Time to lower level in two-orifice tank

Given: tank area $A = 3.2$ m²; two orifices each $a = 8.8$ mm² $= 8.8\times 10^{-6}$ m² at 5 m and 2 m above the bottom; $C_d = 0.62$; level falls from 8 m to 3 m.

The upper orifice (at 5 m) works only while the level is above 5 m. So the problem has two stages:

• Stage 1: level 8 m to 5 m, both orifices discharging.
• Stage 2: level 5 m to 3 m, only the lower orifice (at 2 m) discharging.

$$k = C_d\,a\sqrt{2g} = 0.62\times 8.8\times 10^{-6}\times 4.4294 = 2.4167\times 10^{-5}\ \text{m}^{5/2}/\text{s}$$

Head over the upper orifice $= h - 5$; over the lower orifice $= h - 2$ ($h$ = level above tank bottom).

Stage 1 (8 m to 5 m)

$$-A\,dh = k\left[\sqrt{h - 5} + \sqrt{h - 2}\right]dt$$

Rationalise: $\dfrac{1}{\sqrt{h-5} + \sqrt{h-2}} = \dfrac{\sqrt{h-2} - \sqrt{h-5}}{3}$.

$$\begin{aligned} t_1 &= \frac{A}{k}\int_5^8\frac{\sqrt{h-2} - \sqrt{h-5}}{3}dh = \frac{A}{k}\cdot\frac{2}{9}\left[(h-2)^{3/2} - (h-5)^{3/2}\right]_5^8\\ &= \frac{3.2}{2.4167\times 10^{-5}}\times\frac{2}{9}\,[\ldots] \end{aligned}$$

Evaluating the bracket: at $h = 8$: $6^{3/2} - 3^{3/2} = 14.697 - 5.196 = 9.501$; at $h = 5$: $3^{3/2} - 0 = 5.196$. Difference $= 4.305$.

$$t_1 = \frac{3.2}{2.4167\times 10^{-5}}\times\frac{2}{9}\times 4.305 = 126\,663\ \text{s}$$

Stage 2 (5 m to 3 m, lower orifice only)

$$t_2 = \frac{2A}{k}\left[\sqrt{5-2} - \sqrt{3-2}\right] = \frac{2\times 3.2}{2.4167\times 10^{-5}}\times(1.7321 - 1) = 193\,864\ \text{s}$$

Total

$$t = t_1 + t_2 = 126\,663 + 193\,864 = 320\,527\ \text{s} \approx 89.0\ \text{h}$$

Answer: $t \approx 3.2\times 10^{5}$ s (about 89 hours) for the data as given. The very long time results from the tiny orifice areas (8.8 mm²). If the areas were 8.8 cm², all times would be 100 times smaller (about 3205 s, i.e. 53 min).