Pipe Flow — Reynolds Number & the Navier–Stokes Solution

Push fluid through a pipe — inertia fights viscosity. Their ratio (Reynolds number) sets whether you get smooth Hagen–Poiseuille flow or turbulence.

Visualization
Input
m\mathrm{m}
m/s\mathrm{m/s}
kg/m3\mathrm{kg/m^{3}}
Pas\mathrm{Pa\cdot s}
m\mathrm{m}

Show the analytical velocity profile (Hagen–Poiseuille)

Inertia versus viscosity

Two forces fight to set the personality of any fluid flow. Inertia wants the fluid to keep moving in whatever direction it's moving — it doesn't want to be redirected. Viscosity wants neighboring layers of fluid to stick together and damp out any differences. The ratio of those two is the most famous dimensionless number in fluid mechanics: the Reynolds number.

Re=ρvDμ\mathrm{Re} = \frac{\rho v D}{\mu}

Source: Wikipedia — Reynolds number.

For pipe flow, the practical takeaway is a clean threshold:

  • Re2300\mathrm{Re} \lesssim 2300 — viscosity wins. Flow is laminar: smooth parallel layers, perfectly predictable.
  • 2300Re40002300 \lesssim \mathrm{Re} \lesssim 4000transitional. Tiny disturbances start to grow into eddies. Behavior gets unstable.
  • Re4000\mathrm{Re} \gtrsim 4000 — inertia wins. Flow is turbulent: chaotic eddies, time-averaged but never steady at the small scale.

This is what the slider above is exposing. Try the Water through a thin tube preset (Re ≈ 500): the tracers march in clean parallel lanes. Now switch to the faucet preset (Re ≈ 25,000) — same fluid, same physics, but the tracers thrash. Nothing about the fluid changed; only the regime did.

Hagen–Poiseuille — an exact NS solution

The full Navier–Stokes equations are infamously hard:

ρ ⁣(vt+(v ⁣ ⁣)v)=p+μ2v+f\rho\!\left(\frac{\partial \mathbf{v}}{\partial t} + (\mathbf{v}\!\cdot\!\nabla)\mathbf{v}\right) = -\nabla p + \mu \nabla^{2} \mathbf{v} + \mathbf{f}

Source: Wikipedia — Navier–Stokes equations. General closed-form solutions don't exist; that's literally a Millennium Prize Problem.

But for one specific case — steady, fully developed, axisymmetric flow in a long round pipe — every term you can drop, drops. The convective acceleration (v ⁣ ⁣)v(\mathbf{v}\!\cdot\!\nabla)\mathbf{v} vanishes, the time derivative vanishes, and what's left is a balance between the axial pressure gradient and viscous shear:

dpdx=μ1rddr ⁣(rdudr)\frac{dp}{dx} = \mu \frac{1}{r}\frac{d}{dr}\!\left(r \frac{du}{dr}\right)

Integrate twice with the no-slip boundary condition (u=0u = 0 at r=Rr = R) and you get a perfect parabolic profile:

u(r)=14μ ⁣(dpdx) ⁣(R2r2)u(r) = \frac{1}{4\mu}\!\left(-\frac{dp}{dx}\right)\!(R^{2} - r^{2})

The cross-section average is half the centerline peak, umax=2vˉu_{\max} = 2\bar{v}, and integrating the profile across the pipe yields the Hagen–Poiseuille pressure drop:

ΔP=32μLvˉD2\Delta P = \frac{32 \mu L \bar{v}}{D^{2}}

Source: Wikipedia — Hagen–Poiseuille equation. Open the engineering view above to see the parabolic profile drawn explicitly.

Why doubling the diameter is so powerful

Look at the pressure-drop formula: ΔP1/D2\Delta P \propto 1/D^{2} at fixed vˉ\bar{v}. But the flow rate relationship is even more dramatic. Because Q=vˉA=vˉπR2Q = \bar{v} A = \bar{v} \pi R^{2}, if you instead hold ΔP\Delta P constant and ask how much fluid you can move:

Q=πR4ΔP8μLQ = \frac{\pi R^{4} \Delta P}{8 \mu L}

The fourth power on RR is why your kitchen sink drain is so much more effective than a thin straw, and why narrowing of arteries (which shrinks RR) so dramatically increases the heart's pumping work. Halve the radius and you cut the carrying capacity by a factor of 16.

Why the laminar formula breaks at high Re

Hagen–Poiseuille assumed the velocity profile is steady, axisymmetric, and parallel — every fluid particle moves in a straight line down the pipe. That's a perfectly good solution to NS, but at high Reynolds number it's an unstable one. Tiny disturbances (a wall bump, a vibration, the inlet) get amplified by the inertia of the flow instead of damped by viscosity. The neat parabola breaks up into eddies, the time-averaged profile flattens out into a "plug" with steep wall gradients, and pressure drop now scales roughly as vˉ1.75\bar{v}^{1.75} to vˉ2\bar{v}^{2} instead of linearly. None of the closed-form formulas above apply once you cross the transition.

Show your work
Pipe inner radius
R  =  D/2  =  (0.025)/2  =  0.0125  mR \;=\; D / 2 \;=\; \textcolor{#dc4124}{\left(0.025\right)} / 2 \;=\; 0.0125\;\mathrm{m}
Cross-section area
A  =  πR2  =  π(0.0125)2  =  4.909×104  m2A \;=\; \pi R^{2} \;=\; \pi \textcolor{#dc4124}{\left(0.0125\right)}^{2} \;=\; 4.909 \times 10^{-4}\;\mathrm{m^{2}}
Volumetric flow rate
Q  =  vA  =  (0.1)(4.909×104)  =  4.909×105  m3/sQ \;=\; v \cdot A \;=\; \textcolor{#dc4124}{\left(0.1\right)} \cdot \textcolor{#dc4124}{\left(4.909 \times 10^{-4}\right)} \;=\; 4.909 \times 10^{-5}\;\mathrm{m^{3}/s}
Reynolds number — inertia / viscosity
Re  =  ρvD/μ  =  ρ(0.1)(0.025)/μ  =  2500\mathrm{Re} \;=\; \rho v D / \mu \;=\; \rho \textcolor{#dc4124}{\left(0.1\right)} \textcolor{#dc4124}{\left(0.025\right)} / \mu \;=\; 2500
Centerline velocity in fully-developed laminar flow (= 2× mean)
umax  =  2v  =  2(0.1)  =  0.2  m/su_{\max} \;=\; 2 v \;=\; 2 \textcolor{#dc4124}{\left(0.1\right)} \;=\; 0.2\;\mathrm{m/s}
Hagen–Poiseuille pressure drop (laminar only)
ΔPHP  =  32μLv/D2  =  32μ(1)(0.1)/(0.025)2  =  5.12  Pa\Delta P_{HP} \;=\; 32 \mu L v / D^{2} \;=\; 32 \mu \textcolor{#dc4124}{\left(1\right)} \textcolor{#dc4124}{\left(0.1\right)} / \textcolor{#dc4124}{\left(0.025\right)}^{2} \;=\; 5.12\;\mathrm{Pa}
Wall shear stress in laminar pipe flow
τw  =  8μv/D  =  8μ(0.1)/(0.025)  =  0.032  Pa\tau_{w} \;=\; 8 \mu v / D \;=\; 8 \mu \textcolor{#dc4124}{\left(0.1\right)} / \textcolor{#dc4124}{\left(0.025\right)} \;=\; 0.032\;\mathrm{Pa}
Laminar Darcy friction factor (only meaningful for Re ≲ 2300)
flam  =  64/Re  =  64/(2500)  =  0.0256f_{\text{lam}} \;=\; 64 / \mathrm{Re} \;=\; 64 / \mathrm{\textcolor{#dc4124}{\left(2500\right)}} \;=\; 0.0256
Worked Example

Water at 20 °C (ρ = 1000 kg/m³, μ = 0.001 Pa·s) through a 5 mm tube at v = 0.4 m/s gives Re = 1000·0.4·0.005/0.001 = 2000 — laminar, just under the transition. Hagen–Poiseuille predicts ΔP = 32·0.001·1·0.4/(0.005)² = 512 Pa over 1 m, with centerline velocity u_max = 0.8 m/s and wall shear τ_w = 8·0.001·0.4/0.005 = 0.64 Pa.

QuantityExpectedComputedΔ
Re200020000
u_max0.80.80
dP_HP5125120
tau_wall_HP0.640.640
f_lam0.0320.0320
Worked Example

Tap-water flow at v = 1 m/s through a D = 25 mm pipe (kitchen-sink scale) gives Re = 1000·1·0.025/0.001 = 25,000 — well past the turbulent threshold (~4000). Hagen–Poiseuille no longer applies; the laminar formulas are reported but not physical here.

QuantityExpectedComputedΔ
Re25000250000
Sources
  1. [1] Wikipedia. Reynolds number — definition and pipe-flow critical values, eq. Re = ρvD/μ; for pipe flow Re ≲ 2300 is laminar, transition is roughly 2300–4000, Re ≳ 4000 is fully turbulent. (en.wikipedia.org)
  2. [2] Wikipedia. Hagen–Poiseuille equation — exact NS solution for steady laminar pipe flow, eq. ΔP = 8μLQ / (πR⁴) = 32μLv / D²; u(r) = G/(4μ)·(R² − r²) with G = ΔP/L; u_max = 2·v_avg. (en.wikipedia.org)
  3. [3] Wikipedia. Navier–Stokes equations — momentum conservation for a Newtonian fluid, eq. ρ(∂v/∂t + (v·∇)v) = −∇p + μ∇²v + f. Hagen–Poiseuille is the closed-form solution for steady, axisymmetric, fully-developed flow in a round pipe. (en.wikipedia.org)
Assumptions
  • Steady, fully-developed, axisymmetric flow in a long straight round pipe — Hagen–Poiseuille assumes the velocity profile no longer changes with axial position.
  • Newtonian fluid — viscosity is a single constant, independent of strain rate.
  • No body forces along the flow axis (horizontal pipe, or pressure already accounts for elevation).
  • No-slip boundary at the wall — fluid velocity is zero at r = R.
  • Hagen–Poiseuille formulas are valid only while flow is laminar (Re ≲ 2300). Above that, transitional or turbulent behavior breaks the parabolic-profile assumption and a different friction model is required.