Question

The following table is for a wind turbine:

1. Calculate the Annual Energy production (AEP) using the measured wind data.
2. Calculate the Annual Energy production (AEP) using the Rayleigh distribution.
3. Calculate the error incurred, compared to (a) when using the Rayleigh distribution under (b).

U [m/s] 0.00 power [kW] SNIUI 7.50 1735 2050 2000 17.50 2050 2050 2050 2050 19.50 22.00 Table 2: turbine power production vs. Wind speed

Answer 1

	Power, P_i (kW)	Speed, u_i (m/s)	C	2v/c^2	(v/c)^2	f(V)	sum(p(u)f(v))	Turbine power (kW)
	0	0	11.386	0.000	0.000	0	0	1151.584938
	0	0.5	11.386	0.008	0.002	0.007698241	0
	2	1.5	11.386	0.023	0.017	0.022741192	0.045482385
	9	2.5	11.386	0.039	0.048	0.036750475	0.330754273
	67	3.5	11.386	0.054	0.094	0.049123856	3.291298356
	129	4.5	11.386	0.069	0.156	0.059379819	7.659996618
	226	5.5	11.386	0.085	0.233	0.067187964	15.18447977
	398	6.5	11.386	0.100	0.326	0.072384415	28.80899708
	666	7.5	11.386	0.116	0.434	0.074971532	49.93104005
	1015	8.5	11.386	0.131	0.557	0.075103113	76.22966018
	1397	9.5	11.386	0.147	0.696	0.07305787	102.0618439
	1735	10.5	11.386	0.162	0.850	0.069205015	120.0707012
	1957	11.5	11.386	0.177	1.020	0.063966335	125.182117
	2050	12.5	11.386	0.193	1.205	0.057778928	118.4468032
	2050	13.5	11.386	0.208	1.406	0.051062206	104.6775213
	2050	14.5	11.386	0.224	1.622	0.044191691	90.5929667
	2050	15.5	11.386	0.239	1.853	0.037481039	76.83613005
	2050	16.5	11.386	0.255	2.100	0.031172502	63.90362887
	2050	17.5	11.386	0.270	2.362	0.025435135	52.14202631
	2050	18.5	11.386	0.285	2.640	0.020369318	41.75710197
	2050	19.5	11.386	0.301	2.933	0.0160158	32.8323891
	2050	22	11.386	0.339	3.733	0.008117073	41.59999968
Average	1184.136364	U_avg=10.0909091

2500 2000 Nominal Velocity Power, kW - 15 Velocity, m/s 20 25

(a) The Annual Energy Production (AEP) of the wing turbine by using the nominal method is given below:

The nominal power is the power corresponding to the nominal velocity. The nominal velocity is the velocity at which the power output of the turbine becomes almost constant.

Walker (2011) have given the following formula for calculating the power output of the turbine

(1)

Pturbine = Pnominal X 712 hominal - U1

where as calculated above in the table.

Vaug = 10.09 m/s

Nominal velocity can be obtain from the graph as shown above or table given above which is:

Unominal = 12.5 m/s

and (using table)

Pnominal = 2050 kW

is the initial velocity,

U1=0

Thus,

Pturbine = 2050 x 10.092 - 02 12.52 _ 02= 1335.72 kW

Hence the AEP is given by:

   (2)

AEP = Pturbine x 356 x 24 = Pturbine x 8760 kWh

(3)

AEP = 1335.72 x 8760 = 11700907.2 kWh = 11.7 GWh

which is the required result.

(b) AEP using the Rayleigh method.

The Rayleigh distribution of the average velocity is given by:

(4)

2x el-(92

Where,shape factor in case of Rayleigh distribution  $k=2$ is substituted in weibull distribution,

also, The scale factor, c is given by:

$c=\frac{2U_{avg}}{\sqrt{\pi}}=\frac{2\times10.09}{\sqrt \pi}=11.385$

The Turbine Power output is given by:

(5)

Purbine = _." PW)F(ody = * pv) x el-eild

The value of this power output of turbine is calculated in the table as shown above in the 9th column given by:

Pfurbine = 1151.585 kW

Now using equation (2) we have

AEP = Pturbine x 356 x 24 = 1151.585 x 8760 kWh

AEP = 10087884.6 kWh = 10.09 GWh

Which is the required result.

(C)The error incured in method A compared to Rayleigh method in B is given as:

error = AEPA - AE PRayleigh = 11.7 – 10.09 = 1.61 GWh

Also,

AE P. - AEPRayleigh x 100 = 11.7 – 10.09 - 10.09 error(percent) = 7 15.95 Percent X 100 = AE PRayleigh