Valve opening to double discharge into C

Data. Take the datum at the surface of C: $Z_C = 0$, $Z_B = 20$ m, $Z_A = 20 + 40 = 60$ m. Head losses $h = rQ^2$: $r_{AJ} = 150$, $r_{BJ} = 200$, $r_{CJ} = 300$, and the valve in AJ has $r_v = (400/n)^2$. The discharge into C is twice the discharge into B, so B receives water: $Q_C = 2Q_B$, and by continuity at J,

$$Q_A = Q_B + Q_C = 3Q_B$$

Let $H_J$ be the head at the junction.

Heads at J from the lower reservoirs

• J to B: $H_J - 20 = r_{BJ}Q_B^2 = 200\,Q_B^2$
• J to C: $H_J - 0 = r_{CJ}Q_C^2 = 300(2Q_B)^2 = 1200\,Q_B^2$

Subtracting: $1200Q_B^2 - 200Q_B^2 = 20 \Rightarrow Q_B^2 = 0.02$, so

$$Q_B = 0.1414\ \text{m}^3/\text{s},\quad Q_C = 2Q_B = 0.2828\ \text{m}^3/\text{s},\quad Q_A = 3Q_B = 0.4243\ \text{m}^3/\text{s}$$ $$H_J = 1200(0.02) = 24.0\ \text{m}\quad(\text{check: } 20 + 200(0.02) = 24\ \text{m})$$

Valve opening

Head loss from A to J: $60 - 24.0 = 36.0$ m $= (r_{AJ} + r_v)Q_A^2$:

$$150 + r_v = \frac{36.0}{(0.4243)^2} = \frac{36.0}{0.18} = 200 \;\Rightarrow\; r_v = 50$$ $$\left(\frac{400}{n}\right)^2 = 50 \;\Rightarrow\; n = \frac{400}{\sqrt{50}} = 56.57\ \%$$

Answer: The valve must be about $56.6\%$ open ($Q_A = 424$ l/s, $Q_B = 141$ l/s, $Q_C = 283$ l/s, $H_J = 24$ m above C).