Viscous flow between parallel plates and in pipes

Why it matters

Laminar viscous flow is the one flow regime engineers can solve exactly, and it is everywhere in fluid power and machinery: oil films in bearings, leakage past pistons and spool valves, flow in capillary tubes and small hydraulic lines, lubrication of slideways and the viscometers used to measure oil viscosity. The exact solutions also give the friction factor f = 64/Re and are the reference against which turbulent flow is compared.

Key ideas

Laminar versus turbulent. In laminar flow, fluid moves in smooth layers and momentum is exchanged only by molecular viscosity. In a circular pipe the flow is laminar when Re = ρVD/μ is below about 2000 (transition up to about 4000). Between plates, using the gap as length scale, laminar flow persists to roughly Re ≈ 1000–1400. Leakage paths in hydraulic components have tiny gaps, so their flow is almost always laminar.

Fully developed flow. Near a pipe entry the boundary layers grow until they meet at the centre; after this entrance length the velocity profile stops changing along the pipe. For laminar flow, L_e ≈ 0.06·Re·D. The exact solutions below apply only to fully developed, steady, incompressible laminar flow.

Force balance. In fully developed flow the fluid does not accelerate, so the net pressure force on a fluid element balances the net viscous shear force. Combined with Newton's law of viscosity, this gives a second-order equation for the velocity profile, solved with no-slip at the walls.

Couette flow (one plate moving, no pressure gradient). The velocity rises linearly from 0 at the fixed plate to U at the moving plate; the shear stress is uniform, τ = μU/h. This models a lightly loaded journal bearing.

Plane Poiseuille flow (both plates fixed, pressure-driven). The velocity profile is a parabola, maximum at mid-gap. The mean velocity is two-thirds of the maximum. Shear stress varies linearly from zero at the centre to a maximum at the walls. Flow rate per unit width is proportional to h³ — so leakage past a spool or piston rises with the cube of the clearance. Doubling the clearance increases leakage eightfold, which is why fluid-power components are built with clearances of a few micrometres and why wear quickly reduces volumetric efficiency.

Combined Couette–Poiseuille flow. When one plate moves and a pressure gradient also acts, the two solutions add (the equation is linear). With an adverse pressure gradient, back-flow can appear near the fixed plate.

Hagen–Poiseuille flow (circular pipe). The profile is a paraboloid with maximum at the axis equal to twice the mean velocity. Shear stress varies linearly from zero at the axis to τ_w at the wall. The pressure drop is proportional to the mean velocity (and to Q), to viscosity and to length, and inversely to D² (or D⁴ at a given Q). Expressed in Darcy form, the friction factor is f = 64/Re, independent of wall roughness.

Correction factors. Because the laminar profile is far from uniform, the kinetic-energy correction factor is α = 2 and the momentum correction factor is β = 4/3 for pipes (α = 1.543, β = 1.2 between plates).

Power and viscometry. The power to push fluid through is Δp·Q. Measuring Q and Δp in a capillary tube of known size gives μ (capillary-tube viscometer); other viscometers use falling spheres (Stokes' law) or rotating cylinders.

Formulas

u(y) = (U/h)·y (Couette)

• u = velocity at distance y from fixed plate (m/s), U = moving plate speed (m/s), h = gap (m).

u(y) = (1/(2μ))·(−dp/dx)·(h·y − y²) (fixed parallel plates)

• −dp/dx = pressure drop per unit length (Pa/m), μ = dynamic viscosity (Pa·s), y measured from one plate (m).

u_max = (−dp/dx)·h² / (8μ), V_mean = (2/3)·u_max

q = (−dp/dx)·h³ / (12μ) or Q = b·h³·Δp / (12·μ·L)

• q = flow rate per unit width (m²/s), Q = flow rate (m³/s), b = width (m), L = length of gap (m). For a concentric annular clearance with small radial gap h, use b = πD.

τ_w = (−dp/dx)·h/2 (plates), τ_w = (−dp/dx)·D/4 = Δp·D/(4L) (pipe)

• τ_w = wall shear stress (Pa).

u(r) = (−dp/dx)·(R² − r²) / (4μ), u_max = 2·V_mean (pipe)

• R = pipe radius (m), r = radial position (m).

Δp = 32·μ·V·L / D² = 128·μ·Q·L / (π·D⁴) (Hagen–Poiseuille)

• Δp = pressure drop (Pa), V = mean velocity (m/s), L = length (m), D = diameter (m).

h_f = f·L·V² / (2g·D) with f = 64 / Re (laminar pipe)

• h_f = head loss (m), f = Darcy friction factor (–).

L_e ≈ 0.06·Re·D (laminar entrance length)

Worked examples

Example 1 — oil pipeline (standard). Oil (ρ = 900 kg/m³, μ = 0.1 Pa·s) flows at 2 L/s through a 50 mm pipe 100 m long. Find the pressure drop, head loss, pumping power and wall shear stress.

1. Mean velocity: V = Q/A = 0.002/(π × 0.05²/4) = 1.019 m/s.
2. Re = ρVD/μ = 900 × 1.019 × 0.05/0.1 = 458 — laminar.
3. Δp = 32μVL/D² = 32 × 0.1 × 1.019 × 100/0.0025 = 130.4 kPa.
4. Head loss: h_f = Δp/(ρg) = 130 380/(900 × 9.81) = 14.77 m. Check: f = 64/458 = 0.1396 and fLV²/(2gD) = 14.77 m.
5. Power: P = Δp·Q = 130 380 × 0.002 = 260.8 W.
6. τ_w = Δp·D/(4L) = 130 380 × 0.05/400 = 16.3 Pa. Answer: Δp ≈ 130 kPa, h_f ≈ 14.8 m, P ≈ 261 W, τ_w ≈ 16.3 Pa.

Example 2 — leakage past a hydraulic piston (GATE level). A 50 mm piston with a 40 mm long land runs concentrically in its bore with a radial clearance of 20 μm. The pressure difference across it is 10 MPa and the oil viscosity is 0.03 Pa·s (ρ = 870 kg/m³). Find the leakage flow and check the regime.

1. Because h ≪ D, unroll the annulus into a slot of width b = πD = 0.1571 m.
2. Q = b·h³·Δp/(12·μ·L) = 0.1571 × (2 × 10⁻⁵)³ × 10⁷ /(12 × 0.03 × 0.04).
3. Numerator: 0.1571 × 8 × 10⁻¹⁵ × 10⁷ = 1.257 × 10⁻⁸; denominator: 0.0144. Q = 8.73 × 10⁻⁷ m³/s.
4. In L/min: 8.73 × 10⁻⁷ × 60 000 = 0.0524 L/min.
5. Mean velocity in the gap: Q/(b·h) = 8.73 × 10⁻⁷/(0.1571 × 2 × 10⁻⁵) = 0.278 m/s; Re = ρVh/μ = 870 × 0.278 × 2 × 10⁻⁵/0.03 = 0.16, deeply laminar. Answer: leakage ≈ 8.7 × 10⁻⁷ m³/s (about 0.052 L/min). A 40 μm clearance would leak eight times as much.

Common mistakes

• Using f = 64/Re without first checking that Re < 2000.
• Mixing radius and diameter in the Hagen–Poiseuille formula (128 uses D⁴; 8 uses R⁴).
• Using h/2 where the formula expects the full gap h, or the reverse — check whether y is measured from a wall or from the centreline.
• Forgetting that leakage scales with h³, so small clearance changes have large effects.
• Taking u_max = 1.5·V_mean for a pipe (that is the plate value; for a pipe it is 2).
• Using the Darcy f from a Moody chart with the Fanning f (which is f/4) in the same formula.

For GATE ME

Expect numericals on pressure drop and power in laminar pipe flow, the ratio of maximum to mean velocity, the radius at which local velocity equals the mean (r = R/√2 for a pipe), shear stress distribution, flow between fixed or moving plates, and leakage through small clearances. Conceptual questions test the dependence of Δp on D at constant Q (D⁻⁴), the value f = 64/Re and correction factors α and β. Practise deriving the plate and pipe profiles from a force balance.

Quick check

1. What is the ratio of maximum to mean velocity in laminar pipe flow, and between fixed parallel plates?
2. At fixed flow rate, how does laminar pressure drop change if the pipe diameter is halved?
3. What is the Darcy friction factor at Re = 1600?
4. How does leakage past a spool change if the radial clearance doubles?
5. Where is the shear stress zero in fully developed pipe flow?

Answers: 1. 2 and 1.5; 2. it rises 16 times; 3. 0.04; 4. it increases 8 times; 5. on the pipe axis.