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I am having a bit of difficulty establishing the efflux velocity of an open tank of water in terms of time, i.e. $u(t)$, according to Torricelli's law. I know that for a known height $h$ (where $h = H - c$, and $H$ is the vertical position of the surface of the water level in the tank, while $c$ is the vertical position of the hole in the tank), the velocity $u = \sqrt{2gh}$.

However, since $h$ is decreasing over time, $u$ is not constant and it has a decreasing parabolic shape over $t$. So what is the right formula for $u(t)$?

I tried to follow the explanation here (Equation 14) but this doesn't make sense, because that is just a straight line. A more realistic solution is here, which says that since $u(t) = \sqrt{2gh(t)}$, we just need to find the rate of change of $h(t)$ and plug it in. In the answer $h(t)$ seems to be defined as $h(t) = [\sqrt{h_0} - \frac{A}{a_t} \sqrt{\frac{g}{2}t} ]^2$, where $h_0$ is the initial $h$, $A$ is the area of the hole in the tank, while $a_t$ is the area of the surface of the water in the tank.

But when I plugged it in the above formula for $u(t)$ and tried to plot it, I didn't get the parabolic shape I was expecting. The rate of change seems to start descending with the expected curve, but towards the end where the tank presumably becomes close to empty, the line seems to become a straight one. It then reflects back up again almost as a straight line. You can see it plotted here for $h_0 = 20$, $A = 1$ and $a_t = 10$.

However, if I remember well, it should just continue the parabola (obviously physically this does not happen). Unless I am wrong about what I am expecting, this doesn't seem right.

What is the right formula for $u(t)$ given a height $h$? What am I missing?

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I know that for a known height h (where $h=H−c$, and H is the vertical position of the surface of the water level in the tank, while $c$ is the vertical position of the hole in the tank), the velocity $u = \sqrt{2gh}$

The velocity $u$ is actually an approximation,

The correct relation, (which you can easily derive using Bernouilli and equation of continuity) is:

$u$ = $\sqrt{2gh\big(\dfrac{A_1^{2}}{{A_1}^{2}-{A_2}^{2}}\big)}$

where $A_1$ is area of the open surface and $A_2$ is the area of the hole.

In ideal cases $A_1<<<$ $A_2$ so the above simplifies to $u =\sqrt{2gh}$

Now coming to your problem regarding the velocity as a function of time.

Since we already know $u(t)$, we require $h(t)$ and then plug it in.

So, considering the area of the open surface to be $A_1$ and that of the hole to be $A_2$, height of cylinder to be $H$ (from bottom) and height at ant time $t=t$ to be $h$(from bottom) and approximating $A_2<<<$ $A_1$, we can begin.

By equation of continuity for a non compressible fluid,

$$A_1v_1 = A_2v_2$$ ($v_1$ being the rate at which water level comes down, or say $h$ decreases and $v_2$ being the speed of efflux.)

So, $$A_1 \big(\frac{-dh}{dt}\big)= A_2(\sqrt{2gh})$$

Which is,

$$\frac{-dh}{\sqrt{2gh}} = \big(\frac{A_2}{A_1}\big)dt$$

And integrating with proper limits,

$$\int_H^h \frac{-dh}{\sqrt{2gh}} = \int_0^t \big(\frac{A_2}{A_1}\big)dt$$

Hence, we get,

$$\frac{-1}{\sqrt{2g}}[2\sqrt{h}-2\sqrt{H}] = \frac{A_2}{A_1}t$$

Which on simpifying is

$$\big(\sqrt{H} - \sqrt{\frac{g}{2}}\big(\frac{A_1}{A_2}\big)t\big)^{2} = h$$

So,

$$h(t) = \frac{{A_2}^2 gt^{2}}{2{A_1}^{2}} - \frac{\sqrt{2gH}(A_2 t)}{A_1}+H$$

Hence, we have the relation for $h(t)$ and $u(h)$ so we can get $u(t)$ as:

$$ u(t) = \sqrt{2g \big(\sqrt{H} - \sqrt{\frac{g}{2}}\big(\frac{A_1}{A_2}\big)t\big)^{2}}$$

See if it makes sense to you.

Cheers!

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  • $\begingroup$ Cheers for your answer. I think it does. I think I was confused by the fact that the height reduces parabolically, while the velocity actually reduces linearly. Plotting your formula creates the descending velocity that makes sense. bit.ly/1DhOJNp $\endgroup$ – jbx Jan 31 '15 at 17:34
  • $\begingroup$ Alright! Pleased to help :) $\endgroup$ – ritvik1512 Feb 1 '15 at 6:01
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However, since h is decreasing over time, u is not constant and it has a decreasing parabolic shape over t.

Why do you think the $u(t)$ must have a decreasing parabolic shape over time. I am assuming you want to find $u(t)$.

The correct way to do it is to see that $u(t) = \sqrt{2gh(t)}$ and find $h(t)$ using the fact that it exits from the water hole.

Let $A_b$ be the area of the base of the cylinder and $A_h$ be area of the hole. Let $u(t)$ be the velocity of water which exits from the hole. We have, $$ \frac{dV}{dt} = -u A_h$$ where $V = A_b H$.

Now,$\frac{dH}{dt} = \frac{dh}{dt}$ since $H = h + c$ and c is just a constant.

So,

$$A_b \frac{dh}{dt} = -u A_h$$ Using our original equation for $u(t)$ we get,

$$\frac{A_b u }{g} \frac{du}{dt} = - u A_h$$ So finally $u(t)$ becomes a linear function of time.

$$ u(t) - u(0) = - \frac{A_h g t}{A_b}$$

This formula when used to find $h(t)$ gives

$$h(t)=\sqrt{h_0−\frac{A_h}{A_b}\sqrt{\frac{gt^2}{2}}}$$.

While $$u(t) = \sqrt{2gh(t)}$$ is still valid since we started with it in the first place.If we want to recover it back,the procedure would be to find $t$ in terms of $u(t)$ and substitute it the expression for $h(t)$ while respecting the fact that $u(0) = \sqrt{2gh_0}$.

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  • $\begingroup$ Thanks for your answer. I am not a physicist, I am just trying to model this but I am getting different solutions wherever I look. I thought the velocity over time $u(t)$ should have a parabolic shape, not sure where I read it though. So I plotted your function for $u(t) = \sqrt{2g \sqrt{h_0 - \frac{A_h}{A_b}\sqrt{\frac{gt^2}{2}}}}$ with $h_0 = 20$, $A_h = 1$, $A_b = 10$ and I got this shape: fooplot.com/plot/ih7j2jzdfe . Is this right? I found another one here, and again it defines it different, assuming $\alpha = 1$ for simplicity (its defined as $v(t)$) b.gatech.edu/15KSUXC $\endgroup$ – jbx Jan 26 '15 at 14:24
  • $\begingroup$ So if you look at their graphs here from their experiments: b.gatech.edu/1JOjbAP, the velocity over time is linear (the height is parabolic, I presume from the loss of velocity it goes down more slowly over time, I must have mixed up with that one). However, it is still not matching the shape from your formula... still confused :( $\endgroup$ – jbx Jan 26 '15 at 14:34

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