Popularly: Very general set of equations describing almost any flow of fluids

• Vast span of applications

• Recall: Potential flow theory
  • Inviscid, irrotational, incompressible flow
• The most general form of the Navier-Stokes equations
  • Viscid, rotational, compressible flow
  • Huge impact on the complexity of the flow → Turbulence

• Note: Incompressibility is still a very accurate assumption used in most engineering applications

Turbulence is the most important unsolved problem of classical physics
Eames & Flor (2011)

There is a physical problem that is common to many fields, that is very old, and that has not been solved. It is not the problem of finding new fundamental particles, but something left over from a long time ago—over a hundred years. Nobody in physics has really been able to analyze it mathematically satisfactorily in spite of its importance to the sister sciences. It is the analysis of circulating or turbulent fluids.
Feynman et al. (1965)

The Clay Mathematics Institute has identified the existence and smoothness of solutions to the Navier-Stokes equations as one of the seven Millennium Prize Problems.

• Sketch of the Problem
  Prove or give a counter-example that in 3-D, solutions to the incompressible Navier-Stokes equations always exist and are smooth for all time.
  Fefferman (2022)


• Prize
  $1,000,000 for a correct solution.

The Navier-Stokes equations describes some very complex physics not fully understood and their mathematical properties are still being investigated by mathematicians

• This is a course for engineering programmes
  • We need to able to reach results and within a reasonable time frame
  • So, why this lecture?
• Combined with modern computer power the Navier-Stokes equations
  • Allows for detailed analysis of virtually all flows of engineering interest
  • Are great for investigation of novel concepts — given their generality
  • Are often the basis of faster, more simple design frameworks
• Understanding the underlying physics of the Navier-Stokes equations is key to applying them meaningfully

Examples of Engineering Applications based on Navier-Stokes solvers (CFD):

Promenade flooding

Wakes in wind farm (starts at 2:20)

Air flow in data center

Wind in urban area (starts at 1:13)

Knowledge

• Must have knowledge about stresses in fluids, equation of motion, constitutive models and Navier-Stokes equations.

Skills

• Must be able to describe assumptions and limitations of mathematical models for different types of flows.
• Must be able to apply analytical and semi-empirical methods for mathematical description of fluid dynamic problems.

Objective: Detailed information on flow inside domain of interest

• Local values of flow fields (velocity, pressure, etc.) not obtained with integral methods
• Finite → Infinitesimal control volume (CV)
• Governing equations: Partial differential equations (PDEs) derived from conservation laws on
  • Mass
  • Linear momentum

• System approach (rather than CV approach)
• Development of kinematic relations for infinitesimal system (fluid element)

Conservation of mass for a finite CV (application of Reynolds Transport Theorem) $$\frac{D M_{sys}}{D t} = \frac{d}{dt}\int_{CV} \rho \, dV\llap{--} + \int_{CS} \rho \boldsymbol{V} \cdot \boldsymbol{\hat{n}} \, dA = 0$$

• Goal: Conservation of mass for a infinitesimal CV

Consider an infinitesimal control volume with dimensions $\delta x\,\delta y\,\delta z$

• Fields given in centroid: $\rho, \boldsymbol{V}$

Infinitesimal CV with mass flow rate in $x$ at faces

Reformulation of volume integral and surface integrals $$ \frac{D M_{sys}}{D t} = \textcolor{#cc445b}{\frac{d}{dt}\int_{CV} \rho \, dV\llap{--}} + \textcolor{#0e8563}{\int_{CS} \rho \boldsymbol{V} \cdot \boldsymbol{\hat{n}} \, dA} = 0 $$

Infinitesimal CV with mass flow rate in $x$ at faces

• Approximation of volume integral for infinitesimal CV,
$$ \textcolor{#cc445b}{\frac{d}{dt}\int_{CV} \rho \, dV\llap{--}} \approx \frac{\partial \rho}{\partial t} \, \delta x\, \delta y\, \delta z $$

Reformulation of surface integral $$ \frac{D M_{sys}}{D t} = \frac{\partial \rho}{\partial t} \, \delta x\, \delta y\, \delta z + \textcolor{#0e8563}{\int_{CS} \rho \boldsymbol{V} \cdot \boldsymbol{\hat{n}} \, dA} = 0 $$

Infinitesimal CV with mass flow rate in $x$ at faces

• Mass flux (rate of mass flow per unit area) in $x$ at centroid: $\rho u$
• Mass flux at faces (First order Taylor series expansion): $$\rho u |_{x+\delta x /2} = \rho u + \frac{\partial (\rho u)}{\partial x} \frac{\delta x}{2}$$ $$\rho u |_{x-\delta x /2} = \rho u - \frac{\partial (\rho u)}{\partial x} \frac{\delta x}{2}$$

Reformulation of surface integral $$ \frac{D M_{sys}}{D t} = \frac{\partial \rho}{\partial t} \, \delta x\, \delta y\, \delta z + \textcolor{#0e8563}{\int_{CS} \rho \boldsymbol{V} \cdot \boldsymbol{\hat{n}} \, dA} = 0 $$

• Mass flux (mass flow rate per unit area) in $x$ at centroid: $\rho u$
• Mass flux in $x$ at faces (First order Taylor series expansion): $$\rho u |_{x+\delta x /2} = \rho u + \frac{\partial (\rho u)}{\partial x} \frac{\delta x}{2}$$ $$\rho u |_{x-\delta x /2} = \rho u - \frac{\partial (\rho u)}{\partial x} \frac{\delta x}{2}$$
• Net mass flow rate in $x$: $$(\rho u |_{x+\delta x /2} - \rho u |_{x-\delta x /2}) \, \delta y\, \delta z \textcolor{#FFFFFF}{= \frac{\partial (\rho u)}{\partial x} \, \delta x\, \delta y\, \delta z}$$

Reformulation of surface integral $$ \frac{D M_{sys}}{D t} = \frac{\partial \rho}{\partial t} \, \delta x\, \delta y\, \delta z + \textcolor{#0e8563}{\int_{CS} \rho \boldsymbol{V} \cdot \boldsymbol{\hat{n}} \, dA} = 0 $$

• Mass flux (mass flow rate per unit area) in $x$ at centroid: $\rho u$
• Mass flux in $x$ at faces (First order Taylor series expansion): $$\rho u |_{x+\delta x /2} = \cancel{\rho u} + \frac{\partial (\rho u)}{\partial x} \frac{\delta x}{\cancel{2}}$$ $$\rho u |_{x-\delta x /2} = \cancel{\rho u} - \frac{\partial (\rho u)}{\partial x} \frac{\delta x}{\cancel{2}}$$
• Net mass flow rate in $x$: $$(\rho u |_{x+\delta x /2} - \rho u |_{x-\delta x /2}) \, \delta y\, \delta z = \frac{\partial (\rho u)}{\partial x} \, \delta x\, \delta y\, \delta z$$
• Similarly for net mass flow rate in $y$ and $z$ directions: $$\frac{\partial (\rho v)}{\partial y} \, \delta x\, \delta y\, \delta z \ \ \text{and} \ \ \frac{\partial (\rho w)}{\partial z} \, \delta x\, \delta y\, \delta z$$
• Net mass flow rate (3-D): $$\textcolor{#0e8563}{\int_{CS} \rho \boldsymbol{V} \cdot \boldsymbol{\hat{n}} \, dA} = \left(\frac{\partial (\rho u)}{\partial x} + \frac{\partial (\rho v)}{\partial y} + \frac{\partial (\rho w)}{\partial z}\right) \, \delta x\, \delta y\, \delta z $$

Reformulation of surface integral $$ \frac{D M_{sys}}{D t} = \frac{\partial \rho}{\partial t} \, \delta x\, \delta y\, \delta z + \textcolor{#0e8563}{\int_{CS} \rho \boldsymbol{V} \cdot \boldsymbol{\hat{n}} \, dA} = 0 $$

• Mass flux (mass flow rate per unit area) in $x$ at centroid: $\rho u$
• Mass flux in $x$ at faces (First order Taylor series expansion): $$\rho u |_{x+\delta x /2} = \cancel{\rho u} + \frac{\partial (\rho u)}{\partial x} \frac{\delta x}{\cancel{2}}$$ $$\rho u |_{x-\delta x /2} = \cancel{\rho u} - \frac{\partial (\rho u)}{\partial x} \frac{\delta x}{\cancel{2}}$$
• Net mass flow rate in $x$: $$(\rho u |_{x+\delta x /2} - \rho u |_{x-\delta x /2}) \, \delta y\, \delta z = \frac{\partial (\rho u)}{\partial x} \, \delta x\, \delta y\, \delta z$$
• Similarly for net mass flow rate in $y$ and $z$ directions: $$\frac{\partial (\rho v)}{\partial y} \, \delta x\, \delta y\, \delta z \ \ \text{and} \ \ \frac{\partial (\rho w)}{\partial z} \, \delta x\, \delta y\, \delta z$$
• Net mass flow rate (3-D): $$\textcolor{#0e8563}{\int_{CS} \rho \boldsymbol{V} \cdot \boldsymbol{\hat{n}} \, dA} = \nabla \cdot (\rho \boldsymbol{V}) \, \delta x\, \delta y\, \delta z$$

Reformulation $$ \frac{D M_{sys}}{D t} = \frac{\partial \rho}{\partial t} \, \delta x\, \delta y\, \delta z + \nabla \cdot (\rho \boldsymbol{V}) \, \delta x\, \delta y\, \delta z = 0$$

Reformulation $$ \frac{D M_{sys}}{D t} = \frac{\partial \rho}{\partial t} \, \delta x\, \delta y\, \delta z + \nabla \cdot (\rho \boldsymbol{V}) \, \delta x\, \delta y\, \delta z = 0$$

Corresponds to a zero volumetric dilatation rate

Approach

1. Conservation of linear momentum for an infinitesimal CV
2. Forces acting on an infinitesimal CV
3. Equations of fluid motion
4. Constitutive law for Newtonian fluids
5. The Navier-Stokes equations

Conservation of linear momentum for a finite CV (application of Reynolds Transport Theorem) $$ \boldsymbol{F} = \frac{D \boldsymbol{P_{sys}}}{D t} = \frac{d}{dt}\int_{CV} \boldsymbol{V} \rho \, dV\llap{--} + \int_{CS} \boldsymbol{V}\rho \boldsymbol{V} \cdot \boldsymbol{\hat{n}} \, dA$$

• Similar methodology as used to derive the continuity equation: Reformulation to an infinitesimal CV

Conservation of linear momentum for a finite CV (application of Reynolds Transport Theorem) $$\boldsymbol{F} = \frac{D \boldsymbol{P_{sys}}}{D t} = \textcolor{#cc445b}{\frac{d}{dt}\int_{CV} \boldsymbol{V}\rho \, dV\llap{--}} + \textcolor{#0e8563}{\int_{CS} \boldsymbol{V}\rho \boldsymbol{V} \cdot \boldsymbol{\hat{n}} \, dA} $$

• Similar methodology as used to derive the continuity equation: Reformulation to an infinitesimal CV
• Recall,

Infinitesimal CV

• Approximation of volume integral for infinitesimal CV,
$$ \textcolor{#cc445b}{\frac{d}{dt}\int_{CV} \boldsymbol{V} \rho \, dV\llap{--}} = \frac{\partial}{\partial t} (\rho \boldsymbol{V}) \, \delta x\, \delta y\, \delta z $$

Conservation of linear momentum $$ \boldsymbol{F} = \frac{D \boldsymbol{P_{sys}}}{D t} = \frac{\partial}{\partial t} (\rho \boldsymbol{V}) \, \delta x\, \delta y\, \delta z + \textcolor{#0e8563}{\int_{CS} \boldsymbol{V}\rho \boldsymbol{V} \cdot \boldsymbol{\hat{n}} \, dA} $$

• Similar methodology as used to derive the continuity equation: Reformulation to an infinitesimal CV
• We reuse a result from 3-D mass conservation: Net mass flow rate → net linear momentum flow rate $$\text{Mass cons.:} \ \ \textcolor{#0e8563}{\int_{CS} \rho \boldsymbol{V} \cdot \boldsymbol{\hat{n}} \, dA} = \left(\frac{\partial (\rho u)}{\partial x} + \frac{\partial (\rho v)}{\partial y} + \frac{\partial (\rho w)}{\partial z}\right) \, \delta x\, \delta y\, \delta z = \nabla \cdot (\rho \boldsymbol{V}) \, \delta x\, \delta y\, \delta z $$ $$\text{Momentum cons.:} \ \ \textcolor{#0e8563}{\int_{CS} \boldsymbol{V}\rho \boldsymbol{V} \cdot \boldsymbol{\hat{n}} \, dA} = \left(\frac{\partial (\rho u\boldsymbol{V})}{\partial x} + \frac{\partial (\rho v\boldsymbol{V})}{\partial y} + \frac{\partial (\rho w\boldsymbol{V})}{\partial z}\right) \, \delta x\, \delta y\, \delta z = \nabla \cdot (\rho\boldsymbol{V}\boldsymbol{V}) \, \delta x\, \delta y\, \delta z $$

Conservation of linear momentum for infinitesimal CV $$ \delta \boldsymbol{F} = \frac{\partial}{\partial t} (\rho \boldsymbol{V}) \, \delta x\, \delta y\, \delta z + \nabla \cdot (\rho\boldsymbol{V}\boldsymbol{V}) \, \delta x\, \delta y\, \delta z $$ $$ \Updownarrow $$ $$ \delta \boldsymbol{F} = \left( \frac{\partial}{\partial t} (\rho \boldsymbol{V}) + \nabla \cdot (\rho \boldsymbol{V}\boldsymbol{V}) \right) \, \delta x\, \delta y\, \delta z $$

• We see that $\nabla \cdot (\rho \boldsymbol{V}\boldsymbol{V})$ is the divergence of an outer product (dyadic product) and can be rewritten
$$\nabla \cdot (\rho \boldsymbol{V}\boldsymbol{V}) = \rho (\boldsymbol{V} \cdot \nabla) \boldsymbol{V} + \boldsymbol{V} (\nabla \cdot (\rho \boldsymbol{V}))$$
• Insertion and rearrangement $$ \delta \boldsymbol{F} = \left( \rho \left( \frac{\partial \boldsymbol{V}}{\partial t} + (\boldsymbol{V} \cdot \nabla) \boldsymbol{V} \right) + \boldsymbol{V} \left( \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \boldsymbol{V})\right) \right) \, \delta x\, \delta y\, \delta z $$

Conservation of linear momentum for infinitesimal CV $$ \delta \boldsymbol{F} = \frac{\partial}{\partial t} (\rho \boldsymbol{V}) \, \delta x\, \delta y\, \delta z + \nabla \cdot (\rho\boldsymbol{V}\boldsymbol{V}) \, \delta x\, \delta y\, \delta z $$ $$ \Updownarrow $$ $$ \delta \boldsymbol{F} = \left( \frac{\partial}{\partial t} (\rho \boldsymbol{V}) + \nabla \cdot (\rho \boldsymbol{V}\boldsymbol{V}) \right) \, \delta x\, \delta y\, \delta z $$

• We see that $\nabla \cdot (\rho \boldsymbol{V}\boldsymbol{V})$ is the divergence of an outer product (dyadic product) and can be rewritten
$$\nabla \cdot (\rho \boldsymbol{V}\boldsymbol{V}) = \rho (\boldsymbol{V} \cdot \nabla) \boldsymbol{V} + \boldsymbol{V} (\nabla \cdot (\rho \boldsymbol{V}))$$
• Insertion and rearrangement $$ \delta \boldsymbol{F} = \left( \rho \left( \textcolor{#cc445b}{\frac{\partial \boldsymbol{V}}{\partial t} + (\boldsymbol{V} \cdot \nabla) \boldsymbol{V}} \right) + \cancel{ \boldsymbol{V} \left( \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \boldsymbol{V})\right) } \right) \, \delta x\, \delta y\, \delta z $$
• Applying the definition of the material derivative and the (compressible) continuity equation
$$ \delta \boldsymbol{F} = \rho \left( \frac{D \boldsymbol{V}}{D t} \right) \, \delta x\, \delta y\, \delta z $$
• Conservation of linear momentum on infinitesimal CV $\rightarrow$ differential form of Newton's 2nd law
$$ \delta \boldsymbol{F} = \delta m \ \boldsymbol{a} $$

Next step: Describe the forces $\delta \boldsymbol{F}$ acting on the infinitesimal CV

• Combination of body forces (e.g., gravity) and surface forces
  (e.g. pressure, viscous stresses) $$ \delta \boldsymbol{F} = \textcolor{#cc445b}{\delta \boldsymbol{F_{b}}} + \textcolor{#0e8563}{\delta \boldsymbol{F_{s}}} $$
• Body force due to gravity (3-D): $$ \textcolor{#cc445b}{\delta \boldsymbol{F_{b}}} = \delta m \ \boldsymbol{g} $$

Next step: Describe the forces $\delta \boldsymbol{F}$ acting on the infinitesimal CV

• Combination of body forces (e.g., gravity) and surface forces
  (e.g. pressure, viscous stresses) $$ \delta \boldsymbol{F} = \delta m \ \boldsymbol{g} + \textcolor{#0e8563}{\delta \boldsymbol{F_{s}}} $$
• Surface force: A combination of normal and shear stresses
• Stress expected to vary in flow field
• First order Taylor series expansion on stress state
  (incl. gradients) in centroid
• Net surface force in $x$ direction
$$ \delta \boldsymbol{F_{sx}} = \left( \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \tau_{yx}}{\partial y} + \frac{\partial \tau_{zx}}{\partial z} \right) \ \delta x\ \delta y\ \delta z $$
• Similarly for $y$ and $z$ directions
• Introducing Cauchy's stress tensor $\boldsymbol{\sigma}$ $$ \boldsymbol{\sigma} = \begin{bmatrix} \sigma_{xx} & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \sigma_{yy} & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \sigma_{zz} \end{bmatrix} $$
• The net surface force vector can be expressed as $$ \textcolor{#0e8563}{\delta \boldsymbol{F_{s}}}= \left( \nabla \cdot \boldsymbol{\sigma} \right) \, \delta x\, \delta y\, \delta z $$

Description of the net force $\delta \boldsymbol{F}$ acting on the infinitesimal CV

• Combination of body forces (e.g., gravity) and surface forces
  (e.g. pressure, viscous stresses) $$ \delta \boldsymbol{F} = \delta m \ \boldsymbol{g} + ( \nabla \cdot \boldsymbol{\sigma} ) \, \delta x\, \delta y\, \delta z $$
$$ \Updownarrow $$ $$ \ \delta \boldsymbol{F} = ( \rho \ \boldsymbol{g} + ( \nabla \cdot \boldsymbol{\sigma} )) \, \delta x\, \delta y\, \delta z $$
• We recall $$ \delta \boldsymbol{F} = \delta m \ \boldsymbol{a} $$ $$ \Downarrow $$ $$ ( \rho \ \boldsymbol{g} + ( \nabla \cdot \boldsymbol{\sigma} )) \, \delta x\, \delta y\, \delta z = \rho \left( \frac{D \boldsymbol{V}}{D t} \right)\, \delta x\, \delta y\, \delta z $$

Description of the net force $\delta \boldsymbol{F}$ acting on the infinitesimal CV

• Combination of body forces (e.g., gravity) and surface forces
  (e.g. pressure, viscous stresses) $$ \delta \boldsymbol{F} = \delta m \ \boldsymbol{g} + ( \nabla \cdot \boldsymbol{\sigma} ) \, \delta x\, \delta y\, \delta z $$
$$ \Updownarrow $$ $$ \ \delta \boldsymbol{F} = ( \rho \ \boldsymbol{g} + ( \nabla \cdot \boldsymbol{\sigma} )) \, \delta x\, \delta y\, \delta z $$
• We recall $$ \delta \boldsymbol{F} = \delta m \ \boldsymbol{a} $$ $$ \Downarrow $$ $$ ( \rho \ \boldsymbol{g} + ( \nabla \cdot \boldsymbol{\sigma} )) \, \cancel{\delta x\, \delta y\, \delta z} = \rho \left( \frac{D \boldsymbol{V}}{D t} \right) \, \cancel{\delta x\, \delta y\, \delta z} $$
• The general differential equations of fluid motion,
  or the Cauchy momentum equation, results $$ \rho \ \boldsymbol{g} + ( \nabla \cdot \boldsymbol{\sigma} ) = \rho \left( \frac{\partial \boldsymbol{V}}{\partial t} + (\boldsymbol{V} \cdot \nabla) \boldsymbol{V} \right) $$

Key result: The general differential equations of fluid motion (Cauchy momentum equation): $$ \rho \ \boldsymbol{g} + ( \nabla \cdot \boldsymbol{\sigma} ) = \rho \left( \frac{\partial \boldsymbol{V}}{\partial t} + (\boldsymbol{V} \cdot \nabla) \boldsymbol{V} \right) $$

• Resemblance to $\delta \boldsymbol{F} = \delta m \ \boldsymbol{a}$
• Written out the equations of fluid motion reads: $$ \rho g_x + \frac{\partial \sigma_{xx}}{\partial x} + \frac{\partial \tau_{yx}}{\partial y} + \frac{\partial \tau_{zx}}{\partial z} = \rho \left( \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} + w\frac{\partial u}{\partial z} \right) $$ $$ \rho g_y + \frac{\partial \tau_{xy}}{\partial x} + \frac{\partial \sigma_{yy}}{\partial y} + \frac{\partial \tau_{zy}}{\partial z} = \rho \left( \frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} + w\frac{\partial v}{\partial z} \right) $$ $$ \rho g_z + \frac{\partial \tau_{xz}}{\partial x} + \frac{\partial \tau_{yz}}{\partial y} + \frac{\partial \sigma_{zz}}{\partial z} = \rho \left( \frac{\partial w}{\partial t} + u\frac{\partial w}{\partial x} + v\frac{\partial w}{\partial y} + w\frac{\partial w}{\partial z} \right) $$
• Can we solve the flow field now?
• No, more unknowns (6 stress components, 3 velocity components, pressure and density) than equations → Ill-posed problem

To close the problem, we need to relate stresses to velocities, i.e., elimate unkowns

• Revisit the Cauchy stress tensor: Sum of pressure (isotropic) and viscous (deviatoric) stress tensors $$\boldsymbol{\sigma} = -p \boldsymbol{I_3} + \boldsymbol{\tau} = \begin{bmatrix} -p & 0 & 0 \\ 0 & -p & 0 \\ 0 & 0 & -p \end{bmatrix} + \begin{bmatrix} \tau_{xx} & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \tau_{yy} & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \tau_{zz} \end{bmatrix} $$
• Constitutive relation for an incompressible Newtonian fluid: Viscous stresses are linearly proportional to strain rates $$\boldsymbol{\tau} = \mu \boldsymbol{S} $$
• with the strain rate tensor $$ \boldsymbol{S} = \left[ \begin{matrix} 2 \dot{\epsilon}_{xx} & \dot{\gamma}_{xy} & \dot{\gamma}_{xz} \\ \dot{\gamma}_{yx} & 2 \dot{\epsilon}_{yy} & \dot{\gamma}_{yz} \\ \dot{\gamma}_{zx} & \dot{\gamma}_{zy} & 2 \dot{\epsilon}_{zz} \end{matrix} \right] $$
• Recalling our kinematic relations developed with the fluid element

To close the problem, we need to relate stresses to velocities, i.e., elimate unkowns

• Revisit the Cauchy stress tensor: Sum of pressure (isotropic) and viscous (deviatoric) stress tensors $$\boldsymbol{\sigma} = -p \boldsymbol{I_3} + \boldsymbol{\tau} = \begin{bmatrix} -p & 0 & 0 \\ 0 & -p & 0 \\ 0 & 0 & -p \end{bmatrix} + \begin{bmatrix} \tau_{xx} & \tau_{xy} & \tau_{xz} \\ \tau_{yx} & \tau_{yy} & \tau_{yz} \\ \tau_{zx} & \tau_{zy} & \tau_{zz} \end{bmatrix} $$
• Constitutive relation for an incompressible Newtonian fluid: Viscous stresses are linearly proportional to strain rates $$\boldsymbol{\tau} = \mu \boldsymbol{S} $$
• with the strain rate tensor
$$ \boldsymbol{S} = \left[ \begin{matrix} 2 \dot{\epsilon}_{xx} & \dot{\gamma}_{xy} & \dot{\gamma}_{xz} \\ \dot{\gamma}_{yx} & 2 \dot{\epsilon}_{yy} & \dot{\gamma}_{yz} \\ \dot{\gamma}_{zx} & \dot{\gamma}_{zy} & 2 \dot{\epsilon}_{zz} \end{matrix} \right] $$
• Recalling our kinematic relations developed with the fluid element
• Strain rates were expressed in terms of velocity gradients
$$ \boldsymbol{S} = \nabla \boldsymbol{V} + (\nabla \boldsymbol{V})^{\mathrm{T}} = \left[ \begin{matrix} 2\frac{\partial u}{\partial x} & \left(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\right) & \ \left(\frac{\partial u}{\partial z} + \frac{\partial w}{\partial x}\right) \\ \left(\frac{\partial v}{\partial x} + \frac{\partial u}{\partial y}\right) & 2\frac{\partial v}{\partial y} & \left(\frac{\partial v}{\partial z} + \frac{\partial w}{\partial y}\right) \\ \left(\frac{\partial w}{\partial x} + \frac{\partial u}{\partial z}\right) & \left(\frac{\partial w}{\partial y} + \frac{\partial v}{\partial z}\right) & 2\frac{\partial w}{\partial z} \end{matrix} \right] $$
• Sum of diagonal entries is zero (traceless) due to continuity

Stress tensor for incompressible Newtonian fluid $$ \boldsymbol{\sigma} = -p \boldsymbol{I_3} + \mu (\nabla \boldsymbol{V} + (\nabla \boldsymbol{V})^{\mathrm{T}}) $$

Substituting into the equations of fluid motion $$ \rho \ \boldsymbol{g} + \nabla \cdot \boldsymbol{\sigma} = \rho \left( \frac{\partial \boldsymbol{V}}{\partial t} + (\boldsymbol{V} \cdot \nabla) \boldsymbol{V} \right) $$ $$ \Downarrow $$ $$ \rho \ \boldsymbol{g} + \nabla \cdot (-p \boldsymbol{I_3} + \mu (\nabla \boldsymbol{V} + (\nabla \boldsymbol{V})^{\mathrm{T}})) = \rho \left( \frac{\partial \boldsymbol{V}}{\partial t} + (\boldsymbol{V} \cdot \nabla) \boldsymbol{V} \right) $$

Using vector calculus identities and the continuity equation $$ \nabla \cdot (p \boldsymbol{I_3}) = \nabla p $$ $$ \nabla \cdot (\nabla \boldsymbol{V}) = \nabla^2 \boldsymbol{V} $$ $$ \nabla \cdot ((\nabla \boldsymbol{V})^{\mathrm{T}}) = \nabla(\nabla \cdot \boldsymbol{V}) = 0 \ \ \text{(continuity)} $$

Final form of the incompressible Navier-Stokes equations $$ \rho\left( \frac{\partial \boldsymbol{V}}{\partial t} + (\boldsymbol{V} \cdot \nabla) \boldsymbol{V} \right) = -\nabla p + \mu \nabla^2 \boldsymbol{V} + \rho \boldsymbol{g} $$

Final form of the incompressible Navier-Stokes equations $$ \rho\left( \frac{\partial \boldsymbol{V}}{\partial t} + (\boldsymbol{V} \cdot \nabla) \boldsymbol{V} \right) = -\nabla p + \rho \boldsymbol{g} + \mu \nabla^2 \boldsymbol{V} $$

• Together with the incompressible continuity equation, we have 4 equations for 4 unknowns ($u,v,w,p$) → well-posed problem!
• These are the governing equations of all incompressible Newtonian fluids!
• Now let's solve the equations!

We have now derived the formidable incompressible Navier-Stokes equations $$ \rho\left( \frac{\partial \boldsymbol{V}}{\partial t} + (\boldsymbol{V} \cdot \nabla) \boldsymbol{V} \right) = -\nabla p + \rho \boldsymbol{g} + \mu \nabla^2 \boldsymbol{V} $$

• What do we have here?
  • A set of nonlinear 2nd order partial differential equations
  • The nonlinear "self"-convection of momentum term $(\boldsymbol{V} \cdot \nabla) \boldsymbol{V}$
    • Greatly complicates solutions
    • Source of turbulence and instabilities
  • Only few exact solutions and only for very simple flows
  • No known accurate analytical solutions to most engineering flows
    • Numerical methods for approximate solutions using CFD