Navier-Stokes equations and exact solutions (Couette, Poiseuille)

Why it matters

The Navier–Stokes equations are the governing equations of viscous flow: every CFD code solves them, and boundary-layer theory, pipe-flow friction and lubrication theory all come from simplifying them. A few flows are simple enough to solve exactly. Couette and Poiseuille flows are the classic ones; they give the friction in bearings and seals, the laminar friction factor 64/Re, and benchmarks for checking numerical codes.

Key ideas

What the equations say. The Navier–Stokes (N–S) equations are Newton's second law for a fluid element with viscous stresses included: mass × acceleration (local plus convective) = pressure force + viscous force + body force. They are solved together with continuity (and, for compressible flow, the energy equation and an equation of state). For a Newtonian fluid the stress is linear in the strain rate (Stokes' hypothesis).

Incompressible, constant viscosity form. ρ DV/Dt = −∇p + μ∇²V + ρg, with ∇·V = 0. Dropping the viscous term gives Euler's equation. Four unknowns (u, v, w, p) and four equations.

Why they are hard. The convective term (V·∇)V is nonlinear; at high Reynolds number the solutions become unstable and turbulent. No general analytical solution exists. Exact solutions exist where the nonlinear term vanishes, typically fully developed parallel flows (velocity in one direction, varying only across the channel), so (V·∇)V = 0.

Boundary conditions. No-slip and no-penetration at solid walls (fluid velocity equals wall velocity); specified velocity or pressure at inlets and outlets; zero shear and constant pressure at a free surface.

Couette flow. Fluid between two parallel plates a gap h apart, the upper plate moving at U, no pressure gradient. The N–S x-equation reduces to d²u/dy² = 0, so u = U·y/h, a linear profile with uniform shear stress μU/h. Model for journal bearings with small clearance.

Generalised Couette flow. With a pressure gradient as well, d²u/dy² = (1/μ)dp/dx. The profile is the linear Couette part plus a parabola. An adverse gradient (dp/dx > 0) can produce backflow near the stationary wall; zero net flow occurs when dp/dx = 6μU/h².

Plane Poiseuille flow. Both plates fixed, gap b, flow driven by −dp/dx: u = (1/(2μ))(−dp/dx)·y(b − y), parabolic, u_max = 1.5 × mean.

Hagen–Poiseuille flow (pipe). Fully developed laminar flow in a pipe of radius R: u(r) = (1/(4μ))(−dp/dx)(R² − r²). u_max = 2 × mean velocity. Flow rate Q = πR⁴(−dp/dx)/(8μ) = πD⁴Δp/(128μL). Wall shear τ_w = 8μV̄/D, and the Darcy friction factor is f = 64/Re. Valid for Re below about 2300 and away from the entrance region.

Assumptions for all three. Steady, laminar, incompressible, Newtonian, fully developed (no change along the flow), gravity absorbed into the pressure or negligible.

Formulas

ρ·(∂V/∂t + (V·∇)V) = −∇p + μ·∇²V + ρ·g with ∇·V = 0

• ρ (kg/m³); V (m/s); p (Pa); μ (Pa·s); g (m/s²). Incompressible Newtonian fluid, constant μ.

u = U·y / h, τ = μ·U / h

• Couette flow; U: plate speed (m/s); h: gap (m); y measured from the fixed plate (m).

u = U·y/h + (1/(2μ))·(−dp/dx)·y·(h − y)

• Generalised Couette flow; dp/dx in Pa/m.

u = (1/(2μ))·(−dp/dx)·y·(b − y), u_max = b²·(−dp/dx)/(8μ), q = b³·(−dp/dx)/(12μ)

• Plane Poiseuille; b: gap (m); q: flow per unit width (m²/s).

u = (1/(4μ))·(−dp/dx)·(R² − r²), u_max = R²·(−dp/dx)/(4μ) = 2·V̄

• Hagen–Poiseuille; R: pipe radius (m); r: radial position (m).

Q = π·D⁴·Δp / (128·μ·L), τ_w = 8·μ·V̄ / D, f = 64 / Re

• Q (m³/s); D: diameter (m); Δp: pressure drop over length L (Pa); V̄: mean velocity (m/s).

Worked examples

Example 1 (standard): lubricating film. Given: μ = 0.1 Pa·s, plate speed U = 3 m/s, gap h = 0.6 mm.

1. τ = μ·U/h = 0.1 × 3/0.0006 = 500 Pa.
2. Velocity at mid-gap: u = U/2 = 1.5 m/s. Answer: τ = 500 Pa, uniform across the gap.

Example 2 (standard): laminar oil pipe. Given: oil μ = 0.3 Pa·s, ρ = 900 kg/m³, D = 20 mm, Q = 0.2 L/s.

1. V̄ = Q/A = 0.0002/(π × 0.02²/4) = 0.6366 m/s.
2. Re = ρV̄D/μ = 900 × 0.6366 × 0.02/0.3 = 38.2, laminar.
3. Δp/L = 128·μ·Q/(π·D⁴) = 128 × 0.3 × 0.0002/(π × 1.6×10⁻⁷) = 15 279 Pa/m.
4. τ_w = 8·μ·V̄/D = 8 × 0.3 × 0.6366/0.02 = 76.4 Pa. Check: τ_w = (Δp/L)·R/2 = 15 279 × 0.01/2 = 76.4 Pa.
5. u_max = 2V̄ = 1.27 m/s. Answer: Δp/L ≈ 15.3 kPa/m, τ_w ≈ 76.4 Pa.

Example 3 (GATE level): generalised Couette flow with zero net flow. Given: upper plate speed U = 2 m/s, gap h = 2 mm, μ = 0.05 Pa·s. Find dp/dx for zero net flow.

1. Flow per unit width: q = U·h/2 + h³·(−dp/dx)/(12μ).
2. Set q = 0: dp/dx = 6μU/h² = 6 × 0.05 × 2/(0.002²) = 150 000 Pa/m.
3. The gradient is adverse (pressure rises in the direction of plate motion); flow near the moving plate goes forward and flow near the fixed plate goes backward. Answer: dp/dx = 150 kPa/m.

Common mistakes

• Using u_max = 2V̄ for flow between plates. That is for a pipe; between plates u_max = 1.5V̄.
• Confusing radius and diameter in Q = πR⁴Δp/(8μL) versus πD⁴Δp/(128μL).
• Losing the sign of dp/dx: flow goes from high to low pressure, so −dp/dx is positive for forward flow.
• Applying Hagen–Poiseuille to turbulent flow (Re above about 2300) or to the entrance region.
• Saying the N–S equations express mass and energy conservation. They are the momentum equations; continuity and energy are separate.

For GATE AE

Expect: simplifying N–S for a given parallel flow and integrating with boundary conditions; Couette shear stress and velocity at a point; maximum, mean velocity and flow rate in pipe and channel Poiseuille flow; wall shear stress and f = 64/Re; and the condition for flow reversal in generalised Couette flow. Practise writing the reduced equation, integrating twice and applying both boundary conditions cleanly.

Quick check

1. In plane Poiseuille flow, what is u_max/V̄?
2. Couette flow, μ = 0.002 Pa·s, U = 2 m/s, h = 5 mm. Find τ.
3. What is the laminar Darcy friction factor at Re = 1600?
4. Which term of the N–S equations vanishes in fully developed parallel flow?

Answers: 1. 1.5 2. 0.8 Pa 3. 0.04 4. The convective (nonlinear) acceleration term