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What is mean by Normalized Force on Polyline 1? |
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#1 |
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Qazi Shahzad Ali
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Hi all,
What is “Normalized Force on Polyline 1” in cfx Post? Since cfx help manual does not provide any meaningful explanation to its normalized factor. I have a circular polyline, although it appears to be Normalized force, on the other hand, while on plotting a chart (Normalized Force on Polyline 1 vs Theta) it shows units of force [N] as well? Can somebody tell the real explanation behind this? |
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#2 |
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Glenn Horrocks
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Assuming what you are plotting is the Force variable over a line object - this is not meaningful. The Navier Stokes equations cannot exert a force on a line, only over surfaces (via pressure and shear stresses, both of which require surfaces).
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#3 |
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Qazi Shahzad Ali
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Glenn, thank you very much for the answer. You are absolutely right in your views about force on the line.
As per CFX help, forceNorm_x()@Polyline1 fucntion can find the force per unit width on polyline created by intersecting a locator on a boundary. However, in my case, this circular polyline (red color in enclosed pic) is created by boundary interaction of surface and vertical plane. I want to find flow induced forces (x, y directions so that to further convert into norm and tang comp) along the circular path at polyline location of axial flow (z-axis). The force_x@surface or forceNorm_x()@Polyline1 are not serving my purpose, so that particular reason I used Normalized Force on Polyline 1, which provided few results but I am unaware of its normalized factor and confusion about units [N]. If I am wrong, then kindly guide me how to find normal and tangential force component at particular location in any other possible way? |
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#4 |
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Glenn Horrocks
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Then you probably want to look at the line integral of pressure and the wall shear at the line, not the force variable. You will need to take account of the wall normal direction. Look in the CFX reference manual for all the variables you need to do this.
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#5 |
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Nidhi
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When you say line integral of the pressure and wall shear, do you mean lengthInt from the function commands? The length integral gives me information in terms of N/m or J/m2. Shouldn't the values be in terms of just N?
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#6 |
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Glenn Horrocks
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That is right, line integral of pressure will give N/m. So this is force per unit length, which is useful in interpreting 2D simulations. You multiply the N/m result by the third dimension length to get the total force.
In 3D you want an area integral of pressure to give a force. But easier is just to use the force() function.
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#7 | |
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Nidhi
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