Geoscience Reference
In-Depth Information
erupting Phillipines volcano
Fig. 3.19 In Nature the most spectacular demonstration of positive gaseous work occurs during volcanic eruption, like those illustrated here.
The cylinder in the thought experiment is replaced by a subsurface magma chamber. Upward magma flux is accompanied by rapid degassing
and blows off. Gaseous expansion may be slow or, as in this case, fast, causing magma eruption through a conduit to the surface. Here, work
done includes the total force exerted on magma and crustal rock in the explosion.
3.5
Hydrostatic pressure
3.5.1
Pascal's result
calculation that f AB
f CA
f BC , an exercise we leave to the
reader.
Hydrostatic pressure arises from surface gravity forces.
The units of p (mind your “ p ”s - do not confuse pressure, p ,
a scalar, with momentum, p , a vector) are N m 2 or Pa and
the dimensions are ML 1 T 2 . We emphasize that pressure
intensity is usually measured with respect to a difference in
pressure between the particular fluid and some reference
pressure, often taken as that of the atmosphere at the time
of measurement. When dealing with pressures in the watery
Earth, it is often sufficient to neglect variations in atmos-
pheric pressure, but in meteorology such a course is not
possible. For example, the column of atmosphere over us,
rising over 100 km, exerts a mean pressure of some
1.01
The most important fact about the scalar quantity pressure,
p , at any point in a stationary fluid is that it has the same
magnitude in all directions, a result that was established by
Pascal. The simplest argument for this is that if it were not
the same in all directions, then a net force would exist which
would cause motion - a paradox for a stationary fluid. For a
formal proof of this striking property of static fluids, consider
a diagonal half slice of a solid unit cubic volume, a stationary
prismatic body, in air or completely immersed within sta-
tionary liquid (Fig. 3.20) of constant density. Imagine the
prism has the same density as the fluid, so no buoyant forces
act. We use the property of stationarity to demonstrate that
there are no shearing forces acting on any of the planes AB,
BC, and CA; if there were, then the larger face AB would
have a larger force acting and the prism would rotate. But it
does not. Our second result is a positive one; all forces acting
on the prism must be normal, that is, in the geometrical
sense, acting at right angles to each face. Now we can make
rapid progress using simple vectorial geometry. First we
shrink the prism so that it is infinitesimal. Then we resolve
the forces acting. Call the force acting on each face f AB , f BC ,
and f CA . From definition, BC
10 5 Pa (1 atm) due to its weight; as weather fronts
move past us this weight changes, typically up to
4 percent
(Fig. 3.21). The higher we climb on a mountain, the lesser
pressure gets as the effective thickness of the column resting
on us reduces. At 8.9 km altitude on Everest's crest, the
pressure is only 0.31 atm.
It is worth pondering on the significance of Pascal's result:
Even though static fluid pressure is caused by the downward
action of gravity, the resulting stresses at a point are equal in
all directions and give the pressure, p . If we define three
orthogonal stresses,
CA and from geometry
z as vector components
(appendix), then we have the equality p
x ,
y , and
AB
CA. The tangent of angle ABC is given vecto-
rially by f CA · AC/ f BC · BC. Since the angle is 45
BC
x y z .
It is only when we realize that the gravity force is transmit-
ted onto any surface by the random three-dimensional (3D)
, tan
1,
AC
f BC . It is then easy,
by resolving the cosine of angle ABC, to complete the
BC, and thus f CA
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