Renal Artery Stenosis: Continuity, Bernoulli, and Poiseuille
A 62-year-old man has difficult-to-control hypertension and an abdominal bruit. Doppler ultrasound shows a focal renal artery stenosis. The normal proximal segment has an internal diameter of 6.0 mm and a time-averaged velocity of 0.18 m/s. At the stenosis, the diameter is 2.0 mm and the measured peak velocity is 1.5 m/s. Blood density is 1060 kg/m3 and viscosity 3.5 × 10−3 Pa·s. The stenotic segment is 10 mm long. The radiologist estimates volumetric flow at 300 mL/min. A trainee asks whether the pressure drop across the stenosis can be estimated from the velocity data, why the stenosis resistance rises so sharply, and whether the flow is turbulent. The team is considering a pressure-wire measurement.
1. Use continuity to calculate the velocity in the stenotic segment from the diameter and the stated flow. Compare with the measured 1.5 m/s and comment.
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Convert flow: 300 mL/min = 300 × 10−6 m3/min = 5 × 10−6 m3/s. Normal radius is 3.0 mm = 3.0 × 10−3 m; area = π(3.0 × 10−3)2 = 2.83 × 10−5 m2; velocity = Q/A = 5 × 10−6 / 2.83 × 10−5 = 0.177 m/s, matching the stated 0.18 m/s. Stenotic radius is 1.0 mm; area = π(1.0 × 10−3)2 = 3.14 × 10−6 m2; velocity = 5 × 10−6 / 3.14 × 10−6 = 1.59 m/s. The measured peak velocity of 1.5 m/s is close to this time-averaged estimate. Peak systolic velocity is usually higher than time-averaged velocity, so the agreement is reasonable rather than exact. Continuity assumes steady, incompressible flow and a fixed lumen; in vivo the stenosis may be irregular, and the ultrasound sample volume may not be exactly at the narrowest point. The calculation nonetheless confirms that the diameter reduction produces roughly a ninefold increase in velocity, because velocity scales with the square of diameter.
2. Estimate the Bernoulli pressure drop in mmHg across the stenosis and the ratio of Poiseuille resistance of the stenotic segment to the normal segment for the same length. Which term dominates?
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Using the simplified Bernoulli relation ΔP ≈ 4v2 (mmHg, v in m/s), the pressure drop is about 4 × (1.5)2 = 9 mmHg. More exactly, 0.5ρ(v22 − v12) = 0.5 × 1060 × (1.52 − 0.182) = 0.5 × 1060 × (2.25 − 0.032) = 1176 Pa, or about 8.8 mmHg; using 1.59 m/s gives about 9.9 mmHg. Poiseuille resistance is R = 8ηL/(πr4), so for the same length the resistance ratio is (rnormal/rstenosis)4 = (3/1)4 = 81. The stenotic segment therefore has about 81 times the resistance per unit length. The viscous pressure drop across the 10-mm stenosis is R × Q ≈ 8.9 × 107 Pa·s/m3 × 5 × 10−6 m3/s ≈ 445 Pa, or 3.3 mmHg, smaller than the Bernoulli term but not negligible. The sharp rise in resistance is dominated by the fourth-power dependence on radius; a 3-fold narrowing produces an 81-fold resistance increase.
3. Calculate the Reynolds number at the stenosis. Is the flow laminar? Why might catheter-measured pressure gradient differ from the Doppler-derived estimate, and what does this imply for deciding on intervention?
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Reynolds number at the stenosis: Re = ρvD/η = (1060 × 1.5 × 0.002)/0.0035 ≈ 909. That is below the conventional 2000 threshold for fully turbulent pipe flow, so the core flow is laminar or transitional, but a stenotic lumen with an abrupt change in caliber, vessel curvature, and pulsatility can produce disturbed flow and post-stenotic turbulence at lower Re. A catheter pressure wire measures the actual pressure difference between the aorta and the distal renal artery, integrating viscous, convective, and turbulent losses. The Doppler-derived Bernoulli estimate uses peak velocity and ignores viscous losses, pressure recovery, and the exact geometry; it can therefore underestimate or overestimate the true gradient. For intervention decisions, the pressure-wire measurement during hyperemia is more direct, but it is invasive and itself subject to drift and equalization error. The Doppler velocity is best used as a screening signal, not as the sole basis for angioplasty.