04 // Interactive Formula Compendium // ENGINEERINGQ = (π · R⁴ · ΔP) / (8 · μ · L)
Hagen-Poiseuille Laminar Pipe Flow Law
Poiseuille's law Q = πR⁴ΔP / (8μL) shows that laminar volumetric flow rate grows with the fourth power of tube radius R and linearly with pressure drop ΔP.
Formula Reference
Main Formula
Q = (π · R⁴ · ΔP) / (8 · μ · L)
Rearranged Forms
ΔP = (8 · μ · L · Q) / (π · R⁴)
R_hyd = 8μL / (πR⁴)
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| Q | Volumetric Flow Rate | m³/s |
| R | Tube Inner Radius | m |
| dP | Pressure Drop ΔP | Pa |
| mu | Dynamic Viscosity | Pa·s |
| L | Tube Length | m |
Inputs & Parameters
[m]
[Pa]
[Pa·s]
[m]
Calculated Result (Volumetric Flow Rate)
Q =3.59039e-5m³/s
Formula Notes & Background
- Integrating viscous shear balance (1/r)d/dr(r μ dv/dr) = −ΔP/L with no-slip wall boundary v(R)=0 gives a parabolic velocity profile v(r) = (ΔP/(4μL))(R² − r²).
- Integrating 2πr·v(r)dr from 0 to R yields Q = πR⁴ΔP/(8μL).
Step-by-Step Examples
#1: Microfluidic capillary (R = 2 mm, L = 0.25 m) carrying blood-mimicking fluid (μ = 0.0035 Pa·s) under ΔP = 5,000 Pa.
R = 0.002 mΔP = 5,000 Paμ = 0.0035 Pa·sL = 0.25 m
Q = π × (1.6×10⁻¹¹) × 5000 / (8 × 0.0035 × 0.25) = 2.513×10⁻⁷ / 0.007
→ Q = 3.59 × 10⁻⁵ m³/s (35.9 mL/s)
#2: Effect of a 20% arterial radius narrowing (R drops to 0.8R) at constant pressure drop.
R₂ / R₁ = 0.80
Q₂ / Q₁ = (0.80)⁴ = 0.4096
→ Flow rate drops by 59.0% (to 41.0% of baseline)
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Frequently Asked Questions
Why does radius R have such a dramatic effect in Poiseuille's law?+
Wider tubes both increase cross-sectional area (∝ R²) and increase peak centerline velocity away from wall shear (∝ R²), combining to scale flow as R⁴.
What is the ratio of maximum centerline speed to mean speed?+
In parabolic laminar pipe flow, v_max = 2 · v_mean exactly.
Can Poiseuille's law be used for turbulent water mains?+
No — once Re exceeds ~2,300, turbulent eddies flatten the velocity profile and flow scales roughly with ΔP^(0.5) instead of ΔP¹.