Hagen–Poiseuille Law: Full Derivation

1. Force balance on a fluid cylinder

Consider steady, fully developed laminar flow in a pipe of radius R, driven by pressure drop ΔP over length L. Isolate a coaxial fluid cylinder of radius r. The pressure force pushing it forward is ΔP·πr²; the viscous shear on its lateral surface, τ·2πrL, resists. Equilibrium (no acceleration in fully developed flow):

Shear stress grows linearly from zero at the centreline to a maximum at the wall — before any constitutive law is invoked.

2. Insert Newton's law of viscosity

With τ = −μ du/dr (velocity decreases outward), separate and integrate using the no-slip condition u(R) = 0:

The profile is a parabola: maximum velocity u_max = ΔPR²/4μL on the axis, and the mean velocity works out to exactly half of it.

3. Integrate to the flow rate

Sum the flow through annular rings of area 2πr dr:

The r⁴ scaling compounds two squares: the cross-sectional area grows as r², and the parabolic profile lets the core speed grow as r² again. Halving an artery's radius cuts flow sixteen-fold at fixed pressure — the hemodynamic reason arterial narrowing is so dangerous.

4. Validity

The derivation assumed laminar (Re < 2300), steady, fully developed, incompressible Newtonian flow in a straight circular pipe. The Hagen–Poiseuille calculator checks the Reynolds number of its own answer and warns when the assumption breaks.

5. Extended worked examples

Example 1 — an infusion line. Saline (μ ≈ 1.0×10⁻³ Pa·s) driven at ΔP = 8 kPa (a pressurised bag ≈ 0.8 m head) through a needle of radius 0.3 mm and length 3 cm: Q = π×8 000×(3×10⁻⁴)⁴/(8×10⁻³×0.03) = π×8 000×8.1×10⁻¹⁵/2.4×10⁻⁴ ≈ 8.5×10⁻⁷ m³/s ≈ 51 mL/min. Halve the needle radius and the same pressure delivers 51/16 ≈ 3.2 mL/min — the r⁴ law is why needle gauge dominates every infusion chart.

Example 2 — arterial stenosis, quantified. A plaque narrows an artery to 50% of its radius. At fixed pressure drop, Q ∝ r⁴ falls to (0.5)⁴ = 6.25% of normal. The body compensates by raising ΔP and dilating downstream vessels, but the arithmetic explains why symptoms often stay silent until narrowing is severe, then deteriorate rapidly — the fourth power is flat, then a cliff.

Example 3 — checking the assumption that checks itself. Take Example 1's answer: mean velocity v̄ = Q/πr² = 8.5×10⁻⁷/(π×9×10⁻⁸) = 3.0 m/s; Re = ρv̄D/μ = 998×3.0×6×10⁻⁴/10⁻³ ≈ 1 800 < 2 300 ✓ laminar, self-consistent. Had Re exceeded 2 300, the computed Q would be fiction — always close this loop, exactly as the site's Hagen–Poiseuille calculator does automatically.

Example 4 — capillary viscometry. Run a test fluid through r = 0.5 mm, L = 0.2 m under ΔP = 2 kPa and collect Q = 4.0×10⁻⁸ m³/s. Invert the law: μ = πΔPr⁴/(8LQ) = π×2 000×6.25×10⁻¹⁴/(8×0.2×4×10⁻⁸) = 6.1×10⁻³ Pa·s. Glass capillary viscometers (Ostwald, Ubbelohde) in every rheology lab are this equation run backwards, traceable to national standards.

6. Common misconceptions

"Mean velocity ≈ maximum velocity." For the parabola, v̄ = u_max/2 exactly. Reading a centreline probe and multiplying by area overestimates Q by 100%. (Turbulent profiles are much flatter — v̄ ≈ 0.82u_max — a difference exploited to detect regime.)

"The law holds from the pipe entrance." Fully developed flow needs an entrance length L_e ≈ 0.06 Re D. At Re = 1 800 in Example 3's needle that is 0.06×1 800×0.0006 = 6.5 cm — longer than the needle! Short-tube corrections exist precisely because real capillaries often never reach the parabola; treat computed values as slight overestimates of resistance.

"Doubling pressure doubles velocity, so Re doubles — still fine." It does, and the check must be redone: Poiseuille's law can predict flows that violate its own laminar premise. The formula contains no alarm; you are the alarm.

"Blood in capillaries obeys this law." In vessels a few red-cell diameters wide, blood is not a continuum Newtonian fluid: cells deform single-file, apparent viscosity drops (Fåhræus–Lindqvist effect), and Casson-type rheology takes over. Poiseuille remains the reference against which those corrections are defined.

7. Where this shows up

Microfluidics designs whole labs-on-chips with the electrical analogy this law licenses: ΔP = R_H Q with hydraulic resistance R_H = 8μL/πr⁴, so channel networks compute like resistor circuits — series resistances add, parallel conductances add, and a droplet-sorting chip is an exercise in Kirchhoff's laws. Physiology grades vascular resistance the same way (systemic vascular resistance in clinics is ΔP/Q in Wood units). Filtration, membrane pores, soil percolation (each pore a tiny Poiseuille tube summed into Darcy's law), 3-D printer nozzles, and the ink channels in your ballpoint pen: wherever viscous liquid meets a narrow conduit, this derivation is doing the pricing.

8. Practice problems

P1. Olive oil (μ = 0.084 Pa·s) through r = 2 mm, L = 1 m at ΔP = 20 kPa. Find Q. Solution: Q = π×2×10⁴×1.6×10⁻¹¹/(8×0.084×1) = 1.5×10⁻⁶ m³/s ≈ 1.5 mL/s.

P2. Verify P1's regime. Solution: v̄ = Q/πr² = 0.119 m/s; Re = 910×0.119×0.004/0.084 ≈ 5.2 — deeply laminar.

P3. Two capillaries in series: identical L, radii r and r/2. What fraction of ΔP falls across the narrow one? Solution: resistances ∝ 1/r⁴, ratio 1:16 → 16/17 ≈ 94% of the drop sits on the narrow segment — why one stenosis dominates a whole vessel.

P4. A viscometer must keep Re < 100 measuring μ ≈ 5×10⁻³ Pa·s fluid (ρ = 1 050) in r = 0.4 mm. Max ΔP for L = 0.15 m? Solution: Re = 2ρQ/(πrμ) < 100 → Q < 100πrμ/2ρ = 100π×4×10⁻⁴×5×10⁻³/2 100 = 3.0×10⁻⁷; then ΔP = 8μLQ/πr⁴ < 8×5×10⁻³×0.15×3×10⁻⁷/(π×2.56×10⁻¹⁴) ≈ 22 kPa.

9. Going deeper

Make the driving pressure oscillate — a heartbeat — and the problem becomes Womersley flow. A new dimensionless group α = r√(ω/ν) compares the tube radius with the oscillatory viscous penetration depth. For α ≪ 1 the flow is quasi-steady Poiseuille tracking the pressure in phase; for α ≫ 1 the core moves as a plug lagging the pressure by up to 90°, with the shear confined to thin Stokes layers at the wall. The human aorta operates at α ≈ 15–20: pulsatile, plug-like, decidedly not Poiseuille — while your capillaries, at α ≪ 1, obey these notes beautifully. One derivation, and its own well-marked exits.

10. Historical context

The law has two independent parents. Gotthilf Hagen, a German hydraulic engineer, published measurements on brass tubes in 1839, catching the fourth-power dependence and even noting anomalies we now read as transition to turbulence. Jean Léonard Marie Poiseuille, a Parisian physician, wanted to understand blood flow; unable to control experiments in living vessels, he built capillary rigs of heroic precision (1840–46), varying tube radius, length, pressure and temperature systematically — his data confirm the r⁴ law to fractions of a percent. Neither derived it theoretically: that came via Navier–Stokes solutions by Stokes (1845, unpublished for pipes), and definitively by Hagenbach (1860), who also christened the law. Physiology repaid the debt with interest: the fourth-power sensitivity became the central fact of vascular physiology — the body regulates blood distribution almost entirely by tuning vessel radii, since a 19% radius change already doubles the flow.

11. Another way to see it: the shell balance

Instead of solving the reduced Navier–Stokes equation, balance forces on a fluid cylinder of radius r inside the pipe. Pressure pushes it forward with (ΔP)(πr²); the surrounding fluid drags its lateral surface backward with τ(r)·2πrL. Steady flow means these balance:

— the shear stress grows linearly from zero at the centreline to its maximum at the wall, a result independent of the fluid's constitutive law (it holds for blood, paint and polymer melts too, which is why rheologists love pipe flow). Only now invoke Newton's law τ = −μ du/dr and integrate with u(R) = 0: the parabola, then Q by one more integral. The two-step structure — universal stress distribution, then material-specific velocity profile — is the template for every non-Newtonian pipe formula: substitute the power-law or Bingham constitutive relation at step two and the corresponding design equations (used daily for slurries and drilling muds) fall out of the same balance.

12. Frequently asked questions

Why r⁴ and not r²? Two factors of r from area (more cross-section), two from speed: a wider tube both carries more fluid and lets each parcel move faster (the wall's braking is farther away, and v ∝ r² at the centre). Area alone would give r²; the profile supplies the rest.

Thumb over a garden hose makes water jet farther — doesn't that contradict Q ∝ r⁴? No: the law gives Q at fixed ΔP across the tube in question. The hose's flow is set mostly by the long hose and supply pressure; your thumb adds a short constriction that barely reduces Q but forces that Q through a small area — velocity up, by continuity. Different question, different equation.

Does the law apply to gases? Yes, at low speeds where compressibility is negligible and Kn is small; for long lines with large pressure ratios an integrated compressible version replaces it, and in fine capillaries at low pressure a slip correction (Kn again) raises the flow above the classical value.

What if the tube is an annulus, an ellipse, a rectangle? The same derivation runs with the corresponding Poisson-equation solution; each shape earns its own constant (e.g., a thin slit of width w, gap h: Q = ΔP w h³/12μL — the "cubic law" of fracture flow in hydrogeology).

Series and parallel tubes? Exactly like resistors, via R_H = 8μL/πr⁴: resistances add in series; conductances add in parallel. P3 of §8 was Ohm's-law reasoning in disguise.

13. Further practice

P5. From τ(r) = ΔP r/2L, find the wall shear stress in Example 1's needle. Solution: τ_w = 8 000×3×10⁻⁴/(2×0.03) = 40 Pa — the number a biomedical engineer checks against hemolysis thresholds.

P6. A power-law fluid (τ = K γ̇ⁿ, n = 0.5) flows in a pipe. Using the shell balance, how does Q scale with ΔP? Solution: γ̇ ∝ τ^{1/n} = τ² → u ∝ (ΔP)²; Q ∝ (ΔP)² — shear-thinning fluids reward extra pressure superlinearly, a fact exploited in polymer extrusion.

P7. Two identical syringes, one with water and one with glycerine (μ = 1.4), are pressed with equal force. Flow-time ratio? Solution: t ∝ 1/Q ∝ μ: glycerine takes 1 400× longer — why viscous drugs come in wide-bore syringes.

P8. A capillary viscometer's timing bulb empties under gravity head h ≈ 0.1 m. Show μ ∝ ρt and explain why such instruments report kinematic viscosity. Solution: ΔP = ρgh → μ = πρghr⁴t/(8LV): μ/ρ = ν depends on t alone for a given instrument — the glass measures ν, and labs quote centistokes for exactly this reason.

14. Worked exam problem

Problem. A microfluidic chip channel (treat as circular): r = 25 μm, L = 2 cm, water. The assay needs Q = 0.5 μL/min. Find (a) the driving pressure, (b) the mean velocity, (c) the Reynolds number, and comment.

Solution. (a) Q = 0.5×10⁻⁹/60 = 8.33×10⁻¹² m³/s; ΔP = 8μLQ/πr⁴ = 8×10⁻³×0.02×8.33×10⁻¹²/(π×3.91×10⁻¹⁹) = 1.09 kPa — a hundredth of an atmosphere, deliverable by a small syringe pump, and computable to the pascal because the law is exact here. (b) v̄ = Q/πr² = 8.33×10⁻¹²/1.963×10⁻⁹ = 4.2 mm/s. (c) Re = ρv̄(2r)/μ = 998×4.2×10⁻³×5×10⁻⁵/10⁻³ ≈ 2×10⁻⁴ — ten million times below transition. Microfluidics is Poiseuille's kingdom: the entire chip is designed as a resistor network with R_H = 8μL/πr⁴, exactly as §7 promised.

15. Key takeaways

For steady, laminar, fully developed pipe flow: parabolic profile, v̄ = u_max/2, and Q = πΔPr⁴/8μL — the fourth power making radius the master variable of every conduit from arteries to chips. The shear stress distribution τ = ΔPr/2L is constitutive-law-independent; the flow formula is Newtonian-specific. Always close the loop: check Re, check entrance length, and remember pulsatile flow (Womersley) is a different regime with its own number.

16. Where to go next

The regime check lives in the Reynolds chapter; the momentum equation this solved exactly is Navier–Stokes. The Hagen–Poiseuille calculator computes Q with the laminar check built in, and the Hagen–Poiseuille calculator runs the circuit analogy directly.