2
$\begingroup$

The problem is to find an expression for the time it takes for a cylindrical water tank is half empty. I am given that the water tank has a diameter of $D_0$ and a hight $H$. The water outlet is placed in the bottom of the water tank and has a diameter $D$.

I am using the Bernoulli's equation to find the velocity to have the expression $v=\sqrt{2gH}$, because both pressures are atmospheric.

The method I used to find the expression for the time it took for the water tank to be half empty, is to divide the volume of the tank with the flow rate:

$t=\frac{V}{\dot V}$.

The expression for the volume of half of the water tank (We are supposed to calculate the time it takes for half of the volume to flow out, and therefore only half of the volume is used here):

$V=A\cdot H=\frac{\pi}{4} D_0^2 \frac{1}{2}H$=$\frac{\pi}{8} D_0^2H$

while the expression for the flow rate is:

$\dot V=A\cdot v=\frac{\pi}{4} D^2\sqrt{2gH}$

The expression for the time then becomes:

$t=\frac{\frac{\pi}{8} D_0^2H}{\frac{\pi}{4} D^2\sqrt{2gH}}=(\frac{D_0}{D})^2\frac{H}{2\sqrt{2gH}}=\frac{1}{2}(\frac{D_0}{D})^2\sqrt{\frac{H}{2g}}$.

My book uses that

$dV=\frac{\pi}{4}D^2\sqrt{2gz}dt$

and that $dV$ can also be expressed as:

$dV=A_{tank}(-dz)=-\frac{\pi}{4}D_0^2dz$

These two expressions for $dV$ are then put equal to each other and solved for $dt$. This expression is then integrated and the result is the following:

$t=(\frac{D_0}{D})^2 \frac{1}{\sqrt{2g}}(\sqrt{H}-\sqrt{\frac{H}{2}})$

My question is therefore why the first method does not give the same answer for time as the second method?

$\endgroup$
  • $\begingroup$ Is a $\frac{1}{\sqrt{2g}}$ in the last expression? $\endgroup$ – Snaporaz Jun 2 '15 at 8:55
  • $\begingroup$ Yes, thank you. I have now corrected the last expression. $\endgroup$ – David Jun 2 '15 at 9:23
  • $\begingroup$ I consider this an example of a "good" homework question. Given the volume of "bad" homework questions that the site attracts, it is good to see some people read the FAQ... $\endgroup$ – Floris Jun 2 '15 at 21:44
2
$\begingroup$

The problem with your solution is that you are assuming the flow rate to be constant in time. In reality as the hight of the water lowers, the pressure at the bottom (therefore the flow rate) decreases as well.

In the book's solution you can see how the problem is solved with an integral, where the change in flow rate is considered.

If you look at \begin{equation} dV=\frac{\pi}{2}D^2\sqrt{2gz}dt, \end{equation} you can see that $dV$ for the same time interval $dt$ will have different values for different hights $z$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.