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UDF for an annular disc.

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Old   May 26, 2016, 14:14
Default UDF for an annular disc.
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Alex Skerhut
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Hey I have an annular disc whose radius goes from 5mm to 10mm and I need to write a UDF as a function of angle and radius in cylindrical co-ordinate system. I will use the DEFINE_PROFILE function but I am not able to understand how to in corporate the radius and the angle variation in disc.

Kindly help, I have searched a lot but not able to figure out the solution.
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Old   May 26, 2016, 19:18
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Are you running Fluent with Cartesian coordinates? You can convert between cylindrical to Cartesian coordinates with (x,y) = (r cos(theta),r sin(theta)). Note: you need to account for the displacement of the circle if it's not centred about the origin.
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Old   June 3, 2016, 02:27
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Alex Skerhut
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Yes, I am running fluent with cartesian coordinates. My global coordinate system is 10mm above the centre of the annular disc and I want to define heat flux using DEFINE_PROFILE as a function of radial distance and angular position. Heat flux if of type = r - sin(theta) form multiplied by some constants. I am unable to write UDF for such function in cylindrical coordinates. Also theta is the angle measured from which of the axis. I need to measure theta in anticlockwise sense from the x axis along z axis. Please help me writing the UDF for the above function.

Please help me and thanks in Advance.
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rotary parts, rotating boundary, rotating disc, rotating disk, udf code

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