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-   -   How to derive this simple stress tensor calculation? (https://www.cfd-online.com/Forums/main/77793-how-derive-simple-stress-tensor-calculation.html)

bearcat July 4, 2010 12:33
How to derive this simple stress tensor calculation?
 
Most turbulence models are written in tensor expressions so we have to write out the full formulation before we can implement them. I am not very experienced in difficult tensor analysis. So can anyone explain a little about writing this term in the standard k-e model?

LHS=S_{ij}\frac{\partial U_{i}}{\partial x_{j}}=(\frac{\partial U_{i}}{\partial x_{j}}+\frac{\partial U_{j}}{\partial x_{i}})\frac{\partial U_{i}}{\partial x_{j}}

From other books, the LHS should equals RHS, which reads:

RHS=2[(\frac{\partial u}{\partial x})^{2}+(\frac{\partial v}{\partial y})^{2}+(\frac{\partial w}{\partial z})^{2}]+(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x})^{2}+( \frac{\partial u}{\partial z}+\frac{\partial w}{\partial x})^{2}+(\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z})^{2}

How to get from LHS to RHS? In my understanding, for 3D domain LHS should be a 3x3 matrix as it's a product of two rank-two tensors. But RHS is a scalar. How can this happen?

Thank you very much.

jola July 4, 2010 13:46
Check out: http://www.cfd-online.com/Wiki/Einst...ion_convention

The LHS you have written is also a scalar and uses the Einstein summation convention

bearcat July 4, 2010 14:29
Quote:
Originally Posted by jola (Post 265647)
Check out: http://www.cfd-online.com/Wiki/Einst...ion_convention

The LHS you have written is also a scalar and uses the Einstein summation convention

The difficulty is that the index does not repeat in any individual term, we don't have any summation.

http://topex.ucsd.edu/geodynamics/shearer.pdf

this is about solid mechanics, but those concepts such as strain tensor and rotation tensor also apply to fluid mechanics.

Guillaume_Fournier July 4, 2010 17:03
Actually, (dui/dxj + duj/dxi)*dui/dxj = (dui/dxj)² + duj/dxi*dui/dxj

Now, you have two terms in which the indexes repeat themselves. Therefore, there is a summation on both i and j for each terms

After a few step, you will get the RHS from the LHS in your first post

bearcat July 4, 2010 18:27
yes, Guillaume, you're right from your direction.

On the other hand, isn't that

S_{ij}=\begin{bmatrix} S_{11} &S_{12} &S_{13} \\ S_{21} &S_{22} &S_{23} \\ S_{31} &S_{32} &S_{33} \end{bmatrix}=\frac{\partial U_{i} }{\partial x_{j}}+\frac{\partial U_{j}}{\partial x_{i}} =

\begin{bmatrix} 2\frac{\partial u}{\partial x} &\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x} &\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x} \\ \frac{\partial v}{\partial x}+\frac{\partial u}{\partial y} &2\frac{\partial v}{\partial y} &\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y} \\ ... &... &2\frac{\partial w}{\partial z} \end{bmatrix} (not fully written because of Latex limit here)

and

\frac{\partial U_{i}}{\partial x_{j}}=\begin{bmatrix} \frac{\partial u}{\partial x} &\frac{\partial u}{\partial y} &\frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial x} &\frac{\partial v}{\partial y} &\frac{\partial v}{\partial z} \\ \frac{\partial w}{\partial x} &\frac{\partial w}{\partial y} &\frac{\partial w}{\partial z} \\ \end{bmatrix}

They are all 3x3 matrices, how can their product be a scalar? Anything wrong with the above expressions?

Guillaume_Fournier July 5, 2010 03:45
I think you're right in the way you wrote the expressions of S_{ij} and \frac{\partial u_i}{\partial x_j}

However, you're mistaken in the fact that S_{ij}\frac{\partial u_i}{\partial x_j} is not only a product of matrices when using Einstein's convention: you have to sum over i and j, which corresponds to computing the trace of the matrix.

Let me show you. I won't develop the Stress tensor, and I will call each of its components S11, S12....

Let's start with the Einstein's summation convention

S_{ij}\frac{\partial u_i}{\partial x_j} = S_{11}\frac{\partial u_1}{\partial x_1} + S_{12}\frac{\partial u_1}{\partial x_2} + S_{13}\frac{\partial u_1}{\partial x_3} + S_{21}\frac{\partial u_2}{\partial x_1}+ S_{22}\frac{\partial u_2}{\partial x_2} + S_{23}\frac{\partial u_2}{\partial x_3} + S_{31}\frac{\partial u_3}{\partial x_1} + S_{32}\frac{\partial u_3}{\partial x_2} + S_{33}\frac{\partial u_3}{\partial x_3}

That will eventually lead to the RHS term you wrote in your first message

Without the convention, it would be a classical matrix product between two 3x3 matrices, yielding a 3x3 matrix as well. If we call it M, we have

M_{11} =S_{11}\frac{\partial u_1}{\partial x_1}+ S_{12}\frac{\partial u_2}{\partial x_1} + S_{13}\frac{\partial u_3}{\partial x_1}

M_{12} =S_{11}\frac{\partial u_1}{\partial x_2}+ S_{12}\frac{\partial u_2}{\partial x_2} + S_{13}\frac{\partial u_3}{\partial x_2}

M_{13} =S_{11}\frac{\partial u_1}{\partial x_3}+ S_{12}\frac{\partial u_2}{\partial x_3} + S_{13}\frac{\partial u_3}{\partial x_3}

and so on for the next lines.

We can notice that the RHS term only contains the sum of the diagonal terms of M. Therefore, if you prefer to use the matrix product rather than the Einstein convention to compute and develop your term, you just have to remember that this convention is equivalent to the computation of the trace of the resultant matrix

Here we are. I hope it will help a little

Cheers

Guillaume

bearcat July 5, 2010 13:34
Thank you for your explanation. This is very helpful discussion.

Quote:
Originally Posted by Guillaume_Fournier (Post 265691)
I think you're right in the way you wrote the expressions of S_{ij} and \frac{\partial u_i}{\partial x_j}

However, you're mistaken in the fact that S_{ij}\frac{\partial u_i}{\partial x_j} is not only a product of matrices when using Einstein's convention: you have to sum over i and j, which corresponds to computing the trace of the matrix.

Let me show you. I won't develop the Stress tensor, and I will call each of its components S11, S12....

Let's start with the Einstein's summation convention

S_{ij}\frac{\partial u_i}{\partial x_j} = S_{11}\frac{\partial u_1}{\partial x_1} + S_{12}\frac{\partial u_1}{\partial x_2} + S_{13}\frac{\partial u_1}{\partial x_3} + S_{21}\frac{\partial u_2}{\partial x_1}+ S_{22}\frac{\partial u_2}{\partial x_2} + S_{23}\frac{\partial u_2}{\partial x_3} + S_{31}\frac{\partial u_3}{\partial x_1} + S_{32}\frac{\partial u_3}{\partial x_2} + S_{33}\frac{\partial u_3}{\partial x_3}

That will eventually lead to the RHS term you wrote in your first message

Without the convention, it would be a classical matrix product between two 3x3 matrices, yielding a 3x3 matrix as well. If we call it M, we have

M_{11} =S_{11}\frac{\partial u_1}{\partial x_1}+ S_{12}\frac{\partial u_2}{\partial x_1} + S_{13}\frac{\partial u_3}{\partial x_1}

M_{12} =S_{11}\frac{\partial u_1}{\partial x_2}+ S_{12}\frac{\partial u_2}{\partial x_2} + S_{13}\frac{\partial u_3}{\partial x_2}

M_{13} =S_{11}\frac{\partial u_1}{\partial x_3}+ S_{12}\frac{\partial u_2}{\partial x_3} + S_{13}\frac{\partial u_3}{\partial x_3}

and so on for the next lines.

We can notice that the RHS term only contains the sum of the diagonal terms of M. Therefore, if you prefer to use the matrix product rather than the Einstein convention to compute and develop your term, you just have to remember that this convention is equivalent to the computation of the trace of the resultant matrix

Here we are. I hope it will help a little

Cheers

Guillaume


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