--- title: "Series and parallel pipe network" subject: "Fluid Mechanics" chapter: 8 key: "semester_4/fluid_mechanics_mechanical:series-parallel-pipe-network-discharge" source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/series-parallel-pipe-network-discharge site: BCT Engineering --- # Series and parallel pipe network Fluid Mechanics · Chapter 8: Flow Measurement - Asked in: year not given, 8 marks - Video: [Fluid Mechanics: Topic 9.2 - Introduction to pipe networks (pipes in series, parallel, branching)](https://www.youtube.com/watch?v=ZO18AXE8TZo) — CPPMechEngTutorials - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/series-parallel-pipe-network-discharge - Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-8 ## Question Two reservoirs A and B have water surface elevations of 120 m and 90 m. They are connected by a pipe 1 (length 800 m, diameter 300 mm, $f = 0.02$) from A to a junction J, after which pipes 2 (600 m, 200 mm, $f = 0.025$) and 3 (900 m, 150 mm, $f = 0.025$) run in parallel from J to B. Neglecting minor losses, find the total discharge and the discharge in each parallel pipe. ## Short answer Equivalent resistance of parallel pipes gives K_eq=1982.5. Total head 30 m yields total Q = 0.109 m³/s. Head loss in parallel section 23.5 m gives Q2 = 0.078 m³/s and Q3 = 0.031 m³/s. ## Answer #### 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: $Q = 0.109\ \text{m}^3/\text{s}$; $Q_2 = 0.078\ \text{m}^3/\text{s}$; $Q_3 = 0.031\ \text{m}^3/\text{s}$** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which rules apply to series and parallel pipes? A. Series: same head loss; parallel: same discharge B. Series: same discharge, losses add; parallel: same head loss, discharges add C. Both: discharges add D. Both: losses add Answer: B. Series pipes carry the same Q with additive losses; parallel branches share the same loss with discharges adding. (Key point: Series and parallel rules) 2. How is the head loss expressed in terms of discharge? A. h_f = K Q with K = 8fL/(pi^2 g D^5) B. h_f = K Q^2 with K = 8fL/(pi^2 g D^5) C. h_f = K Q^2 with K = 8fL/(pi^2 g D^2) D. h_f = K sqrt(Q) Answer: B. Substituting V = 4Q/(pi D^2) in Darcy-Weisbach gives h_f = (8fL/(pi^2 g D^5)) Q^2. (Key point: Loss coefficient) 3. How is K_eq of the two parallel pipes found? A. K_eq = K2 + K3 B. 1/K_eq = 1/K2 + 1/K3 C. 1/sqrt(K_eq) = 1/sqrt(K2) + 1/sqrt(K3) D. K_eq = K2 K3 Answer: C. With Q = sqrt(h/K2) + sqrt(h/K3), the equivalent satisfies 1/sqrt(K_eq) = sum of 1/sqrt(K). (Key point: Equivalent of parallel pipes) 4. What is the total head used and the result? A. 120 m; Q = 0.2 m^3/s B. 90 m; Q = 0.19 m^3/s C. 30 m (120 - 90); Q = sqrt(30/(K1 + Keq)) = 0.109 m^3/s D. 30 m; Q = 0.078 m^3/s Answer: C. The available head is the level difference of 30 m, giving Q = sqrt(30/2526.5) = 0.109 m^3/s. (Key point: Total discharge) ## Related in Fluid Mechanics - [Pump head and power between two reservoirs](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Apumped-system-power-between-reservoirs) — Fluid Mechanics, chapter 8 - [Differential U-tube manometer between two pipes](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Adifferential-manometer-pipes) — Fluid Mechanics, chapter 2 - [Force on a reducing pipe bend](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Areducing-bend-force-numerical) — Fluid Mechanics, chapter 4 - [Buckingham pi theorem for pipe pressure drop](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Abuckingham-pi-pipe-pressure-drop) — Fluid Mechanics, chapter 5 - [Laminar oil flow in a pipe](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Alaminar-oil-pipe-pressure-drop) — Fluid Mechanics, chapter 6 - [Laminar flow between fixed parallel plates](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Alaminar-flow-between-fixed-parallel-plates) — Fluid Mechanics, chapter 6 ## Related in other subjects - [Three reservoirs with two parallel pipes](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Athree-reservoirs-five-pipes-parallel) — Hydraulics, chapter 3 - [Siphon with 90 mm pipe, number and length](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Asiphon-90mm-pipe-total-length) — Hydraulics, chapter 2 - [Three reservoirs with 250 mm concrete pipes](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Athree-reservoir-concrete-pipes-flow-rates) — Hydraulics, chapter 3, asked 2× - [Series-parallel pipes with control valve](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aseries-parallel-valve-equal-discharge) — Hydraulics, chapter 2, asked 2×