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#1 |
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Senior Member
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Hi Foamers,
I want to calculate torque from the simulation, can you guys help me? I have used MRFSimpleFoam for my simulation. The fan RPM is 750. please give me some idea. Thanks and regards, Sivakumar |
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#2 |
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Member
Laurent B
Join Date: Jun 2009
Location: Lille, FRANCE
Posts: 59
Rep Power: 5 ![]() |
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#3 |
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Senior Member
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Hi Guys,
After a long struggle, I came to the conclusion regarding torque. To calculate torque we need to use this function in the controlDict file. functions ( forces { type forces; functionObjectLibs ("libforces.so"); outputControl timeStep; outputInterval 1; patches (wallBlade); // pname p; // Uname U; rhoName rhoInf; log true; rhoInf 1.205; CofR (0 0 0); } ); results will be as follows: forces(pressure, viscous)((2162.2 -313.192 35.8823) (12.5865 3.1166 3.97532)) moment(pressure, viscous)((1162.4 6862.37 100.944) (-12.5067 35.4299 -50.9462)) now the the output pressure force is not exact pressure (unit m2/s2) so to get exact torque = obtained Torque (moment) / density Thanks, Sivakumar |
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#4 |
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Member
Laurent B
Join Date: Jun 2009
Location: Lille, FRANCE
Posts: 59
Rep Power: 5 ![]() |
Hi,
Does somebody can explain in details how to calculate torque with "forces" library data ? I try to compare experimental and simulation results but there is a big difference. I calculate the torque as follow : Torque = |moment(pressure) + moment(viscous)| and i obtain twice the experimental value... Thanks Laurent |
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#5 |
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Senior Member
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hi,
Which solver you are using? for any simpleFoam you need to divide it with the density what you are using. Thanks, Sivakumar |
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#6 |
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Member
Laurent B
Join Date: Jun 2009
Location: Lille, FRANCE
Posts: 59
Rep Power: 5 ![]() |
I'm using MRFSimpleFoam. The density is 1300 kg/m3. If i divide by 1300 the difference is bigger...
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#7 |
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Senior Member
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hi,
What kind of simulation you are doing, I am also using MRFSimpleFoam for axial flow fan. In my case MRFSimpleFoam torque results keep on changing for each case Actually, I tried two cases one is with passage (suction side and pressure side of the fan) secondly two blade and quarter of the cylinder see the fig. for passage case I am getting 45 N-m (it should be 70-80 N-m) for 2 blade case I am getting 80 N-m I am totally confused, is there any problem with MRFSimpleFoam or in my settings. I am totally gone. Thanks, Sivakumar Last edited by sivakumar; February 20, 2013 at 05:57. |
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#8 |
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Senior Member
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Hi Guys and Laurentb,
I got the right way to calculate torque. I was doing some wrong calculation, that is why my results were wrong. now I got the right results, with 2% error. To Laurentb, I am sorry man, I think I have miss guided you. Actually we need to multiply the pressure with density to get the exact pressure. to calculate torque: take the z component of the pressure force and viscous. torque = F*r F = ((P *A) *density+ tau *A) A --> area of the patch which you want to calculate torque. I hope it is clear now Thanks, Sivakumar |
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#9 |
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Member
Laurent B
Join Date: Jun 2009
Location: Lille, FRANCE
Posts: 59
Rep Power: 5 ![]() |
Hi,
Finally the torque computed by the forces library fits my experimental data with 10% error. My error comes from that i turn off the turbulence in RASProperties file for laminar flow conditions. If i turn on the turbulence option the force computation works well. Laurent |
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#10 |
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New Member
Baek, Donghae
Join Date: Jan 2013
Location: Seoul
Posts: 7
Rep Power: 2 ![]() |
Hi all
I have something to ask. My case is 2D on x-y plane. Object look like airfoil and it is in wind tunnel. I obtain moment using force function added controldic file. Center of moment is z-axis. From result, moment has minus value. In this case, Is object rotated to clockwise or countclockwise?? |
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#11 | |
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Senior Member
Forrest
Join Date: Jun 2012
Posts: 218
Rep Power: 3 ![]() |
Quote:
Thats cool,Thanks very much.But I still have some problems. 1.How can I get the area of my patch? cuz its not a regular geometry. 2.If my impeller is rotating around y axiz.Should I pick the y component forces? 3.I have many impellers,How can I get the sum torque? 4.As we know,torque is F*radius, is this radius the max radius? 5.I am using MRFInterFoam, and I add this code into controldict. Code:
outputControl timeStep;
outputInterval 1000;
patches (intermigOpposite1);
// pname p;
// Uname U;
rhoName rhoInf;
log true;
rhoInf 1000;
CofR (0 0 0);
}
and I get the forces: Code:
forces output:
forces(pressure,viscous)((-0.111013 9.4727 -0.09283),(0.00142333 -0.00100197 0.0121232))
moment(pressure,viscous)((-0.0196299 0.941766 0.19644),(0.00167602 0.0451228 0.000146112))
forces output:
forces(pressure,viscous)((0.447146 12.7018 -0.170442),(-0.00129838 0.00185303 -0.00024443))
moment(pressure,viscous)((-0.0262434 0.928373 -0.127495),(1.8784e-05 0.0418585 0.000395807))
forces output:
forces(pressure,viscous)((-0.0942824 4.63998 0.16988),(-0.00217494 -0.0359383 0.00219822))
moment(pressure,viscous)((0.0744167 1.1975 0.0769798),(0.00104654 0.0454346 0.00164999))
forces output:
forces(pressure,viscous)((-0.226807 5.13595 -0.159452),(-0.0119485 -0.0779935 -0.00055085))
moment(pressure,viscous)((0.015837 1.36635 0.137174),(4.43451e-05 0.0472514 0.00681036))
forces output:
forces(pressure,viscous)((0.300891 1.99927 0.0561256),(-0.00293573 0.00419335 0.000610194))
moment(pressure,viscous)((0.0233319 9.04241e-06 -0.05736),(0.000348815 0.000324186 0.000962111))
If I want to calculate,for example the first one. Force=(9.4727-0.0010017)*0.033555(my patch area)=0.3178n torque=0.3178*0.145(radius)=0.046Nm. Well, 0.046Nm is too small. There are many problems,Thanks again for your patience. Regards. Last edited by sharonyue; May 6, 2013 at 22:01. |
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#12 |
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Senior Member
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Hi,
To calculate torque, just add the components (moment and viscous) which belongs to the axis of rotation. You no need to think of the area. In your case for the first output moment about y axis: y component of moment + viscous = 0.941766+0.0451228 If you get wrong torque, check your settings and blade angle. if your torque is small, how small it is? what is the experimental torque value? Thanks, Sivakumar |
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#13 | |
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Senior Member
Forrest
Join Date: Jun 2012
Posts: 218
Rep Power: 3 ![]() |
Quote:
If I add them all.and is this the torque? y component of moment + viscous = 0.941766+0.0451228 +0.928373 +0.0418585 +.....=4.6 Well , my experiment is a little different with my CFD case. but the torque should not differ much. if 4.6's unit is Nm, I think its normal. But in you last post, torque = F*r F = ((P *A) *density+ tau *A) A --> area of the patch which you want to calculate torque. Why is this using area? Anyway,you give me lots of hints,Thanks very much!! |
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#14 |
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Senior Member
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Hi,
In the previous post I used force to calculate torque, so OF out put is interns of pressure force and viscous force. Pressure = F/A, so I used area. but, in your case just add the Y component of the pressure moment and viscous moment (this is the normal way). how many % error you are getting in torque? how good is your results with experimental results? Thanks, Sivakumar |
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#15 | |
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Senior Member
Forrest
Join Date: Jun 2012
Posts: 218
Rep Power: 3 ![]() |
Quote:
My experiment is a little different with my case,so I cannot make a comparision between them.But it should not differ much. If my case's torque is about 4 Nm. its normal. If its below 1 or 2Nm its not normal. And in my experiment, its two phase.Thats a gas-liquid mixer.Now I dont know how to calculate two phase's torque in FOAM.So... Well,looks like there are two ways calculating torque,1.use the forces to calculate which would need to know the area.2.use the moment just add them all.But in my case I think the first method is not working... |
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