turbulent flows

What is turbulent flow?

Short exam answer: A turbulent flow is irregular, time-dependent, rotational, and three-dimensional, with a broad range of eddy sizes. It appears chaotic but contains recognizable coherent structures.

Explanation: The Navier–Stokes equations are deterministic, but their nonlinear dynamics allow small differences to develop into very different instantaneous states. Turbulence is therefore often described statistically.

Key terms: chaotic; 3D; rotational; eddies; statistics

Source: Lecture 01; Compendium Ch. 1.1

Name at least four differences between laminar and turbulent flow.

Short exam answer: Turbulent flow has stronger mixing, greater heat and mass transfer, higher wall friction, and greater resistance to separation. Laminar flow is layered and smoother, whereas turbulent flow is irregular and contains transverse motion.

Explanation: The transverse motion in turbulent flow transports mass, heat, and momentum much faster than molecular diffusion alone.

Key terms: mixing; heat transfer; friction; separation

Source: Review Questions – Introduction 1

Why does turbulent flow produce stronger mixing than laminar flow?

Short exam answer: Turbulent fluctuations move fluid parcels across the mean flow and therefore transport mass, heat, and momentum between regions with different values.

Explanation: In laminar flow, transverse transport occurs mainly through molecular diffusion. Turbulent flow adds a much faster macroscopic transport mechanism.

Key terms: transverse transport; fluctuations; momentum exchange

Source: Lecture 01; Review Questions – Introduction

Why is wall friction usually greater in a turbulent boundary layer than in a laminar one?

Short exam answer: Turbulent mixing brings high-speed fluid closer to the wall and increases the wall-normal velocity gradient at the wall. The wall shear stress therefore increases.

Explanation: The turbulent flux is zero at the wall itself, but turbulence changes the velocity profile so that the molecular shear stress at the wall becomes larger.

Formula: τ_w = μ(∂U/∂y)_w

Key terms: wall gradient; shear stress; friction

Source: Review Questions – Introduction 2

Why can a turbulent boundary layer withstand an adverse pressure gradient better than a laminar one?

Short exam answer: Turbulent mixing transports high-momentum fluid from the outer layer toward the wall. The near-wall fluid therefore has more energy and can continue moving against the increasing pressure for longer.

Explanation: The result is usually delayed separation, at the cost of greater skin friction.

Key terms: adverse pressure gradient; momentum; separation

Source: Lecture 01; Review Questions – Introduction

What is the time scale of molecular diffusion?

Short exam answer: Molecular diffusion over a distance l has the time scale t_m ~ l²/ν.

Explanation: Diffusion becomes very slow as the transport distance increases because the time grows with the square of the length.

Formula: t_m ~ l²/ν

Key terms: molecular diffusion; viscosity

Source: Lecture 01; Review Questions – Introduction 3

What is the time scale of turbulent transport?

Short exam answer: Turbulent transport over a distance l has the time scale t_t ~ l/u′.

Explanation: A turbulent eddy moves fluid over the distance l with a characteristic fluctuation velocity u′.

Formula: t_t ~ l/u′

Key terms: turbulent diffusion; fluctuation velocity

Source: Lecture 01; Review Questions – Introduction 3

How are molecular and turbulent diffusion compared quantitatively?

Short exam answer: The ratio of the time scales is t_m/t_t ~ u′l/ν = Re_t. When Re_t ≫ 1, turbulent transport is much faster than molecular diffusion.

Explanation: For l = 1 cm, u′ = 1 m/s, and ν ≈ 10⁻⁵ m²/s, Re_t ≈ 1000.

Formula: t_m/t_t ~ u′l/ν = Re_t

Key terms: turbulent Reynolds number; time scale

Source: Review Questions – Introduction 3

Name three engineering applications that make use of turbulence.

Short exam answer: Examples include dimples on golf balls, vortex generators on wings, swirl injectors in combustion systems, corrugated pipes, and blenders.

Explanation: They exploit increased mixing, increased heat or mass transfer, or delayed flow separation.

Key terms: golf ball; vortex generator; combustion

Source: Review Questions – Introduction 4

What is the energy cascade?

Short exam answer: Kinetic energy is supplied at large scales, transferred through nonlinear interactions to progressively smaller scales, and converted into internal energy at the smallest scales.

Explanation: Large eddies are determined by geometry and boundary conditions, while small eddies are more universal.

Key terms: energy transfer; large scales; small scales

Source: Lecture 01–03; Compendium Ch. 1.3 and 2.3.1

What is vortex stretching, and why is it important?

Short exam answer: Vortex stretching occurs when a vortex is elongated, reducing its cross-sectional area and increasing its rotation rate. The mechanism transfers energy to smaller scales and helps sustain three-dimensional turbulence.

Explanation: It appears in the vorticity transport equation as ω_j∂u_i/∂x_j. In purely two-dimensional flow, the stretching mechanism vanishes.

Formula: Dω_i/Dt = ν∂²ω_i/∂x_j² + ω_j∂u_i/∂x_j

Key terms: vorticity; 3D; energy cascade

Source: Compendium Ch. 1.3; Lecture 01

When and why does the energy cascade stop?

Short exam answer: It stops at the Kolmogorov scales, where the local Reynolds number is of order one and viscous forces are as important as inertial forces.

Explanation: Viscosity then smooths the smallest velocity gradients and converts turbulent kinetic energy into heat.

Formula: Re_η = u_ηη/ν ~ 1

Key terms: dissipation; Kolmogorov scale; Re≈1

Source: Lecture 01–03; Compendium Ch. 1.3

Why are canonical flows used in turbulence research?

Short exam answer: They isolate fundamental mechanisms in simple geometries that can be studied thoroughly using experiments, DNS, and theory.

Explanation: Complex flows can often be understood as combinations of boundary layers, free jets, channel flows, wakes, and separation regions.

Key terms: canonical flow; validation; model development

Source: Review Questions – Free shear flow 1

What is a tensor?

Short exam answer: A tensor describes a physical quantity that is independent of the coordinate system used to represent it.

Explanation: Its components change when the coordinate system is rotated, but the physical quantity remains the same. A scalar is rank 0, a vector rank 1, and a matrix-like tensor rank 2.

Key terms: coordinate independent; rank; components

Source: Exercise 1; Lecture general maths

What does the Einstein summation convention mean?

Short exam answer: An index that appears exactly twice in one term is automatically summed over the spatial directions.

Explanation: For example, a_i b_i = a_1b_1 + a_2b_2 + a_3b_3. An index that appears only once is a free index.

Formula: a_i b_i = Σ_i a_i b_i

Key terms: dummy index; free index; summation

Source: Exercise 1; Lecture general maths

What is the difference between a free index and a summation index?

Short exam answer: A free index specifies the component of the result and must be the same in every term. A summation index appears twice in one term and may be renamed without changing the meaning.

Explanation: In u_j∂u_i/∂x_j, i is free and j is summed.

Key terms: free index; dummy index

Source: Exercise 1; Lecture general maths

How is incompressible continuity written in index notation?

Short exam answer: The divergence of the velocity is zero: ∂u_i/∂x_i = 0.

Explanation: This means that the volume of a fluid element does not change.

Formula: ∂u_i/∂x_i = 0

Key terms: continuity; incompressible; divergence

Source: Exercise 1

How is the convective term in the Navier–Stokes equations written in index notation?

Short exam answer: The convective term is u_j∂u_i/∂x_j.

Explanation: The index j sums the transport contributions from all three directions, while i identifies the momentum component being transported.

Formula: (u·∇)u → u_j∂u_i/∂x_j

Key terms: convection; momentum transport

Source: Exercise 1

How is the Laplacian of velocity written in index notation?

Short exam answer: The viscous diffusion term is ∂²u_i/∂x_k², which is the sum of second derivatives in all spatial directions.

Explanation: The index k is only a summation index and may be replaced by j as long as it does not conflict with other indices in the same term.

Formula: ∇²u_i = ∂²u_i/(∂x_k∂x_k)

Key terms: Laplacian; viscous diffusion

Source: Exercise 1

What is the mean strain-rate tensor?

Short exam answer: S_ij is the symmetric part of the velocity-gradient tensor and describes local deformation.

Explanation: The antisymmetric part describes rigid-body rotation. For a Newtonian fluid, the viscous stress is proportional to S_ij.

Formula: S_ij = 1/2(∂U_i/∂x_j + ∂U_j/∂x_i), τ_ij = 2μS_ij

Key terms: strain rate; symmetric tensor

Source: Exercise 1; Compendium appendix

What is Reynolds decomposition?

Short exam answer: An instantaneous quantity is divided into a mean value and a fluctuation: u_i = ⟨u_i⟩ + u_i′.

Explanation: The mean describes the statistically stable flow, while u_i′ describes the instantaneous deviation from the mean.

Formula: u_i(x,t)=⟨u_i(x,t)⟩+u_i′(x,t)

Key terms: mean value; fluctuation

Source: Lecture 01; Compendium Ch. 2.2

What is the mean value of a Reynolds fluctuation?

Short exam answer: It is zero: ⟨u_i′⟩ = 0.

Explanation: This follows directly from the definition u_i′ = u_i − ⟨u_i⟩.

Formula: ⟨u_i′⟩=0

Key terms: Reynolds averaging rules

Source: Lecture 01; Compendium Ch. 2.2

What is an ensemble?

Short exam answer: An ensemble is the collection of all possible realizations of the same statistical flow.

Explanation: An ensemble average is obtained by averaging many repeated realizations at the same location and time.

Key terms: realizations; expectation value

Source: Review Questions – HIT 3; Lecture 01

How is the ensemble average obtained for a statistically stationary flow?

Short exam answer: If the flow is ergodic, the ensemble average can be replaced by a sufficiently long time average at one measurement point.

Explanation: Stationary means that the statistics do not change with time; ergodic means that the time series samples the relevant states representatively.

Key terms: ergodicity; time average; stationary

Source: Review Questions – HIT 3

How is the ensemble average obtained for a statistically unsteady flow?

Short exam answer: The same experiment must be repeated many times, and the value at the same relative time in each realization must be averaged.

Explanation: A conventional time average would otherwise mix the systematic time development with the turbulent fluctuations.

Key terms: unsteady; repeated experiments

Source: Review Questions – HIT 3

What does the law of large numbers mean in this context?

Short exam answer: As the number of independent realizations increases, estimated statistical quantities approach their true ensemble values.

Explanation: The relevant moments, such as the variance, must be finite.

Key terms: sample size; convergence

Source: Review Questions – HIT 3

What do variance and standard deviation measure?

Short exam answer: Variance measures the mean squared distance from the mean value; standard deviation is its square root and has the same unit as the signal.

Formula: σ_u²=⟨(u−⟨u⟩)²⟩, σ_u=√σ_u²

Key terms: spread; statistical moment

Source: Exercise 2; Lecture 01

What is the RMS value of a velocity fluctuation?

Short exam answer: u′_rms = √⟨u′²⟩, which equals the standard deviation when the fluctuation has zero mean.

Explanation: The RMS value gives a characteristic magnitude of the turbulent velocity fluctuations.

Formula: u′_rms=√⟨u′²⟩=σ_u

Key terms: RMS; fluctuation level

Source: Exercise 2

What is turbulence intensity?

Short exam answer: Turbulence intensity is the RMS fluctuation normalized by the mean velocity.

Formula: I = u′_rms/⟨u⟩

Key terms: normalized fluctuation

Source: Exercise 2

What does covariance measure?

Short exam answer: Covariance measures whether two fluctuations tend to be positive or negative at the same time.

Explanation: Positive covariance means that the signals usually vary in the same direction; negative covariance means that they vary in opposite directions.

Formula: Cov(u,v)=⟨(u−⟨u⟩)(v−⟨v⟩)⟩

Key terms: co-variation; Reynolds stress

Source: Exercise 2

What is the correlation coefficient?

Short exam answer: It is the normalized covariance and lies between −1 and 1.

Explanation: ρ = 1 means perfect positive correlation, ρ = −1 perfect negative correlation, and ρ = 0 no linear correlation.

Formula: ρ_uv = Cov(u,v)/(σ_uσ_v)

Key terms: normalized covariance

Source: Exercise 2

What does skewness measure?

Short exam answer: Skewness is the normalized third central moment and measures the asymmetry of a probability distribution.

Explanation: Positive skewness indicates a long tail toward positive values; negative skewness indicates a long tail toward negative values.

Formula: g₃=⟨(u−⟨u⟩)³⟩/σ³

Key terms: asymmetry; third moment

Source: Exercise 2

What does kurtosis measure?

Short exam answer: Kurtosis is the normalized fourth central moment and measures the strength of the tails and the frequency of extreme values.

Explanation: The course commonly uses excess kurtosis, for which a normal distribution has g₄ = 0.

Formula: g₄=⟨(u−⟨u⟩)⁴⟩/σ⁴−3

Key terms: outliers; fourth moment

Source: Exercise 2

What is the difference between a PDF and a CDF?

Short exam answer: The PDF describes probability density, whereas the CDF gives the probability that the variable is smaller than a specified value.

Explanation: The CDF is the integral of the PDF, and the PDF is the derivative of the CDF.

Formula: F(V)=∫_{−∞}^{V} f(v)dv, f(V)=dF/dV

Key terms: probability; distribution

Source: Compendium Ch. 1.4; Exercise 2

What is autocorrelation?

Short exam answer: Autocorrelation measures how similar a signal is to itself after a time or spatial shift.

Explanation: When the correlation approaches zero, the signal has lost memory of its earlier value. This can be used to define time or length scales.

Formula: R_uu(τ)=⟨u′(t)u′(t+τ)⟩

Key terms: memory; time scale; length scale

Source: Lecture 01–02; Compendium Ch. 1.5

What does a Fourier transform do to a turbulent signal?

Short exam answer: It represents the signal as a sum of waves with different frequencies or wavenumbers and shows how amplitude or energy is distributed among the scales.

Explanation: Small wavenumbers correspond to large spatial scales; large wavenumbers correspond to small scales.

Key terms: frequency space; wavenumber; scales

Source: Lecture 01–03; Exercise 2

What does homogeneous turbulence mean?

Short exam answer: Statistical quantities are unchanged when the coordinate system is translated. The statistics do not depend on absolute position.

Explanation: Homogeneity does not mean that the instantaneous flow is identical everywhere, only that the statistics are the same.

Key terms: translation invariant; position

Source: Review Questions – HIT 1

What does isotropic turbulence mean?

Short exam answer: Statistical quantities are unchanged under rotation and reflection of the coordinate system. No direction is statistically preferred.

Explanation: Isotropy applies to the statistics; it does not mean that the instantaneous velocity points in the same direction everywhere.

Key terms: rotation invariant; reflection; direction

Source: Review Questions – HIT 1

How can homogeneous isotropic turbulence be generated experimentally?

Short exam answer: An approximately uniform mean flow is passed through a regular grid or mesh. Some distance downstream, a region of approximately homogeneous and isotropic turbulence is obtained.

Explanation: Close to the grid, the individual wakes dominate and the flow is inhomogeneous. Farther downstream, the wakes have mixed, although the turbulence also decays with distance.

Key terms: grid turbulence; wind tunnel

Source: Review Questions – HIT 2

Why must the measurement volume for grid-generated HIT be small and sufficiently far downstream?

Short exam answer: Far enough downstream, the wakes from the grid bars have mixed. A small measurement volume allows the slow streamwise change in turbulence level to be neglected locally.

Explanation: The flow can then be treated as locally homogeneous and isotropic.

Key terms: measurement location; locally homogeneous

Source: Review Questions – HIT 2

What is Taylor's frozen-turbulence hypothesis?

Short exam answer: When the mean velocity is much larger than the fluctuation velocity, turbulent structures are assumed to change very little while they are advected past the measurement point.

Explanation: A time series at one point can then be interpreted as a spatial series: a time shift τ corresponds approximately to the distance r = Uτ.

Formula: r ≈ Uτ, ∂()/∂t ≈ −U∂()/∂x

Key terms: frozen turbulence; advection

Source: Lecture 02

What does the two-point correlation describe?

Short exam answer: It measures the statistical relationship between velocity fluctuations at two points separated by the vector r.

Explanation: In homogeneous turbulence, it depends on the separation r, not on the absolute position x.

Formula: R_ij(r)=⟨u_i′(x)u_j′(x+r)⟩

Key terms: spatial correlation; structure size

Source: Lecture 02; Compendium Ch. 2.3.1

What is the longitudinal two-point correlation?

Short exam answer: It is the correlation between velocity components that are both parallel to the separation vector r.

Formula: f(r)=⟨u_L′(x)u_L′(x+r)⟩/⟨u_L′²⟩

Key terms: longitudinal; parallel

Source: Lecture 02

What is the lateral or transverse two-point correlation?

Short exam answer: It is the correlation between velocity components that are perpendicular to the separation vector r.

Formula: g(r)=⟨u_T′(x)u_T′(x+r)⟩/⟨u_T′²⟩

Key terms: transverse; perpendicular

Source: Lecture 02

What is an integral length scale?

Short exam answer: It is the integral of a normalized spatial correlation and provides a characteristic size of the energy-containing eddies.

Explanation: It measures the distance over which velocity fluctuations remain correlated.

Formula: L = ∫₀^∞ f(r)dr

Key terms: large scale; correlation length

Source: Lecture 02; Compendium Ch. 2.3.1

What does the Reynolds-stress tensor look like in fully isotropic turbulence?

Short exam answer: The normal stresses are equal and the cross correlations are zero: ⟨u_i′u_j′⟩ = (2K/3)δ_ij.

Explanation: Isotropy permits neither a preferred direction nor a preferred shear correlation.

Formula: R_ij=(2/3)Kδ_ij

Key terms: isotropic RST; diagonal tensor

Source: Lecture 02–04; Compendium Ch. 2.3.1

Name the three spectral ranges in Kolmogorov's theory.

Short exam answer: The energy-containing range, the inertial range, and the dissipation range.

Explanation: They represent, respectively, energy input and large eddies, conservative energy transfer, and viscous conversion into heat.

Key terms: energy-containing; inertial; dissipation

Source: Review Questions – HIT 4

Which spectral range is non-universal, and why?

Short exam answer: The energy-containing range is non-universal because the large scales are determined by the geometry, boundary conditions, and the way energy is supplied.

Key terms: large scales; global flow

Source: Review Questions – HIT 4

Why are the inertial and dissipation ranges regarded as universal at high Reynolds numbers?

Short exam answer: The small scales largely lose direct information about the geometry. The inertial range is governed mainly by ε and k, while the dissipation range is governed by ε and ν.

Explanation: Laws and models for these ranges can therefore be applied across different flows.

Key terms: local isotropy; universal small scales

Source: Review Questions – HIT 4

What is Kolmogorov's −5/3 law?

Short exam answer: In the inertial range, the energy spectrum is E(k) = C_K ε^(2/3) k^(−5/3).

Explanation: There is no direct energy production or viscous dissipation in this range; energy passes through the scales with an approximately constant flux ε.

Formula: E(k)=C_K ε^(2/3)k^(−5/3)

Key terms: inertial range; spectral slope

Source: Review Questions – HIT 5

How are the exponents 2/3 and −5/3 obtained by dimensional analysis?

Short exam answer: Set E = Cε^a k^b and match the dimensions [E]=L³/T², [ε]=L²/T³, and [k]=1/L. This gives a=2/3 and b=−5/3.

Formula: [E]=L³T⁻² = (L²T⁻³)^a(L⁻¹)^b

Key terms: dimensional analysis

Source: Review Questions – HIT 5

How is the −5/3 exponent identified graphically?

Short exam answer: In a log–log plot of E versus k, a power law becomes a straight line. The slope in the inertial range is −5/3.

Formula: log E = konstant − (5/3)log k

Key terms: log-log; slope

Source: Review Questions – HIT 5

Which terms appear in the spectral energy equation, and where are they active?

Short exam answer: Production acts mainly at small wavenumbers, nonlinear transfer moves energy between wavenumbers, and dissipation acts mainly at large wavenumbers.

Key terms: production; transfer; dissipation

Source: Review Questions – HIT 6

Why is the integral of the spectral transfer term over all wavenumbers equal to zero?

Short exam answer: Because the transfer term only redistributes energy among scales; it neither creates nor destroys total turbulent kinetic energy.

Formula: ∫₀^∞ T(k)dk = 0

Key terms: conservative redistribution

Source: Review Questions – HIT 7

What is a wavenumber triad?

Short exam answer: It is a set of three Fourier modes coupled through the nonlinear convective term and satisfying a vector relation such as k+p+q=0.

Explanation: Triads are the mechanism by which energy is transferred between scales. A numerical model must therefore represent a sufficiently broad wavenumber range.

Formula: k+p+q=0

Key terms: nonlinear coupling; Fourier modes

Source: Review Questions – HIT 8

What is the Kolmogorov length scale?

Short exam answer: It is the smallest characteristic turbulent length scale, determined by ν and ε.

Formula: η=(ν³/ε)^(1/4)

Key terms: smallest scale; viscosity

Source: Lecture 03; Compendium Ch. 2.3.1

What are the Kolmogorov velocity and time scales?

Short exam answer: u_η=(νε)^(1/4) and τ_η=(ν/ε)^(1/2).

Explanation: They describe the characteristic velocity and time of the smallest dissipative eddies.

Formula: u_η=(νε)^(1/4), τ_η=(ν/ε)^(1/2)

Key terms: small scales; dissipation

Source: Lecture 03

What happens to scale separation as the Reynolds number increases?

Short exam answer: The ratio between the large and small scales increases, and the inertial range becomes wider.

Explanation: The Kolmogorov scale becomes smaller relative to the global length scale.

Formula: η/L ~ Re^(−3/4)

Key terms: scale separation; high Re

Source: Lecture 03; Review Questions – HIT 5

What is the difference between Reynolds averaging and Reynolds decomposition of the continuity equation?

Short exam answer: Averaging gives ∂⟨u_i⟩/∂x_i=0. Decomposition substitutes u_i=⟨u_i⟩+u_i′ into the instantaneous equation and gives both ∂⟨u_i⟩/∂x_i=0 and ∂u_i′/∂x_i=0.

Explanation: Averaging is an operation applied to the equation; decomposition is a splitting of the variable.

Formula: ⟨∂u_i/∂x_i⟩=0; ∂(⟨u_i⟩+u_i′)/∂x_i=0

Key terms: averaging; decomposition

Source: Review Questions – Reynolds averaging 1

Why do Reynolds stresses appear when the Navier–Stokes equations are averaged?

Short exam answer: The nonlinear product u_j∂u_i/∂x_j contains products of fluctuations. After averaging, terms of the form ⟨u_i′u_j′⟩ remain.

Explanation: Averaging commutes with linear operations, but the average of a product is not generally the product of the averages.

Formula: ⟨u_i u_j⟩=⟨u_i⟩⟨u_j⟩+⟨u_i′u_j′⟩

Key terms: nonlinearity; unknown correlation

Source: Lecture 04; Compendium Ch. 2.2.1

What is the incompressible RANS momentum equation?

Short exam answer: The mean flow satisfies the Navier–Stokes equations with an additional divergence term arising from the Reynolds stresses.

Formula: ∂U_i/∂t + U_j∂U_i/∂x_j = −(1/ρ)∂P/∂x_i + ν∂²U_i/∂x_j² − ∂⟨u_i′u_j′⟩/∂x_j

Key terms: RANS; Reynolds stress

Source: Lecture 04; Compendium Ch. 2.2.1

What does the Reynolds-stress tensor represent physically?

Short exam answer: R_ij=⟨u_i′u_j′⟩ represents turbulent transport of i-momentum in the j-direction.

Explanation: The diagonal components are fluctuation energies. The off-diagonal components describe correlated transverse transport and act as turbulent shear stresses.

Formula: R_ij=⟨u_i′u_j′⟩

Key terms: momentum flux; tensor

Source: Lecture 04; Compendium Ch. 2.2.3

How many independent components does the Reynolds-stress tensor have in three dimensions?

Short exam answer: Six, because the tensor is symmetric: R_ij=R_ji.

Explanation: There are three normal stresses and three distinct shear stresses.

Key terms: symmetry; six components

Source: Review Questions – Reynolds averaging 2

What is the closure problem in RANS?

Short exam answer: Averaging introduces six unknown Reynolds stresses without providing six new closed equations. The number of unknowns therefore exceeds the number of available governing equations.

Explanation: In the counting used in the review questions, there are eleven unknowns: density, pressure, three mean-velocity components, and six RST components, but only five governing equations.

Key terms: closure; more unknowns than equations

Source: Review Questions – Reynolds averaging 2

What is turbulent kinetic energy K?

Short exam answer: K is one half of the trace of the Reynolds-stress tensor.

Formula: K = 1/2⟨u_i′u_i′⟩ = 1/2(⟨u′²⟩+⟨v′²⟩+⟨w′²⟩)

Key terms: TKE; trace

Source: Lecture 04; Compendium Ch. 2.2.2

What are the main terms in the TKE transport equation?

Short exam answer: Local change, convection, production, dissipation, turbulent transport, pressure diffusion, and molecular diffusion.

Explanation: Production transfers energy from the mean flow to turbulence; dissipation removes TKE as heat. The transport terms redistribute TKE in space.

Key terms: TKE budget

Source: Lecture 04; Review Questions – Reynolds averaging 4

What is the production term in the TKE equation?

Short exam answer: P = −⟨u_i′u_j′⟩∂U_i/∂x_j.

Explanation: It represents the work done by the mean shear against the Reynolds stresses. Positive P means transfer from mean kinetic energy to TKE.

Formula: P = −R_ij∂U_i/∂x_j

Key terms: production; mean shear

Source: Lecture 04; Compendium Ch. 2.2.2

Why is TKE production often positive in a conventional wall shear layer?

Short exam answer: In a wall shear layer, ∂U/∂y>0, while ⟨u′v′⟩ is usually negative. Therefore −⟨u′v′⟩∂U/∂y is positive.

Explanation: The negative correlation is caused by sweeps and ejections that transport streamwise momentum.

Formula: P = −⟨u′v′⟩dU/dy > 0

Key terms: u'v'; sweep; ejection

Source: Lecture 04; channel/boundary-layer lectures

What is dissipation ε?

Short exam answer: Dissipation is the rate at which viscosity converts TKE into internal energy at the small scales.

Formula: ε = ν⟨(∂u_i′/∂x_j)(∂u_i′/∂x_j)⟩

Key terms: viscous conversion; small scales

Source: Lecture 04; Review Questions – Reynolds averaging 4

What does turbulent transport do in the TKE equation?

Short exam answer: The triple correlation transports TKE from regions of high turbulence to other regions.

Explanation: It is a spatial redistribution and does not change the total energy in a closed domain when the boundary flux is zero.

Formula: −1/2 ∂⟨u_i′u_i′u_j′⟩/∂x_j

Key terms: triple correlation; redistribution

Source: Lecture 04; Review Questions – Reynolds averaging 4

What does pressure diffusion do in the TKE equation?

Short exam answer: The correlation between pressure fluctuations and velocity fluctuations transports TKE through space.

Formula: −(1/ρ)∂⟨u_j′p′⟩/∂x_j

Key terms: pressure diffusion; energy flux

Source: Lecture 04; Review Questions – Reynolds averaging 4

Which terms in the TKE equation are unclosed?

Short exam answer: The Reynolds stresses in the production term, the dissipation ε, pressure diffusion, and turbulent transport.

Explanation: They contain unknown second- and third-order correlations or fluctuation gradients that cannot be determined from the mean quantities alone.

Key terms: closure; TKE

Source: Review Questions – Reynolds averaging 4

Why can it be advantageous to solve transport equations for the Reynolds stresses?

Short exam answer: The model can represent anisotropy, flow history, and the effects of curved streamlines better than a scalar eddy-viscosity model.

Explanation: This often gives smaller modeling errors and improved separation prediction.

Key terms: RST transport; anisotropy

Source: Review Questions – Reynolds averaging 3

What are the disadvantages of Reynolds-stress transport models?

Short exam answer: Six additional transport equations must be solved, the equations are themselves unclosed, the model is more expensive and less numerically robust, and it requires a better computational grid.

Key terms: cost; robustness; six equations

Source: Review Questions – Reynolds averaging 3

Why can a conventional RANS calculation not predict a rare instantaneous event such as engine knock?

Short exam answer: RANS computes statistical mean quantities and models all turbulent fluctuations. A rare instantaneous realization is therefore not explicitly represented.

Explanation: LES or another time-resolved method is required when the large instantaneous fluctuations themselves must be observed.

Key terms: rare events; time resolved

Source: Review Questions – Reynolds averaging 5

Which four stability states were introduced?

Short exam answer: Stable, neutrally stable, linearly unstable, and nonlinearly unstable.

Explanation: A small perturbation decays in a stable system, remains constant in a neutrally stable system, grows in a linearly unstable system, and may grow only above a finite amplitude in a nonlinearly unstable system.

Key terms: stability; perturbation

Source: Review Questions – Transition 1

Which stability description best applies to a laminar flow in general?

Short exam answer: A laminar flow is best described as nonlinearly unstable.

Explanation: It may be linearly stable to infinitesimal disturbances, while finite-amplitude disturbances can still trigger transition through nonlinear mechanisms.

Key terms: subcritical transition; finite amplitude

Source: Review Questions – Transition 1

What is the procedure for a linear stability analysis?

Short exam answer: 1) Define the governing equations. 2) Split the flow into a base flow and a perturbation. 3) Subtract the base-flow equation. 4) Linearize the perturbation equation. 5) Apply a normal-mode ansatz. 6) Solve the eigenvalue problem. 7) Interpret the growth rates and construct a stability map.

Key terms: base flow; linearization; eigenvalue

Source: Review Questions – Transition 2

What does it mean to linearize the perturbation equation?

Short exam answer: Products of two or more perturbation quantities are neglected because they are second or higher order in the small amplitude.

Explanation: This produces a linear operator and allows each wave mode to be analyzed separately.

Key terms: small perturbations; second-order terms

Source: Lecture 06; Compendium Ch. 2.1.1

What is a normal-mode ansatz?

Short exam answer: The perturbation is assumed to be a wave with separated spatial and temporal dependence, for example q̂(y)e^{i(αx+βz−ωt)}.

Explanation: This transforms the differential equations into an eigenvalue problem for the amplitude profile and the complex frequency or wavenumber.

Formula: q′=q̂(y)e^{i(αx+βz−ωt)}

Key terms: wave mode; eigenvalue

Source: Lecture 06; Compendium Ch. 2.1.1

How is a complex frequency interpreted in temporal stability analysis?

Short exam answer: When ω=ω_r+iω_i and the wavenumber is real, the perturbation grows if ω_i>0 and decays if ω_i<0.

Formula: e^{-iωt}=e^{-iω_rt}e^{ω_it}

Key terms: temporal growth rate

Source: Lecture 06

How is a complex wavenumber interpreted in spatial stability analysis?

Short exam answer: When α=α_r+iα_i and the frequency is real, the amplitude varies as e^{−α_i x}. With this sign convention, downstream growth therefore requires α_i<0.

Formula: e^{iαx}=e^{iα_rx}e^{−α_ix}

Key terms: spatial growth

Source: Lecture 06

What does Rayleigh's inflection-point theorem state?

Short exam answer: For an inviscid parallel shear flow, an inflection point in the mean velocity profile is a necessary condition for linear instability.

Explanation: Necessary does not mean sufficient: an inflection point does not guarantee instability.

Key terms: inviscid; inflection point

Source: Review Questions – Transition 3

Why does Rayleigh's criterion not apply directly to viscous boundary layers?

Short exam answer: At finite Reynolds number, viscosity can enable Tollmien–Schlichting instability even when the velocity profile has no inflection point.

Explanation: The Blasius profile can therefore become linearly unstable at sufficiently high Reynolds number.

Key terms: viscous instability; TS waves

Source: Review Questions – Transition 3

What is Kelvin–Helmholtz instability?

Short exam answer: It is an inviscid shear-layer instability in which a velocity jump or a profile with an inflection point rolls up into vortices.

Explanation: It is typical of free shear layers, jets, and wakes.

Key terms: free shear layer; roll-up

Source: Lecture 06; Compendium Ch. 2.1.2

What is transient growth?

Short exam answer: It is strong temporary growth of perturbation energy even though all linear eigenmodes decay asymptotically.

Explanation: The linear operator is non-normal, so different decaying modes can reinforce one another temporarily. This can trigger nonlinear transition.

Key terms: non-normal operator; lift-up

Source: Lecture 06

What is the e^N method?

Short exam answer: It is a semi-empirical transition method in which the spatial growth rate of unstable waves is integrated. Transition is assumed when the amplitude amplification reaches e^N.

Explanation: N depends on the disturbance environment; low external turbulence usually allows a larger N before transition.

Formula: N(x)=ln(A/A₀)=∫−α_i dx

Key terms: transition prediction; amplitude

Source: Review Questions – Transition 4; Lecture 06

Name physical parameters that influence transition.

Short exam answer: Surface roughness, wall geometry and curvature, pressure gradient, external turbulence level, and the disturbance spectrum.

Key terms: roughness; disturbance spectrum

Source: Review Questions – Transition 4

What does self-similarity mean in a turbulent flow?

Short exam answer: When velocity and coordinates are normalized using suitable local scales, profiles from different positions collapse onto one universal shape function.

Explanation: The shape is unchanged even though the amplitude and width vary.

Key terms: profile collapse; similarity function

Source: Review Questions – Free shear flow 2

How does the centerline velocity decay in the self-similar region of a round jet?

Short exam answer: It decreases approximately inversely with the virtual distance from the nozzle.

Formula: U₀/U_J = C_u / ((x−x₀)/d)

Key terms: centerline decay; virtual origin

Source: Review Questions – Free shear flow 3

What is the virtual origin x₀ of a jet?

Short exam answer: It is an extrapolated point at which the self-similar solution would mathematically have started.

Explanation: It corrects for the fact that the near field at the nozzle is not self-similar.

Key terms: virtual origin; near field

Source: Lecture 07; Compendium Ch. 2.3.2

How does the width of a round jet develop downstream?

Short exam answer: The half-width r₀.₅ grows approximately linearly with x−x₀.

Explanation: The jet entrains surrounding fluid, so it spreads while the centerline velocity decreases.

Formula: r₀.₅ ∝ x−x₀

Key terms: jet spreading; entrainment

Source: Lecture 07; Compendium Ch. 2.3.2

How is the radial mean-velocity profile normalized in a self-similar round jet?

Short exam answer: Velocity is normalized by the local centerline velocity U₀(x), and radius is normalized by the local half-width r₀.₅(x).

Formula: U(r,x)/U₀(x)=f(r/r₀.₅)

Key terms: radial similarity

Source: Review Questions – Free shear flow 3

What does the eddy-viscosity approach state for the Reynolds shear stress in a round jet?

Short exam answer: It assumes that the Reynolds shear stress is proportional to the radial gradient of the axial mean velocity.

Formula: ⟨u_r′u_x′⟩ = −ν_T ∂U_x/∂r

Key terms: Boussinesq; free shear flow

Source: Review Questions – Free shear flow 4

How is eddy viscosity scaled in a self-similar round jet?

Short exam answer: ν_T is scaled by the product of the local centerline velocity and the local jet width.

Formula: ν_T/(U₀r₀.₅) ≈ 0.03

Key terms: eddy viscosity; similarity

Source: Review Questions – Free shear flow 4

Why is the momentum flux approximately conserved in a free jet?

Short exam answer: There is no wall force in the streamwise direction, and the surrounding fluid is initially at rest. The integrated axial momentum flux therefore remains approximately constant.

Explanation: This links the linear growth of the jet width to the 1/x decay of the centerline velocity.

Key terms: momentum conservation; entrainment

Source: Lecture 07; Compendium Ch. 2.3.2

What is entrainment in a jet?

Short exam answer: Entrainment is the process by which surrounding fluid is drawn into the turbulent jet.

Explanation: It increases the volume flow rate and the jet width downstream, but dilutes the momentum so that the mean velocity decreases.

Key terms: entrainment; volume flow rate

Source: Lecture 07

Name two similarities between laminar and turbulent channel flow.

Short exam answer: Both are driven by a pressure gradient and have a total shear stress that varies linearly from the wall to zero at the center plane.

Explanation: They are also described by the same outer control parameters: ρ, ν, channel half-height δ, and the pressure gradient or τ_w.

Key terms: pressure driven; total stress

Source: Review Questions – Channel 1

Name two differences between laminar and turbulent channel flow.

Short exam answer: Turbulent flow contains Reynolds shear stress and has a flatter mean-velocity profile with a lower maximum velocity at the same volume flow rate.

Explanation: In laminar flow, the entire shear stress is carried by molecular viscosity.

Key terms: Reynolds stress; flat profile

Source: Review Questions – Channel 1

What is the friction velocity u_τ?

Short exam answer: It is a velocity scale constructed from the wall shear stress.

Explanation: It is not the actual fluid velocity at the wall, but a useful scale for near-wall dynamics.

Formula: u_τ=√(τ_w/ρ)

Key terms: wall shear; velocity scale

Source: Review Questions – Channel 2