SIO 210 Problem Set 3 November 15, 2013 Due Nov. 22, 2013 Answerkey

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1 SIO 210 Problem Set 3 November 15, 2013 Due Nov. 22, 2013 Answerkey 1. Dynamics: rotation (a) Draw the direction of an inertial current in (i) the northern hemisphere and (ii) the southern hemisphere and label them. Figure from lecture or DPO: (b) Write down the inertial current balance from the x momentum and y momentum equations. Which terms from the full x and y momentum equations are missing from inertial current balance? (either the word equation or formal equation answer is OK) x-acceleration + Coriolis = 0 u/ t fv = 0 y-accceleration + Coriolis = 0 v/ t + fu = 0 Missing terms: advection, pressure gradient force, viscous terms (c) Over what time scale would an inertial current be important? Assume that the velocity scales in the x and y direction are the same. Calculate the time scale for an inertial current at 40 N. (This means you will be calculating the Coriolis parameter at that latitude why?) Time scale is when rotation becomes important > 1 day Timescale = 1/f from the equations above (Actual period T = 2π/f, as in the figure above) At 40 N, f = 2Ωsin(latitude) = 1.45 x 10-4 sec -1 sin(40 ) = 1.45 x 10-4 sec -1 (0.64) = 9.32 x 10-5 sec -1 Timescale = e+04 sec /(86400 sec/day) = 0.12 days Period = 2π/f = 0.78 days 2. Dynamics: geostrophy

2 The attached figure shows surface height in the Pacific Ocean. (a) Label the Kuroshio and Kuroshio Extension on the figure. (b) Label the Antarctic Circumpolar Current. (c) Label at least TWO high pressure regions, one in the northern hemisphere and one in the southern hemisphere. (d) Label at least TWO low pressure regions, one in the northern hemisphere and one in the southern hemisphere.!"#$%&'$()*+,-%'$-(!"#$%&'$( 9:;( 6786( 6786(.-+/#010(2'#0"34$5/#(2"##,-+( 9:;( (g) Write down the geostrophic balance from the x momentum equation. We will apply this to the northward Kuroshio. You may do this with a word equation or with the full momentum equation. Coriolis = pressure gradient force -fv = -(1/ρ) p/ x Which terms in the full x momentum equation are missing from the geostrophic balance? Acceleration, advection, viscous terms (h) Understanding the non-dimensional Rossby number. Compare the size of the nonlinear advection terms with the Coriolis term to understand why we ignore the advection term. For the Kuroshio, assume a velocity (northward and eastward) scale of 1 m/sec, and a width of 100 km. Use these scales as did for the Reynolds number to calculate first the magnitude of the advection and Coriolis terms, and then form a non-dimensional

3 parameter that is the ratio of the scales of the advective to the Coriolis term. This will be the Rossby number. Under what conditions (large or small Rossby number) is a flow geostrophic? Does the Kuroshio fit this condition of geostrophic flow? Nonlinear terms in the x-momentum equation are, e.g. u u/ x + v u/ y and in y-momentum equation, e.g. u v/ x + v v/ y Sizes of these terms (with x derivatives) are U 2 /L Size of Coriolis terms are fu Ratio of nonlinear to Coriolis terms are (U 2 /L)/(fU) = U/(fL) = Rossby number Geostrophic flow has small Rossby number. Kuroshio has small Rossby number. (e) Calculation 1. From the map, assuming that steric height is very close to physical height, what is the difference in height across the Kuroshio (low onshore to high offshore)? Use this height difference, an average seawater density of 1025 kg/m 3, and hydrostatic balance to calculate the pressure difference across the Kuroshio at an average depth of 10 m: note that the sea surface height on the west and east sides will be higher and lower than the average of 10 m. From the pressure difference, use the geostrophic balance, and latitude 30 N, to calculate the northward speed of the Kuroshio at 10 m. Height difference from the coast to the greatest height just offshore of the coast is about 0.7 m. Pressure at a depth of 10 m. Use expression p = ρgh. g = 9.8 m/sec 2 Onshore: p onshore = (1025 kg/m 3 )(9.8 m/sec 2 )(10 m) = x10 5 kg/(m sec 2 ) Offshore: p offshore = (1025 kg/m 3 )(9.8 m/sec 2 )(10.7 m) = x10 5 kg/(m sec 2 ) Geostrophic balance: -fv = -(1/ρ)( p/ x) f = 2Ωsin(30 ) = x 10-4 sec -1 sin(30 ) = 7.29 x 10-5 sec -1 v = (1/f) (9.8 m/sec 2 )(0.7m)/(100 x 10 3 m) = 0.94 m/sec (f) Calculation 2. Assume now that the seawater density just east of the Kuroshio is 1024 kg/m 3 and density just west of the Kuroshio is 1025 kg/m 3. Now do the same calculation as in part (e), but at a depth of 1000 m. What is the northward speed of the Kuroshio at 1000 m with this density difference across the current? Compare with the surface speed from part (e). Onshore (West): p west = (1025 kg/m 3 )(9.8 m/sec 2 )(1000 m) = x10 7 kg/(m sec 2 ) Offshore (East): p east = (1024 kg/m 3 )(9.8 m/sec 2 )( m) = x10 7 kg/(m sec 2 ) f = 7.29 x 10-5 sec -1 v = (1/7.29 x 10-5 sec -1 ) (1/1024.5kg/m 3 )[ x10 7 kg/(m sec 2 ) x10 7 kg/(m sec 2 )] /(10 5 m) = m/sec

4 My choice of average density, which wasn t very accurate, means that the reduction in northward velocity downward from the surface creates a significant southward velocity at 1000 m. 3. Dynamics: Ekman The attached figure is wind stress at the sea surface (from Chapter 5). (%&!%")*%&'!"#$%&' (a) In the North Pacific, label the Trade Winds and the Westerlies. (b) In the N. Pacific, sketch the direction of Ekman transport in the Trade Winds and Westerlies. (c) Also in the N. Pacific, sketch the direction of Ekman transport along the California coast. (d) Calculate the Ekman transport at 40 N. Use a single estimate of the wind stress from this map (so it won t be exact, but please use an exact expression relating Ekman transport to wind stress). Integrate this very approximately across the full width of the N. Pacific. Southward Ekman transport. Choose a wind stress of τ(x) = 0.1 N/m2 (eastward) Take f from Problem 1 = VEk = -τ(x) /(ρf) = -0.1 N/m2/((1025 kg/m3 )(9.3204e-05/sec)) = N m sec/kg (units 1 N = 1 kg m/sec2) so VEk = m2/sec Integrated across width of Pacific: assume W = 10,000km = 107m WVEk = x 106m3/sec = Sv (e) Calculate Ekman transport also at 15 N, as in (d). Integrate this very approximately across the full width of the N Pacific. Replace f with the f for 15 N. I ll do it by changing f, but one could do this directly in the expression for Vek

5 f = f 40 (sin(15 )/sin(40 )) = f 40 (0.2588/0.6428) = (9.3204e-05/sec)(0.2588/0.6428) = e-05/sec V Ek15 = V Ek40 (f 40 /f 15 ) = (-1.05 m 2 /sec)*(9.3204e-05/sec)/3.7529e-05/sec = -2.61m 2 /sec Using the same width (although it is actually more like 15,000 km at 15 N), the transport is WV Ek15 = x 10 6 m 3 /sec = -26.1Sv (f) Write down the terms in the x- momentum equation that are applicable for Ekman transport and remark on how this differs from the inertial currents of the first problem, and geostrophic balance of the second problem. Coriolis = viscous terms Terms that are not included are acceleration, advection, pressure gradient force For these three balances, the only terms that we haven t invoked are the nonlinear terms. Those are actually important in the Kuroshio Extension for instance.