Suppose a water tank has 1" diameter drain at the bottom and is filled with water up to one meter height above the drain. What time it will take the tank to drain out completely. Now say, the tank is filled up to two meter height above the drain then what time it will take the tank to drain out? Will it be double or less than it? Can we establish a relation between flow rate for the given height of water column?

  • $\begingroup$ Please provide your thoughts on the question. Otherwise there is a chance that this question will be put on hold. $\endgroup$ – Rajath Krishna R Oct 4 '13 at 18:09

Use Bernoulli's equation to derive Torricelli's Law (check any website for this) for the velocity out of the hole; $ v = \sqrt{2 g h(t)} $, where g is gravity and $ h(t) $ is the height of the fluid in the tank at any time.

Write a balance on the mass of the fluid in the tank as:

$$ \text{in - out + gen = accumulation} $$ $$ \rho Q_{in} - \rho Q_{out} = \frac{d(\rho V)}{dt} $$ where the generation term is zero, $\rho$ is the fluid density (constant here) and $Q_{in}$ and $Q_{out}$ is the flow rate in and out of the tank, respectively. $Q_{in}$ is zero so we get: $$ \frac{dV}{dt} = -Q_{out} $$ The flow out is $v A = \sqrt{2 g h(t)} A$, where $A$ is the area of the hole which is calculated by knowing the diameter of the circular hole; given in the problem statement as 1 inch.

The volume of the tank, $V = a_t h(t)$, is the height, $h(t)$ times the area, $a_{t}$.

Putting it all together, we get the separable first order differential equation for the height of the fluid in the tank versus time:

$$ \frac{dh}{dt} = -\frac{A}{a_t}\sqrt{2g}\sqrt{h}$$

Prepare it for Integration $$ \frac{dh}{\sqrt{h}} = -\frac{A}{a_t}\sqrt{2g}{dt}$$

Integrate the equation. The upper bound for $dh$ is $h(t)$. The lower bound is $h(0)=H$ . For $dt$ we integrate from $t$ to $0$: $$ 2 (\sqrt{h(t)} - \sqrt{H}) =-\frac{A}{ a_t}\sqrt{2g} t$$ Solve for $h(t)$ $$ h(t)=[\sqrt{H} -\frac{A}{2 a_t}\sqrt{2g} t]^2 $$

To find the time when the tank empties, set $h$ equal to zero and solve for $t$:

$$ t= \sqrt{\frac{H}{2g}} \frac{2 a_t}{A}$$

The times to empty the same tank for two different starting heights, $H_1$ and $H_2$ is:

$$ \frac{t_1}{t_2} = \sqrt{\frac{H_1}{H_2}} $$

So, finally, if $H_2 = 2 H_1$ as in the problem statement, then the time to empty the tank is not double, but $\sqrt{2}$ times longer.

Make sense?

Paul Safier

| cite | improve this answer | |
  • $\begingroup$ Dear Paul Safier, Thanks to you for explaining whole theory to get down to the correct answer. I am sure it would help me solve the problem where I am trying to establish a correlation between surface area of the tank and the flow rate. $\endgroup$ – Arun Khatri Oct 22 '13 at 16:45
  • $\begingroup$ When I tried to plug in $h(t)$ as defined here into the fomula for $v(t)$ I didn't get the expected parabola. You can see it here for $h=20$, $A=1$ and $a_t=10$. What's wrong? bit.ly/1uhSc9M $\endgroup$ – jbx Jan 19 '15 at 11:52
  • $\begingroup$ I think your eqution $2 (\sqrt{h(t)} - \sqrt{H}) =-\frac{A}{ a_t}\sqrt{\frac{g}{2}} t$ should be $2 (\sqrt{h(t)} - \sqrt{H}) =-\frac{A}{ a_t}\sqrt{2g} t$ $\endgroup$ – Ehegh Nov 7 '15 at 20:06

The issue is, what is the flow velocity?

It is not simple, and it depends on pressure, which is proportional to height.

For low pressure, viscosity will dominate, and velocity will be proportional to pressure.

For high pressure, velocity will be proportional to square root of pressure.

In any case, the geometry of the opening matters. The general subject is Orifice Flow.

| cite | improve this answer | |
  • $\begingroup$ With respect to Mike Dunlavey's reply, shall I assume to take velocity (proportional to pressure for low pressure) of the water flowing through 1" diameter drain having water column of 1 meter above it to be 0.1 m/s since the pressure developed by 1 meter water column would be 0.1 Kg/cm2. And then substitute it in the equation mentioned in the answer # 2 by programing enthusiat to derive the flow rate. $\endgroup$ – Arun Khatri Oct 8 '13 at 6:30
  • $\begingroup$ @Arun: It's not that simple. Programming's answer requires you to know the exit velocity v. The pressure is just the weight of water in a column above a unit of area. You still need to know the relationship between pressure and velocity through the orifice, which is why I gave you a link to orifice flow. An orifice with sharp edges will have less velocity than one with a bell-shaped entrance, for example. $\endgroup$ – Mike Dunlavey Oct 8 '13 at 11:48

This is a very simple problem. The relation is given by using

$$A \frac{dh}{dt} = \frac{-\pi D^2}{4}v;$$ where $D$, $h$, $A$ are the diameter, height and area of the tank and v is velocity.

I will leave the derivation up to you

| cite | improve this answer | |

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