Laminar flow in pipes and Hagen-Poiseuille equation

Why it matters

Viscous liquids — lube oils, polymer solutions, glycerol, heavy crude, many process intermediates — usually flow laminarly in plant pipelines and in small-bore tubing, capillaries and viscometers. For laminar flow the pressure drop can be calculated exactly, with no empirical friction chart, and the same result underlies capillary viscometers and the Darcy and Kozeny–Carman equations for packed beds.

Key ideas

Reynolds number and the laminar regime. In a circular pipe, Re = ρVD/μ, with V the average velocity. Flow is laminar for Re below about 2100 (values of 2000–2300 are quoted in different texts), transitional up to about 4000, and turbulent beyond. Laminar flow can persist to higher Re in very smooth, disturbance-free pipes, but design uses about 2100.

Assumptions of Hagen–Poiseuille flow.

• Steady, incompressible, Newtonian fluid.
• Laminar and fully developed: the velocity profile no longer changes along the pipe, so only the axial velocity u(r) is non-zero.
• Straight pipe of constant circular cross-section, no slip at the wall.
• Gravity is not neglected — it is absorbed by using the piezometric pressure p + ρgz. For a horizontal pipe this is just p.

Derivation by a shell (force) balance. For a cylinder of fluid of radius r and length L, pressure force (Δp)πr² balances shear force τ(2πrL). So τ = (Δp/L)(r/2): shear stress varies linearly from zero at the centre to a maximum at the wall, τ_w = (Δp/L)(R/2). This linear τ distribution holds for laminar and turbulent flow, and for any fluid. Substituting Newton's law τ = −μ du/dr and integrating with u = 0 at r = R gives the parabolic profile.

Velocity profile results.

• u(r) = u_max[1 − (r/R)²], a paraboloid.
• u_max = 2V: the centreline velocity is twice the average.
• The local velocity equals the mean at r = R/√2 ≈ 0.707R.
• Kinetic-energy correction factor α = 2 and momentum correction factor β = 4/3.

Pressure drop and friction factor. Δp = 32μLV/D², so pressure drop is proportional to velocity (and flow rate) to the first power and inversely to D². Written in Darcy–Weisbach form, the Darcy friction factor is f = 64/Re (Fanning f = 16/Re, one quarter of the Darcy value). Laminar friction factor is independent of wall roughness.

Sensitivity to diameter. At constant Δp, Q ∝ D⁴: a 19 % increase in diameter doubles the flow; a small deposit or fouling layer cuts flow sharply.

Entrance length. The profile needs about L_e ≈ 0.05·Re·D to become fully developed; before this the pressure drop is higher than Hagen–Poiseuille predicts.

Non-Newtonian extension. For a power-law fluid the profile is flatter (n < 1) or more pointed (n > 1), and Re is replaced by a generalised (Metzner–Reed) Reynolds number; treated in rheology texts.

Formulas

• Re = ρ·V·D/μ — ρ (kg/m³), V average velocity (m/s), D diameter (m), μ (Pa·s). Laminar if Re < ≈2100.
• τ = (Δp/L)·(r/2), τ_w = Δp·D/(4L) — shear stress (Pa); valid for any fully developed pipe flow.
• u = u_max·[1 − (r/R)²], u_max = 2V — laminar profile (m/s).
• Δp = 32·μ·L·V/D² — Hagen–Poiseuille (Pa); L length (m).
• Q = π·Δp·D⁴/(128·μ·L) = π·Δp·R⁴/(8·μ·L) — flow rate (m³/s).
• f_D = 64/Re, f_F = 16/Re — Darcy and Fanning friction factors for laminar flow.
• h_L = f_D·(L/D)·V²/(2g) — head loss (m), consistent with the above.
• L_e ≈ 0.05·Re·D — laminar entrance length (m).

Worked examples

Example 1 (standard). Oil (ρ = 870 kg/m³, μ = 0.08 Pa·s) flows at 0.5 L/s through a horizontal 25 mm pipe, 30 m long. Find Re, the pressure drop, the friction factor and the pumping power.

1. A = (π/4)(0.025)² = 4.909 × 10⁻⁴ m²; V = Q/A = 0.0005/4.909 × 10⁻⁴ = 1.019 m/s.
2. Re = 870 × 1.019 × 0.025 / 0.08 = 277 → laminar.
3. Δp = 32·μ·L·V/D² = 32 × 0.08 × 30 × 1.019 / (0.025)² = 125 165 Pa ≈ 125.2 kPa.
4. f_D = 64/277 = 0.231. Check: f(L/D)(ρV²/2) = 0.231 × 1200 × 451.3 = 125.2 kPa ✓.
5. Power = Δp·Q = 125 165 × 0.0005 = 62.6 W. Re ≈ 277, Δp ≈ 125 kPa, f_D ≈ 0.231, P ≈ 62.6 W.

Example 2 (GATE level). A viscous liquid (ρ = 950 kg/m³, μ = 0.5 Pa·s) flows through a 50 mm horizontal pipe, 100 m long, under a pressure difference of 200 kPa. Find (a) the mean and maximum velocity, (b) the flow rate, (c) the wall shear stress, (d) the velocity at r = R/2, and (e) confirm the regime.

1. V = Δp·D²/(32·μ·L) = 200 000 × 0.0025 / (32 × 0.5 × 100) = 0.3125 m/s; u_max = 2V = 0.625 m/s.
2. Q = V·A = 0.3125 × 1.9635 × 10⁻³ = 6.14 × 10⁻⁴ m³/s.
3. τ_w = Δp·D/(4L) = 200 000 × 0.05 / 400 = 25 Pa.
4. u(R/2) = 0.625 × (1 − 0.25) = 0.469 m/s.
5. Re = 950 × 0.3125 × 0.05 / 0.5 = 29.7 → strongly laminar; entrance length 0.05 × 29.7 × 0.05 = 0.074 m, negligible compared with 100 m. (a) 0.3125 and 0.625 m/s, (b) 6.14 × 10⁻⁴ m³/s (0.614 L/s), (c) 25 Pa, (d) 0.469 m/s.

Common mistakes

• Using Hagen–Poiseuille without checking Re; many textbook-looking numbers give turbulent flow.
• Mixing radius and diameter forms: π·Δp·R⁴/(8μL) and π·Δp·D⁴/(128μL) are the same; mixing them is a factor-16 error.
• Confusing Darcy (64/Re) and Fanning (16/Re) friction factors.
• Taking u_max = 1.5V (that is the value for flow between parallel plates).
• Thinking gravity invalidates the result; use piezometric pressure for an inclined pipe.

For GATE CH

Expect numericals for Δp, Q or viscosity from Hagen–Poiseuille, ratios (what happens to Q when D or μ changes at fixed Δp), the location where u equals V, wall shear stress, kinetic-energy correction factor, and power required. Laminar flow between parallel plates and in an annulus are common variants. Always state the regime check in your working.

Quick check

1. What is u_max/V for laminar flow in a pipe?
2. At constant Δp, by what factor does Q change if D is doubled?
3. Laminar flow has Re = 800. What is the Darcy friction factor?
4. Where in the pipe is the shear stress zero?

Answers: 1. 2. 2. 16 times. 3. 64/800 = 0.08. 4. On the centreline.