomes $2\rho gd=v_2^2$. I am assuming I need to use the equation of volume rate of flow for time, but then I would need the velocity $v_2$. But how do I solve for time from that? How am I to find Q?

 

[TSny]

Hello. Hmm, you have a factor of 2 in the third term on the left of Bernoulli's equation. Check to make sure that's right.

It would help if you told us where you're picking points 1 and 2.

What is the justification for cancelling the terms that involve h₁ and h₂?

 

[Flipmeister]

Whoooops I should have reread my post. Typos everywhere. I've edited now; thanks for pointing them out.

 

[TSny]

OK, so I presume point 2 is at the little hole at the bottom and that you made the approximation $v_1≈0$. Still looks like a little error (or typo) in your expression for $v_2^2$. Can you find it?

 

[Flipmeister]

Ah the $\rho$ should have canceled out there as well. So now that I know what the velocity is, can I then use ##vA=\frac {\Delta V}{\Delta t}?

 

[haruspex]

I believe, since $p_1$ and $p_2$ are the same, that Bernoulli's equation becomes $2\rho gd=v_2^2$.

Not quite. Check that again.
If $v_2$ is the linear flow rate out of the hole, what is the volume flow rate? What does that then tell you about how fast the depth in the tank changes?

 

[TSny]

Ah the $\rho$ should have canceled out there as well. So now that I know what the velocity is, can I then use $vA=\frac {\Delta V}{\Delta t}$?

Yes, Can you express the volume in terms of the depth $d$? Instead of using finite differences Δ, you will want to use instantaneous rates of change (you're heading towards a differential equation). It might be better to let the depth of the water be denoted by $y$ or $h$ instead of $d$ in case you need to express the rate of change of the depth as a derivative.

 

[Flipmeister]

Alright, so if I have $vA=\frac{dV}{dt}=A\frac{dy}{dt}$, then plug in y from $v=\sqrt{2gy}$ it looks like $\frac{dV}{dt}=A\frac{d}{dt}(\frac{v^2}{2g})$ which I believe gives me...
$$Q=\frac{dV}{dt}=\frac{Av}{g}$$

Can I say that Volume(final) = Volume(initial) + QΔt and solve for Δt? Or is it more complicated than this...

 

[TSny]

Stick with $y$ as the dependent variable and $t$ as the independent variable. Can you find an expression for the rate of change of $y$?

 

[TSny]

Alright, so if I have $vA=\frac{dV}{dt}=A\frac{dy}{dt}$

Careful here. Are the two areas $A$ the same in this equation?