University of Warsaw Advanced Hydrodynamics
Faculty of Physics Selected Topics in Fluid Mechanics
Summer Semester 2019/20
Exercise Sheet 8
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1. Rotating flow systems. The so-called geophysical flows occurring in the atmosphere and oceans are described by an equation of motion in a non-inertial reference system, which rotates with the Earth at constant angular velocity Ω.
(a) Show that the equation of motion for the fluid velocity u relative to the rotating framereads
∂u
∂t + u · ∇u + 2Ω × u + Ω × (Ω × r) = −1
ρ∇p + ν∇2u, ∇ · u = 0, (1) where the familiar expression for the acceleration has been augmented by a Cori- olis term, 2Ω × u, and a centrifugal term, Ω × (Ω × r).
Solution. The equation of motion in the inertial, nonrotating frame is duI
dt = −1
ρ∇p + ν∇2uI, ∇ · uI= 0, (2)
where uIdenotes the fluid velocity seen in the nonrotating frame.
In general, the rate of change of a vector B perceived in an inertial, nonrotating frame is
dB dt
I
= dB
dt + Ω × B, (3)
where dB/dt is the rate of change of vector B seen in the frame rotating with velocity Ω.
Thus we have to do the following substitutions in (2):
duI dt → du
dt + Ω × u, (4)
for the acceleration and
uI→ u + Ω × r, (5)
for the velocity, where u denotes the fluid velocity relative to the rotating frame.
Then (for Ω constant) d
dt(u + Ω × r) + Ω × (u + Ω × r) = du
dt + 2Ω × u + Ω × (Ω × r), (6)
and the equation of motion for the fluid velocity u relative to the rotating frame is
∂u
∂t + u · ∇u + 2Ω × u + Ω × (Ω × r) = −1
ρ∇p + ν∇2u, ∇ · u = 0, (7) with a Coriolis term, 2Ω × u, and a centrifugal term, Ω × (Ω × r).
It is a simple matter to show that spatial gradients are perceived identically in rotating and nonrotating coordinate frames, and that ν∇2uI= ν∇2u.
(b) Show that (Ω being constant)
Ω × (Ω × r) = −∇ 1
2(Ω × r)2
, (8)
and use this identity to define an effective pressure P . Solution. Using
Ω × (Ω × r) = −∇ 1
2(Ω × r)2
, (9)
(an identity which can be straightforwardly demonstrated using index notation) one can define an effective pressure,
P = p − 1
2ρ(Ω × r)2, (10)
and write the governing equation in the form du
dt + 2Ω × u = −1
ρ∇P + ν∇2u. (11)
(c) Neglecting viscosity effects, write the resulting equation of motion in dimension- less form, identify the dimensionless parameter (the so-called Rossby number) and interpret it physically. Estimate typical values of the Rossby number for the following flows: (i) in the atmosphere, for characteristic length scale L ∼ 104km and velocity scale U ∼ 10 m/s; (ii) in the ocean, with L ∼ 102km and U ∼ 1 m/s.
Can the equation of motion be further simplified in these flows?
Solution. We neglect the viscous term and write Eq. (11) in the form du
dt + 2Ω ˆΩ × u = −1
ρ∇P, (12)
where ˆΩ = Ω/Ω is the unit vector in the direction of Ω (assumed to be constant).
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Let U be a typical flow speed and L a typical length scale of the flow. Then one may write
u = u∗U, t = t∗L
U, (13)
where u∗ and t∗ are dimensionless variables. Substitution of the variables above into Eq. (12) yields
U2 L
du∗
dt∗ + 2ΩU ˆΩ × u∗ = −1
ρ∇P. (14)
Dividing by ΩU Rodu∗
dt∗ + 2 ˆΩ × u∗ = −∇∗P∗, (15)
where
∇P = (ρΩU )∇∗P∗, (16)
and
Ro = U
ΩL ∼ |du∗/dt∗|
|2Ω × u|, (17)
is the Rossby number. If Ro 1, than the term du∗/dt∗, in special the non-linear term (u · ∇)u, may be neglected in comparison with the Coriolis term 2Ω × u.
The Earth’s rotation rate is Ω = 7.27 × 10−5 s−1 (once per day). (i) In mid-latitude atmosphere, L ∼ 104km (size of cyclones) and U ∼ 10 m/s, Ro = U/(ΩL) ≈ 0.014;
(ii) in the ocean, with L ∼ 102 km (width of ocean currents) and U ∼ 1 m/s, Ro = U/(ΩL) ≈ 0.14.
In both cases, the Rossby number is small. In the atmosphere flow, the neglect of (u · ∇)u in the governing equation is a better approximation than in the case of ocean flows. In the limit of small Rossby numbers, at leading order there is a balance between the Coriolis term and the pressure gradients. This is called the geostrophic balance.
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