# Introduction to turbulence/Reynolds averaged equations

(Difference between revisions)
 Revision as of 17:32, 25 June 2007 (view source)Jola (Talk | contribs)← Older edit Revision as of 18:05, 25 June 2007 (view source)Jola (Talk | contribs) Newer edit → Line 1: Line 1: - {{Turbulence}} + {{Introduction to turbulence menu}} - ==The Reynolds averaged equations and the turbulence closure problem== + - === The equations governing the instantaneous fluid motions === + == Equations governing instantaneous fluid motion == All fluid motions, whether turbulent or not, are governed by the dynamical equations for a fluid. These can be written using Cartesian tensor notation as: All fluid motions, whether turbulent or not, are governed by the dynamical equations for a fluid. These can be written using Cartesian tensor notation as: - +
:$:[itex] \rho\left[\frac{\partial \tilde{u}_i}{\partial t}+\tilde{u}_j\frac{\partial \tilde{u}_i}{\partial x_j}\right] = -\frac{\partial \tilde{p}}{\partial x_i}+\frac{\partial \tilde{T}_{ij}^{(v)}}{\partial x_j}$ \rho\left[\frac{\partial \tilde{u}_i}{\partial t}+\tilde{u}_j\frac{\partial \tilde{u}_i}{\partial x_j}\right] = -\frac{\partial \tilde{p}}{\partial x_i}+\frac{\partial \tilde{T}_{ij}^{(v)}}{\partial x_j}[/itex] - (2.1)
+
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- + +
:$:[itex] \left[\frac{\partial \tilde{\rho}}{\partial t}+\tilde{u}_j\frac{\partial \tilde{\rho}}{\partial x_j}\right]+ \tilde{\rho}\frac{\partial \tilde{u}_j}{\partial x_j}= 0$ \left[\frac{\partial \tilde{\rho}}{\partial t}+\tilde{u}_j\frac{\partial \tilde{\rho}}{\partial x_j}\right]+ \tilde{\rho}\frac{\partial \tilde{u}_j}{\partial x_j}= 0 [/itex] - (2.2)
+
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- + where $\tilde{u_i}(\vec{x},t)$ represents the i-the component of the fluid velocity at a point in space,$[\vec{x}]_i=x_i$, and time,t. Also where $\tilde{u_i}(\vec{x},t)$ represents the i-the component of the fluid velocity at a point in space,$[\vec{x}]_i=x_i$, and time,t. Also - $\tilde{p}(\vec{x},t)$ represents the static pressure, $\tilde{T}_{ij}^{(v)}(\vec{x},t)$, the viscous(or deviatoric) stresses, and $\tilde\rho$ the fluid density. The tilde over the symbol indicates that an instantaneous quantity is being considered. Also the [[Einstein summation convention]] has been employed. + $\tilde{p}(\vec{x},t)$ represents the static pressure, $\tilde{T}_{ij}^{(v)}(\vec{x},t)$, the viscous(or deviatoric) stresses, and $\tilde\rho$ the fluid density. The tilde over the symbol indicates that an instantaneous quantity is being considered. Also the [[einstein summation convention]] has been employed. - + - In equation 2.1, the subscript $i$ is a free index which can take on the values 1,2 and 3. Thus equation 2.1 is in reality three separate equations. These three equations are just Newton's second law written for a continuum in a spatial(or Eulerian) reference frame. Together they relate the rate of change of momentum per unit mass $(\rho{u_i})$,a vector quantity, to the contact and body forces. + In equation 1, the subscript $i$ is a free index which can take on the values 1,2 and 3. Thus equation 1 is in reality three separate equations. These three equations are just Newton's second law written for a continuum in a spatial(or Eulerian) reference frame. Together they relate the rate of change of momentum per unit mass $(\rho{u_i})$,a vector quantity, to the contact and body forces. - Equation 2.2 is the equation for mass conservation in the absence of sources(or sinks) of mass. Almost all flows considered in this material will be incompressible, which implies that derivative of the density following the fluid material[the term in brackets] is zero. Thus for incompressible flows, the mass conservation equation reduces to: + Equation 2 is the equation for mass conservation in the absence of sources(or sinks) of mass. Almost all flows considered in this material will be incompressible, which implies that derivative of the density following the fluid material[the term in brackets] is zero. Thus for incompressible flows, the mass conservation equation reduces to: - +
:$:[itex] \frac{D \tilde{\rho}}{Dt}=\frac{\partial \tilde{\rho}}{\partial t}+\tilde{u}_j\frac{\partial \tilde{\rho}}{\partial x_j}= 0$ \frac{D \tilde{\rho}}{Dt}=\frac{\partial \tilde{\rho}}{\partial t}+\tilde{u}_j\frac{\partial \tilde{\rho}}{\partial x_j}= 0[/itex] - (2.3)
+
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- From equation 2.2 it follows that for incompressible flows, + From equation 2 it follows that for incompressible flows, - +
:$:[itex] \frac{\partial \tilde{u}_j}{\partial x_j}= 0$ \frac{\partial \tilde{u}_j}{\partial x_j}= 0[/itex] - (2.4)
+
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The viscous stresses(the stress minus the mean normal stress) are represented by the tensor$\tilde{T}_{ij}^{(v)}$. From its definition,$\tilde{T}_{kk}^{(v)}$=0. In many flows of interest, the fluid behaves as a Newtonian fluid in which the viscous stress can be related to the fluid motion by a constitutive relation of the form. The viscous stresses(the stress minus the mean normal stress) are represented by the tensor$\tilde{T}_{ij}^{(v)}$. From its definition,$\tilde{T}_{kk}^{(v)}$=0. In many flows of interest, the fluid behaves as a Newtonian fluid in which the viscous stress can be related to the fluid motion by a constitutive relation of the form. - +
- $\tilde{T}_{ij}^{(v)}= 2\mu[\tilde{s}_{ij}-\frac{1}{3}\tilde{s}_{kk}\delta_{ij}]$ + :$\tilde{T}_{ij}^{(v)}= 2\mu[\tilde{s}_{ij}-\frac{1}{3}\tilde{s}_{kk}\delta_{ij}]$ - (2.5)
+
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The viscosity, $\mu$, is a property of the fluid that can be measured in an independent experiment. $\tilde s_{ij}$ is the instantaneous strain rate tensor defined by The viscosity, $\mu$, is a property of the fluid that can be measured in an independent experiment. $\tilde s_{ij}$ is the instantaneous strain rate tensor defined by - +
- $\tilde{s}_{ij}= \frac{1}{2}\left[\frac{\partial \tilde u_i}{\partial x_j}+\frac{\partial \tilde u_j}{\partial x_i}\right]$ + :$\tilde{s}_{ij}= \frac{1}{2}\left[\frac{\partial \tilde u_i}{\partial x_j}+\frac{\partial \tilde u_j}{\partial x_i}\right]$ - (2.6)
+
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From its definition, $\tilde s_{kk}=\frac{\partial \tilde u_k}{\partial x_k}$. If the flow is incompressible, $\tilde s_{kk}=0$ and the Newtonian constitutive equation reduces to From its definition, $\tilde s_{kk}=\frac{\partial \tilde u_k}{\partial x_k}$. If the flow is incompressible, $\tilde s_{kk}=0$ and the Newtonian constitutive equation reduces to - + +
- $\tilde{T}_{ij}^{(v)}= 2\mu\tilde{s}_{ij}$ + :$\tilde{T}_{ij}^{(v)}= 2\mu\tilde{s}_{ij}$ - (2.7)
+
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Throughout this material, unless explicitly stated otherwise, the density $\tilde\rho=\rho$ and the viscosity $\mu$ will be assumed constant. With these assumptions, the instantaneous momentum equations for a Newtonian Fluid reduce to: Throughout this material, unless explicitly stated otherwise, the density $\tilde\rho=\rho$ and the viscosity $\mu$ will be assumed constant. With these assumptions, the instantaneous momentum equations for a Newtonian Fluid reduce to: - +
:$:[itex] \left[\frac{\partial \tilde{u}_i}{\partial t}+\tilde{u}_j\frac{\partial \tilde{u}_i}{\partial x_j}\right] = -\frac {1}{\tilde\rho}\frac{\partial \tilde{p}}{\partial x_i}+\nu\frac{\partial^2 {\tilde{u}_i}}{\partial x_j^2}$ \left[\frac{\partial \tilde{u}_i}{\partial t}+\tilde{u}_j\frac{\partial \tilde{u}_i}{\partial x_j}\right] = -\frac {1}{\tilde\rho}\frac{\partial \tilde{p}}{\partial x_i}+\nu\frac{\partial^2 {\tilde{u}_i}}{\partial x_j^2}[/itex] - (2.8)
+
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where the kinematic viscosity, $\nu$, has been defined as: where the kinematic viscosity, $\nu$, has been defined as: - + +
- $\nu\equiv\frac{\mu}{\rho}$ + :$\nu\equiv\frac{\mu}{\rho}$ - (2.9)
+
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Note that since the density is assumed constant, the tilde is no longer necessary. Note that since the density is assumed constant, the tilde is no longer necessary. Line 78: Line 78: Sometimes it will be more instructive and convenient to not explicitly include incompressibilty in the stress term, but to refer to the incompressible momentum equation in the following form: Sometimes it will be more instructive and convenient to not explicitly include incompressibilty in the stress term, but to refer to the incompressible momentum equation in the following form: - +
:$:[itex] \rho\left[\frac{\partial \tilde{u}_i}{\partial t}+\tilde{u}_j\frac{\partial \tilde{u}_i}{\partial x_j}\right] = -\frac{\partial \tilde{p}}{\partial x_i}+\frac{\partial \tilde{T}_{ij}^{(v)}}{\partial x_j}$ \rho\left[\frac{\partial \tilde{u}_i}{\partial t}+\tilde{u}_j\frac{\partial \tilde{u}_i}{\partial x_j}\right] = -\frac{\partial \tilde{p}}{\partial x_i}+\frac{\partial \tilde{T}_{ij}^{(v)}}{\partial x_j}[/itex] - (2.10)
+
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This form has the advantage that it is easier to keep track of the exact role of the viscous stresses. This form has the advantage that it is easier to keep track of the exact role of the viscous stresses. - === Equations for the average velocity === + == Equations for the average velocity == - + Although laminar solutions to the equations often exist that are consistent with the boundary conditions, perturbations to these solutions(sometimes even infinitesimal) can cause them to become turbulent. To see how this can happen, it is convenient to analyze the flow in two parts, a mean(or average) component and a fluctuating component. Thus the instantaneous velocity and stresses can be written as: Although laminar solutions to the equations often exist that are consistent with the boundary conditions, perturbations to these solutions(sometimes even infinitesimal) can cause them to become turbulent. To see how this can happen, it is convenient to analyze the flow in two parts, a mean(or average) component and a fluctuating component. Thus the instantaneous velocity and stresses can be written as: - +
:$:[itex] Line 102: Line 101: \tilde T_{ij}^{(v)}=T_{ij}^{(v)}+\tau_{ij}^{(v)} \tilde T_{ij}^{(v)}=T_{ij}^{(v)}+\tau_{ij}^{(v)}$ [/itex] - (2.11)
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+ - Where $U_i$, $P$ and $T_{ij}^{(v)}$ represent the mean motion, and $u_i$, $p$ and $\tau_{ij}^{(v)}$ the fluctuating motions. This technique for decomposing the instantaneous motion is referred to as the '''''Reynolds decomposition.''''' Note that if the averages are defined as ensemble means, they are, in general, time-dependent. For the remainder of this material unless other wise stated, the density will be assumed constant so$\tilde{\rho}\equiv\rho$,and its fluctuation is zero. Where $U_i$, $P$ and $T_{ij}^{(v)}$ represent the mean motion, and $u_i$, $p$ and $\tau_{ij}^{(v)}$ the fluctuating motions. This technique for decomposing the instantaneous motion is referred to as the '''''Reynolds decomposition.''''' Note that if the averages are defined as ensemble means, they are, in general, time-dependent. For the remainder of this material unless other wise stated, the density will be assumed constant so$\tilde{\rho}\equiv\rho$,and its fluctuation is zero. - Substitution of equations 2.11 into equations 2.10 yields + Substitution of equations 11 into equations 10 yields - +
:$:[itex] \rho\left[\frac{\partial (U_i+u_i)}{\partial t}+(U_j+u_j)\frac{\partial (U_i+u_i)}{\partial x_j}\right] = -\frac{\partial (P+p)}{\partial x_i}+\frac{\partial (T_{ij}^{(v)}+\tau_{ij}^{(v)})}{\partial x_j}$ \rho\left[\frac{\partial (U_i+u_i)}{\partial t}+(U_j+u_j)\frac{\partial (U_i+u_i)}{\partial x_j}\right] = -\frac{\partial (P+p)}{\partial x_i}+\frac{\partial (T_{ij}^{(v)}+\tau_{ij}^{(v)})}{\partial x_j}[/itex] - (2.12)
+
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This equation can now be averaged to yield an equation expressing momentum conservation for the averaged motion. Note that the operations of averaging and differentiation commute; i.e., the average of a derivative is the same as the derivative of the average. Also the average of a fluctuating quantity is zero. Thus the equation for the averaged motion reduces to: This equation can now be averaged to yield an equation expressing momentum conservation for the averaged motion. Note that the operations of averaging and differentiation commute; i.e., the average of a derivative is the same as the derivative of the average. Also the average of a fluctuating quantity is zero. Thus the equation for the averaged motion reduces to: - +
:$:[itex] \rho\left[\frac{\partial U_i}{\partial t}+U_j\frac{\partial U_i}{\partial x_j}\right] = -\frac{\partial P}{\partial x_i}+\frac{\partial T_{ij}^{(v)}}{\partial x_j}-\rho\left \langle u_j\frac{\partial u_i }{\partial x_j} \right \rangle$ \rho\left[\frac{\partial U_i}{\partial t}+U_j\frac{\partial U_i}{\partial x_j}\right] = -\frac{\partial P}{\partial x_i}+\frac{\partial T_{ij}^{(v)}}{\partial x_j}-\rho\left \langle u_j\frac{\partial u_i }{\partial x_j} \right \rangle[/itex] - (2.13)
+
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where the remaining fluctuating product term has been moved to the right hand side of the equation. Whether or not the last term is zero like the other fluctuating term depends on the correlation of the terms in the product. In general, these correlations are not zero. where the remaining fluctuating product term has been moved to the right hand side of the equation. Whether or not the last term is zero like the other fluctuating term depends on the correlation of the terms in the product. In general, these correlations are not zero. - The mass conservation equation can be similarly decomposed. In incompressible form, substitution of equations 2.11 into equation 2.4 yields: + The mass conservation equation can be similarly decomposed. In incompressible form, substitution of equations 11 into equation 4 yields: - +
:$:[itex] \frac{\partial (U_j+u_j)}{\partial x_j}=0$ \frac{\partial (U_j+u_j)}{\partial x_j}=0[/itex] - (2.14)
+
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of which average is of which average is - +
:$:[itex] \frac{\partial U_j}{\partial x_j}=0$ \frac{\partial U_j}{\partial x_j}=0[/itex] - (2.15)
+
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- It is clear from the equation 2.15 that the averaged motion satisfies the same form of the mass conservation equation as does the instantaneous motion at least for incompressible flows. How much simpler the turbulence problem would be if the same were true for the momentum! Unfortunately, as is easily seen from the equation 2.13, such is not the case. + It is clear from equation 15 that the averaged motion satisfies the same form of the mass conservation equation as does the instantaneous motion at least for incompressible flows. How much simpler the turbulence problem would be if the same were true for the momentum! Unfortunately, as is easily seen from equation 13, such is not the case. - Equation 2.15 can be subtracted from equation 2.14 to yield an equation for instantaneous motion alone; i.e, + Equation 15 can be subtracted from equation 14 to yield an equation for instantaneous motion alone; i.e, - + +
:$:[itex] \frac{\partial u_j}{\partial x_j}=0$ \frac{\partial u_j}{\partial x_j}=0[/itex] - (2.16)
+
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Again, like the mean, the form of the original instantaneous equation is seen to be preserved. The reason, of course, is obvious: the continuity equation is linear. The momentum equation  , on the other hand, is not; hence the difference. Again, like the mean, the form of the original instantaneous equation is seen to be preserved. The reason, of course, is obvious: the continuity equation is linear. The momentum equation  , on the other hand, is not; hence the difference. - Equation 2.16 can be used to rewrite the last term in equation 2.13 for the mean momentum. Multiplying equation 2.16 by $u_i$ and averaging yields: + Equation 16 can be used to rewrite the last term in equation 13 for the mean momentum. Multiplying equation 16 by $u_i$ and averaging yields: - +
:$:[itex] \left \langle u_i\frac{\partial u_j}{\partial x_j}\right \rangle=0$ \left \langle u_i\frac{\partial u_j}{\partial x_j}\right \rangle=0[/itex] - (2.17)
+
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This can be added to:$\left \langle u_j\frac{\partial u_i}{\partial x_j}\right \rangle$ to obtain: This can be added to:$\left \langle u_j\frac{\partial u_i}{\partial x_j}\right \rangle$ to obtain: - +
:$:[itex] \left \langle u_j\frac{\partial u_i}{\partial x_j}\right \rangle +0=\left \langle u_j\frac{\partial u_i}{\partial x_j}\right \rangle+ \left \langle u_i\frac{\partial u_j}{\partial x_j}\right \rangle =\frac{ \partial}{\partial x_j}{\left \langle u_iu_j\right \rangle}$ \left \langle u_j\frac{\partial u_i}{\partial x_j}\right \rangle +0=\left \langle u_j\frac{\partial u_i}{\partial x_j}\right \rangle+ \left \langle u_i\frac{\partial u_j}{\partial x_j}\right \rangle =\frac{ \partial}{\partial x_j}{\left \langle u_iu_j\right \rangle} [/itex] - (2.18)
+
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Where again the fact that arithmetic and averaging operations commute has been used. Where again the fact that arithmetic and averaging operations commute has been used. - The equation for the averaged momentum, equation 2.13 can now be rewritten as: + The equation for the averaged momentum, equation 13 can now be rewritten as: - +
:$:[itex] \rho\left[\frac{\partial U_i}{\partial t}+U_j\frac{\partial U_i}{\partial x_j}\right] = -\frac{\partial P}{\partial x_i}+\frac{\partial T_{ij}^{(v)}}{\partial x_j}-\frac{ \partial}{\partial x_j}{(\rho\left \langle u_iu_j\right \rangle)}$ \rho\left[\frac{\partial U_i}{\partial t}+U_j\frac{\partial U_i}{\partial x_j}\right] = -\frac{\partial P}{\partial x_i}+\frac{\partial T_{ij}^{(v)}}{\partial x_j}-\frac{ \partial}{\partial x_j}{(\rho\left \langle u_iu_j\right \rangle)}[/itex] - (2.19)
+
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The last two terms on the right hand side are both divergence terms and can be combined; the result is: The last two terms on the right hand side are both divergence terms and can be combined; the result is: - +
:$:[itex] \rho\left[\frac{\partial U_i}{\partial t}+U_j\frac{\partial U_i}{\partial x_j}\right] = -\frac{\partial P}{\partial x_i}+\frac{\partial }{\partial x_j}[T_{ij}^{(v)}-{\rho\left \langle u_iu_j\right \rangle}]$ \rho\left[\frac{\partial U_i}{\partial t}+U_j\frac{\partial U_i}{\partial x_j}\right] = -\frac{\partial P}{\partial x_i}+\frac{\partial }{\partial x_j}[T_{ij}^{(v)}-{\rho\left \langle u_iu_j\right \rangle}][/itex] - (2.20)
+
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:$:[itex] \rho\left[\frac{\partial u_{i}}{\partial t} + U_{j}\frac{\partial u_{i}}{\partial x_{j}} \right] = - \frac{\partial p}{\partial x_{i}} + \frac{\partial\tau^{(v)}_{ij}}{\partial x_{j}} - \rho \left[ u_{j}\frac{\partial U_{i}}{\partial x_{j}} \right] - \left\{ u_{j} \frac{\partial u_{i}}{ \partial x_{j}} - \rho \left\langle u_{j} \frac{\partial u_{i}}{\partial x_{j}} \right\rangle \right\} \rho\left[\frac{\partial u_{i}}{\partial t} + U_{j}\frac{\partial u_{i}}{\partial x_{j}} \right] = - \frac{\partial p}{\partial x_{i}} + \frac{\partial\tau^{(v)}_{ij}}{\partial x_{j}} - \rho \left[ u_{j}\frac{\partial U_{i}}{\partial x_{j}} \right] - \left\{ u_{j} \frac{\partial u_{i}}{ \partial x_{j}} - \rho \left\langle u_{j} \frac{\partial u_{i}}{\partial x_{j}} \right\rangle \right\}$ [/itex] - (3.21)
+
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+ :$+ y_{n+1}= r y_{n} \left(1- y_{n} \right) +$ + (22)
+ + where $n= 1,2..., 0 0$. The idea is that you pick any value for $y_{1}$, use the equation to find $y_{2}$, then insert that value on the right-hand side to find $y_{3}$, and just continue the process as long as you like. Make sure you note any dependece of the final result on the initial value for $y$. + + *First notice what happens if you linearize this equation by disregarding the term in parentheses; i.e., comsider the simpler equation $y_{1+1} = r y_{n}$. My guess is that you won't find this too eciting - unless, of course, you are one of those rare individuals who likes watching grass grow. + + * Now consider the full equation and note what happens for $r<3$ , and especially what happens for very small values of $r$. Run as many iterations as necessary to make sure your answer has converged. Do NOT try to take short-cuts by programming all the steps at once. Do them one at time so you can see what is happening. It will be much easier this way in the long run. + + * Now research carefully what happens when $r=3.1, 3.5,$ and $3.8$. Can you recognize any patterns. + + * Vary $r$ between 3 and 4 to see if you can find the boundaries for what you are observing. + + * Now try values of $r>4$. How do you explain this + + '''Example 2:''' Stretching of a simple vortex. + + Imagine a simple vortex filament that looks about like a strand of spaghetti. Now suppose it is in otherwise steady inviscid incompressible flow. Use the vorticity equation to examine the following: + + * Examine first what happens to it in two-dimensional velocity field. Note particularly whether any new vorticity can be produced; i.e., can the material derivative of the vorticity ever be greater than zero? (Hint: look at the $\omega_{j} \partial u_{i}/ \partial x_{j} +$ - term) + + * Now consider the same vortex filament in a three-dimensional flow. Note particularly the various ways new vorticity can be produced - if you have some to start with! Does all this have anything to do with non-linearities?

## Revision as of 18:05, 25 June 2007

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## Equations governing instantaneous fluid motion

All fluid motions, whether turbulent or not, are governed by the dynamical equations for a fluid. These can be written using Cartesian tensor notation as:

 $\rho\left[\frac{\partial \tilde{u}_i}{\partial t}+\tilde{u}_j\frac{\partial \tilde{u}_i}{\partial x_j}\right] = -\frac{\partial \tilde{p}}{\partial x_i}+\frac{\partial \tilde{T}_{ij}^{(v)}}{\partial x_j}$ (1)
 $\left[\frac{\partial \tilde{\rho}}{\partial t}+\tilde{u}_j\frac{\partial \tilde{\rho}}{\partial x_j}\right]+ \tilde{\rho}\frac{\partial \tilde{u}_j}{\partial x_j}= 0$ (2)

where $\tilde{u_i}(\vec{x},t)$ represents the i-the component of the fluid velocity at a point in space,$[\vec{x}]_i=x_i$, and time,t. Also $\tilde{p}(\vec{x},t)$ represents the static pressure, $\tilde{T}_{ij}^{(v)}(\vec{x},t)$, the viscous(or deviatoric) stresses, and $\tilde\rho$ the fluid density. The tilde over the symbol indicates that an instantaneous quantity is being considered. Also the einstein summation convention has been employed.

In equation 1, the subscript $i$ is a free index which can take on the values 1,2 and 3. Thus equation 1 is in reality three separate equations. These three equations are just Newton's second law written for a continuum in a spatial(or Eulerian) reference frame. Together they relate the rate of change of momentum per unit mass $(\rho{u_i})$,a vector quantity, to the contact and body forces.

Equation 2 is the equation for mass conservation in the absence of sources(or sinks) of mass. Almost all flows considered in this material will be incompressible, which implies that derivative of the density following the fluid material[the term in brackets] is zero. Thus for incompressible flows, the mass conservation equation reduces to:

 $\frac{D \tilde{\rho}}{Dt}=\frac{\partial \tilde{\rho}}{\partial t}+\tilde{u}_j\frac{\partial \tilde{\rho}}{\partial x_j}= 0$ (3)

From equation 2 it follows that for incompressible flows,

 $\frac{\partial \tilde{u}_j}{\partial x_j}= 0$ (4)

The viscous stresses(the stress minus the mean normal stress) are represented by the tensor$\tilde{T}_{ij}^{(v)}$. From its definition,$\tilde{T}_{kk}^{(v)}$=0. In many flows of interest, the fluid behaves as a Newtonian fluid in which the viscous stress can be related to the fluid motion by a constitutive relation of the form.

 $\tilde{T}_{ij}^{(v)}= 2\mu[\tilde{s}_{ij}-\frac{1}{3}\tilde{s}_{kk}\delta_{ij}]$ (5)

The viscosity, $\mu$, is a property of the fluid that can be measured in an independent experiment. $\tilde s_{ij}$ is the instantaneous strain rate tensor defined by

 $\tilde{s}_{ij}= \frac{1}{2}\left[\frac{\partial \tilde u_i}{\partial x_j}+\frac{\partial \tilde u_j}{\partial x_i}\right]$ (6)

From its definition, $\tilde s_{kk}=\frac{\partial \tilde u_k}{\partial x_k}$. If the flow is incompressible, $\tilde s_{kk}=0$ and the Newtonian constitutive equation reduces to

 $\tilde{T}_{ij}^{(v)}= 2\mu\tilde{s}_{ij}$ (7)

Throughout this material, unless explicitly stated otherwise, the density $\tilde\rho=\rho$ and the viscosity $\mu$ will be assumed constant. With these assumptions, the instantaneous momentum equations for a Newtonian Fluid reduce to:

 $\left[\frac{\partial \tilde{u}_i}{\partial t}+\tilde{u}_j\frac{\partial \tilde{u}_i}{\partial x_j}\right] = -\frac {1}{\tilde\rho}\frac{\partial \tilde{p}}{\partial x_i}+\nu\frac{\partial^2 {\tilde{u}_i}}{\partial x_j^2}$ (8)

where the kinematic viscosity, $\nu$, has been defined as:

 $\nu\equiv\frac{\mu}{\rho}$ (9)

Note that since the density is assumed constant, the tilde is no longer necessary.

Sometimes it will be more instructive and convenient to not explicitly include incompressibilty in the stress term, but to refer to the incompressible momentum equation in the following form:

 $\rho\left[\frac{\partial \tilde{u}_i}{\partial t}+\tilde{u}_j\frac{\partial \tilde{u}_i}{\partial x_j}\right] = -\frac{\partial \tilde{p}}{\partial x_i}+\frac{\partial \tilde{T}_{ij}^{(v)}}{\partial x_j}$ (10)

This form has the advantage that it is easier to keep track of the exact role of the viscous stresses.

## Equations for the average velocity

Although laminar solutions to the equations often exist that are consistent with the boundary conditions, perturbations to these solutions(sometimes even infinitesimal) can cause them to become turbulent. To see how this can happen, it is convenient to analyze the flow in two parts, a mean(or average) component and a fluctuating component. Thus the instantaneous velocity and stresses can be written as:

 $\tilde {u}_i=U_i+u_i$ $\tilde p=P+p$ $\tilde T_{ij}^{(v)}=T_{ij}^{(v)}+\tau_{ij}^{(v)}$ (11)

Where $U_i$, $P$ and $T_{ij}^{(v)}$ represent the mean motion, and $u_i$, $p$ and $\tau_{ij}^{(v)}$ the fluctuating motions. This technique for decomposing the instantaneous motion is referred to as the Reynolds decomposition. Note that if the averages are defined as ensemble means, they are, in general, time-dependent. For the remainder of this material unless other wise stated, the density will be assumed constant so$\tilde{\rho}\equiv\rho$,and its fluctuation is zero.

Substitution of equations 11 into equations 10 yields

 $\rho\left[\frac{\partial (U_i+u_i)}{\partial t}+(U_j+u_j)\frac{\partial (U_i+u_i)}{\partial x_j}\right] = -\frac{\partial (P+p)}{\partial x_i}+\frac{\partial (T_{ij}^{(v)}+\tau_{ij}^{(v)})}{\partial x_j}$ (12)

This equation can now be averaged to yield an equation expressing momentum conservation for the averaged motion. Note that the operations of averaging and differentiation commute; i.e., the average of a derivative is the same as the derivative of the average. Also the average of a fluctuating quantity is zero. Thus the equation for the averaged motion reduces to:

 $\rho\left[\frac{\partial U_i}{\partial t}+U_j\frac{\partial U_i}{\partial x_j}\right] = -\frac{\partial P}{\partial x_i}+\frac{\partial T_{ij}^{(v)}}{\partial x_j}-\rho\left \langle u_j\frac{\partial u_i }{\partial x_j} \right \rangle$ (13)

where the remaining fluctuating product term has been moved to the right hand side of the equation. Whether or not the last term is zero like the other fluctuating term depends on the correlation of the terms in the product. In general, these correlations are not zero.

The mass conservation equation can be similarly decomposed. In incompressible form, substitution of equations 11 into equation 4 yields:

 $\frac{\partial (U_j+u_j)}{\partial x_j}=0$ (14)

of which average is

 $\frac{\partial U_j}{\partial x_j}=0$ (15)

It is clear from equation 15 that the averaged motion satisfies the same form of the mass conservation equation as does the instantaneous motion at least for incompressible flows. How much simpler the turbulence problem would be if the same were true for the momentum! Unfortunately, as is easily seen from equation 13, such is not the case.

Equation 15 can be subtracted from equation 14 to yield an equation for instantaneous motion alone; i.e,

 $\frac{\partial u_j}{\partial x_j}=0$ (16)

Again, like the mean, the form of the original instantaneous equation is seen to be preserved. The reason, of course, is obvious: the continuity equation is linear. The momentum equation , on the other hand, is not; hence the difference.

Equation 16 can be used to rewrite the last term in equation 13 for the mean momentum. Multiplying equation 16 by $u_i$ and averaging yields:

 $\left \langle u_i\frac{\partial u_j}{\partial x_j}\right \rangle=0$ (17)

This can be added to:$\left \langle u_j\frac{\partial u_i}{\partial x_j}\right \rangle$ to obtain:

 $\left \langle u_j\frac{\partial u_i}{\partial x_j}\right \rangle +0=\left \langle u_j\frac{\partial u_i}{\partial x_j}\right \rangle+ \left \langle u_i\frac{\partial u_j}{\partial x_j}\right \rangle =\frac{ \partial}{\partial x_j}{\left \langle u_iu_j\right \rangle}$ (18)

Where again the fact that arithmetic and averaging operations commute has been used.

The equation for the averaged momentum, equation 13 can now be rewritten as:

 $\rho\left[\frac{\partial U_i}{\partial t}+U_j\frac{\partial U_i}{\partial x_j}\right] = -\frac{\partial P}{\partial x_i}+\frac{\partial T_{ij}^{(v)}}{\partial x_j}-\frac{ \partial}{\partial x_j}{(\rho\left \langle u_iu_j\right \rangle)}$ (19)

The last two terms on the right hand side are both divergence terms and can be combined; the result is:

 $\rho\left[\frac{\partial U_i}{\partial t}+U_j\frac{\partial U_i}{\partial x_j}\right] = -\frac{\partial P}{\partial x_i}+\frac{\partial }{\partial x_j}[T_{ij}^{(v)}-{\rho\left \langle u_iu_j\right \rangle}]$ (20)

Now the terms in square brackets on the right have the dimensions of stress. The first term is, in fact , the viscous stress. The second term, on the other hand, is not a stress at all, but simply a re-worked version of the fluctuating contribution to the non-linear acceleration terms. The fact that it can be written this way, however, indicates that at least as far as the motion is concerned, it acts as though it were a stress- hence its name, the Reynolds stress. In the succeeding sections the consequences of this difference will be examined.

## The turbulence problem

It is the appearance of the Reynolds stress which makes the turbulence problem so difficult - at least from the engineers perspective. Even though we can pretend it is a stress, the physics which give rise to it are very different from the viscous stress. The viscous stress can be related directly to the other flow properties by constitutive equations, which in turn depend only on the properties of the fluid (as in equation 5 for a Newtonian fluid). The reason this works is that when we make such closure approximations for a fluid, we are averaging over characteristic length and time scales much smaller than those of the flows we are interested in. Yet at the same time, these scales are much larger than the molecular length and time scales which characterize the molecular interactions that are actually causing the momentum transfer. (This is what the continuum approximation is all about).

The Reynolds stress, on the other hand, arises from the flow itself! Worse, the scales of the fluctuating motion which give rise to it are the scales we are interested in. This means that the closure ideas which worked so well for the viscous stress, should not be expected to work too well for the Reynolds stress. And as we shall see, they do not.

This leaves us in a terrible position. Physics and engineering are all about writing equations(and boundary conditions) so we can solve them to make predictions. We don't want to have a build prototype airplanes first to see if they will they fall out of the sky. Instead we want to be able to analyze our designs before building, to save the cost in money and lives if our ideas are wrong. The same is true for dams and bridges and tunnels and automobiles. If we had confidence in our turbulence models, we could even build huge one-offs and expect them to work the first time. Unfortunately, even though turbulence models have improved to the point where we can use them in design, we still cannot trust them enough to eliminate expensive wind tunnel and model studies. And recent history is full of examples to prove this.

The turbulence problem (from the engineers perspective) is then three-fold:

• The averaged equations are not closed. Count the number of unknowns in equation 20 above. Then count the number of equations. Even with the continuity equation we have atleast six equations too few.
• The simple ideas to provide the extra equations usually do not work. And even when we can fix them up for a particular class of flows (like the flow in a pipe, for example), they will most likely not be able to predict what happens in even a slightly different environment (like a bend).
• Even the last resort of compiling engineering tables for design handbooks carries substantial risk. This is the last resort for the engineer who lacks equations or cannot trust them. Even when based on a wealth of experience, they require expensive model testing to see if they can be extrapolated to a particular situation. Often they cannot, so infinitely clever is Mother Nature in creating turbulence that is unique to a particular set of boundary conditions.

Turbulent flows are indeed flows!. And that is the problem.

## Origins of turbulence

Turbulent flows can often be observed to arise from laminar flows as the Reynolds number, (or someother relevant parameter) is increased. This happens because small disturbances to the flow are no longer damped by the flow, but begin to grow by taking energy from the original laminar flow. This natural process is easily visualized by watching the simple stream of water from a faucet (or even a pitcher). Turn the flow on very slow (or pour) so the stream is very smooth initially, at least near the outlet. Now slowly open the faucet (or pour faster) abd observe what happens, first far away, then closer to the spout. The surface begins to exhibit waves or ripples which appear to grow downstream . In fact, they are growing by extracting energy from the primary flow. Eventually they grow enough that the flow breaks into drops. These are capillary instabilities arisiing from surface tension, but regardless of the type of instability, the idea is the same -small (or infinitesimal ) disturbances have grown to disrupt the serenity (and simplicity) of laminar flow.

The manner in which the instabilities grow naturally in a flow can be examined using the equations we have already developed above. We derived them by decomposing the motion into a mean and fluctuating part. But suppose instead we had decomposed the motion into a base flow part (the initial laminar part) and into a disturbance which represents a fluctuating part superimposed on the base flow. The result of substituting such a decomposition into the full Navier-Stokes equations and averaging is precisely that given by equations (13) and (15). But the very important difference is the additional restriction that what was previously identified as the mean (or averaged ) motion is now also the base or laminar flow.

Now if the base flow is really laminar flow (which it must be by our original hypothesis), then our averaged equations governing the base flow must yield the same mean flow as the original laminar flow on which the disturbances was superimposed. But this can happen only if these new averaged equations reduce to exactly the same lamiane flow equations without any evidence of a disturbance. Clearly from equations 13 and 15, this can happen only if all the Reynolds stress terms vanish identically! Obviously this requires that the disturbances be infintesimal so the extra terms can be neglected - hence our interest in infinitesimal disturbances.

So we hypothesized a base flow which was laminar and showed that it is unchanged even with the imposition of infintesimal disturbances on it - but only as long as the disturbances remain infinitesimal! What happens if the disturbance starts to grow? Obviously before we conclude that all laminar flows are laminar forever we better investigate whether or not these infinitesimal disturbances can grow to finite size. To do this we need an equation for the fluctuation itself.

An equation for the fluctuation (which might be an imposed disturbance) can be obtained by subtracting the equation for the mean (or base) flow from that for the instantaneous motion. We already did this for the continuity equation. Now we will do it for the momentum equation. Subtracting equation 13 from equation 11 yields an equation for the fluctuation as:

 $\rho\left[\frac{\partial u_{i}}{\partial t} + U_{j}\frac{\partial u_{i}}{\partial x_{j}} \right] = - \frac{\partial p}{\partial x_{i}} + \frac{\partial\tau^{(v)}_{ij}}{\partial x_{j}} - \rho \left[ u_{j}\frac{\partial U_{i}}{\partial x_{j}} \right] - \left\{ u_{j} \frac{\partial u_{i}}{ \partial x_{j}} - \rho \left\langle u_{j} \frac{\partial u_{i}}{\partial x_{j}} \right\rangle \right\}$ (21)

It is very important to note the type and character of the terms in this equation. First note that the left-hand side is the derivative of the fluctuating velocity following the mean motion. This is exactly like the term which appears on the left-hand side of the equation for the mean velocity, equation 13. The first two terms on the right-hand side are also like those in the mean motion, and represent the fluctuating pressure gradient and the fluctuating viscous stresses. The third term on the right-hand is is new, and will be seen later to represent the primary means by which fluctuations (and turbulence as well!) extract energy from the mean flow, the so-called production terms. The last term is quadratic in the fluctuating velocity, unlike all the otherwhich are linear. Note that all of the terms vanish identically if the equation is averaged, the last because its mean is subtructed from it.

Now we want to examine what happens if the disturbance is small. In the limit as the amplitude of the disturbance (or fluctuation) is infinitesmal, the bracketed term in the equationfor the fluctuation vanishes (since it involves productsof infinitesimals) , and the remaining equation is linear in the disturbance. The study of whether or not such infinitesmal disturbances can grow is called Linear Fluid Dynamic Stability Theory. These linearized equations are very different from those govering turbulence. Unlike the equations for disturbances of finite amplitude, the linearized equations are well-posed (or closed) since the Reynolds stress terms are gone.

The absence of the non-linear terms, however, constrains the validity of the linear analysis to only the initial stage of disturbance growth. This is because as soon as the fluctuations begin to grow, their amplitudes can no longer be assumed infinitesmal and the Reynolds stress (or more properly, the non-linear fluctuating terms) become important. As a result the base flow equations begin to be modified so that the solution to them can no longer be identical to the laminar flow (or base flow) from which it arose. Thus while linear stability theory can predict when many flows become unstable, it can say very little about transition to turbulence since this progress is highly non-linear.

## Importance of non-linearity

We saw in the preceding section that non-linearity was one of essential features of turbulence. When small disturbances grow large enough to interact with each other, we enter a whole new world of complex behavior. Most of the rules we learned for linear system do not apply. Since most of your mathematical training has been for linear equations, most of your mathematical intuition therefore will not apply either. On the other hand, you may surprise yourself by discovering how much your non-mathematical intuition already recognizes non-linear behavior and accounts for it.

Considering the following simple example. Take a long stick with one person holding each end and stand at the corner of a building. Now place the middle of against the building and let each person apply pressure in the same direction so as to bend the stick. If the applied force is small, the stick deflects (or bends) a small amount. Double the force, and the deflection is approximately doubled. Quadruple the force and the deflection is quadrupled. Now you don't need a Ph.D. in Engineering to know what is going to happen if you continue this process. The stick is going to break!

But where in the equations for the deflection of the stick is there anything that predicts this can happen? Now if you are only like engineer, you are probably thinking: he's asking a stupid question. Of course you can't continue to increase the force because you will exceed first the yield stress, then the breaking limit, and of course the stick will break.

But pretend I am the company president with nothing more than MBA. I don't know much about these things, but you have told me in the past that your computers have equations to predict everything. So I repeat: Where in the equations for the deflection of this stick does it tell me this going to happen?

The answer is very simple: There is nothing in the equations that will predict this. And the reason is also quite simple: You lost the ability to predict catasrophes like breaking when you linearized the fundamental equations - which started out as Newton's Law too. In fact, before linearization, they were exactly the same as those for a fluid, only the constitutive equation was different.

If we had NOT linearized these equations and had constituve equations that were more general, then we possibly could apply these equation right to and past the limit. The point of fracture would be a bifurcation point for the solution.

Now the good news is that for things like reasonable deflections of beams linearization work woderfully since we hope most things we build don't deflect too much. Unfortunately, as we noted above, for fluids the disturbances tend to quickly become dominated by non-linear terms. This, of course, means our linear analytical techniques are pretty useless for fluid mechanics, and especially turbulence.

But all is not lost. Just as we learned to train ourselves to anticipate when sticks break, we have to train ourselves to anticipate how non-linear fluid phenomena behave. Toward that end we will consider two simple examples: one from algebra - the logistic map, and one from fluid mechanics - simple vortex streching.

Example 1: An experiment with the logistic map.

Consider the behavior of the simple equation:

 $y_{n+1}= r y_{n} \left(1- y_{n} \right)$ (22)

where $n= 1,2..., 0 and $r > 0$. The idea is that you pick any value for $y_{1}$, use the equation to find $y_{2}$, then insert that value on the right-hand side to find $y_{3}$, and just continue the process as long as you like. Make sure you note any dependece of the final result on the initial value for $y$.

• First notice what happens if you linearize this equation by disregarding the term in parentheses; i.e., comsider the simpler equation $y_{1+1} = r y_{n}$. My guess is that you won't find this too eciting - unless, of course, you are one of those rare individuals who likes watching grass grow.
• Now consider the full equation and note what happens for $r<3$ , and especially what happens for very small values of $r$. Run as many iterations as necessary to make sure your answer has converged. Do NOT try to take short-cuts by programming all the steps at once. Do them one at time so you can see what is happening. It will be much easier this way in the long run.
• Now research carefully what happens when $r=3.1, 3.5,$ and $3.8$. Can you recognize any patterns.
• Vary $r$ between 3 and 4 to see if you can find the boundaries for what you are observing.
• Now try values of $r>4$. How do you explain this

Example 2: Stretching of a simple vortex.

Imagine a simple vortex filament that looks about like a strand of spaghetti. Now suppose it is in otherwise steady inviscid incompressible flow. Use the vorticity equation to examine the following:

• Examine first what happens to it in two-dimensional velocity field. Note particularly whether any new vorticity can be produced; i.e., can the material derivative of the vorticity ever be greater than zero? (Hint: look at the $\omega_{j} \partial u_{i}/ \partial x_{j}$ - term)
• Now consider the same vortex filament in a three-dimensional flow. Note particularly the various ways new vorticity can be produced - if you have some to start with! Does all this have anything to do with non-linearities?