Exercise: Show that � π exp( − Z ) sin Z dz = 1 2 (1 + e − π ) ≈ 1 2 0 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 0 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 � D � ∂u � � ∂u � ∂x + ∂v ∂y + ∂w ∂x + ∂v dz = dz + [ w ( D ) − w (0)] = 0 ∂z ∂y 0 0 9
Next, integrate the continuity equation through the PBL: � D � D � ∂u � � ∂u � ∂x + ∂v ∂y + ∂w ∂x + ∂v dz = dz + [ w ( D ) − w (0)] = 0 ∂z ∂y 0 0 We assume the surface is flat, so that w (0) = 0 . Then � D � ∂u � ∂x + ∂v w ( D ) = − dz ∂y 0 9
Next, integrate the continuity equation through the PBL: � D � D � ∂u � � ∂u � ∂x + ∂v ∂y + ∂w ∂x + ∂v dz = dz + [ w ( D ) − w (0)] = 0 ∂z ∂y 0 0 We assume the surface is flat, so that w (0) = 0 . Then � D � ∂u � ∂x + ∂v w ( D ) = − dz ∂y 0 Now recall the Ekman solution u = u g [1 − e − πz/D cos( πz/D )] , v = u g e − πz/D sin( πz/D ) 9
Next, integrate the continuity equation through the PBL: � D � D � ∂u � � ∂u � ∂x + ∂v ∂y + ∂w ∂x + ∂v dz = dz + [ w ( D ) − w (0)] = 0 ∂z ∂y 0 0 We assume the surface is flat, so that w (0) = 0 . Then � D � ∂u � ∂x + ∂v w ( D ) = − dz ∂y 0 Now recall the Ekman solution u = u g [1 − e − πz/D cos( πz/D )] , 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 � � ∂u g � ∂x + ∂v = ∂v e − πz/D sin( πz/D ) ∂y = ∂y ∂y 9
Next, integrate the continuity equation through the PBL: � D � D � ∂u � � ∂u � ∂x + ∂v ∂y + ∂w ∂x + ∂v dz = dz + [ w ( D ) − w (0)] = 0 ∂z ∂y 0 0 We assume the surface is flat, so that w (0) = 0 . Then � D � ∂u � ∂x + ∂v w ( D ) = − dz ∂y 0 Now recall the Ekman solution u = u g [1 − e − πz/D cos( πz/D )] , 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 � ∂u g � e − πz/D sin( πz/D ) ∂y = ∂y ∂y Subsitituing this into the equation for w ( D ) gives � D � ∂u g w ( D ) = − ∂u g e − πz/D sin( πz/D ) dz = − 1 � D 2 π ∂y ∂y 0 9
Again, � ∂u g � D w ( D ) = − 1 2 π ∂y 10
Again, � ∂u g � D w ( D ) = − 1 2 π ∂y But recall the expression for mass transport � D � M = 1 ρ 0 u g 2 π 10
Again, � ∂u g � D w ( D ) = − 1 2 π ∂y But recall the expression for mass transport � D � M = 1 ρ 0 u g 2 π Combining these, we have ρ 0 w ( D ) = − ∂M ∂y 10
Again, � ∂u g � D w ( D ) = − 1 2 π ∂y But recall the expression for mass transport � D � M = 1 ρ 0 u g 2 π 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
Again, � ∂u g � D w ( D ) = − 1 2 π ∂y But recall the expression for mass transport � D � M = 1 ρ 0 u g 2 π 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 � ∂v g ∂x − ∂u g � = − ∂u g ζ g = ∂y ∂y 10
Again, � ∂u g � D w ( D ) = − 1 2 π ∂y But recall the expression for mass transport � D � M = 1 ρ 0 u g 2 π 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 � ∂v g ∂x − ∂u g � = − ∂u g ζ g = ∂y ∂y Thus, � ∂u g � � � w ( D ) = − 1 D D ∂y = ζ g 2 π 2 π 10
This is the so-called Ekman Pumping formula: � 1 � w ( D ) = Dζ g 2 π 11
This is the so-called Ekman Pumping formula: � 1 � w ( D ) = Dζ g 2 π 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: � 1 � w ( D ) = Dζ g 2 π 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 � Positive � Cyclonic � � � Upward � ⇐ ⇒ ⇐ ⇒ Flow Vorticity Velocity 11
This is the so-called Ekman Pumping formula: � 1 � w ( D ) = Dζ g 2 π 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 � Positive � Cyclonic � � � Upward � ⇐ ⇒ ⇐ ⇒ Flow Vorticity Velocity In the vicinity of High Pressure we have � Antiyclonic � � Negative � � Downward � ⇐ ⇒ ⇐ ⇒ Flow Vorticity Velocity 11
This is the so-called Ekman Pumping formula: � 1 � w ( D ) = Dζ g 2 π 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 � Positive � Cyclonic � � � Upward � ⇐ ⇒ ⇐ ⇒ Flow Vorticity Velocity In the vicinity of High Pressure we have � Antiyclonic � � Negative � � Downward � ⇐ ⇒ ⇐ ⇒ Flow Vorticity Velocity Note on Dines Mechanism to be added later. 11
Magnitude of Ekman Pumping 12
Magnitude of Ekman Pumping The vertical velocity is � 1 � w ( D ) = Dζ g 2 π 12
Magnitude of Ekman Pumping The vertical velocity is � 1 � w ( D ) = Dζ g 2 π Suppose D = 1 km and ζ = 5 × 10 − 5 s − 1 . Then � 1 × 10 3 × (5 × 10 − 5 ) = 5 × 10 − 2 � ≈ 8 mm s − 1 ∼ 1 cm s − 1 w ( D ) = 2 π 2 π 12
Magnitude of Ekman Pumping The vertical velocity is � 1 � w ( D ) = Dζ g 2 π Suppose D = 1 km and ζ = 5 × 10 − 5 s − 1 . Then � 1 × 10 3 × (5 × 10 − 5 ) = 5 × 10 − 2 � ≈ 8 mm s − 1 ∼ 1 cm s − 1 w ( D ) = 2 π 2 π 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 � 1 � w ( D ) = Dζ g 2 π Suppose D = 1 km and ζ = 5 × 10 − 5 s − 1 . Then � 1 × 10 3 × (5 × 10 − 5 ) = 5 × 10 − 2 � ≈ 8 mm s − 1 ∼ 1 cm s − 1 w ( D ) = 2 π 2 π 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: U g = 0 , V g = 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: 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 = ω 2 r 14
For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 ∂p ∂r = ω 2 r ρ 0 15
For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 ∂p ∂r = ω 2 r ρ 0 This can be integrated immediately to give p = p 0 + 1 2 ρ 0 ω 2 r 2 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 ∂p ∂r = ω 2 r ρ 0 This can be integrated immediately to give p = p 0 + 1 2 ρ 0 ω 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
For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 ∂p ∂r = ω 2 r ρ 0 This can be integrated immediately to give p = p 0 + 1 2 ρ 0 ω 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
For steady flow, the centrifugal force is balanced by the pressure gradient force: 1 ∂p ∂r = ω 2 r ρ 0 This can be integrated immediately to give p = p 0 + 1 2 ρ 0 ω 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
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 2 ω , 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 2 ω , 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 2 ω , 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 2 π 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 2 ω , 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 2 π The geostrophic vorticity is given, in cylindrical coordinates, by ∂ ( ωr 2 ) � ∂ ( rV g ) − ∂U g � ζ g = K · ∇ × V g = 1 = 1 = 2 ω r ∂r ∂θ r ∂r 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 ) = 2 π s − 1 × 1 cm = 2 cm s − 1 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 ) = 2 π s − 1 × 1 cm = 2 cm s − 1 π We may compare this to the azimuthal velocity. At r = 5 cm we have V g = ω r = 2 π 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 ) = 2 π s − 1 × 1 cm = 2 cm s − 1 π We may compare this to the azimuthal velocity. At r = 5 cm we have V g = ω r = 2 π 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 � � V inward = 1 ¯ D 2 u g π 20
Sketch of Solution: From the Ekman solution, the mean inward velocity is � � V inward = 1 ¯ D 2 u g π The horizontal inward mass transport is � � M H = ρ 0 ¯ V inward = 1 D 2 ρ 0 u g × 2 πR = ρ 0 u g DR π 20
Sketch of Solution: From the Ekman solution, the mean inward velocity is � � V inward = 1 ¯ D 2 u g π The horizontal inward mass transport is � � M H = ρ 0 ¯ V inward = 1 D 2 ρ 0 u g × 2 πR = ρ 0 u g DR π The vertical mass transport through the top is ρ 0 w ( D ) × πR 2 20
Sketch of Solution: From the Ekman solution, the mean inward velocity is � � V inward = 1 ¯ D 2 u g π The horizontal inward mass transport is � � M H = ρ 0 ¯ V inward = 1 D 2 ρ 0 u g × 2 πR = ρ 0 u g DR π The vertical mass transport through the top is ρ 0 w ( D ) × πR 2 These must be equal, so w ( D ) = ρ 0 u g DR ρ 0 × πR 2 = D πRu g 20
Sketch of Solution: From the Ekman solution, the mean inward velocity is � � V inward = 1 D ¯ 2 u g π The horizontal inward mass transport is � � M H = ρ 0 ¯ V inward = 1 D 2 ρ 0 u g × 2 πR = ρ 0 u g DR π The vertical mass transport through the top is ρ 0 w ( D ) × πR 2 These must be equal, so w ( D ) = ρ 0 u g DR ρ 0 × πR 2 = D πRu g For solid body rotation, u g = ωR and the geostrophic vortic- ity is ζ g = 2 ω , so w ( D ) = D 2 π ζ g 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 dζ g � ∂u ∂x + ∂v � = f ∂w dt = − f ∂y ∂z 21
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 ∂x + ∂v � = f ∂w dt = − f ∂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
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 ∂x + ∂v � = f ∂w dt = − f ∂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
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 � ∂x + ∂v = f ∂w dt = − f ∂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 � 1 dζ g f � dt = − Dζ g ( H − D ) 2 π 21
Assuming H ≫ D , this is dζ g dt = − fD 2 πH ζ g 22
Assuming H ≫ D , this is dζ g dt = − fD 2 πH ζ g If we define the time-scale τ Ekman = 2 πH fD the equation for vorticity may be written dζ g 1 dt = − ζ g τ Ekman 22
Assuming H ≫ D , this is dζ g dt = − fD 2 πH ζ g If we define the time-scale τ Ekman = 2 πH fD the equation for vorticity may be written dζ g 1 dt = − ζ g τ Ekman The solution for the vorticity is ζ g = ζ g (0) exp( − t/τ Ekman ) 22
Assuming H ≫ D , this is dζ g dt = − fD 2 πH ζ g If we define the time-scale τ Ekman = 2 πH fD the equation for vorticity may be written dζ g 1 dt = − 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: fD = 2 π × 10 4 τ Ekman = 2 πH 10 − 4 × 10 3 ≈ 6 × 10 5 s which is about seven days. 22
We may compare this to the time-scale for eddy diffusion. The diffusion equation is ∂t = K∂ 2 u ∂u ∂z 2 23
We may compare this to the time-scale for eddy diffusion. The diffusion equation is ∂t = K∂ 2 u ∂u ∂z 2 If τ Diff is the diffusion time-scale and H the vertical scale for diffusion, we get U = KU H 2 τ Diff 23
We may compare this to the time-scale for eddy diffusion. The diffusion equation is ∂t = K∂ 2 u ∂u ∂z 2 If τ Diff is the diffusion time-scale and H the vertical scale for diffusion, we get U = KU H 2 τ Diff For the values already assumed ( K = 5 m 2 s − 1 and H = 10 km) we get τ Diff = H 2 K ≈ 10 8 5 = 2 × 10 7 s which is of the order of 225 days, about 30 times longer than τ Ekman . 23
We may compare this to the time-scale for eddy diffusion. The diffusion equation is ∂t = K∂ 2 u ∂u ∂z 2 If τ Diff is the diffusion time-scale and H the vertical scale for diffusion, we get U = KU H 2 τ Diff For the values already assumed ( K = 5 m 2 s − 1 and H = 10 km) we get τ Diff = H 2 K ≈ 10 8 5 = 2 × 10 7 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
We may compare this to the time-scale for eddy diffusion. The diffusion equation is ∂t = K∂ 2 u ∂u ∂z 2 If τ Diff is the diffusion time-scale and H the vertical scale for diffusion, we get U = KU H 2 τ Diff For the values already assumed ( K = 5 m 2 s − 1 and H = 10 km) we get τ Diff = H 2 K ≈ 10 8 5 = 2 × 10 7 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 trans- port of heat and momentum through the entire troposphere. 23
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