Effective Depth of Ekman Layer.

Transcription

1 5.5: Ekman Pumping

2 Effective Depth of Ekman Layer. 2

3 Effective Depth of Ekman Layer. Defining γ = f/2k, we derived the solution u = u g (1 e γz cos γz) v = u g e γz sin γz corresponding to the Ekman spiral. 2

4 Effective Depth of Ekman Layer. Defining γ = f/2k, we derived the solution u = u g (1 e γz cos γz) v = u g e γz sin γz corresponding to the Ekman spiral. Let us assume the values f = 10 4 s 1 and K = 5 m 2 s 1. 2

5 Effective Depth of Ekman Layer. Defining γ = f/2k, we derived the solution u = u g (1 e γz cos γz) v = u g e γz sin γz corresponding to the Ekman spiral. Let us assume the values f = 10 4 s 1 and K = 5 m 2 s 1. The effective depth is D = π/γ. 2

6 Effective Depth of Ekman Layer. Defining γ = f/2k, we derived the solution u = u g (1 e γz cos γz) v = u g e γz sin γz corresponding to the Ekman spiral. Let us assume the values f = 10 4 s 1 and K = 5 m 2 s 1. The effective depth is D = π/γ. With f = 10 4 s 1 and K = 5 m 2 s 1 we have D = π γ = π 2K 2 5 f = π = 993 m 1 km 10 4 Thus, the effective depth of the Ekman boundary layer is about one kilometre. 2

7 Effective Depth of Ekman Layer. Defining γ = f/2k, we derived the solution u = u g (1 e γz cos γz) v = u g e γz sin γz corresponding to the Ekman spiral. Let us assume the values f = 10 4 s 1 and K = 5 m 2 s 1. The effective depth is D = π/γ. With f = 10 4 s 1 and K = 5 m 2 s 1 we have D = π γ = π 2K 2 5 f = π = 993 m 1 km 10 4 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 particular value 1 km is more an indication of the scale that a sharp quantitative estimate. 2

8 Remarks on the Ekman Spiral 3

9 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 observations 3

10 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 observations Better agreement can be obtained by coupling the Ekman layer to a surface layer where the wind direction is unchanging and the speed varies logarithmically 3

11 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 observations Better agreement can be obtained by coupling the Ekman 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 z = z z B 3

12 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 observations Better agreement can be obtained by coupling the Ekman 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 z = z z B The solution is then called a modified Ekman spiral. 3

13 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 observations Better agreement can be obtained by coupling the Ekman 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 z = z z B 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

14 Vertical Velocity The Ekman solution implies cross-isobar flow in the Planetary Boundary Layer (PBL). 4

15 Vertical Velocity The Ekman solution implies cross-isobar flow in the Planetary Boundary Layer (PBL). The flow is consistently towards lower pressure. 4

16 Vertical Velocity The Ekman solution implies cross-isobar flow in the Planetary Boundary Layer (PBL). The flow is consistently towards lower pressure. 4

17 For a steady-state solution, the convergence towards lower pressure centres and divergence from higher pressure centres has two implications: 5

18 For a steady-state solution, the convergence towards lower pressure centres and divergence from higher pressure centres has two implications: There must be upward velocity at the top of the PBL in regions of low pressure. 5

19 For a steady-state solution, the convergence towards lower pressure centres and divergence from higher pressure centres 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

20 For a steady-state solution, the convergence towards lower pressure centres and divergence from higher pressure centres 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

21 For a steady-state solution, the convergence towards lower pressure centres and divergence from higher pressure centres 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

22 First, consider a purely zonal geostrophic flow. So the isobars are oriented in an east-west direction. 6

23 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 ρ 0 v (kg m 2 s 1 ). 6

24 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 ρ 0 v (kg m 2 s 1 ). The cross-isober mass transport through a column of unit width extending through the entire PBL is the vertical integral of ρ 0 v through the layer z = 0 to z = D = π/γ: M = D 0 ρ 0 v dz (kg m 1 s 1 ) 6

25 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 ρ 0 v (kg m 2 s 1 ). The cross-isober mass transport through a column of unit width extending through the entire PBL is the vertical integral of ρ 0 v through the layer z = 0 to z = D = π/γ: M = D 0 ρ 0 v dz (kg m 1 s 1 ) Now substitute the Ekman solution for v: v = u g e γz sin(γz) = u g e πz/d sin(πz/d) 6

26 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 ρ 0 v (kg m 2 s 1 ). The cross-isober mass transport through a column of unit width extending through the entire PBL is the vertical integral of ρ 0 v through the layer z = 0 to z = D = π/γ: M = D 0 ρ 0 v dz (kg m 1 s 1 ) Now substitute the Ekman solution for v: v = u g e γz sin(γz) = u g e πz/d sin(πz/d) The result is thus M = D 0 ρ 0 u g exp( πz/d) sin(πz/d) dz 6

27 Since ρ 0 and u g are assumed to be constant, we have M = ρ 0 u g D 0 exp( πz/d) sin(πz/d) dz 7

28 Since ρ 0 and u g are assumed to be constant, we have M = ρ 0 u g D 0 exp( πz/d) sin(πz/d) dz Defining a new vertical variable Z = πz/d, this is ( ) D π M = ρ 0 u g exp( Z) sin Z dz π 0 7

29 Since ρ 0 and u g are assumed to be constant, we have M = ρ 0 u g D 0 exp( πz/d) sin(πz/d) dz Defining a new vertical variable Z = πz/d, this is ( ) D π M = ρ 0 u g exp( Z) sin Z dz π Using a standard integral, this may be written M = ( D π ) ρ 0 u g 7

30 Since ρ 0 and u g are assumed to be constant, we have M = ρ 0 u g D 0 exp( πz/d) sin(πz/d) dz Defining a new vertical variable Z = πz/d, this is ( ) D π M = ρ 0 u g exp( Z) sin Z dz π Using a standard integral, this may be written M = ( D π Here we have used the fact that ) ρ 0 u g e π

31 Exercise: Show that π 0 exp( Z) sin Z dz = 1 2 (1 + e π ) 1 2 8

32 Exercise: Show that Solution: π 0 exp( Z) sin Z dz = 1 2 (1 + e π ) 1 2 Evaluate the integral analytically Consult a Table of Integrals (e.g., GR2.663) Evaluate by numerical integration (MatLab) Use Maple to evaluate it. 8

33 Exercise: Show that Solution: π 0 exp( Z) sin Z dz = 1 2 (1 + e π ) 1 2 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 straightforward. 8

34 Exercise: Show that Solution: π 0 exp( Z) sin Z dz = 1 2 (1 + e π ) 1 2 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 straightforward. 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

35 Next, integrate the continuity equation through the PBL: D ( u x + v y + w ) D ( u dz = z x + v ) dz + [w(d) w(0)] = 0 y 0 0 9

36 Next, integrate the continuity equation through the PBL: D ( u x + v y + w ) D ( u dz = z x + v ) dz + [w(d) w(0)] = 0 y 0 0 We assume the surface is flat, so that w(0) = 0. Then D ( u w(d) = x + v ) dz y 0 9

37 Next, integrate the continuity equation through the PBL: D ( u x + v y + w ) D ( u dz = z x + v ) dz + [w(d) w(0)] = 0 y 0 0 We assume the surface is flat, so that w(0) = 0. Then D ( u w(d) = x + v ) dz y Now recall the Ekman solution u = u g [1 e πz/d cos(πz/d)], 0 v = u g e πz/d sin(πz/d) 9

38 Next, integrate the continuity equation through the PBL: D ( u x + v y + w ) D ( u dz = z x + v ) dz + [w(d) w(0)] = 0 y 0 0 We assume the surface is flat, so that w(0) = 0. Then D ( u w(d) = x + v ) dz y Now recall the Ekman solution u = u g [1 e πz/d cos(πz/d)], 0 v = u g e πz/d sin(πz/d) Since the flow is purely zonal, u g is independent of x so the (horizontal) divergence reduces to ( u x + v ) = v ( ) ug y y = e πz/d sin(πz/d) y 9

39 Next, integrate the continuity equation through the PBL: D ( u x + v y + w ) D ( u dz = z x + v ) dz + [w(d) w(0)] = 0 y 0 0 We assume the surface is flat, so that w(0) = 0. Then D ( u w(d) = x + v ) dz y Now recall the Ekman solution u = u g [1 e πz/d cos(πz/d)], 0 v = u g e πz/d sin(πz/d) Since the flow is purely zonal, u g is independent of x so the (horizontal) divergence reduces to ( u x + v ) = v ( ) ug y y = e πz/d sin(πz/d) y Subsitituing this into the equation for w(d) gives w(d) = u g y D 0 e πz/d sin(πz/d) dz = 1 2 ( Dπ ) ug y 9

40 Again, w(d) = 1 2 ( ) D ug π y 10

41 Again, w(d) = 1 2 ( ) D ug π y But recall the expression for mass transport ( ) M = 1 D 2 ρ π 0 u g 10

42 Again, w(d) = 1 2 ( ) D ug π y But recall the expression for mass transport ( ) M = 1 D 2 ρ π 0 u g Combining these, we have ρ 0 w(d) = M y 10

43 Again, w(d) = 1 2 ( ) D ug π y But recall the expression for mass transport ( ) M = 1 D 2 ρ π 0 u g Combining these, we have ρ 0 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

44 Again, w(d) = 1 2 ( ) D ug π y But recall the expression for mass transport ( ) M = 1 D 2 ρ π 0 u g Combining these, we have ρ 0 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 ( vg ζ g = x u ) g = u g y y 10

45 Again, w(d) = 1 2 ( ) D ug π y But recall the expression for mass transport ( ) M = 1 D 2 ρ π 0 u g Combining these, we have ρ 0 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 ( vg ζ g = x u ) g = u g y y Thus, w(d) = 1 2 ( Dπ ) ug y = ( D2π ) ζ g 10

46 This is the so-called Ekman Pumping formula: w(d) = ( 1 2π ) Dζ g 11

47 This is the so-called Ekman Pumping formula: w(d) = ( 1 2π ) Dζ g It shows that the vertical velocity at the top of the Ekman Layer is proportional to the geostrophic vorticity. 11

48 This is the so-called Ekman Pumping formula: w(d) = ( 1 2π ) Dζ g 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 Positive Flow Vorticity [ ] Upward Velocity 11

49 This is the so-called Ekman Pumping formula: w(d) = ( 1 2π ) Dζ g 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 Positive Flow Vorticity [ ] Upward Velocity In the vicinity of High Pressure we have [ ] [ ] [ ] Antiyclonic Negative Downward Flow Vorticity Velocity 11

50 This is the so-called Ekman Pumping formula: w(d) = ( 1 2π ) Dζ g 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 Positive Flow Vorticity [ ] Upward Velocity In the vicinity of High Pressure we have [ ] [ ] [ ] Antiyclonic Negative Downward Flow Vorticity Velocity Note on Dines Mechanism to be added later. 11

51 Magnitude of Ekman Pumping 12

52 Magnitude of Ekman Pumping The vertical velocity is w(d) = ( ) 1 Dζ g 2π 12

53 Magnitude of Ekman Pumping The vertical velocity is w(d) = ( ) 1 Dζ g 2π Suppose D = 1 km and ζ = s 1. Then ( ) 1 w(d) = 10 3 ( ) = mm s 1 1 cm s 1 2π 2π 12

54 Magnitude of Ekman Pumping The vertical velocity is w(d) = ( ) 1 Dζ g 2π Suppose D = 1 km and ζ = s 1. Then ( ) 1 w(d) = 10 3 ( ) = mm s 1 1 cm s 1 2π 2π 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

55 Magnitude of Ekman Pumping The vertical velocity is w(d) = ( ) 1 Dζ g 2π Suppose D = 1 km and ζ = s 1. Then ( ) 1 w(d) = 10 3 ( ) = mm s 1 1 cm s 1 2π 2π 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

56 Storms in Teacups Standing waves in a tea cup, induced by the propeller rotation of an airoplane. 13

57 Cyclostrophic Balanced Rotation 14

58 Cyclostrophic Balanced Rotation We consider now another flow configuration. We ignore the rotation of the earth. 14

59 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 corresponding velocity components (U, V, W ). 14

60 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 corresponding velocity components (U, V, W ). We consider a cyclostrophically balanced vortex spinning in solid rotation. 14

61 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 corresponding velocity components (U, V, W ). We consider a cyclostrophically balanced vortex spinning in solid rotation. That is, the azimuthal velocity depends linearly on the radial distance: U g = 0, V g = ω r where ω = θ is the constant angular velocity. 14

62 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 corresponding velocity components (U, V, W ). We consider a cyclostrophically balanced vortex spinning in solid rotation. That is, the azimuthal velocity depends linearly on the radial distance: U g = 0, V g = ω r where ω = θ is the constant angular velocity. The centrifugal force is given, as usual, by V 2 r = ω2 r 14

63 For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 p ρ 0 r = ω2 r 15

64 For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 p ρ 0 r = ω2 r This can be integrated immediately to give p = p ρ 0ω 2 r 2 so the surface has the form of a parabola (or more correctly a paraboloid of revolution). 15

65 For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 p ρ 0 r = ω2 r This can be integrated immediately to give p = p ρ 0ω 2 r 2 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 of viscosity. Then, the centrifugal force is insufficient to balance the pressure gradient force. 15

66 For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 p ρ 0 r = ω2 r This can be integrated immediately to give p = p ρ 0ω 2 r 2 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 of 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

67 For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 p ρ 0 r = ω2 r This can be integrated immediately to give p = p ρ 0ω 2 r 2 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 of 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

68 16

69 This secondary circulation is completed by downward flow near the edges of the container. 17

70 This secondary circulation is completed by downward flow near the edges of the container. At the bottom surface, the flow must vanish completely. 17

71 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

72 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

73 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 ( ) 1 w(d) = D ζ g 2π 17

74 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 ( ) 1 w(d) = D ζ g 2π The geostrophic vorticity is given, in cylindrical coordinates, by ζ g = K V g = 1 [ (rvg ) U ] g = 1 (ωr 2 ) r r θ r r = 2ω 17

75 Thus, the Ekman pumping is w(d) = ωd π 18

76 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

77 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 V g = 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 dynamically important. 18

78 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 V g = 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 dynamically important. Exercise: Create a storm in a teacup: Stir your tea (no milk) and observe the leaves. 18

79 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

80 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

81 Sketch of Solution: From the Ekman solution, the mean inward velocity is ( ) V inward = 1 2 u Dπ g 20

82 Sketch of Solution: From the Ekman solution, the mean inward velocity is ( ) V inward = 1 2 u Dπ g The horizontal inward mass transport is ( ) M H = ρ 0 Vinward = 1 2 ρ Dπ 0u g 2πR = ρ 0 u g DR 20

83 Sketch of Solution: From the Ekman solution, the mean inward velocity is ( ) V inward = 1 2 u Dπ g The horizontal inward mass transport is ( ) M H = ρ 0 Vinward = 1 2 ρ Dπ 0u g 2πR = ρ 0 u g DR The vertical mass transport through the top is ρ 0 w(d) πr 2 20

84 Sketch of Solution: From the Ekman solution, the mean inward velocity is ( ) V inward = 1 2 u Dπ g The horizontal inward mass transport is M H = ρ 0 Vinward = 1 2 ρ 0u g ( Dπ ) 2πR = ρ 0 u g DR The vertical mass transport through the top is These must be equal, so ρ 0 w(d) πr 2 w(d) = ρ 0u g DR ρ 0 πr 2 = D πr u g 20

85 Sketch of Solution: From the Ekman solution, the mean inward velocity is ( ) V inward = 1 2 u Dπ g The horizontal inward mass transport is M H = ρ 0 Vinward = 1 2 ρ 0u g ( Dπ ) 2πR = ρ 0 u g DR The vertical mass transport through the top is ρ 0 w(d) πr 2 These must be equal, so w(d) = ρ 0u g DR ρ 0 πr 2 = D πr u g For solid body rotation, u g = ωr and the geostrophic vorticity is ζ g = 2ω, so w(d) = D 2π ζ g 20

86 Spin-Down 21

87 Spin-Down We estimate the characteristic spin-down time of the secondary circulation for a barotropic atmosphere. 21

88 Spin-Down We estimate the characteristic spin-down time of the secondary circulation for a barotropic atmosphere. The barotropic vorticity equation is ( dζ g u dt = f x + v ) = f w y z 21

89 Spin-Down We estimate the characteristic spin-down time of the secondary circulation for a barotropic atmosphere. The barotropic vorticity equation is ( dζ g u dt = f x + v ) = f w y 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

90 Spin-Down We estimate the characteristic spin-down time of the secondary circulation for a barotropic atmosphere. The barotropic vorticity equation is ( dζ g u dt = f x + v ) = f w y 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) dζ g dt = f[w(h) W (D)] 21

91 Spin-Down We estimate the characteristic spin-down time of the secondary circulation for a barotropic atmosphere. The barotropic vorticity equation is ( dζ g u dt = f x + v ) = f w y 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) dζ g dt = f[w(h) W (D)] Assuming w(h) = 0 and substituting the Ekman pumping for W (D) we get ( ) dζ g dt = f 1 Dζ g (H D) 2π 21

92 Assuming H D, this is dζ g dt = fd 2πH ζ g 22

93 Assuming H D, this is dζ g dt = fd 2πH ζ g If we define the time-scale τ Ekman = 2πH fd the equation for vorticity may be written dζ g dt = 1 ζ g τ Ekman 22

94 Assuming H D, this is dζ g dt = fd 2πH ζ g If we define the time-scale τ Ekman = 2πH fd the equation for vorticity may be written dζ g dt = 1 ζ g τ Ekman The solution for the vorticity is ζ g = ζ g (0) exp( t/τ Ekman ) 22

95 Assuming H D, this is dζ g dt = fd 2πH ζ g If we define the time-scale τ Ekman = 2πH fd the equation for vorticity may be written dζ g dt = 1 ζ g τ Ekman The solution for the vorticity is ζ g = ζ g (0) exp( t/τ Ekman ) The size of τ Ekman may be estimated for typical values: τ Ekman = 2πH fd which is about seven days. = 2π s 22

96 We may compare this to the time-scale for eddy diffusion. The diffusion equation is u t = K 2 u z 2 23

97 We may compare this to the time-scale for eddy diffusion. The diffusion equation is u t = K 2 u z 2 If τ Diff is the diffusion time-scale and H the vertical scale for diffusion, we get U = KU τ Diff H 2 23

98 We may compare this to the time-scale for eddy diffusion. The diffusion equation is u t = K 2 u z 2 If τ Diff is the diffusion time-scale and H the vertical scale for diffusion, we get U = KU τ Diff H 2 For the values already assumed (K = 5 m 2 s 1 and H = 10 km) we get τ Diff = H2 K = s which is of the order of 225 days, about 30 times longer than τ Ekman. 23

99 We may compare this to the time-scale for eddy diffusion. The diffusion equation is u t = K 2 u z 2 If τ Diff is the diffusion time-scale and H the vertical scale for diffusion, we get U = KU τ Diff H 2 For the values already assumed (K = 5 m 2 s 1 and H = 10 km) we get τ Diff = H2 K = s which is of the order of 225 days, about 30 times longer than τ Ekman. Thus, in the absence of convective clouds, Ekman spin-down is much more effective than eddy diffusion. 23

100 We may compare this to the time-scale for eddy diffusion. The diffusion equation is u t = u K 2 z 2 If τ Diff is the diffusion time-scale and H the vertical scale for diffusion, we get U = KU τ Diff H 2 For the values already assumed (K = 5 m 2 s 1 and H = 10 km) we get τ Diff = H2 K = s which is of the order of 225 days, about 30 times longer than τ Ekman. Thus, in the absence of convective clouds, Ekman spin-down is much more effective than eddy diffusion. However, cumulonumbus convection can produce rapid transport of heat and momentum through the entire troposphere. 23

101 Ekman Layer in the Ocean Ekman spiral in the ocean. 24

102 25

103 Typical La Niña Pattern Mean sea surface temperature, eastern Pacific Ocean 5 September to 5 October,

104 End of

105 28