Mass Moment of Inertia

[sinu21]

Dear friends

I am trying to find a mass moment of inertia of an wind turbine on its rotor.
No of blades = 3
Weight of blade = 11.5T
COG = 15m

If i use mass x radius 2
i get 2587.5 x 3 = 7762.5 Tm2

i want to find an equivalent box section with provides me same mass moment of inertia in the same vertical axis.

for a box section = (M/12) x (b2 + h2)


If i keep the mass constant 11.5*3= 34.5 T
b=51m and h=14.5m


can anyone advice me how to best represent a 3 blade rotor to a rectangular box.

Thanks in advance.

Regards

 

[rb1957]

i think using m*r^2 is a major under-estimate.

m*r^2 is the inertia of a point mass at a distance r, but you have a rotating blade. to use an analogy, when calculating the moment of inertia of a section you have Ax^2+Io, the 2nd term being the self moment of inertia of the area.

i think you need to divide the blade up into a bunch of small segments so that the self moment term is small enough to be negligible.

 

[johnbridge231]

You may find a solution in a basic reference or book for dynamic analysis, for examnple Chopra's book is sxcellent.

Analysis and Design of arbitrary cross sections
Reinforcement design to all major codes
Moment Curvature analysis

http://www.engissol.com/cross-section-analysis-design.html

 

[BAretired]

Polar moment of inertia:
For rotors, I_p = M(15)² = 225M

For box section, Ip = M(b² +h²)/12

So (b² + h²) = 12*225 = 2700
For a square, b = h = 36.7m
For a rectangle with b = 51m, h = 9.95m (say 10m)


BA

 

[IRstuff]

Why a rectangular box? Why wouldn't a cylinder be more natural?

In any case, you need to perform the integral from 0 to R of mass(r)*r^2 to get the moment of inertia.

TTFN
faq731-376

 

[BAretired]

You do not have to perform any integral if you know the centroid of each blade.

BA

 

[IRstuff]

Agreed, but the centroid is also the result of an integral.

TTFN
faq731-376

 

[rb1957]

maybe he wants something to simulate a blade, like for a test rig.

are you guys sure it's imply m*cg^2 ? particularly with the rectangle solution. i mean the cg^2 for a rectangle will be (b^2+h^2)/4 ... no?

 

[BAretired]

Polar Moment of Inertia is defined as [∑]R²dA where R is the radius of the small area dA from the c.g.

For a rectangle, Ip = Ix + Iy = bh³/12 +hb³/12 = bh(b² + h²)/12

Polar Mass Moment of Inertia is Ip*[ρ] = M(b² + h²)/12 where [ρ] is density per unit area.

BA

 

[rb1957]

aren't the rectangle results above (Ip) about the centroid of the rectangle ? which is why i think there's more to it than m*cg^2 (this'd be the result for a point mass)

it seems to me that we're talking about a blade (maybe it can be simplified to a rectangle) rotating about it's base. so that'd change "/12" to /3 ... ?

 

[IRstuff]

If you know enough to approximate the blades themselves with rectangles, then it would be just as easy, I think, to crank the integral and get a more precise answer.

TTFN
faq731-376

 

[BAretired]

My understanding of the problem is that the OP wants to find the size of a rectangle of mass M centered at the hub of the three blades, each having mass M/3 centered 15m from the hub. The plane of the rectangle is normal to the axis of rotation of the rotor.

I have no idea why he wants to find this, but that is my understanding of the problem. If that is not what the OP is asking, perhaps he could clarify.

BA

 

[IRstuff]

I read that, but it made no sense, since he's ostensibly trading a symmetrical structure for a partially symmetrical structure.

TTFN
faq731-376

 

[rb1957]

i thought he was setting up a test of the generator, and needed to simulate the blades ...
though a round plate, or better yet three concentrated weights, would be "better".

 

[GearsNsparks]

If you have access to the 3D solid model, you can probably use your CAD software to find an accurate value for your MMOI.

 

[BAretired]

If you have access to the 3D solid model, you can probably use your CAD software to find an accurate value for your MMOI.

What a sad commentary on the status of engineering today.

BA