GFD-1 AS509/OC512 Problem Set 3/review problems
version c: additions/typos in red Out: Monday 2 Feb 2009
Back: Monday 9 Feb 2009
Hand in problem 15 and any 3 otherproblems from the following set and study the rest..many will be discussed in class.
Quiz now scheduled for Weds 11 Feb 2009.
First some short questions:
1. Suppose an ice cube melts while floating in a cylinder of water on a rotating table. Will it start to spin? If so which way? We did the experiment: convection from the cold icecube causes downflow beneath it. Water drawn to feed this downflow is horizontally converging…making cyclonic rotation which rotates the ice cube. With very salty water, the fresh melt water is buoyant and should spread out at the surface rather than sink. This horizontal divergence leads to anticyclonic currents…if the water is salty enough. The coldness of the ice cube actually will drive some thermal convection below, so there probably will be layered vortex due to the layers of horiz convergence and divergence.
2. Scale analysis of the MASS conservation equation, for an incompressible fluid, suggests that the ratio of vertical and horizontal velocities is W/U ~ H/L where H is not the fluid depth, but is the length scale in the vertical direction of variations in vertical velocity, and L is the horizontal scale of variations in u and v. Show that for a rotating fluid you can use the vertical vorticity equation to give another estimate of W/U, which is very different. Note that a scale estimate of vertical vorticity ζ is U/L. The vert. vorticity equation is Scale analysis of the size of the respective terms gives
U/LT U2/L2 fW/H UW/LH
divide by fW/H to give (1/fT)(U/W )(H/L) (U/fL )(U/W )(H/L) 1 U/fL
which involves our two
Rossby numbers, Ro = U/fL (1/fT)(U/W )(H/L) Ro(U/W )(H/L) 1 Ro
and 1/fT which doesn’t
seem to have a name. Now, it’s up to us to choose the kind of flow we are interested in. A steady flow
(T=>∞) with Ro <1 leaves just two terms, which therefore must balance: Ro(U/W )(H/L) ~ 1
This gives us the estimate U/W ~ Ro(H/L) which is very much smaller than the estimate H/L obtained from quick scale analysis of MASS conservation: ux + vy + wz = 0 for incompressible fluid. In this case ux + vy nearly sum to zero by themselves: the fluid is ‘kinematically non-horizontally-divergent yet the vorticity equation shows that horiz. divergence is acting and important, so the fluid is ‘dynamically divergent’.
The other choice for time-varying flows balances term 1 and term 3 typically, giving
W/U ~ H/L (1/fT), again < H/L for motions with time scale T > 1/f.
3. Describe the vertical vorticity in a Kelvin wave, and what causes it to change with time. (Recall that without Coriolis effects, long, linear gravity waves have no vertical vorticity.)The Kelvin wave seems to act like a long gravity wave without rotation (because the dispersion relation is σ2 = gHk2 as if there were no Coriolis effects), yet it is trapped to the coast by rotation…it leans on the coast in order that the Coriolis force on the velocities can be balanced by the slope of the free surface η. Unlike an ordinary gravity wave the vorticity equation (see #2 above) tells us that it indeed has vorticity generated by the vertical stretching of planetary vorticity, f. So writing wz = 1/H ∂η/∂t (this is from integrating wz vertically and noting that w(z=η) = Dη/Dt, here ≈ ∂η/∂τ). We now know that ζ = -fη/H (could have done this directly from PV conservation). Hence it is big near the coast where η is biggest. This voriticity is entirely shear of the velocity (which is parallel to the coast), with no streamline curvature effect.
from Gill p 380
4. In geostrophic adjustment models using simple initial conditions as in Vallis’ book or as with a η = cos(kx) initial surface elevation, the geostrophic flow part of the solution (which has straight streamlines) does not change with time. For more general initial conditions, the geostrophic flow will indeed change with time, and that is the more interesting part of the problem.Suppose the initial height field η0(x,y) is a ‘hill’ next to a ‘valley’, so the height contours form two sets of closed circles. If the initial velocities are zero,
•suggest how you think the geostrophically balanced flow will evolve. How will it depend on the initial length scale, L of η0? lThis is a case where the geostrophic flow will be two vortices of opposite sign. They will act like a vortex pair….a dipole…and move across the fluid. I’m not sure if you have worked with this idea in the first fluids course. Two point vortices can advect on another, as a sort of 2 dimensional ‘smoke ring’. We did have a vortex gun in the lab if you recall. The initial scale L matters because, as in our earlier geos. adjustment problems, it tells you how strong the geos. velocity will be; if L/LD >1 most of the energy is PE and there won’t be much released to make KE of a current.
•during the initial adjustment to geostrophic balance, as the waves begin to propagate away, fluid particles will move horizontally as the geostrophic flow develops. How far (as a fraction of L) will they move in the brief time taken to establish geostrophic balance, as a function of the amplitude of the initial conditions? How long does it take for the flow to become nearly geostrophic? [All answers are meant to be approximate, made with scale analysis, and not through some complete solution.]{We have seen that the waves propagate away the part of the initial conditions that has no PV signature, while the PV of the initial conditions is ‘locked’ to fluid particles from the beginning. Vallis shows that the geostrophically balanced final flow has the minimum possible energy of any flow with that PV.} This question relates to the amplitude of the flow and the initial conditions. We did it as a linear problem, so the amplitude merely sets the scale of the velocity and pressure field, but changing it doesn’t change the form of the solution. In real life amplitude matters. You can to the usual scale analysis to say when that will be true…how big the initial η will be to make nonlinear terms important. Withing the confines of our linear solution you can estimate how far the fluid moves in the x-direction (that’s perpendicular to the wave crests in the homework problem on geos.adjustment). The fluid surface is ‘slumping’ so there has to be movement in x-direction,
ux = - ηt /H Think of u as dX/dt, where X is the position of a fluid particle. Thus
Xxt = - ηt /H or ∂X/∂x = - η/H. So if the net change in η during geos. adjustment is δη, then
a fluid particle should move a distance ~ δηL/H where L = k-1 is the length scale of the sine wave. This is a good answer, and we can also substitute for δη because we know it is connected to L/LD. The particular solution for that problem, which was the geos. adjusted steady flow, is
η = η0/(1+Ld2/L2) so you can estimate the change in η as δη = η-η0 = η0/(L2/LD2 +1)l. The answer is that a fluid particle moves a distance η0 L/H if most of the energy in the initial conditions goes into waves (this is obvious from ηt~ Hux in that case or a fluid particle moves a distance much less than this if L>LD in which case most of the initial wave is locked in geostrophic balance.
5. In a 2-layer stratified, rotating fluid, with small Rossby number and small aspect ratio, H/L, how does the vector velocity difference, between the upper layer (1) and lower layer (2) relate to the contours of constant interface height (η)? Done in class: Margules formula for the shear in a two layer fluid is
which shows that the velocity difference points along the contours of constant η (normal to η). More generally, in a continuously stratified fluid, the velocity vector at one level z and a neighboring level z +δz
differ by a velocity change that lies normal to ρ …that is, along height contours of a surface of constant density. The ‘veering’ and ‘backing’ , that is the change in direction of horizontal velocity at different z levels has interesting implications we will see soon…or look at Gill section 7.7.
6 The lower layer of a 2-layer stratified fluid can be at rest while there is circulation above in the upper layer. If the top of the fluid is a free surface at z = η(x,y,t) how does η relate to the interface elevation, ηINT (x,y,t)? {This is called ‘compensation’ and it frequently occurs. You can imagine that it cannot really happen if the upper layer has time-dependent flow, since then ηINT would vary in time and excite the lower layer. But the ocean has a shallow thermocline (thin upper layer), 10 – 100m deep, as well as deep stratification, and quite often that upper layer acts as if the deep layers beneath were at rest. }Discussed in class: in order that the pressure gradient p (horizontal component) vanish in the lower layer we need to cancel the pressure gradient due to the upper surface, ρ0gη with an equal and opposite pressure gradient due to the interface, gΔρ ηINT thus showing that the water surface and the ‘thermocline’ ηINT are opposite in shape and the vertical displacement of the density interface is bigger by a factor ~ ρ/Δρ.
7. We have seen that the energy flux in a linear, long gravity wave with or without rotation is .
Discuss relations between velocity and η fields in such a wave, showing how the energy flux points in the direction of propagation, and contrasting waves propagating in one direction with standing waves{this question suggests looking at the 1-layer energy equation and its energy flux term}.
Show that the u-velocity is in the direction of propagation when the pressure (and η) are higher, and opposite to the direction of propagation when pressure and η are lower…this is true in any gravity wave, short or long. Recall the streak photographs from the waves lab in fluid dynamics, Fall 2008?
Use the x-MOM equation to show that the u velocity is positive beneath the crest (η>0) and negative beneath the trough (η<0), if the wave is propagating to the right, to increasing values of x. This gives us the energy flux correlation.
8. If the ocean tides have elevation ~ 0.5 m in mid-ocean and ~ 1m closer to shore, estimate the potential energy stored in them, using the total area of the oceans. Compare this number with the energy in the geostrophic ocean circulation, assuming that to be a combination of KE and PE, with currents averaging 0.1 m sec-1 in the top 500m of the ocean only, and lateral scale of the currents L ~ 100 km. The Rossby radius of deformation is needed for this estimate: calculate it as LD = c1/f where c1 is the phase speed of long gravity waves on the thermocline, c1 ≈ (g Δρ/ρ H1)1/2 . Use Δρ/ρ ~ 2 x10-3, H1 = 500m. Use ρ g η2 as potential energy PE per unit surface area, multiply by area of oceans, 3.61x1014 m2 (roughly 7/10 of the Earth’s surface, if you need to calculate it). For the PE and KE of ocean currents
use ½ Hu2 where H1 = 500m for KE and PE ≈ (LD/L)2 x KE where L = 100 km.
9. A tsunami is a long gravity wave excited by an earthquake...usually a vertical displacement in the seafloor at a plate boundary. How does the amplitude of the surface elevation wave, η, vary with the ocean depth?
Assume that the energy flux is Ec0, (the product of wave energy density (Joules m-3 )and group velocity c0) is the same in both places, implying that there is no wave reflection and dynamics remain linear, neither of which is really true. Rotation can be neglected for these waves, so co2 = gH.
If the wave has an amplitude of 1 m in the deep sea (5 km deep) what could its amplitude be where it comes into shallow water, (say 25m deep)? {Tsunamis have wavelengths of roughly 200 km and amplitude roughly 1m. in the deep sea.Animation of the Indian Ocean tsunami: Note, be sure to get the fluid depth in mind in calculating energy flux: the flux of KE is ½ (u2+v2) c0 times the area of the cross-section (height x width) that it propagates through. What is the corresponding relation for flux of PE?
As we have pointed out in class, gravity waves without significant Coriolis effects have equipartition of energy, KE=PE. We want to have the same energy flux in deep and shallow water, so equate ρgη2 c0 in the two places. This should be the same as equating ρgH(u2+v2)c0 in the midocean and shallow ocean. Either way the amplitude of the wave η varies like H-1/4growing slowly as the depth decreases.
10. Fluid flows may involve equations that are nonlinear, or linear. With linear equations you can add any two solutions to give a third solution…and so on to Fourier analysis where you add many sine-wave components to build a solution that may be just an isolated flow. Nonlinear equations often have no simple analytical solutions yet can be approached with conservation principles and numerical experiments. Use scale analysis to describe the ratio of typical nonlinear terms to linear terms in a long gravity wave without rotation (e.g. compare ∂η/∂t with u∂η/∂x where η is the free surface elevation). How does the nonlinearity parameter you find relate to observations of η and of the motion of fluid particles? This is just a comparison of the linear MOM term ut and the nonlinear MOM term u ux which gives U/T compared with U2/L. The importance of nonlinearity is thus UT/L which we can think of as U/C, C being the phase speed of the wave. Sometimes it is confusing because we think U must be the same as L/T. They have the same dimensions but it is the size that interests us. L and T are scales of the wave motion and are independent of U. Now, we can think of UT as a particle displacement during a wave period (or really period/2π) . So, the nonlinearity parameter can be writing (particle displacement/wavelength) (or really…wavelength/2π!). This is another nice description of nonlinearity. For short, deep-water surface waves you can see that the parameter also describes the geometric slope of the wave, from peak to trough because the velocity scale is the same for w and for u (not true for long hydrostatic waves!).
11. In the GFD lab we created a flow in a single layer of water with a mountain. Fluid that had been sitting over the mountain was swept off it. Describe what happens using PV (potential voriticy) ideas. There was a free surface. How would the result have differed if the upper surface had been a rigid, horizontal glass sheet?
Done in class; the stretched cyclone has low pressure at its center. This sucks down the fluid surface if it is free to be sucked down. If not, with a rigid lid, there will be more stretching and stronger vorticity.
12. For a single layer with a free surface we looked at both long, high-frequency waves (σ>f) and geostrophically balanced flow. The energy ratio, PE/KE was derived. Make a similar scale analysis of the ratio of the two contributors to PV at low frequency: the vertical vorticity and the ‘stretching’ term (that is the f/h contribution). Take PV to be q = (f+ζ)/h ≈ H-1 (f + ζ - fη/H) where ζ is relative vertical vorticity and η is free surface displacement.
{Note that we found two interrelated equations in this problem: the horizontal divergence (the η-equation for high frequency waves) and the vertical vorticity equation for ζ. At low frequency or with steady geostrophic flow what does our equation for η (formerly the wave equation) tell us?}
We have the vertical VORT equation, ζt = fwz = (f/H) ηt so changes in relative vorticity mirror changes in stretching following a fluid particle, in order to conserve PV. However that doesn’t mean the two components of PV are always the same. For geostrophic adjustment with a very large initial length scale, L > LD, most of the energy is PE rather than KE and most of the PV is stretching, and it stays there after adjustment to geos. balance. To see this look at the η equation below in prob 13
At low frequency 1/fT < 1 so the 1st term can be neglected.This leaves
For nearly geost. balance, η = fψ/g = p/fρ0 to order Ro + 1/fT. The first two terms are the relative vorticity and the stretching part of PV, with sizes ~ ETA c02/L2 and f2ETA where ETA is the scale estimate of η. Their ratio is just f2 L2/c02 ≡ (L/LD)2 which is what we are looking for.
13. (Long waves/geostrophic flow). A one-layer fluid on a rotating f-plane (‘flat Earth’) has a side boundary at x = 0. The boundary oscillates in the x-direction, sinusoidally in time.
Solve for the long waves generated by this oscillatory forcing. The long-wave equation is
where Q0 is the potential vorticity of the fluid (here we choose it to be zero, thinking that the fluid was at rest
before the oscillation began).
The boundary conditions are:
u = A cos σ0t at x = 0.
wave energy propagates away from the boundary toward x = ∞, or if there are no waves,
u => 0 as x => ∞
Note the great dependence of the solution on the forcing frequency σ0.
Discussed in class: the steps are: realize there is no active PV in this problem because there was none in the fluid to start with. This is not stated in words, but in the equation.
• then look for free wave solutions appropriate to the problem. Here, there is no variation of the boundary forcing in the y direction so we expect to find a wave that varies only in x and t.
This can be found by separation of variables or simply ‘by inspection’ as my old professors used to say.
That is, suspecting that η = Real (B exp(ikx – iσt)) is the right form, and furthermore we expect to find that σ = σ0because a linear equation with constant (in time) coefficients cannot turn one frequency into another. More simply, find this by fitting the solution to the boundary conditions (b.c.)
• find the dispersion relation by substituting the assumed wave solution into the equation, giving
σ2 = f2 + c02 k2
which tells us k as soon as we know σ.
• apply the b.c. You need to express the connection between η and u. This is from the MOM equations and was given in the notes: here is found from the pair of equations -iσu –fv = -gηx, -iσv +fu = -gηy There is a tight connection between u and v in the 2d MOM equation. We have no variation in the y direction so -iσv +f u = 0. Even though ∂/∂y of everything is zero, we can still have velocity in the y-direction and (with rotation) we need to. Solving,