Pump Primer 1 $100 - 8 Hours - Technical Learning College

Free Surface Perpendicular to Gravity
When gravity acts, the liquid assumes a free surface perpendicular to gravity, which can be
proved by Thomson's method. A straight cylinder of unit cross-sectional area (assumed only for
ease in the arithmetic) can be used to find the increase of pressure with depth. Indeed, we see
that p2 = p1 + ρgh. The upper surface of the cylinder can be placed at the free surface if
desired. The pressure is now the same in any direction at a point, but is greater at points that lie
deeper. From this same figure, it is easy to prove Archimedes’ Principle that the buoyant force is
equal to the weight of the displaced fluid, and passes through the center of mass of this
displaced fluid.
Geometric Arguments
Ingenious geometric arguments can be used to
substitute for easier, but less transparent arguments
using calculus. For example, the force acting on one
side of an inclined plane surface whose projection is
AB can be found as in the diagram on the previous
page. O is the point at which the prolonged projection
intersects the free surface. The line AC' perpendicular
to the plane is made equal to the depth AC of point A,
and line BD' is similarly drawn equal to BD. The line
OD' also passes through C', by proportionality of
triangles OAC' and OAD'. Therefore, the thrust F on
the plane is the weight of a prism of fluid of crosssection
AC'D'B, passing through its centroid normal to
plane AB. Note that the thrust is equal to the density
times the area times the depth of the center of the
area; its line of action does not pass through the
center, but below it, at the center of thrust. The same
result can be obtained with calculus by summing the
pressures and the moments.
Atmospheric Pressure and its Effects
Suppose a vertical pipe is stood in a pool of water,
and a vacuum pump applied to the upper end. Before
we start the pump, the water levels outside and inside the pipe are equal, and the pressures on
the surfaces are also equal and are equal to the atmospheric pressure.
Now start the pump. When it has sucked all the air out above the water, the pressure on the
surface of the water inside the pipe is zero, and the pressure at the level of the water on the
outside of the pipe is still the atmospheric pressure. Of course, there is the vapor pressure of
the water to worry about if you want to be precise, but we neglect this complication in making
our point. We require a column of water 33.9 ft high inside the pipe, with a vacuum above it, to
balance the atmospheric pressure. Now do the same thing with liquid mercury, whose density at
0 °C is 13.5951 times that of water. The height of the column is 2.494 ft, 29.92 in, or 760.0 mm.
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