GFD 2 Spring 2010 P.B. Rhines Problem set 1-solns out: 5 April back: 12 April

Transcription

1 GFD 2 Spring 2010 P.B. Rhines Problem set 1-solns out: 5 April back: 12 April 1 The Gulf Stream has a dramatic thermal-wind signature : the sloping isotherms and isohalines (hence isopycnals) not only balance the strong current but also represent the water-mass boundary between warm subtropics and cold subpolar oceans: a subpolar front. In this potential temperature section at the end of this document, cold, dense water lies north (left) of warm less dense water. The hydrostatic pressure gradient p/ z is thus larger at a given depth in the north than in the south, and it would look as if this would provide a southward force, - p/ y. But, a southward force would balance the Coriolis force due to a westward current and we know the Gulf Stream is flowing dominantly eastward. Resolve this paradox (the first figure in Lecture 1 will help). To help with this, consider the following question: In the web version of Lecture 1 two possible velocity profiles in balance with the same sloping density field are shown. The first one is the same as the solved geostrophic adjustment problem for an upper ocean mixed layer with a free upper surface, in those notes. How would we change the initial conditions to produce the second velocity field instead of the first? We can tilt the free surface such that there is zero pressure gradient at bottom, and large gradient at top. η x = -Hα. The sloping free surface provides a horizontal pressure gradient which is constant in z, uniform throughout the layer (by the hydrostatic pressure equation). This can cancel the pressure gradient p x due to the stratification at one level (here, at the bottom) while adding to the pressure gradient at other depths. This is very much the same idea as the compensation of a tilted density field and a tilted ocean surface, such that at great depth the pressure gradient vanishes. 2 Coriolis effects arise from living on an accelerating coordinate system because of Earth s rotation. Coriolis theorem relating the rate of change of a vector in rotating and non-rotating reference frames turns F = MA into our GFD momentum equations. Another way see Coriolis effects is to consider a simple ring of fluid centered on the North Pole, and consider its angular momentum in an inertial (non-rotating) reference frame. The angular MOM balance is formed from r F = m r D u / Dt where F is the force acting on the fluid (could be a pressure gradient). Let r be the radial distance from the rotation axis (the cylindrical radius) and m is the mass of the ring, u 1

2 is the east-west (zonal-) velocity and θ and λ are latitude and longitude. The radial (north-south) component tells us that in absence of east-west forces F λ the angular momentum mru is conserved following the fluid ring (or think of spinning a point mass on a string about your head). Suppose r changes by an amount δr in a time δt; find the change in u, δu. Now express this change as seen by a rotating observer by defining u rot = u - Ωr where Ω is the initial angular velocity of the fluid; initially u rot =0 and u = Ωr. Show that this observer sees an acceleration δu rot /δt due to an east-west Coriolis force on the radial velocity δr/δt. {Note, the complementary north-south Coriolis force can be found by considering east-west momentum equation, for a case with constant r but changing u and changing F r. How can this be shown (optional)?} East-west Coriolis force: ru=const. So a small change in u, say u => u + δu δu= -u/r δr. A rotating observer sees this small change in u as δu rot = δu - Ωδr = (-u/r Ω)δr = -2Ω δr for δr/r<<1 since u initially = Ωr. Thus we see a 2Ω and we can further divide by a time interval δt to get δu rot /δt = 2Ω v where v = -δr/δt which says that the rotating observer sees an east-west acceleration due to a Coriolis force {extra note, not part of the problem: to diagnose the North/south Coriolis force: for a steadily spinning mass ( a point mass on a string, or a ring of fluid), the centrifugal force must be balanced by an inward force: mu 2 /r = -F r is the non-rot. radial MOM eqn. If we change u (with a zonal east/west force) while holding the fluid at the same radius, then the radial force changes as δf r = -2mu/r δu; now let u rot = u - Ωr => δf r = m(2ω + u rot /r) δu rot ~ 2Ωδu r + O(Ro) which expresses the change in radial force as a Coriolis force (you see the 2Ω term).} 3 The Friday GFD lab demonstration showed buoyant fluid injected at the center of a rotating cylinder with more dense salty fluid (by about 1%). Say the density of the salty fluid is ρ 0+δρ and the injected fluid has density ρ 0. Suppose the fluid was injected at radius r=a with zero azimuthal ( east-west ) velocity in the rotating frame, u rot (r=a)=0 [that is, u(r=a)=ωr]. As above, conserve inertial-reference-frame angular momentum of a fluid ring as it expands toward larger r. Find u (r). Integrate the radial momentum equation (neglecting any acceleration terms) to find the pressure, p(r ) and free surface elevation η(r) ) (the atmospheric pressure above the water is approximately constant). The bed of salty water was very deep compared with the injected eddy. As a consequence, its velocities were very small; assume they vanished. Calculate the shape of the interface η i (r) using hydrostatic balance in the vertical. If we had taken the u(r) profile above and gone to the rotating reference frame, with the injection radius a large enough that the Rossby number was small, the pressure would be determined by 2Ωu rot = 1 ρ 0 p r along Φ=const where urot = u-ωr. This gives a different pressure field, because it is a geostrophic approximation to the exact pressure: find this approximate solution 2

3 (informational note:) The above solution goes a bit crazy at large r because we assume it to be nearly geostrophic. The experiment started from a state of rest (in the rotating frame). The timedevelopment of the velocity and interface depth, η i (r,t) is a further interesting problem. (1) Non rotating observer: Angular momentum conservation for a non-rotating observer says that ru = constant; with the injected fluid having the same azimuthal velocity as the tank at r= a, we have ru = Ωa 2 or u = Ωa 2 /r The radial momentum balance is centrifugal vs. pressure gradient: u 2 /r = 1/ρ p r => p r /ρ = Ω 2 a 4 /r 3 (which is >0 ) (1) {near r =a this is ~ Ω 2 a(1-3r /a) = Ω 2 (a - 3r ) > 0) i.e., the vortex has a low-pressure center, whichever way it rotates. => p/ρ = -1/2 Ω 2 a 4 /r 2 +Const on z=const. for the non rotating observer. an approximation near r=a is p/ρ ~ const + Ω 2 r 2 = Ω 2 a r + O(r /a) 2 where r = a+r (1) The pressure increases outward (the center is a low pressure region) as it must to balance the u 2 /r outward-pointing centripetal force (2) For the rotating observer, write u rot = u Ωr = Ωa 2 /r - Ωr The radial momentum equation is now, with the geostrophic assumption, 2Ωu rot = p r /ρ Φ {the exact radial MOM equation is (u rot ) 2 /r + 2Ωu rot = p r /ρ} Use the velocity above, the geostrophic approximation gives 2Ωu rot = 2Ω 2 ( a 2 /r - r) = p r /ρ (near r=a this ~ -4Ω 2 r ) {+ or if we keep the centrifugal term, (Ωa 2 /r Ωr) 2 /r = (Ω 2 a 4 /r 3-2Ω 2 a 2 /r +Ω 2 r) the exact pressure gradient is p r /ρ = -2Ω 2 r + Ω 2 a 4 /r 3 +Ω 2 r = -Ω 2 r + Ω 2 a 4 /r 3 Note, p r /ρ < 0 for r>a } (2) near r=a this is approximately~ -Ω 2 (a+r ) +Ω 2 a(1-3r /a) =-4Ω 2 r } The integral of the geostrophic r-mom equation gives the expression p/ρ = 2Ω 2 a 2 ln r/a - Ω 2 r 2 + const [slope of ln(x) is 1 near x=1] whereas using the full radial MOM balance including u 2 /r gives p/ρ = - ½ Ω 2 r 2 ¼ Ω 2 a 4 /r 4 + const. How can it be that the non-rotating observer sees a low pressure at the center of the vortex while a rotating observer sees a high pressure? The answer is that the rotating observer measures pressure variation along the parabolic free surface (where the gravity + centrifugal potential Φ is constant); that is the natural horizontal for rotating coordinate system. The non-rotating observer measures pressure variation along z=constant: the two observers do not agree as to what is horizontal! The newly injected fluid is a dome relative to the z=constant plane yet it is a puddle in a parabola with a low point in the center, for the non-rotating observer as you see in the sketch below. Expressions (1, non-rot observer) and (2 rot observer) for the pressure gradient 3

4 differ by Ω 2 r which is equal to p/ z z/ r, just the term needed to convert p/ r z=const to p/ r η=const where η is the free surface of the fluid (before the flow starts); η is a surface with Φ = constant, where Φ is the laboratory geopotential, Φ = gz - ½ Ω 2 r 2 (draw the little triangle that shows this). The injected fluid is the thin eye-brow region sitting on the parabola, outside r=a) below: Gulf Stream profiles, Pierce & Joyce JGR

5 5