Fluid Dynamics - Differential Relations for Fluid Mechanics

Differential Relations for Fluid Mechanics

• Objective: Obtain partial differential equations (PDEs) for conservation laws.

• Solution: Properties at every point within the domain at any time, $x = f(x, t)$.

Describing Properties

• Property defined on an element (Lagrangian description).

• Property can change with time.

• Eulerian description: Property at a point x occupied by element X at time t.

• Time derivative of properties needed for conservation laws.

  • Physical meaning: Time derivative of a property following an element.

  • Use substantial or material derivative relations:

    • $\frac{Df}{Dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} \frac{dx}{dt}$

    • $\frac{Df}{Dt} = \frac{\partial f}{\partial t} + v \cdot \nabla f$

Differential Relations &

• Differential relation can be derived from integral forms using Reynolds transport theorem.

• Steps:

  1. Use material volume (VM), a control volume enclosing the same fluid elements at all times.

     • Mass inside VM stays constant.

  2. Apply Reynolds transport theorem:

     • $\frac{d}{dt} \int<em>{V(t)} f dV = \int</em>{V(t)} \frac{\partial f}{\partial t} dV + \int<em>{A(t)} f V</em>s \cdot n dA$

       • Where $V_s$ is the velocity of the surface.

     • This theorem accounts for the time derivative of the integral, considering both the change in f with time and the change in volume with time.

  3. For material volume, the boundary moves with the same velocity as the element ($V_s = v$).

  4. Convert area integral to volume integral using Gauss's theorem.

     • Requires the property $f$ to be a continuous function.

     • Gauss's Theorem: $\int<em>{A} f \cdot dA = \int</em>{V} \nabla \cdot f dV$

     • Discontinuities (e.g., interfaces between different fluids, shock waves) invalidate this conversion if the volume encloses the discontinuity.

Conservation of Mass

• Fix and observe a fluid element.

• Mass inside the fluid element remains constant.

• Elemental mass: $dm = \rho dV = constant$

  • \rho = density, dV = volume of the element

• If volume increases, density must decrease, and vice versa.

• Time rate of change of density is proportional to the negative of the time rate of change of volume.

• $\frac{d\rho}{dt} \propto -\frac{dV}{dt}$

• Time rate of change of volume per unit volume: $\frac{1}{V} \frac{dV}{dt} = \nabla \cdot v$

• $\frac{d\rho}{dt} = - \rho (\nabla \cdot v)$

• $\frac{1}{\rho}\frac{d\rho}{dt} = -(\nabla \cdot v)$

• Bring everything to the left-hand side:

  • $\frac{D\rho}{Dt} + \rho (\nabla \cdot v) = 0$

    • Lagrangian version of conservation of mass

• Eulerian version, using substantial derivative:

  • $\frac{\partial \rho}{\partial t} + v \cdot \nabla \rho + \rho (\nabla \cdot v) = 0$

  • $\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho v) = 0$

    • Valid at any point at any time.

• Physical meaning of terms:

  • $\nabla \cdot (\rho v)$ represents the flux of mass.

  • If the derivative is positive, more mass is leaving the point than entering, causing the density at that point to decrease.

• One-dimensional example:

  • $\frac{\partial \rho}{\partial t} = - \frac{\partial (\rho v_x)}{\partial x}$

• The equation describes how the density change at a particular point equals the negative of the mass flux entering and leaving that point.

Conservation of Linear Momentum

• Based on Newton's second law: $F = ma$

• Forces acting on the fluid element:

  • Pressure forces.

  • Viscous forces (stresses).

  • Gravitational force.

• Consider the x component:

  • $\rho \frac{Dv<em>x}{Dt} = \sum F</em>x$

• Pressure force in the x direction, where p is pressure:

  • $p - (p + \frac{\partial p}{\partial x} dx) dy dz = - \frac{\partial p}{\partial x} dx dy dz$

• Viscous forces:

  • $\tau<em>{xx} dy dz$, $\tau</em>{xx} + \frac{\partial \tau_{xx}}{\partial x}dx dy dz$

  • $\tau<em>{yx} dx dz$, $\tau</em>{yx} + \frac{\partial \tau_{yx}}{\partial y}dy dx dz$

• Newton's second law for the element:

  • $\rho dxdydz \frac{Dv_x}{Dt} = \text{Pressure Forces} + \text{Viscous Forces} + \text{Gravitational Forces}$

  • $\rho dxdydz \frac{Dv<em>x}{Dt} = - \frac{\partial p}{\partial x} dxdydz + (\frac{\partial \tau</em>{xx}}{\partial x} + \frac{\partial \tau<em>{yx}}{\partial y} + \frac{\partial \tau</em>{zx}}{\partial z})dxdydz + \rho g_x dxdydz$

• Generalize to three dimensions:

  • Replace $v_x$ with $v$ (velocity vector).

  • Replace $-\frac{\partial p}{\partial x}$ with $\nabla p$ (gradient of pressure).

  • Viscous force term becomes $\nabla \cdot \tau$ (divergence of the stress tensor).

• Conservation of linear momentum:

  • $\rho \frac{Dv}{Dt} = -\nabla p + \nabla \cdot \tau + \rho g$

• Mathematical manipulation to obtain the Eulerian form:

  • Using the substantial derivative:

    • $\frac{Dv}{Dt} = \frac{\partial v}{\partial t} + (v \cdot \nabla)v$

  • $\rho(\frac{\partial v}{\partial t} + (v \cdot \nabla)v) = -\nabla p + \nabla \cdot \tau + \rho g$

• Add zero using conservation of mass: $\nabla \cdot (\rho v) = 0$

  • $v( \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho v)) = 0$

  • $\rho \frac{\partial v}{\partial t} + \rho(v \cdot \nabla)v + v(\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho v)) = \frac{\partial (\rho v)}{\partial t} + (\nabla \cdot (\rho vv))$

  • $\frac{\partial (\rho v)}{\partial t} + \nabla \cdot (\rho vv) = -\nabla p + \nabla \cdot \tau + \rho g$

• Understanding the physical meaning:

  • $\nabla \cdot (\rho vv)$ is the momentum flux.

Conservation of Angular Momentum

• Based on Newton's second law for rotation: $\frac{dL}{dt} = \text{Net Moment}$

• Moments can come from pressure, viscous forces, and gravity.

• Gravity does not produce a moment about the center of mass.

• Shear stresses contribute to the moment.

• Mathematical derivation:

  • $\tau<em>{yx} = \tau</em>{yx} + \frac{\partial \tau_{yx}}{\partial y}dy$

  • $\tau<em>{xy} = \tau</em>{xy} + \frac{\partial \tau_{xy}}{\partial x}dx$

  • $\frac{dL}{dt} = I \frac{d\alpha}{dt}$

• The conservation of angular momentum requires the stress tensor to be symmetric:

  • $\tau<em>{xy} = \tau</em>{yx}$

  • $\tau<em>{xz} = \tau</em>{zx}$

  • $\tau<em>{yz} = \tau</em>{zy}$

• Total stress (including pressure) is a symmetric tensor.

  • $\sigma = -pI + \tau$

Conservation of Energy

• First law of thermodynamics for a closed system:

  • $\frac{dE}{dt} = \dot{W} + \dot{Q}$

  • $\frac{d}{dt} \int_V e \rho dV = \dot{W} + \dot{Q}$

  • $e$ is specific energy.

• Energy of a moving fluid element:

  • Internal energy (u) + Kinetic energy (\frac{1}{2}v^2).

• Power ($\dot{W}$) is due to:

  • Pressure.

  • Viscous forces.

  • Gravity.

• Heat transfer ($\dot{Q}$) is due to:

  • Radiation.

  • Conduction.

• Pressure on x surface:

  • $Force = p \cdot Area = p dy dz$

• Radiative heat transfer is expressed as: $\dot{q} = \rho q$ ; q is heat transfer per unit volume.

• Substitute into the conservation of energy:

  • $\frac{d}{dt} \int_V (u + \frac{1}{2}v^2) \rho dV = \dot{W} + \dot{Q}$

  • $\rho \frac{D}{Dt}(u + \frac{1}{2}v^2) = \dot{q} - \nabla \cdot (pv) + \nabla \cdot (\tau \cdot v) + \rho (g \cdot v)$

• Del operator application:

  • $\nabla \cdot (\tau \cdot v)$: This term describes the work done by viscous forces. Expanding this term fully would lead to a complex expression containing several terms with velocity and stress components. The term can be rewritten as $\tau : (\nabla v)$

• Subtract the kinetic energy (linear momentum) equation from the total energy equation to obtain the conservation of internal energy.

• $\frac{D}{Dt}(\frac{1}{2} v^2) = v \cdot \frac{Dv}{Dt} = v \cdot(-\frac{\nabla p}{\rho} + \frac{\nabla \cdot \tau}{\rho} + g)$

• Conservation of internal energy:

  • $\rho \frac{Du}{Dt} = \dot{q} - p(\nabla \cdot v) + \Phi$

• $\Phi = \tau : \nabla v$ This expression represents viscous dissipation.

• Physical interpretation:

  • Increasing internal energy:

    • Increase temperature through radiation and conduction.

    • Compress the fluid (decrease volume).

    • Viscous heating (viscous dissipation).

Alternative Forms of the Energy Equation

• Express conservation of energy in terms of enthalpy and entropy.

Enthalpy forms

• Enthalpy: h = u + pv, v = 1/p $dh = du + pdv + vdp$ $\rho \frac{Dh}{Dt} = \rho \frac{Du}{Dt} + P\nabla*v + v \frac{DP}{Dt}$

  • Replacing Du/Dt yields
    $\rho \frac{Dh}{Dt} = \dot{q} + \nabla<em>v + \Phi+ \nabla</em>v + v \frac{DP}{Dt}$

• Final Form:
  $\rho \frac{Dh}{Dt} = \dot{q} + \Phi+ v \frac{DP}{Dt}$

Entropy forms

• Start with:$du = Tds - pdv$

  • Substituting for specific volume and multiplying by density yields
    $\frac{du}{Dt} = \rho T\frac{ds}{Dt} -p \rho \frac{dv}{Dt}$

• Final form:
  $\frac{du}{Dt} = \rho T\frac{ds}{Dt} + p \nabla v$

  • Combining with previous statements:
    $\rho T \frac{Ds}{Dt} = \dot{q} + \Phi$

• Useful form if Inviscid and Adiabatic (ds/dt=0):

  • Adiabatic: No heat transfer

  • Inviscid: No Viscosity

• Constitutive relations will be discussed in the next video.

Differential Relations for Fluid Mechanics

Objective: Obtain partial differential equations (PDEs) for conservation laws.

Solution: Properties at every point within the domain at any time, $x = f(x, t)$.

Describing Properties

Property defined on an element (Lagrangian description).

Property can change with time.

Eulerian description: Property at a point x occupied by element X at time t.

Time derivative of properties needed for conservation laws.

• Physical meaning: Time derivative of a property following an element.

• Use substantial or material derivative relations:-

  • $\frac{Df}{Dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x} \frac{dx}{dt}$

  • $\frac{Df}{Dt} = \frac{\partial f}{\partial t} + v \cdot \nabla f$

Differential Relations &

Differential relation can be derived from integral forms using Reynolds transport theorem.

Steps:

1. Use material volume (VM), a control volume enclosing the same fluid elements at all times.

   • Mass inside VM stays constant.

2. Apply Reynolds transport theorem:-

   • $\frac{d}{dt} \int<em>{V(t)} f dV = \int</em>{V(t)} \frac{\partial f}{\partial t} dV + \int<em>{A(t)} f V</em>s \cdot n dA$

   • Where $V_s$ is the velocity of the surface.

   • This theorem accounts for the time derivative of the integral, considering both the change in f with time and the change in volume with time.

3. For material volume, the boundary moves with the same velocity as the element ($V_s = v$).

4. Convert area integral to volume integral using Gauss's theorem.

   • Requires the property $f$ to be a continuous function.

   • Gauss's Theorem: $\int<em>{A} f \cdot dA = \int</em>{V} \nabla \cdot f dV$

   • Discontinuities (e.g., interfaces between different fluids, shock waves) invalidate this conversion if the volume encloses the discontinuity.

Conservation of Mass

Fix and observe a fluid element.

Mass inside the fluid element remains constant.

Elemental mass: $dm = \rho dV = constant$

• \rho = density, dV = volume of the element

If volume increases, density must decrease, and vice versa.

Time rate of change of density is proportional to the negative of the time rate of change of volume.

$\frac{d\rho}{dt} \propto -\frac{dV}{dt}$

Time rate of change of volume per unit volume: $\frac{1}{V} \frac{dV}{dt} = \nabla \cdot v$

$\frac{d\rho}{dt} = - \rho (\nabla \cdot v)$

$\frac{1}{\rho}\frac{d\rho}{dt} = -(\nabla \cdot v)$

Bring everything to the left-hand side:-

$\frac{D\rho}{Dt} + \rho (\nabla \cdot v) = 0$

• Lagrangian version of conservation of mass

Eulerian version, using substantial derivative:-

• $\frac{\partial \rho}{\partial t} + v \cdot \nabla \rho + \rho (\nabla \cdot v) = 0$

• $\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho v) = 0$

  • Valid at any point at any time.

Physical meaning of terms:-

• $\nabla \cdot (\rho v)$ represents the flux of mass.

• If the derivative is positive, more mass is leaving the point than entering, causing the density at that point to decrease.

One-dimensional example:-

• $\frac{\partial \rho}{\partial t} = - \frac{\partial (\rho v_x)}{\partial x}$

The equation describes how the density change at a particular point equals the negative of the mass flux entering and leaving that point.

Conservation of Linear Momentum

Based on Newton's second law: $F = ma$

Forces acting on the fluid element:

Pressure forces.

Viscous forces (stresses).

Gravitational force.

Consider the x component:-

• $\rho \frac{Dv<em>x}{Dt} = \sum F</em>x$

Pressure force in the x direction, where p is pressure:-

• $p - (p + \frac{\partial p}{\partial x} dx) dy dz = - \frac{\partial p}{\partial x} dx dy dz$

Viscous forces:-

• $\tau<em>{xx} dy dz$, $\tau</em>{xx} + \frac{\partial \tau_{xx}}{\partial x}dx dy dz$

• $\tau<em>{yx} dx dz$, $\tau</em>{yx} + \frac{\partial \tau_{yx}}{\partial y}dy dx dz$

Newton's second law for the element:-

• $\rho dxdydz \frac{Dv_x}{Dt} = \text{Pressure Forces} + \text{Viscous Forces} + \text{Gravitational Forces}$

• $\rho dxdydz \frac{Dv<em>x}{Dt} = - \frac{\partial p}{\partial x} dxdydz + (\frac{\partial \tau</em>{xx}}{\partial x} + \frac{\partial \tau<em>{yx}}{\partial y} + \frac{\partial \tau</em>{zx}}{\partial z})dxdydz + \rho g_x dxdydz$

Generalize to three dimensions:-

• Replace $v_x$ with $v$ (velocity vector).

• Replace $-\frac{\partial p}{\partial x}$ with $\nabla p$ (gradient of pressure).

• Viscous force term becomes $\nabla \cdot \tau$ (divergence of the stress tensor).

Conservation of linear momentum:-

• $\rho \frac{Dv}{Dt} = -\nabla p + \nabla \cdot \tau + \rho g$

Mathematical manipulation to obtain the Eulerian form:-

• Using the substantial derivative:-

  • $\frac{Dv}{Dt} = \frac{\partial v}{\partial t} + (v \cdot \nabla)v$

• $\rho(\frac{\partial v}{\partial t} + (v \cdot \nabla)v) = -\nabla p + \nabla \cdot \tau + \rho g$

Add zero using conservation of mass: $\nabla \cdot (\rho v) = 0$

• $v( \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho v)) = 0$

• $\rho \frac{\partial v}{\partial t} + \rho(v \cdot \nabla)v + v(\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho v)) = \frac{\partial (\rho v)}{\partial t} + (\nabla \cdot (\rho vv))$

• $\frac{\partial (\rho v)}{\partial t} + \nabla \cdot (\rho vv) = -\nabla p + \nabla \cdot \tau + \rho g$

Understanding the physical meaning:-

• $\nabla \cdot (\rho vv)$ is the momentum flux.

Conservation of Angular Momentum

Based on Newton's second law for rotation: $\frac{dL}{dt} = \text{Net Moment}$

Moments can come from pressure, viscous forces, and gravity.

Gravity does not produce a moment about the center of mass.

Shear stresses contribute to the moment.

Mathematical derivation:-

• $\tau<em>{yx} = \tau</em>{yx} + \frac{\partial \tau_{yx}}{\partial y}dy$

• $\tau<em>{xy} = \tau</em>{xy} + \frac{\partial \tau_{xy}}{\partial x}dx$

• $\frac{dL}{dt} = I \frac{d\alpha}{dt}$

The conservation of angular momentum requires the stress tensor to be symmetric:-

• $\tau<em>{xy} = \tau</em>{yx}$

• $\tau<em>{xz} = \tau</em>{zx}$

• $\tau<em>{yz} = \tau</em>{zy}$

Total stress (including pressure) is a symmetric tensor.-

• $\sigma = -pI + \tau$

Conservation of Energy

First law of thermodynamics for a closed system:-

• $\frac{dE}{dt} = \dot{W} + \dot{Q}$

• $\frac{d}{dt} \int_V e \rho dV = \dot{W} + \dot{Q}$

• $e$ is specific energy.

Energy of a moving fluid element:-

• Internal energy (u) + Kinetic energy (\frac{1}{2}v^2).

Power ($\dot{W}$) is due to:-

Pressure.

Viscous forces.

Gravity.

Heat transfer ($\dot{Q}$) is due to:-

Radiation.

Conduction.

Pressure on x surface:-

• $Force = p \cdot Area = p dy dz$

Radiative heat transfer is expressed as: $\dot{q} = \rho q$ ; q is heat transfer per unit volume.

Substitute into the conservation of energy:-

• $\frac{d}{dt} \int_V (u + \frac{1}{2}v^2) \rho dV = \dot{W} + \dot{Q}$

• $\rho \frac{D}{Dt}(u + \frac{1}{2}v^2) = \dot{q} - \nabla \cdot (pv) + \nabla \cdot (\tau \cdot v) + \rho (g \cdot v)$

Del operator application:-

• $\nabla \cdot (\tau \cdot v)$: This term describes the work done by viscous forces. Expanding this term fully would lead to a complex expression containing several terms with velocity and stress components. The term can be rewritten as $\tau : (\nabla v)$

Subtract the kinetic energy (linear momentum) equation from the total energy equation to obtain the conservation of internal energy.

$\frac{D}{Dt}(\frac{1}{2} v^2) = v \cdot \frac{Dv}{Dt} = v \cdot(-\frac{\nabla p}{\rho} + \frac{\nabla \cdot \tau}{\rho} + g)$

Conservation of internal energy:-

• $\rho \frac{Du}{Dt} = \dot{q} - p(\nabla \cdot v) + \Phi$

• $\Phi = \tau : \nabla v$ This expression represents viscous dissipation.

Physical interpretation:-

Increasing internal energy:-

• Increase temperature through radiation and conduction.

• Compress the fluid (decrease volume).

• Viscous heating (viscous dissipation).

Alternative Forms of the Energy Equation

Express conservation of energy in terms of enthalpy and entropy.

Enthalpy forms

Enthalpy: h = u + pv, v = 1/p $dh = du + pdv + vdp$ $\rho \frac{Dh}{Dt} = \rho \frac{Du}{Dt} + P\nabla v + v \frac{DP}{Dt}$- Replacing Du/Dt yields

$\rho \frac{Dh}{Dt} = \dot{q} + \nabla v + \Phi+ \nabla*v + v \frac{DP}{Dt}$

Final Form:

$\rho \frac{Dh}{Dt} = \dot{q} + \Phi+ v \frac{DP}{Dt}$

Entropy forms

Start with:$du = Tds - pdv$

• Substituting for specific volume and multiplying by density yields

• $\frac{du}{Dt} = \rho T\frac{ds}{Dt} -p \rho \frac{dv}{Dt}$

Final form:

$\frac{du}{Dt} = \rho T\frac{ds}{Dt} + p \nabla v$

• Combining with previous statements:

• $\rho T \frac{Ds}{Dt} = \dot{q} + \Phi$

Useful form if Inviscid and Adiabatic (