Fully Developed Laminar Flow — Poiseuille's Law

The exact parabolic profile of laminar pipe flow — and why flow rate scales with radius to the fourth power.

Viscous (Internal) Flow: Pipe FlowMechanical EngineeringFree preview
⏱️ About 16 min

A cardiologist threads a slightly thicker catheter into an artery and the infusion rate collapses to a trickle. A plumber ups a pipe from ½ inch to ¾ inch and the delivered water roughly triples. Both shocks come from one equation: Poiseuille's law, which says that in laminar pipe flow the flow rate grows with the fourth power of the radius. Double the radius and you move sixteen times the fluid. Halve it and you keep one sixteenth. That ferocious sensitivity is not a rule of thumb — it is the integral of the exact parabolic profile you derived in the last lesson.

💡
The big idea: For fully developed, steady, laminar flow in a circular tube, the Navier-Stokes equation (re-solved in cylindrical coordinates) gives a paraboloid of revolution: u(r) ∝ (R² − r²), zero at the wall and maximum on the axis. Integrating that profile across the cross-section yields Poiseuille's law, Q = πΔpR⁴/(8μL): flow rate is directly proportional to the pressure drop and to the fourth power of the radius, and inversely proportional to viscosity and length. The corresponding Darcy friction factor is exactly f = 64/Re — a clean, exact result that holds only in the laminar regime and breaks down at the transition.
🎯 By the end, you'll be able to
  • State the parabolic velocity profile and centreline velocity for laminar pipe flow
  • Derive Poiseuille's law Q = πΔpR⁴/(8μL) by integrating the profile across the section
  • Use the laminar friction factor f = 64/Re and state the regime where it is valid
  • Quantify the R⁴ sensitivity of flow rate to pipe radius
📎 Helpful to know first

From flat plates to a round tube

The last lesson solved the reduced Navier-Stokes equation for flow between flat plates and found a parabola. A circular pipe is the same problem wrapped around an axis: solve μ d²u/dy² = dp/dx in cylindrical coordinates, with no-slip at the wall (u(R) = 0) and symmetry on the axis (du/dr|_{r=0} = 0). The result is a paraboloid of revolution — a parabola spun around the pipe axis. Fluid at the wall is stuck fast; fluid on the centreline moves fastest, at twice the mean velocity.

\[ u(r) = \frac{1}{4\mu}\!\left(-\frac{dp}{dx}\right)(R^{2} - r^{2}) \qquad u_{\max} = \frac{\Delta p\, R^{2}}{4\mu L} \]
The exact laminar pipe velocity profile: a paraboloid, zero at the wall (r = R) and maximum on the axis (r = 0). The centreline speed u_max is exactly twice the mean velocity V. Here Δp = (−dp/dx)L is the total pressure drop over length L.
Side-by-side pipe velocity profiles: laminar flow has a sharp parabolic profile peaking at the centreline; turbulent flow has a much flatter, fuller profile. A dashed line marks the mean velocity in each. Laminar — Re < 2300 V = 0.5 u_max flow u_max Turbulent — Re > 4000 V ≈ 0.8 u_max u_max Both pipes carry the same mean velocity V (dashed). Turbulence flattens the profile — the centreline speed u_max is lower relative to V. velocity u →
Laminar pipe flow: the parabolic profile (reproduced from lesson 1) is the cross-section of the paraboloid u(r) ∝ (R² − r²). Integrating this exact profile over the circular area gives Poiseuille's law below — no fitting required.

Integrating the profile: Poiseuille's law

To get the total volume flow rate Q, multiply the local speed by the ring of area 2πr dr at each radius and integrate from the axis to the wall. The integral of r(R² − r²) brings down an R⁴, which is the origin of the famous fourth-power dependence. Collect the constants and the result is Poiseuille's law, sometimes called the Hagen-Poiseuille equation after the two who established it experimentally in the 1840s — and later shown (as you just saw) to be an exact consequence of the Navier-Stokes equations.

\[ Q = \frac{\pi\,\Delta p\, R^{4}}{8\,\mu\, L} \]
Poiseuille's law for fully developed laminar flow in a circular tube. Flow rate is proportional to the pressure drop and to R⁴, and inversely proportional to viscosity and length. Valid only for laminar flow (Re < 2300); it fails once the flow goes turbulent.
✨ The R⁴ sensitivity: why radius dominates everything

Because Q ∝ R⁴, the radius is overwhelmingly the most important variable in laminar flow. Increase the radius by 10% and the flow rate rises by (1.1)⁴ ≈ 1.46 — a 46% gain from a 10% widening. Double the radius and the flow rises by 2⁴ = 16. Conversely, halving a tube's diameter cuts the flow to 1/16. This is why small blood vessels dominate the resistance of the bloodstream (a slight constriction throttles flow), why narrow capillaries need large pressure gradients, and why doubling a pipe size is so much more effective than doubling the pump pressure for laminar service.

\[ f = \frac{64}{Re} \]
The Darcy friction factor for fully developed laminar pipe flow — an exact result (not a correlation). f depends only on Re and is independent of wall roughness, because in laminar flow the viscous sublayer buries any roughness. Valid strictly for Re < 2300.
⚠️ f = 64/Re is a LAMINAR result — do not carry it into turbulence

This is the misconception the whole module is built to dismantle. The clean formula f = 64/Re is exact for laminar flow only. The moment Reynolds number crosses ~2300, the parabolic profile is destroyed and this formula stops applying — yet students routinely plug a turbulent Reynolds number into 64/Re and get a nonsense friction factor. In turbulent flow the friction factor depends on both Re and relative roughness, and you must read it from the Moody chart or the Colebrook/Haaland equations (lesson 4). Always check the regime first, then choose the friction-factor method that belongs to it.

🎮 Interactive: Poiseuille flow rate LIVE
Predict first: If you double the pipe radius, does the flow rate double — or much more? Predict, then drag.

An interactive slider tool computing the laminar volume flow rate from pressure drop, radius, viscosity, and length using Poiseuille's law, in mL/s.

Poiseuille's law Q = πΔpR⁴/(8μL), live. Slide the pressure drop, radius, viscosity, and length. The radius slider drives Q to the fourth power — the lever that matters most in laminar flow.
📝 Worked example: Water (μ = 1.0 × 10⁻³ Pa·s, ρ = 1000 kg/m³) flows through a capillary of radius R = 1.0 mm = 0.001 m and length L = 1.0 m, driven by a pressure drop Δp = 8.0 kPa = 8000 Pa. Find the volume flow rate Q, confirm the flow is laminar, and state the friction factor.
  1. Poiseuille's law: Q = πΔpR⁴/(8μL) = π(8000)(0.001)⁴/(8 × 1.0 × 10⁻³ × 1).
  2. R⁴ = (0.001)⁴ = 1.0 × 10⁻¹²; numerator = π × 8000 × 1.0 × 10⁻¹² = π × 8.0 × 10⁻⁹.
  3. Denominator 8μL = 8.0 × 10⁻³; Q = (π × 8.0 × 10⁻⁹)/(8.0 × 10⁻³) = π × 1.0 × 10⁻⁶ = 3.14 × 10⁻⁶ m³/s.
  4. In mL/s: Q = 3.14 × 10⁻⁶ m³/s × 10⁶ mL/m³ = 3.14 mL/s.
  5. Mean velocity: V = Q/A = (3.14 × 10⁻⁶)/(π × 0.001²) = 1.0 m/s.
  6. Re = ρVD/μ = (1000)(1.0)(0.002)/(1.0 × 10⁻³) = 2000 < 2300 → laminar. ✓
  7. Friction factor: f = 64/Re = 64/2000 = 0.032 (laminar).
✓ Q ≈ 3.14 mL/s (3.14 × 10⁻⁶ m³/s); Re = 2000 (laminar); f = 0.032.
✏️ Practice: The same capillary (R = 1.0 mm, L = 1.0 m, μ = 1.0 × 10⁻³ Pa·s) must carry Q = 1.0 mL/s = 1.0 × 10⁻⁶ m³/s. Find the required pressure drop Δp, in kPa. (First confirm Re stays laminar: V = Q/A ≈ 0.32 m/s gives Re ≈ 640.)
kPa
Solution
  1. Rearrange Poiseuille's law: Δp = 8μLQ/(πR⁴) = (8 × 1.0 × 10⁻³ × 1 × 1.0 × 10⁻⁶)/(π × (0.001)⁴).
  2. Numerator = 8.0 × 10⁻⁹; R⁴ = 1.0 × 10⁻¹²; πR⁴ = π × 1.0 × 10⁻¹².
  3. Δp = 8.0 × 10⁻⁹/(π × 1.0 × 10⁻¹²) = 2546 Pa = 2.55 kPa.
✏️ Practice: A viscous fluid flows laminarly through a tube under a fixed pressure drop. If the tube radius is doubled (R → 2R) with everything else held constant, by what factor does the flow rate Q increase? (Enter the numerical factor; e.g. 2 means twice the flow.)
×
Solution
  1. Poiseuille's law has Q ∝ R⁴, so doubling R multiplies Q by 2⁴.
  2. 2⁴ = 2 × 2 × 2 × 2 = 16.
  3. The flow rate increases by a factor of 16 — the R⁴ sensitivity in one line.

Check your understanding

1. In fully developed laminar pipe flow, the velocity profile is:
Solving the reduced Navier-Stokes equation in cylindrical coordinates gives u(r) ∝ (R² − r²) — a paraboloid. The centreline speed is exactly twice the mean velocity.
2. Poiseuille's law says Q ∝ R⁴. If a pipe's radius is increased by 20% (with Δp, μ, and L unchanged), the flow rate rises by roughly:
(1.2)⁴ = 2.0736, so Q rises by about 107% — call it roughly double. The fourth power makes even modest radius changes dramatic; this is why radius dominates laminar flow design.
3. The friction-factor relation f = 64/Re is:
It is exact, but strictly laminar. In turbulent flow the profile is no longer parabolic and f depends on both Re and roughness (Moody chart). Never plug a turbulent Re into 64/Re.
🔎 Related lessons
Where Poiseuille's Law Comes From: Exact Solutions of Navier-Stokes · soon Turbulent Flow & the Moody Chart · soon
✅ Key takeaways
  • Fully developed laminar pipe flow has a parabolic profile u(r) ∝ (R² − r²); the centreline speed is twice the mean velocity.
  • Poiseuille's law: Q = πΔpR⁴/(8μL) — flow rate scales with the pressure drop and with R⁴, and inversely with viscosity and length.
  • The R⁴ dependence makes radius the dominant variable; doubling the radius gives 16× the flow.
  • The laminar friction factor f = 64/Re is exact but valid only for Re < 2300 — it must never be carried into turbulent flow.
➡️ So long as the flow stays laminar, everything is exact and clean. But push the Reynolds number past 2300 and the orderly parabola shatters into turbulence — the exact theory stops, and friction factor must be read from a chart. That chart is the Moody chart, and it is the subject of the next lesson.
Want to test yourself on this? Try the Mechanical Aptitude test →
🎓 Go deeper: external courses & trusted references

External sites are listed for reference only. This course is independent and has no affiliation with, or endorsement from, the institutions named.