Worked example: a branched distribution main
Written by Haimi Jordaan, MEng (Mechanical), University of Pretoria. Seven years in a specialist engineering analysis and design group across CFD, FEA and DEM. He wrote the solver behind Fluid Network Studio.Published . Last updated .
Most real distribution systems are branched, not looped: a main runs out from the source and sheds smaller branches to each demand, like a tree. The engineering question is rarely whether water reaches the far end, it almost always does, but whether it arrives with enough pressure. Pressure falls along the main with friction, and rises and falls again as the ground does, so the lowest pressure is not always at the point furthest from the source. This worked example shows that interaction: one main, three branches, and three demands at three different ground levels.
The setup
A single reservoir feeds a branched main that supplies three demand points at different elevations. Every pipe has a roughness of 0.15 mm:
- Reservoir at 60 m head.
- Main, first leg: 300 m of 200 mm pipe to the first junction.
- First branch: 200 m of 100 mm pipe to demand one, drawing 0.01 cubic metres per second at 15 m ground level.
- Main, second leg: 250 m of 150 mm pipe to the second junction.
- Second branch: 200 m of 100 mm pipe to demand two, drawing 0.012 cubic metres per second at 25 m ground level.
- Third branch: 150 m of 100 mm pipe to demand three, drawing 0.008 cubic metres per second at 10 m ground level.
The main steps down in diameter as it sheds flow, from 200 mm to 150 mm, which is normal practice. Demand two draws the most flow and sits highest, at 25 m, so it is the obvious candidate for the worst pressure, but only the solve confirms it.
The physics and the method
Each pipe loses head to Darcy-Weisbach friction with a Churchill (1977) friction factor. The flows are set by the three fixed demands, and continuity at each junction fixes how much flows through each leg of the main: the first leg carries all three demands, the second leg carries demands two and three. The head at each junction is the reservoir head minus the friction accumulated to that point. The gauge pressure at each demand is then that head minus the local ground elevation, which is why a high demand can see a lower pressure than a lower one further out.
Fluid Network Studio solves the whole network at once with the Global Gradient Algorithm (Todini and Pilati, 1988), satisfying continuity at every junction and the head-loss relation on every pipe. The conservation residual is reported each solve, and this incompressible mode is cross-checked against EPANET, part of the published verification cases. You can colour the network by pressure to see the low point stand out immediately.
The solved result
The worst pressure is at demand two, 251.7 kPa, and it is not the demand furthest from the source:
| Point | Ground level (m) | Demand (L/s) | Head (m) | Gauge pressure (kPa) |
|---|---|---|---|---|
| Reservoir | - | - | 60.00 | fixed boundary |
| J1, first junction | 0 | - | 58.59 | 573.5 |
| J2, second junction | 0 | - | 56.26 | 550.7 |
| Demand one | 15 | 10 | 54.70 | 388.6 |
| Demand two | 25 | 12 | 50.71 | 251.7 |
| Demand three | 10 | 8 | 54.36 | 434.3 |
| Pipe | Size | Flow (L/s) | Velocity (m/s) | Head loss (m) |
|---|---|---|---|---|
| Main, first leg | 300 m of 200 mm | 30 | 0.955 | 1.410 |
| Branch to demand one | 200 m of 100 mm | 10 | 1.273 | 3.893 |
| Main, second leg | 250 m of 150 mm | 20 | 1.132 | 2.331 |
| Branch to demand two | 200 m of 100 mm | 12 | 1.528 | 5.544 |
| Branch to demand three | 150 m of 100 mm | 8 | 1.019 | 1.897 |
Demand three is further from the reservoir than demand one, yet it sits 45.7 kPa higher (434.3 against 388.6), purely because its ground is 5 m lower. Demand two loses on both counts at once: it draws the most flow, 12 L/s, through the same 100 mm branch as the others, which costs it 5.544 m of friction, and it sits on the highest ground at 25 m. Distance from the source did not decide the low point. The combination of local elevation and branch velocity did, and that is the reason a branched system still needs solving rather than sketching.
Every value above is solver output and it is reproducible. Open the example, press Solve, and compare.
What you learn
Solving the example gives you the flow and velocity in every pipe, the head at each junction, and the gauge pressure at each of the three demands. You can see where the system pressure is lowest and confirm whether elevation or distance is the deciding factor. The instructive experiment is to lift one demand's elevation or increase its draw and re-solve: the pressure there drops, and if it falls too far the negative-pressure advisory flags it. That is the quickest way to find which branch needs a larger diameter.
This example runs on the free plan. The glossary explains head, gauge pressure and the hydraulic grade line, and the how it works page describes the solver method.
Open this example in FNS and raise a demand's elevation to see which branch loses pressure first.