Valve openings for equal discharge to B and C

Data. A = 110 m, B = 95 m, C = 90 m; junction D at elevation 100 m with $p_D = 39.24$ kN/m². Friction only, $h_f = rQ^2$, $r = \dfrac{8fL}{\pi^2gD^5}$ for the pipes and the valve resistances given.

Piezometric head at D

$$H_D = z_D + \frac{p_D}{\gamma} = 100 + \frac{39.24}{9.81} = 100 + 4.00 = 104.00\ \text{m}$$

Since $H_D = 104$ m is below A (110 m) and above B (95 m) and C (90 m), pipe AD supplies D, and D supplies both B and C.

Pipe resistances (s²/m⁵)

Discharges

$$Q_{AD} = \sqrt{\frac{110 - 104}{1692.2}} = 0.05955\ \text{m}^3/\text{s} = 59.55\ \text{l/s}$$

Equal discharge to B and C: $Q_B = Q_C = \dfrac{Q_{AD}}{2} = 0.02977$ m³/s ($29.77$ l/s).

Valve $V_1$ in pipe DB

Head available from D to B: $104 - 95 = 9$ m $= (r_{BD} + r_{V1})Q_B^2$:

$$r_{V1} = \frac{9}{0.02977^2} - 4131.3 = 6021.8 = \left(\frac{6000}{n_1}\right)^2 \;\Rightarrow\; n_1 = \frac{6000}{\sqrt{6021.8}} = 77.3\ \%$$

Valve $V_2$ in pipe DC

Head available from D to C: $104 - 90 = 14$ m $= (r_{CD} + r_{V2})Q_C^2$:

$$r_{V2} = \frac{14}{0.02977^2} - 6197.0 = 9596.8 = \left(\frac{8000}{n_2}\right)^2 \;\Rightarrow\; n_2 = \frac{8000}{\sqrt{9596.8}} = 81.7\ \%$$

Answer: $n_1 \approx 77.3\ \%$ and $n_2 \approx 81.7\ \%$ (each with $Q = 29.8$ l/s to B and to C).