Three reservoirs ZA 65, ZB 70, ZC 40

Data. $Z_A = 65$ m, $Z_B = 70$ m, $Z_C = 40$ m; AD: 600 m, 25 mm; BD: 900 m, 100 mm; DC: 150 m, 150 mm; $f = 0.025$. Only friction, $h_f = rQ^2$, $r = \dfrac{8fL}{\pi^2gD^5}$.

Method

Let $H_D$ be the piezometric head at D. Flows are directed from the higher to the lower head. C (40 m) is the lowest, so DC always carries water to C. B (70 m) is the highest, so B supplies. A (65 m) supplies if $H_D < 65$ m and receives water if $H_D > 65$ m.

Trial with $H_D = 65$ m (no flow in AD): $Q_{BD} = \sqrt{5/185910} = 0.0052$ m³/s in, $Q_{DC} = \sqrt{25/4080.3} = 0.0783$ m³/s out. The outflow is larger than the inflow, so $H_D$ must be lowered below 65 m, and A also supplies D: $Q_{AD} + Q_{BD} = Q_{DC}$.

The balance is at $H_D = 40.69$ m.

Discharges

$$Q_{AD} = \sqrt{\frac{65 - 40.69}{126914853}} = 0.000438\ \text{m}^3/\text{s} = 0.44\ \text{l/s}\quad(\text{A to D})$$ $$Q_{BD} = \sqrt{\frac{70 - 40.69}{185910}} = 0.01256\ \text{m}^3/\text{s} = 12.56\ \text{l/s}\quad(\text{B to D})$$ $$Q_{DC} = \sqrt{\frac{40.69 - 40}{4080.3}} = 0.01299\ \text{m}^3/\text{s} = 12.99\ \text{l/s}\quad(\text{D to C})$$

Check: $0.438 + 12.556 = 12.994 \approx 12.99$ l/s.

Answer: $Q_{AD} = 0.44$ l/s (A to D), $Q_{BD} = 12.56$ l/s (B to D), $Q_{DC} = 12.99$ l/s (D to C); $H_D = 40.69$ m.