---
title: "Pressure drop in a pipe network"
subject: "Applied Computational Fluid Dynamics"
chapter: 6
key: "semester_7/applied_computational_fluid_dynamics:pipe-network-pressure-drop-numerical"
source: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/pipe-network-pressure-drop-numerical
site: BCT Engineering
---

# Pressure drop in a pipe network

Applied Computational Fluid Dynamics · Chapter 6: CFD Applications

- Asked in: year not given, 5 marks
- Video: [Piping Networks and Pressure Losses](https://www.youtube.com/watch?v=yKgZWZYBD8Y) — JoVE (Journal of Visualized Experiments)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/pipe-network-pressure-drop-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/applied_computational_fluid_dynamics/chapter-6

## Question

Water ($\rho = 998$ kg/m³, $\mu = 1.0\times10^{-3}$ Pa·s) flows at 0.030 m³/s through a 300 m long, 150 mm diameter pipe, which then splits into two parallel pipes that rejoin: pipe B (100 mm diameter, 200 m long) and pipe C (80 mm diameter, 200 m long). All pipes have roughness 0.046 mm. Neglecting minor losses, find the flow in each parallel branch and the total pressure drop. Explain how CFD is used for pressure drop prediction in pipe networks.

## Short answer

Using Darcy–Weisbach and Colebrook friction factors, iterating equal pressure drop in parallel pipes gives Q_B = 0.0193 m³/s, Q_C =0.0107 m³/s. Total pressure drop including series pipe A is about 160 kPa.

## Answer

#### Method
Darcy-Weisbach: $\Delta p=f\dfrac LD\dfrac{\rho V^2}2$, with $f$ from the Colebrook equation
$$
\frac1{\sqrt f}=-2\log_{10}\left(\frac{\varepsilon/D}{3.7}+\frac{2.51}{Re\sqrt f}\right)
$$
For parallel pipes: the pressure drop is the same, and $Q_B+Q_C=Q$. Iterate on the split.

#### Pipe A (series, full flow)
$V=\dfrac{4Q}{\pi D^2}=\dfrac{4\times0.03}{\pi\times0.15^2}=1.698$ m/s; $Re=\dfrac{998\times1.698\times0.15}{10^{-3}}=2.54\times10^5$; $\varepsilon/D=3.07\times10^{-4}$.

Colebrook gives $f=0.0173$.
$$
\Delta p_A=0.0173\times\frac{300}{0.15}\times\frac{998\times1.698^2}{2}=4.98\times10^4\ \text{Pa}
$$

#### Parallel pipes B and C
Trial splits are adjusted until $\Delta p_B=\Delta p_C$. The converged split is

| Pipe | $D$ (m) | $Q$ (m³/s) | $V$ (m/s) | $Re$ | $f$ | $\Delta p$ (kPa) |
|---|---|---|---|---|---|---|
| B | 0.10 | 0.01930 | 2.457 | $2.45\times10^5$ | 0.0183 | 110.1 |
| C | 0.08 | 0.01070 | 2.129 | $1.70\times10^5$ | 0.0195 | 110.1 |

$Q_B+Q_C=0.0300$ m³/s, as required. Pipe B (larger) carries 64% of the flow.

#### Total
$$
\Delta p_{total}=\Delta p_A+\Delta p_{parallel}=49.8+110.1=159.9\ \text{kPa}
$$
(about 16.3 m of water head).

#### Use of CFD for pipe networks
- A full 3D CFD model of a part (bends, tees, valves, manifolds) gives loss coefficients $K$ and velocity maps, including flow maldistribution at junctions.
- Typical set-up: velocity inlet, pressure outlet, no-slip walls with roughness, RANS k-$\varepsilon$ or k-$\omega$ SST, prism layers on walls; pressure drop = difference of area-averaged total pressure.
- For large networks, 1D network solvers (using $f$ and $K$ from CFD or correlations) are coupled with 3D CFD at critical components.
- Results are checked against Darcy-Weisbach and experiments.

**Answer:** $Q_B=0.0193$ m³/s, $Q_C=0.0107$ m³/s; total $\Delta p\approx160$ kPa.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What conditions govern the flow split in parallel pipes B and C?
   A. Equal flows in both, and equal velocities
   B. Equal pressure drop across both, and Q_B + Q_C = Q
   C. Equal Reynolds numbers, and Q_B = Q_C
   D. Equal friction factors only
   Answer: B. Both branches connect the same two junctions, so they share the same pressure drop and their flows sum to the total. (Key point: Parallel pipe conditions)

2. How is the friction factor f found for these turbulent flows?
   A. f = 64/Re
   B. From the Colebrook equation using Re and relative roughness (iterated)
   C. f = 0.5 always
   D. f is independent of roughness
   Answer: B. Re is about 10^5, so the Colebrook equation with relative roughness is used; 64/Re is laminar only. (Key point: Friction factor)

3. What is the total pressure drop?
   A. 110.1 kPa, parallel section only
   B. 49.8 kPa, pipe A only
   C. About 160 kPa: 49.8 (pipe A, series) + 110.1 (parallel section)
   D. About 220 kPa, adding both parallel drops
   Answer: C. Series drops add, but parallel pipes share one drop: 49.8 + 110.1 = 159.9 kPa. (Key point: Total pressure drop)

4. How is CFD used for pipe networks?
   A. It replaces Darcy-Weisbach for all straight pipes
   B. To obtain loss coefficients K and flow maps of components like bends and tees, often coupled with 1D network solvers
   C. It cannot predict pressure drop
   D. It only gives the friction factor for smooth pipes
   Answer: B. 3D CFD of components gives K and maldistribution; 1D network solvers handle the rest, with checks against correlations and experiments. (Key point: Role of CFD)

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