Fluid structure interaction of Wind Turbine

Problem statement : The blade is 43.2 meters long and starts with a cylindrical shape at the root and then transitions to the airfoils S818, S825 and S826 for the root, body and tip, respectively. This blade also has pitch to vary as a function of radius, giving it a twist and the pitch angle at the blade tip is 4 degrees. This blade was created to be similar in size to a GE 1.5XLE turbine.The blade is made out of an orthotropic composite material, it has a varying thickness and it also has a spar inside the blade for structural rigidity. The turbulent wind flows towards the negative z-direction (into the page on the above diagram) at 12 m/s which is a typical rated wind speed for a turbine this size. This incoming flow is assumed to make the blade rotate at an angular velocity of -2.22 rad/s about the z-axis. ts density is 1.225 kg/m^3 and its viscosity is 1.7894e-05 kg/(m*s).

orthotropic material properties

Density (kg/m^3)	1550
Young's Modulus-X (Pa)	1.1375E+11
Young's Modulus-Y (Pa)	7.583E+09
Young's Modulus-Z (Pa)	7.583E+09
Poisson's Ratio-XY	0.32
Poisson's Ratio-YZ	0.37
Poisson's Ratio-XZ	0.35
Shear Modulus-XY (Pa)	5.446E+09
Shear Modulus-YZ (Pa)	2.964E+09
Shear Modulus-XZ (Pa)	2.964E+09

To find :

¨A) Plot the pressure distribution in the y-z plane at x=-20m and at x=-40m, with the negative x-axis direction pointing into the page (eg. you are looking towards the blade tip). Export these plots out of FLUENT and sketch in:

¨The z and y axis, superimposed on the blade

¨The chordline

¨The relative wind vector ( Urel ) along with its component in both the radial (coming from the rotation of the blade), called Urad, and incoming wind direction (the one passing through the rotor at 12 m/s), called U.

¨The angle of attack ( α )

¨The lift and drag vectors (L and D)

¨The resulting tangential force, T, (which means it is parallel to the y-axis) and normal force, N, (which means it is parallel to the z-axis), coming from the lift and drag force vectors.

¨Try to draw the vectors with appropriate magnitudes.

¨B) Use trigonometry to derive the equations for the normal and tangential forces as functions of the angle of attack. What is the effect of the tangential force? What about the normal force?

¨C) From the relevant forces on the blade, derive

¨the expression for the torque acting on the blade,

¨use this to find the expression for the total torque acting on the rotor,

¨and finally use this to determine the expression of the aerodynamic power

¨D) Thoroughly explain the difference in results between the cross-sections at x=-20 and at x=-40 (i.e pressure distribution, direction and magnitude of the vectors, etc.) Also, from these results, explain why more power is produced at x=-40m.

Step 1] : Hand Calculations

1. Velocity at the tip of blade

V=r*w

= 44.2*2.22

= 98.12m/s

2. Power Coefficient (Cp)

GE1.5XLE Wind turbine (Reference)

Cp=Prated/Pwind=Prated/0.5ρAV^3

=Prate/0.5(1.225kgm^3)(π*(82.5)^2/4)* (11.5m/s)^3=0.30

Our Cp calculations

Protor= n*T*2.22rad/s

= 3*140903*2.22

= 938413.98KW

Pwind =0.5ρAV^3=0.5*1.225* π*(88.4)^2/4*(12)^3

= 6.4*10^6

Cp=Prated/Pwind=0.1466=14.66%

Betz limit=59.2% of wind kinetic energy can be absorbed

Cp should lie within 59.2%