# Torque Calculation of 2D rotating disk

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 July 2, 2012, 01:57 Torque Calculation of 2D rotating disk #1 New Member   playmaster Join Date: Aug 2010 Posts: 13 Rep Power: 6 Dear all I have simulated a rotating disk floating in water by a 2D model in CFX using symmetry boundary condition. after solution, I want to determine torque on this disk. would anyone help me how to calculate it? Thank you in advance, playmaster

 July 2, 2012, 08:44 #2 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,824 Rep Power: 85 Use the function calculator in CFD-Post, with the force_x/y/z or torque_x/y/z functions.

 July 3, 2012, 14:34 #3 New Member   playmaster Join Date: Aug 2010 Posts: 13 Rep Power: 6 thank you ghorrocks, actually I did, but I expected that it would be about 11000 N.m, when I saw the calculated value (about 450 N.m). It seems that torque of 2D antisymmetric flow with swirl competent need another calculations by writing an expression.

 July 3, 2012, 19:06 #4 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,824 Rep Power: 85 Have you taken into account the lack of a z dimension in 2D models? So your expected value will actually be of units Nm/m (where the /m comes from per metre of length in the z dimension), and your CFX model will have a torque over whatever extrusion length you did in the z direction, so you need to convert that to Nm/m also.

 July 6, 2012, 02:44 #5 New Member   playmaster Join Date: Aug 2010 Posts: 13 Rep Power: 6 Mesh between rotor and stator was generated in X-Y plane. Then , in CFX, the domain has been considered as a rotating domain around X-axis. In CFD post, torque was claculated around x-axis. it is also interesting that I expect that torque value around Y and Z axis be zero, but its value around Z- axis is greater than X-axis. Confusing!!!

 July 6, 2012, 06:27 #6 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,824 Rep Power: 85 Confusing - yes definitely, I did not understand a word you said. But my point (which I suspect I did not make clearly enough last post) is that if you make the z extrusion 1mm it will give you half the torque of if you made it 2mm. The simulation is not normalised for 2D, you have to do that yourself.

 July 26, 2012, 23:55 #7 Member   Khayyamian Join Date: Dec 2010 Posts: 46 Rep Power: 6 Hey Glenn; I had the same problem. you are telling the output value of the solver is actually in Nm/m. is it right? the problem is that there is no extrusion in z direction. actually it is a disk rotating around x axis? So I dont understand that how output value should be converted in real 3D case? thanks in advance for your reply.

 July 27, 2012, 06:25 #8 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,824 Rep Power: 85 No. The output of the solver in simply Nm on the body as modelled. You have to convert that to a 2D result yourself if that is of relevance. I do not understand your second sentence, can you post an image?

 July 27, 2012, 13:23 #9 Member   Khayyamian Join Date: Dec 2010 Posts: 46 Rep Power: 6 Hey Glenn; thanks for reply. I would like to model a disk like in this figure: http://www.cfd-online.com/Forums/att...1-geometry.jpg for simplicity I used 2d axisymmetric swirl. so the computational domain and geomtry in 2d is like: http://www.cfd-online.com/Forums/att...nal-domain.jpg the output torque is much lower than expected torque! Do you mean that 450 Nm torque which I obtained from solver has to be changed? Please note that I expect final torque to be about 10K Nm. Regards.

 July 28, 2012, 08:03 #10 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,824 Rep Power: 85 In your case, from the torque returned on the disk you will have to double it (for symmetry) then scale it up to 360 degrees - so it you model a 2 degree slice you will need to multiply it by 180. Also this type of flow generates 3D features when the rotational velocity is high enough, and if you are in this regime your 2D model will be inaccurate. Your model is only applicable in the 2D flow regime.

 July 29, 2012, 14:45 #11 Member   Khayyamian Join Date: Dec 2010 Posts: 46 Rep Power: 6 Glenn; I think I could not explain my model well enough. So I will ask in another way: If you see Tutorial Guide: Using a Single Rotating Reference Frame ( 2D, axisymmetric, co-rotating disk cavity system) there are two rotating disks. in this model there is no 2 degree slice of the disk. it is just a section of the disk. and so it is a line (2D). Can you tell me, for example in that example, how you calculate fluid viscouse torque on the right rotating wall? will you just use the value returned by going to Report>Force>Moment around x-axis? Thanks for your reply!

 July 29, 2012, 19:42 #12 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,824 Rep Power: 85 Which example are you referring to?

 July 31, 2012, 16:50 #13 Member   Khayyamian Join Date: Dec 2010 Posts: 46 Rep Power: 6 See below, please: https://www.sharcnet.ca/Software/Flu...tg/node144.htm

 July 31, 2012, 19:08 #14 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 10,824 Rep Power: 85 That is a fluent example. CFX is a bit different. Fluent has a true 2D solver, but CFX does not. hadikhayyamian and zhaitb like this.

 Tags cfx, rotating disk, torque

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