Hydrostatics is one of the sections of hydraulics that studies the equilibrium state of a fluid and the pressure that appears in a liquid resting on different surfaces.
Hydrostatic pressure is a fundamentalconcept in hydrostatics. Let us consider an arbitrary volume of liquid that is in equilibrium. Inside this volume, mark the point A and mentally divide it in half by a plane passing through point A. On this plane, select the area with area S and center at point A. We remove one half of the volume and replace the force with which it acted on the remaining volume by counterbalancing force F. Thus, the liquid in the second half will still be at rest.
Now we begin to reduce the area S so thatpoint A was always inside it. With sufficient reduction, the point A coincides with the area S. And the pressure at the point A will be determined by the formula P (A) = lim dF / dS for dS tending to zero.
Then the pressure exerted on the site S will beis equal to the sum of the pressures exerted on all points belonging to this surface. That is, in other words: p = F / S. The hydrostatic pressure is a value equal to the quotient of dividing the force F by the area S.
The reason for the hydrostatic pressure are: the weight of the liquid itself and the pressure that is applied to the surface of the liquid. Thus, the pressure caused by the weight of the liquid itself and the external pressure are the kinds of hydrostatic pressure. If the liquid is placed in the piston, and some force is applied to it, then naturally the pressure inside the liquid will rise. Under normal conditions, the liquid is pressurized by atmospheric pressure. If the pressure on the surface of the liquid is below atmospheric pressure, then this pressure is called gauge pressure.
The liquid is in equilibrium if all the pressure forces acting on any sufficiently small volume of liquid are balanced with each other.
Let us consider the hydrostatic pressure and its properties:
Let us prove this property: let us assume that the angle at which the force is applied to a certain area is not direct. We represent the force F as P (normal), P (tangential). Suppose that the tangent component is not equal to zero, then under its influence the liquid must flow along an inclined one, but it rests at a point. Hence the conclusion suggests that the tangent is zero and the effect of pressure occurs perpendicular to the area. The property is proved.
Let us prove this property of hydrostatic pressure: in an arbitrary volume of liquid, we select a tetrahedron whose two planes coincide with the coordinate planes, and the third is chosen arbitrarily. At the bottom we get a right triangle. The action of the liquid on each face is denoted by: X * (P), Y * (P), Z * (P) The liquid is in equilibrium, therefore the total result of the action of all forces is 0.
E * (x) = 0
X * (P) dz -E * (P) de sin a = 0,
E * (y) = 0, E * (z) = 0
Z * (P) dx -E * (P) de cos a = 0
It is obvious that dz = de sin a, dx = de cos a
from this: X * (P) = E * (P), Z * (P) = E * (P)
output: X * (P) = Y * (P) = Z * (P) = E * (P)
The property is proved. Since the face was chosen arbitrarily, this equality is valid for any case.
Any point of the fluid, which is in equilibrium,corresponds to the following equation: j + p / g = j (o) + p (o) / g = H, where j is the coordinate of the given point, j (O) is the fluid surface coordinate, p and p (o) g - specific gravity of the liquid, H - hydrostatic head.
As a result of the transformations, we obtain p = p (o) + g [j (0) -j] or p = p (o) + gh
where h is the immersion depth of a given point, and gh -is nothing other than the weight of a column of liquid equal in height h and having a unit in the base area. This property of hydrostatic pressure is called Pascal's Law.
