Water Tank Drainage

Classical Mechanics Level 2

The flow rate of water out of a valve at the bottom of an open cylindrical tank is given by

$\frac{dV}{dt} = C A \sqrt{2 g y}$

where $C$ is a constant, $C \lt 1$ , $A$ is the valve cross-sectional area, $g$ is the gravitational acceleration, and $y$ is the vertical height of the water level above the valve.

Let $C = 0.9 , A = \pi r^2$ , with $r = 1.5 cm$ .

The water in the tank is drained through the valve starting at $t = 0$ .

At $t = 1 \text{min}$ , the water level has dropped by $30$ cm from its original value. What was the height $H$ (in $\text{cm}$ ) of the water level (above the valve) at $t = 0$ ?

Take the cross-sectional area of the tank to be $0.5 m^2$ , and $g = 9.81 m/s^2$ .

The answer is 94.4.

1 solution

A Former Brilliant Member
Sep 4, 2020

The flow rate equation can be rewritten as

$a\dfrac{dy}{dt}=-CA\sqrt {2gy}$ ,

where $a$ is the cross-sectional area of the cylinder.

So, $y^{-\frac 12}dy=-\dfrac {CA}{a}\sqrt {2g}dt$

Integrating we get

$\sqrt y=\sqrt {y_0}-\dfrac {CAt}{a}\sqrt {\dfrac g2}$ ,

where $y_0$ is the initial height of the water level above the valve.

Substituting values we get

$\sqrt {y_0-0.3}\approx \sqrt {y_0}-0.169$

$\implies y_0\approx \left (\dfrac {0.3+0.209}{0.338}\right ) ^2\approx \boxed {94.4}$ cm.

Water Tank Drainage

The answer is 94.4.

1 solution

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