Negative power and torque(Axial Turbine analysis)

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 December 18, 2011, 13:10 Negative power and torque(Axial Turbine analysis) #1 New Member   Join Date: Jul 2010 Posts: 23 Rep Power: 9 Sponsored Links Dear all, I have modeled an axial turbine in CFX. I used the macro calculator to get the torque and power and it gives me negative torque and power! What does it mean? If a negative torque is just mentioning the direction of rotation so why should I get a negative power(since P=T.Omega)? Thanks

 December 18, 2011, 17:43 #2 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 13,651 Rep Power: 105 It means you have not hit the steady state operating condition. For instance, if you run a turbine at too high a rotational velocity you will get negative net torque - this means the rotor is running too fast and will decelerate. If you run it too slow it will generate positive torque and accelerate. The steady state operating point is where the net torque is zero.

 December 18, 2011, 18:05 #3 Super Moderator   Sijal Join Date: Mar 2009 Location: Islamabad Posts: 4,329 Blog Entries: 6 Rep Power: 45 What is the value of total pressure (stationary reference frame) at inlet and outlet? What type of boundary conditions you are applying at inlet and outlet.

December 19, 2011, 03:00
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Thank you both.
The rotor speed is -1000 rpm, the input BC is a mass flow of 250 kg/s and the output BC is static pressure of 1 atm.
The results are attached.
Attached Images
 results.png (16.0 KB, 117 views)

 December 19, 2011, 04:51 #5 Super Moderator   Sijal Join Date: Mar 2009 Location: Islamabad Posts: 4,329 Blog Entries: 6 Rep Power: 45 1. cant you apply the total pressure at inlet? 2. Why efficiency is more than 100%, it seems that either rotation direction is wrong (working as compressor) or solution is not converged. 3. Did you model the NGV before the rotor, if no how you are specifying the velocity components in axial and tangential dirction along with correct sign. (obviously r component is zero).

 December 19, 2011, 05:10 #6 Super Moderator   Sijal Join Date: Mar 2009 Location: Islamabad Posts: 4,329 Blog Entries: 6 Rep Power: 45 did you specify the correct rotation axis in macro calcular panel?

December 19, 2011, 05:36
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Quote:
 Originally Posted by Far 1. cant you apply the total pressure at inlet? 2. Why efficiency is more than 100%, it seems that either rotation direction is wrong (working as compressor) or solution is not converged. 3. Did you model the NGV before the rotor, if no how you are specifying the velocity components in axial and tangential dirction along with correct sign. (obviously r component is zero).
1.Unfortunately not.
2.I know that, the rotation direction is correct but the interesting point is that even changing it will not lead into a positive magnitude.
3.I have specified mass flow rate at inlet in turbo mode.

Quote:
 Originally Posted by Far did you specify the correct rotation axis in macro calcular panel?
Yes
Attached Images
 BC.png (14.0 KB, 64 views) geometry.png (76.7 KB, 120 views)

 December 19, 2011, 06:48 #8 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 13,651 Rep Power: 105 Your mesh looks very coarse. Have you read this FAQ: http://www.cfd-online.com/Wiki/Ansys..._inaccurate.3F Especially the bit about mesh resolution?

 December 20, 2011, 02:10 #9 New Member   Join Date: Jul 2010 Posts: 23 Rep Power: 9 @ghorrocks: I decreased the mesh size to 0.003 but no lock... it is still negative. and the solver manager shows this for every step in the out file. +--------------------------------------------------------------------+ | ****** Notice ****** | | A wall has been placed at portion(s) of an OUTLET | | boundary condition (at 40.1% of the faces, 43.3% of the area) | | to prevent fluid from flowing into the domain. | | The boundary condition name is: R1 Outlet. | | The fluid name is: Water. | | If this situation persists, consider switching | | to an Opening type boundary condition instead. | +--------------------------------------------------------------------+ The solver reaches an rms value of 1e-4 at about 50 steps. Last edited by mak86; December 20, 2011 at 02:30.

 December 20, 2011, 05:00 #10 Super Moderator   Sijal Join Date: Mar 2009 Location: Islamabad Posts: 4,329 Blog Entries: 6 Rep Power: 45 1. decrease static pressure at oultet 2. From the Picture of your geomtry, why geomtry converged to single line on hub side, is this a design feature.

December 20, 2011, 05:10
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Quote:
 Originally Posted by Far 1. decrease static pressure at oultet 2. From the Picture of your geomtry, why geomtry converged to single line on hub side, is this a design feature.
1.Static pressure is zero.
2. That is where hub ends.

 December 20, 2011, 05:38 #12 Super Moderator   Sijal Join Date: Mar 2009 Location: Islamabad Posts: 4,329 Blog Entries: 6 Rep Power: 45 can you please attach the ccl file?

December 20, 2011, 06:09
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Here it is.
Attached Files
 AxialTurbine.zip (3.5 KB, 47 views)

December 20, 2011, 06:36
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Quote:
 I decreased the mesh size to 0.003 but no lock... it is still negative.
Look a bit deeper. Graph the torque versus mesh size. Is it trending in the right direction? Is it all over the place? This can tell you if you are close or not.

You outlet is showing lots of back flow and the suggests your outlet is too close to the blades. You will need to extend your domain further downstream.

 December 20, 2011, 06:40 #15 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 13,651 Rep Power: 105 And please show some images of the flow field - steamlines would be nice.

 December 20, 2011, 07:49 #16 Super Moderator   Sijal Join Date: Mar 2009 Location: Islamabad Posts: 4,329 Blog Entries: 6 Rep Power: 45 I guess you are making some common/basic mistake in setting-up the problem in CFX. Are you clear that the rpm are -1000 or 1000. Could you please show some pics of CFX pre with axis visible. What about reference pressure, is it also equal to 0?

 December 20, 2011, 10:03 #17 New Member   Join Date: Feb 2011 Posts: 13 Rep Power: 8 by default in CFX pre if you give negetive speed(- sign) then rotor rotates anticlockwise. find the enthalpy drop ,for turbine the enthalpy will reduce it from inlet to outlet.

 December 20, 2011, 10:16 #18 Super Moderator   Sijal Join Date: Mar 2009 Location: Islamabad Posts: 4,329 Blog Entries: 6 Rep Power: 45 not necessary, it depends on the geometry and also whether it is turbine or compressor. Just take an example of twin spool turbofan engine.

December 20, 2011, 15:33
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First of all, thank you all.
Quote:
 Originally Posted by ghorrocks Look a bit deeper. Graph the torque versus mesh size. Is it trending in the right direction? Is it all over the place? This can tell you if you are close or not. You outlet is showing lots of back flow and the suggests your outlet is too close to the blades. You will need to extend your domain further downstream.
Decreasing the mesh size did not influence the torque.
Extending the domain at downstream increased the backflow area to 90 percent.
Quote:
 Originally Posted by ghorrocks And please show some images of the flow field - steamlines would be nice.
Quote:
 Originally Posted by Far I guess you are making some common/basic mistake in setting-up the problem in CFX. Are you clear that the rpm are -1000 or 1000. Could you please show some pics of CFX pre with axis visible. What about reference pressure, is it also equal to 0?
Omega equals -1000rpm and as I said even changing it to 1000 did not influence the negative torque.
Attached Images
 Pressure.jpg (36.9 KB, 64 views) Streamline.jpg (56.3 KB, 73 views) VelocityVector.jpg (44.9 KB, 57 views)

December 20, 2011, 17:08
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Glenn Horrocks
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Quote:
 Decreasing the mesh size did not influence the torque.
Sounds extremely unlikely. What meshes did you check?

Quote:
 Extending the domain at downstream increased the backflow area to 90 percent.
Then you need to go further downstream.

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