Turbulent flows intensive

What characterizes turbulent flow?

Short exam answer
Turbulent flow is irregular, time-dependent, rotational, and three-dimensional. It contains a broad range of eddy sizes and is best described statistically, even though the governing equations are deterministic.

Explanation
Turbulence appears chaotic, but it still contains coherent structures and repeatable statistical behavior.

Key terms
irregular; 3D; rotational; eddies; statistical description

Source: Lecture 01; Compendium Ch. 1.1

Name four important differences between laminar and turbulent flow.

Short exam answer
Compared with laminar flow, turbulent flow has stronger mixing, greater heat and mass transfer, higher wall friction, and greater resistance to flow separation. Laminar flow is layered and smoother.

Explanation
Transverse turbulent motion transports momentum, heat, and material much faster than molecular diffusion alone.

Key terms
mixing; heat transfer; friction; separation

Source: Review Questions – Introduction 1

Why is wall friction usually higher in turbulent flow?

Short exam answer
Turbulent mixing brings high-speed fluid closer to the wall, which increases the wall-normal mean-velocity gradient. Because wall shear stress is proportional to this gradient, friction increases.

Explanation
The turbulent stress itself vanishes at the wall; turbulence changes the velocity profile so that the molecular shear stress at the wall becomes larger.

Formula / key relation
τ_w = μ(∂U/∂y)_w

Key terms
wall gradient; wall shear stress

Source: Review Questions – Introduction 2

Why does turbulence usually delay flow separation?

Short exam answer
Turbulent mixing transports high-momentum fluid from the outer flow toward the wall. The near-wall fluid can therefore oppose an adverse pressure gradient for longer, so separation is delayed.

Explanation
The benefit is delayed separation; the cost is increased skin friction.

Key terms
adverse pressure gradient; momentum transport; separation

Source: Lecture 01; Review Questions – Introduction

Compare molecular and turbulent diffusion using their time scales.

Short exam answer
Molecular diffusion has t_m ~ l²/ν, while turbulent transport has t_t ~ l/u′. Their ratio is t_m/t_t ~ u′l/ν = Re_t. For Re_t ≫ 1, turbulent transport is much faster.

Explanation
The official example gives Re_t ≈ 1000 for air with l≈1 cm and u′≈1 m/s.

Formula / key relation
t_m ~ l²/ν; t_t ~ l/u′; t_m/t_t ~ Re_t

Key terms
diffusion time scale; turbulent Reynolds number

Source: Review Questions – Introduction 3

Explain the turbulent energy cascade and where it ends.

Short exam answer
Energy is supplied to large eddies, transferred by nonlinear interactions to progressively smaller eddies, and finally dissipated into internal energy at the smallest scales.

Explanation
The cascade ends when the local Reynolds number is of order one, so viscous and inertial effects are comparable.

Formula / key relation
Re_η ~ 1

Key terms
energy cascade; dissipation; Kolmogorov 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
When a vortex is stretched, its cross-sectional area decreases and its rotation rate increases. Vortex stretching transfers energy toward smaller scales and helps sustain three-dimensional turbulence.

Explanation
The stretching term vanishes in purely two-dimensional flow, which is why the three-dimensional nature of turbulence is fundamental.

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

Key terms
vorticity; 3D; cascade

Source: Compendium Ch. 1.3; Lecture 01

What is Reynolds decomposition?

Short exam answer
An instantaneous quantity is split into a mean and a fluctuation: u_i = ⟨u_i⟩ + u′_i. By definition, the mean fluctuation is zero: ⟨u′_i⟩ = 0.

Explanation
This decomposition is the starting point for Reynolds averaging and the derivation of the RANS equations.

Formula / key relation
u_i = ⟨u_i⟩ + u′_i; ⟨u′_i⟩=0

Key terms
instantaneous; mean; fluctuation

Source: Lecture 01–02; Compendium Ch. 2.2.1

What is an ensemble average, and how is it obtained for steady and unsteady flows?

Short exam answer
An ensemble contains all realizations of a turbulent flow. For a statistically steady and ergodic flow, the ensemble average can be replaced by a sufficiently long time average. For an unsteady flow, it is obtained by averaging many identical experiments at the same phase or time.

Explanation
The law of large numbers means that estimated statistics approach the true values as the number of realizations increases.

Key terms
ensemble; ergodicity; time average; repeated experiments

Source: Review Questions – HIT 3

What is the difference between homogeneous and isotropic turbulence?

Short exam answer
Homogeneous means that statistical properties are invariant under translation in space. Isotropic means that statistical properties are invariant under rotation and reflection.

Explanation
Homogeneity concerns position; isotropy concerns direction.

Key terms
translation; rotation; reflection

Source: Review Questions – HIT 1; Lecture 02

How is homogeneous isotropic turbulence generated experimentally, and where should it be measured?

Short exam answer
A steady flow is passed through an approximately uniform grid. Measurements should be taken sufficiently far downstream and within a sufficiently small control volume.

Explanation
Close to the grid, the flow still remembers individual grid bars; over a large volume, residual inhomogeneity can become visible.

Key terms
grid turbulence; downstream distance; small control volume

Source: Review Questions – HIT 2; Lecture 02

What is Taylor's frozen-turbulence hypothesis?

Short exam answer
When the mean convection velocity is much larger than the fluctuation velocity, turbulent structures are treated as nearly frozen while they pass a fixed probe. Temporal changes can then be converted into spatial changes.

Explanation
It allows a time signal at one point to be interpreted approximately as a spatial signal.

Formula / key relation
∂(·)/∂t ≈ −Ū ∂(·)/∂x

Key terms
frozen turbulence; probe; convection

Source: Lecture 02; Compendium Ch. 2.3.1

What do two-point correlation and the integral length scale describe?

Short exam answer
Two-point correlation measures how similar velocity fluctuations are at two positions separated by r. The integral length scale is the area under the correlation curve and represents the characteristic size of the large energy-containing eddies.

Explanation
Correlation is close to one for small separation and decreases as the points become dynamically independent.

Formula / key relation
L ~ ∫₀^∞ f(r)dr

Key terms
spatial correlation; large eddies; length scale

Source: Lecture 02; Compendium Ch. 2.3.1

Name the three spectral ranges and state which are universal.

Short exam answer
The energy-containing range is non-universal and depends on geometry and forcing. The inertial range is universal and depends mainly on ε and k. The dissipation range is also universal at high Reynolds number and depends on ε and ν.

Explanation
The inertial and dissipation ranges are especially useful for modeling because their small-scale behavior is less dependent on the global flow.

Key terms
energy-containing; inertial; dissipation; universality

Source: Review Questions – HIT 4

What is Kolmogorov's −5/3 law, and how are the exponents obtained?

Short exam answer
In the inertial range, E(k)=C_K ε^(2/3)k^(−5/3). The exponents follow from dimensional analysis using only ε and k.

Explanation
In a log–log spectrum, the exponent −5/3 appears as the slope of a straight inertial-range segment.

Formula / key relation
E(k)=C_K ε^(2/3)k^(−5/3)

Key terms
energy spectrum; inertial range; dimensional analysis

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 low wavenumbers in the energy-containing range. Nonlinear transfer moves energy through wavenumber space. Dissipation acts mainly at high wavenumbers in the dissipation range.

Explanation
Transfer redistributes energy between scales; it does not create or destroy total energy.

Key terms
production; transfer; dissipation; wavenumber

Source: Review Questions – HIT 6

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

Short exam answer
Because nonlinear transfer only redistributes energy between wavenumbers. Energy lost by one part of the spectrum is gained by another, so the net integral is zero.

Explanation
Only production adds turbulent energy and viscous dissipation removes it.

Formula / key relation
∫₀^∞ T(k)dk = 0

Key terms
conservative redistribution; spectral transfer

Source: Review Questions – HIT 7

What is a wavenumber triad?

Short exam answer
The nonlinear term couples three Fourier modes whose wavevectors satisfy a triad relation. Energy exchange at one wavenumber therefore depends on interactions with other wavenumbers.

Explanation
A numerical model must represent the relevant wavenumber interactions, not only an isolated scale.

Formula / key relation
k + p + q = 0

Key terms
nonlinear interaction; Fourier modes

Source: Review Questions – HIT 8

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

Short exam answer
Decomposition inserts u_i=⟨u_i⟩+u′_i into ∂u_i/∂x_i=0. Reynolds averaging takes the average of the equation and gives ∂⟨u_i⟩/∂x_i=0.

Explanation
For incompressible flow, both the mean field and the fluctuation field are divergence-free.

Formula / key relation
∂(⟨u_i⟩+u′_i)/∂x_i=0; ∂⟨u_i⟩/∂x_i=0

Key terms
decomposition; averaging; continuity

Source: Review Questions – Reynolds averaging 1

Why do Reynolds stresses appear, and what do they represent physically?

Short exam answer
They appear because averaging the nonlinear convection term produces correlations of velocity fluctuations, ⟨u′_i u′_j⟩. Physically, they represent turbulent transport of momentum.

Explanation
The diagonal components are fluctuation intensities; the off-diagonal components are turbulent shear correlations.

Formula / key relation
R_ij = ⟨u′_i u′_j⟩

Key terms
nonlinearity; momentum flux; Reynolds stress

Source: Lecture 04; Compendium Ch. 2.2.1

Explain the closure problem in RANS.

Short exam answer
Averaging introduces the six independent Reynolds-stress components as new unknowns, but the RANS equations do not provide enough additional equations to determine them. The system is therefore not closed.

Explanation
The review solution counts eleven unknown parameters but only five governing equations, so turbulence modeling is required.

Key terms
unknowns; equations; closure; modeling

Source: Review Questions – Reynolds averaging 2

What is turbulent kinetic energy?

Short exam answer
Turbulent kinetic energy is half the trace of the Reynolds-stress tensor: K = 1/2⟨u′_i u′_i⟩. It measures the kinetic energy contained in velocity fluctuations per unit mass.

Explanation
It combines the fluctuation energy in all three velocity directions.

Formula / key relation
K = 1/2(⟨u′²⟩+⟨v′²⟩+⟨w′²⟩)

Key terms
TKE; trace; fluctuation energy

Source: Lecture 04; Compendium Ch. 2.2.2

Name and explain the main terms in the TKE transport equation.

Short exam answer
The equation contains local change, mean convection, production, dissipation, turbulent transport, pressure diffusion, and viscous diffusion.

Explanation
Production transfers energy from the mean flow to turbulence; dissipation converts TKE to internal energy; the diffusion and transport terms redistribute TKE in space.

Key terms
production; dissipation; transport; diffusion

Source: Lecture 04; Compendium Ch. 2.2.2

What is TKE production in a simple wall shear flow?

Short exam answer
Production is the transfer of kinetic energy from the mean velocity gradient to turbulent fluctuations. In a simple wall shear flow, P = −⟨u′v′⟩ dU/dy, which is usually positive.

Explanation
Typically dU/dy>0 and ⟨u′v′⟩<0, so the product with the leading minus sign is positive.

Formula / key relation
P = −⟨u′v′⟩ dU/dy

Key terms
mean shear; Reynolds shear stress; positive production

Source: Lecture 04; Compendium Ch. 2.2.2

What is dissipation ε?

Short exam answer
Dissipation is the viscous destruction of turbulent kinetic energy at the smallest scales, where velocity gradients are large. The lost mechanical energy becomes internal energy.

Explanation
Dissipation is always non-negative in the TKE balance.

Formula / key relation
ε = ν⟨(∂u′_i/∂x_j)(∂u′_i/∂x_j)⟩

Key terms
viscosity; smallest scales; heat

Source: Lecture 04; Compendium Ch. 2.2.2

Which terms in the TKE equation are unclosed?

Short exam answer
The Reynolds stresses in production, the dissipation rate, pressure diffusion, and turbulent transport are unclosed.

Explanation
They contain correlations or gradients of fluctuating quantities that cannot be expressed directly using only the mean variables.

Key terms
RST; ε; pressure diffusion; triple correlation

Source: Review Questions – Reynolds averaging 4

What are the advantages and disadvantages of solving Reynolds-stress transport equations?

Short exam answer
Advantages: better representation of anisotropy, curvature, and complex strain, with generally smaller modeling error. Disadvantages: six additional transport equations, greater cost, stricter grid requirements, and lower numerical robustness.

Explanation
The RST equations also contain unclosed terms, so they still require modeling.

Key terms
anisotropy; six equations; accuracy; robustness

Source: Review Questions – Reynolds averaging 3; RSM 2

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

Short exam answer
RANS predicts statistical mean quantities, not individual instantaneous realizations. A rare event can disappear in the averaging process.

Explanation
A scale-resolving method such as LES is required when the transient fluctuation history is essential.

Key terms
mean flow; rare event; LES

Source: Review Questions – Reynolds averaging 5

Give key similarities and differences between laminar and turbulent channel flow.

Short exam answer
Both are driven by a streamwise pressure gradient, have a total shear stress that varies linearly across the half-channel, and use the same control parameters. Turbulent flow additionally contains Reynolds shear stress and has a fuller mean-velocity profile with a lower centerline maximum for the same flow rate.

Explanation
The fuller profile follows from stronger transverse momentum transport.

Key terms
pressure gradient; total stress; fuller profile

Source: Review Questions – Turbulent channel flow 1

Define friction velocity, viscous length, wall units, and friction Reynolds number.

Short exam answer
u_τ=√(τ_w/ρ), l⁺=ν/u_τ, y⁺=u_τy/ν, u⁺=U/u_τ, and Re_τ=u_τδ/ν=δ/l⁺.

Explanation
These variables scale the near-wall flow using wall shear stress and viscosity.

Formula / key relation
u_τ=√(τ_w/ρ); l⁺=ν/u_τ; y⁺=u_τy/ν; Re_τ=u_τδ/ν

Key terms
wall units; inner scaling

Source: Review Questions – Turbulent channel flow 2

How is total shear stress distributed in a fully developed channel?

Short exam answer
The total shear stress varies linearly from τ_w at the wall to zero at the channel centerline. It is the sum of molecular and turbulent shear stress.

Explanation
Near the wall molecular stress dominates; farther from the wall Reynolds shear stress dominates.

Formula / key relation
τ_tot = μ dU/dy − ρ⟨u′v′⟩

Key terms
linear total stress; molecular; turbulent

Source: Lecture 08; Compendium Ch. 2.3.3

State the universal law-of-the-wall regions and their approximate limits.

Short exam answer
Viscous sublayer: y⁺<5. Buffer layer: 5<y⁺<30. Logarithmic layer: y⁺>30 while y/δ<0.3. Outer velocity-defect region: approximately y/δ>0.3.

Explanation
The viscous sublayer and log layer have simple universal velocity laws; the buffer and outer layers are less universal.

Key terms
viscous sublayer; buffer; log layer; outer layer

Source: Review Questions – Turbulent channel flow 3

What are the linear and logarithmic laws of the wall?

Short exam answer
In the viscous sublayer, u⁺=y⁺. In the logarithmic layer, u⁺=(1/κ)ln(y⁺)+A.

Explanation
The linear law reflects molecular-viscous dominance; the log law follows from an overlap region where neither inner nor outer scaling alone is sufficient.

Formula / key relation
u⁺=y⁺; u⁺=(1/κ)ln(y⁺)+A

Key terms
linear law; log law; universal velocity

Source: Review Questions – Turbulent channel flow 3; Boundary layer 3

How do velocity fluctuations behave very close to a no-slip wall, and why is this hard to model?

Short exam answer
The streamwise and spanwise fluctuations scale approximately as u′~y and w′~y, while the wall-normal fluctuation scales as v′~y². The turbulence therefore becomes strongly anisotropic and nearly two-dimensional near the wall.

Explanation
A scalar eddy viscosity cannot easily reproduce this directional behavior.

Formula / key relation
u′~y; w′~y; v′~y²

Key terms
anisotropy; no-slip; impermeability

Source: Review Questions – Turbulent channel flow 4

What is the main difference between a fully developed channel flow and a boundary layer?

Short exam answer
A fully developed channel flow is statistically independent of the streamwise coordinate, whereas a boundary layer grows downstream and its mean profiles depend on x. A boundary layer also matches an outer free-stream velocity U_∞.

Explanation
Because the boundary-layer thickness changes with x, continuity requires a nonzero mean wall-normal velocity V even in a two-dimensional boundary layer.

Key terms
streamwise development; free stream; U and V

Source: Lecture 08–09; Compendium Ch. 2.3.3–2.3.4

To what extent is the law of the wall universal in a turbulent boundary layer?

Short exam answer
The viscous sublayer and logarithmic layer are approximately universal under suitable conditions. The buffer layer and outer velocity-defect region are not strictly universal and are influenced by pressure gradient, flow history, free-stream turbulence, roughness, and three-dimensionality.

Explanation
The inner wall laws are more universal than the outer-layer description.

Key terms
inner layer; outer layer; pressure gradient; history

Source: Review Questions – Boundary layer 3; Lecture 09

What mesh criteria distinguish DNS, wall-resolved RANS, and wall-modeled RANS near a wall?

Short exam answer
DNS must resolve the Kolmogorov scale with several cells. Wall-resolved RANS places the first cell center around y⁺≈1. Wall-modeled RANS places the first cell in the log layer, typically y⁺>30, and uses a wall function.

Explanation
These approaches trade computational cost against the amount of near-wall physics that is modeled.

Key terms
Kolmogorov scale; y+=1; y+>30; wall function

Source: Review Questions – Boundary layer 1

How does DNS cost scale with Reynolds number, and what happens if L is halved while U is quadrupled?

Short exam answer
The Reynolds number doubles. Using t_comp~Re^2.25, the cost increases by about 2^2.25≈4.8, i.e. roughly a factor of five.

Explanation
The estimate follows from the shrinking Kolmogorov length scale and the increasing number of three-dimensional grid cells.

Formula / key relation
t_comp ~ Re^2.25

Key terms
DNS cost; Reynolds scaling

Source: Review Questions – Boundary layer 2

What is the role of hairpin or horseshoe vortices, sweeps, and ejections?

Short exam answer
Near-wall vortices connect the outer flow with the near-wall region. Sweeps carry high-speed fluid toward the wall; ejections carry low-speed fluid away from the wall. Together they sustain turbulence and momentum exchange.

Explanation
This structural transport increases near-wall energy and wall shear.

Key terms
hairpin vortex; sweep; ejection; momentum transport

Source: Review Questions – Boundary layer 4; Compendium Ch. 2.3.5

Give the causal chain for delayed separation in a turbulent boundary layer.

Short exam answer
Turbulent structures increase cross-stream momentum transport → high-speed outer fluid is brought toward the wall → the near-wall fluid gains momentum → it resists the adverse pressure gradient longer → separation is delayed.

Explanation
The same mechanism also increases wall friction.

Key terms
momentum transport; energetic boundary layer; delayed separation

Source: Lecture 01 and 09; Compendium Ch. 2.3.4–2.3.5

Compare DNS, LES, and RANS in terms of what is resolved and modeled.

Short exam answer
DNS resolves all turbulent scales. LES resolves the large scales and models the small subgrid scales. RANS resolves the mean flow and models all turbulent fluctuations.

Explanation
Computational cost and instantaneous detail decrease from DNS to LES to RANS.

Key terms
DNS; LES; RANS; resolved scales; modeled scales

Source: Review Questions – Zero- and one-equation models 1

What is the physical analogy behind the eddy-viscosity hypothesis?

Short exam answer
Molecular viscosity represents momentum exchange by molecular motion. Eddy viscosity represents momentum exchange by turbulent fluctuations. The anisotropic Reynolds stress is modeled as proportional to the mean strain-rate tensor.

Explanation
This is a modeling analogy; eddy viscosity is not a material constant.

Formula / key relation
⟨u′_i u′_j⟩ − (2/3)Kδ_ij = −2ν_T S_ij

Key terms
Boussinesq hypothesis; turbulent momentum transport

Source: Review Questions – Zero- and one-equation models 2

What is the difference between molecular viscosity and eddy viscosity?

Short exam answer
Molecular viscosity is a material property determined by the fluid and thermodynamic state. Eddy viscosity is a property of the flow and turbulence model, so it varies with position and flow conditions.

Explanation
ν_T can be many times larger than ν, but it has no universal value for a fluid.

Key terms
material property; flow property

Source: Review Questions – Zero- and one-equation models 2

What is the fundamental limitation of a linear eddy-viscosity model?

Short exam answer
It assumes that the Reynolds-stress anisotropy is instantaneously aligned with and proportional to the local mean strain through one scalar ν_T. This is generally not true in complex, curved, rotating, separating, or history-dependent flows.

Explanation
The model therefore often predicts anisotropy incorrectly.

Key terms
linear relation; scalar ν_T; anisotropy

Source: Review Questions – Two-equation models 3

How does the mixing-length model estimate eddy viscosity, and why is a universal mixing length difficult?

Short exam answer
In a simple shear flow, ν_T=l_m²|dU/dy|, often with l_m=κy near a wall. A universal l_m is difficult because the relevant turbulent length scale depends on geometry, boundaries, separation, and flow history.

Explanation
Using wall distance directly links the turbulence model to a particular boundary condition.

Formula / key relation
ν_T=l_m²|dU/dy|; l_m=κy

Key terms
Prandtl mixing length; length-scale problem

Source: Review Questions – Zero- and one-equation models 3

What is the idea of algebraic two-layer models, and why can they fail in a separated diffuser?

Short exam answer
They use different empirical eddy-viscosity formulas in inner and outer regions and connect them algebraically. They can fail in separation because the turbulence is nonlocal and history-dependent, while the algebraic model reacts only to local quantities.

Explanation
Excessive modeled momentum transport can make the boundary layer too energetic and delay separation too much.

Key terms
Cebeci–Smith; Baldwin–Lomax; separation

Source: Lecture 10; Compendium Ch. 4.2

What are the main advantage and weakness of the Spalart–Allmaras model?

Short exam answer
Advantage: it transports a viscosity-like variable directly, avoiding an explicit mixing-length formula, and often works well for calibrated aerodynamic boundary layers. Weakness: the transported eddy-viscosity variable is not a fundamental conserved quantity, so its transport equation is mainly empirical.

Explanation
It is a practical one-equation model rather than a first-principles turbulence equation.

Key terms
one-equation model; aerodynamic flows; empirical

Source: Review Questions – Zero- and one-equation models 4

What is the main advantage of two-equation models?

Short exam answer
They transport two turbulence variables, typically K and a scale-setting quantity such as ε or ω. This allows ν_T to be calculated without prescribing a mixing length.

Explanation
The second equation provides a turbulent time or length scale.

Formula / key relation
ν_T~K²/ε or ν_T~K/ω

Key terms
two equations; turbulence scale

Source: Review Questions – Two-equation models 1

Compare the k–ε and k–ω models.

Short exam answer
k–ε transports K and ε and is generally strong in free-shear flows but less reliable very near walls and in separation. k–ω transports K and ω, is robust near walls, but can be sensitive to the specified outer-flow value of ω.

Explanation
Their eddy viscosities scale as ν_T~K²/ε and ν_T~K/ω, respectively.

Formula / key relation
ν_T~K²/ε; ν_T~K/ω

Key terms
free shear; near wall; boundary conditions

Source: Review Questions – Two-equation models 2

What is the fundamental difference between Reynolds-stress models and eddy-viscosity models?

Short exam answer
Eddy-viscosity models use a scalar ν_T to relate Reynolds stresses to mean strain. Reynolds-stress models solve transport equations for the individual Reynolds-stress components and can therefore represent anisotropy more directly.

Explanation
RSMs are usually more accurate for strongly curved or anisotropic flows, but are more expensive and less robust.

Key terms
RSM; EVM; anisotropy; cost

Source: Review Questions – RSM 1–2

Why is wall-bounded turbulence difficult to model, and what wall treatments are used?

Short exam answer
Very large gradients occur on very small scales, and the Reynolds stresses become strongly anisotropic near the wall. The main strategies are fine wall-resolved grids with y⁺≈1, wall functions with the first cell in the log layer, or simplified high-Re approaches that neglect part of the direct wall influence.

Explanation
Any wall treatment must be consistent with its mesh placement and boundary conditions.

Key terms
near-wall gradients; anisotropy; wall function; y+

Source: Review Questions – RSM and wall modeling 3

Why are canonical flows useful, and what does self-similarity mean?

Short exam answer
Canonical flows are simple configurations that can be studied thoroughly and used to develop and validate models. Self-similarity means that profiles at different positions collapse onto one curve when normalized with suitable local velocity and length scales.

Explanation
A successful similarity description reduces many downstream profiles to one universal function.

Key terms
canonical flow; normalized profile; model validation

Source: Review Questions – Free shear flow 1–2

How do centerline velocity and jet width behave in the self-similar region of a round jet?

Short exam answer
The centerline velocity decays approximately as 1/(x−x₀), while the jet half-width grows approximately linearly with x−x₀.

Explanation
The virtual origin x₀ shifts the idealized self-similar solution upstream or downstream of the physical nozzle.

Formula / key relation
U₀/U_J = C_u /[(x−x₀)/d]; r₀.₅∝(x−x₀)

Key terms
centerline decay; spreading; virtual origin

Source: Review Questions – Free shear flow 3; Compendium Ch. 2.3.2

How are the radial velocity profile and eddy viscosity normalized in a self-similar round jet?

Short exam answer
The mean velocity collapses as ⟨u⟩/U₀=f(r/r₀.₅). The eddy viscosity scales with U₀r₀.₅, so ν_T/(U₀r₀.₅) is approximately constant over a substantial region.

Explanation
The local centerline velocity is the velocity scale and the half-width is the length scale.

Formula / key relation
⟨u⟩/U₀=f(r/r₀.₅); ν_T/(U₀r₀.₅)≈0.03

Key terms
similarity variables; eddy viscosity

Source: Review Questions – Free shear flow 3–4

What are entrainment and momentum conservation in a free jet?

Short exam answer
Entrainment is the drawing of surrounding fluid into the jet by turbulent mixing. With negligible external axial force, the integral axial momentum flux is approximately conserved even though the jet widens and its centerline velocity decreases.

Explanation
Entrainment explains why the volume flow rate grows downstream.

Key terms
entrainment; momentum flux; jet spreading

Source: Lecture 07; Compendium Ch. 2.3.2

Which stability states were introduced, and which best describes laminar flow?

Short exam answer
The states are stable, neutrally stable, unstable, and nonlinearly unstable. The review solution describes a laminar flow in general as nonlinearly unstable.

Explanation
Small disturbances may decay, while disturbances above a finite threshold can trigger transition.

Key terms
stable; neutral; unstable; nonlinear instability

Source: Review Questions – Transition 1

Give the procedure for a linear stability analysis.

Short exam answer
Define the governing equations and base flow; add a small perturbation; subtract the base-flow equations; neglect nonlinear perturbation products; apply a normal-mode ansatz; solve the eigenvalue problem; interpret growth rates and map the stability region.

Explanation
The analysis asks whether infinitesimal disturbances grow or decay.

Key terms
base flow; perturbation; linearization; eigenvalue

Source: Review Questions – Transition 2

State Rayleigh's inflection-point theorem and explain the viscous–inviscid difference.

Short exam answer
For an inviscid parallel shear flow, an inflection point in the mean velocity profile is a necessary condition for instability. At finite Reynolds number, viscous boundary layers without an inflection point can still become unstable.

Explanation
Tollmien–Schlichting instability is a key example of a viscous instability.

Key terms
inflection point; inviscid; viscous; TS wave

Source: Review Questions – Transition 3

What are transient growth and the e^N method, and which physical factors influence transition?

Short exam answer
Transient growth is temporary amplification caused by non-normal disturbance interactions even when all eigenmodes are asymptotically stable. The e^N method integrates linear disturbance growth until an empirical amplification threshold is reached. Transition is influenced by roughness, wall geometry, pressure gradient, and the disturbance spectrum.

Explanation
These ideas explain why transition location depends strongly on the environment and not only on one critical Reynolds number.

Key terms
non-modal growth; amplification factor; roughness; disturbances

Source: Review Questions – Transition 4; Lecture 06