0
$\begingroup$

If we have a cylindrical tank of area $A$ which has a small outlet of area $a$ at the bottom of the surface . And the container starts filling with a constant rate $k [m^3/sec]$ . Now the max. level of water will be $$h = k^2/(2a^2g)$$ as by using Bernoulli's equation . But how will find the time after which level of water become $h$ ? I got it till $$\frac{dM}{dt} = pk -pa\sqrt{2gh}$$

$\endgroup$
  • $\begingroup$ So you are close, as you can see at steady-state $d_tM=0$ you get your maximum height back from the rate equation. Now to get the time; like i said in my answer, $M=\rho V\left(t\right)$ where the volume $V$ of liquid in the tank is related to the tank area $A$ and liquid height $h\left(t\right)$. You then have a rate equation for the liquid height; solve it as a function of time, plug in the maximum height and solve for the time to reach it. $\endgroup$ – nluigi Dec 18 '15 at 10:45
  • $\begingroup$ I just realized something and am going to spoil the answer for you; so the maximum height you calculate is the height at steady-state $d_tM=0$. But it mathematically takes an infinite amount of time to reach a steady-state, i.e. $\Delta t\rightarrow\infty$. So unfortunately the mathematical answer to your question is an infinite amount of time! You could however solve for the case $0.99h_{max}$ which would give you the finite time until it has reached $99\%$ of the steady-state height. $\endgroup$ – nluigi Dec 18 '15 at 10:49
0
$\begingroup$

Relate the rate of change of fluid mass in the tank to the amount flowing in and the amount flowing out: $$\frac{dM}{dt} = \phi_{in} - \phi_{out}\left(t\right)$$ where $\phi_{in}=\rho k$ and $\phi_{out}\left(t\right)=\rho a v_{out}$ needs the outflow velocity $v_{out}$ to be determined from the Bernoulli equation in terms of the height $h\left(t\right)$ of the fluid. The mass in the tank: $$M\left(t\right) = \rho V\left(t\right)$$ where the volume $V\left(t\right)$ of the liquid in the tank is related to the area of the tank and the height $h\left(t\right)$ of the water. From there you should be able to get a rate equation for the height of the liquid in the tank.

$\endgroup$
  • $\begingroup$ How the mass in the tank is related to area and height $\endgroup$ – user101522 Dec 16 '15 at 17:02
  • $\begingroup$ @user101522: $M=\rho V$, where $V$ is the volume of the liquid in the tank, can you guess how the volume is related to the tank area and the height of the fluid? $\endgroup$ – nluigi Dec 16 '15 at 17:04
  • $\begingroup$ dm/pK= dt then integrate it so I am getting hA/K = t . Is it correct $\endgroup$ – user101522 Dec 16 '15 at 17:16
  • $\begingroup$ @user101522 in your first equation you are not accounting for the fluid flow out of the tank, $\rho a v_{out}$, just an inflow. If there is only inflow the level in the tank will increase indefinitely right? can you think of a way to calculate $v_{out}$? Hint: the name of the equation is in your question $\endgroup$ – nluigi Dec 16 '15 at 19:00
  • $\begingroup$ I think the vol. out should be Av/a $\endgroup$ – user101522 Dec 16 '15 at 19:06

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.