University of Warsaw Advanced Hydrodynamics
Faculty of Physics Selected Topics in Fluid Mechanics
Summer Semester 2019/20
Exercise Sheet 10
Questions, comments and corrections: e-mail togustavo.abade@fuw.edu.pl
1. Consider two ideal, imiscible fluids of density ρ1 and ρ2 (withρ2 > ρ1) separated by a sharp interface as shown in the figure. For the fluids in equilibrium in gravitational field the interface is plane and horizontal. We then introduce a “small” perturbation of the interface described byy = ξ(x, t) with
ξ(x, t) = a ei(kx−ωt), (1)
the real part being understood. Here “small” meansa λ = 2π/k. This perturbation will induce gravity waves, that propagate without either growth or decay (provided ρ2 > ρ1).
ρ1 y
x
g
ρ2
y = ξ(x, t)
The velocity u and pressurep fields after perturbation satisfy the incompressible Euler equations,
∂u
∂t + u · ∇u = −1
ρ∇p − gˆy, ∇ · u = 0. (2)
For our analysis we decompose (u, p) into the sum of a steady base state (¯u, ¯p) (that corresponds to fluids in equilibrium in gravitational field) and a time-dependent pertur- bation (u0, p0),
u = ¯u + u0, p = ¯p + p0, (3)
where the perturbations in velocity and pressure have the form
u0(x, y, t) = ˆu0(y)ei(kx−ωt), p0(x, y, t) = ˆp0(y)ei(kx−ωt). (4)
Our task is to characterize the linear gravity wave propagation by the determinig the relation between the frequency ω of oscilations and the wavenumber k, the so-called dispersion relationω(k).
Due to the sharp discontinuity in density we separate the solution (u, p) into upper (indexed 1) and lower (indexed 2) parts,
(u, p) = (u1, p1) z > ξ,
(u2, p2) z ≤ ξ. (5)
The same holds for the base state and perturbations.
(a) Obtain the pressure distribution in the base state.
Solution. For equilibrium
¯uj = 0, d¯pj
dy = −ρjg, j = 1, 2. (6)
Integration of the hydrostatic balance equation for pressure yields,
¯
pj = −ρjgy, j = 1, 2, (7)
where we have used the continuity of ¯p across the unperturbed interface at y = 0.
(b) Obtain the linearized equations for the perturbations (u0, p0). What are the bound- ary conditions?
Solution. Substitution of (3) into (2) and neglecting quadratic and higher order terms in the perturbation variables yields
∂u0
∂t = −1
ρ∇p0, ∇ · u0= 0. (8)
The perturbations (u0, p0) must decay to zero far from the interface (y → ±∞), and the total pressure p = ¯p + p0must be continuous across the interface (if one neglects surface tension). These are the dynamic boundary conditions.
The kinematic boundary condition assumes that fluid particles initially forming the in- terface remain on it. Then the vertical velocity component v at the interface equals the material time derivative of the interface elevation ξ,
Dξ Dt = ∂ξ
∂t + u0∂ξ
∂x = v0, for y = ξ(x, t). (9)
For small u0and ξ, this boundary condition may be linearized,
∂ξ
∂t = v0, for y = 0, (10)
where we have neglected the higher order terms in the perturbation variables (u0, ξ) and considered the boundary condition on the unperturbed surface y = 0 rather than on the unknown interface elevation y = ξ(x, t).
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(c) Obtain the pressure perturbation at the interface elevationξ(x, t) and the disper- sion relationω(k). Neglect surface tension at the interface.
Solution. We start from the distribution of the pressure perturbation. Using the incom- pressibility condition in Eq. (8) one may show that the pressure perturbation is harmonic,
∇2p0j = 0, j = 1, 2. (11)
Using the wave form suggested in (4) and applying the boundary conditions
p01 → 0 as y → ∞, p02→ 0 as y → −∞ (12)
we obtain
p01 = e−kyei(kx−ωt), p02 = e+kyei(kx−ωt). (13)
Substitution of (4) into the linearized equations (8) yields (for the upper part)
−iωu01 = −1
ρ1[ikp01ˆx − kp01ˆy]. (14)
For the vertical component,
−iωv10 = k
ρ1p01, (15)
and
p01 = −iρ1ω
kv10. (16)
From (1) and the kinematic boundary condition (10) at the interface
v10 = −iωξ, (17)
and the pressure perturbation at the interface reads p01 = −ρ1
ω2
k ξ. (18)
Performing similarly for the lower part yields p02 = +ρ2
ω2
k ξ, (19)
where we have used the continuity of the vertical velocity perturbation (v10 = v20) across the interface.
Applying the continuity of the pressure field at the interface,
¯
p1+ p01= ¯p2+ p02 at y = ξ, (20)
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yields
−ρ1gξ − ρ1
ω2
k ξ = −ρ2gξ + ρ2
ω2
k ξ, (21)
or
(ρ1+ ρ2)ω2− (ρ2− ρ1)gk = 0, (22)
which is the frequency equation for the dispersion relation ω(k).
Then, finally ω(k) = ±p
∆(k), (23)
with
∆(k) = ρ2− ρ1
ρ1+ ρ2gk. (24)
Note that ω is real-valued provided ρ2 > ρ1.
(d) Surface tension. Surface tension creates an extra pressure difference across the interface that is proportional to the curvature of the interface,
p2− p1 = −γκ, (25)
whereγ is the surface tension coefficient and κ = ∂2ξ
∂x2
"
1 + ∂ξ
∂x
2#−3
, (26)
is the curvature.
Show that inclusion of surface tension effects yields a dispersion relation of the form (23) with
∆(k) = ρ2 − ρ1
ρ1+ρ2
gk + γ
ρ1+ρ2
k3. (27)
Solution. The pressure boundary condition (20) must be modified to account for the difference (25) due to surface tension,
¯
p1+ p01= ¯p2+ p02+ γk2ξ at y = ξ, (28) where we have used (1) to estimate the curvature κ ≈ ∂2ξ/∂x2 = −k2ξ. This adds an extra term to the frequency equation (22),
(ρ1+ ρ2)ω2− (ρ2− ρ1)gk − γk3 = 0, (29)
yielding a dispersion relation of the form (23) with
∆(k) = ρ2− ρ1
ρ1+ ρ2gk + γ
ρ1+ ρ2k3. (30)
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