S

The distance hB - hA (which is the same as the difference between l>R and So- -j) js therefore about 0.03'k of r:(, so the assumption you made m part (a), that /iH and r() - were equal, was a reasonable one.

(ii) U7ib - hA = 0.3 m, tan 6 - 03IL = 0.3/1 (X) x 101 = 0.000003, which (typically) is a very small gradient indeed. The angle 0 is correspondingly small, at ~ 0.000 ! 7° = 0.000 T.

(c) In the situation described here, as in (he example given in the text, p,^ > pB so ihc isobar />, slopes up towards B and the horizontal pressure gradient force acts from B to A. in the Southern Hemisphere, the Coriolis force acts to the left of (he direction of current How. so in conditions of geostrophic equilibrium the horizontal pressure gradient force must be acting to the right of the direction of flow. In this case, therefore, the current must be flowing out of the page, i.e towards the south (as station B is due east of station A). For the diagram showing this, see Figure 3.17(a). Note that the arrow representing the pressure gradient force is horizontal.

ANTlCYCLONIC WINDS IN SOUTHERN HEMISPHERE

Question 3.3 (a) The contours in Figure 3.21 show the dynamic topography of ihe sea-surface, assuming that the 1500dbar isobane surface is horizontal. For a more accurate picture, any topography in ihis 1500 isobaric surface would have to he 'added on' to thai shown

(h> You should have had no difficulty in identifying the Gulf Stream and the Antarctic Circumpolar Current.

(c) The stronger the geo atrophic current, the closer logether are the contours ol dynamic height. Therefore, according to Figure 3.21 the fastest part Of the Antarctic Circumpolar Current is to the east of the southern tip of Africa, at about 60° E.

id) By inspecting the contour values to the north of the 4.4 dynamic metre closed contour, we can .see that it represents a depression of the sea-surface, rather than a hill In the Southern Hemisphere, isobaric surfaces slope up lo the left of the current direction and so. given that the Current is flowing clockwise, we would indeed expect ihe sea-surface to slope down towards the middle of the region defined by the closed contour.

Question 3 9 The conditions that would lead to convergence and sinking of surface water in the Southern Hemisphere are illustrated in Figure A3(a) andib)

surface convergence ir downwetlmg 11

(tiermocline

Ekman wind surface transport current

Figure A3 Answer lo Question 3 9

Question 3.10 (a I Kinetic energy is proportional to the square of currenl speed. So. if current speeds associated with eddies are (0 times those associated with the mean, the energy must he 10- = 100 times bigger.

ib> According to Figure 3.32, a severe thunderstorm is the atmospheric phenomenon which has a kinetic energy closest to that of a mesoscale eddy,

(ci The Antarctic Circumpolar Currenl has ihe greatest kinetic energy. This is not surprising as. acted upon by prevailing westerlies, this broad current can How unimpeded around the globe.

Question 3,11 (a) I Wind stress is an external force (i). as it acts on the upper surface of the ocean.

2 Viscous forces are internal fnctional forces (ii). However, frietional forces act to oppose motion and so you would also have been justified m choosing (iii).

3 The tides are produced by the gravitational attraction of the Sun and the Moon, and so ihe tide-producing forces are external forces (i).

4 Horizontal pressure gradient forces that result from sea-surface slopes caused by variations in atmospheric pressure (i.e. when conditions arc barolropic) could be argued to be external forces (i). However, it could also be argued that in baroclmic conditions, the horizontal pressure gradient force is partly an internal force I ii). as with increasing depth it is determined more and more by ihe distribution of mass (density) within the body of ocean waler.

(It is also possible to argue thai as all horizontal pressure gradient forces resull from lateral differences in the pressure of overlying water, and these pressures derive ultimately from the gravitational attraction of the water to the mass of the Earth, horizontal pressure gradient forces arc themselves external forces. Do not worry if you did not think of this.)

5 The Coriolis force is a secondary force (iii). as it only acts on water that is already moving. Note also that the Coriolis force is an apparent secondary force. In effect, it had lo be "invented' to explain movement relative to the Earth that results from the rotation of the Earth itself

(b) (i > In geographic flow, the horizontal pressure gradient force is balancing the Coriolis force.

(ii| The mean flow of the Ekman layer at right angles to the wind is the result of the wind stress balancing the Coriolis force.

Question 3.12 At the end of Section 3.1.2. we stated thai the Ekinan transport - the total volume transport in the wind-driven layer - may be calculated by multiplying it. the depth mean current, by the thickness of the wind-driven layer, D From Equation 3.4. ¡7 is given by X/Dpf. Multiplying this by D gives i/pf. A does not appear in this expression, and so we can say that the Ekrnan transport resulting from a given w ind stress is independent of the eddy viscosity. This is an important result because it means that we can calculate Ekrnan transports without needing to know an appropriate value for eddy viscosity which, as you might imagine, is very hard to estimate.

Question 3.13 Inertia currents are indeed manifestations of the Coriolis force tn action, because they are what results when the Coriolis force acts unopposed on already-moving water (ignoring frictiotial forces, which will eventually cause an inertia current to die away), (This obviously contrasts with the situation in a geostrophic current, where the Coriolis force is balanced by a horizontal pressure gradient force, and motion is at right angles to both.)

Question 3,14 (a) Because the water is well mixed, there will be no lateral variations in density, and so conditions will be barotropie.

(b) Flow is to the east, so the Coriolis force to the right of the flow, i.e to the south, will be balanced by the horizontal pressure gradient force acting towards the left, i.e. to the north. In other words, the mean sea-level must be higher on the southern (French) side of (he Straits. Your sketch should therefore look something like Figure A4. with isobars parallel to the sea-surface, and horizontal arrows showing the Coriolis force balanced by the horizontal pressure gradient force, ( if isopycnals had been drawn, they would have been parallel to the isobars.)

<c) If the slope of the sea-surface, and all other isobars, is given by tan from the gradient equation (Equation 3.11);

Taking Cl as 7.29 x I0"V. and $ as 5 T N (so that sin 0 = 0.778).

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