Dependence of pressure in liquid on depth

Consider a fluid in a gravity field. The gravity field acts on the liquid and tries to compress it, but the liquid is very weakly compressed, since it is not compressible, and under any action the density of the liquid is always the same. This is a serious difference between a liquid and a gas, therefore the formulas that we will consider refer to an incompressible liquid and are not applicable in a gaseous environment.

See also online calculator - Fluid Pressure Calculator.

Consider an object with a liquid of area S = 1, height h, density of the liquid ρ, which is in a gravity field with an acceleration of gravity g. Above the pressure of the liquid P0 and below the pressure Ph, since the object is in a state of equilibrium, the sum of the forces acting on it will be equal to zero. The force of gravity will be equal to the density of the fluid on the acceleration of gravity and on the volume Ft = ρ g V, since V = h S, and S = 1, then we get Ft = ρ g h.

The total pressure force is equal to the pressure difference multiplied by the cross-sectional area, but since we have it equal to one, then P = Рh - Р0

Since this object does not move with us, these two forces are equal to each other Ft = P.

We get the dependence of fluid pressure on depth or the law of hydrostatic pressure. Pressure at depth h differs from pressure at zero depth by ρ g h: Рh = Р0 + (ρ g h).

The magnitude of pressure does not depend on the direction of the normal to the surface on which this pressure is located, that is, the pressure distribution is isotropic (the same) in all directions.

This law was established experimentally. Suppose that in some liquid there is a rectangular prism, one of the legs of which is located vertically, and the other - horizontally. The pressure on the vertical wall will be equal to P2, the pressure on the horizontal wall will be P3, the pressure on an arbitrary wall will be P1.

Three sides form a right-angled triangle, the pressure forces acting on these sides are directed along the normal to these surfaces. Since the allocated volume is in a state of equilibrium, at rest, does not move anywhere, therefore, the sum of the forces acting on it is equal to zero. The force normal to the hypotenuse is proportional to the surface area, that is, equal to the pressure times the surface area.

The forces acting on the vertical and horizontal walls are also proportional to the areas of these surfaces and are also directed perpendicularly. That is, the force acting on the vertical is directed horizontally, and the force acting on the horizontal is directed vertically. These three forces add up to zero, therefore, they form a triangle, which is completely similar to this triangle.