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 g y j HEAVIER FLUID ∂φ 2 y = η ( x , t ) ρ 2 , p 2 , φ 2 ∂ y x i LIGHTER FLUID ∂φ 1 ∂ y ρ 1 , p 1 , φ 1

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 ∂ 2 φ 1 ∂ x 2 + ∂ 2 φ 1 ∂ y 2 = 0 ∂ 2 φ 2 ∂ x 2 + ∂ 2 φ 2 ∂ y 2 = 0 y = 0 : ∂φ 1 ∂ y ( x , 0 , t ) = ∂η ∂ t ( x , t ) y = 0 : ∂φ 2 ∂ y ( x , 0 , t ) = ∂η ∂ t ( x , t ) y = 0 : ρ 1 [ ∂φ 1 ∂ t ( x , 0 , t )+ g η ( x , t )] = ρ 2 [ ∂φ 2 ∂ t ( x , 0 , t )+ g η ( x , t )]

Form of Solution and Dispersion Relation φ 1 ( x , y , t ) = F 1 ( y ) exp [ i ( kx − ω t )] φ 2 ( x , y , t ) = F 2 ( y ) exp [ i ( kx − ω t )] � kg ( ρ 2 − ρ 1 ) ω = ± i ρ 1 + ρ 2

Solution and Analysis � k ( ρ 1 + ρ 2 ) g ( ρ 2 − ρ 1 )( e β t − e − β t ) Re ( η ) = A 1 cos( kx ) This is an unstable exponentially growing standing wave if ρ 2 > ρ 1 , and is a stable standing wave if ρ 1 > ρ 2

Rayleigh-Taylor Instability with Surface Tension g y j HEAVIER FLUID ∂φ 2 y = η ( x , t ) T ∂ 2 η ρ 2 , p 2 , φ 2 ∂ y ∂ x 2 x i LIGHTER FLUID ∂φ 1 ∂ y ρ 1 , p 1 , φ 1

Instability of Fluids with Interfacial Tension Net upward force per unit area due to interfacial tension T ∂ 2 y ∂ x 2 ∂ 2 φ 1 ∂ x 2 + ∂ 2 φ 1 ∂ y 2 = 0 ∂ 2 φ 2 ∂ x 2 + ∂ 2 φ 2 ∂ y 2 = 0 y = 0 : ∂φ 1 ∂ y ( x , 0 , t ) = ∂η ∂ t ( x , t ) y = 0 : ∂φ 2 ∂ y ( x , 0 , t ) = ∂η ∂ t ( x , t ) p 1 ( x , 0 , t ) + T ∂ 2 y y = 0 : ∂ x 2 = p 2 ( x , 0 , t )

Form of Solution and Dispersion Relation The form of solution: η ( x , t ) = η 0 exp [ i ( kx − ω t )] φ 1 ( x , y , t ) = F 1 ( y ) exp [ i ( kx − ω t )] φ 2 ( x , y , t ) = F 2 ( y ) exp [ i ( kx − ω t )] Dispersion Relation: � k ( ρ 2 + ρ 1 )( − g ( ρ 2 − ρ 1 ) + Tk 2 ) ω = ±

Solution and Analysis Stable if k 2 > ( ρ 2 − ρ 1 ) g T � � 4 π 2 T T λ < ( ρ 2 − ρ 1 ) g = 2 π ( ρ 2 − ρ 1 ) g .

Kelvin-Helmholtz Instability g y j v 2 y = η ( x , t ) ρ, p 2 , φ 2 x i v 1 ρ, p 1 , φ 1

Kelvin-Helmholtz Instability = V 1 + ∂φ 1 = ∂φ 1 V (1) V (1) ∂ x , x y ∂ y = V 2 + ∂φ 2 = ∂φ 2 V (2) V (2) ∂ x , x y ∂ y ∂ 2 φ 1 ∂ x 2 + ∂ 2 φ 1 ∂ 2 φ 2 ∂ x 2 + ∂ 2 φ 2 ∂ y 2 = 0; ∂ y 2 = 0 , ∂φ 1 ∂ y ( x , 0 , t ) = ∂η ∂η ∂ t + V 1 ∂ y . ∂φ 2 ∂ y ( x , 0 , t ) = ∂η ∂η ∂ t + V 2 ∂ y . ∂φ 1 ∂ x + ∂φ 1 ∂φ 2 ∂ x + ∂φ 2 V 1 ∂ t = V 2 ∂ t .

Form of Solution and Dispersion Relation The form of solution: η ( x , t ) = η 0 exp [ i ( kx − ω t )] φ 1 ( x , y , t ) = F 1 ( y ) exp [ i ( kx − ω t )] φ 2 ( x , y , t ) = F 2 ( y ) exp [ i ( kx − ω t )] Dispersion Relation: ω = k ( V 2 + V 1 ) ± ik ( V 1 − V 2 ) 2

Solution and Analysis The perturbation solution is � � kx − k � − k � η ( x , t ) = η 0 exp 2( V 2 + V 1 ) t 2( V 1 − V 2 ) t + i � � kx − k � + k � η 0 exp 2( V 2 + V 1 ) t 2( V 1 − V 2 ) t i � kx − k � � � − k � Re [ η ( x , t )] = η 0 cos 2( V 2 + V 1 ) t exp 2( V 2 + V 1) t � − k �� + exp 2( V 1 − V 2) t This is unstable for V 1 < V 2 and V 2 < V 1 .

Kelvin-Helmholtz and Rayleigh-Taylor Instability with Interfacial Tension y j v 2 p 2 T ∂ 2 η ρ 2 , p 2 ∂ x 2 x i v 1 ρ 1 , p 1 p 1

Kelvin-Helmholtz and Rayleigh-Taylor Instability with Interfacial Tension = V 1 + ∂φ 1 = ∂φ 1 V (1) V (1) ∂ x , x y ∂ y = V 2 + ∂φ 2 = ∂φ 2 V (2) V (2) ∂ x , x y ∂ y ∂ 2 φ 1 ∂ x 2 + ∂ 2 φ 1 ∂ 2 φ 2 ∂ x 2 + ∂ 2 φ 2 ∂ y 2 = 0; ∂ y 2 = 0 , ∂φ 1 ∂ y ( x , 0 , t ) = ∂η ∂η ∂ t + V 1 ∂ y . ∂φ 2 ∂ y ( x , 0 , t ) = ∂η ∂η ∂ t + V 2 ∂ y . p 1 ( x , 0 , t ) + T ∂ 2 y y = 0 : ∂ x 2 = p 2 ( x , 0 , t )

Form of Solution and Dispersion Relation The form of solution: η ( x , t ) = η 0 exp [ i ( kx − ω t )] φ 1 ( x , y , t ) = F 1 ( y ) exp [ i ( kx − ω t )] φ 2 ( x , y , t ) = F 2 ( y ) exp [ i ( kx − ω t )] k = ρ 1 V 1 + ρ 2 V 2 ± √ Q ω ρ 1 + ρ 2 where Q = − ρ 1 ρ 2 ( V 1 − V 2 ) 2 + ( ρ 1 + ρ 2 )( Tk − g / k ( ρ 2 − ρ 1 ))

Solution and Analysis ( ρ 1 + ρ 2 )( Tk − g It is unstable when ( V 1 − V 2 ) 2 > k ( ρ 2 − ρ 1 )) ρ 1 ρ 2 � g k > T ( ρ 2 − ρ 1 ) is the first necessary condition for stability

Graphical Analysis for ρ 2 > ρ 1 f ( k ) ( v 1 − v 2 ) 2 k f ( k ) = ( ρ 1 + ρ 2 )( Tk − g k ( ρ 2 − ρ 1 )) ρ 1 ρ 2

Graphical Analysis for ρ 1 > ρ 2 f ( k ) ( v 1 − v 2 ) 2 k f ( k ) = ( ρ 1 + ρ 2 )( Tk + g k ( ρ 1 − ρ 2 )) ρ 1 ρ 2

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

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: ρ Dv Dt = −∇ p + µ ∇ 2 v + ρ g Energy Eqn: DT Dt = κ ∇ 2 T +viscous 2nd order terms ρ (1 − α ( T − ˜ Equation of State: ρ = ˜ T ))

Benard Problem - Unperturbed State v 0 = 0 T = T 0 ( z ) ρ = ρ 0 ( z ) p = p 0 ( z ) Simplified Constitutive Eqns: dp 0 dz = − ρ 0 g κ d 2 T 0 dz 2 = 0 ρ (1 − α ( T 0 − ˜ ρ 0 = ˜ T ))

Benard Problem - Unperturbed State Solving yields: T 0 ( z ) = T L − ∆ Tz d ρ (1 − α ( T L − ∆ Tz − ˜ ρ 0 ( z ) = ˜ T )) d ρ z ( α ( T L − ∆ Tz − ˜ p 0 ( z ) = C + g ˜ T ) − 1) 2 d

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 + v 1 ( x , y , z , t ) T ( x , y , z , t ) = T 0 ( z ) + T 1 ( x , y , z , t ) ρ ( x , y , z , t ) = ρ 0 ( z ) + ρ 1 ( x , y , z , t ) p ( x , y , z , t ) = p 0 ( z ) + p 1 ( x , y , z , t )

New Constitutive Equations We still have incompressibility: ∇ · v = ∇ · v 1 = 0 ρ∂ v 1 ∂ t = −∇ p 1 + µ ∇ 2 v 1 + ρ 1 g ˜ ∂ T 1 ∆ T ∂ t − v ( z ) = κ ∇ 2 T 1 1 d ρ 1 = − ˜ ρα T 1

Form of Solution and Eigenvalue Problem We want to find v ( z ) to analyse the stability of the system. 1 We assume the form v ( z ) = w ( z ) f ( x , y ) e st 1 We obtain an eigenvalue problem with eigenfunction w(z) �� � z − a 2 ) − α a 2 g dT 0 κ ( D 2 z − a 2 ) − s ν ( D 2 z − a 2 ) − s ( D 2 � � � w ( z ) dz = 0

Boundary Conditions and Form of Solution w (0) = w ( d ) = 0 as v ( z ) would be 0 at the boundary, as the 1 boundaries are not moving. d 2 w dz 2 (0) = d 2 w dz 2 ( d ) = 0 d 4 w dz 4 (0) = d 4 w dz 4 ( d ) = 0 To satisfy the boundary conditions, we let w ( z ) = sin( n π z d ); n = 1 , 2 , 3 ..

Stability for Negative Temperature Gradient To analyse stability we need to focus on the e st factor. Solving the eigenvalue problem for w(z) yields an equation for s. Solving ( ν − κ ) 2 a 6 ∗ + 4 α a 2 g ∆ T > 0 for a yields that in this d case, s is always real However, for stability, we need s < 0. This is true when α gd 3 ∆ T < 27 π 4 νκ 4 Where Rayleigh’s number is defined as R = α gd 3 ∆ T νκ

Stability for Positive Temperature Gradient ∗ − 4 α a 2 g ∆ T Solving ( ν − κ ) 2 a 6 > 0 for a yields that in this d case, s is not always real. We require α g ∆ Td 3 ( ν − κ ) 2 > 27 π 4 16 in order for s to be real, otherwise we get an oscillating velocity, where Mason’s number is defined as M = α g ∆ Td 3 ( ν − κ ) 2 > 27 π 4 16 If s is real, for stability, we need s < 0. This is true when α gd 3 ∆ T > − 27 π 4 νκ 4 Which is always true, as the Rayleigh number is always positive

Convective Unstable Fluid T L − ∆ T cold z = d z = 0 hot T L