Emptying time of cylinder-cone vessel

Given (assumed from the figure): cylinder 2 m diameter, 2 m high on top of a conical frustum 4 m high tapering from 2 m diameter (top) to 1 m diameter (bottom); orifice $d = 0.10$ m at the bottom; $C_d = 0.6$. Initially full, so the head above the orifice is $H = 4 + 2 = 6$ m.

   |<--2 m-->|
   |         |  2 m
   |_________|
    \       /
     \     /  4 m (frustum)
      \___/
       o  <- orifice 10 cm (bottom dia 1 m)

Orifice area $a = \dfrac{\pi}{4}(0.1)^2 = 0.007854$ m².

$$k = C_d\,a\sqrt{2g} = 0.6\times 0.007854\times 4.4294 = 0.020873\ \text{m}^{5/2}/\text{s}$$

Rate equation: $-A(h)\,dh = k\sqrt{h}\,dt\;\Rightarrow\; t = \dfrac{1}{k}\displaystyle\int\dfrac{A(h)}{\sqrt h}dh$.

Time to lower 6 m to 4 m (cylinder, $A = \pi$ m²)

$$t_1 = \frac{A}{k}\cdot 2\left(\sqrt{6} - \sqrt{4}\right) = \frac{3.1416\times 2\times 0.4495}{0.020873} = 135.3\ \text{s}$$

Time to lower 4 m to 0 (frustum)

At height $h$ above the orifice, the diameter is $D = 1 + h/4$ (1 m at $h = 0$, 2 m at $h = 4$), so

$$A(h) = \frac{\pi}{4}\left(1 + \frac{h}{4}\right)^2 = \frac{\pi}{4}\left(1 + 0.5h + 0.0625h^2\right)$$ $$\begin{aligned} t_2 &= \frac{\pi}{4k}\int_0^4\left(h^{-1/2} + 0.5h^{1/2} + 0.0625h^{3/2}\right)dh\\ &= \frac{\pi}{4k}\left[2h^{1/2} + \frac{1}{3}h^{3/2} + 0.025h^{5/2}\right]_0^4\\ &= \frac{\pi}{4k}\left[4 + 2.667 + 0.8\right] = \frac{0.7854\times 7.467}{0.020873} = 280.9\ \text{s} \end{aligned}$$

Total time

$$t = t_1 + t_2 = 135.3 + 280.9 = 416.3\ \text{s} \approx 6.94\ \text{min}$$

Answer: time of emptying $\approx 416$ s (about 6 min 56 s).