## Objective

In this article, you will learn how to predict linear and exponent coefficients for a power law porous media using experimental data.

## Overview

In our simulations, often times we have to model a structure as a porous media. These structures can be as simple as perforated plates and as complex as radiators. The objective, however, is the same: save mesh cells while still accounting for local pressure losses.

In the Power Law formulation, pressure losses are given by:

$$\Delta P = C_0 \cdot \rho \cdot L \cdot U^{C_1}$$

- ΔP is the pressure drop across the porous media (Pa);
- ρ is the density of the fluid (kg/m³);
- L is the thickness of the porous zone (m);
- U is the fluid velocity (m/s);
- C0 is the linear coefficient and;
- C1 is the exponent coefficient.

*Note*: with the power law formulation, pressure drop will always be isotropic within the medium. If your media is anisotropic, make sure to check out the Darcy-Forchheimer approach. For compressible flows, check the Fixed Coefficient model.

## Power curve fitting approach

Using experimental data for pressure *versus* velocity, we can extract the coefficients.

For example, let’s take a 0.01 meters thick perforated plate and water as fluid. The following case was simulated on SimScale at 6 different velocities.

The table below shows the results from the simulations:

\begin{array} {|r|r|} \hline Velocity (m/s) & ΔP (Pa) \\ \hline 0.5 & 960 \\ \hline 0.7 & 1840 \\ \hline 1 & 3720 \\ \hline 1.2 & 5390 \\ \hline 1.5 & 8240 \\ \hline 2 & 14690 \\ \hline \end{array}

By fitting this data set with a power curve, we get the following:

The expression below is extracted from the plot:

$$\Delta P = 3736.2 \cdot U^{1.9685}$$

From that, we obtain C0 and C1:

$$C_0 = {3736.2 \over \rho \cdot L} = {3736.2 \over 997.33 \cdot 0.01} = 374.62$$

$$C_1 = 1.9685$$

In the platform, you can readily create a porous media and assign it to a volume.

Now to compare experimental data with simulation results:

The power law model proves to be accurate in accounting for pressure losses.

Note

If none of the above suggestions did solve your problem, then please post the issue on our forum or contact us.