Physical Oceanography, MSCI 3001 Oceanographic Processes, MSCI Dr. Katrin Meissner Ocean Dynamics.

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1 Physical Oceanography, MSCI 3001 Oceanographic Processes, MSCI 5004 Dr. Katrin Meissner Ocean Dynamics The Equations of Motion d u dt = 1 ρ Σ F dt = 1 ρ ΣF x dt = 1 ρ ΣF y dw dt = 1 ρ ΣF z Horizontal Equations: Acceleration = Pressure Gradient Force + Coriolis dt = 1 ρ + fv dt = 1 ρ dy fu Or, for a Barotropic Ocean: dt dt = g dη + fv = g dη dy fu Vertical Equation: Pressure Gradient force = Gravitational Force 1

2 If the only force acting on a water parcel is the Coriolis force, the Navier Stokes equations can be simplified to: dt = fv dt = fu Northern Hemisphere: f > 0 What does this mean? u > 0 u < 0 y v > 0 v < 0 x If the only force acting on a water parcel is the Coriolis force, the Navier Stokes equations can be simplified to: dt = fv dt = fu Northern Hemisphere: f > 0 What does this mean? u > 0 u < 0 y v > 0 v < 0 x 2

3 If the only force acting on a water parcel is the Coriolis force, the Navier Stokes equations can be simplified to: dt = fv dt = fu Southern Hemisphere: f < 0 What does this mean? u > 0 u < 0 y v > 0 v < 0 x Horizontal Equations: Acceleration = Pressure Gradient Force + Coriolis dt = 1 ρ dt = 1 ρ + fv dy fu If pressure gradients are small: dt = fv dt = fu Inertia currents the water flows around in a circle with frequency f. T=2π/f T(Sydney) = 21 hours 27 minutes 3

4 Scaling arguments: 1. If there are no surface slopes or horizontal density differences then there will be no pressure gradient force (i.e. left with Coriolis, inertia currents) What forces are important in a bathtub? What kind of speeds will the water get up to? What kind of accelerations? What surface slopes? dt dt = g dη + fv = g dη dy fu Barotropic! Scaling arguments: What forces are important in a bathtub? What kind of speeds will the water get up to? What kind of accelerations? What surface slopes? Magnitude of the pressure gradient force: dη = 0.1m /1m = 0.1 g dη =10ms 2 (0.1) =1ms 2 Magnitude of the Coriolis force: fu = ms 2 = ms 2 dt dt = g dη + fv = g dη dy fu Coriolis << Pressure Gradient, so we can neglect rotation effects Acceleration in the bathtub is driven by pressure differences (e to changes in surface slopes) 4

5 Scaling arguments: 1. If there are no surface slopes or horizontal density differences then there will be no pressure gradient force (i.e. left with /dt=fv, /dt=-fu, inertia currents): dt = fv dt = fu 2. If you are sitting in your bathtub, you are in a barotropic environment AND the Coriolis force can be neglected: dt dt = g dη = g dη dy But the ocean is not a bathtub. We will conct a scaling analysis on our equations of motion... to find further simplifications for motions with a period greater than ~10 days dt = 1 Scaling Analysis: ρ + fv T~10 days = 8.64 x 10 5 s ~ 10 6 s dt = 1 ρ dy fu u,v ~ U ~ 1cms -1-1ms -1 f~ 10-4 s -1 Acceleration << Coriolis 5

6 Geostrophic Balance Acceleration is much smaller than Coriolis and Pressure Gradient Force (this is true almost everywhere in the ocean) The ocean is in Geostrophic Balance (= balance between Pressure Gradient and Coriolis Forces) Ocean is in Steady State (no acceleration) /dt is negligible dt = 1 ρ dt = 1 ρ + fv dy fu 1 ρ 1 ρ = fv dy = fu Geostrophic Balance What does the geostrophic balance mean physically? Suppose we have a difference in sea-level height. Water will want to move from the region of high pressure towards the region of low pressure. 6

7 Geostrophic Balance As the water starts to move, the Coriolis effect (rotation) deflects the water to the right (NH) or left (SH). The water keeps getting deflected until the force e to the pressure difference balances the Coriolis force. This balance is called a geostrophic balance and the resulting current is referred to as a geostrophic current. 13 Which direction is the Geostrophic wind? (f <0 SH) y x PG CF V 7

8 Geostrophy Problem 1 In the NH: Which way does the current flow if sea level height is increasing towards the South? You can use the equations, or just think about the forces NH: f>0 dη/dy<0 fv = g dη fu = g dη dy C P u = g f u=(-)(+)(+)(-)>0 East! y dη dy Geostrophy Problem 2 Which direction does the water flow around this pressure feature if it is in the Northern Hemisphere? 8

9 Geostrophy Problem 3 Which direction does the water flow around this pressure feature if it is in the Southern Hemisphere? Anticyclone Cyclone Northern Hemisphere clockwise counterclockwise Southern Hemisphere 9

10 Anticyclonic circulation: ALWAYS around a high pressure system. clockwise in the Northern Hemisphere counter-clockwise in the Southern Hemisphere Cyclonic circulation: ALWAYS around a low pressure system. counter-clockwise in the Northern Hemisphere clockwise in the Southern Hemisphere Geostrophy Problem 4 A certain ocean current has a height change of 1.1 m (increasing to the east) over its width of 100 km at 45 N. How fast is the current flowing? fv = g dη g Δη Δx f = 2Ωsin(φ) = 1x10-4 s -1 g = 10 ms -2 Δη = 1.1m Δx = 100x1000 m V=1 m/s Summary: Geostrophy is the balance between pressure gradient forces and the Coriolis force. All major current systems in the ocean can essentially be considered geostrophic. Geostrophy does not work over short periods of time or small distances (other forces become dominant). Geostrophy also fails in regions where friction becomes important. 10

11 Forces on a Parcel of Water d u dt = 1 ρ ( F g + F C + F P + F f +...) Gravity Coriolis Pressure Friction dt = 1 ρ F x dt = 1 ρ F y dw = 1 dt ρ F z d u dt = 1 ρ ( F g + F C + F P + F f +...) The last force to consider is friction. This is only important at continental boundaries, at the bottom of the ocean, and at the surface (e to wind). What will the friction term look like? We know that friction always tries to retard motion. 11

12 Effects of Friction A simple model for the frictional force at the sea floor in the x and y direction is: F x = -ru F y = -rv Rayleigh frictional dissipation, r is a coefficient (r ~ 10-7 s -1 ) Hence the equations of motion become dt = 1 + fv ru ρ dt = 1 fu rv ρ dy 12

13 Henry Stommel (1948) F x = -ru F y = -rv Coriolis + pressure gradient forces Harald Sverdrup (1947) wind forcing Coriolis + pressure gradient forces wind forcing To examine the effects of friction, consider the simple balance: dt = ru A solution is: u(t) = u o e rt So that at t=0, u(0)=u o U i.e no pressure gradient forces and no coriolis force. Motion is just decelerated because of friction. 1/r represents the time it takes for the speed to drop to 1/e (~1/3) of its initial value. Velocity decreases with time because of friction. U o 1/3U o U o e -1 1/r Time 26 13

14 8/27/ (Monday) Mid Semester Break (no class) (Monday) Dr. Caroline Ummenhofer (and Dr. Matthew England) (Monday) Dr. Caroline Ummenhofer (Monday) Dr. Laura Ciasto (Monday) Labour Day (no class) (Wednesday!) Me again on a WEDNESDAY! Thermal Wind Balance So far we have assumed that density ρ is constant (barotropic) Small horizontal changes in ρ can result in large vertical changes in current/ wind e.g. near fronts and eddies 1 = fv ρ Starting with the geostrophic balance 1 = fu ρ dy We can differentiate the equations with regard to depth (z) 1 v= ρf 1 u= ρf dy = ρg dz d 1 = dz dz ρf d 1 = dz dz ρf dy = ρg dz 14

15 Thermal Wind Balance So far we have assumed that density ρ is constant (barotropic) Small horizontal changes in ρ can result in large vertical changes in current/ dz = 1 d ( ρf ρg) dz = g dρ ρf wind e.g. near fronts and eddies Starting with the geostrophic balance dz = 1 dz = g dρ ρf dy ρf dz dz We can differentiate the equations with regard to depth (z), then substitute for /dz v = 1 ρf u = 1 ρf dz = ρg dy dz = d 1 dz ρf dz = d dz 1 ρf dy dz = ρg u dρ << ρ dz = ρg dz = 1 ρf dz = 1 ρf dz = ρg d dz d dz dy dz = ρg dz = 1 ρf dz = 1 ρf dz = ρg d dy ρg ( ) d dz d dy dz For geostrophic conditions: Thermal Wind Balance The vertical structure of u and v is related to the horizontal density gradients dz = g ρf dz = g ρf dρ dρ dy i.e. horizontal density gradients in temperature (T) and salinity (S) can explain the change in horizontal velocity with depth (vertical profile of horizontal velocity). 15

16 Thermal Wind Balance dz = g ρf dρ > 0 dρ NH: f > 0, so, dz > 0 Velocity is into the page and increases with depth i.e. v gets more positive with increasing depth Thermal Wind Balance: Cold Core Eddy Example 16

17 Thermal Wind Balance: Cold Core Eddy Example y cold/saline core x Potential Density through an Eddy near the Gulf Stream. How does the geostrophic current velocity in the eddy change with depth? dz = g ρf Thermal Wind Balance: Cold Core Eddy Example dρ g ρf y Δρ Δx x dz = g ρf dz = g ρf dρ dρ dy Estimate the density gradient at x=40km: ρ changes from to kgm -3 over 25 km Δρ = Δx 25km dz = g ρf = s 1 =

18 dz = g ρf Thermal Wind Balance: Cold Core Eddy Example dρ g ρf y Δρ Δx x This means that at z = 500m, dz = s 1 let s see how much v varies over 100m of depth at z=500m (Δz=100m, Δv=?): Δv = s 1 100m = 0.15ms 1 = 15cms 1 i.e the velocity shear ~ -15 cm/s per 100m depth increase This means that v (velocity into page) is getting more negative as we get deeper Thermal Wind Balance: Cold Core Eddy Example y x To work out the actual velocities, we need one more piece of information: at depth (say 2000m) the velocity is 0. This is called the depth of no motion

19 Going down in depth, the density surfaces flatten out. We can assume a level of no motion where there is no longer a change in density. Hence we can calculate the change in velocity up through the water column. Note that on the other side of the eddy, the density surfaces slope the other way, so the circulation must also be in the opposite sense. V= V= V=0+.15 V=0 V is positive into the page 100m 100m 100m Thermal Wind Balance: Cold Core Eddy Example A physical explanation NH f > 0 Step 1: denser in the middle so the surface will be depressed v = 1 ρf 19

20 Thermal Wind Balance: Cold Core Eddy Example NH f > 0 P Step 2: Start by figuring out the pressure force e to the surface slope: Pressure increases moving away from centre barotropic component Remember the fishtank experiment Thermal Wind Balance: Cold Core Eddy Example NH f > 0 Step 3: Forget the barotropic component. What happens at greater depth? Density is higher at the centre than further away. Remember the fishtank experiment ρ 1 < ρ 2 20

21 Thermal Wind Balance: Cold Core Eddy Example NH f > 0 Step 3: Forget the barotropic component. What happens at greater depth? Density is higher at the centre than further away. So the pressure force, just e to density would be in the positive x-direction (on the right hand side of the gyre). The density gradient causes a pressure force that increases with depth. Opposite happens on the left side of the gyre. Remember the fishtank experiment baroclinic component ρ 1 < ρ 2 Thermal Wind Balance: Cold Core Eddy Example NH f > 0 Step 4: Add up barotropic component (black) and baroclinic component (blue). Keep repeating this down the water column, until there is no density gradient (i.e. at the bottom of the cold core eddy) A smaller pressure gradient force means a smaller geostropic velocity with depth v = 1 ρf 21

22 Thermal Wind Balance: Cold Core Eddy Example NH f > 0 Step 5: If we add up the forces e to the surface slope and e to the density gradient we get a pressure gradient force that decreases with depth. Thermal Wind Balance: Cold Core Eddy Example NH f > 0 C P P C Finally we need to use the geostrophic relationship: If we have a pressure gradient in the x direction, it will create a geostrophic velocity in the y direction, proportional in strength to the pressure gradient. v = 1 ρf 22

23 Example: Derive the rotation of these cold and warm water eddies, in the SH, using the thermal wind balance. Given that u & v = 0 at 2000m (depth of no motion) estimate the surface current. 23

24 Thermal Wind Balance For geostrophic flow (i.e. pressure is balanced by Coriolis): Geostrophic flow in the presence of horizontal density gradients Horizontal density gradients (T,S) can explain vertical velocity changes To know absolute velocity we need extra information (i.e. we need to know absolute velocity at some depth) dz = g ρf dz = g ρf dρ - Can figure out surface velocity from surface heights. - Often we assume that at a certain depth (e.g. 2000m) velocities are zero this is called the depth of no motion. - Once we know the velocity at the surface or the depth of no motion we can calculate velocity at all other depths using the thermal wind equation. dρ dy Summary: Ocean Dynamics Most of the motion in the ocean can be understood in terms of Newton s Law that the acceleration of a parcel of water (how fast its velocity changes with time /dt) is related to the sum of forces acting on that parcel of water. We can split the forces, velocities and accelerations into south-north (y,v), west-east (x,u) and up-down (z,w) components. In the vertical direction the acceleration is related to the difference between the water weight and the bouyancy (or pressure) force. When there is a vertical density gradient this leads to oscillations (Brunt Väisälä frequency N). The density gradient tries to inhibit vertical motion (and mixing). These vertical accelerations are generally very weak, so we get the hydrostatic equation. If the hydrostatic equation is integrated over depth, it just says that the pressure at a point just equals the weight of water above that point. Acceleration in the horizontal can be driven by a number of different forces: (1) The pressure gradient force. This exists whenever there is a surface slope and/ or a horizontal density gradient. (2) Coriolis (because we live on a rotating planet). It is very weak, so we only feel its effect over long times (> few days) and large distances (> 10s of km). Coriolis only affects moving fluids, deflecting to the right in the NH and to the left in SH. (3) Friction. Also only acts on moving water. Always acts to slow down motion. Important at the boundaries of the ocean. 24

25 Summary: Ocean Dynamics dt = 1 + fv ru ρ dt = 1 fu rv ρ dy Or for a constant density (barotropic ocean): dt dt dη = g + fv ru dη = g fu rv dy Over much of the ocean, the flow is steady (i.e. /dt=/dt=0) and friction is negligible, so we are left with the geostrophic balance i.e. pressure gradient forces balance coriolis. The current moves at right angles to the pressure gradient. P C 1 ρ = fv, 1 ρ dy = fu When there is a horizontal density gradient the velocity changes with depth. This can be calculated using the thermal wind equations dz = g ρf dρ, dz = g ρf dρ dy 25