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April 19, 2013, 11:15 |
wind turbine
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#1 |
Senior Member
Francisco
Join Date: Mar 2012
Location: Spain
Posts: 273
Rep Power: 15 |
Hello everybody!
I'm working on a three blade horizontal axis wind turbine of 29m of radius. I have a 120º sector cylinder of 5R of raidus, 5R upstream and 15R downstream. My mesh is a 4 mill cells hexa mesh with rather good quality. The target is to measure the torque generated for 15m/s and 10rpm. To achieve this I'm increasing the windspeed from 5m/s to 10m/s and 15m/s, with 1rpm, 2rpm and 3rpm of angular speed. I'm using single precision (I've also tested double seeing no differences), SIMPLE, least squares for gradients, 2nd order for pressure and 2nd order for momentum. Residuals started at 1e-7, then I've reduced them to 1e-5 and 1e-4 to speed up convergence. I've first tried the steady laminar case, but with no success. For 5m/s I got reversed flow at pressure outlet, and for 10 and 15m/s I got very low torque coefficient, almost 10 times lower than the experimental data. At least flow vectors seemed to be correct. Then I've tried the transient laminar case. I've used 1st order time scheme to use the adpative timestepping: 1e-7 to 1e-3 s. Torque coefficient looks better but reverse flow appears and vectors looks the opposite way they should be! I've even doubted if the sens of rotation is okay and have tried the opposite. Results are the same. How can I avoid the reverse flow at pressure outlet? This can be due to not large enough domain, but I've checked literature and mine is pretty okay... It can also be due to bad boundary conditions, but I'm using the usual: velocity inelt, periodic, symmetry and pressure outlet. Can anybody tell me how reverse flow can be avoid? And if it is the main cause of such a poor predicition? Is my rotating reference frame okay? or it should be the opposite? See pictures. Any comment or suggestion is highly appreciated. Thanks a lot! |
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April 20, 2013, 11:53 |
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#3 |
Senior Member
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reversed flow in farfield and in few cells should not be the problem. Try some turbulence model as you have to decide it based on your reynolds number and keeping in view dia of your wind turbine and relative velocity, I don't think flow is laminar.
Last edited by Far; April 25, 2013 at 00:27. |
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April 24, 2013, 06:29 |
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#4 |
Senior Member
Francisco
Join Date: Mar 2012
Location: Spain
Posts: 273
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I know it's not a laminar flow, but I wanted to test this case as a upper limit of torque prediction as I had read.
I've run the case with k-omega SST model (inlet default values). No change in the solution. Reversed flow appears in the same amount and torque coefficient es still very low. When I check the flow patern it's complety wrong. Absolute velocity vectors go the opposite way and relative vectors too! (see pictures) What's more, turbulent viscosity ratio limited to 1e5 is starting in a few cells! Why is the flow going the opposite way? Is the rotating reference frame right as I exposed in the first post http://www.cfd-online.com/Forums/flu...tml#post421762 Or is it due to the reversed flow? I'm stuck! Please help! Thanks |
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April 24, 2013, 06:46 |
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#5 |
Senior Member
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Looks like you have wrong boundary conditions and flow angles.
How did you determine them? You know in wind turbine incoming flow is higher angle (towards 90) and relative flow is less than stall angle. Whereas in your case I see relative velocity at right angle to turbine? and free stream velocity is coming from trailing edge to leading edge !!! Absolute velocity should remain same Why this is so? R you running your case at very low RPM? This is also because of weired flow angles (but still it should not make freestream air coming from rear side). If so the don't nt expect solution and convergence at such off design condition and massive separation is expected and you are trying to get solution in stall regime. Moreover with laminar flow your flow gets more worsen as laminar flow has more tendency for separation. PS: are you sure solution is fully converged? |
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April 24, 2013, 07:32 |
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#6 | ||
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Francisco
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Location: Spain
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Quote:
Incident velocity is okay, but relative due to rotating ref frame is the opposite. I've also tried the other sense of rotation but it looks the same. Quote:
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April 24, 2013, 07:45 |
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#8 | |
Senior Member
Francisco
Join Date: Mar 2012
Location: Spain
Posts: 273
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Quote:
Relative flow should be parallel to incident flow coming from inlet as it is rotating with the ref frame. So absolute one, observed from outside the rotating ref frame, should look like the real flow. Right? |
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April 24, 2013, 22:18 |
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#9 |
Senior Member
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You should be able to obtain a steady solution with some turbulence model. When you chose a turbulence model, re-initialize the flow from the inlet b.c. You may want to change the symmetric b.c. to velocity inlet b.c. at first. The reversed flow may appear at the outlet for first tens or hundreds steps but will disappear thereafter. Make sure that the origin and axis for your rotation frame is set right. Never use laminar flow model unless you are a hundred percent sure what you are doing. Do not turn on the unsteady solver because it is unnecessary here. You may need to reduce the URF or use the coupled solver with a smaller Courant number with your final inlet velocity and rpm but you do not have to increase them gradually.
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April 25, 2013, 07:11 |
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#11 | ||||
Senior Member
Francisco
Join Date: Mar 2012
Location: Spain
Posts: 273
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Quote:
Should I try non shear stress wall b.c.? Would that help to avoid the reversed flow at outlet? Quote:
1) Rotating reference frame clockwise, relative flow should be anti-clockwise? (see last post picture) 2) Rotating reference frame anti-clockwise 3) Rotating mesh clockwise In all cases there's reversed flow in the same amount (23k cells in outlet) and vectors for absolute flow goes the opposite way! Which setup should I use? Rotating reference frame in the same sense as the turbine? Opposite to it? Which difference is there with rotating mesh? Quote:
Quote:
Yes, the blade is twisted but its center line is stragith from hub to the tip. Thanks! |
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April 25, 2013, 07:48 |
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#12 | |
Senior Member
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Quote:
Also be careful in giving tangential flow angle. Sign is important. For Axial flow direction I am trying to digg my memory but cannot recall ... Look into fluent help |
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April 25, 2013, 07:58 |
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#13 |
Senior Member
Francisco
Join Date: Mar 2012
Location: Spain
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I am applying the right hand rule, what I don't know is how the relative flow-frame movement is.
What do you mean by tangential flow angle? The angle of attack? It is determined by the axial flow speed (inlet) and the angular velocity. I don't get your point here. |
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April 25, 2013, 13:48 |
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#14 |
Senior Member
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Ok. Let me explain:
1. Wrap you figures in the rotation direction. You thumb will give you - or + sign. If it is pointing towards positive z then it is +1 if it is in -ve z then you have -1. 2. Lets say you rotation axis is -Z and now if you incoming air is going towards +z direction then you will give the -ve component of angle in axial direction. If it is totally axial flow then you will have the -axial component (i.e. -cos0 = -1). If incoming flow is at some angle then you will have to specify the tangential component as well. Again if it is in the direction of rotation then it is +ve otherwise -ve. While specifying the boundary conditions you don't need to consider the relative flow velocity and angle. It is taken care by solver. But if it is different than the expected direction and magnitude then your boundary conditions must be wrong. |
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April 26, 2013, 05:10 |
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#16 | |
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Francisco
Join Date: Mar 2012
Location: Spain
Posts: 273
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I've tried using wall bc with zero shear stress instead of symmetry condition and reversed flow has disappeared, at least for the laminar case U15m/s-30rpm. Residuals cannot reach 1e-5 (see picture 1).
Torque coefficient is still low CT=0.008, and it oscillates too much, from 0 to 0.015 (see picture 2). Relative flow looks the right way (see picture 3). I'm trying now the k-omega SST case, I'll share my results. Quote:
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April 26, 2013, 05:14 |
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#17 |
Senior Member
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How you define the coefficient of torque? did you make a custom field function?
Did you get the point about setting the axis of rotation and flow direction? Why you want to have the 1e-05 as convergence criteria? it is too tight convergence criteria in Fluent |
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April 26, 2013, 05:35 |
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#18 | ||
Senior Member
Francisco
Join Date: Mar 2012
Location: Spain
Posts: 273
Rep Power: 15 |
Quote:
Quote:
I don't know, I just wanted to get it very converged. Which will be an appropiate value? |
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April 26, 2013, 05:46 |
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#19 | |||
Senior Member
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Quote:
Maximum (theoretical) torque is = 0.5 * density * Area * V1^2 Quote:
Quote:
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April 26, 2013, 06:09 |
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#20 | |
Senior Member
Francisco
Join Date: Mar 2012
Location: Spain
Posts: 273
Rep Power: 15 |
Quote:
I know it's independent, but when I checked the resulting flow it was going the opposite way and I think it's due to reversed flow in the outlet that modiefies all the flow pattern. I'll check that once I have good results. Thanks. |
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Tags |
hawt, reverse flow, srf, torque, wind turbine |
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