Climate Physics and Chemistry - Problem Set 3 - Solutions

Climate Physics and Chemistry Ocean and Climate: Problem Set Solution Comment: I paid little attention to numerical values, only looking to see if you understood the basic idea. 1. The pressure field at depth } = 1000m in the ocean is found to follow the rule, ¡ ¢ ¡ ¢ s = s0 cos {@104 cos |@104 = The origin of | is taken to be 30 N and {> | are measured in meters. What are the northward and eastward components of geostrophic velocity at { = 104 @2> | = 0? If the fluid density is approximated as uniform,  = 0 = 1=03 × 103 kg/m3 > how much water (mass) is moving northward between the seasurface and 1000m, between { = 0 and { = 104 @2? For a numerical answer, let s0 = 238N/m2 = The general expression for northward geostrophic flow is iy =  ¢ ¡ ¢ ¡  Cs = s0 4 sin {@104 cos |@104 > C{ 10 which at { = 104 @2> | = 0 produces y= s0 s0  = = 4= 194 2 × 1033 s0 = 1m/s 4 35 i 0 × 10 (7= 272 2 × 10 ) 1=03 × 103 × 104 To get the total mass transport, at fixed }> Z 0 104 @2 Z 0 yg{ =  104 @2 0 ¢ ¡ ¢ ¡ s0  sin {@104 cos |@104 g{ = 4 10 i ¯ 4 ¢ ¡ ¢¯10 @4 s0 ¡ ¢ s0 = cos {@104 cos |@104 ¯¯ (0  1) = 3=3 × 106 m2 /s kg/m3 i i 0 ¡ (i = 7 = 272 2 × 10 35 ) To get the vertical transport, just multiply by the depth range of 1000m, as by assumption, there is no depth dependence in y= 2. A ship measures the temperature and salinity in the ocean at { = 0> and { = 50km at a latitude of 45 N. When converted to density, the two profiles are found to be closely approximated as, ¡ ¡ ¢¢  ({ = 0> }) = 1=03 × 103 kg/m3 1  }@ 2 × 104 > £ ¡ ¢¤  ({ = 50km,}) = 1=03 × 103 kg/m3 1  (} + 1 × 1037 } 2 )@ 2 × 104 where } is in meters. Compute and plot the northward velocity as a function of } for 0  }  3000m under the assumption that } = 2000m is a level of no motion. What is dierent at 10 N? Take gravity, j = 10m/s2 = 1 The thermal wind relation for the flow in the northward direction, Z j } C y (}) =  g} + e i }r C{ Choose }0 = 2000 so that e = 0= Also, ¢ ¡ ¢ ¡ 1=03 × 103 kg/m 1 × 1037 } 2 @ 2 × 104 C  = =  50 × 103 m C{ { (  = 1=03 × 103 > almost exactly. Hence, ¡ ¢¢ ¡ Z } 1=03 × 103 1 × 1037 } 2 @ 2 × 104 j y (}) =  g} 50 × 103 i  32000 = 4= 583 5 × 10312 } 3  3= 666 8 × 1032 0.08 0.06 0.04 0.02 -3000 -2500 -2000 -1500 z -1000 -500 0 -0.02 Velocity (vertical axis) against depth (horizontal axis) in m/s. One can take  outside the integral in the last equation, as its inclusion makes no observable dierence (part of the “Boussinesq approximation”). At 10  N, the velocity will be larger in the ratio of sin 30 @ sin 10 . 3. A uniform wind blows towards the northeast such that the windstress on the ocean is  = 0 (1> 1). Using the equations C2x C} 2 C2y ix = D 2 C} iy = D which govern the Ekman layer, find x> y as a function of }= (Hint: multiply the second equation by l and add to the first equation. Solve this equation for the complex quantity x + ly=) Note that the implied density is 0 = 1> which for seawater implies cgs units. Alternatively, one can define D0 = D@= This problem is worked out in almost every textbook. It can be reduced to the textbook calcula­ tions by defining new axes so that {0 and velocity x0 are oriented to the northeast, perpendicular 2 to the windstress. Then | 0 > y 0 point towards the northwest. The equations in the new coordinate system are the same as those written above. Multiplying the second one by l and adding to the first produces, ¢ ¡ ¢ C2 ¡ i y 0  lx0 = D 2 x0 + ly 0 C} which is the same as ¢ ¡ ¢ C2 ¡ li x0 + ly 0 = D 2 x0 + ly 0 C} an ordinary dierential equation for the variable X 0 = x0 + ly 0 g2 X 0 li 0  X = 0> g} 02 D (1) subject to the boundary conditions ¯ s CX 0 ¯¯ D 0¯ = 20 C} }=0 (in the new coordinate system, the wind field is aligned with the with x> {) and that |Xs0 | $ 0 as } $ 4= The square root of 2 appears because the vector 0 (1> 1) has magnitude 2 and direction in the new {0 sense. Dierential equations like Eq. (1) are usually solved by substitution of a trial solution X 0 = F exp (s} 0 ) where D> s are constants to be determined. Substituting this solution into (1) produces μ ¶ li 02 F s  =0 D The only way to avoid a solution F = 0 (which would not satisfy the boundary condition) is to have li s02  =0 D or, r s s 2 li 0 s =± [1 + l2] I = ± lI = ± D 2 So that there are two possible solutions, Ãs μ ¶ ! à s μ ¶ ! 2 i 1@2 2 i 1@2 0 X = F exp (1 + l) } + G exp  (1 + l) } 2 D 2 D As we go into the ocean away from the surface, } becomes very large and negative. The second solution would become exponentially large, an unphysical solution for one driven by the wind at } = 0= So we set G = 0= It remains only to find F> and we do that by using the boundary condition, ! #¯ " Ãs μ ¶ ¯ s g 2 i 1@2 ¯ D F exp (1 + l) } ¯ = 20 ¯ g} 2 D }=0 s μ ¶1@2 s 2 i DF (1 + l) = 20 2 D 3 whose solution is, F = (1  l) D Hence, X0 is, X 0 = (1  l) 0 (i D)1@2 0 0 ³ ´1@2 = (1  l) s iD i D "à s μ ¶ ! # 2 i 1@2 exp (1 + l) } 2 D The real part is x0 > and the imaginary part is y 0 = "à s μ ¶ ! # 0 2 i 1@2 0 0 exp (1 + l) } x + ly = (1  l) 2 D (iD)1@2 I " (" s μ ¶1@2 s μ ¶1@2 # s μ ¶1@2 s μ ¶1@2 #) 2 i 1@2 2 i 2 i 2 i 2 i 0 h 2 ( D ) } = cos } + sin } + l  cos } + sin } 1@2 2 D 2 D 2 D 2 D (iD) At the surface, x0 = 0 @ (i D)1@2 > y0 = 0 @ (iD)1@2 > that is, 45  to the right of the wind. As } 0 decreases (becomes negative) below the seasurface, a diminishing spiral is seen. 4