SIO 210 Problem Set 2 October 17, 2011 Due Oct. 24, 2011

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1 SIO 210 Problem Set 2 October 17, 2011 Due Oct. 24, The Pacific Ocean is approximately 10,000 km wide. Its upper layer (wind-driven gyre*) is approximately 1,000 m deep. Consider a west-to-east cross-section at about 30 N across the whole width of the Pacific, from Asia to North America, for this layer. Assume that there is a narrow western boundary current and a very broad interior flow across most of the section. For the following questions, assume that velocity does not vary with depth within this layer. * wind-driven means that the wind stress, that acts directly on the top 50 m, creates flow there ( Ekman transport ) that is convergent or divergent, which drives the interior gyre circulation, which is geostrophic; the wind-driven gyre is surface-intensified, decaying with depth, and mostly vanishes by about 1500 m in the N. Pacific. (a) If the water in that cross-section is moving southward at 1 cm/sec, calculate the total southward volume transport, in MKS units. (Ignore the western boundary current for this calculation.) Calculate velocity times cross-sectional area being careful about units. Assume that the Pacific is about 10,000 km wide (many options for choosing a width, but it should be of this order of magnitude): V = v*width*depth = [(1 cm/sec) * (10-2 m/cm)] * [(10,000 km) (10 3 m/km)] * 1,000 m = 10 8 m 3 /sec = 100 x 10 6 m 3 /sec = 100 Sv (b) If this same amount of water is returning northward in a western boundary current that is 100 km wide (and still 1 km deep), calculate the average northward velocity of the western boundary current. v = V/area = [100 x 10 6 m 3 /sec]/[(100 km)*(10 3 m/km)*(1 km)*(10 3 m/km)] = 1 m/sec = 100 cm/sec (c) If the average oxygen content of the northward flow in the western boundary current is 150 µmol/kg, calculate the net northward transport of oxygen in the western boundary current, in units of µmol/sec. Use the information from (b) to calculate. Oxygen transport = (oxygen concentration)*density*transport = (150µmol/kg)*(1025 kg/m 3 )*100x10 6 m 3 /sec = 1.54 x µmol/sec = 1.54 x 10 7 mol/sec. (d) Suppose this circulation is transporting 1 PW of heat northward. If all of the northward flow is of one temperature and all of the southward flow is of another temperature, what is the temperature difference between the northward and southward flow? Use typical (uniform) values for density and specific heat, as given in class or in a textbook. Heat transport divergence = density*specific heat*(temperature difference)*volume transport So (temperature difference) = Heat transport divergence /( density*specific heat*(temperature difference)*volume transport)

2 1 PW = 1x10 15 W = 1x10 15 J/sec = (1025 kg/m 3 )*(3850 J/ C kg)*( ΔT)*(100x10 6 m 3 /sec) ΔT = (1x10 15 J/sec)/[ (1025 kg/m 3 )*(3850 J/ C kg)*(100x10 6 m 3 /sec)] = 2.5 C (e) Explain why I asked you to calculate a temperature difference in (d), rather than the actual temperature. The net heat transport through this coast-to-coast section, across which volume (or mass) transport must be balanced, is due to northward flow of warm water, cooling to the north of the closed section, and return of cooler water southward. (f) Calculate the average air-sea heat flux between this section and the northern edge of the Pacific Ocean (Japan/Russia/U.S./Canada). Use very simplified assumptions about the width and length of this region (i.e. don t worry about calculating the exact dimensions, just approximate it). Ignore the Bering Strait assume there is no leakage out of this large box. This heat transport through the section is equal to the total heat loss through the surface of the ocean between this section and northern boundary of the N. Pacific. The air-sea heat flux is therefore Air-sea heat flux = Heat transport/surface area. Calculation of the surface area can be VERY crude for this exercise. I would use 10,000 km for the width, and 3,000 km for the north-south extent. The exact number you get depends on your estimate for the surface area of course. Air-sea heat flux = (1 x W)/[(10,000 km * 10 3 m/km)*(3,000 km * 10 3 m/km)] = 0.33 x10 2 W/m 2 = 33 W/m Satellite sensors have revolutionized the way we can observe the ocean, in comparison with observations made prior to about (a) Name a property associated with ocean circulation that can be measured from satellites, and what satellite instrument measures it, with a very very brief description of the instrument. Sea surface height; altimeter; measures distance from satellite to sea surface by sending out EM pulse with a specific frequency (further details in Chapter S16 and various sources, such as NASA online materials) Sea surface temperature; infrared sensors that measure infrared radiation from sea surface (see Chapter S16 and various sources, such as NASA online materials) Wind speed and direction; scatterometer; measures scatter of radiation by sending out EM pulse and sensing what comes back (further details in Chapter S16 and various sources, such as NASA online materials) Other choices OK too. (b) Explain how this satellite measurement relates to ocean circulation Altimetry: sea surface height is related to the surface pressure distribution that is associated with geostrophic flow

3 SST: surface temperature distribution is related to the circulation since advection moves temperature around, upwelling creates low surface temperature. Wind speed and direction: forcing for the surface layer and hence waves and overall circulation (c) Describe how that measurement is made in situ (what kinds of instruments might be used that are located in the actual water) Sea surface height: tide gauges are the reliable method SST: temperature sensors such as thermistors Wind speed and direction: various sensors that can be mounted on island stations or on ocean buoys (d) Discuss whether the measurement can also be made using an autonomous in situ instrument (and briefly describe the instrument) SSH: cannot really be measured autonomously within the ocean itself instantaneously. However, average distributions of sea surface height can be derived from surface drifter velocities if the velocities are averaged and the direct wind-driven component removed, with the residual due to the geostrphic circulation. The geostrophic velocity distribution at the sea surface can then be used to calculated the geostrophic streamfunction which is equivalent to the sea surface height. (This is beyond the information that we have presented thus far in the class.) SST: yes, this is one of the most basic in situ observations, measured autonomously from drifters, subsurface floats, moored instruments, gliders, etc. Wind speed and direction: can be measured from moored buoys that have weather instrumentation. (e) Discuss (briefly) advantages and disadvantages of satellite observations versus ship-based observations. Advantages: Satellite observations have very broad geographical coverage, and can sample every several days; with a constellation of satellites there can be near-continuous coverage, although this is not the case for the altimeters at present. Disadvantage is that relating surface properties to subsurface ones is complicated if there are no in situ observations; if very high temporal resolution is needed (say, for tides or waves), in situ measurements are necessary. Calibration of satellite data could be an issue, but once a satellite instrument has been in service for a long time, the team working on the satellite data set will be providing good calibrations, which might require in situ sampling networks. 3. Diffusivity κ has units of (length) 2 / time. (a) Describe very briefly (1-2 sentences) what we mean by "eddy diffusivity. Eddy diffusivity is the effective or turbulent diffusivity arising from macroscale motions of water, rather than arising from its molecular properties. Eddy diffusivity for a given scale of motion arises from motions at smaller scales - for large scale circulation, it arises from mesoscale eddies and submesoscale motions. For mesoscale eddies, it can arise from frontal structure and

4 intrusions, and microstructure (cm scale variability) associated with turbulence. For waves it would arise from the smallest scales of turbulence. (b) The molecular diffusivity of temperature in water is cm 2 /sec. Approximately how long would it take for temperature to diffuse 50 meters? (You may round the diffusivity to ) Please make this a very simple argument just use scale arguments, don t solve the diffusion equation, and don t worry about the temperature dependence of the diffusivity. Since κ has units of (length) 2 / time, the time scale T = L 2 /κ = [(50 m)*(100 cm/m)] 2 /( cm 2 /sec) = (25x10 6 /0.0014) sec = 1.78x10 10 sec = x10 3 years = 567 years. (c) Is eddy diffusivity larger or smaller than molecular diffusivity? Larger. (d) Approximately how large is vertical eddy diffusivity? Compare this with molecular diffusivity of temperature and with horizontal eddy diffusivity. From class notes, vertical eddy diffusivity is observed to be about 0.1 cm 2 /sec, which is much larger than the cm 2 /sec listed above. Horizontal eddy diffusivity can have a very large range; from the Ledwell et al. (Nature, 1993) experiment with dye release (in the class powerpoint), the horizontal eddy diffusivity was 10 4 to 10 8 cm 2 /sec. (e) Approximately how long would it take for temperature to diffuse 50 meters if it diffuses through eddy diffusivity? T = L 2 /κ = [(50 m)*(100 cm/m)] 2 /(0.1 cm 2 /sec) = 0.025x10 10 sec = 2.5x10 8 sec = 7.9 years vertically (I used the vertical diffusivity here), or [(50 m)*(100 cm/m)] 2 /(10 4 cm 2 /sec) = 2.5x10 3 sec = 42 minutes for the lower horizontal eddy diffusivity. 4. This figure shows contours of sea surface height (with respect to the geoid) in a Southern Hemisphere ocean basin. The contour spacing is 3 cm and the basin is 1000 m deep. Indicate on the figure the direction of flow for each gyre. Estimate the transport of the western boundary current at 20 S assuming the velocity at all depths is the same as at the surface and that the western boundary current is 100 km wide. (Express your answer in Sv.) The difference in SSH across the western boundary current is 24 cm, or 0.24 m. This is almost exactly equal to a pressure difference of 0.24 dbar across the current, so I will not do a full conversion of depth to dbar. From geostrophic balance using the x-momentum equation, the geostrophic velocity is the horizontal pressure gradient force across the current divided by the Coriolis parameter for 20 S (and by density). v = (1/fρ)(dp/dx) (note that the current depth is not needed for this part of the calculation) Coriolis parameter f = 2Ωsin(latitude) = (1.4x10-4 /sec)(0.3) = 0.42x10-4 /sec Convert the pressure difference to mks units: 0.24 dbar = 0.24x10 4 N/m 2 = 0.24 x10 4 (kg m/s 2 )/m 2 = 0.24 x10 4 kg/s 2 m. Convert the distance to m: 100 km = 10 5 m. Calculate velocity: v = 0.56 m/sec = 56 cm/sec

5 Transport: V = v*h*w = 0.56 m/sec * 1000 m * 100 km *(1000 m/km) = 56 x 10 6 m 3 /sec = 56 Sv. I can t draw on this diagram in word, so I will just describe the arrows: In the top panel (negative SSH values), the arrows should go clockwise around the low in the middle. In the bottom panel (positive SSH values), the arrows go counterclockwise around the high in the middle. (Southern Hemisphere). 10S 20S 30S S 3 50S 9 3 6

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