Physical Oceanography, MSCI 3001 Oceanographic Processes, MSCI Dr. Katrin Meissner Week 5.

Transcription

1 Physical Oceanography, MSCI 3001 Oceanographic Processes, MSCI 5004 Dr. Katrin Meissner Week 5 Ocean Dynamics Transport of Volume, Heat & Salt Flux: Amount of heat, salt or volume that flows through a unit area per unit time (e.g. in Jm -2 s -1 (for heat), kgm -2 s -1 (for salt) or m 3 m -2 s -1 =ms -1 (for volume)). Volume transport: The volume of moving water measured between two points of reference and expressed in cubic meters per second or in Sverdrup (Sv). 1 Sv = 1 million cubic meters per second = 10 6 m 3 /s Harald Sverdrup 1

2 The amount of salt in a parcel of water moving through the ocean remains relatively constant (unless the parcel is in contact with the surface, then river discharge, evaporation, precipitation, sea ice, will change it!) Salinity is conservative. The flux of salt into the cube includes an aective contribution (ρsu) and a diffusive contribution (-Kρds/dx) Diffusive flux (turbulent and molecular) of salt has a negative sign because it is directed in the opposite direction to the salinity gradient The conversation equation for salt per unit volume is given by Aective Flux Diffusive Flux But don t panic... In most instances the aective flux will dominate, so the diffusive part can be ignored. We are also usually interested in 1D flow (e.g. flow through a channel or across a section) Volume transport of an ocean current with velocity u through an area A is given by: Volume transport = u. A = u.h.l u A Volume swept out 2

3 Transport of Volume, Heat & Salt If we know the heat (or salt) per m 3 we can convert volume flux into heat (or salt) flux. Heat per m 3 = ρc p T Salt per m 3 = ρs So, multiplying by ua Q=ρc p TuA S f =ρsua ρ is density, c p ~ 4000 J/Ckg, salinity is in absolute units, 35kg/ 1000kg UNITS heat flux Watts (1 Watt = 1 J/s). Salt flux - kg/s Total Heat and Salt Fluxes V T = u L H Q T = ρc p T u L H S ft = ρs u L H m 3 /s J/s kg/s T, S 3

4 Example Calculate the total volume, heat and salt fluxes a) Past Tasmania b) Through the Drake Passage. Assume the ocean currents persist over the upper 1000m only. c) Why might the volume flux be different? The Equations of Motion Dynamics = how things move and why things move. The fundamental equations describing this motion is Newton s second law: F = force i.e. F = m a or a = 1 m Σ F m = mass a = acceleration ρ = ρ 0 (1-αT+βS) 4

5 Forces Slope force component of weight acting down the slope Friction force only acts on moving car The car will accelerate /dt>0 until the slope force is balanced by friction. At that point the net force will be zero and the car will continue at a steady speed. Just because the net force is 0 doesn t mean that the car won t be moving. Newton s Laws of Motion An object will continue to move in a straight line and at a constant speed unless acted on by a NET force The change in the velocity (speed and/or direction) of an object (e.g. a bit of water) is proportional to the force and inversely proportional to the mass of the object. a = 1 m Σ F but m = ρv so, a = 1 ρv Σ F Oceanographers willconsider force per unit volume. Acceleration a = d u dt So, dt = 1 ρ ΣF x dt = 1 ρ ΣF y dw dt = 1 ρ ΣF z u,x: west to east v,y: south to north w,z: up to down 5

6 Ocean Dynamics What are the forces that drive the Ocean? Gravity (earth, sun and moon) and centrifugal force Pressure forces Friction predominantly at boundaries* Wind only at surface boundary Seismic Forces occasional impulse only Coriolis* - e to the earth s rotation Buoyancy e to vertical T,S differences (related to heat and water fluxes at the ocean surface) * These forces are only experienced by fluid in motion 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 6

7 Newton s Laws of Motion Vertical direction: P ρ ΔV Δz What are the forces acting in the vertical direction? The boxes weight is acting downwards (mg) The pressure at the top of the box is also trying to force the box downwards. But the pressure at the bottom of the box is trying to force it upwards (the difference in the pressure forces is just the buoyancy force discussed for Archimedes) Δx Δy P=F/A P+ΔP ma z = ΣF z ρδva z = mg + PΔA (P + ΔP)ΔA Weight (mg) Surface: ΔA=ΔxΔy Volume: ΔV=ΔxΔyΔz Mass: m=ρδv=ρδxδyδz ρ(δxδyδz)a z = ρ(δxδyδz)g + P(ΔxΔy) (P + ΔP)(ΔxΔy) a z = g 1 ΔP ρ Δz But acceleration in the z direction is just: dw/dt, so Newton s Laws of Motion Vertical direction Vertical acceleration Buoyancy force Weight We can normally make this even simpler: We can us SCALE ANALYSIS to evaluate the size of each term, compare their magnitudes, and neglect the small terms. e.g. Suppose A = B + C If we know that B= and C=10, then to a good approximation we could simplify this equation to A C e.g. Suppose D = E + F If D= and E=10, then then to a good approximation we could simplify this equation to E F 7

8 Exercise Note that the Equatorial Pacific is ~ km wide and the upwelling is concentrated in a 50 km wide band at the equator. Once the winds start blowing it takes about a day for the upwelling to spin up What is dw/dt? How does it compare in size to gravity? Exercise Note that the Equatorial Pacific is ~ km wide and the upwelling is concentrated in a 50 km wide band at the equator. Once the winds start blowing it takes about a day for the upwelling to spin up What is dw/dt? How does it compare in size to gravity? Upwelling =100,000,000m 3 /s. This is coming up through an area of: A=10,000,000x50,000=5x10 11 m 2. So (remember volume transport): w= (100,000,000m 3 /s)/(5x10 11 m 2 ) = m/s (very small!) 8

9 Exercise What is dw/dt? How does it compare in size to gravity? w= (100,000,000m 3 /s)/(5x10 11 m 2 ) = m/s (very small!) If the wind were to stop and then start again it would take about a day to start upwelling at that rate again i.e. It takes 1 day to reach a velocity of w=0.0002m/s if we start from w 0 =0m/s. So we can approximate our acceleration to dw dt w w 0 T = ( )ms s = ms 2 << g = 10ms -2 In general over the ocean vertical acceleration is much smaller than g. This means that in the vertical equation g and must be of similar magnitude In general over the ocean vertical acceleration is much smaller than g. This means that in the vertical equation g and must be of similar magnitude This is called the hydrostatic equation We can integrate this equation since density and g are essentially constant. h h dp dz dz = ρgdz 0 0 Or simply Which you are hopefully familiar with already! 9

10 Newton s Laws of Motion Horizontal direction P Just like vertical pressure gradients can create a force on the water, so can horizontal gradients ρ ΔV Δz Surface Δx Δy P+ΔP We can now forget the vertical we ve done that already. What about forces in the horizontal Bottom The pressure on the right of the box will be larger than the pressure on the left of the box There will be a net force from right to left. Let s zoom in Newton s Laws of Motion Horizontal direction ma x = ΣF x ρδva x = P 1 ΔA P 2 ΔA ρ (ΔxΔyΔz)a x = (P)ΔA (P + ΔP)ΔA ρ (ΔxΔyΔz)a x = (P)ΔyΔz (P + ΔP)ΔyΔz ρ P Δx Δz Δy P+ ΔP ρ a x Δx = P P ΔP a x = dt = 1 ΔP ρ Δx dt = 1 ρ dp dx Surface: ΔA=ΔyΔz Volume: ΔV=ΔxΔyΔz Mass: m=ρδv=ρδxδyδz And, doing the same in the y-direction: dt = 1 dp ρ dy 10

11 Barotropic and Baroclinic Motion Remember, p = ρgz, and 5cm 5cm 1cm ρ 0 =1027 ρ 1 =1026 < ρ 2 =1028 kgm -3 Barotropic and Baroclinic Motion Remember, p = ρgz p1=1027*10*0.01=102.7kg/m/s 2 or Pa p2=1027*10*0.03=308.1 Pa p3=513.5 Pa p1=205.4 Pa p2=410.8 Pa p3=616.2 Pa cm 1cm ρ 0 = dt 2 dt 3 dt dt = 1 p ρ x = 1 Δp ρ Δx = = = = 2m /s 2 = 2m /s 2 = 2m /s 2 11

12 Barotropic and Baroclinic Motion dt = 1 p ρ x = 1 Δp Remember, p = ρgz, and ρ Δx p1=1026*10*0.02=205.2pa p2=1026*10*0.04=410.4pa p3=615.6pa p1=205.6pa p2=411.2pa p3=616.8pa 1 /dt=-(1/1027)*( )/0.05 = ms -2 2 /dt=-(1/1027)*( )/0.05 = ms -2 3 /dt= ms -2 5cm ρ 1 =1026 < ρ 2 =1028 kgm -3 Barotropic and Baroclinic Motion Remember, p = ρgz, and dt = 1 p ρ x Barotropic velocity is constant with depth Baroclinic velocity changes with depth ρ 0 ρ 1 < ρ 2 Motion e to surface slopes Motion e to density differences 12

13 Barotropic and Baroclinic motion Barotropic Currents move the whole water column. Can be inced by a horizontal tilt of the water surface, which proce a pressure gradient throughout the whole water column. Baroclinic Currents vary with depth. Can be inced by horizontal density gradients. This effect is known as the thermal wind balance (more later). BAROTROPIC BAROCLINIC Barotropic and Baroclinic Motion Remember, p = ρgz, and Mixed situation ρ 1 < ρ 2 Motion e to density differences 13

14 Barotropic: isobaric (equal pressure) surfaces are parallel to isopycnic (equal density) surfaces Baroclinic: isobaric and isopycnic surfaces are NOT parallel Barotropic: Isobaric surfaces are parallel to isopycnic surfaces Baroclinic: isobaric and isopycnic surfaces are NOT parallel Where are we likely to find barotropic conditions in the ocean? Well-mixed surface layers Shallow shelf seas (particularly where shelf waters are well mixed by tidal currents) Deep ocean (below the permanent thermocline) Where are we likely to find baroclinic conditions in the ocean? Regions of fast surface currents 14

15 Barotropic Ocean For a constant density ocean, we can write the pressure gradient in an easier way. Remember, p = ρgz, and η2 η1 d h1 h2 P1=h1ρg h1=d+η1 P2=h2ρg Δx h2=d+η2 So we are left with 15

16 Barotropic Motion For a constant density ocean, the acceleration of water just depends on what is happening at the surface i.e. the slope of the sea surface Δη ( ) 1 = = Δx = 10( ) = 2ms 2 dt 5 5cm 1cm Same answer as we got before (using pressure) It s magic! ρ0=1027kgm-3 Pressure Gradients 16

17 Summary: Start with Newton s Law: the acceleration of a parcel of water is related to the sum of forces acting on that water parcel. We can break down the acceleration and forces into those acting in the x,y and z directions In the z-direction, vertical velocities/accelerations are small, which leaves us with the hydrostatic equation In the x-direction For a barotropic (constant density) situation we can simplify the equation to: Similarly in the y-direction This just means that the weight of water is balanced by the vertical pressure gradient (in other words the buoyancy force) The water acceleration is driven by horizontal differences in pressure (these differences may be related to horizontal differences in surface height (barotropic) or in density (baroclinic)) But there are other forces out there The Coriolis Force a = 1 m Σ F dt = 1 p + Coriolis force +... ρ x dt = 1 p + Coriolis force +... ρ y p z = ρg 17

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19 Earth s rotation Observations of moving parcels are judged relative to coordinates that are fixed to the Earth If we were somewhere on a spaceship we would see the atmosphere s independent motion and the Earth turning out from under the moving parcels of air Deflection to the right in the Northern Hemisphere Northern hemisphere Deflection to the left in the Southern Hemisphere Southern hemisphere 19

20 The Coriolis Force The strength of the Coriolis force varies with latitude It is proportional to the Coriolis parameter f=2ωsin(φ), where Φ is latitude And Ω is the angular velocity (in radians per second) It is maximum at the poles, zero at the equator and changes sign from NH to SH Ω= 2π / 24 hours = 7.272x10-5 rad/s The Coriolis Force Earth spins counter clockwise looking at it from the north pole 20

21 The Coriolis Force The strength of the Coriolis force varies with latitude It is proportional to the Coriolis parameter f=2ωsin(φ), where Φ is latitude And Ω is the angular velocity (in radians per second) It is maximum at the poles, zero at the equator and changes sign from NH to SH The acceleration e to the Coriolis force is: f=2ωsin(φ) fv - in the x-direction (east-west), and -fu - in the y-direction (north-south) What is the value of Ω (the earth rotates 360º every day)? What is f at the north pole, the equator, 30N and 30S? What is the acceleration of water flowing at 1m/s at 30S? What pathway will the water trace out? The Coriolis Force Summary Acceleration e to the Coriolis force is fv (x direction) and -fu (y direction) acts only if water/air is moving acts at right angles to the direction of motion causes water/air to move to the right in the northern hemisphere causes water/air to move to the left in the southern hemisphere 21

22 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 dp ρ dx + fv dt = 1 dp ρ dy fu Or, for a Barotropic Ocean: dt dt = g dη dx + fv = g dη dy fu Vertical Equation: Pressure Gradient Force = Gravitational Force dt = 1 ρ dt = 1 ρ dp dx + fv dp dy fu If the only force acting on a body is Coriolis, the Navier Stokes equations can be simplified to: What does this mean? dt = fv dt = fu u > 0 u < 0 y v > 0 v < 0 x 22

23 If the only force acting on a body is Coriolis, 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 body is Coriolis, 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 23

24 If the only force acting on a body is Coriolis, 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 dp dx + fv dp 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 24