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Bollonga April 19, 2013 10:15

wind turbine
 
3 Attachment(s)
Hello everybody!

I'm working on a three blade horizontal axis wind turbine of 29m of radius. I have a 120 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!

blackmask April 20, 2013 01:14

How could it be laminar?

Far April 20, 2013 10:53

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.

Bollonga April 24, 2013 05:29

2 Attachment(s)
Quote:

Originally Posted by blackmask (Post 421884)
How could it be laminar?

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

Far April 24, 2013 05:46

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?

Bollonga April 24, 2013 06:32

1 Attachment(s)
Quote:

Originally Posted by Far (Post 422794)
Looks like you have wrong boundary conditions and flow angles.

How did you determine them?

Well, I upload a picture of the wanted flow angles that I guess helps to understand the case. Inlet velocity is 5m/s for this case with an angular velocity of 1.034 rad/s.
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:

Originally Posted by Far (Post 422794)
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?

I guess flow coming from trailing to lead edge is due to reversed flow disturbing all the flow. Or maybe for bad rotation of the reference frame. That's exactly what I want to solve.

Far April 24, 2013 06:36

well we need to find out the root cause of this behavior. I suspect the wrong case set up at some point. Where it is I dont know!!!

Bollonga April 24, 2013 06:45

Quote:

Originally Posted by Far (Post 422806)
well we need to find out the root cause of this behavior. I suspect the wrong case set up at some point. Where it is I dont know!!!

Yes, there must be some set up error. :(

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?

blackmask April 24, 2013 21:18

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.

Far April 24, 2013 23:30

One basic question: Do you have twist and camber in blade?

Bollonga April 25, 2013 06:11

Quote:

Originally Posted by blackmask (Post 422965)
You may want to change the symmetric b.c. to velocity inlet b.c. at first.

To inlet b.c.? Would that be realistic? and when should I change that back to symmetry?
Should I try non shear stress wall b.c.? Would that help to avoid the reversed flow at outlet?

Quote:

Originally Posted by blackmask (Post 422965)
Make sure that the origin and axis for your rotation frame is set right.

I'm sure about the center of rotation, I moved the mesh to make it the (0,0,0). I'm not pretty sure about the rotating axis: (0,0,-1) or (0,0,1). I want the turbine to rotate clockwise. I've tried several options:
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:

Originally Posted by blackmask (Post 422965)
Do not turn on the unsteady solver because it is unnecessary here.

Thats's what I thought about the steady solution. This flow doesn't seem to be very time-dependent for an optimun angle of attack.

Quote:

Originally Posted by blackmask (Post 422965)
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.

No need to increase rpm-windspeed gradually? That's what I read at fluent user guide.

Quote:

Originally Posted by Far (Post 422971)
One basic question: Do you have twist and camber in blade?

Yes, the blade is twisted but its center line is stragith from hub to the tip.

Thanks!

Far April 25, 2013 06:48

Quote:

I'm not pretty sure about the rotating axis: (0,0,-1) or (0,0,1).
Use right hand rule to determine + or - z .

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 :mad:... Look into fluent help :o

Bollonga April 25, 2013 06:58

Quote:

Originally Posted by Far (Post 423078)
Use right hand rule to determine + or - z .

I am applying the right hand rule, what I don't know is how the relative flow-frame movement is.

Quote:

Originally Posted by Far (Post 423078)
Also be careful in giving tangential flow angle. Sign is important.

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.

Quote:

Originally Posted by Far (Post 423078)
For Axial flow direction I am trying to digg my memory but cannot recall :mad:... Look into fluent help :o

I don't get your point here.

Far April 25, 2013 12:48

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.

blackmask April 26, 2013 01:32

Hi, Bollonga, could you please post the result of the TUI command
/report/summay yes "report.sum"
so that we could put the discussion on a more firm basis.

Bollonga April 26, 2013 04:10

4 Attachment(s)
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:

Originally Posted by blackmask (Post 423236)
Hi, Bollonga, could you please post the result of the TUI command
/report/summay yes "report.sum"
so that we could put the discussion on a more firm basis.

You have the report in the zip file.

Far April 26, 2013 04:14

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

Bollonga April 26, 2013 04:35

Quote:

Originally Posted by Far (Post 423269)
How you define the coefficient of torque? did you make a custom field function?

No, I'm measuring moment coefficient of every wall surface for center (0,0,0) and axis (0,0,-1)

Quote:

Originally Posted by Far (Post 423269)
Did you get the point about setting the axis of rotation and flow direction?

Yes, without reversed flow the rotation seems to be the right way.

Quote:

Originally Posted by Far (Post 423269)
Why you want to have the 1e-05 as convergence criteria? it is too tight convergence criteria in Fluent

I don't know, I just wanted to get it very converged. Which will be an appropiate value?

Far April 26, 2013 04:46

Quote:

No, I'm measuring moment coefficient of every wall surface for center (0,0,0) and axis (0,0,-1)
Torque coefficient = torque you get from CFD / Maximum torque (V4 = 0)

Maximum (theoretical) torque is = 0.5 * density * Area * V1^2

Quote:

Yes, without reversed flow the rotation seems to be the right way.
Reverse flow should not the your rotation axis criteria. :rolleyes:

Quote:

I don't know, I just wanted to get it very converged. Which will be an appropriate value?
Very converged : I think 1e-03 or 1e-04 is good convergence. You can compare them to find out the convergence error. You should also monitor some other value of your interest such as power coefficient or torque coefficient.

Bollonga April 26, 2013 05:09

Quote:

Originally Posted by Far (Post 423278)
Torque coefficient = torque you get from CFD / Maximum torque (V4 = 0)

Maximum (theoretical) torque is = 0.5 * density * Area * V1^2

I know, I was talking about moment coefficient = torque I get from CFD

Quote:

Originally Posted by Far (Post 423278)
Reverse flow should not the your rotation axis criteria. :rolleyes:

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.

Quote:

Originally Posted by Far (Post 423278)
Very converged : I think 1e-03 or 1e-04 is good convergence. You can compare them to find out the convergence error. You should also monitor some other value of your interest such as power coefficient or torque coefficient.

I'll check that once I have good results. Thanks.


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