# I need to find out the force exerted on an internal wall by the fluid.Could you explain how the "force" is calculated? For example, for a coordinate-aligned wall, the area integration of pressure should equal the force normal to the wall? I checked but "areaInt(Pressure)@Face" does not seem to equal "force_x()@Face".

 There are different methods to get a result for the force on a wall, internal or external. They should all give similar results, but there will be slight differences as they are calculated in different ways.1).Look at the momentum flows at the end of the .out file. for example:+--------------------------------------------------------------------+| W-Mom+--------------------------------------------------------------------+Boundary : Domain 1 Default -2.8533E-02Boundary : in 7.4036E-01Boundary : internalwallup -1.3078E-02Boundary : internalwallup Other -5.5560E-03Boundary : out -6.9320E-01Domain Imbalance : -5.8413E-06Domain Imbalance, in %: -0.0008 %The force in the Z-direction (because I'm looking at W-Mom) is equal and opposite to the change in momentum. The force on internalwallup is therefore 1.3078E-2 N, so that's 0.013 N in the positive Z-direction.2).Look at the Pressure and Viscous Force on the Walls at the end of the .out file:+--------------------------------------------------------------------+| Pressure Force On Walls |+--------------------------------------------------------------------+ X-Comp.Y-Comp. Z-Comp.Domain Group: Domain 1Domain 1 Default 6.4206E-04 -1.9441E-04 0.0000E+00internalwallup 0.0000E+00 0.0000E+00 1.3078E-02internalwallup Other Side 0.0000E+00 0.0000E+00 5.5560E-03 ----------- ----------- -----------Domain Group Totals : 6.4206E-04 -1.9441E-04 1.8633E-02+--------------------------------------------------------------------+| Viscous Force On Walls+--------------------------------------------------------------------+ X-Comp. Y-Comp. Z-Comp.Domain Group: Domain 1Domain 1 Default 8.5505E-06 -7.6983E-06 2.8533E-02internalwallup 1.0876E-06 -1.5087E-06 -5.4205E-11internalwallup Other Side -1.7727E-06 2.7746E-08 6.8420E-11 ----------- ----------- -----------Domain Group Totals : 7.8654E-06 -9.1792E-06 2.8533E-02In this case my wall lies in the X-Y plane, so I would probably be interested in the Z-direction force. The sum of the Viscous and Pressure forces should equal the change in momentum. Here, the Viscous Force is negligible so the Pressure force is the same as the force from 1).3).In CFX-Post, calculate force_z()@Face using the Function Calculator. This will read Momentum Flow data from the results file if it is available and will therefore give you the same answer as 1). Momentum Flow data should be available for all the wall boundary conditions. If you just create an internal plane in CFX-Post and then calculate force_z on that plane, you will get an approximate force based on the advection of momentum through that plane.4).In CFX-Post, calculate the areaInt(Pressure)@Face using the Function Calculator. This will exclude viscous forces. CFX-Post will integrate the Pressure values over the plane. This should give you a similar result to 2), but since CFX-Post is calculating this rather than reading it from the results file it could be slightly different.

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