MISG 2018: Instability in Fluids Yilun Wang, Nolwazi Nkomo, Shina D - - PowerPoint PPT Presentation

MISG 2018: Instability in Fluids

Yilun Wang, Nolwazi Nkomo, Shina D Oloniiju, Jessica Ihesie, Williams Chukwu, Keegan Anderson, Mojalefa Nchupang, Saul Hurwitz Supervisor: Prof David Mason

January 13, 2018

Rayleigh-Taylor Instability

y j x i y = η(x, t)

∂φ1 ∂y ∂φ2 ∂y

HEAVIER FLUID ρ2, p2, φ2 LIGHTER FLUID ρ1, p1, φ1 g

Rayleigh-Taylor Instability

The problem proposed is a situation with 2 fluids, one atop the other with different densities. Between them is the interface η(x, t) which is a perturbation across y = 0. Some assumptions are made: The vorticity is 0 (it is irrotational) so ∇ × v = 0 It is incompressible, meaning the volume is constant. This results in ∇ · v = 0

Equations of State and Boundary Conditions

∂2φ1 ∂x2 + ∂2φ1 ∂y2 = 0 ∂2φ2 ∂x2 + ∂2φ2 ∂y2 = 0 y = 0 : ∂φ1 ∂y (x, 0, t) = ∂η ∂t (x, t) y = 0 : ∂φ2 ∂y (x, 0, t) = ∂η ∂t (x, t) y = 0 : ρ1[∂φ1 ∂t (x, 0, t)+gη(x, t)] = ρ2[∂φ2 ∂t (x, 0, t)+gη(x, t)]

Form of Solution and Dispersion Relation

φ1(x, y, t) = F1(y)exp[i(kx − ωt)] φ2(x, y, t) = F2(y)exp[i(kx − ωt)] ω = ±i

• kg(ρ2 − ρ1)

ρ1 + ρ2

Solution and Analysis

Re(η) = A1 cos(kx)

• k(ρ1 + ρ2)

g(ρ2 − ρ1)(eβt − e−βt) This is an unstable exponentially growing standing wave if ρ2 > ρ1, and is a stable standing wave if ρ1 > ρ2

Rayleigh-Taylor Instability with Surface Tension

y j x i y = η(x, t)

∂φ1 ∂y ∂φ2 ∂y

T ∂2η

∂x2

HEAVIER FLUID ρ2, p2, φ2 LIGHTER FLUID ρ1, p1, φ1 g

Instability of Fluids with Interfacial Tension

Net upward force per unit area due to interfacial tension T ∂2y ∂x2 ∂2φ1 ∂x2 + ∂2φ1 ∂y2 = 0 ∂2φ2 ∂x2 + ∂2φ2 ∂y2 = 0 y = 0 : ∂φ1 ∂y (x, 0, t) = ∂η ∂t (x, t) y = 0 : ∂φ2 ∂y (x, 0, t) = ∂η ∂t (x, t) y = 0 : p1(x, 0, t) + T ∂2y ∂x2 = p2(x, 0, t)

Form of Solution and Dispersion Relation

The form of solution: η(x, t) = η0exp[i(kx − ωt)] φ1(x, y, t) = F1(y)exp[i(kx − ωt)] φ2(x, y, t) = F2(y)exp[i(kx − ωt)] Dispersion Relation: ω = ±

• k

(ρ2 + ρ1)(−g(ρ2 − ρ1) + Tk2)

Solution and Analysis

Stable if k2 > (ρ2 − ρ1)g T λ <

• 4π2T

(ρ2 − ρ1)g = 2π

• T

(ρ2 − ρ1)g .

Kelvin-Helmholtz Instability

y j x i y = η(x, t) ρ, p2, φ2 ρ, p1, φ1 g v2 v1

Kelvin-Helmholtz Instability

V (1)

x

= V1 + ∂φ1 ∂x , V (1)

y

= ∂φ1 ∂y V (2)

x

= V2 + ∂φ2 ∂x , V (2)

y

= ∂φ2 ∂y ∂2φ1 ∂x2 + ∂2φ1 ∂y2 = 0; ∂2φ2 ∂x2 + ∂2φ2 ∂y2 = 0, ∂φ1 ∂y (x, 0, t) = ∂η ∂t + V1 ∂η ∂y . ∂φ2 ∂y (x, 0, t) = ∂η ∂t + V2 ∂η ∂y . V1 ∂φ1 ∂x + ∂φ1 ∂t = V2 ∂φ2 ∂x + ∂φ2 ∂t .

Form of Solution and Dispersion Relation

The form of solution: η(x, t) = η0exp[i(kx − ωt)] φ1(x, y, t) = F1(y)exp[i(kx − ωt)] φ2(x, y, t) = F2(y)exp[i(kx − ωt)] Dispersion Relation: ω = k(V2 + V1) ± ik(V1 − V2) 2

Solution and Analysis

The perturbation solution is η(x, t) = η0 exp

• i
• kx − k

2(V2 + V1)t

• − k

2(V1 − V2)t

• +

η0 exp

• i
• kx − k

2(V2 + V1)t

• + k

2(V1 − V2)t

• Re[η(x, t)] = η0 cos
• kx − k

2(V2 + V1)t exp

• −k

2(V2 + V 1)t

• + exp
• −k

2(V1 − V 2)t

• This is unstable for V1 < V2 and V2 < V1.

Kelvin-Helmholtz and Rayleigh-Taylor Instability with Interfacial Tension

y j x i p1 p2 T ∂2η

∂x2

ρ2, p2 ρ1, p1 v2 v1

Kelvin-Helmholtz and Rayleigh-Taylor Instability with Interfacial Tension

V (1)

x

= V1 + ∂φ1 ∂x , V (1)

y

= ∂φ1 ∂y V (2)

x

= V2 + ∂φ2 ∂x , V (2)

y

= ∂φ2 ∂y ∂2φ1 ∂x2 + ∂2φ1 ∂y2 = 0; ∂2φ2 ∂x2 + ∂2φ2 ∂y2 = 0, ∂φ1 ∂y (x, 0, t) = ∂η ∂t + V1 ∂η ∂y . ∂φ2 ∂y (x, 0, t) = ∂η ∂t + V2 ∂η ∂y . y = 0 : p1(x, 0, t) + T ∂2y ∂x2 = p2(x, 0, t)

Form of Solution and Dispersion Relation

The form of solution: η(x, t) = η0exp[i(kx − ωt)] φ1(x, y, t) = F1(y)exp[i(kx − ωt)] φ2(x, y, t) = F2(y)exp[i(kx − ωt)] ω k = ρ1V1 + ρ2V2 ± √Q ρ1 + ρ2 where Q = −ρ1ρ2(V1 − V2)2 + (ρ1 + ρ2)(Tk − g/

k(ρ2 − ρ1))

Solution and Analysis

It is unstable when (V1 − V2)2 >

(ρ1+ρ2)(Tk− g

k (ρ2−ρ1))

ρ1ρ2

k >

• g

T (ρ2 − ρ1) is the first necessary condition for stability

Graphical Analysis for ρ2 > ρ1

(v1 − v2)2 k f (k) f (k) = (ρ1 + ρ2)(Tk − g

k (ρ2 − ρ1))

ρ1ρ2

Graphical Analysis for ρ1 > ρ2

(v1 − v2)2 k f (k) f (k) = (ρ1 + ρ2)(Tk + g

k (ρ1 − ρ2))

ρ1ρ2

Benard Problem

z = d z = 0 cold hot k j i TL − ∆T TL VISCOUS FLUID AT REST

Benard Problem

We wish to study the instability in the fluid owing to heat transfer - that is, when does the heat transfer mechanism switch from conduction to convection Temperature Gradient: dT

dz = − ∆T d

Navier-Stokes Momentum Eqn: ρDv

Dt = −∇p + µ∇2v + ρg

Energy Eqn: DT

Dt = κ∇2T+viscous 2nd order terms

Equation of State: ρ = ˜ ρ(1 − α(T − ˜ T))

Benard Problem - Unperturbed State

v0 = 0 T = T0(z) ρ = ρ0(z) p = p0(z) Simplified Constitutive Eqns: dp0 dz = −ρ0g κd2T0 dz2 = 0 ρ0 = ˜ ρ(1 − α(T0 − ˜ T))

Benard Problem - Unperturbed State

Solving yields: T0(z) = TL − ∆Tz d ρ0(z) = ˜ ρ(1 − α(TL − ∆Tz d − ˜ T)) p0(z) = C + g ˜ ρz(α(TL − ∆Tz 2d − ˜ T) − 1)

Perturbation and Boussinesq Approximation

We make the Boussinesq Approximation: we take ρ to be constant and approximately ˜ ρ unless it gives rise to buoyancy forces in the Navier-Stokes eqn We perturb the state by making it no longer at rest: v(x, y, z, t) = 0 + v1(x, y, z, t) T(x, y, z, t) = T0(z) + T1(x, y, z, t) ρ(x, y, z, t) = ρ0(z) + ρ1(x, y, z, t) p(x, y, z, t) = p0(z) + p1(x, y, z, t)

New Constitutive Equations

We still have incompressibility: ∇ · v = ∇ · v1 = 0 ˜ ρ∂v1 ∂t = −∇p1 + µ∇2v1 + ρ1g ∂T1 ∂t − v(z)

1

∆T d = κ∇2T1 ρ1 = −˜ ραT1

Form of Solution and Eigenvalue Problem

We want to find v(z)

1

to analyse the stability of the system. We assume the form v(z)

1

= w(z)f (x, y)est We obtain an eigenvalue problem with eigenfunction w(z)

• κ(D2

z − a2) − s

ν(D2

z − a2) − s

• (D2

z − a2) − αa2g dT0

dz

• w(z)

= 0

Boundary Conditions and Form of Solution

w(0) = w(d) = 0 as v(z)

1

would be 0 at the boundary, as the boundaries are not moving. d2w dz2 (0) = d2w dz2 (d) = 0 d4w dz4 (0) = d4w dz4 (d) = 0 To satisfy the boundary conditions, we let w(z) = sin(nπz d ); n = 1, 2, 3..

Stability for Negative Temperature Gradient

To analyse stability we need to focus on the est factor. Solving the eigenvalue problem for w(z) yields an equation for s. Solving (ν − κ)2a6

∗ + 4αa2g ∆T d

> 0 for a yields that in this case, s is always real However, for stability, we need s < 0. This is true when αgd3∆T νκ < 27π4 4 Where Rayleigh’s number is defined as R = αgd3∆T

νκ

Stability for Positive Temperature Gradient

Solving (ν − κ)2a6

∗ − 4αa2g ∆T d

> 0 for a yields that in this case, s is not always real. We require αg∆Td3 (ν − κ)2 > 27π4 16 in order for s to be real, otherwise we get an oscillating velocity, where Mason’s number is defined as M = αg∆Td3

(ν−κ)2 > 27π4 16

If s is real, for stability, we need s < 0. This is true when αgd3∆T νκ > −27π4 4 Which is always true, as the Rayleigh number is always positive

Convective Unstable Fluid

z = d z = 0 cold hot TL − ∆T TL