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Volume Flow Rate and Area * Velocity values different

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Old   July 9, 2013, 00:55
Default Volume Flow Rate and Area * Velocity values different
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Hello,

I have a fluent model, where the fluid is of constant density.
Structured mesh, well converged.

But the volume flow rate what fluent shows is different when I multiply the area and area weighted average of velocity magnitude.

Did not understand ? Why ?

(Volume Flow Rate) Fluent != (Area) fluent * (Area weighted average of velocity magnitude) fluent

Thanks in advance
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Old   July 10, 2013, 03:41
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Is it multiphase?
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Old   July 10, 2013, 04:41
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No .... the fluid is air
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Old   July 10, 2013, 04:46
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That sounds strange... How much do they differ?
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Old   July 10, 2013, 04:49
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Ahh maybe this:
What do you integrate for the "Area weighted average of velocity magnitude"? You need to integrate only the perpendicular (to the surface) component of the velocity vector - not the length of the complete velocity vector!
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Old   July 10, 2013, 04:54
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(Volume Flow Rate) Fluent == (Area) fluent * (Area weighted average of normal component of velocity) fluent

Note the difference between <vector u> dot <vector A> and |magnitude u|*|magnitude A|.
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Old   July 10, 2013, 04:54
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I tried with Velocity magnitude, X, Y and Z velocities .... still they differ.
When I calculate velocity = (volume flow rate/area) fluent and
area weighted avg of (velocity magnitude/X-vel/Y-vel/Z-vel), both differ by almost 4-5m/s
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Old   July 10, 2013, 04:57
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But this isn't what I wrote:
You need the perpendicular velocity component. If your surface is no "simple" cartesian surface, the cartesian (x,y,z) components won't help you.
What you need is "A dot v"!
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Old   July 10, 2013, 05:10
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It is a simple rectangular surface. So don't you think the velocity magnitude is normal to that ?
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Old   July 10, 2013, 05:12
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Not necessarily... Why don't you post pictures?
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Old   July 10, 2013, 05:29
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Let the normal of the rectangle be [nx, ny, nz], and the area-weighted average of x-,y-,z- velocity be [U, V, W], then (U*nx + V*ny + W*nz) should not be differ much from the volumetric flow rate divided by the area.
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Old   July 10, 2013, 11:27
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Thanks !
Will do that.
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