# MRF Rotation

#7

As one blade will be ‘extracting’ wind energy and symmetric blade will be ‘wasting’ the wind energy. you need to ADD those forces in order to obtain resulting NET force.

That net force from Savonius turbine is, as you expect already, will be rotation angle bound.

#8

So , is my angular velocity value setted in mrf going to affect my force plot?

#9

the same logic effects the helical savonius or not ? one blade each time and calculate resultant .
https://www.simscale.com/projects/ahmedna8/vawt_1_test_1/

#10

i have validated what @Retsam suggest. The output of the turbine that I obtained through simulation is very close to the experiment one. That indicated that the simulation is successful. For your question, I think the answer is ‘yes’ , as long as negative torque exists , then you should measure the forces separately

Regarding to the question you inbox me, you suppose to post it in the forum , but i can answer your question here, you use pressure force x,y,z in your force plot and calculated the magnitude of them to get your resultant force by the wind turbine. Viscous force can be ignored as we are dealing with low viscosity fluid (air)

Cp equals to wind turbine output divided by wind power available , wind turbine output is Force x radius of your turbine x angular velocity
Wind power available is 0.5xdensity of air x rotor area x wind velocity cube , the calculation of rotor area for hawt and vawt is different, i dont know if helical savonius is the same as conventional savonius or not, you better find out before assuming anything

#11

Initially I thought the force plot already give us the resultant value , thats why I am stressing out for so much… thank you sir, this really helped me by a lot! I am very close to completing my project now!

#12

one more question .
if i calculated it separately. how can i calculate the resultant force?
i need the equation

#13

this what i am thinking of .
i add x components together .so do y and z
then F=((Fx)^2+(Fy)^2+(Fz)^2)^1/2

is this right?

#14

I agree, but helical Savonius, in case it has 360 degree twist should not have negative torque. In most cases, the twist is only 45 - 90 degrees which makes rotation smoother, but still oscillating.

As you both are interested in Savonius turbine, I can cite an interesting study and simulation (ANSYS Fluent was used), but that study could help to better understand the complexity of that simple Savonius setup.

https://www.sciencedirect.com/science/article/pii/S1110016819300171#f0010

#15

I question your ability to use that formula, in this instance, to determine any form of VAT output power…

The ‘forces and moments’ output of the simulation run, are steady state forces at a snapshot in time, when the VAT is rotating at -5.91 rad/s in a 5 mps wind. The snapshot in time that you used is when the blades are close to the rotation angle where you would expect to generate the greatest torque about the axis of rotation.

Also, it looks to me that the ‘Center of Rotation’ used for the forces and moments output is well outside the extremes of the VAT geometry (see geometry bounding box max,mins and ranges). This does not make sense to me. If you are trying to actually calculate the torque that is required to be applied to the VAT rotation shaft in order to hold the blades at that rotation angle and rotational speed, then either you must translate the force and moment result values that you have, to a point that lies on the VAT rotation axis or, much more simply, put the Center of Rotation somewhere on the VAT rotation axis (I would put it the Y location of one of the VAT rotation shaft bearings) and then re-run the sim…

Even after all that, you only have one data point of the infinite number of VAT angular positions,which could can be analysed in a steady state (each of those positions would generate a different VAT rotation shaft torque that is required to hold the VAT in that position …)

Because of symmetry you could maybe take 18 sim runs from 0 to 170 degrees of rotation and then maybe take the average of those torque (MomentY) results as some form of output power estimation (you may even have chosen a rotational speed that is higher than what would be attainable without drawing ANY power at all from the shaft, in that case you could estimate how big a motor you would have to ADD to the shaft to get it to that rotational speed, in that wind )

What an I missing here???

#16

CAN I USE AMI instead of mrf to get transient forces?

#17

I did 8 simruns , 0-180 blade angle

#18

why moment y ? or should i take force ?

#19

Once you have the ‘Forces and Moments’ center of rotation anywhere on the axis of rotation, then the only ‘Forces and Moments’ variables that have anything to do with output power are the moments about the rotation axis.

#20

as example this my simulation about savanous helical rotor with 180 degree twist at 135 degree of rotation
https://www.simscale.com/projects/ahmedna8/vawt_1_test_135/

moment or forces that i should take to put in the equation of power coefficient?

Contra-Rotating Windmill Simulation
#21
1. You will have to learn how to calculate Cp for your design, lots of info about that is out there.
2. With SimScale, you MAY be able to obtain a very rough guess as to how much power you can extract from your design at different wind speeds and rpms… This power that you calculate will be used in Cp calculation.
3. Power is a function of torque (moment) and rpm. Formulas are out there. Rpm is what you input in MRF (I think you may have the incorrect direction for that in your simulation). Torque for each blade is in your ‘Forces and Moments’ results (both pressure and viscous moments need to be added together).
4. In order to have any faith at all in the values of forces and moments results, those values should be stable within a 1% range of the value, over the last 500 iterations… Your values are not considered stable.
5. Again, be careful that you do not spin the MRF faster than the no load rpm of that design in the wind speed of the sim run. If you do, then you will be INPUTing power to the windmill to get it to the rpm you have set the MRF to (you will see that by the sign of the moments once you are sure you have the correct rotation direction in the MRF parms).
6. You should be working with coarser meshes until you zero in on having a sim setup that works.

#22

thank you so much

#23

i have one more question
for example
pressure moment x : -0.255 Pm y: 0.555 Pm z; -0.424
Pmx; 0.455 Pmy: 0.623 Pmz; -0.546
how i get resultant?
because i have two opinions and i am confused

#24

y is rotation axis, so pressure momentY sum is 0.555+0.623 Nm (edit, sorry decimal location was incorrect)

#25

if i have Pmy 1: -0.256
Pmy 2: 0.4
the resultant is 0.144 not 0.656

#26

thank you sir
you helped me alot