What is Poiseuille's Law: Laminar Flow Through Cylindrical Pipes?
The Hagen-Poiseuille equation (derived independently by Gotthilf Hagen in 1839 and Jean Léonard Marie Poiseuille in 1840) is the exact analytical solution for steady, incompressible, laminar flow through a long, straight, rigid pipe with constant circular cross-section. It is one of the few exact solutions to the Navier-Stokes equations. The equation reveals that flow rate is proportional to the fourth power of radius — making pipe diameter the overwhelmingly dominant factor in flow capacity.
Mathematical Foundation
Hagen-Poiseuille Equation
= Volumetric flow rate (m³/s) — the volume of fluid passing through the pipe per unit time.
= Internal radius of the pipe (m). Flow rate scales with r⁴ — doubling the radius increases flow 16×.
= Pressure difference between the two ends of the pipe (Pa = N/m²).
= Dynamic viscosity of the fluid (Pa·s). Higher viscosity means more resistance to flow.
= Length of the pipe (m). Longer pipe → more friction → less flow at the same pressure.
Laws & Principles
- The r⁴ Power Law: This is the most critical engineering insight. Doubling pipe radius increases flow rate by 2⁴ = 16×. A 10% increase in radius gives 1.1⁴ = 1.46× more flow (46% increase). This is why even small amounts of arterial plaque cause dramatic blood flow reduction — a 25% radius reduction cuts flow to (0.75)⁴ = 31.6% of normal.
- Laminar Flow Only: Poiseuille's law is valid ONLY for laminar flow (Re < 2,000). In turbulent flow, the pressure-flow relationship changes from linear (Q ∝ ΔP) to approximately Q ∝ ΔP^0.5. Using Poiseuille's equation for turbulent flow vastly overestimates flow rate.
- Parabolic Velocity Profile: In Poiseuille flow, fluid at the center moves at exactly twice the mean velocity, while fluid at the wall has zero velocity (no-slip condition). The velocity profile is v(r) = v_max(1 - r²/R²) — a perfect parabola.
Step-by-Step Example Walkthrough
" Water (μ = 0.001 Pa·s) flows through a 5mm radius pipe that is 2 meters long, with a pressure drop of 500 Pa. "
- 1. Convert: r = 5mm = 0.005 m, L = 2 m, ΔP = 500 Pa, μ = 0.001 Pa·s.
- 2. Q = π × (0.005)⁴ × 500 / (8 × 0.001 × 2).
- 3. Q = π × 6.25×10⁻¹⁰ × 500 / 0.016 = π × 3.125×10⁻⁷ / 0.016.
- 4. Q = π × 1.953×10⁻⁵ = 6.14 × 10⁻⁵ m³/s = 61.4 mL/s.
Final Result: Flow rate = 6.14 × 10⁻⁵ m³/s (61.4 mL/s ≈ 3.68 L/min). Verify: mean velocity v = Q/(πr²) = 0.782 m/s → Re = ρvD/μ = 1000×0.782×0.01/0.001 = 7,820 — this exceeds Re 2,000, so Poiseuille's law would NOT be valid here. A smaller pipe or lower ΔP is needed for laminar conditions.