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Fluid Statics

Field: Fluid Mechanics

The study of fluids at rest.

Sequence of Expressions

In the context of a continuous fluid medium Ω⊂R3\Omega \subset \mathbb{R}^3 at rest (i.e., v⃗=0⃗\vec{v} = \vec{0}), the pressure PP is defined via the Cauchy stress tensor σ\boldsymbol{\sigma}. For an incompressible, static fluid, the stress tensor is isotropic and purely normal, σ=−PI\boldsymbol{\sigma} = -P \mathbf{I}, where I\mathbf{I} is the identity tensor. The pressure PP is mathematically defined as the scalar field representing the normal stress component acting on a surface element dAdA due to the fluid's internal energy and external body forces. Specifically, considering the equilibrium of a differential volume element dVdV, the Cauchy momentum equation reduces to: \n\n∇⋅σ+ρg⃗=0⃗\nabla \cdot \boldsymbol{\sigma} + \rho \vec{g} = \vec{0}\n\nSubstituting the hydrostatic stress assumption σ=−PI\boldsymbol{\sigma} = -P \mathbf{I} into the equilibrium equation yields: \n\n∇(−PI)+ρg⃗=0⃗\nabla (-P \mathbf{I}) + \rho \vec{g} = \vec{0}\n\nThis simplifies to the fundamental differential equation for pressure: \n\n∇P=−ρg⃗\nabla P = -\rho \vec{g}\n\nIntegrating this gradient yields the hydrostatic pressure equation, which defines PP at any point (x,y,z)(x, y, z) relative to a reference point (x0,y0,z0)(x_0, y_0, z_0): \n\nP(x,y,z)−P0=−ρg⃗⋅(r⃗−r⃗0)P(x, y, z) - P_0 = -\rho \vec{g} \cdot (\vec{r} - \vec{r}_0) \n\nWhere ρ\rho is the constant fluid density, g⃗\vec{g} is the acceleration due to gravity, and r⃗\vec{r} and r⃗0\vec{r}_0 are position vectors.
Let Ω⊂R3\Omega \subset \mathbb{R}^3 be a continuous spatial domain occupied by the fluid, and let m(Ω)m(\Omega) be the total mass contained within Ω\Omega. The fluid density ρ(x⃗,t)\rho(\vec{x}, t) is defined as the material density function, which is a scalar field ρ:Ω×[t1,t2]→R+\rho: \Omega \times [t_1, t_2] \to \mathbb{R}^+. Formally, for any point x⃗∈Ω\vec{x} \in \Omega, the density ρ(x⃗,t)\rho(\vec{x}, t) is the limit of the ratio of mass mVm_{V} contained in a small volume VV surrounding x⃗\vec{x} to the volume VV itself, as VV approaches the zero volume limit: \nρ(x⃗,t)=lim⁡V→0mVV\rho(\vec{x}, t) = \lim_{V \to 0} \frac{m_{V}}{V} \nAlternatively, in the context of continuum mechanics, the mass mm contained within an arbitrary volume V⊂ΩV \subset \Omega is given by the volume integral of the density function: \nm=∭Vρ(x⃗,t) dVm = \iiint_{V} \rho(\vec{x}, t) \, dV \nFurthermore, the density is related to the fluid's equation of state, P=P(ρ,T)P = P(\rho, T), where PP is pressure, ρ\rho is density, and TT is temperature.
Consider a control volume V\mathcal{V} through which an incompressible fluid of constant density ρ0\rho_0 flows steadily. Let v(x,y,z)\textbf{v}(x, y, z) be the velocity field, and A(x)A(x) be the cross-sectional area perpendicular to the xx-axis. The principle of continuity, derived from the conservation of mass, mandates that the net volumetric flow rate QQ must be constant along the flow axis xx. Mathematically, this is expressed as:\nddx(A(x)v(x))=0\frac{\text{d}}{\text{d}x} \left( A(x) v(x) \right) = 0 \nIntegrating this differential equation yields the fundamental statement of continuity:\nA1v1=A2v2=⋯=QconstA_1 v_1 = A_2 v_2 = \dots = Q_{\text{const}} \nwhere AiA_i and viv_i are the cross-sectional area and the average fluid velocity, respectively, at any two arbitrary sections ii and jj along the conduit, provided the fluid remains incompressible (ρ=constant\rho = \text{constant}). This formulation is a direct consequence of the divergence-free condition for the velocity field: ∇⋅v=0\nabla \cdot \textbf{v} = 0.
The hydrostatic pressure P(h)P(h) at a depth hh within a fluid of constant density ρ\rho under gravitational acceleration g⃗\vec{g} is derived from the condition of mechanical equilibrium. Consider a fluid element of height dhdh and cross-sectional area AA. The pressure gradient ∇P\nabla P must balance the body force per unit volume, f⃗b=ρg⃗\vec{f}_b = \rho \vec{g}. Since the fluid is static, the pressure gradient is purely vertical: ∂P∂h=−ρg\frac{\partial P}{\partial h} = -\rho g. Integrating this differential equation from the surface (h=0h=0) to an arbitrary depth hh, and assuming the surface pressure P0P_0 is the reference pressure, yields the scalar relationship for the gauge pressure Pgauge(h)P_{gauge}(h): \n\nPgauge(h)=∫0hρg dh′=ρghP_{gauge}(h) = \int_{0}^{h} \rho g \, d h' = \rho g h
The Equation of Fluid Statics is derived from the principle of mechanical equilibrium applied to a differential volume element dV\mathrm{d}\mathcal{V} within a continuous fluid domain Ω⊂R3\Omega \subset \mathbb{R}^3 at rest. Let ρ\rho be the fluid density, PP be the thermodynamic pressure, and g⃗\vec{g} be the gravitational acceleration vector. The condition for static equilibrium requires that the net force F⃗net\vec{F}_{net} acting on dV\mathrm{d}\mathcal{V} vanishes.\n\nFormally, the governing equation is expressed as the balance of forces per unit volume:\n\n∇P+ρg⃗=0⃗\nabla P + \rho \vec{g} = \vec{0} \n\nThis vector equation implies three scalar partial differential equations (PDEs) in Cartesian coordinates (x,y,z)(x, y, z): \n\n∂P∂x+ρgx=0  ⟹  ∂P∂x=−ρgx(x-component)\frac{\partial P}{\partial x} + \rho g_x = 0 \quad \implies \quad \frac{\partial P}{\partial x} = -\rho g_x \quad \text{(x-component)}\n∂P∂y+ρgy=0  ⟹  ∂P∂y=−ρgy(y-component)\frac{\partial P}{\partial y} + \rho g_y = 0 \quad \implies \quad \frac{\partial P}{\partial y} = -\rho g_y \quad \text{(y-component)}\n∂P∂z+ρgz=0  ⟹  ∂P∂z=−ρgz(z-component)\frac{\partial P}{\partial z} + \rho g_z = 0 \quad \implies \quad \frac{\partial P}{\partial z} = -\rho g_z \quad \text{(z-component)}\n\nIntegrating these partial derivatives yields the fundamental hydrostatic pressure relation:\n\nP(z)=P0+∫z0zρgzdz′P(z) = P_0 + \int_{z_0}^{z} \rho g_z \mathrm{d}z'
The relationship is derived from the fundamental principle of fluid statics, which states that in a fluid at equilibrium, the pressure gradient must balance the body forces. Assuming a fluid of density ρ\rho and constant gravitational acceleration g⃗=−gk^\vec{g} = -g\hat{k} (where k^\hat{k} is the unit vector in the vertical direction, and zz is the vertical coordinate increasing upwards), the governing differential equation is:\∇P=−ρg⃗\nabla P = -\rho \vec{g}. Since ρ\rho and gg are assumed constant, and the pressure PP is a function of position r⃗=(x,y,z)\vec{r} = (x, y, z), we have:\∂P∂x=0\frac{\partial P}{\partial x} = 0, \∂P∂y=0\frac{\partial P}{\partial y} = 0, and \∂P∂z=−ρg\frac{\partial P}{\partial z} = -\rho g. Integrating the partial derivative with respect to zz yields the hydrostatic pressure profile:\dPdz=−ρg\frac{dP}{dz} = -\rho g. Integrating this differential equation from the surface (z=0z=0) to an arbitrary depth zz gives the pressure P(z)P(z) at depth zz: \∫PatmP(z)dP=−ρg∫0zdz\int_{P_{atm}}^{P(z)} dP = -\rho g \int_{0}^{z} dz. \\Therefore, the pressure P(z)P(z) at depth zz is given by the equation:\P(z)=Patm+ρgh\mathbf{P(z) = P_{atm} + \rho g h},where, where h = -zisthedepthmeasuredpositivelydownwards,and is the depth measured positively downwards, and P_{atm}istheatmosphericpressureatthesurface( is the atmospheric pressure at the surface (z=0).If). If hisdefinedasthedepth,therelationshipis:$P(h)=Patm+ρgh is defined as the depth, the relationship is:\$\mathbf{P(h) = P_{atm} + \rho g h}.
Consider a fluid F\mathcal{F} occupying a volume V\mathcal{V} bounded by a surface ∂V\partial\mathcal{V}. Let P\mathbf{P} be the Cauchy stress tensor within the fluid. In the state of fluid statics, the fluid is in mechanical equilibrium, meaning the net force on any differential volume element dVd\mathcal{V} is zero. The governing equation is the Cauchy momentum equation: ∇⋅P+ρg=0\nabla \cdot \mathbf{P} + \rho \mathbf{g} = \mathbf{0}. For a Newtonian fluid at rest, the stress tensor P\mathbf{P} simplifies to P=−PI+2μE\mathbf{P} = -P \mathbf{I} + 2\mu \mathbf{E}, where PP is the pressure, I\mathbf{I} is the identity tensor, μ\mu is the dynamic viscosity, and E\mathbf{E} is the strain rate tensor. Since the fluid is static, E=0\mathbf{E} = \mathbf{0}. Thus, the equilibrium condition reduces to ∇PI+ρg=0\nabla P \mathbf{I} + \rho \mathbf{g} = \mathbf{0}. This implies that the pressure gradient must balance the body forces: ∇P=−ρg\nabla P = -\rho \mathbf{g}. Pascal's Law asserts that the pressure PP at any point x∈V\mathbf{x} \in \mathcal{V} depends only on the depth hh (or the vertical coordinate zz) and is independent of the horizontal coordinates (x,y)(x, y), provided the fluid is incompressible and the container is rigid. Mathematically, this means that the pressure field P(x)P(\mathbf{x}) must satisfy the condition of path independence for the pressure change, leading to the differential form: \begin{equation} \frac{\partial P}{\partial x} = 0 \quad \text{and} \quad \frac{\partial P}{\partial y} = 0 \end{equation} in a horizontal plane, which is consistent with the full hydrostatic equation: \begin{equation} P(z) = P_0 + \rho g (z_0 - z) \end{equation} where P0P_0 is the pressure at the surface z0z_0, ρ\rho is the fluid density, and gg is the acceleration due to gravity.
Let Ω⊂R3\Omega \subset \mathbb{R}^3 be the continuous domain occupied by the fluid, and let V⊂ΩV \subset \Omega be the volume of the object submerged in the fluid. Assume the fluid is in hydrostatic equilibrium, characterized by a pressure field p(r⃗)p(\vec{r}) such that ∇p=−ρ(r⃗)g⃗\nabla p = -\rho(\vec{r}) \vec{g}, where ρ(r⃗)\rho(\vec{r}) is the fluid density and g⃗\vec{g} is the gravitational acceleration vector. The buoyant force F⃗B\vec{F}_B acting on the object is the net force exerted by the pressure differential over the object's surface ∂V\partial V. Mathematically, this force is given by the surface integral:\n\nF⃗B=−∯∂Vpn⃗dS\vec{F}_B = -\oiint_{\partial V} p \vec{n} dS \n\nBy applying the Divergence Theorem and the hydrostatic pressure gradient relation, the buoyant force can be expressed as the volume integral of the pressure force density over the submerged volume VV:\n\nF⃗B=∭V(−∇p)dV=∭Vρg⃗dV\vec{F}_B = \iiint_{V} (-\nabla p) dV = \iiint_{V} \rho \vec{g} dV \n\nSince ρ\rho and g⃗\vec{g} are assumed constant throughout the fluid domain, we obtain:\n\nF⃗B=ρVg⃗\vec{F}_B = \rho V \vec{g} \n\nThis result is equated to the weight of the displaced fluid, WdispW_{disp}. The mass of the displaced fluid is mdisp=ρVm_{disp} = \rho V, and its weight is Wdisp=mdispg=ρVgW_{disp} = m_{disp} g = \rho V g. Thus, the principle states the vector equality:\n\nF⃗B=Wdisp  ⟹  ρVg⃗=ρVg⃗\vec{F}_B = W_{disp} \implies \rho V \vec{g} = \rho V \vec{g}
Let Ω\Omega be a continuous, connected region in R3\mathbb{R}^3 representing the fluid domain, and let ρ(x⃗)\rho(\vec{x}) be the density function. For a differential fluid element dVdV located at x⃗∈Ω\vec{x} \in \Omega to be in static equilibrium, the net force acting on it must vanish. This condition is mathematically expressed by the balance of forces, which states that the gradient of the pressure field pp must exactly counteract the body force per unit volume f⃗b\vec{f}_b. Specifically, the governing equation is:\n\n∇p+f⃗b=0⃗\nabla p + \vec{f}_b = \vec{0}\n\nIf the body force is solely due to gravity, f⃗b=ρg⃗\vec{f}_b = \rho \vec{g}, leading to the fundamental differential equation of fluid statics:\n\n∇p=−ρg⃗\nabla p = -\rho \vec{g}\n\nFurthermore, the condition of static equilibrium requires that the velocity field v⃗\vec{v} is identically zero throughout Ω\Omega, i.e., v⃗=0⃗\vec{v} = \vec{0}, ensuring that the inertial terms vanish.
Consider a fluid domain Ω⊂R3\Omega \subset \mathbb{R}^3 in static equilibrium under a constant gravitational field g⃗=−gk^\vec{g} = -g \hat{\mathbf{k}}. The pressure PP must satisfy the momentum balance equation (Cauchy's equation of motion for a fluid at rest): \n\n∇P=−ρg⃗\nabla P = -\rho \vec{g} \n\nSubstituting the gravitational vector yields:\n\n∇P=ρgk^\nabla P = \rho g \hat{\mathbf{k}} \n\nThis vector equation implies that the pressure PP is independent of the horizontal coordinates xx and yy. Formally, the partial derivatives must vanish:\n\n∂P∂x=0 and ∂P∂y=0\frac{\partial P}{\partial x} = 0 \text{ and } \frac{\partial P}{\partial y} = 0 \n\nConsequently, the pressure PP is a function solely of the vertical coordinate zz (or depth hh): P=P(z)P = P(z). Integrating the differential equation yields the fundamental hydrostatic relation:\n\nP(z)=P0+ρgzP(z) = P_0 + \rho g z