A Thermal stresses in the stress tensor

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The discussion centers on incorporating thermal stresses into the mechanical stress tensor for viscous flow. The original equation presented is deemed appropriate for incompressible fluids, but corrections are suggested for compressible fluids, particularly regarding thermal expansion effects. The correct formulation for a compressible viscous fluid includes terms for both pressure and shear stress, with thermal effects typically considered negligible. However, the user seeks to couple temperature changes directly to the Navier equations, indicating a focus on thermal stresses rather than expansion. The conversation highlights the complexities of modeling thermal effects in fluid dynamics, especially in applications like sintering.
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How to I include thermal stresses in the stress tensor
Suppose I have a mechanical stress tensor \sigma. Say I have the stress tensor for viscous flow:
\boldsymbol{\sigma}=-p\mathbf{I}+\frac{1}{2}(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})
If the thermal flux is given by \boldsymbol{\sigma}_{th}=\alpha T\mathbf{I}, so I have a total flux as:
\boldsymbol{\sigma}=-p\mathbf{I}+\frac{1}{2}(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})+\alpha T\mathbf{I}
Is this correct?
 
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hunt_mat said:
TL;DR Summary: How to I include thermal stresses in the stress tensor

Suppose I have a mechanical stress tensor \sigma. Say I have the stress tensor for viscous flow:
\boldsymbol{\sigma}=-p\mathbf{I}+\frac{1}{2}(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})
If the thermal flux is given by \boldsymbol{\sigma}_{th}=\alpha T\mathbf{I}, so I have a total flux as:
\boldsymbol{\sigma}=-p\mathbf{I}+\frac{1}{2}(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})+\alpha T\mathbf{I}
Is this correct?
Your original equation is for an incompressible fluid.
 
The correct equation for a compressible viscous fluid without thermal expansion is \boldsymbol{\sigma}=-(p+\frac{2\mu}{3 }\nabla \centerdot \mathbf u)\mathbf{I}+\mu(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})
With thermal expansion, this becomes \boldsymbol{\sigma}=-(p+\frac{2\mu}{3 }(\nabla \centerdot \mathbf u-\alpha \frac{D T}{Dt}))\mathbf{I}+\mu(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})The thermal expansion term is usually considered negligible in determining the stress.
 
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Chestermiller said:
The correct equation for a compressible viscous fluid without thermal expansion is \boldsymbol{\sigma}=-(p+\frac{2\mu}{3 }\nabla \centerdot \mathbf u)\mathbf{I}+\mu(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})
With thermal expansion, this becomes \boldsymbol{\sigma}=-(p+\frac{2\mu}{3 }\nabla \centerdot \mathbf u-\alpha \frac{\partial T}{\partial t})\mathbf{I}+\mu(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})The thermal expansion term is usually considered negligible in determining the stress.
Hi, thanks for your reply. I'm not interested in expansion, but thermal stresses within a material. I want temperature to be coupled to Navier's equations. I would include the \partial_{t}T term as part of the stress tensor to fully couple the derivative?

I'm thinking of sintering with this application, and how thermal expansion affects everything.
 
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