# Calculating the Cp (Power Coefficient ) for a cross flow vertical wind turbine

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 February 12, 2018, 04:46 Calculating the Cp (Power Coefficient ) for a cross flow vertical wind turbine #1 Senior Member   Join Date: Mar 2013 Location: Germany Posts: 357 Rep Power: 14 Hi All I am doing a simulation for vertical axis wind turbine where the flow is cross flow . I have the input details like RPM 409 and wind speed or 15m/sec and I have 20 blades of each 2m length. But for simulation am just using a small part of it 20cm. So now my doubt is Will the effect of 20cm simulation result be same for the 2m blade in real ? Then how can I calculate the Cp or power coefficient ? I calculate the power using the Power = torque_y* omega Omega = RPM*Pi/30 But for Cp it is Rated Power /Power right ? How can i calculate this ? And what is rated power and does cfx directly give me this ? I want to increase the RPM and study how the efficiecny increases or decreases !! Thanks in advance

 February 12, 2018, 04:58 #2 Senior Member   Gert-Jan Join Date: Oct 2012 Location: Europe Posts: 1,835 Rep Power: 27

 February 12, 2018, 05:20 #3 Senior Member   M Join Date: Dec 2017 Posts: 643 Rep Power: 12 The torque should be heavily influenced by the outer regions of the blade because it depends on radius. /edit: deleted my comment on periodic BC

 February 12, 2018, 05:29 #4 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,727 Rep Power: 143 You cannot use rotational boundary conditions to reduce the model size when you are modelling cross flow. The cross flow is not periodic. The most complete way is to just model the entire thing, then you get wake effects and everything - along with a dirty big simulation. You can simplify it down to a single blade and run a series of angles of attack, but then to reconstruct that into torque over the full rotation can be challenging. __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.

 February 12, 2018, 05:51 #5 Senior Member   Gert-Jan Join Date: Oct 2012 Location: Europe Posts: 1,835 Rep Power: 27 I think the deviation of your simplification will depend on the exact geometry and orientation of your blades. If they are straight, then the deviation could be limited. So if you can shed a light, that could be helpful.

 February 12, 2018, 10:40 #6 Senior Member   Join Date: Mar 2013 Location: Germany Posts: 357 Rep Power: 14 Thanks a lot for your replies. Well my geometry is 2m long and its similar to vertical axis wind turbine but kept 90 degree so its like a cross flow. so now I decided to take a small part of it and simulate and then compare it with the experiment results, else I can take the half and then give the symmetry bc'n as well right ?

 February 12, 2018, 16:49 #7 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,727 Rep Power: 143 If the geometry and flow is symmetric then a symmetry boundary makes sense. Regarding what you are modelling - an image would help. Also a key question is how the blades align to the wind flow direction, or if there is no alignment and they are fixed and just go around the main axis. __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.

February 13, 2018, 06:56
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If you see in the image the flow is in the positive x direction ! and this is just a slice of the turbine. both sides are the plates but I gave it as interface
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 February 13, 2018, 07:08 #9 Senior Member   Join Date: Mar 2013 Location: Germany Posts: 357 Rep Power: 14 Also my concern is how to calculate the Power Extracted from the Shaft ? I have the RPM as 400 and Wind Velocity as 15m/sec In CFX I can calculate the torque along the axis of rotor. So omega is 2*pi*RPM /60 but this RPM is Revolution per minute right ?

 February 13, 2018, 17:16 #10 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,727 Rep Power: 143 In this case you definitely need to model all blades. The blades are going to experience massively difference conditions as they rotate, so you need to model the whole thing to capture that. But not a lot happens in the z direction. So I think you could consider a 2D planar model for initial work. The 3D model would include end effects but will be a much bigger simulation. How to calculate power? torque * rotation speed Be careful with omega. You specify it as rpm, but CFX then converts it to a standard unit (radians per second) which means you do not need to convert rpm to radians per second, it has already been done. __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.

 February 14, 2018, 02:30 #11 Senior Member   Join Date: Mar 2013 Location: Germany Posts: 357 Rep Power: 14 Thanks for your reply. So when I do a 2D simulation how about the interfaces on the side walls of the rotor ? There wornt be any interface right ? So in principle I will have only the cylindrical part (in 2D the circle) will be the interface ? Second thing regarding the Power calculation In CFX am getting power in Joules so am making some mistake. !! I just used RPM = 400 Moment = Torque_z()Rotor Power =RPM*Pi*Moment/30 So I dint use any units either for RPM or Moment !! Do I have to use is the question, if yes what should I. My main concern is about this RPM !! Kindly let me know

 February 14, 2018, 03:14 #12 Senior Member   Gert-Jan Join Date: Oct 2012 Location: Europe Posts: 1,835 Rep Power: 27 I always use: N = 400 [rev/min] Torque = abs(torque_z()@Rotor) ShaftPower = abs(N)*2*pi*Torque/60[s/min]/360[degree]

 February 14, 2018, 03:37 #13 Senior Member   Join Date: Mar 2013 Location: Germany Posts: 357 Rep Power: 14 Perfect this works perfectly , thanks a lot !!

 February 14, 2018, 16:57 #14 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,727 Rep Power: 143 You should not need to do the conversion from rpm to rad/s. The conversion is done internally and the variable is automatically stored in rad/s even when you specify it as rpm. So: N = 400 [rev/min] Torque = abs(torque_z()@Rotor) ShaftPower = abs(N)*Torque Should work. Why does your function including the conversion from rpm to rad/s work? Because the 2*pi/60[s/min]/360[degree] evaluates to 1.0 __________________ Note: I do not answer CFD questions by PM. CFD questions should be posted on the forum.