Close a branching flow balance without guessing the split
Volume flow rate Q is volume transported per time, not the speed of an individual fluid element. For a section of area A and mean normal speed v, Q=Av. One litre is 0.001 m³ and one minute is 60 s, so 6 L/min is 0.0001 m³/s. Converting area requires squaring the length conversion: 1 cm² is 0.0001 m².
Learning outcome: Compare two areas at the same prescribed split, then compare two splits at one area. Retain volume and mass residuals. State explicitly why continuity verifies an imposed split but does not predict a pressure-driven split.
Starting recap
Declare density, pressure reference, flow regime and admissible force constraints. Convert area and volume units before balancing forces, mass or energy.
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At first-branch flow 3 L/min and second area 1 cm², second mean speed is 0.5 m/s and mass flow 0.045 kg/s. Doubling the area at that imposed flow halves the mean speed.
Open the experimentPredict
For steady flow with changing density, the conserved flow quantity is:
Experiment
Compare two areas at the same prescribed split, then compare two splits at one area. Retain volume and mass residuals. State explicitly why continuity verifies an imposed split but does not predict a pressure-driven split.
Close a branching volume and mass balance
Worked teaching model · synthetic values · Steady incompressible liquid density 900 kg/m³ enters a junction at 6 L/min. No storage or leakage. Selected first-branch flow is independently stipulated, not predicted by a pipe-resistance model.
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Second branch speed at fixed inlet flow
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: no-storage balance (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Second branch speed at fixed inlet flow
no-storage balance
| first branch flow (L/min) | second branch mean speed (m/s) |
|---|---|
| 0 | 0.5 |
| 0.15 | 0.4875 |
| 0.3 | 0.475 |
| 0.45 | 0.4625 |
| 0.6 | 0.45 |
| 0.75 | 0.4375 |
| 0.9 | 0.425 |
| 1.05 | 0.4125 |
| 1.2 | 0.4 |
| 1.35 | 0.3875 |
| 1.5 | 0.375 |
| 1.65 | 0.3625 |
| 1.8 | 0.35 |
| 1.95 | 0.3375 |
| 2.1 | 0.325 |
| 2.25 | 0.3125 |
| 2.4 | 0.3 |
| 2.55 | 0.2875 |
| 2.7 | 0.275 |
| 2.85 | 0.2625 |
| 3 | 0.25 |
| 3.15 | 0.2375 |
| 3.3 | 0.225 |
| 3.45 | 0.2125 |
| 3.6 | 0.2 |
| 3.75 | 0.1875 |
| 3.9 | 0.175 |
| 4.05 | 0.1625 |
| 4.2 | 0.15 |
| 4.35 | 0.1375 |
| 4.5 | 0.125 |
| 4.65 | 0.1125 |
| 4.8 | 0.1 |
| 4.95 | 0.0875 |
| 5.1 | 0.075 |
| 5.25 | 0.0625 |
| 5.4 | 0.05 |
| 5.55 | 0.0375 |
| 5.7 | 0.025 |
| 5.85 | 0.0125 |
| 6 | 0 |
At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.
| Quantity | Value | Unit |
|---|---|---|
| Inlet volume flow | 0.0001 | m³/s |
| First branch volume flow | 0.0000333333 | m³/s |
| Second branch volume flow | 0.0000666667 | m³/s |
| Second branch mean speed | 0.333333 | m/s |
| Inlet mass flow | 0.09 | kg/s |
| Second branch mass flow | 0.06 | kg/s |
| Junction volume residual | 0 | m³/s |
Continuity closes the imposed flow split but does not determine it from pipe areas alone.
Q=A v for mean section velocity; a litre is .001 m³ and a minute is 60 s. In steady incompressible flow, incoming Q equals the sum of outgoing Q. With varying density conserve rho Q instead. Pressure, resistance and boundary conditions are needed to predict an unknown branch split.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
Continuity closes the imposed flow split but does not determine it from pipe areas alone.
Q=A v for mean section velocity; a litre is .001 m³ and a minute is 60 s. In steady incompressible flow, incoming Q equals the sum of outgoing Q. With varying density conserve rho Q instead. Pressure, resistance and boundary conditions are needed to predict an unknown branch split.
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What additional evidence would be needed to infer an unknown junction flow split?
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