Viscosity, Reynolds number, and Navier–Stokes equations

Start with a moving plate

A liquid fills a 2 mm gap between two flat plates. Hold the lower plate still and move the upper one right at 0.020 m/s. Assume the liquid follows each wall without slipping. What speed lies halfway across the gap, and what contact force must each plate exert on the liquid? If pressure also falls to the right, can the moving upper plate receive power from the liquid?

Choose x to the right and y from the lower plate upward. Write the fluid velocity as v = (u, v_y, v_z); our gap cases have u = u(y,t) and v_y = v_z = 0. The total gap is h = 0.002 m. Supply constant density ρ = 1000 kg/m³ and dynamic viscosity μ = 0.001 Pa·s. Density is mass per volume. Viscosity will relate a local deformation rate to stress; these assigned properties do not identify a measured fluid.

First follow the three gap profiles, their contact-force balance, the Reynolds scale table and the decaying profile. Vectors and vector spaces and Derivatives, Taylor series, and local approximation provide useful preparation. Gradient, divergence, curl, and vector calculus and Integrals and multivariable calculus are optional return paths for the final matrix and integral derivations. The essential terms and conditions are introduced here.

Connect viscosity to deformation rate

Neighbours at different heights can have different horizontal velocities. Their engineering shear rate is γ̇ = ∂u/∂y in this planar motion, with units (m/s)/m = s⁻¹. A Newtonian fluid has shear stress τ_xy = μγ̇. Stress is force per area, measured in pascals: 1 Pa = 1 N/m². Dynamic viscosity therefore has units Pa·s.

The general local deformation-rate tensor is D_ij = (∂v_i/∂x_j + ∂v_j/∂x_i)/2. Its diagonal terms describe extension rates and its off-diagonal terms describe shearing. Here D_xy = γ̇/2; engineering shear rate and tensor shear component differ by a factor of two. For our incompressible isotropic Newtonian model, total stress is σ_ij = −pδ_ij + 2μD_ij, where p is pressure and δ_ij is 1 for equal indices and 0 otherwise.

Viscosity responds to deformation rate, rather than to absolute velocity. Uniform translation has no velocity gradient and no viscous stress in this model. Rigid rotation has a skew velocity gradient but D = 0, so its viscous stress is also zero. Pressure can still be present; this statement concerns only the viscous part of stress.

Kinematic viscosity is ν = μ/ρ, with units m²/s. For the supplied gap, ν = 0.000001 m²/s. It sets a momentum-diffusion timescale when the constant-property momentum equation takes a diffusion form. μ converts deformation rate into stress; ν compares that stress response with mass density. Interchanging their units would change the equation.

Balance mass and local momentum

A velocity field must conserve mass. The local equation is ∂ρ/∂t + ∇·(ρv) = 0. Here ∇· means divergence, the net outward flux per local volume. With uniform constant density, it reduces to ∇·v = 0: the motion does not change a material volume. Our planar gap fields satisfy it because ∂u/∂x = 0 and the other velocity components vanish.

Newton’s law for a moving fluid parcel is ρ Dv/Dt = ∇·σ + ρb. The body-force acceleration b is force per mass, such as gravity; it is zero in our supplied gap cases. The divergence of stress gives net contact force per volume. A large stress alone does not determine acceleration: stress on opposite faces can cancel, while a spatial stress gradient leaves a net force.

For constant μ and ρ in the stated incompressible Newtonian model, this becomes ρ[∂v/∂t + (v·∇)v] = −∇p + μ∇²v + ρb. These are the incompressible Navier–Stokes equations together with ∇·v = 0. The Laplacian ∇² adds second spatial derivatives. Every momentum term is force per volume, N/m³, equivalently Pa/m.

The material acceleration Dv/Dt follows a parcel. Its first part, ∂v/∂t, measures change at a fixed position; its second, (v·∇)v, measures change encountered as a parcel moves through a spatially varying field. A steady flow can still accelerate along a curved or changing-speed path. In these fully developed gap fields the advective part is exactly zero: u has no x dependence, and there is no velocity across y. That exact cancellation does not require small Reynolds number.

Compare wall and pressure drives

Use infinite parallel plates with no slip and no penetration, and assume a steady fully developed profile u(y). Define G = −∂p/∂x: positive G means pressure falls to the right and pushes the fluid right. Let upper-wall speed be U and lower-wall speed zero. There is no body force and no pressure gradient across the gap. The streamwise balance reduces to 0 = G + μ d²u/dy².

Integrate this equation twice and impose u(0) = 0 and u(h) = U. The result is u(y) = Uy/h + G y(h−y)/(2μ). The first term is the moving-wall contribution, called plane Couette flow. The second is the pressure contribution, called plane Poiseuille flow. They add here because the reduced equation is linear with constant μ and the same wall geometry.

Compare U = 0.020 m/s, G = 0; U = 0, G = 20 Pa/m; and U = 0.020 m/s, G = 20 Pa/m. The moving-wall profile is linear. The pressure-only profile is parabolic and has its maximum at the centre. The combined profile reaches 0.0225 m/s at y = 1.5 mm, faster than the upper wall. No slip fixes velocity at the walls, not an upper bound throughout the fluid.

Select a bookkeeping strip of width w = 0.100 m and streamwise length ℓ = 0.200 m in this infinite-plate model. These dimensions supply area and total-flow factors; they do not introduce physical sidewalls, an entrance region or a stability prediction. The mean is ū = U/2 + Gh²/(12μ), and volume flow is Q = whū. The three means are 0.010000, 0.006667 and 0.016667 m/s, rounded to six decimals.

Read contact forces with the outward normal

Define t_i = σ_ij n_j as contact traction exerted on the fluid at its outward unit normal n. The first index gives the force component and the second selects the normal component. On the upper wall n = +e_y, so t_x = τ_xy(h). On the lower wall n = −e_y, so t_x = −τ_xy(0). Fluid forces on the walls are the opposites of these contact forces.

Differentiate the profile: τ_xy(y) = μU/h + G(h/2−y). In the moving-wall-only case, τ_xy = 0.010 Pa throughout the gap. Its gradient is zero, so the two wall forces cancel and the fluid has zero acceleration despite nonzero shear stress. In the pressure-only case, τ_xy falls from +0.020 Pa to −0.020 Pa. The viscous force density dτ_xy/dy = −G balances the pressure force density +G.

Across the strip, pressure drop is Δp = Gℓ and net pressure force is Δp wh. Multiply upper and lower tractions by wall area wℓ. In each steady case the total streamwise force is zero. The profile is identical at both bookkeeping ends, so their momentum fluxes are equal; a through-flow does not require a nonzero net force on the strip.

For the combined drives, pressure supplies +0.8 mN while the lower and upper walls supply −0.6 and −0.2 mN. The upper wall moves right, but the fluid immediately below it moves faster. Its contact force on the fluid therefore points left. Read the normal, gradient and relative neighbouring motion together, rather than assigning a force direction from wall speed alone.

Follow signed power and viscous dissipation

Mechanical power supplied by a wall is its contact force on the fluid dotted with wall velocity. A stationary wall can exert force while supplying zero mechanical power. For the moving-wall-only strip, the upper wall supplies (0.0002 N)(0.020 m/s) = 0.000004 W = 4 μW. The lower wall supplies zero.

Pressure supplies Δp Q. In the combined case, Δp = 4 Pa and Q = 0.000003333333… m³/s, so pressure power is 13.333333… μW. Upper-wall power is (−0.0002 N)(0.020 m/s) = −4 μW: the wall receives mechanical power. Net supplied mechanical power is 9.333333… μW.

Positive viscosity dissipates mechanical energy into internal energy. For our planar field, the local dissipation rate per volume is Φ = μ(du/dy)² ≥ 0, with units W/m³. Integrating over the strip gives exactly the net supplied drive power: 4, 5.333333… and 9.333333… μW for the three cases. The velocity profile is unchanged along x, so net kinetic-energy flux through the bookkeeping ends is zero.

This is a mechanical energy account. Determining temperature would also require an energy equation, thermal properties and heat-transfer conditions. Negative power from one wall does not mean negative viscous dissipation; pressure can supply enough power both to the wall and to internal-energy production.

Choose scales before reading Reynolds number

Choose a positive characteristic speed U and length L for a particular comparison. Reynolds number is Re = ρUL/μ = UL/ν, a dimensionless ratio of advective-inertial and viscous force-density scales. Those scales are ρU²/L and μU/L². State the choices: using top-wall speed and total gap is different from using mean speed and a hydraulic length.

Two associated times are t_adv = L/U and t_diff = L²/ν. Their ratio t_diff/t_adv equals Re. A larger ν smooths velocity differences faster over the same distance. All three supplied comparisons have t_adv = 0.10 s, while their diffusion times differ by four orders of magnitude. The table compares characteristic scales, not measured transitions.

Small Re suggests weak advection relative to viscosity under those scales, but rapid externally imposed time variation can still require ∂v/∂t. Large Re can make viscosity small in much of a flow while thin wall regions retain strong gradients. Geometry, disturbances, boundary conditions and other forcing scales matter. No universal Reynolds threshold decides whether every flow is laminar, steady or turbulent.

Laminar motion can be time dependent. Navier–Stokes governs the instantaneous velocity in both smooth laminar motion and turbulence within its continuum and constitutive assumptions; averaging turbulent motion introduces additional correlations that need their own treatment. Boundary layers, lift, and drag, Fluid instabilities and convection, and Turbulence develop those mechanisms separately.

Watch momentum diffuse in time

Now keep both plates stationary, set G = 0 and prescribe the initial profile u(y,0) = U₀ sin(πy/h), with U₀ = 0.020 m/s and the same h, ρ and μ. The sine profile vanishes at both walls and is smooth. No new drive is imposed. This is a separate unsteady initial-value problem, not the steady moving-wall profile after an unspecified stopping event.

Advection is still exactly zero, but local acceleration is not. The momentum equation becomes ∂u/∂t = ν ∂²u/∂y². Velocity differences diffuse across the gap. Since the second derivative of sin(πy/h) is −(π/h)² times the same sine, u(y,t) = U₀ sin(πy/h) exp(−νπ²t/h²) satisfies the equation and both wall values.

The centre amplitude is A(t) = U₀ exp(−t/τ), where τ = h²/(π²ν) = 4/π² s ≈ 0.405285 s. This e-folding time is derived from the supplied mode, not an arbitrary damping constant. Momentum and mean velocity have ratio exp(−t/τ); kinetic energy, proportional to velocity squared, has ratio exp(−2t/τ). The mean speed is 2A(t)/π.

At the lower and upper walls, τ_xy is respectively +μπA/h and −μπA/h. Their outward normals make both wall contact forces on the fluid point left while the fluid moves right. Both stationary walls supply zero mechanical power. Integrated viscous dissipation accounts for the loss of kinetic energy. The figure compares actual profile snapshots and velocity/energy ratios with their distinct meanings.

Decide which model is closed

A differential equation needs a domain, initial data when time dependent, and suitable boundary and driving conditions. No slip means fluid velocity matches the contacting wall; no penetration matches its normal motion. These are the assumed continuum wall conditions here, not universal rules for every rarefied gas or microscopic interface. Our pressure gradient is supplied; its constant level is not determined by these velocity balances.

Constant μ and ρ make the displayed equations simple. Temperature-dependent viscosity requires retaining its spatial variation inside the stress divergence. Compressible flow requires variable-density mass balance and a constitutive law including volumetric response, together with thermodynamic closure. A non-Newtonian material needs an appropriate stress law instead of simply inserting one fixed μ; Rheology and non-Newtonian matter provides a published comparison.

The parallel-plate solutions are useful local models for sheared gaps, pressure-driven layers and combinations of both drives. They do not supply entry-flow corrections, sidewall effects, instability onset, thermal limits or safe loads for a device. An inviscid Euler approximation can be useful where viscous terms are negligible under the chosen conditions, but it cannot by itself impose a viscous no-slip layer at a wall. Potential flow and boundary-layer models address different parts of that approximation.

Optional return: stress, scales and energy

Return after the profiles and balances for the index and scaling steps. With (grad v)_ij = ∂v_i/∂x_j, the constant-μ stress divergence is ∂σ_ij/∂x_j = −∂p/∂x_i + μ∂²v_i/∂x_j² + μ∂(∇·v)/∂x_i, summing repeated j. Incompressibility removes the last term. This is why μ∇²v is valid here; it is not a general substitution for variable viscosity.

For U > 0 and L > 0, choose x = Lx*, t = (L/U)t*, v = Uv*, pressure variation p = ρU²p* and b = (U²/L)b*. Divide momentum balance by ρU²/L to obtain ∂v*/∂t* + (v*·∇*)v* = −∇*p* + (1/Re)∇*²v* + b*, with ∇*·v* = 0. A different pressure or time scale moves the coefficients; choosing a shorter imposed forcing time introduces another dimensionless comparison.

For incompressible Newtonian deformation, Φ = 2μD:D, where D:D means ∑_ij D_ij². In planar shear, the two equal off-diagonal entries are u′/2, so Φ = μ(u′)². Integrating u′ = U/h + G(h−2y)/(2μ) gives strip dissipation μwℓ[U²/h + G²h³/(12μ²)]. The cross term integrates to zero. Pressure and wall powers sum to this value even when the wall contribution is negative.

For the decaying sine mode, kinetic energy per wall area is E = (ρ/2)∫₀ʰ u²dy = ρhA²/4. Its derivative is dE/dt = −2(νπ²/h²)E. Integrating Φ gives μA²π²/(2h), the positive magnitude of this energy loss. Multiplying the diffusion equation by u and integrating by parts yields the same relation; the boundary power term vanishes because u = 0 at both walls.

Worked examples

Try it yourself

1. In the pressure-only case, double μ to 0.002 Pa·s while keeping ρ, h, G, strip dimensions and stationary walls unchanged. Find centre and mean speeds, Q, both wall forces, dissipation and Re using U = mean speed and L = h.

   Show the answer and reasoning

   Centre speed is 0.005 m/s, mean speed 0.003333333… m/s and Q = 0.000000666666… m³/s. Both wall forces on the fluid remain −0.4 mN. Dissipation is 2.666666… μW and Re = 10/3 ≈ 3.333333.

   At fixed G and gap, the pressure profile and Q scale as 1/μ, while wall stresses ±Gh/2 are fixed by force balance. Pressure power Δp Q therefore halves. Re falls by a factor of four: the chosen mean speed halves while μ doubles.

2. Keep the moving-wall-only speed 0.020 m/s, μ = 0.001 Pa·s and ρ = 1000 kg/m³, but halve the gap to 0.001 m. Find γ̇, D_xy, shear stress and Re with U = top-wall speed and L = new gap.

   Show the answer and reasoning

   γ̇ = 20 s⁻¹, D_xy = 10 s⁻¹, τ_xy = 0.020 Pa and Re = 20.

   The same speed change across half the distance doubles the gradient and stress. Re is proportional to the selected length and therefore halves from 40 to 20. Larger wall stress and smaller Re are consistent because they measure different comparisons.

3. For the supplied sine-mode transient, compare velocity amplitude and kinetic-energy ratios after one and two e-folding times. Is the energy e-folding time the same as the velocity e-folding time?

   Show the answer and reasoning

   At t = τ, ratios are e⁻¹ ≈ 0.367879 for velocity and e⁻² ≈ 0.135335 for energy. At t = 2τ, they are e⁻² ≈ 0.135335 and e⁻⁴ ≈ 0.018316. Energy e-folding time is τ/2.

   The whole profile scales with A, so momentum is proportional to A and kinetic energy to A². Squaring exp(−t/τ) doubles the exponent. These exact mode ratios do not define a decay law for an arbitrary initial flow.

4. Prescribe local rigid motion v = (−Ωy, Ωx, 0), Ω = 3 s⁻¹, in the same incompressible Newtonian stress model. Find D and viscous stress. Compare with ω_z = ∂v_y/∂x − ∂u/∂y; no pressure or boundary dynamics is supplied.

   Show the answer and reasoning

   D = 0 and 2μD = 0, while ω_z = 6 s⁻¹.

   The off-diagonal velocity derivatives are −3 and +3 s⁻¹. Their symmetric sum is zero, while curl subtracts the negative derivative. Rotation without local deformation produces no viscous stress in this model. The result does not specify the pressure needed for a complete rotating-flow solution.

5. Check units of ρ∂v/∂t, ∇p, μ∇²v and ρb. A proposed equation replaces μ with ν while leaving all other terms unchanged; is it dimensionally consistent?

   Show the answer and reasoning

   All four original terms have units kg/(m²·s²) = N/m³ = Pa/m. Replacing μ by ν in that force-density equation is inconsistent.

   Velocity acceleration has units m/s², pressure gradient Pa/m, and velocity Laplacian 1/(m·s). Multiplication by ρ, μ and ρ respectively gives the same force-density unit. ν has units m²/s; it belongs after dividing the constant-density equation by ρ, giving acceleration terms, not unchanged force-density terms.

What to carry forward

• Newtonian viscous stress responds to local deformation rate: σ = −pI + 2μD. Engineering shear rate is twice D_xy in the stated planar flow.
• Dynamic viscosity μ has units Pa·s; kinematic viscosity ν = μ/ρ has units m²/s. A stress gradient, rather than stress alone, supplies net contact force density.
• Navier–Stokes combines material momentum balance with the Newtonian stress law. Incompressibility, material properties, initial data and boundary conditions must be specified.
• Actual wall traction on the fluid is σn. A moving wall can supply negative mechanical power while total drive power equals positive viscous dissipation.
• Reynolds number needs explicit U and L. It compares chosen inertial and viscous scales without supplying a universal transition or steadiness criterion.
• Unsteady momentum diffusion can persist with exactly zero advection. A supplied sine mode fixes its own decay time; velocity and kinetic energy decay at different rates.