### Pressure Potential with SOLIDWORKS Flow Simulation

###### September 22, 2015 • Michael S.

Not only will we discuss the Pressure Potential checkbox that seen in General Settings (including Project Wizard) and the Boundary Condition, but also how it affects the calculation of hydrostatic pressure.

For example, let’s take a look at a cylindrical column of water. Creating this in Flow Simulation is easy (draw a cylinder and shell it).

Notice the location of the global coordinate system (GCC). This will come in to play later.

For an internal steady state analysis, the Thermodynamic Parameters in the general settings provide the initial conditions (since we’re solving time-dependent equations) and default settings for creating new boundary conditions.

As shown in the screenshot of the Boundary Conditions PropertyManager, we have Pressure Potential enabled. We’ll apply 1 atm (101325 Pa) to the top inside surface. Since Pressure Potential is enabled, what static pressure will actually be defined at this surface?

Before we click solve, let’s do some hand calculations. What do we know?

The height (h) of the column is 1 meter.

We’re using water. Density (r) is 998.16 kg/m^3 at 293.16 K.

Gravity (g) is 9.81 m/s^2 (defined in Analysis Type in General Settings).

The pressure difference between the top and bottom surfaces is r*g*h. This gives us 9791.95 Pa.

Since Pressure Potential is enabled, Flow simulation is automatically changing this applied pressure to account for its location.

Think of it this way. The pressure we define in the property manager is applying that pressure at Y=0m. Since Y=1m in this case, we need to subtract 9791.95 Pa from 101325 Pa. This gives us 91533.05 Pa (0.90 atm). So, how do our results look in Flow Simulation?

Spot on!

Now, what if we turn off Pressure Potential in the boundary condition?

What should our pressure be at the bottom of the column?

With Pressure Potential off, our applied static pressure is unchanged. So, we reverse our calculation to add 0.1 atm to the 1 atm we applied. We get 1.1 atm at the bottom.

The image looks to same, but notice the legend (color bar). Again, we’re spot on. We have our pressure at the bottom of the column at 1.1 atm.

Here’s your homework. Keeping Pressure Potential on and using the same surface for the boundary condition, how can you get the same result that we get when Pressure Potential was off (0.9 atm at top, 1.0 atm at bottom)? Tip: there are two possible answers.

Lesson to remember: The Pressure Potential option modifies the applied static pressure of the boundary condition.

Important Note: Since we only have one boundary condition, it is expected that you will receive the warning that a vortex crosses the boundary condition. Sometimes, this can significantly impact the results accuracy. As we’ve shown in this simple case, there is no impact. However, be aware of this risk. If you are getting unexpected results, you may want to add a second boundary condition. For this simple column, we would create two pressure openings (one at the top and one at the bottom) with the same properties (Pressure Potential enabled). If you have any questions, please let us know.

Michael S.

Senior Application Engineer

support@qintegration.com

800-370-3750