# Missing factor of S/s in hydrodynamics question

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A cylindrical vessel of height H and base area S is filled with water. An orifice of surface area $s \ll S$ is opened in the bottom of the vessel. Neglecting the viscosity of water, determine how soon all the water will pour out of the vessel.

Attempt :

Let at any instant height of water be h.

Let the speed from the bottom be $v$.

$v^2 = 2gh$

Differentating,

$2v \dfrac{dv}{dt}= 2g\dfrac{dh}{dt}$

$\implies v\dfrac{dv}{dt}= g\dfrac{dh}{dt}$

Now, $\dfrac{dh}{dt}= -v$

Substituting then integrating we get,

$v_i = gt$

$v_i = \sqrt{2gH}$

$\implies t = \sqrt{\dfrac{2H}{g}}$

But answer is: $t = \dfrac{S}{s}\sqrt{\dfrac{2H}{g}}$

Why am I missing that extra $S/s$ factor?

Please let me know my error. If my error is $v_2 \ne \dfrac{-dh}{dt}$, then please explain why.

edited Jul 21, 2018

1 vote

The velocity of the water leaving through the hole is $v=\sqrt{2gh}$. Your mistake is that this is not equal to $-\frac{dh}{dt}=V$ which is the velocity of the water level in the vessel.
Using the Continuity Equation the volume flow rate is the same at the water level and the hole. So the speeds $v, V$ are related by $$SV=sv$$ So $$\sqrt{2gh}=v=\frac{S}{s} V=-\frac{S}{s}\frac{dh}{dt}$$ $$dt=-\frac{S}{s}\frac{1}{\sqrt{2g}}\frac{dh}{\sqrt{h}}$$ $$t=\frac{S}{s}\frac{2\sqrt{H}}{\sqrt{2g}}=\frac{S}{s}\sqrt{\frac{2H}{g}}$$
If water is flowing through a pipe which gets narrower, the speed of the water is slow when the pipe is wide and fast when the pipe is narrow. $dh/dt$ is the speed in the wide part (the cylinder), the speed of the water level, which is slow. This is different from the speed at the narrow part (the hole), which is faster.