The Ekman Spiral We consider now an elegant method of describing the wind in the boundary layer, due originally to V. Walfrid Ekman.
The Ekman Spiral We consider now an elegant method of describing the wind in the boundary layer, due originally to V. Walfrid Ekman. The momentum equations will be taken in the form du dt − fv + 1 ∂x + ∂ ∂p � � w ′ u ′ = 0 ρ 0 ∂z dv dt + fu + 1 ∂y + ∂ ∂p � � w ′ v ′ = 0 ρ 0 ∂z
The Ekman Spiral We consider now an elegant method of describing the wind in the boundary layer, due originally to V. Walfrid Ekman. The momentum equations will be taken in the form du dt − fv + 1 ∂x + ∂ ∂p � � w ′ u ′ = 0 ρ 0 ∂z dv dt + fu + 1 ∂y + ∂ ∂p � � w ′ v ′ = 0 ρ 0 ∂z When the eddy fluxes are parameterized in terms of the mean flow, as indicated in the previous lecture, the mo- mentum equations become ∂x − K∂ 2 u − fv + 1 ∂p ∂z 2 = 0 ρ ∂y − K∂ 2 v + fu + 1 ∂p ∂z 2 = 0 ρ
Exercise: The Ekman Spiral 2
Exercise: The Ekman Spiral For steady, horizontally homogeneous, incompressible flow the momentum equations for the atmospheric boundary layer may be written ∂x − K∂ 2 u − fv + 1 ∂p ∂z 2 = 0 ρ ∂y − K∂ 2 v + fu + 1 ∂p ∂z 2 = 0 ρ where K and f may be assumed to be constant. 2
Exercise: The Ekman Spiral For steady, horizontally homogeneous, incompressible flow the momentum equations for the atmospheric boundary layer may be written ∂x − K∂ 2 u − fv + 1 ∂p ∂z 2 = 0 ρ ∂y − K∂ 2 v + fu + 1 ∂p ∂z 2 = 0 ρ where K and f may be assumed to be constant. � Defining γ = f/ 2 K and assuming that the motion vanishes at z = 0 and tends to the zonal geostrophic value V = ( u g , 0) in the free atmosphere, derive the equations u = u g (1 − e − γz cos γz ) v = u g e − γz sin γz corresponding to the Ekman spiral. 2
Solution: 3
Solution: There are two alternatives for solving this problem: 3
Solution: There are two alternatives for solving this problem: • Eliminate one of the velocities, say v , in favour of the other. This yields a fourth-order o.d.e. for u . 3
Solution: There are two alternatives for solving this problem: • Eliminate one of the velocities, say v , in favour of the other. This yields a fourth-order o.d.e. for u . • Introduce a complex velocity w = u + iv . This seems curi- ous, but it yields a more convenient second-order o.d.e. 3
Solution: There are two alternatives for solving this problem: • Eliminate one of the velocities, say v , in favour of the other. This yields a fourth-order o.d.e. for u . • Introduce a complex velocity w = u + iv . This seems curi- ous, but it yields a more convenient second-order o.d.e. We will choose the second alternative. 3
Solution: There are two alternatives for solving this problem: • Eliminate one of the velocities, say v , in favour of the other. This yields a fourth-order o.d.e. for u . • Introduce a complex velocity w = u + iv . This seems curi- ous, but it yields a more convenient second-order o.d.e. We will choose the second alternative. Therefore, let us define w = u + iv 3
Solution: There are two alternatives for solving this problem: • Eliminate one of the velocities, say v , in favour of the other. This yields a fourth-order o.d.e. for u . • Introduce a complex velocity w = u + iv . This seems curi- ous, but it yields a more convenient second-order o.d.e. We will choose the second alternative. Therefore, let us define w = u + iv We define the components of the geostrophic velocity as u G = − 1 ∂p v G = + 1 ∂p fρ ∂y fρ ∂x and the corresponding complex geostrophic velocity as w G = u G + iv G 3
Now we may write the equations of motion as − fv + fv G − K∂ 2 u ∂z 2 = 0 + fu − fu G − K∂ 2 v ∂z 2 = 0 4
Now we may write the equations of motion as − fv + fv G − K∂ 2 u ∂z 2 = 0 + fu − fu G − K∂ 2 v ∂z 2 = 0 Now multiply the first equation by i and subtract it from the second: + ifv − ifv G + iK∂ 2 u ∂z 2 + fu − fu G − K∂ 2 v ∂z 2 = 0 4
Now we may write the equations of motion as − fv + fv G − K∂ 2 u ∂z 2 = 0 + fu − fu G − K∂ 2 v ∂z 2 = 0 Now multiply the first equation by i and subtract it from the second: + ifv − ifv G + iK∂ 2 u ∂z 2 + fu − fu G − K∂ 2 v ∂z 2 = 0 Rearranging terms we get fw − fw G + iK∂ 2 w ∂z 2 = 0 4
Now we may write the equations of motion as − fv + fv G − K∂ 2 u ∂z 2 = 0 + fu − fu G − K∂ 2 v ∂z 2 = 0 Now multiply the first equation by i and subtract it from the second: + ifv − ifv G + iK∂ 2 u ∂z 2 + fu − fu G − K∂ 2 v ∂z 2 = 0 Rearranging terms we get fw − fw G + iK∂ 2 w ∂z 2 = 0 We re-write this as ∂ 2 w � if � � if � ∂z 2 − w = − w G K K 4
Again: ∂ 2 w � if � � if � ∂z 2 − w = − w G K K 5
Again: ∂ 2 w � if � � if � ∂z 2 − w = − w G K K As usual, we solve this inhomogeneous equation in two steps 5
Again: ∂ 2 w � if � � if � ∂z 2 − w = − w G K K As usual, we solve this inhomogeneous equation in two steps 1. Find a Particular Integral (PI) of the inhomogeneous equation. 5
Again: ∂ 2 w � if � � if � ∂z 2 − w = − w G K K As usual, we solve this inhomogeneous equation in two steps 1. Find a Particular Integral (PI) of the inhomogeneous equation. 2. Find a Complementary Function (CF), a general solution of the homogeneous part of the equation. 5
Again: ∂ 2 w � if � � if � ∂z 2 − w = − w G K K As usual, we solve this inhomogeneous equation in two steps 1. Find a Particular Integral (PI) of the inhomogeneous equation. 2. Find a Complementary Function (CF), a general solution of the homogeneous part of the equation. Particular Integral : Clearly, one solution of the inhomoge- neous equation is obtained by assuming that w is indepen- dent of z . This reduces the equation to � if � � if � − w = − w G K K with the solution w = w G . 5
Complementary Function : The homogeneous version of the equation is ∂ 2 w � if � ∂z 2 − w = 0 K 6
Complementary Function : The homogeneous version of the equation is ∂ 2 w � if � ∂z 2 − w = 0 K If we seek a solution of the form w = A exp( λz ) , we get λ 2 = if K 6
Complementary Function : The homogeneous version of the equation is ∂ 2 w � if � ∂z 2 − w = 0 K If we seek a solution of the form w = A exp( λz ) , we get λ 2 = if K Thus, there are two possible values of λ : � � λ + = 1 + i f λ − = − 1 − i f √ √ and K K 2 2 6
Complementary Function : The homogeneous version of the equation is ∂ 2 w � if � ∂z 2 − w = 0 K If we seek a solution of the form w = A exp( λz ) , we get λ 2 = if K Thus, there are two possible values of λ : � � λ + = 1 + i f λ − = − 1 − i f √ √ and K K 2 2 We define the quantity γ as � f γ = 2 K (Check that γ has the dimensions of an inverse length L − 1 .) 6
Complementary Function : The homogeneous version of the equation is ∂ 2 w � if � ∂z 2 − w = 0 K If we seek a solution of the form w = A exp( λz ) , we get λ 2 = if K Thus, there are two possible values of λ : � � λ + = 1 + i f λ − = − 1 − i f √ and √ K K 2 2 We define the quantity γ as � f γ = 2 K (Check that γ has the dimensions of an inverse length L − 1 .) Now we have λ + = (1 + i ) γ and λ − = ( − 1 − i ) γ 6
The general solution of the homogeneous equation is w = A exp λ + z + B exp λ − z = A exp(1 + i )( γz ) + B exp( − 1 − i )( γz ) = A exp( γz ) exp( iγz ) + B exp( − γz ) exp( − iγz ) where A and B are arbitrary constants, which must be determined by imposing boundary conditions. 7
The general solution of the homogeneous equation is w = A exp λ + z + B exp λ − z = A exp(1 + i )( γz ) + B exp( − 1 − i )( γz ) = A exp( γz ) exp( iγz ) + B exp( − γz ) exp( − iγz ) where A and B are arbitrary constants, which must be determined by imposing boundary conditions. The boundary conditions are as follows: 7
The general solution of the homogeneous equation is w = A exp λ + z + B exp λ − z = A exp(1 + i )( γz ) + B exp( − 1 − i )( γz ) = A exp( γz ) exp( iγz ) + B exp( − γz ) exp( − iγz ) where A and B are arbitrary constants, which must be determined by imposing boundary conditions. The boundary conditions are as follows: • w → w G as z → ∞ ; thus, the solution must remain finite as z → ∞ 7
The general solution of the homogeneous equation is w = A exp λ + z + B exp λ − z = A exp(1 + i )( γz ) + B exp( − 1 − i )( γz ) = A exp( γz ) exp( iγz ) + B exp( − γz ) exp( − iγz ) where A and B are arbitrary constants, which must be determined by imposing boundary conditions. The boundary conditions are as follows: • w → w G as z → ∞ ; thus, the solution must remain finite as z → ∞ • The velocity must be zero at the earth’s surface. That is, w = 0 at z = 0 . 7
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