Lab 1: Pipe networks

Students solve four steady incompressible pipe problems by hand (a branching junction, parallel pipes, a multi-loop network, and a pumped distribution system), then run each in OpenFLUME and explain any difference. Two are the shipped Three-pipe junction and Water distribution network examples; the other two come from the incompressible hydraulics validation report.

Learning objectives

By the end of the lab, students should be able to:

1. Compute the flow in a pipe from a pressure drop with the Darcy–Weisbach equation and an explicit (Swamee–Jain) friction factor.
2. Solve a branching (three-pipe) problem by iterating on the junction pressure until continuity closes.
3. Solve a parallel-pipe problem from the common-ΔP condition.
4. Set up the Hardy-Cross loop corrections for a network with two fixed-pressure boundaries, and iterate them to convergence.
5. Find a pump operating point and read a pressure profile (the hydraulic grade line minus elevation) along a flow path.

Prerequisites

• Fluid mechanics through internal flow: Darcy–Weisbach, the Moody chart, and continuity.
• A spreadsheet or a short script for the iterative parts (Parts A and C).
• No OpenFLUME experience. The Getting started page helps but is optional.

Time

About 60 minutes of pre-lab hand calculation and 2 hours in the lab.

Theory

All four problems use water as an incompressible liquid with the solver's water preset: ρ = 998 kg/m³, μ = 0.001 Pa·s. Gravity is g = 9.80665 m/s².

Pipe law. A pipe branch applies Darcy–Weisbach friction plus a hydrostatic term (user manual §4.3.1):

ΔP = P_from − P_to = f (L/D) ρ V|V| / 2 + ρ g Δz
V  = ṁ / (ρ A),   A = π D² / 4,   Re = ρ V D / μ

Δz is the pipe's Elevation Change, positive when the to end is higher (manual §1.4). Entrance and fitting losses must be added as separate components.

Friction factor. The code (src/core/components/frictionFactor.ts) uses f = 64/Re below Re = 2300, the Swamee–Jain approximation of Colebrook at and above Re = 4000,

f = 0.25 / [ log10( ε/(3.7 D) + 5.74 / Re^0.9 ) ]²

and a smooth blend between the two. All flows in this lab are well above Re = 4000, so only Swamee–Jain applies. A pipe set to Constant f uses that value at every Reynolds number.

Valves and resistances. A valve with area A, discharge coefficient Cd, and position p obeys ΔP = ṁ|ṁ| / (2ρ (Cd A p)²) (manual §4.3.6). A resistance obeys ΔP = K ṁ|ṁ| / (2ρA²) (manual §4.3.5). Both have the form ΔP = R ṁ|ṁ|.

Pump. A pump adds a pressure rise interpolated linearly from its curve of volumetric flow Q = ṁ/ρ against rise (manual §4.3.10). The operating point is where the pump rise equals the system's demand.

Networks. Every internal node conserves mass, and every branch relates its end pressures to its flow. Around any closed loop the signed pressure drops sum to zero. A path between two fixed-pressure boundaries is a pseudo-loop whose drops sum to the boundary pressure difference.

Hardy-Cross. Start from any set of flows that satisfies continuity. For each loop, compute a flow correction and add it, with sign, to every branch in the loop:

ΔQ = − [ Σ s_j R_j Q_j|Q_j| − H ] / [ Σ 2 R_j |Q_j| ]

s_j = +1 when branch j is drawn in the loop's direction of travel and −1 otherwise, and H is the imposed pressure difference (zero for a closed loop, P_HP − P_LP for a pseudo-loop). Each correction is added around a closed path, so continuity always holds. Here Q is mass flow in kg/s and R is in Pa/(kg/s)².

OpenFLUME solves the nodal mass balances and branch momentum equations together with Newton–Raphson (manual §4.6.1). It converges to the same solution as Hardy-Cross, so the two can be compared directly.

Pre-lab hand calculation

Part A: Three-pipe junction (branching)

Water enters at 300 kPa, flows through pipe 1 to a junction, and splits into pipe 2 (to an outlet at 200 kPa) and pipe 3 (to an outlet at 150 kPa). All pipes are horizontal with roughness ε = 1×10⁻⁵ m.

Pipe	From → to	L [m]	D [mm]
1	Inlet (300 kPa) → Junction	2	30
2	Junction → Out 1 (200 kPa)	3	20
3	Junction → Out 2 (150 kPa)	4	15

1. Write the continuity condition at the junction in terms of the junction pressure P_j.
2. Write a procedure that returns ṁ for one pipe given its ΔP. Guess f = 0.02, compute V = sqrt(2 ΔP D / (f ρ L)), then Re, update f with Swamee–Jain, and repeat until f stops changing (two or three passes).
3. Iterate on P_j (bisection between 150 and 300 kPa works) until the inflow equals the sum of the outflows to within 0.1 %. Report P_j, the three mass flows, and each pipe's Re and f.
4. Predict, without calculating, how P_j and the three flows change if pipe 3's diameter is increased from 15 mm to 20 mm.

Part B: Parallel pipes

Three pipes connect the same two reservoirs, at 150 kPa and 100 kPa (ΔP = 50 kPa). Each has a constant Darcy friction factor. This is case 3 of the validation report.

Pipe	D [mm]	L [m]	f
p1	40	8	0.020
p2	60	12	0.025
p3	25	6	0.030

Compute each pipe's mass flow from ṁ = A sqrt(2 ρ ΔP D / (f L)) and the total. Which pipe carries the most flow per unit of cross-sectional area, and why?

Part C: Hardy-Cross on a loop network

The network in Figure 1 is case 4 of the validation report. Two boundaries, HP at 500 kPa and LP at 100 kPa, are joined through six internal nodes by twelve square-law resistances, all with A = 0.001 m².

Figure 1. Multi-loop network (validation report, case 4).

Branch	From → to	K	Branch	From → to	K
b1	HP → N1	10	b7	N3 → N5	10
b2	N1 → N2	5	b8	N5 → N6	5
b3	N2 → LP	10	b9	N6 → LP	10
b4	HP → N3	8	b10	N1 → N3	2
b5	N3 → N4	6	b11	N2 → N4	2
b6	N4 → LP	8	b12	N4 → N6	2

1. Compute R = K / (2ρA²) for each branch.

2. Count the unknowns. There are 12 branch flows and 6 internal nodes, so you need 12 − 6 = 6 loop equations: 5 closed loops (12 branches − 8 nodes + 1) and 1 pseudo-loop from HP to LP. Use these:

   Loop	Path	Branches with sign
   L1	HP → N1 → N3 → HP	+b1, +b10, −b4
   L2	N1 → N2 → N4 → N3 → N1	+b2, +b11, −b5, −b10
   L3	N2 → LP → N4 → N2	+b3, −b6, −b11
   L4	N3 → N4 → N6 → N5 → N3	+b5, +b12, −b8, −b7
   L5	N4 → LP → N6 → N4	+b6, −b9, −b12
   P	HP → N1 → N2 → LP, H = 400 kPa	+b1, +b2, +b3

3. Start from this guess, which satisfies continuity at every node (kg/s): b1 = 6, b2 = 5, b3 = 4, b4 = 8, b5 = 5, b6 = 5, b7 = 4, b8 = 4, b9 = 5, b10 = 1, b11 = 1, b12 = 1. Check continuity at N3 and N4 yourself.

4. Do one full sweep by hand: apply the corrections in the order L1, L2, L3, L4, L5, P, and update the flows after each loop before you compute the next one. Report the six corrections.

5. Put the procedure in a spreadsheet and iterate until every correction is below 0.001 kg/s. Report the converged flows, the total flow from HP to LP, and the pressures at N1 to N6.

Part D: Pumped network (estimate only)

This is the shipped Water distribution network example (see the examples gallery). A pump draws from a 200 kPa supply, and the flow passes through a header, three parallel legs (each a pipe plus a valve), and two return lines that discharge to boundaries at 150 kPa and 120 kPa. The pump curve (volumetric flow, rise) is (0, 300 kPa), (0.01 m³/s, 250 kPa), (0.02 m³/s, 150 kPa), (0.03 m³/s, 50 kPa).

Branch	Type	Parameters
Leg1 pipe / valve	pipe + valve	L = 8 m, D = 30 mm, Δz = +3 m; valve A = 0.001 m², Cd = 0.6, position 0.5
Leg2 pipe / valve	pipe + valve	L = 8 m, D = 30 mm, Δz = 0; position 0.8
Leg3 pipe / valve	pipe + valve	L = 8 m, D = 30 mm, Δz = −3 m; position 1.0

1. Rank the three legs by the flow you expect them to carry, and give your reasons.
2. Write down how you would check, from solver output alone, that the pump is operating on its curve.

Procedure in OpenFLUME

Set Units ▾ to SI before you start. Every value in this lab is then displayed in the units used above.

Part A: Three-pipe junction

1. Open the example: Three-pipe junction. (Or choose it from Examples ▾ → Applications.) Opening an example replaces the current model.
2. Click each pipe and check its Length, Diameter, and Roughness in the property panel against the table in Part A. Click each boundary node and check its Pressure.
3. Click Run. The status pill should read Converged.
4. Click the Junction node. The Results block at the bottom of the property panel shows its solved pressure. Click each pipe to read its mass flow and Reynolds number. Record them next to your hand values.
5. Change Pipe 3's Diameter to 0.02 m and click Run again. Compare the result with your prediction from Part A, step 4.
6. Click Undo in the toolbar to restore the 15 mm diameter, select Pipe 3, and delete it. The Out 2 node is now unconnected; delete it too. Run again. You now have two pipes in series. Record the flow and the junction pressure.

Part B: Parallel pipes

1. Click New in the toolbar. A new model is steady-state incompressible water.
2. From the left rail, place two Boundary nodes. Set their Pressure to 150000 and 100000 Pa and their Temperature to 300 K.
3. Drag from the first node's handle to the second. In the Connect with chooser, pick Pipe. Repeat twice more so that three pipes join the same two nodes.
4. For each pipe, set Length, Diameter, and Roughness (1e-5), set Friction to Constant f, and enter the Darcy friction factor f from the Part B table.
5. Click Run and read each pipe's mass flow.

If drawing three branches between the same two nodes is awkward on the canvas, open Text view, replace the text with the model below, and click Apply:

// Fluid Network config v2
network "Lab 1 parallel pipes" {
settings: {"mode":"steady","tolerance":1e-9,"maxIterations":500,"relaxation":0.9}
fluid: {"model":"incompressible","preset":"water"}
node "A" boundary @ (0, 0) data: {"pressure":150000,"temperature":300}
node "B" boundary @ (400, 0) data: {"pressure":100000,"temperature":300}
branch "p1": "A" -> "B" pipe data: {"length":8,"diameter":0.04,"roughness":0.00001,"frictionFactor":0.02}
branch "p2": "A" -> "B" pipe data: {"length":12,"diameter":0.06,"roughness":0.00001,"frictionFactor":0.025}
branch "p3": "A" -> "B" pipe data: {"length":6,"diameter":0.025,"roughness":0.00001,"frictionFactor":0.03}
}

Part C: Loop network

Building eight nodes and twelve branches by hand takes about 20 minutes and is good practice with the canvas. To save time, click New, open Text view, replace the text with the model below, and click Apply:

// Fluid Network config v2
network "Lab 1 Hardy-Cross network" {
settings: {"mode":"steady","tolerance":1e-9,"maxIterations":500,"relaxation":0.8}
fluid: {"model":"incompressible","preset":"water"}
node "HP" boundary @ (0, 150) data: {"pressure":500000,"temperature":300}
node "N1" internal @ (200, 0) data: {"pressure":300000,"temperature":300}
node "N2" internal @ (400, 0) data: {"pressure":300000,"temperature":300}
node "N3" internal @ (200, 150) data: {"pressure":300000,"temperature":300}
node "N4" internal @ (400, 150) data: {"pressure":300000,"temperature":300}
node "N5" internal @ (200, 300) data: {"pressure":300000,"temperature":300}
node "N6" internal @ (400, 300) data: {"pressure":300000,"temperature":300}
node "LP" boundary @ (600, 150) data: {"pressure":100000,"temperature":300}
branch "b1": "HP" -> "N1" resistance data: {"k":10,"area":0.001}
branch "b2": "N1" -> "N2" resistance data: {"k":5,"area":0.001}
branch "b3": "N2" -> "LP" resistance data: {"k":10,"area":0.001}
branch "b4": "HP" -> "N3" resistance data: {"k":8,"area":0.001}
branch "b5": "N3" -> "N4" resistance data: {"k":6,"area":0.001}
branch "b6": "N4" -> "LP" resistance data: {"k":8,"area":0.001}
branch "b7": "N3" -> "N5" resistance data: {"k":10,"area":0.001}
branch "b8": "N5" -> "N6" resistance data: {"k":5,"area":0.001}
branch "b9": "N6" -> "LP" resistance data: {"k":10,"area":0.001}
branch "b10": "N1" -> "N3" resistance data: {"k":2,"area":0.001}
branch "b11": "N2" -> "N4" resistance data: {"k":2,"area":0.001}
branch "b12": "N4" -> "N6" resistance data: {"k":2,"area":0.001}
}

1. Click Run. The status pill shows the number of Newton iterations. Compare it with the number of Hardy-Cross sweeps you needed.
2. Click each branch and node to read the solved flows and pressures in the property panel's Results block (hovering a row in the project outline also shows them). Compare them with your converged spreadsheet.
3. Check one loop by hand from the solver's numbers: the signed sum of the branch pressure drops around L2 should be zero.

Part D: Water distribution network

1. Open the example: Water distribution network.
2. Click Run.
3. Click the Pump branch and record its mass flow and pressure drop (the pressure drop is negative because a pump raises pressure). Convert the mass flow to Q = ṁ/ρ and check that the rise lies on the pump curve.
4. Record the mass flow in each leg pipe. Check continuity at the Return merge node: the three legs in, the two returns out.
5. For Leg1, compute the pipe pressure drop (friction plus ρ g Δz) and the valve pressure drop by hand from the solved flow. Their sum should equal P(Header) − P(Return merge).
6. Open the Results tab, click + Plot, choose Station along path as the x axis, pick a path that runs through Leg1 in the Path selector, and add the node pressures. This is the pressure along the path. Adding ρ g z at each node would give the hydraulic grade line. Identify the largest single pressure loss.
7. Back on the Model tab, set Leg1 valve Position to 1.0 and run again. Record the new flows in all three legs and the new pump operating point.

Questions

1. In Part A, how close did your hand calculation come to the solver? If the two disagree by more than 0.1 %, find the cause (friction factor, units, or an unconverged iteration).
2. Opening pipe 3 (step 5 of Part A) raises the total inflow but lowers the flow through pipe 2. Explain why with the junction pressure.
3. In Part B, the solver's result agrees with the closed form to round-off (the validation report shows errors below 10⁻¹² %). In Part A it agrees with your hand calculation only as well as your iteration converged. What is the difference between the two problems that explains this?
4. How many Hardy-Cross sweeps did you need to reach 0.001 kg/s? How many Newton iterations did OpenFLUME report? Why does Newton converge in fewer iterations, and what does each iteration cost?
5. In Part D, Leg3 carries the most flow. Separate the reasons: how much of the difference between Leg1 and Leg3 comes from elevation and how much from valve position?
6. In Part D, the Discharge low pipe shows a negative pressure drop even though water flows through it in the drawn direction. Explain.
7. Opening Leg1's valve fully (Part D, step 7) increases Leg1's flow but reduces the flow in Legs 2 and 3. Explain using the pump curve.

Extensions

• Friction-factor sensitivity. In Part A, set every pipe to Constant f = 0.02 and rerun. How much does the junction pressure move? The shift is the effect of the Swamee–Jain correction.
• Elevation. In Part D, raise Leg1 pipe's Elevation Change from 3 m to 6 m and predict the change in Leg1 flow before running. The user manual (§7.4) states that raising a leg must reduce its flow. Check this.
• Sweep. Use the Sweep tab to vary Leg1 valve position from 0.1 to 1.0 in 10 variants. Plot Leg1 flow against position. Why is the curve not linear?
• Hardy-Cross ordering. Repeat Part C with simultaneous corrections (compute all six from the same flows, then apply them together). Does it converge faster or slower than the sequential sweep?

Answer key

All values below were computed independently with a Python implementation of the same equations (Swamee–Jain with the code's blend, bisection, and the Hardy-Cross sweep described above). They were cross-checked against the validation report where it covers the case, and against OpenFLUME 0.3.1 run locally. Values marked solver are OpenFLUME 0.3.1 results. Hand and solver results agree to at least five significant figures wherever both exist.

Part A: three-pipe junction

Quantity	Pipe 1	Pipe 2	Pipe 3
ṁ [kg/s]	3.6004	2.3802	1.2202
V [m/s]	5.10	7.59	6.92
Re	1.53×10⁵	1.52×10⁵	1.04×10⁵
f	0.01857	0.01945	0.02102

• Junction pressure P_j = 283.91 kPa. Continuity: 3.6004 = 2.3802 + 1.2202.
• With a first guess of P_j = 250 kPa, the inflow (6.55 kg/s) far exceeds the outflow (2.85 kg/s), so P_j must rise. Bisection on [150, 300] kPa reaches 0.1 % continuity closure in about 12 steps.
• Pipe 3 at D = 20 mm (step 5): P_j = 273.21 kPa; ṁ₁ = 4.719, ṁ₂ = 2.215, ṁ₃ = 2.504 kg/s. The junction pressure falls, so pipe 1 carries more flow and pipe 2 carries less.
• Series (pipe 3 removed, step 6): P_j = 291.88 kPa, ṁ = 2.4968 kg/s.

Part B: parallel pipes (validation report, case 3)

Pipe	ṁ [kg/s]	Re
p1	6.276899	2.00×10⁵
p2	12.632016	2.68×10⁵
p3	1.827549	9.3×10⁴
Total	20.736463	—

Mass flux ṁ/A = sqrt(2ρΔP D/(fL)) grows with D/(fL): it is about 4470 kg/(m²·s) for p2, 4990 for p1, and 3720 for p3. Per unit area, p1 carries the most flow: its D/(fL) is 0.25 m⁻¹, against 0.20 m⁻¹ for p2 and 0.139 m⁻¹ for p3. The solver matches the closed form to 6×10⁻¹³ % (report, case 3) because, with a constant f, the closed form is the solver's own equation. With Swamee–Jain in Part A the hand answer is only as exact as the iteration.

Part C: Hardy-Cross

R = K/(2ρA²) with 2ρA² = 1.996×10⁻³: R = 5010.0 (K = 10), 4008.0 (K = 8), 3006.0 (K = 6), 2505.0 (K = 5), 1002.0 (K = 2) Pa/(kg/s)².

First sequential sweep (kg/s): ΔQ_L1 = +0.5952, ΔQ_L2 = +0.2333, ΔQ_L3 = +0.2610, ΔQ_L4 = +0.5611, ΔQ_L5 = +0.4131, ΔQ_P = +0.1667. Flows after the sweep: b1 6.762, b2 5.400, b3 4.428, b4 7.405, b5 5.328, b6 5.152, b7 3.439, b8 3.439, b9 4.587, b10 1.362, b11 0.972, b12 1.148.

Sequential sweeps reach |ΔQ| below 0.001 kg/s after 13 sweeps, and below 10⁻¹⁰ after 51. The converged solution matches the validation report's independent Newton reference and OpenFLUME (report case 4 tables; solver 0.3.1 identical to the digits shown):

Branch	ṁ [kg/s]	Branch	ṁ [kg/s]
b1	6.607218	b7	3.272332
b2	5.610671	b8	3.272332
b3	4.521599	b9	4.463610
b4	7.403880	b10	0.996547
b5	5.128094	b11	1.089072
b6	5.025888	b12	1.191278

Total flow HP → LP = b1 + b4 = 14.0111 kg/s. Node pressures (kPa): N1 281.286, N2 202.429, N3 280.291, N4 201.241, N5 226.643, N6 199.819.

OpenFLUME 0.3.1 reports 28 Newton iterations with the model above (relaxation 0.8, tolerance 10⁻⁹). The report's own undamped reference Newton solve took 18. Each Newton iteration solves all twelve flows and six pressures together with a full Jacobian, while each Hardy-Cross correction only updates one loop. The Newton method converges quadratically near the solution, and Hardy-Cross converges linearly.

L2 loop check (solver ΔP, kPa): b2 78.857 + b11 1.189 − b5 79.050 − b10 0.995 = 0.000.

Part D: water distribution network (solver, OpenFLUME 0.3.1)

Branch	ṁ [kg/s]	ΔP [kPa]
Pump	13.9965	−209.755
Main header	13.9965	40.766
Leg1 pipe / valve	3.5588	92.327 / 70.504
Leg2 pipe / valve	4.8323	112.053 / 50.778
Leg3 pipe / valve	5.6053	119.105 / 43.726
Return low / Discharge low	8.8808	63.056 / −6.898
Return high / Discharge high	5.1157	22.334 / 63.824

Node pressures (kPa): Pump out 409.755, Header 368.989, Return merge 206.158, Return low 143.102, Return high 183.824.

• Pump check: Q = 13.9965/998 = 0.0140245 m³/s. On the segment between 0.01 and 0.02 m³/s, rise = 250 − (0.0040245/0.01)×100 = 209.755 kPa, which matches the solver.
• Continuity at Return merge: 3.5588 + 4.8323 + 5.6053 = 13.9965 = 8.8808 + 5.1157.
• Leg1 by hand from ṁ = 3.5588 kg/s: V = 5.045 m/s, Re = 1.51×10⁵, f = 0.01859. Friction = 62.97 kPa, plus ρgΔz = 998 × 9.80665 × 3 = 29.36 kPa, gives a pipe ΔP of 92.33 kPa. Valve: 3.5588² / (2 × 998 × (0.6 × 0.001 × 0.5)²) = 70.50 kPa. The sum, 162.83 kPa, equals 368.989 − 206.158.
• The largest single loss along the Leg1 path is the Leg1 pipe (92.3 kPa, about a third of it hydrostatic), followed by the Leg1 valve (70.5 kPa).
• Discharge low: friction is +42.04 kPa, but the pipe falls 5 m, so ρgΔz = −48.94 kPa. The net ΔP is −6.90 kPa, so pressure rises along the pipe because of the elevation drop.
• Leg1 valve at position 1.0: Leg1 4.4155, Leg2 4.6464, Leg3 5.4228, total 14.4847 kg/s, pump rise 204.86 kPa. The total flow rises, so the pump moves down its curve and delivers less rise. The Header pressure falls (361.36 kPa), and Legs 2 and 3 lose flow.
• Question 5: with every valve at the same position, the only difference between the legs would be ρgΔz. Leg1 then has 2 × 29.36 = 58.7 kPa more to overcome than Leg3. Leg1's half-open valve has four times the resistance of a fully open valve, because (Cd A p)² is quartered. Students can separate the two effects with the valve-position runs: at position 1.0, Leg1 still carries less than Leg3 (4.416 against 5.423 kg/s), which is the elevation effect alone.
• Extension, Leg1 Δz = 6 m: Leg1 flow falls from 3.5588 to 3.2060 kg/s, and the total falls to 13.7920 kg/s.