What conditions govern the flow split in parallel pipes B and C?

Equal pressure drop across both, and Q_B + Q_C = Q

Equal flows in both, and equal velocities

Equal Reynolds numbers, and Q_B = Q_C

How is the friction factor f found for these turbulent flows?

From the Colebrook equation using Re and relative roughness (iterated)

What is the total pressure drop?

About 160 kPa: 49.8 (pipe A, series) + 110.1 (parallel section)

110.1 kPa, parallel section only

About 220 kPa, adding both parallel drops

How is CFD used for pipe networks?

To obtain loss coefficients K and flow maps of components like bends and tees, often coupled with 1D network solvers

It replaces Darcy-Weisbach for all straight pipes

It cannot predict pressure drop

It only gives the friction factor for smooth pipes

Practice · 5 marks

Pressure drop in a pipe network

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.

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.

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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