Required field
Required field
Required field
Required field

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

In this article, you will learn how to predict linear and exponent coefficients for a power law porous media modeling 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:

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

where:

• $$\Delta P$$ is the pressure drop across the porous media $$[Pa]$$;
• $$\rho$$ is the density of the fluid $$[kg/m³]$$;
• $$L$$ is the thickness of the porous zone $$[m]$$;
• $$U$$ is the fluid velocity $$[m/s]$$;
• $$C_0$$ is the linear coefficient and;
• $$C_1$$ is the exponent coefficient.

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 in SimScale at 6 different velocities.

Instead of simulating the original geometry of the perforated plate, we will model it as a porous medium. In the platform, you can readily create a porous media and assign it to a volume or a geometry primitive such as the cartesian box shown below:

The table below shows the results from the simulations:

By fitting this data set with a power curve, we get the following: Figure 3: Fitting the data set with a power curve. y-axis represents the pressure drop and x-axis represents velocity.

The expression below is extracted from the plot:

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

From that, we obtain $$C_0$$ and $$C_1$$:

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

$$C_1 = 1.9685$$

Now to compare experimental data with simulation results: