The setup of the stress tensor for non-Newtonian fluids, particularly power law fluids, involves defining shear stress as K(du/dy)^n, where K is the consistency index and n is the flow behavior index. In symmetric pipes or rectangular ducts, the normal stresses are represented by pressure, while other stress components may be zero. For purely viscous non-Newtonian fluids, the viscosity is expressed as a function of the second invariant of the rate of deformation tensor, paralleling the approach used for Newtonian fluids as detailed in "Transport Phenomena" by Bird, Stewart, and Lightfoot.
PREREQUISITESFluid mechanics engineers, researchers in rheology, and professionals involved in the design of systems handling non-Newtonian fluids will benefit from this discussion.
For a purely viscous non-Newtonian fluid (not viscoelastic), you use exactly the same form of equation as for a Newtonian fluid (see Bird, Stewart, and Lightfoot), except that you represent the viscosity as a function of the 2nd invariant of the rate of deformation tensor. There is a form of this functionality that gives the same behavior as a power law fluid for simple shear flows.ccrook said:How does one setup the stress tensor for a non-Newtonian fluid? I know that for any fluid the normals should be the pressure and for a power law fluid the shear stress in the direction of flow is related by K(du/dy)^n. Does this mean that all other components are 0 for a symmetric pipe or rectangular duct?