1. Fluid Equations and Linear Wave Theory Flashcards
Density Definition
-let Q denote the amount of a quantity in a volume V
-the density of Q, q, is defined as:
Q = ∫ q(x_,t) dV
-this also applies to vectors
Mass Density
Q = mass q = ρ
Kinetic Energy Density
Q = KE q = 1/2 ρu²
Volume Density
Q = V q = 1
Linear Momentum Density
Q_ = linear momentum q = ρu_
Flux Definition
-flux of a quantity Q through a surface S is the rate at which Q passes through S
flux = ∫ q (u_ . n_) dS
Mass Flux
mass flux = ∫ ρ (u_ . n_) dS
Momentum Flux
flux = ∫ (ρu_) (u_ . n_) dS
Conservation Laws
-the evolution of any quantity Q in a volume V is governed by a conservation law of universal form
dQ/dt = d/dt ∫qdV
= ∫qu_.(-n_)dS + ∫K.(-n_)dS + ∫adV
-first term is flux through S, second term includes any other surface sources of Q and the third term is for body sources
Mass Conservation
-substitute q=ρ into the universal conservation law
-physics tells us that the second and third terms are 0
=>
d/dt ∫ρ dV = ∫ρu_.(-n_) dS
-put under single integarl using divergence theorem
∫∂ρ/∂t + ∇.(ρu_) dV = 0
-this is true for any V, thus:
∂ρ/∂t + ∇.(ρu_) = 0
Momentum Conservation
q_ = ρu_
-sub into universal conservation law:
d/dt ∫ρ dV = ∫ρu_.(-n_) dS + ∫ρu_(u_.(-n_)) dS + ∫F_ dV
-put into suffix notation to put all terms under one volume integral, set equal to 0
=>
ρ Du_/Dt = -∇p + F_
-this is essentially Newton’s 2nd Law
Equations of State
- physics => p = f(ρ,T)
- where f is material dependent
Ideal Gas Law
p = RρT
- doubling the density will double the frequency of particle collisions so pressure would be twice as great
- doubling the temperature doubles the energy so you would get twice the momentum transfer per collision and thus double the pressure
Linear Approximation to the Ideal Gas Law
-when we make a small perturbation from the base state we can use a linear approximation, a Taylor expansion of (3) about the base state (ρ,T,p) = (ρo,To,po): p = po + Aρ (ρ-ρo) + At (T-To) -where: Aρ = ∂f/∂ρ|ρo,To At = ∂f/∂T|ρo,To -this is commonly written as: ρ = ρo [1 + αp(p-po) + αt(T-To)] -where αp is the coefficient of compressibility and αt is the thermal expansion coefficient
Fluid Mechanical Equations for a Compressible or Incompressible Fluid
∂ρ/∂t + ∇.(ρu_) = 0
ρ Du_/Dt = -∇p + F_
p = F(ρ), assuming T is constant
Boundary Conditions
Moving Rigid Boundary
u_.n_ = U_.n_
-where u_ is the fluid velocity and U_ is the boundary velocity
Boundary Conditions
Fluid-Fluid Interface
u1_.n_ = u2.n_ -also need stress to be continuous -for an invicid fluid this implies: p1=p2 on S -generally: σ1_.n_ = σ2_.n_
Incompressibility
-many situations are incompressible, where αp~0
-if αp->0, we lose the relation between ρ and p, and:
Dρ/Dt = 0
-then mass conservation =>
∇.u_ = 0
Fluid Mechanical Equations for an Incompressible Fluid
Dρ/Dt = 0
ρ Du_/Dt = -∇p + F_
∇.u_ = 0
αp in the Atmosphere
αp~10^(-2) /Pa
=> αp(p-p0) ~ o(1)
-of order 1, this is significant so we would not model the atmosphere as incompressible
αp in the Ocean
αp~10^(-11) /Pa -in the deep ocean p-p0~10^8Pa = 1000bar -so: αp(p-p0) ~ 10^(-3) << 1 -so this term is negligible and we can model the ocean as incompressible
Does incompressibiltiy mean that ρ is constant?
- NO
- only that Dρ/Dt=0
Static States and Hydrostatic Pressure
u_=0
F_ = -ρg ez^ 0_ = -∇p + F_ => ∂p/∂x = ∂p/∂y = 0 ∂p/∂z = -ρg
Static States and Hydrostatic Pressure
u_=0, incompressible, ρ const.
p = po - ρgz
-pressure gets larger the deeper you go as there is a greater weight above e.g. the ocean
Static States and Hydrostatic Pressure
u_=0, incompressible, ρ variant
ρ = ρo(z)
=>
p = -g ∫ρdz ez^ + po
Static States and Hydrostatic Pressure
u_=0, compressible, ρ(z) unknown
-use appropriate equation of state for ρ
Static States and Hydrostatic Pressure
Atmosphere
p = RρT => RT ∂ρ/∂z = -ρg => ρ = ρo e^(-g/RT z)
Archimedes Principle
-object immersed in hydrostatic pressure experiences a net pressure force
-this buoyancy force is equal to the weight of the displaced fluid:
∫pn_ dS = ∫ρgdV ez^
Wave
Definition
- a propagating disturbance that leaves the medium unchanged
- needs a restoring force / mechanism
Particle in 1D
x'' = f(x) -suppose x=xo is the neutral state such that f(xo)=0 -consider a small perturbation x~ from xo: x = xo + x~ x~'' = f(xo + x~) -Taylor Expansion -remove higher order terms to linearise x~'' = -α² x~ -simple harmonic motion -try solution of form: x~ = Ae^(-iωt) => ω=±α -general solution: x~ = Ae^(-iαt) + Be^(iαt)
Linear Wave Analysis
Steps
1) define the neutral / background state from which perturbations are made
2) linearisation
3) wave ansatz (a guess that works), for waves:
u~ = Re[ u^ e^(i(kx-ωt))]
Dispersion Relation
Definition
-the relation between frequency, ω and wave vector, k
Non-Dispersive
Definition
-if ω(k)∝k, then the wave is non-dispersive
Completeness Principle
- the most general superposition of linear wave modes IS the general solution
- obtain this by integrating over all values of k
- or summing if there are restrictions on the values that k can take
Complex Amplitude and Phase
Definitions
-in the wave ansatz: u~ = Re[ u^ e^(i(kx-ωt))] -the complex amplitude is u^ -define phase as: θ = kx-ωt -and let: u^ = |u^| e^(iθo) -where θo=arg(u^) -then the wave ansatz becomes: u~ = |u^| Re[e^(i(kx-ωt+θo))] = |u^|cos(kx-ωt+θo) -here |u^| is the amplitude and θo is the phase shift
Phase Velocity
Definition
-factorise k out of the exponent in the ansatz:
u~ = Re[ u^ e^(ik(x-Cp(k)t))]
-then Cp(k)=ω(k)/k, the phase velocity
-note that the ansatz is only constant when the exponent is constant, x-Cp(k)t=const.
-therefore crest/troughs etc. all move at the phase velocity
Superposition
-by setting A(k) equal to a delta function in the general solution you can recover the superposition of two wave solutions
Conditions for Equations to Support Waves
- to support waves, an equation must have a solution that allows both k and ω to be real at the same time
- this is only the case if all derivatives have the same parity
- i.e. they are all odd or all even
Travelling Wave Solutions
-if a wave is non-dispersive, then it preserves its shape
-can then use a faster method, transformation of coordinates, to solve it
e.g.
ζ = x-Ut
τ = t
Sound Waves
-apply linear wave analysis to the full compressible fluid mechanical equations with no body forces and a background state of rest
The Wave Equation
∂²ρ~/∂t² - c²∇²ρ~ = 0
-ρ~, p~ and u_~ all obey the wave equation, just with different phases etc.
3D Wave Modes
-introduce the wave vector: k_ = (k,l,m) -then the 3D ansatz is given by: u~ = Re[ u^ e^(i(k_.x_-ωt))] -this is now constant when k_.x_-ωt=const. corresponding to planes moving in the k_ direction -the 3D phase velocity is then given by: Cp_ = k_/|k_|² * ω
Standing Waves
-introduce a boundary, then to meet the boundary condition, u~=0 at x=0, the ansatz becomes a superposition of incident and reflected waves:
u~ = Re[ u^ e^(i(kx-ωt)) + u^ e^(-i(kx-ωt))]
-introducing two boundaries (with waves inbetween) restricts the k to a discrete range of possible values:
u~ = 2|u^|sin(kx)cos(ckt)
-for sound
