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The explanation of shear rate in laminar flow is straightforward: We imagine small layers of fluid that glide on each other. Now, in turbulent flow, this does not work as there are no layers. If I want to know the apparent viscosity of a shear thinning (or other non-Newtonian) liquid, I need to know the shear rate. I've asked about this before here, and received an answer. However, I can't solve the Navier Stokes equation, so someone has to walk me through it or present me with an answer.

Fluid may be assumed to be a power-law fluid.

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marked as duplicate by Brandon Enright, tpg2114, BebopButUnsteady, dmckee Dec 16 '13 at 0:29

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I thought that the shear rate was the square root of the sum of the squares of the entries of the deformation matrix divided by 2, that is $$ \dot\gamma = \left(\frac12 \sum_{i=1}^3 \sum_{j=1}^3 \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right)^2 \right)^{1/2} .$$ This matches the usual formula for a uniform shear, and is invariant under change of coordinates. Notice that this is a function of space, and so varies from point to point in a non-laminar flow.

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