Series and parallel pipe network

Rules for pipe combinations

• Series: the same discharge flows in each pipe, and the head losses add.
• Parallel: the head loss between the two junctions is the same in each branch, and the discharges add.

Loss coefficients

The Darcy-Weisbach equation in terms of discharge:

$$h_f = \frac{8 f L}{\pi^2 g D^5}Q^2 = K Q^2$$ $$\begin{aligned} K_1 &= \frac{8 \times 0.02 \times 800}{\pi^2 \times 9.81 \times (0.3)^5} = 544.0\ \text{s}^2/\text{m}^5 \\ K_2 &= \frac{8 \times 0.025 \times 600}{\pi^2 \times 9.81 \times (0.2)^5} = 3873\ \text{s}^2/\text{m}^5 \\ K_3 &= \frac{8 \times 0.025 \times 900}{\pi^2 \times 9.81 \times (0.15)^5} = 24\,482\ \text{s}^2/\text{m}^5 \end{aligned}$$

Equivalent of the parallel pipes

For equal head loss $h_p = K_2Q_2^2 = K_3Q_3^2$, $Q = Q_2 + Q_3$:

$$\frac{1}{\sqrt{K_{eq}}} = \frac{1}{\sqrt{K_2}} + \frac{1}{\sqrt{K_3}} = 0.016068 + 0.006391 = 0.022459$$ $$K_{eq} = 1982.5\ \text{s}^2/\text{m}^5$$

Total discharge

The total head available is $120 - 90 = 30\ \text{m}$:

$$30 = (K_1 + K_{eq})Q^2 = (544.0 + 1982.5)Q^2$$ $$Q = \sqrt{\frac{30}{2526.5}} = 0.1090\ \text{m}^3/\text{s}$$

Head losses and branch flows

$$\begin{aligned} h_1 &= 544.0 \times (0.1090)^2 = 6.46\ \text{m} \\ h_p &= 30 - 6.46 = 23.54\ \text{m} \end{aligned}$$ $$Q_2 = \sqrt{\frac{23.54}{3873}} = 0.0780\ \text{m}^3/\text{s}, \qquad Q_3 = \sqrt{\frac{23.54}{24\,482}} = 0.0310\ \text{m}^3/\text{s}$$

Check: $Q_2 + Q_3 = 0.0780 + 0.0310 = 0.1090\ \text{m}^3/\text{s}$. This agrees.

Answer: $Q = 0.109\ \text{m}^3/\text{s}$; $Q_2 = 0.078\ \text{m}^3/\text{s}$; $Q_3 = 0.031\ \text{m}^3/\text{s}$