Hydrostatic pressure is the static pressure exerted at a point of interest by the weight of a fluid column above the point.
Background Due to the fundamental nature of fluids, a fluid cannot remain at rest under the presence of a shear stress. However, fluids can exert pressure normal to any contacting surface. If a point in the fluid is thought of as an infinitesimally small cube, then it follows from the principles of equilibrium that the pressure on every side of this unit of fluid must be equal. If this were not the case, the fluid would move in the direction of the resulting force. Thus, the pressure on a fluid at rest is isotropic; i.e., it acts with equal magnitude in all directions. This characteristic allows fluids to transmit force through the length of pipes or tubes; i.e., a force applied to a fluid in a pipe is transmitted, via the fluid, to the other end of the pipe. This principle was first formulated, in a slightly extended form, by Blaise Pascal, and is now called Pascal's law.
Formulation In a fluid at rest, all frictional and inertial stresses vanish and the state of stress of the system is called hydrostatic. When this condition of V = 0 is applied to the Navier–Stokes equations for viscous fluids or Euler equations (fluid dynamics) for ideal inviscid fluid, the gradient of pressure becomes a function of body forces only. The Navier-Stokes momentum equations are:
By setting the flow velocity : u = 0 {\displaystyle \mathbf {u} =\mathbf {0} } , they become simply:
0 = − ∇ p + ρ g {\displaystyle \mathbf {0} =-\nabla p+\rho \mathbf {g} }
or:
∇ p = ρ g {\displaystyle \nabla p=\rho \mathbf {g} }
This is the general form of Stevin's law: the pressure gradient equals the body force force density field. Let us now consider two particular cases of this law. In case of a conservative body force with scalar potential : ϕ {\displaystyle \phi } :
ρ g = − ∇ ϕ {\displaystyle \rho \mathbf {g} =-\nabla \phi }
the Stevin equation becomes:
∇ p = − ∇ ϕ {\displaystyle \nabla p=-\nabla \phi }
That can be integrated to give:
Δ p = − Δ ϕ {\displaystyle \Delta p=-\Delta \phi }
So in this case the pressure difference is the opposite of the difference of the scalar potential associated to the body force. In the other particular case of a body force of constant direction along z:
g = − g ( x , y , z ) k ^ {\displaystyle \mathbf {g} =-g(x,y,z){\hat {k}}}
the generalised Stevin's law above becomes:
∂ p ∂ z = − ρ ( x , y , z ) g ( x , y , z ) {\displaystyle {\frac {\partial p}{\partial z}}=-\rho (x,y,z)g(x,y,z)}
That can be integrated to give another (less-) generalised Stevin's law:
p ( x , y , z ) − p 0 ( x , y ) = − ∫ 0 z ρ ( x , y , z ′ ) g ( x , y , z ′ ) d z ′ {\displaystyle p(x,y,z)-p_{0}(x,y)=-\int _{0}^{z}\rho (x,y,z')g(x,y,z')dz'}
where:
p {\displaystyle p} is the hydrostatic pressure (Pa),
ρ {\displaystyle \rho } is the fluid density (kg/m3),
g {\displaystyle g} is gravitational acceleration (m/s2),
z {\displaystyle z} is the height (parallel to the direction of gravity) of the test area (m),
0 {\displaystyle 0} is the height of the zero reference point of the pressure (m)
p 0 {\displaystyle p_{0}} is the hydrostatic pressure field (Pa) along x and y at the zero reference point
Simplification for liquids For water and other liquids, this integral can be simplified significantly for many practical applications, based on the following two assumptions. Since many liquids can be considered incompressible, a reasonable good estimation can be made from assuming a constant density throughout the liquid. The same assumption cannot be made within a gaseous environment. Also, since the height Δ z {\displaystyle \Delta z} of the fluid column between z and z0 is often reasonably small compared to the radius of the Earth, one can neglect the variation of g. Under these circumstances, one can transport out of the integral the density and the gravity acceleration and the law is simplified into the formula
Δ p ( z ) = ρ g Δ z , {\displaystyle \Delta p(z)=\rho g\Delta z,}
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