cee 262a h ydrodynamics lecture 18 surface ekman layer

CEE 262A

HYDRODYNAMICS

Lecture 18

Surface Ekman layer

Another exact solution: Coriolis Viscous Stress: The Ekman Layer

A rotational boundary layer in an infinite ocean: flow driven by a wind stress at the surface (x3 = 0), acting in the x1 direction

x1x2

x3

u1u2

0x3 = 0

Assumptions: Steady flow, uniform density, constant viscosity, no pressure gradients

Steady Linear

Constant density, only hydrostatic pressure with no motion

Coriolis – Friction Balance

The governing equations reduce to (f-plane)2

12 2

3

22

1 23

ufu

x

ufu

x

With the BCs at x3 = 0:201

*3 0

2

3

0

uu

x

u

x

Again, we wish to make the momentum equations dimensionless, and again, we look at the surface BC

unknown velocity scale

unknown length scale

These imply that

Let’s define * * *1 1 2 2 3 3u Uu u Uu x Zx

2 * 2 *2* *1 1

2 22 *2 *23 3

2 * 2 *2* *2 2

1 12 *2 *23 3

u uU ZfUu f u

Z x x

u uU ZfUu f u

Z x x

Thus if we choose

2 ** 1

2 *23

2 ** 2

1 *23

uu

x

uu

x

the momentum equations are parameter free

The Ekman layer thickness

These must be solved subject to the dimensionless BCs:

*1*

3

*2

3

1

0

u

x

u

x

Both at x3*=0

We also require that the flow decay to zero as *

3x

This completes the physics part of the problem; what remains is the mathematical problem.

To solve this system of two equations, we follow Ekman and define

* *1 2u iu

In terms of , the two real o.d.e.s become one complex equation:

which has the solution

where A,B, 1, and 2 are all complex

Using the ode itself, we see that for either i

2i i

The two roots are given by

1 2

1 11 1

2 2i i

With

1 2

1 11 1

2 2i i

the condition that the flow disappear at great depth implies that B=0 since

* * *2 3 3 3exp exp 2 as B x x x

The condition on the water surface

1

1 11 1 1

2 2A A i A i

Thus, 3* *

1 2

3 3 3

11exp

2 2

1exp cos sin

2 2 2 2

i xiu i u

x x xii

Taking the real part to find u1*

* 3 3 31

3 3

1exp cos sin

2 2 2 2

exp sin42 2

x x xu

x x

(after using some trigonometric identities)

* 3 32 exp sin

42 2

x xu

Likewise, u2* is found from the imaginary part of and a

little more trigonometry to be

Thus, the velocity vector decays and rotates with depth – aka the “Ekman Spiral”

The vertical structure of the Ekman spiral, as originally plotted by Ekman

At the surface (x3=0):

E

x2

x1

Friction + wind:

Wind stress

motion

Friction (stress on bottom of parcel)

Friction + wind + Coriolis:

Wind stress

motion

Friction (stress on bottom of parcel)

Coriolis

In order to balance Coriolis and wind stress, motion must be at some angle to the wind

Finally, imagine we go deep enough into then ocean to have the stress be zero on the bottom of our slab

Thus, we see that in an integral sense the motion we produce must be at right angles to the wind - this motion is known as Ekman drift.To see this more formally, we integrate the dimensional momentum equations: 00 0

21 12 3 3 *

3 3 3

00 02 2

1 3 33 3 3

0

du dudf u dx dx u

dx dx dx

du dudf u dx dx

dx dx dx

Wind stress

motion

Coriolis

0 2*

2 2 3

0

1 1 3 0

uq u dx

f

q u dx

Thus the net transports are found to be

i.e., to the right of the wind in the Northern hemisphere!

California

Wind from north

Upwelling (downwelling): The effects of variations in Ekman transport

Ekman drift

Upwelling from depth

California

Wind from south

Ekman drift

Downwelling from surface

Why are California coastal waters so COLD? Upwelling!

Upwelling-favorable winds duringspring/summer.