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I am familiar with the known formula to find Reynold's number in a circular pipe for newtonian fluids:

$$\text{Re} = \frac{\rho \bar{v}D}{\mu}$$

where $\rho$ is the density, $\bar{v}$ is the average velocity, and $\mu$ is the viscosity of the newtonian fluid; and $D$ is the inner diameter of the pipe.

At first, I presumed finding the $\text{Re}$ value for a non-Newtonian fluid is pretty straightforward: replace $\mu$ with whatever is equivalent for a non-newtonian fluid. I know that newtonian fluids hold the property that the viscous shear stress on any infinitesimal surface $dS$ is proportional to the velocity gradient along the normal of $dS$:

$$\tau \propto \frac{dv}{dz}$$

where $\tau$ is the viscous shear stress. Of course, the proportionality constant here is $\mu$.

For non-newtonian fluids, the shear stress is not proportional to velocity gradient, but still remains a function of it:

$$\tau = f\left(\frac{dv}{dz}\right)$$

Various types of non-Newtonian fluids have been experimentally discovered, such as pseudoplastics and dilatants, with their shear-stress and shear rate (velocity gradient) relations:

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I thought we must use the "viscosity-equivalent" for non-Newtonian fluids:

$$\mu_{\text{equiv}}=\frac{\tau}{\gamma}$$

where

$$\gamma = \frac{dv}{dz}$$

This becomes:

$$\mu_{\text{equiv}}=\frac{f(\gamma)}{\gamma}$$

This "viscosity equivalent" becomes a function of shear rate. This is where I hit a roadblock:

  • In a circular pipe, the shear-rate at different distances from center are different, and thus, the "viscosity equivalent" is different at every point in the pipe. What should I substitute $\mu$ with in the newtonian formula to get my result for any non-newtonian fluid?

  • If my entire premise was wrong, I would like to know how I may obtain the Reynold's number for a non-Newtonian fluid.

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  • $\begingroup$ Are you trying to do this to get the pressure drop vs flow rate relationship for non-Newtonian laminar flow in a tube, or are you trying to use a turbulent friction factor correlation? $\endgroup$ – Chet Miller May 26 at 17:43
  • $\begingroup$ @ChetMiller Ideally I'd prefer for both. I know that there are empirical relations that relate $\text{Re}$ with $f$ (fanning friction factor) under laminar and turbulent flows. I believe finding pressure drop vs. flow rate can be found from $\text{Re}$ vs f relations (since wall shear can be obtained from $f$, and fluid velocity from $\text{Re}$). $\endgroup$ – Pritt Balagopal May 26 at 18:40
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I am no expert on non Newtonian fluids, but to me it looks like (unless you have a special type of viscosity) you can't. A non Newtonian fluid is more complicated than a Newtonian one and can not be described by a single dimensionless parameter. I will illustrate this with a specific example of 'shear stress'-'velocity gradient' relation (I guess the 'dilatant' fluid):

$$ f(\gamma) = \mu \gamma + \nu \gamma^2 \, .$$

Making everything dimensionless as for the Newtonian fluid, you can define $ Re = {\rho \bar{v} D}/{\mu}$, as before. You are however stuck with the $\nu$ parameter, which provides a second dimensionless number. For example you can define

$$Re_{nN} = \frac{D^2 \rho}{\nu \bar{v}}\, .$$

I use the index ${nN}$ to signify that that $Re_{nN}$ is a dimensionless number (just like $Re$) that quantifies the non Newtonian viscosity. If $Re_{nN} \gg 1$, ($\nu$ is small and) the non Newtonian viscosity is not important. The opposite happens when $Re_{nN}\ll 1$.

There are however two numbers. The fluid is approximatively Newtonian if $Re \ll Re_{nN}$, and it is far from being Newtonian when $Re \gg Re_{nN}$. I guess that you need $Re \gg 1$ and $Re_{nN} \gg 1$ if you want to reach a turbulent regime and that the flow will be laminar if either $Re \ll 1$ or $Re_{nN} \ll 1$, but we are now entering into the specifics of non Newtonian fluids of which I know nothing. The takeaway message is (unfortunately) that non Newtonian fluids are complicated and depend on more than one parameter.

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If the flow is laminar, then there is no problem, because you don't need to know the Reynolds number. You just integrate the velocity profile for the desired viscosity parameterization. This is particularly easy for a power-law fluid. To check for the onset of turbulence, you can use the viscosity evaluated at the wall shear rate to get the Reynolds number.

For turbulent flow, I suggest researching with Google. However, here is a reference to get you started for the case of a power law fluid:

  1. Dodge, D.W.; Metzner, A.B. Turbulent flow of non-Newtonian systems. AIChE J. 1959, 5, 189–204.

I just briefly looked with Google, and found lots of references covering methodology for turbulent flow.

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  • $\begingroup$ To check for the onset of turbulence, you can use the viscosity evaluated at the wall shear rate to get the Reynolds number. Oh, so do I just replace the $\mu$ in $$\frac{\rho \bar{v}D}{\mu}$$ with the "viscosity-equivalent" I mentioned at the wall region? Would that give me a Reynold's number to adequately test for turbulence? $\endgroup$ – Pritt Balagopal May 28 at 14:03
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    $\begingroup$ The onset of turbulence is not an exact thing, even for a Newtonian fluid. So this is a judgment call. I would use the viscosity evaluated at the wall shear rate, as calculated for laminar flow of the non-Newtonian fluid. $\endgroup$ – Chet Miller May 28 at 14:41

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