Classical Mechanics · Upper Division

Fluid Mechanics

Fluids are continuous media described by field equations — the Navier-Stokes equations — that govern everything from blood flow to ocean currents to atmospheric weather, and remain one of the great unsolved problems of mathematics.

PrerequisitesNewtonslaws(Ch.3)Vectorsandcalculus(Ch.2122)PartialdifferentialequationsNewton's laws (Ch. 3) \cdot Vectors and calculus (Ch. 21-22) \cdot Partial differential equations
Learning Goals
  • Distinguish Eulerian and Lagrangian descriptions and write the material derivative D/Dt in terms of partial derivatives.
  • Derive thecontinuityequationforacompressiblefluidandstatetheincompressibilityconditione continuity equation for a compressible fluid and state the incompressibility condition \nabla
  • Apply Bernoulli's equation to find flow velocities and pressure differences in steady, inviscid, incompressible flow.
  • Define the Reynolds number and use it to classify flow regimes from creeping (Stokes) to fully turbulent.
  • ExplainPrandtlsboundarylayerconceptandtheδ(\nux/U)scalingforaflatplateExplain Prandtl's boundary layer concept and the \delta \propto \sqrt(\nux/U) scaling for a flat plate.

FM.1 Kinematics and the Continuity Equation

A fluid is described by the velocity field v(r, t) — the velocity of the fluid element at position r at time t. Two frames of description:

Eulerian: fixed position r, varying t. What velocity do we observe at a fixed point in space? The relevant time derivative is ∂/∂t (local).

Lagrangian: follow a fluid parcel. The material derivative(rate of change following the fluid) is:

DDt=t+v(material/substantial derivative)\frac{D}{Dt} = \frac{\partial}{\partial t} + \mathbf{v}\cdot\nabla \qquad \text{(material/substantial derivative)}(FM.1)

Mass conservation gives the continuity equation:

ρt+(ρv)=0(continuity, compressible)\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v}) = 0 \qquad \text{(continuity, compressible)}(FM.2)

For an incompressible fluid (ρ = const): ∇·v = 0 — the flow is divergence-free. This holds for liquids and subsonic gas flows.

FM.2 Euler and Navier-Stokes Equations

Newton's second law for a fluid element, including pressure and viscosity:

ρ(vt+(v)v)=P+η2v+ρg(Navier-Stokes)\rho\left(\frac{\partial \mathbf{v}}{\partial t}+(\mathbf{v}\cdot\nabla)\mathbf{v}\right) = -\nabla P + \eta\nabla^2\mathbf{v} + \rho\mathbf{g} \qquad \text{(Navier-Stokes)}(FM.3)

For inviscid flow (η = 0): Euler's equation. The nonlinear term (v·∇)v is the inertial term — it causes turbulence, vortex stretching, and makes Navier-Stokes brutally difficult. Whether smooth solutions always exist in 3D is one of the Millennium Prize Problems (unsolved, $1M prize).

Definition FM.1Reynolds Number
TheReynoldsnumberRe=\rhovL/η=vL/ν(ν=η/ρiskinematicviscosity)measurestheratioThe Reynolds number Re = \rhovL/\eta = vL/\nu (\nu = \eta/\rho is kinematic viscosity) measures the ratio of inertial to viscous forces:Re=(inertialforce)/(viscousforce)=\rhov2L2/(\etavL)=\rhovL/ηRe = (inertial force) / (viscous force) = \rhov^{2}L^{2} / (\etavL) = \rhovL/\etaRe ≪ 1: viscous dominates (Stokes flow — creeping, reversible, spermswimming).Re1:inertialdominates(turbulent,mixing,aircraft).Transition:Reperm swimming). Re ≫ 1: inertial dominates (turbulent, mixing, aircraft). Transition: Re \approx

FM.3 Bernoulli's Equation

For inviscid, steady, incompressible flow along a streamline:

P+12ρv2+ρgh=constant(Bernoulli’s equation)P + \frac{1}{2}\rho v^2 + \rho gh = \mathrm{constant} \qquad \text{(Bernoulli's equation)}(FM.4)

This is energy conservation for fluid elements. Bernoulli's principle — faster flow, lower pressure — explains airfoil lift (wing), carburetor operation, and the Venturi effect. It is derived from Euler's equation by integration along a streamline.

Example FM.1Torricelli's Law

A large tank has a small hole at depth h below the surface. Find the exit velocity.

Apply Bernoulli:Atsurface:P=Patm,v0(largetank),height=h.Athole:P=Patm,height=0At surface: P = P_{atm}, v \approx 0 (large tank), height = h. At hole: P = P_{atm}, height = 0.
Bernoulli:Patm+0+\rhogh=Patm+12\rhove2xit+0P_{atm} + 0 + \rhogh = P_{atm} + \frac{1}{2}\rhov^{2}_exit + 0
Result:vexit=(2gh)(Torricellislaw,1643v_{exit} = \sqrt(2gh) \qquad (Torricelli's law, 1643
Same as projectile:This is the velocity a ballacquiresfallingfreelyaheighthasifthewaterfellfreely.Dischargerate:Q=ball acquires falling freely a height h — as if the water fell freely. Discharge rate: Q =le(2ghle \sqrt(2gh.

FM.4 Vorticity and Potential Flow

The vorticity ω = ∇ × v measures local rotation of the fluid. For irrotational flow (ω = 0): v = ∇φ for a velocity potential φ. Combined with incompressibility (∇·v = 0):

2ϕ=0(Laplace’s equation for potential flow)\nabla^2\phi = 0 \qquad \text{(Laplace's equation for potential flow)}(FM.5)

Potential flow is solved by the same methods as electrostatics! A cylinder in uniform flow U_∞ has solution φ = U_∞ r(1 + R²/r²) cos θ — D'Alembert's paradox: no drag. Real fluids have viscous boundary layers that separate, creating drag — potential flow misses this entirely. Adding circulation Γ (rotation around the cylinder) gives lift:

L=ρUΓ(Kutta-Joukowski theorem)L = \rho U_\infty \Gamma \qquad \text{(Kutta-Joukowski theorem)}(FM.6)
Example FM.2Stokes Drag on a Sphere

For very viscous flow (Re ≪ 1) past a sphere of radius R moving at velocity U, the drag force is:

Stokes flow solution:IntheStokeslimit,droptheinertialterminNS:η2v=\nablaP.SolvewithnoslipBCatspIn the Stokes limit, drop the inertial term in N-S: \eta\nabla^{2}v = \nablaP. Solve with no-slip BC at sphere surface.
Stokes drag formula:Fdrag=6π\etaRUF_{drag} = 6\pi\etaRU
Terminal velocity:Forasphereofdensityρsinfluidρf:weightbuoyancy=drag(4/3)\piR3(ρsρf)gFor a sphere of density \rho_s in fluid \rho_f: weight - buoyancy = drag \to (4/3)\piR^{3}(\rho_s - \rho_f)g = 6π\etaRUUterminal=2R2(ρsρf)g/(9η6\pi\etaRU \to U_{terminal} = 2R^{2}(\rho_s-\rho_f)g/(9\eta
Example:RaindropR=1mm,ρwater=1000,ρair=1.2kg/m3,ηair=1.8×105Pa.U=2(103)2×99Rain drop R=1mm, \rho_water = 1000, \rho_air = 1.2 kg/m^{3}, \eta_air = 1.8\times10^{-5} Pa\cdots. U = 2(10^{-3})^{2}\times999×9.8/(9×1.8×105)12m/s.(Re=\rhovR/η800Stokesisnotvalidhere;theactualterm9\times9.8/(9\times1.8\times10^{-5}) \approx 12 m/s. (Re = \rhovR/\eta \approx 800 — Stokes is not valid here; the actual terminal velocity with form drag correction is ~9 m/s.)
Definition FM.2Common Traps
  • Eulerian and Lagrangian are viewpoints: fixed-point derivatives and parcel-following derivatives are not the same.
  • Incompressible does not mean densityless: itmeansdensityofeachparcelisconstant,giving\cdotv=0it means density of each parcel is constant, giving \nabla\cdotv = 0
  • Bernoulli has assumptions: steady, inviscid, incompressible flow along a streamline.
  • High Reynolds number does not mean no viscosity: viscosity may dominate thin boundary layers.
  • Potential flow misses drag: real drag often comes from viscosity, separation, and wake formation.
Exercises — FM.1–FM.4 Fluid Mechanics
1.
A Venturi meter has upstream radius 5 cm and throat radius 2 cm. The pressure difference is 1000 Pa. Find the flow velocity and volume flow rate Q (L/s) for water.
L/s
Straightforward
2.Describethedynamicsoftwoparallellinevorticeswithcirculations±ΓseparatedbydistDescribe the dynamics of two parallel line vortices with circulations \pmΓ separated by distance d. What happens if both have the same sign? Opposite signs?
Intermediate
3.DescribethePrandtlboundarylayer.HowdoesthethicknessδscalewithdistancexalongDescribe the Prandtl boundary layer. How does the thickness \delta scale with distance x along a flat plate? When does the boundary layer separate?
Intermediate
4.State Kolmogorov's theory of turbulence and the -5/3 power law for the energy spectrum. What determines the smallest and largest scales? Why is turbulence computationally intractable?
Challenging
Key Takeaways
  • Materialderivative:D/Dt=/\partialt+v(rateofchangefollowingfluidparcelMaterial derivative: D/Dt = \partial/\partialt + v\cdot\nabla (rate of change following fluid parcel.
  • Continuity:ρ/\partialt+(\rhov)=0.Incompressible:\cdotv=0Continuity: \partial\rho/\partialt + \nabla\cdot(\rhov) = 0. Incompressible: \nabla\cdotv = 0.
  • NavierStokes:ρDv/Dt=\nablaP+η2v+\rhog.Nonlinear,unsolvedin3DNavier-Stokes: \rho Dv/Dt = -\nablaP + \eta\nabla^{2}v + \rhog. Nonlinear, unsolved in 3D.
  • ReynoldsnumberRe=\rhovL/η:Re1laminar(Stokes),Re1turbulentReynolds number Re = \rhovL/\eta: Re≪1 laminar (Stokes), Re≫1 turbulent.
  • Bernoulli(inviscid,steady):P+12\rhov2+\rhogh=constalongstreamlineBernoulli (inviscid, steady): P + \frac{1}{2}\rhov^{2} + \rhogh = const along streamline.
  • Potentialflow:2ϕ=0=electrostatics.KuttaJoukowski:L=\rhoUΓPotential flow: \nabla^{2}\phi = 0 = electrostatics. Kutta-Joukowski: L = \rhoU\inftyΓ.