Engineering & Technology · Fluid Mechanics, Flow Systems & Machines
Internal flow, losses, pipes and flow networks
Two branches between the same junctions do not normally divide flow equally. Each path offers a different relation between flow and required mechanical-energy change. A network finds a distribution that satisfies both conservation and every path’s resistance.
Review force vectors, energy and power, control volumes and steady-flow thermodynamics, SI units and measurement uncertainty. Be comfortable with ratios, areas, square roots and reading graphs. Symbols are defined where used: ρ is density, μ dynamic viscosity, Q volume flow and ṁ mass flow. For the constructed liquid cases, use g = 9.81 m/s² where stated. Prior physics study of fluids is helpful but not required.
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Connect pipe losses with flow regime and solve a simple network using continuity and compatible head changes.
A velocity profile develops inside the pipe
A viscous fluid near an ordinary stationary wall satisfies the no-slip approximation: its velocity relative to the wall is zero. Across a pipe, local speed therefore differs from the area mean U used in Q = AU. Near an inlet, the profile changes with distance; fully developed flow means its shape no longer changes along a straight, uniform pipe. It does not mean zero friction or zero pressure gradient.
For fully developed laminar flow of an incompressible Newtonian fluid in a circular pipe, the Darcy friction factor is fD = 64/Re, with Re = ρUD/μ. Turbulent friction depends on Reynolds number and relative roughness, among other details. Transition is sensitive to disturbances and geometry; a regime estimate should not be treated as a universal sharp switch.
Head loss represents mechanical-energy dissipation
The Darcy–Weisbach model gives straight-pipe head loss hL = fD(L/D)U²/(2g). Here L is pipe length, D diameter, and hL is energy dissipated per unit weight, expressed in metres of the flowing fluid. For equal elevation and equal velocity at the two ends, pressure decrease is ρghL. A pressure decrease caused by rising elevation is not itself an irreversible loss.
Fittings can add terms K Uref²/(2g), with each coefficient K tied to a stated reference velocity, geometry and flow conditions. “Minor loss” names the category, not a promise that it is small. Darcy and Fanning friction factors differ by a factor of four: fD = 4fF. Mixing a chart from one convention with the other equation gives a substantial error. Quadratic head-flow curves assume approximately fixed coefficients; they are not universal across regimes.
Join paths with conservation and compatibility
At a steady junction, total incoming mass flow equals total outgoing mass flow. For a constant-density liquid this becomes a volume-flow balance. Parallel branches between the same two network nodes share the same node-to-node total-head change; their flows can differ. Series elements carry the same flow when there is no intervening branch or storage, and their losses add. Closed-loop head changes must also be compatible, including any machine head.
Constructed pipe loss and parallel-flow check
For a hypothetical straight pipe plus fittings, let L = 5 m, D = 0.020 m, U = 1 m/s, assumed Darcy fD = 0.040, and combined K = 2 referenced to this same U. With g = 9.81 m/s², hL = [0.040(5/0.020) + 2]/19.62 = 0.612 m. For ρ = 1,000 kg/m³, the corresponding loss in pressure-equivalent terms is 6,000 Pa. Q = πD²U/4 = 0.314 L/s. The chosen fD is illustrative, not a prediction for an unspecified real pipe.
Separately, the plotted fictional parallel branches obey h₁ = 2Q₁² and h₂ = 8Q₂², using Q in L/s and h in m. At a common head change 2 m, Q₁ = √(2/2) = 1 L/s and Q₂ = √(2/8) = 0.5 L/s. Total is 1.5 L/s. Assigning half the total to each branch would fail the required equal head change.
Try the reasoning
Does a horizontal constant-diameter pipe lose total energy when friction acts?
Show the answer and reasoning
Energy is conserved, while organized mechanical energy is irreversibly converted to internal energy and possibly transferred as heat. The mechanical head decreases.
Why can doubling flow produce more or less than four times the loss?
Show the answer and reasoning
The U² factor gives four times only if friction and fitting coefficients remain fixed. Reynolds number, regime and coefficients can change with flow.
If branch 2’s resistance rises while node heads remain fixed, what changes?
Show the answer and reasoning
Its flow falls according to its head-flow relation. The other branch keeps its flow if its own relation and both node heads truly stay fixed; total flow decreases.
What to carry forward
Use the correct friction convention and flow conditions, then enforce both junction balances and path head changes.
References
Teaching resources and research supporting this lesson’s explanations and examples.
- University Physics Volume 1, 14.7: Viscosity and TurbulenceWilliam Moebs, Samuel J. Ling and Jeff Sanny; OpenStaxTeaching resource
- Chapter 11: Applications of Viscous Flows Through Pipes—RecapNPTEL, Fluid Mechanics teaching course 112104118Teaching resource
- Lecture 21: Minor Loss, Kinetic Energy Correction FactorNPTEL, teaching course 103104043Teaching resource
- Lecture 15: Pipe Network AnalysisNPTEL, Water Supply Network Design teaching course 105104102Teaching resource