Fuel tank draining [closed]

A cylindrical fuel tank is being drained from the bottom as in this picture : Conservation of flow rate : $v_A = \frac{S_B}{S_A}v_B = \alpha v_B$

Assuming that $z_B = 0$, Bernoulli's theorem states that :

$$p_{atm} + \rho g h(t) + \frac{1}{2}\rho (\alpha v_B)^2 = (p_{atm} - \rho g h(t)) + \frac{1}{2} \rho g {v_B}^2$$

$$\Leftrightarrow v_B = 2\sqrt{\frac{gh(t)}{1-\alpha^2}}$$

and

$$v_A = \alpha v_B = 2\alpha \sqrt{\frac{gh(t)}{1-\alpha^2}}$$

With $v_A = \frac{dh}{dt}$

We have the differential equation :

$$h'(t) - 2\alpha \sqrt{\frac{gh(t)}{1-\alpha^2}} = 0$$

Solving this, we get :

$$h(t) = \frac{g \alpha^2}{1-\alpha^2}t^2 + C \alpha \sqrt{\frac{g}{1-\alpha^2}}t + \frac{C^2}{4}$$

Now, $h(t = 0) \Rightarrow C = 2\sqrt{H}$

So finally :

$$h(t) = \frac{g \alpha^2}{1-\alpha^2}t^2 + 2 \alpha \sqrt{\frac{gH}{1-\alpha^2}}t + H$$

closed as off-topic by David Z♦Nov 17 '14 at 8:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – David Z
If this question can be reworded to fit the rules in the help center, please edit the question.

Hint: If you divide by $\sqrt {h(t)}$ the equation separates.