Hagen–Poiseuille Discharge

Q = \frac{π r ^{4} Δ p}{8 μL}

Exact laminar, fully developed, Newtonian flow in a circular tube. Q is extremely sensitive to radius (∝ r⁴). Mean velocity is V = Δp r² / (8μL) and the centerline is 2V. Invalid once the flow is transitional or turbulent.

$Q$ (m³/s): Discharge
$r$ (m): Pipe inner radius
$Δ p$ (Pa): Pressure drop over L
$μ$ (Pa·s): Dynamic viscosity
$L$ (m): Length

Step-by-step solved example

r = 2.0 mm, Δp = 20 kPa, μ = 0.0010 Pa·s, L = 4.0 m. Find Q.

Laminar pipe: Δp = 32μLV/D², parabolic profile.

1. r⁴
r = 0.0020 m → r⁴ = 1.60×10⁻¹¹ m⁴.
2. Discharge
Q = π(1.60e−11)(20000) / (8×0.0010×4.0) = 3.14×10⁻⁵ m³/s.
$Q = 3.14 \times 1 0^{- 5} m^{3} /s$

Answer: 3.14×10⁻⁵ m³/s

10 practice questions

0/10 correct

1.Hagen–Poiseuille Q is proportional to

2.Doubling r (same Δp, L, μ) multiplies Q by

3.The law assumes

4.Δp is proportional to

5.Mean velocity V equals

6.Laminar-pipe centerline speed is

7.Increasing μ

8.Units of r⁴ Δp / (μ L) are

9.r = 1.0 mm, Δp = 8.0 kPa, μ = 0.0010 Pa·s, L = 1.0 m. Q ≈

10.Hagen–Poiseuille is invalid when