1. Fluid Equations and Linear Wave Theory

Question 1

Density Definition

Answer 1

-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

Question 2

Mass Density

Answer 2

Q = mass
q = ρ

Question 3

Kinetic Energy Density

Answer 3

Q = KE
q = 1/2 ρu²

Question 4

Volume Density

Answer 4

Q = V
q = 1

Question 5

Linear Momentum Density

Answer 5

Q_ = linear momentum
q = ρu_

Question 6

Flux Definition

Answer 6

-flux of a quantity Q through a surface S is the rate at which Q passes through S
flux = ∫ q (u_ . n_) dS

Question 7

Mass Flux

Answer 7

mass flux = ∫ ρ (u_ . n_) dS

Question 8

Momentum Flux

Answer 8

flux = ∫ (ρu_) (u_ . n_) dS

Question 9

Conservation Laws

Answer 9

-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

Question 10

Mass Conservation

Answer 10

-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

Question 11

Momentum Conservation

Answer 11

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

Question 12

Equations of State

Answer 12

• physics => p = f(ρ,T)

- where f is material dependent

Question 13

Ideal Gas Law

Answer 13

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

Question 14

Linear Approximation to the Ideal Gas Law

Answer 14

-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

Question 15

Fluid Mechanical Equations for a Compressible or Incompressible Fluid

Answer 15

∂ρ/∂t + ∇.(ρu_) = 0
ρ Du_/Dt = -∇p + F_
p = F(ρ), assuming T is constant

Question 16

Boundary Conditions

Moving Rigid Boundary

Answer 16

u_.n_ = U_.n_

-where u_ is the fluid velocity and U_ is the boundary velocity

Question 17

Boundary Conditions

Fluid-Fluid Interface

Answer 17

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_

Question 18

Incompressibility

Answer 18

-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

Question 19

Fluid Mechanical Equations for an Incompressible Fluid

Answer 19

Dρ/Dt = 0
ρ Du_/Dt = -∇p + F_
∇.u_ = 0

Question 20

αp in the Atmosphere

Answer 20

αp~10^(-2) /Pa
=> αp(p-p0) ~ o(1)
-of order 1, this is significant so we would not model the atmosphere as incompressible

Question 21

αp in the Ocean

Answer 21

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

Question 22

Does incompressibiltiy mean that ρ is constant?

Answer 22

• NO

- only that Dρ/Dt=0

Question 23

Static States and Hydrostatic Pressure

u_=0

Answer 23

F_ = -ρg ez^
0_ = -∇p + F_
=>
∂p/∂x = ∂p/∂y = 0
∂p/∂z = -ρg

Question 24

Static States and Hydrostatic Pressure

u_=0, incompressible, ρ const.

Answer 24

p = po - ρgz

-pressure gets larger the deeper you go as there is a greater weight above e.g. the ocean

Question 25

Static States and Hydrostatic Pressure

u_=0, incompressible, ρ variant

Answer 25

ρ = ρo(z)
=>
p = -g ∫ρdz ez^ + po

Question 26

Static States and Hydrostatic Pressure

u_=0, compressible, ρ(z) unknown

Answer 26

-use appropriate equation of state for ρ

Question 27

Static States and Hydrostatic Pressure

Atmosphere

Answer 27

p = RρT
=>
RT ∂ρ/∂z = -ρg
=>
ρ = ρo e^(-g/RT z)

Question 28

Archimedes Principle

Answer 28

-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^

Question 29

Wave

Definition

Answer 29

• a propagating disturbance that leaves the medium unchanged
• needs a restoring force / mechanism

Question 30

Particle in 1D

Answer 30

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)

Question 31

Linear Wave Analysis

Steps

Answer 31

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))]

Question 32

Dispersion Relation

Definition

Answer 32

-the relation between frequency, ω and wave vector, k

Question 33

Non-Dispersive

Definition

Answer 33

-if ω(k)∝k, then the wave is non-dispersive

Question 34

Completeness Principle

Answer 34

• 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

Question 35

Complex Amplitude and Phase

Definitions

Answer 35

-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

Question 36

Phase Velocity

Definition

Answer 36

-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

Question 37

Superposition

Answer 37

-by setting A(k) equal to a delta function in the general solution you can recover the superposition of two wave solutions

Question 38

Conditions for Equations to Support Waves

Answer 38

• 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

Question 39

Travelling Wave Solutions

Answer 39

-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

Question 40

Sound Waves

Answer 40

-apply linear wave analysis to the full compressible fluid mechanical equations with no body forces and a background state of rest

Question 41

The Wave Equation

Answer 41

∂²ρ~/∂t² - c²∇²ρ~ = 0

-ρ~, p~ and u_~ all obey the wave equation, just with different phases etc.

Question 42

3D Wave Modes

Answer 42

-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

Question 43

Standing Waves

Answer 43

-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