Three reservoirs: pipe AEC length and level B

Data missing from the question. The figure gives only the levels (A = 40 m, outlet D = 10 m) and the pipe table; no discharge is given, and two quantities are unknown ($L_{AEC}$ and $Z_B$). As the paper says "assume suitable data", the following standard conditions are assumed:

1. The two parallel pipes AEC and AFC (which together join A to C) carry equal discharges.
2. Reservoir B neither supplies nor takes water (no flow in BC), so its level equals the piezometric head at C. If B supplies a discharge $Q_B$ the level would be $Z_B = H_C + r_{BC}Q_B^2$.

Friction losses only; $h_f = rQ^2$ with $r = \dfrac{8fL}{\pi^2gD^5}$.

Length of pipe AEC

Parallel pipes have the same head loss: $\dfrac{f_1L_1Q_1^2}{D_1^5} = \dfrac{f_2L_2Q_2^2}{D_2^5}$. With $f$ equal and $Q_1 = Q_2$:

$$L_{AEC} = L_{AFC}\left(\frac{D_{AEC}}{D_{AFC}}\right)^5 = 1500\left(\frac{0.60}{0.75}\right)^5 = 1500(0.32768) = 491.5\ \text{m}$$

(General case: $L_{AEC} = L_{AFC}\left(\dfrac{D_{AEC}}{D_{AFC}}\right)^5\left(\dfrac{Q_{AFC}}{Q_{AEC}}\right)^2$.)

Discharge from A to D

Resistance of AFC: $r_{AFC} = 15.67$ s²/m⁵. Since AEC has the same resistance (equal $L/D^5$ ratio and flow), the pair in parallel has $r_{par} = r_{AFC}/4 = 3.917$ s²/m⁵. Resistance of CD: $r_{CD} = \dfrac{8(0.02)(3000)}{\pi^2(9.81)(0.4)^5} = 484.1$ s²/m⁵.

Between the levels of A and D:

$$40 - 10 = (r_{par} + r_{CD})Q^2 \;\Rightarrow\; Q = \sqrt{\frac{30}{3.917 + 484.1}} = 0.2479\ \text{m}^3/\text{s}$$

Each parallel pipe carries 0.1240 m³/s.

Elevation of reservoir B (no flow in BC)

Head loss in the parallel pair $= r_{par}Q^2 = 0.24$ m; in CD $= r_{CD}Q^2 = 29.76$ m. The piezometric head at the junction C is the head at D plus the loss in CD:

$$H_C = 10 + 29.76 = 39.76\ \text{m}$$ $$Z_B = H_C = 39.76\ \text{m}$$

Answer: $L_{AEC} \approx 491.5$ m (about 492 m) and $Z_B \approx 39.8$ m, with $Q \approx 0.248$ m³/s in CD under the stated assumptions. (For a B that supplies water, $Z_B$ is higher than this by $r_{BC}Q_B^2$, with $r_{BC} = 126.9$ s²/m⁵.)