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:

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