Rotating tube with quadrant, speed and head

Reading of the figure: $B$ is on the axis, $C$ is on the axis 250 mm below $B$, and the quadrant $CD$ (radius $R=0.25$ m, centre $B$) curves from $C$ up to $D$, which is level with $B$ and 0.25 m from the axis. The water surface in $A$ is 0.3 m above $C$ (on the axis), and $A$ is open.

   A  o   <- water surface, 0.3 m above C
      |
      |
   B  +------- D (r = 0.25 m)
      |     /
      |   /  quadrant, centre B
   C  +--

For rigid-body rotation, the pressure head at a point at radius $r$ and height $z$ above $C$ is

$$h_p=h_C+\frac{\omega^{2}r^{2}}{2g}-z,\qquad h_C=0.3\ \text{m}$$

(a) Speed for $h_D=h_C$

At $D$: $r=0.25$ m and $z=0.25$ m.

$$\frac{\omega^{2}(0.25)^{2}}{2g}=0.25\ \Rightarrow\ \omega^{2}=\frac{2(9.81)}{0.25}=78.48$$ $$\omega=8.859\ \text{rad/s},\qquad N=\frac{60\omega}{2\pi}=84.6\ \text{rpm}$$

(b) Maximum head in the curved part

Let a point on the quadrant make angle $\theta$ at $B$ measured from $BC$. Then $r=R\sin\theta$, $z=R(1-\cos\theta)$. With $\dfrac{\omega^{2}R^{2}}{2g}=R$:

$$h_p=0.3+R\sin^{2}\theta-R(1-\cos\theta)=0.3+R(\cos\theta-\cos^{2}\theta)$$ $$\frac{dh_p}{d\theta}=0\ \Rightarrow\ \sin\theta(2\cos\theta-1)=0\ \Rightarrow\ \cos\theta=0.5,\ \theta=60^\circ$$ $$h_{p,max}=0.3+0.25(0.5-0.25)=0.3625\ \text{m of water}$$

Position: at $60^\circ$ from $C$ along the arc, i.e. 0.2165 m from the axis and 0.125 m above $C$.

Answer: (a) 8.86 rad/s (84.6 rpm); (b) maximum head 0.3625 m of water, at $60^\circ$ from C on the arc.