Check flow regimes and dissipative pressure work

A useful equation needs a regime check

A calculation can return a number even when its assumptions fail. Try radius 4 mm at 100 Pa and then 150 Pa. The laminar candidate still exists algebraically in both cases. Predict why the accepted result and velocity profile might disappear in the second case.

Compare inertia with viscosity

For pipe flow, Re = ρv̄D/μ. Re is the dimensionless Reynolds number; ρ is density in kg/m³, v̄ is mean speed in m/s, D is pipe diameter in metres and μ is dynamic viscosity in Pa·s. Use diameter D=2r, not radius, and use mean rather than center speed. The units cancel.

In the 2 mm radius, 100 Pa case, v̄ = 0.050 m/s and D = 0.004 m. Therefore Re = 1000 × 0.050 × 0.004 / 0.001 = 200. The investigation’s conservative teaching screen accepts a laminar prediction only when candidate Re is below 2000.

This screen is a model-use rule. Transition depends on disturbances, geometry and history; 2000 is not a universal switch that proves the behavior of every real flow. A low Reynolds number also does not establish that an entrance region is fully developed. The model assumes the measured 1 m section is already fully developed and excludes entrance, exit and fitting losses.

Follow the boundary case

At radius 4 mm and 100 Pa, candidate mean speed is 0.200 m/s and diameter is 0.008 m. Re = 1600, inside the screen. Candidate volume flow is approximately 10.053 mL/s.

Increase pressure difference to 150 Pa. The laminar formula gives mean speed 0.300 m/s and Re = 2400. It fails the screen. The ledger deliberately distinguishes candidate calculations from accepted predictions, and removes the profile. The candidate is not a turbulent-flow solution. Inferring the actual flow would require an appropriate resistance relation and more information.

The OpenStax viscosity and turbulence reference discusses these distinct flow regimes. In this investigation the exact screen and synthetic controls are teaching choices, not measured transition data.

Account for work without accelerating the flow

Return to radius 2 mm and 100 Pa. The pipe has constant area and its fully developed profile is unchanged from one section to the next. The mean speeds are equal. Yet a pressure difference is needed to balance viscous effects.

Pressure work enters at rate p₁Q and leaves at rate p₂Q. Their difference is P = (p₁ − p₂)Q = ΔpQ. P is mechanical dissipation power in watts. With Q = 6.283 × 10⁻⁷ m³/s and Δp = 100 Pa, P is 6.283 × 10⁻⁵ W, or 0.06283 mW.

The kinetic-energy flow is the same at the two fully developed sections, as is elevation. The pressure-work difference is converted from organized mechanical energy into internal energy by viscous dissipation. Total energy is conserved. Do not insert this pressure drop into the lossless Bernoulli equation and conclude the result is impossible: the missing term is the dissipative loss.

A real apparatus may carry the resulting energy away with the fluid or through the pipe walls. This fixed-property model supplies neither a temperature rise nor electrical pump consumption. Those would require thermal properties, heat-transfer conditions and pump efficiency. Its dissipation value is conditional on the accepted flow model.

Check the force balance

For a section of length L, pressure produces downstream force Δpπr². Wall shear stress τ acts over area 2πrL in the opposite direction. At steady fully developed flow, τ = Δp r/(2L) makes these forces equal.

In the default case τ = 100 × 0.002 / 2 = 0.100 Pa. Both forces have magnitude approximately 0.001257 N. Fluid can move steadily with zero net force on the section: balanced forces do not imply zero speed or zero dissipation.

At zero Δp, the solution is rest with zero Re and zero dissipation. At fixed radius within the accepted regime, doubling Δp doubles Q but quadruples dissipation, because ΔpQ contains two factors that each double. Test that claim using 100 and 200 Pa at radius 2 mm.

Transfer and defend a model choice

A candidate liquid flow has density 900 kg/m³, viscosity 0.003 Pa·s, mean speed 0.25 m/s and diameter 0.006 m. Calculate Re. If pressure difference is 600 Pa and accepted flow is 7.0 × 10⁻⁶ m³/s, find dissipation power. Explain what still has to be established before calling these a physical prediction.

Check the regime and energy transfer

Re = 900 × 0.25 × 0.006 / 0.003 = 450. It passes the teaching screen. Power is 600 × 7.0 × 10⁻⁶ = 0.0042 W = 4.2 mW. The stated flow is rounded consistently with the diameter and mean speed. Passing the screen does not verify a straight rigid circular section, steady Newtonian behavior, constant properties, a fully developed profile or compatible pressure boundaries. Measurements and their uncertainty would be needed to compare this prediction with an apparatus.