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# How to Predict Linear and Exponent Coefficients for a Power Law Porous Media?

## 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