# UDF for an annular disc.

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 May 26, 2016, 14:14 UDF for an annular disc. #1 New Member   Alex Skerhut Join Date: Mar 2016 Posts: 10 Rep Power: 3 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.

 May 26, 2016, 19:18 #2 Senior Member   Join Date: Mar 2015 Posts: 800 Rep Power: 11 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.

 June 3, 2016, 02:27 #3 New Member   Alex Skerhut Join Date: Mar 2016 Posts: 10 Rep Power: 3 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.

 December 30, 2017, 07:12 #4 Member   Join Date: Apr 2017 Location: india Posts: 34 Rep Power: 2 DID u find a solution for it @alexskerhut?

 Tags rotary parts, rotating boundary, rotating disc, rotating disk, udf code

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