5.5: Ekman Pumping Effective Depth of Ekman Layer. 2 Effective - - PowerPoint PPT Presentation

§5.5: Ekman Pumping

Effective Depth of Ekman Layer.

2

Effective Depth of Ekman Layer.

Defining γ =

• f/2K, we derived the solution

u = ug(1 − e−γz cos γz) v = uge−γz sin γz corresponding to the Ekman spiral.

2

Effective Depth of Ekman Layer.

Defining γ =

• f/2K, we derived the solution

u = ug(1 − e−γz cos γz) v = uge−γz sin γz corresponding to the Ekman spiral. Let us assume the values f = 10−4 s−1 and K = 5 m2s−1.

2

Effective Depth of Ekman Layer.

Defining γ =

• f/2K, we derived the solution

u = ug(1 − e−γz cos γz) v = uge−γz sin γz corresponding to the Ekman spiral. Let us assume the values f = 10−4 s−1 and K = 5 m2s−1. The effective depth is D = π/γ.

2

Effective Depth of Ekman Layer.

Defining γ =

• f/2K, we derived the solution

u = ug(1 − e−γz cos γz) v = uge−γz sin γz corresponding to the Ekman spiral. Let us assume the values f = 10−4 s−1 and K = 5 m2s−1. The effective depth is D = π/γ. With f = 10−4 s−1 and K = 5 m2s−1 we have D = π γ = π

• 2K

f = π

• 2 × 5

10−4 = 993 m ≈ 1 km Thus, the effective depth of the Ekman boundary layer is about one kilometre.

2

Effective Depth of Ekman Layer.

Defining γ =

• f/2K, we derived the solution

u = ug(1 − e−γz cos γz) v = uge−γz sin γz corresponding to the Ekman spiral. Let us assume the values f = 10−4 s−1 and K = 5 m2s−1. The effective depth is D = π/γ. With f = 10−4 s−1 and K = 5 m2s−1 we have D = π γ = π

• 2K

f = π

• 2 × 5

10−4 = 993 m ≈ 1 km Thus, the effective depth of the Ekman boundary layer is about one kilometre. Note that D depends on the values of f and K so the par- ticular value 1 km is more an indication of the scale that a sharp quantitative estimate.

2

Remarks on the Ekman Spiral

3

Remarks on the Ekman Spiral

• The Ekman theory predicts a cross-isobar flow of 45◦ at

the lower boundary. This is not in agreement with obser- vations

3

Remarks on the Ekman Spiral

• The Ekman theory predicts a cross-isobar flow of 45◦ at

the lower boundary. This is not in agreement with obser- vations

• Better agreement can be obtained by coupling the Ek-

man layer to a surface layer where the wind direction is unchanging and the speed varies logarithmically

3

Remarks on the Ekman Spiral

• The Ekman theory predicts a cross-isobar flow of 45◦ at

the lower boundary. This is not in agreement with obser- vations

• Better agreement can be obtained by coupling the Ek-

man layer to a surface layer where the wind direction is unchanging and the speed varies logarithmically

• This can be done by taking a boundary condition

V

• ∂V

∂z @ z = zB

3

Remarks on the Ekman Spiral

• The Ekman theory predicts a cross-isobar flow of 45◦ at

the lower boundary. This is not in agreement with obser- vations

• Better agreement can be obtained by coupling the Ek-

man layer to a surface layer where the wind direction is unchanging and the speed varies logarithmically

• This can be done by taking a boundary condition

V

• ∂V

∂z @ z = zB

• The solution is then called a modified Ekman spiral.

3

Remarks on the Ekman Spiral

• The Ekman theory predicts a cross-isobar flow of 45◦ at

the lower boundary. This is not in agreement with obser- vations

• Better agreement can be obtained by coupling the Ek-

man layer to a surface layer where the wind direction is unchanging and the speed varies logarithmically

• This can be done by taking a boundary condition

V

• ∂V

∂z @ z = zB

• The solution is then called a modified Ekman spiral.

The modified Ekman Layer is discussed on Holton (§5.3.6). We will not discuss it here.

3

Vertical Velocity

The Ekman solution implies cross-isobar flow in the Planetary Boundary Layer (PBL).

4

Vertical Velocity

The Ekman solution implies cross-isobar flow in the Planetary Boundary Layer (PBL). The flow is consistently towards lower pressure.

4

Vertical Velocity

The Ekman solution implies cross-isobar flow in the Planetary Boundary Layer (PBL). The flow is consistently towards lower pressure.

4

For a steady-state solution, the convergence towards lower pressure centres and divergence from higher pressure cen- tres has two implications:

5

For a steady-state solution, the convergence towards lower pressure centres and divergence from higher pressure cen- tres has two implications:

• There must be upward velocity at the top of the PBL in

regions of low pressure.

5

For a steady-state solution, the convergence towards lower pressure centres and divergence from higher pressure cen- tres has two implications:

• There must be upward velocity at the top of the PBL in

regions of low pressure.

• There must be downward velocity at the top of the PBL

in regions of high pressure.

5

For a steady-state solution, the convergence towards lower pressure centres and divergence from higher pressure cen- tres has two implications:

• There must be upward velocity at the top of the PBL in

regions of low pressure.

• There must be downward velocity at the top of the PBL

in regions of high pressure. These implications follow from consideration of the conservation of mass.

5

For a steady-state solution, the convergence towards lower pressure centres and divergence from higher pressure cen- tres has two implications:

• There must be upward velocity at the top of the PBL in

regions of low pressure.

• There must be downward velocity at the top of the PBL

in regions of high pressure. These implications follow from consideration of the conservation of mass. We will now calculate the vertical velocity at the top of the Ekman layer.

5

First, consider a purely zonal geostrophic flow. So the isobars are oriented in an east-west direction.

6

First, consider a purely zonal geostrophic flow. So the isobars are oriented in an east-west direction. The cross-isober mass transport per unit area in the planetary boundary layer (PBL) is just ρ0v (kg m−2s−1).

6

First, consider a purely zonal geostrophic flow. So the isobars are oriented in an east-west direction. The cross-isober mass transport per unit area in the planetary boundary layer (PBL) is just ρ0v (kg m−2s−1). The cross-isober mass transport through a column of unit width extending through the entire PBL is the vertical in- tegral of ρ0v through the layer z = 0 to z = D = π/γ: M = D ρ0v dz (kg m−1s−1)

6

First, consider a purely zonal geostrophic flow. So the isobars are oriented in an east-west direction. The cross-isober mass transport per unit area in the planetary boundary layer (PBL) is just ρ0v (kg m−2s−1). The cross-isober mass transport through a column of unit width extending through the entire PBL is the vertical in- tegral of ρ0v through the layer z = 0 to z = D = π/γ: M = D ρ0v dz (kg m−1s−1) Now substitute the Ekman solution for v: v = uge−γz sin(γz) = uge−πz/D sin(πz/D)

6

First, consider a purely zonal geostrophic flow. So the isobars are oriented in an east-west direction. The cross-isober mass transport per unit area in the planetary boundary layer (PBL) is just ρ0v (kg m−2s−1). The cross-isober mass transport through a column of unit width extending through the entire PBL is the vertical in- tegral of ρ0v through the layer z = 0 to z = D = π/γ: M = D ρ0v dz (kg m−1s−1) Now substitute the Ekman solution for v: v = uge−γz sin(γz) = uge−πz/D sin(πz/D) The result is thus M = D ρ0ug exp(−πz/D) sin(πz/D) dz

6

Since ρ0 and ug are assumed to be constant, we have M = ρ0ug D exp(−πz/D) sin(πz/D) dz

7

Since ρ0 and ug are assumed to be constant, we have M = ρ0ug D exp(−πz/D) sin(πz/D) dz Defining a new vertical variable Z = πz/D, this is M = ρ0ug D π π exp(−Z) sin Z dZ

7

Since ρ0 and ug are assumed to be constant, we have M = ρ0ug D exp(−πz/D) sin(πz/D) dz Defining a new vertical variable Z = πz/D, this is M = ρ0ug D π π exp(−Z) sin Z dZ Using a standard integral, this may be written

M = 1

2

D π

• ρ0ug

7

Since ρ0 and ug are assumed to be constant, we have M = ρ0ug D exp(−πz/D) sin(πz/D) dz Defining a new vertical variable Z = πz/D, this is M = ρ0ug D π π exp(−Z) sin Z dZ Using a standard integral, this may be written

M = 1

2

D π

• ρ0ug

Here we have used the fact that e−π ≈ 0.0432 ≪ 1

7

Exercise: Show that

π exp(−Z) sin Z dz = 1

2(1 + e−π) ≈ 1 2

8

Exercise: Show that

π exp(−Z) sin Z dz = 1

2(1 + e−π) ≈ 1 2

Solution:

• Evaluate the integral analytically
• Consult a Table of Integrals (e.g., GR2.663)
• Evaluate by numerical integration (MatLab)
• Use Maple to evaluate it.

8

Exercise: Show that

π exp(−Z) sin Z dz = 1

2(1 + e−π) ≈ 1 2

Solution:

• Evaluate the integral analytically
• Consult a Table of Integrals (e.g., GR2.663)
• Evaluate by numerical integration (MatLab)
• Use Maple to evaluate it.

Note that the analytical evaluation of the integral is straight- forward.

8

Exercise: Show that

π exp(−Z) sin Z dz = 1

2(1 + e−π) ≈ 1 2

Solution:

• Evaluate the integral analytically
• Consult a Table of Integrals (e.g., GR2.663)
• Evaluate by numerical integration (MatLab)
• Use Maple to evaluate it.

Note that the analytical evaluation of the integral is straight- forward. For example, it can be done by means of integration by parts (twice), or by expressing the sin-function in terms of complex exponentials.

8

Next, integrate the continuity equation through the PBL: D ∂u ∂x + ∂v ∂y + ∂w ∂z

• dz =

D ∂u ∂x + ∂v ∂y

• dz + [w(D) − w(0)] = 0

9

Next, integrate the continuity equation through the PBL: D ∂u ∂x + ∂v ∂y + ∂w ∂z

• dz =

D ∂u ∂x + ∂v ∂y

• dz + [w(D) − w(0)] = 0

We assume the surface is flat, so that w(0) = 0. Then w(D) = − D ∂u ∂x + ∂v ∂y

• dz

9

Next, integrate the continuity equation through the PBL: D ∂u ∂x + ∂v ∂y + ∂w ∂z

• dz =

D ∂u ∂x + ∂v ∂y

• dz + [w(D) − w(0)] = 0

We assume the surface is flat, so that w(0) = 0. Then w(D) = − D ∂u ∂x + ∂v ∂y

• dz

Now recall the Ekman solution u = ug[1 − e−πz/D cos(πz/D)] , v = uge−πz/D sin(πz/D)

9

Next, integrate the continuity equation through the PBL: D ∂u ∂x + ∂v ∂y + ∂w ∂z

• dz =

D ∂u ∂x + ∂v ∂y

• dz + [w(D) − w(0)] = 0

We assume the surface is flat, so that w(0) = 0. Then w(D) = − D ∂u ∂x + ∂v ∂y

• dz

Now recall the Ekman solution u = ug[1 − e−πz/D cos(πz/D)] , v = uge−πz/D sin(πz/D) Since the flow is purely zonal, ug is independent of x so the (horizontal) divergence reduces to ∂u ∂x + ∂v ∂y

• = ∂v

∂y = ∂ug ∂y

• e−πz/D sin(πz/D)

9

Next, integrate the continuity equation through the PBL: D ∂u ∂x + ∂v ∂y + ∂w ∂z

• dz =

D ∂u ∂x + ∂v ∂y

• dz + [w(D) − w(0)] = 0

We assume the surface is flat, so that w(0) = 0. Then w(D) = − D ∂u ∂x + ∂v ∂y

• dz

Now recall the Ekman solution u = ug[1 − e−πz/D cos(πz/D)] , v = uge−πz/D sin(πz/D) Since the flow is purely zonal, ug is independent of x so the (horizontal) divergence reduces to ∂u ∂x + ∂v ∂y

• = ∂v

∂y = ∂ug ∂y

• e−πz/D sin(πz/D)

Subsitituing this into the equation for w(D) gives w(D) = −∂ug ∂y D e−πz/D sin(πz/D) dz = −1

2

• D

π

∂ug ∂y

9

Again, w(D) = −1

2

D π ∂ug ∂y

10

Again, w(D) = −1

2

D π ∂ug ∂y But recall the expression for mass transport M = 1

2

D π

• ρ0ug

10

Again, w(D) = −1

2

D π ∂ug ∂y But recall the expression for mass transport M = 1

2

D π

• ρ0ug

Combining these, we have ρ0 w(D) = − ∂M ∂y

10

Again, w(D) = −1

2

D π ∂ug ∂y But recall the expression for mass transport M = 1

2

D π

• ρ0ug

Combining these, we have ρ0 w(D) = − ∂M ∂y This says that the mass flux out of the boundary layer is equal to the cross-isobar mass transport in the layer.

10

Again, w(D) = −1

2

D π ∂ug ∂y But recall the expression for mass transport M = 1

2

D π

• ρ0ug

Combining these, we have ρ0 w(D) = − ∂M ∂y This says that the mass flux out of the boundary layer is equal to the cross-isobar mass transport in the layer. We now note that the geostrophic vorticity is given by ζg = ∂vg ∂x − ∂ug ∂y

• = −∂ug

∂y

10

Again, w(D) = −1

2

D π ∂ug ∂y But recall the expression for mass transport M = 1

2

D π

• ρ0ug

Combining these, we have ρ0 w(D) = − ∂M ∂y This says that the mass flux out of the boundary layer is equal to the cross-isobar mass transport in the layer. We now note that the geostrophic vorticity is given by ζg = ∂vg ∂x − ∂ug ∂y

• = −∂ug

∂y Thus, w(D) = −1

2

• D

π

∂ug

∂y =

• D

2π

• ζg

10

This is the so-called Ekman Pumping formula:

w(D) = 1 2π

• Dζg

11

This is the so-called Ekman Pumping formula:

w(D) = 1 2π

• Dζg

It shows that the vertical velocity at the top of the Ekman Layer is proportional to the geostrophic vorticity.

11

This is the so-called Ekman Pumping formula:

w(D) = 1 2π

• Dζg

It shows that the vertical velocity at the top of the Ekman Layer is proportional to the geostrophic vorticity. In the vicinity of Low Pressure we have Cyclonic Flow

• ⇐

⇒ Positive Vorticity

• ⇐

⇒ Upward Velocity

• 11

This is the so-called Ekman Pumping formula:

w(D) = 1 2π

• Dζg

It shows that the vertical velocity at the top of the Ekman Layer is proportional to the geostrophic vorticity. In the vicinity of Low Pressure we have Cyclonic Flow

• ⇐

⇒ Positive Vorticity

• ⇐

⇒ Upward Velocity

• In the vicinity of High Pressure we have

Antiyclonic Flow

• ⇐

⇒ Negative Vorticity

• ⇐

⇒ Downward Velocity

• 11

This is the so-called Ekman Pumping formula:

w(D) = 1 2π

• Dζg

It shows that the vertical velocity at the top of the Ekman Layer is proportional to the geostrophic vorticity. In the vicinity of Low Pressure we have Cyclonic Flow

• ⇐

⇒ Positive Vorticity

• ⇐

⇒ Upward Velocity

• In the vicinity of High Pressure we have

Antiyclonic Flow

• ⇐

⇒ Negative Vorticity

• ⇐

⇒ Downward Velocity

• Note on Dines Mechanism to be added later.

11

Magnitude of Ekman Pumping

12

Magnitude of Ekman Pumping

The vertical velocity is w(D) = 1 2π

• Dζg

12

Magnitude of Ekman Pumping

The vertical velocity is w(D) = 1 2π

• Dζg

Suppose D = 1 km and ζ = 5 × 10−5 s−1. Then w(D) = 1 2π

• × 103 × (5 × 10−5) = 5 × 10−2

2π ≈ 8 mm s−1 ∼ 1 cm s−1

12

Magnitude of Ekman Pumping

The vertical velocity is w(D) = 1 2π

• Dζg

Suppose D = 1 km and ζ = 5 × 10−5 s−1. Then w(D) = 1 2π

• × 103 × (5 × 10−5) = 5 × 10−2

2π ≈ 8 mm s−1 ∼ 1 cm s−1 This is a relatively small value for vertical velocity, but it is important as it may extend over a large area and persist for a long time.

12

Magnitude of Ekman Pumping

The vertical velocity is w(D) = 1 2π

• Dζg

Suppose D = 1 km and ζ = 5 × 10−5 s−1. Then w(D) = 1 2π

• × 103 × (5 × 10−5) = 5 × 10−2

2π ≈ 8 mm s−1 ∼ 1 cm s−1 This is a relatively small value for vertical velocity, but it is important as it may extend over a large area and persist for a long time. If it is sufficient to lift air to its LCL, then latent heat release allows stronger updrafts within the convective clouds.

12

Storms in Teacups

Standing waves in a tea cup, induced by the propeller rotation of an airoplane.

13

Cyclostrophic Balanced Rotation

14

Cyclostrophic Balanced Rotation

We consider now another flow configuration. We ignore the rotation of the earth.

14

Cyclostrophic Balanced Rotation

We consider now another flow configuration. We ignore the rotation of the earth. We use cylindrical polar coordinates (r, θ, z) and correspond- ing velocity components (U, V, W).

14

Cyclostrophic Balanced Rotation

We consider now another flow configuration. We ignore the rotation of the earth. We use cylindrical polar coordinates (r, θ, z) and correspond- ing velocity components (U, V, W). We consider a cyclostrophically balanced vortex spinning in solid rotation.

14

Cyclostrophic Balanced Rotation

We consider now another flow configuration. We ignore the rotation of the earth. We use cylindrical polar coordinates (r, θ, z) and correspond- ing velocity components (U, V, W). We consider a cyclostrophically balanced vortex spinning in solid rotation. That is, the azimuthal velocity depends linearly on the ra- dial distance: Ug = 0 , Vg = ω r where ω = ˙ θ is the constant angular velocity.

14

Cyclostrophic Balanced Rotation

We consider now another flow configuration. We ignore the rotation of the earth. We use cylindrical polar coordinates (r, θ, z) and correspond- ing velocity components (U, V, W). We consider a cyclostrophically balanced vortex spinning in solid rotation. That is, the azimuthal velocity depends linearly on the ra- dial distance: Ug = 0 , Vg = ω r where ω = ˙ θ is the constant angular velocity. The centrifugal force is given, as usual, by V 2 r = ω2r

14

For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 ρ0 ∂p ∂r = ω2r

15

For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 ρ0 ∂p ∂r = ω2r This can be integrated immediately to give p = p0 + 1

2ρ0ω2r2

so the surface has the form of a parabola (or more correctly a paraboloid of revolution).

15

For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 ρ0 ∂p ∂r = ω2r This can be integrated immediately to give p = p0 + 1

2ρ0ω2r2

so the surface has the form of a parabola (or more correctly a paraboloid of revolution). Near the bottom boundary, the flow is slowed by the effect

• f viscosity.

Then, the centrifugal force is insufficient to balance the pressure gradient force.

15

For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 ρ0 ∂p ∂r = ω2r This can be integrated immediately to give p = p0 + 1

2ρ0ω2r2

so the surface has the form of a parabola (or more correctly a paraboloid of revolution). Near the bottom boundary, the flow is slowed by the effect

• f viscosity.

Then, the centrifugal force is insufficient to balance the pressure gradient force. As a result, there is radial inflow near the bottom. By continuity of mass, this must result in upward motion near the centre.

15

For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 ρ0 ∂p ∂r = ω2r This can be integrated immediately to give p = p0 + 1

2ρ0ω2r2

so the surface has the form of a parabola (or more correctly a paraboloid of revolution). Near the bottom boundary, the flow is slowed by the effect

• f viscosity.

Then, the centrifugal force is insufficient to balance the pressure gradient force. As a result, there is radial inflow near the bottom. By continuity of mass, this must result in upward motion near the centre. Furthermore, outflow must occur in the fluid above the boundary layer.

15

16

This secondary circulation is completed by downward flow near the edges of the container.

17

This secondary circulation is completed by downward flow near the edges of the container. At the bottom surface, the flow must vanish completely.

17

This secondary circulation is completed by downward flow near the edges of the container. At the bottom surface, the flow must vanish completely. An analysis similar to the case of zonal flow again gives a solution in the form of an Ekman spiral. But now the quantity corresponding to the Coriolis parameter is 2ω, so we have γ =

• ω/K.

17

This secondary circulation is completed by downward flow near the edges of the container. At the bottom surface, the flow must vanish completely. An analysis similar to the case of zonal flow again gives a solution in the form of an Ekman spiral. But now the quantity corresponding to the Coriolis parameter is 2ω, so we have γ =

• ω/K.

The boundary layer depth is again D = π/γ.

17

This secondary circulation is completed by downward flow near the edges of the container. At the bottom surface, the flow must vanish completely. An analysis similar to the case of zonal flow again gives a solution in the form of an Ekman spiral. But now the quantity corresponding to the Coriolis parameter is 2ω, so we have γ =

• ω/K.

The boundary layer depth is again D = π/γ. The vertical velocity at the top of the boundary layer is w(D) = 1 2π

• D ζg

17

This secondary circulation is completed by downward flow near the edges of the container. At the bottom surface, the flow must vanish completely. An analysis similar to the case of zonal flow again gives a solution in the form of an Ekman spiral. But now the quantity corresponding to the Coriolis parameter is 2ω, so we have γ =

• ω/K.

The boundary layer depth is again D = π/γ. The vertical velocity at the top of the boundary layer is w(D) = 1 2π

• D ζg

The geostrophic vorticity is given, in cylindrical coordinates, by ζg = K · ∇ × Vg = 1 r ∂(rVg) ∂r − ∂Ug ∂θ

• = 1

r ∂(ωr2) ∂r = 2ω

17

Thus, the Ekman pumping is w(D) = ωD π

18

Thus, the Ekman pumping is w(D) = ωD π Let us suppose D = 1 cm and ω = 1 c.p.s. Then w(D) = 2π s−1 × 1 cm π = 2 cm s−1

18

Thus, the Ekman pumping is w(D) = ωD π Let us suppose D = 1 cm and ω = 1 c.p.s. Then w(D) = 2π s−1 × 1 cm π = 2 cm s−1 We may compare this to the azimuthal velocity. At r = 5 cm we have Vg = ω r = 2π s−1 × 5 cm ≈ 30 cm s−1 Thus, the secondary circulation is relatively weak compared to the primary (solid rotation) circulation. But it is dynam- ically important.

18

Thus, the Ekman pumping is w(D) = ωD π Let us suppose D = 1 cm and ω = 1 c.p.s. Then w(D) = 2π s−1 × 1 cm π = 2 cm s−1 We may compare this to the azimuthal velocity. At r = 5 cm we have Vg = ω r = 2π s−1 × 5 cm ≈ 30 cm s−1 Thus, the secondary circulation is relatively weak compared to the primary (solid rotation) circulation. But it is dynam- ically important.

Exercise: Create a storm in a teacup:

Stir your tea (no milk) and observe the leaves.

18

Exercise:

• Calculate the mass influx through the sides of a cyclone.
• Equate this to the upward flux through the top of the

boundary layer.

• Deduce an expression for the vertical velocity.

19

Exercise:

• Calculate the mass influx through the sides of a cyclone.
• Equate this to the upward flux through the top of the

boundary layer.

• Deduce an expression for the vertical velocity.

19

Sketch of Solution:

From the Ekman solution, the mean inward velocity is ¯ Vinward = 1

2ug

• D

π

• 20

Sketch of Solution:

From the Ekman solution, the mean inward velocity is ¯ Vinward = 1

2ug

• D

π

• The horizontal inward mass transport is

MH = ρ0 ¯ Vinward = 1

2ρ0ug

• D

π

• × 2πR = ρ0ugDR

20

Sketch of Solution:

From the Ekman solution, the mean inward velocity is ¯ Vinward = 1

2ug

• D

π

• The horizontal inward mass transport is

MH = ρ0 ¯ Vinward = 1

2ρ0ug

• D

π

• × 2πR = ρ0ugDR

The vertical mass transport through the top is ρ0w(D) × πR2

20

Sketch of Solution:

From the Ekman solution, the mean inward velocity is ¯ Vinward = 1

2ug

• D

π

• The horizontal inward mass transport is

MH = ρ0 ¯ Vinward = 1

2ρ0ug

• D

π

• × 2πR = ρ0ugDR

The vertical mass transport through the top is ρ0w(D) × πR2 These must be equal, so w(D) = ρ0ugDR ρ0 × πR2 = D πRug

20

Sketch of Solution:

From the Ekman solution, the mean inward velocity is ¯ Vinward = 1

2ug

• D

π

• The horizontal inward mass transport is

MH = ρ0 ¯ Vinward = 1

2ρ0ug

• D

π

• × 2πR = ρ0ugDR

The vertical mass transport through the top is ρ0w(D) × πR2 These must be equal, so w(D) = ρ0ugDR ρ0 × πR2 = D πRug For solid body rotation, ug = ωR and the geostrophic vortic- ity is ζg = 2ω, so w(D) = D 2π ζg

20

Spin-Down

21

Spin-Down

We estimate the characteristic spin-down time of the secondary circulation for a barotropic atmosphere.

21

Spin-Down

We estimate the characteristic spin-down time of the secondary circulation for a barotropic atmosphere. The barotropic vorticity equation is dζg dt = −f ∂u ∂x + ∂v ∂y

• = f ∂w

∂z

21

Spin-Down

We estimate the characteristic spin-down time of the secondary circulation for a barotropic atmosphere. The barotropic vorticity equation is dζg dt = −f ∂u ∂x + ∂v ∂y

• = f ∂w

∂z We integrate this through the free atmopshere, that is, from z = D to z = H, where H is the height of the tropopause.

21

Spin-Down

We estimate the characteristic spin-down time of the secondary circulation for a barotropic atmosphere. The barotropic vorticity equation is dζg dt = −f ∂u ∂x + ∂v ∂y

• = f ∂w

∂z We integrate this through the free atmopshere, that is, from z = D to z = H, where H is the height of the tropopause. The result is (H − D) dζg dt = f[w(H) − W(D)]

21

Spin-Down

We estimate the characteristic spin-down time of the secondary circulation for a barotropic atmosphere. The barotropic vorticity equation is dζg dt = −f ∂u ∂x + ∂v ∂y

• = f ∂w

∂z We integrate this through the free atmopshere, that is, from z = D to z = H, where H is the height of the tropopause. The result is (H − D) dζg dt = f[w(H) − W(D)] Assuming w(H) = 0 and substituting the Ekman pumping for W(D) we get dζg dt = − f (H − D) 1 2π

• Dζg

21

Assuming H ≫ D, this is dζg dt = − fD 2πH ζg

22

Assuming H ≫ D, this is dζg dt = − fD 2πH ζg If we define the time-scale τEkman = 2πH fD the equation for vorticity may be written dζg dt = − 1 τEkman ζg

22

Assuming H ≫ D, this is dζg dt = − fD 2πH ζg If we define the time-scale τEkman = 2πH fD the equation for vorticity may be written dζg dt = − 1 τEkman ζg The solution for the vorticity is ζg = ζg(0) exp(−t/τEkman)

22

Assuming H ≫ D, this is dζg dt = − fD 2πH ζg If we define the time-scale τEkman = 2πH fD the equation for vorticity may be written dζg dt = − 1 τEkman ζg The solution for the vorticity is ζg = ζg(0) exp(−t/τEkman) The size of τEkman may be estimated for typical values: τEkman = 2πH fD = 2π × 104 10−4 × 103 ≈ 6 × 105 s which is about seven days.

22

We may compare this to the time-scale for eddy diffusion. The diffusion equation is ∂u ∂t = K∂2u ∂z2

23

We may compare this to the time-scale for eddy diffusion. The diffusion equation is ∂u ∂t = K∂2u ∂z2 If τDiff is the diffusion time-scale and H the vertical scale for diffusion, we get U τDiff = KU H2

23

We may compare this to the time-scale for eddy diffusion. The diffusion equation is ∂u ∂t = K∂2u ∂z2 If τDiff is the diffusion time-scale and H the vertical scale for diffusion, we get U τDiff = KU H2 For the values already assumed (K = 5 m2 s−1 and H = 10 km) we get τDiff = H2 K ≈ 108 5 = 2 × 107 s which is of the order of 225 days, about 30 times longer than τEkman.

23

We may compare this to the time-scale for eddy diffusion. The diffusion equation is ∂u ∂t = K∂2u ∂z2 If τDiff is the diffusion time-scale and H the vertical scale for diffusion, we get U τDiff = KU H2 For the values already assumed (K = 5 m2 s−1 and H = 10 km) we get τDiff = H2 K ≈ 108 5 = 2 × 107 s which is of the order of 225 days, about 30 times longer than τEkman. Thus, in the absence of convective clouds, Ekman spin-down is much more effective than eddy diffusion.

23

We may compare this to the time-scale for eddy diffusion. The diffusion equation is ∂u ∂t = K∂2u ∂z2 If τDiff is the diffusion time-scale and H the vertical scale for diffusion, we get U τDiff = KU H2 For the values already assumed (K = 5 m2 s−1 and H = 10 km) we get τDiff = H2 K ≈ 108 5 = 2 × 107 s which is of the order of 225 days, about 30 times longer than τEkman. Thus, in the absence of convective clouds, Ekman spin-down is much more effective than eddy diffusion. However, cumulonumbus convection can produce rapid trans- port of heat and momentum through the entire troposphere.

23

Ekman Layer in the Ocean

Ekman spiral in the ocean.

24

25

Typical La Ni˜ na Pattern Mean sea surface temperature, eastern Pacific Ocean 5 September to 5 October, 1998.

26

End of §5.5

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