text:
User talk:Egm6936.f09/Kolmogorov time scale

= Current discussion page =

= Issues to resolve (reverse chronological order) =

What to do in a wiki article for the equation renumbering to work
See this doc.

Quotations with references
Quotation with the wiki commands and :

Interesting references
Statistical Theory and Modeling for Turbulent Flows by P. A. Durbin and B. A. Pettersson Reif (Paperback - Feb. 13, 2001). amazon

An Introduction to Turbulent Flow by Jean Mathieu and Julian Scott (Paperback - June 26, 2000). amazon no reviews !

Terminologies to define
Fully-developed flows are flows in which the velocity statistics are stationary. For example in channel flows, near the entry of the channel, the velocity statistics are changing along the axis $$x$$ of the channel; this region is called the flow-development region (Pope 2000, p.265). Further away from the channel entry, where the velocity statistics are no longer a function of $$x$$ coordinate along the channel axis, but only on the transverse coordinates in the channel cross section, we have the fully-developed flow region.

Derivation of the vorticity equation
We will first derive the Navier Stokes (NS) equation from the balance of the linear momentum equation, which is

 * {| style="width:100%" border="0"

$$  \displaystyle \rho \frac {D \mathbf u}  {Dt} =  {\rm div} \boldsymbol \tau +  \mathbf b   \Longrightarrow \rho \frac {D u_j} {Dt} \mathbf e_j =  \frac {\partial \tau_{ij}} {\partial x_i} \mathbf e_j +  b_j \mathbf e_j $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

For constant property Newtonian fluid

 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol \tau =  - P   \mathbf I   + 2 \mu \boldsymbol \gamma (\mathbf u) $$ ||  (2)
 * style="width:95%" |
 * style="width:95%" |
 * }

Here, $$\displaystyle \boldsymbol \gamma $$ is the strain rate. Note about the minus sign in front of the pressure term. $$\displaystyle p>0 $$, since stress is compressive and thus negative. The stress tensor in component form can be written as

 * {| style="width:100%" border="0"

$$  \displaystyle \tau_{ij} =     -P \delta_{ij} +      \mu \left[ \frac{\partial u_i} {\partial x_j} + \frac{\partial u_j} {\partial x_i} \right ] $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

The stress tensor can be decomposed into the spherical (isotropic) part, denoted as $$\displaystyle \mathbf s $$, and the deviatoric part, $$\displaystyle \mathbf d $$ as

 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol \tau =  \mathbf s +    \mathbf d $$ (4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

and

 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf s = \bar \tau  \mathbf I $$ (5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

where $$\displaystyle \bar \tau $$ is the mean stress defined as

 * {| style="width:100%" border="0"

$$ \displaystyle \frac{1}{3}  \tau_{ii} = \frac{1}{3} Tr (\boldsymbol \tau) $$ (6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

 * {| style="width:100%" border="0"

$$ \displaystyle Tr (\boldsymbol \tau) = \tau_{ii} = -P \underbrace{\delta_{ii}}_{=3} + 2 \mu \underbrace{\frac{\partial u_i} {\partial x_i}}_{= {\rm div} \mathbf u} $$ (7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

For an incompressible flow: $$\displaystyle {\rm div} \mathbf u = \mathbf 0 $$. Then,

 * {| style="width:100%" border="0"

$$\displaystyle Tr (\boldsymbol \tau) = -3P $$ (8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle P = -\frac{1}{3} Tr (\boldsymbol \tau) $$ (9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Thus $$\displaystyle p $$ is the mean stress $$\displaystyle \bar \tau $$ with a minus sign. Hence,

 * {| style="width:100%" border="0"

$$\displaystyle \mathbf s = -P \mathbf I $$ (10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle \mathbf d = \boldsymbol \tau - \mathbf s = 2 \mu   \boldsymbol \gamma (\mathbf u) $$ (11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Body force $$\displaystyle \mathbf b $$ : Assume $$\displaystyle \mathbf b $$ is the gradient of a potential $$\displaystyle \Psi $$ (like gravitational potential), then $$\displaystyle \mathbf b = - \rho \, {\rm grad} \Psi $$. NS becomes

 * {| style="width:100%" border="0"

$$  \displaystyle \rho \frac {D \mathbf u}  {Dt} =  {\rm div} (-P \mathbf I)  + 2 \mu \, {\rm div} \boldsymbol \gamma (\mathbf u) - \rho \, {\rm grad} \Psi $$ (12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

But,

 * {| style="width:100%" border="0"

$$  \displaystyle {\rm div} (P \mathbf I) = \left ( \frac{\partial } {\partial x_i} P \, \delta_{ij} \right ) \mathbf e_j = \frac{\partial P} {\partial x_j} \mathbf e_j = {\rm grad}\, P  $$ (13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now, let’s focus on the $$\displaystyle {\rm div} \boldsymbol \gamma (\mathbf u) $$ term.

 * {| style="width:100%" border="0"

$$  \displaystyle {\rm div} \boldsymbol \gamma (\mathbf u) = \frac{\partial } {\partial x_i} \gamma_{ij} \, \mathbf e_j = \frac{1}{2} \frac{\partial } {\partial x_i} \left ( \frac{\partial u_i} {\partial x_j} + \frac{\partial u_j} {\partial x_i} \right) \mathbf e_j = \frac{1}{2} \left [ \frac{\partial } {\partial x_j} \underbrace{ \left(\frac{\partial u_i} {\partial x_i} \right) }_{ {\rm div} \mathbf u = \mathbf 0} + \frac{\partial^2 u_j} {\partial x_i \partial x_i}
 * style="width:95%" |
 * style="width:95%" |

\right] \mathbf e_j = \frac{1}{2} \nabla^2 \mathbf u

$$ (14)
 *

 * }

Therefore, for an incompressible flow and assuming constant properties

 * {| style="width:100%" border="0"

$$  \displaystyle \rho \frac {D \mathbf u}  {Dt} =  -{\rm grad} \,  p     + \mu \nabla^2 \mathbf u $$ (15)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

where the modified pressure is defined as $$\displaystyle p = P + \rho \Psi $$. We will take the curl of the NS equation to arrive to the vorticity transport equation, given as Eq.(2.60) (Pope, 2000, p.126)

 * {| style="width:100%" border="0"

$$  \displaystyle \frac{D \boldsymbol \omega}{Dt} = \boldsymbol \omega \cdot \nabla \mathbf u + \nu \nabla^2 \boldsymbol \omega $$ (16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Here, $$\displaystyle \boldsymbol \omega $$ is the vorticity, defined as $$\displaystyle \nabla \times \mathbf u = e_{ijk} \partial_i u_j \mathbf e_k $$. Note that our definition of the gradient is the transpose of Pope’s, the vorticity equation in terms of our gradient is

 * {| style="width:100%" border="0"

$$  \displaystyle \frac{D \boldsymbol \omega}{Dt} = \nabla \mathbf u \cdot \boldsymbol \omega + \nu \nabla^2 \boldsymbol \omega $$ (17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The curl of the NS becomes

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times \frac {D \mathbf u}  {Dt} =  \nabla \times \left( -\frac {1}{\rho} {\rm grad} \, p \right) +  \nabla \times \left(  \mu \nabla^2 \mathbf u \right) $$ (18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Let’s first focus on the left hand side, using the definition of the material time derivative

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times \frac {D \mathbf u}  {Dt} =  \nabla \times \left( \nabla \mathbf u \cdot \mathbf  u \right) +  \nabla \times \frac{\partial \mathbf u} {\partial t}
 * style="width:95%" |
 * style="width:95%" |

$$ (19)
 *

 * }

Here, the last terms is

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times \frac{\partial \mathbf u} {\partial t} = \frac{\partial \boldsymbol \omega} {\partial t}
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

$$ (20)
 *

 * }

Note that

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \mathbf u = \frac{\partial u_i} {\partial x_j} \mathbf e_i \otimes \mathbf e_j $$ (21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

and

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \mathbf u \cdot \mathbf u = \frac{\partial u_i} {\partial x_j} \mathbf u_j \mathbf e_i $$ (22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

So, the second term coming from the material time derivative is

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times \left( \nabla \mathbf u \cdot \mathbf  u \right) = e_{ijk} \partial_i \left[ (\partial_p u_j) u_p \right] \mathbf e_k = e_{ijk} \left[ (\partial_i \partial_p u_j) u_p + (\partial_p u_j)( \partial_i u_p) \right] \mathbf e_k $$ (23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Let

 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol \alpha = e_{ijk} (\partial_i \partial_p u_j) u_p  \mathbf e_k = u_p \partial_p \underbrace{ ( e_{ijk} \partial_i u_j) \mathbf e_k}_{\nabla \times \mathbf u = \boldsymbol \omega} $$ (24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

which gives

 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol \alpha = \nabla \boldsymbol \omega \cdot \mathbf u
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

$$ (25)
 *

 * }

Let

 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol \beta = e_{ijk} (\partial_p u_j)( \partial_i u_p) \mathbf e_k $$ (26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

To facilitate the writing, let’s define Jacobian as

 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf J = {\rm grad} \mathbf u = \frac{\partial u_j} {\partial x_i} \mathbf e_j\otimes \mathbf e_i $$ (27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Or, in component form

 * {| style="width:100%" border="0"

$$  \displaystyle J_i^j = \frac{\partial u_j} {\partial x_i} $$ (28)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Here, the superscript represents the row, and the subscript the column index. Also, let

 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf A = A_i^j \mathbf e_j\otimes \mathbf e_i = \mathbf J^2 $$ (29)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then, $$\displaystyle \boldsymbol \beta $$ can be rewritten as

 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol \beta = e_{ijk} A_i^j \mathbf e_k = \mathbf e_1 (A_2^3 - A_3^2) + \mathbf e_2 (A_3^1 - A_1^3) + \mathbf e_3 (A_1^2 - A_2^1) $$ (30)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We can write the first component of $$\displaystyle \boldsymbol \beta $$ in terms of the Jacobian as

 * {| style="width:100%" border="0"

$$  \displaystyle A_2^3 = \frac{\partial u_3} {\partial x_p} \frac{\partial u_p} {\partial x_2} = J_p^3 J_2^p = J_1^3 J_2^1 + J_2^3 J_2^2 + J_3^3 J_2^3 $$ (31)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle A_3^2 = \frac{\partial u_3} {\partial x_p} \frac{\partial u_p} {\partial x_2} = J_p^2 J_3^p = J_1^2 J_3^1 + J_2^2 J_3^2 + J_3^2 J_3^3 $$ (32)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then,

 * {| style="width:100%" border="0"

$$  \displaystyle A_2^3 - A_3^2 = \frac{\partial u_3} {\partial x_p} \frac{\partial u_p} {\partial x_2} = J_1^3 J_2^1 - J_3^1 J_1^2 + (J_2^2 + J_3^3)( J_2^3 - J_3^2) $$ (33)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Similar expressions can be written for the other components of $$\displaystyle \boldsymbol \beta $$. Now, we want to relate $$\displaystyle \boldsymbol \beta $$ to $$\displaystyle \boldsymbol \gamma = \nabla \mathbf u \cdot \boldsymbol \omega$$ to obtain the vorticity equation. Let’s first write the vorticity in ters of the Jacobian, then we will obtain $$\displaystyle \boldsymbol \gamma $$.

 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol \omega = \nabla \times \mathbf u = \underbrace{ (\partial_ 2 u_3 - \partial_3 u_2)}_{ J_2^3 - J_3^2} \mathbf e_1 + \underbrace{ (\partial_3 u_1 - \partial_1 u_3)}_{ J_3^1 J_1^3}\mathbf e_2 + \underbrace{ (\partial_1 u_2 - \partial_2 u_1) }_{ J_1^2 J_2^1} \mathbf e_3 $$ (34)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol \gamma = \nabla \mathbf u \cdot \boldsymbol \omega = \left( \frac{\partial u_i} {\partial x_j} \mathbf e_i\otimes \mathbf e_j \right) \cdot (\omega_k \mathbf e_k) = \frac{\partial u_i} {\partial x_j} \omega_j \mathbf e_i = \underbrace{ \frac{\partial u_i} {\partial x_j}}_{J_j^i} \left(e_{klj} \partial_k u_l \right) \mathbf e_i = a_i \mathbf e_i $$ (35)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The components of $$\displaystyle \boldsymbol \gamma $$ can be calculated using matrix algebra as

 * {| style="width:100%" border="0"

$$  \displaystyle \begin{vmatrix} J_1^1 & J_2^1 & J_3^1 \\ J_1^2 & J_2^2 & J_3^2 \\ J_1^3 & J_2^3 & J_3^3 \\ \end{vmatrix} \begin{vmatrix} J_2^3 - J_3^2 \\ J_3^1 - J_1^3 \\ J_1^2 - J_2^1 \\ \end{vmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$ (36)
 *

 * }

Let’s compute the first component of $$\displaystyle \boldsymbol \gamma $$, $$\displaystyle a_1 $$

 * {| style="width:100%" border="0"

$$  \displaystyle a_1 = J_1^1 (J_2^3 - J_3^2) + J_2^1 (J_3^1 - J_1^3) + J_3^1 (J_1^2 - J_2^1) $$ (37)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

But, $$\displaystyle J_1^1 = -(J_2^2+J_3^3) $$ as $$\displaystyle J_i^i = 0 = {\rm div} \mathbf u $$. Therefore,

 * {| style="width:100%" border="0"

$$  \displaystyle a_1 = -(J_2^2+J_3^3) (J_2^3 - J_3^2) + J_3^1 J_1^2 - J_2^1 J_1^3 $$ (38)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now, we can relate the components of $$\displaystyle \boldsymbol \beta $$ to those of $$\displaystyle \boldsymbol \gamma $$ as

 :{| style="width:100%" border="0" $$  \displaystyle A_2^3 - A_3^2 = - a_1 $$ (39)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Similarly

 * {| style="width:100%" border="0"

$$  \displaystyle A_3^1 - A_1^3 = - a_2 $$ (40)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

and

 * {| style="width:100%" border="0"

$$  \displaystyle A_1^2 - A_2^1 = - a_3 $$ (41)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Hence,

 :{| style="width:100%" border="0" $$  \displaystyle \boldsymbol \beta = - \boldsymbol \gamma = -\nabla \mathbf u \cdot \boldsymbol \omega $$ (42)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

whose component form is given in Eq.(35). Now, we will work with the right hand side of the NS. The first terms is the pressure one

 :{| style="width:100%" border="0" $$  \displaystyle \nabla \times (\nabla p) = \mathbf 0 $$ (43)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

and as the density is constant

 :{| style="width:100%" border="0" $$  \displaystyle \nabla \times (\frac {1}{\rho} \nabla p) = \mathbf 0 $$ (44)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Next,

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times (\nabla^2 \mathbf u) = {\rm curl} ({\rm div} ({\rm grad^T}\mathbf u)) = e_{ijk} \partial_i (\partial_p \partial_p u_j) \mathbf e_k = \partial_p \partial_p \underbrace{(e_{ijk} \partial_i u_j \mathbf e_k )}_{ \nabla \times \mathbf u = \boldsymbol \omega} = \nabla^2 \boldsymbol \omega $$ (45)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

As the viscosity and the density are constant

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times (\frac{\mu}{\rho} \nabla^2 \mathbf u) = \nu \nabla^2 \boldsymbol \omega $$ (46)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Combining all terms, we obtain the evolution equation for the vorticity for incompressible flows as

 * {| style="width:100%" border="0"

$$  \displaystyle \frac{D \boldsymbol \omega}{Dt} = \nabla \mathbf u \cdot \boldsymbol \omega + \nu \nabla^2 \boldsymbol \omega $$ (47)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

The vorticity equation in component form is

 * {| style="width:100%" border="0"

$$  \displaystyle \frac{D \omega_i}{Dt} = \frac{\partial u_i}{\partial x_j} \omega_j + \nu \nabla^2 \omega_i $$ (48)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Here, the first term is the time rate of change of vorticity and it represents the vorticity production when it is positive. $$\displaystyle \nabla \mathbf u \cdot \boldsymbol \omega $$ is the vorticity production due to vortex-line stretching and $$\displaystyle  \nu \nabla^2 \boldsymbol \omega $$ is the vorticity diffusion.

Derivation of the vorticity equation: An alternative method
The Jacobian $$\displaystyle \mathbf J $$ can be decomposed as

 :{| style="width:100%" border="0" $$  \displaystyle J_j^i = \underbrace{\frac{1}{2}(J_j^i + J_i^j)}_{S_j^i} - \underbrace{\frac{1}{2}(J_i^j - J_j^i)}_{\Omega_j^i} $$ (49)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

So,

 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf J = \mathbf S + \boldsymbol \Omega
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

$$ (50)
 *

 * }

where $$\displaystyle \mathbf S $$ is a small strain tensor and $$\displaystyle  \boldsymbol  \Omega$$ is a small rotation tensor. The vorticity can be expressed as

 :{| style="width:100%" border="0" $$  \displaystyle \omega = \nabla \times \mathbf u = e_{ijk} \underbrace{\partial_i u_j}_{J_i^j} \mathbf e_k = \underbrace{(J_2^3 - J_3^2)}_{2 \Omega_2^3} \mathbf e_1 + \underbrace{(J_3^1 - J_1^3)}_{2 \Omega_3^1} \mathbf e_2 + \underbrace{(J_1^2 - J_2^1)}_{2 \Omega_1^2} \mathbf e_3 = 2 \Omega_i^j \mathbf e_k $$ (51)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Let’s define an extraction operator, $$\displaystyle \zeta$$. When it operates on the Jacobian it extracts the pseudo vector $$\displaystyle \boldsymbol \omega $$ as

 :{| style="width:100%" border="0" $$  \displaystyle \zeta \mathbf J = e_{ijk} J_i^j \mathbf e_k = \boldsymbol \omega $$ (52)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 :{| style="width:100%" border="0" $$  \displaystyle \zeta \mathbf J = \zeta \mathbf S + \zeta \boldsymbol \Omega = \zeta \boldsymbol \Omega = \boldsymbol \omega $$ (53)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Note that as $$\displaystyle \mathbf  S$$  is symmetric $$\displaystyle  \zeta \mathbf S = \mathbf  0$$. Our goal is to find an expression for $$\displaystyle \boldsymbol \beta  $$, which is expressed as

 :{| style="width:100%" border="0" $$  \displaystyle \boldsymbol \beta = e_{ijk} \underbrace{(\partial_p u_j)}_{J_p^j} \underbrace{(\partial_i u_p)}_{J_i^p} \mathbf e_k = \zeta \mathbf J^2 $$ (54) and
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 :{| style="width:100%" border="0" $$  \displaystyle \mathbf J^2 = (\mathbf S + \boldsymbol \Omega) (\mathbf S + \boldsymbol \Omega) = \mathbf S^2 + \boldsymbol \Omega^2 + \boldsymbol \Omega \mathbf S + \mathbf S \boldsymbol \Omega $$ (55)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Here,

 :{| style="width:100%" border="0" $$  \displaystyle (\mathbf S^2)^T = \mathbf S^T \mathbf S^T = \mathbf S^2 $$ (56)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 :{| style="width:100%" border="0" $$  \displaystyle (\boldsymbol \Omega^2)^T = \boldsymbol \Omega^T \boldsymbol \Omega^T = (-\boldsymbol \Omega) (-\boldsymbol \Omega) = \boldsymbol \Omega^2 $$ (57)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

As both $$\displaystyle \mathbf S^2 $$ and $$\displaystyle \boldsymbol \Omega^2 $$ are symmetric

 :{| style="width:100%" border="0" $$  \displaystyle \zeta \mathbf S^2 = \zeta \boldsymbol \Omega^2 = \mathbf 0 $$ (58)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 :{| style="width:100%" border="0" $$  \displaystyle (\boldsymbol \Omega \mathbf S)^T = \mathbf S^T \boldsymbol \Omega^T = \mathbf S (-\boldsymbol \Omega) = - \mathbf S \boldsymbol \Omega \Longrightarrow \mathbf S \boldsymbol \Omega = - (\boldsymbol \Omega \mathbf S)^T $$ (59)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Eq.(59) implies

 :{| style="width:100%" border="0" $$  \displaystyle \boldsymbol \Omega \mathbf S + \mathbf S \boldsymbol \Omega = \boldsymbol \Omega \mathbf S - (\boldsymbol \Omega \mathbf S)^T = \mathbf S \boldsymbol \Omega - (\mathbf S \boldsymbol \Omega)^T $$ (60)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now, we can use the extraction operator

 :{| style="width:100%" border="0" $$  \displaystyle \zeta (\mathbf S \boldsymbol \Omega) = e_{ijk}S_p^j \underbrace{\Omega_i^p}_{\frac{1}{2}e_{ipq}\omega_q} \mathbf e_k = \frac{1}{2}\underbrace{e_{ijk}e_{ipq}}_{\delta_{jp}\delta_{kq}-\delta_{jq}\delta_{kp}} S_p^j \omega_q \mathbf e_k = \frac{1}{2}\left[\underbrace{S_j^j}_{=0}\omega_k - \underbrace{S_k^j}_{=S_j^k}\omega_j \right] \mathbf e_k = -\frac{1}{2} \mathbf S \cdot \boldsymbol \omega $$ (61)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 :{| style="width:100%" border="0" $$  \displaystyle \zeta (\mathbf S \boldsymbol \Omega)^T = \underbrace{ e_{ijk}}_{-e_{jik}} S_p^i \Omega_j^p \mathbf e_k = -\frac{1}{2}\underbrace{e_{jik}e_{jpq}}_{\delta_{ip}\delta_{kq}-\delta_{iq}\delta_{kp}} S_p^i \omega_q \mathbf e_k = -\frac{1}{2}\left[\underbrace{S_i^i}_{=0}\omega_k - \underbrace{S_k^i}_{=S_i^k}\omega_i \right] \mathbf e_k = \frac{1}{2} \mathbf S \cdot \boldsymbol \omega $$ (62)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Collecting the terms,

 :{| style="width:100%" border="0" $$  \displaystyle \beta = \zeta \mathbf J^2 = \zeta \left[ (\mathbf S \boldsymbol \Omega) - (\mathbf S \boldsymbol \Omega)^T \right] = -\mathbf S \cdot \Omega $$ (63)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

But, $$\displaystyle \mathbf S \cdot \boldsymbol \Omega = \underbrace{(\mathbf S + \boldsymbol \Omega)}_{\mathbf J}\cdot \boldsymbol \omega $$ as $$\displaystyle  \boldsymbol \Omega \boldsymbol \omega = \boldsymbol \omega \times \boldsymbol \omega = \mathbf 0$$. Therefore,

 * {| style="width:100%" border="0"

$$  \displaystyle \beta = -\mathbf J \cdot \omega = -\nabla \mathbf u \cdot \boldsymbol \omega $$ (64)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

This completes the proof.

Derivation of the vorticity equation using vector identities
Here, we use the vector calculus identities to derive the vorticity equation.

 Note: The gradient used in the vector calculus identities, which is $$\displaystyle \nabla {\mathbf u} = \frac{\partial u_i}{\partial x_j} {\mathbf e_j} \otimes {\mathbf e_i}$$, is the transpose of our gradient (see Gradient of a vector: Two tensor conventions), which is $$\displaystyle \nabla {\mathbf u} = \frac{\partial u_i}{\partial x_j} {\mathbf e_i} \otimes {\mathbf e_j}$$. Therefore, for all properties we use from vector calculus identities where there is a gradient of a vector field, we put a right arrow on the nabla operator, i.e. $$\displaystyle \overset{\rightarrow}{\nabla} \mathbf u = \frac{\partial u_i}{\partial x_j} {\mathbf e_j} \otimes {\mathbf e_i}$$. --EGM6936.f09.ku 17:22, 6 October 2010 (UTC)

The curl of the Navier-Stokes equation, i.e. Eq.(15), gives

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times  \frac {D \mathbf u}  {Dt} = \nabla \times \left( -\frac{1}{\rho} \nabla p \right) + \nabla \times \left(\nu \nabla^2 \mathbf u \right) $$     (65)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Let’s first focus on the left hand side

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times  \frac {D \mathbf u}  {Dt} = \nabla \times  \left(\frac{\partial \mathbf u}{\partial  t}\right) + \nabla \times \left(\mathbf u \cdot \overset{\rightarrow}{\nabla} \mathbf u \right) $$     (66)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

But, the partial derivative in Eq.(66) can be written as

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times  \left(\frac{\partial \mathbf u}{\partial  t}\right) = \frac{\partial}{\partial t} \underbrace{\left (\nabla \times \mathbf u \right)}_{ \boldsymbol \omega} $$     (67)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Also, using one of the vector dot product identities

 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{u} \cdot (\overset{\rightarrow}{\nabla} \mathbf{u}) = \frac{1}{2}\nabla \left(\mathbf u \cdot \mathbf u \right) - \mathbf u \times \underbrace{\left (\nabla \times \mathbf u \right)}_{ \boldsymbol \omega} $$     (68)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now, using one of the vector cross product identities

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times \left(\mathbf u \times \boldsymbol \omega \right) = \mathbf u \underbrace{\left( \nabla \cdot \boldsymbol \omega \right)}_{0} - \boldsymbol \omega \underbrace{\left( \nabla \cdot \mathbf u \right)}_{0} + \boldsymbol \omega \cdot \overset{\rightarrow}{\nabla} \mathbf u - \mathbf u \cdot \overset{\rightarrow}{\nabla}\boldsymbol \omega $$     (69)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The first term on the right of Eq.(69) is zero because it is the divergence of a curl, and the second term is zero because of the incompressibility condition. As the curl of the gradient of a scalar field vanishes

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times \left( \frac{1}{2}\nabla \left(\mathbf u \cdot \mathbf u \right) \right) = \mathbf 0 $$     (70)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Grouping all terms on the left hand side

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times  \frac{D \mathbf u}{Dt} = \frac{\partial \boldsymbol \omega}{\partial t} - \boldsymbol \omega \cdot \overset{\rightarrow}{\nabla} \mathbf u + \mathbf u \cdot \overset{\rightarrow}{\nabla}\boldsymbol \omega = \frac{D \boldsymbol \omega}{Dt} - \boldsymbol \omega \cdot \overset{\rightarrow}{\nabla} \mathbf u $$ (71)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The first term on the right hand side of Eq.(65) vanishes as in Eq.(70), i.e.,

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times \left( -\frac{1}{\rho} \nabla p \right) = \mathbf 0 $$     (72)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Finally, the last term in Eq.(65) is

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla \times \left(\nu \nabla^2 \mathbf u \right) = \nu \nabla^2 \left(\nabla \times \mathbf u \right) = \nu \nabla^2 \boldsymbol \omega $$     (73)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Combining the left and the right hand sides, we obtain

 * {| style="width:100%" border="0"

$$  \displaystyle \frac{D \boldsymbol \omega}{Dt} - \boldsymbol \omega \cdot \overset{\rightarrow}{\nabla} \mathbf u = \nu \nabla^2 \boldsymbol \omega $$  (74)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Meaning of vortex-line stretching
Vortex-line is a line along the vorticity vector $$\displaystyle \mathbf \omega $$ at a point $$\displaystyle x $$. In a uniform flow field, $$\displaystyle \nabla \mathbf u = \mathbf 0$$, therefore two points on an infinitesimally small vorticity line have the same velocity and the line stays undistorted. However, in a non-uniform filed the vortex line is stretched as different velocities are assumed at different points of the line. Therefore, $$\displaystyle \nabla \mathbf u \cdot \boldsymbol \omega $$ represents the vorticity production by vortex-line stretching.

Evolution equation for an infinitesimally small line element
The mapping at time $$\displaystyle t $$ between the spatial position $$\displaystyle \mathbf x $$ and the material point $$\displaystyle \mathbf X$$ is given with

 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf x = \phi (\mathbf X,t) = \phi_t (\mathbf X) $$ (75)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then,

 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf x + \mathbf h = \phi_t(\mathbf X+ \mathbf H) = \phi_t(\mathbf X) + \frac{\partial \phi(\mathbf X)}{\partial X_k} H_k + \rm HOT = \underbrace{ \phi_t(\mathbf X)}_{ \mathbf x} + \underbrace{\left [\rm grad_\mathbf X \phi_t(\mathbf X) \right] \cdot  \mathbf  H(\mathbf X)}_{\mathbf  h} $$ (76)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac{D \mathbf h(\mathbf x, t)}{Dt} = \left. \frac{d \mathbf h(\mathbf x, t)}{dt} \right|_{X \ fixed} = \left[ \rm grad_\mathbf X \underbrace{ \frac{\partial \phi (\mathbf X, t) }{\partial t } }_{\boldsymbol \mathcal V(\mathbf X, t)= \mathbf V(\mathbf x, t) } \right] $$     (77)
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \frac{\partial V}{\partial x_i} = \frac{\partial \mathcal V}{\partial X_k} \frac{\partial X_k}{\partial x_i} $$     (78)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \rm grad_\mathbf X \mathbf \mathcal V = \frac{\partial \mathcal V_i}{\partial X_j} \mathbf e_i \otimes \mathbf e_j $$     (79)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \frac{\partial \mathcal V_i}{\partial X_j} = \underbrace{\frac{\partial \mathcal V_i}{\partial x_p}}_{\frac{\partial V_i}{\partial x_j}} \underbrace{\frac{\partial x_p}{\partial X_j}}_{F_j^p} $$     (80)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \rm grad_\mathbf X \mathbf \mathcal V = \rm grad_\mathbf x \mathbf V \cdot \mathbf F $$ (81)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf \mathcal V = \mathcal V_i \mathbf E_i = \mathbf V= V_i \mathbf e_i $$     (82)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf e_i = \mathbf E_i \, \ i=1,2,3 $$     (83)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle V_i(\mathbf x,t) = \mathcal V_i (\mathbf X,t) $$     (84)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf F = F_j^i \mathbf e_i \otimes \mathbf E_j $$     (85)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla_\mathbf X \mathbf \mathcal V = \frac{\partial \mathcal V_i}{\partial X_j} \mathbf E_i \otimes \mathbf E_j $$     (86)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla_\mathbf x \mathbf V = \frac{\partial V_i}{\partial x_j} \underbrace{ \mathbf e_i }_{\mathbf E_i} \otimes \; \mathbf e_j $$     (87)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \nabla_\mathbf X \mathbf \mathcal V \cdot \mathbf H = \left(\nabla_\mathbf x \mathbf V \cdot \mathbf F \right) \cdot \mathbf H = \nabla_\mathbf x(\mathbf V) \cdot \underbrace{\left(\mathbf F \cdot \mathbf H \right) }_{\mathbf h} $$ (88)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \frac{D \mathbf h}{Dt} = \nabla_\mathbf x \mathbf V \cdot \mathbf h $$ (89)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Pope (2000, p.14, Eq.(2.16)) is

 * {| style="width:100%" border="0"

$$  \displaystyle \frac{d \mathbf s}{dt} = \mathbf s \cdot \nabla \mathbf u $$ (90)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Pope (2000, p.13, Eq.(2.12)) is

 * {| style="width:100%" border="0"

$$  \displaystyle \underbrace{\mathbf s \cdot \nabla \mathbf u}_{\nabla \mathbf u \cdot \mathbf s \ \rm our \; \rm notation} = s_i \frac{ \partial u_j}{ \partial x_i} \mathbf e_j $$     (91)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

-->

Condition for incompressible flows
Here, we obtain the condition for using incompressible flow model starting with the gas dynamical equation. By assuming that the flow velocity is very small compared to the speed of sound, and that the gradient of the velocity is bounded by some finite number, then the divergence of the velocity field is negligible, i.e., approaching the incompressibility condition; Malvern 1969, p.443, Eq.(7.2.33).

The gas dynamical equation

$$\displaystyle c^2 {\rm div \,} \mathbf u - \mathbf u \cdot {\rm grad \,} \mathbf u \cdot \mathbf u = 0$$

or

$$\displaystyle {\rm div \,} \mathbf u = \frac{\mathbf u \cdot {\rm grad \,} \mathbf u \cdot \mathbf u}{c^2}$$

thus

$$\displaystyle \parallel \mathbf u \parallel \ll c {\ \rm and \ } \parallel {\rm grad \,} \mathbf u \parallel < M \Rightarrow {\rm div \,} \mathbf u \approx 0$$

for some finite number $$\displaystyle M$$.

The downside of this approach is that the condition on the velocity field (being less than the speed of sound) had to be assumed ahead. Instead, Landau and Lifshitz 19xx and by Batchelor 1967 took different starting points to arrive at the velocity condition for the validity of using incompressible-flow model.

 Need to show how to obtain the gas dynamical equation from Navier-Stokes. What are the assumptions. This gas dynamics equation may be related to Eq.(D10) in When to use incompressibility flow model: $$v = grad \phi$$ and $$c^2 div(grad \phi) = c^2 div v$$. Eml5526.s11 13:56, 18 January 2011 (UTC)

An alternative derivation for incompressibility conditions
Here, we consider the criterion under which a fluid may be treated as incompressible. The analysis is treated on page 245 of the Fluid Mechanics book by Landau and Lifshitz.

A sound wave is an oscillatory motion of small amplitude in a compressible fluid. The pressure and the density may be written as

 * {| style="width:100%" border="0"

$$  \displaystyle p = p_0 + p^ \prime $$  (D1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

and

 * {| style="width:100%" border="0"

$$  \displaystyle \rho = \rho_0 + \rho^ \prime $$  (D2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle p_0 $$ and $$\displaystyle  \rho_0 $$ represent the undisturbed/equilibrium pressure and density and $$\displaystyle  p^ \prime $$ and $$\displaystyle  \rho^ \prime $$ are the small changes/disturbances to the equilibrium quantities. The continuity equation

 * {| style="width:100%" border="0"

$$  \displaystyle \frac {\partial \rho} {\partial t} + \rm div (\rho \mathbf v) = 0 $$  (D3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

becomes

 * {| style="width:100%" border="0"

$$  \displaystyle \frac {\partial \rho^ \prime } {\partial t} + \rho_0 \rm div \mathbf v = 0 $$  (D4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Navier Stokes equation

 * {| style="width:100%" border="0"

$$  \displaystyle \frac {\partial \mathbf v} {\partial t} + (\mathbf v \cdot \rm grad) \mathbf v = -\frac{1}{\rho} \rm grad p + \nu \nabla^2 \mathbf v $$ (D5a)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

for an inviscid fluid becomes Euler’s equation, which is

 * {| style="width:100%" border="0"

$$  \displaystyle \frac {\partial \mathbf v} {\partial t} + (\mathbf v \cdot \rm grad) \mathbf v = -\frac{1}{\rho} \rm grad p $$ (D5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

omitting higher order terms, reduces to

 * {| style="width:100%" border="0"

$$  \displaystyle \frac {\partial \mathbf v} {\partial t} + \frac{1}{\rho_0} \rm grad p^ \prime = 0 $$  (D6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For an ideal fluid (deformation of fluid elements is an isentropic process, i.e. adiabatic and reversible), the small changes in pressure and density are related to each other as

 * {| style="width:100%" border="0"

$$  \displaystyle p^ \prime = \left( \frac {\partial p} {\partial \rho_0} \right)_S \rho^ \prime $$  (D7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then, the continuity equation, i.e. Eq.(D4) becomes

 * {| style="width:100%" border="0"

$$  \displaystyle \frac {\partial p^ \prime } {\partial t} + \rho_0 \left( \frac {\partial p} {\partial \rho_0} \right)_S \rm div \mathbf v = 0 $$  (D8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For an ideal fluid, the velocity field may be expressed as a potential $$\displaystyle \mathbf v = \rm grad \phi $$. Then, the Euler’s equation, Eq.(D5) becomes

 * {| style="width:100%" border="0"

$$  \displaystyle p^ \prime = -\rho_0 \frac {\partial \phi} {\partial t} $$ (D9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Continuity, i.e. Eq.(D8) becomes
 * {| style="width:100%" border="0"

$$  \displaystyle \frac {\partial^2 \phi } {\partial t^2} - c^2 \rm div (\rm grad \phi) = 0 $$  (D10) which is called a wave equation. Here, $$\displaystyle c := \sqrt{(\partial p/\partial \rho_0)_S} $$. Now, let’s consider a wave equation in one dimension as
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \frac {\partial^2 \phi} {\partial x^2} - \frac {1} {c^2} \frac {\partial^2 \phi} {\partial t^2} = 0 $$  (D11) which has a solution of the form
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \phi = f_1(x-ct) + f_2(x+ct) $$  (D12) This form of the solution is also valid for $$\displaystyle (p^\prime, \rho^\prime, \mathbf v) $$. The function $$\displaystyle f_1 $$ implies that the value of $$\displaystyle \phi $$ at $$\displaystyle t=0 $$ is also observed after time $$\displaystyle t $$ at a distance $$\displaystyle ct $$ from the previous point. Therefore $$\displaystyle c $$ represents the speed of the sound and the function $$\displaystyle f_1 $$ is called a travelling plane wave. Note that $$\displaystyle f_2 $$ propagates in the opposite direction. For a travelling wave, as $$\displaystyle \phi = f(x-ct) $$, $$\displaystyle v_x = v = f^\prime (x-ct) $$ and $$\displaystyle p^\prime = -\rho_0 \partial \phi/ \partial t = \rho_0 c f^\prime(x-ct) $$. Therefore,
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="0"

$$  \displaystyle v = p^\prime/(\rho_0 c) $$ (D13) Substituting Eq.(D7) into Eq.(D13), with the definition of $$\displaystyle c$$ yields
 * style="width:95%" |
 * style="width:95%" |
 *

 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{\rho^\prime}{\rho_0} = \frac{v}{c} $$  (D14) The fluid may be treated as incompressible if $$\displaystyle \rho^\prime/\rho_0 << 1 $$ or
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle v/c << 1 $$     (D15)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

This last quantity is also known as Mach number. At $$\displaystyle 25^{\circ}{\rm C} $$, the speed of sound in air is $$\displaystyle 346.1 \rm {m/s} $$ and it is $$\displaystyle 1497\rm {m/s} $$  in water. Therefore, for all practical purposes, the liquids are assumed to be incompressible and for gases, one needs to check the Mach number.

For an unsteady flow, a second criteria needs to be satisfied

 * {| style="width:100%" border="0"

$$  \displaystyle \tau >> l/c $$     (D16) where $$\displaystyle \tau $$ and $$\displaystyle l$$  are the time and length scales over which the fluid goes significant changes. The above equation is obtained when the $$\displaystyle \partial \mathbf v/\partial t$$ and $$\displaystyle (1/\rho) \rm grad p$$ in Euler’s equation are comparable. The order of magnitude of each term are $$\displaystyle v/\tau \sim \Delta p/(l \rho)$$. Then, $$\displaystyle \Delta \rho \sim l\rho v /(\tau c^2)$$. When the $$\displaystyle \partial \rho / \partial t$$ and $$\displaystyle \rho \rm{div} \mathbf v$$ terms in the continuity equation are compared, $$\displaystyle \partial \rho / \partial t$$ can be neglected if $$\displaystyle \Delta \rho/\tau << \rho v/l$$, which yields Eq.(D16). Note that, for steady flow, $$\displaystyle \tau = \infty$$, and the second criteria, Eq.(D16) is always fulfilled.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

discussion on 2010.04.30
Length, Pope, p.129, (5.152)

$$\eta ={\rm Re}_{0}^{-3/4}\hat{\varepsilon }^{-1/4}r_{1/2}$$

Velocity, Pope, p.129, (5.154)

$$u_{\eta }={\rm Re}_{0}^{-1/4}\hat{\varepsilon }^{+1/4}U_{0}$$

Thus

$$\eta u_{\eta }=\operatorname{Re}_{0}^{-1}r_{1/2}U_{0}=\nu $$

Since the Reynolds number for the round jet is defined as follows (Pope 2000, p.100, Eq.(5.9)):

$$\operatorname{Re}_{0}=\frac{r_{1/2}U_{0}}{\nu }$$

Hence, we have the following first fundamental relation for the Kolmogorov scales

$$\frac{\eta u_{\eta }}{\nu }=1$$

Recall that the Reynolds number is defined as

$$\displaystyle {\rm Re} = \frac{LV}{\nu}$$

where   $$\displaystyle L$$   is length,   $$\displaystyle V$$   velocity,   $$\displaystyle \nu$$   kinematic viscosity,  $$\displaystyle \mu$$  the dynamic viscosity, and  $$\displaystyle \rho$$  the mass density.

Thus, at the Kolmogorov length scale, the Reynolds number is equal to one.

The Reynolds number has the dimension of the ratio of inertia force  $$\displaystyle F_I$$  over viscous force  $$\displaystyle F_v$$, and can be seen as follows. From the relation of shear stress $$\displaystyle \tau$$ and strain rate $$\displaystyle \dot \gamma$$

$$\displaystyle \tau = \mu \dot \gamma$$

we have the dimensional relation

$$\displaystyle [\tau] = \frac{[F_v]}{L^2} = [\mu] [\dot \gamma] = [\mu] \frac{1}{T}$$

Thus

$$\displaystyle [{\rm Re}] = \frac{[\rho] L [V^2]}{[\mu][V]} = \frac{[\rho][V^2]}{[\mu]\frac{1}{T}} = \frac{[\rho][V^2]}{[F_v] / L^2} = \frac{[\rho] L^2 [V^2]}{[F_v]} = \frac{[M] [V^2] / L}{[F_v]} = \frac{[F_I]}{[F_v]}$$

since

$$\displaystyle [M] [V^2] = [K] = [F_I] L$$

where $$\displaystyle K$$ is the kinetic energy.

Another way to see the definition of the Reynolds number as coming from the ratio of the inertia force over the viscous force is to relate the quantity  $$\displaystyle V^2 / L$$ to the inertia force per unit mass    $$\displaystyle F_I / M$$, and to relate the quantity   $$\displaystyle \nu V / L^2$$ with the viscous force per unit mass  $$\displaystyle F_v / M$$:

$$\displaystyle [K] = M [V^2] = [F_I] [L] \Rightarrow \frac{[F_I]}{M} =  \frac{[V^2]}{L}$$

$$\displaystyle [\tau] = \frac{[F_v]}{L^2} = \frac{\mu}{T} \Rightarrow \frac{[F_v]}{M} = \frac{\mu L^2}{T M} = \frac{\mu [V] L}{M} = \frac{\mu [V] L}{\rho L^3} = \frac{\nu [V]}{L^2}$$

Then the ratio of inertia force over viscous force defines the Reynolds number:

$$\displaystyle \frac{[F_I]}{[F_v]} = \frac{[F_I]/M}{[F_v] / M} = \frac{[V^2]/L}{\nu [V]/L^2} = \frac{[V]L}{\nu} =: [{\rm Re}] \Rightarrow {\rm Re} := \frac{LV}{\nu}$$

Thus, at the Kolmogorov length scale, the inertia force is at the same order as the viscous force, since the Reynolds number is equal to one.

Since the dimensional relation for the turbulence dissipation $$\displaystyle \epsilon$$ is

$$\displaystyle [\epsilon] = \frac{[K]}{MT} = \frac{M [V^2]}{MT} = \frac{L^2 T^{-2}}{T} = L^2 T^{-3} = (LT^{-1})^3 / L = [V^3] / L$$

we can select the length $$\displaystyle L$$  as  $$\displaystyle r_{1/2}$$  and velocity  $$\displaystyle V$$  as  $$\displaystyle U_0$$ to non-dimensionalize the turbulence dissipation

$$\hat{\varepsilon }=\varepsilon \frac{r_{1/2}}{U_{0}^{3}}$$

The Kolmogorov length scale $$\eta $$ can then be expressed in terms of  $$\hat{\epsilon }$$ as follows

$$\displaystyle \eta =\left( \frac{\nu ^{3}}{\epsilon } \right)^{1/4}=\left( \frac{\nu ^{3}r_{1/2}}{\hat{\epsilon }U_{0}^{3}} \right)^{1/4}=\left( \frac{\nu ^{3}r_{1/2}}{\hat{\epsilon }({\rm Re}_{0}\nu r_{1/2}^{-1})^{3}} \right)^{1/4}=\operatorname{Re}_{0}^{-3/4}\hat{\epsilon }^{-1/4}r_{1/2}$$

Note that since both $$\displaystyle {\rm Re}_0$$  and  $$\displaystyle \hat \epsilon$$  are dimensionless, it can be verified that the dimensions on both sides of above relation are the same.

The ratio    $$\displaystyle \frac{\eta u_\eta}{\nu}$$  is the Reynolds number at the smallest scale of turbulence, and is equal to the ratio of inertia force over the viscous force as in the usual definition of the Reynolds number.

discussion on 2010.05.07
Length (Pope, 2000, p.128, Eq.5.149)

 * {| style="width:100%" border="0"

$$  \displaystyle \eta :=  \left(      \frac{\nu^3}{\epsilon}   \right)^{1/4} $$     (D50)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Time (Pope, 2000, p.129, Eq.5.150)
 * {| style="width:100%" border="0"

$$  \displaystyle \tau_\eta :=  \left(      \frac      {\nu}      {\epsilon}   \right)^{1/2} $$     (D51)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Velocity (Pope, 2000, p.129, Eq.5.151)

 * {| style="width:100%" border="0"

$$  \displaystyle u_\eta :=  (\nu \epsilon)^{1/4} $$     (D52)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using Eqs.(D50)-(D52), we obtain

 * {| style="width:100%" border="0"

$$  \displaystyle \frac {u_\eta} {\eta} =  \nu^{-1/2} \epsilon^{1/2} = \frac{1} {\tau_\eta} $$     (D53)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Eq.(D53) indicates that $$\displaystyle u_\eta/\eta$$ is independent of the kinematic viscosity, $$\displaystyle\nu$$. Clearly, $$\displaystyle \eta $$, $$\displaystyle \tau_\eta $$, $$\displaystyle u_\eta $$, and $$\displaystyle \epsilon $$, each depends on $$\displaystyle \nu $$. Eq.(D53) provides a consistent characterization of the velocity gradients of the disspative eddies (smallest scale, i.e., Kolmogorov scale) as stated in Pope (2000, p.185, below Eq.6.4).

Pope (2000, p.129) refers to $$\displaystyle u_\eta/\eta$$ as velocity gradient even though it is actually change in velocity with respect to space, i.e., it is not exactly derivative with respect to space. Two fluids of different kinematic viscosity, e.g., water and oil, would have different length, time, and velocity scales. However, the ratio of velocity to length scale would always give the inverse of the time scale as shown by Eq.(D53).

The dimension of $$\displaystyle u_\eta/\eta$$ is

$$\displaystyle [u_\eta] = L/T$$

$$\displaystyle [\eta] = L$$

Therefore,

$$\displaystyle [u_\eta/\eta] = 1/T$$.

Using Eq.(D53)

 * {| style="width:100%" border="0"

$$  \displaystyle \nu \left(\frac {u_\eta} {\eta} \right )^2 =  \nu \nu^{-1} \epsilon = \epsilon = \frac{\nu}{\tau_\eta^2} $$     (D54) On the other hand, the turbulence dissipation is defined as (Kolmogorov time scale section, Eq.(136.b))
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

$$\displaystyle \epsilon := 2 \nu \langle \mathbf s {\mathbf :} \mathbf s \rangle $$

Using the definition of the turbulence dissipation and Eq.(D54)

 * {| style="width:100%" border="0"

$$  \displaystyle \langle \mathbf s {\mathbf :} \mathbf s \rangle = \frac {1} {2 \tau_\eta^2} $$     (D55)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 ''' Is $$\displaystyle \epsilon $$ independent of $$\displaystyle  \nu $$? ''' Pope (2000, p.129) wrote this statement after presenting Eq.(5.156), which is Eq.(D54) in this text. The turbulence dissipation being independent of the kinematic viscosity is a very deep subject and cannot be stated from Eq.(D54). Check Eq.(6.4) in Pope (2000, p.185), which shows that the dissipation depends on the kinematic viscosity. We believe he wanted to only say that $$\displaystyle u_\eta/\eta$$ is independent of $$\displaystyle\nu$$.

The strain rate is a purely kinematic quantity, which does not involve any constitutive relation or a material property. Therefore, it is independent of $$\displaystyle \nu $$, i.e., you may have two fluids of different kinematic viscosity to have the same strain rate. Thus, $$\displaystyle \epsilon $$ has to depend on $$\displaystyle  \nu $$ since $$\displaystyle  \langle \mathbf s {\mathbf :} \mathbf s \rangle $$ cannot cancel out $$\displaystyle \nu $$.

Then, what happened to Taylor’s assumption (PRSA, 1935) that $$\displaystyle \epsilon $$ is independent of $$\displaystyle  \nu $$ in the paper by Lee et al (2008) ? This is still a research topic (See Lee et al (2008), Pope (2000, p.186, Kolmogorov’s second similarity hypothesis) , Frish (1996) , and other recent researches. $$\displaystyle \epsilon $$ is independent of $$\displaystyle  \nu $$ is an approximation at high Reynolds numbers.

discussion on 2010.06.25
What is $$\displaystyle EI $$ in $$\displaystyle l_{EI} $$? (Pope, 2000, p.184)

$$\displaystyle l > l_{EI} $$ : anisotropic large eddies (?)

$$\displaystyle l < l_{EI} $$ : isotropic small eddies (?)

$$\displaystyle l > l_{EI} $$ : anisotropic large eddies (?)

Eddy: no clear definition, coherent (?)(Pope, 2000, p.183) Turbulent region of size $$\displaystyle l $$.

Eddy $$\displaystyle \ne $$ whorl (or vortex or maelstrom)

See our plans.

discussion on 2010.07.07
Energy cascade concept from Richardson (1922)

Eddies of the largest scale are of lengthscale $$\displaystyle l_0$$, velocity scale $$\displaystyle u_0$$, and time scale $$\displaystyle \tau_0$$. Those large eddies are unstable and breakup into smaller ones. These new eddies breakup into even smaller ones until a smallest scale is reached. This scale is called Kolmogorov scale and the lengthscale is represented with $$\displaystyle \eta$$. Note that "eddy" (Pope, p.183) has no clear definition.

Energy density is $$\displaystyle u_0^2$$ and $$\displaystyle \tau_0=l_0/u_0$$. Rate of energy transfer is

$$\displaystyle \epsilon= \frac{u_0^2}{\tau_0}= \frac{ u_0^3}{l_0} $$

High Re flows $$\displaystyle \Rightarrow $$ viscosity is negligible.

Energy cascade $$\displaystyle \Rightarrow $$ $$\displaystyle \epsilon \propto \frac {u_0^3}{l_0}$$, which is independent of $$\displaystyle \nu $$.

Pope (2000, p.185, Eq.6.4)

$$\displaystyle \epsilon= \nu \left( \frac{u_\eta}{\eta} \right)^2 = \frac{ \nu}{\tau_\eta^2} $$

Low Re flows, nothing indicates that $$\displaystyle \epsilon $$ is independent of $$\displaystyle \nu $$. Actually, it is more reasonable that $$\displaystyle \epsilon $$ depends on $$\displaystyle \nu $$.

H1) Kolmogorov’s Hypothesis 1: Local isotropy

At small scales, flow statistics are independent of flow geometry and boundary conditions. $$\displaystyle \Rightarrow \exists $$ a statistically universal range (state) of turbulence at small scale. Q: What are the governing parameters? $$\displaystyle \Rightarrow $$ the second hypothesis ( First similarity hypothesis. Why similarity?)

H2) Kolmogorov’s first Similarity Hypothesis

At high Re, small scale statistics depend on $$\displaystyle \epsilon $$ and $$\displaystyle \nu $$.

Derive Kolmogorov’s scales $$\displaystyle \eta, u_\eta, \tau_\eta $$ see our wiki discussion Intuitive Meaning of Kolmogorov scales.

Dimensionless quantities (Pope, 2000, Eqs.6.5-6.6) :

$$\displaystyle \mathbf y= \frac{\mathbf x - \mathbf x_0}{\eta} = \frac{x - x_0}{\eta} $$

$$\displaystyle \mathbf W(\mathbf y)= \frac{\left[ \mathbf U(x,t_0) - \mathbf U(x_0,t_0) \right]} {u_\eta} $$

Why $$\displaystyle \epsilon $$ and $$\displaystyle \nu $$?

$$\displaystyle \epsilon = $$ rate of energy transfer = due to energy cascade introduced by Richardson.

$$\displaystyle \nu = $$ viscosity = again due to Richardson’s energy cascade. At the end of the cascade, i.e., at the smallest scale $$\displaystyle \eta $$, energy of turbulence is dissipated. Thus, we need viscosity (which is the source of energy dissipation) as another parameter.

Back to the meaning of similarity (see the wiki discussion): $$\displaystyle  \left[\nu \right] = L^2 T^{-1} $$ and $$\displaystyle \left[ \epsilon \right] = L^2 T^{-3} $$ $$\displaystyle \Rightarrow $$ cannot form nondimensional quantities out of $$\displaystyle (\epsilon, \nu)$$.

Since $$\displaystyle \left[\mathbf W(\mathbf y) \right]=1 \Rightarrow \mathbf W $$ cannot depend on $$\displaystyle (\epsilon, \nu)$$.

Note: $$\displaystyle (\eta, u_\eta, \tau_\eta) $$ depend on $$\displaystyle (\epsilon, \nu)$$. However, dimensionless velocity $$\displaystyle \mathbf W $$ is independent of $$\displaystyle (\epsilon, \nu)$$.

Example: 2 fluids $$\displaystyle (\epsilon_1, \nu_1)$$ and $$\displaystyle (\epsilon_2, \nu_2)$$, which gives

2 different large scale eddies $$\displaystyle l_{01}$$ and $$\displaystyle  l_{02}$$

2 different large scale velocities $$\displaystyle u_{01}$$ and $$\displaystyle  u_{02}$$

Therefore,

$$\displaystyle \epsilon_1 \propto \frac {u_{01}^3}{l_{01}}$$  and $$\displaystyle  \epsilon_2 \propto \frac {u_{02}^3}{l_{02}}$$

2 different Kolmogorov scales $$\displaystyle (\eta_1, u_{\eta 1}, \tau_{\eta 1}) $$ and $$\displaystyle  (\eta_2, u_{\eta 2}, \tau_{\eta 2}) $$

$$\displaystyle \mathbf W_i(\mathbf y_i)= \frac{\left[ \mathbf U_i(x,t_0) - \mathbf U_i (x_0,t_0) \right]} {u_{\eta i}} $$ where $$\displaystyle i = 1,2 $$. $$\displaystyle \mathbf W_i $$ depend only on $$\displaystyle  \mathbf y_i $$. Since $$\displaystyle \mathbf W_i $$ is independent of $$\displaystyle (\epsilon_i, \nu_i)$$, we can remove index $$\displaystyle i$$ and write $$\displaystyle  \mathbf W $$ only.

So $$\displaystyle \mathbf W $$ is a kind of universal  function. How about $$\displaystyle \mathbf y_i $$? Compare $$\displaystyle \mathbf W(y_1) $$ and $$\displaystyle  \mathbf W(y_2) $$. Only $$\displaystyle \eta_1 $$ and $$\displaystyle  \eta_2 $$ differ. Consider

$$\displaystyle \mathbf  {\hat y} =  \underbrace{\mathbf  y_1}_{\frac{x_1-x_0}{\eta_1}} = \underbrace{\mathbf  y_2}_{\frac{x_2-x_0}{\eta_2}} \Rightarrow $$

$$\displaystyle \mathbf W(\mathbf  {\hat y}) =  \mathbf W(\mathbf  y_1) = \mathbf W(\mathbf  y_2) $$ $$\displaystyle \Rightarrow $$

$$\displaystyle \frac{x_1-x_0}{x_2-x_0}= \frac{\eta_1}{\eta_2} $$

At two different points $$\displaystyle x_1 \rm{and} x_2$$, we have the same dimensionless velocity $$\displaystyle   \mathbf W(\mathbf  {\hat y}) $$. So, if we scale the geometry such that $$\displaystyle \frac{x_1-x_0}{x_2-x_0}= \frac{\eta_1}{\eta_2} $$, we have the same dimensionless velocity.

A turbulent flow at small scales is characterized by 2 parameters $$\displaystyle (\epsilon, \nu)$$ (Second Kolmogorov hypothesis $$\displaystyle \equiv$$First Kolmogorov similarity hypothesis)

Turbulent flow 1 $$\displaystyle \equiv(\epsilon_1, \nu_1)$$

Turbulent flow 2 $$\displaystyle \equiv(\epsilon_2, \nu_2)$$

$$\displaystyle x_i = $$ point in physical space of turbulent flow $$\displaystyle i $$ where $$\displaystyle i = 1,2$$. if we scale $$\displaystyle x_1 $$ and $$\displaystyle x_2 $$ such that $$\displaystyle \frac{x_1-x_0}{x_2-x_0}= \frac{\eta_1}{\eta_2} $$, we have the same dimensionless velocity profile $$\displaystyle   \mathbf W $$. That is the meaning of similarity. At this point we reached the end of the second hypothesis.

Motivation for the third hypothesis

The length, velocity, and time scales are given in Pope (2000, Eqs.6.1-6.3) as:

$$  \displaystyle \eta :=  \left(      \frac{\nu^3}{\epsilon}   \right)^{1/4} $$

$$  \displaystyle u_\eta :=  (\nu \epsilon)^{1/4} $$

$$  \displaystyle \tau_\eta :=  \left(      \frac      {\nu}      {\epsilon}   \right)^{1/2} $$

Also, $$ \displaystyle \epsilon \propto \frac {u_0^3}{l_0}$$  (Pope, p. 183). Keeping in mind that $$ \displaystyle Re = \frac {l_0 u_0}{\nu}$$, we obtain equations 6.7-6.9 in Pope (2000, p.186) as

$$  \displaystyle \eta/l_0 \propto Re^{-3/4} $$

$$  \displaystyle u_\eta/u_0 \propto Re^{-1/4} $$

$$  \displaystyle \tau_\eta/\tau_0 \propto Re^{-1/2} $$

Note that as $$ \displaystyle Re\to\infty$$

$$ \displaystyle \eta/l_0 \to 0$$

$$ \displaystyle u_\eta/u_0 \to 0$$

$$ \displaystyle \tau_\eta/\tau_0 \to 0$$

Q: How do $$ \displaystyle u(l)$$ and $$ \displaystyle \tau(l)$$ change when $$ \displaystyle l \to 0$$? Pope (2000, p.184) : Assuming $$ \displaystyle Re = \frac {l u(l)}{\nu}$$ to decrease with l is not sufficient to determine the trend of $$ \displaystyle u(l)$$ and $$ \displaystyle \tau(l)$$. That is why we need Kolmogorov’s third hypothesis, which assume that as $$ \displaystyle l \to 0 $$, there is an intermediate range $$ \displaystyle \eta \ll l \ll l_0$$ in which $$ \displaystyle Re$$ is still large compared to $$ \displaystyle 1=\frac{\eta u_\eta}{\nu}$$, and thus viscosity is negligible.

Reviews of Pope's book
In this derivation of the Kolmogorov scales, we will primarily follow the exposition introduced in the book Turbulent Flows by Stephen B. Pope. Therefore, we will first introduce some of the reviews about Pope’s book on turbulence. Then, give some information about the Turbulence video prepared by the National Committee for Fluid Mechanics.

Reviews about the book: Turbulent Flows

First, a couple of words about Stephen B. Pope who is a professor at Cornell University in the Sibley School of Mechanical and Aerospace Engineering. Pope was elected to the National Academy of Engineering and the quotation on the press release of February 17th, 2010 was as follows:

Joel Henry Ferziger, who is famous for his book Computational Methods for Fluid Dynamics, died in 2004, commented on the Turbulent Flows at the web site of Cambridge as follows:

Note: The book by Tennekes and Lumley is called A First Course in Turbulence

Stewart Cant wrote a review in 2001 about Turbulent Flows on the Combustion and Flame journal (125 (4), pp. 1361-1362). Some quotations from the review:

Jean-Luc Thiffeault who works in the area of Dynamical Systems wrote a review on for the Geophysical and Astrophysical Fluid Dynamics. Some quotations from his review:

= Issues resolved (chronological order) =

Turbulence Video by the National Committee for Fluid Mechanics
The National Committee for Fluid Mechanics (NCFM) has prepared some educational videos on fluid mechanics and MIT has made some of these films available online (Realplayer is required) through the iFluids program. The film notes are also published online. One of these movies, prepared by Robert W. Stewart (His biography is available through the Bibliographical Memoirs of Fellows of The Royal Society) from University of British Columbia, is called “Turbulence”.

In this video, the physics of the turbulence is explained. The syndromes of turbulence are given as disorder, efficient mixing, and vorticity. No matter how carefully one tries to reproduce the turbulence, it is never possible. However, the average speeds are well-defined and they are reproducible. Water waves example was given to show that disorder alone is not enough for turbulence. Mixing is another requirement. Turbulent motion occurs in three-dimensions. In turbulent flows, there is appreciable cross-stream motion, which makes the gradients smoother. Once the flow becomes turbulent, the Reynolds number is not important for the large scale motion. However, the small scale motion is affected by the Reynolds number, which is explained through energy dissipation and energy cascades. The large scale motions, smaller than the main flow scales, are unstable and break into smaller scales. At some small scale, the Reynolds number becomes very small and viscosity starts playing role. At that scale, the energy is dissipated via viscosity and these scales are stable. The Reynolds number at which turbulence starts can be delayed or advanced by using another type of instability. Here, the author suggests the use of buoyancy. In channel flow, if the liquid has a stable density stratification, i.e., a light liquid on top of a heavy liquid is composed by flowing cold liquid to underlie the hot liquid, the turbulence occurs at a much larger Reynolds number.

Reference to Pope
Keep the google characteristics of the google books here, not in the main article, so to keep the article clean.

The Kolmogorov time scale is one of the three characteristic scales of the smallest turbulent motions, named after the Russian mathematician A.N. Kolmogorov (Pope, 2000, p.128)

Intuitive meaning of Kolmogorov scales
It is useful to develop some intuitive feel for the Kolmogorov scales before diving into lengthy detailed derivation.

For a giving flow setting, turbulence occurs when the flow Reynolds number $$\displaystyle {\rm Re}$$ becomes larger than the critical Reynolds number $$\displaystyle {\rm Re}_{cr}$$ (Barenblatt 1996, p.253), when in general, inertia force dominates over viscous force, which becomes negligible with respect to the inertia force at this large scale. On the other hand, the large scale is unstable, and breaks down to a cascade of smaller scales (Richardson energy cascade); at the smallest scale where stability sets in, the inertia force is at the same order as the viscous force, which provides a mechanism to dissipate energy. One can use dimensional analysis to find the relation for this smallest length scale at which viscous force, represented by the kinematic viscosity $$\displaystyle \nu$$, plays a key role in the energy dissipation, represented by the turbulence dissipation  $$\displaystyle \epsilon$$ (c.f. Tennekes & Lumley 1972, p.20) .

The dimensional relation for the kinematic viscosity $$\displaystyle \nu$$ is

$$\displaystyle [\nu] = \frac{[\mu]}{[\rho]} = \frac{[\tau]/[\dot \gamma]}{M / L^{3}} = \frac{(MLT^{-2}/L^2)(T^{-1})}{M L^{-3}} = L^2 T^{-1} = L [V]$$ The turbulence dissipation $$\displaystyle \epsilon$$ represents the time rate of change of kinetic energy per unit mass, and thus

$$\displaystyle [\epsilon] = \frac{[K]}{MT} = \frac{M [V^2]}{MT} = \frac{L^2 T^{-2}}{T} = L^2 T^{-3} = [V^3] / L$$

Thus to obtain a length scale from $$\displaystyle [\nu]$$  and  $$\displaystyle [\epsilon]$$, we need to cancel out the time dimension   $$\displaystyle T$$ by raising  $$\displaystyle [\nu]$$ to power 3, and divide by  $$\displaystyle [\epsilon]$$ (to obtain  $$\displaystyle L^4$$), then take the quartic root; hence the Kolmogorov length scale

$$\displaystyle \eta := \left( \frac{\nu^3}{\epsilon} \right)^{1/4}$$

Similarly, the Kolmogorov time scale is obtained from $$\displaystyle [\nu]$$  and  $$\displaystyle [\epsilon]$$ by canceling the length dimension  $$\displaystyle L$$ with the ratio   $$\displaystyle [\nu] / [\epsilon]$$, then take the square root, i.e.,

$$\displaystyle \tau_\eta = \left( \frac{\nu}{\epsilon} \right)^{1/2}$$

Again, the Kolmogorov velocity scale is obtained from $$\displaystyle [\nu]$$  and  $$\displaystyle [\epsilon]$$ with the product     $$\displaystyle [\nu] [\epsilon]$$, then take the quartic root, i.e.,

$$\displaystyle u_\eta = \left( {\nu}{\epsilon} \right)^{1/4}$$

We can now merge this section and the section on the turbulence videos into a single section, and incorporate it into the article.

Kolmo length scale
Kolmogorov, in his 1941 paper, defined a transformation of coordinate using a length scale denoted by $$\displaystyle \eta$$; he subsequently related $$\displaystyle \eta$$ to  $$\displaystyle \nu$$  and $$\displaystyle \epsilon$$  as shown in Eq.(1), then changed this length scale notation to  $$\displaystyle \lambda$$, which is Greek for (lowercase) L, and is more mnemonic for “length”. It is not clear why subsequent authors to present day decided to preserve the notation $$\displaystyle \eta$$, instead of the more mnemonic notation $$\displaystyle \lambda$$, which Kolmogorov selected himself, for the Kolmogorov length scale.

Opening section
Turbulence has four main observable characteristics: (1) disorder, (2) irreproducible details, (3) efficient mixing and transport, (4) irregular distribution of vorticities; see the highly informative video and film notes by R.W. Stewart 1969.

Stewart characterizes the vorticities in turbulent flows as strictly occurring in 3-D, since the first term in the right-hand side of the vorticity equation

 * {| style="width:100%" border="0"

$$  \displaystyle \frac{D \boldsymbol \omega}{Dt} = ({\rm grad} \, {\boldsymbol U}) \cdot {\boldsymbol \omega} + {\rm div} \, {\rm grad} \, {\boldsymbol \omega} = (\nabla {\boldsymbol U}) \cdot {\boldsymbol \omega} + \nabla^2 {\boldsymbol \omega}
 * style="width:95%" |
 * style="width:95%" |

$$  (B1)
 *

 * }

i.e.,

 * {| style="width:100%" border="0"

$$  \displaystyle (\nabla {\boldsymbol U}) \cdot {\boldsymbol \omega} = \left( \frac{\partial U_i}{\partial x_j} \omega_j  \right) \mathbf e_i $$  (B2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

which corresponds to vorticity production by vortex-line stretching, would vanish in 2-D, even though he did say that “something like turbulence can occur in 2-D”, such as large scale weather systems. Pope 2000, p.22, on the other hand, refers to “2-D turbulence (which can occurs in special circumstances)”.

It is useful to develop some intuitive feel for the Kolmogorov scales before diving into lengthy detailed derivation.

For a giving flow setting, turbulence occurs when the flow Reynolds number $$\displaystyle {\rm Re}$$ becomes larger than the critical Reynolds number $$\displaystyle {\rm Re}_{cr}$$ (Barenblatt 1996, p.253), when in general, inertia force dominates over viscous force, which becomes negligible with respect to the inertia force at this large scale. On the other hand, the large scale is unstable, and breaks down to a cascade of smaller scales (Richardson energy cascade); at the smallest scale where stability sets in, the inertia force is at the same order as the viscous force, which provides a mechanism to dissipate energy. One can use dimensional analysis to find the relation for this smallest length scale at which viscous force, represented by the kinematic viscosity $$\displaystyle \nu$$, plays a key role in the energy dissipation, represented by the turbulence dissipation  $$\displaystyle \epsilon$$ (c.f. Tennekes & Lumley 1972, p.20) .

The dimensional relation for the kinematic viscosity $$\displaystyle \displaystyle \nu$$   is

 * {| style="width:100%" border="0"

$$  \displaystyle \displaystyle [\nu] = \frac{[\mu]}{[\rho]} = \frac{[\tau]/[\dot \gamma]}{M / L^{3}} = \frac{(MLT^{-2}/L^2)(T^{-1})}{M L^{-3}} = L^2 T^{-1} = L [V] $$  (B5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The turbulence dissipation $$\displaystyle \displaystyle \epsilon$$   represents the time rate of change of kinetic energy per unit mass, and thus

 * {| style="width:100%" border="0"

$$  \displaystyle \displaystyle [\epsilon] = \frac{[K]}{MT} = \frac{M [V^2]}{MT} = \frac{L^2 T^{-2}}{T} = L^2 T^{-3} = [V^3] / L $$ (B6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Thus to obtain a length scale from $$\displaystyle [\nu]$$  and  $$\displaystyle [\epsilon]$$, we need to cancel out the time dimension   $$\displaystyle T$$ by raising  $$\displaystyle [\nu]$$ to power 3, and divide by  $$\displaystyle [\epsilon]$$ (to obtain  $$\displaystyle L^4$$), then take the quartic root; hence the Kolmogorov length scale

 * {| style="width:100%" border="0"

$$  \displaystyle \eta := \left( \frac{\nu^3}{\epsilon} \right)^{1/4} $$  (B7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Similarly, the Kolmogorov time scale is obtained from $$\displaystyle [\nu]$$  and $$\displaystyle [\epsilon]$$   by canceling the length dimension $$\displaystyle L$$    with the ratio $$\displaystyle [\nu] / [\epsilon]$$  , then take the square root, i.e.,

 * {| style="width:100%" border="0"

$$  \displaystyle \tau_\eta = \left( \frac{\nu}{\epsilon} \right)^{1/2} $$  (B8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Again, the Kolmogorov velocity scale is obtained from $$\displaystyle [\nu]$$   and $$\displaystyle [\epsilon]$$   with the product  $$\displaystyle [\nu][\epsilon]$$ , then take the quartic root, i.e.,

 * {| style="width:100%" border="0"

$$  \displaystyle u_\eta = \left( {\nu}{\epsilon} \right)^{1/4} $$  (B9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 Check out the licensing issue at Steven G. Johnson, Guide to the dual-license, Template:DualLicenseWithCC-BySA-to-3.0. Perhaps dual licensing is already established as indicated in Wikipedia:Copyrights Egm6341.s10 11:05, 25 December 2009 (UTC)

 I lightened the citation style throughout the article, e.g., write (Pope, 2000, p.128), instead of (Pope (2000), p.128), following the APA citation style. Egm6341.s10 11:05, 25 December 2009 (UTC)

Kinetic energy
We follow primarily the exposition in Pope, 2000, Chaps 3, 5 ...

 I added the above sentence; see further below for the reason.

Decomposition of kinetic energy density

 Change “kinetic energy” to “kinetic energy density” throughout this subsection. Egm6341.s10 01:31, 31 December 2009 (UTC)

Energy flux

 I changed "Eq. $$(55)_1$$" to "Eq. (55)" as there is no subparts of that equation. Also, now the writing is consistent with the previous equations. --EGM6936.f09.ku 10:06, 21 December 2009 (UTC)

Actually, "Eq. $$(55)_1$$" intended to mean the first equality in Eq.(55), to distinguish from the second equality in Eq.(55). So it is best to keep the subscript 1. We can discuss more on this issue; for now, I restore the subscript 1.

Also, don't have a blank space between "Eq." and the equation number "(55)" to avoid having "Eq." on one line and "(55)" on the next line. Egm6341.s10 01:11, 25 December 2009 (UTC)

 I changed "incompressibility condition $$(53)_1$$" to "incompressibility condition stated in Eq. (53)". --EGM6936.f09.ku 10:06, 21 December 2009 (UTC)

Actually, the "incompressibility condition $$(53)_1$$" was used to mean the left-hand side of $$\Longrightarrow$$. Saying the "incompressibility condition stated in Eq. (53)" is not precise, since the right-hand side of $$\Longrightarrow$$ is not the incompressibility condition. We can discuss on this issue; for now, I revert to the previous expression, for lack of a better alternative. Egm6341.s10 01:16, 25 December 2009 (UTC)

Constant-density flows

 I suspected that Pope made some kind of assumption similar to Eq.(81), and just found out about the assumption of constant-density flows in Pope, and added the above section. Egm6341.s10 01:30, 25 December 2009 (UTC)

Time rate of Jacobian

 I just noticed the 3rd equality in Pope, and added the above sentence. Egm6341.s10 01:42, 25 December 2009 (UTC)

 I used openoffice to create the above table very fast. Egm6341.s10 23:06, 25 December 2009 (UTC)

Evolution of kinetic energy of mean field
Alternative derivation

We are primarily following Exercise (5.19) in Pope (2000, p.125) and adopted the solution given by Liu and Pope.

Inserting Eq.(61) into Eq.(57) and dividing by $$\displaystyle \rho  $$ yields

 * {| style="width:100%" border="0"

$$  \displaystyle \frac {D U_j} {Dt} = -\frac {1}  {\rho} \frac {\partial p}  {\partial x_j} + 2 \nu \frac {\partial S_{ij}} {\partial x_i} $$     (139)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

known as Navier-Stokes Equation. Here, $$\displaystyle S $$ is the strain rate given by (73). Taking the mean of (139) and assuming $$\displaystyle \rho$$ is not a random variable

 * {| style="width:100%" border="0"

$$  \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\bigg\langle \frac {D U_j} {Dt} \bigg\rangle = -\frac {1}  {\rho} \frac {\partial \langle p \rangle} {\partial x_j} + 2 \nu \frac {\partial \bar S_{ij}} {\partial x_i} $$     (140)
 *

 * }

where $$\displaystyle \bar S_{ij} $$ is defined by (127) and (131). The left hand side of (140) can be written as

 * {| style="width:100%" border="0"

$$  \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\bigg\langle \frac {D U_j} {Dt} \bigg\rangle = \frac {\partial \langle U_j \rangle} {\partial t} + \frac {\partial \langle U_i U_j \rangle } {\partial x_i} $$     (141) and
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \langle U_i U_j \rangle = \langle U_i \rangle  \langle U_j  \rangle + \langle u_i u_j \rangle
 * style="width:95%" |
 * style="width:95%" |

$$     (142) Inserting (142) into (141)
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\bigg\langle \frac {D U_j} {Dt} \bigg\rangle = \frac {\partial \langle U_j \rangle} {\partial t} + U_i \frac {\partial \langle U_j \rangle} {\partial x_i} + \frac {\partial \langle u_i u_j \rangle} {\partial x_i} $$     (143)
 *

 * }

with $$\displaystyle \frac {\partial \langle U_j \rangle} {\partial x_i} = 0 $$ from Eq.(114). It should be noted that the first two terms form the mean material derivative defined in (112). Then, Eq.(140) becomes

 * {| style="width:100%" border="0"

$$  \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac {\partial \langle U_j \rangle} {\partial t} + \langle U_i \rangle \frac {\partial \langle U_j \rangle} {\partial x_i} + \frac {\partial \langle u_i u_j \rangle} {\partial x_i} = -\frac {1}  {\rho} \frac {\partial \langle p \rangle} {\partial x_i} + 2 \nu \frac {\partial \bar S_{ij}} {\partial x_i} $$     (144)
 *

 * }

Multiply Eq.(144) with $$\displaystyle \langle U_j \rangle $$ to obtain

 * {| style="width:100%" border="0"

$$  \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac {\bar D \bar E}  {\bar D t } + \langle U_j \rangle \frac {\partial \langle u_i u_j \rangle} {\partial x_i} = -\frac {\langle U_j \rangle} {\rho} \frac {\partial \langle p \rangle} {\partial x_i} + \langle U_j \rangle 2 \nu \frac {\partial \bar S_{ij}} {\partial x_i} $$     (145)
 *

 * }

The last term in (145) can be written as

 * {| style="width:100%" border="0"

$$  \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\langle U_j \rangle 2 \nu \frac {\partial \bar S_{ij}} {\partial x_i} = 2 \nu \frac {\partial } {\partial x_i} \bigg (  \langle U_j \rangle \frac   {\partial \bar S_{ij}}   {\partial x_i} \bigg) - \bar \epsilon $$     (146)
 *

 * }

where $$\displaystyle \bar \epsilon $$ defined in Eq.(136a) is $$\displaystyle 2\nu \bar S_{ij}   \frac {\partial    \langle U_j \rangle} {\partial x_i} $$ in indicial notation.Then, Eq.(145) becomes

 * {| style="width:100%" border="0"

$$  \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac {\bar D \bar E}  {\bar D t } + \frac {\partial } {\partial x_i} \bigg (  \langle U_j \rangle   \langle u_i u_j  \rangle  +  \frac  {\langle p \rangle}  {\rho}  \langle U_j \rangle  -  \langle U_j \rangle  2 \nu  \frac   {\partial \bar S_{ij}}   {\partial x_i} \bigg )

- \frac {\partial \rho } {\partial x_i} \bigg [ \langle p \rangle \langle U_j \rangle - 2\mu\langle U_j \rangle \frac {\partial \bar S_{ij}} {\partial x_i} \bigg ] = - \mathcal P - \bar \epsilon

$$     (147)
 *

 * }

Here, $$\displaystyle \mathcal P = -    \langle u_i u_j  \rangle \frac {\partial    \langle U_j \rangle} {\partial x_i} $$ is called production by Pope (2000, p.125).

Evolution of k
$$\left ( \mathbf a \otimes \mathbf b \right ) \mathbf c = \mathbf a \left ( \mathbf b \cdot \mathbf c \right )$$

so

$$\left ( \mathbf u \otimes \mathbf u \right ) \mathbf u = \mathbf u \left ( \mathbf u \cdot \mathbf u \right )$$

Also

$$\mathbf c \cdot \left ( \mathbf a \otimes \mathbf b \right ) = \left ( \mathbf c \cdot \mathbf a \right ) \mathbf b$$

and thus

$$\mathbf u \cdot \left ( \mathbf u \otimes \mathbf u \right ) = \left ( \mathbf u \cdot \mathbf u \right ) \mathbf u$$

Note:

$$\left ( \mathbf e_k \otimes \mathbf e_l \right ) \cdot \mathbf e_j = \mathbf e_k \left ( \mathbf e_l \cdot \mathbf e_j \right ) = \mathbf e_k \delta_{lj} $$

Averaging

 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf u (\mathbf U) := \mathbf U - \langle \mathbf U \rangle $$  (C1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \langle \mathbf U \cdot \left[ \mathbf u (\mathbf U) \cdot \mathbf u (\mathbf U) \right] \rangle = \int \mathbf V \cdot \left[ \mathbf u (\mathbf V) \cdot \mathbf u (\mathbf V) \right] f(\mathbf V) d \mathbf V = \int \mathbf V \cdot \left[ \left( \mathbf V - \langle \mathbf V \rangle \right) \cdot \left( \mathbf V - \langle \mathbf V \rangle \right) \right] f(\mathbf V) d \mathbf V $$ (C2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

As an example of applying the definition of taking the mean, let’s take Eq.(147) from Kolmogorov Time Scale section

 * {| style="width:100%" border="0"

$$  \displaystyle \langle \mathbf U \otimes \mathbf U \rangle = \int \left [ \mathbf V \otimes \mathbf V \right ] f(\mathbf V) d \mathbf V $$ (C3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using the velocity decomposition, one gets

 * {| style="width:100%" border="0"

$$  \displaystyle \langle \mathbf U \otimes \mathbf U \rangle = \int \left[ \langle \mathbf V \rangle \otimes \langle \mathbf V \rangle + \langle \mathbf V \rangle \otimes \mathbf v + \mathbf v \otimes \langle \mathbf V \rangle + \mathbf v \otimes \mathbf v \right] f(\mathbf V) d \mathbf V $$ (C5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

As $$\displaystyle \langle \mathbf V \rangle  $$ is not a random variable, it can be taken outside of the integral. Also, it should be noted that $$\displaystyle \langle \mathbf u \rangle = \mathbf 0 $$. Therefore, Eq.(C5) is simplified to

 * {| style="width:100%" border="0"

$$  \displaystyle \langle \mathbf U \otimes \mathbf U \rangle = \langle \mathbf U \rangle \otimes \langle \mathbf U \rangle + \langle \mathbf u \otimes \mathbf u \rangle $$  (C6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We recovered Eq.(148) of Kolmogorov Time Scale section by applying the definition of the averaging.


text:
User:Danaemerritt

=English Composition=

Statement of Intent
I intend to write several character studies from a television show that I obsess over. I plan to write a page-page and a half a week and cover a fair amount of depth with them. I want to be able to explain the person and have the character speak for itself while also displaying my own style of writing, and, hopefully, my skill. As these characters are very dear to me and I take them very seriously, I will do my best and make use of my considerable knowledge of them to bring them to life to an ausiance, that I will assume, has no previous understanding of them. I want everything in my studies to be clear and understandable, easy to fall into, avoiding techinical speak and assumptions.

Overview
I want to create around six character profiles, in depth, about five-six pages each, depending on which character I'm currently doing. Some may be longer, others shorter.

Rationale
I hope to gain a better understanding of the ins-and-outs of English/writing, of the small details.

Publication/Presentation

Schedule

 * week 2
 * week 3
 * week 4
 * etc.

Anticipated Problems
I may misjudge the amount of information I wish to go into and deal with. There may be more than I anticipate. Also, I may end up digging in too far into a particular subject, excluding others, etc.

Week 6 Project Review

Not so good. My second half of the character study was without an aim or a goal, so it unconsciously turned into a summary. Had I realized that at the time, I would have made it a lot better, as I am actually good at writing summaries. Unfortunately, I didn’t and it was well below my capabilities. As that was not my intended direction, I’m going to switch characters and move on and try to find my way again. I’m fairly sure of my goal, so it should turn out better than the first did.

Reader's Report

I read lots of fanfiction of the TV show and its characters therein. Daily. This occurs on the internet, as there are no published, available books on these subjects as much as I wish there were. Usually revolving around the character of Mr. Gray in my first study guide as he is my favorite. I love to look at all the other writers' thoughts and ideas in fiction writing. Its so very inspiring to me, some are truly breathtaking. It is also benificial to me, as a writer, who also roleplays in her spare time with this same show and character sets. I use it to explore other people's imaginations and bring about different ideas in my own writing from their example.

Project

Example of a subpage
character study #1

character study #2

character study #3

character study #4

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Danaemerritt


text:
User:Egm6341.s10.Team4.yunseok

 *  Welcome, I am Yunseok Im

 * Homework1 due by 27 January 2010.
 * Homework2 due by 3 February 2010.
 * Homework3 due by 17 February 2010.
 * Homework4 due by 3 March 2010.

 * Homework5

 *  Team Homework
 * Team Homework

 *  Team members
 * Egm6341.s10.Team4.andy
 * Egm6341.s10.Team4.riherd
 * Egm6341.s10.Team4.nimaa&m
 * Egm6341.s10.Team4.roni
 * Egm6341.s10.team4.anandankala

 *  Course EGM6341 Numerical method 1
 * [EGM6341] Numerical method 1

title: User:Egm6341.s10.Team4.yunseok