Connect pipe geometry to viscous flow

Compare two pressure-driven pipes

Imagine two long straight pipes carrying the same liquid. Both have the same length and pressure difference. One has twice the other’s radius. Predict the volume collected in a fixed time before changing the control. More area matters, but it is not the whole explanation: the speed distribution also changes.

This model specifies a 1 m fully developed section of horizontal circular pipe. The liquid is incompressible and Newtonian: shear stress is proportional to the local velocity gradient. Its dynamic viscosity μ is 0.001 Pa·s and density ρ is 1000 kg/m³. These are fixed teaching properties, not measurements of a particular sample. Temperature is constant. The rigid walls are stationary and impose no slip: liquid speed at the wall is zero.

Let the pressure difference determine the flow

In the preceding continuity experiment, the supply prescribed volume flow and a narrower section had to carry that same flow. Here the control prescribes the pressure difference Δp between two sections. Flow is an output. Different controlled conditions answer different questions; their radius comparisons must not be mixed.

For fully developed laminar flow, the hydraulic resistance is R = 8μL/(πr⁴). L is section length in metres, r is radius in metres and μ is dynamic viscosity in Pa·s. R has units Pa·s/m³. The volume flow is Q = Δp/R, in m³/s. In words: divide the driving pressure difference by the resistance to find the passing volume per second.

The OpenStax viscosity reference gives the assumptions and radius dependence of this relation. The following values and exercises are original teaching examples. This equation does not describe arbitrary turbulent, branching, flexible or non-Newtonian flows.

Work through one case

Set radius to 2 mm and pressure difference to 100 Pa. First convert radius: 2 mm = 0.002 m. Its fourth power is 1.6 × 10⁻¹¹ m⁴. Resistance is 0.008/(π × 1.6 × 10⁻¹¹), approximately 1.592 × 10⁸ Pa·s/m³.

Q = 100/R is approximately 6.283 × 10⁻⁷ m³/s, or 0.628 mL/s. There are one million millilitres per cubic metre. Pipe area πr² is approximately 1.257 × 10⁻⁵ m². The mean speed Q/(πr²) is 0.050 m/s. Speed and volume flow have different units and meanings.

A hypothetical collection over 20 seconds would contain about 12.57 mL, provided these steady conditions persisted. This is a model prediction, not recorded experimental evidence.

Read the profile rather than assuming uniform speed

Fully developed laminar flow has a parabolic velocity profile. At distance s from the center, v(s) = Δp(r² − s²)/(4μL). Here 0 ≤ s ≤ r; s and r are measured in metres. The plotted diameter uses signed positions on either side of the center; speed depends on their squares.

At the wall s=r, speed is zero. At the center s=0, speed is greatest. The center speed is twice the area-average speed: 0.100 m/s in the default case. Q equals area times the mean, not area times the center speed. The mean is weighted by cross-sectional area; averaging the two endpoint speeds or a few equally spaced positions is not a general way to calculate it.

The solid curve shows local speed, the dashed line marks the mean and horizontal lines mark the walls. Both axes use fixed scales when you change radius or pressure. A larger plotted pipe does not imply a longer pipe.

Test separate scaling claims

At the same 100 Pa, radius 1 mm gives Q approximately 0.0393 mL/s and mean speed 0.0125 m/s. Doubling radius to 2 mm increases Q sixteenfold, area fourfold and mean speed fourfold. Resistance falls by sixteen. Capture both cases and calculate these ratios from the ledger.

At fixed radius, doubling Δp doubles Q and mean speed while resistance remains unchanged. If a different experiment instead maintained Q, doubling radius would reduce the required Δp by sixteen and reduce mean speed by four. State the controlled quantity with every claim.

At zero pressure difference the model is at rest. Its resistance remains finite, but flow, shear and pressure-work dissipation vanish. Only pressure difference is specified: the model does not infer absolute pressure.

Know when this calculation is only a candidate

The model computes a candidate flow and checks its Reynolds number before accepting its profile. Above the teaching screen it keeps candidate numbers visible for diagnosis but supplies no accepted flow. Even a passing screen assumes fully developed laminar conditions. The next lesson explains that check and the energy lost through viscosity.

Entrance regions, fittings, rough disturbances, temperature changes and flexible walls can require additional models or evidence. A laboratory comparison would need measured radius, liquid properties, pressure difference, collected volume and timing, with uncertainty and repeated trials. Do not call a saved simulated collection a physical measurement.

Transfer to a different length

A supported laminar setup gives 0.80 mL/s at a given radius and pressure difference. The same liquid passes through a section twice as long and with radius 1.5 times as large. Keep Δp fixed. Predict the flow ratio and new flow, then identify an assumption you would recheck before applying it.

Check the resistance transfer

Q scales as r⁴/L, so the ratio is 1.5⁴/2 = 2.53125. The candidate flow is 2.025 mL/s. Mean speed changes by 2.53125/1.5² = 1.125, and diameter changes by 1.5, so Reynolds number rises by 1.6875. Check the flow regime and fully developed conditions again; a previously valid case does not guarantee this candidate is valid.