We consider now an elegant method of describing the ? The Ekman Spiral We consider now an elegant

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The Ekman SpiralWe consider now an elegant method of describing the windin the boundary layer, due originally to V. Walfrid Ekman.The Ekman SpiralWe consider now an elegant method of describing the windin the boundary layer, due originally to V. Walfrid Ekman.The momentum equations will be taken in the formdudt fv + 10px+z(wu)= 0dvdt+ fu +10py+z(wv)= 0The Ekman SpiralWe consider now an elegant method of describing the windin the boundary layer, due originally to V. Walfrid Ekman.The momentum equations will be taken in the formdudt fv + 10px+z(wu)= 0dvdt+ fu +10py+z(wv)= 0When the eddy fluxes are parameterized in terms of themean flow, as indicated in the previous lecture, the mo-mentum equations becomefv + 1pxK2uz2= 0+fu +1pyK2vz2= 0Exercise: The Ekman Spiral2Exercise: The Ekman SpiralFor steady, horizontally homogeneous, incompressible flowthe momentum equations for the atmospheric boundary layermay be writtenfv + 1pxK2uz2= 0+fu +1pyK2vz2= 0where K and f may be assumed to be constant.2Exercise: The Ekman SpiralFor steady, horizontally homogeneous, incompressible flowthe momentum equations for the atmospheric boundary layermay be writtenfv + 1pxK2uz2= 0+fu +1pyK2vz2= 0where K and f may be assumed to be constant.Defining =f/2K and assuming that the motion vanishesat z = 0 and tends to the zonal geostrophic value V = (ug, 0)in the free atmosphere, derive the equationsu = ug(1 ez cos z)v = ugez sin zcorresponding to the Ekman spiral.2Solution:3Solution:There are two alternatives for solving this problem:3Solution:There are two alternatives for solving this problem: Eliminate one of the velocities, say v, in favour of theother. This yields a fourth-order o.d.e. for u.3Solution:There are two alternatives for solving this problem: Eliminate one of the velocities, say v, in favour of theother. 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.3Solution:There are two alternatives for solving this problem: Eliminate one of the velocities, say v, in favour of theother. 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.3Solution:There are two alternatives for solving this problem: Eliminate one of the velocities, say v, in favour of theother. 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 definew = u + iv3Solution:There are two alternatives for solving this problem: Eliminate one of the velocities, say v, in favour of theother. 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 definew = u + ivWe define the components of the geostrophic velocity asuG = 1fpyvG = +1fpxand the corresponding complex geostrophic velocity aswG = uG + ivG3Now we may write the equations of motion asfv + fvG K2uz2= 0+fu fuG K2vz2= 04Now we may write the equations of motion asfv + fvG K2uz2= 0+fu fuG K2vz2= 0Now multiply the first equation by i and subtract it fromthe second:+ifv ifvG + iK2uz2+ fu fuG K2vz2= 04Now we may write the equations of motion asfv + fvG K2uz2= 0+fu fuG K2vz2= 0Now multiply the first equation by i and subtract it fromthe second:+ifv ifvG + iK2uz2+ fu fuG K2vz2= 0Rearranging terms we getfw fwG + iK2wz2= 04Now we may write the equations of motion asfv + fvG K2uz2= 0+fu fuG K2vz2= 0Now multiply the first equation by i and subtract it fromthe second:+ifv ifvG + iK2uz2+ fu fuG K2vz2= 0Rearranging terms we getfw fwG + iK2wz2= 0We re-write this as2wz2(ifK)w = (ifK)wG4Again:2wz2(ifK)w = (ifK)wG5Again:2wz2(ifK)w = (ifK)wGAs usual, we solve this inhomogeneous equation in two steps5Again:2wz2(ifK)w = (ifK)wGAs usual, we solve this inhomogeneous equation in two steps1. Find a Particular Integral (PI) of the inhomogeneousequation.5Again:2wz2(ifK)w = (ifK)wGAs usual, we solve this inhomogeneous equation in two steps1. Find a Particular Integral (PI) of the inhomogeneousequation.2. Find a Complementary Function (CF), a general solutionof the homogeneous part of the equation.5Again:2wz2(ifK)w = (ifK)wGAs usual, we solve this inhomogeneous equation in two steps1. Find a Particular Integral (PI) of the inhomogeneousequation.2. Find a Complementary Function (CF), a general solutionof 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(ifK)w = (ifK)wGwith the solution w = wG.5Complementary Function: The homogeneous version of theequation is2wz2(ifK)w = 06Complementary Function: The homogeneous version of theequation is2wz2(ifK)w = 0If we seek a solution of the form w = A exp(z), we get2 =ifK6Complementary Function: The homogeneous version of theequation is2wz2(ifK)w = 0If we seek a solution of the form w = A exp(z), we get2 =ifKThus, there are two possible values of :+ =1 + i2fKand =1 i2fK6Complementary Function: The homogeneous version of theequation is2wz2(ifK)w = 0If we seek a solution of the form w = A exp(z), we get2 =ifKThus, there are two possible values of :+ =1 + i2fKand =1 i2fKWe define the quantity as =f2K(Check that has the dimensions of an inverse length L1.)6Complementary Function: The homogeneous version of theequation is2wz2(ifK)w = 0If we seek a solution of the form w = A exp(z), we get2 =ifKThus, there are two possible values of :+ =1 + i2fKand =1 i2fKWe define the quantity as =f2K(Check that has the dimensions of an inverse length L1.)Now we have+ = (1 + i) and = (1 i)6The general solution of the homogeneous equation isw = A exp +z + B exp z= A exp(1 + i)(z) + B exp(1 i)(z)= A exp(z) exp(iz) + B exp(z) exp(iz)where A and B are arbitrary constants, which must bedetermined by imposing boundary conditions.7The general solution of the homogeneous equation isw = A exp +z + B exp z= A exp(1 + i)(z) + B exp(1 i)(z)= A exp(z) exp(iz) + B exp(z) exp(iz)where A and B are arbitrary constants, which must bedetermined by imposing boundary conditions.The boundary conditions are as follows:7The general solution of the homogeneous equation isw = A exp +z + B exp z= A exp(1 + i)(z) + B exp(1 i)(z)= A exp(z) exp(iz) + B exp(z) exp(iz)where A and B are arbitrary constants, which must bedetermined by imposing boundary conditions.The boundary conditions are as follows: w wG as z ; thus, the solution must remainfinite as z 7The general solution of the homogeneous equation isw = A exp +z + B exp z= A exp(1 + i)(z) + B exp(1 i)(z)= A exp(z) exp(iz) + B exp(z) exp(iz)where A and B are arbitrary constants, which must bedetermined by imposing boundary conditions.The boundary conditions are as follows: w wG as z ; thus, the solution must remainfinite as z The velocity must be zero at the earths surface.That is, w = 0 at z = 0.7The general solution of the homogeneous equation isw = A exp +z + B exp z= A exp(1 + i)(z) + B exp(1 i)(z)= A exp(z) exp(iz) + B exp(z) exp(iz)where A and B are arbitrary constants, which must bedetermined by imposing boundary conditions.The boundary conditions are as follows: w wG as z ; thus, the solution must remainfinite as z The velocity must be zero at the earths surface.That is, w = 0 at z = 0.The term multiplied by A grows exponentially with z andso must be rejected. The physically acceptable solution isthusw = B exp(z) exp(iz)7So, the complete solution (PI + CF) isw = wG + B exp(z) exp(iz)8So, the complete solution (PI + CF) isw = wG + B exp(z) exp(iz)Setting z = 0 this gives0 = wG + B8So, the complete solution (PI + CF) isw = wG + B exp(z) exp(iz)Setting z = 0 this gives0 = wG + BThus, B = wG and the complete solution isw = wG [1 exp(z) exp(iz)]8So, the complete solution (PI + CF) isw = wG + B exp(z) exp(iz)Setting z = 0 this gives0 = wG + BThus, B = wG and the complete solution isw = wG [1 exp(z) exp(iz)]Expanding this into real and imaginary parts, we haveu + iv = (uG + ivG) [1 exp(z) cos(z) + i exp(z) sin(z)]8So, the complete solution (PI + CF) isw = wG + B exp(z) exp(iz)Setting z = 0 this gives0 = wG + BThus, B = wG and the complete solution isw = wG [1 exp(z) exp(iz)]Expanding this into real and imaginary parts, we haveu + iv = (uG + ivG) [1 exp(z) cos(z) + i exp(z) sin(z)]For simplicity, we now assume that the geostrophic wind ispurely zonal, so that vG = 0. Then, separating the real andimaginary components of w, we haveu = uG [1 exp(z) cos(z)]v = uG [ + exp(z) sin(z)]8Horizontal axis: u. Vertical axis: v. Geostrophic wind: uG = 10m s1.9Description of the solution in qualitative terms.10Description of the solution in qualitative terms. There is cross-isobar flow towards low pressure.10Description of the solution in qualitative terms. There is cross-isobar flow towards low pressure. The velocity vanishes at the lower boundary.10Description of the solution in qualitative terms. There is cross-isobar flow towards low pressure. The velocity vanishes at the lower boundary. The velocity tends to the geostrophic flow at high levels.10Description of the solution in qualitative terms. There is cross-isobar flow towards low pressure. The velocity vanishes at the lower boundary. The velocity tends to the geostrophic flow at high levels. For small z, we have u uG(z) and v uG(z). Thus,the flow near the surface is 45 to the left of the limitinggeostrophic flow (purely zonal).10Description of the solution in qualitative terms. There is cross-isobar flow towards low pressure. The velocity vanishes at the lower boundary. The velocity tends to the geostrophic flow at high levels. For small z, we have u uG(z) and v uG(z). Thus,the flow near the surface is 45 to the left of the limitinggeostrophic flow (purely zonal). The hodograph of the velocity against height is a clock-wise spiral converging to (uG, 0).10Description of the solution in qualitative terms. There is cross-isobar flow towards low pressure. The velocity vanishes at the lower boundary. The velocity tends to the geostrophic flow at high levels. For small z, we have u uG(z) and v uG(z). Thus,the flow near the surface is 45 to the left of the limitinggeostrophic flow (purely zonal). The hodograph of the velocity against height is a clock-wise spiral converging to (uG, 0). The velocity reaches a maximum at the first zero of v,which is at z = .10Description of the solution in qualitative terms. There is cross-isobar flow towards low pressure. The velocity vanishes at the lower boundary. The velocity tends to the geostrophic flow at high levels. For small z, we have u uG(z) and v uG(z). Thus,the flow near the surface is 45 to the left of the limitinggeostrophic flow (purely zonal). The hodograph of the velocity against height is a clock-wise spiral converging to (uG, 0). The velocity reaches a maximum at the first zero of v,which is at z = . The flow is super-geostrophic at this point.10Description of the solution in qualitative terms. There is cross-isobar flow towards low pressure. The velocity vanishes at the lower boundary. The velocity tends to the geostrophic flow at high levels. For small z, we have u uG(z) and v uG(z). Thus,the flow near the surface is 45 to the left of the limitinggeostrophic flow (purely zonal). The hodograph of the velocity against height is a clock-wise spiral converging to (uG, 0). The velocity reaches a maximum at the first zero of v,which is at z = . The flow is super-geostrophic at this point. The height where this occurs may be taken as the ef-fective height of the Ekman layer. The wind is close togeostrophic above this height.10Effective depth of the boundary layer.11Effective depth of the boundary layer.We assume the values f = 104 s1 and K = 10m2s1.11Effective depth of the boundary layer.We assume the values f = 104 s1 and K = 10m2s1.The effective height is z0 = /. With f = 104 s1 andK = 10m2s1 we havez0 == 2Kf= 2 10104 1400 mThus, the effective depth of the Ekman boundary layer isabout 1.4 km.11Remarks on the Ekman Spiral12Remarks on the Ekman Spiral The Ekman theory predicts a cross-isobar flow of 45 atthe lower boundary. This is not in agreement with obser-vations12Remarks on the Ekman Spiral The Ekman theory predicts a cross-isobar flow of 45 atthe 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 isunchanging and the speed varies logarithmicaly12Remarks on the Ekman Spiral The Ekman theory predicts a cross-isobar flow of 45 atthe 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 isunchanging and the speed varies logarithmicaly This can be done by taking a boundary conditionV Vz@ z = zB12Remarks on the Ekman Spiral The Ekman theory predicts a cross-isobar flow of 45 atthe 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 isunchanging and the speed varies logarithmicaly This can be done by taking a boundary conditionV Vz@ z = zB The solution is then called a modified Ekman spiral.12MatLab Exercise:Write a program to calculate the wind speed as a function ofaltitude. Assume the values f = 104 s1 and K = 10m2s1.13End of 5.414

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