Worked example: a looped water distribution network
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 .
A ring main does not behave like a tree of branches. Water reaching a demand can take more than one path, and how the flow divides between those paths is exactly what a network solver works out. This worked example is the most complete of the built-in networks: a pump station lifting from a low supply into two parallel mains, cross-linked into loops, feeding three demands and balanced against an elevated storage tower.
The setup
A supply reservoir sits at 8 m of head. A pump lifts from it into the pump-station node, with a head-flow curve running from 65 m at shut-off to 50 m at 60 L/s and 25 m at 120 L/s, and an efficiency curve entered through 60 per cent at 30 L/s, 78 per cent at 70 L/s and 68 per cent at 110 L/s. From the pump station two mains run in parallel, a top row and a bottom row, cross-linked by three rungs so the network forms a ladder of loops rather than a tree:
- The top main carries a 90 degree elbow (loss coefficient 0.9) and two draw-offs, 20 L/s and 15 L/s.
- The bottom main carries a throttle valve (loss coefficient 2.5, open) and a 20 L/s draw-off.
- Both mains converge on an elevated storage tower held at 55 m.
Pipes step from DN300 on the pump-station legs down to DN200 on the branches, all at 0.5 mm roughness. The three draw-offs total 55 L/s, and whatever the demands do not take is balanced by the tower.
The physics and the method
Fluid Network Studio solves the whole network at once with the global-gradient (Todini-Pilati) method. Because the mains are cross-linked into loops, the solver splits the flow between the parallel paths until the head loss agrees around every loop and continuity holds at every junction, at the same time as it places the pump on its own head-flow curve. There is no assumed flow direction: the solver finds which way water moves through each rung from the pressures it computes.
The throttle valve is the interesting element. Closing it (raising its loss coefficient) does not simply cut one branch, it re-routes flow through the rungs into the other main, the way isolating a section of a real ring main pushes supply around the loop. The tower floats on the network, filling when supply exceeds demand and draining when it does not.
The solved result
The pump settles at 62.89 L/s and 49.03 m, which is 7.89 L/s more than the demands take, so the tower fills:
| Quantity | Value |
|---|---|
| Pump duty flow | 62.89 L/s |
| Pump head at duty | 49.03 m |
| Efficiency at duty | 76.8 % |
| Hydraulic power | 30.18 kW |
| Shaft power | 39.27 kW |
| Best efficiency point | 75.71 L/s at 78.3 % |
| Duty as a fraction of the best-efficiency flow | 0.831, inside the preferred operating region |
| NPSHa at the pump suction | 18.11 m |
| Head at the pump station | 57.03 m, which is 558.2 kPa gauge |
| Flow into the tower, top main | 2.791 L/s |
| Flow into the tower, bottom main | 5.098 L/s |
The three draw-offs and the three cross-links:
| Element | Value |
|---|---|
| Pressure at the 20 L/s draw-off on the top main | 543.2 kPa |
| Pressure at the 15 L/s draw-off on the top main | 538.6 kPa |
| Pressure at the 20 L/s draw-off on the bottom main | 540.8 kPa |
| Head loss across the open throttle valve, K = 2.5 | 0.0754 m |
| Head loss across the 90 degree elbow, K = 0.9 | 0.0286 m |
| First rung (A1 to B1) | 5.678 L/s, running from the bottom main up to the top |
| Second rung (A3 to B3) | 6.067 L/s, running from the top main down to the bottom |
| Third rung (A4 to B4) | 5.132 L/s, running from the bottom main up to the top |
The rungs are the point of the example. Nobody set their directions, and they do not all run the same way: the first and third carry water from the bottom main up into the top, while the second sends it back down. That is what a ring main does, and it is why a branch-by-branch hand calculation cannot reproduce this network. The pressure spread across the whole grid is small, 543.2 kPa at the best-served draw-off against 538.6 kPa at the worst, a difference of under half a metre of head, so the loops are doing their job of sharing the load. Meanwhile the pump delivers 62.89 L/s against 55 L/s of demand, and the surplus 7.89 L/s flows into the elevated tower, split 2.791 L/s through the top main and 5.098 L/s through the bottom. At this demand the tower is filling, not supporting the network.
Every figure comes from the solver. Open the example, press Solve, and check the balance at the pump station for yourself.
What you learn
Solving the example gives the flow and direction in every pipe, the pressure at each draw-off, and the pump operating point read off against its efficiency curve. You can see at a glance whether the tower is filling or draining at this demand, and whether the worst-served draw-off still has enough pressure. Raise the throttle valve's loss and re-solve to watch the flow redistribute through the loops. The wider workflow, from sizing the pump to checking residual pressures, is set out on the pump system design page, and the simpler looped network example isolates the flow-splitting behaviour on its own.
Open this example in FNS and close the throttle valve to see the flow re-route around the loop.