Turbulence

Start with a fluctuating water velocity

Imagine placing a small velocity sensor in moving water. Its downstream readings fluctuate around 2 m/s, while the water also moves sideways. A mean of 2 m/s describes the average downstream motion at that point, but it cannot tell us whether fast-moving fluid is carrying momentum toward a wall or away from it. What information disappears when we keep only the average?

We will first compare four assigned velocity states and calculate their momentum transport. Then we will follow kinetic energy from a maintained large-scale motion to small-scale viscous dissipation. The covariance comparison, energy-path diagram and scale/spectrum graph form a first-reading path. The indexed equations and third-order law are an optional return path after the physical account.

The assigned states below are an arithmetic model, not measurements of a particular water flow. Four changing readings would not establish turbulence: a vibrating pump, a regular wave or instrument noise could also produce a changing signal. Turbulence involves irregular interacting motion over a range of scales; its interpretation needs the flow, the measurements and their spatial and temporal resolution.

Published preparation includes Vectors and vector spaces, Derivatives, Taylor series, and local approximation, Conservation of mass, momentum, and energy, and Vorticity and circulation. The neighboring Viscosity, Reynolds number, and Navier–Stokes equations develops the instantaneous viscous equations; Boundary layers, lift, and drag treats wall forces, while Fluid instabilities and convection treats onset. Their essential terms are defined here, so a reader can follow the first path without completing every derivation.

Identify the motion and the regime

A steady laminar flow has an orderly velocity field that does not change in time. Laminar flow can also be unsteady: an oscillating but orderly viscous flow remains a counterexample to ‘changing means turbulent’. A single rotating fluid region is not enough either. Vorticity measures local fluid rotation; interacting, stretching and deforming vortical structures are part of three-dimensional turbulence, but vorticity can exist in a laminar flow.

For a speed scale U, length scale L and kinematic viscosity ν, the Reynolds number is Re = UL/ν. Kinematic viscosity has units m²/s and equals dynamic viscosity divided by density: ν = μ/ρ. A large Re indicates that advection can compete strongly with viscous diffusion on those selected scales. It does not eliminate viscosity from the smallest scales.

There is no transition number that applies to every geometry and disturbance. Wall roughness, incoming disturbances, boundary conditions and the chosen speed and length all matter. Instability describes growth of disturbances to a specified base state; its initial growth is different from a sustained nonlinear turbulent regime. The preceding instability lesson ends at that distinction.

Here the operative continuum model is a constant-density, incompressible Newtonian fluid with constant ν. Density-changing, shock-bearing, rotating, stratified and magnetised regimes may have additional energy pathways and scaling constraints. We will identify those boundaries rather than transfer one three-dimensional model to every fluid.

Separate a mean from fluctuations

Write the instantaneous velocity as u = U + u′, where U is the chosen average and u′ is the departure from it. The prime means ‘fluctuation’, not a derivative. At one location the components are downstream u, transverse v and spanwise w; their means are U, V and W.

For the four equally weighted states in case A, the mean is (2, 0, 0) m/s. Subtract that vector from every state. The downstream departures are +0.2, −0.2, +0.2 and −0.2 m/s, so their mean is zero. Their squares are positive: the mean square is 0.040 m²/s² and the root-mean-square departure is 0.200 m/s.

Streamwise intensity here is I_u = √⟨u′²⟩/|U| = 0.200/2.000 = 0.100, or 10%. Angle brackets mean the stated average. Intensity is dimensionless; its definition needs a component and reference speed. This version is undefined when U = 0 even if the fluctuations have a perfectly meaningful finite strength.

In experiments, an ensemble average compares repeated realisations; a time average uses a record at one location. Equating them requires appropriate statistical assumptions. A finite record gives an estimate, and slow drift, insufficient duration or a probe that misses small scales changes what it can support. The exact arithmetic of our four assigned states carries no claim that four observations suffice to characterise a real turbulent flow.

Look for correlated momentum transport

The product u′v′ combines a downstream momentum departure with cross-stream motion. In case A, faster-than-average water has v′ < 0 and slower-than-average water has v′ > 0. Both combinations give u′v′ < 0. Their average is −0.020 m²/s² even though ⟨u′⟩ = ⟨v′⟩ = 0.

Across a plane with normal +y, the fluctuating advective flux of x momentum is ρ⟨u′v′⟩. With assigned density ρ = 1000 kg/m³, it is −20 Pa: net x-momentum transport in the −y direction. The averaged momentum equation can move this flux to the stress side and define the turbulent stress contribution as τᵀ_xy = −ρ⟨u′v′⟩ = +20 Pa. The minus sign changes the accounting convention, not the physical transport.

Now make case B by changing only the transverse states to −0.1, +0.1, +0.1 and −0.1 m/s. The downstream and transverse means and variances stay the same. The four u′v′ values become −0.020, −0.020, +0.020 and +0.020 m²/s², whose average is zero. Equal intensity does not guarantee equal momentum transport.

This covariance is the missing information. Define R_ij = ⟨u′_i u′_j⟩, with units m²/s². Its diagonal entries are component variances; off-diagonal entries describe correlations. We use τᵀ_ij = −ρR_ij throughout. Other references call R_ij or +ρR_ij ‘Reynolds stress’, so inspect the sign and density convention before comparing a formula. This is transport by moving fluid, not a new molecular contact force.

Read the averaged momentum equation

For a constant-density incompressible fluid, Reynolds averaging gives ∂U_i/∂t + U_j∂U_i/∂x_j = −(1/ρ)∂P/∂x_i + ν∇²U_i − ∂R_ij/∂x_j + F_i, with ∂U_j/∂x_j = 0. P is mean pressure and F is mean body force per mass. A repeated index is summed over the three spatial directions. The last correlation term is the divergence of the momentum flux we just calculated.

This equation assumes an averaging operation that is linear, removes the mean fluctuation and commutes with the derivatives being used. An ensemble average with adequate regularity provides the clean formal setting. Applying the same algebra exactly to an arbitrary short sliding window is not automatic; the mean and its departure need not obey all Reynolds averaging rules for that filter.

Averaging has not removed nonlinear effects. The averaged product satisfies ⟨u_i u_j⟩ = U_iU_j + R_ij, not merely U_iU_j. Knowing the mean velocity therefore leaves unknown correlations. An evolution equation for R_ij introduces further correlations, including third-order velocity products and pressure–velocity terms. The need to supply additional relations is the closure problem.

A constant R_xy at one location does not specify a mean acceleration. The equation needs its spatial gradient and the other terms. Our two cases expose what mean speed misses; they are not complete spatial solutions for a pipe, jet or wall layer.

Ask what an eddy-viscosity model assumes

A common closure represents part of the turbulent stress as proportional to the mean strain rate. In this incompressible convention, τᵀ_ij ≈ 2ρν_t S_ij − (2/3)ρKδ_ij, where S_ij = (∂U_i/∂x_j + ∂U_j/∂x_i)/2 and K = ⟨u′_i u′_i⟩/2. The symbol δ_ij is one when i = j and zero otherwise. K is turbulent kinetic energy per mass, measured in m²/s²; ν_t is a modelled eddy viscosity in m²/s.

For a local parallel shear U = U(y), this gives −⟨u′v′⟩ ≈ ν_t dU/dy. Supply dU/dy = 10 s⁻¹ and the case A covariance. Matching this one component gives ν_t = 0.002 m²/s. With ν = 0.000001 m²/s, that diagnostic ratio is 2000. At density 1000 kg/m³ the molecular shear is μ dU/dy = 0.010 Pa, compared with the assigned 20 Pa turbulent stress contribution.

Eddy viscosity is not a property of the water alone. It summarises a specified flow and model, can vary in space and time, and is not a replacement for molecular viscosity in the small-scale heating law. Matching one shear component also fails to validate the full stress tensor: our assigned component variances are unequal, whereas this simple parallel-shear closure gives equal diagonal R_ii = 2K/3.

Linear eddy-viscosity models can be useful for suitable mean-flow predictions, but strong anisotropy, rotation, curvature, separation or rapid changes can challenge the alignment and local-response assumptions. Reynolds-stress models transport the separate covariance components and model other terms instead. A more elaborate closure still requires verification and validation for the intended conditions.

Track turbulent kinetic energy

For both finite-state cases, K = (0.0400 + 0.0100 + 0.0025)/2 = 0.02625 m²/s². This measures departures from the chosen mean, not total kinetic energy. The mean motion itself has energy |U|²/2 = 2 m²/s². Changing the averaging convention can change the division between those parts.

Shear can transfer energy from the mean motion into fluctuations. The production rate per mass is P_K = −R_ij∂U_i/∂x_j. For the assigned local gradient and case A, P_K = −(−0.020)(10) = 0.200 m²/s³. Positive production is a transfer between mean and fluctuating kinetic energy, not creation of total energy. The corresponding term has the opposite sign in the mean kinetic-energy balance.

Let ε be mean viscous dissipation of the fluctuating motion per mass. For this incompressible Newtonian model, ε = 2ν⟨s′_ij s′_ij⟩ ≥ 0, with s′ the fluctuating strain-rate tensor. Its units are m²/s³, equivalently W/kg. Viscosity converts kinetic energy into internal energy. Molecular viscosity remains essential even when transport by fluctuations is much larger than molecular momentum transport at a large scale.

A local energy balance also includes transport by fluctuating motion, pressure and viscosity. Schematically D K/Dt = P_K − ε − ∇·T for the conditions detailed in the return path. If, as a separate diagnostic assumption, transport is zero and ε is assigned 0.100 m²/s³, the local material rate is +0.100 m²/s³. We must not call this case stationary: production exceeds dissipation. A real stationary shear flow can instead balance their difference through energy transport.

Follow a three-dimensional energy cascade

A pump, moving boundary, buoyant forcing or another source can maintain large-scale fluid motion. Turbulence can also decay after its driving is removed: ongoing forcing is needed for a sustained stationary energy balance, not for every turbulent state. In a common high-Reynolds-number three-dimensional regime, nonlinear interactions transfer kinetic energy to progressively smaller scales, on average. ‘Scale’ means a size of spatial variation, not a separate species of fluid. Eddies are a useful picture, but they are not rigid little wheels that split on a schedule.

The cascade transfers energy between scales; it is different from viscous dissipation into internal energy. Strong deformation makes small-scale velocity gradients large enough for molecular viscosity to act efficiently. Three-dimensional vortex stretching is one important part of the dynamics: the surrounding strain can amplify vorticity aligned with an extending direction. Pressure also couples regions of the incompressible flow; this is not a sequence of isolated rotating objects.

For a statistically stationary, appropriately homogeneous region forced at large scales, a mean forward energy flux across an intermediate scale range can be approximately constant and equal to the mean dissipation ε. ‘Stationary’ means statistics do not drift with time; it does not mean each fluid velocity is constant. ‘Homogeneous’ means the statistics do not depend on position within the modelled region; ‘isotropic’ means they do not depend on direction.

These are conditions, not properties granted by the word turbulence. Near a wall, in a rapidly changing flow or in strong rotation or stratification, different terms and directional preferences may remain important. Even within the ordinary forward-cascade regime, individual transfers can run in both directions. A positive average flux is not a claim that every local event sends energy only downward in size.

Compare outer and viscous scales

At an intermediate separation ℓ in the classical three-dimensional picture, a velocity-difference scale is δu_ℓ of order (εℓ)¹ᐟ³. The turnover-time estimate is ℓ/δu_ℓ, and δu_ℓ³/ℓ has the units and scale of the energy flux. These are similarity estimates under the cascade assumptions; they do not assign the exact velocity of an individual eddy.

Combining ν and mean ε gives the Kolmogorov scales: η = (ν³/ε)¹ᐟ⁴ for length, τ_η = (ν/ε)¹ᐟ² for time, and u_η = (νε)¹ᐟ⁴ for velocity. They satisfy u_ηη/ν = 1 by construction. This expresses competition between nonlinear and viscous effects at a characteristic small scale, rather than a solid cutoff below which every motion vanishes.

For a separate assigned cascade reference, set ν = 0.000001 m²/s, ε = 0.100 m²/s³ and outer length L = 0.100 m. Then η ≈ 0.056234 mm, τ_η ≈ 3.162278 ms and u_η ≈ 17.782794 mm/s. The outer estimate u_L = (εL)¹ᐟ³ ≈ 0.215443 m/s uses an explicitly chosen coefficient of one. It is not the 2 m/s mean velocity of our finite-state exercise.

This reference gives Re_L = u_LL/ν ≈ 21544.35 and L/η ≈ 1778.28. With that coefficient choice, L/η = Re_L³ᐟ⁴. The arithmetic illustrates scale separation; it does not prove that a real flow has a clean inertial interval or determine a simulation grid. Intermittent extreme gradients can require scales smaller than this mean-dissipation estimate.

Reducing ν does not necessarily drive the mean turbulent dissipation toward zero in a maintained high-Reynolds-number family of flows. Increasingly strong gradients can compensate for the smaller viscosity. This observed finite-dissipation behavior is often called the dissipation anomaly. It is a statement about a limit of turbulent flows, not permission to erase viscosity from a finite-flow heating calculation or a proof that every inviscid solution loses energy.

If ε is representative of a supplied volume V = 0.001 m³ at density 1000 kg/m³, the associated fluctuating viscous heating power is ρVε = 0.100 W. Matching it to external input requires the appropriate whole-system balance, including boundary fluxes, mean-flow dissipation and any storage. It is not automatically the required pump power.

Read a spectrum with its measure

A spectrum organises kinetic energy by spatial wavenumber. Let q be the magnitude of an angular wavenumber, with units m⁻¹, and use the isotropic three-dimensional convention K = ∫₀∞E(q)dq. Then E has units m³/s². Angular wavenumber corresponds to sinusoidal wavelength λ = 2π/q; the dimensional scale ℓ = 1/q is smaller by 2π. State which one an axis represents.

The Kolmogorov inertial-range reference is E(q) ≈ C_K ε²ᐟ³q⁻⁵ᐟ³ for 1/L ≪ q ≪ 1/η, with adequate scale separation and the statistical conditions already stated. If E depends only on ε and q, matching dimensions m³/s² = (m²/s³)ᵃ(m⁻¹)ᵇ gives a = 2/3 and b = −5/3. That dependence is the physical hypothesis; dimensional algebra then supplies the powers. The coefficient depends on the spectrum convention and is not fixed by this argument. A slope on its own does not demonstrate turbulence or determine the direction of energy transfer.

The graph shows only a declared reference curve from q = 100 to 1000 m⁻¹, using the assigned C_K = 1.5. At q = 100 m⁻¹, E = 0.000150 m³/s²; at 1000 m⁻¹, E ≈ 0.000003231652 m³/s². Increasing q tenfold reduces E by 10⁻⁵ᐟ³ ≈ 0.0215443. This curve is not a measured spectrum, a fitted constant or a validated prediction of the four-state exercise.

The area under E against q gives energy per mass. A log–log plot makes the power-law slope readable, but its ordinary screen area is not ∫E dq. Energy per logarithmic wavenumber interval is qE(q), because dq = q d(ln q). Our reference-band integral is about 0.0176525 m²/s², not a total K obtained by extending q⁻⁵ᐟ³ down to zero or up through the dissipation range.

A probe record often gives a frequency spectrum or a one-dimensional component spectrum, not this shell-integrated three-dimensional spectrum. Converting time variation into spatial structure with a frozen-advection approximation needs a suitable mean advection and slowly evolving structures over the sampling time. Instrument resolution, record length, noise and that assumption must be checked before comparing coefficients or claiming a scaling interval.

Keep bursts visible

An average dissipation rate does not tell us how evenly dissipation is distributed. Compare eight assigned equal-weight samples: one set is 0.1 in every sample; the other is zero in seven samples and 0.8 in one. Both means are 0.1 m²/s³. Their mean squares are 0.01 and 0.08 m⁴/s⁶, respectively. The second set has much stronger variation despite the same mean.

This simple comparison explains why a mean can hide bursts. Real small-scale intermittency concerns uneven, sometimes intense velocity gradients and increments across space and time, with statistics that depend on scale. The eight samples are a teaching illustration, not a measured intermittency distribution, a universal probability law or an intermittency exponent.

Consequently, similarity based only on mean ε and ν does not determine every high-order moment or extreme event. Large-scale anisotropy can also persist into smaller-scale statistics. Approximate success of an energy-spectrum slope does not establish exact isotropy, Gaussian fluctuations or all-moment universality. The Kolmogorov scales remain useful reference quantities with those limits.

For mixing, kinetic stirring stretches a concentration field into finer structures, while molecular diffusion eliminates concentration differences. A visibly stirred region need not already be molecularly uniform. Scalar diffusivity supplies an additional scale comparison, so the velocity dissipation scale alone does not set every concentration or temperature scale. Detailed reacting, biological and geophysical consequences belong to their neighboring subject.

Choose what to resolve and what to model

Reynolds-averaged Navier–Stokes methods predict chosen mean fields and model the influence of velocity correlations. They do not reconstruct each turbulent fluctuation merely by converging to a mean solution. An unsteady RANS calculation can represent variation in the mean while still modelling the turbulence beneath that averaging description.

Large-eddy simulation filters the velocity field, resolving larger motions and modelling the stress associated with unresolved scales. Filtering a nonlinear product again leaves a residual: filtering u_i u_j is not the same as multiplying the filtered velocities. The cutoff, wall treatment, numerical dissipation and model all affect the result. LES filtering is a scale separation, whereas Reynolds averaging is a statistical separation; they are related ideas with different retained information.

Direct numerical simulation resolves the dynamically relevant turbulent scales of the chosen governing model without a turbulence closure. It still needs adequate domain, mesh, time step, boundary/initial data and statistical sampling. It is a numerical solution of a physical model, not an exact representation of every molecular process or automatic validation of a real experiment.

A rough three-dimensional spatial-count estimate is (L/η)³ of order Re_L⁹ᐟ⁴ under the conditioned outer-to-inner scaling. This is not a cost quotation: actual grid constants, boundaries, time integration and intermittent extremes add requirements. Report which quantities were checked for resolution and sampling convergence. Verification asks whether the chosen equations are solved as intended; validation asks how well they represent the relevant physical observations. Neither a fine-looking image nor agreement with a single mean speed settles both questions.

Connect applications with regime limits

In a pipe or a cooling passage, turbulent transport affects momentum and heat transfer; the pressure and thermal requirements depend on the wall and flow conditions. Around a body, a turbulent boundary layer may increase skin friction yet delay separation, changing pressure drag. Total drag therefore requires both accounts, as developed in Boundary layers, lift, and drag.

Jets and wakes exchange momentum with their surroundings and entrain surrounding fluid. Predicting their spreading or scalar concentrations needs the actual geometry, forcing, density effects and transport description. A covariance at one point or a Kolmogorov length does not supply that complete prediction.

Three-dimensional forward energy transfer is not a universal direction rule. In idealised two-dimensional incompressible turbulence, energy can transfer toward larger scales while enstrophy, a measure based on squared vorticity, transfers toward smaller scales. Forcing, boundaries and friction modify these regimes. A two-dimensional simulation may represent the mean geometry of a three-dimensional experiment without resolving its three-dimensional turbulent dynamics.

Rotation and stratification can introduce waves, anisotropy and competing transfer pathways. Compressible flow adds density and internal-energy coupling, including effects near shocks; ordinary constant-density Reynolds averaging then needs revision, commonly with density-weighted averaging. Compressible flow and shock waves and Microfluidic, reacting, biological, and geophysical flows are the permanent neighboring homes for those developments. The present lesson supplies the statistical and scale reasoning they will need.

Return path: derive the missing product and energy terms

Start from ∂u_i/∂t + ∂(u_i u_j)/∂x_j = −(1/ρ)∂p/∂x_i + ν∇²u_i + f_i with ∂u_j/∂x_j = 0. Write u_i = U_i + u′_i and p = P + p′. Expanding and averaging u_i u_j leaves U_iU_j + ⟨u′_i u′_j⟩; the mixed terms vanish under the stated Reynolds averaging rules. Moving the covariance divergence to the right yields the averaged equation above.

The covariance matrix is symmetric and positive semidefinite: for any real vector a, a_i R_ij a_j = ⟨(a_i u′_i)²⟩ ≥ 0. Thus component variances cannot be negative, and |R_12| ≤ √(R_11R_22). Case A reaches equality: |−0.020| = √(0.040 × 0.010). Its assigned transverse departure is exactly −u′/2. That extreme correlation is intentional for arithmetic clarity, not a typical turbulence statistic.

Subtract the averaged momentum equation from the instantaneous equation, multiply by u′_i and average. With constant density/ν, incompressibility, a nonfluctuating body force and the same averaging rules, the resulting local budget is ∂K/∂t + U_j∂K/∂x_j = −R_ij∂U_i/∂x_j − ε − ∂T_j/∂x_j. Define s′_ij = (∂u′_i/∂x_j + ∂u′_j/∂x_i)/2, ε = 2ν⟨s′_ij s′_ij⟩ and T_j = ⟨u′_i u′_i u′_j⟩/2 + ⟨p′u′_j⟩/ρ − 2ν⟨u′_i s′_ij⟩. The transport flux has units m³/s³; its divergence has the same W/kg units as production and dissipation.

This form keeps the physical strain-based dissipation and its matching viscous transport together. Alternative local rearrangements use ν⟨(∂u′_i/∂x_j)²⟩ and a different viscous flux; the divergence terms must be reconciled before substituting one local definition into the other form. For suitable homogeneous or boundary-integrated settings the corresponding averages agree. With fluctuating body forcing, add its covariance ⟨u′_i f′_i⟩ to the energy input.

For a maintained homogeneous stationary cascade with no mean shear, energy is instead supplied by fluctuating forcing; a stationary balance can then give ⟨u′·f′⟩ = ε. In a shear-driven flow, production comes from the mean and spatial transport may remain important. These balances explain why ‘everywhere P_K = ε’ is an extra local-equilibrium assumption, not a general definition of turbulence.

Return path: increments and the four-fifths law

Choose a unit direction e and a positive separation r. Define the longitudinal increment δu_L(r) = [u(x + re) − u(x)]·e. A structure function compares velocities at two separated locations, rather than comparing one velocity with a mean. For example, S_2(r) = ⟨[δu_L(r)]²⟩ and the signed S_3(r) = ⟨[δu_L(r)]³⟩. The signed third moment is different from ⟨|δu_L|³⟩.

For the classical homogeneous, isotropic, incompressible stationary three-dimensional setting with large-scale forcing, the inertial asymptotic gives S_3(r) = −(4/5)εr for η ≪ r ≪ L. Viscous, time-development, forcing-scale and inhomogeneous contributions need appropriate treatment outside that limit. The coefficient and sign arise from the scale-by-scale energy balance; dimensional analysis alone supplies neither.

With the convention above and positive mean ε, S_3 is negative. This is compatible with a mean forward cascade, not negative kinetic energy. For the assigned ε = 0.100 m²/s³ and r = 0.010 m, the conditional law gives S_3 = −0.000800 m³/s³. Reversing the definition of the signed increment changes the sign convention, so write it before comparing data.

The law does not say every increment is negative or every event transports energy forward. Nor does it imply exact S_p proportional to rᵖᐟ³ for every order p. Intermittency can alter higher-order scaling while the conditioned third-order energy-balance result retains its distinct role. A finite measured value at one separation cannot verify the asymptotic law, its averaging convergence or all of its conditions.

Check the common misreadings

‘The fluctuations average to zero, so they have no effect’ misses the products that survive averaging. ‘Both flows have 10% intensity, so they transport the same momentum’ misses correlation. ‘A large Reynolds number means viscosity is irrelevant’ misses the strong velocity gradients at small length scales and the conversion into internal energy.

‘A cascade destroys energy at every step’ confuses transfer with dissipation. ‘The Kolmogorov length is the smallest possible disturbance’ turns a characteristic mean-dissipation scale into a strict cutoff. ‘A straight −5/3 line proves our model’ ignores spectrum conventions, finite ranges, other mechanisms and independent transport evidence.

‘Turbulence always mixes everything uniformly’ ignores scalar diffusion, barriers and spatial variability. ‘DNS has no assumptions’ ignores the governing fluid model, resolution and sampling. A useful conclusion names the flow regime, average, measured or supplied inputs, transport convention and conditions under which the answer changes.

Worked examples

Try it yourself

1. A velocity probe records one clean sinusoid at one location. Does that alone demonstrate turbulence?

   Show the answer and reasoning

   No. A regular wave or periodic laminar motion can give that signal.

   The classification needs the underlying flow and resolved multiscale interactions; time variation by itself is insufficient.

2. For variances 0.0400, 0.0100 and 0.0025 m²/s², calculate K. Is K determined by the covariance R_12 alone?

   Show the answer and reasoning

   K = 0.02625 m²/s²; R_12 alone does not determine it.

   K is half the trace of the covariance matrix, so it uses the three component variances. An off-diagonal correlation instead affects directional transport and production with the appropriate mean gradient.

3. A source defines Rstress_xy = +ρ⟨u′v′⟩. For case A what is its value, and what is the value of this lesson’s τᵀ_xy?

   Show the answer and reasoning

   Rstress_xy = −20 Pa and τᵀ_xy = +20 Pa.

   The two definitions differ by a minus sign. The same physical advective flux appears with corresponding signs on opposite sides of the averaged momentum balance.

4. Matching ν_t = 0.002 m²/s reproduces case A’s shear stress at dU/dy = 10 s⁻¹. Does that prove the stated linear eddy-viscosity model also reproduces its three normal variances?

   Show the answer and reasoning

   No. For this parallel shear the model gives equal R_ii = 2K/3 = 0.0175 m²/s², unlike the assigned 0.0400, 0.0100 and 0.0025.

   All mean normal strain rates vanish in this local parallel-shear diagnostic. Only the isotropic part remains on the diagonal; fitting one off-diagonal stress does not establish the full anisotropic covariance.

5. Double ν while retaining the assigned mean ε. By what factors do η and τ_η change?

   Show the answer and reasoning

   η increases by 2³ᐟ⁴, about 1.68179; τ_η increases by √2, about 1.41421.

   The definitions give η proportional to ν³ᐟ⁴ and τ_η proportional to ν¹ᐟ² at fixed ε. Keeping ε fixed is part of the question, not a claim that changing a real fluid’s viscosity leaves its flow unchanged.

6. For angular wavenumber q = 200 m⁻¹, find λ and ℓ = 1/q. May they be labelled as the same length?

   Show the answer and reasoning

   λ ≈ 31.4159 mm and ℓ = 5 mm; they are different by 2π.

   A sinusoidal phase changes by 2π over one wavelength, so λ = 2π/q. Scaling estimates often use 1/q only as a characteristic length.

7. With δu_L = [u(x + re) − u(x)]·e, ε = 0.100 m²/s³ and r = 0.010 m, what does the conditioned four-fifths law give? Does it make kinetic energy negative?

   Show the answer and reasoning

   S_3 = −0.000800 m³/s³; kinetic energy remains nonnegative.

   The signed third moment records asymmetry of increments. K is a mean square and therefore nonnegative. The law requires its homogeneous/isotropic stationary inertial setting and is not a statement about every increment.

8. A steady RANS calculation reproduces a measured mean velocity. Does this establish the frequency distribution of turbulent fluctuations, and can the same evidence validate a mixing prediction?

   Show the answer and reasoning

   No to both conclusions from that evidence alone.

   RANS models the turbulent influence on the mean; it does not reconstruct a fluctuation time history. A mixing prediction also needs the scalar model, boundary conditions and relevant independent physical comparisons.

What to carry forward

• Turbulence is a multiscale nonlinear regime; an unsteady signal, a single vortex or linear instability alone does not establish it.
• Zero mean fluctuation can coexist with nonzero variance and correlated momentum transport; intensity alone does not specify that transport.
• Averaging leaves Reynolds correlations, so a mean-flow prediction needs closure and an explicit stress convention.
• Mean-shear production, spatial energy transport, transfer across scales and viscous dissipation perform different roles in the energy balance.
• The classical three-dimensional cascade, Kolmogorov scales and −5/3 spectrum have stated statistical and regime conditions; intermittent extremes and alternative regimes limit their use.
• Spectra and signed structure functions retain different information, and a power-law line is not a complete validation.
• RANS, LES, DNS and experiments each require evidence suited to their retained information, resolution and model scope.