Time to discharge pressurised tank

Given: tank $D = 2$ m, so $A = \dfrac{\pi}{4}(2)^2 = 3.1416$ m²; orifice $d = 0.1$ m, $a = \dfrac{\pi}{4}(0.1)^2 = 0.007854$ m²; $h_0 = 3$ m; air pressure 450 kPa absolute, atmospheric 100 kPa; $C_d = 1$ (no friction); air pressure kept constant.

Pressure head of air above the water (gauge):

$$h_p = \frac{450 - 100}{9.81} = 35.678\ \text{m}$$

The effective head at the orifice is $h + h_p$, so the exit velocity is $V = \sqrt{2g(h + h_p)}$.

Rate equation: $-A\dfrac{dh}{dt} = a\sqrt{2g(h + h_p)}$.

$$t = \frac{A}{a\sqrt{2g}}\int_{h_2}^{h_1}\frac{dh}{\sqrt{h + h_p}} = \frac{2A}{a\sqrt{2g}}\left[\sqrt{h_1 + h_p} - \sqrt{h_2 + h_p}\right]$$ $$\frac{a\sqrt{2g}}{2A} = \frac{0.007854\times 4.4294}{2\times 3.1416} = 0.0055368\ \text{m}^{1/2}/\text{s}$$

(i) Time for half the water to discharge

The water level falls from 3 m to 1.5 m (half the volume, since the tank is cylindrical).

$$\begin{aligned} \sqrt{3 + 35.678} &= 6.2192,\qquad \sqrt{1.5 + 35.678} = 6.0974\\ t &= \frac{6.2192 - 6.0974}{0.0055368} = 22.0\ \text{s} \end{aligned}$$

(ii) Level after 10 s

$$\begin{aligned} \sqrt{h + h_p} &= 6.2192 - 0.0055368\times 10 = 6.1638\\ h + h_p &= 37.992\\ h &= 37.992 - 35.678 = 2.31\ \text{m} \end{aligned}$$

Answer: (i) $t \approx 22.0$ s; (ii) water level after 10 s $\approx 2.31$ m above the outlet.