How do I calculate Reynold's number for a non-newtonian fluid?

Question

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:

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.

Answer 1

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.

Answer 2

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:

28. 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.

Answer 3

We need to define effective viscosity of the fluid for a given a set of fluid flow conditions to determine Reynolds numbers of a non-Newtonian fluid. Please refer to the paper below for more information.

Shende, T., Niasar, V.J. & Babaei, M. Effective viscosity and Reynolds number of non-Newtonian fluids using Meter model. Rheol Acta (2020). https://doi.org/10.1007/s00397-020-01248-y

Answer 4

Similar to Steven Mathey, I am not an expert in non-Newtonian flows, but if your fluid can be modeled as a power law fluid (let's say using the Oswald-de Waele equation)

$$\tau(r)=K(T)\left(\frac{du}{dr}\right)^n$$

you can work out a version of the Reynolds number commonly used for non-Newtonian flows.

• For $n<1$: the fluid is pseudoplastic (shear-thinning)
• For $n>1$: the fluid is dilatant (shear-thickening)

As Piya mentioned, the effective or apparent viscosity should also be defined as $$\mu_\text{app}=K(T)\left(\frac{du}{dr}\right)^{n-1}$$ which will be more helpful towards the middle of my answer.

I will detail the derivation described in this article for non-Newtonian flow in a circular duct.

1. Start with a force balance on a cylindrical differential volume within the duct. The result will be $\frac{dp}{dz}=\frac{2\tau}{r}$ where $p$ is the pressure, $z$ is the direction of the flow, and $r$ is the local radius. Extending this radius to the diameter of the duct gives $\frac{dp}{dz}=\frac{4\tau_w}{D}$.
2. Using the Oswald-de Waele equation, solve for the velocity profile and find the average velocity $\bar{V}=\frac{2}{R^2}\int_0^Rru(r)\,dr$. You can do some algebra and find the wall shear stress to be $\tau_w=K_n\left(\frac{8\bar{V}}{D}\right)^n$ where $K_n=\left(\frac{3n+1}{4n}\right)^nK$. Note that for $n=1$ you get back to Newtonian flows, so $K_1=\mu_\text{app}=\mu$ of the Newtonian fluid.
3. The Darcy-Weisbach friction friction factor is defined as $f=\frac{8\tau_w}{\rho\bar{V}^2}$. The article I cited has a 2 in the numerator, but I have seen an 8 more often for laminar flows. You should know for laminar flows that $f=\frac{64}{{Re}_D}$ (${Re}_D<2300$). From other first principles, you can then calculate what is known as the Metzner-Reed Reynolds number, which is... $${Re}_\text{MR}=\frac{\rho D^n\bar{V}^{2-n}}{K_n}=\frac{K_1}{K_n}\left(\frac{D}{\bar{V}}\right)^{n-1}{Re}_D$$

I recommend reading the article I cited above and Chapter 7 of this Chemical Engineering text which brought me up to speed for a project I am working on.