Series Pipe Rules
For pipes in series, the same discharge passes through each pipe and the total head loss is the sum of the individual losses.
$$Q_1=Q_2=Q_3$$$$h_L=h_{L1}+h_{L2}+h_{L3}$$
Parallel Pipe Rules
For parallel pipes between two common junctions, head loss is the same in each branch and total flow is the sum of branch flows.
$$h_{L1}=h_{L2}=h_{L3}$$$$Q=Q_1+Q_2+Q_3$$
Darcy-Weisbach Head Loss
$$h_f=f\frac{L}{D}\frac{V^2}{2g}$$$$V=\frac{4Q}{\pi D^2}$$
Use Darcy-Weisbach when $f$ is given; use Hazen-Williams only when the coefficient $C$ is specified.
Parallel Flow Split Check
Two parallel pipes carry 0.10 m3/s total. If branch A carries 0.065 m3/s, find branch B and the head-loss condition.
$$Q_B=Q-Q_A=0.10-0.065=0.035\text{ m}^3/\text{s}$$$$h_{fA}=h_{fB}$$
Answer: Branch B carries 0.035 m3/s, and both branches must have equal head loss between the same junctions.
Series Pipes Total Head Loss
Water flows at 0.050 m3/s through two pipes in series. Pipe 1 is 200 m long and 200 mm in diameter. Pipe 2 is 150 m long and 150 mm in diameter. Take $f=0.020$ for both pipes and neglect minor losses. Find the total head loss.
$$V_1=\frac{Q}{A_1}=\frac{0.050}{\pi(0.20)^2/4}=1.59\text{ m/s}$$$$V_2=\frac{0.050}{\pi(0.15)^2/4}=2.83\text{ m/s}$$$$h_{f1}=0.020\frac{200}{0.20}\frac{1.59^2}{2(9.81)}=2.58\text{ m}$$$$h_{f2}=0.020\frac{150}{0.15}\frac{2.83^2}{2(9.81)}=8.12\text{ m}$$$$h_L=2.58+8.12=10.70\text{ m}$$
Answer: The total head loss is 10.70 m.
Parallel Pipes Same Diameter Flow Division
Two parallel pipes have the same diameter and friction factor. Branch A is 100 m long and branch B is 400 m long. The total flow is 0.120 m3/s. Determine the flow in each branch.
$$h_{fA}=h_{fB},\quad h_f\propto LQ^2\text{ for same }D\text{ and }f$$$$L_AQ_A^2=L_BQ_B^2$$$$100Q_A^2=400Q_B^2\Rightarrow Q_A=2Q_B$$$$Q_A+Q_B=0.120\Rightarrow 3Q_B=0.120$$$$Q_B=0.040\text{ m}^3/\text{s},\quad Q_A=0.080\text{ m}^3/\text{s}$$
Answer: Shorter branch A carries 0.080 m3/s; branch B carries 0.040 m3/s.
Problem: Equivalent Pipe for Three Pipes in Series
Three pipes are connected in series between two reservoirs with a total head difference of 25 m. Pipe 1: L = 500 m, D = 300 mm, f = 0.020. Pipe 2: L = 400 m, D = 250 mm, f = 0.022. Pipe 3: L = 300 m, D = 200 mm, f = 0.018. Neglect minor losses. Find the discharge through the system.
Express friction head loss for each pipe in terms of Q using $h_f = \frac{8fLQ^2}{\pi^2 g D^5}$:
$$K_i = \frac{8f_i L_i}{\pi^2 g D_i^5}$$
$$K_1 = \frac{8(0.020)(500)}{\pi^2(9.81)(0.30)^5} = \frac{80}{9.81\pi^2(0.00243)} = 337.9$$
$$K_2 = \frac{8(0.022)(400)}{\pi^2(9.81)(0.25)^5} = \frac{70.4}{9.81\pi^2(0.000977)} = 744.5$$
$$K_3 = \frac{8(0.018)(300)}{\pi^2(9.81)(0.20)^5} = \frac{43.2}{9.81\pi^2(0.000320)} = 1399.5$$
$$\sum h_f = (K_1 + K_2 + K_3)Q^2 = (337.9 + 744.5 + 1399.5)Q^2 = 2481.9Q^2$$
$$25 = 2481.9Q^2 \Rightarrow Q = \sqrt{\frac{25}{2481.9}} = 0.1004 \text{ m}^3/\text{s}$$
Answer: Discharge is 0.100 m³/s (100 L/s). The smallest pipe (200 mm) contributes the most friction loss, controlling the system capacity.
Problem: Three Unequal Parallel Pipes — Flow Distribution
Three parallel pipes connect junction J1 to junction J2 with a head loss of 8.0 m between them. All have f = 0.020. Branch 1: L = 600 m, D = 250 mm. Branch 2: L = 400 m, D = 200 mm. Branch 3: L = 300 m, D = 150 mm. Find the discharge in each branch and the total discharge.
For each branch: $h_f = K_i Q_i^2 = 8.0$ m, so $Q_i = \sqrt{8.0/K_i}$.
$$K_1 = \frac{8(0.020)(600)}{\pi^2(9.81)(0.25)^5} = \frac{96}{9.81\pi^2(0.000977)} = 1014.8$$
$$K_2 = \frac{8(0.020)(400)}{\pi^2(9.81)(0.20)^5} = \frac{64}{9.81\pi^2(0.000320)} = 2065.2$$
$$K_3 = \frac{8(0.020)(300)}{\pi^2(9.81)(0.15)^5} = \frac{48}{9.81\pi^2(0.0000759)} = 6539.6$$
$$Q_1 = \sqrt{\frac{8.0}{1014.8}} = 0.0888 \text{ m}^3/\text{s}$$
$$Q_2 = \sqrt{\frac{8.0}{2065.2}} = 0.0622 \text{ m}^3/\text{s}$$
$$Q_3 = \sqrt{\frac{8.0}{6539.6}} = 0.0350 \text{ m}^3/\text{s}$$
$$Q_{total} = 0.0888 + 0.0622 + 0.0350 = 0.1860 \text{ m}^3/\text{s}$$
Answer: Q1 = 88.8 L/s, Q2 = 62.2 L/s, Q3 = 35.0 L/s. Total = 186 L/s. The 250 mm pipe carries the most flow even though it is the longest, because its large diameter (raised to the 5th power) dominates.
Problem: Hazen-Williams Formula — Pipe Sizing
A water main must carry 0.085 m³/s over a length of 800 m with a maximum allowable head loss of 12 m. Using the Hazen-Williams formula $V = 0.8492 C R^{0.63} S^{0.54}$ with C = 120 and $R = D/4$ for a full circular pipe, determine the required pipe diameter in mm. Round up to the next standard size.
$$S = \frac{h_f}{L} = \frac{12}{800} = 0.015$$
$$Q = VA = 0.8492C\left(\frac{D}{4}\right)^{0.63} S^{0.54} \cdot \frac{\pi D^2}{4}$$
$$0.085 = 0.8492(120)\left(\frac{D}{4}\right)^{0.63}(0.015)^{0.54}\frac{\pi D^2}{4}$$
$$0.085 = 101.9(0.01488)(D/4)^{0.63}\frac{\pi D^2}{4}$$
Solve by trial: try D = 0.25 m:
$$Q_{trial} = 0.8492(120)\left(\frac{0.25}{4}\right)^{0.63}(0.015)^{0.54}\frac{\pi(0.25)^2}{4}$$
$$= 101.9(0.0625)^{0.63}(0.01488)(0.04909)$$
$$= 101.9(0.1633)(0.01488)(0.04909) = 0.1215 \text{ m}^3/\text{s}$$
Too large. Try D = 0.20 m: Q ≈ 0.068 m³/s (too small). Interpolate: required D ≈ 0.225 m.
Answer: Required diameter is approximately 225 mm. Use the next standard size of 250 mm to ensure adequate capacity with margin.