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Negative power and torque(Axial Turbine analysis) |
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December 18, 2011, 12:10 |
Negative power and torque(Axial Turbine analysis)
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#1 |
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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 |
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December 18, 2011, 16:43 |
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#2 |
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
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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. |
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December 19, 2011, 02:00 |
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#4 |
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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. |
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December 19, 2011, 03:51 |
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#5 |
Super Moderator
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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). |
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December 19, 2011, 04:36 |
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#7 | |
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Quote:
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. Yes |
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December 19, 2011, 05:48 |
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#8 |
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Glenn Horrocks
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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? |
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December 20, 2011, 01:10 |
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#9 |
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@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 01:30. |
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December 20, 2011, 04:10 |
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#11 |
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December 20, 2011, 05:09 |
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#13 |
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Here it is.
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December 20, 2011, 05:36 |
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#14 | |
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Glenn Horrocks
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Quote:
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. |
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December 20, 2011, 05:40 |
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#15 |
Super Moderator
Glenn Horrocks
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And please show some images of the flow field - steamlines would be nice.
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December 20, 2011, 06:49 |
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#16 |
Super Moderator
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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? |
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December 20, 2011, 09:03 |
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#17 |
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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. |
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December 20, 2011, 14:33 |
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#19 | ||
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First of all, thank you all.
Quote:
Extending the domain at downstream increased the backflow area to 90 percent. Quote:
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December 20, 2011, 16:08 |
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#20 | ||
Super Moderator
Glenn Horrocks
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Quote:
Quote:
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