## Info

in GFD Lab V. (Note that in this figure, the rotation is anticlockwise, meaning that Q > 0, like that of the northern hemisphere viewed from above the north pole; for the southern hemisphere, the effective sign of rotation is reversed; see Fig. 6.17.)

In the absence of any other forces acting on it, a fluid parcel would accelerate as

With the signs shown, the parcel would turn to the right in response to the Coriolis force (to the left in the southern hemisphere). Note that since, by definition, (Q x u) ■ u = 0, the Coriolis force is workless: it does no work, but merely acts to change the flow direction.

To breathe some life in to these acceleration terms, we will now describe experiments with a parabolic rotating surface.

6.6.4. GFD Labs IV and V: Experiments with Coriolis forces on a parabolic rotating table

QFD Lab TV: Studies of parabolic equipotential surfaces

We fill a tank with water, set it turning, and leave it until it comes into solid body rotation, which is, the state in which fluid parcels have zero velocity in the rotating frame of reference. This is easily determined by viewing a paper dot floating on the free surface from a co-rotating camera. We note that the free surface of the water is not flat; it is depressed in the middle and rises up to its highest point along the rim of the tank, as sketched in Fig. 6.11. What's going on?

In solid body rotation, u = 0, F = 0, and so Eq. 6-29 reduces to pVp + V< = 0 (a generalization of hydrostatic balance to the rotating frame). For this to be true, p

- + < = constant p everywhere in the fluid (note that here we are assuming p = constant). Thus on surfaces where p = constant, < must be constant too; meaning that p and < surfaces must be coincident with one another.

At the free surface of the fluid, p = 0. Thus, from Eq. 6-30

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