WT-Perf Compare to FAST

Wind & Water Computer-Aided Engineering Software Tools
1

Dear NREL Team

I made the performance curves by WT-Perf using same blade aerodynamic input file from FAST(AeroDyn).
I run FAST for specific wind speed and I measured the Output Power, Shaft Speed and Pitch angle. Then I calculated the Power Coefficient, TSR for this operation point.
But when I use simillar TSR and PITCH angle to interpolate the relevant CP from WT-Perf CP curve I got quite different value. Something around 15% difference.
What do you think about the reason for this difference ?

Best Regards
Mahmoud

2

Dear Mahmoud,

Can you explain how you obtained the CP from FAST? Did you output RotCp or something else? Did you run FAST until a steady state condition was reached before outputting?

Best regards,

3

Dear Mr. Jonkman

I measured Cp from fast in 2 ways.
Method 1: Direct measurment from fast by RotCp channel. For 200 Sec simulation with constant wind speed of 8 [m/s] CP=0.483 TSR=10.13 BPITCH=0;
METHOD 2: From an observer,
I have designed an observer according to dynamic model of dtive train, The estimated aerodynamic torque is

                              Tr=(Ir+NG^2*Ig)*d/dt(Wr)+(Br)*Wr+(Bdt)*(Wr-Wg/NG)+Kdt*int(Wr-Wg/NG)+NG*(Wg*Bg+Tg).

The description and magnetude for each parameters according to FAST are :

  • Tr : Aerodynamic Torque [Unknown]
  • Wr : Low Speed Shaft speed [rad/s]
  • Wg : High speed shaft speed [rad/s]
  • Ir : Rotor Inertia : Ir=3Ib+Ih=312266982.6129+115926=419423713;[Kh.m^2]
  • Br : Rotor Damping Coef. = 0
  • Kdt : Low Speed shaft tortional stiffenes 867637000 [N.m/Rad]
  • Bdt : Low Speed shaft Damping coef 6215000 [N.m/(Rad.s)]
  • Bg : Generator damping coefficinet 0 [N.m/(Rad.s)]
  • Ig : Generator Inertia 534.116 [kg.m^2]
  • NG : GearBox Ratio
  • int() : Integral Operator
  • d/dt() : Derivative Operator

Aerodynamic power will be :
Pr=Tr*Wr
now one can simply calculate the power coefficients from :

                   Cp=Pr/(0.5*1.225*pi*63^2*Ur^3)

where Ur is the incoming wind speed.
But the magnetude of Power and Power coefficient from this estimationn is increidebly high. Almost 150 times bigger.

As a summary the value of CP for each method are
Method 1 : Fast Output Cp=0.47 [-]
Method 2 : From torque observer CP= 81 [-] ?
Method 3 : WT-Perf CP Curve Cp=0.41

Why method 2 gives such a big value. What is the reason for difference between method 1 and 3.

Do you think this observe is correct for fast.

Best Regards
Mahmoud

4

Dear Mahmoud,

The last time I compared FAST and WT_Perf was back in 2004 when I was developing the NREL 5-MW model. The comparison was very good. FAST predicted CpMax = 0.482 at TSR = 7.55 and Pitch = 0.0. WT_Perf predicted CpMax = 0.487 at TSR = 7.69 and Pitch = 0.0. I’m not sure why your comparison (Method 1 versus Method 3) does not match as well. I suspect you’ve made a mistake in one of your models.

I’m not sure I understand how you obtained your equation in Method 2. I would think that a “Br” term would be included in “Tr” and a “Bg” term would be included in “Tg” (althoug you’ve assumed these to be zero?). Also, I’m not sure why you include the generator inertia and torque in your equation. The aerodynamic torque could be calculated as follows (using your nomenclature):

Tr = Ird/dt(Wr)+Bdt(Wr-Wg/NG)+Kdt*int(Wr-Wg/NG)

The sum of the last two terms equals the output “LSShftTq” in FAST, so, you could simplify the equation to:

Tr = Ir*d/dt(Wr)+LSShftTq

Most values in your post appear to be the values defined for the NREL 5-MW turbine except for the rotor inertia. I’m not sure where you got the numbers for the rotor inertia, but the inertia calculated by FAST and written to the FAST summary file (.fsm file) for the NREL 5-MW turbine is Ir = 38759228 kgm^2, which is over an order of magnitude less than the number you reported. Regardless, for a simulation that reaches steady state, the inertia term should tend to zero, meaning that the LSShftTq output by FAST equals the applied aerodynamic torque.

I hope that helps.

Best regards,