A narrower tube can produce faster flow at one location or reduce the total flow through a system. Those statements can both be correct because they answer different questions. One compares locations within a specified steady flow; the other changes a system’s resistance while its driving pressure is held fixed.
The AAMC content outline explicitly includes continuity, Bernoulli’s equation, viscosity and Poiseuille flow. This article uses original pipe examples to separate the assumptions behind those relationships.
Distinguish volume per time from distance per time
Volumetric flow rate Q measures volume crossing a section per unit time, such as m³/s. Mean speed v measures distance per time, such as m/s. They are related by Q = Av, where A is the cross-sectional area and v is the cross-sectional average speed.
For steady incompressible flow through an unbranched tube, with no leakage or accumulation, A₁v₁ = A₂v₂. At a branch, incoming flow equals the sum of outgoing flows; each daughter branch need not carry the entire original flow. These distinctions follow from the continuity treatment in OpenStax.
Suppose a pipe carries Q = 100 cm³/s, or 1.00 × 10⁻⁴ m³/s. Its area narrows from 2.00 cm² to 0.500 cm²:
- A₁ = 2.00 × 10⁻⁴ m², giving v₁ = 0.500 m/s.
- A₂ = 5.00 × 10⁻⁵ m², giving v₂ = 2.00 m/s.
The mean speed increases fourfold while volumetric flow stays the same. No extra water is created at the narrow section.
The conversion matters: 1 cm² is 10⁻⁴ m², whereas 1 cm³ is 10⁻⁶ m³. Converting both with the same power of ten destroys the units even if the later algebra is correct.
Bernoulli adds a mechanical-energy comparison
For steady incompressible flow along a streamline, neglecting viscosity and external mechanical work between the points:
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂.
The terms have pressure units. The simplified pipe calculation also treats the section speeds as representative of the streamline speeds, rather than resolving a nonuniform velocity profile. OpenStax derives the relationship from mechanical energy.
Use the narrowing pipe above, fluid density ρ = 1000 kg/m³, and initially equal heights. Then:
P₁ − P₂ = ½(1000)(2.00² − 0.500²) = 1875 Pa.
The faster section has lower static pressure in this ideal, equal-height comparison. The phrase “faster always means lower pressure” would be too broad: height, added mechanical work, losses and the points being compared all matter.
Now place point 2 at 0.100 m above point 1, with the same prescribed flow and areas. Taking g = 9.80 m/s² adds a gravitational term:
P₁ − P₂ = 1875 + (1000)(9.80)(0.100) = 2855 Pa.
The upstream pressure must support both the greater speed and the greater elevation. If point 2 were instead 0.300 m lower, the gravitational term would be −2940 Pa. Then P₁ − P₂ = −1065 Pa: the faster downstream section would have the higher pressure. That counterexample identifies the missing condition in the memorized shortcut.
Poiseuille describes a different idealization
A horizontal, straight, rigid circular tube carrying steady, fully developed laminar flow of an incompressible Newtonian fluid, with no slip at the wall, has:
Q = πr⁴ΔP/(8ηL).
Here ΔP is upstream minus downstream pressure, η is dynamic viscosity and L is tube length. Unlike the lossless Bernoulli calculation, viscosity is essential. OpenStax presents the fourth-power radius dependence and its laminar-flow limits.
Consider a separate model tube with radius 1.00 mm, length 0.200 m, viscosity 1.00 × 10⁻³ Pa·s, and ΔP = 100 Pa. Substitution gives:
Q = 1.96 × 10⁻⁷ m³/s = 0.196 mL/s.
The mean speed is Q/(πr²) = 0.0625 m/s. For density 1000 kg/m³, the Reynolds number ρv(2r)/η is 125, consistent with the intended laminar regime. A low Reynolds number alone does not establish the other assumptions, such as a sufficiently developed velocity profile.
At the same pressure difference, doubling radius predicts sixteen times the flow and four times the mean speed: Q scales with r⁴, while area scales with r². The prediction remains conditional on the altered system still satisfying the flow model.
If instead the pump maintains the same flow, doubling radius requires only one-sixteenth of the pressure difference, and mean speed falls to one-quarter. The fixed variable changes the answer.
Why equal-area pipe flow can still lose pressure
In a horizontal uniform tube, continuity gives the same mean speed at both ends. Lossless Bernoulli would then predict equal pressures. Poiseuille predicts a pressure drop needed to overcome viscous resistance.
There is no algebraic contradiction. One model neglects a mechanism that the other retains. For real flow with meaningful losses, the mechanical-energy balance needs a loss term; applying the lossless equation and the viscous equation simultaneously to the same segment without acknowledging that term is inconsistent.
Biological vessels add further complications: pulsation, changing radius, compliant walls, branching and blood properties. A passage may deliberately idealize them. Use the stated model for the question while recognizing what it leaves out.
Transfer exercise: two experiments, one radius change
A tube’s radius is halved while length and viscosity remain constant. Assume Poiseuille conditions continue to hold.
Experiment A: the pressure difference stays fixed. What happens to Q and mean speed?
Experiment B: the flow rate stays fixed. What happens to the required pressure difference and mean speed?
In A, Q becomes 1/16 of its original value. Area becomes 1/4, so mean speed becomes (1/16)/(1/4) = 1/4 of its original value.
In B, maintaining Q requires 16 times the pressure difference. With one-quarter the area, mean speed becomes 4 times the original value.
The answer “speed always quadruples” silently assumes fixed flow. The answer “speed always falls” silently assumes fixed pressure. Neither is a complete response to an unspecified system change.
Make your next solution state its conditions
Before calculating, identify the comparison: two positions in one flow, or two differently configured systems. Mark the held-constant quantities and choose the model that describes the relevant mechanism.
After calculating, check that the units match the requested quantity and that your model assumptions remain plausible. A correct substitution into the wrong physical model is still the wrong explanation.