Topic 3: surface ocean circulation


E. What is the impact of Ekman Transport on the surface ocean circulation?



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E. What is the impact of Ekman Transport on the surface ocean circulation?

1. Generalized global wind regimes (Fig. 8)



  • Two wind regimes have the greatest impact on surface ocean currents

  • Trade winds are the strong easterly winds between 0º and 30° in both hemispheres

  • Westerlies are the strong westerly winds between 30º and 60° in both hemispheres

2. Because the average water transport in the Ekman layer is perpendicular to the wind direction (to right in NH and to left in SH) then we can expect the following patterns of Ekman transport (Fig. 8)



  • net poleward and westward surface water flow away from the equator in the Trade Wind regime (0° to 30°)

  • net equatorward and eastward surface water flow in the Westerlies regime (30° to 60°)

  • Remember: the water flow directly caused by wind stress occurs in a surface layer which is <100m deep (Ekman layer)

3. This large scale (global) pattern of Ekman transport implies that, generally, there is a net surface transport of water towards 30°, away from 60° and away from the equator. (Look at the directions of the black arrows in Fig. 8 to see this pattern).


4. The Ekman transport resulting from steady Trade and Westerly winds distorts the sea surface height causing slight “hills” near 30° and slight “valleys” at 0º and near 60° in both hemispheres
5. These sea surface height distortions (from a level or flat ocean surface) have a very important impact on ocean circulation by producing geostrophic currents, which dominate the large-scale circulation patterns in the upper ocean.
III. GEOSTROPHIC CURRENTS
A. Horizontal Pressure Gradients
1. Under steady winds, Ekman transport causes surface water to pile up in certain regions of the ocean and be removed from other regions (Fig. 9).

  • Thus the winds cause small elevation distortions from a level sea surface

  • the total range in sea surface height is small, at ~2.5 meters

  • these small height changes occur over 1000s of kilometers of horizontal distances and are much too small to be detected by sight

-however, they can be detected and accurately measured by satellites
2. What prevents the water from continuing to pile up in certain regions (e.g. gyres) as winds continue to blow?

- A steady-state condition is reached where the rate of piling up surface water via Ekman transport (resulting from winds) is exactly balanced by the loss of water via surface currents away from the region.

( Likewise, in regions where Ekman transport is removing surface water (e.g. equator), this loss rate of water is exactly balanced by the flow of water via surface currents input to the region.)
3. What causes water to leave the regions where surface water is being piled up?


  • increased pressure caused by the piling up of surface water increases the pressure in these regions and causes water to flow away

  • Remember: Water, like air, moves away from regions of high pressure towards regions of low pressure

4. Pressure is a force that is exerted downward (it is a force per unit horizontal area)



  • Pressure at any depth equals the mass of water above the depth times gravitational acceleration per unit area

- Remember: Force = mass*acceleration

- In this case, Pressure = mass of the water column * gravitational acceleration


5. The mass of the seawater column above a chosen depth depends on the height of the seawater column and the average density of the seawater above the chosen depth

6. Thus Pressure (P) equals the gravitational acceleration * density*column height (Fig. 10)

- thus P = g * r * z, where r is the density of water (kg/m3), g is gravitational acceleration constant on earth (g = 9.8 m/s2) and z is depth in meters

- since Pressure is a force per unit area, the units are:

P = Force / area = mass acceleration /area = (kg * m/s2 )/(m2)

- Remember: the units of Force are Newtons, where 1 Newton (N) is the force required to accelerate 1 kg of mass by 1 m/sec2

- thus the units of Pressure are Newtons/m2

(Note: A negative sign is sometimes used to express pressure (P = -r*g*z) when the depths in the ocean are expressed as negative numbers.)


7. If you consider a column of seawater as a series of layers then the pressure change across one of these layers is

  • P = r*g*z (Fig. 10)

  • since the seawater density usually increase with depth then you usually have to calculate the P over z increments using the average density (ρ) for each layer

  • thus the total pressure at a chosen depth equals sum (Σ) of all the individual pressure above that depth (P =  g*r*z or Ptotal = g*ρ1*Δz1 + g*ρ2*Δz2 + g*ρ3*Δz3 + ….. until you reach the depth of interest

8. By the way, the relationship between pressure and depth (P = r*g*z) is how oceanographers calculate depth from a CTD’s measurement of pressure

-this depth calculation requires a calculation of density for the water column above the chosen depth, which is calculated from the CTD’s continuous measurements of temperature and salinity and pressure.

9. In regions of the ocean where surface water is piled up by winds (as a result of Ekman Transport), there is higher pressure than in regions where less or no piling of surface water has occurred because the sea surface height (ΔZ) is slightly greater (Fig. 11)


10. These slight differences in pressure, as a result of slight differences in sea surface height, between various locations in the ocean yield significant horizontal pressure gradients (HPG)

-an HPG is the difference in pressure at two sites in the ocean divided by the distance between the two sites (P/x) (Fig. 11)

Pa at A = ρa*g*z

Pb at B = ρb*g*(z+Δz)

-where ΔZ is the difference in sea surface height between sites A and B

11. The Horizontal Pressure Gradient (HPG) between A and B is (Fig. 11):

ΔP/Δx = (Pb - Pa) / (Xb - Xa)

ΔP/Δx = [(ρb*g*z) - (ρa*g*(z+Δz))] / Δx

-if we assume for now that density at the two sites is the same (ρb = ρa),

-then ΔP/Δx = ρb*g*Δz/Δx,


12. For example: if ρ = 1025 kg/m3 , Z = 2m and X = 10° latitude

-then ΔP/Δx = g * ρ * Δz/Δx

=9.8 m/s2 * 1025 kg/m3 * 2m/(10° *110km/° *1000m/km)

= 0.018 m/s2 *kg/m3 = 0.018 Newtons/m3

13. If θ represents the angle of the linear sea surface slope (Fig. 11)

-then ΔP/Δx = g* ρ * Δz/Δx

ΔP/Δx = g* ρ * tan θ (because tan θ = Δz/Δx)
14. HPG is a force exerted per unit volume, i.e., ΔP/Δx = (ΔF/Area)/ Δx = ΔF/(Area*Δx) = ΔF/ΔV

-that is, pressure is a force per unit area and it is divided by length (Δx)

-thus, HPG represents a Force/volume (Newtons/m3)
15. The convert the force/volume exerted by the horizontal pressure gradient to force/mass, one divides the HPG by density:

• HPG/ ρ = (Force/volume) / ρ = Force/(volume* ρ) = Force/mass (because mass = volume* ρ)

• HPG Force / ρ = (Force/volume) / ρ = (1/ ρ)*(ΔP/Δx)

• HPG Force / ρ = (1/ ρ)*(g * ρ *Δz/Δx) = g * Δz/Δx = g*tan θ (see Fig. 11)

-the units are m/s2 (units of acceleration)

16. The HPG Force/mass is important because it equals acceleration



-Remember: Force = mass * acceleration

-therefore, Acceleration = Force/mass

17. The greater the HPG/mass, then the greater the potential acceleration rate of horizontal currents

- thus, the greater the horizontal gradient in sea surface height (z/x), the greater the HPG/mass, the greater the potential acceleration, and the greater the potential geostrophic current velocity.

-that is, the greater the horizontal gradient in sea surface height (ΔZ/Δx), the faster the resulting geostrophic current velocity

18. Summary: In the upper ocean, Ekman transport of surface water caused by winds produces horizontal pressure gradients (ΔP/Δx) because winds pile up surface water in some regions and remove surface water from other regions



  • these horizontal pressure gradients (HPGs) are the forces that initiate horizontal water movement and accelerate currents

  • HPGs in the ocean, generally, produce currents that move water away from the regions of piled up water (higher pressure) towards regions where surface water has been removed (low pressure).

  • The water transported by these currents moving away from regions of high pressure exactly offsets the surface water transported towards these regions by wind-induced Ekman transport

  • This water volume balance in the upper ocean prevents the “hills” in the sea surface from continually growing and the “valleys” from continually deepening while winds continue to blow over the ocean.


B. Geostrophic Currents
1. Water flows away from high pressure and towards low pressure, so horizontal gradients in pressure cause water parcels to move.
2. However, water parcels moving away from high pressure are influenced by Coriolis Force (Fig. 12)

  • water parcels moving away from high pressure will be deflected to the right in the northern hemisphere and to the left in the southern hemisphere because of Coriolis Force

3. To see the effect of the Coriolis Force on currents caused by HPGs, let’s start with a level ocean surface and no motion. Then let winds blow over the ocean and pile up water in the east that, in turn, produces a horizontal pressure gradient force towards the west (Fig. 12). Let the winds continue to blow at the same speed and in the same direction.

What is the ocean’s response? (Fig. 12)


  • Initially, the HPG force causes water to move westward in the same direction as the HPG force is exerted.

  • However, the current is deflected towards the right (in the Northern Hemisphere) due to the Coriolis Force.

  • The HPG is a force which accelerates the current, so the current speeds up. The Coriolis Force keeps turning the current towards the right in the northern hemisphere (to the left in the Southern Hemisphere) and increasing in magnitude as the current speed increases.

  • The current speed increases until it reaches a velocity where the Coriolis Force on the current equals (and opposes in direction) the HPG force. (The Coriolis Force is proportional to current speed.)

  • This force balance occurs when the current is moving perpendicular to the HPG force because the Coriolis Force acts at a right angle to the direction of current flow.

  • At a force balance situation, the current speed remains constant over time

- a force balance occurs because the HPG force is exactly offset by the Coriolis Force

- under conditions of a force balance, the net force = 0 and, as a result, acceleration = 0

- as a result the current speed remains constant
4. Currents that result from a force balance between HPG force and Coriolis Force are called Geostrophic Currents


  • Geostrophic currents flow at right angles (perpendicular) to the HPG force (Fig. 12)

  • one can determine the direction of geostrophic currents by assuming that the Coriolis Force opposes the HPG force and the Coriolis Force is oriented perpendicular to current direction

  • thus Geostrophic Currents run in a circle around a high pressure dome (Fig. 13)

    • the currents run clockwise in the northern hemisphere and counterclockwise in the Southern Hemisphere

    • this hypothetical situation resembles the circular pattern of flow around the subtropical gyres observed in the ocean

  • in the Northern Hemisphere, the high pressure is to the right of the current flow direction (and in the Southern Hemisphere the higher pressure is to the left) as one looks in the direction of the current flow.

  • use Fig. 13 to help visualize the relationship between the directions of the geostrophic current and HPG force in both hemispheres.

5. The general pattern of surface current flow in the ocean is a result of geostrophic currents that respond to the horizontal pressure gradients caused by differences in sea surface height

-thus geostrophic currents flow perpendicular to HPGs and, thus, flow along the contour lines of sea surface height (look at the arrow directions along contours of sea surface height in Fig. 14)
6. Note: Frictional force is assumed to be negligible for Geostrophic Currents


  • it is assumed that in the interior of the ocean the magnitude of the frictional force is small (negligible) compared to the magnitude of the Coriolis and HPG forces

  • in contrast, at the edges (or interfaces) of the ocean (near sea floor or at the air-sea boundary) frictional forces must be accounted for (as they are in the Ekman layer)

  • velocity shear (Δvel/Δx or Δvel/Δz) is greatest near boundaries (air-sea, basin walls and bottom) which increases frictional forces

  • this simplification of a frictionless geostrophic force balance is reasonable for most of the ocean, i.e., generally, the predicted pattern of geostrophic current velocities (away from the ocean boundaries) agrees reasonably well with observed surface current patterns





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